Chapter Two Exercise Solutions Engineering Circuit Analysis (6th Edition, 2001) Hayt Solution Manual

CHAPTER TWO SOLUTIONS

1. (a) 12 µs (d) 3.5 Gbits (g) 39 pA

(b) 750 mJ (e) 6.5 nm (h) 49 kΩ

(c) 1.13 kΩ (f) 13.56 MHz (i) 11.73 pA

CHAPTER TWO SOLUTIONS

2. (a) 1 MW (e) 33 µJ (i) 32 mm

(b) 12.35 mm (f) 5.33 nW

(c) 47. kW (g) 1 ns

(d) 5.46 mA (h) 5.555 MW

CHAPTER TWO SOLUTIONS

3. Motor power = 175 Hp

(a) With 100% efficient mechanical to electrical power conversion,

(175 Hp)[1 W/ (1/745.7 Hp)] = 130.5 kW

(b) Running for 3 hours,

Energy = (130.5×103 W)(3 hr)(60 min/hr)(60 s/min) = 1.409 GJ

(c) A single battery has 430 kW-hr capacity. We require

(130.5 kW)(3 hr) = 391.5 kW-hr therefore one battery is sufficient.

CHAPTER TWO SOLUTIONS

4. The 400-mJ pulse lasts 20 ns.

(a) To compute the peak power, we assume the pulse shape is square:

Then P = 400×10-3/20×10-9 = 20 MW.

(b) At 20 pulses per second, the average power is

Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = 8 W.

400

Energy (mJ)

t (ns)

20

CHAPTER TWO SOLUTIONS

5. The 1-mJ pulse lasts 75 fs.

(c) To compute the peak power, we assume the pulse shape is square:

Then P = 1×10-3/75×10-15 = 13.33 GW.

(d) At 100 pulses per second, the average power is

Pavg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW.

1

Energy (mJ)

t (fs)

75

CHAPTER TWO SOLUTIONS

6. The power drawn from the battery is (not quite drawn to scale):

(a) Total energy (in J) expended is

[6(5) + 0(2) + 0.5(10)(10) + 0.5(10)(7)]60 = 6.9 kJ.

(b) The average power in Btu/hr is

(6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr.

5 7 17 24

P (W)

10

6

t (min)

CHAPTER TWO SOLUTIONS

7. Total charge q = 18t2 – 2t4 C.

(a) q(2 s) = 40 C.

(b) To find the maximum charge within 0 ≤ t ≤ 3 s, we need to take the first and

second derivitives:

dq/dt = 36t – 8t3 = 0, leading to roots at 0, ± 2.121 s

d2q/dt2 = 36 – 24t2

substituting t = 2.121 s into the expression for d2q/dt2, we obtain a value of

–14.9, so that this root represents a maximum.

Thus, we find a maximum charge q = 40.5 C at t = 2.121 s.

(c) The rate of charge accumulation at t = 8 s is

dq/dt|t = 0.8 = 36(0.8) – 8(0.8)3 = 24.7 C/s.

(d) See Fig. (a) and (b).

(a) (b)

CHAPTER TWO SOLUTIONS

8. Referring to Fig. 2.6c,

0 A, 3 2-

)( 3

5

1





>+

<+

=−

te

ti t

t

Thus,

(a) i1(-0.2) = 6.155 A

(b) i1 (0.2) = 3.466 A

(c) To determine the instants at which i1 = 0, we must consider t < 0 and t > 0 separately:

for t < 0, - 2 + 3e-5t = 0 leads to t = -0.2 ln (2/3) = +2.027 s (impossible)

for t > 0, -2 + 3e3t = 0 leads to t = (1/3) ln (2/3) = -0.135 s (impossible)

Therefore, the current is never negative.

(d) The total charge passed left to right in the interval 0.08 < t < 0.1 s is

q(t) = ∫−

1.0

08.0 1)( dtti

=

[][]

∫∫ +−++−

−

−1.0

0

3

0

08.0

532 32 dtedte tt

= 1.0

0

3

0

0.08-

532 32 tt ee +−++− −

= 0.1351 + 0.1499

= 285 mC

CHAPTER TWO SOLUTIONS

9. Referring to Fig. 2.28,

(a) The average current over one period (10 s) is

iavg = [-4(2) + 2(2) + 6(2) + 0(4)]/10 = 800 mA

(b) The total charge transferred over the interval 1 < t < 12 s is

∫

=12

1

total )( dttiq = -4(2) + 2(2) + 6(2) + 0(4) –4(2) = 0 C

(d) See Fig. below

2 4 6

8

10 12

16

q (C)

t(s)

-16

8

16

-8

14

CHAPTER TWO SOLUTIONS

10. (a) Pabs = (+3.2 V)(-2 mA) = -6.4 mW (or +6.4 mW supplied)

(b) Pabs = (+6 V)(-20 A) = -120 W (or +120 W supplied)

(d) Pabs = (+6 V)(2 ix) = (+6 V)[(2)(5 A)] = +60 W

(e) Pabs = (4 sin 1000t V)(-8 cos 1000t mA)| t = 2 ms = +12.11 W

CHAPTER TWO SOLUTIONS

11. i = 3te-100t mA and v = [6 – 600t] e-100t mV

(a) The power absorbed at t = 5 ms is

Pabs =

()

[

]

W 3 6006 5

100100

µ

mst

tt teet =

−− ⋅−

= 0.01655 µW = 16.55 nW

(b) The energy delivered over the interval 0 < t < ∞ is

()

∫∫

∞∞

−

−=

00

200 J 60063

µ

dtettdtP t

abs

Making use of the relationship

1

0

!

+

∞−=

∫n

axn

a

n

dxex where n is a positive integer and a > 0,

we find the energy delivered to be

= 18/(200)2 - 1800/(200)3

= 0

CHAPTER TWO SOLUTIONS

12. (a) Pabs = (40i)(3e-100t)| t = 8 ms =

[]

2

8

100

360 mst

t

e=

− = 72.68 W

(b) Pabs =

[]

W36.34- 180- 2.0 2

8

100 ==











=

−mst

t

ei

dt

di

(c) Pabs =

()

mst

t

teidt

8

100

03 20 30

=

−











+

∫

=

mst

t

ttt etdee

8

100

0

100100 60 3 90

=

−

′

−− 









+

′

∫ = 27.63 W

CHAPTER TWO SOLUTIONS

13. (a) The short-circuit current is the value of the current at V = 0.

Reading from the graph, this corresponds to approximately 3.0 A.

(b) The open-circuit voltage is the value of the voltage at I = 0.

Reading from the graph, this corresponds to roughly 0.4875 V, estimating the curve as

hitting the x-axis 1 mm behind the 0.5 V mark.

(c) We see that the maximum current corresponds to zero voltage, and likewise, the

maximum voltage occurs at zero current. The maximum power point, therefore,

occurs somewhere between these two points. By trial and error,

Pmax is roughly (375 mV)(2.5 A) = 938 mW, or just under 1 W.

CHAPTER TWO SOLUTIONS

14. Note that in the table below, only the –4-A source and the –3-A source are actually

“absorbing” power; the remaining sources are supplying power to the circuit.

Source Absorbed Power Absorbed Power

2-V source (2 V)(-2 A) - 4 W

8-V source (8 V)(-2 A) - 16 W

-4-A source (10 V)[-(-4 A)] 40 W

10-V source (10 V)(-5 A) - 50 W

-3-A source (10 V)[-(-3 A)] 30 W

The 5 power quantities sum to –4 – 16 + 40 – 50 + 30 = 0, as demanded from

conservation of energy.

CHAPTER TWO SOLUTIONS

15. We are told that Vx = 1 V, and from Fig. 2.33 we see that the current flowing through

the dependent source (and hence through each element of the circuit) is 5Vx = 5 A.

We will compute absorbed power by using the current flowing into the positive

reference terminal of the appropriate voltage (passive sign convention), and we will

compute supplied power by using the current flowing out of the positive reference

terminal of the appropriate voltage.

(a) The power absorbed by element “A” = (9 V)(5 A) = 45 W

(b) The power supplied by the 1-V source = (1 V)(5 A) = 5 W, and

the power supplied by the dependent source = (8 V)(5 A) = 40 W

(c) The sum of the supplied power = 5 + 40 = 45 W

The sum of the absorbed power is 45 W, so

yes, the sum of the power supplied = the sum of the power absorbed, as we

expect from the principle of conservation of energy.

CHAPTER TWO SOLUTIONS

16. We are asked to determine the voltage vs, which is identical to the voltage labeled v1.

The only remaining reference to v1 is in the expression for the current flowing through

the dependent source, 5v1.

This current is equal to –i2.

Thus,

5 v1 = -i2 = - 5 mA

Therefore v1 = -1 mV

and so vs = v1 = -1 mV

CHAPTER TWO SOLUTIONS

17. The battery delivers an energy of 460.8 W-hr over a period of 8 hrs.

(a) The power delivered to the headlight is therefore (460.8 W-hr)/(8 hr) = 57.6 W

(b) The current through the headlight is equal to the power it absorbs from the battery

divided by the voltage at which the power is supplied, or

I = (57.6 W)/(12 V) = 4.8 A

CHAPTER TWO SOLUTIONS

18. The supply voltage is 110 V, and the maximum dissipated power is 500 W. The fuses

are specified in terms of current, so we need to determine the maximum current that

can flow through the fuse.

P = V I therefore Imax = Pmax/V = (500 W)/(110 V) = 4.545 A

If we choose the 5-A fuse, it will allow up to (110 V)(5 A) = 550 W of power to be

delivered to the application (we must assume here that the fuse absorbs zero power, a

reasonable assumption in practice). This exceeds the specified maximum power.

If we choose the 4.5-A fuse instead, we will have a maximum current of 4.5 A. This

leads to a maximum power of (110)(4.5) = 495 W delivered to the application.

Although 495 W is less than the maximum power allowed, this fuse will provide

adequate protection for the application circuitry. If a fault occurs and the application

circuitry attempts to draw too much power, 1000 W for example, the fuse will blow,

no current will flow, and the application circuitry will be protected. However, if the

application circuitry tries to draw its maximum rated power (500 W), the fuse will

also blow. In practice, most equipment will not draw its maximum rated power

continuously- although to be safe, we typically assume that it will.

CHAPTER TWO SOLUTIONS

19. (a) Pabs = i2R = [20e-12t] 2 (1200) µW

= [20e-1.2] 2 (1200) µW

= 43.54 mW

(b) Pabs = v2/R = [40 cos 20t] 2 / 1200 W

= [40 cos 2] 2 / 1200 W

= 230.9 mW

(c) Pabs = v i = 8t 1.5 W

= 253.0 mW

keep in mind we

are using radians

CHAPTER TWO SOLUTIONS

20. It’s probably best to begin this problem by sketching the voltage waveform:

(a) vmax = +10 V

(b) vavg = [(+10)(20×10-3) + (-10)(20×10-3)]/(40×10-3) = 0

(c) iavg = vavg /R = 0

(d) R

v

pabs

2

max

max = = (10)2 / 50 = 2 W

(e) 









⋅

−

+⋅

+

=20

)10(

20

)10(

40

1

22

RR

pavg

abs = 2 W

60

40

20 t (ms)

v (V)

+10

-10

CHAPTER TWO SOLUTIONS

21. We are given that the conductivity

σ

of copper is 5.8×107 S/m.

(a) 50 ft of #18 (18 AWG) copper wire, which has a diameter of 1.024 mm, will have

a resistance of l/(

σ

A) ohms, where A = the cross-sectional area and l = 50 ft.

Converting the dimensional quantities to meters,

l = (50 ft)(12 in/ft)(2.54 cm/in)(1 m/100 cm) = 15.24 m

and

r = 0.5(1.024 mm)(1 m/1000 mm) = 5.12×10-4 m

so that

A =

π

r2 =

π

(5.12×10-4 m)2 = 8.236×10-7 m2

Thus, R = (15.24 m)/[( 5.8×107)( 8.236×10-7)] = 319.0 mΩ

(b) We assume that the conductivity value specified also holds true at 50oC.

The cross-sectional area of the foil is

A = (33 µm)(500 µm)(1 m/106 µm)( 1 m/106 µm) = 1.65×10-8 m2

So that

R = (15 cm)(1 m/100 cm)/[( 5.8×107)( 1.65×10-8)] = 156.7 mΩ

A 3-A current flowing through this copper in the direction specified would

lead to the dissipation of

I2R = (3)2 (156.7) mW = 1.410 W

CHAPTER TWO SOLUTIONS

22. Since we are informed that the same current must flow through each component, we

begin by defining a current I flowing out of the positive reference terminal of the

voltage source.

The power supplied by the voltage source is Vs I.

The power absorbed by resistor R1 is I2R1.

The power absorbed by resistor R2 is I2R2.

Since we know that the total power supplied is equal to the total power absorbed,

we may write: Vs I = I2R1 + I2R2

or Vs = I R1 + I R2

Vs = I (R1 + R2)

By Ohm’s law,

I = 2

R

V/ R2

so that

Vs =

()

21

2

2RR

R

VR+

Solving for 2

R

Vwe find

()

21

2

s

2RR

R

VVR+

= Q.E.D.

CHAPTER TWO SOLUTIONS

23. (a)

(b) We see from our answer to part (a) that this device has a reasonably linear

characteristic (a not unreasonable degree of experimental error is evident in the data).

Thus, we choose to estimate the resistance using the two extreme points:

Reff = [(2.5 – (-1.5)]/[5.23 – (-3.19)] kΩ = 475 Ω

Using the last two points instead, we find Reff = 469 Ω, so that we can state with some

certainty at least that a reasonable estimate of the resistance is approximately 470 Ω.

(c)

CHAPTER TWO SOLUTIONS

24. Top Left Circuit: I = (5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW

Top Right Circuit: I = -(5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW

Bottom Left Circuit: I = (-5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW

Bottom Right Circuit: I = -(-5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW

CHAPTER TWO SOLUTIONS

25. The voltage vout is given by

vout = -10-3 vπ (1000)

= - vπ

Since vπ = vs = 0.01 cos 1000t V, we find that

vout = - vπ = -0.001 cos 1000t V

CHAPTER TWO SOLUTIONS

26. 18 AWG wire has a resistance of 6.39 Ω / 1000 ft.

Thus, we require 1000 (53) / 6.39 = 8294 ft of wire.

(Or 1.57 miles. Or, 2.53 km).

CHAPTER TWO SOLUTIONS

27. We need to create a 470-Ω resistor from 28 AWG wire, knowing that the ambient

temperature is 108oF, or 42.22oC.

Referring to Table 2.3, 28 AWG wire is 65.3 mΩ/ft at 20oC, and using the equation

provided we compute

R2/R1 = (234.5 + T2)/(234.5 + T1) = (234.5 + 42.22)/(234.5 + 20) = 1.087

We thus find that 28 AWG wire is (1.087)(65.3) = 71.0 mΩ/ft.

Thus, to repair the transmitter we will need

(470 Ω)/(71.0 × 10-3 Ω/ft) = 6620 ft (1.25 miles, or 2.02 km).

Note: This seems like a lot of wire to be washing up on shore. We may find we don’t

have enough. In that case, perhaps we should take our cue from Eq. [6], and try to

squash a piece of the wire flat so that it has a very small cross-sectional area…..

CHAPTER TWO SOLUTIONS

28. (a) We need to plot the negative and positive voltage ranges separately, as the positive

voltage range is, after all, exponential!

(b) To determine the resistance of the device at V = 550 mV, we compute the

corresponding current:

I = 10-6 [e39(0.55) – 1] = 2.068 A

Thus, R(0.55 V) = 0.55/2.068 = 266 mΩ

(c) R = 1 Ω corresponds to V = I. Thus, we need to solve the transcendental equation

I = 10-6 [e39I – 1]

Using a scientific calculator or the tried-and-true trial and error approach, we find that

I = 325.5 mA

CHAPTER TWO SOLUTIONS

29. We require a 10-Ω resistor, and are told it is for a portable application, implying that

size, weight or both would be important to consider when selecting a wire gauge. We

have 10,000 ft of each of the gauges listed in Table 2.3 with which to work. Quick

inspection of the values listed eliminates 2, 4 and 6 AWG wire as their respective

resistances are too low for only 10,000 ft of wire.

Using 12-AWG wire would require (10 Ω) / (1.59 mΩ/ft) = 6290 ft.

Using 28-AWG wire, the narrowest available, would require

(10 Ω) / (65.3 mΩ/ft) = 153 ft.

Would the 28-AWG wire weight less? Again referring to Table 2.3, we see that the

cross-sectional area of 28-AWG wire is 0.0804 mm2, and that of 12-AWG wire is

3.31 mm2. The volume of 12-AWG wire required is therefore 6345900 mm3, and that

of 28-AWG wire required is only 3750 mm3.

The best (but not the only) choice for a portable application is clear: 28-AWG wire!

CHAPTER TWO SOLUTIONS

30. Our target is a 100-Ω resistor. We see from the plot that at ND = 1015 cm-3, µn ~ 2x103

cm2/V-s, yielding a resistivity of 3.121 Ω-cm.

At ND = 1018 cm-3, µn ~ 230 cm2/ V-s, yielding a resistivity of 0.02714 Ω-cm.

Thus, we see that the lower doping level clearly provides material with higher

resistivity, requiring less of the available area on the silicon wafer.

Since R = ρL/A, where we know R = 10 Ω and ρ = 3.121 Ω-cm, we need only define

the resistor geometry to complete the design.

We typically form contacts primarily on the surface of a silicon wafer, so that the

wafer thickness would be part of the factor A; L represents the distance between the

contacts. Thus, we may write

R = 3.121 L/(250x10-4 Y)

where L and Y are dimensions on the surface of the wafer.

If we make Y small (i.e. a narrow width as viewed from the top of the wafer), then L

can also be small. Seeking a value of 0.080103 then for L/Y, and choosing Y = 100

µm (a large dimension for silicon devices), we find a contact-to-contact length of L =

8 cm! While this easily fits onto a 6” diameter wafer, we could probably do a little

better. We are also assuming that the resistor is to be cut from the wafer, and the ends

made the contacts, as shown below in the figure.

Design summary (one possibility): ND = 1015 cm-3

L = 8 cm

Y = 100 µm

250 µm

8 cm contact

100

µ

m

CHAPTER THREE SOLUTIONS

1.

CHAPTER THREE SOLUTIONS

2. (a) six nodes; (b) nine branches.

CHAPTER THREE SOLUTIONS

3. (a) Four nodes; (b) five branches; (c) path, yes – loop, no.

CHAPTER THREE SOLUTIONS

4. (a) Five nodes; (b) seven branches; (c) path, yes – loop, no.

CHAPTER THREE SOLUTIONS

5. (a) 3 A; (b) –3 A; (c) 0.

CHAPTER THREE SOLUTIONS

6. By KCL, we may write: 5 + iy + iz = 3 + ix

(a) ix = 2 + iy + iz = 2 + 2 + 0 = 4 A

(b) iy = 3 + ix – 5 – iz

iy = –2 + 2 – 2 iy

Thus, we find that iy = 0.

(c) This situation is impossible, since ix and iz are in opposite directions. The only

possible value (zero), is also disallowed, as KCL will not be satisfied ( 5 ≠ 3).

CHAPTER THREE SOLUTIONS

7. Focusing our attention on the bottom left node, we see that ix = 1 A.

Focusing our attention next on the top right node, we see that iy = 5A.

CHAPTER THREE SOLUTIONS

8. (a) vy = 1(3vx + iz)

vx = 5 V and given that iz = –3 A, we find that

vy = 3(5) – 3 = 12 V

(b) vy = 1(3vx + iz) = –6 = 3vx + 0.5

Solving, we find that vx = (–6 – 0.5)/3 = –2.167 V.

CHAPTER THREE SOLUTIONS

9. (a) ix = v1/10 + v1/10 = 5

2v1 = 50

so v1 = 25 V.

By Ohm’s law, we see that iy = v2/10

also, using Ohm’s law in combination with KCL, we may write

ix = v2/10 + v2/10 = iy + iy = 5 A

Thus, iy = 2.5 A.

(b) From part (a), ix = 2 v1/ 10. Substituting the new value for v1, we find that

ix = 6/10 = 600 mA.

Since we have found that iy = 0.5 ix, iy = 300 mA.

(c) no value – this is impossible.

CHAPTER THREE SOLUTIONS

10. We begin by making use of the information given regarding the power generated by

the 5-A and the 40-V sources. The 5-A source supplies 100 W, so it must therefore

have a terminal voltage of 20 V. The 40-V source supplies 500 W, so it must therefore

provide a current of 12.5 A. These quantities are marked on our schematic below:

(1) By KVL, -40 – 110 + R(5) = 0

Thus, R = 30 Ω.

(2) By KVL, -VG – (-110) + 40 = 0

So VG = 150 V

Now that we know the voltage across the unknown conductance G, we need only to

find the current flowing through it to find its value by making use of Ohm’s law.

KCL provides us with the means to find this current: The current flowing into the “+”

terminal of the –110-V source is 12.5 + 6 = 18.5 A.

Then, Ix = 18.5 – 5 = 13.5 A

By Ohm’s law, Ix = G · VG

So G = 13.5/ 150 or G = 90 mS

CHAPTER THREE SOLUTIONS

11. (a) -1 + 2 + 10i – 3.5 + 10i = 0

Solving, i = 125 mA

(b) +10 + 1i - 2 + 2i + 2 – 6 + i = 0

Solving, we find that 4i = -4 or i = - 1 A.

CHAPTER THREE SOLUTIONS

12. (a) By KVL, -2 + vx + 8 = 0

so that vx = -6 V.

(b) By KCL at the top right node,

IS + 4 vx = 4 - vx/4

So IS = 29.5 A.

(c) By KCL at the top left node,

iin = 1 + IS + vx/4 – 6

or iin = 23 A

(d) The power provided by the dependent source is 8(4vx) = -192 W.

CHAPTER THREE SOLUTIONS

13. (a) Working from left to right,

v1 = 60 V

v2 = 60 V

i2 = 60/20 = 3 A

i4 = v1/4 = 60/4 = 15 A

v3 = 5i2 = 15 V

By KVL, -60 + v3 + v5 = 0

v5 = 60 – 15 = 45 V

v4 = v5 = 45

i5 = v5/5 = 45/5 = 9 A

i3 = i4 + i5 = 15 + 9 = 24 A

i1 = i2 + i3 = 3 + 24 = 27

(b) It is now a simple matter to compute the power absorbed by each element:

p1 = -v1i1 = -(60)(27) = -1.62 kW

p2 = v2i2 = (60)(3) = 180 W

p3 = v3i3 = (15)(24) = 360 W

p4 = v4i4 = (45)(15) = 675 W

p5 = v5i5 = (45)(9) = 405 W

and it is a simple matter to check that these values indeed sum to zero as they should.

v1 = 60 V i1 = 27 A

v2 = 60 V i2 = 3 A

v3 = 15 V i3 = 24 A

v4 = 45 V i4 = 15 A

v5 = 45 V i5 = 9 A

CHAPTER THREE SOLUTIONS

14. Refer to the labeled diagram below.

Beginning from the left, we find

p20V = -(20)(4) = -80 W

v1.5 = 4(1.5) = 6 V therefore p1.5 = (v1.5)2/ 1.5 = 24 W.

v14 = 20 – v1.5 = 20 – 6 = 14 V therefore p14 = 142/ 14 = 14 W.

i2 = v2/2 = v1.5/1.5 – v14/14 = 6/1.5 – 14/14 = 3 A

Therefore v2 = 2(3) = 6 V and p2 = 62/2 = 18 W.

v4 = v14 – v2 = 14 – 6 = 8 V therefore p4 = 82/4 = 16 W

i2.5 = v2.5/ 2.5 = v2/2 – v4/4 = 3 – 2 = 1 A

Therefore v2.5 = (2.5)(1) = 2.5 V and so p2.5 = (2.5)2/2.5 = 2.5 W.

I2.5 = - IS, thefore IS = -1 A.

KVL allows us to write -v4 + v2.5 + vIS = 0

so VIS = v4 – v2.5 = 8 – 2.5 = 5.5 V and pIS = -VIS IS = 5.5 W.

A quick check assures us that these power quantities sum to zero.

CHAPTER THREE SOLUTIONS

15. Sketching the circuit as described,

(a) v14 = 0. v13 = v43 = 8 V

v23 = -v12 – v34 = -12 + 8 = -4 V

v24 = v23 + v34 = -4 – 8 = -12 V

(b) v14 = 6 V. v13 = v14 + v43 = 6 + 8 = 14 V

v23 = v13 – v12 = 14 – 12 = 2 V

v24 = v23 + v34 = 2 – 8 = -6 V

(c) v14 = -6 V. v13 = v14 + v43 = -6 + 8 = 2 V

v23 = v13 – v12 = 2 – 12 = -10 V

v24 = v23 + v34 = -10 – 8 = -18 V

CHAPTER THREE SOLUTIONS

16. (a) By KVL, -12 + 5000ID + VDS + 2000ID = 0

Therefore, VDS = 12 – 7(1.5) = 1.5 V.

(b) By KVL, - VG + VGS + 2000ID = 0

Therefore, VGS = VG – 2(2) = -1 V.

CHAPTER THREE SOLUTIONS

17. Applying KVL around this series circuit,

-120 + 30ix + 40ix + 20ix + vx + 20 + 10ix = 0

where vx is defined across the unknown element X, with the “+” reference on top.

Simplifying, we find that 100ix + vx = 100

To solve further we require specific information about the element X and its

properties.

(a) if X is a 100-Ω resistor,

vx = 100ix so we find that 100 ix + 100 ix = 100.

Thus ix = 500 mA and px = vx ix = 25 W.

(b) If X is a 40-V independent voltage source such that vx = 40 V, we find that

ix = (100 – 40) / 100 = 600 mA and px = vx ix = 24 W

(c) If X is a dependent voltage source such that vx = 25ix,

ix = 100/125 = 800 mA and px = vx ix = 16 W.

(d) If X is a dependent voltage source so that vx = 0.8v1,

where v1 = 40ix, we have 100 ix + 0.8(40ix) = 100

or ix = 100/132 = 757.6 mA and px = vx ix = 0.8(40)(0.7576)2 = 18.37 W.

(e) If X is a 2-A independent current source, arrow up,

100(-2) + vx = 100

so that vx = 100 + 200 = 300 V and px = vx ix = -600 W

CHAPTER THREE SOLUTIONS

18. (a) We first apply KVL:

-20 + 10i1 + 90 + 40i1 + 2v2 = 0

where v2 = 10i1. Substituting,

70 + 70 i1 = 0

or i1= -1 A.

(b) Applying KVL, -20 + 10i1 + 90 + 40i1 + 1.5v3 = 0 [1]

where v3 = -90 – 10i1 + 20 = -70 – 10 i1

alternatively, we could write v3 = 40i1 + 1.5v3 = -80i1

Using either expression in Eq. [1], we find i1 = 1 A.

(c) Applying KVL, -20 + 10i1 + 90 + 40i1 - 15 i1 = 0

Solving, i1 = - 2A.

CHAPTER THREE SOLUTIONS

19. Applying KVL, we find that

-20 + 10i1 + 90 + 40i1 + 1.8v3 = 0 [1]

Also, KVL allows us to write v3 = 40i1 + 1.8v3

v3 = -50i1

So that we may write Eq. [1] as

50i1 – 1.8(50)i1 = -70

or i1 = -70/-40 = 1.75 A.

Since v3 = -50i1 = -87.5 V, no further information is required to determine its value.

The 90-V source is absorbing (90)(i1) = 157.5 W of power and the dependent source

is absorbing (1.8v3)(i1) = -275.6 W of power.

Therefore, none of the conditions specified in (a) to (d) can be met by this circuit.

CHAPTER THREE SOLUTIONS

20. (a) Define the charging current i as flowing clockwise in the circuit provided.

By application of KVL,

-13 + 0.02i + Ri + 0.035i + 10.5 = 0

We know that we need a current i = 4 A, so we may calculate the necessary resistance

R = [13 – 10.5 – 0.055(4)]/ 4 = 570 mΩ

(b) The total power delivered to the battery consists of the power absorbed by the

0.035-Ω resistance (0.035i2), and the power absorbed by the 10.5-V ideal battery

(10.5i). Thus, we need to solve the quadratic equation

0.035i2 + 10.5i = 25

which has the solutions i = -302.4 A and i = 2.362 A.

In order to determine which of these two values should be used, we must recall that

the idea is to charge the battery, implying that it is absorbing power, or that i as

defined is positive. Thus, we choose i = 2.362 A, and, making use of the expression

developed in part (a), we find that

R = [13 – 10.5 – 0.055(2.362)]/ 2.362 = 1.003 Ω

(c) To obtain a voltage of 11 V across the battery, we apply KVL:

0.035i + 10.5 = 11 so that i = 14.29 A

From part (a), this means we need

R = [13 – 10.5 – 0.055(14.29)]/ 14.29 = 119.9 mW

CHAPTER THREE SOLUTIONS

21. Drawing the circuit described, we also define a clockwise current i.

By KVL, we find that

-13 + (0.02 + 0.5 + 0.035)i + 10.5 – 0.05i = 0

or that i = (13 – 10.5)/0.505 = 4.950 A

and Vbattery = 13 – (0.02 + 0.5)i = 10.43 V.

CHAPTER THREE SOLUTIONS

22. Applying KVL about this simple loop circuit (the dependent sources are still linear

elements, by the way, as they depend only upon a sum of voltages)

-40 + (5 + 25 + 20)i – (2v3 + v2) + (4v1 – v2) = 0 [1]

where we have defined i to be flowing in the clockwise direction, and

v1 = 5i, v2 = 25i, and v3 = 20i.

Performing the necessary substition, Eq. [1] becomes

50i - (40i + 25i) + (20i – 25i) = 40

so that i = 40/-20 = -2 A

Computing the absorbed power is now a straightforward matter:

p40V = (40)(-i) = 80 W

p5W = 5i2 = 20 W

p25W = 25i2 = 100 W

p20W = 20i2 = 80 W

pdepsrc1 = (2v3 + v2)(-i) = (40i + 25i) = -260 W

pdepsrc2 = (4v1 - v2)(-i) = (20i - 25i) = -20 W

and we can easily verify that these quantities indeed sum to zero as expected.

CHAPTER THREE SOLUTIONS

23. We begin by defining a clockwise current i.

(a) i = 12/(40 + R) mA, with R expressed in kΩ.

We want i2 · 25 = 2

or 2 25

40

12 2

=⋅













+R

Rearranging, we find a quadratic expression involving R:

R2 + 80R – 200 = 0

which has the solutions R = -82.43 kΩ and R = 2.426 kΩ. Only the latter is a

physical solution, so

R = 2.426 kΩ.

(b) We require i · 12 = 3.6 or i = 0.3 mA

From the circuit, we also see that i = 12/(15 + R + 25) mA.

Substituting the desired value for i, we find that the required value of R is R = 0.

(c)

CHAPTER THREE SOLUTIONS

24. By KVL, –12 + (1 + 2.3 + Rwire segment) i = 0

The wire segment is a 3000–ft section of 28–AWG solid copper wire. Using Table

2.3, we compute its resistance as

(16.2 mΩ/ft)(3000 ft) = 48.6 Ω

which is certainly not negligible compared to the other resistances in the circuit!

Thus, i = 12/(1 + 2.3 + 48.6) = 231.2 mA

CHAPTER THREE SOLUTIONS

25. We can apply Ohm’s law to find an expression for vo:

vo = 1000(–gm vπ)

We do not have a value for vπ, but KVL will allow us to express that in terms of vo,

which we do know:

–10×10–3 cos 5t + (300 + 50×103) i = 0

where i is defined as flowing clockwise.

Thus, vπ = 50×103 i = 50×103 (10×10–3 cos 5t) / (300 + 50×103)

= 9.940×10–3 cos 5t V

and we by substitution we find that

vo = 1000(–25×10–3)( 9.940×10–3 cos 5t )

= –248.5 cos 5t mV

CHAPTER THREE SOLUTIONS

26. By KVL, we find that

–3 + 100 ID + VD = 0

Substituting ID = 3×10–6(eVD

/ 27×10–3 – 1), we find that

–3 + 300×10–6(eVD

/ 27×10–3 – 1) + VD = 0

This is a transcendental equation. Using a scientific calculator or a numerical

software package such as MATLAB®, we find

VD = 246.4 mV

Let’s assume digital assistance is unavailable. In that case, we need to “guess” a value

for VD, substitute it into the right hand side of our equation, and see how close the

result is to the left hand side (in this case, zero).

GUESS RESULT

0 –3

1 3.648×1012

0.5 3.308×104

0.25 0.4001

0.245 –0.1375

0.248 0.1732

0.246 –0.0377

oops

better

At this point, the error is

getting much smaller, and

our confidence is increasing

as to the value of VD.

CHAPTER THREE SOLUTIONS

27. Define a voltage vx, “+” reference on the right, across the dependent current source.

Note that in fact vx appears across each of the four elements. We first convert the 10

mS conductance into a 100–Ω resistor, and the 40–mS conductance into a 25–Ω

resistor.

(a) Applying KCL, we sum the currents flowing into the right–hand node:

5 – vx / 100 – vx / 25 + 0.8 ix = 0 [1]

This represents one equation in two unknowns. A second equation to introduce at this

point is

ix = vx /25 so that Eq. [1] becomes

5 – vx / 100 – vx / 25 + 0.8 (vx / 25) = 0

Solving for vx, we find vx = 277.8 V. It is a simple matter now to compute the power

absorbed by each element:

P5A = –5 vx = –1.389 kW

P100Ω = (vx)2 / 100 = 771.7 W

P25Ω = (vx)2 / 25 = 3.087 kW

Pdep = –vx(0.8 ix) = –0.8 (vx)2 / 25 = –2.470 kW

A quick check assures us that the calculated values sum to zero, as they should.

(b) Again summing the currents into the right–hand node,

5 – vx / 100 – vx / 25 + 0.8 iy = 0 [2]

where iy = 5 – vx/100

Thus, Eq. [2] becomes

5 – vx / 100 – vx / 25 + 0.8(5) – 0.8 (iy) / 100 = 0

Solving, we find that vx x = 155.2 V and iy = 3.448 A

So that

P5A = –5 vx = –776.0 W

P100Ω = (vx)2 / 100 = 240.9 W

P25Ω = (vx)2 / 25 = 963.5 W

Pdep = –vx(0.8 iy) = –428.1 W

A quick check assures us that the calculated values sum to 0.3, which is reasonably

close to zero (small roundoff errors accumulate here).

CHAPTER THREE SOLUTIONS

28. Define a voltage v with the “+” reference at the top node. Applying KCL and

summing the currents flowing out of the top node,

v/5,000 + 4×10–3 + 3i1 + v/20,000 = 0 [1]

This, unfortunately, is one equation in two unknowns, necessitating the search for a

second suitable equation. Returning to the circuit diagram, we observe that

i1 = 3 i1 + v/2,000

or i1 = –v/40,000 [2]

Upon substituting Eq. [2] into Eq. [1], Eq. [1] becomes,

v/5,000 + 4×10–3 – 3v/40,000 + v/20,000 = 0

Solving, we find that

v = –22.86 V

and

i1 = 571.4 µA

Since ix = i1, we find that ix = 571.4 µA.

CHAPTER THREE SOLUTIONS

29. Define a voltage vx with its “+” reference at the center node. Applying KCL and

summing the currents into the center node,

8 – vx /6 + 7 – vx /12 – vx /4 = 0

Solving, vx = 30 V.

It is now a straightforward matter to compute the power absorbed by each element:

P8A = –8 vx = –240 W

P6Ω = (vx)2 / 6 = 150 W

P8A = –7 vx = –210 W

P12Ω = (vx)2 / 12 = 75 W

P4Ω = (vx)2 / 4 = 225 W

and a quick check verifies that the computed quantities sum to zero, as expected.

CHAPTER THREE SOLUTIONS

30. (a) Define a voltage v across the 1–kΩ resistor with the “+” reference at the top node.

Applying KCL at this top node, we find that

80×10–3 – 30×10–3 = v/1000 + v/4000

Solving, v = (50×10–3)(4×106 / 5×103) = 40 V

and P4kΩ = v2/4000 = 400 mW

(b) Once again, we first define a voltage v across the 1–kΩ resistor with the “+”

reference at the top node. Applying KCL at this top node, we find that

80×10–3 – 30×10–3 – 20×10–3 = v/1000

Solving, v = 30 V

and P20mA = v · 20×10–3 = 600 mW

(c) Once again, we first define a voltage v across the 1–kΩ resistor with the “+”

reference at the top node. Applying KCL at this top node, we find that

80×10–3 – 30×10–3 – 2ix = v/1000

where ix = v/1000

so that 80×10–3 – 30×10–3 = 2v/1000 + v/1000

and v = 50×10–3 (1000)/3 = 16.67 V

Thus, Pdep = v · 2ix = 555.8 mW

(d) We note that ix = 60/1000 = 60 mA. KCL stipulates that (viewing currents

into and out of the top node)

80 – 30 + is = ix = 60

Thus, is = 10 mA

and P60V = 60(–10) mW = –600 mW

CHAPTER THREE SOLUTIONS

31. (a) To cancel out the effects of both the 80-mA and 30-mA sources, iS must be set to

iS = –50 mA.

(b) Define a current is flowing out of the “+” reference terminal of the independent

voltage source. Interpret “no power” to mean “zero power.”

Summing the currents flowing into the top node and invoking KCL, we find that

80×10-3 - 30×10-3 - vS/1×103 + iS = 0

Simplifying slightly, this becomes

50 - vS + 103 iS = 0 [1]

We are seeking a value for vS such that vS · iS = 0. Clearly, setting vS = 0 will achieve

this. From Eq. [1], we also see that setting vS = 50 V will work as well.

CHAPTER THREE SOLUTIONS

32. Define a voltage v9 across the 9-Ω resistor, with the “+” reference at the top node.

(a) Summing the currents into the right-hand node and applying KCL,

5 + 7 = v9 / 3 + v9 / 9

Solving, we find that v9 = 27 V. Since ix = v9 / 9, ix = 3 A.

(b) Again, we apply KCL, this time to the top left node:

2 – v8 / 8 + 2ix – 5 = 0

Since we know from part (a) that ix = 3 A, we may calculate v8 = 24 V.

(c) p5A = (v9 – v8) · 5 = 15 W.

CHAPTER THREE SOLUTIONS

33. Define a voltage vx across the 5-A source, with the “+” reference on top.

Applying KCL at the top node then yields

5 + 5v1 - vx/ (1 + 2) – vx/ 5 = 0 [1]

where v1 = 2[vx /(1 + 2)] = 2 vx / 3.

Thus, Eq. [1] becomes 5 + 5(2 vx / 3) – vx / 3 – vx / 5 = 0

or 75 + 50 vx – 5 vx – 3 vx = 0, which, upon solving, yields vx = -1.786 V.

The power absorbed by the 5-Ω resistor is then simply (vx)2/5 = 638.0 mW.

CHAPTER THREE SOLUTIONS

34. Despite the way it may appear at first glance, this is actually a simple node-pair

circuit. Define a voltage v across the elements, with the “+” reference at the top node.

Summing the currents leaving the top node and applying KCL, we find that

2 + 6 + 3 + v/5 + v/5 + v/5 = 0

or v = -55/3 = -18.33 V. The power supplied by each source is then computed as:

p2A = -v(2) = 36.67 W

p6A = -v(6) = 110 W

p3A = -v(3) = 55 W

We can check our results by first determining the power absorbed by each resistor,

which is simply v2/5 = 67.22 W for a total of 201.67 W, which is the total power

supplied by all sources.

CHAPTER THREE SOLUTIONS

35. Defining a voltage Vx across the 10-A source with the “+” reference at the top node,

KCL tells us that 10 = 5 + I1Ω, where I1Ω is defined flowing downward through the

1-Ω resistor.

Solving, we find that I1Ω = 5 A, so that Vx = (1)(5) = 5 V.

So, we need to solve Vx = 5 = 5(0.5 + Rsegment)

with Rsegment = 500 mΩ.

From Table 2.3, we see that 28-AWG solid copper wire has a resistance of 65.3

mΩ/ft. Thus, the total number of miles needed of the wire is

miles 101.450

ft/mi) /ft)(5280m (65.3

m 500 3-

×=

Ω

CHAPTER THREE SOLUTIONS

36. Since v = 6 V, we know the current through the 1-Ω resistor is 6 A, the current

through the 2-Ω resistor is 3 A, and the current through the 5-Ω resistor is 6/5

= 1.2 A, as shown below:

By KCL, 6 + 3 + 1.2 + iS = 0 or iS = -10.2 A.

CHAPTER THREE SOLUTIONS

37. (a) Applying KCL, 1 – i – 3 + 3 = 0 so i = 1 A.

(b) The rightmost source should be labeled 3.5 A to satisfy KCL.

Then, looking at the left part of the circuit, we see 1 + 3 = 4 A flowing into the

unknown current source, which, by virtue of KCL, must therefore be a 4-A current

source. Thus, KCL at the node labeled with the “+” reference of the voltage v gives

4 – 2 + 7 – i = 0 or i = 9 A

CHAPTER THREE SOLUTIONS

38. (a) We may redraw the circuit as

Then, we see that v = (1)(1) = 1 V.

(b) The current source at the far right should be labeled 3.5 A, or KCL is violated.

In that case, we may combine all sources to the right of the 1-Ω resistor into a single

7-A current source. On the left, the two 1-A sources in sereies reduce to a single 1-A

source.

The new 1-A source and the 3-A source combine to yield a 4-A source in series with

the unknown current source which, by KCL, must be a 4-A current source.

At this point we have reduced the circuit to

Further simplification is possible, resulting in

From which we see clearly that v = (9)(1) = 9 V.

CHAPTER THREE SOLUTIONS

39. (a) Combine the 12-V and 2-V series connected sources to obtain a new 12 – 2 = 10 V

source, with the “+” reference terminal at the top. The result is two 10-V sources in

parallel, which is permitted by KVL. Therefore,

i = 10/1000 = 10 mA.

(b) No current flows through the 6-V source, so we may neglect it for this calculation.

The 12-V, 10-V and 3-V sources are connected in series as a result, so we replace

them with a 12 + 10 –3 = 19 V source as shown

Thus, i = 19/5 = 3.8 A.

CHAPTER THREE SOLUTIONS

40. We first combine the 10-V and 5-V sources into a single 15-V source, with the “+”

reference on top. The 2-A and 7-A current sources combine into a 7 – 2 = 5 A current

source (arrow pointing down); although these two current sources may not appear to

be in parallel at first glance, they actually are.

Redrawing our circuit,

we see that v = 15 V (note that we can completely the ignore the 5-A source here,

since we have a voltage source directly across the resistor). Thus,

p16Ω = v2/16 = 14.06 W.

CHAPTER THREE SOLUTIONS

41. We can combine the voltage sources such that

i = vS/ 14

(a) vS = 10 + 10 – 6 – 6 = 20 –12 = 8

Therefore i = 8/14 = 571.4 mA.

(b) vS = 3 + 2.5 – 3 – 2.5 = 0 Therefore i = 0.

(c) vS = -3 + 1.5 – (-0.5) – 0 = -1 V

Therefore i = -1/14 = -71.43 mA.

CHAPTER THREE SOLUTIONS

42. We first simplify as shown, making use of the fact that we are told ix = 2 A to find the

voltage across the middle and right-most 1-Ω resistors as labeled.

By KVL, then, we find that v1 = 2 + 3 = 5 V.

CHAPTER THREE SOLUTIONS

43. We see that to determine the voltage v we will need vx due to the presence of the

dependent current soruce. So, let’s begin with the right-hand side, where we find that

vx = 1000(1 – 3) × 10-3 = -2 V.

Returning to the left-hand side of the circuit, and summing currents into the top node,

we find that

(12 – 3.5) ×10-3 + 0.03 vx = v/10×103

or v = -515 V.

CHAPTER THREE SOLUTIONS

44. (a) We first label the circuit with a focus on determining the current flowing through

each voltage source:

Then the power absorbed by each voltage source is

P2V = -2(-5) = 10 W

P4V = -(-4)(4) = 16 W

P-3V = -(-9)(-3) = 27 W

For the current sources,

So that the absorbed power is

P

-5A = +(-5)(6) = -30 W

P-4A = -(-4)(4) = 16 W

P3A = -(3)(7) = -21 W

P12A = -(12)(-3) = 36 W

A quick check assures us that these absorbed powers sum to zero as they should.

(b) We need to change the 4-V source such that the voltage across the –5-A source

drops to zero. Define Vx across the –5-A source such that the “+” reference terminal is

on the left. Then, -2 + Vx – Vneeded = 0

or Vneeded = -2 V.

CHAPTER THREE SOLUTIONS

45. We begin by noting several things:

(1) The bottom resistor has been shorted out;

(2) the far-right resistor is only connected by one terminal and therefore does

not affect the equivalent resistance as seen from the indicated terminals;

(3) All resistors to the right of the top left resistor have been shorted.

Thus, from the indicated terminals, we only see the single 1-kΩ resistor, so that

Req = 1 kΩ.

CHAPTER THREE SOLUTIONS

46. (a) We see 1Ω || (1 Ω + 1 Ω) || (1 Ω + 1 Ω + 1 Ω)

= 1Ω || 2 Ω || 3 Ω

= 545.5 mΩ

(b) 1/Req = 1 + 1/2 + 1/3 + … 1/N

Thus, Req = [1 + 1/2 + 1/3 + … 1/N]-1

CHAPTER THREE SOLUTIONS

47. (a) 5 kΩ = 10 kΩ || 10 kΩ

(b) 57 333 Ω = 47 kΩ + 10 kΩ + 1 kΩ || 1kΩ || 1kΩ

(c) 29.5 kΩ = 47 kΩ || 47 kΩ + 10 kΩ || 10 kΩ + 1 kΩ

CHAPTER THREE SOLUTIONS

0 V 1 A

5 Ω 7 Ω

5 Ω

1 A 2.917 Ω

48. (a) no simplification is possible using only source and/or resistor combination

techniques.

(b) We first simplify the circuit to

and then notice that the 0-V source is shorting out one of the 5-Ω resistors, so a

further simplification is possible, noting that 5 Ω || 7 Ω = 2.917 Ω:

CHAPTER THREE SOLUTIONS

49. Req = 1 kΩ + 2 kΩ || 2 kΩ + 3 kΩ || 3 kΩ + 4 kΩ || 4 kΩ

= 1 kΩ + 1 k Ω + 1.5 kΩ + 2 kΩ

= 5.5 kΩ.

CHAPTER THREE SOLUTIONS

Req →

100 Ω

100 Ω 250 Ω

100 Ω

10 Ω

20 Ω

5 Ω

14.4 Ω

Req →

2 Ω

15 Ω 10 Ω

50 Ω

8 Ω 20 Ω 30 Ω

2 Ω 16.67 Ω

8 Ω 50 Ω

50. (a) Working from right to left, we first see that we may combine several resistors as

100 Ω + 100 Ω || 100 Ω + 100 Ω = 250 Ω, yielding the following circuit:

Next, we see 100 Ω + 100 Ω || 250 Ω + 100 Ω = 271.4 Ω,

and subsequently 100 Ω + 100 Ω || 271.4 Ω + 100 Ω = 273.1 Ω,

and, finally,

Req = 100 Ω || 273.1 Ω = 73.20 Ω.

(b) First, we combine 24 Ω || (50 Ω + 40 Ω) || 60 Ω = 14.4 Ω, which leaves us with

Thus, Req = 10 Ω + 20 Ω || (5 + 14.4 Ω) = 19.85 Ω.

(c) First combine the 10-Ω and 40-Ω resistors and redraw the circuit:

We now see we have (10 Ω + 15 Ω) || 50 Ω = 16.67 Ω. Redrawing once again,

where the equivalent resistance is seen to be 2 Ω + 50 Ω || 16.67 Ω + 8 Ω = 22.5 Ω.

CHAPTER THREE SOLUTIONS

51. (a) Req = [(40 Ω + 20 Ω) || 30 Ω + 80 Ω] || 100 Ω + 10 Ω = 60 Ω.

(b) Req = 80 Ω = [(40 Ω + 20 Ω) || 30 Ω + R] || 100 Ω + 10 Ω

70 Ω = [(60 Ω || 30 Ω) + R] || 100 Ω

1/70 = 1/(20 + R) + 0.01

20+ R = 233.3 Ω therefore R = 213.3 Ω.

(c) R = [(40 Ω + 20 Ω) || 30 Ω + R] || 100 Ω + 10 Ω

R – 10 Ω = [20 + R] || 100

1/(R – 10) = 1/(R + 20) + 1/ 100

3000 = R2 + 10R – 200

Solving, we find R = -61.79 Ω or R = 51.79 Ω.

Clearly, the first is not a physical solution, so R = 51.79 Ω.

CHAPTER THREE SOLUTIONS

52. (a) 25 Ω = 100 Ω || 100 Ω || 100 Ω

(b) 60 Ω = [(100 Ω || 100 Ω) + 100 Ω] || 100 Ω

(c) 40 Ω = (100 Ω + 100 Ω) || 100 Ω || 100 Ω

CHAPTER THREE SOLUTIONS

53. Req = [(5 Ω || 20 Ω) + 6 Ω] || 30 Ω + 2.5 Ω = 10 Ω

The source therefore provides a total of 1000 W and a current of 100/10 = 10 A.

P

2.5Ω = (10)2 · 2.5 = 250 W

V

30Ω = 100 - 2.5(10) = 75 V

P

30Ω = 752/ 30 = 187.5 W

I

6Ω = 10 – V30Ω /30 = 10 – 75/30 = 7.5 A

P

6Ω = (7.5)2 · 6 = 337.5 W

V

5Ω = 75 – 6(7.5) = 30 V

P

5Ω = 302/ 5 = 180 W

V

20Ω = V5Ω = 30 V

P

20Ω = 302/20 = 45 W

We check our results by verifying that the absorbed powers in fact add to 1000 W.

CHAPTER THREE SOLUTIONS

- vx +

9 A

14 Ω 6 Ω

4 Ω 6 Ω

54. To begin with, the 10-Ω and 15-Ω resistors are in parallel ( = 6 Ω), and so are the

20-Ω and 5-Ω resistors (= 4 Ω).

Also, the 4-A, 1-A and 6-A current sources are in parallel, so they can be combined

into a single 4 + 6 – 1 = 9 A current source as shown:

Next, we note that (14 Ω + 6 Ω) || (4 Ω + 6 Ω) = 6.667 Ω

so that vx = 9(6.667) = 60 V

and ix = -60/10 = -6 A.

CHAPTER THREE SOLUTIONS

13.64 mS

100 mS

22.22 mS

Gin →

55. (a) Working from right to left, and borrowing x || y notation from resistance

calculations to indicate the operation xy/(x + y),

Gin = {[(6 || 2 || 3) + 0.5] || 1.5 || 2.5 + 0.8} || 4 || 5 mS

= {[(1) + 0.5] || 1.5 || 2.5 + 0.8} || 4 || 5 mS

= {1.377} || 4 || 5

= 0.8502 mS = 850.2 mS

(b) The 50-mS and 40-mS conductances are in series, equivalent to (50(40)/90 =

22.22 mS. The 30-mS and 25-mS conductances are also in series, equivalent to 13.64

mS. Redrawing for clarity,

we see that Gin = 10 + 22.22 + 13.64 = 135.9 mS.

CHAPTER THREE SOLUTIONS

56. The bottom four resistors between the 2-Ω resistor and the 30-V source are shorted

out. The 10-Ω and 40-Ω resistors are in parallel (= 8 Ω), as are the 15-Ω and 60-Ω

(=12 Ω) resistors. These combinations are in series.

Define a clockwise current I through the 1-Ω resistor:

I = (150 – 30)/(2 + 8 + 12 + 3 + 1 + 2) = 4.286 A

P1Ω = I2 · 1 = 18.37 W

To compute P10Ω, consider that since the 10-Ω and 40-Ω resistors are in parallel, the

same voltage Vx (“+” reference on the left) appears across both resistors. The current I

= 4.286 A flows into this combination. Thus, Vx = (8)(4.286) = 34.29 V and

P10Ω = (Vx)2 / 10 = 117.6 W.

P13Ω = 0 since no current flows through that resistor.

CHAPTER THREE SOLUTIONS

57. One possible solution of many:

The basic concept is as shown

If we use 28-AWG soft copper wire, we see from Table 2.3 that 9-Ω would require

138 feet, which is somewhat impractical. Referring to p. 4-48 of the Standard

Handbook for Electrical Engineers (this should be available in most

engineering/science libraries), we see that 44-AWG soft copper wire has a resistance

of 2590 Ω per 1000 ft, or 0.08497 Ω/cm.

Thus, 1-Ω requires 11.8 cm of 44-AWG wire, and 9-Ω requires 105.9 cm. We decide

to make the wiper arm and leads out of 28-AWG wire, which will add a slight

resistance to the total value, but a negligible amount.

The radius of the wiper arm should be (105.9 cm)/

π

= 33.7 cm.

9-Ω wire

se

g

ment

1-Ω wire

se

g

ment

external contacts

CHAPTER THREE SOLUTIONS

58. One possible solution of many:

vS = 2(5.5) = 11 V

R

1 = R2 = 1 kΩ.

CHAPTER THREE SOLUTIONS

59. One possible solution of many:

iS = 11 mA

R

1 = R2 = 1 kΩ.

CHAPTER THREE SOLUTIONS

60. p15Ω = (v15)2 / 15×103 A

v15 = 15×103 (-0.3 v1)

where v1 = [4 (5)/ (5 + 2)] · 2 = 5.714 V

Therefore v15 = -25714 V and p15 = 44.08 kW.

CHAPTER THREE SOLUTIONS

61. Replace the top 10 kΩ, 4 kΩ and 47 kΩ resistors with 10 kΩ + 4 kΩ || 47 kΩ =

13.69 kΩ.

Define vx across the 10 kΩ resistor with its “+” reference at the top node: then

vx = 5 · (10 kΩ || 13.69 kΩ) / (15 kΩ + 10 || 13.69 kΩ) = 1.391 V

ix = vx/ 10 mA = 139.1 µA

v15 = 5 – 1.391 = 3.609 V and p15 = (v15)2/ 15×103 = 868.3 µW.

CHAPTER THREE SOLUTIONS

62. We may combine the 12-A and 5-A current sources into a single 7-A current source

with its arrow oriented upwards. The left three resistors may be replaced by a 3 +

6 || 13 = 7.105 Ω resistor, and the right three resistors may be replaced by a 7 + 20 || 4

= 10.33 Ω resistor.

By current division, iy = 7 (7.105)/(7.105 + 10.33) = 2.853 A

We must now return to the original circuit. The current into the 6 Ω, 13 Ω parallel

combination is 7 – iy = 4.147 A. By current division,

ix = 4.147 . 13/ (13 + 6) = 2.837 A

and px = (4.147)2 · 3 = 51.59 W

CHAPTER THREE SOLUTIONS

63. The controlling voltage v1, needed to obtain the power into the 47-kΩ resistor, can be

found separately as that network does not depend on the left-hand network.

The right-most 2 kΩ resistor can be neglected.

By current division, then, in combination with Ohm’s law,

v1 = 3000[5×10-3 (2000)/ (2000 + 3000 + 7000)] = 2.5 V

Voltage division gives the voltage across the 47-kΩ resistor:

V 0.9228

16.67 47

7)0.5(2.5)(4

20 ||100 47

47

5.0 1=

+

=

+

v

So that p47kΩ = (0.9928)2 / 47×103 = 18.12 µW

CHAPTER THREE SOLUTIONS

64. The temptation to write an equation such as

v1 = 2020

20

10 +

must be fought!

Voltage division only applies to resistors connected in series, meaning that the same

current must flow through each resistor. In this circuit, i1 ≠ 0 , so we do not have the

same current flowing through both 20 kΩ resistors.

CHAPTER THREE SOLUTIONS

65. (a) )]R (R || [R R

)R (R || R

V

4321

432

S2 ++

+

=v

=

()

4324321

432432

SR R R)R (R R R

R R R)R (R R

V ++++

+++

=

()

)R (RR R R RR

)R (R R

V

4324321

432

S++++

+

(b) )]R (R || [R R

R

V

4321

1

S1 ++

=v

=

()

4324321

1

SR R R)R (R R R

R

V ++++

=

()

)R (RR R R RR

)R R (R R

V

4324321

4321

S++++

++

(c) 













++













=

432

2

1

4R R R

R

v

i

=

()

[]

)R R )(RR (RR )R R (RRR

R R R R R

V

43243243211

24321

S++++++

++

=

()

)R (RR R R RR

R

V

4324321

2

S++++

CHAPTER THREE SOLUTIONS

66. (a) With the current source open-circuited, we find that

v1 = V 8-

6000 || 3000 500

500

40 =

+

−

(b) With the voltage source short-circuited, we find that

i2 =

()

mA 400

1/6000 1/3000 500/1

1/3000

103 3=

++

×−

i3 =

()

mA 600

6000 || 3000 500

500

103 3=

+

×−

CHAPTER THREE SOLUTIONS

67. (a) The current through the 5-Ω resistor is 10/5 = 2 A. Define R as 3 || (4 + 5)

= 2.25 Ω. The current through the 2-Ω resistor then is given by

25.5

I

R) (2 1

1

I S

S=

++

The current through the 5-Ω resistor is

A 2

9 3

3

25.5

I

S=













+

so that IS = 42 A.

(b) Given that IS is now 50 A, the current through the 5-Ω resistor becomes

A 2.381

9 3

3

25.5

I

S=













+

Thus, vx = 5(2.381) = 11.90 V

(c) 0.2381

I

9 3

3

5.25

5I

I

S

=























+

=

x

v

CHAPTER THREE SOLUTIONS

68. First combine the 1 kΩ and 3 kΩ resistors to obtain 750 Ω.

By current division, the current through resistor Rx is

IRx = 750 R 2000

2000

1010

x

3

++

×−

and we know that Rx · I

Rx = 9

so

x

R 2750

R 20

9 +

=

9 Rx + 24750 = 20 Rx or Rx = 2250 W. Thus,

PRx = 92/ Rx = 36 mW.

CHAPTER THREE SOLUTIONS

69. Define R = R3 || (R4 + R5)

Then 













+

=

2

SR R R

R

V v

=

()













++++

+++

2543543

543543

SR R R R)R (R R

R R R)R (R R

V

=













++++

+

)R (RR )R R(R R

)R (R R

V

54354 32

543

S

Thus,













+

=

54

5

R5 R R

R

vv

= 













++++ )R (RR )R R(R R

R R

V

54354 32

53

S

CHAPTER THREE SOLUTIONS

70. Define R1 = 10 + 15 || 30 = 20 Ω and R2 = 5 + 25 = 30 Ω.

(a) Ix = I1 . 15 / (15 + 30) = 4 mA

(b) I1 = Ix . 45/15 = 36 mA

(c) I2 = IS R1 / (R1 + R2) and I1 = IS R2 / (R1 + R2)

So I1/I2 = R2/R1

Therefore I1 = R2I2/R1 = 30(15)/20 = 22.5 mA

Thus, Ix = I1 . 15/ 45 = 7.5 mA

(d) I1 = IS R2/ (R1 + R2) = 60 (30)/ 50 = 36 A

Thus, Ix = I1 15/ 45 = 12 A.

CHAPTER THREE SOLUTIONS

71. vout = -gm vπ (100 kΩ || 100 kΩ) = -4.762×103 gm vπ

where vπ = (3 sin 10t) · 15/(15 + 0.3) = 2.941 sin 10t

Thus, vout = -56.02 sin 10t V

CHAPTER THREE SOLUTIONS

72. vout = -1000gm vπ

where vπ = V 10sin 2.679

0.3 3) || (15

3 || 15

10sin 3 tt =

+

therefore

vout = -(2.679)(1000)(38×10-3) sin 10t = -101.8 sin 10t V.

CHAPTER FOUR SOLUTIONS

1. (a) 0.1 -0.3 -0.4 v1 0

-0.5 0.1 0 v2 = 4

-0.2 -0.3 0.4 v3 6

Solving this matrix equation using a scientific calculator, v2 = -8.387 V

(b) Using a scientific calculator, the determinant is equal to 32.

CHAPTER FOUR SOLUTIONS

2. (a) 1 1 1 vA 27

-1 2 3 vB = -16

2 0 4 vC -6

Solving this matrix equation using a scientific calculator,

vA = 19.57

vB = 18.71

vC = -11.29

(b) Using a scientific calculator,

1 1 1

-1 2 3 = 16

2 0 4

CHAPTER FOUR SOLUTIONS

3. The bottom node has the largest number of branch connections, so we choose that as

our reference node. This also makes vP easier to find, as it will be a nodal voltage.

Working from left to right, we name our nodes 1, P, 2, and 3.

NODE 1: 10 = v1/ 20 + (v1 – vP)/ 40 [1]

NODE P: 0 = (vP – v1)/ 40 + vP/ 100 + (vP – v2)/ 50 [2]

NODE 2: -2.5 + 2 = (v2 – vP)/ 50 + (v2 – v3)/ 10 [3]

NODE 3: 5 – 2 = v3/ 200 + (v3 – v2)/ 10 [4]

Simplifying,

60v1 - 20vP = 8000 [1]

-50v1 + 110 vP - 40v2 = 0 [2]

- vP + 6v2 - 5v3 = -25 [3]

-200v2 + 210v3 = 6000 [4]

Solving,

vP = 171.6 V

CHAPTER FOUR SOLUTIONS

4. The logical choice for a reference node is the bottom node, as then vx will

automatically become a nodal voltage.

NODE 1: 4 = v1/ 100 + (v1 – v2)/ 20 + (v1 – vx)/ 50 [1]

NODE x: 10 – 4 – (-2) = (vx – v1)/ 50 + (vx – v2)/ 40 [2]

NODE 2: -2 = v2 / 25 + (v2 – vx)/ 40 + (v2 – v1)/ 20 [3]

Simplifying,

4 = 0.0800v1 – 0.0500v2 – 0.0200vx [1]

8 = -0.0200v1 – 0.02500v2 + 0.04500vx [2]

-2 = -0.0500v1 + 0.1150v2 – 0.02500vx [3]

Solving, vx = 397.4 V.

CHAPTER FOUR SOLUTIONS

5. Designate the node between the 3-Ω and 6-Ω resistors as node X, and the right-hand

node of the 6-Ω resistor as node Y. The bottom node is chosen as the reference node.

(a) Writing the two nodal equations, then

NODE X: –10 = (vX – 240)/ 3 + (vX – vY)/ 6 [1]

NODE Y: 0 = (vY – vX)/ 6 + vY/ 30 + (vY – 60)/ 12 [2]

Simplifying, -180 + 1440 = 9 vX – 3 vY [1]

10800 = - 360 vX + 612 vY [2]

Solving, vX = 181.5 V and vY = 124.4 V

Thus, v1 = 240 – vX = 58.50 V and v2 = vY – 60 = 64.40 V

(b) The power absorbed by the 6-Ω resistor is

(vX – vY)2 / 6 = 543.4 W

CHAPTER FOUR SOLUTIONS

6. Only one nodal equation is required: At the node where three resistors join,

0.02v1 = (vx – 5 i2) / 45 + (vx – 100) / 30 + (vx – 0.2 v3) / 50 [1]

This, however, is one equation in four unknowns, the other three resulting from the

presence of the dependent sources. Thus, we require three additional equations:

i2 = (0.2 v3 - vx) / 50 [2]

v1 = 0.2 v3 - 100 [3]

v3 = 50i2 [4]

Simplifying,

v1 – 0.2v3 = -100 [3]

– v3 + 50 i2 = 0 [4]

–vx + 0.2v3 – 50 i2 = 0 [2]

0.07556vx – 0.02v1 – 0.004v3 – 0.111i2 = 33.33 [1]

Solving, we find that v1 = -103..8 V and i2 = -377.4 mA.

CHAPTER FOUR SOLUTIONS

7. If v1 = 0, the dependent source is a short circuit and we may redraw the circuit as:

At NODE 1: 4 - 6 = v1/ 40 + (v1 – 96)/ 20 + (v1 – V2)/ 10

Since v1 = 0, this simplifies to

-2 = -96 / 20 - V2/ 10

so that V2 = -28 V.

20 Ω 10 Ω

40 Ω

+

v1 = 0

-

.

CHAPTER FOUR SOLUTIONS

8. We choose the bottom node as ground to make calculation of i5 easier. The left-most

node is named “1”, the top node is named “2”, the central node is named “3” and the

node between the 4-Ω and 6-Ω resistors is named “4.”

NODE 1: - 3 = v1/2 + (v1 – v2)/ 1 [1]

NODE 2: 2 = (v2 – v1)/ 1 + (v2 – v3)/ 3 + (v2 – v4)/ 4 [2]

NODE 3: 3 = v3/ 5 + (v3 – v4)/ 7 + (v3 – v2)/ 3 [3]

NODE 4: 0 = v4/ 6 + (v4 – v3)/ 7 + (v4 – v2)/ 4 [4]

Rearranging and grouping terms,

3v1 – 2v2 = -6 [1]

-12v1 + 19v2 – 4v3 – 3v4 = 24 [2]

–35v2 + 71v3 – 15v4 = 315 [3]

-42v2 – 24v3 + 94v4 = 0 [4]

Solving, we find that v3 = 6.760 V and so i5 = v3/ 5 = 1.352 A.

CHAPTER FOUR SOLUTIONS

9. We can redraw this circuit and eliminate the 2.2-kΩ resistor as no current flows

through it:

At NODE 2: 7×10-3 – 5×10-3 = (v2 + 9)/ 470 + (v2 – vx)/ 10×10-3 [1]

At NODE x: 5×10-3 – 0.2v1 = (vx – v2)/ 10×103 [2]

The additional equation required by the presence of the dependent source and the fact

that its controlling variable is not one of the nodal voltages:

v1 = v2 – vx [3]

Eliminating the variable v1 and grouping terms, we obtain:

10,470 v2 – 470 vx = –89,518

and 1999 v2 – 1999 vx = 50

Solving, we find vx = –8.086 V.

↓

9 V 7 mA

5 mA

0.2 v1

10 kΩ

470 Ω + v1 - vx

v2

CHAPTER FOUR SOLUTIONS

10. We need concern ourselves with the bottom part of this circuit only. Writing a single

nodal equation, -4 + 2 = v/ 50

We find that v = -100 V.

CHAPTER FOUR SOLUTIONS

11. We choose the center node for our common terminal, since it connects to the largest

number of branches. We name the left node “A”, the top node “B”, the right node

“C”, and the bottom node “D”. We next form a supernode between nodes A and B.

At the supernode: 5 = (VA – VB)/ 10 + VA/ 20 + (VB – VC)/ 12.5 [1]

At node C: VC = 150 [2]

At node D: -10 = VD/ 25 + (VD – VA)/ 10 [3]

Our supernode-related equation is VB – VA = 100 [4]

Simplifiying and grouping terms,

0.15 VA + 0.08 VB - 0.08 VC – 0.1 VD = 5 [1]

VC = 150 [2]

-25 VA + 35 VD = -2500 [3]

- VA + VB = 100 [4]

Solving, we find that VD = -63.06 V. Since v4 = - VD,

v4 = 63.06 V.

CHAPTER FOUR SOLUTIONS

12. Choosing the bottom node as the reference terminal and naming the left node “1”, the

center node “2” and the right node “3”, we next form a supernode about nodes 1 and

2, encompassing the dependent voltage source.

At the supernode, 5 – 8 = (v1 – v2)/ 2 + v3/ 2.5 [1]

At node 2, 8 = v2 / 5 + (v2 – v1)/ 2 [2]

Our supernode equation is v1 - v3 = 0.8 vA [3]

Since vA = v2, we can rewrite [3] as v1 – v3 = 0.8v2

Simplifying and collecting terms,

0.5 v1 - 0.5 v2 + 0.4 v3 = -3 [1]

-0.5 v1 + 0.7 v2 = 8 [2]

v1 - 0.8 v2 - v3 = 0 [3]

(a) Solving for v2 = vA, we find that vA = 25.91 V

(b) The power absorbed by the 2.5-Ω resistor is

(v3)2/ 2.5 = (-0.4546)2/ 2.5 = 82.66 mW.

CHAPTER FOUR SOLUTIONS

13. Selecting the bottom node as the reference terminal, we name the left node “1”, the

middle node “2” and the right node “3.”

NODE 1: 5 = (v1 – v2)/ 20 + (v1 – v3)/ 50 [1]

NODE 2: v2 = 0.4 v1 [2]

NODE 3: 0.01 v1 = (v3 – v2)/ 30 + (v3 – v1)/ 50 [3]

Simplifying and collecting terms, we obtain

0.07 v1 – 0.05 v2 – 0.02 v3 = 5 [1]

0.4 v1 – v2 = 0 [2]

-0.03 v1 – 0.03333 v2 + 0.05333 v3 = 0 [3]

Since our choice of reference terminal makes the controlling variable of both

dependent sources a nodal voltage, we have no need for an additional equation as we

might have expected.

Solving, we find that v1 = 148.2 V, v2 = 59.26 V, and v3 = 120.4 V.

The power supplied by the dependent current source is therefore

(0.01 v1) • v3 = 177.4 W.

CHAPTER FOUR SOLUTIONS

14. At node x: vx/ 4 + (vx – vy)/ 2 + (vx – 6)/ 1 = 0 [1]

At node y: (vy – kvx)/ 3 + (vy – vx)/ 2 = 2 [2]

Our additional constraint is that vy = 0, so we may simplify Eqs. [1] and [2]:

14 vx = 48 [1]

-2k vx - 3 vx = 12 [2]

Since Eq. [1] yields vx = 48/14 = 3.429 V, we find that

k = (12 + 3 vx)/ (-2 vx) = -3.250

CHAPTER FOUR SOLUTIONS

15. Choosing the bottom node joining the 4-Ω resistor, the 2-A current sourcee and the

4-V voltage source as our reference node, we next name the other node of the 4-Ω

resistor node “1”, and the node joining the 2-Ω resistor and the 2-A current source

node “2.” Finally, we create a supernode with nodes “1” and “2.”

At the supernode: –2 = v1/ 4 + (v2 – 4)/ 2 [1]

Our remaining equations: v1 – v2 = –3 – 0.5i1 [2]

and i1 = (v2 – 4)/ 2 [3]

Equation [1] simplifies to v1 + 2 v2 = 0 [1]

Combining Eqs. [2] and [3, 4 v1 – 3 v2 = –8 [4]

Solving these last two equations, we find that v2 = 727.3 mV. Making use of Eq. [3],

we therefore find that i1 = – 1.636 A.

CHAPTER FOUR SOLUTIONS

16. We first number the nodes as 1, 2, 3, 4, and 5 moving left to right. We next select

node 5 as the reference terminal. To simplify the analysis, we form a supernode from

nodes 1, 2, and 3.

At the supernode,

-4 – 8 + 6 = v1/ 40 + (v1 – v3)/ 10 + (v3 – v1)/ 10 + v2/ 50 + (v3 – v4)/ 20 [1]

Note that since both ends of the 10-Ω resistor are connected to the supernode, the

related terms cancel each other out, and so could have been ignored.

At node 4: v4 = 200 [2]

Supernode KVL equation: v1 – v3 = 400 + 4v20 [3]

Where the controlling voltage v20 = v3 – v4 = v3 – 200 [4]

Thus, Eq. [1] becomes -6 = v1/ 40 + v2/ 50 + (v3 – 200)/ 20 or, more simply,

4 = v1/ 40 + v2/ 50 + v3/ 20 [1’]

and Eq. [3] becomes v1 – 5 v3 = -400 [3’]

Eqs. [1’], [3’], and [5] are not sufficient, however, as we have four unknowns. At this

point we need to seek an additional equation, possibly in terms of v2. Referring to the

circuit, v1 - v2 = 400 [5]

Rewriting as a matrix equation,













=

























400

400-

4

0 1- 1

5- 0 1

20

1

50

1

40

1

3

2

1

v

Solving, we find that

v1 = 145.5 V, v2 = -254.5 V, and v3 = 109.1 V. Since v20 = v3 – 200, we find that

v20 = -90.9 V.

CHAPTER FOUR SOLUTIONS

17. We begin by naming the top left node “1”, the top right node “2”, the bottom node of

the 6-V source “3” and the top node of the 2-Ω resistor “4.” The reference node has

already been selected, and designated using a ground symbol.

By inspection, v2 = 5 V.

Forming a supernode with nodes 1 & 3, we find

At the supernode: -2 = v3/ 1 + (v1 – 5)/ 10 [1]

At node 4: 2 = v4/ 2 + (v4 – 5)/ 4 [2]

Our supernode KVL equation: v1 – v3 = 6 [3]

Rearranging, simplifying and collecting terms,

v1 + 10 v3 = -20 + 5 = -15 [1]

and v1 - v3 = 6 [2]

Eq. [3] may be directly solved to obtain v4 = 4.333 V.

Solving Eqs. [1] and [2], we find that

v1 = 4.091 V and v3 = -1.909 V.

CHAPTER FOUR SOLUTIONS

18. We begin by selecting the bottom node as the reference, naming the nodes as shown

below, and forming a supernode with nodes 5 & 6.

By inspection, v4 = 4 V.

By KVL, v3 – v4 = 1 so v3 = -1 + v4 = -1 + 4 or v3 = 3 V.

At the supernode, 2 = v6/ 1 + (v5 – 4)/ 2 [1]

At node 1, 4 = v1/ 3 therefore, v1 = 12 V.

At node 2, -4 – 2 = (v2 – 3)/ 4

Solving, we find that v2 = -21 V

Our supernode KVL equation is v5 - v6 = 3 [2]

Solving Eqs. [1] and [2], we find that

v5 = 4.667 V and v6 = 1.667 V.

The power supplied by the 2-A source therefore is (v6 – v2)(2) = 45.33 W.

4 A

2 A

1 V

4 V

3 V

4 Ω

3 Ω

2 Ω

1 Ω

v2

v1

v3 v4 v5 v6

CHAPTER FOUR SOLUTIONS

19. We begin by selecting the bottom node as the reference, naming each node as shown

below, and forming two different supernodes as indicated.

By inspection, v7 = 4 V and v1 = (3)(4) = 12 V.

At node 2: -4 – 2 = (v2 – v3)/ 4 or v2 -v3 = -24 [1]

At the 3-4 supernode:

0 = (v3 – v2)/ 4 + (v4 – v5)/ 6 or -6v2 + 6v3 + 4v4 – 4v5 = 0 [2]

At node 5:

0 = (v5 – v4)/ 6 + (v5 – 4)/ 7 + (v5 – v6)/ 2 or -14v4 + 68v5 – 42v6 = 48 [3]

At the 6-8 supernode: 2 = (v6 – v5)/ 2 + v8/ 1 or -v5 + v6 + 2v8 = 4 [4]

3-4 supernode KVL equation: v3 - v4 = -1 [5]

6-8 supernode KVL equation: v6 – v8 = 3 [6]

Rewriting Eqs. [1] to [6] in matrix form,













=

























3

1-

4

48

0

24-

1- 1 0 0 0 0

0 0 0 1- 1 0

2 1 1- 0 0 0

0 42- 68 14- 0 0

0 0 4- 4 6 6-

0 0 0 0 1- 1

8

6

5

4

3

2

v

Solving, we find that

v2 = -68.9 V, v3 = -44.9 V, v4 = -43.9 V, v5 = -7.9 V, v6 = 700 mV, v8 = -2.3 V.

The power generated by the 2-A source is therefore (v8 – v6)(2) = 133.2 W.

v1

v2 v3 v4

v5 v6

v7

v8

Voltages in

volts.

Currents in

amperes.

Resistances

in ohms.

CHAPTER FOUR SOLUTIONS

20. With the reference terminal already specified, we name the bottom terminal of the

3-mA source node “1,” the right terminal of the bottom 2.2-kΩ resistor node “2,” the

top terminal of the 3-mA source node “3,” the “+” reference terminal of the 9-V

source node “4,” and the “-” terminal of the 9-V source node “5.”

Since we know that 1 mA flows through the top 2.2-kΩ resistor, v5 = -2.2 V.

Also, we see that v4 – v5 = 9, so that v4 = 9 – 2.2 = 6.8 V.

Proceeding with nodal analysis,

At node 1: -3×10-3 = v1/ 10x103 + (v1 – v2)/ 2.2×103 [1]

At node 2: 0 = (v2 – v1)/ 2.2×103 + (v2 – v3)/ 4.7×103 [2]

At node 3: 1×103 + 3×103 = (v3 – v2)/ 4.7×103 + v3/3.3×103 [3]

Solving, v1 = -8.614 V, v2 = -3.909 V and v3 = 6.143 V.

Note that we could also have made use of the supernode approach here.

CHAPTER FOUR SOLUTIONS

21. Moving from left to right, we name the bottom three meshes, mesh “1”, mesh “2,”

and mesh “3.” In each of these three meshes we define a clockwise current. The

remaining mesh current is clearly 8 A. We may then write:

MESH 1: 12 i1 - 4 i2 = 100

MESH 2: -4 i1 + 9 i2 - 3 i3 = 0

MESH 3: -3 i2 + 18 i3 = -80

Solving this system of three (independent) equations in three unknowns, we find that

i2 = ix = 2.791 A.

CHAPTER FOUR SOLUTIONS

22. We define four clockwise mesh currents. The top mesh current is labeled i4. The

bottom left mesh current is labeled i1, the bottom right mesh current is labeled i3, and

the remaining mesh current is labeled i2. Define a voltage “v4A” across the 4-A current

source with the “+” reference terminal on the left.

By inspection, i3 = 5 A and ia = i4.

MESH 1: -60 + 2i1 – 2i4 + 6i4 = 0 or 2i1 + 4i4 = 60 [1]

MESH 2: -6i4 + v4A + 4i2 – 4(5) = 0 or 4i2 - 6i4 + v4A = 30 [2]

MESH 4: 2i4 – 2i1 + 5i4 + 3i4 – 3(5) – v4A = 0 or -2i1 + 10i4 - v4A = 15 [3]

At this point, we are short an equation. Returning to the circuit diagram, we note that

i2 – i4 = 4 [4]

Collecting these equations and writing in matrix form, we have













=

























4

15

20

60

0 1- 1 0

1- 10 0 2-

1 6- 4 0

0 4 0 2

A4

4

2

1

v

i

Solving, i1 = 16.83 A, i2 = 10.58 A, i4 = 6.583 A and v4A = 17.17 V.

Thus, the power dissipated by the 2-Ω resistor is

(i1 – i4)2 • (2) = 210.0 W

CHAPTER FOUR SOLUTIONS

23. We begin our analysis by defining three clockwise mesh currents. We will call the top

mesh current i3, the bottom left mesh current i1, and the bottom right mesh current i2.

By inspection, i1 = 5 A [1] and i2 = -0.01 v1 [2]

MESH 3: 50 i3 + 30 i3 – 30 i2 + 20 i3 – 20 i1 = 0

or -20 i1 – 30 i2 + 100 i3 = 0 [3]

These three equations are insufficient, however, to solve for the unknowns. It would

be nice to be able to express the dependent source controlling variable v1 in terms of

the mesh currents. Returning to the diagram, it can be seen that KVL around mesh 1

will yield - v1 + 20 i1 – 20 i3 + 0.4 v1 = 0

or v1 = 20 i1/ 0.6 – 20 i3/ 0.6 or v1 = (20(5)/ 0.6 - 20 i3/ 0.6 [4]

Substituting Eq. [4] into Eq. [2] and then the modified Eq. [2] into Eq. [3], we find

-20(5) – 30(-0.01)(20)(5)/0.6 + 30(-0.01)(20) i3/ 0.6 + 100 i3 = 0

Solving, we find that i3 = (100 – 50)/ 90 = 555.6 mA

Thus, v1 = 148.1 V, i2 = -1.481 A, and the power generated by the dependent voltage

source is 0.4 v1 (i2 – i1) = -383.9 W.

CHAPTER FOUR SOLUTIONS

24. We begin by defining four clockwise mesh currents i1, i2, i3 and i4, in the meshes of

our circuit, starting at the left-most mesh. We also define a voltage vdep across the

dependent current source, with the “+” on the top.

By inspection, i1 = 2A and i4 = -5 A.

At Mesh 2: 10 i2 - 10(2) + 20 i2 + vdep = 0 [1]

At Mesh 3: - vdep + 25 i3 + 5 i3 –5(-5) = 0 [2]

Collecting terms, we rewrite Eqs. [1] and [2] as

30 i2 + vdep = 20 [1]

30 i3 – vdep = -25 [2]

This is only two equations but three unknowns, however, so we require an additional

equation. Returning to the circuit diagram, we note that it is possible to express the

current of the dependent source in terms of mesh currents. (We might also choose to

obtain an expression for vdep in terms of mesh currents using KVL around mesh 2 or

3.)

Thus, 1.5ix = i3 - i2 where ix = i1 – i2 so -0.5 i2 - i3 = -3 [3]

In matrix form,













=

























3-

25-

20

0 1- 0.5-

1- 30 0

1 0 30

3

2

dep

v

i

Solving, we find that i2 = -6.333 A so that ix = i1 – i2 = 8.333 A.

CHAPTER FOUR SOLUTIONS

25. We define a clockwise mesh current i1 in the bottom left mesh, a clockwise mesh

current i2 in the top left mesh, a clockwise mesh current i3 in the top right mesh, and a

clockwise mesh current i4 in the bottom right mesh.

MESH 1: -0.1 va + 4700 i1 – 4700 i2 + 4700 i1 – 4700 i4 = 0 [1]

MESH 2: 9400 i2 – 4700 i1 – 9 = 0 [2]

MESH 3: 9 + 9400 i3 – 4700 i4 = 0 [3]

MESH 4: 9400 i4 – 4700 i1 – 4700 i3 + 0.1 ix = 0 [4]

The presence of the two dependent sources has led to the introduction of two

additional unknowns (ix and va) besides our four mesh currents. In a perfect world, it

would simplify the solution if we could express these two quantities in terms of the

mesh currents.

Referring to the circuit diagram, we see that ix = i2 (easy enough) and that

va = 4700 i3 (also straightforward). Thus, substituting these expressions into our

four mesh equations and creating a matrix equation, we arrive at:













=

























0

9-

9

0

9400 4700- 0.1 4700-

4700- 9400 0 0

0 0 9400 4700-

4700- 470- 4700- 9400

4

3

2

1

i

Solving,

i1 = 239.3 µA, i2 = 1.077 mA, i3 = -1.197 mA and i4 = -478.8 µA.

CHAPTER FOUR SOLUTIONS

26. We define a clockwise mesh current i3 in the upper right mesh, a clockwise mesh

current i1 in the lower left mesh, and a clockwise mesh current i2 in the lower right

mesh.

MESH 1: -6 + 6 i1 - 2 = 0 [1]

MESH 2: 2 + 15 i2 – 12 i3 – 1.5 = 0 [2]

MESH 3: i3 = 0.1 vx [3]

Eq. [1] may be solved directly to obtain i1 = 1.333 A.

It would help in the solution of Eqs. [2] and [3] if we could express the dependent

source controlling variable vx in terms of mesh currents. Referring to the circuit

diagram, we see that vx = (1)( i1) = i1, so Eq. [3] reduces to

i3 = 0.1 vx = 0.1 i1 = 133.3 mA.

As a result, Eq. [1] reduces to i2 = [-0.5 + 12(0.1333)]/ 15 = 73.31 mA.

CHAPTER FOUR SOLUTIONS

27. (a) Define a mesh current i2 in the second mesh. Then KVL allows us to write:

MESH 1: -9 + R i1 + 47000 i1 – 47000 i2 = 0 [1]

MESH 2: 67000 i2 – 47000 i1 – 5 = 0 [2]

Given that i1 = 1.5 mA, we may solve Eq. [2] to find that

mA 1.127 mA

67

47(1.5) 5

2=

+

=i

and so

3-

101.5

47(1.127) 47(1.5) - 9

×

+

=R = -5687 Ω.

(b) This value of R is unique; no other value will satisfy both Eqs. [1] and [2].

CHAPTER FOUR SOLUTIONS

28. Define three clockwise mesh currents i1, i2 and i3. The bottom 1-kΩ resistor can be

ignored, as no current flows through it.

MESH 1: -4 + (2700 + 1000 + 5000) i1 – 1000 i2 = 0 [1]

MESH 2: (1000 + 1000 + 4400 + 3000) i2 – 1000 i1 – 4400 i3 + 2.2 – 3 = 0 [2]

MESH 3: (4400 + 4000 + 3000) i3 - 4400 i2 – 1.5 = 0 [3]

Combining terms,

8700 i1 – 1000 i2 = 4 [1]

–1000 i1 + 9400 i2 – 4400 i3 = 0.8 [2]

– 4400 i2 + 11400 i3 = 1.5 [3]

Solving,

i1 = 487.6 mA, i2 = 242.4 mA and i3 = 225.1 mA.

The power absorbed by each resistor may now be calculated:

P5k = 5000 (i1)2 = 1.189 mW

P2.7k = 2700 (i1)2 = 641.9 µW

P1ktop = 1000 (i1 – i2)2 = 60.12 µW

P1kmiddle = 1000 (i2)2 = 58.76 µW

P1kbottom = 0 = 0

P4.4k = 4400 (i2 – i3)2 = 1.317 µW

P3ktop = 3000 (i3)2 = 152.0 µW

P4k = 4000 (i3)2 = 202.7 µW

P3kbottom = 3000 (i2)2 = 176.3 µW

Check: The sources supply a total of

4(487.6) + (3 – 2.2)(242.4) + 1.5(225.1) = 2482 µW.

The absorbed powers add to 2482 µW.

CHAPTER FOUR SOLUTIONS

29. (a) We begin by naming four mesh currents as depicted below:

Proceeding with mesh analysis, then, keeping in mind that Ix = -i4,

MESH 1: (4700 + 300) i1 - 4700 i2 = 0 [1]

MESH 2: (4700 + 1700) i2 – 4700 i1 – 1700 i3 = 0 [2]

Since we have a current source on the perimeter of mesh 3, we do not require a KVL

equation for that mesh. Instead, we may simply write

i3 = -0.03 vπ [3a] where vπ = 4700(i1 – i2) [3b]

MESH 4: 3000 i4 – 3000 i3 + 1 = 0 [4]

Simplifying and combining Eqs. 3a and 3b,

5000 i1 – 4700 i2 = 0

–4700 i1 + 6400 i2 – 1700 i3 = 0

–141 i1 + 141 i2 – i3 = 0

– 3000 i3 + 3000 i4 = –1

Solving, we find that i4 = -333.3 mA, so Ix = 333.3 µA.

(b) At node “π” : 0.03 vπ = vπ / 300 + vπ / 4700 + vπ /1700

Solving, we find that vπ = 0, therefore no current flows through the dependent source.

Hence, Ix = 333.3 µA as found in part (a).

(c) Vs/ Ix has units of resistance. It can be thought of as the resistance “seen” by the

voltage source Vs…. more on this in Chap. 5….

CHAPTER FOUR SOLUTIONS

30. We begin by naming each mesh and the three undefined voltage sources as shown

below:

MESH 1: –Vz + 9i1 – 2i2 – 7i4 = 0

MESH 2: –2i1 + 7i2 – 5i3 = 0

MESH 3: Vx – 5i2 + 8i3 – 3i4 = 0

MESH 4: Vy – 7i1 – 3i3 + 10i4 = 0

Rearranging and setting i1 – i2 = 0, i2 – i3 = 0, i1 – i4 = 0 and i4 – i3 = 0,

9i1 - 2i2 -7i4 = Vz

-2i1 + 7i2 - 5i3 = 0

-5i2 + 8i3 – 3i4 = - Vx

-7i1 -3i3 + 10i4 = - Vy

Since i1 = i2 = i3 = i4, these equations produce:

V

z = 0

0 = 0

-V

x = 0

-V

y = 0

This is a unique solution. Therefore,

the request that nonzero values be

found cannot be satisfied.

CHAPTER FOUR SOLUTIONS

31. The “supermesh” concept is not required (or helpful) in solving this problem, as there

are no current sources shared between meshes. Starting with the left-most mesh and

moving right, we define four clockwise mesh currents i1, i2, i3 and i4. By inspection,

we see that i1 = 2 mA.

MESH 2: -10 + 5000i2 + 4 + 1000i3 = 0 [1]

MESH 3: -1000i3 + 6 + 10,000 – 10,000i4 = 0 [2]

MESH 4: i4 = -0.5i2 [3]

Reorganising, we find

5000 i2 + 1000 i3 = 6 [1]

9000 i3 – 10,000 i4 = -6 [2]

0.5 i2 + i4 = 0 [3]

We could either subtitute Eq. [3] into Eq. [2] to reduce the number of equations, or

simply go ahead and solve the system of Eqs. [1-3]. Either way, we find that

i1 = 2 mA, i2 = 1.5 mA, i3 = -1.5 mA and i4 = -0.75 mA.

The power generated by each source is:

P2mA = 5000(i1 – i2)(i1) = 5 mW

P4V = 4 (-i2) = -6 mW

P6V = 6 (-i3) = 9 mW

PdepV = 1000 i3 (i3 – i2) = 4.5 mW

PdepI = 10,000(i3 – i4)(0.5 i2) = -5.625 mW

CHAPTER FOUR SOLUTIONS

32. This circuit does not require the supermesh technique, as it does not contain any

current sources. Redrawing the circuit so its planar nature and mesh structure are

clear,

MESH 1: -20 + 2 i1 – 2 i2 + 2.5 iA = 0 [1]

MESH 2: 2 i2 + 3 i2 – 3 i3 + 2 i2 – 2 i1 = 0 [2]

MESH 3: -2.5 iA + 7 i3 – 3 i2 = 0 [3]

Combining terms and making use of the fact that iA = - i3,

2 i1 – 2 i2 – 2.5 i3 = 20 [1]

-2 i1 + 7i2 – 3 i3 = 0 [2]

–3 i2 + 9.5 i3 = 0 [3]

Solving, i1 = 18.55 A, i2 = 6.129 A, and i3 = 1.935 A. Since iA = - i3,

iA = -1.935 A.

20 V 2.5 iA

2 Ω

4 Ω

2 Ω

iA →

i2

i3

i1

•

3 Ω

CHAPTER FOUR SOLUTIONS

33. Define four mesh currents

By inspection, i1 = -4.5 A.

We form a supermesh with meshes 3 and 4 as defined above.

MESH 2: 2.2 + 3 i2 + 4 i2 + 5 – 4 i3 = 0 [1]

SUPERMESH: 3 i4 + 9 i4 – 9 i1 + 4 i3 – 4 i2 + 6 i3 + i3 – 3 = 0 [2]

Supermesh KCL equation: i4 - i3 = 2 [3]

Simplifying and combining terms, we may rewrite these three equations as:

7 i2 – 4 i3 = -7.2 [1]

-4 i2 + 11 i3 + 12 i4 = -37.5 [2]

- i3 + i4 = 2 [3]

Solving, we find that i2 = -2.839 A, i3 = -3.168 A, and i4 = -1.168 A.

The power supplied by the 2.2-V source is then 2.2 (i1 – i2) = -3.654 W.

i1 i2

i4

i3

CHAPTER FOUR SOLUTIONS

34. We begin by defining six mesh currents as depicted below:

• We form a supermesh with meshes 1 and 2 since they share a current source.

• We form a second supermesh with meshes 3 and 4 since they also share a current

source.

1, 2 Supermesh:

(4700 + 1000 + 10,000) i1 – 2200 i5 + (2200 + 1000 + 4700) i2 – 1000 i3 = 0 [1]

3, 4 Supermesh:

(4700 + 1000 + 2200) i3 – 1000 i2 – 2200 i6 + (4700 + 10,000 + 1000) i4 = 0 [2]

MESH 5: (2200 + 4700) i5 – 2200 i2 + 3.2 – 1.5 = 0 [3]

MESH 6: 1.5 + (4700 + 4700 + 2200) c – 2200 i3 = 0 [4]

1, 2 Supermesh KCL equation: i1 – i2 = 3×10-3 [5]

3, 4 Supermesh KCL equation: i4 – i3 = 2×10-3 [6]

We can simplify these equations prior to solution in several ways. Choosing to retain

six equations,

15,700 i1 + 7900 i2 - 1000 i3 -2200 i5 = 0 [1]

- 1000 i2 + 7900 i3 + 15,700 i4 -2200 i6 = 0 [2]

- 2200 i2 + 6900 i5 = -1.7 [3]

- 2200 i3 + 11,600 i6 = -1.5 [4]

i1 – i2 = 3×10-3 [5]

- i3 + i4 = 2×10-3 [6]

Solving, we find that i4 = 540.8 mA. Thus, the voltage across the 2-mA source is

(4700 + 10,000 + 1000) (540.8×10-6) = 8.491 V

i5 i6

i3 i2

i1 i4

CHAPTER FOUR SOLUTIONS

35. We define a mesh current ia in the left-hand mesh, a mesh current i1 in the top right

mesh, and a mesh current i2 in the bottom right mesh.

The left-most mesh can be analysed separately to determine the controlling voltage va,

as KCL assures us that no current flows through either the 1-Ω or 6-Ω resistor.

Thus, -1.8 + 3 ia – 1.5 + 2 ia = 0, which may be solved to find ia = 0.66 A. Hence,

va = 3 ia = 1.98 V.

Forming one supermesh from the remaining two meshes, we may write:

-3 + 2.5 i1 + 3 i2 + 4 i2 = 0

and the supermesh KCL equation: i2 – i1 = 0.05 va = 0.05(1.98) = 99×10-3

Thus, we have two equations to solve:

2.5 i1 + 7 i2 = 3

-i1 + i2 = 99×10-3

Solving, we find that i1 = 242.8 mA and the voltage across the 2.5-Ω resistor

(arbitrarily assuming the left terminal is the “+” reference) is 2.5 i1 = 607 mV.

CHAPTER FOUR SOLUTIONS

36. UNDEFINED RESISTOR VALUE IN FIGURE. Set to 10 mΩ.

There are only three meshes in this circuit, as the botton 22-mΩ resistor is not

connected connected at its left terminal. Thus, we define three mesh currents, i1, i2,

and i3, beginning with the left-most mesh.

We next create a supermesh from meshes 1 and 2 (note that mesh 3 is independent,

and can be analysed separately).

Thus, -11.8 + 10×10-3 i1 + 22×10-3 i2 + 10×10-3 i2 + 17×10-3 i1 = 0

and applying KCL to obtain an equation containing the current source,

i1 – i2 = 100

Combining terms and simplifying, we obtain

27×10-3 i1 + 32×10-3 i2 = 11.8

i1 – i2 = 100

Solving, we find that i1 = 254.2 A and i2 = 154.2 A.

The final mesh current is easily found: i3 = 13×103/ (14 + 11.6 + 15) = 320.2 A.

CHAPTER FOUR SOLUTIONS

37. MESH 1: -7 + i1 – i2 = 0 [1]

MESH 2: i2 – i1 + 2i2 + 3i2 – 3i3 = 0 [2]

MESH 3: 3i3 – 3i2 + xi3 +2i3 – 7 = 0 [3]

Grouping terms, we find that

i1 – i2 = 7 [1]

-i1 + 6i2 – 3i3 = 0 [2]

-3i2 + (5 + x)i3 = 7 [3]

This, unfortunately, is four unknowns but only three equations. However, we have not

yet made use of the fact that we are trying to obtain i2 = 2.273 A. Solving these “four”

equations, we find that

x = (7 + 3 i2 – 5 i3)/ i3 = 4.498 Ω.

CHAPTER FOUR SOLUTIONS

7 V

7 A

1 Ω

2 Ω

3 Ω

300 mΩ

2 Ω

i2 i3

i1

38. We begin by redrawing the circuit as instructed, and define three mesh currents:

By inspection, i3 = 7 A.

MESH 1: -7 + i1 – i2 = 0 or i1 – i2 = 7 [1]

MESH 2: (1 + 2 + 3) i2 – i1 –3(7) = 0 or -i1 + 6i2 = 21 [2]

There is no need for supermesh techniques for this situation, as the only current

source lies on the outside perimeter of a mesh- it is not shared between meshes.

Solving, we find that i1 = 12.6 A, i2 = 5.6 A and i3 = 7 A.

CHAPTER FOUR SOLUTIONS

39. (a) We are asked for a voltage, and have one current source and one voltage source.

Nodal analysis is probably best then- the nodes can be named so that the desired

voltage is a nodal voltage, or, at worst, we have one supernode equation to solve.

Name the top left node “1” and the top right node “x”; designate the bottom node as

the reference terminal. Next, form a supernode with nodes “1” and “x.”

At the supernode: 11 = v1/ 2 + vx/ 9 [1]

and the KVL Eqn: v1 – vx = 22 [2]

Rearranging, 11(18) = 9 v1 + 2 vx [1]

22 = v1 – vx [2]

Solving, vx = 0

(b) We are asked for a voltage, and so may suspect that nodal analysis is preferrable;

with two current sources and only one voltage source (easily dealt with using the

supernode technique), nodal analysis does seem to have an edge over mesh analysis

here.

Name the top left node “x,” the top right node “y” and designate the bottom node as

the reference node. Forming a supernode from nodes “x” and “y,”

At the supernode: 6 + 9 = vx / 10 + vy/ 20 [1]

and the KVL Eqn: vy – vx = 12 [2]

Rearranging, 15(20) = 2 vx + vy [1]

and 12 = - vx + vy [2]

Solving, we find that vx = 96 V.

(c) We are asked for a voltage, but would have to subtract two nodal voltages (not

much harder than invoking Ohm’s law). On the other hand, the dependent current

source depends on the desired unknown, which would lead to the need for another

equation if invoking mesh analysis. Trying nodal analysis,

0.1 vx = (v1 – 50) / 2 + vx / 4 [1]

referring to the circuit we see that vx = v1 – 100. Rearranging so that we may

eliminate v1 in Eq. [1], we obtain v1 = vx + 100. Thus, Eq. [1] becomes

0.1 vx = (vx + 100 – 50)/ 2 + vx / 4

and a little algebra yields vx = -38.46 V.

CHAPTER FOUR SOLUTIONS

Both ends of the

resistor are

connected to the

supernode, so we

could actually just

ignore it…

(a) (b)

Ref.

v1

40.

(a) We begin by noting that it is a voltage that is required; no current values are

requested. This is a three-mesh circuit, or a four-node circuit, depending on your

perspective. Either approach requires three equations…. Except that applying the

supernode technique reduces the number of needed equations by one.

At the 1, 3 supernode:

0 = (v1 – 80)/ 10 + (v1 – v3)/ 20 + (v3 – v1)/ 20 + v3/ 40 + v3/ 30

and v3 - v1 = 30

We simplify these two equations and collect terms, yielding

0.1 v1 + 0.05833 v3 = 8

- v1 + v3 = 30

Solving, we find that v3 = 69.48 V

(b) Mesh analysis would be straightforward, requiring 3 equations and a (trivial)

application of Ohm’s law to obtain the final answer. Nodal analysis, on the other

hand, would require only two equations, and the desired voltage will be a nodal

voltage.

At the b, c, d supernode: 0 = (vb – 80)/ 10 + vd/ 40 + vc/ 30

and: vd – vb = 30 vc – vd = 9

Simplify and collect terms: 0.1 vb + 0.03333 vc + 0.025 vd = 80

-vb + vd = 30

vc - vd = 9

Solving, vd (= v3) = 67.58 V

(c) We are now faced with a dependent current source whose value depends on a

mesh current. Mesh analysis in this situation requires 1 supermesh, 1 KCL equation

and Ohm’s law. Nodal analysis requires 1 supernode, 1 KVL equation, 1 other nodal

equation, and one equation to express i1 in terms of nodal voltages. Thus, mesh

analysis has an edge here. Define the left mesh as “1,” the top mesh as “2”, and the

bottom mesh as “3.”

Mesh 1: -80 + 10 i1 + 20 i1 – 20 i2 + 30 i1 – 30 i3 = 0

2, 3 supermesh: 20 i2 – 20 i1 – 30 + 40 i3 + 30 i3 – 30 i1 = 0

and: i2 - i3 = 5 i1

Rewriting, 60 i1 – 20 i2 – 30 i3 = 80

-50 i1 + 20 i2 + 70 i3 = 30

5 i1 – i2 + i3 = 0

Solving, i3 = 4.727 A so v3 = 40 i3 = 189 V.

CHAPTER FOUR SOLUTIONS

41. This circuit consists of 3 meshes, and no dependent sources. Therefore 3 simultaneous

equations and 1 subtraction operation would be required to solve for the two desired

currents. On the other hand, if we use nodal analysis, forming a supernode about the

30-V source would lead to 5 – 1 – 1 = 3 simulataneous equations as well, plus several

subtraction and division operations to find the currents. Thus, mesh analysis has a

slight edge here.

Define three clockwise mesh currents: ia in the left-most mesh, ib in the top right

mesh, and ic in the bottom right mesh. Then our mesh equations will be:

Mesh a: -80 + (10 + 20 + 30) ia – 20 ib – 30 ic = 0 [1]

Mesh b: -30 + (12 + 20) ib – 12 ic – 20 ia = 0 [2]

Mesh c: (12 + 40 + 30) ic – 12 ib – 30 ia = 0 [3]

Simplifying and collecting terms,

60 ia – 20 ib – 30 ic = 80 [1]

-20 ia + 32 ib – 12 ic = 30 [2]

-30 ia – 12 ib + 82 ic = 0 [3]

Solving, we find that ia = 3.549 A, ib = 3.854 A, and ic = 1.863 A. Thus,

i1 = ia = 3.549 A and i2 = ia – ic = 1.686 A.

CHAPTER FOUR SOLUTIONS

42. Approaching this problem using nodal analysis would require 3 separate nodal

equations, plus one equation to deal with the dependent source, plus subtraction and

division steps to actually find the current i10. Mesh analysis, on the other hand, will

require 2 mesh/supermesh equations, 1 KCL equation, and one subtraction step to find

i10. Thus, mesh analysis has a clear edge. Define three clockwise mesh currents: i1 in

the bottom left mesh, i2 in the top mesh, and i3 in the bottom right mesh.

MESH 1: i1 = 5 mA by inspection [1]

SUPERMESH: i1 – i2 = 0.4 i10

i1 – i2 = 0.4(i3 – i2)

i1 – 0.6 i2 – 0.4 i3 = 0 [2]

MESH 3: -5000 i1 – 10000 i2 + 35000 i3 = 0 [3]

Simplify: 0.6 i2 + 0.4 i3 = 5×10-3 [2]

-10000 i2 + 35000 i3 = 25 [3]

Solving, we find i2 = 6.6 mA and i3 = 2.6 mA. Since i10 = i3 – i2, we find that

i10 = -4 mA.

CHAPTER FOUR SOLUTIONS

43. For this circuit problem, nodal analysis will require 3 simultaneous nodal equations,

then subtraction/ division steps to obtain the desired currents. Mesh analysis requires

1 mesh equation, 1 supermesh equation, 2 simple KCL equations and one subtraction

step to determine the currents. If either technique has an edge in this situation, it’s

probably mesh analysis. Thus, define four clockwise mesh equations: ia in the bottom

left mesh, ib in the top left mesh, ic in the top right mesh, and id in the bottom right

mesh.

At the a, b, c supermesh: -100 + 6 ia + 20 ib + 4 ic + 10 ic – 10 id = 0 [1]

Mesh d: 100 + 10 id – 10 ic + 24 id = 0 [2]

KCL: - ia + ib = 2 [3]

and - ib + ic = 3 i3 = 3 ia [4]

Collecting terms & simplifying,

6 ia + 20 ib + 14 ic – 10 id = 100 [1]

-10 ic + 34 id = -100 [2]

- ia + ib = 2 [3]

-3 ia – ib + ic = 0 [4]

Solving,

ia = 0.1206 A, ib = 2.121 A, ic = 2.482 A, and id = -2.211 A. Thus,

i3 = ia = 120.6 mA and i10 = ic – id = 4.693 A.

CHAPTER FOUR SOLUTIONS

44. With 7 nodes in this circuit, nodal analysis will require the solution of three

simultaneous nodal equations (assuming we make use of the supernode technique)

and one KVL equation. Mesh analysis will require the solution of three simultaneous

mesh equations (one mesh current can be found by inspection), plus several

subtraction and multiplication operations to finally determine the voltage at the

central node. Either will probably require a comparable amount of algebraic

manoeuvres, so we go with nodal analysis, as the desired unknown is a direct result of

solving the simultaneous equations. Define the nodes as:

NODE 1: -2×10-3 = (v1 – 1.3)/ 1.8×103 → v1 = -2.84 V.

2, 4 Supernode:

2.3×10-3 = (v2 – v5)/ 1x103 + (v4 – 1.3)/ 7.3×103 + (v4 – v5)/ 1.3×103 + v4/ 1.5×103

KVL equation: -v2 + v4 = 5.2

Node 5: 0 = (v5 – v2)/ 1x103 + (v5 – v4)/ 1.3x103 + (v5 – 2.6)/ 6.3x103

Simplifying and collecting terms,

14.235 v2 + 22.39 v4 – 25.185 v5 = 35.275 [1]

-v2 + v4 = 5.2 [2]

-8.19 v2 – 6.3 v4 + 15.79 v5 = 3.38 [3]

Solving, we find the voltage at the central node is v4 = 3.460 V.

v1 v2

v3

v4 v5

v6

CHAPTER FOUR SOLUTIONS

45. Mesh analysis yields current values directly, so use that approach. We therefore

define four clockwise mesh currents, starting with i1 in the left-most mesh, then i2, i3

and i4 moving towards the right.

Mesh 1: -0.8ix + (2 + 5) i1 – 5 i2 = 0 [1]

Mesh 2: i2 = 1 A by inspection [2]

Mesh 3: (3 + 4) i3 – 3(1) – 4(i4) = 0 [3]

Mesh 4: (4 + 3) i4 – 4 i3 – 5 = 0 [4]

Simplify and collect terms, noting that ix = i1 – i2 = i1 – 1

-0.8(i1 – 1) + 7 i1 – 5(1) = 0 yields i1 = 677.4 mA

Thus, [3] and [4] become: 7 i3 – 4 i4 = 3 [3]

-4 i3 + 7 i4 = 5 [4]

Solving, we find that i3 = 1.242 A and i4 = 1.424 A. A map of individual branch

currents can now be drawn:

↑

677.4 mA

→

-322.6 mA

↓ -242.0 mA

↓

↑

182.0 mA

↑

-1.424 A

CHAPTER FOUR SOLUTIONS

46. If we choose to perform mesh analysis, we require 2 simultaneous equations (there are

four meshes, but one mesh current is known, and we can employ the supermesh

technique around the left two meshes). In order to find the voltage across the 2-mA

source we will need to write a KVL equation, however. Using nodal analysis is less

desirable in this case, as there will be a large number of nodal equations needed.

Thus, we define four clockwise mesh currents i1, i2, i3 and i4 starting with the left-

most mesh and moving towards the right of the circuit.

At the 1,2 supermesh: 2000 i1 + 6000 i2 – 3 + 5000 i2 = 0 [1]

and i1 – i2 = 2×10-3 [2]

by inspection, i4 = -1 mA. However, this as well as any equation for mesh

four are unnecessary: we already have two equations in two unknowns and i1 and i2

are sufficient to enable us to find the voltage across the current source.

Simplifying, we obtain 2000 i1 + 11000 i2 = 3 [1]

1000 i1 - 1000 i2 = 2 [2]

Solving, i1 = 1.923 mA and i2 = -76.92 µA.

Thus, the voltage across the 2-mA source (“+” reference at the top of the source) is

v = -2000 i1 – 6000 (i1 – i2) = -15.85 V.

CHAPTER FOUR SOLUTIONS

47. Nodal analysis will require 2 nodal equations (one being a “supernode” equation), 1

KVL equation, and subtraction/division operations to obtain the desired current. Mesh

analysis simply requires 2 “supermesh” equations and 2 KCL equations, with the

desired current being a mesh current. Thus, we define four clockwise mesh currents

ia, ib, ic, id starting with the left-most mesh and proceeding to the right of the circuit.

At the a, b supermesh: -5 + 2 ia + 2 ib + 3 ib – 3 ic = 0 [1]

At the c, d supermesh: 3 ic – 3 ib + 1 + 4 id = 0 [2]

and ia - ib = 3 [3]

ic - id = 2 [4]

Simplifying and collecting terms, we obtain

2 ia + 5 ib – 3 ic = 5 [1]

-3 ib + 3 ic + 4 id = -1 [2]

ia - ib = 3 [3]

ic - id = 2 [4]

Solving, we find ia = 3.35 A, ib = 350 mA, ic = 1.15 A, and id = -850 mA. As i1 = ib,

i1 = 350 mA.

CHAPTER FOUR SOLUTIONS

48. Define a voltage vx at the top node of the current source I2, and a clockwise mesh

current ib in the right-most mesh.

We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement ib = 1 A.

Applying nodal analysis to the circuit,

I

1 + I2 = (vx – v1)/ 6 = 1

so our requirement is I1 + I2 = 1. There is no constraint on the value of v1 other than

we are told to select a nonzero value.

Thus, we choose I1 = I2 = 500 mA and v1 = 3.1415 V.

CHAPTER FOUR SOLUTIONS

49. Inserting the new 2-V source with “+” reference at the bottom, and the new 7-mA

source with the arrow pointing down, we define four clockwise mesh currents i1, i2, i3,

i4 starting with the left-most mesh and proceeding towards the right of the circuit.

Mesh 1: (2000 + 1000 + 5000) i1 – 6000 i2 – 2 = 0 [1]

2, 3 Supermesh:

2 + (5000 + 5000 + 1000 + 6000) i2 – 6000 i1 + (3000 + 4000 + 5000) i3 – 5000 i4

= 0 [2]

and i2 - i3 = 7×10-3 [3]

Mesh 4: i4 = -1 mA by inspection [4]

Simplifying and combining terms,

8000 i1 – 6000 i2 = 2 [1]

1000 i2 – 1000 i3 = 7 [4]

-6000 i1 + 17000 i2 + 12000 i3 = -7 [2]

Solving, we find that

i1 = 2.653 A, i2 = 3.204 A, i3 = -3.796 A, i4 = -1 mA

CHAPTER FOUR SOLUTIONS

50. Define node 1 as the top left node, and node 2 as the node joining the three 2-Ω

resistors. Place the “+” reference terminal of the 2-V source at the right. The right-

most 2-Ω resistor has therefore been shorted out. Applying nodal analysis then,

Node 1: -5 i1 = (v1 – v2)/ 2 [1]

Node 2: 0 = (v2 – v1)/ 2 + v2/ 2 + (v2 – 2)/ 2 [2]

and, i1 = (v2 – 2)/ 2 [3]

Simplifying and collecting terms,

v1 + v2 = 10 [1]

-v1 + 3 v2 = 2 [2]

Solving, we find that v1 = 3.143 V and v2 = 1.714 V.

Defining clockwise mesh currents ia, ib, ic, id starting with the left-most mesh and

proceeding right, we may easily determine that

ia = -5 i1 = 714.3 mA

ib = -142.9 mA

ic = i1 – 2 = -2.143 A

id = 3 + ic = 857.1 mA

CHAPTER FOUR SOLUTIONS

51. Hand analysis:

Define three clockwise mesh currents: i1 in the bottom left mesh, i2 in the top mesh,

and i3 in the bottom right mesh.

MESH 1: i1 = 5 mA by inspection [1]

SUPERMESH: i1 – i2 = 0.4 i10

i1 – i2 = 0.4(i3 – i2)

i1 – 0.6 i2 – 0.4 i3 = 0 [2]

MESH 3: -5000 i1 – 10000 i2 + 35000 i3 = 0 [3]

Simplify: 0.6 i2 + 0.4 i3 = 5×10-3 [2]

-10000 i2 + 35000 i3 = 25 [3]

Solving, we find i2 = 6.6 mA and i3 = 2.6 mA. Since i10 = i3 – i2, we find that

i10 = -4 mA.

PSpice simulation results:

Summary: The current entering the right-hand node of the 10-kΩ resistor R2 is

equal to 4.000 mA. Since this current is –i10, i10 = -4.000 mA as found by hand.

i10

→

CHAPTER FOUR SOLUTIONS

52. Hand analysis:

Define the nodes as:

NODE 1: -2×10-3 = (v1 – 1.3)/ 1.8×103 → v1 = -2.84 V.

2, 4 Supernode:

2.3×10-3 = (v2 – v5)/ 1x103 + (v4 – 1.3)/ 7.3×103 + (v4 – v5)/ 1.3×103 + v4/ 1.5×103

KVL equation: -v2 + v4 = 5.2

Node 5: 0 = (v5 – v2)/ 1x103 + (v5 – v4)/ 1.3x103 + (v5 – 2.6)/ 6.3x103

Simplifying and collecting terms,

14.235 v2 + 22.39 v4 – 25.185 v5 = 35.275 [1]

-v2 + v4 = 5.2 [2]

-8.19 v2 – 6.3 v4 + 15.79 v5 = 3.38 [3]

Solving, we find the voltage at the central node is v4 = 3.460 V.

PSpice simulation results:

Summary: The voltage at the center node is found to be 3.460 V, which is in

agreement with our hand calculation.

v1 v2

v3

v4 v5

v6

CHAPTER FOUR SOLUTIONS

53. Hand analysis:

At the 1,2 supermesh: 2000 i1 + 6000 i2 – 3 + 5000 i2 = 0 [1]

and i1 – i2 = 2×10-3 [2]

by inspection, i4 = -1 mA. However, this as well as any equation for mesh

four are unnecessary: we already have two equations in two unknowns and i1 and i2

are sufficient to enable us to find the voltage across the current source.

Simplifying, we obtain 2000 i1 + 11000 i2 = 3 [1]

1000 i1 - 1000 i2 = 2 [2]

Solving, i1 = 1.923 mA and i2 = -76.92 µA.

Thus, the voltage across the 2-mA source (“+” reference at the top of the source) is

v = -2000 i1 – 6000 (i1 – i2) = -15.85 V.

PSpice simulation results:

Summary: Again arbitrarily selecting the “+” reference as the top node of the

2-mA current source, we find the voltage across it is –5.846 – 10 = -15.846 V, in

agreement with our hand calculation.

CHAPTER FOUR SOLUTIONS

54. Hand analysis:

Define a voltage vx at the top node of the current source I2, and a clockwise mesh

current ib in the right-most mesh.

We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement ib = 1 A.

Applying nodal analysis to the circuit,

I

1 + I2 = (vx – v1)/ 6 = 1

so our requirement is I1 + I2 = 1. There is no constraint on the value of v1 other than

we are told to select a nonzero value.

Thus, we choose I1 = I2 = 500 mA and v1 = 3.1415 V.

PSpice simulation results:

Summary: We see from the labeled schematic above that our choice for I1, I2 and

V1 lead to 1 A through the 6-Ω resistor, or 6 W dissipated in that resistor, as desired.

CHAPTER FOUR SOLUTIONS

55. Hand analysis:

Define node 1 as the top left node, and node 2 as the node joining the three 2-Ω

resistors. Place the “+” reference terminal of the 2-V source at the right. The right-

most 2-Ω resistor has therefore been shorted out. Applying nodal analysis then,

Node 1: -5 i1 = (v1 – v2)/ 2 [1]

Node 2: 0 = (v2 – v1)/ 2 + v2/ 2 + (v2 – 2)/ 2 [2]

and, i1 = (v2 – 2)/ 2 [3]

Simplifying and collecting terms,

v1 + v2 = 10 [1]

-v1 + 3 v2 = 2 [2]

Solving, we find that v1 = 3.143 V and v2 = 1.714 V.

Defining clockwise mesh currents ia, ib, ic, id starting with the left-most mesh and

proceeding right, we may easily determine that

ia = -5 i1 = 714.3 mA

ib = -142.9 mA

ic = i1 – 2 = -2.143 A

id = 3 + ic = 857.1 mA

PSpice simulation results:

Summary: The simulation results agree with the hand calculations.

CHAPTER FOUR SOLUTIONS

56. (a) One possible circuit configuration of many that would satisfy the requirements:

At node 1: -3 = (v1 – 5)/ 100 + (v1 – v2)/ 50 [1]

At node 2: 2 vx = (v2 – v1)/ 50 + v2/ 30 [2]

and, vx = 5 – v1 [3]

Simplifying and collecting terms,

150 v1 – 100 v2 = -14750 [1]

2970 v1 + 80 v2 = 15000 [2]

Solving, we find that v1 = 1.036 V and v2 = 149.1 V.

The current through the 100-Ω resistor is simply (5 – v1)/100 = 39.64 mA

The current through the 50-Ω resistor is (v1 – v2)/ 50 = -2.961 A,

and the current through the 20-Ω and 10-Ω series combination is v2/ 30 = 4.97 A.

Finally, the dependent source generates a current of 2 vx = 7.928 A.

(b) PSpice simulation results

Summary: The simulated results agree with the hand calculations.

↑

5 V 3 A 2 vx

100 Ω

50 Ω 20 Ω

10 Ω

+ vx -

CHAPTER FOUR SOLUTIONS

+ 5 V -

57. One possible solution of many:

Choose R so that 3R = 5; then the voltage across the current source will be 5 V, and

so will the voltage across the resistor R.

R = 5/3 Ω. To construct this from 1-Ω resistors, note that

5/3 Ω = 1 Ω + 2/3 Ω = 1 Ω + 1 Ω || 1Ω || 1Ω + 1Ω || 1Ω || 1Ω

* Solution to Problem 4.57

.OP

V1 1 0 DC 10

I1 0 4 DC 3

R1 1 2 1

R2 2 3 1

R3 2 3 1

R4 2 3 1

R5 3 4 1

R6 3 4 1

R7 3 4 1

.END

**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C

******************************************************************************

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE

( 1) 10.0000 ( 2) 7.0000 ( 3) 6.0000 ( 4) 5.0000

VOLTAGE SOURCE CURRENTS

NAME CURRENT

V1 -3.000E+00

CHAPTER FOUR SOLUTIONS

58. We first name each node, resistor and voltage source:

We next write an appropriate input deck for SPICE:

And obtain the following output:

We see from this simulation result that the voltage v5 = 2.847 V.

1 2 3

4

5

0

R1

R2

R3

R4

R5

R6

R7

R8

R9

R10

V1

* Solution to Problem 4.58

.OP

V1 1 0 DC 20

R1 1 2 2

R2 2 0 3

R3 2 3 4

R4 2 4 10

R5 3 0 5

R6 3 4 6

R7 3 5 11

R8 4 0 7

R9 4 5 8

R10 5 0 9

.END

CHAPTER FOUR SOLUTIONS

v1

v2

v3

All resistors are

1 Ω, except R1,

which represents

5 1-Ω resistors

in series.

59. One possible solution of many:

Verify: v1 = 9(4/9) = 4 V

v2 = 9(3/9) = 3 V

v3 = 9(2/9) = 2 V

SPICE INPUT DECK:

* Solution to Problem 4.59

.OP

V1 1 0 DC 9

R1 1 2 5

R2 2 3 1

R3 3 4 1

R4 4 5 1

R5 5 0 1

.END

R1 R2

R3

R4

R5

**** 07/29/01 21:36:26 *********** Evaluation PSpice (Nov 1999) **************

* Solution to Problem 4.59

**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C

***********************************************************************

*******

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE

VOLTAGE

( 1) 9.0000 ( 2) 4.0000 ( 3) 3.0000 ( 4) 2.0000

( 5) 1.0000 1 2

3

4

5

0

R1

R4

R5

R2

R3

CHAPTER FOUR SOLUTIONS

60. (a) If only two bulbs are not lit (and thinking of each bulb as a resistor), the bulbs

must be in parallel- otherwise, the burned out bulbs, acting as short circuits, would

prevent current from flowing to the “good” bulbs.

(b) In a parallel connected circuit, each bulb “sees” 115 VAC. Therefore, the

individual bulb current is 1 W/ 115 V = 8.696 mA. The resistance of each “good”

bulb is V/I = 13.22 kΩ. A simplified, electrically-equivalent model for this circuit

would be a 115 VAC source connected in parallel to a resistor Req such that

1/Req = 1/13.22×103 + 1/13.22×103 + …. + 1/13.22×103 (400 – 2 = 398 terms)

or Req = 33.22 Ω. We expect the source to provide 398 W.

(c) The inherent series resistance of the wire connections leads to a voltage drop

which increases the further one is from the voltage source. Thus, the furthest bulbs

actually have less than 115 VAC across them, so they draw slightly less current and

glow more dimly.

* Solution to Problem 4.60

.OP

V1 1 0 AC 115 60

R1 1 0 33.22

.AC LIN 1 60 60

.PRINT AC VM(1)IM(V1)

.END

**** 07/29/01 21:09:32 *********** Evaluation PSpice (Nov 1999) **************

* Solution to Problem 4.60

**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C

******************************************************************************

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE

( 1) 0.0000

VOLTAGE SOURCE CURRENTS

NAME CURRENT

V1 0.000E+00

TOTAL POWER DISSIPATION 0.00E+00 WATTS

**** 07/29/01 21:09:32 *********** Evaluation PSpice (Nov 1999) **************

* Solution to Problem 4.60

**** AC ANALYSIS TEMPERATURE = 27.000 DEG C

******************************************************************************

FREQ VM(1) IM(V1)

6.000E+01 1.150E+02 3.462E+00

This calculated power is not the value

sought. It is an artifact of the use o

f

an ac source, which requires that we

perform an ac analysis. The supplied

power is then separately computed as

(1.15×102)(3.462) = 398.1 W.

CHAPTER FIVE SOLUTIONS

1. Define percent error as 100 [ex – (1 + x)]/ ex

x 1 + x ex % error

0.001 1.001 1.001 5×10-5

0.005 1.005 1.005 1×10-3

0.01 1.01 1.010 5×10-3

0.05 1.05 1.051 0.1

0.10 1.10 1.105 0.5

0.50 1.50 1.649 9

1.00 2.00 2.718 26

5.00 6.00 148.4 96

Of course, “reasonable” is a very subjective term. However, if we choose x < 0.1, we

ensure that the error is less than 1%.

CHAPTER FIVE SOLUTIONS

2. iA, vB “on”, vC = 0: ix = 20 A

iA, vC “on”, vB = 0: ix = -5 A

iA, vB, vC “on” : ix = 12 A

so, we can write ix’ + ix” + ix”’ = 12

ix’ + ix” = 20

ix’ + ix”’ = - 5

In matrix form,













=













′′′

′′

′













5-

20

12

1 0 1

0 1 1

1 1 1

x

i

(a) with iA on only, the response ix = ix’ = 3 A.

(b) with vB on only, the response ix = ix” = 17 A.

(c) with vC on only, the response ix = ix”’ = -8 A.

(d) iA and vC doubled, vB reversed: 2(3) + 2(-8) + (-1)(17) = -27 A.

CHAPTER FIVE SOLUTIONS

3. One source at a time:

The contribution from the 24-V source may be found by shorting the 45-V source and

open-circuiting the 2-A source. Applying voltage division,

vx’ = V 10

182010

20

24

30||452010

20

24 =

++

=

++

We find the contribution of the 2-A source by shorting both voltage sources and

applying current division:

vx” = V 8.333

182010

10

220 =













++

Finally, the contribution from the 45-V source is found by open-circuiting the 2-A

source and shorting the 24-V source. Defining v30 across the 30-Ω resistor with the

“+” reference on top:

0 = v30/ 20 + v30/ (10 + 20) + (v30 – 45)/ 45

solving, v30 = 11.25 V and hence vx”’ = -11.25(20)/(10 + 20) = -7.5 V

Adding the individual contributions, we find that vx = vx’ + vx” + vx”’ = 10.83 V.

CHAPTER FIVE SOLUTIONS

4. The contribution of the 8-A source is found by shorting out the two voltage sources

and employing simple current division:

i3' = A 5-

3050

50

8=

+

−

The contribution of the voltage sources may be found collectively or individually. The

contribution of the 100-V source is found by open-circuiting the 8-A source and

shorting the 60-V source. Then,

i3" = A 6.25

30||60||)3050(

100 =

+

The contribution of the 60-V source is found in a similar way as i3"' = -60/30 = -2 A.

The total response is i3 = i3' + i3" + i3"' = -750 mA.

CHAPTER FIVE SOLUTIONS

5. (a) By current division, the contribution of the 1-A source i2’ is

i2’ = 1 (200)/ 250 = 800 mA.

The contribution of the 100-V source is i2” = 100/ 250 = 400 mA.

The contribution of the 0.5-A source is found by current division once the 1-A source

is open-circuited and the voltage source is shorted. Thus,

i2”’ = 0.5 (50)/ 250 = 100 mA

Thus, i2 = i2’ + i2” + i2”’ = 1.3 A

(b) P1A = (1) [(200)(1 – 1.3)] = 60 W

P200 = (1 – 1.3)2 (200) = 18 W

P100V = -(1.3)(100) = -130 W

P50 = (1.3 – 0.5)2 (50) = 32 W

P0.5A = (0.5) [(50)(1.3 – 0.5)] = 20 W

Check: 60 + 18 + 32 + 20 = +130.

CHAPTER FIVE SOLUTIONS

6. We find the contribution of the 4-A source by shorting out the 100-V source and

analysing the resulting circuit:

4 = V1' / 20 + (V1' – V')/ 10 [1]

0.4 i1' = V1'/ 30 + (V' – V1')/ 10 [2]

where i1' = V1'/ 20

Simplifying & collecting terms, we obtain 30 V1' – 20 V' = 800 [1]

-7.2 V1' + 8 V' = 0 [2]

Solving, we find that V' = 60 V. Proceeding to the contribution of the 60-V source, we

analyse the following circuit after defining a clockwise mesh current ia flowing in the

left mesh and a clockwise mesh current ib flowing in the right mesh.

30 ia – 60 + 30 ia – 30 ib = 0 [1]

ib = -0.4 i1" = +0.4 ia [2]

Solving, we find that ia = 1.25 A and so V" = 30(ia – ib) = 22.5 V.

Thus, V = V' + V" = 82.5 V.

'

"

'

"

CHAPTER FIVE SOLUTIONS

7. (a) Linearity allows us to consider this by viewing each source as being scaled by

25/ 10. This means that the response (v3) will be scaled by the same factor:

25 iA'/ 10 + 25 iB'/ 10 = 25 v3'/ 10

∴ v3 = 25v3'/ 10 = 25(80)/ 10 = 200 V

(b) iA' = 10 A, iB' = 25 A → v4' = 100 V

iA" = 10 A, iB" = 25 A → v4" = -50 V

iA = 20 A, iB = -10 A → v4 = ?

We can view this in a somewhat abstract form: the currents iA and iB multiply

the same circuit parameters regardless of their value; the result is v4.

Writing in matrix form, 











=

























50-

100

b

a

10 25

25 10 , we can solve to find

a = -4.286 and b = 5.714, so that 20a – 10b leads to v4 = -142.9 V

CHAPTER FIVE SOLUTIONS

8. With the current source open-circuited and the 7-V source shorted, we are left with

100k || (22k + 4.7k) = 21.07 kΩ.

Thus, V3V = 3 (21.07)/ (21.07 + 47) = 0.9286 V.

In a similar fashion, we find that the contribution of the 7-V source is:

V

7V = 7 (31.97) / (31.97 + 26.7) = 3.814 V

Finally, the contribution of the current source to the voltage V across it is:

V

5mA = (5×10-3) ( 47k || 100k || 26.7k) = 72.75 V.

Adding, we find that V = 0.9286 + 3.814 + 72.75 = 77.49 V.

CHAPTER FIVE SOLUTIONS

9. We must find the current through the 500-kΩ resistor using superposition, and then

calculate the dissipated power.

The contribution from the current source may be calculated by first noting that

1M || 2.7M || 5M = 636.8 kΩ. Then,

i60µA = A 43.51

0.636830.5

3

1060 6

µ

=













++

×−

The contribution from the voltage source is found by first noting that 2.7M || 5M =

1.753 MΩ. The total current flowing from the voltage source (with the current source

open-circuited) is –1.5/ (3.5 || 1.753 + 1) µA = -0.6919 µA. The current flowing

through the 500-kΩ resistor due to the voltage source acting alone is then

i1.5V = 0.6919 (1.753)/ (1.753 + 3.5) mA = 230.9 nA.

The total current through the 500-kΩ resistor is then i60µA + i1.5V = 43.74 µA and the

dissipated power is (43.74×10-9)2 (500×103) = 956.6 µW.

CHAPTER FIVE SOLUTIONS

10. We first determine the contribution of the voltage source:

Via mesh analysis, we write: 5 = 18000 I1' – 17000 Ix'

-6 I

x' = -17000 Ix' + 39000 Ix'

Solving, we find I1' = 472.1 mA and Ix' = 205.8 mA, so V' = 17×103 (I1' - Ix')

= 4.527 V. We proceed to find the contribution of the current source:

Via supernode: -20×10-3 = Vx"/ 22×103 + V"/ 0.9444×103 [1]

and V" – Vx" = 6Ix" or V" – Vx" = 6 Vx"/ 22×103 [2]

Solving, we find that V" = -18.11 V. Thus, V = V' + V" = -13.58 V.

The maximum power is V2/ 17×103 = V2/ 17 mW = 250 mW, so

V = 13.58 - V 65.19 )250)(17( ′

== . Solving, we find max

V′= 78.77 V.

The 5-V source may then be increased by a factor of 78.77/ 4.527, so that its

maximum positive value is 87 V; past this value, and the resistor will overheat.

'

↑

I1'

'

Vx"

V"

CHAPTER FIVE SOLUTIONS

11. It is impossible to identify the individual contribution of each source to the power

dissipated in the resistor; superposition cannot be used for such a purpose.

Simplifying the circuit, we may at least determine the total power dissipated in the

resistor:

Via superposition in one step, we may write

i = mA 195.1

2.12

2.1

2 -

1.22

5=

++

Thus,

P2Ω = i2 . 2 = 76.15 mW

i →

CHAPTER FIVE SOLUTIONS

12. We will analyse this circuit by first considering the combined effect of both dc

sources (left), and then finding the effect of the single ac source acting alone (right).

1, 3 supernode: V1/ 100 + V1/ 17×103 + (V1 – 15)/ 33×103 + V3/ 103 = 20 IB [1]

and: V1 – V3 = 0.7 [2]

Node 2: -20 IB = (V2 – 15)/ 1000 [3]

We require one additional equation if we wish to have IB as an unknown:

20 IB + IB = V3/ 1000 [4]

Simplifying and collecting terms,

10.08912 V1 + V3 – 20×103 IB = 0.4545 [1]

V

1 - V3 = 0.7 [2]

V2 + 20×103 IB = 15 [3]

-V3 + 21×103 IB = 0 [4]

Solving, we find that IB = -31.04 µA.

To analyse the right-hand circuit, we first find the Thévenin equivalent to the left of

the wire marked iB', noting that the 33-kΩ and 17-kΩ resistors are now in parallel. We

find that VTH = 16.85 cos 6t V by voltage division, and RTH = 100 || 17k || 33k =

99.12 Ω. We may now proceed:

20 iB' = vx' / 1000 + (vx' – 16.85 cos 6t)/ 99.12 [1]

20 iB' + iB'' = vx'/ 1000 [2]

Solving, we find that iB' = 798.6 cos 6t mA. Thus, adding our two results, we find the

complete current is

iB = iB' + IB = -31.04 + 798.6 cos 6t µA.

IB

V1

V2

V3 vx'

CHAPTER FIVE SOLUTIONS

13.

We first consider the effect of the 2-A source separately, using the left circuit:

Vx' = V 1.765

143

3

2 5 =













+

Next we consider the effect of the 6-A source on its own using the right circuit:

Vx" = V 15.88

89

9

6 5 =













+

Thus, Vx = Vx' + Vx" = 17.65 V.

(b) PSpice verification (DC Sweep)

The DC

sweep

results

below

confirm

that Vx' =

1.765 V

CHAPTER FIVE SOLUTIONS

14.

(a) Beginning with the circuit on the left, we find the contribution of the 2-V source to

Vx:

50

2V

100

V

V4 xx

x

−

′

+

′

=

′

−

which leads to Vx' = 9.926 mV.

The circuit on the right yields the contribution of the 6-A source to Vx:

50

V

100

V

V4 xx

x

′′

+

′′

=

′′

−

which leads to Vx" = 0.

Thus, Vx = Vx' + Vx" = 9.926 mV.

(b) PSpice verification.

'

As can be seen from the two

separate PSpice simulations,

our hand calculations are

correct; the pV-scale voltage in

the second simulation is a result

of numerical inaccuracy.

CHAPTER FIVE SOLUTIONS

15.

V 2.455 V so

0

2

15V

3

V

1

12V

x

xxx

=

+

++

−

V 0.5455 V so

0

2

10V

3

V

1

6V

x

xxx

=

′

=

+

′

+

′

+

−

′

V 1.909 V so

0

2

5V

3

V

1

6V

x

xxx

=

′′

=

+

′′

+

′′

+

−

′′

Adding, we find that Vx' + Vx" = 2.455 V = Vx as promised.

CHAPTER FIVE SOLUTIONS

16. (a) [120 cos 400t] / 60 = 2 cos 400t A. 60 || 120 = 40 Ω.

[2 cos 400t] (40) = 80 cos 400t V. 40 + 10 = 50 Ω.

[80 cos 400t]/ 50 = 1.6 cos 400t A. 50 || 50 = 25 Ω.

(b) 2k || 3k + 6k = 7.2 kΩ. 7.2k || 12k = 4.5 kΩ

(20)(4.5) = 90 V.

25 Ω

1.6 cos 400t A

4.5 kΩ

3.5 kΩ

8 kΩ

CHAPTER FIVE SOLUTIONS

17. We can ignore the 1-kΩ resistor, at least when performing a source transformation on

this circuit, as the 1-mA source will pump 1 mA through whatever value resistor we

place there. So, we need only combine the 1 and 2 mA sources (which are in parallel

once we replace the 1-kΩ resistor with a 0-Ω resistor). The current through the 5.8-

kΩ resistor is then simply given by voltage division:

mA 1.343

5.84.7

4.7

103 3- =

+

×=i

The power dissipated by the 5.8-kΩ resistor is then i2 . 5.8×103 = 10.46 mW.

CHAPTER FIVE SOLUTIONS

18. We may ignore the 10-kΩ and 9.7-kΩ resistors, as 3-V will appear across them

regardless of their value. Performing a quick source transformation on the 10-kΩ

resistor/ 4-mA current source combination, we replace them with a 40-V source in

series with a 10-kΩ resistor:

I = 43/ 15.8 mA = 2.722 mA. Therefore, P5.8Ω = I2. 5.8×103 = 42.97 mW.

Ω

I →

CHAPTER FIVE SOLUTIONS

19. (100 kΩ)(6 mA) = 0.6 V

470 k || 300 k = 183.1 kΩ

(-3 – 0.6)/ 300×103 = -12 µA

(183.1 kΩ)(-12 µA) = -2.197 V

Solving, 9 + 1183.1×103 I – 2.197 = 0, so I = -5.750 µA. Thus,

P1MΩ = I2 . 106 = 33.06 µW.

0.6 V

183.1

-2.197 V

← I

CHAPTER FIVE SOLUTIONS

20. (1)(47) = 47 V. (20)(10) = 200 V. Each voltage source “+” corresponds to its

corresponding current source’s arrow head.

Using KVL on the simplified circuit above,

47 + 47×103 I1 – 4 I1 + 13.3×103 I1 + 200 = 0

Solving, we find that I1 = -247/ (60.3×103 – 4) = -4.096 mA.

CHAPTER FIVE SOLUTIONS

21. (2 V1)(17) = 34 V1

Analysing the simplified circuit above,

34 V1 – 0.6 + 7 I + 2 I + 17 I = 0 [1] and V1 = 2 I [2]

Substituting, we find that I = 0.6/ (68 + 7 + 2 + 17) = 6.383 mA. Thus,

V1 = 2 I = 12.77 mV

34

← I

CHAPTER FIVE SOLUTIONS

22. 12/ 9000 = 1.333 mA. 9k || 7k = 3.938 kΩ. → (1.333 mA)(3.938 kΩ) = 5.249 V.

5.249/ 473.938×103 = 11.08 µA

473.9 k || 10 k = 9.793 kΩ. (11.08 mA)(9.793 kΩ) = 0.1085 V

I

x = 0.1085/ 28.793×103 = 3.768 µA.

5.249 V

3.938

473.9 kΩ 10 kΩ

11.08 µA

0.1085 V 17 kΩ

11.793 kΩ

CHAPTER FIVE SOLUTIONS

23. First, (-7 µA)(2 MΩ) = -14 V, “+” reference down. 2 MΩ + 4 MΩ = 6 MΩ.

+14 V/ 6 ΜΩ = 2.333 µA, arrow pointing up; 6 M || 10 M = 3.75 MΩ.

(2.333)(3.75) = 8.749 V. Req = 6.75 MΩ

∴ Ix = 8.749/ (6.75 + 4.7) µA = 764.1 nA.

2.333

3.75 MΩ

CHAPTER FIVE SOLUTIONS

24. To begin, note that (1 mA)(9 Ω) = 9 mV, and 5 || 4 = 2.222 Ω.

The above circuit may not be further simplified using only source transformation

techniques.

9

15

2.222 Ω

CHAPTER FIVE SOLUTIONS

25. Label the “-” terminal of the 9-V source node x and the other terminal node x'. The

9-V source will force the voltage across these two terminals to be –9 V regardless of

the value of the current source and resistor to its left. These two components may

therefore be neglected from the perspective of terminals a & b. Thus, we may draw:

7.25 A 2 Ω

CHAPTER FIVE SOLUTIONS

26. Beware of the temptation to employ superposition to compute the dissipated power- it

won’t work!

Instead, define a current I flowing into the bottom terminal of the 1-MΩ resistor.

Using superposition to compute this current,

I = 1.8/ 1.840 + 0 + 0 µA = 978.3 nA.

Thus,

P1MΩ = (978.3×10-9)2 (106) = 957.1 nW.

CHAPTER FIVE SOLUTIONS

27. Let’s begin by plotting the experimental results, along with a least-squares fit to part

of the data:

We see from the figure that we cannot draw a very good line through all data points

representing currents from 1 mA to 20 mA. We have therefore chosen to perform a

linear fit for the three lower voltages only, as shown. Our model will not be as

accurate at 1 mA; there is no way to know if our model will be accurate at 20 mA,

since that is beyond the range of the experimental data.

Modeling this system as an ideal voltage source in series with a resistance

(representing the internal resistance of the battery) and a varying load resistance, we

may write the following two equations based on the linear fit to the data:

1.567 = Vsrc – Rs (1.6681×10-3)

1.558 = Vsrc – Rs (12.763×10-3)

Solving, Vsrc = 1.568 V and Rs = 811.2 mΩ. It should be noted that depending on the

line fit to the experimental data, these values can change somewhat, particularly the

series resistance value.

Least-squares fit results:

Voltage (V) Current (mA)

1.567 1.6681

1.563 6.599

1.558 12.763

CHAPTER FIVE SOLUTIONS

28. Let’s begin by plotting the experimental results, along with a least-squares fit to part

of the data:

We see from the figure that we cannot draw a very good line through all data points

representing currents from 1 mA to 20 mA. We have therefore chosen to perform a

linear fit for the three lower voltages only, as shown. Our model will not be as

accurate at 1 mA; there is no way to know if our model will be accurate at 20 mA,

since that is beyond the range of the experimental data.

Modeling this system as an ideal current source in parallel with a resistance Rp

(representing the internal resistance of the battery) and a varying load resistance, we

may write the following two equations based on the linear fit to the data:

1.6681×10-3 = Isrc – 1.567/ Rp

12.763×10-3 = Isrc – 1.558/ Rp

Solving, Isrc = 1.933 A and Rs = 811.2 mΩ. It should be noted that depending on the

line fit to the experimental data, these values can change somewhat, particularly the

series resistance value.

Least-squares fit results:

Voltage (V) Current (mA)

1.567 1.6681

1.563 6.599

1.558 12.763

CHAPTER FIVE SOLUTIONS

29. Reference terminals are required to avoid ambiguity: depending on the sources with

which we begin the transformation process, we will obtain entirely different answers.

Working from left to right in this case,

2 µA – 1.8 µA = 200 nA, arrow up.

1.4 MΩ + 2.7 MΩ = 4.1 MΩ

An additional transformation back to a voltage source yields (200 nA)(4.1 MΩ) =

0.82 V in series with 4.1 MΩ + 2 MΩ = 6.1 MΩ, as shown below:

Then, 0.82 V/ 6.1 MΩ = 134.4 nA, arrow up.

6.1 MΩ || 3 MΩ = 2.011 MΩ

4.1 µA + 134.4 nA = 4.234 mA, arrow up.

(4.234 µA) (2.011 MΩ) = 8.515 V.

±

6.1

0.82 V

±

8.515 V

2.011

CHAPTER FIVE SOLUTIONS

30. To begin, we note that the 5-V and 2-V sources are in series:

Next, noting that 3 V/ 1 Ω = 3 A, and 4 A – 3 A = +1 A (arrow down), we obtain:

By voltage division, the voltage across the

5-Ω resistor in the circuit to the right is:

(-1) 2 5||2

5||2

+ = -0.4167 V.

Thus, the power dissipated by the 5-Ω resistor is (-0.4167)2 / 5 = 34.73 mW.

3

The left-hand resistor and the current

source are easily transformed into a

1-V source in series with a 1-Ω

resistor:

±

-1 V

CHAPTER FIVE SOLUTIONS

31. (a) RTH = 25 || (10 + 15) = 25 || 25 = 12.5 Ω.

V

TH = Vab = 











++

+













++ 251015

1015

100

251510

25

50 = 75 V.

(b) If Rab = 50 Ω,

P

50Ω = W72

50

1

12.550

50

75

2

=



































+

(c) If Rab = 12.5 Ω,

P

12.5Ω = W112.5

12.5

1

12.55.21

12.5

75

2

=



































+

CHAPTER FIVE SOLUTIONS

32. (a) Removing terminal c, we need write only one nodal equation:

0.1 = 15

5V

12

2V bb −

+

−, which may be solved to

yield Vb = 4 V. Therefore, Vab = VTH = 2 – 4

= -2 V.

RTH = 12 || 15 = 6.667 Ω. We may then

calculate IN as IN = VTH/ RTH

= -300 mA (arrow pointing upwards).

(b) Removing terminal a, we again find RTH = 6.667 Ω, and only need write a single

nodal equation; in fact, it is identical to that written for the circuit above, and we once

again find that Vb = 4 V. In this case, VTH = Vbc = 4 – 5 = -1 V, so IN = -1/ 6.667

= -150 mA (arrow pointing upwards).

CHAPTER FIVE SOLUTIONS

33. (a) Shorting out the 88-V source and open-circuiting the 1-A source, we see looking

into the terminals x and x' a 50-Ω resistor in parallel with 10 Ω in parallel with

(20 Ω + 40 Ω), so

RTH = 50 || 10 || (20 + 40) = 7.317 Ω

Using superposition to determine the voltage Vxx' across the 50-Ω resistor, we find

Vxx' = VTH = 











++

+













++

+

)10||50(2040

40

10)||(1)(50

)]4020(||50[10

)4020(||50

88

=











++

+













333.82040

40

(1)(8.333)

27.37

27.27

88 = 69.27 V

(b) Shorting out the 88-V source and open-circuiting the 1-A source, we see looking

into the terminals y and y' a 40-Ω resistor in parallel with [20 Ω + (10 Ω || 50 Ω)]:

RTH = 40 || [20 + (10 || 50)] = 16.59 Ω

Using superposition to determine the voltage Vyy' across the 1-A source, we find

Vyy' = VTH = (1)(RTH) + 











+













+4020

40

27.2710

27.27

88

= 59.52 V

CHAPTER FIVE SOLUTIONS

34. (a) Select terminal b as the reference terminal, and define a nodal voltage V1 at the

top of the 200-Ω resistor. Then,

0 = 200

V

100

VV

40

20V 1TH11 +

−

+

− [1]

1.5 i1 = (VTH – V1)/ 100 [2]

where i1 = V1/ 200, so Eq. [2] becomes 150 V1/ 200 + V1 - VTH = 0 [2]

Simplifying and collecting terms, these equations may be re-written as:

(0.25 + 0.1 + 0.05) V1 – 0.1 VTH = 5 [1]

(1 + 15/ 20) V1 – VTH = 0 [2]

Solving, we find that VTH = 38.89 V. To find RTH, we short the voltage source and

inject 1 A into the port:

0 = 200

V

40

V

100

VV 11in1++

− [1]

1.5 i1 + 1 = 100

VV 1in − [2]

i1 = V1/ 200 [3]

Combining Eqs. [2] and [3] yields 1.75 V1 – Vin = -100 [4]

Solving Eqs. [1] & [4] then results in Vin = 177.8 V, so that RTH = Vin/ 1 A = 177.8 Ω.

(b) Adding a 100-Ω load to the original circuit or our Thévenin equivalent, the

voltage across the load is

V100Ω = V 14.00

177.8100

100

VTH =













+, and so P100Ω = (V100Ω)2 / 100 = 1.96 W.

1 A

+

Vin

-

V1

Ref.

CHAPTER FIVE SOLUTIONS

35. We inject a current of 1 A into the port (arrow pointing up), select the bottom terminal

as our reference terminal, and define the nodal voltage Vx across the 200-Ω resistor.

Then, 1 = V1/ 100 + (V1 – Vx)/ 50 [1]

-0.1 V1 = Vx/ 200 + (Vx – V1)/ 50 [2]

which may be simplified to

3 V1 – 2 Vx = 100 [1]

16 V1 + 5 Vx = 0 [2]

Solving, we find that V1 = 10.64 V, so RTH = V1/ (1 A) = 10.64 Ω.

Since there are no independent sources present in the original network, IN = 0.

CHAPTER FIVE SOLUTIONS

36. With no independent sources present, VTH = 0.

We decide to inject a 1-A current into the port:

Node ‘x’: 0.01 vab = vx/ 200 + (vx – vf)/ 50 [1]

Supernode: 1 = vab/ 100 + (vf – vx) [2]

and: vab – vf = 0.2 vab [3]

Rearranging and collecting terms,

-2 vab + 5 vx – 4 vf = 0 [1]

vab – 2 vx + 2 vf = 100 [2]

0.8 vab - vf = 0 [3]

Solving, we find that vab = 192.3 V, so RTH = vab/ (1 A) = 192.3 Ω.

Ref.

vx

vf

•

A

CHAPTER FIVE SOLUTIONS

37. We first find RTH by shorting out the voltage source and open-circuiting the current

source.

Looking into the terminals a & b, we see

RTH = 10 || [47 + (100 || 12)]

= 8.523 Ω.

Returning to the original circuit, we decide to perform nodal analysis to obtain VTH:

-12×103 = (V1 – 12)/ 100×103 + V1/ 12×103 + (V1 – VTH)/ 47×103 [1]

12×103 = VTH/ 10×103 + (VTH – V1)/ 47×103 [2]

Rearranging and collecting terms,

0.1146 V1 - 0.02128 VTH = -11.88 [1]

-0.02128 V1 + 0.02128 VTH = 12 [2]

Solving, we find that VTH = 83.48 V.

CHAPTER FIVE SOLUTIONS

38. (a) RTH = 4 + 2 || 2 + 10 = 15 Ω.

(b) same as above: 15 Ω.

CHAPTER FIVE SOLUTIONS

39. For Fig. 5.78a, IN = 12/ ~0 → ∞ A in parallel with ~ 0 Ω.

For Fig. 5.78b, VTH = (2)(~∞) → ∞ V in series with ~∞ Ω.

CHAPTER FIVE SOLUTIONS

40. With no independent sources present, VTH = 0.

Connecting a 1-V source to the port and measuring the current that flows as a result,

I = 0.5 Vx + 0.25 Vx = 0.5 + 0.25 = 0.75 A.

R

TH = 1/ I = 1.333 Ω.

The Norton equivalent is 0 A in parallel with 1.333 Ω.

+

-

← I

CHAPTER FIVE SOLUTIONS

41. Performing nodal analysis to determine VTH,

100×10-3 = Vx/ 250 + Voc/ 7.5×103 [1]

and Vx – Voc = 5 ix

where ix = Vx/ 250. Thus, we may write the second equation as

0.98 Vx – Voc = 0 [2]

Solving, we find that Voc = VTH = 23.72 V.

In order to determine RTH, we inject 1 A into the port:

V

ab/ 7.5×103 + Vx/ 250 = 1 [1]

and Vx – Vab = 5 ix = 5Vx/ 250 or

-V

ab + (1 – 5/ 250) Vx = 0 [2]

Solving, we find that Vab = 237.2 V. Since RTH = Vab/ (1 A), RTH = 237.2 Ω.

Finally, IN = VTH/ RTH = 100 mA.

CHAPTER FIVE SOLUTIONS

42. We first note that VTH = Vx, so performing nodal analysis,

-5 V

x = Vx/ 19 which has the solution Vx = 0 V.

Thus, VTH (and hence IN) = 0. (Assuming RTH ≠ 0)

To find RTH, we inject 1 A into the port, noting that RTH = Vx/ 1 A:

-5 Vx + 1 = Vx/ 19

Solving, we find that Vx = 197.9 mV, so that RTH = RN = 197.9 mV.

CHAPTER FIVE SOLUTIONS

43. Shorting out the voltage source, we redraw the circuit with a 1-A source in place of

the 2-kΩ resistor:

Noting that 300 Ω || 2 MΩ ≈ 300 Ω,

0 = (vgs – V)/ 300 [1]

1 – 0.02 vgs = V/ 1000 + (V – vgs)/ 300 [2]

Simplifying & collecting terms,

vgs – V = 0 [1]

0.01667 vgs + 0.00433 V = 1 [2]

Solving, we find that vgs = V = 47.62 V. Hence, RTH = V/ 1 A = 47.62 Ω.

1 A

Ref.

V

CHAPTER FIVE SOLUTIONS

44. We replace the source vs and the 300-Ω resistor with a 1-A source and seek its

voltage:

By nodal analysis, 1 = V1/ 2×106 so V1 = 2×106 V.

Since V = V1, we have Rin = V/ 1 A = 2 MΩ.

1 A

+

V

-

Ref.

V1 V2

CHAPTER FIVE SOLUTIONS

45. Removing the voltage source and the 300-Ω resistor, we replace them with a 1-A

source and seek the voltage that develops across its terminals:

We select the bottom node as our reference terminal, and define nodal voltages V1

and V2. Then,

1 = V1 / 2×106 + (V1 – V2)/ rπ [1]

0.02 vπ = (V2 – V1)/ rπ + V2/ 1000 + V2/ 2000 [2]

where vπ = V1 – V2

Simplifying & collecting terms,

(2×106 + rπ) V1 – 2×106 V2 = 2×106 rπ [1]

-(2000 + 40 rπ) V1 + (2000 + 43 rπ) V2 = 0 [2]

Solving, we find that V1 = V = 













++×

+

×

π

r 14.33 666.7 102

r 14.33 666.7

102 6

6.

Thus, RTH = 2×106 || (666.7 + 14.33 rπ) Ω.

1 A

+

V

- Ref.

CHAPTER FIVE SOLUTIONS

46. Such a scheme probably would lead to maximum or at least near-maximum power

transfer to our home. Since we pay the utility company based on the power we use,

however, this might not be such a hot idea…

CHAPTER FIVE SOLUTIONS

47. We need to find the Thévenin equivalent resistance of the circuit connected to RL, so

we short the 20-V source and open-circuit the 2-A source; by inspection, then

RTH = 12 || 8 + 5 + 6 = 15.8 Ω

Analyzing the original circuit to obtain V1 and V2 with RL removed:

V

1 = 20 8/ 20 = 8 V; V2 = -2 (6) = -12 V.

We define VTH = V1 – V2 = 8 + 12 = 20 V. Then,

P

RL|max = W6.329

4(15.8)

400

R 4

V

L

2

TH ==

+ VTH -

V1 V2

Ref.

CHAPTER FIVE SOLUTIONS

48. (a) RTH = 25 || (10 + 15) = 12.5 Ω

Using superposition, Vab = VTH = 50

1015

100

251015

25

50 +

+

++ = 75 V.

(b) Connecting a 50-Ω resistor,

P

load = W90

50 12.5

75

R R

V2

loadTH

2

TH =

+

=

+

(c) Connecting a 12.5-Ω resistor,

P

load =

()

W112.5

12.5 4

75

R 4

V2

TH

2

TH ==

CHAPTER FIVE SOLUTIONS

49. (a) By inspection, we see that i10 = 5 A, so

VTH = Vab = 2(0) + 3 i10 + 10 i10 = 13 i10 = 13(5) = 65 V.

To find RTH, we connect a 1-A source between terminals a & b:

5 = V1/ 10 + (V2 – V)/ 2 [1] → V

1 + 5 V2 – 5 V = 50 [1]

1 = (V – V2)/ 2 [2] → -V2 + V = 2 [2]

and V2 – V1 = 3 i10 [3]

where i10 = V1/ 10 → -13 V1 + 10 V2 = 0 [3]

Solving, we find that V = 80 V, so that RTH = V / 1 A = 80 Ω.

(b) Pmax = W13.20

4(80)

65

R 4

V2

TH

2

TH ==

Ref.

V1 V2 V

•

CHAPTER FIVE SOLUTIONS

50.

(a) Replacing the resistor RL with a 1-A source, we seek the voltage that develops across

its terminals with the independent voltage source shorted:

11 1

11

1

10 20 40 0 [1] 30 20 0 [1]

and 1 [2] 1 [2]

Solving, 400mA

So 40 16V and 16

1A

−+ + = ⇒ + =

−= ⇒ −=

=

== ==Ω

xx

TH

iii ii

ii ii

i

V

Vi R

(b) Removing the resistor RL from the original circuit, we seek the resulting open-circuit

voltage:

1

10 50

0[1]

20 40

50

where 40

150 50

so [1] becomes 0 20 2 40 40

50

020 80

04 50

550

or 10V

−−

=+

−

=

−−



=− +





−

=+

=+−

=

TH TH

TH

TH TH TH

TH TH

TH

ViV

V

i

VV V

VV

V

Thus, if

16 ,

5V

2

==Ω

===

+

L

LTH

RTH

LTH

RR

RV

VV

RR

+

VTH

-

CHAPTER FIVE SOLUTIONS

51.

(a) 2.5A=

N

I

2

20 80

2A

=

i

By current division,

2 2.5 20

=+

N

R

Solving, 80==Ω

NTH

RR

Thus, 2.5 80 200V==×=

TH OC

VV

(b)

22

max

200 125W

4480

== =

×

TH

V

PR

(c) 80==Ω

LTH

RR

RN 20 Ω

↓

2 A

CHAPTER FIVE SOLUTIONS

52.

By Voltage ÷, =+N

RN

N

R

II

RR

So

Solving, 1.7 A=

N

I and 33.33=Ω

N

R

(a) If

(b) If

(c) If

is a maximum,

33.33

1.7 850mA

33.33 33.33

33.33 28.33V

== Ω

=× =

+

==

LL

LN

L

LL

vi

RR

i

vi

is a maximum

; max when 0

So 1.7A

0V

==Ω

+

=

L

N

LN L

NL

L

i

R

ii R

RR

i

v

0.2 [1]

250

0.5 [2]

80

=+

N

R

IR

R

IR

10W to 250 corresp to 200mA.

20W to 80 corresp to 500mA.

Ω

RN R

IN

is a maximum

()

So is a maximum when is a maximum, which occurs at .

Then 0 and 1.7 56.66V

=

=∞

==×=

L

LNNL

LNL L

LLN

v

VIRR

vRR R

ivR

CHAPTER FIVE SOLUTIONS

53. There is no conflict with our derivation concerning maximum power. While a dead

short across the battery terminals will indeed result in maximum current draw from

the battery, and power is indeed proportional to i2, the power delivered to the load is

i2RLOAD = i2(0) = 0 watts. This is the minimum, not the maximum, power that the

battery can deliver to a load.

CHAPTER FIVE SOLUTIONS

54. Remove RE: =

TH E in

RRR

bottom node: 3

3

1 3 10 [1]

300 70 10

−ππ

π

−−

−× = + ×

Vv Vv

v

at other node: 33

010 10 300 70 10

ππ π

−−

=++

××

vvVvV

[2]

Simplifying and collecting terms,

210×105 = 70×103 V + 300V + 63000 vπ – 70x103 vπ - 300 vπ

35

33

or 70.3 10 7300 210 10 [1]

0 2100 70 10 70 10 300 300

or 69.7 10 72.4 10 0 [2]

solving, 331.9 So 331.9

π

ππ π

π

×− =×

=+×−×+−

−× +× =

==Ω

TH E

Vv

vvVvV

Vv

VVRR

Next, we determine vTH using mesh analysis:

33

12

70.3 10 70 10 0 [1]−+ × − × =

s

vii

33

213

80 10 70 10 0 [2]×−×+=

E

iiRi

and: 3

32

310 [3]

−π

−=×ii v

or 33

32 2

3 10 (10 10 )

−

−=× ×ii i

or 32 2

30−=ii i

or

23

31 0 [3]−+=ii

Solving :

33

1

33

2

3

70.3 10 70 10 0

70 10 80 10 0

03110



×−×  



 

−× × =



 



 

− 



s

E

iv

Ri

i

We seek i3:

3

363

21.7 10

7.24 10 21.79 10

−×

=×+ ×

s

E

v

iR

So 3

363

21.7 10

7.24 10 21.79 10

−×

== = ×+ ×

E

OC TH E s

E

R

VVRi v

R

22

32

82

63

62

2

632

21.7 10 8

88 7.24 10 21.79 10 331.9

331.9

11.35 10 (331.9 )

(7.24 10 21.79 10 )

Ω

 

−×

==

 

+×+×



 



+



×+

=×+ ×

TH E

TH E E

E

VRvs

PRR

R

Rvs

R

This is maximized by setting .=∞

E

R

V

1 A

CHAPTER FIVE SOLUTIONS

55. Thévenize the left-hand network, assigning the nodal voltage Vx at the free end of

right-most 1-kΩ resistor.

A single nodal equation: 3

3

40 10 710

−

×=

×

xoc

V

So 280V==

TH x oc

VV

R

TH = 1 k + 7 k = 8 kΩ

Select R1 = RTH = 8 kΩ.

CHAPTER FIVE SOLUTIONS

56.

6

65

1

56

26

66

36

1 850 0.1 851.1 10

10 10 117.5

10 850 10 99.87

851.1 10

850 10 10 998.7

851.1 10

=++=+ += ×

×

== =Ω

××

== = Ω

×

××

== = Ω

×

ABC

AB

BC

CA

DR R R

RR

RDD

RR

Rk

D

RR

Rk

D

1 MΩ

100 kΩ

850 MΩ

CHAPTER FIVE SOLUTIONS

57.

12 23 31

2

3

1

0.1 0.4 0.4 0.9 0.9 0.1

0.49

1.225

544.4m

4.9

=++

=×+×+×

=Ω

== Ω

==Ω

A

B

C

NRR RRRR

N

RR

N

RR

N

RR

R1 R2

R3

CHAPTER FIVE SOLUTIONS

58.

1

2

:16310

61 63 31

0.6, 1.8 , 0.3

10 10 10

:5 1 4 10

51 14 54

0.5 , 0.4 , 2

10 10 10

1.8 2 0.5 4.3

0.3 0.6 0.4 1.3

1.3 4.3 0.9982

0.9982 0.6 2 3.598

3.598 6 2.249

∆++=Ω

×××

===

∆++=Ω

×× ×

===

++ = Ω

++=Ω

=Ω

++= Ω

=Ω

CHAPTER FIVE SOLUTIONS

59.

2

622336 36

36 36 36

6 , 18 , 12

62 3

12 4 3 , 6 12 4

4 3 18 25

18

3 2.16

25

18

4 2.88

25

3

4 0.48

25

9.48 2.16 9.48 2.88 2.88 2.16 54

54 54

18.75 5.696

2.88 9.48

54 25

2.16

75 18.75 15

100 25 20

(15 20) 5.696

×+×+×= Ω

=Ω = Ω = Ω

=Ω Ω=Ω

++ = Ω

×= Ω

×+×+×=Ω

=Ω =Ω

=Ω

+=4.899

5 4.899 9.899

Ω

∴=+ = Ω

in

R

96

CHAPTER FIVE SOLUTIONS

60. We begin by converting the ∆-connected network consisting of the 4-, 6-, and 3-Ω

resistors to an equivalent Y-connected network:

1

2

3

64313

64 1.846

13

43 923.1m

13

36 1.385

13

=++= Ω

×

===Ω

×

=== Ω

×

===Ω

AB

BC

CA

D

RR

RD

RR

RD

RR

RD

Then network becomes:

Then we may write

12 [13.846 (19.385 6.9231)]

7.347

=+

=Ω

in

R

RA RB

RC

1.846 Ω 0.9231 Ω

1.385 Ω

CHAPTER FIVE SOLUTIONS

0.5 0.5

0.25

61.

Next, we convert the Y-connected network on the left to a ∆-connected network:

2

1 0.5 0.5 2 2 1 3.5×+×+×=Ω

3.5 7

0.5

3.5 1.75

2

3.5 3.5

1

==Ω

A

B

C

R

After this procedure, we have a 3.5-Ω resistor in parallel with the 2.5-Ω resistor. Replacing

them with a 1.458-Ω resistor, we may redraw the circuit:

This circuit may be easily analysed to find:

1

2

3

112 4

12 1

42

21 1

42

11 0.25

4

++ = Ω

×

==Ω

×

==Ω

×

==Ω

R

12 1.458 5.454V

1.75 1.458

0.25 1.458 1.75

1.045

×

==

+

=+

=Ω

oc

TH

V

R

7

1.75 0.25

1.458

CHAPTER FIVE SOLUTIONS

62. We begin by converting the Y-network to a ∆-connected network:

2

1.1 1.1 1.1 3

33

1

33

1

33

1

=++=Ω

==Ω

A

B

C

N

R

Next, we note that 1 3 0.75=Ω, and hence have a simple ∆-network. This is easily

converted to a Y-connected network:

1

2

3

0.75 3 3 6.75

0.75 3 0.3333

6.75

33 1.333

6.75

3 0.75 0.3333

6.75

++= Ω

×

==Ω

×

==Ω

×

==Ω

R

1.333 0.3333

1.667

1/3

11/3 1 1/3

11

131 5

0.2A

200mA

=+

=Ω

==×++

==

++

=

N

NSC

R

II

33

3 Ω

1 A

Analysing this final

circuit,

CHAPTER FIVE SOLUTIONS

63. Since 1 V appears across the resistor associated with I1, we know that I1 = 1 V/ 10 Ω

= 100 mA. From the perspective of the open terminals, the 10-Ω resistor in parallel

with the voltage source has no influence if we replace the “dependent” source with a

fixed 0.5-A source:

Then, we may write: -1 + (10 + 10 + 10) ia – 10 (0.5) = 0

so that ia = 200 mA.

We next find that VTH = Vab = 10(-0.5) + 10(ia – 0.5) + 10(-0.5) = -13 V.

To determine RTH, we first recognise that with the 1-V source shorted, I1 = 0 and

hence the dependent current source is dead. Thus, we may write RTH from inspection:

RTH = 10 + 10 + 10 || 20 = 26.67 Ω.

0.5 A

CHAPTER FIVE SOLUTIONS

64. (a) We begin by splitting the 1-kΩ resistor into two 500-Ω resistors in series. We then

have two related Y-connected networks, each with a 500-Ω resistor as a leg.

Converting those networks into ∆-connected networks,

Σ = (17)(10) + (1)(4) + (4)(17) = 89×106 Ω2

89/0.5 = 178 kΩ; 89/ 17 = 5.236 kΩ; 89/4 = 22.25 kΩ

Following this conversion, we find that we have two 5.235 kΩ resistors in parallel,

and a 178-kΩ resistor in parallel with the 4-kΩ resistor. Noting that 5.235 k || 5.235 k

= 2.618 kΩ and 178 k || 4 k = 3.912 kΩ, we may draw the circuit as:

We next attack the Y-connected network in the center:

Σ = (22.25)(22.25) + (22.25)(2.618) + (2.618)(22.25) = 611.6×106 Ω2

611.6/ 22.25 = 27.49 kΩ; 611.6/ 2.618 = 233.6 kΩ

Noting that 178 k || 27.49 k = 23.81 kΩ and 27.49 || 3.912 = 3.425 kΩ, we are left

with a simple ∆-connected network. To convert this to the requested Y-network,

Σ = 23.81 + 233.6 + 3.425 = 260.8 kΩ

(23.81)(233.6)/ 260.8 = 21.33 kΩ

(233.6)(3.425)/ 260.8 = 3.068 kΩ

(3.425)(23.81)/ 260.8 = 312.6 Ω

178 kΩ

233.6 kΩ

27.49 kΩ 27.49 kΩ 3.912 kΩ

312.6 Ω

21.33 kΩ 3.068 kΩ

(b)

CHAPTER FIVE SOLUTIONS

65. (a) Although this network may be simplified, it is not possible to replace it with a

three-resistor equivalent.

(b) See (a).

CHAPTER FIVE SOLUTIONS

66. First, replace network to left of the 0.7-V source with its Thévenin equivalent:

15

20 2.609V

100 15

100 15 13.04

=× =

+

==Ω

TH

V

Rkkk

Redraw:

Analysing the new circuit to find IB, we note that IC = 250 IB:

3

4

2.609 13.04 10 0.7 5000( 250 ) 0

2.609 0.7 1.505 A

13.04 10 251 5000

250 3.764 10 A

376.4 A

−

−+×++ + =

−

==µ

×+×

==×

=µ

BBB

B

CB

III

I

II

2.609 V

13.04 kΩ

CHAPTER FIVE SOLUTIONS

67. (a) Define a nodal voltage V1 at the top of the current source IS, and a nodal voltage

V2 at the top of the load resistor RL. Since the load resistor can safely dissipate 1 W,

and we know that

PRL = 1000

V2

2

then V 31.62 V max

2=. This corresponds to a load resistor (and hence lamp) current

of 32.62 mA, so we may treat the lamp as a 10.6-Ω resistor.

Proceeding with nodal analysis, we may write:

IS = V1/ 200 + (V1 – 5 Vx)/ 200 [1]

0 = V2/ 1000 + (V2 – 5 Vx)/ 10.6 [2]

Vx = V1 – 5 Vx or Vx = V1/ 6 [3]

Substituting Eq. [3] into Eqs. [1] and [2], we find that

7 V1 = 1200 IS [1]

-5000 V1 + 6063.6 V2 = 0 [2]

Substituting V 31.62 V max

2=into Eq. [2] then yields V1 = 38.35 V, so that

IS| max = (7)(38.35)/ 1200 = 223.7 mA.

(b) PSpice verification.

The lamp current does not exceed 36 mA in the range of operation allowed

(i.e. a load power of < 1 W.) The simulation result shows that the load will dissipate

slightly more than 1 W for a source current magnitude of 224 mA, as predicted by

hand analysis.

CHAPTER FIVE SOLUTIONS

68. Short out all but the source operating at 104 rad/s, and define three clockwise mesh

currents i1, i2, and i3 starting with the left-most mesh. Then

608 i1 – 300 i2 = 3.5 cos 104 t [1]

-300 i1 + 316 i2 – 8 i3 = 0 [2]

-8 i2 + 322 i3 = 0 [3]

Solving, we find that i1(t) = 10.84 cos 104 t mA

i2(t) = 10.29 cos 104 t mA

i3(t) = 255.7 cos 104 t µA

Next, short out all but the 7 sin 200t V source, and and define three clockwise mesh

currents ia, ib, and ic starting with the left-most mesh. Then

608 ia – 300 ib = -7 sin 200t [1]

-300 ia + 316 ib – 8 ic = 7 sin 200t [2]

-8 ib + 322 ic = 0 [3]

Solving, we find that ia(t) = -1.084 sin 200t mA

ib(t) = 21.14 sin 200t mA

ic(t) = 525.1 sin 200t µA

Next, short out all but the source operating at 103 rad/s, and define three clockwise

mesh currents iA, iB, and iC starting with the left-most mesh. Then

608 iA – 300 iB = 0 [1]

-300 iA + 316 iB – 8 iC = 0 [2]

-8 iB + 322 iC = -8 cos 104 t [3]

Solving, we find that iA(t) = -584.5 cos 103 t µA

iB(t) = -1.185 cos 103 t mA

iC(t) = -24.87 cos 103 t mA

We may now compute the power delivered to each of the three 8-Ω speakers:

p1 = 8[i1 + ia + iA]2 = 8[10.84×10-3 cos 104 t -1.084×10-3 sin 200t -584.5×10-6 cos 103 t]2

p2 = 8[i2 + ib + iB]2 = 8[10.29×10-3 cos 104 t +21.14×10-3 sin 200t –1.185×10-3 cos 103 t]2

p3 = 8[i3 + ic + iC]2 = 8[255.7×10-6 cos 104 t +525.1×10-6 sin 200t –24.87×10-3 cos 103 t]2

CHAPTER FIVE SOLUTIONS

69. Replacing the DMM with a possible Norton equivalent (a 1-MΩ resistor in parallel

with a 1-A source):

We begin by noting that 33 Ω || 1 MΩ ≈ 33 Ω. Then,

0 = (V1 – Vin)/ 33 + V1/ 275×103 [1]

and

1 - 0.7 V1 = Vin/ 106 + Vin/ 33×103 + (Vin – V1)/ 33 [2]

Simplifying and collecting terms,

(275×103 + 33) V1 - 275×103 Vin = 0 [1]

22.1 V1 + 1.001 Vin = 33 [2]

Solving, we find that Vin = 1.429 V; in other words, the DMM sees 1.429 V across its

terminals in response to the known current of 1 A it’s supplying. It therefore thinks

that it is connected to a resistance of 1.429 Ω.

1 A

Vin

CHAPTER FIVE SOLUTIONS

70. We know that the resistor R is absorbing maximum power. We might be tempted to

say that the resistance of the cylinder is therefore 10 Ω, but this is wrong: The larger

we make the cylinder resistance, the small the power delivery to R:

PR = 10 i2 = 10

2

10

120













+

cylinder

R

Thus, if we are in fact delivering the maximum possible power to the resistor from the

120-V source, the resistance of the cylinder must be zero.

This corresponds to a temperature of absolute zero using the equation given.

CHAPTER FIVE SOLUTIONS

71. We note that the buzzer draws 15 mA at 6 V, so that it may be modeled as a

400-Ω resistor. One possible solution of many, then, is:

Note: construct the 18-V source from 12 1.5-V batteries in series, and the two 400-Ω

resistors can be fabricated by soldering 400 1-Ω resistors in series, although there’s

probably a much better alternative…

CHAPTER FIVE SOLUTIONS

72. To solve this problem, we need to assume that “45 W” is a designation that

applies

when 120 Vac is applied directly to a particular lamp. This corresponds to a current

draw of 375 mA, or a light bulb resistance of 120/ 0.375 = 320 Ω.

Original wiring scheme New wiring scheme

In the original wiring scheme, Lamps 1 & 2 draw (40)2 / 320 = 5 W of power each,

and Lamp 3 draws (80)2 / 320 = 20 W of power. Therefore, none of the lamps is

running at its maximum rating of 45 W. We require a circuit which will deliver the

same intensity after the lamps are reconnected in a ∆ configuration. Thus, we need a

total of 30 W from the new network of lamps.

There are several ways to accomplish this, but the simplest may be to just use one

120-Vac source connected to the left port in series with a resistor whose value is

chosen to obtain 30 W delivered to the three lamps.

In other words,

30

320

213.3Rs

213.3

06

2

320

213.3Rs

213.3

120

22

=













+













+

Solving, we find that we require Rs = 106.65 Ω, as confirmed by the PSpice

simulation below, which shows that both wiring configurations lead to one lamp with

80-V across it, and two lamps with 40 V across each.

3

CHAPTER FIVE SOLUTIONS

73.

• Maximum current rating for the LED is 35 mA.

• Its resistance can vary between 47 and 117 Ω.

• A 9-V battery must be used as a power source.

• Only standard resistance values may be used.

One possible current-limiting scheme is to connect a 9-V battery in series with a

resistor Rlimiting and in series with the LED.

From KVL,

ILED =

LEDlimiting R R

9

+

The maximum value of this current will occur at the minimum LED resistance, 47 Ω.

Thus, we solve

35×10-3 = 47 R

9

limiting +

to obtain Rlimiting ≥ 210.1 Ω to ensure an LED current of less than 35 mA. This is not a

standard resistor value, however, so we select

Rlimiting = 220 Ω.

CHAPTER SIX SOLUTIONS

1. The first step is to perform a simple source transformation, so that a 0.15-V source in

series with a 150-Ω resistor is connected to the inverting pin of the ideal op amp.

Then, vout =

()

V 2.2- 15.0

150

2200

=−

CHAPTER SIX SOLUTIONS

2. In order to deliver 150 mW to the 10-kΩ resistor, we need vout =

V. 38.73 )1010)(15.0( 3=× Writing a nodal equation at the inverting input,

we find

1000

5

R

5

0 out

v−

+=

Using vout = 38.73, we find that R = 148.2 Ω.

CHAPTER SIX SOLUTIONS

3. Since the 670-Ω switch requires 100 mA to activate, the voltage delivered to it by our

op amp circuit must be (670)(0.1) = 67 V. The microphone acts as the input to the

circuit, and provides 0.5 V. Thus, an amplifier circuit having a gain = 67/0.5 = 134 is

required.

One possible solution of many: a non-inverting op amp circuit with the microphone

connected to the non-inverting input terminal, the switch connected between the op

amp output pin and ground, a feedback resistor Rf = 133 Ω, and a resistor R1 = 1 Ω.

CHAPTER SIX SOLUTIONS

4. We begin by labeling the nodal voltages v- and v+ at the inverting and non-inverting

input terminals, respectively. Since no current can flow into the non-inverting input,

no current flows through the 40-kΩ resistor; hence, v+ = 0. Therefore, we know that

v- = 0 as well.

Writing a single nodal equation at the non-inverting input then leads to

22000

) - (

100

) - (

0 out-S- vvvv +=

or

22000

-

100

-

0 outSvv +=

Solving,

vout = -220 vS

CHAPTER SIX SOLUTIONS

5. We first label the nodal voltage at the output pin Vo. Then, writing a single nodal

equation at the inverting input terminal of the op amp,

17000

V-4

1000

3-4

0 o

+=

Solving, we find that Vo = 21 V. Since no current can flow through the 300-kΩ

resistor, V1 = 21 as well.

CHAPTER SIX SOLUTIONS

6. A source transformation and some series combinations are well worthwhile prior to

launching into the analysis. With 5 kΩ || 3 kΩ = 1.875 kΩ and (1 mA)(1.875 kΩ) =

1.875 V, we may redraw the circuit as

This is now a simple inverting amplifier with gain – Rf/ R1 = -75.33/ 1.975 = -38.14.

Thus, V2 = -38.14(3.975) = -151.6 V.

Ω

V2

CHAPTER SIX SOLUTIONS

7. This is a simple inverting amplifier, so we may write

()()

V 3sin 14- 3sin22

1000

2000-

out ttv +=+=

vout(t = 3 s) = -5.648 V.

CHAPTER SIX SOLUTIONS

8. We first combine the 2 MΩ and 700 kΩ resistors into a 518.5 kΩ resistor.

We are left with a simple non-inverting amplifier having a gain of

1 + 518.5/ 250 = 3.074. Thus,

vout = (3.074) vin = 18 so vin = 5.856 V.

CHAPTER SIX SOLUTIONS

9. This is a simple non-inverting amplifier circuit, and so it has a gain of 1 + Rf/ R1.

We want vout = 23.7 cos 500t V when the input is 0.1 cos 500t V, so a gain of 23.7/0.1

= 237 is required.

One possible solution of many: Rf = 236 kΩ and R1 = 1 kΩ.

CHAPTER SIX SOLUTIONS

10. Define a nodal voltage V- at the inverting input, and a nodal voltage V+ at the non-

inverting input. Then,

At the non-inverting input: -3×10-6 = 6

101.5

V

×

+ [1]

Thus, V+ = -4.5 V, and we therefore also know that V- = -4.5 V.

At the inverting input:

7

out-

6

-

R

V - V

R

V

0 += [2]

Solving and making use of the fact that V- = -4.5 V,

vout =

()

V 1

R

4.5- 4.5 - 5.4

R

6

7

6

7











+=−

CHAPTER SIX SOLUTIONS

11. (a) B must be the non-inverting input: that yields a gain of 1 + 70/10 = 8 and an

output of 8 V for a 1-V input.

(b) R1 = ∞, RA = 0. We need a gain of 20/10 = 2, so choose R2 = RB = 1 Ω.

(c) A is the inverting input since it has the feedback connection to the output pin.

CHAPTER SIX SOLUTIONS

12. It is probably best to first perform a simple source transformation:

(1 mA)(2 kΩ) = 2 V.

Since no current can flow into the non-inverting input pin, we know that V+ = 2 V,

and therefore also that V- = 2 V. A single nodal equation at the inverting input yields:

13000

-2

1000

3-2

0 out

v

+=

which yields vout = -11 V.

2 V

3 V

vout

13 kΩ

1 kΩ

2 kΩ V+

V-

CHAPTER SIX SOLUTIONS

13. We begin by find the Thévenin equivalent to the left of the op amp:

Vth = -3.3(3) v

π

= -9.9 v

π

= 1100

1000

9.9 S

v

− = -9 vS

R

th = 3.3 kΩ, so we can redraw the circuit as:

which is simply a classic inverting op amp circuit with gain of -100/3.3 = -30.3.

Thus, vout = (-30.3)( -9 vS) = 272.7 vS

For vS = 5 sin 3t mV, vout = 1.364 sin 3t V, and vout(0.25 s) = 0.9298 V.

vout

-9 vS

3.3 kΩ

100 kΩ

CHAPTER SIX SOLUTIONS

14. We first combine the 4.7 MΩ and 1.3 kΩ resistors: 4.7 MΩ || 1.3 kΩ = 1.30 kΩ.

Next, a source transformation yields (3×10-6)(1300) = 3.899 mV which appears in

series with the 20 mV source and the 500-Ω resistor. Thus, we may redraw the circuit

as

Since no current flows through the 1.8 kΩ resistor, V+ = 23.899 mV and hence

V- = 23.899 mV as well. A single nodal equation at the inverting input terminal yields

3

out

-3-3

1037.7

- 1023.899

500

1023.899

0 ×

×

+

×

=v

Solving, vout = 1.826 V

23.899 mV

-6 V

370 Ω

37.7 kΩ

1.8 kΩ

vout

500 Ω

CHAPTER SIX SOLUTIONS

15. We first combine the 4.7 MΩ and 1.3 kΩ resistors: 4.7 MΩ || 1.3 kΩ = 1.30 kΩ.

Next, a source transformation yields (27×10-6)(1300) = 35.1 mV which appears in

series with the 20 mV source and the 500-Ω resistor. Thus, we may redraw the circuit

as

Since no current flows through the 1.8 kΩ resistor, V+ = 55.1 mV and hence

V- = 55.1 mV as well. A single nodal equation at the inverting input terminal yields

3

out

-3-3

1037.7

- 101.55

500

101.55

0 ×

×

+

×

=v

Solving, vout = 4.21 V

55.1 mV

-6 V

370 Ω

37.7 kΩ

1.8 kΩ

vout

500 Ω

CHAPTER SIX SOLUTIONS

16. The 3 mA source, 1 kΩ resistor and 20 kΩ resistor may be replaced with a –3 V

source (“+” reference up) in series with a 21 kΩ resistor. No current flows through

either 1 MΩ resistor, so that the voltage at each of the four input terminals is

identically zero. Considering each op amp circuit separately,

V 14.29

21

100

(-3)-

AMLEFTOP

out ==

P

v

V 50-

10

100

(5)-

AMPOP RIGH

out ==v

vx = PP vv AMOP RIGH

out

AMLEFTOP

out - = 14.29 + 50 = 64.29 V.

CHAPTER SIX SOLUTIONS

17. A general summing amplifier with N input sources:

1. va = vb = 0

2. A single nodal equation at the inverting input leads to:

N

Na

2

2a

1

1a

f

a

R

...

R

0 vvvvvvvv out −

++

−

+

−

+

−

=

Simplifying and making use of the fact that va = 0, we may write this as

∏∏∏∏ ====

+++=











−N

1

i

N

1

i

2

N

1

i

1

N

1

i

f

R

... R

R

1

iii

out

i

vvv

v

or simply

N

2

1

fR

...

R

vvvvout +++=−

Thus,

vout = ∑

=

N

1

fR

R-

ii

i

v

v1

v2

vN

vout

RN

R1

R2

Rf

va

vb

CHAPTER SIX SOLUTIONS

18. A general difference amplifier:

Writing a nodal equation at the inverting input,

f

a

1

1a

R

-

R

-

0 out

vvvv +=

Writing a nodal equation at the non-inverting input,

2

2b

3

b

R

-

R

0 vvv +=

Simplifying and collecting terms, we may write

(R

f + R1) va – R1 vout = Rf v1 [1]

(R

2 + R3) vb = R3 v2 [2]

From Eqn. [2], we have vb = 2

32

3

RR

Rv

+

Since va = vb, we can now rewrite Eqn. [1] as

()

2

32

31f

1f1 RR

RRR

R R vvvout +

+

−=−

and hence

2

32

1f

1

3

1

f

RR

R

- vvvout 













+

+=

v1

v2

R1

R2

R3

R4

vout

CHAPTER SIX SOLUTIONS

19. In total darkness, the CdS cell has a resistance of 100 kΩ, and at a light intensity L of

6 candela it has a resistance of 6 kΩ. Thus, we may compute the light-dependent

resistance (assuming a linear response in the range between 0 and 6 candela) as RCdS =

-15L + 100 Ω.

Our design requirement (using the standard inverting op amp circuit shown) is that the

voltage across the load is 1.5 V at 2 candela, and less than 1.5 V for intensities greater

than 2 candela.

Thus, vout(2 candela) = -RCdS vS/ R1 = -70 VS/ R1 = 1.5 (R1 in kΩ).

Pick R1 = 10 kΩ. Then vS = -0.2143 V.

CHAPTER SIX SOLUTIONS

20. We want Rf/ Rinstrument = 2K, and Rf/ Rvocal = 1K, where K is a constant not specified.

Assuming K = 1, one possible solution of many is:

vout

vocals

microphone

instruments

microphone

Rvocal = 1 Ω

Rinstruments = 2 Ω

Rf = 2 Ω

CHAPTER SIX SOLUTIONS

21. One possible solution of many:

vS

vout

2 V

1 kΩ

99 kΩ

CHAPTER SIX SOLUTIONS

22. vout of stage 1 is (1)(-20/ 2) = -10 V.

vout of stage 2 is (-10)(-1000/ 10) = 1000 V

Note: in reality, the output voltage will be limited to a value less than that used to

power the op amps.

CHAPTER SIX SOLUTIONS

23. We have a difference amplifier as the first amplifier stage, and a simple voltage

follower as the second stage. We therefore need only to find the output voltage of the

first stage: vout will track this voltage. Using voltage division, then, we vind that the

voltage at the non-inverting input pin of the first op amp is













+32

3

2RR

R

V

and this is the voltage at the inverting input terminal also. Thus, we may write a

single nodal equation at the inverting input of the first op amp:

























+

























+

=1

out

32

3

2

f

1

32

3

2

1

V -

RR

R

V

R

1

V -

RR

R

V

R

1

0 Stage

which may be solved to obtain:

1

f

2

32

3

1

f

1

outout V

R

- V

RR

R

1

R

V V +











+== Stage

CHAPTER SIX SOLUTIONS

24. The output of the first op amp stage may be found by realising that the voltage at the

non-inverting input (and hence the voltage at the inverting input) is 0, and writing a

ingle nodal equation at the inverting input:

7

3 - 0

1

2 - 0

47

V - 0

0 1 stage

out ++= which leads to 1 steage

out

V = -114.1 V

This voltage appears at the input of the second op amp stage, which has a gain of

–3/ 0.3 = 10. Thus, the output of the second op amp stage is –10(-114.1) = 1141 V.

This voltage appears at the input of the final op amp stage, which has a gain of

–47/ 0.3 = -156.7.

Thus, the output of the circuit is –156.7(1141) = -178.8 kV, which is completely and

utterly ridiculous.

CHAPTER SIX SOLUTIONS

25. The output of the top left stage is –1(10/ 2) = -5 V.

The output of the middle left stage is –2(10/ 2) = -10 V.

The output of the bottom right stage is –3(10/ 2) = -15 V.

These three voltages are the input to a summing amplifier such that

Vout =

()

10 15105

100

R

=−−−−

Solving, we find that R = 33.33 Ω.

CHAPTER SIX SOLUTIONS

26. Stage 1 is configured as a voltage follower: the output voltage will be equal to the

input voltage. Using voltage division, the voltage at the non-inverting input (and

hence at the inverting input, as well), is

V 1.667

50100

50

5=

+

The second stage is wired as a voltage follower also, so vout = 1.667 V.

CHAPTER SIX SOLUTIONS

27.

(a) va = vb = 1 nV ∴ vd = 0 and vout = 0. Thus, P8Ω = 0 W.

(b) va = 0, vb = 1 nV ∴ vd = -1 nV

vout = (2×105)(-1×10-9)875

8

+ = -19.28 µV. Thus, P8Ω = 8

2

out

v = 46.46 pW.

(c) va = 2 pV, vb = 1 fV ∴ vd = 1.999 pV

vout = (2×105)(1.999×10-12)875

8

+ = 38.53 nV. Thus, P8Ω = 8

2

out

v = 185.6 aW.

(c) va = 50 µV, vb = -4 µV ∴ vd = 54 µV

vout = (2×105)(54×10-6)875

8

+ = 1.041 V. Thus, P8Ω = 8

2

out

v = 135.5 mW.

CHAPTER SIX SOLUTIONS

28.

Writing a nodal equation at the “-vd” node,

f

outd

1

Sd

in

d

R

- -

R

V - -

R

-

0 vvvv ++= [1]

or (R1Rf + RinRf + RinR1) vd + RinR1vout = -RinRfVS [1]

Writing a nodal equation at the “vout” node,

f

dout

o

dout

R

)(- -

R

A - -

0 vvvv += [2]

Eqn. [2] can be rewritten as:

()

out

fo

of

d

AR - R

R R-

vv +

= [2]

so that Eqn. [1] becomes:

vout =

()

ino1o1infin1f1in

Sofin

RR RR RR RR RR RAR

V R - ARR

- +++++

where for this circuit, A = 106, Rin = 10 TΩ, Ro = 15 Ω, Rf = 1000 kΩ, R1 = 270 kΩ.

(a) –3.704 mV; (b) 27.78 mV; (c) –3.704 V.

CHAPTER SIX SOLUTIONS

29. vout = Avd =

()

V 2 sin 1080

R16

R

A15

i

it×

+

(a) A = 105, Ri = 100 MΩ, Ro value irrelevant. vout = 8 sin 2t nV

(b) A = 106, Ri = 1 TΩ, Ro value irrelevant. vout = 80 sin 2t nV

CHAPTER SIX SOLUTIONS

30.

(a) Find vout/ vin if Ri = ∞, Ro = 0, and A is finite.

The nodal equation at the inverting input is

100

- -

1

- -

0 outdindvvvv += [1]

At the output, with Ro = 0 we may write vout = Avd so vd = vout/ A. Thus, Eqn. [1]

becomes

100

100A

A

0 outout

in

out vv

v

v+++=

from which we find

A 101

100A-

in

out

+

=

v

v [2]

(b) We want the value of A such that vout/ vin = -99 (the “ideal” value would be –100

if A were infinite). Substituting into Eqn. [2], we find

A = 9999

CHAPTER SIX SOLUTIONS

31. (a)

δ

= 0 V ∴ vd = 0, and P8Ω = 0 W.

(b)

δ

= 1 nV, so vd = 5 – (5 + 10-9) = -10-9 V

Thus,

vout = (2×105)vd 758

8

+ = -19.28 µV and P8Ω = (vout)2/ 8 = 46.46 pW.

(c)

δ

= 2.5 µV, so vd = 5 – (5 + 2.5×10-6) = -2.5×10-6V

Thus,

vout = (2×105)vd 758

8

+ = -48.19 mV and P8Ω = (vout)2/ 8 = 290.3 µW.

CHAPTER SIX SOLUTIONS

32.

Writing a single nodal equation at the output, we find that

o

dout

i

inout

R

A -

R

-

0 vvvv += [1]

Also, vin – vout = vd, so Eqn. [1] becomes

0 = (vout – vin) Ro + (vout – Avin + Avout) Ri

and

vout =

()

in

v

A

R )1(R

AR R

io

++

+

To within 4 significant figures (and more, actually), when vin = -16 mV, vout = -16 mV

(this is, after all, a voltage follower circuit).

AD549

CHAPTER SIX SOLUTIONS

33. The ideal op amp model predicts a gain vout/ vin = -1000/ 10 = -100, regardless of the

value of vin. In other words, it predicts an input-output characteristic such as:

From the PSpice simulation result shown below, we see that the ideal op amp model

is reasonably accurate for |vin| × 100 < 15 V (the supply voltage, assuming both have

the same magnitude), but the onset of saturation is at ±14.5 V, or |vin| ~ 145 mV.

Increasing |vin| past this value does not lead to an increase in |vout|.

vout (V)

vin (V)

-100

1

CHAPTER SIX SOLUTIONS

34. Positive voltage supply, negative voltage supply, inverting input, ground, output pin.

CHAPTER SIX SOLUTIONS

35. This op amp circuit is an open-loop circuit; there is no external feedback path from

the output terminal to either input. Thus, the output should be the open-loop gain

times the differential input voltage, minus any resistive losses.

From the simulation results below, we see that all three op amps saturate at a voltage

magnitude of approximately 14 V, corresponding to a differential input voltage of 50

to 100 µV, except in the interest case of the LM 324, which may be showing some

unexpected input offset behavior.

op amp onset of

negative

saturation

negative

saturation

voltage

onset of

positive

saturation

positive

saturation

voltage

µA 741 -92 µV -14.32 V 54.4 mV 14.34 V

LM 324 41.3 µV -14.71 V 337.2 mV 13.87 V

LF 411 -31.77 µV -13.81 V 39.78 mV 13.86 V

CHAPTER SIX SOLUTIONS

36. This is a non-inverting op amp circuit, so we expect a gain of 1 + 1000/4.7 = 213.8.

With ±15 V DC supplies, we need to sweep the input just above and just below this

value divided by the gain to ensure that we see the saturation regions. Performing the

indicated simulation and a DC sweep from –0.1 V to +0.1 V with 0.001 V steps, we

obtain the following input-output characteristic:

Using the cursor tool, we see that the linear region is in the range of

–68.2 mV < Vin < 68.5 mV.

The simulation predicts a gain of 7.103 V/ 32.87 mV = 216.1, which is reasonably

close to the value predicted using the ideal op amp model.

CHAPTER SIX SOLUTIONS

37. Referring to the detailed model of an op amp, shorting the input terminals together

shorts the dependent source to ground. Therefore, any 1-V source connected to the

output through a 1-Ω resistor should “see” 1 Ω + Ro. For the µA 741, we expect 1 +

75 = 76 Ω. For the LF 411, we expect ~1 + 1 Ω or ~ 2 Ω.

Simulating the µA741 circuit, we find:

Supply voltages Current into output Total resistance Output resistance

±15 V -42.5 mA -23.53 Ω -22.53 Ω

±5 V -40.55 mA -24.66 Ω -24.66 Ω

±2 V -40.55 mA -24.66 Ω -24.66 Ω

0 V 579.2 mA 1.727 Ω 727 mΩ

Conclusion: as we might expect from previous experience in determining Thévenin

equivalent resistances, we must short out the voltage supplies to the op amp when

performing such an experiment (hence the negative resistance values obtained above).

However, we obtained 0.727 Ω instead of the expected 75 Ω, which leads to two

possible conclusions: (1) The PSpice model is not designed to represent the op amp

behavior accurately in such a circuit configuration or (2) such an experimental

connection is not adequate for measuring the output resistance.

Simulating the LF411 circuit, we find:

Supply voltages Current into output Total resistance Output resistance

±15 V 25.46 mA 39.28 Ω 38.28 Ω

±5 V 25.43 mA 39.32 Ω 38.32 Ω

±2 V 25.48 mA 39.24 Ω 28.24 Ω

0 V 1000 mA 1 Ω 0 Ω

Conclusion: as we might expect from previous experience in determining Thévenin

equivalent resistances, we must short out the voltage supplies to the op amp when

performing such an experiment. However, we obtained ~0 Ω instead of the expected 1

Ω, which leads to two possible conclusions: (1) The PSpice model is not designed to

represent the op amp behavior accurately in such a circuit configuration or (2) such an

experimental connection is not adequate for measuring the output resistance.

However, it is interesting that PSpice did predict a much lower output resistance for

the LF 411 than the µA 741, as we would expect.

CHAPTER SIX SOLUTIONS

38. Based on the detailed model of the LF 411 op amp, we can write the following nodal

equation at the inverting input:

o

6

dd

4

dx

in

d

R10

- A

10

-

R

-

0 +

++= vvvvv

Substituting values for the LF 411 and simplifying, we make appropriate

approximations and then solve for vd in terms of vx, finding that

199.9

-

10199.9

10-

x

6

d

v

vv =

×

=

With a gain of –1000/10 = -100 and supply voltage magnitudes of 15 V, we are

effectively limited to values of |vx| < 150 mV.

For vx = -10 mV, PSpice predicts vd = 6 µV, where the hand calculations based on the

detailed model predict 50 µV, which is about one order of magnitude larger. For the

same input voltage, PSpice predicts an input current of -1 µA, whereas the hand

calculations predict 99.5vx mA = -995 nA (which is reasonably close).

CHAPTER SIX SOLUTIONS

(a) The gain of the inverting amplifier is –1000. At a sensor voltage of –30 mV, the

predicted output voltage (assuming an ideal op amp) is +30 V. At a sensor voltage of +75

mV, the predicted output voltage (again assuming an ideal op amp) is –75 V. Since the op

amp is being powered by dc sources with voltage magnitude equal to 15 V, the output

voltage range will realistically be limited to the range

–15 < Vout < 15 V.

(b) The peak input voltage is 75 mV. Therefore, 15/ 75×10-3 = 200, and we should set

the resistance ratio Rf/ R1 < 199 to ensure the op amp does not saturate.

CHAPTER SIX SOLUTIONS

39. (a)

We see from the simulation result that negative saturation begins at Vin = –4.72 V,

and positive saturation begins at Vin = +4.67 V.

(b) Using a 1 pΩ resistor between the output pin and ground, we obtain an output

current of 40.61 mA, slightly larger than the expected 35 mA, but not too far off.

CHAPTER SIX SOLUTIONS

41. We assume that the strength of the separately-broadcast chaotic “noise” signal is

received at the appropriate intensity such that it may precisely cancel out the chaotic

component of the total received signal; otherwise, a variable-gain stage would need to

be added so that this could be adjusted by the user. We also assume that the signal

frequency is separate from the “carrier” or broadcast frequency, and has already been

separated out by an appropriate circuit (in a similar fashion, a radio station

transmitting at 92 MHz is sending an audio signal of between 20 and 20 kHz, which

must be separated from the 92 MHz frequency.)

One possible solution of many (all resistances in ohms):

CHAPTER SIX SOLUTIONS

41. One possible solution of many:

This circuit produces an output equal to the average of V1, V2, and V3, as shown in the

simulation result: Vaverage = (1.45 + 3.95 + 7.82)/ 3 = 4.407 V.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1. (a)

()

pF

d

A

C954.6

10100

1054.7810854.8

6

612 =

×

××

== −

−−

ε

(b)

()

kV

C

E

VCVEEnergy 959.16

10954.6

10122

2

1

,12

3

2=

×

==∴= −

−

(c)

()

pF

V

E

CCVE 500

100

105.222

2

1

2

6

2

2=

×

==∴= −

()()

()

1

6

612 .62.636

1054.78

1010010500 −

−

−− =

×

××

==∴= mpF

A

Cd

d

A

C

ε

9.71

10854.8

1062.636

,_Re 12

12

0

=

×

=∴ −

−

ε

typermittivilative

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2. (a) For VA = -1V,

()

()()

()

157.0

101106.1

10854.88.112

2

2419

12

0+

××

×

=−= −

−

Abi

sVV

qN

K

W

ε

m

9

10281.45 −

×=

(

)

(

)

fFC j307.2

10281.45

10110854.88.11

9

1212 =

×

××

=−

−−

(b) For VA = -5V,

()

()()

()

557.0

101106.1

10854.88.112

2

2419

12

0+

××

×

=−= −

−

Abi

sVV

qN

K

W

ε

m

9

10289.85 −

×=

(

)

(

)

fFC j225.1

10289.85

10110854.88.11

9

1212 =

×

××

=−

−−

(c) For VA = -10V,

()

()()

()

1057.0

101106.1

10854.88.112

2

2419

12

0+

××

×

=−= −

−

Abi

sVV

qN

K

W

ε

m

9

10491.117 −

×=

(

)

(

)

aFC j239.889

10491.117

10110854.88.11

9

1212 =

×

××

=−

−−

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3. We require a capacitor that may be manually varied between 100 and 1000 pF by rotation

of a knob. Let’s choose an air dielectric for simplicity of construction, and a series of 11

half-plates:

Constructed as shown, the half-plates are in parallel, so that each of the 10 pairs must

have a capacitance of 1000/ 10 = 100 pF when rotated such that they overlap completely.

If we arbitrarily select an area of 1 cm2 for each half-plate, then the gap spacing between

each plate is d = εA/C = (8.854×10-14 F/cm)(1 cm2)/ (100×10-12 F) = 0.8854 mm. This is

tight, but not impossible to achieve. The final step is to determine the amount of overlap

which corresponds to 100 pF for the total capacitor structure. A capacitance of 100 pF is

equal to 10% of the capacitance when all of the plate areas are aligned, so we need a pie-

shaped wedge having an area of 0.1 cm2. If the middle figure above corresponds to an

angle of 0o and the case of perfect alignment (maximum capacitance) corresponds to an

angle of 180o, we need to set out minimum angle to be 18o.

Side view with no

overlap between plates

Top view Side view with a small

overlap between plates.

fixed

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4. (a) Energy stored JdteeC

dt

dv

Cv

tt

t

µ

080.1

5

3

.3. 5

102

05

3

0

−=











−== −

×−

∫∫ −

(b) Vmax = 3V

Max. energy at t=0, JEmJCV

µ

5.499%3735.1

2

1max

2=∴==

V at 37% Emax = 1.825V

()

sstetv

t

2486.23825.1 5⇒≈=∴== −

(c) Ae

dt

dv

Ci

µ

593.141

5

3

10300 5

2.1

6−=











−×== −

−

(d)

(

)

WviP

µ

6.24210658.120011.2 6−=×−== −

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5. (a)

()

mV

C

v421.33101

2

14159.3

.

1047

1

101

2

.

12

3

6

2

3=×

×

=×= −

−

π

(b)

()

mV

C

v421.33101

2

14159.3

.

1047

1

0101

2

.

12

3

6

2

3=×

×

=











+×= −

−

π

(c)

()() ()

mV

C

v132.50101

4

3

.

1047

1

101

4

101

2

.

12

3

6

2

3

2

3=











×

×

=











×+×= −

−

−−

πππ

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6. C

t

C

idt

C

V

ms

ms 426.0

cos

10711 200

0

3

200

0=























×

−== −

∫

π

()

FC

C

CVE

µ

30181

1032

10086.181

2

10086.181

103

2

1

6

99

62 =

×

=∴

×

=×== −

−−

−

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7.

(a)

(b) 2722 7

max

11

2 10 (5 3cos 200 ) 10 64 6.4 J

22

cc c

wcv tw

µ

−−

==×× + ∴=×=

(c) 6 100 3 3 100 100

0

110 8 10 10 40( 0.01)( 1) 400(1 )V

0.2

−− −

=× × =×− −= −

∫tttt

c

vedt e e

(d) 100

500 400 V

t

c

ve

−

=−

0.12sin 400 mA∴=−

c

it

26

0.2 F, 5 3cos 200 V; 0.2 10 (3)( 2)200sin 200 cos 200

−

==+ ∴=×−

cc

cv ti tt

µ

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

0.1

0

0.2

0.1

0.9

22

(0) 250V, 2mF ( ) (0.1) 250 500 5

(0.1) 500V; (0.2) 500 10 1000V

(0.6) 1750V, (0.9) 2000V

0.9 1: 2000 500 10 2000 5000( 0.9)

2100 2000 5000( 0.9) 0.92 0.9 0

== =+

∴= = =

∴= =

∴<< = + = + −

∴= = + − ∴= ∴ <<

∫

cc

t

c

vcav dt

vv dt

vv

t v dt t

vttt.92s

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9.

(a)

(b)

2 6 2 2 1000 500

6 500 500

11

C 10 2 10 200 V

22

C 10 ( 200)( 500) 0.1

200

R2

0.1

−−− −

−−−

==×=× ∴=±

′

== ± − =

−

∴= = = Ω

m

tt

c

tt

wv v ev e

iv e e

vk

i

2 1000 1000

1000 1000

R0

0

P R 0.01 2000 20 W

W 20 0.02 0.02J

−−

∞−−∞

==× =

∴= =− =

∫

tt

R

tt

iee

edt e

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10. (a) Left circuit:

By Voltage division,

()

V

kk

k

VC877.05

17.4

1=

+

=

Right circuit:

()

VV 3

2

2//11

1==

By Voltage Division, VVVV C3

1

3

1

2−=∴=

(b)

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

(a)

(b) max

P P ( 100)( 5) 500W at 40 ms

−

=∴ =− −= =

L

LLi L

vt

(c) min

P 100( 5) 500W at 20 and 40 ms

++

=−=− =

Lt

(d) 22

11

W L W (40ms) 0.2( 5) 2.5J

22

=∴ =×−=

L

LL

i

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12.

3 100 100

-100t 100 1

max

100

max

100 100

L 50 10 , 0: 0; 0 80 mA 0.08 A

i =0.08e 8 0.08 8 , , 0.01 , 0.08 0.01

0.2943mA; 0.05 (0.004 0.4 )

( 0.4) 100 (0.004 0.4 ) 0.4 0.4 40 ,

−−−

−−

−

−−

=× < = > = =

′

∴−∴===×

′

∴= = = −

′

∴= − − − ∴− = −

tt

t

m

t

tt

t i t i te te

te t t s i e

iviet

ve e t t

2

max

0.8 0.02

40

(0.004 0.008) 0.5413mV this is minimum 0.004Vat t=0

−

==

=−=− ∴=

ts

ve v

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13.

(a) 22

0: 0.4 A 10 5 4 4 V

′

>=∴=+=+

sinss

ti tv iitt

(b) 2

0

1

0.1 40 5 4 4 5A

5

′=+ +=++

∫t

in s

iv tdttt

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14. 20cos1000 V, L 25mH, (0) 0===

LL

vti

(a) 0

40 20 cos 1000 0.8sin 1000 A 8sin 2000 W==∴=

∫t

L

i tdt t p t

(b) 32 2

125 10 0.64sin 1000 8sin 1000 mJ

2

wtt

−

=× × × =

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

(a) 0

0 10 ms: 2 5 100 2 500 (10ms) 3A, (8ms) 2A<< =−+ =−+ ∴ = =

∫t

LLL

t i dt t i i

(b)

(c) If the circuit has been connected for a long time, L appears like short circuit.

()

VVV 80100

82

8

8=

+

=

Ω

A

V

I10

2

20

2=

Ω

=

Ω

A

V

ix1

80

80 =

Ω

=∴

0.02 4

0.01

40.024

0.01

2

(0) 0 (10ms) 500 0.01 5A (20ms) 5 5 10 (0.02 )

(20ms) 5 5 10 (0.02 0.5 ) 5 5 10 (0.0002 0.00015) 7.5A

10.2 7.5 5.625J

2

LL L

L

ii i tdt

itt

w

=∴ = × = ∴ =+ −

∴ =+× − =+× − =

∴=×× =

∫

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16. 2

L 5H,V 10( )V, (0) 0.08A

−−

==− =

tt

LL

ee i

(a) 12

(1) 10( ) 2.325 V

L

vee

−− +

=−=

(b)

(c) ( ) 1.08A

L

i∞=

22

0

22

0.08 0.2 10( ) 0.08 2( 0.5 )

0.08 2( 0.5 1 0.5) 1.08 2 (1) 0.4796A

−− − −

−− −−

=+ − =+−+

=+−+ +−=+−∴=

∫ttt t tt

L

tt tt

L L

ieedtee

iee eei

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

(a)

(b)

40 12

120 40 5

12 20 40 12 20 40

200 100 100V

33

=× +××

++ ++

=+=

x

v

15 12

120 15 40 40 5

12 15 60 15 60 15 12 60

120 1 6.667

40 200

12 12 5 66.667

40 20 60V

=××+×

++ +

=××+

+

=+=

x

v

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18.

(a) 2

15 1.6 6.4J

2

L

w=×× =

(b) 62

120 10 100 0.1J

2

c

w−

=× × × =

(c) Left to right (magnitudes): 100, 0, 100, 116, 16, 16, 0 (V)

(d) Left to right (magnitudes): 0, 0, 2, 2, 0.4, 1.6, 0 (A)

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19.

(a)

(b)

(c) 0.4

24 45

0

4 , P 100 16 1600 320 0.4 3.277J==×∴= =×=

∫

RR R

it tw tdt

2

52

400 V, 0; (0) 0.5A; 0.4

1

400 0.16 64V, 10 64 20.48mJ

2

−

=>==

=× = ××=

sL

cc

vtti ts

vw

0.4 23

0

2

1

0.5 0.1 400 0.5 40 0.4 1.3533A

3

110 1.3533 9.1581J

2

=+ =+×× =

∴=×× =

∫

L

itdt

w

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20. (a)

()

W

R

V

PWP 4.0

10

2

;0 2

2

107==== ΩΩ

(b) PSpice verification

We see from the PSpice

simulation that the

voltage across the 10-Ω

resistor is –2 V, so that

it is dissipating 4/10

= 400 mW.

The 7-Ω resistor has

zero volts across its

terminals, and hence

dissipates zero power.

Both results agree with

the hand calculations.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.













+

+≡

µµ

µ

10

1

10

1

10

equiv

Cin series with

µ

10 in series with













+

µµ

µ

10

1

10

1

10

F

µ

286.4

≡

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22.

()()()()

pHpppppppLequiv 6.1797777//77777777//77 &

=++++≡

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23. (a) Assuming all resistors have value R, all inductors have value L, and all capacitors

have value C,

(b) At dc, 20µF is open circuit; 500µH is short circuit.

Using voltage division,

()

V

kk

k

Vx6.39

1510

10 =

+

=

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24. (a) Assuming all resistors have value R, all inductors value L, and all capacitors value C,

(b) VVx0= as L is short circuit at dc.

+

Vx

-

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25. Cequiv = { [(100 n + 40 n) || 12 n] + 75 n} || {7 µ + (2 µ || 12 µ)}

nFCequiv 211.85≡

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26. Lequiv = {[ (17 p || 4 n) + 77 p] || 12 n} + {1 n || (72 p + 14 p)}

pHLequiv 388.172≡

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

()

FCC xT

µ

171100161477 =++++=−

()()()

JVCCE xTCC xT

µµ

375.5345.2171

2

1

2

12

2==−=

−

()

nJJEEE xTTx CCCC 425375.5348.534 =−=−= −

µ

()

nF

n

CVCnE xxCx136

5.2

2425

2

1

425 2

2==⇒==∴

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28.

(a) For all L = 1.5H, HLequiv 75.2

5.1

1

5.1

1

5.1

1

5.1

1

5.1

1

5.1 =













++

+













+

+=

(b) For a general network of this type, having N stages (and all L values equiv),

∑

=−

=n

NN

N

equiv NL

L

11

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29.

(a) HLequiv 3

3

1

3

1

3

1

2

1

2

1

1=













++

+













+

+=

(b) For a network of this type having 3 stages,

() ()

()

() ()

()

2

32

22 33

3

22

2

1

3

1

3

33

1

2

22

1

1++=

+

+=

equiv

L

Extending for the general case of N stages,

N

1

N

1

3

1

3

1

3

1

2

1

2

1

1 Κ

Κ+

++

+

+=

equiv

L

N

N(1/N)

1

)3/1(3

1

)2/1(2

1

1 =++++=Κ

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30.

()( )

pF

pp

Cequiv 231.0

25.03

25.03 =

+

=

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31.

()()

nH

n

nn

Lequiv 6291.0

6.2

3.03.2 &

&

&& ==

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32. (a) Use 2 x 1µH in series with 4 x 1µH in parallel.

(b) Use 2 x 1µH in parallel, in series with 4 x 1µH in parallel.

(c) Use 5 x 1µH in parallel, in series with 4 x 1µH in parallel.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

(a)

(b) L 10H L 11.379H=∴=

eq

(c)

10 10 55

R 10 :10 10 10 , 10 10 10

33 3

55

R 30 11.379

3

=Ω = ++ =

∴= = Ω

eq

1

C 10F : 5.4545

1/ 30 1/10 1/ 20

10

C 5.4545 8.788F

3

==

++

∴= +=

eq

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34.

(a)

(b)

17 1

:1 , 1.3125F

1/4 1/2 3 3/7 1/2

15 5

: ,C 4 4.833F

1/5 1 6 6

+== =

++

==+=

+

eq

oc c

sc

:L 6 1 3 3.857H

: L (3 2 1) 4 2.2 4 1.4194H

=+=

=+= =

eq

oc

sc

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35.

(a)

(b)

(c)

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36. 200

1

60 mA, (0) 20mA

t

s

ie i

−

==

(a)

(b)

(c) 200 200 200

21

60 24 4 36 4mA( 0)

tt t

s

iii e e e t

−− −

=−= − += + >

200

6 4 2.4H L 2.4 0.06( 200)

or 28.8 V

−

′

=∴= =×−

=−

t

eq s

t

vi e

ve

200 200

1

200

14.8

28.8 0.02 ( 1) 0.02

6200

24 4mA( 0)

−−

−

=− + = −+

=−>

∫ttt

o

t

iedt e

et

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37. 80

1

100 , (0) 20V

t

s

veVv

−

==

(a) 680 380

C 0.8 10 ( 80)100 6.4 10 A

−− −−

′

==×− =−×

tt

eq s

iv e e

(b)

(c)

6 3 80 80

1

80

1

6400

10 ( 6.4 10 ) 20 ( 1) 20

80

80 60V

−− −

−

=−× += −+

∴= −

∫ttt

o

t

v e dt e

ve

6380 80

2

80

10 1600

( 6.4 10 ) 80 ( 1) 80

480

20 60V

−− −

−

−× += −+

=+

∫ttt

o

t

v e dt e

e

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38.

(a)

(b)

6

3

510 0

20 10

120

10 8 10

−

−−

′

+× + =

−++=

×∫

cs cL

c

t

Lc L

o

vv vv

v

vv vdt

20 20

6

3

20

6

1

20 ( ) 12

510

1( ) 12 10 8 10 0

510

−

+−+=

×

′

−−++× =

×

∫

t

Ls

o

t

LLL

o

iiidtv

iidt i i

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

3

333

( ): 30mA: 0.03 20 0.6V, 0.6V

9V: 9V, 20mA: 0.02 20 0.4V

0.04cos10 : 0

( ) 9.2V

( ): 30mA,20mA,

9V: 0; 0.04cos10 : 0.06 0.04( 1000)sin10 2.4sin10 V

×= =

==−×=

=

∴=

==−×−=

cc

c

L

LL

vt v

vv

tv

vt

vtv tt

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40. We begin by selecting the bottom node as the reference and assigning four nodal

voltages:

1, 4 Supernode: 20×10-3 e-20t =

()

∫′

−×+ −

tttde

0

20

4

3

21 40V100.02

50

V-V [1]

and: V1 – V4 = 0.2 Vx or 0.8V1 + 0.2 V2 – V4 = 0 [2]

Node 2: dt

de t

2

6-

20

212 V

10

100

40 - V

50

V - V

0 ++= − [3]

Ref.

V1

V2 V3

V4

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41. (a)

R,R0,A 0 C

also 0 R 0 RC

1

RA 0,

11A

AAR

1111A

AARCA

1A 1A

AorA0

RC RC

′

=∞ = =∞∴ = ∴ =

′

++=∴=−

−+ − = = +

−+

=− ∴ = ∴ =

+

∴= − =− + −

++

′′ ′ ′

∴=−− + +=

∫

∫∫

io i s

oo s

iis i

oiio i

o

soo

so oo os

viv

iv v v

viv v idtv

c

vvvvi v

v

v idt v v dt

c

vv vv vv

(b)

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42. Place a current source in parallel with a 1-MΩ resistor on the positive input of a buffer

with output voltage, v. This feeds into an integrator stage with input resistor, R2, of

1-MΩ and feedback capacitor, Cf, of 1 µF.

sec

10602.1 19 ions

dt

dv

Ci f

c

f××== −

sec

10602.1

101101

019

66

ions

VV

dt

dv

C

VV a

c

f

af−

×+

×

−

=+

×

−

=

sec

10602.1

101

019

6

2

ionsV

dt

dv

C

R

Vf

c

f−

×+

×

−

=+

−

=

Integrating current with respect to t,

()

0'

1

0

2ff cc

t

fVVCvdt

R−=

∫

f

cf VC

R

ions =

×× −

2

19

10602.1

ions

C

Vions

CR

R

VVVV

f

out

f

outoutacf×××

−

=⇒×××

−

=⇒−= −− 1919

2

110602.1

1

10602.1

R1 = 1 MΩ, Cf = 1µF

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43. R 0.5M , C 2 F, R , R 0, cos10 1V=Ω= =∞== −

ioo

vt

µ

(a)

(b) Let A 10sin10 V

s

vt=∞∴ =

11

Eq. (16) is: 1 (0)

ARCA

11 1 11

1 1 ( 10sin10 ) 1 cos10

ARCAA A A

111

1 10sin10 cos10 Let A 2000

AAA

10.005sin10 0.0005 0.0005c



+=− +−

  

  



   

′

∴ + =− + ∴ + − =− + −

   

   





∴=+ + − =





∴= + −

∫to

osc

o

os s

s

v

v v dt v

v

vv tv t

vtt

vt os10t

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44. Create a op-amp based differentiator using an ideal op amp with input capacitor C1 and

feedback resistor Rf followed by inverter stage with unity gain.

min/

1

60

1rpm

mV

dt

dvs

CR

R

Vfout ×=+=

R

fC1=60 so choose Rf = 6 MΩ and C1 = 10 µF.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45. (a) ∫−

+=

f

outa

R

VV

vdt

L

1

0

'

1

,0 0dtv

L

R

V

R

V

dtv

L

VV t

s

f

out

f

out

La ∫∫ −

=⇒=∴==

(b) In practice, capacitors are usually used as capacitor values are more readily

available than inductor values.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46.

(a)

(b)

(c)

20 20

6

3

20

6

1

20 ( ) 12

510

1( ) 12 10 8 10 0

510

−

+−+=

×

′

−−++× =

×

∫

t

cs

o

t

ccc

o

vvvdti

vvdt v v

6

3

510 0

20 10

120

10 8 10

−

−−

′

+× + =

−++=

×∫

Ls Lc

L

t

cL c

o

ii ii

i

ii idt

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49.

(a)

(b) “Let is = 100e-80t A and i1 (0) = 20 A in the circuit of (new) Fig. 7.62.

(a) Determine v(t) for all t.

(b) Find i1(t) for t ≥ 0.

(c) Find v2(t) for t ≥ 0.”

(c) (a)

(b) 6 3 80 80 80

11

6400

( ) 10 6.4 10 20 ( ) ( 1) 80 60A

80

−− − −

=−× +∴= −= −

∫tttt

o

it e dt it e e

(c) 80

212

() () () 20 60A

t

s

it i it it e

−

=− ∴ = +

680

-80t

L 1 4 0.8 H ( ) L 0.8 10 100( 80) V

v(t) 6.43 mV

−−

′

== ∴= =× × −

∴=−

t

eq eq s

vt i r

µ

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50.

In creating the dual of the original circuit, we have lost both vs and vout. However, we

may write the dual of the original transfer function: iout/ is. Performing nodal analysis,

)V - (V G V

L

1

21in

01

1

S+

′

=∫ttdi [1]

iout = Aid = GfV2 + Gin (V2 – V1) [2]

Dividing, we find that

)V - (V G V

L

1

V G )V - (V G

21in

01

1

2f12in

S

out

+

′

+

=∫ttd

i

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



51. PSpice verification

w = ½ Cv2 = 0.5 (33×10-6)[5 cos (75×10-2)]2 = 220.8 µJ. This is in agreement with the

PSpice simulation results shown below.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



52. PSpice verification

w = ½ Li2 = 0.5 (100×10-12)[5 cos (75×10-2)]2 = 669.2 pJ. This is in agreement with

the PSpice simulation results shown below.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



53. ∫

+

−

=dtv

LR

VV

f

L

sa 1

0

1

,0== ba VV ∫

+

−

=dtv

LR

V

f

L

s1

0

1

dt

dVs

R

L

VVVV outoutaLf1

0=−=−=

()

LetRLtA

dt

d

R

L

dt

dVs

R

L

Vf

ff

out

π

_

;2102cos 1

3

11

=⇒−=−= R = 1 Ω and L = 1 H.

PSpice Verification: clearly, something rather odd is occuring in the simulation of this

particular circuit, since the output is not a pure sinusoid, but a combination of several

sinusoids.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



54. PSpice verification

w = ½ Cv2 = 0.5 (33×10-6)[5 cos (75×10-2) - 7]2 = 184.2 µJ. This is in reasonable

agreement with the PSpice simulation results shown below.

CHAPTER SEVEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



55. PSpice verification

w = ½ Li2 = 0.5 (100×10-12)[5 cos (75×10-2) - 7]2 = 558.3 pJ. This is in agreement with

the PSpice simulation results shown below.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1.

(a)

(b) 4

(0.01) 2 36.63mA

L

ie

−

==

(c) 11

400 400

1

2 1, 2, 1.7329ms

−===

tt

eet

80 /0.2

400

100

(0) 2A ( ) 2

50

2A,0

−

==∴=

=>

t

LL

t

iite

et

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2.

(a) 12

(0 ) 60 30mA, (0 ) 30 20mA

23

Lx

ii

−−

=× = =× =

(b) (0 ) 30mA, (0 ) 30mA

Lx

ii

++

==−

(c)

250 /0.05 5000

1.5

( ) 30 30 mA, (0.3ms)

30 6.694mA

tt

LL

x

it e e i

ei

−−

−

==

== =−

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3.

(a)

= 1.289 mA.

(b) iSW( 5 µs) = 9 – 1.289 = 7.711 mA.

6

36

3

10 /4 1.25

10 10

(0) 4.5mA, R/L 410 4

4.5 mA (5 ) 4.5

−

−−

===

×

∴= ∴ =

L

t

LL

i

ie ise

µ

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4.

(a) Since the inductor current can’t change instantaneously, we simply need to find iL while

the switch is closed. The inductor is shorting out both of the resistors, so iL(0+) = 2 A.

(b) The instant after the switch is thrown, we know that 2 A flows through the inductor. By

KCL, the simple circuit must have 2 A flowing through the 20-Ω resistor as well. Thus,

v = 4(20) = 80 V.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5. (a) Prior to the switch being thrown, the 12-Ω resistor is isolated and we have a simple

two-resistor current divider (the inductor is acting like a short circuit in the DC circuit,

since it has been connected in this fashion long enough for any transients to have

decayed). Thus, the current iL through the inductor is simply 5(8)/ (8 + 2) = 4 A.

The voltage v must be 0 V.

(b) The instant just after the switch is thrown, the inductor current must remain the same,

so iL = 4 A. KCL requires that the same current must now be flowing through the 12-Ω

resistor, so v = 12(-4) = -48 V.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6. For t < 0, we have a current divider with iL(0-) = ix(0-) = 0.5 [ 10 (1/ (1 + 1.5)] mA

= 2 mA. For t > 0, the resistor through which ix flows is shorted, so that ix(t > 0) = 0.

The remaining 1-kΩ resistor and 1-mH inductor network exhibits a decaying current such

that iL(t) = 2e-t/

τ

mA where

τ

= L/R = 1 µS.

(a)

(b)

t (µs)

ix (mA)

2

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7.

(a)

(b)

/II

, , 10 10 2.303;

I

II

100, 4.605; 1000, 6.908

−

== =∴==

== = =

ll

too

o

oo

it t

en n

ii

tt

ii

τ

ττ

// 1

11

(/I) ()

,; at /1,

I(/) ()

Now, ( 1) ( 1) , I

At 0, ( 1) 2 / 2

−−

==−==−



=−+=− −+ = =





=−=∴=∴=

tt

o

diid

eete

dt d

ti

ymx b ex e x y

yex ex t

ττ

τ

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8. Reading from the graph current is at 0.37 at 2 ms

A 10I

ms 2

0=

=∴

τ

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9. w = ½ Li2, so an initial energy of 15 mJ in a 10-mH inductor corresponds to an initial

inductor current of 1.732 A. For R = 1 kΩ,

τ

= L/R = 10 µs, so iL(t) = 1.732 e–0.1t A.

For R = 10 kΩ,

τ

= 1 µs so iL(t) = 1.732 e-t. For R = 100 kΩ,

τ

= 100 ns or 0.1 µs so

iL(t) = 1.732 e-10t A. For each current expression above, it is assumed that time is

expressed in microseconds.

To create a sketch, we first realise that the maximum current for any of the three cases

will be 1.732 A, and after one time constant (10, 1, or 0.1 µs), the current will drop to

36.79% of this value (637.2 mA); after approximately 5 time constants, the current will

be close to zero.

Sketch based on hand analysis Circuit used for PSpice

verification

As can be seen by

comparing the two plots,

which probably should

have the same x-axis scale

labels for easier

comparison, the PSpice

simulation results obtained

using a parametric sweep

do in fact agree with our

hand calculations.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10.

(a) 12

6

6103.3

101

103.3 −

−×=

×

=

τ

(b)

()

A 1.121.55

A 1.5

103.3

10432

..

2

1

6126 103.3/105101

6

0

2

0

==

=

×

××

=

−− ××××−

−

epsi

I

IL

ω

(c)

From the PSpice

simulation, we see that

the inductor current is

1.121 A at t = 5 ps, in

agreement with the hand

calculation.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11. Assume the source Thévenin resistance is zero, and assume the transient is measured to

5τ. Then,

9

6

7

L5L

5 100 10 secs

RR

(5)(125.7)10

R10

−

τ=∴τ==×

∴> so R must be greater than 6.285 kΩ.

(If 1τ assumed then 6.285

R 125.7

5

>=Ω)

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



The film acts as an intensity integrator. Assuming that we may model the intensity as a simple

decaying exponential,

φ(t) = φo e-t/ τ

where the time constant τ represents the effect of the Thévenin equivalent resistance of

the equipment as it drains the energy stored in the two capacitors, then the intensity of the

image on the film Φ is actually proportional to the integrated exposure:

Φ = dte t

∫−

timeexposure

0

/

o

K

τ

φ

where K is some constant. Solving the integral, we find that

Φ =

[

]

1 - K - time)/ osure(exp

o

τ

τφ

−

e

The maximum value of this intensity function is –Kφo

τ

.

With 150 ms yielding an image intensity of approximately 14% of the maximum

observed and the knowledge that at 2 s no further increase is seen leads us to estimate

that 1 - e-150x10-3/

τ

= 0.14, assuming that we are observing single-exponential decay

behavior and that the response speed of the film is not affecting the measurement. Thus,

we may extract an estimate of the circuit time constant as

τ

= 994.5 ms.

This estimate is consistent with the additional observation that at t = 2 s, the image

appears to be saturated.

With two 50-mF capacitors connected in parallel for a total capacitance of 100 mF, we

may estimate the Thévenin equivalent resistance from

τ

= RC as Rth =

τ

/ C

= 9.945 Ω.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13.

(a)

(b) 125

0.1 18.421ms

t

et

−

=∴=

3000 / 24 125

30

(0) 8(50 200) 192V

50

( ) 192 192 V

c

tt

c

v

vt e e

−−

=×=

==

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14.

(a) 64 4

10 /100 10 10

80 80 V, 0; 0.5 69.31

−− −

== >=∴=

tt t

c

ve e t e t s

µ

(b) 2 20,000 2

11

80 80 34.66

24

−

==∴=

t

c

wCe C t s

µ

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

46

4

/ 2 10 2 10

25 25 25

3

2

0: ( ) 0, 10 5000 10 mA

3

20

( ) 6.667V

3

0: 0 ( ) 6.667

6.667

( ) 6.667 V ( ) 0.3333 mA

20 10

−

−×××

−−−

<==+∴=

∴==

>=∴=

−

∴= ∴= =

×

csss

c

t

sc

ttt

cc

tit i ii

vt

ti vt e

vt e it e e

vC(t)

iC(t)

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16.

()

Amsi

Vemsv

Amsi

Vemsv

Ai

Vv

03

1203

05.1

5.4205.1

1.00

200

63

102050/103

102050/105.1

=

==

=

==

=

−−

×××−

+

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

(a)

(b)

500

0.4

(0)4A ()4 A(0 1ms)

(0.8ms) 4 2.681A

t

LL

L

iitet

ie

−

=∴ = ≤≤

==

0.5

250( 0.001)

0.25

(1ms) 4 2.426A

( ) 2.426

(2ms) 2.426 1.8895 A

L

t

L

ie

it e

ie

−

−−

==

∴=

∴= =

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18.

(a) 50,000 50,000 1

40 mA 10 40 , 27.73

−−

=∴=∴=

tt

L

ie e t s

µ

(b)

6

0.5

(1000 )50

(1000 )5 10

33

(10 ) 40 24.26mA

24.26 ( 10 )

10 24.26 2.426 0.8863

0.25(1000 R)10 ,1000 R 0.8863 4 10 R 2545

LL

Rt

R

ise i

ets

en

µ

−

−+

−+×

−+

== ∴

=>

∴= ∴ =

=+ +=××∴=Ω

l

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19.

(a)

(b) 0.75 1.5

(15 ) 40 45 28.94Vvs e e

µ

−−

=+=

(c)

12

50000 100000

(0) 20mA, (0) 15mA

( ) 40 45 V (0) 85V

−−

==

∴= + ∴ =

tt

ii

vt e e v

50000 100000 . 50000

2

50000

85 40 45 Let

10

45 40 8.5 0

40 1600 1530 0.17718, 0

90

0.17718, 34.61

ttt

t

ee ex

xx

x

ets

µ

−− −

−

=+ =

∴+−=

−± +

∴= = <

∴= =

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20.

22

2

12 1

12 1 2

50 50

112

12 12

12

50 /1000

22

1

2R R 2R

0: , (0)

R+R R R

2R 2R R

0: ( ) RR R+R

2R R

(0 ) 10 R R 5 . Also, (1ms)

R+R

5 10 0.05R 0.6931 R 13.863

111

R 7.821

13.863 R 5

−−

+

−

<= ↓=

+

>= ∴=

+

∴== ∴=Ω

== ∴ = ∴ = Ω

∴+=∴=Ω

RL

Rt Rt

LR

RR

R

tv i

tit e v e

vv

e

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

(a) 750

24

(0) 0.4A ( ) 0.4 A, 0

60

−

== ∴ = >

t

LL

iitet

(b)

750

524 20V, 0

6

3

(0 ) 50 0.4 7.5V

8

() 7.5 V, 0

+

−

=× = <

=× ×=

∴= >

x

t

x

vt

v

vt e t

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22.

25 /0.5 50

320 10 25

4

25 10 10 A, 0

−−

=×+ =

∴=Ω∴= = >

L

in L L

tt

in

in L

L

i

vii

v

vieet

i

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23.

= 2.5 A

24 /8 3

3

11

64 40

(0) 5A

440848

55A

( ) 2.5 A, 0; ( 0.1)

(0.03) 2.285 A, (0.1) 1.

8

−−

−

=×=

+

∴= =

∴= > −

==

L

tt

L

t

i

ie e

it e t i

ii

52 A

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24.

(a)

(b)

100

1.5

(0) 4A 4 A, 0 15ms

(15ms) 4 0.8925 A

−

−+

=∴= <<

∴==

t

LL

L

iiet

ie

20( 0.015)

0.3

15ms: 0.8925 A

(30ms) 0.8925 0.6612A

t

L

ti e

ie

+− −

+−

>=

∴= =

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25.

(a) 11 2 2

(0 ) (0 ) 10A, (0 ) (0 ) 20A (0 ) 30Aii ii i

+− + − +

== ==∴=

(b) 0.08 5

L / R ms 1.6667ms;

48 3

eq eq

τ

====

(c) 12

(0 ) 10A, (0 ) 20Aii

−−

==; 600

( ) 30 A

t

it e

−

=

(d) 600

48 1440 V

t

vi e

−

=− =−

(e)

(f)

600 600 600

10

0

600

20

600 600

0

10( 440) 10 24 10 24 14A

2.5( 1440) 20

6 20 6 14A

−−−

−

−−

=− += += −

=− +

=+=+

∫

ttttt

tt

tt t

iedtee

iedt

ee

22

L

2 1200

R00

11

W (0) 0.1 10 0.4 20 5 80 85J

22

11

W ( ) 0.1 14 0.4 14 9.8 39.2 49J

22

900 48

W 48 900 48 ( 1) 36J

1200

49 36 85 checks

L

t

idt e dt

∞∞

−

=× × +× × =+ =

∞= × × + × × = + =

×

==× =−=

−

∴+=

∫∫

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26.

(a)

(b)

(c)

τ

= 400 s, so 1.2

τ

= 480 s. vC(1.2

τ

) = 33.33 e–1.2 = 10.04 V.

Using Ohm’s law, we find that i1(1.2

τ

) = vC(1.2

τ

)/ 4000 = 2.51 mA.

(d) PSpice Verification:

We see from the DC analysis of the circuit that our initial value is correct; the Probe

output confirms our hand calculations, especially for part (c).

1

c1

22 1002

(0) 100 33.33V; (0 ) 16.667mA

223 223

v (9:59) 33.33V, (9:59) 16.667mA

c

vi

i

−

=× ×= = ×=

++

∴= =

/ 400 300/ 400

1

( ) 33.33 , 10:00 (10:05) 33.33

15.745

15.745 V, (10:05) 3.936mA

4000

t

cc

vt e t v e

i

−−

+

=>∴=

===

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

25

0: 1.25 34 100(1.25 0.8 ) 25 0.2A

20

xxxxxxx

i

ti iiiii>=∴= −++∴=

(a) (0 ) (1.25 0.8 1)0.2 0.290A

s

i−=−+=

(b) (0 ) 0.2A

x

i−=

(c) 5

( ) 25 0.2 5 V (0 ) 0.05A

100

tt

cx

vt e e i

−− +

=× = ∴ = =

(d) 34 20 33.2

0.8 (0 ) 0.04A (0 ) 0.04 0.2767A

120 120 120

xs

ii

++

=∴=−×==

(e) 0.4

1

(0.4) 5 0.03352A

100

x

ie

−

=× =

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28.

(a) 6

10 /(10 50 200) 20000

(0) 10V ( ) 10 10 V

tt

cc

vvte e

−+ −

=∴ = =

(b)

(c) PSpice Verification.

From the DC simulation, we see that PSpice verifies our hand calculation of iA = 50 mA.

The transient response plotted using Probe indicates that at 100 µs, the current is approximately

5.46 mA, which is within acceptable round-off error compared to the hand calculated value.

2

10

( 100 ) (0 ) 50mA

200

150

(100 ) 10 5.413mA

10 40 250

AA

A

isi

ise

µ

−

−===



==



+



CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29.

(a) 1

12

() 8( 1) 6mA( 0)

12 4

it t=− =− <

+

(b)

63

10 / 5 2 10 100

100

4 12 6 2 , (0) 48V

( ) 48 48 V, 0

() 12 mA, 0

−×× −

−

=Ω =

∴= = >

∴= >

c

tt

c

t

kv

vt e e t

it e t

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30.

(a)

(b)

66

10 /8 10 /0.8

1,250,000 125,000

(0) 20V, (0) 80V

20 , 80

80 20 V, 0

−−

==

∴= =

∴=−= − >

CLeft CRIGHT

tt

CL CR

tt

out CR CL

vv

veve

vvv e e t

1.25 0.125

6.25 0.625

(0 ) 60V; (1 ) 80 20 5.270V

(5 ) 80 20 10.551V

+−−

−−

==−=

=− =−

out out

out

vvsee

vse e

µ

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31. (a)

(b) vC(3 µs) = 9.447 V

(c) PSpice verification. Note that the switch parameters had to be changed in order to

perform this simulation.

As can be seen from the simulation results, our hand calculations are accurate.

6

10 /4 250,000

0.25 40

0: 0 20V( 0)

5104

1 0.25

0: Apply 1V 0.1 0.25A

5

1

R4

0.25

( ) 20 20 V( 0)

−−

<++=∴=<

−

>=∴+−=

∴= =Ω

∴= = >

c ccc c

cin

eq

tt

c

v vvv

tvt

tv i

vt e e t

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32.

6

63

10 /( 1000) 500/( 1000)

1000/ 2742.4

10 ( 10 )

1

10

0: (0) 60V

50

01ms: 60 60

R500 1.2 0.18232 2 5.4848, 1742.4

R 1000 500

(1ms) 60 41.67V

1ms : 41.67 /(1742.4 R 1000)

25 41.67

oo

c

tR R

c

oo

o

c

t

c

tv

tve e

nR

ve

tv e

e

−

−+ −+

−

−−

−

<=

<< = ∴

∴==∴+==Ω

+

∴= =

>= +

∴=

l

00() 1

1

3

11

1

.1000

0.5108 ,1742.4 R 1000

1742.4 R 1000

11

1957.6, R 1000 215.2 10 R 274.2

R 215.2

−

∴= +

+

==+=∴=Ω

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

(a) With the switch closed, define a nodal voltage V1 at the top of the 5-kΩ resistor.

Then,

0 = (V1 – 100)/ 2 + (V1 – VC)/ 3 + V1/ 5 [1]

0 = VC/ 10 + (VC – V1)/ 3 + (VC – 100) [2]

Solving, we find that VC = vC(0-) = 99.76 V.

(b) 7

10 /3939 2539

0: R 10 6.5 3.939 87.59 87.59 V ( 0)

−−

>==Ω∴= = >

tt

eq c

tkveet

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. t < 0:

39 6

111 11 1

111 1

/75 10 2 10 10 /150 6667

6667

14

12 4 20 0.5mA (0) 6 20 26

(0) 13V

0: Apply 1mA 1 0.6 2.5mA; 30 75V R 75

( ) 13 13 13

( ) 0.4333 mA ( 0)

310

−

−××× − −

−

=+ ∴= ∴ =+ =

=

>←∴+=∴=±==∴=Ω

∴= = =

∴= = >

×

c

in eq

ttt

c

t

o

iii v iii

v

tiiivik

vt e e e

v

it e t

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35.

(a) 12

(0 ) 100V. (0 ) 0, (0 ) 0

R

vvv

−−−

===

(b) 12

(0 ) 100V. (0 ) 0, (0 ) 100V

R

vvv

−++

===

(c) 64 2

20 5 10 2 10 8 10

20 5 s

τ

−−

×

=××=×

+

(d) 12.5

( ) 100 V, 0

t

R

vt e t

−

=>

(e) 12.5

4

()

() 5 mA

210

t

R

vt

it e

−

==

×

(f)

(g)

63

3 12.5 12.5 12.5

1

12.5 12.5 12.5

2

10 10

( ) 5 10 100 100 20 80V

20 50

1000

( ) 5 0 80 0 80 80V

5

−− − −

−−−

=−× += +=− +

=+=−+=−+

∫

ttttt

o

ttttt

o

vt e dt e e

62 62

12

62

12

625 4 4 6

11

( ) 20 10 80 64mJ, ( ) 5 10 80 16mJ

22

1

(0) 20 10 100 100mJ, (0) 0

2

25

25 10 2 10 2 10 ( 1)10 20mJ

25

64 16 20 100 checks

−−

−

∞−− −

∞= × × × = ∞ ×× × =

=× × × = =

=× ××=××− =

−

++ =

∫

cc

t

Ro

ww

wedt

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36.

(a) 0: 1mA (0) 10V, (0) 1mA (0) 10V, 0

sc L x

ti v i v t<=∴± =↓=−∴= <

(b)

49

/10 20 10 5000

3 3 /0.1 3 10000 10000

L

5000 10000

0: ( ) 10 10 V

i() 10 10 10 A () V, 0

( ) 10 V, 0

tt

c

tt t

L

tt

xcL

tvt e e

te e vtet

vvvt e e t

−

−×× −

−−− −

−−

>= =

=− − =− ∴± = >

∴=− = − >

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37.

(a) 0: 20V 20V, 20mA ( ) 20mA, 0<=∴= = ∴= <

scL x

tv v i it t

(b)

84

10000 / 2 10 10 5000

8 5000 5000

10000 5000 10000 5000

0: 0 ( ) 0.02 A; ( ) 20 20 V

( ) 2 10 20( 5000) 2 mA

( ) ( ) ( ) 0.02 0.002 A 20 2 m

A

tt t

sL c

tt

c

tttt

xLc

tv it e vt e e

it e e

it it it e e e e

−

−−×−

−−−

−−−−

>=∴= = =

↓=××− =−

=+= − = −

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38.

PSpice verification: Note that the switch properties were changed.

We see from the simulation result that the current through the fuse (R3) is 869 mA, in

agreement with our hand calculation.

()

blow.not does fuse thems, 100 prior toA 1 than less todropped hascurrent thesince

A .868501.11.0

A )( :0

A 1.1

909.0

1

R

V

)0(

1.0363.2

363.2

=∴

==

=>

===

×−

−

t

esi

etit

i

L

t

L

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39. (a)

1

11

300 ( 1)V, 120 ( 1)V; 3 ( )A

100

1.5: ( 1.5) 3 1A

300

120

0.5: ( 0.5) 1 0.6A;

300

120 300 120

0.5: 0.4A; 1.5: 0.6A

300 300 300

=− =−+=−

=− − = × =

−

=−=+=

= =−=− = =−=

AB c

vutv utiut

ti

ti ti

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40.

600 ( 1) , 600( 1) ( )V, 6( 1) ( 1)A

AB c

vtutvv tutitut

↑

=+=+ =−−

(a)

(b)

11

1

1.5: 0; 0.5: 600( 0.5) /300 1A

600(0.5) 600(1.5)

0.5: 4A

300 300

600(1.5) 600(2.5) 1

1.5: 6 0.5 3 5 1 9A

300 300 3

=− = =− = − =−

== + =

== + +××=++=

titi

ti

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

(a) 2 ( 1) 3 (1) 4 (3) 3 4 1uuu−− + =−+=

(b)

(c) (1) 1

4 (1) 4 1.4715

u

eu e

−−+

==

[5 (2)] [2 (1)] [1 ( 1)]

43112

uuu−+−−

=××=

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42.

(a)

(b) t < 0: The voltage source is shorting out the 30-Ω resistor, so ix = 0.

t > 0: ix = 60/ 30 = 2 A.

100 20

0: 0 10 6A

50 50

60

0: 0 0 2A

30

x

ti

<=++×=

>=++=

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43. 0.5:t=− 200 1/50 1

50 25 16.667, 2 3 2.5A

66.67 1/50 1/ 25 1/ 50 2

x

i==− =−=

++

0.5:t= 200 3A

66.67

x

i==

1.5:t= 100 1

3 2.5A

66.67 3

x

i=− ×=

2.5:t= 200 100 2A

50

x

i−

==

3.5:t= 100 2A

50

x

i=− =−

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44. () 4 16() 20( 4) 6( 6)Vvt ut ut ut=− + − − −

t (s)

v(t) (V)

4

-12

8

2

0 4 6

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45. (a)

(b) Resistor of value 2Ω

7()0.2()8( 2)3

(1) 9.8 volts

−+−+

=

ut ut t

v

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46.

(a) 200000

() (2 2 ) ()

m

t

L

it e ut

−

=− mA

(b)

33

200000 200000

() L 15 10 10 ( 2)

( 200000 ) ( ) 6 ( )V

−−

′

==×× −

−=

LL

tt

vt i

euteut

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47.

(a) 2.5 1

( ) 2 2(1 ) ( )A ( 0.5) 2A

t

L

it e ut i

−

=+ − ∴− =

(b) 1.25

(0.5) 2 2(1 ) 3.427A

L

ie

−

=+ − =

(c) 3.75

(1.5) 2 2(1 ) 3.953A

L

ie

−

=+ − =

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48.

Ai

Aii

ii

ns

R

L

R

L

th

3

1

33

1

3

6

3

33

102.3

1010

1010.107.4.

156

102.3

10500

102.3

107.41010

−

×=∴

×=+

×=×

=

×

==

Ω×=

×+×

×××

=

τ

(a)

(b)

(c)

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49.

(a)

(b) 1000

1( ) (100 80 ) ( )V

t

vt e ut

−

=−

20 /0.02

1000

() (4 4 ) ()

( ) 4(1 ) ( )A

t

L

t

L

it e ut

−

=−

∴=−

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50. (a) 0 W

(b) The total inductance is 30 || 10 = 7.5 mH. The Thévenin equivalent resistance is

12 || 11 = 5.739 kΩ. Thus, the circuit time constant is L/R = 1.307 µs. The final value of

the total current flowing into the parallel inductor combination is 50/12 mA = 4.167 mA.

This will be divided between the two inductors, so that i(∞) = (4.167)(30)/ (30 + 10) =

3.125 mA.

We may therefore write i(t) = 3.125[1 – e-106t/ 1.307] A. Solving at t = 3 µs,

we find 2.810 A.

(c) PSpice verification

We see from the Probe output that our hand calculations are correct by verifying using

the cursor tool at t = 3 µs.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



51.

(a) ( ) 10 , 0=<

L

it At

(b)

5/0.5

10

() 8 2

() 8 2 A, 0

t

L

t

L

it e

it e t

−

=+

∴=+ >

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



52.

(a) ( ) 2A, 0

L

it t=>

(b)

4 /0.1

40

() 5

() 5 3 A, 0

t

L

t

L

it e

it e t

−

=−

∴=− >

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



53.

(a)

(b)

(c) 25

10 10 ( )

Q( ) 2 2 ( ) 0.16 0.08 A, 0

5

−

+

==+∴=− >

t

ut

tutiet

(d)

P Q, Q , R 125 , L 5H

L LP LQ LP 5P R 125 P 25

−−

+= = + = Ω=

∴=∴===∴=

∫

Pt Pt Pt

di iieedtAe

dt

di i

dt

25 25 25 25 25 25

25

10 2

Q( ) 2 2 A A

L25

21022

A , (0) A 0 0.08A

25 125 25 25

−−− −

−

==∴= + = × +

∴= + = = ∴ = ∴= =

∫t

tt tt tt t

o

t

tieedteeee

iei i

25 25 25 25

25

10 ( ) 2

Q( ) 2 ( ) 2 A A

525

2

(0) 0 A ( ) 0.08(1 )A, 0

25

−−−

−

==∴= +=+

=∴ =− ∴ = − >

∫t

tt t t

o

t

ut

tutieedtee

iitet

25 25 25

25

25 25

22

25

25 25

10 ( )cos50

Q( ) 2 ( )cos50 2cos50

5

2 (25cos50 50sin 50 ) A

50 25

1

2 (25cos50 50sin50 ) 25 A

3125 3125

242

cos50 sin50

125 125 125

−−

− −

==∴= ×+



∴= + +



+





=+−×+





=+−

∫t

ttt

o

t

tt

o

t

t t

ut t

tuttietedtAe

e

ie t t e

e

ette

tte

25 25

25

A

22

(0) 0 0 A A 0

125 125

( ) 0.016cos50 0.032sin50 0.016 A, 0

−−

−

+

=∴= − + ∴ =

∴= + − >

tt

t

e

i

it t t e t

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



54.

(a) 0, 0

(b) 0, 200V

(c) 1A, 100V

(d)

34000

4000

11

50 10 1 ms 1(1 ) ( )A, (0.2ms) 0.5507A

200 4

( ) (100 100 ) ( )V, (0.2ms) 144.93V

t

LL

t

ieuti

vt e ut v

τ

−−

−

×

==∴=− =

=+ =

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



55.

(a) 100 100

( ) 15A, 0

20 5

L

it t=−=− <

(b) (0 ) (0 ) 15A

LL

ii

+−

==−

(c) 100

() 5A

20

L

i∞= =

(d) 40

() 5 20 A, 0

t

L

it e t

−

=− >

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



56.

L

9000

18 1

(0 ) 0.1A (0 ) 0.1A

60 30 2

i ( ) 0.1 0.1 0.2A

( ) 0.2 0.1 A, 0

( ) 0.1 ( ) (0.2 0.1 ) ( )A

or, ( ) 0.1 (0.1 0.1 ) ( )A

−+

−

=×=∴=

+

∞= + =

∴=− >

∴= −+−

=+ −

LL

t

L

t

L

t

L

ii

it e t

it u t e ut

it e ut

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



57.

(a) 30 3

(0 ) 3A, (0 ) 4A

7.5 4

xL

ii

−−

=×= =

(b) (0 ) (0 ) 4A

xL

ii

++

==

(c)

10 /0.5 20

0.8

() () 3A

() 3 1 3 A (0.04)

3 3.449A

−−

−

∞= ∞=

∴=+ =+ ∴

=+ =

xL

tt

xx

ii

it e e i

e

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



58.

(a) 30

(0 ) (0 ) 3A

10

xL

ii

−−

===

(b)

(c)

30 30 15

(0 ) 3 2.4A

30 7.5 40 10 15

+=×+×=

++

x

i

6 /0.5

12 0.48

30 30

() 3A () 30.6

7.5 40

3 0.6 (0.04) 3 0.6 2.629A

t

xx

t

x

iite

ei e

−

−−

∞= × = ∴ = −

=− ∴ =− =

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



59.

52 /0.2 260

260

OC: 0, 4 ( )V

0.2

SC : 0.1 ( ) ,12 ( ) 0.6 2

40 60

12 ( ) 12 ( ) ( )

2.6 60 2.6 60 13

4() ()

413 52 (1 )() (1 )()

52 13

60 4.615 (1 ) ( )V

−−

+−

==

−

=+ =+

∴= ∴= = =

×

∴=×=Ω∴= − = −

∴= = −

xoc

xxx xx

x

xab

tt

th L

t

xL

vvut

vvv

ut ut v v

vut ut ut

vi

ut ut

Rieuteut

vi eut

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



60.

(a)

(b)

11 1

1

40

OC: 100 30 20 0, 2A

80 ( )V

20 10

SC: 10A, 10 20A

20

R 4 () 20(1 ) ()A

−

−++ = =

∴=

×

=↓=+ =

∴=Ω∴ = −

oc

sc

t

th L

iii

vut

ii

it e ut

40 40

1

0.12040()80()

100 ( ) 80 ( )

() 10 8 ()A

10

tt

L

tt

veuteut

ut e ut

it e ut

−−

=×× =

−

∴= =−

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



61. Unfortunately, PSpice will not allow us to use negative time values. Thus, we must

perform the simulation starting from t = 0, and manually shift the results if needed to

account for sources that change value prior to t = 0.

Voltage source

vB(t) from

Exercise 39.

Current source

iC(t) from

Exercise 39.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



62. (a)

(b) PSpice verification.

As can be seen from the plot above, the PSpice simulation results confirm our hand

calculations of vC(t < 0) = 2 V and vC(t = 2 µs) = -3.06 V

93

10 / 2 10 500000

1

6

(0 ) 3 2V (0 )

9

6

() 26(27) 6V

7

() 68 68 V, 0

( 2 ) (0 ) 2V, (2 ) 6 8 3.057V

−+

−× −

−−

=×= =

∞= − =−

∴=−+ =−+ >

−= = =−+=−

cc

c

tt

c

cc c

vv

v

vt e e t

vsv vs e

µµ

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



63.

83 5

10 /10 10

10

(0 ) 2.5mA, ( ) 10mA

4

10 7.5

(0) 7.5V (0 ) 17.5mA

11

10 7.5 10 7.5 mA, 0, 2.5mA 0

−

+

−−

== ∞=

=∴ =+=

=+ =+ > = <

AA

cA

tt

AA

ii

vi

ie etit

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



64.

83

10 /1.75 10 57140

10

(0 ) 10mA, ( ) 2.5mA, (0) 0

1

10 1

(0 ) 1.4286mA 10mA, 0

1.75 4

2.5 (1.4286 2.5) 2.5 1.0714 mA, 0

AAc

AA

tt

A

iiv

iit

ie et

−

+

−× −

== ∞= =

=× ∴= <

=+ − =− >

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



65.

(a)

(b)

37 3

500

12 ( ) 24 ( )V

0: (0 ) 8V (0 ) 8V

2

0: ( ) 24 16V

3

200

RC 10 3 10 2 10

30

( ) 16 24 V, 0

( ) 8 ( ) (16 24 ) ( )

s

cc

c

t

c

t

c

vutut

tv v

tv

vt e t

vt u t e ut

−+

−−

−

=− − +

<=−∴=−

>∞=×=

=×××=×

∴=− >

∴=−−+−

500

12 24 8

(0 ) 0.4mA, (0 ) 3.2mA

30 10

24

( ) 0.8mA

30

( ) 0.4 ( ) (0.8 2.4 ) ( )mA

in in

in

t

in

ii

i

it ut e ut

−+

−

−+

==− = =

∞= =

=− + +

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



66.

(a) iin (-1.5) = 0

(b) iin (1.5) = 0

(0 ) 10V (0 ), (0 ) 0

(0 ) 0 ( ) 0 for all

−+−

+

== =

=∴ =

ccin

in in

vvi

iit t

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



67.

2.5

0.375 20( 0.5) 7.5( 0.5)

7.5(0.3)

0 0.5 : 10(1 )V

(0.4) 6.321V, (0.5) 7.135V

20 10 5 5 50 8

0.5: A ( ) 10 8 V, 4 8

12 6 6 3 3

50 50

( ) 7.135 16.667 9.532 V

33

(0.8) 16.667 9.532 15.

t

c

cc

c

tt

c

tsv e

vv

tv

vt e e

ve

−

−×− −−

−

<< = −

==

−

>=∴∞=++==Ω



=+ − = −





∴=− =662V

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



68.

6

/10 /33.33

30,000

3

OC: 0 1, 3 1 2V

100 100 100

SC : 3V 0.06A

100 100

R / 2/ 0.06 33.33

(1 ) 2(1 )

2(1 ) V, 0

th

xx x xoc

xx

xsc

th oc sc

tRC t

coc

t

vv v vv

vv

vi

vv e e

et

−−

−

−−

−+ =∴= =−=

=∴= + =

∴= = = Ω

∴= − = −

=− >

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



69.

(a) 0: 8(10 20) 240V ( ) 80V, 0

R

tvtt<+===<

(b)

(c) 0: ( ) 80V

R

tvt<=

(d)

6

/10 10 100000

100000

0: ( ) 8 30 240V (0 ) 240V

1

( ) : ( ) 8(10 10) 80V

2

( ) 80 160 80 160 V

( ) 80 160 V, 0

cc

c

tt

c

t

R

tvt v

tv

vt e e

vt e t

−

+

−× −

−

<=×=∴=

=∞ ∞= × + =

∴=+ =+

∴=+ >

6

/50 10 20000

20000

(0 ) 80V, ( ) 240V ( ) 240 160 240 160 V

20 80

(0 ) 80V, (0 ) 8 10 10 32 16 48V

30 20 50

() 80V () 8032 V, 0

tt

cc c

RR

t

RR

vv vte e

vv

vvtet

−

−−×−

−+

−

= ∞= ∴ =− =−

==×+×=+=

+

∞= ∴ = − >

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



70.

(a) For t < 0, there are no active sources, and so vC = 0.

For 0 < t < 1, only the 40-V source is active. Rth = 5k || 20 k = 4 kΩ and hence

τ

= Rth C = 0.4 s. The “final” value (assuming no other source is ever added) is found

by voltage division to be vC(∞) = 40(20)/(20 + 5) = 32 V. Thus, we may write

vC(t) = 32 + [0 – 32] e–t/ 0.4 V = 32(1 – e-2.5t) V.

For t > 1, we now have two sources operating, although the circuit time constant remains

unchanged. We define a new time axis temporarily: t' = t – 1. Then vC(t' = 0+) =

vC(t = 1) = 29.37 V. This is the voltage across the capacitor when the second source

kicks on. The new final voltage is found to be vC(∞) = 40(20)/ (20 + 5) +

100(5)/ (20 + 5) = 52 V.

Thus, vC(t') = 52 + [29.37 – 52] e-2.5t' = 52 – 22.63 e-2.5(t – 1) V.

(b)

(c)

For t < 0,

vC = 0.

We see from the simulation results that

our hand calculations and sketch are

indeed correct.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



71.

2000

,

1000

,

,,

2000 1000

1000 1000 2

1000

200 V

100(1 )V

0

200 100 100

100 200( ) 100 0,

100 10,000 80,000 0.25 0.75

400

0.5, 0.6931ms

−

−−

−

=

=−

=−=

∴=−

∴+ −=

−± +

==−±

∴==

t

xL

t

xc

xxLxc

tt

t

ve

vv v

ee

e

et

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



72.

61

11

62

10 /( 100)

1000/( 100) 1000/( 100)

1

10 /( 100) 3 1000

2

0: 0

01ms:9(1 )

1

89(1 ),

9

1000 2.197, R 355.1

R 100

1ms : 8 , 10 1 8 (R 100)

1000 2.079, R 480.9 100 380.9

R 100

c

tR

c

RR

tR

c

tv

tve

ee

tve tt e

−+

−+ −+

′

−+ − −

<=

<< = −

∴= − =

∴= =Ω

+

′

>= =−∴− +

∴= =−=Ω

+

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



73. For t < 0, the voltage across all three capacitors is simply 9 (4.7)/ 5.7 = 7.421 V. The

circuit time constant is

τ

= RC = 4700 (0.5455×10-6) = 2.564 ms.

When the circuit was first constructed, we assume no energy was stored in any of the

capacitors, and hence the voltage across each was zero. When the switch was closed, the

capacitors began to charge according to ½ Cv2. The capacitors charge with the same

current flowing through each, so that by KCL we may write

dt

dv

C

dt

dv

C

dt

dv

C3

3

2

1

1 ==

With no initial energy stored, integration yields the relationship C1v1 = C2v2 = C3v3

throughout the charging (i.e. until the switch is eventually opened). Thus, just prior to

the switch being thrown at what we now call t = 0, the total voltage across the capacitor

string is 7.421 V, and the individual voltages may be found by solving:

v1 + v2 + v3 = 7.421

10

-6 v1 – 2×10-6 v2 = 0

2×10-6 v2 – 3×10-6 v3 = 0

so that v2 = 2.024 V.

With the initial voltage across the 2-uF capacitor now known, we may write

v(t) = 2.024 e-t/ 2.564×10-3 V

(a) v(t = 5.45 ms) = 241.6 mV.

(b) The voltage across the entire capacitor string can be written as 7.421 e-t/ 2.564×10-3 V.

Thus, the voltage across the 4.7-kΩ resistor at t = 1.7 ms = 3.824 V and the dissipated

power is therefore 3.111 mW.

(c) Energy stored at t = 0 is ½ Cv2 = 0.5(0.5455×10-6)(7.421)2 = 15.02 µJ.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



74.

()

W0.02

V 3.31000//900107

W0.08

V 6.3900107.

W0.00110001.00

2

3

2

3

322

==

=ΩΩ××=

==

=××==

=×==<

−

R

V

P

V

R

V

P

RIV

RItP

final

init

8

6

4

2

Power (W)

07

Time (ms)

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



75. voltage follower 2

() ()

o

vt vt∴=

PSpice verification:

From the simulation results, we see that modeling the operation of this circuit using an

ideal op amp model does not provide an accurate accounting for the operation of the

actual circuit.

2

0.1

(0 ) 5(0.25 1) 1 ( )V

v ( ) 0, 1.25 8 10

() ()V

+

−

==

∞= = × =

∴=

t

o

vut

s

vt e ut

τ

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



76. voltage follower 2

() ()

o

vt vt∴=

6

2

10 /0.5 200

10,000

( ) 1.25 ( )V ( )

( ) 1.25 ( )

1.25 ( )V

−×

−

==

=

o

x

t

vt ut vt

vt e ut

eut

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



77. (a)

And we may write vo(t) = -0.2[1 + e-20×103t]u(t) V.

(b) PSpice verification:

We can see from the simulation result that our ideal op amp approximation is not

providing a great deal of accuracy in modeling the transient response of an op amp in this

particular circuit; the output was predicted to be negative for t > 0.

20,000

14

7 20,000 20,000

4

20,000

3 20,000

20,000 20,000

4

0 (virtual gnd) ( )A

10

4

10 0.2

10

( ) 0.2(1 ) ( )

() 10 () 0.4 ()V

( ) ( ) ( ) ( 0.2 0.2 0.4 ) ( )

t

tttt

co

o

t

c

t

R

tt

ocR

vieut

vedte

vt e ut

vt it e ut

vt vt vt e e ut

−

−−

−

−−

=∴=

∴= =−

∴= −

∴= =

∴=−− =−+ −

∫

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



78. For t < 0, the current source is an open circuit and so i1 = 40/ 50 = 0.8 A.

The current through the 5-Ω resistor is [40 – 10(0.8)]/ 5 = 7.2 A, so the inductor current

is equal to – 7.2 A

PSpice Simulation

From the PSpice simulation, we see that our t < 0 calculation is indeed correct, and find

that the inductor current at t = 50 ms is 7.82 A.

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



79. Assume at least 1 µA required otherwise alarm triggers.

Add capacitor C.

6

1

10 6

(1) 1 volt

1000

(0) .1.5 1.496 volts

1002.37

1

We have 1 1.496 or C 2.48 F

10 (1.496)

−

=

==

∴= = =

µ

l

c

C

v

en

CHAPTER EIGHT SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



80. One possible solution of many: implement a capacitor to retain charge; assuming the

light is left on long enough to fully charge the capacitor, the stored charge will run the

lightbulb after the wall switch is turned off. Taking a 40-W light bulb connected to 115

V, we estimate the resistance of the light bulb (which changes with its temperature) as

330.6 Ω. We define “on” for the light bulb somewhat arbitrarily as 50% intensity, taking

intensity as proportional to the dissipated power. Thus, we need at least 20 W (246 mA

or 81.33 V) to the light bulb for 5 seconds after the light switch is turned off.

The circuit above contains a 1-MΩ resistor in parallel with the capacitor to allow current

to flow through the light bulb when the light switch is on. In order to determine the

required capacitor size, we first recognise that it will see a Thevenin equivalent resistance

of 1 MΩ || 330.6 Ω = 330.5 Ω. We want vC(t = 5s) = 81.33 = 115e-5/

τ

, so we need a

circuit time constant of t = 14.43 s and a capacitor value of

τ

/ Rth = 43.67 mF.

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1.

11

12

22 22

1

22

L10, 6 , 8

6 , 8 adding,

14 2 7

1

6 7 49 48 , 6.928

LC

rad/s 6.928L 10, L 1.4434H,

11

C 14.434mF, 7 R 4.949

48L 2RC

o

oo

oo o

ssss

s

ω

ααω ααω

αα

ωω ω

−−

−

= Ω =− =−

∴− = + − − =− − −

−=−∴=

∴− =− + − ∴ = =

∴==

== =∴= Ω

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2. 100 200

40 30 mA, C 1mF, (0) 0.25V

−−

=− = =−

tt

c

ie e v

(a)

(b)

(c)

100 200

1

( ) 0.25 (40 30 ) 0.25

C

( ) 0.4( 1) 0.15( 1) 0.25

( ) 0.4 0.15 V

tt

c

oo

tt

vt idt e e dt

vt e e

−−

=−= − −

∴=− −+ −−

∴=− +

∫∫

22 22

12

3

22

100 200

R

100 , 200

300 2 , 150 1

1 500

150 ,R 3.333 Also,

2R10 150

200 150 22500 20000

1 100

20000 , L 0.5H

LC L

i ( ) 0.12 0.045 A

R

−

−−

=− =−α+ α −ω =− =−α− α −ω

∴− =− α α = −

∴+ = = Ω

−=−− −ω∴ω=

∴== =

∴== +

oo

tt

ss

s

v

tee

100 200

( ) ( ) ( ) (0.12 0.04) ( 0.045 0.03)

( ) 80 15 mA, 0

tt

Rc

tt

it i t i t e e

it e e t

−−

=− − = − + − +

∴= − >

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3. Parallel RLC with ωo = 70.71 × 1012 rad/s. L = 2 pH.

(a)

(b)

(c) α is the neper frequency: 5 Gs-1

(d)

(e) 95

12

510 7.071 10

70.71 10

−

α×

ζ= = = ×

ω×

o

2122

12 2 12

1(70.71 10 )

1

So 100.0aF

(70.71 10 ) (2 10 )

−

ω= = ×

==

××

oLC

C

91

10 18

1510

2

1

So 1 M

(10 )(100 10 )

−

α= = ×

==Ω

×

s

RC

R

22 9 121

1

22 9 121

2

5 10 70.71 10

−

=−α+ α −ω =− × + ×

=−α− α −ω =− × − ×

o

Sjs

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4. Given: 21

4, 2

=α=LRC RC

Show that 12

() ( )

−α

=+

t

vt e At A is a solution to

2

11

0[1]++=

dv dv

Cv

dt R dt L

112

11 2

2

11 2 1

2

1211

121

() ( )

() [2]

()()

()

(2 ) [3]

−α −α

−α

−α −α

−α

=−α+

=−α−α

=−α−α−α −α

=−α −α + −α

=−α −α −α

tt

t

tt

t

dv eA eAtA

dt

AAtAe

dv AAtA e Ae

dt

AAAAte

AAAte

Substituting Eqs. [2] and [3] into Eq. [1], and using the information initially provided,

2

1121

12 12

22

11 1

(2 ) ( ) ( )

22

11

() ()

24

0

−α −α −α

−α −α



−+++





−++ +

=

ttt

tt

Ae At Ae Ae

RC RC RC

At A e At A e

RC R C

Thus, 12

() ( )

−α

=+

t

vt e At A is in fact a solution to the differential equation.

Next, with 2

(0) 16==vA

and 121

0

()(16)4

=

=−α=−α=

t

dv AA A

dt

we find that 1416=+ αA

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5. Parallel RLC with ωo = 800 rad/s, and α = 1000 s-1 when R = 100 Ω.

2

1 so 5 F

2

1 so 312.5 mH

α= = µ

ω= =

o

C

RC

L

LC

Replace the resistor with 5 meters of 18 AWG copper wire. From Table 2.3, 18 AWG soft solid

copper wire has a resistance of 6.39 Ω/1000ft. Thus, the wire has a resistance of

100cm 1in 1ft 6.39

(5m) 1m 2.54cm 12in 1000ft

0.1048 or 104.8m

 

Ω

 

 

=Ω Ω

(a) The resonant frequency is unchanged, so 800 rad/sω=

o

(b) 31

1954.0 10

2

−

α= = × s

RC

(c)

Define the percent change as 100

ζ

−

ζ

×

ζ

new old

old

100

95300%

α−α

=×

α

=

new old

old

α

ζ=

ω

α

ζ=

ω

old

o

new

o

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6. 5H, R 8 , C 12.5mF, (0 ) 40V

+

==Ω= =Lv

(a)

(b)

2

28

1,2 1 2

12

12 12 2 2 1

1 1000 1

(0 ) 8A: 5, 16,

2RC 2 8 12.5 LC

4 5 25 16 2, 8 ( ) A A

1000 40

40 A A (0 ) (0 ) 80( 8 5) 1040

12.5 8

/ 2A 8A 520 A 4A 3A 480, A 160,A 120

( ) 120

o

tt

o

L

i

svtee

vi

vs

vt

+

−−

++

=α= = =ω==

××

ω= =−± − =− −∴ = +



′

∴=+ = − − = −−=−





=− − ∴− =− − ∴− =− = =−

∴=− 28

160 V, 0

tt

eet

−−

+>

28

34

34 34

28

443

(0 ) 40

(0 ) 8A Let ( ) A A ; (0 ) 5A

R8

(0 ) A A (0 ) (0 ) 8 5 13A;

40

(0 ) 2A 8A 8 A / 4 A 4A

5

3A 13 4, A 3, A 16 ( ) 16 3 A, 0

+

+−−+

+++

+

−−

==+ ===

∴=+=− − =−−=−

=− − = = ∴ =− −

∴− =− + = =− ∴ =− + >

tt

cR

Rc

tt

v

iiteei

iii

is

it e e t

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7. 1

(0) 40A, (0) 40V, L H, R 0.1 , C 0.2F

80

====Ω=iv

(a)

(b) i(t) = – v/ R – C dt

dv = tt-tt eeee 401040 10 -40)(0.2)(60)( 10)0.2(-20)(- 600 200 −−− −−−

= tt ee 40 10 120 160 −− − A

2

1,2

10 40

12 12

12

12 2 2 1

10 40

180

25, 400,

2 0.1 0.2 0.2

20, 25 625 400 10, 40

( ) A A 40 A A ;

1(0)

(0 ) 10A 40A (0 ) (0) 2200

CR

A 4A 220 3A 180 A 60, A 20

( ) 20 60 V,

−−

++

−−

α= = ω = =

××

ω= =− ± − = −

∴= + ∴=+



′′

=− − = − =−





∴−− =−∴− =−∴= =−

∴=− + >

o

tt

s

vt e e

v

vvi

vt e e t 0

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

663

26

3

1,2

2000 6000

12 1

3

12 12 2

100

(0) 2A, (0) 100V

50

10 3 10

4000, 12 10

2 50 2.5 100 2.5

16 12 10 200, 4000 2000

() A A , 0 A A 2

10 3

(0 ) 100 3000 2000A 6000A 1.5 A 3A 0.5 2A

100

Lc

o

tt

Ls

L

iv

w

s

it e e t

i

+

−−

+

== =

×

α= = = = ×

×× ×

−× = =− ±

∴= + >∴+=

−×

′=×=−=−−∴−=−−∴=−

2000 6000

21

2000 6000

A 0.25, A 2.25 ( ) 2.25 0.25 A, 0

0: ( ) 2A ( ) 2 ( ) (2.25 0.25 ) ( )A

tt

L

tt

LL

it e e t

tit itut e e ut

−−

∴=− = ∴ = − >

>=∴=−+ −

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9.

2

21

1,2

50 450

12 12 12

12 2 2 1

12

(0) 2A, (0) 2V

51

1000 1000 45

= 250, 22500

212 2

250 250 22500 50, 450

A A A A 2; (0 ) 45( 2) 50A 450A

A 9A 1.8 8A 0.2 A 0.025, A 2.025(A)

( ) 2.025

−

−− +

−

== =

+×

== =

××

=− ± − =− −

′

∴= + ∴ + = = −=− −

∴+ =∴− = ∴ =− =

∴=

Lc

o

tt

LL

L

iv

ss

ie e i

it e

αω

50 450

0.025 A, 0

−

−>

tt

et

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10.

(a)

(b)

(c)

Solving using a scientific calculator, we find that ts = 1.181 s.

2

436

1,2 1 2

12

1212 22 1

436

1 1440 1440

20, 144

2RC 72 10

20 400 144 4, 36: A A

118

(0) 18 , (0) 1440 0

236

0 4A 36A A 9A 18 8A , A 2.25, A 20.25

( ) 20.25 2.25 V, 0

o

tt

t

svee

vAAv

vt e e t

αω

+

−−

=== ==

=− ± − =− − = +



′

==+ = − =





∴=− − =− − =∴ =− =− =

∴= − >

436 4 36

436

1

( ) 0.5625 0.0625 0.05625 0.05625

36 1440

( ) 0.50625 0.00625 A, 0

tt t t

tt

v

it v e e e e

it e e t

−− − −

−−

′

=+ = − − +

∴= − >

436

max max

at 0 18V 0.18 20.25 2.25

ss

tt

vt v e e

−−

=∴ = ∴ = −

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11. L = 1250 mH

so Since α > ωo, this circuit is over damped.

The capacitor stores 390 J at t = 0−:

2

1

2

So (0 ) 125 V (0 )

+

=

===

cc

c

cc

WCv

W

vv

C

The inductor initially stores zero energy,

so

Thus, 82

() −−

=+

tt

v t Ae Be

Using the initial conditions, (0) 125 [1]==+vAB

38 2

3

(0 )

(0 ) (0 ) (0 ) 0 (0 ) 0

2

(0 ) 125

So (0 ) 62.5 V

22

50 10 [ 8 2 ]

(0 ) 62.5 50 10 (8 2 ) [2]

+

+++ +

+

−− −

+−

++=++=

=− =− =−

==×− −

=− =− × +

LRc c

c

tt

c

v

iii i

v

i

dv

iC Ae Be

dt

iAB

Solving Eqs. [1] and [2], A = 150 V

B = −25 V

Thus, 82

( ) 166.7 41.67 , 0

−−

=− >

tt

vt e e t

1

14rad/s

15

2

−

ω= =

α= =

oLC

s

RC

22

1,2

(0 ) (0 ) 0

53 8, 2

−+

==

=−α± α −ω =− ± =− −

LL

o

ii

S

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12. (a) We want a response 46−−

=+

tt

vAe Be

1

22 2

1

22 2

2

15

2

4525

6525

−

α= =

=−α+ α −ω =− =− + −ω

=−α− α −ω =− =− − −ω

oo

s

RC

S

Solving either equation, we obtain ωo = 4.899 rad/s

Since

2

11

, 833.3 mHω= = =

ω

o

L

LC C

(b) If (0 ) 10 A and (0 ) 15 A, find A and B.

++

==

Rc

ii

with

34 6

3

46

(0 ) 10 A, (0 ) (0 ) (0 ) 20 V

(0) 20 [1]

50 10 ( 4 6 )

(0 ) 50 10 ( 4 6 ) 15 [2]

Solving, 210 V, 190 V

Thus, 210 190 , 0

++++

−− −

+−

−−

====

=+=

==×− −

=× −− =

==−

=− >

RR c

tt

c

tt

ivvv

vAB

dv

i C Ae Be

dt

iAB

AB

ve et

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13. Initial conditions: 50

(0) (0) 0 (0) 2 A

25

−+ +

== ==

LL R

ii i

(a) (0 ) (0 ) 2(25) 50 V

+−

===

cc

vv

(b) (0 ) (0 ) (0 ) 0 2 2 A

+++

=− − = − =−

cLR

iii

(c) t > 0: parallel (source-free) RLC circuit

1

14000

2

13464 rad/s

−

α= =

ω= =

o

s

RC

LC

Since α >

ω

0, this system is overdamped. Thus,

Solving, we find A = −25 and B = 75

so that 2000 6000

( ) 25 75 , 0

−−

=− + >

tt

c

vt e e t

(d)

(e) 2000 6000

25 75 0 274.7

−−

−+ =⇒=

µ

tt

ee t s

using a scientific calculator

(f) max 25 75 50 V=− + =

c

v

So, solving | 2000 6000

25 75

−−

−+

ss

tt

ee| = 0.5 in view of the graph in part (d),

we find ts = 1.955 ms using a scientific calculator’s equation solver routine.

2000 6000

6 2000 6000

()

(5 10 )( 2000 6000 )

(0 ) 0.01 0.03 2 [1]

and (0 ) 50 [2]

−−

−− −

+

=+

==× − −

=− − =−

=+=

tt

c

tt

c

v t Ae Be

dv

i C Ae Be

dt

iAB

vAB

22

2000, 6000

=−α± α −ω

=− −

o

s1,2

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14. Due to the presence of the inductor, (0 ) 0

−=

c

v. Performing mesh analysis,

12

21 2

12

92 2 0 [1]

22370 [2]

and

−+ − =

−+ + =

−=

A

ii

iiii

ii i

Rearranging, we obtain 2i1 – 2i2 = 0 and –4i1 + 6 i2 = 0. Solving, i1 = 13.5 A and i2 = 9 A.

(a) 12 2

(0 ) 4.5 A and (0 ) 9 A

−−

=−= = =

AL

iii ii

(b) t > 0:

(c) (0 ) 0

−=

c

v due to the presence of the inductor.

(d)

(e)

To find the second equation required to determine the coefficients, we write:

=− −

LcR

cA

iii

dv

Ci

dt

3

(0 ) 9 25 10 [ 1.881(6 ) 4.785(6 )]

+−

==−× − − −−

L

iABAB

or 9 = -0.7178A – 0.2822B [2]

Solving Eqs. [1] and [2], A = −20.66 and B = +20.66

So that 4.785 1.881

( ) 20.66[ ]

−−

=−

tt

A

it e e

1

22

1,2

13.333

2

13 rad/s

1.881, 4.785

−

α= =

ω= =

=−α± α −ω =− −

o

s

RC

LC

S

4.444 H

→i2

→i1

(0 ) 7 (0 ) 3 (0 ) 2 (0 ) 0

so, (0 ) 0

++++

+

−+ − + =

=

cAAA

A

viii

i

around left mesh:

4.444 H

1 A

73(1)20

6

6 V 6

1

−+− +=

=∴ ==Ω

LC

LC TH

v

vR

1.881 4.785

()

(0 ) 0 [1]

−−

+

=+

==+

tt

A

i t Ae Be

iAB

Thus,

3 1.881 4.785

1.881 4.785

25 10 1.881(6 ) 4.785(6 )

−− −

−−



−× − −



−

tt

Ae Be

=

-

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15. Diameter of a dime: approximately 8 mm. Area = 22

0.5027cmπ=r

Capacitance

4H

=

µ

L

179.89Mrad/sω= =

oLC

For an over damped response, we require α > ωo.

Thus,

or 159.8<ΩR

*Note: The final answer depends quite strongly on the choice of εr.

14 2

(88)(8.854 10 F/cm)(0.5027cm )

0.1cm

39.17pF

−

εε ×

==

=

ro

A

d

6

12 6

179.89 10

2

1

2(39.17 10 )(79.89 10 )

−

>×

<××

RC

R

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16.

crit. damp. (a) 2326

100

L 4R C 10 4R 10 R 57.74

3

−−

==×=×∴=Ω

(b)

31

3464

12

2

65

121

3464 5

1

10 / 2.5 3464

30

( ) (A A ) (0) 100V

100

(0) 1.7321A 100 A

57.74

10 100

(0 ) 1.7321 0 A 3464A A 3.464 10

2.5 57.74

( ) (3.464 10 100)V, 0

o

t

cc

L

c

t

c

s

vt e t v

i

v

vt e t t

ωα

−

+

−

== × =

∴= + =

== ∴=



′=−==−∴=×





∴= ×+ >

R

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

crit. damp. (a) 23

L4RC41210 8mH

−

==×××=

(b)

(c)

max

250

max

: (250 2) 0, 1 250 2, 0 No!

0, 2A 0.02 (250 2); SOLVE: 23.96ms

s

Lm mm

t

mL s s

it tt

ti e t t

−

+= = + <

∴= = ∴ = + =

250

12

250

1

3 3 1.25

1

1 1000 250 (A A )

2RC 2 1 2

(0) 2A, (0) 2V (A 2)

Then 8 10 (0 ) 2 8 10 (A 500), (1.25 2) 0.9311A

t

oL

t

Lc L

L

ie t

iv iet

ie

αω

−

−+ − −

===∴= +

××

==∴=+

′

×=−=×−=+=

L

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18. 8

L 5mH, C 10 F, crit. damp. (0) 400V, (0) 0.1A

−

== =− =vi

(a) 2328

L 4R C 5 10 4R 10 R 353.6

−−

==×= ∴=Ω

(b)

(c) max (0) 0.1Aii∴= =

8141,420

12

141,421 3

21

11

141,421

10 141,420 (A A )

2 353.6

A 0.1 (A 0.1), 5 10

(A 141,420 0.1) 400 A 65,860

( 65,860 0.1). 0

( 65860) 141,420 ( 65,860 0.1) 0

8.590 ( )

t

tt

m

mm

ie t

et

ie t i

eet

tsit

αα

α

µ

−

−−

−

−−

==∴= +

×

∴=∴= + ×

−×=−∴=−

′

∴= − + =

∴+ + − +=

∴= ∴ = 6

141,420 8.590 10

6

max

( 65,860 8.590 10 0.1) 0.13821A

i ( ) 0.13821A

m

e

it

−

−××

−

−××+=−

∴= =

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19. Diameter of a dime is approximately 8 mm. The area, therefore, is πr2 = 0.5027 cm2.

The capacitance is

14

(88)(8.854 10 )(0.5027)

0.1

39.17 pF

−

εε ×

=

ro

A

d

with 1

4 H, 79.89 Mrad/s=µ ω= =

o

LLC

For critical damping, we require 1

2=ω

o

RC

or 1159.8

2

==Ω

ωo

RC

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20. Critically damped parallel RLC with 31

10 , 1M

−−

α= = ΩsR .

We know 3

3

6

110

10 , so 500 F

2210

−

===

µ

×

C

RC

Since α = ωo,

If

29

2

72

9

29 22 9

13

210

50 turns 1 m

(4 10 H/m) . (0.5cm) . .

cm 100 cm

So

210

(4 10 )(50) (0.5) 2 10

So 8.106 10 cm

−

µ

==×



 

π× π

 



 



=×

π× = ×

=×

NA

LS

s

3

6

110

1

or 10

so 2 GH (!)

−

ω= =

=

oLC

LC

L

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

663

27

66

4000

12

6

110 110

4000, 2 10

2RC 100 2.5 LC 50

20 10 16 10 2000

(B cos 2000 B sin 2000 )

(0) 2A, (0) 0 (0 ) 2A; (0 ) (0 ) (0 )

11 1 210

(0 ) (0) (0 ) 0 (0 )

L R RC 125

o

d

t

c

Lcc cLR

ccc c

ie t t

ivi iii

ivv i

αω

ω

+

−

++++

+++

== = ===×

×

=×−×=

∴= +

′

′′

==∴=− =−−

×

′′

∴=− − =− =

∴6

122

4000

210

B 2A, 16,000 2000B ( 2)( 4000) B 4

125

( ) ( 2cos2000 4sin 2000 )A, 0

t

c

it e t t t

−

×

=− = = + − − ∴ =

∴= − + >

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22.

2

12

14 1413

1, 26, 26 1 5

2RC 2 2 LC 2

() ( cos5 sin5)

od

t

c

vt e B t B t

αω ω

−

×

===== = =−=

×

∴= +

(a) (0 ) (0) 4A

LL

ii

+==

(b) (0 ) (0) 0

cc

vv

+==

(c) 1

(0 ) (0 ) 0

L

Lc

iv

++

′==

(d) (0 )

1

(0 ) [ (0 ) (0 )] 4 4 4( 4 0) 16

2

c

cLR

v

vii

c

+

+++



′=− − =−− =−+=−



 V/s

(e)

(f)

11 2 2

2

( )0 1(B ) B 0, ( ) B sin5 , (0 ) B (5) 16

B 3.2, ( ) 3.2 sin5 V, 0

t

cc

t

c

evtetv

vt e t t

−+

−

′

∴=∴= = = =−

∴=− =− >

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23.

(a)

(b)

93 9

28

22 6 6

5000 4 4

12

5000 4 4

2

9

110 110

5000, 1.25 10

2RC 2 20 5 LC 1.6 5

125 10 25 10 10,000

() (Bcos10 B sin10 )

(0) 200V, (0) 10mA ( ) (200cos10 B sin10 )

110

(0 ) (0 ) (0)

5

−

++

α= = ω== =×

×× ×

ω= ω−α= × − × =

∴= +

==∴= +

′== −

o

do

t

c

t

cL c

c

cc L

vt e t t

vi vte tt

v

vi i

c

924

2

5000 4 4

2

(0)

20,000

10 200

10 0 10 B 200(5000)

5 20,000

B 100V ( ) (200cos10 100sin10 ) V, 0

−









=−==−





∴= ∴ = + >

t

c

vt e t t t

2

5000 4

500 6 6 5000 4

5000 9 6 5000 4

5

1

10 , C

R

[10 ( 200sin 100cos] 5000(200cos 100sin)]

[10 ( 2sin 0.5cos)] 2.5 10 sin10 /

1(200cos 100sin) 5 10 2.5 10 sin10

20,000

−

−−

−−−

−

′

=− = +

′=−+− +

=−−=−×



∴= + −× × ×





=

sw L L c c

t

c

tt

t t

L

iiivv

ve

eetvs

ie e t

e000 4 4

5000 4 4

(0.01cos10 0.0075sin10 )A

10 (10cos10 7.5sin10 ) mA, 0

−

∴=− − >

t

sw

tt

ie t tt

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24.

(a)

(b)

(c)

222

2

1 100 1 100

8, , 36 64

2RC 12.5 LC L

100

100 L 1H

L

odo

o

α= = = ω = = ω = =ω −

∴ω = = ∴ =

812

8

12

22

8

0: ( ) 4A; 0: ( ) (B cos6 B sin 6 )

(0) 4A B 4A, (4cos6 B sin 6 ) (0) 0

(0 ) (0 ) 0 6B 8(4) 0, B 16/ 3

( ) 4 ( ) (4cos6 5.333sin6 ) ( )A

−

++

−

<=>= +

=∴= = + =

′==∴−==

∴=−+ +

t

LL

t

LL c

Lc

t

L

tit tite t t

iiettv

itv

it u t e t tut

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25.

63 12 6

44

100 12

100

12

66

1

110 1

100 , 1.01 10

2RC 2 5 LC

60

101 10 10 100; (0) 6mA

10

(0) 0 ( ) (A cos1000 A sin1000 ), 0

A 0, ( ) A sin1000

11

(0) (0) 10[ (0) (0)] 10

C 5000

(610

−−

−

++ + +

−

α= = = ω = = ×

×

∴ω = × − = = =

=∴ = + >

∴= =

′==−− =

−×

o

dL

t

cc

t

c

cc c

s

i

vvte t tt

vt e t

vi i v

3

22

100 14

4 100

100

1

) 6000 1000A A 6

1

( ) 6 sin1000 V, 0 ( ) 10

( ) 10 ( 6) sin1000 A

( ) 0.6 sin1000 mA, 0

−

−−

−

=− = ∴ =−

∴=− >∴=−

=− −

∴= >

t

c

t

c

t

vt e t t it

vt e t

it e t t

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26.

(a)

(b)

66

2

22

20

12

1

20

2

22

6

1 10 1 1.01 10

20, 40,400

2RC 2000 25 LC 25

40,400 400 200

(A cos200 A sin 200 )

(0) 10V, (0) 9mA A 10V

(10cos 200 A sin 200 ) V, 0

1

(0 ) 200A 20 10 200(A 1) (0 )

C

10

2

o

do

t

L

t

o

ve t t

vi

ve t t t

vi

−

++

×

α= = = ω = = =

×

ω= ω−α= − =

∴= +

==∴=

∴= + >

′=−×= −=

=3

2

20

( 10 ) 40 A 1 0.2 0.8

5

( ) (10cos 200 0.8sin 200 ) V, 0

t

vt e t t t

−

−=−∴=−=

∴= + >

20

10.032 cos (200 4.574 )V

2

T3.42ms

200

−

=−°

==

t

ve t

π

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

11

1

21 1

/

112

/

2

1

(0) 0; (0) 10A

(Acos Bsin ) A 0,

Bsin

[Bsin Bcos ]0

1

tan , tan

1T;

2

Bsin B

sin ; let

V

mmd

d

t

dd

t

d

t

dd d

dd

dm

d

mm dm

d

tt

mdmm

m

dm

m

vi

ve t t

ve t

ve t t

tt

tt t

ve tv e

v

te

α

−α

−

−α −α −απ ω

−απ ω

==

=ω+ω∴=

=ω

′=−αω+ω ω=

ωω

∴ω= =

αωα

π

=+ =+

ω

=ω=−

ω∴=− 2

1

/

222

0

2

1

2

1

100

121

100, 100; ,

2RC R

1 21 100

6 6 441/ R 6R 441

LC R R

21

R 1/6 441 10.3781 To keep

100

0.01, chose R 10.3780 (0 )

21 0

BB6 4R10

10.378 10.

d

m

d

md

m

v

en

n

vv

v

απ ω

+

=

ω

∴ = α= α= =

π

ω= = ∴ω= − ∴ −

π



π



∴= + = Ω











′

<=Ω=ω



=− = +





l

2

2.02351

1

21

B 1.380363

3780

21 21

2.02351; 6 1.380363

10.378 10.378

304.268 sin 1.380363 0.434 ,

71.2926 Computed values show

2.145sec; 0.7126 0.01

d

t

m

sm m

ve tvts

vv

tv v

−



∴=







α= = ω = − =





∴= =

=

==<

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28. We replace the 25-Ω resistor to obtain an underdamped response:

LC

1

and

2RC

1

0==

ωα

; we require

α

<

ω

0.

Thus, 3464

R 1010

1

6<

×− or R > 34.64 mΩ.

For R = 34.64 Ω (1000× the minimum required value), the response is:

v(t) = e-

α

t (A cos

ω

dt + B sin

ω

dt) where

α

= 2887 s-1 and

ω

d = 1914 rad/s.

iL(0+) = iL(0-) = 0 and vC(0+) = vC(0-) = (2)(25) = 50 V = A.

iL(t) = dt

dv

dt

dv CL L L =

=

()()

[

]

tBtAettBttAe dd

t

dddd

t

ωωαωωωω αα

sin cos - cos sin L ++− −−

iL(0+) = 0 =

[]

A - B

3

1050

d

3

αω

−

×, so that B = 75.42 V.

Thus, v(t) = e-2887t (50 cos 1914t + 75.42 sin 1914t) V.

From PSpice the settling time using R = 34.64 Ω is approximately 1.6 ms.

Sketch of v(t). PSpice schematic for t > 0 circuit.

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29. (a,b) For t < 0 s, we see from the circuit below that the capacitor and the resistor are

shorted by the presence of the inductor. Hence, iL(0-) = 4 A and vC(0-) = 0 V.

When the 4-A source turns off at t = 0 s, we are left with a parallel RLC circuit such that

α

= 1/2RC = 1 s-1 and

ω

0 = 5.099 rad/s. Since

α

<

ω

0, the response will be underdamped

with

ω

d = 5 rad/s. Assume the form iL(t) = e-

α

t (C cos

ω

dt + D sin

ω

dt) for the response.

With iL(0+) = iL(0-) = 4 A, we find C = 4 A. To find D, we first note that

vC(t) = vL(t) = L dt

diL

and so vC(t) = (2/13) [e-

α

t (-C

ω

d sin

ω

dt + D

ω

d cos

ω

dt) -

α

e-

α

t (C cos

ω

dt + D sin

ω

dt)]

With vC(0+) = 0 = (2/13) (5D – 4), we obtain D = 0.8 A.

Thus, iL(t) = e-t (4 cos 5t + 0.8 sin 5t) A

(c) Using the cursor tool, the settling

time is approximately 4.65 s.

We see that the simulation

result confirms our hand

analysis; there is only a

slight difference due to

numerical error between

the simulation result and

our exact expression.

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30. (a) For t < 0 s, we see from the circuit that the capacitor and the resistor are shorted by

the presence of the inductor. Hence, iL(0-) = 4 A and vC(0-) = 0 V.

When the 4-A source turns off at t = 0 s, we are left with a parallel RLC circuit such that

α

= 1/2RC = 0.4 s-1 and

ω

0 = 5.099 rad/s. Since

α

<

ω

0, the response will be

underdamped with

ω

d = 5.083 rad/s. Assume the form iL(t) = e-

α

t (C cos

ω

dt + D sin

ω

dt)

for the response.

With iL(0+) = iL(0-) = 4 A, we find C = 4 A. To find D, we first note that

vC(t) = vL(t) = L dt

diL

and so vC(t) = (2/13) [e-

α

t (-C

ω

d sin

ω

dt + D

ω

d cos

ω

dt) -

α

e-

α

t (C cos

ω

dt + D sin

ω

dt)]

With vC(0+) = 0 = (2/13) (5.083D – 0.4C), we obtain D = 0.3148 A.

Thus, iL(t) = e-0.4t (4 cos 5.083t + 0.3148 sin 5.083t) A and iL(2.5) = 1.473 A.

(b)

α

= 1/2RC = 4 s-1 and

ω

0 = 5.099 rad/s. Since

α

<

ω

0, the new response will still be

underdamped, but with

ω

d = 3.162 rad/s. We still may write

vC(t) = (2/13) [e-

α

t (-C

ω

d sin

ω

dt + D

ω

d cos

ω

dt) -

α

e-

α

t (C cos

ω

dt + D sin

ω

dt)]

and so with vC(0+) = 0 = (2/13) (3.162D – 4C), we obtain D = 5.06 A.

Thus, iL(t) = e-4t (4 cos 3.162t + 5.06 sin 3.162t) A and iL(.25) = 2.358 A.

(c)

We see from the simulation

result below that our hand

calculations are correct; the

slight disagreement is due to

numerical inaccuracy.

Changing the step ceiling

from the 10-ms value

employed to a smaller value

will improve the accuracy.

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31. Series:

2

4

12

1

22

4

214

4, 20, 20 16 2

2L 1/ 2 LC 0.2

(A cos 2 A sin 2 ); (0) 10A, (0) 20V

1

A 10; (0 ) (0 ) 4(20 20) 0

L

(0 ) 2A 4 10 A 20

( ) (10cos 2 20sin 2 )A, 0

od

t

LLc

LL

L

t

L

R

ie t ti v

iv

i

it e t t t

αω ω

−

++

+

−

== = === = −=

∴= + = =

′

∴= = = − =

′

∴=−×∴=

∴= + >

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32.

22

20

12 1

20

2

20

22

R80

(0)50802210V,(0)0, 20

2L 4

100 500: 500 20 10

2

( ) (A cos10 A sin10 ) A 210V

1

( ) (210cos10 A sin10 ); (0 ) (0 ) 0

C

0 10A 20(210), A 420 ( ) (210cos10

cL

od

t

c

t

ccc

t

c

vi

vt e t t

vt e t t v i

vt e t

α

ωω

−

−++

−

=+×= = = = =

== = −=

∴= + ∴=

′

∴= + = =

∴= − = ∴ = +

0.8

20

12

22

20

420sin10 )

(40ms) (210cos0.4 420sin 0.4) 160.40V

Also, (B cos10 B sin10 ),

11 1

(0 ) (0 ) [0 (0 )] 210

L2 2

(0 ) 105 10B B 10.5

( ) 10.5 sin10 A, 0

( ) 80 840 sin

c

t

L

LL c

L

t

L

t

RL

t

ve

ie t t

iv v

i

it e t t

vt i e

−

++ +

+

−

∴= + =

=+

==−=×

′

∴=−=∴=

∴=− >

∴==

0.8

20

0.8

10 V

(40ms) 840 sin 0.4 146.98V

() () () ()

(40ms) 160.40 146.98 13.420V

[check: ( 210cos 420sin 840sin)

( 210cos10 420sin10 ) V, 0

(40ms) ( 210cos 420sin 840

R

LccRL

t

L

t

L

t

ve

vt vt vt vt v

ve

ettt

ve

−

∴=− =−

=− − − ∴

=− + =−

=− − +

=− + >

∴=−−+ 20

0.8

sin)

( 210cos10 420sin10 )V, 0

V (40ms)

(420sin 0.4 210cos0.4) 13.420VChecks]

t

L

e

ttt

e

−

=

−+ >

∴=

−=−

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33. “Obtain an expression for vc(t) in the circuit of Fig. 9.8 (dual) that is valid for all t′′.

66

27

66

1,2

2000 6000

12

3

R 0.02 10 10 3

4000, 1.2 10

2L 2 2.5 2.5 10

4000 16 10 12 10 2000, 6000

1

( ) A A ; (0) 100 2V

50

1

(0) 100A 2 A A , (0 ) C

3

( (0)) 10 100 3000 /

100

3000 200A

o

tt

cc

Lc

L

s

vt e e v

iv

ivs

αω

−−

+

××

== = = =×

××

∴=− ± ×−× =− −

∴= + =×=

′

=∴=+ =

−=−××=−

∴− =− 12 12

21

200 6000

600A , 1.5 A 3A

0.5 2A , 0.25, A 2.25

( ) (2.25 0.25 ) ( ) 2 ( )V (checks)

tt

c

vt e e ut u t

−−

−−=−−

∴=− =− =

∴= − +−

A

Ω

µF

mF

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. (a)

(b)

(c)

,max 0.2489V

c

v=

2

22 2

2

46 4

10000

12

10000

21

6

1

R11

crit. damp; L R C

4L LC 4

1 200

L 4 10 0.01H, 10

4 0.02

( ) (A A ); (0) 10V, (0) 0.15A

1

A10,() (A0);(0)

C

(0) 10 ( 0.15) 150,000

Now, (0 ) A

−

−+

+

===∴=

∴=×× = = = =

∴= + =− =−

′

∴=− = − =−

=− − =

′=

o

t

ccL

t

cc

L

c

vt e t v i

vt e t v

i

v

αω

5

1

10,000

10 150,000 A 50,000

( ) (50,000 10) V, 0

−

+= ∴=

∴= − >

t

c

vt e t t

10,000

33

max

( ) [50,000 10,000(50,000 10)]

15

5 50,000 10 0.3ms

50,000

( ) (15 10) 5 0.2489V

(0) 10V 10V

t

c

mm

cm

cc

vt e t

tt

vt e e

vv

−

−−

′=− −=∴

=−∴==

∴= −==

=− ∴ =

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35. (a)

(b)

6

2

22

1,2

10 40

12

111

11

R250 1 10

25, 400

2L 10 LC 2500

25 15 10, 40

A A , (0) 0.5A, (0) 100V

11

0.5 A A , (0 ) (0 )

55

(100 25 100) 5 A / 10A 40A

5 10A 40 (0.5 A ) 10A 40

A20 30A

o

tt

LLc

LL

s

ie ei v

iv

s

αω

ααω

−−

++

== = == =

=− ± − =− ± =− −

∴= + = =

′

∴=+ = =

− − =− =− −

∴= + − = −

+∴− 12

10

15, A 0.5, A 0

( ) 0.5 A, 0

t

L

it e t

−

=− = =

∴= >

10 40

34 34

6

34 443

10

A A 100 A A ;

110

(0 ) ( 0.5) 1000

500

10A 40A 1000 3A 0, A 0, A 100

( ) 100 V 0

tt

c

cc

t

c

ve e

vi

c

vt e t

−−

+

−

=+∴=+

′

′=−=−

∴− − =− ∴− = = =

∴= >

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36. (a)

(b)

222

12

2

R2 1

1, 5, 2

2L 2 LC

(B cos2 B sin2 ), (0) 0, (0) 10V

B0, B sin2

1

(0) (0 ) (0 ) V (0 ) 0 10 2B

1

B5 5sin2A,0

odo

t

LLc

t

L

LR c

t

L

ie t ti v

iet

ivv

iett

αωωωα

−

+++

−

=== == = −=

∴= + = =

∴= =

==−=−=

∴=∴=− >

11

22

2max

max

5[ (2cos2 sin2 )] 0

2cos2 sin 2 ,tan 2 2

0.5536 , ( ) 2.571A

2 2 0.5536 , 2.124,

( ) 0.5345 2.571A

and 0.5345A

−

′=− − =

∴= =

∴= =−

=× + =

=∴=

=

t

L

LL

L

ie tt

ttt

tsit

tt

it i

i

π

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37. We are presented with a series RLC circuit having

α = R/2L = 4700 s-1 and

ω

o = 1/ LC = 447.2 rad/s; therefore we expect an

overdamped response with s1 = -21.32 s-1 and s2 = -9379 s-1.

From the circuit as it exists for t < 0, it is evident that iL(0-) = 0 and vC(0-) = 4.7 kV

Thus, vL(t) = A e–21.32t + B e-9379t [1]

With iL(0+) = iL(0-) = 0 and iR(0+) = 0 we conclude that vR(0+) = 0; this leads to vL(0+) =

-vC(0-) = -4.7 kV and hence A + B = -4700 [2]

Since vL = L dt

di , we may integrate Eq. [1] to find an expression for the inductor current:

iL(t) = 









−− tt ee 937932.21

9379

B

-

21.32

A

-

L

1

At t = 0+, iL = 0 so we have 0

9379

B

-

21.32

A

-

10500

1

3- =













× [3]

Simultaneous solution of Eqs. [2] and [3] yields A = 10.71 and B = -4711. Thus,

vL(t) = 10.71e-21.32t - 4711 e-9379t V, t > 0

and the peak inductor voltage magnitude is 4700 V.

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38. Considering the circuit as it exists for t < 0, we conclude that vC(0-) = 0 and iL(0-) = 9/4 =

2.25 A. For t > 0, we are left with a parallel RLC circuit having α = 1/2RC = 0.25 s-1 and

ω

o = 1/ LC = 0.3333 rad/s. Thus, we expect an underdamped response with

ω

d =

0.2205 rad/s:

iL(t) = e-αt (A cos

ω

dt + B sin

ω

dt)

iL(0+) = iL(0-) = 2.25 = A

so iL(t) = e–0.25t (2.25 cos 0.2205t + B sin 0.2205t)

In order to determine B, we must invoke the remaining boundary condition. Noting that

vC(t) = vL(t) = L dt

diL

= (9)(-0.25)e-0.25t (2.25 cos 0.2205t + B sin 0.2205t)

+ (9) e-0.25t [-2.25(0.2205) sin 0.2205t + 0.2205B cos 0.2205t]

vC(0+) = vC(0-) = 0 = (9)(-0.25)(2.25) + (9)(0.2205B)

so B = 2.551 and

iL(t) = e-0.25t [2.25 cos 0.2205t + 2.551 sin 0.2205t] A

Thus, iL(2) = 1.895 A

This answer is borne out by PSpice simulation:

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39. Considering the circuit at t < 0, we note that iL(0-) = 9/4 = 2.25 A and vC(0-) = 0.

For a critically damped circuit, we require α =

ω

o, or LC

1

RC2

1=, which, with

L = 9 H and C = 1 F, leads to the requirement that R = 1.5 Ω (so α = 0.3333 s-1).

The inductor energy is given by wL = ½ L [iL(t)]2, so we seek an expression for iL(t):

iL(t) = e-αt (At + B)

Noting that iL(0+) = iL(0-) = 2.25, we see that B = 2.25 and hence

iL(t) = e-0.3333t (At + 2.25)

Invoking the remaining initial condition requires consideration of the voltage across the

capacitor, which is equal in this case to the inductor voltage, given by:

vC(t) = vL(t) = dt

diL

L = 9(-0.3333) e-0.3333t (At + 2.25) + 9A e-0.3333t

vC(0+) = vC(0-) = 0 = 9(-0.333)(2.25) + 9A so A = 0.7499 amperes and

iL(t) = e-0.3333t (0.7499t + 2.25) A

Thus, iL(100 ms) = 2.249 A and so wL(100 ms) = 22.76 J

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40. With the 144 mJ originally stored via a 12-V battery, we know that the capacitor has a

value of 2 mF. The initial inductor current is zero, and the initial capacitor voltage is 12

V. We begin by seeking a (painful) current response of the form

ibear = Aes1t + Bes2t

Using our first initial condition, ibear(0+) = iL(0+) = iL(0-) = 0 = A + B

di/dt = As1 es1t + Bs2 es2t

vL = Ldi/dt = ALs1 es1t + BLs2 es2t

vL(0+) = ALs1 + BLs2 = vC(0+) = vC(0-) = 12

What else is known? We know that the bear stops reacting at t = 18 µs, meaning that the

current flowing through its fur coat has dropped just below 100 mA by then (not a long

shock).

Thus, A exp[(18×10-6)s1] + B exp[(18×10-6)s2] = 100×10-3

Iterating, we find that Rbear = 119.9775 Ω.

This corresponds to A = 100 mA, B = -100 mA, s1 = -4.167 s-1 and s2 = -24×106 s-1

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

2

4

12,

4

,12

4

11 2

R2 1

(0) 0, (0) 0, 4, 4 5 20

2L 0.5 LC

20 16 2 ( ) (A cos2 A sin 2 )

10A ( ) 10 (A cos2 A sin 2 )

010A,A 10,()10 (Asin2 10cos2)

1

(0) (0) 4 0 0 (0)

L

cL o

t

dL Lf

t

Lf L

t

L

LL L

vi

it e t t i

iitett

it e t t

iv i

αω

ω

−

++ +

=======×=

∴= −=∴ = + +

=∴=+ +

∴= + =− = + −

==×=∴ 22

02A 40,A 20== + =−

i

L(t) = 10 - e-4t (20 sin 2t + 10 cos 2t) A, t > 0

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42. (a)

(b) 500 (3000 6) 3; by SOLVE, 3.357ms

o

t

oo

et t

−+= =

6

2

500

12 22

1

5000 500

1

R 100

Series, driven: 500,

2L 0.2

11010250,000

LC 40

Crit. damp ( ) 3(1 2) 3,

(0) 3, (0) 300V

3(AA)33A,A6A

1

(0 ) A 300 [ (0) (0 )] 0

L

A 3000 ( ) 3

o

L

Lc

t

L

LcR

t

L

if

iv

iet

ivv

eit e

α

ω

−

++

−−

== =

×

== =

∴=−=−

==

∴=−+ + ∴=−+ =

=− = − =

∴= ∴ =−+

500

(3000 6), 0

( ) 3 ( ) [ 3 (3000 6)] ( )A

t

L

tt

it u t e t ut

−

+>

∴=−+−+ +

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43.

663

26

22

400

,, 1 2

4000 12 1

4000

2

110 110

4000, 20 10

2RC 100 2.5 LC 50

2000, (0) 2A, (0) 0

0, ( 0) (A cos2000 A sin 2000 )

work with : ( ) (B cos2000 B sin 2000 ) B 0

B sin 2000

o

do L c

t

cf cf c

t

cc

t

c

iv

iv ie t t

vvt e t t

ve t

+

−

α= = = ω = = = ×

×

∴ω = ω −α = = =

==∴= +

=+∴=

∴= 65

54000

22

6 4000

6 3 3 4000

4000

110

,(0) (0) (21)810

C 2.5

8 10 2000B , B 400, 400 sin 2000

( ) C 2.5 10 400 ( 4000sin 200 2000cos 200 )

10 ( 4sin 2000 2cos2000 )

(2cos2000 4sin 2000

cc

t

c

tt

cc

t

vi

ve t

it v e t

ett

++

−

−−

−++ −

−

′==×=×

∴× = = =

′

∴= =×× − +

=−+

=−

)A, 0t>

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44.

6

2

1,2

,

10 40

12

R 250 1 10

25, 400

2L 10 LC 2500

25 625 400 10, 40

(0) 0.5A, (0) 100V, 0.5A

( ) 0.5 A A A

0 : (0 ) 100 50 1 200 0.5 50V 50 5 (0 )

(0 ) 10 10 10A 40A , 0.5 0.5

o

LcLf

tt

L

L L

L

s

ivi

it e e

tv i

i

−−

++ +

+

α= = = ω = = =

=− ± − =− −

===−

∴=−+ +

′

==−×−×=−∴−=

′

∴=−∴−=−− =−12

12 2 1 1 1 2

10

AA

A A 1 10 10A 40( 1+A ) 50A 40, A 1,A 0

( ) 0.5 1 A, 0; ( ) 0.5A, 0

t

LL

it e t it t

−

++

∴+ =∴−=− − − =− + = =

∴=−+ > = >

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45.

2

,

1

11

R1 1

1, 1 crit. damp

2L 1 LC

5

(0) 12 10V, (0) 2A, 12V

6

11

( ) 12 (A 2); (0 ) (0 ) (0 ) 1

C2

1A 2;A 1 ()12 ( 2)V, 0

−+++

−

α= = = ω = = ∴

=× = = =

′

∴=+ − = =× =

∴= + =−∴ = − + >

o

cLcf

t

cccL

t

c

viv

vt e t v i i

vt e t t

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46. (a)

So vc(t) = e-1000t (8 cos 1000t + 1.6 sin 1000t) V, t > 0

(b)

66

26

3

,

1000

12

6

1

22

1 8 10 8 10 13

1000, 26 10

2RC 2 4 10 4

26 1 10 5000, (0) 8V

(0) 8mA, 0

(A cos1000 A sin5000 )

18

A 8; (0 ) (0 ) 8 10 (0.01 0.008) 0

C 4000

5000A 1000 8 0, A 1.6

o

dc

Lcf

t

c

cc

v

iv

ve t t

vi

−

++

×××

α= = = ω = = ×

××

∴ω = − × = =

==

∴= +

′

∴= = =× − − =

∴−×==

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47. (a)

(b)

6

266

1,2

500 1500

,12

66 4

12

4

11

110

( ) 10 ( )V: 1000

2RC 1000

1103 3

1000 10 10 500, 1500

LC 4 4

0 A A , (0) 10V, (0) 0

10

10 A A , 10 (0 ) 10 0 2 10

500

2 10 500A 1500A 40 A

s

o

tt

cf c c L

cc

vt u t

s

vveev i

vi

−−

+

=−α= = =

×

ω= = ∴ =− ± − × =− −

=∴ = + = =



′

∴= + = = − =−×





∴− × =− − ∴ = 2221

500 1500

6 500 1500

500 1500

3A 30 2A , A 15, A 5

515V,0 C

10 (2500 22,500 )

2.5 22.5 mA, 0

tt

cscc

tt

s

tt

ve e t iiv

iee

eet

−−

−− −

−−

+∴= = =−

′

∴=− + >∴==

∴= −

=− >

,

500 1500

34 34

66 4

34

34 4 4 3

500 1500

6

() 10 ()V 10V, (0) 0, (0) 0

10 A A A 10

10

(0 ) 10 (0 ) 10 0 2 10 500A 1500A

500

A 3A 40, add: 2A 30, A 15, A 5,

10 5 15 V,

10 (

scfcL

tt

c

cc

tt

csc

vt ut v v i

vAe e

vi

veeii

−−

++

−−

−

=∴= ==

∴= + + ∴ + =−



′==+=×=−−





∴− − = − = =− =

=+ − ==

500 1500 500 1500

2500 22,500 ) 25 22.5 mA, 0

tttt

eeeet

−−−−

−+ =+ >

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48. (a)

(b)

500 1500

,34

6

34

6

34

34 4 4 3

500 1500

10 ( )V, 10, 10 A A ,

(0) 0, (0) 0 A A 10V, (0 ) 2 10

[ (0) (0 )] 2 10 (0 0) 0 500A 1500A

A 3A 0, add: 2A 10, A 5 A 15

( ) 10 15 5 V, 0

()

tt

scfc

cL c

LR

tt

c

R

vutv v e e

vi v

ii

vt e e t

it

−−

+

−−

===++

′

==∴+=− =×

−=×−==−−

∴− − = − =− = ∴ = −

∴=− + >

∴500 1500

10 15 5 mA, 0

tt

ee t

−

=− + >

6

26

1,2

500 1500

12

66

12

110

10 ( ) V : 1000

2RC 2000 0.5

1 2 10 3 0.75 10 500, 1500

LC 8

A A , (0) 10V, (0) 10mA

A 10, (0 ) 2 10 [ (0) (0 )] 2 10

10

0.01 0 500A 1500A 0,

1000

A

s

o

tt

coL

cLR

vut

s

ve ev i

Av ii

−−

++

=− α= = =

×

××

ω= = = × ∴ =− −

∴= + = =

′

∴+ = =× − =×



−=∴−− =





−12 2 2 1

500 1500

3A 0; add: 2A 10, A 5, A 15

( ) 15 5 V 0

( ) 15 5 mA, 0

tt

c

tt

R

vt e e t

it e e t

−−

−= − = =− =

∴= − >

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 

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49. (a) vS(0-) = vC(0-) = 2(15) = 30 V

(b) iL(0+) = iL(0-) = 15 A

Thus, iC(0+) = 22 – 15 = 7 A and vS(0+) = 3(7) + vC(0+) = 51 V

(c) As t → ∞, the current through the inductor approaches 22 A, so vS(t→ ∞,) = 44 A.

(d) We are presented with a series RLC circuit having α = 5/2 = 2.5 s-1 and

ω

o = 3.536

rad/s. The natural response will therefore be underdamped with

ω

d = 2.501 rad/s.

iL(t) = 22 + e-αt (A cos

ω

dt + B sin

ω

dt)

iL(0+) = iL(0-) = 15 = 22 + A so A = -7 amperes

Thus, iL(t) = 22 + e-2.5t (-7 cos 2.501t + B sin 2.501t)

vS(t) = 2 iL(t) + dt

di

i

dt

di L

L

L 2 L += = 44 + 2e-2.5t (-7cos 2.501t + Bsin 2.501t)

– 2.5e-2.5t (-7cos 2.501t + Bsin 2.501t) + e-2.5t [7(2.501) sin 2.501t + 2.501B cos 2.501t)]

vS(t) = 51 = 44 + 2(-7) – 2.5(-7) + 2.501B so B = 1.399 amperes and hence

vS(t) = 44 + 2e-2.5t (-7cos 2.501t + 1.399sin 2.501t)

-2.5e-2.5t (-7cos 2.501t + 1.399sin 2.501t) + e-2.5t [17.51sin 2.501t + 3.499cos 2.501t)]

and vS(t) at t = 3.4 s = 44.002 V. This is borne out by PSpice simulation:

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 

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50. Considering the circuit at t < 0, we see that iL(0-) = 15 A and vC(0-) = 0.

The circuit is a series RLC with α = R/2L = 0.375 s-1 and

ω

0 = 1.768 rad/s. We therefore

expect an underdamped response with

ω

d = 1.728 rad/s. The general form of the

response will be

vC(t) = e-αt (A cos

ω

dt + B sin

ω

dt) + 0 (vC(∞) = 0)

vC(0+) = vC(0-) = 0 = A and we may therefore write vC(t) = Be-0.375t sin (1.728t) V

iC(t) = -iL(t) = C dt

dvC= (80×10-3)(-0.375B e-0.375t sin 1.728t

At t = 0+, iC = 15 + 7 – iL(0+) = 7 = (80×10-3)(1.728B) so that B = 50.64 V.

Thus, vC(t) = 50.64 e–0.375t sin 1.807t V and vC(t = 200 ms) = 16.61 V.

The energy stored in the capacitor at that instant is ½ CvC2 = 11.04 J

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



51. It’s probably easiest to begin by sketching the waveform vx:

(a) The source current ( = iL(t) ) = 0 at t = 0-.

(b) iL(t) = 0 at t = 0+

(c) We are faced with a series RLC circuit having α = R/2L = 2000 rad/s and

ω

0 = 2828

rad/s. Thus, an underdamped response is expected with

ω

d = 1999 rad/s.

The general form of the expected response is iL(t) = e-αt (A cos

ω

dt + B sin

ω

dt)

iL(0+) = iL(0-) = 0 = A so A = 0. This leaves iL(t) = B e-2000t sin 1999t

vL(t) = L dt

diL= B[(5×10-3)(-2000 e-2000t sin 1999t + 1999 e-2000t cos 1999t)]

vL(0+) = vx(0+) – vC(0+) – 20 iL(0+) = B (5×10-3)(1999) so B = 7.504 A.

Thus, iL(t) = 7.504 e-2000t sin 1999t and iL(1 ms) = 0.9239 A.

(d) Define t' = t – 1 ms for notational convenience. With no source present, we expect a

new response but with the same general form:

iL(t') = e-2000t' (A' cos 1999t' + B' sin 1999t')

vL(t) = L dt

diL, and this enables us to calculate that vL(t = 1 ms) = -13.54 V. Prior to the

pulse returning to zero volts, -75 + vL + vC + 20 iL = 0 so vC(t' = 0) = 69.97 V.

iL(t' = 0) = A' = 0.9239 and –vx + vL + vC + 20 iL = 0 so that B' = -7.925.

Thus, iL(t') = e-2000 t' (0.9239 cos 1999t' – 7.925 sin 1999t') and

hence iL(t = 2 ms) = iL(t' = 1 ms) = -1.028 A.

1 2 3 4 t (s)

vx (V)

75

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



52. For t < 0, we have 15 A dc flowing, so that iL = 15 A, vC = 30 V, v3Ω = 0 and vS = 30

V. This is a series RLC circuit with α = R/2L = 2.5 s-1 and

ω

0 = 3.536 rad/s. We

therefore expect an underdamped response with

ω

d = 2.501 rad/s.

vC(t) = e-αt (A cos

ω

dt + B sin

ω

dt)

vC(0+) = vC(0-) = 30 = A so we may write vC(t) = e-2.5t (30 cos 2.501t + B sin 2.501t)

C=

dt

dv -2.5e-2.5t(30 cos 2.501t + B sin 2.501t)

+ e-2.5t [-30(2.501)sin 2.501t + 2.501B cos 2.501t]

iC(0+) =

0

C

+

=t

dt

dv

C= 80×10-3[-2.5(30) + 2.501B] = -iL(0+) = -iL(0-) = -15 so B = -44.98 V

Thus, vC(t) = e-2.5t (30 cos 2.501t – 44.98 sin 2.501t) and

iC(t) = e-2.5t (-15 cos 2.501t + 2.994 sin 2.501t).

Hence, vS(t) = 3 iC(t) + vC(t) = e-2.5t (-15 cos 2.501t – 36 sin 2.501t)

Prior to switching, vC(t = 1) = -4.181 V and iL(t = 1) = -iC(t = 1) = -1.134 A.

t > 2: Define t' = t – 1 for notational simplicity. Then, with the fact that vC(∞) = 6 V,

our response will now be vC(t') = e-αt' (A' cos

ω

dt' + B' sin

ω

dt') + 6.

With vC(0+) = A' + 6 = -4.181, we find that A' = -10.18 V.

iC(0+) =

0

C

+

=

′

′t

td

dv

C= (80×10-3)[(-2.5)(-10.18) + 2.501B')] = 3 – iL(0+) so B' = 10.48 V

Thus, vC(t') = e-2.5t (-10.18 cos 2.501t' + 10.48 sin 2.501t') and

iC(t') = e-2.5t (4.133 cos 2.501t' – 0.05919 sin 2.501t').

Hence, vS(t') = 3 iC(t') + vC(t') = e-2.5t (2.219 cos 2.501t' + 10.36 sin 2.501t')

0 < t < 1

We see that our hand

calculations are supported by

the PSpice simulation.

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



53. The circuit described is a series RLC circuit, and the fact that oscillations are detected

tells us that it is an underdamped response that we are modeling. Thus,

iL(t) = e-αt (A cos

ω

dt + B sin

ω

dt) where we were given that

ω

d = 1.825×106 rad/s.

ω

0 = LC

1 = 1.914×106 rad/s, and so

ω

d2 =

ω

02 – α2 leads to α2 = 332.8×109

Thus, α = R/2L = 576863 s-1, and hence R = 1003 Ω.

Theoretically, this value must include the “radiation resistance” that accounts for the

power lost from the circuit and received by the radio; there is no way to separate this

effect from the resistance of the rag with the information provided.

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



54. The key will be to coordinate the decay dictated by α, and the oscillation period

determined by

ω

d (and hence partially by α). One possible solution of many:

Arbitrarily set

ω

d = 2π rad/s.

We want a capacitor voltage vC(t) = e-αt (A cos 2πt + B sin 2πt). If we go ahead and

decide to set vC(0-) = 0, then we can force A = 0 and simplify some of our algebra.

Thus, vC(t) = B e-αt sin 2πt. This function has max/min at t = 0.25 s, 0.75 s, 1.25 s, etc.

Designing so that there is no strong damping for several seconds, we pick α = 0.5 s-1.

Choosing a series RLC circuit, this now establishes the following:

R/2L = 0.5 so R = L and

ω

d =

2

02

1

- 











ω

= 39.73 rad/s = LC

1

Arbitrarily selecting R = 1 Ω, we find that L = 1 H and C = 25.17 mF. We need the first

peak to be at least 5 V. Designing for B = 10 V, we ∴need iL(0+) = 2π(25.17×10-3)(10) =

1.58 A. Our final circuit, then is:

And the operation is verified by a simple PSpice simulation:

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



55. For t < 0, iL(0-) = 3 A and vC(0-) = 25(3) = 75 V. This is a series RLC circuit with α =

R/2L = 5000 s-1 and

ω

0 = 4000 rad/s. We therefore expect an overdamped response with

s1 = -2000 s-1 and s2 = -8000 s-1. The final value of vC = -50 V.

For t > 0, vC(t) = A e-2000t + B e-8000t - 50

vC(0+) = vC(0-) = 75 = A + B – 50

so A + B = 125 [1]

dt

dvC= -2000 Ae-2000t – 8000 Be-8000t

iC(0+) =

+

=0

C

t

dt

dv

C= 3 – 5 – iL(0-) = -5 = -25×10-6 (2000A + 8000B)

Thus, 2000A + 8000B = 5/25×10-6 [2]

Solving Eqs. [1] and [2], we find that A = 133.3 V and B = -8.333 V. Thus,

vC(t) = 133.3 e-2000t – 8.333 e-8000t – 50

and vC(1 ms) = -31.96 V. This is confirmed by the PSpice simulation shown below.

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



56. α = 0 (this is a series RLC with R = 0, or a parallel RLC with R = ∞)

ω

o2 = 0.05 therefore

ω

d = 0.223 rad/s. We anticipate a response of the form:

v(t) = A cos 0.2236t + B sin 0.2236t

v(0+) = v(0-) = 0 = A therefore v(t) = B sin 0.2236t

dv/dt = 0.2236B cos 0.2236t; iC(t) = Cdv/dt = 0.4472B cos 0.2236t

iC(0+) = 0.4472B = -iL(0+) = -iL(0-) = -1×10-3 so B = -2.236×10-3 and thus

v(t) = -2.236 sin 0.2236t mV

In designing the op amp stage, we first write the differential equation:

)0( 0 2 10

10

13-

0=+=++

′

∫LC

tii

dt

dv

tdv

and then take the derivative of both sides:

v

dt

vd

20

1

- 2

2=

With 43

0

105)10236.2)(2236.0( −−

=

×−=×−=

+

t

dt

dv , one possible solution is:

PSpice simulations are very sensitive to parameter values; better results were obtained

using LF411 instead of 741s (both were compared to the simple LC circuit simulation.)

Simulation using 741 op amps Simulation using LF411 op amps

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



57.

(a)

v

dt

dv

dt

dvv

3.3

1

-

or

0 103.3

1000

3-

=

=×+

(b) One possible solution:

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



58. α = 0 (this is a series RLC with R = 0, or a parallel RLC with R = ∞)

ω

o2 = 50 therefore

ω

d = 7.071 rad/s. We anticipate a response of the form:

v(t) = A cos 7.071t + B sin 7.071t, knowing that iL(0-) = 2 A and v(0-) = 0.

v(0+) = v(0-) = 0 = A therefore v(t) = B sin 7.071t

dv/dt = 7.071B cos 7.071t; iC(t) = Cdv/dt = 0.007071B cos 7.071t

iC(0+) = 0.007071B = -iL(0+) = -iL(0-) = -2 so B = -282.8 and thus

v(t) = -282.8 sin 7.071t V

In designing the op amp stage, we first write the differential equation:

)0( 0 10 2

20

13-

0=+=++

′

∫LC

tii

dt

dv

tdv

and then take the derivative of both sides:

v

dt

vd 05- 2

2=

With 2178)8.282)(071.7(

0

−=−=

+

=t

dt

dv , one possible solution is:

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



59. (a) vR = vL

20(-iL) = 5 dt

diL or L

L4- i

dt

di =

(b) We expect a response of the form iL(t) = A e-t/ τ where τ = L/R = 0.25.

We know that iL(0-) = 2 amperes, so A = 2 and iL(t) = 2 e-4t

+

=0

L

t

dt

di = -4(2) = -8 A/s.

One possible solution, then, is

1 MΩ

4 kΩ

1 kΩ

1 Ω

1 µF

8 V

i

↓

CHAPTER NINE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



and

+

=0 t

dt

di = 0

60. We see either a series RLC with R = 0 or a parallel RLC with R = ∞; either way, α = 0.

ω

02 = 0.3 so

ω

d = 0.5477 rad/s (combining the two inductors in parallel for the

calculation). We expect a response of the form i(t) = A cos

ω

dt + B sin

ω

dt.

i(0+) = i(0-) = A = 1×10-3

di/dt = -A

ω

d sin

ω

dt + B

ω

d cos

ω

dt

vL = 10di/dt = -10A

ω

d sin

ω

dt + 10B

ω

d cos

ω

dt

vL(0+) = vC(0+) = vC(0-) = 0 = 10B(0.5477) so that B = 0

and hence i(t) = 10-3 cos 0.5477t A

The differential equation for this circuit is

v

dt

vd

dt

dv

tvdtvd

tt

3.0

or

0 2

2

1

10

1

2

0

3-

0

−=

=+

′

++

′∫∫

One possible solution is:

1 Ω

↓

i

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1.

(a)

(b)

(c)

3

33

3

rad rad

210

T 4(7.5 2.1)10 21.6 10 , 290.9 rad/s

21.6

( ) 8.5sin (290.9 ) 0 8.5sin (290.9 2.1 10 )

0.6109 2 5.672 or 325.0

( ) 8.5sin (290.9 325.0 )

−−

−

π

=− =×ω= =

∴= +Φ∴= ××+Φ

∴Φ = − + π = °

∴= + °

t

ft t

8.5sin (290.9 325.0 ) 8.5

cos(290.9 235 ) 8.5cos(290.9 125 )

t

tt

+°=

+°= −°

8.5cos( 125 )cos 8.5sin125

sin 4.875 cos 290.9 6.963sin 290.9

+

−° ω+ °

ω=− +

t

ttt

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2.

(a)

(b)

(c)

(d)

10cos 4sin ACos( ), A 0, 180 180

A 116 10.770, Acos 10, Asin 4 tan 0.4, 3 quad

21.80 201.8 ,too large 201.8 360 158.20

d

tt wt−ω+ω+ +Φ>−°<Φ≤°

== Φ=− Φ=−∴Φ=

∴Φ = °= ° ∴Φ = °− °=− °

200cos(5 130 ) Fcos5 Gsin5 F 200cos130 128.56

G 200sin130 153.21

+°= + ∴= °=−

=− °=− °

ttt

sin10 5

( ) 5cos10 3sin10 0, 0 1 , 10 1.0304,

cos10 3

0.10304 ; also, 10 1.0304 , 0.4172 ; 2 : 0.7314

=−=≤≤∴==

==+π=π

t

it t t t s t

t

tst ts s

0 10ms, 10cos100 12sin100 ; let 10cos100 t =12sin100 t

10

tan100 t = , 100 0.6947 2.211ms 0 2.211ms

12

<< π≥ π π π

∴π π= ∴= ∴<<

ttt

tt t

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3.

(a)

(b) ( ) leads ( ) by 149.04 15.255 133.78

°− °= °

ft gt

( ) 50cos 30sin 58.31cos( 149.04 )

( ) 55cos 15sin 57.01cos( 15.255 )

ampl. of ( ) 58.31, ampl. of ( ) 57.01

=− ω − ω = ω + °

=ω−ω= ω+°

∴= =

ft t t t

gt t t t

ft gt

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4.

222 222

( ) Acos( ), L( / ) R V cos

L[ Asin ( )] RA cos( ) V cos

LA sin cos LA cos sin

RAcos cos RAsin sin

V cos LAcos RAsin and sin RAcos V

LL R

tan LA RA

RL RL

m

m m

it t di dt i t

ttt

tt

tLA

R

=ω−θ +=ω

∴ −ω ω −θ + ω −θ = ω

∴−ω ω θ+ω ω θ

+ωθ+ωθ=

ω∴ω θ= θ ω θ+ θ=

ωω

∴θ= ∴ω +

+ω +ω

22 2 222

222

V

VLR

AV R LAV.A RL

m

mm

=



ω

∴+ =∴+ω==

 +ω



CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5. f = 13.56 MHz so

ω

= 2

π

f = 85.20 Mrad/s.

Delivering 300 W (peak) to a 5-Ω load implies that 300

5

V2

m= so Vm = 38.73 V.

Finally, (85.2×106)(21.15×10-3) +

φ

= n

π

, n = 1, 3, 5, …

Since (85.2×106)(21.15×10-3) = 1801980, which is 573588+

π

, we find that

F = 573587

π

- (85.2×106)(21.15×10-3) = -3.295 rad = -188.8o.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6. (a) -33 sin(8t – 9o) → -33∠(-9-90)o = 33∠81o

12 cos (8t – 1o) → 12∠-1o

(b) 15 cos (1000t + 66o) → 15 ∠ 66o

-2 cos (1000t + 450o) → -2 ∠ 450o = -2 ∠90o = 2 ∠ 270o

(c) sin (t – 13o) → 1∠-103o

cos (t – 90o) → 1 ∠ -90o

(d) sin t → 1 ∠ -90o

cos (t – 90o) → 1 ∠ -90o

These two waveforms are in phase. Neither leads the other.

33∠81o

12∠-1o

-33 sin(8t – 9o) leads 12 cos (8t – 1o) by

81 – (-1) = 82o.

15∠66o

2∠270o

15 cos (1000t + 66o) leads -2 cos (1000t + 450o)

by 66 – -90 = 156o.

1∠-103o 1 ∠ -90o

cos (t – 90o) leads sin (t – 13o)

by 66 – -90 = 156o.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7. (a) 6 cos (2

π

60t – 9o) → 6∠-9o

-6 cos (2

π

60t + 9o) → 6∠189o

(b) cos (t - 100o) → 1 ∠ -100o

-cos (t - 100o) → -1 ∠ -100o = 1 ∠80o

(c) -sin t → -1∠-90o = 1∠90o

sin t → 1∠ -90o

(d) 7000 cos (t –

π

) → 7000 ∠ -

π

= 7000 ∠ -180o

9 cos (t – 3.14o) → 9 ∠ -3.14o

6∠-9o

6∠189o

-6 cos (2

π

60t + 9o) lags 6 cos (2

π

60t – 9o)

by 360 – 9 – 189 = 162o.

1∠80o

1∠-100o

-cos (t - 100o) lags cos (t - 100o) by 180o.

1∠90o

1 ∠ -90o

-sin t lags sin t by 180o.

7000 ∠ -180o

9 ∠ -3.14o

7000 cos (t –

π

) lags 9 cos (t – 3.14o)

b

y

180 – 3.14 = 176.9o.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8. v(t) = V1 cos

ω

t - V2 sin

ω

t [1]

We assume this can be written as a single cosine such that

v(t) = Vm cos (

ω

t +

φ

) = Vm cos

ω

t cos

φ

- Vm sin

ω

t sin

φ

[2]

Equating terms on the right hand sides of Eqs. [1] and [2],

V

1 cos

ω

t – V2 sin

ω

t = (Vm cos

φ

) cos

ω

t – (Vm sin

φ

) sin

ω

t

yields

V1 = Vm cos

φ

and V2 = Vm sin

φ

Dividing, we find that

φ

tan

cos V

sin V

V

m

1

2== and

φ

= tan-1(V2/ V1)

Next, we see from the above sketch that we may write Vm = V1/ cos

φ

or

2

11

1

mV VV

V

V +

= = 2

2

1V V +

Thus, we can write v(t) = Vm cos (

ω

t +

φ

) = 2

2

1V V + cos [

ω

t + tan-1(V2/ V1)].

φ

V2

V1

2

1V V +

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9. (a) In the range 0 ≤ t ≤ 0.5, v(t) = t/0.5 V.

Thus, v(0.4) = 0.4/0.5 = 0.8 V.

(b) Remembering to set the calculator to radians, 0.7709 V.

(c) 0.8141 V.

(d) 0.8046 V.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10. (a) Vrms = 2

1

T

0

2

m cos

T

V









∫dtt

ω

= 2

1

T

0

2

m

T

2

cos

T

V









∫dt

t

π

= 2

1

T

0

2

m

T

4

cos 1

2T

V



















+

∫dt

t

π

= 2

1

T

0

2

m

T

0

2

m

T

4

cos

2T

V

2T

V









+∫∫ dt

t

dt

π

= 2

1

4

0

2

m

2

m cos

8

V

T

2T

V









+

π

u

=

2

Vm

(b) Vm = V 155.6 2 110 =, V 162.6 2 115 =, V 169.7 2 201 =

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

1

22

3

At : 15cos500 V

4

R 5 20 60 20

V 15V,R 20 , L 10

15 10

cos 500 tan 0.6708cos(500 26.57)A

20

20 10

−

−==

=+ = Ω

∴= =Ωω=Ω



∴= − = −





+

oc s

th

m

L

xxv v t

it t

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12.

(a)

(b)

At : R 80 20 16

80

0.4(15 85) cos500

85

4.8cos500 V

th

oc

xx

vt

−==Ω

=−

∴=

1

22

4.8 10

cos 500 tan 15

16 10

0.2544cos(500 32.01 )A

−



=−





+

=−°

L

it

t

L 0.02 0.02544( 500)

sin (500 32.01 ) 2.544sin (500 32.01 ) V

2.544cos(500 57.99 ) V,

31.80cos(500 57.99 ) mA

LL

Lx

vi

tt

vti

t

′

==× −

−°=− −°

∴= + °

=+°

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13.

(a)

(b)

(c)

55

22

5

100 800

cos 10 0.10600cos(10 57.99 )A

500

500 800

57.99

0 when 0 10 , 25.83

180 2

R

it t

pit ts



=−=−°





+

°π

==∴−π==µ

355

5

55

5

L 8 10 0.10600( 10 ) sin(10 57.99 )

84.80sin (10 57.99 )

8.989sin (10 57.99 )

cos(10 57.99 ) 4.494 sin (2 16 115.989 )

0 when 2 10 115.989 0 , 180 ,

10.121 or 25.83

L

LL

L

vi t

vt

pvi t

tt

pt

ts

−

′

±= =× × − − °

∴=− − °

∴==− − °

−°=− ×− °

∴= × − °=° °

∴=

µ

55

5

10.600cos10 cos(10 57.99 )

0 when 10 , 15.708 and also 25.83

2

== − °

π

∴= = =

µ

=

µ

ssL

s

pvi t t

pttsts

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14.

55

5

51 5

22

3cos10 V, 0.1cos10 A

in series with 30 0.1cos10 A 30

Add, getting 0.2cos10 A 30

change to 6 cos 10 Vin series with 30 ; 30 20 50

610

cos 10 tan 0.11767cos(10 11.310

50

50 10

−

==

Ω→ Ω

Ω

ΩΩ+Ω=Ω



∴= − = − °





+

ss

s

L

vti t

vt

t

it t

5rad

rad

)A

At 10 , 10 1 0.1167cos(1 11.310 ) 81.76mA

0.11767 10cos(1 11.30 90 ) 0.8462V

=µ =∴= − °=

∴= × − °+°=−

L

tsti

v

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

1

22

cos500 V

cos500 100(0.8 )

1cos500 A R 80

80

1 150

cos 500 tan 80

80 150

5.882cos(500 61.93 )mA

−

∴=

∴= ∴ = =Ω



∴= −





+

=−°

oc

sc

oc

sc th

sc

L

vt

ti

v

it

i

it

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16.

(a)

(b) ,

10.2383 0.11916J

2

ω=× =

Lav

12

1

22

V 120cos120 V

120 120

2A, 1A, 2 1 3A, 60 120 40

60 12

3 40 120V, L 12 37.70

120 37.70

cos 120 tan 40

40 37.70

2.183cos(120 43.30 ) A

ss

L

vt

it

t

−

== π

==+= =Ω

×= ω=π= Ω



∴= π−





+

=π−°

22

2

10.1 2.183 cos (120 43.30 )

2

0.2383cos (120 43.30 ) J

∴ω = × × π − °

=π−°

Lt

t

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

12

22 22

120cos400 V, 180cos200 V

120 180

2A, 1.5A, 60 120 40

60 120

2 40 80V, 1.5 40 60V

80 60

cos(400 45 ) cos(200 26.57 ) A

40 40 40 20

or 1.4142cos(400 45 ) 1.3416cos(200 26.57 )A

ss

L

vtvt

it t

↑= =

== =Ω

×= ×=

=−°+−°

++

=−°+−°

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18.

11

1

11

1

11

L

R , R 0, A , ideal, R C R

Vcos ,

RR

(Vcos)C

R

L

Vcos RC R

For RL circuit, V cos L R

L

Vcos R

By c

io

m out

upper lower

c upper lower out m out

m out out out out

R

mr

mRR

tv

ii

i

ii i v t v

tv v v v

dv

tv dt

tv v

ω

=∞ = =∞ =

=− =

′

∴= + = − =−

′′

∴=+=+



=+ 



′

∴=+

omparison, Rout

vv=

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19.

(a)

(b)

1

V cos R (ignore I.C)

C

1

Vsin R

=+

′

∴− = +

∫

m

ti idt

ti i

C

ω

ωω

()

1

Assume Acos( )

A

Vsin R Asin( ) cos( )

C

AA

V sin R Acos sin R Asin cos cos cos sin sin

CC

Equating terms on the left and right side,

A1

[1] R Asin cos tan so tan 1 CR ,

CCR

−

=+Φ

∴− = − +Φ + +Φ

∴− = − Φ Φ + Φ− Φ

Φ= Φ∴ Φ= Φ=

m

it

tt t

ttttt

ω

ωωωω ω

ωωω ωω ω ω ω

ωω

ω

222 22 2

222 222

222 22 2

1

222

and

CR A 1

[2] V R A C

1CR 1CR

CVAR C1 A

V1CRA

CC

1CR 1CR

CV 1

cos tan CR

1CR

−

−=− −

++



+

∴= = + ∴=



++





∴= +



+

m

it

ω

ωω

ωωω

ω

-

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20. (a) 7 ∠ -90o = -j 7

(b) 3 + j + 7 ∠ -17o = 3 + j + 6.694 – j 2.047 = 9.694 – j 1.047

(c) 14ej15o = 14 ∠ 15o = 14 cos 15o + j 14 sin 15o = 13.52 + j 3.263

(d) 1 ∠ 0o = 1

(e) –2 (1 + j 9) = -2 – j 18 = 18.11 ∠ - 96.34o

(f) 3 = 3 ∠ 0o

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21. (a) 3 + 15 ∠ -23o = 3 + 13.81 – j 5.861 = 16.81 – j 5.861

(b) (j 12)(17 ∠ 180o) = (12 ∠ 90o)(17 ∠ 180o) = 204 ∠ 270o = -j 204

(c) 5 – 16(9 – j 5)/ (33 ∠ -9o) = 5 – (164 ∠ -29.05o)/ (33 ∠ -9o)

= 5 – 4.992 ∠ -20.05o = 5 – 4.689 – j 1.712 = 0.3109 + j 1.712

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22. (a) 5 ∠ 9o – 9 ∠ -17o = 4.938 + j 0.7822 – 8.607 + j 2.631 = -3.669 – j 1.849

= 4.108 ∠ -153.3o

(b) (8 – j 15)(4 + j 16) – j = 272 + j 68 – j = 272 + j 67 = 280.1 ∠ 13.84o

(c) (14 – j 9)/ (2 – j 8) + 5 ∠ -30o = (16.64 ∠-32.74o)/ (8.246 ∠ - 75.96o) + 4.330 – j 2.5

= 1.471 + j 1.382 + 4.330 – j 2.5 = 5.801 – j 1.118 = 5.908 ∠ -10.91o

(d) 17 ∠ -33o + 6 ∠-21o + j 3 = 14.26 – j 9.259 + 5.601 – j 2.150 + j 3

= 19.86 – j 8.409 = 21.57 ∠ -22.95o

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23. (a) ej14o + 9 ∠ 3o – (8 – j 6)/ j2 = 1 ∠ 14o + 9 ∠ 3o – (8 – j 6)/ (-1)

= 0.9703 + j 0.2419 + 8.988 + j 0.4710 + 8 – j 6 = 17.96 – j 5.287 = 18.72 ∠ -16.40o

(b) (5 ∠ 30o)/ (2 ∠ -15o) + 2 e j5o/ (2 – j 2)

= 2.5 ∠ 45o + (2 ∠ 5o)/ (2.828 ∠ -45o) = 1.768 + j 1.768 + 0.7072 ∠ 50o

= 1.768 + j 1.768 + 0.4546 + j 0.5418

= 2.224 + j 2.310 = 3.207 ∠ 46.09o

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24.

(a) 5 110 1.7101 4.698∠− °=− − j

(b) 160

6 5.638 2.052

°=− +

j

ej

(c) (3 6)(2 50 ) 5.336 12.310

+∠°=−+

jj

(d) 100 40 107.70 158.20−− = ∠− °j

(e) 2 50 3 120 1.0873 101.37∠°+∠− °= ∠− °

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25.

(a) 40 50 18 25 39.39 76.20∠− °− ∠ °= ∠− °

(b) 22 5

3 4.050 69.78

12

j

jj

−

++ = ∠− °

+

(c) 3

(2.1 25 ) 9.261 75 2.397 8.945+

∠°= ∠°= +j

(d) 0.3 rad

0.7 0.7 0.3 0.6687 0.2069=∠ = +

j

ej

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26.

(40 30 ) (40 30 )

(40 53

20 A 100 20

50 , 10 A

(20 10) , 40 0.08(20 10)

(32 64) V (32 64 50)

34.93

+° +°

−

=∴=

=− =−

∴= − = × −

∴= + ∴= + −

∴=

∫

tjt

cc

jt jt

cR

jt jt

LL

jt jt

Ls

jt

s

ie v e dt

vje ije

ijevj je

vje vjje

ve

.63 ) V

°

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

(10 25 )

(10 25 ) (10 25 )

(10 25 )

(10 25 ) (10 25 )

(10 125.62 )

20 A

0.2 [20 ] 40

80

(80 40) , 0.08(80 40) 10

( 32 64) ( 12 64)

65.12 A

jt

L

jt t

L

jt

R

jt jt

sc

jt jt

cs

jt

s

ie

d

veje

dt

ve

vjei jje

ijeije

ie

+°

+° =°

+°

+° +°

+°

=

==

=

=+ = +

∴=−+ ∴=−+

∴=

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28. 80cos(500 20 )V 5cos(500 12 )A−° → +°tt

(a) 40cos(500 10 ) 2.5cos(500 42 )A

s out

vti t=+°∴= +°

(b)

(c)

(d)

40sin (500 10 ) 40cos(500 80 )

2.5cos(500 48 )A

=+°=−°

∴= −°

s

out

vt t

it

(500 10 )

(500 42 )

40 40cos(500 10 )

40sin (500 10 ) 2.5 A

+°

==+°

++°∴=

jt

s

jt

out

ve t

jtie

500 21.80 500

(500 53.80 )

(50 20) 53.85

3.366 A

+°+

+°

=+ =

∴=

jt j jt

s

jt

out

vje e

ie

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29.

(a) 12sin (400 110 )A 12 20 A+°→∠°t

(b)

(c)

(d)

(e)

7sin800 3cos800 7 3

3 7 7.616 113.20 A

−− →−

=− + = ∠ °

ttj

j

4cos(200 30 ) 5cos(200 20 )

4 30 5 20 3.910 108.40 A

−°− +°

→∠− °−∠ °= ∠− °

tt

3rad

600, 5ms : 70 30 V

70cos(600 5 10 30 ) 64.95 V

−+

ω= = ∠ °

→××+°=−

t

rad

600, 5ms : 60 40 V 72.11 146.3

72.11cos(3 146.31 ) 53.75 V

+

ω= = + = ∠ °

→+°=

tj

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30. 4000, 1mstω= =

(a)

(b)

(c)

(d)

(e)

rad

I5 80A

4cos(4 80 ) 4.294A

=∠− °

∴= − °=−

x

i

rad

I 4 1.5 4.272 159.44 A

4.272cos(4 159.44 ) 3.750 A

−

=− + = ∠ °

∴= + °=

x

j

i

( ) 50sin (250 40 )

50cos(250 130 ) V 50 103 V

=−°

=−°→=∠−°

x

vt t

t

20cos108 30sin108

20 30 36.06 56.31 V

=−

→+ = ∠ °

x

vtt

j

33cos(80 50 ) 41cos(80 75 )!

33 50 41 75 72.27 63.87 V

=−°+−°

→∠−°+∠−°= ∠− °

x

vt t

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31.

12

3rad

rad rad

V 10 90mV, 500;V 8 90mV,

1200, M by 5, 0.5ms

( 5) [10cos(500 0.5 10 90 )

8cos(1.2 0.5 90 )]

50sin 0.25 40sin 0.6 34.96mV

out

t

v−

=∠° ω= =∠°

ω= − =

=− × × + °

+×+°

=+=

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32. Begin with the inductor:

(2.5 ∠40o) (j500) (20×10-3) = 25∠130o V across the inductor and the 25-Ω resistor.

The current through the 25-Ω resistor is then (25∠130o) / 25 = 1∠130o A.

The current through the unknown element is therefore 25∠130o + 1∠130o = 26∠130o;

this is the same current through the 10-Ω resistor as well. Thus, KVL provides that

Vs = 10(26∠130o) + (25 ∠ -30o) + (25∠130o) = 261.6 ∠128.1o

and so vs(t) = 261.6 cos (500t + 128.1o) V.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

ω

= 5000 rad/s.

(a) The inductor voltage = 48∠ 30o = j

ω

L IL = j(5000)(1.2×10-3) IL

So IL = 8∠-60o and the total current flowing through the capacitor is

10 ∠ 0o - IL = 9.165∠49.11o A and the voltage V1 across the capacitor is

V1 = (1/j

ω

C)(9.165∠49.11o) = -j2 (9.165∠49.11o) = 18.33∠-40.89o V.

Thus, v1(t) = 18.33 cos (5000t – 40.89o) V.

(b) V2 = V1 + 5(9.165∠49.11o) + 60∠120o = 75.88∠79.48o V

(c) V3 = V2 – 48∠30o = 75.88 ∠79.48o – 48∠30o = 57.70∠118.7o V

2( ) 75.88cos(5000 79.48 ) V∴= + °vt t

3( ) 57.70cos(5000 118.70 )V∴= + °vt t

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. VR = 1∠0o V, Vseries = (1 + j

ω

–j/

ω

)(1∠0o)

V

R = 1 and Vseries =

()

1/ - 1 2

ωω

+

We desire the frequency w at which Vseries = 2VR or Vseries = 2

Thus, we need to solve the equation

()

4 1/ - 1 2=+

ωω

or 0 1 - 3 -

2=

ωω

Solving, we find that

ω

= 2.189 rad/s.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35. With an operating frequency of

ω

= 400 rad/s, the impedance of the 10-mH inductor is

j

ω

L = j4 Ω, and the impedance of the 1-mF capacitor is –j/

ω

C = -j2.5 Ω.

V 2 40 ( 2.5) 5 50 A

I 3 2 40 1.9513 41.211 A

V 4 1.9513 90 4.211 7.805 48.79 V

V V V 7.805 48.79 5 50

V 9.892 78.76 V, 9.892cos(400 78.76 ) V

+

∴=∠°− =∠−°

∴=−∠°= ∠− °

∴=× ∠°− °= ∠ °

∴=−= ∠ °−∠−°

∴= ∠ ° = + °

c

L

xLc

xx

j

vt

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36.

2

12

out 1 2

If I 2 20 A, I 3 30 A V 80 10 V

II440AV 9030V

Now let I 2.5 60 A and I 2.5 60 A

Let V AI BI 80 10 A(2 20 ) B(3 30 )

90 30

and 90 30 (A B)(4 40 ) A B 12.45 20.21

440

si s out

s s out

ss

j

jj

+

=∠ ° =∠− ° → = ∠ °

==∠°→ =−

=∠−° =∠°

=+∴∠°=∠°+∠−°

−

−=+ ∠°∴+= = −

∠°

80 10 3 30

A B A 40 10 B(1.5 50 )

220 220

12.415 20.21 B 40 10 B(1.5 50 )

12.415 20.21 40 10 B(1 1.5 50 )

B(1.1496 88.21 )

30.06 153.82

B 10.800 23.81

1.1496 88.21

26.15 65.61

j

+

∠° ∠− °

∴=+ ∴=∠−°−∠−°

∠° ∠

∴ − − = ∠− °− ∠− °

∴−−∠−°=−∠−°

=∠−°

∠− °

∴= = −

∠− °

=∠−°

∴A 12.415 20.21 10.800 23.81

3.952 65.87

V (3.952 65.87 )(2.5 60 )

(26.15 65.61 ) (2.5 60 ) 75.08 4.106 V

out

jj

+

−

=−−+

=∠°

∴= ∠ ° ∠−°

+∠−°∠°=∠−°

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37.

(a)

(b)

800: 2 F 625, 0.6H 480

300( 625) 600( 480)

Z300 625 600 480

478.0 175.65

in

jj

j

ω=

µ

→− →

−

∴= +

−+

=+ Ω

300( 312.5)

1600: Z 300 312.5

600( 960) 587.6 119.79

600 960

in

j

jj

j

−

ω= = −

+=+Ω

+

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38.

(a)

(b)

SC , : 20 10 6.667, (6.667 5) 10

50 66.67 150 200 30 40 4 3

6.667 5 20 15 4 3 4 3

(1.2 1.6)(4 3) 9.6 2.8

in in

ab j j

jjjj

jj jj

ZZ j j j

=−

+++−

===×

++ +−

=∴ + − =+ Ω

50 50 10 10 2 1

(10 10) ( 5) 10 5 2 1 2 1

26 Z226

−−−

+−= =

++−

=− Ω∴ − − Ω

in

jjj

jj jjj

jj

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

120 625

I478.0 175.65 300 625

or I 0.2124 45.82 A

−

∴= ×

+−

=∠−°

j

jj

Thus, i(t) = 212.4 cos (800t – 45.82o) mA.

800: 2 F 625, 0.6H 480

300( 625) 600( 480)

Z300 625 600 480

478.0 175.65

in

jj

j

ω=

µ

→− →

−

∴= +

−+

=+ Ω

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40.

(a) 3 2mH : V (3 20 ) (3 4) 15,000 33.13 V

Ω=∠−°+=∠°

j

(b) 3 125 F: V (3 20 ) (3 4) 15,000 73.3 VjΩ

µ

=∠− ° − = ∠− °

(c) 3 2mH 125 F: V (3 20 ) 3 9,000 20 V

Ω

µ

=∠−°= ∠−°

(d)

same: 4000 V (3 20 ) (3 8 2)

V (3 20 ) (3 6) 20.12 43.43 V

ω= ∴ = ∠− ° + −

∴=∠−° + = ∠ °

jj

j

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

(a)

(b)

(c)

6

C 20 F, 100

11

Z11 0.005 0.01 0.002

1000 20 10

200 1000

1

Z 196.12 11.310

0.005 0.001

−

=

µ

ω=

==

−+

++××

∴= = ∠− °Ω

+

in

jj

j

222

3

11

100 Z 125

0.005 0.001 1000C 0.008

0.005 (100C 0.001) 0.008 100C 0.001

6.245 10 , C 72.45 F

in jj

−−

ω= ∴ = = ∠=

−+ ∠

∴+− =∴−=

±−× = µ

5

22

25 5 5

55525

55

11

C 20 F Z 100

0.0005 0.1/ 2 10 0.01

0.1 0.1

0.005 2 10 0.0001, 2 10 7.5 10

0.01

2 10 866.0 10 0 2 10 866.0 10 0.1 0

866.0 10 7.5 10 8 10

use sign:

in jj

−

−−−

−−−−

−−−

=µ∴ = = ∠=

−ω+×ω ∠



∴+×ω−= ×−=×



ωω



∴×− × =∴×ω ×ω−=

ω

×± ×+×

−ω=

mm

6

5

556

5

444.3 and 0

410

866.0 10 7.5 10 8 10

use + sign: 11.254 and <0

410

=11.254 and 444.3rad/s

−

−−−

−

=<

×

−×±×+×

ω= =

×

∴ω

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42.

(a)

(b)

(c)

(d)

2

1111

25 0.0016

11 0.04 900

30

X 45.23 0.002W, 2261rad/s

x

jx

== ∴ +=

+

∴= Ω= ω=

1

11 30

Y25 of tan

30

64.34 0.02 , 3217rad/s

−−



∠=−°=∠ − =





∴= = ωω=

in jxx

x

2

22

30( 0.02 ) 30 0.092 0.012 18

Z30 0.02 30 0.02 900 0.0004

0.012 25 (900 0.0004 )

0.012 0.01 22,500, 3354rad/s

ω− ω ω+ω

=× =

+ω−ω + ω

∴ω= + ω

∴ω=ω+ ω=

in

jj j

jj

22

26

66

18 10(900 0.0004 ),0.004 18 9000 0,

4500 2.25 10 0

4500 20.25 10 9 10 4500 3354 572.9, 3927rad/s

22

ω= + ω ω − ω+ =

ω− ω+ × =

±×−× ±

ω= = =

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43. With an operating frequency of

ω

= 400 rad/s, the impedance of the 10-mH inductor is

j

ω

L = j4 Ω, and the impedance of the 1-mF capacitor is –j/

ω

C = -j2.5 Ω.

V240(2.5)5 50A

I 3 2 40 1.9513 41.211 A

∴=∠°− =∠−°

∴=−∠°= ∠− °

c

L

j

2

1

2

1

2

22

1

2 40 (R 2.5)

IR4

2 40 (R 2.5)

R4

1.9513 41.21

1.0250 81.21 ( 2.5)

R (1.0250 81.21 ) 2.562 8.789

0.15662R 1.0130R 2.532 0.3915

R 2.532 0.

L

j

Rj

jj

∠° −

=+

∠° −

∴+= ∠− °

=∠°−

=∠°+∠−°

=++−

∴= + 22

21

15662R , 4 1.0130R 0.395

R 4.335 , R 3.211

−

+

=−

∴= Ω = Ω

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44.

ω

= 1200 rad/s.

(a)

(b)

2

(200 80) (80 200 )[200 ( 80)]

Z200 (80 ) 40,000 6400 160

X 0 40,000 80 6400 0

1

46,400 80 , 580 C 14.368 F

1200

−× + − + −

==

+− +−+

=∴− + − =

∴==Ω=∴=

µ

in

jjxjxjx

jx xx

xx x

xx c

22

2

222

2

80X 200X

Z Z 100

200 (80 X)

6400X 40,000X 10,000

40,000 6400 160X X

0.64X 4X X 160X 46,400

3.64X 160 46,400 0,

160 25,600 675,600 160 837.4

X7.28 7.28

1

X 93.05 ( 0) C 8.956 F

1200C

in in

j

X

−

==

+−

+

∴=

+−+

∴+=−+

∴+−=

−± + −±

==

∴= > = ∴= µ

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45. At

ω

= 4 rad/s, the 1/8-F capacitor has an impedance of –j/

ω

C = -j2 Ω, and the 4-H

inductor has an impedance of j

ω

L = j16 Ω.

(a)

(b)

(8 16)(2 2) 16(3 1)

abOC : Z 10 14 10 14

2.378 1.7297

in

jj j

jj

j

+− +

==

++

=− Ω

16 32

abSC: Z (8 2) (2 16) 82216

2.440 1.6362

−

=−+ = +

−+

∴= − Ω

in

jj

jj jj

Zj

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46. f = 1 MHz,

ω

= 2

π

f = 6.283 Mrad/s

2 µF → -j0.07958 Ω = Z1

3.2 µH → j20.11 Ω = Z2

1 µF → -j0.1592 Ω = Z3

1 µH → j6.283 Ω = Z4

20 µH → j125.7 Ω = Z5

200 pF → -j795.8 Ω = Z6

The three impedances at the upper right, Z3, 700 kΩ, and Z3 reduce to –j0.01592 Ω

Then we form Z2 in series with Zeq: Z2 + Zeq = j20.09 Ω.

Next we see 106 || (Z2 + Zeq) = j20.09 Ω.

Finally, Zin = Z1 + Z4 + j20.09 = j26.29 Ω.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47.

2H 2,1F 1Let I 1 0 A

V 2V I I 0.5 1 1

V2(11)(1)11

10 1 1 1

V 0.5 0.55

V1111

1

Now 0.5 2 , 0.5S 2H

2

∈

→→− =∠°

∴= ∴=+ =+

∴=++ −=+

∠° −

∴= = =−

+−

→Ω− = →

LcinL

in

jj

jvj

jjjj

jj

sj

j

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48.

(a)

(b) ,

106

R 21.2

5

==Ω

in ab

(c)

500,Z 510159

159 9

Y Y 500C

5 9 106 106

9

C 169.81 F

53,000

inRLC

inRLC c

jj j

j

ω= = + − = +

−

∴= = ∴==

+

∴= =

µ

,

91

1000 Y 53 5 20 0.5

0.012338 0.12169S or 0.12232 84.21 S

ω= ∴ = + +−

=+ ∠°

in ab jjj

j

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49.

(a)

(b)

(c)

(d)

62

462

62 6

462

42 6 42

42

0.1

R 550 : Z 500 100 0.001

50,000 0.6 100 0.001

Z100 0.001 100 0.001

5 10 0.0006 (60 50 )

Z10 10

5 10 0.006

R 550 5.5 10

10 10

5.5 10 5 10 10

0.5 10

in in

in

j

jj

j

−

−−

−

ω

=Ω =++ω

+ω − ω

∴= ×

+ω−ω

×+ ω+ ω−ω

∴= +ω

×+ ω

∴= =∴×

+ω

+× ω=×× ω

∴× ω= 6 2 10 5

0.5 10 , 10 , 10 rad/s×ω= ω=

642

462

2510

51010

55

10

X 50 0.5 10 0.5 10 10

10 10

0, 2 10 10 0

210 410 410 10 10 rad/s

2

−

ω

=Ω= = × + × ω−ω

+ω

=ω−× ω+ =

×±× −×

∴ω= = ∴ω=

in

3

642

82

642

3

82

642 6 62

662

100 0.001 50,000 0.6

G 1.8 10 : Y 50,000 0.6 50,000 0.6

510 610 (50 6)

25 10 0.36

510 610

1.8 10 25 10 0.36

5 10 6 10 4.5 10 648 10

0.5 10 48 10 102.06

−

−−

−

+ω −ω

=× = ×

+ω −ω

×+× ω+ ω−ω

=×+ ω

×+× ω

∴× = ×+ ω

∴× +× ω= × + × ω

∴× =× ω∴ω=

in in

jj

j

Krad/s

4

82

462

62 4

6

10

B 1.5 10 25 10 0.36

10 37.5 10 54 10

54 10 10 37.5 10 0,

100 81

10 52.23 and 133.95krad/s

108 10

−

−ω

=× =×+ ω

∴ω= × + × ω

∴× ω−ω+ × =

−

ω= ± =

×

in

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50.

(a) 1

11

3

1

I 0.1 30

V 20 23.13 V 20V

Y(34)10

−

∠°

== =∠− °∴=

+j

(b) 21 2

VVV 20V=∴ =

(c)

(d)

3

222

312

3

33

3

I Y V (5 2)10 20 23.13 0.10770 1.3286 A

I I I 0.1 30 0.10770 1.3286 0.2 13.740 A

I 0.2 13.740

V 44.72 77.18 V V 44.72V

Y(24)10

j

−

==+ ×∠−°= ∠−°

∴=+= ∠°+ ∠− °= ∠ °

∠°

∴= = = ∠ °∴ =

−

13

V V V 20 23.13 44.72 77.18 45.60 51.62

V 45.60V

=++∠− °+ ∠ °= ∠ °

∴=

in

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



51.

(a)

(b)

1

11

50 F 20 Y 0.1 0.05

1 1000 1

YR 84

1000 C 0.1 0.05

RC

1

R 8 and C 250 F

4

in

jj

j

µ

ω

→− Ω∴ = +

=∴−==−

+

−

∴=Ω = =

1

111

1

2000: 50 F 10 Y 0.1 0.1 500

RC

500

R 5 5 R 5 , C 100 F

C

in

jj

j

jj

ωµ

µ

=→−Ω∴=+=

−

∴− =−∴=Ω =

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



52.

(a)

ω Gin Bin

0

1

2

5

10

20

∞

0

0.0099

0.0385

0.2

0.5

0.8

1

0

0.0099

0.1923

0.4

0.5

0.4

0

2

22

10 10

Z1

10

Y10 10

10

Y100

10

G,B

100 100

in

in in

j

jj

j

ω

ωω

ω

ωω

+

=+ =

−

∴×

+−

+

∴= +

==

++

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



53.

11212 11212

12

21 21 2

21212 1 2

2

VV V

5 , 75 5V 3V 3V 5V 5V

353

(5 2)V 2V 75

VVVV10

356

10V 10V 6V 6V 5V 300 4V (5 4)V 300

52 75

4 300 1500 600 300 1200

V52 2 17 30 8

454

vv

jjjjjj

jj

jj j

v

jj

jjjj j j

jj

jj j

jj

−−

−=+ + −=+−−+

−

∴− + =−

−−

++=

−

−++−+=∴+− =

−−

−− −

∴= = =

−−+

−

600 34.36 23.63 V

25 30

j

j=∠°

−

(1)

(2)

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



54.

3I 5(I I ) 0 2I 5I 0

3(I 5) 5(I I ) 6(I 10) 0

5I (9 5)I 60 15

05

60 15 9 5 75 300

I25 15 18

59 5

13.198 154.23 A

BBD BD

DDBD

BD

B

jj j

jj

jj j

j

jj j

jj

−−=∴−+=

+− −+ +=

∴+− =−−

−− − −+

==

−−

−

=∠°

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



55.

12

20cos1000 V, 20sin1000 V

V200V,V 20V

0.01H 10 , 0.1mF 10

20 20 0, 0.04 2 2 0,

10 25 10

V 25(2 2) 70.71 45 V

( ) 70.71cos(1000 45 )V

ss

xxx x

x

vtvt

j

jj

vvvj vj

jj

j

vt t

==

∴=∠° =−

→Ω →−Ω

−+

∴++ = +−=

−

=−=∠−°

∴= −°

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



56.

(a)

(b)

32 2

1

3

Assume V 1V V 1 0.5V, I 1 0.5mA

V 1 0.5 (2 0.5)( 0.5) 0.75 1.5V

I 0.75 1.5mA, I 0.75 1.5 2 0.5 2.75 2mA

V 0.75 1.5 1.5 (2.75 2)( 0.5)

100

0.25 2.875V V 34.65 94.

0.25 2.875

in

jj

jjj j

jjjj

jj jj

jjj

+

= ∴ =− =−

∴=− +− − = −

∴= − ∴ = − +− = −

∴=−−+ −−

=− − ∴ = = ∠

−− 97 V°

33

22 12

22 2

11

223 233

2

0.5 Assume V 1V I 1A,

V 1 X, I 1 X, I 2 X

V 1 X(2 X)( X)1X 3X,I 1X 3X,I 3X 4

V1X 3X4X X 3X15X (X6X)X6X0

X 6, X 6, Z 2.449K

in

c

jjx

jj j

jjj j j jX

jjj j

j

−→− =∴=

=− =− → = −

∴=− +− − =− − =− − =− −

∴=−− − + − =− + − ∴−=

∴= = =− Ω

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



57. Define three clockwise mesh currents i1, i2, i3 with i1 in the left mesh, i2 in the top right

mesh, and i3 in the bottom right mesh.

Mesh 1: 10∠0o + (1 + 1 – j0.25)I1 – (-j0.25)I2 = 0

Mesh 2: – I1 + (1 + 1 + j4)I2 – I3 = 0

Mesh 3: (-j0.25 + 1 + 1)I3 – I2 – (-j0.25I1) = 0

2 0.25 1 10

1240

0.25 1 0

I2 0.25 1 0.25

124 1

0.25 1 2 0.25

10(1 1 0.5)

I0.25(2 0.5) ( 2 0.25 0.25) (2 0.25)(4 1 0.5 8 1)

20 5 I 1.217 75.96 A, ( ) 1.2127cos(100 75.96 )A

815

x

xx

j

jj

j

jj

j

jj jj j jj

jit t

j

−−

−+

−

=−−

−+−

−−

+−

∴= −+−+ + +− +−+−

−

=∴=∠−°= −°

+

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



58.

11 12

12

212 2

12

V 10 0.25V 0.25V V V 0

(2 0.25) V V 0.25V 10

VVVV 4V0

V(2 4)VV 0

0.25V 0.25V V V V

0.25V V (2 0.25)V 0

2 0.25 1 10

1240

0.25 1 0

V0.25 1 0.25

124 1

0.25 1 2

x

xxx

x

jj

j

jj

j

jj

j

jj

−− + +− =

∴− − + =

−+−+ =

−++ − =

−+++−

∴−+− =

−−

−+

−

=−−

−+ −

−−

0.25

10(1 1 0.5)

0.25(2 0.5) ( 2 0.25 0.25) (2 0.25)(4 1 0.5 8 1)

20 5 1.2127 75.96 V

815

1.2127cos(100 75.96 ) V

x

j

jj jj j jj

j

vt

+−

=−+−+ + +− +−+−

−

==∠−°

+

∴= − °

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



59.

(a)

(b)

1

11

1

R,R0,AV/V0

VAV

IC(VV)

R

V(1 A CR ) CR V

V

VAV (1A CR) CRV

A

CR A

VV

As A , C R

V1A CR V

ooi

isi

f

iffs

o

oi f fs

f

oo

f

sfs

j

jj

j

ω

ωω

ω

=∞ = =− >>

+

==−

∴++ =

=− ∴− + + =

∴=− →∞ →−

++

11

1

R

1

RC 11CR

CR

(V AV )

I(1CR)(VV)C,VAV

R

V(1 A)(1 C R ) V CR CR V,

V [(1 A) (1 C R ) C R ] C R V

V[(1 A)(1 C R ) C R ] C R V

A

CR A

V

V(1A)(1

f

ff

f

iff s i o i

f

iffsffi

iffffs

off f fs

f

o

s

j

jj

jjj

j

ω

ωω

ωωω

ω

==

+

=+=− =−

∴+ + = −

++ + =

∴− + + + =

−

∴=

++

1

CR

V

As A ,

CR) CR V 1 CR

f

o

ff f s ff

j

jj

ω

ωω ω

−

→∞ →

++

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



60. Define the nodal voltage v1(t) at the junction between the two dependent sources.

The voltage source may be replaced by a 3∠-3o V source, the 600-µF capacitor by a

–j/ 0.6 Ω impedance, the 500-µF capacitor by a –j2 Ω impedance, and the inductor by a

j2 Ω impedance.

5V2 + 3V2 = 2-

) - (

6.0/100

3-3 - 21

o

1

jj

VVV +

−

∠ [1]

-5V2 =

()

2

221

jj

VVV +

−

− [2]

Simplifying and collecting terms,

j2 V1 + (960.1∠ -90.95o) V2 = 6 ∠ 87o [1]

-j2 V1 + 20 V2 = 0 [2]

Solving, we find that V1 = 62.5 ∠ 86.76o mV and V2 = 6.25 ∠ 176.8o mV.

Converting back to the time domain,

v2(t) = 6.25 cos (103t + 176.8o) mV

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 

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61. Define three clockwise mesh currents: i1(t) in the left-most mesh, i2(t) in the bottom right

mesh, and i3(t) in the top right mesh. The 15-µF capacitor is replaced with a –j/ 0.15 Ω

impedance, the inductor is replaced by a j20 Ω impedance, the 74 µF capacitor is

replaced by a –j1.351 Ω impedance, the current source is replaced by a 2∠0o mA source,

and the voltage source is replaced with a 5∠0o V source.

Around the 1, 2 supermesh: (100 + j20) I1 + (13000 – j1.351) I2 – 5000 I3 = 0

and

-I1 + I2 = 2×10-3

Mesh 3: 5∠0o + (5000 – j6.667) I2 – 5000 I3 = 0

Solving, we find that I1 = 1.22∠179.9o mA. Converting to the time domain,

i1(t) = 1.22 cos (104t + 179.9o) mA

Thus, P1000 = [i1(1 ms)]2 • 1000

= (1.025×10-6)(1000) W = 1.025 W.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



62. We define an additional clockwise mesh current i4(t) flowing in the upper right-hand

mesh. The inductor is replaced by a j0.004 Ω impedance, the 750 µF capacitor is

replaced by a –j/ 0.0015 Ω impedance, and the 1000 µF capacitor is replaced by a –j/ 2

Ω impedance. We replace the left voltage source with a a 6 ∠ -13o V source, and the

right voltage source with a 6 ∠ 0o V source.

(1 – j/ 0.0015) I1 – I3 = 6 ∠ -13o [1]

(0.005 + j/ 0.0015) I1 + j0.004 I2 - j0.004 I4 = 0 [2]

-I1 + (1 – j/ 2) I3 + j0.5 I4 = -6 ∠ 0o [3]

-j0.004 I2 + j0.5 I3 + (j0.004 – j0.5) I4 = 0 [4]

Solving, we find that

I1 = 0.00144∠ -51.5o A, I2 = 233.6 ∠ 39.65o A, and I3 = 6.64 ∠ 173.5o A.

Converting to the time domain,

i1(t) = 1.44 cos (2t – 51.5o) mA

i2(t) = 233.6 cos (2t + 39.65o) A

i3(t) = 6.64 cos (2t + 173.5o) A

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



63. We replace the voltage source with a 2115 ∠0o V source, the capacitor with a

–j/ 2

π

C1 Ω impedance, and the inductor with a j0.03142 Ω impedance.

Define Z such that Z-1 = 2

π

C1 - j/0.03142 + 1/20

By voltage division, we can write that 6.014 ∠85.76o = 2115 20 +Z

Z

Thus, Z = 0.7411 ∠ 87.88o Ω. This allows us to solve for C1:

2

π

C1 – 1/0.03142 = -1.348 so that C1 = 4.85 F.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



64. Defining a clockwise mesh current i1(t), we replace the voltage source with a

2115 ∠0o V source, the inductor with a j2

π

L Ω impedance, and the capacitor with a

–j1.592 Ω impedance.

Ohm’s law then yields

()

o

108.132

592.1220

2115

∠=

−+

=Lj

π

I

Thus, 20 =

()

2

2592.12 20 −+ L

π

and we find that L = 253.4 mH.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



65. (a) By nodal analysis:

0 = (V

π

– 1)/ Rs + V

π

/ RB + V

π

/ r

π

+ j

ω

C

π

V

π

+ (V

π

– Vout) j

ω

C

µ

[1]

-gmV

π

= (Vout – V

π

) j

ω

Cµ + Vout / RC + Vout / RL [2]

Simplify and collect terms:

()

S

out

BS R

1

C - CC

r

1

R

1

R

1=











++











++ VV

µπµπ

π

ωω

jj [1]

(-gm + j

ω

C

µ

) V

π

- (j

ω

C

µ

+ 1/RC + 1/RL) Vout = 0 [2]

Define

π

r

1

R

1

R

1

R

1

BS

S

++=

′ and ′

L

R = R

C || RL

Then

()













′

+

′

+

+++

′′

=∆

SL

22

LS R

C

R

CC

C - CC 2C

R R

1-

µπµ

µπµµ ωω

m

gj

And Vout =

()













′

+

′

+

+++

′′

−

SL

22

LS

SSm

R

C

R

CC

C - CC 2C

R R

1-

RCRg

µπµ

µπµµ

µ

ωω

ω

m

gj

j

Therefore, ang(Vout) = 











−

−

2

1

tan

SmRg

Cj

µ

ω

-

()













++

′′













′

+

′

+

+−

−

πµµ

µπµ

µ

ω

CC 2C

R R

1-

R

C

R

CC

C

tan

22

LS

SL

1

m

g

(c) The output is ~180o out of phase with the input for f < 105 Hz; only for f = 0 is it

exactly 180o out of phase with the input.

(b)

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



66.

,

V 100 V

OC: 0.02V 0

20 10

10

10 (0.05 0.1 0.02) V ,V 0.07 0.1

V 67.11 46.98

V 100 V 32.89 46.98 57.35 55.01 V

100

SC:V 100 I 0.02 100 7A

20

57.35 55.01

Z 4.698 6.711

7

xx

x

xx

x

ab oc x

xSC

th

j

jj j

j

−

−+ − =

−

=++ =

+

∴= +

∴=−= − =∠−°

=∴↓= ×+ =

∠− °

∴= = − Ω

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



67.

V 210 0.5V

1

V(1 ) 2

1

12

V1

Z12

1

At 1, Z 1 1 2 1

1

Y 0.5 0.5

11

LL

in

jj

j

jj j

j

ω

ωω

ω

=∴ =

∴=+ +

=+ +

∴= =+ +

==−+=+

∴= =+

+

R = 500 mΩ, L = 500 mH.

so Yin =

()

12 2−+

ωω

ω

j

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



68.

(a)

(b) Is:

1

21 2

2 1 0.8 0.4 V

1212

0.8 0.4 10 20

25 11.785 135 V

1 1 0.8 0.4 1.8 0.6

jj

jjj j

+

−

==+∴

+−

+−+

===∠°

−+ + −

so v1(t) = 11.79 cos (1000t + 135o) V.

1

11

(1 1)1 2 1 3 1 15

V: V 0.6 0.2

2121 5 20.60.2

V 5 90 ( ) 5cos(1000 90 )V

s

jjj j

jj j j

vt t

−+− −

×=∴= ×−

−+ +−

∴=∠°∴ = +°

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



69.

,

OC:V 0 V 1 0 V

SC : I V 2I 1 0 1[0.25( 2I ) I ] 2I

1 (0.5 2)I (0.5 1)I

I1

I 0.4 0.8 0.4 0.8

0.5 1 1 0

111 1

R 2.5 , 0.8, L 1.25H

0.4 L L 0.8

I 0.4 0.8 0.8944 63.43 A

Laboc

NL N NN N

NN

N

NN

N

jjjj

jj j

jY j

j

jj

j

ω

=∴ =∠°

↓∴ = ∴∠°=− + +

∴= − + = +

∴= =− ∴= =−

+∠°

∴= =Ω = =− = =

=− = ∠− °

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



70.

,200

,100

1211

V 2( 1) ( 2) 2 1 1

11 1111

V 1.4142cos(200 45 ) V

1

100:V , 0.5cos(100 90 )V

2

L

LL

jj

jj j j

jjj

t

jv t

ω

−

=+− =+ = =+

++−

∴= +°

== = +°

so vL(t) = 1.414 cos (200t + 45o) + 0.5 cos (100t + 90o) V

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



71.

100

Use superposition. Left: V 100 100 300

300

50 0 V Right: V 100 150V

300 100

V 50 150 158.11 108.43 V

30,000

Z 100 300 150

200

ab

th

j

jj

j

jj

j

jj j

j

=−

−

=− ∠ ° = =

−+

∴=−+ = ∠ °

=−= =Ω

−

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



72. This problem is easily solved if we first perform two source transformations to yield a

circuit containing only voltage sources and impedances:

Then I = 4131073

45-2.920 90-0.240 175 oo

jj

o

−++

∠+∠+∠

= (6.898∠ -7.019o)/ (83.49 ∠ 6.189o) = 82.62 ∠ -13.21 mA

Converting back to the time domain, we find that

i(t) = 82.62 cos (103t – 13.21o) mA

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



73.

(a) There are a number of possible approaches: Thévenizing everything to the left of the

capacitor is one of them.

VTH = 6(j2)/ (5 + j2) = 2.228 ∠ 68.2o V

ZTH = 5 || j2 = j10/ (5 + j2) = 1.857 ∠ 68.2o Ω

Then, by simple voltage division, we find that

V

C = (2.228 ∠ 68.2o) 7 3/ - 2.68857.1

3/

ojj

j

+∠

−

= 88.21 ∠-107.1o mV

Converting back to the time domain, vC(t) = 88.21 cos (t – 107.1o) mV.

(b) PSpice verification.

Running an ac sweep at the

frequency f = 1/2

π

= 0.1592 Hz,

we obtain a phasor magnitude of

88.23 mV, and a phasor angle of

–107.1o, in agreement with our

calculated result (the slight

disagreement is a combination

of round-off error in the hand

calculations and the rounding

due to expressing 1 rad/s in Hz.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



74. (a) Performing nodal analysis on the circuit,

Node 1: 1 = V1/ 5 + V1/ (-j10) + (V1 – V2)/ (-j5) + (V1 – V2)/ j10 [1]

Node 2: j0.5 = V2/ 10 + (V2 – V1)/ (-j5) + (V2 – V1)/ j10 [2]

Simplifying and collecting terms,

(0.2 + j0.2) V1 – j0.1 V2 = 1 [1]

-j V1 + (1 + j) V2 = j5 [2]

Solving, we find that V2 = VTH = 5.423 ∠ 40.60o V

ZTH = 10 || [(j10 || -j5) + (5 || -j10)] = 10 || (-j10 + 4 – j2) = 5.882 – j3.529 Ω.

FREQ VM($N_0002,0) VP($N_0002,0)

1.592E+01 4.474E+00 1.165E+02

FREQ VM($N_0005,0) VP($N_0005,0)

1.592E+01 4.473E+00 1.165E+02

(b)

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



75. Consider the circuit below:

Using voltage division, we may write:

Vout = Vin CjR

Cj

ω

/1

+, or RCj

ω

+

=1

1

in

out

V

The magnitude of this ratio (consider, for example, an input with unity magnitude and

zero phase) is

()

2

in

out

1

RC

ω

+

=

V

As

ω

→ 0, this magnitude → 1, its maximum value.

As

ω

→ ∞, this magnitude → 0; the capacitor is acting as a short circuit to the ac signal.

Thus, low frequency signals are transferred from the input to the output relatively

unaffected by this circuit, but high frequency signals are attenuated, or “filtered out.”

This is readily apparent if we plot the magnitude as a function of frequency (assuming R

= 1 Ω and C = 1 F for convenience):

Cj

ω

1

Vin

Vout

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



76. Consider the circuit below:

Using voltage division, we may write:

Vout = Vin CjR

R

ω

/1+, or RCj

RCj

ω

+

=1

in

out

V

The magnitude of this ratio (consider, for example, an input with unity magnitude and

zero phase) is

()

2

in

out

1

RC

ω

+

=

V

As

ω

→ ∞, this magnitude → 1, its maximum value.

As

ω

→ 0, this magnitude → 0; the capacitor is acting as an open circuit to the ac signal.

Thus, high frequency signals are transferred from the input to the output relatively

unaffected by this circuit, but low frequency signals are attenuated, or “filtered out.”

This is readily apparent if we plot the magnitude as a function of frequency (assuming R

= 1 Ω and C = 1 F for convenience):

Vin

Vout

1/j

ω

C

R

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



77. (a) Removing the capacitor temporarily, we easily find the Thevenin equivalent:

Vth = (405/505) VS and Rth = 100 || (330 + 75) = 80.2 Ω

(b) Cj

Cj

ω

/12.80

1/

505

405

Sout +

=VV so

ω

12

S

out

10532.21

1

505

405

−

×+













=jV

V

and hence 224

S

out

10411.61

0.802

ω

−

×+

=

V

(c)

S

505

405 V

80.2 Ω

31.57 fF +

Vout

-

Both the MATLAB plot of the

frequency response and the PSpice

simulation show essentially the

same behavior; at a frequency of

approximately 20 MHz, there is a

sharp roll-off in the transfer

function ma

g

nitude.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



78. From the derivation, we see that

µ

ω

)CR||(R1

)CR||(R )R || (Rg-

LC

LCLCm

in

out

j

+

=

V

so that

2

1

2

LC

2

LC

2

LC

2

m

in

out

C

RR

1

C

RR

g

























+













+













+

=

µ

ω

V

This function has a maximum value of gm (RC || RL) at

ω

= 0. Thus, the capacitors reduce

the gain at high frequencies; this is the frequency regime at which they begin to act as

short circuits. Therefore, the maximum gain is obtained at frequencies at which the

capacitors may be treated as open circuits. If we do this, we may analyze the circuit

of Fig. 10.25b without the capacitors, which leads to

()

BBS

B

LC

m

BS

B

LC

m

frequency low

S

out

Rr)Rr(R

Rr

RR

g-

R|| rR

R|| r

RR

g-

ππ

π

++













+

=

+













+

=

V

The resistor network comprised of rπ, RS, and RB acts as a voltage divider, leading to a

reduction in the gain of the amplifier. In the situation where rπ || RB >> RS, then it has

minimal effect and the gain will equal its “maximum” value of –gm (RC || RL).

(b) If we set RS = 100 Ω, RL = 8 Ω, RC | max = 10 kΩ and rπgm = 300, then we find that

B

m

S

t

ou

R || r100

R || r

(7.994) g-

π

+

=

V

We seek to maximize this term within the stated constraints. This requires a large value

of gm, but also a large value of rπ || RB. This parallel combination will be less than the

smaller of the two terms, so even if we allow RB → ∞, we are left with

ππ

π

r100

2398-

r100

rg

(7.994) - m

S

t

ou

+

=

+

≈

V

Considering this simpler expression, it is clear that if we select rπ to be small, (i.e.

r

π << 100), then gm will be large and the gain will have a maximum value of

approximately –23.98.

(c) Referring to our original expression in which the gain Vout/ Vin was computed, we

see that the critical frequency

ω

C = [(RC || RL) Cµ]-1. Our selection of maximum RC,

R

B → ∞, and rπ << 100 has not affected this frequency.

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



79. Considering the

ω

= 2×104 rad/ s source first, we make the following replacements:

100 cos (2×104t + 3o) V → 100 ∠3o V

33 µF → -j1.515 Ω 112 µH → j2.24 Ω 92 µF → -j0.5435 Ω

Then

(V1´ – 100 ∠ 3o)/ 47×103 + V1´/ (-j1.515) + (V1´ – V2´)/ (56×103 + j4.48) = 0 [1]

(V2´ – V1´)/ (56×103 + j4.48) + V2´/ (-j0.5435) = 0 [2]

Solving, we find that

V1´ = 3.223 ∠ -87o mV and V2´ = 31.28 ∠ -177o nV

Thus, v1´(t) = 3.223 cos (2×104t – 87o) mV and v2´(t) = 31.28 cos(2×104t – 177o) nV

Considering the effects of the

ω

= 2×105 rad/ s source next,

100 cos (2×105t - 3o) V → 100 ∠-3o V

33 µF → -j0.1515 Ω 112 µH → j22.4 Ω 92 µF → -j0.05435 Ω

Then

V1"/ -j0.1515 + (V1" – V2")/ (56×103 + j44.8) = 0 [3]

(V2" – V1")/ (56×103 + j44.8) + (V2" – 100 ∠ 3o)/ 47×103 + V2"/ (-j0.05435) = 0 [4]

Solving, we find that

V1" = 312.8 ∠ 177o pV and V2" = 115.7 ∠ -93o µV

Thus,

v1"(t) = 312.8 cos (2×105t + 177o) pV and v2"(t) = 115.7 cos(2×105t – 93o) µV

Adding, we find

v1(t) = 3.223×10-3 cos (2×104t – 87o) + 312.8×10-12 cos (2×105t + 177o) V and

v2(t) = 31.28×10-9 cos(2×104t – 177o) + 115.7×10-12 cos(2×105t – 93o) V

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



80. For the source operating at

ω

= 4 rad/s,

7 cos 4t → 7∠ 0o V, 1 H → j4 Ω, 500 mF → -j0.5 Ω, 3 H → j12 Ω, and 2 F → -j/ 8 Ω.

Then by mesh analysis, (define 4 clockwise mesh currents I1, I2, I3, I4 in the top left, top

right, bottom left and bottom right meshes, respectively):

(9.5 + j4) I1 – j4 I2 – 7 I3 - 4 I4 = 0 [1]

-j4 I1 + (3 + j3.5) I2 – 3 I4 = -7 [2]

-7 I1 + (12 – j/ 8) I3 + j/ 8 I4 = 0 [3]

-3 I2 + j/ 8 I3 + (4 + j11.875) I4 = 0 [4]

Solving, we find that I3 = 365.3 ∠ -166.1o mA and I4 = 330.97 ∠ 72.66o mA.

For the source operating at

ω

= 2 rad/s,

5.5 cos 2t → 5.5∠ 0o V, 1 H → j2 Ω, 500 mF → -j Ω, 3 H → j6 Ω, and 2 F → -j/ 4 Ω.

Then by mesh analysis, (define 4 clockwise mesh currents IA, IB, IC, ID in the top left, top

right, bottom left and bottom right meshes, respectively):

(9.5 + j2) IA – j2 IB – 7 IC – 4 ID = 0 [1]

-j2 IA + (3 + j) IB – 3 ID = -7 [2]

-7 IA + (12 – j/ 4) IC + j/ 4 ID = 0 [3]

-3 I2 + j/ 4 IC + (4 + j5.75) ID = 0 [4]

Solving, we find that IC = 783.8 ∠ -4.427o mA and ID = 134 ∠ -25.93o mA.

V1´ = -j0.25 (I3 – I4) = 0.1517∠131.7o V and V1" = -j0.25(IC – ID) = 0.1652∠-90.17o V

V2´ = (1 + j6) I4 = 2.013∠155.2o V and V2" = (1 + j6) ID = 0.8151∠54.61o V

Converting back to the time domain,

v

1(t) = 0.1517 cos (4t + 131.7o) + 0.1652 cos (2t - 90.17o) V

v

2(t) = 2.013 cos (4t + 155.2o) + 0.8151 cos (2t + 54.61o) V

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



81.

(a) 100 100(2 1)

I 57.26 76.76 (2.29 )

22.5 3

2.5 21

L

jin

j

jj

−

===∠−°

−+

+−

1

I (57.26 76.76 ) 25.61 140.19 (1.02 )

21

R

jin

j

−

=∠−° =∠−°

−

2

I (57.26 76.76 ) 51.21 50.19 (2.05 )

21

cin

j

=∠−° =∠−°

−

V 2.5 57.26 90 76.76 143.15 13.24 (2.86 )

Lin=× ∠°− °= ∠ °

V 2 25.61 140.19 51.22 140.19 (1.02 )

Rin= × ∠− °= ∠− °

V 51.21 140.19 (1.02 )

cin=∠− °

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



82.

(a)

(b)

(c)

1

2

3

120

I 3 30 A

40 30

120

I 2.058 30.96 A

50 30

120

I 2.4 53.13 A

30 40

j

==∠−°

∠°

==∠°

−

==∠−°

+

123

IIII

6.265 22.14 A

s=++

=∠−°

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



83.

12

12 1 2

12

22 2

I5A,I7A

I I 10 0 , I lags V,I leads V

I lags I . Use 2.5A /

[Analytically: 5 7 10

5cos 5sin 7cos 7sin

sin 1.4sin

5 1 1.4 sin 7 1 1sin 10

By SOLVE, 40.54 27.66 ]

in

jj

αβ

αα

ββ

αβ

ββ

αβ

==

+=∠°

∠+∠=

=+ ++

∴=−

∴− +− =

=− ° = °

CHAPTER TEN (Phasor Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



84. V1 = 100∠0o V, |V2| = 140 V, |V1 + V2| = 120 V.

Let 50 V = 1 inch. From the sketch, for ∠V2 positive,

V

2 = 140∠122.5o. We may also have V2 = 140∠-122.5o V

[Analytically: |100 + 140∠

α

| = 120

so | 100 + 140 cos

α

+ j140 sin

α

| = 120

Using the “Solve” routine of a scientific calculator,

α

= ±122.88o.]

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1.

6

22

50( 80)

10 80 , 42.40 32.01

500 25 50 80

84.80 32.01 V, 1.696 32.01 A

1.0600 57.99 A

( / 2ms) 84.80cos(45 32.01 )2cos 45 116.85W

50 1.696 cos (45 32.01 ) 136.55W

84.80cos(45 32

c

R

c

s

R

c

j

jj

p

π

−

==−Ω =∠−°Ω

×−

∴ = ∠− ° = ∠− °

=∠°

=°−°°=

=× °− °=

=°−

Z

VI

I

.01 ) 1.060cos(45 57.99 ) 19.69W°= °+ °=−

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2.

(a)

(b)

2242

42 4 2

11

4H: 2 1 4(4 ) 16 , 4(4 4 1)

22

8 8 2 (3) (1) 83 83 2 81812 576 J

L

LLL

it vLi t tw Li t t

wtt w w

′

=−∴== = = =× −+

∴ = − +∴ − = × −× + −×+×− =

233

11

1222

0.2F: (2 1) 25 25 5 1 2

0.2 3 3 3

10 10 61 61

(2) 8 10 5 2 V P (2) 7 142.33W

333 3

t

c

cc

vtdttt tt

v

 

=−+=−+=−−−+

 

 

∴=×−−++=∴=×=

∫

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3. 2

1,2

R1

(0) 2V, (0) 4A, 2, 3, 2 1 1, 3

2L LC

co

vi s

αω

=− = =====−±=−−

(a)

(b) 0.2 0.6 0.2 0.6

P (0.2) (3 5 )( 5 ) 0.5542W

ceeee

−− −

=− −+=−

(c) 0.4 1.2 1.2 0.4

P (0.4) (3 5 )(5 ) 0.4220W

ceeee

−−−−

=− −=

3

333

3

1

AB AB4;(0) (0)(442)14

1

A38 14B5,A 1, 5 A

3( 5 ) 23( 5 ) 2 35 52

35 P(0)(35)(15)8W

tt

L

tt

ttt ttt t t

co

o

tt

cc

ie e i v

ie e

veedtee ee

ve e

−− + +

−−

−− −− − −

−− +

=+∴+= = =−××+=−

∴− − =− ∴ = =− =− +

∴+ = − + − = − − = − − + −

∴= − ∴ =− −+=−

∫

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4. We assume the circuit has already reached sinusoidal steady state by t = 0.

2.5 kΩ → 2.5 kΩ, 1 H → j1000 Ω, 4 µF → -j250 Ω, 10 kΩ → 10 kΩ

Zeq = j1000 || -j250 || 10000 = 11.10 – j333.0 Ω

Veq = V 54.50631.2

0.33310.112500

)0.33310.11)(3020( o

∠=

−+

−∠

j

I10k = mA 50.54- 2631.0

10000

o

eq ∠=

V I1 H = mA 140.5- 631.2

1000

o

eq ∠=

j

V

I

4 µF = mA 46.93 52.10

250

o

eq ∠=

−j

V V2.5k = V 55.3774.19

0.33310.112500

)2500)(3020( o

∠=

−+

∠

j

Thus, P2.5k =

[]

mW 97.97

2500

55.37cos74.19 2

o=

P

1 H =

()

[

]

[]

mW 3.395- )5.140cos(102.631 54.50cos631.2 o-3

o=−×−

P

4 µF =

(

)

[

]

[

]

mW 13.58 )cos(39.461010.52 54.50cos631.2 o-3o =×−

P

2.5k =

()

[]

W 279.6

10000

54.50cos631.2 2

o

µ

=

−

FREQ IM(V_PRINT1) IP(V_PRINT1)

1.592E+02 7.896E-03 3.755E+01

FREQ VM(R2_5k,$N_0002)VP(R2_5k,$N_0002)

1.592E+02 1.974E+01 3.755E+01

FREQ IM(V_PRINT2) IP(V_PRINT2)

1.592E+02 2.628E-03 -1.405E+02

FREQ VM(L,0) VP(L,0)

1.592E+02 2.629E+00 -5.054E+01

FREQ IM(V_PRINT11) IP(V_PRINT11)

1.592E+02 1.052E-02 3.946E+01

FREQ IM(V_PRINT12) IP(V_PRINT12)

1.592E+02 2.629E-04 -5.054E+01

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5.

rad rad

,

8

40 53.13

50A,C 4, 8(3 4) 11 4

3.417 33.15 17.087 33.15 ,

17.087cos(25 33.15 )V

P (0.1) 17.087cos(2.5 33.147 ) 5cos2.5 23.51 W

17.087 cos(25 33.15 )

8

(0.1) 2.136c

sin

s

s abs

ijj

j

vt

it

i

∠− °

→∠° →− Ω = − = −

=∠−°∴= ∠−°

=−°∴

=− − ° × =−

=−°∴

=

Z

V

rad

2

8,

3

rad

3

2

3,

os(2.5 33.15 ) 0.7338 A

P 0.7338 8 4.307 W;

17.087 33.15 3.417 19.98 A

34

(0.1) 3.417cos(2.5 19.98 ) 3.272A

P 3.272 3 32.12 W

4(3.417 19.983 ) 13.67 70.02 ,

(0.1) 13.67

abs

abc

c

j

i

j

v

−°=−

∴= ×=

∠− °

==∠°

−

∴= +°=− ∴

=×=

=− ∠ ° = ∠− °

=

I

V

rad

,

0cos(2.5 70.02 ) 3.946V

P 3.946( 3.272) 12.911 W ( 0)

cabc

−°=

∴= − =− Σ=

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6.

,

2

4,

10

2

10,

5(10 5)

4 4 2.5 5 6.5 5

10

100 12.194 37.57 A

6.5 5

1

P 100 12.194cos37.57 483.3W

2

1

P (12.194) 4 297.4W,

2

P0

100 5 6.097 52.43 so

6.5 5 10

1

P (6.097) 10 185.87

2

in

s

s abs

abs

cabs

abs

jj jj

j

−

=+ =+ + = + Ω

∴= = ∠− °

+

∴=−×× °=−

==

=

==∠°

+

=×=

Z

I

W

P 0 ( 0)

L=Σ=

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7.

10,

2

820

40 30

(10 10) 52.44 69.18 V

550 8 20

1

P 10 52.44cos69.18 93.20 W

2

1

P 10 52.44cos(90 69.18 ) 245.08 W

2

1 52.44

P 8cos( 20 ) 161.51 W ( )

28

gen

jgen

abs gen abs

j

∠−

∠°

=+ = ∠ °

∠°+∠− °

=×× °=

=×× °− °=



=−°=Σ=Σ





V

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

1

331343

0.1 0.3

25 529

Ignore 30 on , 5 ,

68 10

R

sR R

jj

j

=+ =++ =+ Ω

−

+

°= =

+

Z

VI I

(a)

2

3

1529

P 3 10.875 W

210

Ω



=×=





(b)

,

(2 5)(4 3)

V 5 0 13.463 51.94 V

68

1

P 13.463 5cos 51.94 20.75 W

2

s

sgen

jj

j

++

=∠° = ∠ °

+

∴=× × °=

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9.

10 5

10 10 10

10

2

10

50

PP0,

50 50 0

10 10 5

( 0.1 0.1 0.2) 5 10 0

79.06 16.57 V

1 79.06

P 312.5 W;

210

79.06 161.57 50 12.75 78.69 A

10

1

P 50 12.748cos78.69 62.50 W

2

79.06 161.57

jj

V

j

jj

jjj

j

−

Ω

==

−−

++ =

−

∴−++ ++=

∴= ∠ °

==

∠°−

==∠°

∴=×× °=

∠°

=

VVV

V

I

50

50 15.811 7.57 :

5

1

P 50 15.811cos (90 71.57 ) 375.0 W

2

j

−=∠−°

−

=× × °+ °=−

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10.

20 2,

23

3602212

514 60, 0

32

22 30,2(23) 0

60 14

02 3 120 180 9.233 83.88 V

514

10 15 28

22 3

560

20 5.122 140.9 V

18 15

1

P 9.233 2 5

2

xxc

c

xxcc

cx c

xc

cx c x c

x

c

gen

j

jj

j

−−

+=

−+ − =

−

∴− = + =

−

∴−+ =−++ =

−

++

== =∠−°

−+−

−+

−

==∠−°∴

−+

=× ××

VVV

V

VVVV

VV V

VV

VV V V V

V

.122cos( 83.88 140.19 ) 26.23 W−°+ °=

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

(a) X 0 R 0

in L th j=∴ = +Z

(b) R ,X independent R X

LL L th th th

j

∗

∴= = −ZZ

(c)

2

22

V

1

R fixed P R R X

2(R R ) (X X )

th

LL LLLth

th L th L

j∴= × ∴ = −

+++ Z

(d)

(e) 22

X0R RX

L L th th th

=∴ = + =Z

222

22

2

22

222 2

22 2 2

2P RL

X fixed, Let X X (R R )

RR 2R(RR)

0

(R R )

R 2RR R 2RR 2R 0

RR R(XX)

L

LLth

th L

th

th L L th L

Lth L

th th L L th L L

Lth ththL

af a

a

df

dR a

a

+=∴= = ++

++− +

==



++



+++−==

∴= += + +

V

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12. 10

120 107.33 116.57 V

10 5

10(10 15) 814

10 5

th

j

jj j

j

−

==∠−°

+

−+

==−Ω

+

V

Z

(a) 8 15

Lj∴=+ΩZ

(b)

2

,max

107.33 116.57

16

1 107.33

P 8 180 W

216

L

∠− °

=∴



=×=





I

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13.

22

2

22

R R 8 14 16.125

1 107.33

P 16.125 119.38W

2(8 16.125) 14

Lth L

L

=∴=+= Ω

=×=

++

Z

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14.

9.6 4.8I 1.92 I 4.8I

9.6

I5

1.92

V (0.6 5)8 24V

1

P 24 1.6 5 96 W ( )

2

xxx

x

o

jj

gen

−=− − −+

∴= =

∴= × =

∴=×××=

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

(a)

(b)

max

480 80 60

80 60 80 60 80 60

28.8 38.4 28.8 38.4

th

L

jj

jjj

jj

−

==

+−

=+ Ω∴ =− Ω

Z

22

,max 2

5(28.8 38.4) 144 192V,

144 192

2 28.8

1 144 192

and P 28.8 250 W

24 28.8

th

L

jj

j

=+=+

+

∴= ×

+

=×=

×

V

I

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16. Zeq = (6 – j8) || (12 + j9) = 8.321 ∠ -19.44o W

Veq = (5 ∠-30o) (8.321 ∠ -19.44o) = 41.61 ∠ -49.44o V

P

total = ½ (41.61)(5) cos (-19.44o) = 98.09 W

I6-j8 = Veq / (6 – j8) = 4.161 ∠ 3.69o A

I4+j2 = I8+j7 = Veq/ 12+j9 = 2.774 ∠ -86.31o A

P

6-j8 = ½ (41.61)(4.161) cos (-49.44o – 3.69o) = 51.94 W

P

4+j2 = ½ (2.774)2 (4) = 15.39 W

P

8+j7 = ½ (2.774)2 (8) = 30.78 W

Check: Σ = 98.11 W (okay)

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

22

,max 2

10(20)

10

100 20 40, 4 8

20 10 20 10

R R 8.944

12040

P 8.944 38.63W

2(4 8.944) 64

th th

Lth L

L

j

jjj

jj

==+==+Ω

++

∴= ∴= Ω

+

∴= ×=

++

VZ

Z

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18. We may write a single mesh equation: 170 ∠0o = (30 + j10) I1 – (10 – j50)(-λI1)

Solving,

I1 =

λλ

50101030

0170 o

jj −++

∠

(a) λ = 0, so I1 = A 18.43-5.376

1030

0170 o

o∠=

+

∠

j and, with the same current flowing

through both resistors in this case,

P

20 = ½ (5.376)2 (20) = 289.0 W

P

10 = ½ (5.376)2 (10) = 144.5 W

(b) λ = 1, so I1 = A 543.005

4040

0170 o

o∠=

−

∠

j

P

20 = ½ (3.005)2 (20) = 90.30 W

The current through the 10-Ω resistor is I1 + λI1 = 2 I1 = 6.01 ∠ 45o so

P

10 = ½ (6.01)2 (10) = 180.6 W

(c)

(a)

FREQ IM(V_PRINT3) IP(V_PRINT3)

6.000E+01 5.375E+00 -1.846E+01

FREQ IM(V_PRINT4) IP(V_PRINT4)

6.000E+01 5.375E+00 -1.846E+01

(b)

FREQ IM(V_PRINT3) IP(V_PRINT3)

6.000E+01 6.011E+00 4.499E+01

FREQ IM(V_PRINT4) IP(V_PRINT4)

6.000E+01 3.006E+00 4.499E+01

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19. (a) Waveform (a): Iavg = A 667.1

3

)1(0)1)(5()1)(10( =

+−+

Waveform (b): Iavg = A 5

2

)1(0)1)(20(

2

1

=

+

Waveform (c):

I

avg =

()

3

310

0

3

10

0 33

102

cos

2

104

108-

104

2

8sin

101

1

−

−











×











×

×=

×× −

−

−− ∫t

dt

t

π

=

()

A

16

10

16

ππ

=−−

(b) Waveform (a): 22

avg A 41.67

3

(0)(1)(25)(1)(100)(1)

I =

++

=

Waveform (b): i(t) = -20×103 t + 20

i2(t) = 4×108 t2 – 8×105 t + 400

()

dttt

∫+××

×

=-3

10

0

528

3-

2

avg 400 108 - 104

102

1

I

=

() () ()

2

3-

3

2

3

5

3

8

3- A 66.67

102

0.1333

10400 10

2

108

- 10

3

104

102

1=

×

=











+

××

×

−−−

Waveform (c):

()

2

3

10

0

3

10

0 3

2

3

2

avg

A 32

102

sin

2

10

1064

102

10sin

-

2

1064

104

2

64sin

101

1

I

3

=













×

−×=













×

×=

××

=

−

−−

−

∫

π

ππ

tt

dt

t

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20. At

ω

= 120π, 1 H → j377 Ω, and 4 µF → -j663.1 Ω

Define Zeff = j377 || -j663.1 || 10 000 = 870.5 ∠ 85.01o Ω

V2.5k =

(

)

V 27.61- 520.4

85.01 870.5 2500

2500 92400 o

o

o∠=

∠+

−∠

V

10k =

(

)

()

V 40.75 181.2

85.01 870.5 2500

85.01 870.5 92400 o

o

oo ∠=

∠+

∠−∠

Thus, P2.5k = ½ (520.4)2 / 2 500 = 54.16 W

P

10k = ½ (181.2)2 / 10 000 = 1.642 W

P

1H = 0

P

4µF = 0 (A total absorbed power of 55.80 W.)

To check, the average power delivered by the source:

Isource = A 27.61- 0.2081

85.01870.52500

92400 o

o

o∠=

∠+

−∠

and Psource = ½ ( 2400 )(0.2081) cos (-9o + 27.61o) = 55.78 W (checks out).

FREQ IM(V_PRINT1) IP(V_PRINT1)

6.000E+01 2.081E-01 -2.760E+01

FREQ VM(R2_5k,$N_0002) VP(R2_5k,$N_0002)

6.000E+01 5.204E+02 -2.760E+01

FREQ IM(V_PRINT2) IP(V_PRINT2)

6.000E+01 4.805E-01 -3.260E+01

FREQ VM(L,0) VP(L,0)

6.000E+01 1.812E+02 5.740E+01

FREQ IM(V_PRINT11) IP(V_PRINT11)

6.000E+01 2.732E-01 1.474E+02

FREQ IM(V_PRINT12) IP(V_PRINT12)

6.000E+01 1.812E-02 5.740E+01

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

(a)

(b) 222

1

F (10 20 10 ) 150 12.247

4

eff =++==

(c) Favg = 10

4

40

4

)1)(10()1)(20()1)(10( ==

++

10 9cos100 6sin100

11

V 100 81 36 158.5 12.590V

22

eff

vtt=+ +

∴ = +× +× = =

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22.

(a) g(t) = 2 + 3cos100t + 4cos(100t – 120o)

3 ∠0 + 4∠-120o = 3.606 ∠-73.90o so Geff = 2

606.3

4

2

+ = 3.240

(b)

(c)

22

( ) 2 3cos100 4cos(101 120 )

11

H 2 3 4 16.5 4.062

22

eff

ht t t=+ + − °

∴=++ = =

0.1 62

0

63

1

( ) 100 , 0 0.1 F 10

0.3

10 1

10 10 33.33

33

eff

ft t t tdt

−

=<<∴=

=×××=

∫

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23. 2

( ) (2 3cos100 )ft t=−

(a)

(b) 22 2

11

F 8.5 12 4.5 12.43

22

eff =+×+×=

2

( ) 4 12cos100 9cos 100

( ) 4 12cos100 4.5 4.5cos200 F 4 4.5 8.5

av

ft t t

=− +

∴=− ++ ∴=+=

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24. (a) ieff =

()

A 6.455 0)5(10

3

12

1

22 =











+−+

(b) ieff =

[]

A 2.236 5 0 2020

2

12

1

0 ==











+











+−

∫dtt

(c) ieff = A 2.257

2

cos

2

8-

4

2

sin8

1

1 1

0

2

1

0 =



































=



































∫t

dtt

π

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25.

(a)

(b)

(c)

(d)

(e) 2

10

// 10 P 55.18 80.18 W

4

av

dc+∴= +=

o

2

A B 10V, C D 0 10 0 10 45 18.48 22.50

11

P 18.48 42.68 W

24

= = = = ∴ ∠ °+ ∠− °= ∠−

∴=×× =

22

A C 10V, B D 0, 10cos10 10cos40 ,

110 110

P25 W

24 24

s

vtt== == = +

=+=

o

2

10cos10 10sin (10 45 ) 10 10 45 7.654 67.50

1 7.654

P 7.322 W

24

s

vtt=− +°→−∠−°=∠

∴= =

o

22

10cos10 10sin (10 45 ) 10cos40 ;

10 0 10 45 18.48 22.50

1111

P 18.48 10 55.18 W

2424

vt t t=++°+

∠°+ ∠− °= ∠−

∴=× ×+× ×=

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26. Zeq = R || j0.3

ω

=

ω

RjR

Rj

3.0

+. By voltage division, then, we write:

V100mH =

ωω

ω

Rj

jR

Rj

j

4.003.0

1.00.03-

0120

3.0

1.0

0120 2

2

+−

+

∠=

+

∠

V

300mH = Rω.jω.

Rj

jR

Rj

j

jR

Rj

40030

36

0120

3.0

1.0

3.0

0120 2+−

∠=

+

∠

ω

(a) We’re interested in the value of R that would lead to equal voltage magnitudes, or

()

R0.10.03- (120) 36 2

ωωω

jRj +=

Thus, 36R

ω

= 224 14496.12 R

ωω

+ or R = 0.1061

ω

(b) Substituting into the expression for V100mH, we find that V100mH = 73.47 V,

independent of frequency.

To verify with PSpice, simulate the circuit at 60 Hz, or

ω

= 120π rad/s, so R = 40 Ω.

We also include a miniscule (1 pΩ) resistor to avoid inductor loop warnings. We see

from the simulation results that the two voltage magnitudes are indeed the same.

FREQ VM($N_0002,$N_0003)VP($N_0002,$N_0003)

6.000E+01 7.349E+01 -3.525E+01

FREQ VM($N_0001,$N_0002)VP($N_0001,$N_0002)

6.000E+01 7.347E+01 3.527E+01

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

(a) ,1

V30V

av =

,2

1

V (10 30 50) 30V

3

av =++=

(b)

(c) PSpice verification for Sawtooth waveform of Fig. 11.40a:

32

,1 0

222

,2

111

V (20 ) 400 27 1200 34.64V

333

11

V (10 30 50 ) 3500 34.16V

33

eff

tdt

==×××==

=++=×=

∫

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28. Zeff = R || 6

66

103

10

3

10

jR

jRj

−

=











−

ωω

ISRC =

()

666

6

66 1010310

10-R3120

103

1010

0120

RjjRj

j

jR

R

jj

ωω

−−−

=

−

−−

∠

I3µF = ISRC

ω

3

106

jR

R

−

(a) For the two current magnitudes to be equal, we must have 1

3

106=

−

ω

jR

R. This is

only true when R = ∞; otherwise, current is shunted through the resistor and the two

capacitor currents will be unequal.

(b) In this case, the capacitor current is

A )90 cos( 90 or A, 90

3

1010

1

0120 o

66

µωωµω

ωω

+=

−−

∠tj

jj

(c) PSpice verification: set f = 60 Hz, simulate a single 0.75-µF capacitor, and include a

100-MΩ resistor in parallel with the capacitor to prevent a floating node. This should

resit in a rms current amplitude of 33.93 mA, which it does.

FREQ IM(V_PRINT3) IP(V_PRINT3)

6.000E+01 3.393E-02 9.000E+01

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29.

0.5( 3)

22

1, 0

5(3) 3 5

2, 3

3

22

23

( ) 10 [ ( ) ( 2)] 16 [ ( 3) ( 5)] V

Find eff. value separately

120

V 100 8 7.303

53

1 256

V 256 ( ) 6.654

55

V 7.303 6.654 9.879

1

V 100 256

5

t

eff

tt

eff

t

eff

vt tut ut e ut ut

tdt

edt ee

tdt ee

−−

−− −

−

=−−+ −−−

==×=

==−=

∴= + =

=+

∫

25

03

33 5

2

1 100 8 256 ( )

53

1 800 256(1 ) 9.879VOK

53

dt

ee e

e

−−

−









=×+ −







=+−=





∫∫

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30. The peak instantaneous power is 250 mW. The combination of elements yields

Z = 1000 + j1000 Ω = 1414 ∠45o Ω.

Arbitrarily designate V = Vm ∠0 , so that I = 1414

45V

0V o

mm −∠

=

∠

Z A.

We may write p(t) = ½ Vm Im cos φ + ½ Vm Im cos (2

ω

t + φ) where φ = the angle of the

current (-45o). This function has a maximum value of ½ VmIm cos φ + ½ VmIm.

Thus, 0.250 = ½ VmIm (1 + cos φ) = ½ (1414) Im2 (1.707)

and Im = 14.39 mA.

In terms of rms current, the largest rms current permitted is 14.39 / 2 = 10.18 mA rms.

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31. 435Arms=∠ °

I

(a) ,

20 80 35 Vrms, P 80 10cos35 655.3 W

sgen

=+∠° =× °=VI

(b) 2

P R 16 20 320 W

R==×=I

(c) P 655.3 320 335.3 W

Load =−=

(d) ,

AP 80 10 800 VA

sgen =×=

(e) AP P 320VA

RR

==

(f)

(g)

10 0 4 35 7.104 18.84 A rms

AP 80 7.104 568.3 VA

L

=∠°−∠°= ∠− °

∴=× =

I

P335.3

PF cos 0.599

AP 568.3

since I lags V, PF is lagging

L

LL

L

LL

θ

====

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32.

(a)

(b) P 120 9.214 0.8969 991.7W

s=× × =

(c)

(d) PSpice verification

FREQ VM($N_0003,0) VP($N_0003,0)

6.000E+01 1.200E+02 0.000E+00

FREQ IM(V_PRINT1) IP(V_PRINT1)

6.000E+01 9.215E+00 -2.625E+01 ; (a) and (b) are correct

Next, add a 90.09-

µ

F capacitor in parallel with the source:

FREQ IM(V_PRINT1) IP(V_PRINT1)

6.000E+01 8.264E+00 -9.774E-05 ;(c) is correct (-9.8×10-5 degrees

is essentially zero, for unity PF).

120

I 9.214 26.25 A rms

192

412 16

PF cos26.25 0.8969lag

s

j

==∠−°

++

∴= =

22

48 1

Z 4 4 (192 144)

34 25

11.68 5.76

Z 11.68 5.76 , Y 11.68 5.76

5.76

120 C , C 90.09 F

11.68 5.76

L

LL

jj

j

πµ

=+ =+ +

+

−

∴= + Ω = +

∴= =

+

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

1

2

5 2 , 20 10 , 10 30 8.660 5

10 60 5 8.660

200 20 10

0 33.66 13.660 7265 22.09 15.11 3.908 A rms

25 8 20 10 480.9 26.00

20 10 33.66 13.660

25 8 200

20 10 0 200

480.9 26.00

AB c

D

jj j

j

jj

j

=+ Ω = − Ω = ∠°Ω= + Ω

=∠−°=− Ω

−+

−∠°

===∠°

−−+ ∠− °

−+ −

−

−+

==

∠− °

ZZ Z

Z

I

22

1

22

12

2

22

21

1

(20 10) 9.300 0.5681 A rms

480.9 20.00

AP 15.108 29 1229 VA

AP 5.881 10 5 773.5VA

AP 2 9.3 10 86.49VA

AP 9.3 10 864.9VA

AP 200 200 15.108 3022VA

AA

BB

CC

D

S

j−=∠− °

∠°

== =

=− = × =

==×=

IZ

IIZ

IZ

I

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. Perhaps the easiest approach is to consider the load and the compensation capacitor

separately. The load draws a complex power Sload = P + jQ. The capacitor draws a

purely reactive complex power SC = -jQC.

θ

load = tan-1(Q/P), or Q = P tan θload

Q

C = SC = Vrms rms

(/C)j

ω

−

V = 2

Crms

ω

V = 2

CVrms

ω

Stotal = Sload + SC = P + j(Q – QC)

θnew = ang(Stotal) = 1C

Q-Q

tan P

−







, so that Q – QC = P tan θnew

Substituting, we find that QC = P tan θload – P tan θnew

or

2

CVrms

ω

= P (tan θload – tan θnew)

Thus, noting that θold = θload,

()

old new

2

rms

P tan - tan

C = V

θθ

ω

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35. 12

30 15 , 40 40= ∠ °Ω = ∠ °ΩZZ

(a)

(b)

30 15 40 40 68.37 29.31

PF cos 29.3 0.8719 lag

tot =∠°+∠°= ∠ °Ω

∴= °=

Z

1

68.37 29.31 59.62 33.48

PF 0.9 lag

cos 0.9 25.84

X

tan 25.84 0.4843 X 28.88

59.62

1

33.48 28.88,

100 C

C 691.8 F

tot

new

j

θ

π

µ

−

=∠°=+

=

∴= = °

°= = ∴ = Ω

∴− =

=

Z

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36. θ1 = cos-1(0.92) = 23.07o, θ 2 = cos-1 (0.8) = 36.87o, θ 3 = 0

S

1 = VA 42.59 100

92.0

23.07 100 o

j+=

∠

S

2 = VA 5.871 250

8.0

78.63 250 o

j+=

∠

S

3 = VA 500

1

0 500 o=

∠

Stotal = S1 + S2 + S3 = 500 + j230.1 VA = 550.4 ∠24.71o VA

(a) Ieff = rmsA 4.786

115

550.4

V

S

eff

total ==

(b) PF of composite load = cos (24.71o) = 0.9084 lagging

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37.

2

1

AP 10,000 VA, PF 0.8lag, 40A rms

Let 40 0 A rms; P 10,000 0.8 8000 W

8000

Let R X R 5

40

cos 0.8lag cos 0.8 36.87

X 5tan 36.87 3.75 , 5 3.75, 5.2 3.75

40(5.2 3.75) 256.4

LLL

LL

LL LL

LL

LLtot

s

j

jj

j

θθ

−

===

=∠° = × =

=+ ∴= =Ω

=∴= =°

∴= °= Ω =+ =+ Ω

∴= + =

I

Z

ZZ

V1

35.80 V; 5.2 3.75

0.12651 0.09124S, 0.12651 (120 C 0.09124),

PF 0.9 lag, 25.84 tan 25.84 0.4843

0.09124 120 C

0.12651

C 79.48 F

tot

new

new new

j

jj

π

θ

π

µ

∠° =

+

=− =+ −

==°∴°=

−

=∴

=

Y

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38. Zeff = j100 + j300 || 200 = 237 ∠54.25o. PF = cos 54.25o = 0.5843 lagging.

(a) Raise PF to 0.92 lagging with series capacitance

Znew = j100 + jXC + j300 || 200 = 138.5 + j(192.3 + XC) Ω

o1-

C

123.07 0.92 cos

138.5

X 192.3

tan ==











+

−

Solving, we find that XC = -133.3 Ω = -1/

ω

C, so that C = 7.501 µF

(b) Raise PF to 0.92 lagging with parallel capacitance

Znew = j100 || jXC + j300 || 200 = )X100(

X 100

C

+

−

j+138.5 + j92.31 Ω

= 138.5 + 













+

C

X100

31.92j Ω

o1-

C

123.07 0.92 cos

138.5

X100

100X

92.31

tan ==













+

−

Solving, we find that XC = -25 Ω = -1/

ω

C, so that C = 40 µF

General circuit for simulations. Results agree with hand calculations

FREQ IM(V_PRINT1) IP(V_PRINT1) θ PF

With no compensation: 1.592E+02 4.853E-01 -5.825E+01 54.25o 0.5843 lag

With series compensation: 1.592E+02 7.641E-01 -2.707E+01 23.07o 0.9200 lag

With parallel compensation: 1.592E+02 7.641E-01 -2.707E+01 23.07o 0.9200 lag

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

(a) ,

P 20 25 0.8 30 0.75 70 kW

stot =+×+× =

(b)

(c) 3

70,000

PF 0.8604lag

81,360

==

1

2

1o

22

33

1

33

20,000 80 0 A rms

250

25,000 / 250 100 Arms

cos 0.8 36.87 100 36.87 A rms

30,000 40,000

AP 40,000 VA, 160Arms

0.75 250

cos 0.75 41.41 160 41.41 A rms

80 0 100 36.87 160

s

−

==∠°

==

∠=− =− ∴= ∠−

== ==

∠=− =− °∴ = ∠− °

∴=∠°+ ∠− °+ ∠−

I

II

I

II

I41.41 325.4 30.64 A rms

AP 250 325.4 81,360 VA

s

°= ∠− °

∴=× =

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40. 200 kW average power and 280 kVAR reactive result in a power factor of

PF = cos (tan-1 (280/200) = 0.5813 lagging, which is pretty low.

(a) 0.65 peak = 0.65(200) = 130 kVAR

Excess = 280 – 130 = 150 kVAR, for a cost of (12)(0.22)(150) = $396 / year.

(b) Target = S = P + j0.65 P

θ = tan-1(0.65P/P) = 33.02o, so target PF = cos θ = 0.8385

(c) A single 100-kVAR increment costs $200 to install. The excess kVAR would then be

280 – 100 – 130 = 50 kVAR, for an annual penalty of $332. This would result in a

first-year savings of $64.

A single 200-kVAR increment costs $395 to install, and would remove the entire excess

kVAR. The savings would be $1 (wow) in the first year, but $396 each year thereafter.

The single 200-kVAR increment is the most economical choice.

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

20

2

20

20(1 2)

10 10.769 3.846 11.435 19.65

32

100 8.745 19.65

11.435 19.654

100 8.745 19.65 823.5 294.1VA

10 20

8.745 19.65 5.423 49.40

30 20

20 5.432 588.2 0 VA

in

s

sss

j

jj

j

+

∗

+

=− + = − = ∠− °Ω

+

∴= = ∠ °

∠− °

∴=− =− × ∠− °=− +

+

=∠°× =∠°

+

∴=× = +

Z

I

SVI

I

S

10

2

10

2

20

2

10

20 5.423 49.40 4.851 14.04

10 20

10 4.851 235.3 0 VA

20 4.851 470.6 VA,

10 8.745 764.7 VA, 0

j

jj

−

×∠

==∠−°

+

=× = +

=× =

=− × = − Σ=

I

S

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42.

1

1.

2

6,

4,

100 100 0

6 4 10 5

1 100

0.1 0.2 20

64 64

53.35 42.66 V

100 53.35 42.66 9.806 64.44 A

64

1100 9.806 64.44 211.5 4423VA

2

16 9.806 288.5 0VA

2

1

xxx

x

gen

abs

jabs

j

jj

j

−

−−

++ =

+−



∴++=+



++



∴= ∠ °

−∠°

∴= = ∠− °

+

∴=××∠°= +

=×× = +

=

VVV

V

I

S

S2

2

5

2,

10,

( 4)9.806 0 192.3VA

2

100 53.35 42.66 14.99 121.6 ,

5

15 14.99 561.5 0VA

2

1( 100)14.99 121.57 638.4 392.3VA

2

1 53.35 ( 10) 0 142.3VA 142.3 0 VA 0

210

abs

gen

j abs

jj

j

jj

−

=+

−∠°

==∠°

=×× = +

=∠−°=−



=−=−=∠Σ=





I

S

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43.

(a)

(b)

(c)

1

500VA, PF 0.75 lead

500 cos 0.75 375 330.7 VAj

−

=∴

=∠− = −S

1

500W, PF 0.75 lead

500

500 sin (cos 0.75) 500 441.0 VA

.075 j

j

−

=∴

=− = −S

1

500 VAR, PF 0.75(lead) cos 0.75 41.41

P500/ tan 41.41 566.9W,

566.9 500 VAj

θ

−

−=∴=−=−°

∴°=

=−S

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44. 1600 500VA(gen)

sj=+S

(a)

(b) 11833.3

PF cos tan 0.6575 lag

1600

L

−+



==





(c) 1600 500 1676 17.35 VA PF cos17.35 0.9545 lag

ss

j=+=∠°∴= °=S

1600 500 4 1.25 4 1.25

400

400 3.333A rms 4 1.25 3.333

120

4 4.583A rms

400(4 4.583) 1600 1833 VA

ss

cLsc

L

jjj

j

jj

∗+

==+∴=−

= = ∴=−=− −

−

∴=− ∴

=+ =+

II

IIII

I

S

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45. 11

(cos 0.8 36.87 , cos 0.9 25.84 )

−−

=° =°

(a)

(b) PF cos23.245 0.9188

s=°=

(c) 3300 1417 VAj=+S

1200 36.87 1600 25.84 900

960 720 1440 697.4 900

3300 1417.4 3592 23.25 VA

3591.5 15.62 A rms

230

tot

s

jj

j

=∠°+∠°+

=+ + + +

=+ =∠°

∴= =

S

I

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46. V = 339 ∠-66o V,

ω

= 100π rad/ s, connected to Z = 1000 Ω.

(a) Veff = rms V 239.7

2

339 =

(b) pmax = 3392 / 1000 = 114.9 W

(c) pmin = 0 W

(d) Apparent power = Veff Ieff = VA 57.46

1000

V

1000

2

339

2

339 2

eff ==

























(e) Since the load is purely resistive, it draws zero reactive power.

(f) S = 57.46 VA

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47. V = 339 ∠-66o V,

ω

= 100π rad/s to a purely inductive load of 150 mH (j47.12 Ω)

(a) I = A 156- 7.194

12.47

66-339

o

o∠=

∠

=jZ

V

so Ieff = rmsA 5.087

2

194.7 =

(b) p(t) = ½ VmIm cos φ + ½ VmIm cos(2

ω

t + φ)

where φ = angle of current – angle of voltage

pmax = ½ VmIm cos φ + ½ VmIm = (1 + cos(-90o)) (339)(7.194)/ 2 = 1219 W

(c) pmin = ½ VmIm cos φ - ½ VmIm = -1219 W

(d) apparent power = Veff Ieff =

()

VA 1219 087.5

2

339 =

(e) reactive power = Q = Veff Ieff sin (θ – φ) = 1219 VA

(f) complex power = j1219 VA

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48. 1 H → j Ω, 4 µF → -j250 kΩ

Zeff = j || -j250×103 || 103 Ω = 1 ∠89.99o Ω

V10k = V 89.97 0.002

)89.99 (12500

)89.99 (1 )05( o

o

o∠=

∠+

∠∠

(a) pmax = (0.002)2 / 10×103 = 400 pW

(b) 0 W (purely resistive elements draw no reactive power)

(c) apparent power = VeffIeff = ½ VmIm = ½ (0.002)2 / 10000 = 200 pW

(d) Isource = A 0.02292- 0.002

99.8912500

05 o

o∠=

∠+

∠

S = ½ VmIm ∠(89.99o + 0.02292o) = 0.005 ∠90.01o VA

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49. (a) At

ω

= 400 rad/s, 1 µF → -j2500 Ω, 100 mH → j40 Ω

Define Zeff = -j2500 || (250 + j40) = 256 ∠ 3.287o Ω

IS = rmsA 3.049- 43.48

287.325620

012000 o

o∠=

∠+

∠

Ssource = (12000)(43.48) ∠ 3.049o = 521.8 ∠3.049o kVA

S20Ω = (43.48)2 (20) ∠0 = 37.81 ∠0 kVA

Veff = rms V 0.2381 11130

287.325620

)287.3256)(012000( o

∠=

∠+

∠∠

o

I1µF = rmsA 90.24 4.452

0250-

o

eff ∠=

j

V

so S1µF = (11130)(4.452) ∠-90o = 49.55 ∠-90o kVA

V100mH = rms V 15.18 1758

40250

)40)(2381.011130( o

∠=

+

∠

j

o

I

100mH = rmsA 852.8- 43.96

40

o

100mH ∠=

j

V

so S100µΗ = (1758)(4.43.96) ∠90o = 77.28 ∠90o kVA

V250Ω = rms V 852.8 10990

40250

)250)(2381.011130( o

−∠=

+

∠

j

o

so S250Ω = (10990)2 / 250 = 483.1 ∠0o kVA

(b) 37.81 ∠0 + 49.55 ∠-90o +77.28 ∠90o + 483.1 ∠0o = 521.6 ∠3.014o kVA,

which is within rounding error of the complex power delivered by the source.

(c) The apparent power of the source is 521.8 kVA. The apparent powers of the passive

elements sum to 37.81 + 49.55 + 77.28 + 483.1 = 647.7 kVA, so NO! Phase angle is

important!

(d) P = Veff Ieff cos (ang VS – ang IS) = (12000)(43.48) cos (3.049o) = 521 kW

(e) Q = Veff Ieff sin (ang VS – ang IS) = (12000)(43.48) sin (3.049o) = 27.75 kVAR

CHAPTER 11 SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50. (a) Peak current = 28 2 = 39.6 A

(b) θload = cos-1(0.812) = +35.71o (since lagging PF). Assume ang (V) = 0o.

p(t) =

(

)

(

)

)7135120(cos)120(co 239.60 22300 o

.πt - πts

at t = 2.5 ms, then, p(t) = 147.9 kW

(c) P = Veff Ieff cos θ = (2300)(28) cos (35.71o) = 52.29 kW

(d) S = Veff Ieff ∠θ = 64.4 ∠ 35.71o kVA

(e) apparent power = |S| = 64.4 kVA

(f) |Zload| = |V/ I| = 2300/28 = 82.14 Ω. Thus, Zload = 82.14 ∠ 35.71o Ω

(g) Q = Veff Ieff sin θ = 37.59 kVAR

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1. Vbc = Vbe + Vec = 0.7 – 10 = -9.3 V

V

eb = - Vbe = -0.7 V

V

cb = Vce + Veb = 10 – 0.7 = 9.3 V

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2. (a) Vgd = Vgs + Vsd = -1 – 5 = -6 V

(b) Vsg = Vsd + Vdg = -4 – 2.5 = -6.5 V

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3. (a) positive phase sequence

Van = |Vp| ∠ 0o Vdn = |Vp| ∠ -180o

Vbn = |Vp| ∠ -60o Ven = |Vp| ∠ -240o

Vcn = |Vp| ∠ -120o Vfn = |Vp| ∠ -300o

(b) negative phase sequence

Van = |Vp| ∠ 0o Vdn = |Vp| ∠ 180o

Vbn = |Vp| ∠ 60o Ven = |Vp| ∠ 240o

Vcn = |Vp| ∠ 120o Vfn = |Vp| ∠ 300o

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4. (a) Vyz = Vyx + Vxz = -110 ∠20o + 160 ∠ -50o

= -103.4 – j37.62 + 102.8 – j122.6 = -0.6 – j160.2

= 160.2 ∠ -90.21o V

(b) Vaz = Vay + Vyz = 80 ∠130o + 160.2 ∠ -90.21o

= -51.42 + j61.28 -0.6 – j160.2 = -52.02 – j98.92

= 111.8 ∠ -117.7o V

(c) o

o

xy

zx 1101.455

20110

301160

20110

50-160-

∠=

∠

=

∠

=

V

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5. (a) V25 = V24 + V45 = -80 ∠ 120o + 60 ∠ 75o

= 40 – j69.28 + 15.53 + j57.96 = 55.53 – j11.32

= 56.67 ∠ -11.52o V

(b) V13 = V12 + V25 + V53 = 100 + 55.53 – j11.32 + j120

= 155.53 + j108.7

= 189.8 ∠ 34.95o V

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6.

230/ 460V rms Z : S 10 40 kVA; Z : 8 10 kVA;

10,000 40

Z : 4 80 kVA Let V 230 0 V S V I ,I 43.48 40 A

230

4000 80

I 43.48 40 A, S V I I 8.696 80 , I 8.696 80 I I I

460

I 43.

∗∗

=∠° ∠°

∠°

∠− ° = ∠ ° ∴ = = = ∠ °

∠− °

∴=∠−° = ∴= =∠−°=∠°∴=+

∴=

AN NB

AB AN AN AN AN AN

AN AB AB AB AB AB aA AN AB

aA 48 40 8.696 80 39.85 29.107 I 39.85 A

8000 10

I 34.78 10 , I 34.78 10 A

230

I 34.78 10 8.696 80 35.85 175.96 , I 35.85 A

I 43.48 40 34.78 10 21.93 87.52 , I 21.93A

−−

∗

++

∠°+ ∠°= ∠− °∴ =

∠°

==∠°=∠−°

∴=− ∠−°− ∠°= ∠− °∴=

=− ∠− °+ ∠− °= ∠ ° =

aA

NB NB

bB bB

nN nN

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7. (a) InN = 0 since the circuit is balanced.

IAN = 12 ∠0 IAB = o

36.9- 12

1216

0240 ∠=

+

∠

j

IaA = IAN + IAB = 12 + 9.596 – j7.205 = 22.77 ∠ -18.45o A

(b)

The voltage across the 16-Ω resistor and j12-Ω impedance has not changed, so IAB

has not changed from above.

IaA = IAN + IAB = 24 ∠ 0o + 12 ∠ -36.9o = 34.36 ∠ -12.10o A

IbB = IBN - IAB = -12 ∠ 0o - 12 ∠ -36.9o = 7.595 ∠ -108.5o A

InN = IBN – IAN = -12 – 24 = 36 ∠180o A

IAN = 24 ∠ 0o A

IBN = -12 ∠ 0o A

InN = -12 ∠ 0o A

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

(a)

(b)

(c) 222

,

P 126.06 1 130.65 1 4.730 10 15.891 17.069 0.224 33.18kW

tot

ω

= ×+ ×+×=++=

(d) ,

P 720 126.06 cos 6.479 720 130.65 cos5.968 90.18 93.56 183.74kW

gen tot −

=× °+× °= + =

21 3 10 10 3

10 19 2 8 2 (21 3)(674 167 60 32)

10 3 8 2 36 5

10( 360 50 74 44) (10 3)(80 20 184 77)

5800 1995 6127 18.805

720 10 10 3

720 19 2 8 2 720(614 135 434 94) 7

082365

+−−−

∆= − + − − = + + − −

−− −− +

+− − −− − + + + +

∴∆= + = ∠ °

−−−

+−−= +++=

−− +

jj

jj jj j

jj j

jj jj j

j

jj j j

jj

20 1072.7 12.326

720 1072.7 12.326

I 126.06 6.479 A

6127 18.805

×∠°

∴= = ∠− °

∠°

aA

21 3 720 10 3 720(1084 247)

10 720 8 2 720(1084 247) I 130.65 5.968 A

6127 18.805

10 3 0 36 5

I 130.65 5.968 126.06 6.479 4.730 7.760 A

−

+−− +

−−−=+∴= =∠−°

∠°

−− +

∴= ∠− °− ∠− °= ∠ °

Bb

nN

jj j

jj

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9. V 220Vrms, 60Hz

AN =

(a)

(b) 6

I 377 91.47 10 440 15.172A VA 440 15.172 6.676kVA

AB −

=× ××= ∴=× =

220 0

PF 1 I 40.85 21.80 A; I 377C 440

52

I 40.85cos21.80 (377C440 40.85sin 21.80 )

40.85sin 21.80

C 91.47 F

377 440

+

∠°

=∴ = = ∠− ° = ×

+

∴= °+ − °

°

∴= =

×

AN AB

aA

j

µ

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10. (a) IaA = IAN + IAB =

AB

R

0400

312

0200 ∠

+

∠

j = 15.69 – j3.922 +

AB

R

400

Since we know that |IaA| = 30 A rms = 42.43 A,

2

AB

3.922

R

400

69.1543.42 +











+=

or RAB = 15.06 Ω

(b) IaA = IAN + IAB =

ABAB X

400

922.3- 15.69

X-

0400

312

0200 j

j

jj +=

∠

+

∠

In order for the angle of IaA to be zero,

AB

X

400 = 3.922, so that XAB = 102 Ω capacitive.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11. seq. V 120 60 V rms, R 0.6 P 5kVA, 0.6 lag

BC w load

+=∠° =Ω=

(a)

(b)

2

120 5000 5000

V 150 V S 0.8 0.6

33

3

120

S 150 I I 24.06 113.13 A

3

I 24.06 113.13 P 3 24.06 0.6 1041.7W

AN AN

AN aA aA

aA wire

j

∗∗

=∠°∴= ×+

∴=∠°∴= ∠− °

∴= ∠ °∴ =× × =

V 0.6 24.06 113.13 14.434 113.13 V

120

V V V 14.434 113.13 158 81.29 143.88 V

3

aA

an aA AN

=× ∠ °= ∠ °

∴=+= ∠°+∠°=∠°

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12. rms

V 2300 0 V , R 2 , ., S 100 30kVA

an w tot

seq j↑= ∠° =Ω+ =+

(a) 1(100,000 30,000) 2300I I 15.131 16.699 A

3aA aA

j∗

+=∴=∠−°

(b) V 2300 2 15.131 16.699 2271 0.2194 V

AN =−× ∠− °=∠ °

(c) 2271 0.2194

Z V / I 143.60 43.67

15.131 16.699

pANaA j

∠°

== =+Ω

∠− °

(d) 143.60

trans. eff. 0.9863, or98.63%

145.60

==

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13. Z 12 5 , I 20 0 A rms, +seq., PF 0.935

pbB

j↑=+Ω =∠° =

(a) 15

cos 0.935 20.77 tan 20.77 , R 1.1821

12 R w

w

θ

−

==°∴=°=Ω

+

(b)

(c) V 3 V / V 150 450.3 172.62 V

AB BN BN

=∠+°=∠°

(d)

V I Z 20(12 5) 240 100V V 20(13.1821 5) 281.97 20.77 V==+=+ ∴= += ∠°

BN bB p bn

jj j

S 3 V I 3 281.97 20.77 (20)

15.819 6.000kVA

∗

==×∠−°

=−

source Bn bB

j

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14. 125 mH → j(2π)(60)(0.125) = j47.12 Ω 75 Ω → 75 Ω

55 µF → -j/(2π)(60)(55×10-6) = -j48.23 Ω

The per-phase current magnitude |I| is then I =

)23.4812.47(75

125

22 =

−+ 1.667 A.

The power in each phase = (1.667)2 (75) = 208.4 W, so that the total power taken by the

load is 3(208.4) = 625.2 W.

The power factor of the load is 1.000

75

23.4812.47

cos =











−

This isn’t surprising, as the impedance of the inductor and the impedance of the

capacitor essentially cancel each other out as they have approximately the same

magnitude but opposite sign and are connected in series.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

Bal.,+ seq. Z 8 6 , Z 12 16 , Z 5 0, V 120 0 V rms

120 0 120 120 120 120

R 0.5 ( ) I 6.803 83.86 A

8.5 6 12.5 16 5.5

I 6.803 96.14 A rms

AN BN CN AN

wnN

nN

jj j

ajj

↑=+Ω=−Ω=+=∠°

∠° ∠− ° ∠ °

=Ω −= + + = ∠ °

+−

∴= ∠− °

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16. Working on a per-phase basis, the line current magnitude is simply

()

2

w105R

40

++

=I

(a) RW = 0

Then A 578.3

1025

40

2=

+

=I, and the power delivered to each phase of the load is

(3.578)2(5) = 64.01 W. The total power of the load is therefore 3(64.01) = 192.0 W.

(b) RW = 3 Ω

Then A 123.3

1064

40

2=

+

=I, and the power delivered to each phase of the load is

(3.123)2(5) = 48.77 W. The total power of the load is therefore 3(48.77) = 146.3 W.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17. Z 75 25 25 F, V 240 0 V rms, 60Hz, R 2

pan w

µ

↑=∠°Ω = ∠° =Ω

(a)

(b) 2

P 3(2.968) 2 52.84 W

w=×=

(c) 2

P 3(2.968) 75.34 1990.6W

load ==

(d) PF cos16.989 0.9564 lead

source =°=

6

10 75 25 ( 106.10)

Z 106.10 Z 75.34 23.63

377 25 75 25 106.10

240

Z 77.34 23.63 I 2.968 16.989 A

77.34 23.63

+

∠°−

=− =− Ω∴ = = − Ω

×∠°−

∴= − ∴= = ∠ °

−

cap p

pw aA

j

jj j

j

jj

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18. Working on a per-phase basis and noting that the capacitor corresponds to a –j6366-Ω

impedance,

-j6366 || 100 ∠ 28o = 89.59 + j46.04 Ω so that the current flowing through the combined

load is

rmsA 362.2

04.4690.59

240

22 =

+

=I

The power in each phase is (2.362)2 (90.59) = 505.4 W, so that the power deliverd to the

total load is 3(505.4) = 1.516 kW.

The power lost in the wiring is (3)(2.362)2 (1) = 16.74 W.

Simulation Result:

FREQ IM(V_PRINT1) IP(V_PRINT1)

5.000E+01 1.181E+00 -2.694E+01

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19. Bal., R 0, Z 10 5 , 60 Hz

wp jf↑ = =+Ω =

(a) 10 5 11.180 26.57 PF cos 26.57 0.8944j

+= ∠ °∴= °=

(b)

(c)

1

PF 0.93 lag, 21.57 , Y 0.08 0.04S

11.180 26.57

377C 0.04

Y 0.08 (377C 0.04) tan 21.57 0.3952

0.08

377C 0.04 0.08 0.3952 0.00838 C 22.23 F

pj

θ

µ

==°= =−

∠°

−

′=+ − ∴ =− °=−

∴=−× = ∴=

6

,

10 440 / 3

V 440Vrms, Z 119.30 ,I 2.129A

120 22.23 119.30

440

VAR 2.129 540.9 VAR ( .)

3

−

===−Ω==

∴=×=

L load c c

jj

cap

π

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20. Working from the single-phase equivalent,

Van rms = 











∠

2

0115

3

1o

= 46.9 ∠0o V rms

1.5 H → j565 Ω, 100 µF → -j26.5 Ω and 1 kΩ → 1 kΩ.

These three impedances appear in parallel, with a combined value of 27.8 ∠ -88.4o Ω.

Thus, |Irms| = 46.9/ 27.8 = 1.69 A rms

Zload = 27.8 ∠88.4o = 0.776 – j 27.8 Ω, so Pload = (3)(1.69)2 (0.776) = 2.22 W.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

R 0, V 200 60 V rms. S 2 1 kVA seq.

wan p

j==∠° =− +

(a) V 220 3 30 346.4 30 V

bc = ∠−°= ∠−°

(b)

(c)

S 2000 1000 V I 346.4 30 I

I 6.455 3.435 , I 6.455 3.435

200 3 30

Z 53.67 26.57 48 24

6.455 3.435

BC BC BC BC

BC BC

p

j

∗∗

∗− −

−

=− = =∠−°

∴= ∠ ° = ∠− °

∠− °

∴= = ∠− °=− Ω

∠− °

I I I 6.455 120 3.43 6.455 120 3.43 11.180 86.57 A rms

aA AB CA −−

= − = ∠ °− °− ∠− °− °= ∠ °

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22. 15kVA, 0.8lag, +seq., V 180 30 V rms, R 0.75

BC w

↑=∠°=Ω

(a)

(b)

1

V 180 30 V 180 150 V, S 5000 cos 0.8 5000 36.87 180 30 I

I 27.78 6.87 and I 27.78 113.13 A I I I

I 27.78(1 6.87 1 113.13 ) 48.11 36.87 A V 0.75(I I )

V 0.75 48.11(1

−∗

=∠°∴ =∠° = ∠ = ∠ °=∠°

∴= ∠− ° = ∠ °∴=−

∴= ∠− °−∠ °= ∠− °∴ = −

∴= × ∠−

BC AB p BC

BC AB bB BC AB

bB bC bB cC

bC 36.87 1 156.87 ) 180 30 233.0 20.74 V°− ∠− ° + ∠ °= ∠ °

2

P 3 48.11 0.75 5208W

S 5208 15,000 36.87 17.208 9.000kVA

wire

gen j

=× × =

=+ ∠°= +

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23. Bal., S 3 1.8 kVA, S 3.45 1.8 kVA, R 5

Lgen w

jj↑=+ =+ =Ω

(a) 2

1

P 450 W 450 I 5 I 5.477A rms

3

=∴×=×∴=

waAaA

(b) 1

I 5.477 3.162A rms

3

AB =× =

(c)

1

Assume I 3.162 0 and +seq. (3000 1800) (3.162 0 )

3

V 368.8 30.96 V V V V V V

V 5I 5 5.477 30 27.39 30 ,V 27.39 150

V 27.39 30 27.39 150 368.8 30.96 V (1

∗

=∠° ∴ + = = ∠°

∴= ∠ °∴=+−+

==×∠−°=∠−°=∠−°

∴ = ∠− °− ∠− °+ ∠ °+ ∠−

AB AB AB AB

AB an aA AB bB bn

aA aA bB

an an

jVIV

120 )

27.39 30 27.39 150 368.8 30.96

V 236.8 2.447 236.8V rms

1 1 120

°

∠− °− ∠− °+ ∠ °

∴= = ∠− °∴=

−∠− °

an an

V

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24. If a total of 240 W is lost in the three wires marked Rw, then 80 W is lost in each 2.3-Ω

segment. Thus, the line current is rmsA 898.5

3.2

80 =. Since this is a D-connected load,

the phase current is 1/ 3 times the line current, or 3.405 A rms.

In order to determine the phase voltage of the source, we note that

Ptotal = 













=⋅⋅ 2

2

(5.898) 3 PF 3 linelineline VIV = 1800

where |Vline| =

()()

V 249.2

(5.898) 3 2

21800 =

This is the voltage at the load, so we need to add the voltage lost across the wire, which

(taking the load voltage as the reference phase) is

()

W

1R

2

1

cos898.5 





















−∠ −

= 13.57 ∠-45o V. Thus, the line voltage magnitude of the source is

|249.2 ∠ 0o + 13.57 ∠ -45o| = 259.0 V rms.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25. Bal., +seq.

(a)

(b) I 41.57 20.78 30 60.47 170.10 A rms

bB =− − ∠ °= ∠− °

(c) I 20.78 120 41.57 36.00 30 A rms

cC =∠−°+ =∠−°

(d)

120 3 30

V 120 0 V 120 3 30 , ., I 20.78 30 A

10

120 3 90 120 3 150

I 41.57 ; I 20.78 120 A

510

I I I 20.78(1 30 1 120 ) 40.15 45 Arms

∠°

=∠°∴= ∠° = = ∠°

∠− ° ∠ °

==−==∠−°

−

=−= ∠°−∠−°= ∠°

an ab AB

BC CA

aA AB CA

etc

A

jj

S V I V I V I 120 3 30 20.78 30 120 3 90 ( 41.57)

120 3 150 20.78 120 4320 0 0 8640 0 4320 4320 4320VA

tot AB AB BC BC CA CA

jj j j

∗∗∗

=++=∠°×∠−°+∠−°−+

∠°× ∠°= +++ +− = +

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26. IAB = A 18.4- 21.1

4.1849.9

0200

30 ||10

0200 o

o∠=

∠

=

∠

j

|IA| = A 36.5 I 3 AB =

The power supplied by the source = (3) |IA|2 (0.2) + (3) (200)2 / 10 = 12.8 kW

Define transmission efficiency as η = 100 × Pload/ Psource. Then η = 93.8%.

IA leads IAB by 30o, so that IA = 36.5 ∠ 11.6o.

V 11.6 7.3 )11.6 (0.2)(36.5 oo

RW∠=∠=V

With VAN = o

30

3

200 ∠, and noting that Van = VAN +

W

R

V= 122 ∠ 28.9o, we may now

compute the power factor of the source as

PF = cos (ang(Van) – ang(IA)) = cos (28.9o – 11.6o) = 0.955.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27. Bal., V 140 0 V , seq., R 0, S 15 9 kVA

an rms w L j↑=∠°+==+

(a) V V 3 140 30 242.5 30 V

−

== ∠°= ∠°

ab AB

(b) V I 5000 3000 242.5 30 I I 24.05 0.9638 A rms

AB AB AB AB

j

∗−∗−

=+=∠°∴=∠−°

(c) I I I 24.05 0.9638 24.05 119.03 41.65 30.96 A rms

aA AB CA −−−

=−= ∠− °− ∠ °= ∠− °

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28. 15 mH → j5.65 Ω, 0.25 mF → -j10.6 Ω

VAB = V 30 3120 o

∠

VBC = V 90 3120 o

−∠

VCA = V 012 3120 o

−∠

Defining three clockwise mesh currents I1,

I2 and I3 corresponding to sources VAB,

VBC and VCA, respectively, we may write:

VAB = (10 + j5.65) I1 – 10 I2 + j5.65 I3 [1]

VBC = -10 I1 + (10 – j10.6) I2 + j10.6 I3 [2]

VCA = - j5.65 I1 + j10.6 I2 + (j5.65 – j10.6) I3 [3]

Solving using MATLAB or a scientific calculator, we find that

I

1 = 53.23 ∠ -5.873o A, I2 = 40.55 ∠ 20.31o A, and I3 = 0

(a) VAN = j5.65(I1 – I3) = 300.7 ∠ 84.13o V, so VAN = 300.7 V

(b) VBN = 10(I2 – I1) = 245.7 ∠ 127.4o V, so VBN = 245.7 V

(c) VCN = -j10.6 (-I2) = 429.8 ∠ 110.3o V, so VCN = 429.8 V

PSpice Simulation Results

(agree with hand calculations)

FREQ VM(A,N) VP(A,N)

6.000E+01 3.007E+02 8.410E+01

FREQ VM(B,N) VP(B,N)

6.000E+01 2.456E+02 1.274E+02

FREQ VM(C,N) VP(C,N)

6.000E+01 4.297E+02 1.103E+02

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29. 1↑=Ω

line

R

(a)

1

207.8[1 30 (70 40) 1( 10 45)] 21.690 34.86

I 33.87 45.20 I

630 115 630 115 aA

jj j

jj

∠° + + − − ∠ °

∴= = = ∠ °=

−−

2

12 1 30 10

1 1 5 207.8

10 0 10 5 207.8[ 1 30 ( 10 45) 1(20 60)]

I630 115 630 115

16,136 162.01 25.20 172.36 A

630 115

∠° −

−− −

−− −∠ °− − − −

∴= =

−−

∠°

==∠°

−

jj

jjj j

jj

j

(b) I 25.20 7.641 A

cC

∴= ∠− °

(c) I I I 33.87 45.20 25.20 7.641 53.03 157.05 A rms

bB aA CC

∴=−−=−∠°−∠−°=∠− °

(d)

S 120 3 30 (33.87 45.20 ) 120 3 90 (25.20 7.641 )

6793 1846.1 696.3 5190.4 6096 3344VAjjj

=∠°∠−°+∠°∠°

=− − + =+

1

207.8 30 1 10 1 30 1 10

207.8 90 2 5 5 207.8 1 2 5 5

05105 05105

120 3 207.8 I 12 1 10 12(70 40) ( 10 45) 10(20 55)

12 5 5

10 5 10 5

j

jj j jj

jj jj

jjj j

jj

∠° − − ∠° − −

∠− ° + − − + −

−− −−

== =

−− ++−−− +

−+ −

−− −

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30. |Vline| = 240 V. Set Vab = 240∠0o V. Then Van = o

30

3

240 −∠ .

IA2 = A 1.068.23

35

30

3

240

o

−∠=

+

−∠

j

I

A1B1 =

()

mA 76.40.20

1021

0

3

240

o

3

o

−∠=

×+

∠

j

I

phase leads Iline by 30o, so

IA1 = mA 8.3434.6 mA 8.34320 oo −∠=−∠

Ia = IA1 + IA2 = 11.5 – j20.8 + 28.4 – j19.7 mA = 56.9∠-45.4o mA

The power factor at the source = cos (45.4o – 30o) = 0.964 lagging.

The power taken by the load = (3)(20×10-3)2 (12×103) + (3)(23.8×10-3)2 (5000) = 22.9 W.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31. Define I flowing from the ‘+’ terminal of the source. Then,

I = o

o74.2941.12

29.7416.12

0200

)20||10(10

0200 −∠=

∠

=

+

∠

j

(a) Vxy = 10 I = 124.1 ∠-29.74o V. Thus, Pxy = (12.41)(124.1) = 1.54 kW

(b) Pxz = (200)(12.41) cos (29.74o) = 2.155 kW

(c) Vyz = 200 ∠0 – 124.1 ∠-29.74o = 110.9 ∠ 33.72o V

Thus, Pyz = (110.9)(12.41) cos (33.72o + 29.74o) = 614.9 W

No reversal of meter leads is required for any of the above measurements.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32. 1 H → j377 Ω, 25 µF → -j106 Ω

I1 =

()

[]

A 2186.1

10610037750

0440

o

∠=

+

∠

-j||j

IC = I A 41.32.43

106100377

377 o

∠=

−+ jj

j

V2 = (106∠-90o)(2.43∠-41.3o) = 257∠-48.7o V

P

measured = (257)(1.86) cos (21o + 48.7o) = 166 W.

No reversal of meter leads is needed. PSpice verification:

FREQ VM($N_0002,0) VP($N_0002,0)

6.000E+01 2.581E+02 -4.871E+01

FREQ IM(V_PRINT1) IP(V_PRINT1)

6.000E+01 1.863E+00 2.103E+01

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33. 2.5 A peak = 1.77 A rms. 200 V peak = 141 V rms. 100 µF → -j20 Ω.

Define the clockwise mesh current I1 in the bottom mesh, and the clockwise mesh current

I2 in the top mesh. IC = I1 – I2.

Since I2 = -177∠-90o, we need write only one mesh equation:

141∠0o = (20 - j40o) I1 + (-20 + j20) I2

so that I1 = A78.474.023

40-20

)90-20)(1.77 (-20 0141 o

o∠=

∠++∠

j

and IC = I1 – I2 = 2.361 ∠ 63.43o A. Imeter = -I1 = 4.023∠-105.2o

Vmeter = 20 IC = 47.23 ∠63.43o V

Thus, Pmeter = (47.23)(4.023)cos(63.43o + 105.2o) = -186.3 W.

Since this would result in pegging the meter, we would need to swap the potential leads.

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. (a) Define three clockwise mesh currents I1, I2 and I3 in the top left, bottom left and right-

hand meshes, respectively. Then we may write:

100 ∠0 = (10 – j10) I1 - (10 – j10) I3

50 ∠90o = (8 + j6) I2 – (8 + j6) I3

0 = -(10 – j10) I1 – (8 + j6) I2 + (48 + j6) I3

Solving, we find that I1 = 10.12∠ 32.91o A, I2 = 7.906 ∠ 34.7o and I3 = 3.536 ∠ 8.13o A.

Thus, PA = (100)(10.12) cos (-32.91o) = 849.6 W

and PB = (5)(7.906) cos (90o – 34.7o) = 225.0 W

(b) Yes, the total power absorbed by the combined load (1.075 kW) is the sum of the

wattmeter readings.

PSpice verification:

FREQ IM(V_PRINT1) IP(V_PRINT1)

6.280E+00 1.014E+01 6.144E-02

FREQ IM(V_PRINT2) IP(V_PRINT2)

6.280E+00 4.268E-01 1.465E+02

FREQ VM($N_0002,$N_0006) VP($N_0002,$N_0006)

6.280E+00 1.000E+02 0.000E+00

FREQ VM($N_0004,$N_0006) VP($N_0004,$N_0006)

6.280E+00 5.000E+01 - 9.000E+01

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35. This circuit is equivalent to a Y-connected load in parallel with a ∆-connected load.

For the Y-connected load, Iline = A 6062.4

3025

30

3

200

o

−∠=

∠

−∠

P

Y =

()

kW 1.386 30cos62.4

3

200

)3( o=













For the ∆-connected load, Iline = A 604

6050

0200

o

∠=

−∠

∠

P

∆ = (3)(200)(4 cos 60o) = 1.2 kW

P

total = PY + P∆ = 2.586 kW

P

wattmeter = Ptotal / 3 = 862 W

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36. We assume that the wire resistance cannot be separated from the load, so we measure

from the source connection:

(a)

(b)

CHAPTER TWELVE SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37. We assume that the wire resistance cannot be separated from the load, so we measure

from the source connection:

(a)

(b)

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1. 1 and 3, 2 and 4

1 and 4, 2 and 3

3 and 1, 2 and 4

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2. 12

4 A, 10 A==

SS

iti t

(a) 20 4 4 10 120 V=×+×=

AG

v

(b) 4 6 24 V=− × =−

CG

v

(c) 3 10 4 4 6 4 30 16 24 22 V=× +×−×= + − =

BG

v

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3.

(a)

(b)

2

22

50 2000

113

P 50 0.4745 5.630 W, P 2000 0.4745 4.769 W

2 2 20 5

−

=× × = =× × × =

+

j

(c) 0 each

(d) 0

12 21

21

1

11

,

100 (50 200) I 300I ,(2000 500)I 300I 0

3 900

I , 100 50 200

20 5 20 5

900 4250

100 I I 0.4745 64.01 A

20 5

1

P 100 0.4745cos 64.01 10.399 W

2

=+ + + + =



−

∴= = + +



++



+

∴= ∴= ∠− °

+

∴=−×× °=−

SabS

jj jj

jjI

jj

j

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4.

KVL Loop 1 100 ∠0 = 2(I1 – I2) + j

ω

3 (I1 – I3) + j

ω

2 (I2 – I3)

KVL Loop 2 2(I2 – I1) + 10I2 + j

ω

4 (I2 – I3) + j

ω

2 (I1 – I3) = 0

KVL Loop 3 5I3 + j

ω

3 (I3 – I1) + j

ω

2 (I3 – I2) + j

ω

4 (I3 – I2) + j

ω

2 (I3 – I1) = 0

∴LINEAR EQUATIONS











∠

=

























+−

−++−

−+−+

0

0100

11525

641222

52232 1

3

2

I

jjj

ωω

ωωω

Since

ω

= 2πf = 2π(50) = 314.2 rad/s, the matrix becomes











∠

=

























+−

−++−

−+−+

0

0100

345654.6281571

18851257124.6282

15714.62826.9422 1

3

2

I

jjj

Solving using a scientific calculator or MATLAB, we find that

I1 = 278.5 ∠ -89.65o mA, I2 = 39.78 ∠ -89.43o mA, I3 = 119.4 ∠ -89.58o mA.

Returning to the time domain, we thus find that

i1(t) = 278.5 cos (100πt – 89.65o) mA, i2(t) = 39.78 cos (100πt – 89.43o) mA, and

i3(t) = 119.4 cos (100πt – 89.58o) mA.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5.

(a)

(b)

,

12 2 1

12 22

,2

100

V ( 300) 145.52 165.96 V

50 200

100 (50 200) I 300I , 500I 300I 0

55

I I , 100 (50 200) 300 I I 1.1142 158.199 A

33

145.52 165.96

ZV/I 1.1142 158.

=−=∠−°

+

=+ + + =





∴=− = + − + ∴ = ∠ °









∠− °

∴= = ∠

ab oc

SC SC

SC SC SC

th ab bc SC

j

jjj j

jj

130.60 35.84 105.88 76.47

199 =∠°=+ Ω

°j

2

max

145.52

Z 105.88 76.47 I 0.6872 A

2 105.88

1

P 0.6872 105.88 25.00 W

2

=−Ω∴= =

×

∴=× × =

LL

L

j

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6.

(a) 11 2 11 2 2 2 1

() L M , () L M L M

′′ ′′ ′′

=− =−+ −

AB

vt i ivt i i i i

(b) V1(j

ω

) = j

ω

L1 IA + j

ω

M(IB + IA)

V2(j

ω

) = j

ω

L2 (IB + IA) + j

ω

MIA

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7.

22

222

64 4

222

10 ( ) 1000

0.01 ( )

0.01 0.01

15 1500

0.015 ( ), 100 ( )

0.01 0.01

15 ( 0.01)2 2

100 10 10 ( ) 15 10 ( )

0.01 ( 0.01)

−− −

′′

==∴=

++

′

== =

++

 +−×

′

∴= × = =×



++



sSS

xS x

Cx

tut t

viiut

tt

vi utv ut

tt

dt t ttt

iv ut ut

dt t t

40.02

15 10 (

−

∴=×

C

t

it22 22

30

() A,

0.01) ( 0.01)

∴=

µ

++

C

t

it t

t > 0

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

(a)

Thus, v(t) = 126 cos (100πt + 90o) V

(b) Define V2 across the 2-H inductor with + reference at the dot, and a clockwise

currents I1 and I2, respectively, in each mesh. Then,

V = -V2 and we may also write

V2 = j

ω

L2 I2 + j

ω

MI1 or -V = j

ω

L2 10

V + j

ω

M

Solving for V,

V =

()()

21001

0.4 100

π

j

+

− = o

o

179.1- 2.000

89.0962.84

90-125.7

83.621

907.125 ∠=

∠

=

+

−∠

j

o

Thus,

v(t) = 2 cos (100πt – 179.1o) V.

(c) Define V1 across the left inductor, and V2 across the right inductor, with the “+”

reference at the respective dot; also define two clockwise mesh currents I1 and I2.

Then,

1

12

2

10

Now and

4

and 10

∠−

==−

=

out

V

IVV

V

I

1

11

1

2

11

1

2

1

10 EQN 1

410

10 EQN 2

10 4

10

1410 4

10

1

410 4

1 39 12.6 3

31.4 1 62.8

∠−



⇒=ω +ω





∠−



−=ω +ω





ω−ω ω∠



−





=



ωω ω∠





−+





−− 



=



−+



out

jL jM

jL jM jL

jM jL jM

jj

VV

V

VV

V

9.3

31.4







j

Solving, we find that Vout (= V) = 1.20 ∠ -2.108o V and hence

v(t) = 1.2 cos (100πt – 2.108o) V.

2

0.4 1 0

100 0.4 1 0

−=ω ∠

=− π× × ∠

j

V

V= 126 ∠ 90o V

111 2

222 1

=ω +ω

jL jM

VII

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9.

(a) 100 = j5

ω

(I1 – I2) + j3

ω

I2 + 6(I1 – I3) [1]

(4 + j4

ω

)I2 + j3

ω

(I1 – I2) + j2

ω

(I3 – I2) + j6

ω

(I2 – I3) – j2

ω

I2 + j5

ω

(I2 – I1)

– j3

ω

I2 = 0 [2]

6 (I3 – I1) + j6

ω

(I3 – I2) + j2

ω

I2 + 5 I3 = 0 [3]

Collecting terms,

(6 + j5

ω

) I1 – j2

ω

I2 – 6 I3 = 100 [1]

-j2

ω

I1 + (4 + j5

ω

) I2 – j4

ω

I3 = 0 [2]

-6 I1 - j4

ω

I2 + (11 + j6

ω

) I3 = 0 [3]

(b) For

ω

= 2 rad/s, we find

(6 + j10) I1 – j4 I2 – 6 I3 = 100

-j4 I1 + (4 + j10) I2 – j8 I3 = 0

-6 I1 – j8 I2 + (11 + j12) I3 = 0

Solving, I3 = 4.32 ∠ -54.30o A

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10.

(a)

1 1

2 2

=ω +ω =

=ω +ω =−

aab a

bba b

jL jM

VI I II

VII II

111

11 1

11 11 2

=+

=+ω+ω

=+ω−ω

a

ab

R

RjL jM

VI V

III

222

22 2

22 22 1

=−

=−ω−ω

=+ω−ω

b

ba

R

RjL jM

VI V

III

(b) Assuming that the systems connecting the transformer are fully isolated.

11

22

=ω +ω =−

aab a

bba b

jL jM

VII II

11

111 2

=−

=−ω−ω

=+ω+ω

a

ab

R

RjL jM

VI V

III

22

22 2

22 22 1

=+

=− +ω +ω

=− − ω − ω

bb

ba

R

RjL jM

VVI

III

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

(a)

22

22 22

2222

22

(0.2)

2 0.1 5 0.5

5 (0.2) 0.5 (0.2)

2 0.1 5 ( 0.5) 5 ( 0.5)

0.2 0.02

2 0.1

25 0.25 25 0.25

ω

=

+ω + +ω

ωωω

=

+ω + −

+ω +ω



ωω

=

++ω−



+ω +ω



jj

j

(b)

(c) Zin(j

ω

) at

ω

= 50 is equal to 2 + 0.769 + j(50)(0.023) = 2.77 + j1.15 Ω.

Z

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12.

22

11

22

3

22 32

3

232232

22 322

3

232 232

50 10 81010

81010

50 10 8 ( 10 10 ) 8 ( 10 10 )

81010

50 10

8 ( 10 10 ) 8 ( 10 10 )

−

−−

−

−−

ω

=+

ω

=ω × ++ω ×

ωω×ω

⇒=ω× + −

+ω × +ω ×



ω×ω

=+ω×−



+ω × +ω ×



in

M

jj

Mj M

j

MM

j

ZZ Z

Z

In this circuit the real power delivered by the source is all consumed at the speaker, so

2

222

232

22 2

232

1

20 8

3.2 8 ( 10 10 )

2

820

8 ( 10 10 ) 2 3.2

−

 ω

=⇒= ×

 ω×



ω

⇒=

+ω × ×

rms

VM

PR

M

= 62.5 W

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13. 12

2cos10 A, 1.2cos10 A==

SS

iti t

(a)

(b)

(c) 12

11

P 27.34 2cos69.44 9.601 W, P 23.64 1.2cos66.04 5.760 W

22

=× × °= =× × °=

SS

1

0.6( 20sin10 ) 0.2( 12sin10 ) 0.5( 32sin10 ) 9.6cos10

9.6cos10 25.6sin10 27.34cos(10 69.44 ) V

=− −− +− +

∴= − = + °

vtttt

vtt t

2

0.8( 12sin10 ) 0.2( 20sin10 ) 16sin10 9.6cos10

9.6cos10 21.6sin10 23.64cos(10 66.04 ) V

=− −− − +

∴= − = + °

vtttt

vtt t

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14.

84

10 4 10 5

65

=ω +ω

=ω +ω =ω +ω

=ω +ω

aab

bbabc

ccb

jj

jjjj

jj

VII

*V I I I I

VII

Also =− =− =

abc

II II

Now examine equation *.

10 4 10 5−ω −ω =−ω + ω c

jj jjII II

∴ the only solution to this circuit is I = and hence

( ) 120cos

∴

=ω

vt tV.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

12

21

2

2 1

11

100 10I 15I

0 200I 15I 15I

0 (5 10) I 15I

510 12 12

I I I 0 200 15 I 15I

15 3 3

400 200 118.33 66.67

0 15 I 15I I I

33 15

2

100 (66.67 118.33) 5 10 I (39.

3

=−

=−−

=+ −



++ +

∴= = ∴= − −





+



∴= − + − ∴ =







∴= + −− =





L

LL L

LL

L

jj

jjj

jj

jj j

jjj

jj j

j

jj j j

jj44 68.89) I

I 1.2597 60.21 A

+

∴= ∠− °

L

j

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16. 2cos10 A, 0==

s

itt

(a) 22

11

O.C. (0) 5 2 4 2 10 8 18 J

22

− ∴ =××+××=+=ab w

(b)

22 2

2

1

S.C. 10, I 2 0 A, M 12 3 H

2

20 3

( 30 5) I 10 3 2, I 1.1390 9.462 A 1.1390cos(10 9.462 ) A

530

1

(0) 1.1235 (0) 10 8 3 2 1.1235 3 1.1235 16.001 J

2

−

−ω==∠°==

+− ×∴= = ∠ °∴= + °

+

∴= ∴=+−×× +×× =

S

ab

j

jj i t

j

iw

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

1221

12 2

2

22 2

22

24 2

V 12 0 V rms, 100 rad/s

12 (6 20) I 100(0.4K) I , (24 80) I 40KI 0

310 310

I I 12 (6 20) 40K I

5K 5K

18 200 60 60 200K 60K

12 I I

5K 182 200K 120

60 K 24

P(200K 1

=∠° ω=

=+ + + + =



++

∴= ∴ = + +



−−



−+++ −

∴= ∴=

−−++

∴= −

s

jj jj

jj

jj j

jj

22

22 4 2 4 2

86,400 K 2.16K W

82) 120 40,000K 72,800K 47,524 K 1.82K 1.1881

==

+−+−+

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18.

12

=M

kLL

250 /ω= k rad s

6

12 28010

12.6 H

−

==××

=µ

MLL

Zin 22 22

22 22

11 22 22

22 22 22 22

ω−ω

=+ +

++

n

MR jM X

RX RX

Z

36

11 22

36

22

250 10 2 10 2

0.5 (250 10 )(80 10 )

20

−

=× × ×× =Ω

==××

=

jR

jX

Z

Thus, Zin = j0.5 + 19.8/404 – j198/ 404

= 0.049 + j0.010 Ω.

Zin →

2 Ω

j10 Ω

M

• •

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19. 100 rad/sω=

(a)

(b) 2

20

P 0.1144 20 0.2617 W=×=

(c) P 100 0.5833cos88.92 1.1039 W=× °=

gen

12

123

21

31

3121 1

11 2

K 50,K 20,1H j100

100 j200 I 50I 20I

0 (10 100) I 50I

0 (20 100) I 20I

25 52

II,II102052I

2 10 1 10 1 10 2 10

25 4

10 20 I I 0.5833 88.92 A, I

110210

→Ω →Ω→ Ω

=−−

=+ −



∴= = ∴= − −



++ ++





∴= + + ∴= ∠− ° =



++



jj

jj jj

jj j

jj jj

jjj

2

310

0.2902 83.20 A,

I 0.11440 77.61 A P 0.2902 10 0.8422 W

Ω

∠− °

=∠−°∴=×=

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20.

(a)

(b)

(c) The total energy stored at t = 0.

12

22

11 22 1212

41

11

22

11

5 16 1.8 1 1.2 4 1

22

40 0.9 4.8

36.1J

==

=++

=×× +× ×− ××

=+ −

=

AA

W total L L M

II

IIII

12 3

231

510

510 410

−−

+=

⇒=−

=× −×

tt

II I

III

12

0.4 5 1.8

1.2H

=

⇒= ×

=

M

kLL

M

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

12 1 1 2 2

11 122

12 1 2 2

2

12

62

1212

2

12

2

K 1000K L L , L 1000L , L 1000L

V (2 1000L ) I 1000K L L I

0 1000K L L I (40 1000L ) I

40 1000L

1000 rad/s I I

1000K L L

(2 1000L )(40 1000L ) 10 K L L

VI

1000K L L

I8

→→→

∴=+ −

=− + +

+

ω= ∴ =

+++

∴=

s

jjj

jj

j

jj

j

62

1212

12

2

62

s12 12

0 40,000L 2000L 10 L L (1 K )

40,000K L L

V

V 80 10 L L (1 K ) (40,000L 2000L )

++−−

∴=−−+ +

jj

j

(a)

(b)

(c) 2

12

V 40,000 5 200,000

L 1, L 25, K 1 2.222 0.05093

V 80 0 90,000 80 90,000

×

== =∴= = =∠ °

−+ +

s

jj

33

2

12

V405 200

L 10, L 2510, K 1 1.660941.63

V 80 0 (40 50) 80 90

−− ×

==×=∴= = =∠°

−+ + +

s

jj

2

12 62

3

2

V 40,000 0.99 5

L 1, L 25, K 0.99 V 80 25 10 (1 0.99 ) (40,000 50,000)

V 198,000 0.3917 79.74

V 80 497,500 90,000

××

== =∴=

−× − + +

∴= = ∠− °

−+

S

j

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22.

(a)

(b)

,,

,

12

2

L 10 mH, L 5 mH

L8 mH

L 10 mH, L 5 mH, 8 10 M M (5 M) (mH)

M(5 M)

8 10 M , 5M (10 8)5 5M M M 3.162 mH ( 10)

5

3.162

K K 0.4472

50

==

=

∴= = =−+ −

−

∴= − + ∴ = − + − ∴ = =

∴= ∴=

AB CDOC CD ABOC

AB CDSC

1

33 32 3

22

2222 2

2

Dots at A and D, 5 A, 100 mJ

11

100 10 10 10 25 5 10 10 5 10

22

210 40 40

100 125 2.5 5 10 2 10 10 0, 10

2

3.162 A

−− − −

==

∴× =×× ×+×× − ××

±−

=+ − ∴− += = =

∴=

tot

iw

ii

iiii i

i

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23. Define coil voltages v1 and v2 with the “+” reference at the respective dot.

Also define two clockwise mesh currents i1 and i2. We may then write:

12

11

21

22 260 /

=+

=+ ω=π

dd

vL M

dt dt

dd

v L M rad s

dt dt

II

or, using phasor notation,

111 2

222 1

=ω +ω

jL jM

VII

111 2

222 1

100 0 50

25

∠= +ω +ω

−=ω+ω

jL jM

II I

Rearrange:

or

We can solve for I2 and V2 = −25I2:

V2 = 1 LL

658.1

21 +

−k

j

11 2

122

[50 ] 100 0

[25 ] 0

+ω +ω = ∠

ω−+ω =

jL jM

jM jL

II

11

22

50 100 0

25 0

+ω ω ∠





=





ω−+ω 



jL jM

jM jL

I

12

=

MkLL

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24.

1max

2222

max

2cos500 A W at 0

111

42 62 52 32

222

8121012 42 J

==

∴=××+××+××+×

=+ + + =

it t

w

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25.

(a) 12

All DC: L 2 1 1 H

−=−=

(b) 12

AB SC: L 1 2 8 0.6 H

−=− + =

(c) 12

BC SC: L 2 ( 1) 9 2 9 /8 0.875 H

−=+− =− =

(d) 12

AC SC: L (2 1) (1 2) 1 3 0.750 H

−=− += =

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26.

(a) 













+

++

=

ω

ωω

ω

320

2

320

)20(2

315

1

S

L

j

jj

j

V

I

=

ωω

ω

14511300

2

2j

j

+−

(b)

2

1,2

2.57 10.61

145 145 13,200

( ) 100 ( ), (0) 0, (0) 0, 2.570, 10.612

22

,0, A B,0AB

100 15 5 2 , 0 20 3 2 At 0 : 100 0 5 (0 ) 2 (0 ) and

003(0)2

−−

+++

+

−± −

==== =−−

=+ =∴= + ∴=+

′′ ′′ ′ ′

=+− = +− = =+ −

′′

=+ −

ssL

tt

LLfLnLf L

ssL LLs s L

L

vt ut i i s

ii ii i e e

ii i iii t i i

ii

2.57 10.612

(0 ) (0 ) 1.5 (0 ) 100 7.5 (0 ) 2 (0 ) 5.5 (0 )

(0 ) 18.182 A/s 18.182 2.57A 10.61B 2.57A 10.61A 8.042A

A 2.261, B 2.261, ( ) 2.261( ) A, 0

++ + ++ +

+

−−

′′ ′′ ′

∴= ∴= − =

′

∴= ∴ =−− =−+ =

∴= =− = − >

ss L LL L

L

tt

L

ii ii i

i

it e e t

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

(a) Open-Circuit

4

×

=ω Ω

TA

oc

TB

oc

jM

Z

(b) Short-Circuit

4810

××

==−ωΩ+ωωΩ

TA TB

SS SS jM j j MZZ

(c) If the secondary is connected in parallel with the primary

4108

26 12 8

×

=− ω − ω + ω Ω

=ω ω −ω Ω

TA

in

TB

in

jj jM

Z

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28. Define three clockwise mesh currents I1, I2, and I3 beginning with the left-most mesh.

Vs = j8

ω

I1 – j4

ω

I2

0 = -4j

ω

I1 + (5 + j6

ω

) I2 – j2

ω

I3

0 = -j2

ω

I2 + (3 + j

ω

) I3

Solving, I3 = j

ω

/ (15 + j17

ω

). Since Vo = 3 I3,

ω

1715

3

S

o

j

+

=

V

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29. Leq = 2/ 3 + 1 + 2 + 6/5 = 4.867 H

Z(j

ω

) = 10 j

ω

(4.867)/ (10 + j

ω

4.867)

= j4.867

ω

/ (1 + j0.4867

ω

) Ω.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30. 100 rad/s

V 100 0 V rms

ω=

=∠°

s

(a)

(b)

,

100( 400)

V 39.99 1.146 V rms

20 1000

400(20 600) 240,000 8,000

Z , V 0 200 200 40.19 85.44

20 1000 20 1,000

==∠°

+

+−+

==− + =− + = ∠ °Ω

++

OC cd

incd S

j

jj j

jj

400(10 200) 80,000 4,000

Z 20 600 20 600

10 200 10 200

−−+

=+ + =+ +

++

ina b

jj j

jj

= 210.7∠73.48o V and Voc = 0.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31. 12

L 1 H, L 4 H, K 1, 1000 rad/s===ω=

(a) 6

10 1 4

Z 1000 Z 1000 24.98 0.6246

4000 100

××

=Ω∴= + = + Ω

+

Lin

jj

j

(b) 6

410

Z 1000 0.1 Z 1000 24.39

4000 100

×

=×∴=+ = Ω

+

Lin

jj j

jj

(c) 6

L

410

Z 100 Z 1000 25.46

4000 100

×

=− ∴ = + =− Ω

−

in

jj j

jj

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32.

12

,

,,

,

L 6 H, L 12 H, M 5 H

#1, L 6 H

#2, L 12 H

#3, L 1 7 5 3.917 H

#4, L 7 5 1 7.833 H

#5, L 7 1 8 H

#6, L 7 1 5 5.875 H

#7, L 11 17 28 H

#8, L

== =

=

=+ =

=+=

inAB CDOC

inCD ABOC

inAB CDSC

inCD ABSC

inAC BDSC

inAB ACSC BDSC

inAD BCSC

inA ,5 11/17 1.6786 H=− + =

BADSC

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

22

11

22 22

22

312

22 22

312 12

2222

2

11

31.83 314 /

31.83

ie. a 50Hz system

20 100 10 2 31.83

2 31.83

20 100 10 2 31.83 2 31.83

493 7840

20 31.4 1020 1020

20

−

ω

=++

=⇒ω= =

ω×

ω

=+ω× +

−

ωω

=+ω× + −

++



=+ + −





=+

in

M

RjX

rad s

CC

kLL

jj

kLL j kLL

j

jjk

j

ZZ

Z

2

31.4 [0.483 7.69]

( 0) 20 31.4

( 0.5) 20.2 27.6

( 0.9) 20.4 24.5

( 1.0) 20.5 23.7

+−

==+ Ω

== + Ω

in

jk

kj

Z

(a)

(b)

(c)

(d)

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. 12

L 125 H, L 20 H, K 1, M 2500 50 H, M 5000

↑→ → =∴= = ω= Ω

jj

(a)

(b)

6

5000(10 3000)

Z 20 7500 10 2000

15 10 50,000

20 7500 82.499 0.2170

10 2000

82.498 0.3125 V 0

−

=+ + +

×+

=+ + = ∠ °Ω

+

=+ Ω=

ina b

OC

jj

j

jj

j

,

100( 5000)

V 39.99995 0.09167 V rms

20 12,500

5000(20 7500)

Z , V 0 3000 3.19999 0.00512

20 12,500

==∠°

+

==− + = + Ω

+

OC cd

incd S

j

jj

j

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35.

280 2 0.438A

1280

1000 2 1.56A

1280

∴=×=

=×=

a

b

I

1

2

3

1.56A

5 1.56 7.8A

1.5 7.8 11.7A

∴=

⇒=× =

A

I

2

23

22

1

22

2

22

3

(1 )

0.438 1 10

192W

(30 ) (1.56) 30

73W

(1 ) 7.8 1

60.8W

(4 ) 11.7 4

548W

⇒=

=××

=

⇒Ω== ×

=

⇒Ω= = ×

=

⇒Ω= = ×

=

a

Pk IR

PIR

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36.

(a)

(b)

21 1

21 2 1

1111

11

11 1 11 11

2

21

R 100

VV3V

II/4, V4 VI 40 40

3V I 3V 4V

100 10 I V ,

40 4 40 100

100

I 0.46V 100 10(0.46V 0.075V ) V 4.85 V V 4.85

400 82.47

V 4V 82.47 V P 68.02 W

4.85 100

=Ω

−

==∴==



∴= + = +





∴= ∴ = − + = ∴ =

∴= = = ∴= =

L

X

L

2

max

R sees 10 4 160 use R 160

100

P10250 W

20

×= Ω∴ = Ω



=×=





LL

L

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37.

22

2112

22

2

VV

II,V5V

840

100 300(C 0.025)V 5V

100

V12.5 300C

=∴= =

∴= + +

∴= +

(a) 2

2

8

C0 V 8V P 8 W

8

=∴ = ∴ = =

L

(b)

2

100 100 1

C 0.04 V P 2.082 W (neg. )

24.5 24.5 8



=∴= ∴= =





Lfdbk

(c) 2

2

100 200

C 0.04 V 200 V P 5000 W (pos. )

0.5 8

=− ∴ = = ∴ = =

Lfdbk

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38.

2

22

1

Apply V 1 V I 0.05 A, V 4 V

41

4 60 I 20 0.05 I 0.05 A

60

I 0.2 A I 0.25 A R 4 , V 0

=∴= =

−

∴= + × ∴ = =

∴= ∴ = ∴ =Ω =

ab x

in th th

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

100

1

2

25

P 1000 W, P 500 W

500

I 5 A, V 100 5 V

100

1000

I 10 A V 100 40 60 V

100

Now, P 1000 500 10 4 100W I 2 A; also

25

==

∴= = =

== ∴=−=

=−−×= ∴= =

gen

LL

S

X

130

Around center mesh: 60 2 25 100 5 0.8944

=× + ∴=

aa

05

60 =

Ix = 8944.0

5

2

,25 === bb

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40.

(a)

(b) 21

I 3I 3.297 0 A== ∠°

(c) 3

4

I 3.297 4.396 180 A

3

=− × = ∠ °

(d) 2

25

P 25 1.0989 30.19 W=× =

(e) 2

2

P 3.297 2 21.74 W=×=

(f) 2

3

P 4.396 3 57.96 W=×=

2

1

41616 2222

3,2,(3)66

333 33

100

66 25 91 1.0989 0 A I

91



×=Ω+=Ω =Ω





+=Ω = ∠°=

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

12125022

2

222

22222

222

2

V 2.5 V , I 0.4I , I I 0.1 V

60 2.5 V

60 40(0.4 I ) 2.5V I 16

Also, 60 50(I 0.1 V ) V 50I 6V

60 2.5 V

60 50 6 V 187.5 (7.8125 6) V

16

60 187.5

V 9.231 V

13.8125

===+

+

∴= − ∴=

=+ +=+

+



∴= + = + +





−

∴= =−

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42.

2

1

2

4

2

48

2

400

400 16 , 16 48 12 , 12 4 16

5

16 10

4 I 2 A P 4 W

241

21 A P 4 W, 10 2 1 8 V

2

8 2 16 V, 16 4 1 12 V, 12 / 48 3 W P , 12 5 60 V

60

P9 W

400

=Ω =Ω +=Ω

=Ω∴ = = ∴ =

+

=∴= −×=

×= −×= = = ×=

==

s

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43.

122 2

22

2

I 2I , 2I I I I I 2I 0

1

100 3I (4I 20I 20I )

2

10I 3I 12I 100

100 3 I 5I 20I 20I

25I 3I 20I 100

012

100 3 12

100 3 20 0 100( 26) 100( 18)

I11 2 1(60 36) 10( 20 6) 25( 1

10 3 12

25 3 20

==+∴+−=

=+ + −

∴−−=−

=−+ −

∴−− =−

−

−−−

−−− +−−−

∴= =

−− − −−+ −

−−

sx xs

sx

xs

sx x

xs

X

800 4.819 A

2 6) 166

−

==

−−

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44.

(a)

(b) Specify 3 A and 4 A in secondaries

2

10 10

II4

25 25

II3 (I4)(I3)

33

2I 7, I 3.5 A

25 25

V V ( 3.5 4) V

36

25 1

P P 1.7361 W

610

AB f

CD b f f

ff

AB CD

=+

=− − ∴ + = − −

∴=− =−

∴= =−+=



∴= = =





2

10

2

10

25 25 100

50 10 V 1 4 V

333

100 1 1000

P 111.11 W

3109

25 25

V 1 3 25 V, P 62.5 W

310

AB

CD CD

=Ω∴ =××=



∴= = =





=×× = = =

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45. Corrections required to the problem text: both speakers that comprise the load

are 4-Ω devices. We desire a circuit that will connect the signal generator (whose

Thévenin resistance is 4 Ω) to the individual speakers such that one speaker receives

twice the power delivered to the other. One possible solution of many:

We can see from analysing the above circuit that the voltage across the right-most

speaker will be 2

225.1

732.1 or times that across the left speaker. Since power is

proportional to voltage squared, twice as much power is delivered to the right

speaker.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46. (a) We assume Vsecondary = 230∠0o V as a phasor reference. Then,

I

unity PF load = A 034.8 0

230

8000 oo ∠=∠ and

I

0.8 PF load =

()

A 9.3665.2 0.8 cos

230

15000 o1 −∠=−∠ −

Thus, Iprimary =

()

oo 36.9-65.2 08.34

2300

230 ∠+∠

= 0.1 (86.9 – j39.1) = 9.5 ∠-24.3o A

(b) The magnitude of the secondary current is limited to 25×103/230 = 109 A.

If we include a new load operating at 0.95 PF lagging, whose current is

I0.95 PF load = | I0.95 PF load | ∠ (-cos-1 0.95) = | I0.95 PF load | ∠ -18.2o A,

then the new total secondary current is

86.9 – j39.1 + | I0.95 PF load | cos 18.2o – j | I0.95 PF load | sin 18.2o A.

Thus, we may equate this to the maximum rated current of the secondary:

()()

2

o

load PF 0.95

2

o

load PF 0.95 18.2sin || 39.1 18.2 cos || 86.9 109 II +++=

Solving, we find

2

)2800)(4(189189-

||

2

load PF 0.95

+±

=I

So, |I0.95 PF load | = 13.8 A (or –203 A, which is nonsense).

This transformer, then, can deliver to the additional load a power of

13.8×0.95×230 = 3 kW.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47. After careful examination of the circuit diagram, we (fortunately or unfortunately)

determine that the meter determines individual IQ based on age alone. A simplified

version of the circuit then, is simply a 120 V ac source, a 28.8-kΩ resistor and a

(242)RA resistor all connected in series. The IQ result is equal to the power (W)

dissipated in resistor RA divided by 1000.

P = A

2

A

3576R

576R 108.28

120 ×













+×

Thus, IQ = Age576

Age576 108.28

120

1000

12

3××













×+×

(a) Implementation of the above equation with a given age will yield the “measured”

IQ.

(b) The maximum IQ is achieved when maximum power is delivered to resistor RA,

which will occur when 576RA = 28.8×103, or the person’s age is 50 years.

(c) Well, now, this arguably depends on your answer to part (a), and your own sense

of ethics. Hopefully you’ll do the right thing, and simply write to the Better Business

Bureau. And watch less television.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48. We require a transformer that converts 240 V ac to 120 V ac, so that a turns ratio of

2:1 is needed. We attach a male european plug to the primary coil, and a female US

plug to the secondary coil. Unfortunately, we are not given the current requirements

of the CD writer, so that we will have to over-rate the transformer to ensure that it

doesn’t overheat. Checking specifications on the web for an example CD writer, we

find that the power supply provides a dual DC output: 1.2 A at 5 V, and 0.8 A at 12 V.

This corresponds to a total DC power delivery of 15.6 W. Assuming a moderately

efficient ac to DC converter is being used (e.g. 80% efficient), the unit will draw

approximately 15.6/0.8 or 20 W from the wall socket. Thus, the secondary coil

should be rated for at least that (let’s go for 40 W, corresponding to a peak current

draw of about 333 mA). Thus, we include a 300-mA fuse in series with the

secondary coil and the US plug for safety.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49. You need to purchase (and wire in) a three-phase transformer rated at

(

)

()()

kVA. 3.6 102083 = The turns ratio for each phase needs to be 400:208 or

1.923.

CHAPTER THIRTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50. (a) The input to the left of the unit will have the shape:

and the output voltage will be:

We need to reduce the magnitude from 115-V (rms) to a peak voltage of 5 V. The

corresponding peak voltage at the input will be 2115 = 162.6 V, so we require a

transformer with a turns ratio of 162.6:5 or about 32.5:1, connected as shown:

(b) If we wish to reduce the “ripple” in the output voltage, we can connect a capacitor

in parallel with the output terminals. The necessary size will depend on the maximum

allowable ripple voltage and the minimum anticipated load resistance. When the input

voltage swings negative and the output voltage tries to reduce to follow, current will

flow out of the capacitor to reduce the amount of voltage drop that would otherwise

occur.

±

115 V

rms ac

a = 1/ 32.5

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1. (a) s = 0;

(b) s = ± j9 s-1;

(c) s = -8 s-1;

(d) s = -1000 ± j1000 s-1;

(e) v(t) = 8 + 2 cos 2t mV cannot be attributed a single complex frequency. In a circuit

analysis problem, superposition will need to be invoked, where the original function v(t)

is expressed as v(t) = v1(t) + v2(t), with v1(t) = 8 mV and v2(t) = 2 cos 2t mV. The

complex frequency of v1(t) is s = 0, and the complex frequency of v2(t) is s = ± j2 s-1.

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2. (a) (6 – j)* = 6 + j

(b) (9)* = 9

(c) (-j30)* = +j30

(d) (5 e-j6)* = 5 e+j6

(e) (24 ∠ -45o )* = 24 ∠ 45o

(f) o

o

94.19 5.303

16.72- 477.3

77.47 18.44

33.3

184

*

33.3

184 ∠=

∠

=













−

+

=













+

−

j

(g)

()

oo

o

oo

60.36 .62020

*

60.36 .62020

*

26.60062.8

1.0 5

*

74

1.0 5 −∠=∠=













−∠

∠

=













−

∠

j

(h) (4 – 22 ∠ 92.5o)* = (4 + 0.9596 – j21.98)* = (4.9596 – j21.98)* = 4.9596 + j21.98

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3. Re ( ) ( )=it it. No units provided.

(a)

(b)

(c)

(d)

( 3 15) 3 15 3 (15 60.26 )

3

( ) (4 7) (8.062 60.26 ) 8.062

( ) Re ( ) 8.062 cos(15 60.26 )

−+ − − − °

−

=− = ∠− ° =

∴= = − °

jt tjt tjt

x

t

xx

it j e ee ee

it it e t

3 3 15 60.26

3

( ) (4 7) (cos15 sin15 ) 8.062

( ) 8.062 cos(15 60.26 )

−−−+°

−

=+ − =

∴= − °

ttjtj

y

t

y

it j e t j t e e

it e t

( 1.5 12) 57.99 1.5 12 1.5 (125 57.99 )

0.6

( ) (5 8) 9.434 9.434

Re (0.4) 9.434 cos(4.8 57.99 ) 4.134

−+ − °− − − °

−

=− = =

∴= −°=−

tj t j t jt t j

A

rad

A

it je e e e e e

ie

( 1.5 12) 57.99 1.5 12 1.5 (12 57.99 )

( ) (5 8) 9.434 9.434

Re (0.4) 4.134

−+ °− − − − − °

=+ = =

∴=−

jt j t jt t jt

B

it je e e e e e

i

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4. (a)

ω

= 279 Mrad/s, and

ω

= 2

π

f. Thus, f =

ω

/2

π

= 44.4 MHz

(b) If the current i(t) = 2.33 cos (279×106 t) fA flows through a precision 1-TΩ resistor,

the voltage across the resistor will be 1012 i(t) = 2.33 cos (279×106 t) mV. We may write

this as 0.5(2.33) cos (279×106 t) + j (0.5)2.33 sin (279×106 t) + 0.5(2.33) cos (279×106 t)

- j (0.5)2.33 sin (279×106 t) mV

= 1.165 e j279×106 t + 1.165 e -j279×106 t mV

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5. (a) vs(0.1) = (20 – j30) e(-2 + j50)(0.1) = (36.06 ∠ -56.31o) e(-0.2 + j5)

= 36.06e-0.2 ∠ [-56.31o + j5(180)/

π

] = 29.52 ∠230.2o V (or 29.52 ∠-129.8o V).

(b) Re{ vs } = 36.06 e-2t cos (50t – 56.31o) V.

(c) Re{ vs(0.1) } = 29.52 cos (230.2o) = -18.89 V.

(d) The complex frequency of this waveform is s = -2 + j50 s-1

(e) s* = (-2 + j50)* = -2 – j50 s-1

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6. (a) s = 0 + j120

π

= + j120

π

(b) We first construct an s-domain voltage V(s) = 179 ∠ 0o with s given above.

The equation for the circuit is

v(t) = 100 i(t) + L dt

di = 100 i(t) + 500×10-6

dt

di

and we assume a response of the form Iest.

Substituting, we write (179 ∠ 0o) est = 100 Iest + sL Iest

Supressing the exponential factor, we may write

6-

o

10500100

0179

×+

∠

=s

I =

()

o

6-

o

108.0100

0179

10500120100

0179

∠

=

×+

∠

π

j = 1.79 ∠ -0.108o A

Converting back to the time domain, we find that

i(t) = 1.79 cos (120

π

t – 0.108o) A.

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7.

(a)

(b) 2

( ) 0.3536 cos(10 105 ) A

t

x

it e t

−

=−°

2

10 cos(10 30 )V 2 10, V 10 30 V

10 5 1 5 5 25 ( 25 125)/ 26

Z,Z5

2 10 1 5 26 26 ( 5 25 130) / 26

25 125 1 5

Z 5 1 Z 5 0.5( 2 10) 1 4 4

125 25 5 1

10 30 ( 5 25)/ 26

I445(

t

ss

cc

cin

x

ve t s j

jj j

jj

jjjj

jj

j

−

=+°∴=−+=∠°

−− −− − −

== = =

−+ −+ −− +

−− −−

∴= = =−∴=+−+−=+Ω

−−

∠° −−

∴= ×

++−

10 30 5 25 5 30 5 25 1 30 1 5

5 25)/ 26 4 4 130 5 25 2 2 125 25 2 2 5 1

130

I ( 1) 0.3536 105 A

2245

x

jjj

jjjjjjj

j

∠° −− ∠°−− ∠°−−

===

−+−−+−+−

∠°

∴= − = ∠− °

∠°

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8. (a) s = 0 + j100

π

= + j100

π

(b) We first construct an s-domain voltage V(s) = 339 ∠ 0o with s given above.

The equation for the circuit is

v(t) = 2000 i(t) + vC(t) = 2000 dt

dvC

C + vC(t) = dt

dvC

2.0 + vC(t)

and we assume a response of the form VCest.

Substituting, we write (339 ∠ 0o) est = 0.2s VCest + VCest

Supressing the exponential factor, we may write

s

V2.01

0393

o

C+

∠

= =

()

o

oo

09.8984.62

0339

2.01001

0339

∠

=

+

∠

π

j = 5.395 ∠ -89.09o A

Converting back to the time domain, we find that

vC(t) = 5.395 cos (100

π

t – 89.09o) V.

and so the current is i(t) = dt

dvC

C = -0.1695 sin(100

π

t) A = 169.5 sin (100

π

t) mA.

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9. 33

12

20 cos4 A, 30 sin 4 A

tt

SS

ie tie t

−−

==

(a)

(b) 3

( ) 185.15 cos(4 47.58 ) V

t

x

vt e t

−−

=−°

12

I 20 0 , I 30, 3 4

10 3 4

Z 0.4( 3 4) 1.2 1.6, Z 6 8

3434

5(7.2 6.4) 6 8 ( 6 8)(3.8 1.6)

V20 30

2.2 6.4 7.2 6.4 2.2 6.4

600 800 30( 22.8 12.8 30.4 9.6) 600

2.2 6.4

SS

c L

x

js j

jjj j

jj

jj jj

j

jj j

jj j j

j

=∠° =− =−+

−−

∴= = −− =−− =−+

−+ −−

+−+ −+−

∴= × −

−+ −+ −+

−+ − − + + + −+

==

−+

800 30( 10 40)

2.2 6.4

600 1200 1000 600 1000 185.15 47.58 V

2.2 6.4 2.2 6.4

jj j

j

jj

−

−−+

−+

−+ + +

===∠−°

−+ −+

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10. (a) If v(t) = 240 2e

-2t cos 120πt V, then V = 240 2∠0o V where s = -2 + j120

π

.

Since R = 3 mΩ, the current is simply I = 3

o

103

0 2240

−

×

∠ = 113.1 ∠0o kA. Thus,

i(t) = 113.1e-2t cos 120πt kA

(b) Working in the time domain, we may directly compute

i(t) = v(t) / 3×10-3 = (240 2e

-2t cos 120πt ) / 3×10-3 = 113.1e-2t cos 120πt kA

(c) A 1000-mF capacitor added to this circuit corresponds to an impedance

Ω

+

=

×+

=

1202-

1

)10)(1000120(-2

1

3-

ππ

jjCs in parallel with the 3-mΩ

resistor. However, since the capacitor has been added in parallel (it would have been

more interesting if the connection were in series), the same voltage still appears across its

terminals, and so

i(t) = 113.1e-2t cos 120πt kA as before.

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

{}

∞

−

∞−

∞−−

==== ∫∫∫ 0

000 )( )( )( --

tttt e

K

dteKdttueKdttuKetuK ssss

s

L

= 











+











−−

→

−

∞→

t

te

K

e

Kss

ss lim lim 0

If the integral is going to converge, then

(

)

finite). bemust (i.e. 0 lim

s

s=

−

∞→

t

teThis leads

to the first term dropping out (l’Hospital’s rule assures us of this), and so

{}

s

K

tuK )( =L

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12. (a)

{}

∞

−

∞−

∞−−

==== ∫∫∫ 0

000

3

3 )(3 )(3 )( 3 --

tttt edtedttuedttuetu ssss

s

L

= 











+











−−

→

−

∞→

t

tee ss

ss

3

lim

3

lim 0

If the integral is going to converge, then

(

)

finite). bemust (i.e. 0 lim

s

s=

−

∞→

t

teThis leads

to the first term dropping out (l’Hospital’s rule assures us of this), and so

{}

s

3

)( 3 =tuL

(b)

{}

∞

−

∞−

∞−−

==−=− ∫∫ 3

30

3

3 )3(3 )3( 3 -

ttt edtedttuetu sss

s

L

= 











+











−−−

∞→

ss

3

lim ee t

t

If the integral is going to converge, then

(

)

finite). bemust (i.e. 0 lim

s

s=

−

∞→

t

teThis leads

to the first term dropping out (l’Hospital’s rule assures us of this), and so

{}

s

3

)3( 3 −

=− etuL

(c)

{}

[]

∞

−

∞

−

∞−

−

−−

=

=−−=−− ∫∫∫

03

030

3

-

3

3- 3 3)3(3 3)3( 3 --

tt

ttt

ee

dtedtedtetutu

ss

sss

ss

L

Based on our answers to parts (a) and (b), we may write

{}

3

3)3( 3 3

ss

s−=−− −

etuL =

()

1 -

33s

s

−

e

(d)

{}

() ()

1

3

1

3

3 )3(3 )3( 3

33

3

0

3

00 -

--

ss

sss

ss

s

−−

∞−

−=−

−

=

−

==−=− ∫∫

ee

edtedttuetu ttt

L

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13. (a)

{}

[]

∞

−

∞−

∞−−

==+=+ ∫∫ 0

00

5

5 )(32 )( 32 -

ttt edtedttuetu sss

s

L

= 











+











−−

→

−

∞→

t

tee ss

ss

5

lim

5

lim 0

If the integral is going to converge, then

(

)

finite). bemust (i.e. 0 lim

s

s=

−

∞→

t

teThis leads

to the first term dropping out (l’Hospital’s rule assures us of this), and so

{}

s

5

)( 32 =+ tuL

(b)

{}

∞

+−

∞+−

∞−

−

+

−

=== ∫∫

0

)8(

0

)8(

0

8-8-

8

3

3 3 3 --

ttttt edtedteee sss

s

L

= 











+













+

−+−

→

+−

∞→

t

tee )8(

0

)8(

8

3

lim

8

3

lim ss

ss = 0 + 8

3

+s = 8

3

+s

(c)

{}

0 )0( )( )( )( -

---

0

00 ==−=−=− ∫∫∫ −−

∞−dtedttuedttuetu ttt sss

L

(d)

{}

∞

−

∞−

∞−−

==== ∫∫∫ 0

000 --

tttt e

K

dteKdteKdtKeK ssss

s

L

= 











+











−−

→

−

∞→

t

te

K

e

Kss

ss lim lim 0

If the integral is going to converge, then

(

)

finite). bemust (i.e. 0 lim

s

s=

−

∞→

t

teThis leads

to the first term dropping out (l’Hospital’s rule assures us of this), and so

{}

s

K

K =L

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14. (a) The frequency-domain representation of the voltage across the resistor is (1)I(s)

where I(s) =

{}

A

1

4

)(4 +

=s

t ue-t

L. Thus, the voltage is V

1

4

+s.

(b)

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15. (a)

{}

[]

∞

−

∞

−

∞−

+

−

==

−−=−−

∫∫

∫

20

0

5

5 - 5

)2( 5)( 5 )2( 5)( 5 -

tttt

t

eedtedte

dtetutututu

ssss

s

ss

L

= 











+











−−

→

−

∞→

t

tee ss

ss

5

lim

5

lim 0 + 











+











−−−

∞→

ss

2

5

lim ee t

t

If the integral is going to converge, then

(

)

finite). bemust (i.e. 0 lim

s

s=

−

∞→

t

teThis leads

to the first and third terms dropping out (l’Hospital’s rule assures us of this), and so

{}

()

s

2

1

5

)2( 5)( 5 −

+=−− etutuL

(b) The frequency domain current is simply one ohm times the frequency domain voltage,

or

()

s

2

1

5

−

+e

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16.

(a) f(t) ()

0

1F() (1) 0

−

∞−σ+ω

=+∴ = + ∴σ>

∫jt

tstedt

(b) ()

0

() ( 1) () F() ( 1) 0

−

∞−σ+ω

=+ ∴ = + ∴σ>

∫jt

ft t ut s t e dt

(c) 50 50 ( )

0

() () F() 50

−

∞−σ+ω

=∴= ∴σ>

∫

ttjt

ft e ut s e e dt

(d) 50 50 ( )

0

() (5)F() (5) 50

−

∞−σ+ω

=−∴= − ∴σ>

∫

ttjt

ft e ut s e ut e dt

(e) 50 50 ( )

0

() (5)F() (5) 0

−

∞

−−−σ+ω

=−∴= − ∴σ>

∫

ttjt

ft eut s eut e dt

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

(a)

(b)

(c)

3

22

3

03

(2 ) ( 2 ) 6 3 6 3

30

2(2 )

(2) 33

(2 ) () 63

(2)

() 8 [ ( 3) ( 3)] F( ) 8

88

F( ) 8 8 [1 ] [1 ]

22

F() 8 8

88

8F[][1

22

−−

−ω

−

−ω −−ω + ω −− ω

−

−− −

−−

−− −+ −

=−−−∴ω=

∴ω= + = − + −

−ω +ω

==

+∴=−+−

−+

∫

∫∫

tt

jt

jt jt j j

tst st

st s s

ft e ut ut j e e dt

j e dt e dt e e

jj

s e dt e dt

edt e e

ss

63

3(2 ) 63

0

]

8

F( ) 8 [1 ]

2

−

−− −−

==−

+

∫

s

st s

s e dt e

s

3

2(2)

3

63 63 (2)

(2) 3

63 63 (2 )

0

63 632

() 8 [ ( 3) ( 3)]F( ) 8

8[]F()8

2

8[],F()8

2

88

[1] [1]

22

−

−ω

−

−ω −+ω −

−

∞

−+ − −

−−

=+−−ω=

=− =

−ω

=−=

−

=−=−

−−

∫

tjt

jj st

ss st

s

ft e ut ut j e dt

ee s edt

j

ee s edt

s

ee

ss

2

3(2 ) 6 3 6 3

3

3(2 ) 6 3 6 3

(2) 3

(2 ) 63

00

() 8 [ ( 3) ( 3)] F( ) ()

8

F( ) 8 [ ]

2

8

F() () 8 [ ]

2

8

F( ) ( ) 8 [1 ]

2

−

∞

−ω

−∞

+ω + ω −− ω

−

∞−−+ +−−

∞−

∞∞

−−+ −−

=+−−∴ω=

∴ω= = −

+ω

===−

+

===−

+

∫

∫∫

tjt

jt j j

st s t s s

st s t s

ft e ut ut j fte dt

jedt ee

j

s f t e dt e dt e e

s

sftedtedt e

s

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18. (a) ss

1

1=























1-

LL

(b)

{}

ssss

3

111

)]([)(1 2=++=++ tutuL

(c)

{}

ss

31

3)( 2−=−tutL

(d)

{}

ss

s

2

1

)2()1()( -1 −− −+−=−−−+ eettt

δδδ

L

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19. (a) f(t) = e-3t u(t)

(b) f(t) =

δ

(t)

(c) f(t) = t u(t)

(d) f(t) = 275 u(t)

(e) f(t) = u(t)

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20.

{}

[]

{} {}

)( )(

)( )( )( )( )( )(

21

02

01

02121 ---

tftf

dtetfdtetfdtetftftftf ttt

LL

L

+=

+=+=+ ∫∫∫ ∞−

∞−

∞−sss

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

(a)

(b) 224

( ) 2 ( 2) F( ) 2 , F(1 2) 2 0.17692 0.2048

−−−

=δ − ∴ = + = =− +

sj

ft t s e j e e j

(c)

2

22

24

22

( ) 2 ( 2) F( ) 2 ; 1 2

2

F(1 2) 0.04655 0.11174

12

∞

∞−−

−− +

−

=−∴= + = =+

∴+ = = +

+

∫st st s

j

ft ut s e dt e e s j

ss

jee j

j

4

(1) (1) 22

22

22 3

11

() ( 2) F() 11

1

F(1 2) (0.4724 6.458)10

22

∞

−∞

−−+−+−−

−−− −

=−∴= = =

−+ +

∴+ = = +

+

∫

tststs

j

ft e ut s e dt e e

ss

jeee j

j

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22. (a) 7.671- 15sin 8 )1( 5 sin8 =×=−

∫∞

∞− dttt

δ

(b)

() ( )

9 52 )2(5 2 2 =−=−−

∫∞

∞− dttt

δ

(c) 1.840 5 )10333.3(5 )10333.3(300043000 4==×− −

×−

∞

∞−

−−

∫edtte t

δ

(d) )2( KdttK =−

∫∞

∞−

δ

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23.

(a)

(b) 2

2

4

() 4 ( 2) F() 4 ∞−−

=−∴= =

∫st s

f t u t s e dt e

s

(c)

(d) 222

02

() 4 ( 2) F() 4 ( 2) 4 ( 2) 4

+

−

∞−− −

=δ − ∴ = δ − = δ − =

∫∫

st s s

f t t s t e dt e t dt e

(e)

0

( ) 5 ( )sin (10 0.2 ) F( ) 5 ( ) [sin 0.2 ] X 1 5sin36

F( ) 2.939

+

−

=δ + π∴ = δ π = °

∴=

∫

ft t t s t dt

s

0

5

525

22

() [ (5 )][ ( 2)] (), F() [ (5 )][ ( 2)] ()

11

F( ) ( )

−

∞−

−−−−

=− − ∴ − −

∴= =− = −

∫

st

st st s s

f t u t ut ut s u t ut ut e dt

s e dt e e e

ss

3(3)(3)

22

26

4

() 4 ( 2) F() 4 3

4

F( ) 3

∞

−−+−+

−−

−

=−∴= =

+

∴=

+

∫

tstst

s

ft e ut s e dt e

s

se

s

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24. (a) 1 0500 cos )( 500 cos =×=

∫∞

∞− dttt

δ

(b)

() ( )

32 2 )2( 5 5 ==−

∫∞

∞− dttt

δ

(c)

()

0.9197 52. )1000(5.2 1000001.0001.0 ==− −

∞

∞−

−

∫edtte t

δ

(d) - )( 22 KdtctK =−−

∫∞

∞−

δ

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25.

(a) f(t) = 2 u(t – 1) u(3 – t) u(t3)

F(s) =

()

-

2

- 3

3

1

3

1

ssss

ss

−−−− ==

∫eeedte tt

(b) 44

4

22

( ) 2 ( 4) F( ) 2 (0 )

∞−−−

−

=−∴= = −=

∫st s s

ft ut s e dt e e

ss

(c) 2(2)48

4

3

() 3 ( 4) F() 3 2

∞

−−+−−

=−∴= =

+

∫

tsts

ft e ut s e dt e

s

(d) 5

0

() 3 ( 5) F() 3 ( 5) 3

−

∞−−

=δ − ∴ = δ − =

∫st s

f t t s t e dt e

(e)

0

() 4 ( 1)[cos sin ]

F( ) 4 ( 1) [cos sin ] F( ) 4

−

∞−−

=δ − π− π

∴=δ− π−π ∴=−

∫st s

ft t t t

st ttedtse

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26. (a) f(t) = 5 u(t) – 16

δ

(t) + e-4.4t u(t)

(b) f(t) =

δ

(t) + u(t) + t u(t)

(c) F(s) = 1

6

88

7

5

+

++

+ssss

ba

where a = 3.4-

1

17

6-

=

+=

s

s and b = 3.4

6

17

1-

=

+=

s

s.

Thus, f(t) = 5 e-7t u(t) + 88 u(t) –3.4 e-6t u(t) + 3.4 e-t u(t)

Check with MATLAB:

EDU» T1 = '5/(s+7)';

EDU» T2 = '88/s';

EDU» T3 = '17/(s^2 + 7*s + 6)';

EDU» T = symadd(T1,T2);

EDU» P = symadd(T,T3);

EDU» p = ilaplace(P)

p =

5*exp(-7*t)+88-17/5*exp(-6*t)+17/5*exp(-t)

EDU» pretty(p)

5 exp(-7 t) + 88 - 17/5 exp(-6 t) + 17/5 exp(-t)

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27. If V(s) = s

5, then v(t) = 5 u(t) V. The voltage at t = 1 ms is then simply 5 V, and the

current through the 2-kΩ resistor at that instant in time is 2.5 mA.

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28. I(s) = pA,

10

5

+s so i(t) = 5 e-10t u(t) pA. The voltage across the 100-MΩ resistor is

therefore 500 e-10t u(t) µV.

(a) The voltage as specified has zero value for t < 0, and a peak value of 500 µV.

(b) i(0.1 s) = 1.839 pA, so the power absorbed by the resistor at that instant = i2R

= 338.2 aW. (A pretty small number).

(c) 500 e-10t1% = 5

Taking the natural log of both sides, we find t1% = 460.5 ms

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29.

(a) 12 12

F() 1 () () 2 ()

11

−

+

=+=++↔δ++

++

t

s

stuteut

ss ss

(b) 22

F( ) ( 1) 2 1 ( 2) 2 ( 1) ( )

−−−

=+=++↔δ−+δ−+δ

sss

se e e t t t

(c) (1) 1 2 1

F( ) 2 2 2 ( 1)

−+ − − −

== +δ−

ss

se ee et

(d) F(s) = 2e-3s cosh 2s = e-3s (e2s + e-2s) = e-s + e-5s ↔

δ

(t – 1) +

δ

(t – 5)

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30. N(s) = 5s.

(a) D(s) = s2 – 9 so

()()()()

3

b

3

a

3- 3

5

9-

5

)(

2−

+

=

+

== ssss

s

sD

sN

where

()

2.5

6-

15-

3

5

a

3-

==

−

=

=s

s

s and

()

2.5

6

15

3

5

b

3

==

+

=

=s

s

s. Thus,

f(t) = [2.5 e-3t + 2.5 e3t] u(t)

(b) D(s) = (s + 3)(s2 + 19s + 90) = (s + 3)(s + 10)(s + 9) so

()()()

9

c

10

b

3

a

)9)(10)(3(

5

)(

+

=

+++

=ssssss

s

sD

sN

()()

0.3571,-

(7)(6)

15-

901

5

a

3-

==

++

=

=s

ss

s

()()

7.143-

(-7)(-1)

50-

93

5

b

10-

==

++

=

=s

ss

s

()( )

7.5.

(-6)(1)

45-

103

5

c

9-

==

++

=

=s

ss

s ∴ f(t) = [-0.3571 e-3t - 7.143 e-10t + 7.5 e-9t] u(t)

(c) D(s) = (4s + 12)(8s2 + 6s + 1) = 32(s + 3)(s + 0.5)(s + 0.25) so

()()()

25.0

c

5.0

b

3

a

)25.0)(5.0)(3(32

5

)(

+

=

+++













=ssssss

s

sD

sN

()( )

0.06818,

25.05.032

5

a

3-

−=

++













=

=s

ss

s

()( )

0.125

25.0332

5

b

0.5-

=

++













=

=s

ss

s

()( )

0.05682-

5.0332

5

c

-0.25-

=

++













=

=s

ss

s

∴ f(t) = [-0.06818 e-3t + 0.125 e-0.5t – 0.05682e-0.25t] u(t)

(d) Part (a): Part (b): Part (c):

EDU» N = [5 0];

EDU» D = [1 0 -9];

EDU» [r p y] = residue(N,D)

r =

2.5000

p =

3

-3

y =

[]

EDU» N = [5 0];

EDU» D = [1 22 147 270];

EDU» [r p y] = residue(N,D)

r =

-7.1429

7.5000

-0.3571

p =

-10.0000

-9.0000

-3.0000

y =

[]

EDU» N = [5 0];

EDU» D = [32 120 76 12];

EDU» [r p y] = residue(N,D)

r =

-0.0682

0.1250

-0.0568

p =

-3.0000

-0.5000

-0.2500

y =

[]

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31.

(a) 5

F( ) 5 ( )

1

−

=↔

+

t

seut

s

(b) 4

52

F( ) (5 2 ) ( )

14

−−

=− ↔ −

++

tt

seeut

ss

(c) 4

18 6 6

F( ) 6( ) ( )

(1)(4) 1 4

−−

==−↔−

++ + +

tt

seeut

ss s s

(d) 4

18 6 24

F( ) 6(4 ) ( )

(1)(4) 1 4

−−

−

==+↔−

++ + +

tt

s

seeut

ss s s

(e) 24

18 6 96

F( ) 18 18 ( ) 6( 16 ) ( )

(1)(4) 1 4

−−

==+−↔δ+−

++ + +

tt

s

steeut

ss s s

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32. N(s) = 2s2.

(a) D(s) = s2 – 1 so

()()()()

1

b

1

a

1- 1

2

1-

2

)(

)( 2

2

−

+

=

+

== ssss

s

sD

sN + 2

where

()

1-

2-

2

1

2

a

1-

2==

−

=

=s

s

s and

()

1

2

1

2

b

1

2==

+

=

=s

s

s. Thus,

f(t) = [2

δ

(t) + e-t + et] u(t)

(b) D(s) = (s + 3)(s2 + 19s + 90) = (s + 3)(s + 10)(s + 9) so

()()()

9

c

10

b

3

a

)9)(10)(3(

2

)(

)( 2

+

=

+++

=ssssss

s

sD

sN

()()

0.4286,

(7)(6)

18

901

2

a

3-

2==

++

=

=s

ss

s

()()

28.57

(-7)(-1)

200

93

2

b

10-

2==

++

=

=s

ss

s

()( )

27.-

(-6)(1)

162

103

2

c

9-

2==

++

=

=s

ss

s ∴ f(t) = [0.4286 e-3t + 28.57 e-10t - 27 e-9t] u(t)

(c) D(s) = (8s + 12)(16s2 + 12s + 2) = 128(s + 1.5)(s + 0.5)(s + 0.25) so

()()()

25.0

c

5.0

b

5.1

a

)25.0)(5.0)(5.1(128

2

)(

)( 2

+

=

+++













=ssssss

s

sD

sN

()( )

.02813,0

25.05.0128

2

a

1.5-

2

=

++













=

=s

ss

s

()( )

0.01563-

25.05.1128

2

b

0.5-

2

=

++













=

=s

ss

s

()()

0.003125

5.05.1128

2

c

0.25-

2=

++













=

=s

ss

s

∴ f(t) = 0.02813 e-1.5t – 0.01563 e-0.5t + 0.003125e-0.25t] u(t)

(d) Part (a): Part (b): Part (c):

EDU» N = [2 0 0];

EDU» D = [1 0 -1];

EDU» [r p y] = residue(N,D)

r =

-1.0000

1.0000

p =

-1.0000

1.0000

y =

2

EDU» N = [2 0 0];

EDU» D = [1 22 147 270];

EDU» [r p y] = residue(N,D)

r =

28.5714

-27.0000

0.4286

p =

-10.0000

-9.0000

-3.0000

y =

[]

EDU» N = [2 0 0];

EDU» D = [128 288 160 24];

EDU» [r p y] = residue(N,D)

r =

0.0281

-0.0156

0.0031

p =

-1.5000

-0.5000

-0.2500

y =

[]

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

(a) 23

F( ) 1

=−

+

sss so f(t) = 2 u(t) – 3 e-t u(t)

(b) 3

210 4

F( ) 2 2 ( ) 4 ( )

33

−

+

==+↔δ+

++

t

s

steut

ss

(c) 0.8

F( ) 3 3 ( 0.8)

−

=↔δ−

s

se t

(d) 26

12 3 3

F( ) 3( ) ( )

(2)(6) 2 6

−−

==−↔−

++ + +

tt

seeut

ss s s

(e)

22

226

2

12 3 A 0.75

F( ) (2)(6)(2) 2 6

12 3 A 0.75

Let 0 A 0.75

46 4 2 6

3 0.75 0.75

F( ) (3 0.75 0.75 ) ( )

(2) 2 6

−−−

==++

++ + ++

=∴ =+ + ∴ =−

×

∴= − + ↔ − +

+++

ttt

sss s ss

s

steeeut

sss

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. F(s) = 2 5 4

1

- 2 23 +++

+ssss

π

=

)954.71)(954.71)(2(

1

- 2 jj ++−++

+ssss

π

= )954.71(

*

)954.71(

)2(

1

- 2 j

b

j

ba

++

+

−+

+

+ssss

where 0.04888

)954.71)(954.71(

2-

=

++−+

=

s

ss jj

a

π

003073.0 0.02444-

)954.71)(2(

954.71-

j

b

j

+=

+++

=

+=s

ss

π

and hence b* = -0.02444 – j0.003073

Thus, we may write

f(t) = 2

δ

(t) – u(t) + 0.04888 e-2t u(t) + [(-0.02444 + j0.003073) e(-1 + j7.954)t

+ (-0.02444 + j0.003073) e(-1 + j7.954)t ] u(t)

This may be further simplified by expressing (-0.02444 + j0.003073) e(-1 + j7.954)t

as 0.02463 e j172.83o e(-1 + j7.954)t . This term, plus its complex conjugate above, add to the

purely real expression 0.02463 e-t cos (7.954t + 172.8o).

Thus, f(t) = 2

δ

(t) – u(t) + 0.04888 e-2t u(t) + 0.02463 e-t cos (7.954t + 172.8o).

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35. (a) F(s) = )3(

)3(

)2)(1(

+

+=

+

++

ssss

ss ba

3

2

)3(

)2)(1(

0

=

+

++

=

s

ss

a and 3

2

-

3-

(-2)(-1)

)2)(1(

3-

==

++

=

=s

s

ss

b

so

f(t) =

()

)( 1

3

2

)(

3

2

- )(

3

233 tuetuetu tt −− −=

(b) F(s) = )2(

*

)2(

)4(

)2(

2 2 2 j

c

j

cba

−

+

++=

+

ssssss

s

0.5

4

2

)4(

)2(

0

2 ==

+

=

=s

s

a

25.0

4

)4(

)2(2)4(

)4(

)2(

2

0

2 2

2

0

2 ==













+

+−+

=













+

=

=s

ss

sss

s

sd

d

b

51200.125

)4(4

22

)2(

j2

2 .j

j

c+=

−

=

−

+

=

−=s

ss

s = 0.1768∠45o (c* = 0.1768∠-45o)

so

f(t) = 0.5 t u(t) + 0.25 u(t) + 0.1768 ej45o e-j2t u(t) + 0.1768 e-j45o ej2t u(t)

The last two terms may be combined so that

f(t) = 0.5 t u(t) + 0.25 u(t) + 0.3536 cos (2t - 45o)

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36.

(a) 5[sI(s) – i(0-)] – 7[sI(s) – si(0-) – i'(0-)] + 9I(s) = s

4

(b) m[sP(s) – sp(0-) – p'(0-)] + µf [sP(s) – p(0-)] + kP(s) = 0

(c) [s ∆

∆∆

∆Np(s) – ∆np(0-)] =

()

s

sN L

G

p+

∆

−

τ

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37.

4/3

15 () 4 () 8 () 6 (), (0) 3

15 15 4 15 4

48F()6F()18 F()(6 8)18

22 15 15/8

F( ) ( ) (1.875 5.542 ) ( )

6 ( 4/30) 4/3

−

′

−δ = + =−

−−

∴−= + += ∴ +=+

−+

∴= = ∴= −

++

t

ut t f t f t f

ss

sss ss

sss

s

sfteut

ss s

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38.

(a) -5 u(t – 2) + 10 iL(t) + 5 dt

diL = 0

(b) 0 )](0 - )([ 5 )( 10

5-

LLL

2=++

−−ie ssIsI

s

IL(s) = 105

)(0 5

5-

L

2

+

−

s

sie =

()

2

105

-32

+

×+

−

ss

s

e

(c) IL(s) = 2

105

2

-3

2

+

×

+













+

−

sss

sba

e

where a = 2

1

2

1

0

=

+=s

s, and b = 2

1

-

1

2-

=

=s

s, so that we may write

IL(s) = 2

105

2

1

2

1

-3

2

+

×

+













+

−

sss

s

e

Thus, iL(t) =

()

[]

)(105 )2( )2(

2

12-3-22 t uetuetu tt ×+−−− −−

=

()

[]

)(105 )2( 1

2

12-3-2 2 t uetue tt ×+−− −−

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

(a) (0 ) 50V, (0 ) 50 V

cc

vv

−+

==

(b)

3

0.1 0.2 0.1( 20) 0

2

0.1 0.3 2, 0.1 V 5 0.3V

25 2

V (0.1 0.3) 5

5 2 20/3 130/ 3 20 130

V( ) () () V

(0.1 0.3) 3 3 3

−

′++ −=

′

∴+ = −+ =

+

∴+=+=

+

∴= =+ ∴=+



++



cc c

cc c c

c

t

cc

vv v

vv s s

s

sss

s

svteut

ss s s

(c)

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40.

(a) 5 u(t) -5 u(t – 2) + 10 iL(t) + 5 dt

diL = 0

(b) 0 )](0 - )([ 5 )( 10

5

5-

LLL

2=++− −ie ssIsI

ss

s

IL(s) = 105

)(0 5

5

-

5-

L

2

+

−

s

ss

sie =

()

2

105 5

-32

+

×+−

−

ss

s

e

(c) IL(s) = 2

2

+

++













+

−

ssss

sdcba

e where

a = 2

1

2

1

0

=

+=s

s, b = 2

1

-

1

2-

=

=s

s, c = 5.2

2

5

-

2

5105

0

3−==

+

−×

=

−

s

s, and

d = 505.2

2-

5-1010-

5105 3

2-

3=

×

=

−× −

=

−

s

s,

so that we may write

IL(s) = ssss

s2.505

2

5.2

2

1

2

1

2+

+

−













+

−

e

Thus, iL(t) =

()

[]

)( 2.505)(2.5 )2( )2(

2

12-22 tut uetuetu tt +−−−− −−

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

22 2 2

22

0.15

2

12

12 ( ) 20 ( ) 3 (0 ) 2 20 F 20(2) 3F

12 12 40 2 0.6

40 (20 3) F F ( ) ( 0.15)

42

F( ) (4 2 ) ()

0.15

−

′

=+=∴=−+

++

∴+= + = ∴ = +

∴=− ↔−

+

t

ut f t f s

s

ss

ssss

seut

ss

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42. (a) f(t) = 2 u(t) - 4

δ

(t)

(b) f(t) =

(

)

t 99 cos

(c) F(s) = 5 -

2

3

5 -

65

1

2−

+

−

=

++ ssss

ba

where 1

2

1

3

=

−

=

=s

s

a and 1-

3

1

2

=

−

=

=s

s

b

Thus,

f(t) = e-3t u(t) – e-2t u(t) – 5

δ

(t)

(d) f(t) =

δ

'(t) (a “doublet”)

(e) f(t) =

δ

'"(t)

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43.

2

22

222

2

2 ( ), 2 3 8 ( ), (0 ) 5, (0 ) 8

281225Y

X5Y ,Y82X3Y X 5Y

4102Y 8 2 418

Y3Y 8 Y 3 8

32 4188 8 1842 6

Y,Y()

( 1)( 2) 1

−−

′′

+= − + = = =



−+ = −− + = ∴ = +− = + −







∴+−−+ =+∴ ++=++







++ ++ ++

=+=++

 ++ +



xy uty x y utx y

ss

sssssss

ss

sss s sss

ss ss s s

s

ss sssss

0

2

11

() (2 6 ) (); () [ 3 8 ()] 1.5 4 ()

22

1

() [ 6 ()] 1.5[2 6 ] () 4 ()

2

() 6 () () (6 1) ()

−

−−

+

′′

∴=+ = +− =+ −

∴=− + + −

∴= −= −

t

tt

s

yt e ut xt y y ut y y ut

xt e ut e ut ut

xt e ut ut e ut

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44. (a) F(s) = s

s8

88 ++ , with f(0-) = 0. Thus, we may write:

f(t) = 8

δ

(t) + 8 u(t) + 8

δ

' (t)

(b) F(s) = 2 -

)2(

2+

+s

s

s.

f(t) =

δ

' (t) - 2

δ

(t) + 4e-2t u(t) -

δ

' (t) + 2

δ

(t) = 4e-2t u(t)

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45.

(a) 40 100

(0 ) 0, (0) 100 V, (0 ) 0.6 A

100

−+

−

==∴= =−

cc c

iv i

(b) 0

40 100 50 100

−

∞

=+ +

∫

cc

iidt

(c)

0.5

60 50

100 I ( ) I ( )

6 10 5 6 0.6

I , I ( ) ( ) 0.6 ( )

10 5 0.5

−

−= +

+−

∴= = = ↔ =

++

cc

t

cc c

ss

ssiteut

ss ss

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46. (a) 4 cos 100t ↔ 22 100

4

+s

s

(b) 2 sin 103t – 3 cos 100t ↔ 62

3

10

102

+

×

s - 22 100

3

+s

(c) 14 cos 8t - 2 sin 8o ↔ 64

14

2+s

s - s

o

8sin2

(d)

δ

(t) + [sin 6t ]u(t) ↔ 1 + 36

6

2+s

(e) cos 5t sin 3t = ½ sin 8t + ½ sin (-2t) = ½ (sin 8t – sin 2t) ↔ 64

4

2+s - 4

1

2+s

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47. 5

0

100 ( ) A; 4 3 −

−′

==++

∫t

t

ss

ieutivvvdt

(a) 0

11 1

C;R,C1F,L H

RL 4 3

−

′

=+ + =Ω = =

∫t

s

v

ivvdt

(b)

2

35

100 3

V( ) 4V( ) V( )

5

3 4 3 100 100

V( ) 4 V( ) , V( )

45(1)(3)(5)

12.5 75 62.5

V( ) , ( ) (75 12.5 62.5 ) ( ) V

135

−−−

=++

+

++



++ = = =

 ++++



−

∴=+− =−−

+++

tt t

ss s s

ss

ss s

ss s s

ss sss

svteeeut

sss

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48.

(a) V(s) = ss

s2

7−

+eV

(b) V(s) = V

1

2

+

−

s

e

(c) V(s) = 48e-s V

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49.

0

2

4 ( ) 10 4 [ 0.5 ( )] 0

410 10 424

II4I2, I5 2

2 4 1.6

I 0.4

510 2

( ) 0.4 ( ) 1.6 ( ) A

−

∞

−

++ + − δ =

−



∴+ + + = + =−+





−

∴= = −

++

∴+δ−

∫

ccc

cccc

c

t

c

ut i idt i t

s

ss s ss

s

ss

it t e ut

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50.

0

22

2

22

22 2

23(

2

6 9 ( ) 24( 2) ( 2), (0) 0

9169(3)

V( ) 0 6 V( ) V( ) 24 V( ) V( )

1 1/9 1/9 1/3

V( ) 24 24

(3) 3(3)

8/3 8 8 8

V( ) [ ( 2)

3( 3) 3

−

−−

′′

++ = − − =

++ +

∴−+ + = = =



∴= = −−



+++





∴= −− ↔ −−



++



∫t

s

ss

s

vv vzdz t ut v

ss s

ss s s e s s

sss s

s

se e

ss s s s

se ut e

ss s

2)

3( 2) 3( 2) 3( 2)

(2)]

88

8(2) (2) () 8(2) (2)

33

−

−− −− −−

−



−− −∴ =− −− −





t

ttt

ut

te ut vt e te ut

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



51.

(a)

(b)

(c)

22

33

2

3

0

5 ( 1) 5 ( 1)

F( ) (0 ) 5

(1) 1

5( 1)

( ) , but 1 pole in RHP indeterminate

1

+

→∞

→

++

=∴= =

++

+

∞= ∴

+

l

im

s

im

s

sss

sf

ss

fs

22

34

5( 1) 5 ( 1)

F( ) (0 ) 0

16 16

( ) is indeterminate since poles on axis

+

→∞

++

=∴= =

++

∞ω

lim

s

sss

sf

ss

fj

44

22

(1)(1 ) (1)(1 )

F( ) (0 ) 1

22

( ) is indeterminate since poles on axis

−−

+

→∞

++ ++

=∴= =

++

∞ω

l

ss

im

s

se sse

sf

ss

fj

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



52. (a) f(0+) =

[]

2

2 5

6 2

lim )( lim 2

2

=













++

+

=∞→∞→ ss

s

sFs ss

f(∞) =

[]

3

2

6

2 5

6 2

lim )( lim 2

2

0 0 ==













++

+

=→→ ss

s

sFs ss

(b) f(0+) =

[]

0

3

2

lim )( lim =













+

=−

∞→∞→ s

s

sFs s

ss

e

f(∞) =

[]

0

3

2

lim )( lim 0 0 =













+

=−

→→ s

s

sFs s

ss

e

(c) f(0+) =

[]

()

∞=













+

=∞→∞→

5

1

lim )( lim 2

2

s

ss

sFs ss

f(∞) : This function has poles on the j

ω

axis, so we may not apply the final value

theorem to determine f(∞).

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



53.

(a) 22

33

5( 1) 5 ( 1)

F( ) (0 ) 5

(1) (1)

+

→∞

++

=∴= =

++

lim

s

sss

sf

ss

(b)

(c)

22

33

2

3

0

5( 1) 5( 1)

F( ) (0 ) 0

(1) (1)

5( 1)

( ) 5 (pole OK)

(1)

+

→∞

→

++

=∴= =

++

+

∞= =

+

l

im

s

im

s

ss

sf

ss s

s

fs

33

2

32

00

(1 ) 1

F( ) (0 ) 0

111

( ) 1 1 3 9 ... 3 (no poles)

22

−−

+

−

→∞

→→

−−

=∴= =

−

∞= = −+ − × + =





l

ll

ss

s

im

s

im im

ss

ee

sf

ss

e

fss

s

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



54.

1

() ( ) ()

at bt

ft e e ut

t

−

=−

(a)

(b)

2

00

1 1 ... 1

()()

1/

1/( ) 1/( ) ( )

Use Hopital. F( ) 1/ ( )( )

−−

++

→→

→∞ →∞

−+−+

−= =−

++−+

=

+



+− + −

′∴= =− =−



−++



ll

l

at bt

im im

tt

im im

ss

im im

ss

at bt

eeut ba

tt

sb nsb nsa

sn

sa s

sb sa ab

ss s ba

ssbsa

111

Now, () F() () , ()

111

(()

∞−−

∞∞

∞

−−

↔∴↔−↔−

++

+++



∴− ↔ − = = =



++ + + +



∫

∫lll

at bt

s

at bt

sss

ft sds e ut e ut

tsasb

sa sa sb

e e u t ds n n n

tsasbsbsbsa

CHAPTER FOURTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



55.

(a)

(b)

(c)

(d)

2

22

2

22

0

82

F( ) (0 ) F( ) 0

( 2) ( 1)( 6 10)

(8 2) 10 (pole OK)

( 2) ( 1)( 6 10)

+

→∞

→

−

=∴==

++++

−

∞= =

++++

l

im

s

im

s

sfs

ssss

ss

fssss

22

2

0

82 (82)

F( ) (0 ) 8

610 610

(8 2) 6 36 40

( ) 0 poles: , LHP, OK

610 2

+

→∞

→

−−

=∴= =

++ ++



−−±−

∞= = = ∴



++



l

im

s

im

s

sss

sf

ss ss

ss

fs

ss

32 32

32 2

32

2

0

235 235

F( ) (0 )

610 610

235

( ) 0.5 (poles OK)

610

+

→∞

→

−−− −−−

=∴= =∞

++ ++

−−−

∞= =−

++

l

im

s

im

s

ss s ss s

sf

ss s ss

ss s

fss

22

2

0

82 (82)

F( ) (0 ) 8

610 610

(8 2) 6 36 40

( ) , RHP indeterminate

610 2

+

→∞

→

−−

=∴= =

−+ −+

−±−

∞= = ∴

−+

l

im

s

im

s

sss

sf

ss ss

ss

fs

ss

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1. Note that iL(0+) = 12 mA. We have two choices for inductor model:

= 0.032s Ω

= 384 µV

0.032s Ω 12

smA

CHAPTER FIFTEEN SOLUTIONS

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

2. iL(0-) = 0, vC(0+) = 7.2 V (‘+’ reference on left). There are two possible circuits, since

the inductor is modeled simply as an impedance:

73 Ω 1

0.002s Ω

0.03s Ω

7.2 V

s+

V(s)

-

73 Ω

1

0.002s Ω

0.03s Ω

14.4 mA +

V(s)

-

CHAPTER FIFTEEN SOLUTIONS

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

3.

(a)

(b) Zin(-80) = - 10.95 Ω

(c) 128,000 880,000 200,000

( 80) 8.095 54.43

6400 56,000 100,000

in

j

jj

−+ +

==∠°Ω

−+ +

Z

(d) 1 10 200

20 20

RL

+

=+=

s

Yss

(e) 1 500

0.001

2 1000

RC

+

=+ =

s

Ys

(f)

22

2 2000 / 20 1000

() 20 0.1 2 1000 / 200 500

20 10,000 1000 200,000 20 11, 000 200,000

700 100,000 700 100,000

m=+ =+

++ ++

+++ ++

==

++ ++

sss

Zs ssss

sss ss

ss ss

2

200 0.5 0.001 200 10 0.02

20

(200) 0.001 0.7 100

(0.001 0.5)

20

20 11,000 200,000

( )

700 100,000

RL RC

s

+++

++++

==

+++

+

++

==

++

ss

YY sss

s

YY ss

s

ss Zs

ss

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4.

Zin = 33

11

20 || 40

210 210

−−



++



××



ss

= (20 + 500s-1) || (40 + 500s-1)

=

80 3000 +25000

6 100

+

2

ss

3

1

210

−

×s3

1

210

−

×s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5. (a)

2

50 16(0.2 ) 50 16 16 50 4000

16 0.2 80 80

in

++

=+ =+ =

++

+

ssss

Zssss ss

(b) 1024 4000 400

( 8) 0.15842 4.666

64 640

in

j

jj

j

−+ +

==−Ω

−+

Z

(c) 16(4 36 24) 100 300 4000

( 2 6) 6.850 114.3

32 24 160 480

in

jj

jjj

−− − + +

−+ = = ∠− °Ω

−− − +

Z

(d)

(e)

R1

=Ω

2

0.2 R

50 0.2R 10 50R ,

R 0.2 0.2 R

5R 50 50

( 5) 55R 50, R 0.9091

55R

in

R

++

=+ =

++

−+

−= ∴ = = Ω

−

sss

Zssss

Z

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6. 2 mF → 3

1

210

−

×sΩ, 1 mH → 0.001s Ω,

Zin = (55 + 500/ s) || (100 + s/ 1000) =

267

25 5

500

55 100 + 55 + 5.5005 10 5 10

1000 =

500 + 5 10 + 1.55 10

155 1000

 

+

  ×+×

  ××

++

s

ss

ssss

s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7. We convert the circuit to the s-domain:

Defining Zπ = RB || rπ || (1/sCπ) = B

BB

rR

r + R + r R C

π

πππ

s and

ZL = RC || RL = RCRL/ (RC + RL), we next connect a 1-A source to the input and write two

nodal equations:

1 = Vπ/ Zπ + (Vπ – VL)Cµ s [1]

-g

mVπ = VL/ ZL + (VL – Vπ)Cµ s [2]

Solving,

Vπ =

()

BL

2

LB mLB B B L LB B

r R 1 + C

r R C C + (g r R C + r R C + r R C + r C + R C ) + r + R

πµ

ππµ πµπππµ πµ µ π

Zs

ZsZ ZZs

Since we used a 1-A ‘test’ source, this is the input impedance. Setting both capacitors to

zero results in rπ || RB as expected.

Vπ

1/sCπ Ω

1/sCµ Ω

Vπ

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

V(s) =

6

2460 10

4700 4700 0.115

−

+×

+

ss = 2.162 9400

+

0.115 4700 (0.115 + 4700)+

sss

=

18.8 81740

+

40870 ( + 40870)+sss

= 18.8 a b

+ +

40870 + 40870+sss

where a =

= 0

81740 = 2

40870+s

s and b =

= -40870

81740 = -2

s

Thus, V(s) = 18.8 2 2

+ -

40870 + 40870+sss

. Taking the inverse transform of each term,

v(t) = [16.8 e-40870t + 2] u(t) V

V

I

2

s

460 µV

0.115s Ω

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9. v(0-) = 4 V

I(s) =

-6

6

94

5 4.545 10

= =

303030 +0.2755

1.1 10 303030

1.1 10

−×

×+

+×

ss s

s

Taking the inverse transform, we find that i(t) = 4.545 e-0.2755t u(t) µA

9

s

303030

s Ω

4 V

s

I(s)

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10. From the information provided, we assume no initial energy stored in the inductor.

(a) Replace the 100 mH inductor with a 0.1s-Ω impedance, and the current source with a

6

25 10 A

−

×

ssource.

(b) V(s) =

6-6-5

25 10 2 (0.1 ) 5 10 5 10

= = V

2 + 0.1 0.1 2 20

−

×××



 ++



s

ssss

Taking the inverse transform,

v(t) = 50 e-20t mV

The power absorbed in the resistor R is then p(t) = 0.5 v2(t) = 1.25 e-40t nW

V

25

s

0.1s Ω

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11. v(t) = 10e-2t cos (10t + 30o) V

cos (10t + 30o) ⇔

oo

22

cos30 -10sin30 0.866 - 5

=

+100 + 100

ss

Λ {f(t)e-at} ⇔ F(s + a), so

V(s) =

()

2

0.866 2 - 5 8.66 - 16.34

10 = + 100

2 + 100

+

ss

s

The voltage across the 5-Ω resistor may be found by simple voltage division. We first

note that Zeff = (10/s) || 5 = 50

510+sΩ. Thus,

V5Ω =

()()

s

ss

2

50 50 50

510 = =

50 0.5 + 5 5 + 10 + 50 2.5 30 100

0.5 + 5 + 510





+

 ++





+



VVV

s

sss

ss

s

(a) Ix = eff

5

V =

() () ()

222

2

0.866 - 3.268 34.64 - 130.7

40 =

2 100 12 + 40 2 100 6 + 100

 



++ + ++ +



 

ss

ssss s

(b) Taking the inverse transform using MATLAB, we find that

ix(t) = e-6t [0.0915cos 2t - 1.5245 sin 2t] - e-2t [0.0915 cos10t - 0.3415 sin 10t] A

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12.

Node 1: 0 = 0.2 (V1 – 3/ s) + 0.2 V1 s + 0.5 (V1 – V2) s

Node 2: 0 = 0.5 (V2 – V1) s + 0.125 V2 s + 0.1 (V2 + 5/ s)

Rewriting, (3.5 s 2 + s) V1 + 2.5 s 2 V2 = 3 [1]

-4 s 2 V1 + (4 s 2 + 0.8 s + 1) V2 = -4 [2]

Solving using MATLAB or substitution, we find that

()( )()

2

1432

2

20 + 16 + 20

( ) =

40 + 68 + 43 + 10

1 20 + 16 + 20

=

40 + 0.5457 - 0.3361 + 0.5457 + 0.3361 + 0.6086jj

−







ss

Vs ssss

ss

ss s s

which can be expanded:

*

1( ) = + + +

+ 0.5457 - 0.3361 + 0.5457 + 0.3361 + 0.6086

ab b c

jj

Vs ss s s

Using the method of residues, we find that

a = 2, b = 2.511 ∠101.5o, b* = 2.511∠-101.5o and c = -1.003.

Thus,taking the inverse transform,

v1(t) = [2 – 1.003 e-0.6086t + 5.022 e-0.5457t cos (0.3361t – 101.5o)] u(t) V

VC1

V1 V2

3

s 5 V

s

5 Ω

s

2 Ω

s

8s Ω

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13. With zero initial energy, we may draw the following circuit:

Define three clockwise mesh currents I1, I2, and I3 in the left, centre and right meshes,

respectively.

Mesh 1: -3/s + 5I1 + (5/s)I1 – (5/s)I2 = 0

Mesh 2: -(5/s)I1 + (8s + 7/s)I2 – 8s I3 = 0

Mesh 3: -8sI2 + (8s + 10)I3 – 5/s = 0

Rewriting,

(5s + 5) I1 – 5 I2 = 3 [1]

-5 I1 + (8s2 +7) I2 – 8s2 I3 = 0 [2]

- 8s2 I2 + (8s2 + 10s) I3 = 5 [3]

Solving, we find that

I2(s) =

2

32

20 + 32 + 15

40 + 68 + 43 + 10

ss

sss=

()( )( )

2

1 20 + 32 + 15

40 + 0.6086 + 0.5457 - 0.3361 + 0.5457 + 0.3361

jj







ss

ss s

=

()()()

*

+ 0.6086 + 0.5457 - 0.3361 + 0.5457 + 0.3361

ab b

jj

++

ss s

where a = 0.6269, b = 0.3953∠-99.25o, and b* = 0.3955∠+99.25o

Taking the inverse tranform, we find that

i2(t) = [0.6271e-0.6086t + 0.3953e-j99.25o e(-0.5457 + j0.3361)t + 0.3953ej99.25o e(-0.5457 - j0.3361)t ]u(t)

= [0.6271e-0.6086t + 0.7906 e-0.5457t cos(0.3361t + 99.25o)] u(t)

8 s Ω

5 s-1 Ω

2 s-1 Ω

3 V

s

5 V

s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14. We choose to represent the initial energy stored in the capacitor with a current source:

Node 1:

1

112

3

-

1.8 = + + ( - )

552

Vss

sVVV

Node 2:

2

21 2

5

+

1

0 = ( - ) + +

2810

V

ss

VV V

s

Rewriting, (5s2 + 4s) V1 – 5s2 V2 = 18s + 6 [1]

-4s2 V1 + (4s2 + 0.8s + 1)V2 = -4 [2]

Solving, we find that V1(s) =

32

360 + 92 + 114 + 30

(40 + 68 + 43 + 10)

ss s

ss s s

=

*

+ + +

+ 0.6086 + 0.5457 - 0.3361 + 0.5457 + 0.3361

ab c c

jjss s s

where a = 3, b = 30.37, c = 16.84 ∠136.3o and c* = 16.84 ∠-136.3o

Taking the inverse transform, we find that

v

1(t) = [3 + 30.37e-0.6086t + 16.84 ej136.3o e-0.5457t ej0.3361t

+ 16.84 e-j136.3o e-0.5457t e-j0.3361t ]u(t) V

= [3 + 30.37e-0.6086t + 33.68e-0.5457t cos (0.3361t + 136.3o]u(t) V

1.8 A

↑

5Ω

s

2Ω

s

8s Ω

3 V

s

5 V

s

V1 V2

•

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15. We begin by assuming no initial energy in the circuit and transforming to the s-domain:

(a) via nodal analysis, we write:

()

1

1x

2

20 + 60 = - +

( + 3) + 16 10 5

Vss

VV

s [1] and

()

x

x1

2

120 = +

( + 3) + 16 2 10 −

Vs

VV

ss [2]

Collecting terms and solving for Vx(s), we find that

Vx(s)

(b) Using the method of residues, this function may be rewritten as

with a = 92.57 ∠ -47.58o, a* = 92.57 ∠ 47.58o, b = 43.14 ∠106.8o, b* = 43.14 ∠-106.8o

Taking the inverse transform, then, yields

vx(t) = [92.57 e-j47.58o e-3t ej4t + 92.57 ej47.58o e-3t e-j4t

+ 43.14ej106.8o e-1.25t ej1.854t + 43.14e-j106.8o e-1.25t e-j1.854t] u(t)

= [185.1 e-3t cos (4t - 47.58o) + 86.28 e-1.25t cos (1.854t + 106.8o)] u(t)

2

+ 3

20 A

( + 3) + 16

s

s 2s Ω Vx

10 Ω

s

V1

2

4

30 A

( + 3) + 16s

()( )( )( )

2

432

2

200 ( + 9 + 12)

2 + 17 + 90 + 185 + 250

200 ( + 9 + 12)

= + 3 - 4 + 3 + 4 + 1.25 - 1.854 + 1.25 + 1.854jj j j

=ss s

sss s

ss s

ss s s

()()()()

**

+ 3 - 4 + 3 + 4 + 1.25 - 1.854 + 1.25 + 1.854

aa b b

jj j j

++ +

ss s s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16. We model the initial energy in the capacitor as a 75-µA independent current source:

First, define Zeff = 106/s || 0.005s || 20 = -6 2

10 + 0.005 + 200 Ω

s

ss

Then, writing a single KCL equation, -6

22

eff

(s) 1 162.6

75 10 = + ( ) -

20 + 4

π



×



Vs

Zs

which may be solved for V(s):

V(s) =

()

25

44382 6 9

75 + 1.084 10 + 39.48

+ 5.5 10 + 2 10 + 2.171 10 + 7.896 10

×

×× × ×

ss s

sss s

=

()

()()()()

25

75 + 1.084 10 + 12.57

+ 51085 + 3915 - 6.283 + 6.283jj

×ss s

ssss

(NOTE: factored with higher-precision denominator coefficients using MATLAB to

obtain accurate complex poles: otherwise, numerical error led to an exponentially

growing pole i.e. real part of the pole was positive)

=

()()()()

*

+ 51085 + 3915 - 2 + 2

abcc

jj

ππ

+++

ssss

where a = -91.13, b = 166.1, c = 0.1277∠89.91o and c* = 0.1277∠-89.91o.

Thus, consolidating the complex exponential terms (the imaginary components cancel),

v(t) = [-91.13e-51085t + 166.1e-3915t + 0.2554 cos (2πt + 89.91o)] u(t) V

(b) The steady-state voltage across the capacitor is V = [255.4 cos(2πt + 89.91o)] mV

This can be written in phasor notation as 0.2554 ∠89.91o V. The impedance across

which this appears is Zeff = [j

ω

C + 1/j

ω

L + 1/20]-1 = 0.03142 ∠89.91o Ω, so

Isource = V/ Zeff = 8.129∠-89.91o A.

Thus, isource = 8.129 cos 2πt A.

(c) By phasor analysis, we can use simple voltage division to find the voltage division to

find the capacitor voltage:

VC(j

ω

) =

()

o

162.6 0 0.03142 89.91 = 0.2554 89.92 V

20 + 0.03142 89.91

∠∠ ∠

∠ which agrees with

our answer to (a), assuming steady state. Dividing by 0.03142 ∠89.91o Ω, we find

isource = 8.129 cos 2πt A.

↑

22

162.6 V

+ 4

π

s

s V

75 µA

0.005s Ω

6

10 Ω

s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17. Only the inductor appears to have initial energy, so we model that with a voltage source:

Mesh 1: 123

2

5.846 + 2.699 1333 1333

= 2 + - - 2

+ 4







sIII

sss

Mesh 2: 0 = 0.005I1 – 0.001 + (0.001s + 1333/s) I2 – (1333/s)I1 – 0.001sI4

Mesh 3: 0 = (2 + 1000/s)I3 – 2I1 – (1000/s)I4 + 2

6

4+

s

Mesh 4: 0 = (0.001s + 1000/s) I4 - 0.001sI2 – (1000/s)I3 + 0.001

Solving, we find that I1 = 2

154 2699

0.2 4

−

−+

s

s and

I2 =

47310213 14

453929

154 - 7.378 10 - 1.912 10 - 4.07 10 + 7.196 10

0.001 2333 + 6.665 10 + 1.333 10 + 5.332 10

××××

×××

ss ss

sss

oo

-5

0.4328 166.6 0.4328 166.6

+

+ 142.8 + 742 + 142.8 - 742

135.9 96.51 135.9 96.51

+ + + 6.6 10

- 2 + 2

jj

∠− ∠+

=

∠− ∠+ ×

ss

Taking the inverse transform of each,

i1(t) = 271.7 cos (2t – 96.51o) A and

i2(t) = 0.8656 e-142.8t cos (742.3t + 166.6o) + 271.8 cos (2t – 96.51o) + 6.6×10-5

δ

(t) A

Verifying via phasor analysis, we again write four mesh equations:

6∠-13o = (2 – j666.7)I1 + j667I2 – 2I3

0 = (0.005 + j666.7)I1 + (j2x10-3 – j666.7)I2 – j2×10-3I4

-6∠0 = -2I1 + (2 – j500)I3 + j500I4

0 = -j2×10-3I2 + j500I3 + (j2×10-3 – j500)I4

Solving, we find I1 = 271.7∠-96.5o A and I2 = 272∠-96.5o A. From the Laplace analysis,

we see that this agrees with our expression for i1(t), and as t → ∞, our expression for i2(t)

→ 272 cos (2t – 96.5o) in agreement with the phasor analysis.

I1

I2

I3

I4

0.001s Ω

1 mV

2

5.846 + 2.699 V

+ 4

s

1333/s Ω 1000/s Ω

2

6V

+ 4

s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18. With no initial energy storage, we simply convert the circuit to the s-domain:

Writing a supermesh equation,

11332

24

1 1 2000 2000

= 100 + + + 0.002 -

610

−

×

IIIsII

ssss

we next note that I2 = -5V2 = -5(0.002s)I3 = -0.01sI3

and I3 – I1 = 3V2 = 0.006sI3, or I1 = (1 – 0.006s)I3, we may write

332

1

=

0.598 + 110 + 3666−

Iss s

V2(s) = I3/ 0.002s = 43 2

1

0.0012 + 0.22 + 7.332−ss s

= 53 3

2

7.645 10 4.167 10 4.091 10 0.1364

+

212.8 28.82

−− −

×× ×

−−+

−+ss ss

Taking the inverse transform,

v2(t) = -7.645×10-5 e212.8t + 4.167×10-3 e-28.82t – 4.091×10-3 + 0.1364 t] u(t) V

(a) v2(1 ms) = -5.58×10-7 V

(b) v2(100 ms) = -1.334×105 V

(c) v2(10 s) = -1.154×10920 V. This is pretty big- best to start running.

0.002s Ω

2000/s Ω

1667/s Ω

s-2

V2

I1

I2

I3

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19. We need to write three mesh equations:

Mesh 1: 13

2

5.846 + 2.699 1333

= 2 + - 2

+ 4







sII

ss

Mesh 3: 0 = (2 + 1000/s)I3 – 2I1 – (1000/s)I4 + 2

6

4+

s

Mesh 4: 0 = (0.001s + 1000/s) I4 – (1000/s)I3 + 10-6

Solving,

I1 =

()

362 8 9

453 26 9

154 - 2.925 10 + 1.527 10 - 2.699 10

0.001 2333 + 6.665 10 + 1.333 109 + 2.666 10 + 5.332 10

×××

−×× ××

ss s

sssss

oo

-5

0.6507 12.54 0.6507 12.54

+

+ 142.8 742.3 + 142.8 + 742.3

0.00101 6.538 0.00101 6.538

+ + 6.601 10

2 2

jj

∠∠−

=−

∠− ∠ −×

−+

ss

which corresponds to

i1(t) = 1.301 e-142.8t cos (742.3t + 12.54o) + 0.00202 cos (2t – 6.538o) – 6.601×10-5 δ(t) A

and

I

3 =

()

()( )

4638212 6

22 5 9

154 + 3.997 10 + 1.547 10 + 3.996 10 - 2.667 10

0.001 + 4 2333 + 6.665 10 + 1.333 10

××××

−××

sss s

ss s

oo

0.7821 33.56 0.7821 33.56

+

+ 142.8 742.3 + 142.8 + 742.3

1.499 179.9 1.499 179.9

+ +

2 2

jj

∠− ∠

=−

∠∠−

−+

ss

which corresponds to

i3(t) = 1.564 e-142.8t cos (742.3t – 33.56o) + 2.998 cos (2t + 179.9o) A

The power absorbed by the 2-Ω resistor, then, is 2

13

2() ()it it



−



or

p(t) = 2[1.301 e-142.8t cos (742.3t + 12.54o) + 0.00202 cos (2t – 6.538o) – 6.601×10-5 δ(t)

- 1.564 e-142.8t cos (742.3t – 33.56o) - 2.998 cos (2t + 179.9o)]2 W

CHAPTER FIFTEEN SOLUTIONS

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20. (a) We first define Zeff = RB || rπ || (1/sCπ) = B

BB

rR

r + R rRC

π

πππ

+s. Writing two nodal

equations, then, we obtain:

0 = (Vπ – VS)/ RS + Vπ (rπ + RB + rπ RBCπs)/rπRB + (Vπ – Vo)Cµ s

and

-gmVπ = Vo(RC + RL)/RCRL + (Vo – Vp) Cµ s

Solving using MATLAB, we find that

2

o

BCLm sBCL sBC sBC

s

sBL sBL BCL sCL

sBCL ms BCL

= rRRR(-g + C ) [RrRRRCC (RrRRC RrRRC

+ R r R R C R r R R C r R R R C R r R R C

R R R R C g R r R R R C )

πµππµπππµ

πππµπ µπµ

µπ µ

++

+++

++

Vss

V

s

1

BCsCsBC BLsLsBL

+ r R R R r R R R R r R R R r R R R R ]

ππ ππ

−

++ +++

(b) Since we have only two energy storage elements in the circuit, the maximum number

of poles would be two. The capacitors cannot be combined (either series or in parallel),

so we expect a second-order denominator polynomial, which is what we found in part (a).

CHAPTER FIFTEEN SOLUTIONS

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21.

(a)

(b) ZTH = (5 + 0.001s) || (500/ s) = 2

2500 + 0.5

0.001 + 5 + 500 Ω

s

ss

VTH = (3/ s)ZTH =

()

6

25

7.5 10 + 1500 V

+ 5000 + 5 10

×

s

ss s

(c) V1Ω =

()

6

TH

2

TH

2

1 7.5 10 + 1500

= 2500 + 0.5

1 + + 505000 1 + 0.001 + 5 + 500

×







s

Vs

Zss ss

oo

6

-3

2.988 10.53 -89.92 10.53 +89.92

+ +

+ 2.505 10 + 710.6 710.6

2.956 2.967 10

+ +

+ 0.1998

jj

∠∠

=− ×−

×

sss

ss

Thus, i1Ω = v1Ω(t) = [-2.988 e-2.505×106t + 2.956 e-0.1998t + 2.967×10-3

+ 21.06 cos(710.6t + 89.92o)] u(t)

3/s

0.001s Ω

500/s Ω

I IC

CHAPTER FIFTEEN SOLUTIONS

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22.

(a)

(b) ZTH = 0, VTH = 20/ s V so IN = ∞

(c) IC =

20

0.04 A

500







=







s

. Taking the inverse transform, we obtain a delta function:

iC(t) = 40δ(t) mA.

This “unphysical” solution arises from the circuit above

attempting to force the voltage across the capacitor to change in zero time.

20/s

0.001s Ω

500/s Ω

I IC

±

CHAPTER FIFTEEN SOLUTIONS

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

23. Beginning with the source on the left (10/s V) we write two nodal equations:

12

11

-6

10 1

- + + = 0

47000 30303 56 + 336 10

′′

−



′′

 ×



VVs

VV

ss

221

2-6

+ + = 0

47000 10870 56 + 336 10

′′′

−

′×

VVVsVs

Solving,

13 11 2

110 3 15 2 18 18

303030(0.3197 10 + 0.1645 10 + 98700 )

= (0.4639 10 + 0.7732 10 + 0.5691 10 + 0.1936 10 )

××

′××××

ss

Vss s s

18

210 3 15 2 18 18

0.9676 10

= (0.4639 10 + 0.7732 10 + 0.5691 10 + 0.1936 10 )

×

′××××

Vss s s

Shorting out the left source and activating the right-hand source (5 – 3/s) V:

12

11 -6

1 + + = 0

47000 30303 56 + 336 10

′′ ′′

−

′′ ′′ ×

VVs

VV s

221

2-6

3

- 5 +

+ + = 0

47000 10870 56 + 336 10

′′ ′′ ′′

−

′′ ×

VVVs

sVs

Solving,

17

110 3 15 2 18 18

0.9676 10 (5 3)

= (0.4639 10 + 0.7732 10 + 0.5691 10 + 0.1936 10 )

×−

′′ ××××

s

Vss s s

312214 14

210 3 15 2 18 18

7609(705000 + 0.1175 10 + 0.6359 10 - 0.3819 10 )

= (0.4639 10 + 0.7732 10 + 0.5691 10 + 0.1936 10 )

×××

′′ ××××

sss

Vss s s

Adding, we find that

13 13 2

110 3 15 2 18 18

30303(0.2239 10 + 0.1613 10 + 98700 )

= (0.4639 10 + 0.7732 10 + 0.5691 10 + 0.1936 10 )

××

××××

ss

Vss s s

3 12 2 14 14

210 3 15 2 18 18

7609(705000 + 0.1175 10 + 0.6359 10 + 0.8897 10 )

= (0.4639 10 + 0.7732 10 + 0.5691 10 + 0.1936 10 )

×××

××××

sss

Vss s s

(b) Using the ilaplace() routine in MATLAB, we take the inverse transform of each:

v1(t) = [3.504 + 0.3805×10-2 e-165928t – 0.8618 e-739t – 2.646 e-0.3404t] u(t) V

v2(t) = [3.496 – 0.1365×10-2 e-165928t + 0.309 e-739t – 2.647 e-0.3404t] u(t) V

CHAPTER FIFTEEN SOLUTIONS

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24. (10/ s)(1/47000) = 2.128×10-4/ s A

(5 – 3/s)/ 47000 = (1.064 – 0.6383/ s)×10-4 A

ZL = 47000 || (30303/ s) =

9

1.424 10

47000 + 30303

×Ω

s

ZR = 47000 || (10870/ s) =

8

5.109 10

47000 + 10870

×Ω

s

Convert these back to voltage sources, one on the left (VL) and one on the right (VR):

VL = (2.128×10-4/ s )

9

1.424 10

47000 + 30303



×





s =

()

5

3.0303 10 V

47000 + 30303

×

ss

VR = (1.064 – 0.6383/ s)×10-4

8

5.109 10

47000 + 10870



×





s

=

()

54360 32611

-

47000 + 10870 47000 + 10870sss

Then, I56Ω = LR

-6

LR

+ + 336 10 +56

−

×

VV

ZZ s

=

()

9 2 10 9

9 3 14 2 17 17

2.555 10 - 1.413 10 - 4.282 10

6250 4.639 10 + 7.732 10 + 5.691 10 + 1.936 10

×××

−××××

ss

ss s s

18

5

-5 18 4

0.208 0.0210 1.533 10

+ 1.659 10 + 739 + 0.6447

2.658 10 2.755 10 1.382 10

+ + 0.3404 + 0.2313

−

−−

×

=−−

×

×× ×

++

sss

ss s

Thus,

i56Ω(t) = [0.208 exp(-1.659×105t) – 0.0210 exp(-739t) – 1.533×10-18 exp(-.06447t)

+ 2.658×10-5 exp(-0.3404t) + 2.755×10-18 exp(-0.2313t) + 1.382×10-4] u(t) A.

The power absorbed in the 56-Ω resistor is simply 56 [i56Ω(t)]2 or

56 [0.208 exp(-1.659×105t) – 0.0210 exp(-739t) – 1.533×10-18 exp(-.06447t)

+ 2.658×10-5 exp(-0.3404t) + 2.755×10-18 exp(-0.2313t) + 1.382×10-4]2 W

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25. (a) Begin by finding ZTH = ZN:

ZTH = 47000 + (30303/ s) || [336×10-6 s + 56 + (10870/ s) || 47000]

=

3 14 2 17 17

3 10 2 13 12

4.639 109 + 7.732 10 + 5.691 10 + 1.936 10

98700 + 1.645 10 + 1.21 10 + 2.059 10

××××

×××

sss

sss Ω

To find the Norton source value, define three clockwise mesh currents I1, I2 and I3 in the

left, centre and right hand meshes, such that IN(s) = -I1(s) and the 10/s source is replaced

by a short circuit.

(47000 + 30303/ s) I1 - (30303/ s) I2 = 0

(10870/ s + 56 + 336×10-6 s + 30303/ s) I2 - (30303/ s) I1 – (10870/ s)I3 = 0

(47000 + 10870/ s) I3 - (10870/ s)I2 = -5 + 3/ s

Solving,

IN = -I1 =

12

9 3 14 2 17 17

2.059 10 (5 - 3)

(4.639 10 + 7.732 10 + 5.691 10 + 1.936 10 )

×

×× ××

s

ss s s

(b) Isource = (10/ s) (1/ ZTH) - IN(s)

13 6 19 5 24 4 27 3

29 2 29 28 3 9 2

12

0.001(0.4579 10 + 0.1526 10 + 0.1283 10 + 0.1792 10

+ 0.6306 10 + 0.3667 10 + 0.5183 10 )[ (4639 + 0.7732 10

+ 0.5691 1012 + 0.1936 10 )(0.4639 10

=× × × ×

××× ×

×××

ssss

ss sss

s10 3 15 2 18

18 -1

+0.7732 10 + 0.5691 10

+ 0.1936 10 )]

××

×

ss s

Taking the inverse transform using the MATLAB ilaplace() routine, we find that

isource(t) = 0.1382×10-3 + 0.8607×10-8 exp(-165930t) + 0.8723×10-7exp(-739t)

+ 0.1063×10-3 exp(-0.3403t) – 0.8096×10-7 exp(-165930t)

+ 0.1820×10-4 exp(-739t) – 0.5×10-4 exp(-0.3404t)

isource(1.5 ms) = 2.0055×10-4 A = 200.6 µA

CHAPTER FIFTEEN SOLUTIONS

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26. We begin by shorting the 7 cos 4t source, and replacing the 5 cos 2t source with 2

5

4+

s

s.

(a) Define four clockwise mesh currents I1, I2, I3 and Ix in the top left, top right, bottom

left and bottom right meshes, respectively. Then,

2

5

4+

s

s = (12 + 1/2s) I3 – 7 I1 – (1/ 2s) Ix [1]

0 = -4 Ix + (9.5 + s) I1 – s I2 – 7 I3 [2]

0 = (3 + s + 2/ s) I2 – s I1 – 3 Ix [3]

0 = (4 + 3s + 1/2s) Ix – 3 I2 – (1/2s) I3 [4]

1

′

V = (I3 – Ix) (2s) [5]

Solving all five equations simultaneously using MATLAB, we find that

1

′

V=

33 2

654 3 2

20 (75 + 199 + 187 + 152)

1212 + 3311 + 7875 + 15780 + 12408 + 10148 + 1200

ss s s

ssssss

Next we short the 5 cos 2t source, and replace the 7 cos 4t source with 2

7

16+

s

s.

Define four clockwise mesh currents I1, I2, I3 and Ix in the bottom left, top left, top right

and bottom right meshes, respectively (note order changed from above). Then,

0 = (12 + 1/2s) I1 – 7 I2 – (1/ 2s) Ix [1]

0 = -4 Ix + (9.5 + s) I2 – s I3 – 7 I1 [2]

2

7

16

−+

s

s = (3 + s + 2/ s) I3 – s I2 – 3 Ix [3]

0 = (4 + 3s + 1/2s) Ix – 3 I3 – (1/2s) I1 [4]

1′′

V = (I1 – Ix) (2s) [5]

Solving all five equations simultaneously using MATLAB, we find that

1′′

V=

()

42

65 4 3 2

-56 (21 - 8 - 111)

1212 + 3311 + 22420 + 55513 + 48730 + 40590 + 4800

sss

ss s s s s

The next step is to form the sum V1(s) = 1

′

V + 1′′

V, which is accomplished in MATLAB

using the function symadd(): V1 = symadd(V1prime, V1doubleprime);

V1(s) =

()( )

3543 2

265432

4 (81 + 1107 + 7313 + 17130 + 21180 + 12160)

+ 4 1212 + 3311 + 22420 + 55513 + 48730 + 40590 + 4800

ss s s s s

sssssss

(b) Using the ilaplace() routine from MATLAB, we find that

v1(t) = [0.2673 δ(t) + 6.903×10-3 cos 2t – 2.403 sin 2t – 0.1167 e-1.971t

– 0.1948 e-0.3315t cos 0.903t + 0.1611 e-0.3115t sin 0.903t – 0.823×10-3 e-0.1376t

+ 3.229 cos 4t + 3.626 sin 4t] u(t) V

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 

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27. (a) We can combine the two sinusoidal sources in the time domain as they have the same

frequency. Thus, there is really no need to invoke source transformation as such to find

the current.

65 cos 103t ⇔ 26

65

+ 10

s

s, and 13 mH → 0.013s Ω

We may therefore write

I(s) =

()

26 26

65 1 5000

=

+ 10 83 + 0.013 + 10 + 6385

 

 

 

ss

=

()

() ()

oo

33

0.7643 0.3869 8.907 0.3869 8.907

+ 6385 - 10 + 10jj

∠− ∠

−+ +

sss

(b) Taking the inverse transform,

i(t) = [-0.7643 e-6385t + 0.7738 cos (103t – 8.907o)] u(t) A

(c) The steady-state value of i(t) is simply 0.7738 cos (103t – 8.907o) A.

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28.

(a)

(b)

(c) -1

Poles: same; = -0.1615, -1.239 ss

(d) 1

Zeros: same; 0.7 0.7141 sj−

=− ±s

2 1

1.4 1.96 0.8

Poles: 1.4 0.2 0, 0.1615, 1.239s

2

1.4 1.96 4

Zeros: 1.4 1 0, 0.7 0.7141s

2j

−

−± −

++== =− −

−± −

++== =−±

ss s

2

22

2

5

5(25)

(5 5)(2 5 ) 25 35 10

575/ 575 575

575

() 25 35 10

in



++

 ++ ++



===

++ ++ ++

++

∴=

++

sssss

s

Zss

ss ss

ss

Ys ss

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29. (a) Regarding the circuit of Fig. 15.45, we replace each 2-mF capacitor with a 500/ s Ω

impedance. Then,

Zin(s) =

500 500

20 + 40 + ( + 25)( + 12.5)

= 13.33

100 ( + 1.667)

60 +





 ss

ss

s

Reading from the transfer function, we have

zeros at s = -25 and -12.5 s-1, and

poles at s = 0 and s = -1.667 s-1.

(b) Regarding the circuit of Fig. 15.47, we replace the 2-mF capacitor with a 500/ s Ω

impedance and the 1-mH inductor with a 0.001s-Ω impedance. Then,

Zin(s) =

()

5

500 500

55 + 100 + 0.001 ( + )( + 10 )

55

= 55

500 + 1.55 10 ( + 3.226)

155 + + 0.001







×

sss

s

ss

s

Reading from the transfer function, we have

zeros at s = -9.091 and -105 s-1, and

poles at s = -1.55×105 and s = -3.226 s-1.

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 

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30. 1

( ) : zeros at 0; 10; at 5, 20 s ; ( ) 12 S as poles −

=− =−− → →∞Ys s s Ys s

(a)

(b) ( 10) 6.788 45 Sj

−= ∠−°Y

(c) 12( 15)( 5)

( 15) 18 S

( 10)5

−−

−= =−

−

Y

(d)

2

K ( 10)

() ,K 12

( 5)( 20)

12 ( 10) 12 120

() ( 5)( 20) 25 100

1200 1200

( 10) 4.800 4.800 6.788 45 S

100 250 100

j

jj

j

+

==∴

++

==

++ ++

−+

∴= =+=∠°

−+ +

ss

Ys ss

ss s s

Ys ss ss

Y

2

22

2

-1 -1

245 245 68(500)

12 120 17 245 500

5()5 ,

( 5)( 20) 34

25 100

Zeros: 2.461 and 11.951 s ; Poles: 5, 20 s

−± −

+++

+=+ = =

++

=− − =− −

ssss

Ys s

ss

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31.

(a) 1 1 0.2(6 9) 5( 1)( 4)

4 5 5 (4 )(1 ) 6( 1.5)

in in

+++

=+ = ∴=

++ ++ +

sss

YZ

ssss s

(b) -1

Poles: 1.5, ; Zeros: -1, -4 s=− ∞ =ss

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32. 2

+ 2

( ) = ( + 5)( + 6 + 25)

s

Hs sss

(a) d(t) ⇔ 1, so the output is 2

+ 2

( + 5)( + 6 + 25)

s

sss

(b) e-4t u(t) ⇔ 1 / (s + 4), so the output is 2

+ 2

( + 4)( + 5)( + 6 + 25)

s

ssss

(c) 2 cos 15t u(t) ⇔ 2

2

+ 225

s

s, so the output is

()

22

2 + 2

( + 225)( + 5)( + 6 + 25)

ss

ssss

(d) t e-t u(t) ⇔ 1/ (s + 1), so the output is 2

+ 2

( + 1)( + 5)( + 6 + 25)

s

ssss

(e) poles and zeros of each:

(a): zero at s = -2, poles at s = -5, -3 ± j4

(b): zero at s = -2, poles at s = -4, -5, -3 ± j4

(c): zeros at s = 0, -2, poles at s = ± j15, -5, -3 ± j4

(d): zero at s = -2, poles at s = -1, -5, -3 ± j4

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33. h(t) = 5 [u(t) – u(t – 1)] sin πt x(t) = 2[u(t) – u(t – 2)]

y(t) =

()

0() hxt d

λλλ

−

∞−

∫

t < 0: y(t) = 0

0 < t < 1: y(t) =

()

00

10 10

10sin = - cos = 1 cos

t

tdt

πλ λ πλ π

λπ

−

∫

1 < t < 2: y(t) = 1

0

20

10sin = d

πλ λ π

∫

2 < t < 3: y(t) =

[]

1

22

10 10

10sin = - cos = - 1 cos( 2 )

tt

dt

πλ λ πλ π π

ππ

−−

−− −

∫

= (10/

π

) (1 + cos

π

t)

t > 3: y(t) = 0

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. f1(t) = e-5t u(t), f2(t) = (1 – e-2t) u(t)

(a) f1 * f2 =

() ( )

-12

0 fftd

λλλ

∞−

∫

t < 0: f1 * f2 = 0

t > 0: f1 * f2 =

()( )

--

522 523

00

1 =

tt

eed eeed

λλ λ λ

λλ

−− −−−

−−

∫∫

=

523 52

00

11 121

+ = ( )

53 5153

tt

ttt

eee eeut

λλ

−−− −−



−+−





(b) F1(s) = 1/ (s + 5), F2(s) = 1/s – 1/ (s + 2)

F1(s) F2(s) =

()()()

11abc

= + +

+ 5 + 5 + 2 + 2 + 5

−

ss s s s s s

Where a = 0.2, b = -1/3, and c = -1/5 + 1/3 = 2/15.

Taking the inverse transform, we find that f1 * f2 = 52

12 1

()

515 3

tt

eeut

−−



+−





CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 

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35. The impulse response is vo(t) = 4u(t) – 4u(t – 2) V,

so we know that h(t) = 4u(t) – 4u(t – 2). vi(t) = 2u(t - 1), and vo(t) = h(t) * vi(t).

Thus, vo(t) =

()

0()

i

hvt d

λλλ

−

∞−

∫

t < 0: vo = 0

1 < t < 3: vo =

1

08 = 8 8

tdt

λ

−

−−

∫

t > 3: vo =

2

08 = 16d

λ

∫

t (s)

vo (V)

16

1 3

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36. h(t) = 2e-3t u(t), x(t) = u(t) – δ(t)

(a) y(t) =

(b)

Taking the inverse transform, we find that y(t) = 3

28

() ()

33

t

ut e ut

−

[]

0

333

00

33 3

()( )

0: ( ) 0

1

0 : ( ) 2 1 ( ) = 2 - ( ) - ( )

3

228

(1 ) ( ) 2 ( ) = ( )

333

t

tt

tt t

hxt d

tyt

tyt e td eut eut

eut eut e ut

λλ

λλλ

δλλ

−

∞

−−−

−− −

−

<=



>= −− 







=− − −





∫

()

21

( ) = ( ) = - 1

+ 3

21 - 21 8 1

, ( ) = = -

+ 3 3 3 + 3

thus







Hs Xs

ss

s

Ys ss s s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37. h(t) = 5 u(t) - 5 u(t – 2), so H(s) = -2

5 - 5e s

s

(a) vin(t) = 3δ(t), so Vin(s) = 3

Vout(s) = Vin(s) H(s) = -2

15 - 15e s

s. vout(t) = Λ-1{Vout(s)} = 15 u(t) – 15 u(t - 2)

(b) vin(t) = 3u(t), so Vin(s) = 3

s

Vout(s) = Vin(s) H(s) = -2 -2

2

3 5 15 15

- 5e = - e

 

 

 

ss

ss s

s.

vout(t) = Λ-1{Vout(s)} = 15 t u(t) – 15 u2(t - 2) = 15 t u(t) – 15 u (t - 2)

(c) vin(t) = 3u(t) – 3u(t – 2), so Vin(s) = -2

3 - 3e s

s

Vout(s) = Vin(s) H(s) = -2 -2 -2 4

2

3 5 15 30

- 3e - 5e = - e 15e−

 +





ss ss

ss s

s.

vout(t) = Λ-1{Vout(s)} = 15 t u(t) – 30 u2(t - 2) + 15 u2 (t – 4)

= 15 t u(t) – 30 u (t - 2) + 15 u(t)

(d) vin(t) = 3 cos 3t, so Vin(s) = 2

3

+ 9

s

Vout(s) = Vin(s) H(s) = -2

22

15 15

- e

+ 9 + 9

s

ss .

vout(t) = Λ-1{Vout(s)} = 5 sin 3t u(t) – 15 cos [3(t – 2)] u(t - 2)

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38.

2

22

2

22

10 20(20 10 / )

10 10

20 20 20 40 10 /

40

10 40 20 40 60 10 40 60 10

41 4

20 2 2 ;

10 41 40 60 10

40

10 4 1 4

20 4614

in in

in

in in in

top in in in

out in top in

s

==

+



+

++

 +



+

== =

+++ ++

+++

∴= = =

+++

+

=+= +

++ +

VV

Is

ss

VV ss

V

sss ss

ss ss

ss

II I V

sss

ss

VIIV

ss s

22 2

22

61

4 4 1 0.25 ( 0.5)

() ( 0.19098)( 1.3090)

4 6 1 1.5 0.25

zeros: 0.5, 0.5; poles: 1.3090, 0.19098

out

in

s



∴



+



++ ++ +

== = = ∴

++

++ + +

=− =− =− −

s

Vss ss s

Hs Vss

ss s s

ss

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

(a)

(b)

2

22 2

4

( ) 8.5 (17 ) 4

Hj ω+

ω= −ω + ω

(c)

21

2

() ()/ (), (0) 1

K( 2) K( 2)

() (1 4)(1 4) 217

K

1 2 , so K=8.5

17

8.5( 2)

, ( ) 217

8.5( 2)

Let 0 ( ) 217

jj

Thus

==

++

∴= =

++ +− ++

=

+

=++σ+

ω= ∴ σ =σ+σ+

Hs V s V s H

ss

Hs ss ss

s

Hs ss

H

max

By trial & error: ( ) 4.729 at 4.07 rad/sjω= ω=

H

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40. (a) pole-zero constellation (b) elastic-sheet model

(c) H(s) =

() ()

22

2

22

+ 1 + 1

= + + 1

33

+ 0.5 + + 0.5 -

22

+ 2 + 1

= = 1 +

+ + 1 + + 1

jj







ss

ss s

ss ss

We can implement this with a 1-Ω resistor in series with a network having the impedance

given by the second term. There are two energy storage elements in that network (the

denominator is order 2). That network impedance can be rewritten as

2

1

+ + 1 1

=++

s

ss ss

, which can be seen to be equal to the parallel combination of a 1-Ω

resistor, a 1-H inductor, and a 1-F capacitor.

1 Ω

1 Ω 1 H 1 F

O

-1

(2 zeros)

X

3

2

3

2

−

σ

j

ω

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41. 22

( ) (10 55 75) /( 16)=++ +Hs s s s

(a)

()( )

( + 3)( + 2.5)

( ) = 10 + 4 4jj−

ss

Hs ss. Critical frequencies: zeros at –3, -2.5; poles at ± j4.

(b) 75

(0) 4.688, ( ) 10

16

== ∞=HH

(c)

(d)

(0) 4.679 K 3, so K 0.64

90 75 165 0.64

( 3) 0.64 15 165 15.15

77

j

jjcm

===

−+ +

∴= =−+=

H

X 4

-4

O O

3 2 1

X

σ

j

ω

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42. (a)

()()

()( )

2

+ 0.5 + 0.3873 + 0.5 - 0.3873

5 + 5 + 2

( ) = =

5 + 15 + 2 + 2.86 + 0.1399

jjss

ss

Ys ss s s

Zeros: s = -0.5 ± j0.3873

Poles: s = -2.86, s = -0.1399

(b) elastic sheet model

(c) lattitude 5o5’2”, longitude 5o15’2” puts it a little off the coast of Timbuktu.

σ

j

ω

+1

-1

-3 -2 -1

X X

O

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43. 0

( ) ; ( 2) 6

M

=−=

I

Hs H

I

(a)

(b) 30

(0) 2.308, ( )

13

=− =− ∞ =∞HH

(c)

232

22

(1)(1)(3)

() K

(3 2)(3 2)

(3)(1)K 3K

(2) 6 K 10,

(1 2)(1 2) 5

( 1)( 3) 10 30 10 30

, ( ) 10 613 613

jj

Thus

−++

=++ +−

−−

−== = ∴=

+−

−+ + −−

==

++ ++

sss

Hs ss

H

ss sss

Hs ss ss

1: ( 1) ( 2 1) 2.236 116.57

1:( 1)(21)2.23663.43

3 : ( 3) 2 3 3.606 33.69

3 2 : 2 3 2 5.000 53.13

32:23230

sj

jj j

−= −= ∠ °

−+=+= ∠ °

−+=+= ∠°

−− + + = ∠ °

−+ + − =∠°

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44.

: zero at 10 0; 20 : zero at 3.6 0

R/ C R 1/ C 5 5 / RC 1/ C

555

R1/SC CR1 1/RC 1/RC

5( 1/ RC 1/5C)

1/ RC

11

Thus, using the fact that = 0 at = -10, we may write 10

RC 5C

Also,

AA

A

jj

=− + + =− +

++

∴=+ =+ =+ =

++++

++

∴= +

+=

Zs Z s

ss

Zss s

s

Zs

Z

11

25 1 25

25

1/C RC 25C

RC C

25 1

1/ RC 1/ RC

RC

11 4

3.6 or 6.4,

RC 25C 25C

1

C25 mF,

40

40 40 40

10, 2, so R 20

R5 R

B



++

++ 



=+ = =

++ +

∴+ = =

==

+= = =Ω

s

ss s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45. 2

( ) 100( 2) /( 2 5)=+ ++Hs s s s

(a) zero at s = -2, poles at 2420 12

2j

−± −

==−±s

(b) 2

100(2 )

()(5 ) 2

j

jj

+ω

ω= −ω + ω

H

(c)

2

42

4

( ) 100 625

jω+

ω= ω−ω+

H

(d)

(e)

22

24223

42

42 2 2 42 42 42

2

() ()

4 ( 6 25)2 ( 4)(4 12 )

,

10 000 625

6 25 ( 4)(2 6), 6 25 2 2 24, 8 49 0

8 64 196 4.062 2.016 rad/s, H( 2.016) 68.61

2mar

jdj

detc

j

ωω

ω+ ω−ω+ ω−ω+ ω− ω

==

ω

ω−ω+

∴ω−ω+ =ω+ ω− ω−ω+ =ω+ω− ω+ω− =

−± +

∴ω = = ∴ω = =

HH

X j2

-j2

O

3 2 1

X

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46. 520

()

2

in

+

=Ω

+

s

Zs s

(a)

(b)

2

5( 4)

(0) 25V; ( ) ,

2

5( 4)

( ) , single pole at 2 ( ) 25 V, 0

2

ab in ab in in

t

ab

v

vt e t

−

+

== =

+

∴= =−∴ = >

+

s

Zs V ZI

s

Hs s

s

4

12

(0) 3A ( ) single poleat 4

5( 4)

() 3 A, 0

sab

ab ab

in in in

t

ab

i

it e t

−

+

=∴= ∴ === =−

+

∴= >

VI s

IHs s

ZVZs

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47. 2

( ) 5( 4 20) /( 1)

in =++ +Zs s s s

(a)

(b)

6

2

160 V 160 V, 6

160( 1) 32( 5) 5A ( ) 5 A (all )

32420

5( 4 20)

t

ab ab

t

ab

a a

in

ve

it e t

b

−

=∴==−

+−

== = =−∴=−

−+

++

Vs

IZss

6

2

62

11

160 ( ), (0) 0, (0) 32 A/s ( ) 5( 4 20)

41680 2 4 ( ) 5 (A cos 4 Bsin 4 ) 0 5 A, A 5

2

(0) 32 30 10 4B B 3 ( ) [ 5 (5cos 4 3sin 4 )] ( ) A

ta

ab a a

sin

tt

a

tt

aa

veuti i

jit ee t t

iiteettut

−

−−

+

′

===∴===

++

−± −

==−±∴=−++∴=−+=

′==−+ ∴=∴ =− + +

Is

Hs VZ ss

s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48.

(a) 2

0.5 250

() / 0.5 0.002 500 / 250 25 000

cs

== =

++ ++

s

Hs I I ss

ss

(b) 61

1( 250 62 500 10 ) 125 484.1s

2j−

=− ± − =− ±s

(c) -1 6 -1 4 -1

od

R 0.5

= = 125 s , ω= 10 /4= 500 s , ω= 25×10 -15,625 = 484.1 s

2L 0.004

α

=

(d) 1, 0 I 0 0

sccf

i==∴=∴=Is

(e) 125

,( cos 484 Bsin 484 )

t

cn

ieA t t

−

=+

(f) 3

1

(0) 0 (0 ) 0, (0) 0 1 2 10 (0 ) 0 (0 ) 250 A/s

2

Lcc

iiv ii

+−++

=∴ = =∴×=× +∴ =

(g) 125

A 0, 484B 250, B 0.5164 ( ) (0.5164 sin 484.1 ) ( ) A

t

c

it e tut

−

∴= = = ∴ =

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49.

(a)

(b) Note that the element labeled 6 H should be an inductor, as is suggested by the context of

the text (i.e. initial condition provided). Convert to s-domain and define a clockwise mesh

current I2 in the right-hand mesh.

Mesh 1: 0 = -500/ s + (50 + 6s) Iin – 30 - 6s I2 [1]

Mesh 2: 0 = 30 + (20 + 10s) I2 – 6s Iin – 8 [2]

Solving, we find that

where a = 10, b = -2.115 and c = -0.8855. Thus, we may write

iin(t) = [10 – 2.115 e-24.10t – 0.885 e-1.729t] u(t) A

2

2-1

1 1 10 20

() / 6 (4 20) 24 620 1000

50 10 20

1( 620 620 96,000) 1.729 and 24.10 s

48

in in

in

+

=== =

+++

++

∴= − ± − =− −

s

Hs I V ss

Zss

s

6s Ω 4s Ω

-30 V -8 V

500 V

s

Iin

()

()()

22

in 2

42 + 1400 + 2500 7 + 233.3 + 416.7

= = + 24.10 + 1.729

6 + 155 + 250

+ 24.10 + 1.729

ab c

=+ +

ss s s

Iss s

ss s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50.

(a) 50(1000 / ) 1000

() 50 (1000 / ) 20

s

== =

++

Vs

Hs Iss

(b)

() ()

020

-20

2 2 1000 2000 b

= so ( ) = = +

20 20 20

2000 2000

100; b = 100

20

100 100

, ( ) = - and ( ) = 100 1 - e ( ) V

20

s

t

a

s

aa

Thus v t u t

s

==−



 =



+++



== ==−

+





+

ss

IVs

ssssss

ss

Vs s

(c) This function as written is technically valid for all time (although that can’t be possible

physically). Therefore, we can’t use the one-sided Laplace technique we’ve been

studying. We can, however, use simple s-domain/ complex frequency analysis:

(d) 4e-10t u(t) ⇔ 4

+ 10s, so V(s) = 4 1000 a b

= +

+ 10 + 20 + 10 + 20







ss s s

a = 400 and b = -400, so v(t) = 400 [e-10t – e-20t] u(t) V

10

1000

4 A 4 A, 10 4 ( 10) 4 400 V

10

() 400 V (all )

t

ss

t

ie

vt e t

−

=∴==∴=−=×=∴

=

IsVH

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



51.

(a)

(b) No initial energy stored in either capacitor. With vs = u(t), Vs(s) = 1

s, so

VC2 =

()( )

2.5

+ 6.357 + 0.3933ss s = + +

+ 6.357 + 0.3933

ab c

ss s

Where a = 1, b = 0.06594 and c = -1.066. Thus,

vC2(t) = [1 + 0.06594 e-6.357t – 1.066 e-0.3933t ] u(t) V

2

100 25

100 20 (20 100)25

() / 50

100 125 (20 125)

20 20

2500 (20 125)

() (20 125) 1000 6250 500 2500

2.5

() 6.75 2.5

c

s



+





+



== × +



+



++

+

∴= ++++

∴=

++

Vs

ss

s

Hs Vss

ss

Hs ss s s s

Hs ss

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



52.

22

11

() 1 0.05

0.1 0.025 0.1 0.025

20 (80 / ) 4

4 40( 4) 40( 4)

(2)(8)

0.025 0.25 0.4 10 16

in ==

++ ++

++

+++

===Ω

++

++ ++

Zs s

ss

sss

ss

ss ss

20u(t) 20

⇔s, so Vin(s) = 20 40( 4)

(2)(8)



+







++





s

ss s = a + +

28

bc

++ss s

a = 200, b = -133.3 and c = -66.67, so vin(t) = [200 – 133.3 e-2t – 66.67 e-8t] u(t) V

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



53.

1

() f

=−Z

Hs Z

(a)

(b)

38 8 5

38

1

10 10 / 1000 10 R 10

5000, 10 10 / ( ) 5000 5000 5

f

+++

= = + ∴ =− =− =−

ss

ZZ sHs ss

(c)

8

3

188

5

10 5000 5000

10 , 5000 ( ) 1000 (10 / ) 1000 10

5

() 10

f

s

=+ = ∴ =− =−

++

−

∴=

+

s

ZZHs

sss

s

Hs

48 4 8 5

38 48

1885

10 10 / 10 10 10 10

10 10 / , 10 10 / ( ) 1000 10 / 1000 10 10

f

+++

=+ =+ ∴ =− =− =−

+++

ss s

ZsZ sHs ss s

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



54.

1

11

4

1

11

1

R20 k,() RC RC

1

() 2 10C RC

out

ff

in



=Ω ==− +







∴=−× +





V

Hs s

V

Hs s

(a)

4

11

1

210

( ) 50 C 0, 50, R 400

R

×

=− ∴ = = = ΩHs

(b)

(c)

34 4 4 3

11

4

11

9

1

() 10( 10) 210C 210C 10

RC

1

C 50nF; 10 , so R 2 k

50 10 R

− −

−



=− + =− × + ∴ × =





∴= = = Ω

×

Hs s s

44 44

111

11

28

25

2

1

( ) 10 ( 1000) 2 10 C 2 10 C =10 , C 5 nF

RC

( ) 200 A B , (0) 0 0 200 A B

(0 ) 20 0.025 (0 ) 0.025( 2A 8B) A 4B 400 3B 200

200 400 400 200

B , A ( ) 200

33 33

tt

in in

in

t

in

vt e e v

iv

vt e e

− −

−−

++

−



=−+=−× + ∴× =





∴=+ + =∴=++

== = −− ∴−−= ∴−=

′

=− ∴ =− ∴ = − −

Hs s s

8()V

tut

−







CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



55.

(a)

(b)

(c)

(d)

1

3

1

( ) 50,

1/ R C

( ) , R 20 k

1/ R C

R20 10

set C 0 50 R 400

R50

out

in

f

ff

f

==−

−

==Ω

+

×

=∴−=− ∴ = = Ω

V

Hs V

Hs s

1

11

8 9

1

1/R C

1000 1

( ) 10 000

10 000 1/ 20 000C 20 000 C

11

C 5 nF We may then find R : 1000 R 200 kΩ

2 10 5 10 R

f

ff

f−

=− = ∴ =

++

== = ∴=

××

Hs ss

1

9

1

1/ R C

10 000 1

( ) 1000 C 50 nF

1000 1/ 20 000 C 20 000

11000, R 200 kΩ

510R

f

ff

C

−

=− = ∴ = =

++

==

×

Hs ss

5

1A fA 1B fB 1A fA fB

1B

fA fA fB fB fA fA

fB

1B 1A fA

5

fA fA

100

()

+ 10

11 1

RC RC RC R

= - - = - -

11 1

R

s + s + s +

RC RC RC

R

We may therefore set 100

RRC

1

and 10 . Arbitraril

RC

out

in

==

















=

V

Hs Vs

fA fA

fB

1B 1A

y choosing R = 1 k , we find that C 10 nF.

Arbitrarily selecting R = 100 , we may complete the design by choosing

R = R = 10 k

Ω=

Ω

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



56.

()

4

C

o

i

1A

1A f a

11

[ K ][ K ( 100)]

10 ( 100)

() K

1000

Let ( ) K . Choose inverting op amp with parallel RC network at inverting input.

V

-V

0 = 1+ C -

RR

() 1 R C =

AB

AA

fA fA

AAA

AA

RR

−−−+

−+

==

+

−



+



=−

∴=−+ −

ss

Hs s

s

Hs s

s

Hs s

()

11A

4

11

1

R C K . Set R = . Then

R C 10 C

R

Same configuration for ( ) ( ) K ( 100) = 1 R C

For the last stage, choose an inverting op amp circuit with a parallel RC circui

fA A A

fB

BB B BB

B

R

−=− ∞

−=−

∴=−+ −+

ss

Hs Hs s s

()

1

t in the feedback loop.

R

11

Let ( ) K

1000 1 R C

fC

CC

CFCFC

R

=− = −

++

Hs ss

Cascading these three tranfer functions, we find that

HA HB HC = fC

fB

1fB1B

1B 1 fc fc

R

R1

RC RC + RRRC + 1

fA A

C







−− −





 





ss s

Choosing all remaining resistors to be 10 kΩ, we compare this to our desired transfer function.

(Rfc Cfc)-1 = 1000 so Cfc = 100 nF

Next, fB

1B fB 1B

R = 100

R R C so C1B = 1 µF.

Finally, RfAC1ARfBC1BRfC (R1CRfCCfC) = 10-4, so C1A = 1 nF

Cfc

C1a

R1a

Rfa

C1b

R1b

Rfb

R1c

Rfc

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



57. Design a Wien-bridge oscillator for operation at 1 kHz, using only standard resistor

values. One possible solution:

ω

= 2πf = 1/RC, so set (2πRC)-1 = 1000

If we use a 1-µF capacitor, then R = 159 Ω. To construct this using standard resistor

values, connect a 100-Ω, 56-Ω and 3-Ω in series.

To complete the design, select Rf = 2 kΩ and R1 = 1 kΩ.

PSpice verification:

The feedback resistor was set to

2.05 kΩ to initiate oscillations

in the simulation. The output

waveform shown below exhibits

a frequency of 1 kHz as desired.

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



58. Design a Wien-bridge oscillator for operation at 60 Hz. One possible solution:

ω

= 2πf = 1/RC, so set (2πRC)-1 = 60

If we use 10-nF capacitors, then R = 265.3 kΩ.

To complete the design, select Rf = 200 kΩ and R1 = 100 kΩ.

PSpice verification:

The simulated output of the circuit

shows a sinusoidal waveform

having period 54.3 ms – 37.67 ms =

0.01663 ms, which corresponds to a

frequency of 60.13 Hz, as desired.

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



59. Design a Wien-bridge oscillator for operation at 440 Hz, using only standard resistor

values. One possible solution:

ω

= 2πf = 1/RC, so set (2πRC)-1 = 440

If we use 100-nF capacitors, then R = 3.167 kΩ. To construct this using standard resistor

values, connect a 3.6-kΩ, 16-Ω and 1-Ω in series. (May not need the 1-Ω, as we’re using

5% tolerance resistors!). This circuit will produce the musical note, ‘A.’

To complete the design, select Rf = 2 kΩ and R1 = 1 kΩ.

PSpice verification:

Simulation results show a sinusoidal

output having a period of approximately

5.128 – 2.864 = 2.264 ms, or a frequency

of approximately 442 Hz. The error is

likely to uncertainty in cursor placement;

a higher-resolution time simulation

would enable greater precision.

CHAPTER FIFTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



60. Design a Wien-bridge oscillator for 440 Hz:

ω

= 2πf = 1/RC, so set (2πRC)-1 = 440

If we use 100-nF capacitors, then R = 3.167 kΩ.

Design a Wien-bridge oscillator for 220 Hz:

ω

= 2πf = 1/RC, so set (2πRC)-1 = 220

If we use 100-nF capacitors, then R = 7.234 kΩ.

Using a summing stage to add the two waveforms together:

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1. We have a parallel RLC with R = 1 kΩ, C = 47 µF and L = 11 mH.

(a) Qo = R(C/L)½ = 65.37

(b) fo =

ω

o/ 2

π

= (LC)-½ / 2

π

= 221.3 Hz

(c) The circuit is excited by a steady-state 1-mA sinusoidal source:

The admittance Y(s) facing the source is Y(s) = 1/R + 1/sL + sC

= C(s2 + s/RC + 1/LC)/ s so Z(s) = (s/C) / (s2 + s/RC + 1/LC) and

Z(j

ω

) = (1/C) (j

ω

) / (1/LC –

ω

2 + j

ω

/RC).

Since V = 10-3 Z, we note that |V| > 0 as

ω

→ 0 and also as

ω

→ ∞.

10-3∠

∠∠

∠0o A j

ω

ωω

ω

L -j/

ω

ωω

ω

C

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2. (a) R = 1000 Ω and C = 1 µF.

Q

o = R(C/L)½ = 200 so L = C(R/ Qo)2 = 25 µH

(b) L = 12 fH and C = 2.4 nF

R = Qo (L/ C)½ = 447.2 mΩ

(c) R = 121.7 kΩ and L = 100 pH

C = (Qo / R)2 L = 270 aF

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3. We take the approximate expression for Q of a varactor to be

Q ≈

ω

CjRp/ (1 +

ω

2 Cj2 Rp Rs)

(a) Cj = 3.77 pF, Rp = 1.5 MΩ, Rs = 2.8 Ω

(b) dQ/d

ω

= [(1 +

ω

2 Cj2 Rp Rs)(Cj Rp) -

ω

CjRp(2

ω

Cj2 Rp Rs)]/ (1 +

ω

2 Cj2 RpRs)

Setting this equal to zero, we may subsequently write

CjRp (1 +

ω

2 Cj2 Rp Rs) -

ω

CjRp(2

ω

Cj2 Rp Rs) = 0

Or 1 –

ω

2 Cj2 Rp Rs = 0. Thus,

ω

o = (Cj2 RpRs)–½ = 129.4 Mrad/s = 21.00 MHz

Q

o = Q(

ω

=

ω

o) = 366.0

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4. Determine Q for (dropping onto a smooth concrete floor):

(a) A ping pong ball: Dropped twice from 121.1 cm (arbitrarily chosen).

Both times, it bounced to a height of 61.65 cm.

Q = 2

π

h1/ (h1 – h2) = 12.82

(b) A quarter (25 ¢). Dropped three times from 121.1 cm.

Trial 1: bounced to 13.18 cm

Trial 2: bounced to 32.70 cm

Trial 3: bounced to 16.03 cm. Quite a bit of variation, depending on how it struck.

Average bounce height = 20.64 cm, so

Qavg = 2

π

h1/ (h1 – h2) = 7.574

(c) Textbook. Dropped once from 121.1 cm. Didn’t bounce much at all- only 2.223 cm.

Since the book bounced differently depending on angle of incidence, only one trial was

performed.

Q = 2

π

h1/ (h1 – h2) = 6.4

All three items were dropped from the same height for comparison purposes. An

interesting experiment would be to repeat the above, but from several different heights,

preferrably ranging several orders of magnitude (e.g. 1 cm, 10 cm, 100 cm, 1000 cm).

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5.

22

80Np/s, 1200 rad/s, Z( 2 ) 400

1200 80 1202.66 rad/s Q 7.517

2

()() ()(2)

Now, Y( ) C Y( 2 ) C 2

80( 80 2400)

Y( 160 1200) C Y( 160 1200)

160 1200

dd

o

oo

dd d

d

j

sjsj j

sj

j

jj

j

αω αω

ω

ωα

αω αω αα ω

αω αω

== −+=Ω

=+= ∴==

+− ++ − −+

=∴−+=

−+

−−+

∴− + = ∴ − + =

−+

2

1130

80C

400 2 15

1 229 1 1

C 15.775 F; L 43.88mH; R 396.7

32,000 901 C 2

o

j

C

µωα

−

−+

=−+

∴= = = = = = Ω

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6.

2

262 226

35 3 2

1 1 2 0.1

Y 0.2 0.2

2 0.1 1 1000/ 4 0.01 1000 10

2 0.1 1000 0.1 1000

0.2 0

4 0.1 10 4 0.01 10

0.1 10 4000 10 9.9 96,000 98.47 rad/s

−

=++ = ++

+++ +

−+−

=++ ∴ +=

++++

∴+= +∴ = ∴=

in

jj

jj j

jj

ωω

ωωω

ωω ω ω ω

ωωωω

ωω ωω ω ω

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7. 66

Parallel: R 10 , L 1, C 10 , I 10 0 A

s

µ

−

=== =∠°

(a) 366

11000 rad/s; Q RC 10 1000

LC

ooo

ωω

+−

== = = =

(b)

ω V

995

996

997

998

999

1000

1001

1002

1003

1004

1005

999.5

1000.5

0.993

1.238

1.642

2.423

4.47

10.0

4.47

2.428

1.646

1.243

0.997

7.070

7.072

66 533

22

2

6

1 I 1000

Y 10 10 , V 10 /10 10

Y 1000

10 10

V,V

1000 1000

0.001 10

1000 1000

jj

j

ω

ωω

−− −−−

−−

−



  

=+ − == + −

  



  



∴= =

 

+− +−

 

 

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

(a)

(b) 2

226

102000

Z ( ) 2 2.294

400 10

o

in o

oo

ω

ωωω

=++=Ω

++

226

426 2 2

226

5(100/ ) 0.1

Z2

5 (100 / ) 10 0.01

500 10 100 10 100(20 ) 10 (1000 )

22 2

100 5 1000 20 1000 400 10

100 10 0 10 100 40,000, 99 960,000

400 10

960,000/99 9

=++

++

−−

=++=++= ++

+++++ +

−

∴+=∴+=+ =

++

∴= =

in

o

jj

jjjjj

jjjj

ωω

ωωωω

ωωωωω ω

ωω ωω ω

ωω

ω

8.47 rad/s

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9.

(a)

(b) 46

11

Y 10 10 , 1000 Z 9997 1.4321

0.9975 Y

j

ωω

ω

−−



=+ − = ∴== ∠ °Ω





11222

66

2

50 , 1000 1,002,500 1001.249

110 110

L0.9975H;R10

C 1,002,500 2 C 100

dod o

o

ss

k

αω ωαω ω

ωα

−−

+

==∴=+= ∴=

== = ===Ω

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10.

min max

o

33

max min

2

14

max min max 3

33

535kHz, 1605kHz, Q 45 at one end and

Q 45 for 535 1605 kHz

11

1/ 2 LC 535 10 ,1605 10

2LC 2LC

1

L / L 3; L C 8.8498 10

2 535 10

RC 45,535 10 1605 10 . Use

2

o

f

ff

f

πππ

π

ω

ωπ

−

===

≤≤≤

=∴×= ×=



∴== =×



××



≤×≤≤× max

33

14 max

max min

12

2 1605 10 20 10 C 45 C 223.1pF

L

8.8498 10

L 397.6 H, L 44.08 H

223.1 10 9

o

ω

π

µµ

−

∴× × ×× = ∴=

×

∴= = = =

×

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

(a)

(b)

4

4548

3

82 4

48

82 4

Apply 1V. I 10 A

1

Y I 10 (1 [10 ( 10 )])10

4.4 10

1000 48.4 10 4.4 10 1000

Y101110

4.4 4.4

1000 48.4 10 4.4 10

Y( ) 4.4

−

−−−

−

−−

±∴=−

∴== + +− −

×

×+×+

∴= + +× =

−× +×

∴=

R

in in

in

s

ss

s

ss

j

jj

ωω

82

1

4

At , 1000 48.4 10 , 45.45 krad/s

4.4 10

Z( ) 10

4.4

−−

−

==× =



×

==Ω





ooo

o

in o

o

j

jk

j

ωω ωω

ω

ωω

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12. 1000 rad/s, Q 80, C 0.2 F

oo

ω

µ

===

(a) 6

26 36

110 80

L 5H, Q RC R 400

C 0.2 10 10 0.2 10

oo

o

k

ω

−

== = = ∴= =Ω

×××

(b)

2

3

B / Q 1000 /80 12.5

1B 6.25 rad/s

2

Z R / 1 400 10 / 1

B/ 2 6.25

oo

j

ω

ωω ωω

== =

∴=

−−



∴= + = × +





CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13.

12

212

o

3

2

1

103rad/s, 118,

Z( 105) 10

103 118

110.245

110.245 , B 118 103 15 rad/s, Q 7.350

B15

7.350 1 1

7.350 RC RC 66.67 10 ,LC

110.245 12,154

11

Y( 105) 0.1 105C 15C 105C

R 105L

o

j

jj j

ωω

ωωω

ω

ωω

+

−

+

==

=Ω

==×

∴= = − = = = =

∴= ∴= =× ==



==+ − = +





12,154 C 18.456C

105

0.1 1 1

C 5.418mF, R C 12.304 , L 15.185 mH

18.456 15 12,154C

−



−=





∴= = = = Ω = =

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14. 30 krad/s, Q 10, R 600 ,===Ω

oo

ω

(a) B 3 krad/s

Q

o

ω

==

(b) 28 30

N 1.3333

B/ 2 1.5

o

ωω

−−

== =−

(c) Zin(j28 000) = 600 / (1 – j1.333) = 360 ∠ 53.13o Ω

(d)

(e)

1

Q11 10

Z ( 28,000) 28,000C ,C

600 28,000L R 30,000 600

R 600 1 30,000 10 1 28 10 30 10

L,Z

Q 30,000 10 L 600 600 30 600 28 600

600

Z 351.906 54.0903

28 30

110

30 28

−



=+ − ==



×



× 



== = ∴=+×−





×





==∠°Ω



+−





o

in

o

in

oo

in

jjj

j

ω

approx-true 360 351.906

magnitude: 100% 100% 2.300%

true 351.906

53.1301 54.0903

angle: 100% 1.7752%

54.0903

−

==

°− °=−

°

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15. 3

400Hz, Q 8, R 500 , I 2 10 A B 50Hz

oo S

f−

===Ω=×∴=

(a)

(b)

32 2 400

V 2 10 500 / 1 N 0.5 1 N 4, N 3 50/ 2

400 25 3 443.3 and 356.7Hz

f

−−

=× × + = ∴+ = =± =

∴= ± =

322

2

11

I 0.5 10 1 N 4, N 15, N 15

R 500

1N

400 25 15 496.8 and 303.2 Hz

R

v

f

−

== × = × ∴+ = = =±

+

∴= ± =

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16. 63

10 , Q 10, R 5 10 , . .===×

oo pr

ω

(a) 3

6

R510

Q L 0.5mH

L1010

o

ω

×

=∴= =

×

(b)

(c)

6

2

6

2

222 2

22

422

10

Approx: 2 5/ 1 N N 2.291 1.1146 Mrad/S

10 / 20

11

Exact: Y 1 Q 0.5 0.2 1 100 ( in Mrad/S)

R

11

6.25 1 100( 2 1/ ), 2 0.0525, 2.0525

1

2.0525 1 0, 2

2

−

=+∴= = ∴=



 

=+ − ∴= + −



 





∴ =+ −+ −+= +=

−+==

o

j

ωω

ωω ωω

ωω ω

ωωω ω

ωω

ωωω

()

2

.0525 2.0525 4 1.2569, 1.1211 Mrad/s+−==

ω

1

2

1

Approx: Y 30 tan N 30 , N 0.5774 , 1.0289 Mrad/s

1/20

11 1

Exact: Y 1 10 (in Mrad/s) tan30 0.5774 10

5000

1 0.05774 0.05774 4

0.05774, 0.05774 1 0, 1.0293 Mrad/s

2

−−

∠= °∴ = ° = = =



 

=+− ∴°==−

 



 



++

∴− = − −= = =

j

ωω

ωωωω

ω

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

(a) 6

48

1

C 3 7 10nF 10 rad/s

10 10

o

ω

−−

=+= ∴ = =

(b)

(c) 3

3

1

3

15 10 15 90

15 10 N 1.5 V 8.321 33.69 V

10 10 1 1.5

oj

ωω

×∠°

−=× ∴= = ∴= = ∠ °

×+

68 3

969

3

33

1,0

QCR1010551050

B/Q20 krad/s

10

Parallel current source is 3 10 At , I 10 3

Z

V3105101590V

oo

os

jj

j

ω

ωω

−

−−

−

== ×=

==

∠°

=× = ×

∴=×××=∠°

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18.

(a)

(b) 25 10,000

Z ( ) (5 100) (5 100) 1002.5

10

in o

jjj

ω

+

=+ − = = Ω

66

626

246

26

62

626

(5 0.01 )(5 10 / ) (5 0.01 )(5 10 )

Z() 10 0.01 10 / 0.01 10 10

0.05 25 10 5 10

Z() 0.01 10 10

5 10 0.05 10,025

Z( ) 10 0.01 10

10,025 10

At , 5 10 0.05 10 0.01

++ + +

==

++ ++

++ +×

=++

×− +

∴=−+

==

×− −

in

oo

o

ss ss

sss ss

sss

j

jj

ωω

ωωω

ωω

ωω ω

9272

2

29

, 10.025 10 100.25 5 10 0.5

99.75 9.975 10 , 10,000 rad/s

×− =×−

∴=×=

oo

o

oo

ωω

ω

ωω

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19.

,

1000Hz, Q 40, Z ( ) 2 B 25Hz

ooino

fjk

ω

===Ω∴=

(a) Zin(j

ω

) =

Z

in = 2000 / (1 + j0.8) = 1562 ∠ -38.66o Ω

(b) 0.9 1.1 900 1100 Hz

oo

ff f f<< ∴ <<

2000 1000

, N , 1010, N 0.8

1 N 12.5

−

==∴=

+

ff

j

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20. Taking 2–½ = 0.7, we read from

Fig. 16.48a: 1.7 kHz – 0.6 kHz = 1.1 kHz

Fig. 16.48b: 2×107 Hz – 900 Hz = 20 MHz

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

(a)

(b)

2

20A 6 , 3 6 2, 40V in series with 2 1 3

L160

10 rad/s, Q 20

R3

LC

10 1

B 0.5, B 0.25, V ( ) 40Q 800V

20 2

10

V ( ) 800 / 1 0.25

o

oo

out o o

out

j

ω

Ω= +=Ω

== ===Ω

== = = =

−



∴=+





9 rad/s

800

(Approx: V ( 9) 194.03V

17

40 600

Exact: V 3 (6 600 / )

24,000

V ( 9) 204.86 13.325 V

9[3 (54 66.67)]

out

j

jj

ω

ωωω

−

=

==

=×

+−

∴= =∠−

+−

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22. 7

Series: R 50 , L 4mH, C 10−

=Ω= =

(a) 37

1/ 4 10 50 krad/s

o

ω

−−

=× =

(b) 3

50 10 / 2 7.958kHz

o

f

π

=× =

(c) 33

L5010410

Q4

R50

o

ω

−

×××

== =

(d) 3

B / Q 50 10 / 4 12.5 krad/s

oo

ω

==×=

(e) 2

11 (1/ 2 ) 1/ 2 50 1 1/ 64 1/8 44.14 krad/s

ooo

QQ

ωω



=+ − =+−=



(f) 250 65/ 64 1/8 56.64 krad/s

ω



=+=



(g) 73

Z ( 45,000) 50 (180 10 / 45) 50 42.22 65.44 40.18

in jj j

−

=+ − =− = ∠− °Ω

(h) 7

45,000

Z / Z 10 / 45,000 50 4.444

cR j=×=

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23. Apply 1 A, in at top. V 10V

R

∴=

(a)

(b) 33

L 346.4 10

Q 34.64

R10

o

ω

−

×

== =

88

33

383 8

211

10 1.2 10

V Z 10 10 (0.5 10 1) 10 10

5

Z ( ) 10 (10 1.2 10 / ) 10 1.2 10 /

1.2 10 , 346.4 krad/s

in in

in o o

oo

ss

jj

ωωωωω

ωω

−−

×

== ++ ×+= ++

=+ −× ∴ = ×

∴=× =

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24. Find the Thévenin equivalent seen by the inductor-capacitor combination:

1

111

1

6

max

V

SC : 1.5 V 10 0.105V V 50 V

125

50

I 0.4A

125

1.5

OC:V 0 V 1.5V R 3.75

0.4

1000 4

1/ 4 0.25 10 1000,Q 1066.7

3.75

1000 1

B / Q 0.9375, B 0.4688 rad/s

1066.7 2

V Q V 1066.7 1.5 1600V

−



=+ − ∴=





∴↓ = =

=∴ = ∴ = = Ω

×

∴= × × = = =

=== =

== ×=

SC

OC th

oo

Coth

ω

Therefore, keep your hands off!

To generate a plot of |VC| vs. frequency, note that VC(j

ω

) =

C

j

Lj

C

j

ω

−+

−

75.3

5.1

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25. ,0

Series, 500Hz, Q 10, X 500

ooL

f===Ω

(a)

(b)

22

,0

12

500 L 2 (500)L L 0.15915 H, C 0.6366 F

L (2 500)

X500

Q10 R50

RR

+

== ∴= = = =

×

== = ∴=Ω

o

L

o

π

ωπ

µ

ωπ

6

1 10 0.5 250,000

1 I 50 2 I 50

22

10 0.5

I 1/ 50 ( 250,000 / ), V I

2

250,000/

V V (2 450) 4.757V

50 ( 250,000/ )

V (2 500) 10,000V V (2 550) 4.218V



×

=+×− =+−





×

∴= + − =

−

=∴×=

+−

×= ×=

c

Cc

cc

jf j jfj

ff

jf f jf

jf

jf f

π

πππ

π

ππ

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26.

14

828

8

1

X : 0, ,0: 20,000 80,000 , Z ( 10 ) 20 0 SERIES

1

20,000, 80,000 (64 4)10 82,462 rad/s, 68 10

LC

R R 1 L 68 10 1

20,000 40,000, 170,000; Z( ) R L

2L L LC R 40,000 C

1

20 R 10,000L 10,00

−

= ∞ =− ± − =− + Ω ∴

==∴=+= ==×

×

== ∴ = ×= = =+ +

∴− = − −

in

do o

ss js j

αω ω ω

ασσ

σ

1 170,000

R R R R 1.2308

0C 4 10,000

1.2308 1

L 30.77 H, C 4.779 F

40,000 170,000 1.2308

=− − ∴= Ω

∴= = = =

×

µµ

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

= 4.287 ∠ 59.04o kΩ

53

37 5

2

57

3

10

1/ 10 10 rad/s, Q 100, R 10,000

1

Q 500, R 500 0.2 50,000

10 0.2

50 10 8.333 Q CR 10 8333 83.33

100,000

B 1200 rad/s, Z ( ) 8333

83.33

(99 100)10

99,000 N 1.6667,Z

600

−

−−

−

ω====Ω

===×=Ω

×

=Ω∴=ω=×=

== ω=Ω

−

ω= ∴ = =−

B

oLPL

cPC

oo

in o

i

k

j

8.333

( 99,000) 1 1.667

=−

njj

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28. Req = Qo/

ω

o C = 50 / 105-7 = 5000 Ω.

Thus, we may write 1/5000 = 1/8333 + 1/Rx so that

R

x = 12.5 kΩ.

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29.

35

34 2

34

43

2

6

4

1

3mH 1.5mH 1mH, 2 F 8 F 10 F, 10 krad/s

10

310 10

Q 100, R 100 0.3 3

0.3

1.5 10 10

Q 60, R 60 0.25 900

0.25

692.3

900 3000 692.3 Q 69.23

10

692.3

R 0.14444

69.23

10

Q 125, R

10 0.1 8

−−

−

=µ+µ=µ∴ω= =

××

===×=Ω

××

===×=Ω

=Ω∴= =

∴= = Ω

==

××

o

p

L

LS

pc

k

2

45

2

,min

125 0.1 1562.5 10 F

1562.5

Q 10 10 15625 156.25 R 0.064

(156.25)

R 0.14444 0.064 0.2084 Z , 10 krad/s

−

=×= Ωµ

∴= × × = ∴ = = Ω

∴= + = Ω= ω=

cSC

S tot in o

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30.

(a)

(b)

3

2

3

1/ 2 0.2 10 50 rad/s

Q 50 2.5/ 2 62.5, 2 62.5 7812.5

50 10

Q 50, 10 50 25

10

1000

Q 100, 100 1 10 , R 7.8125 25 10 3731

50 0.2 1

50 1

Q 50 3731 0.2 10 37.31; B 1.3400, B 0.6700

37.31 2

V10

−

ω××=

=× = × = Ω

×

==×=Ω

==×=Ω= =Ω

××

=× × × = = = =

∴=

B

o

leftL

rightL

cp

o

k

33731 3.731V

−×=

3

V 10 [(2 125) (10 500) (1 100)]

10 3.7321 0.3950 V

111

2 125 10 500 1 100

−

−+

=+ + −

==∠−°

++

++−

jjj

3.731 V

2.638 V

50

↔

↔↔

↔

1.34 rad/s

|V| (volts)

ω

(rad/s)

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31.

(a)

(b)

63

4

2

,

1000 2000 rad/s, Q 2000 2 10 25 10 100

0.25

R2010

R 25,000/100 2.5 ; Q 40

L 2000 0.25

20,000

R 12.5 R 12.5 2.5 15

1600

2000 0.25 1

Q 33.33 V 1 33.33 16.667V

15 2

−

ω= =××××=

×

∴= =Ω== =

ω×

∴= =Ω∴=+=Ω

×

∴= = ∴ =× ×=

B

oc

CS L

o

LS tot

ox

20,000 500

20,000 500 12,4922 499.688

20,000 500

25,000( 250)

25,000 250 2.4998 249.975

25,000 250

Z 12.4922 2.4998 499.688 250 249.975 14.9920 0.2870

I 1/ 14.9920 0.2870 66.6902mA V 250

×

==+Ω

+

−

−= = −

−

∴= + + − − = − Ω

∴= − = ∴ = ×

in

x

j

jj

j

jj

j

jjj j

j3

66.6902 10 16.6726V

−

×=

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32.

(a)

(b) same ordinate; divide numbers on abscissa by 50

3

6

50 20 10

K 0.5 K 0.02

100 10

1 0.5

9.82 H 0.5 9.82 24.55 H, 31.8 H 31.8 795 H

0.02 0.02

2.57

2.57nF 257nF

0.5 0.02

×

== = =

∴

µ

→× × =

µµ

→×=

µ

→=

×

mf

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

(a)

(b) 2( /5 10) 0.1( 50)

K2,K5Z()20( / 5 5) 25

++

==∴→ =

++

mf in

ss

sss

(c) 11

0.1 0.2 , 0.2 0.4 , 0.5F 0.05F, 0.5I 0.5IΩ→ Ω Ω→ Ω → →

11

Apply 1 V I 10A 0.5 I 5A ; 5A 0.2 can be replaced by 1 V in series with 0.2

1 ( 1) 2 4 20 20( 5) 10

I10 10 Z()

0.2 2 / 0.2 2 0.2 2 10 20( 5)

∴= ∴ = ↓ Ω Ω

−− + + +

∴→=+ =+ = = ∴ =

++++ +

in in

ss s s

s

ssss s

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34.

(a)

(b)

(c) 6

610

10 rad/s, Q stays the same, B 21.40 krad/s

46.73

ω= ∴ = =

oo

36 4

4

434

,8 , 2

43

2 2

, ,

46

1/ (2 8)10 10 10 rad/s

10

Q 10 /8 10 10 125 R 0.64

125

10 10 10

2 8 10mH Q 156.25

0.64

1

R 0.64 156.25 15.625 ; Q 100, R 100 1 10

10 10

R 20 15.625 10 4.673 Q 10 10 4

−−

−

ω+ =

=× =∴==Ω

××

+= ∴ = =

∴=× = Ω= = =×=Ω

×

∴= = Ω∴=× ×

B

o

LLS

L

LP C CP

Po

kk

k3

.673 10 46.73×=

64

K 10 /10 100, K 1 R s stay the same; 2 mH 20 H, 8mH 80 H,1 F 10nF

′

===∴ →µ→µµ→

fm

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35.

(a)

(b)

3

0.1

K 250, K 400 0.1F 1 F

250 400

2 250

5 1250 , 2H 1.25H, 4I 10 I

400

==∴→ =µ

×

Ω→ Ω → = →

mf

xx

36

1250

3

6363

3

33

1

10 . Apply 1 V I 10 , I 1250

110

1000 I 10 I 1.25

1 0.8 0.8

I 10 (1 10 ) 10 ; 10

1250

0.8 10 1 1000

I 10 0.2 10 Z 5 V 0

I 0.2

−

−−−

−

−−

ω= ∴ = ↓ =

−

∴=∴→=

∴= + + − = + =

×

∴= + = × ∴ = = =− Ω =

x

xL

in

in th oc

in

s

ss

ssssj

ss

jj jk

jj

1 µF

1.25 H

1250 Ω 103

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36.

(a) I 2 0 A, 50 V 60 25 V=∠° ω= ∴ = ∠ °

s out

(b) I 2 40 A, 50 V 60 65 V=∠ ° ω= ∴ = ∠ °

sout

(c) I 2 40 A, 200, OTSK=∠ ° ω= ∴

s

(d) K 30, I 2 40 A, 50 V 1800 65 V==∠°ω=∴=∠°

m S out

(e) K 30, K 4, I 2 40 A, 200 V 1800 65 V

===∠°ω=∴=∠°

m f s out

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37.

(a) H /( ) 0.2 H 20 log 0.2 13.979dB=∴= =−

dB

s

(b) H( ) 50 H 20 log 50 33.98dB=∴= =

dB

s

(c)

(d) 37.6 / 20

H 37.6dB H( ) 10 75.86=∴==

dB s

(e) 8/20

H 8dB H( ) 10 0.3981

−

=− ∴ = =

dB s

(f) 0.01/ 20

H 0.01dB H( ) 10 1.0012=∴==

dB s

12 26 6 13 292 380

H( 10) H 20 log 20 log 6.451dB

2 10 20 10 1 5 10 5 60 220

+

=+ ∴= += =

++ ++ −+

dB

j

jjj jj j

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38. (d) MATLAB verification- shown adjacent to Bode plots below.

(a) 20( 1) 0.2(1 )

H( ) , 0.2 14dB

100 1 /100

++

== →−

++

ss

sss

(b) 22

2000( 1) 0.2 (1 )

H( ) , 0.2 14dB

( 100) (1 /100)

++

== →−

++

ss ss

sss

(c)

2

200 45 200 ( 5)( 40) 200(1 / 5)(1 / 40)

H( ) 45 , 200 46dB

++ ++ + +

=+ + = = = →

ss ss s s

1 10

100

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

2

V (20 2 )(182 200/ ) 200/

H( ) 202 2 200/ 182 200/

400( 10) 200(10 )

2( 101 100) (1 )(100 )

20(1 /10)

H( ) , 20 26dB

(1 )(1 /100)

++

== ×

++ +

++

==

++ + +

+

=→

++

C

R

sss

sIss s

ss

ss s s

s

sss

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40.

(a) 8

33

5 10 ( 100) 2.5 (1 /100)

H( ) , 2.5 8dB

( 20)( 1000) (1 / 20)(1 /1000)

×+ +

== →

++ + +

ss s s

sss s s

(b)

(c)

29

33

Corners: 20, 34dB;

100, 34dB;

1000, 54dB

Intercepts: 0dB, 2.5 1, 0.4

2.5 ( /100) 2.5 (20)10

1, 8dB; 0dB, 1 22,360 rad/s

( / 20)( /1000) 100

ω=

ω= ω=

ωω ω

ω= = = ∴ω=

ωω ωω

2

223

Corners: 20, 31.13dB

1 ( /100)

100, 36.69dB H 20 log 2.5 [1 ( / 20) ][1 ( /1000) ]

1000, 44.99dB

ω=

+ω

ω= = ω +ω +ω

ω=

dB

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

(a)

8

33

5 10 ( 100) 2.5 (1 /100)

H( ) ,

( 20)( 1000) (1 / 20)(1 /1000)

×+ +

==

++ + +

ss s s

sss s s

(b)

(c)

2: 90

10

10 : 90 45 45 log 58.5

20

100 100

100 : 90 45 45 log 45 45 log 58.5

20 100

200 200

200 : 90 90 45 45 log 3 45 45 log 17.9

100 100

1000 : 90 9

ω

=∠=°



=∠=°−°+° = °







= ∠=°− °+° + °+° = °







=∠=°−°+°+° −°+° =°





=∠=°− 1000

0 90 3 45 45 log 45

1000

10,000 : 90 90 90 3 90 180

ω



°+ °− °+ ° =− °





=∠=°−°+°−×°=−°

111

11 1

1

2 : 90 tan 0.02 tan 0.1 3tan 0.002 85.09

10 : 90 tan 0.1 tan 0.5 3tan 0.01 67.43

100 : 90 tan 1 tan 5 3tan 0.1 39.18

200 : 90 tan 2 tan 10 3tan 0.2 35.22

1000 : 90 tan 10 t

ω

−−−

−− −

−

=∠=°+ − − = °

=∠=°+ − − =°

=∠=°+ −

11

11 1

an 50 3tan 1 49.56

10,000 : 90 tan 100 tan 500 3tan 10 163.33

ω

−−

−− −

−=−°

=∠=°+− − =−°

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42.

(a)

(b)

(c)

2

22

2

20 400 20 400

H( ) 1

1 2 0.5( / 20) ( / 20)

400

20, 0.5

20 log 400 52dB

Correction at is 20 log 2 0 dB

o

ss

sss s

ss

s

ωζ

++

=+ + =

+× +

=

∴= =

=

5: H 52 2 20 log 5 24.0dB(plot)

H 20 log 1 16 4 23.8dB (exact)

100: H 0dB (plot)

H 20 log 1 0.04 0.2 0.170 dB (exact)

dB

j

ω

==−× =

=−+=

==

=−+=−

Hdb

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43.

(a)

(c) 0.5

20, H( 20) H 15.68 dB H( 20) 80.54

1 4 0.5 dB

j

jj

j

ω

== ∴=−∠=−°

−+

2

5

V 25 25 0.025

H( ) V 10 25 1000/ 10 25 1000 1

12

810 10

1

10, 1/8 correction 20 log 2 12 dB

8

0.025 32 dB

== = =

++ + +



++







∴= = ∴ =− ×=





→−

R

o

ss

sssss ss

ωζ

(b)

HdB ang(H)

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44.

36

12 3 36

1/(50 10 10 ) 20

1st two stages, H ( ) H ( ) 10; H ( ) 1/(200 10 10 ) 5

20 400

H( ) ( 10)( 10) 51/5

400 52 dB

ss s

ss

sss

−

−×× −

==− = =

+×× +

−−



∴=−− =



++



−→

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45.

(a)

20log10(0.1) = -20 dB

(b) (c)

5

11 1

1

55

1

56

1st stage: C 1 F, R , R 10 H (S) R C 0.1

1/R C

2nd stage: R 10 , R 10 , C 1 F H ( ) 1/R C

1/(10 10 ) 10

H() 1/(10 10 ) 10

3rd stage: same as 2nd

1

H( ) ( 0.1 )

AAfA AfAA

BfB

BfBfB B

fB fB

B

ss

sss

ss

µ

−

==∞=∴=−=−

−

===∴=

+

×

∴= =−

+× +

−

∴=− 2

0 10 0.1

10 10 (1 /10)

s

ss s

−



=−



++ +



CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46. An amplifier that rejects high-frequency signals is required. There is some ambiguity in

the requirements, as social conversations may include frequencies up to 50 kHz, and

echolocation sounds, which we are asked to filter out, may begin below this value.

Without further information, we decide to set the filter cutoff frequency at 50 kHz to

ensure we do not lose information. However, we note that this decision is not necessarily

the only correct one.

Our input source is a microphone modeled as a sinusoidal voltage source having a peak

amplitude of 15 mV in series with a 1-Ω resistor. Our output device is an earphone

modeled as a 1-kΩ resistor. A voltage of 15 mV from the microphone should correspond

to about 1 V at the earphone according to the specifications, requiring a gain of 1000/15

= 66.7.

If we select a non-inverting op amp topology, we then need 65.7 1- 66.7

1

==

R

Rf

Arbitrarily choosing R1 = 1 kΩ, we then need Rf = 65.7 kΩ. This completes the

amplification part. Next, we need to filter out frequencies greater than 50 kHz.

Placing a capacitor across the microphone terminals will “short out” high frequencies.

We design for

ω

c = 2πfc = 2π(50×103) =

filtermicCR

1. Since Rmic = 1 Ω, we require

Cfilter = 3.183 µF.

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47. We choose a simple series RLC circuit. It was shown in the text that the “gain” of the

circuit with the output taken across the resistor is

()

[]

2

1

222

2

2 -1

RC

CRLC

AV

ωω

ω

+

=.

This results in a bandpass filter with corner frequencies at

LC

LCCR-RC

L

c2

4

22 ++

=

ω

and LC

LCCRRC

H

c2

4

22 ++

=

ω

If we take our output across the inductor-capacitor combination instead, we obtain the

opposite curve- i.e. a bandstop filter with the same cutoff frequencies. Thus, we want

2π(20) = LC

LCCR-RC

2

4

22 ++ and 2π(20×103) = LC

LCCRRC

2

4

22 ++

Noting that

H

c

ω

–

L

c

ω

= R/L = 125.5 krad/s, we arbitrarily select R = 1 kΩ, so that L =

7.966 mH. Returning to either cutoff frequency expression, we then find C = 7.950 µF

PSpice verification. The circuit

performs as required, with a

lower corner frequency of about

20 Hz and an upper corner

frequency of about 20 kHz.

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48. We choose a simple RC filter topology:

Where RC1

1

V

in

out

ω

j+

= and hence

()

2

in

out

RC1

1

V

ω

+

=. We desire a cutoff

frequency of 1 kHz, and note that this circuit does indeed act as a low-pass filter (higher

frequency signals lead to the capacitor appearing more and more as a short circuit).

Thus,

()

2

1

RC1

1

2

c

=

+

=

ω

where

ω

c = 2πfc = 2000π rad/s.

A small amount of algebra yields 1 + [2π(1000)RC]2 = 2 or 2000πRC = 1. Arbitrarily

setting R = 1 kΩ, we then find that C = 159.2 nF. The operation of the filter is verified in

the PSpice simulation below:

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49. We are not provided with the actual spectral shape of the noise signal, although the

reduction to 1% of its peak value (a drop of 40 dB) by 1 kHz is useful to know. If we

place a simple high-pass RC filter at the input of an op amp stage, designing for a pole at

2.5 kHz should ensure an essentially flat response above 25 kHz, and a 3 dB reduction at

2.5 kHz. If greater tolerance is required, the 40 dB reduction at 1 kHz allows the pole to

be moved to a frequency even closer to 1 kHz. The PSpice simulation below shows a

filter with R = 1 kΩ (arbitrarily chosen) and C = nF 63.66

)1000)(105.2(2

1

3=

×

π

.

At a frequency of 25 kHz, the filter shows minimal gain reduction, but at 1 kHz any

signal is reduced by more than 8 dB.

We therefore design a simple non-inverting op amp circuit such as the one below, which

with Rf = 100 kΩ and R1 = 1 kΩ, has a gain of 100 V/V. In simulating the circuit, a gain

of approximately 40 dB at 25 kHz was noted, although the gain dropped at higher

frequencies, reaching 37 dB around 80 kHz. Thus, to completely assess the suitability of

design, more information regarding the frequency spectrum of the “failure” signals would

be required.

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50. We select a simple series RLC circuit with the output taken across the resistor to serve as

a bandpass filter with 500 Hz and 5000 Hz cutoff frequencies. From Example 16.12, we

know that

(500)2 4LC CR

2LC

1

2L

R

- 22

πω

=++=

L

c

and

(5000)2 4LC CR

2LC

1

2L

R

22

H

πω

=++=

c

With - LH cc

ωω

= 2p(5000 – 500) = R/L, we (arbitrarily) select R = 1 kΩ, so that L

= 35.37 mH. Substituting these two values into the equation for the high-frequency

cutoff, we find that C = 286.3 nF. We complete the design by selecting R1 = 1 kΩ and Rf

= 1 kΩ for a gain of 2 (no value of gain was specified). As seen in the PSpice simulation

results shown below, the circuit performs as specified at maximum gain (6 dB or 2 V/V),

with cutoff frequencies of approximately 500 and 5000 KHz and a peak gain of 6 dB.

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



51. For this circuit, we simply need to connect a low-pass filter to the input of a non-

inverting op amp having Rf/R1 = 9 (for a gain of 10). If we use a simple RC filter, the

cutoff frequency is

(3000)2

RC

1

πω

==

c

Selecting (arbitrarily) R = 1 kΩ, we find C = 53.05 nF. The PSpice simulation below

shows that our design does indeed have a bandwidth of 3 kHz and a peak gain of 10 V/V

(20 dB).

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



52. We require four filter stages, and choose to implement the circuit using op amps to isolate each filter sub-

circuit. Selecting a bandwidth of 1 rad/s (no specification was given) and a simple RLC filter as suggested

in the problem statement, a resistance value of 1 Ω leads to an inductor value of 1 H (bandwidth for this

type of filter =

ω

H –

ω

L = R/L). The capacitance is found by designing each filter’s respective resonant

frequency ( LC1) at the desired “notch” frequency. Thus, we require CF1 = 10.13 µF, CF2 = 2.533 µF,

CF3 = 1.126 µF and CF4 = 633.3 nF.

The Student Version of PSpice® will not permit more than 64 nodes, so that the total solution must be

simulated in two parts. The half with the filters for notching out 50 and 100 Hz components is shown

below; an additional two op amp stages are required to complete the design.

CHAPTER SIXTEEN SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



53. Using the series RLC circuit suggested, we decide to design for a bandwidth of 1 rad/ s

(as no specification was provided). With

ω

H –

ω

L = R/ L, we arbitrarily select R = 1 Ω so

that L = 1 H. The capacitance required is obtained by setting the resonant frequency of

the circuit ( LC1) equal to 60 Hz (120π rad/s). This yields C = 7.04 µF.

1 H

1 Ω

7.04 µF

vin vout

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1.

3

17 8 3

8 17 4 17(273) 8( 148) 3(83) 3208

3417

−−

∆=− −= + − − = Ω

−−

Z

(a) 2

11

3208 100

Z 11.751 P 851.0 W

273 11.751

∆

== = Ω∴= =

∆Z

in

(b) 2

22

3208 100

Z 11.457 P 872.8 W

280 11.457

∆

== = Ω∴= =

∆Z

in

(c) 2

33

3208 100

Z 14.258 P 701.4

225 14.258

∆

== = Ω∴= = Ω

∆Z

in

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2.

3

0.35 0.1 0.2

0.1 0.5 0.15 0.35(0.3525) 0.1( 0.105) 0.2(0.115) 0.089875 S

0.2 0.15 0.75

−−

∆=− − = + − − =

−−

Y

(a) 2

11

0.089875 10

Y 0.254965 P 392.2 W

0.3525 0.254965

Y

in

∆

== = ∴= =

∆

(b) 2

22

0.089875 10

Y 0.403933 P 247.6 W

0.2225 0.403933

Y

in

∆

== = ∴= =

∆

(c) 33

0.089875 100

Y 0.544697 S P 183.59 W

0.165 0.544697

in == ∴==

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3.

4

11

3120 413 120 120

14 13

[R] ()325212522413

22 52 326 326 326

0326

3[4(34) 2(12) 3( 13)] [ 1(34) 2( 12) 3( 4)] 2[ 1(12) 4( 12) 3( 6)]

161

3(73) ( 22) 2(18) 161 R 2.205

73

R

in +

−−

 −− −−



−



=Ω=+−



−−− −− −−



−−



= − −− +− −− −− =− −− −−

∆

=+−−=Ω∴=== Ω

∆

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4. Define a counter-clockwise current I2 in the left-most mesh, and a counter-clockwise

current I1 flowing in the right-most mesh. Then,

12 1 2

11 22 12

2122

12

2

11

2

V 4I 0.2V 0.8I

V I 5( I 0.8 I I ) ( 5) I I

Also, I (2 4) 5(I 0.8I I ) 0

or 0 5I (5 2 )I

( 5)(5 2 ) 5 2 15 20, 5 2

21520

Z25

in

Z

th

ss

s

ssss s

ss

s

=∴ =

=+ + −=+ −

+− + − =

=− + +

∴∆ = + + − = + + ∆ = +

++

∴= +

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5. Define a clockwise mesh current I1 flowing in the bottom left mesh, a clockwise mesh

current I2 flowing in the top mesh, and a clockwise mesh current I3 flowing in the

bottom right mesh. Then,

(a)

(b)

12 2 1 2

213 1 23

232 23

V 10(I I ) 0.6 8I 10I 14.8I

0 50I 10I 12I 10I 50I 12I

0 4.8I 17I 12I 7.2I 17I

10 14.8 0 5120

10 50 12 10(763.6) 10( 251.6) 5120 Z 6.705

763.6

0 7.2 17

in

Zin +

=−−×=−

=−−=−+−

=+−=−+

−

∴∆ = − − = + − = ∴ = = Ω

−

112 12

221 2 12

21 12

V 0.6VVV

I 0.13571 V 0.03571V 0.06V

28 10

V 0.6VVV V

0 0.03571V 0.31905V 0.05V

28 12 5

V VVV

0 0.05V 0.05V 0.175V

820

0.13571 0.03571 0.06

0.03571 0.31905 0.05

0.05 0.05 0.1

x

in x

xx

xx x

y−

−−

=+ = − −

−−

=+ +=− + −

−−

=− + =− + −

−−

∴∆ = − −

−−

11

0.13571( 0.05583 0.0025) 0.03571(0.00625 0.003)

75

0.05(0.00179 0.01914) 0.00724

0.007954

0.007954, 0.05333 Y 0.14926 S

0.05333

1

Z 6.705

0.14926

y

ym

in +

=−++ +

−+=−

∆−

∴∆ = ∆ =− ∴ = = =

∆−

∴= = Ω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6.

VV

0.1 (V 1) 0

210 5

V (0.5 0.3 ) 0.1 0.2

0.2 0.1

V0.3 0.5

0.2 0.1 0.1 0.4

I (1 V ) 1 0.2 0.2

5 0.3 0.5 0.3 0.5

(0.1 0.4) 1.5 2.5 15 25

YI ,Z

1.5 2.5 (0.1 0.4) ( 4)

xx x

x

out out

ss

s

ss s

ss

ss s s

sssss

+−+ −=

∴+=+

+

∴= +

++



∴= − = − =



++



+++

∴== = =

+++

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7.

44

V 1 V, V 0 V V 0, V 1 V

V1

I ;210I 210I 0

RR

1

II R V/I

R

in i x in x

x

xinx

xx

inx inininx

x

R

==∴+==−

==− × +× =

∴=−= ∴ =− =−

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

(a) Assume 1 V at input. Since Vi = 0 at each op-amp input, 1 V is present between R2 and

R3, and also C and R4.

44

3

3434

2

23 12 22

34 34

13412 2

1

1134 2

11 1

VR 1

RCCR

11 1

I11

RCRCRR

1R

II V1RI1

CR R CR R

RRR1V R 1

IIZC

RCRRR I R



∴= + =+







∴= −− =−





∴==− ∴ =+ =−

−

== =∴==

in in

in

jj

j

ωω

ω

(b)

333310

1234

10 6 3

R410, R1010, R1010, R10, C210

4101

Z 2 10 10 0.8 10 (L 0.8 mH)

10

−

=× = × = × = =×

××

∴= × × = ×Ω =

in in

jj

ωω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9. Define a clockwise mesh current I1 in the left-most mesh, a clockwise mesh current Ix in

the center mesh, and a counter-clockwise mesh current I2 in the right-most mesh. Then,

112

1

2

121

22

12

112

11 12

V 13I 10I

V100

03520

V2022

0 10I 35I 20I I 13 10 0

10 35 20

02022

V 20I 22I

V (370) V ( 200) 37 20

I V V

13(370) 10( 220) 261 261

37 20

141.76 mS, 76.63 mS

261 261

=−

−

=− + + ∴ = −

−

=+

+−

∴= = −

+−

−

∴= = = =−

x

yy

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10.

11 2

212 112 2

12112

2212 2121

11

10 5

[ ] (mS) I 0.01V 0.005V ,

50 20

I 0.05V 0.02V , 100 25I V , V 100I

100 0.25V 0.125V V 1.25V 0.125V

5

I 0.01V 0.05V 0.02V 0.03V 0.05V V V

3

0.625 4.375

100 1.25V V V

32

y−



=∴=−





=+ =+=−

∴= − += −

=− = + ∴− = ∴ =−

∴= + = 11 2 1

300 5

V 68.57 V, V V 114.29 V

4.375 3

∴= = =− =−

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

12

112

2

21 11 2 1 2

11 12 21 22

VV

I 0.04V 0.04V

25

V

I 2I I I 0.01V 0.04V 0.03V

100

0.04S, 0.04S, 0.04S, 0.03 Syy y y

−

==−

= + −=+ = −

∴= =− = =−

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12.

111 1 1

22 21 12

21 22 1 2

V 100(I 0.5I ) 50I I 0.02 V

V 300I 200(I 0.5I ) 100I 500I

V 2V 500I , I 0.004V 0.002V

0.02 0

[] (S)

0.004 0.002

y

∴= − = ∴=

=+ +=+

∴= + =− +



∴=



−



CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13.

0.1 0.0025

[] (S)

8 0.05

y−



=

−



(a)

(b)

11 221 2

122

2212 2121

21 111 12 11 1

21 11

I 0.1V 0.0025V , I 8V 0.05V

1 2I V, V 5I

I 0.2V 8V 0.05V 0.25V 8V , V / V 32

I 8V 0.05 32V , I 0.1V 0.0025 32V I 6.4V , I 0.02V

6.4

I / I 320, V / I 50

0.02

=− =−+

=+ =−

∴=− =− + ∴ = =

=− + × = − × ∴ =− =

−

∴==− =Ω

111 1 22 1 2

111 21 2

12 2 2 2 2

2

V 2I , I 0.1V 0.0025V , I 8V 0.05V

I 0.5V 0.1V 0.0025V 0.6V 0.0025V

11

V V / 240, I 8 V / 240 V V

20 60

V60

I

=− = − =− +

∴=− = − ∴ =

∴= =−× + =

∴=Ω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14. 10 5

[] (mS)

20 2

y−



=

−



(a)

(b)

11 22 1 2

111

111 121 2

212 22 2

I 0.01V 0.005V , I 0.02V 0.002V

V100IV

V V1 100I I 0.01V1 I 0.005V I 0.005V1 0.0025V

I 0.02V1 2I 0.002V 0.02V1 0.01V1 0.005V 0.002V 0.01V1 0.003V

0.005 0.0025

[] 0.01 0.0

new

y

=− =−+

′=+

∴= − ∴= −− ∴= −

=− + + =− + − + =− −

−

∴=

−− (S)

03







22 2 2

21 2

21 21

11 2 1 1

11 2

V2 100I V , V V2 100I

I 0.02V 0.002V2 0.2I

11

1.2I 0.02V 0.002V2 I V V2

60 600

11

I 0.01V 0.005(V2 100I ) 0.01V 0.005V2 0.5 V V2

60 600

11 1 1

I V V

100 120 200 1200

=+∴=−

∴=− + −

∴=− + ∴=− +



=− −=− +−+





 

∴= − − −

 

  12

11

V V

600 240

=−

1/ 600 1/ 240

[] (S)

1/ 60 1/ 600

new

y−



∴=



−



CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

1

VS 2

VS 1

I 2

I

Exp #1 100 V 50 V 5 A -32.5 A

Exp #2 50 110 -20 -5

Exp #3 20 0 4 -8

Exp #4 -8.333 -22.22 5 0

Exp #5 -58.33 -55.56 5 15

1 11 1 12 2

2 21 1 22 2

11 12 21 22

11 12, 21 22 21 22

11 12 12 12

IVV

Use 1st 2 rows to find y's

5 100 50 , 32.5 100 50

20 50 100 5 50 100 10 100 200

40 100 200 Substracting, 150 45 0.3 S

5 100

=+

∴= + − = +

−= + −= + →∴−= +

∴− = + =− ∴ =−

∴=

yy

yy yy

yy yy y y

yy y y

y11 11 22

22 21 21

12

15 0.2 S Subtracting 22.5 150

0.2 0.3

0.15 S 32.5 100 7.5 0.4 S [ ] (S)

0.4 0.15

Completing row 3: I 0.2 20 4 A, I 0.4 20 8 A

Completing row 4: 5 0.2V 0.3V , 0 0

−∴ = =

−



∴= ∴− = +∴=− ∴=



−



=×= =−×=−

=− =−

SS

yy

yyyy

1221

11 11 2

12 1 2

1

8

.4V 0.15V V V

3

50

5 0.2V 0.8V 0.6V V 8.333 V, V 22.22 V

6

Completing row 5: 5 0.2V 0.3V , 15 0.4V 0.15V

5 0.3

15 0.15 0.75 4.5 5.25

V 58.33

0.2 0.3 0.03 0.12 0.09

0.4 0.15

+∴=

∴ = − =− ∴ =− =− =−

=− =−+

−

+

∴= = = =−

−−−

−

SSSS

SS SS S

SS S S

S2

0.2 5

0.4 15

V, V 55.56 V

0.09

−

==−

−

S

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16.

1

2

61 63 31

:1 6 3 10 0.6, 1.8, 0.3

10 10 10

51 14 54

: 5 1 4 10 0.5, 0.4, 2

10 10 10

1.8 2 0.5 4.3 , 0.3 0.6 0.4 1.3

1.3 4.3 0.99821 , 0.9982 0.6 2 3.598

3.598 6 2.249

×××

∆++=Ω→ = = =

×× ×

∆++=Ω→ = = =

++ = Ω + + = Ω

=Ω++=Ω

=Ω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

2

622336 36

36/ 6 6, 36/ 2 18, 36 / 3 12

12 4 3, 6 12 4

4 3 18 25

43

3 18/ 25 2.16, 4 18/ 25 2.88, 0.48

25

9.48 2.16 9.48 2.88 2.88 2.16 54

54 54 54

18.75, 25, 5.6962, 75 18.75 15, 100 25 20

2.88 2.16 9.48

(15 20) 5.696 4.

×+×+×= Ω

== =

==

++ = Ω

×

×= ×= =

×+×+×=Ω

=== ==

+=899 R 5 4.899 9.899

in

∴=+ = Ω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18.

:6 431

24 12 18

24, 12, 18, 18 18 0(S.C)

111

ignore 12, 6

24 12 12

12 12 Z

∆−++=

−

=− = =− − =

∴−

−+=−

−=∞∴=∞

in

jjjj

jjjjj

jjj

jj

jj j

jj

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19. 0.4 0.002

[] (S)

5 0.04

y−



=

−



(a)

(b) 2

122221

2

0.05V

I 0.4(0.018)V 0.002V 0.0052V G I / I 9.615

0.0052V

I+

−

=−=∴===−

(c) G G G 55.56 9.615 534.2

pVI +

=− = × =

(d) 12 111

1

I 0.0052V 0.0052 55.56V Z V / I 3.462

0.0052 55.56

in

==×∴== =Ω

×

(e)

11 221 22 2 11

2212 21 21

I 0.4V 0.002V , I 5V 0.04V ,V 20I , V V 2I

500

I 0.05V 5V 0.4V 0.09V 5V G V / V 55.56

9

S

V

=− =−+ =− =+

=− =− + ∴− =− ∴ = = =

11 1 1 1 21 2

222222

0.002

V 2I , V 0 I 0.5V 0.4V 0.002V V V

0.9

0.002

I 5 V 0.04V 0.02889V Z V / I 34.62

0.9

S

out

=− = ∴ =− = − ∴ =



=− + = ∴ = = Ω





CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20. 0.1 0.05

[] (S)

0.5 0.2

y−



=

−



(a)

(b) 1

21

1

( 0.5 0.2 1.25)V

G I / I 6.667

(0.1 0.005 1.25)V

I

−+×

== =−

−×

(c) G 1.25 6.667 8.333

P=× =

(d) 1111

I (0.1 0.05 1.25)V Z V / I 26.67

in

=−× ∴= = Ω

(e)

(f) ,12

G V / V 0.25

Vrev==

(g)

11 2

2121122

212 21

I 0.1V 0.05V

I 0.5V 0.2V , 1 10I V , I 0.2V

0.2V 0.5V 0.2V G V / V 1.25

V

=−

=− + = + =−

∴− =− + ∴ = =

111 11 2

122 22 2 22

V 0, V 10I I 0.1V 0.1V 0.05V

V 0.25V , I 0.05(0.25V ) 0.2V 0.075V Z V / I 13.333

==−∴=−=−

∴= ∴=− + = ∴ = + Ω

S

out

11

22

11 2 2

2

With 2 port: 1 10I 26.67I

6.667 1

1 36.67 I , I 1/ 36.67 I 0.15182 P I 5 2.5(0.15182) 0.08264 W

36.67 2

1 1 0.08264

Without 2 port: P 5 0.011111 W G 7.438

2 15 0.011111

L

Lins

=+

−

∴= = ∴ = =− ∴ = × = =



=×= ∴= =





CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

(a)

(b) 2 in :

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22.

(a) 22

12 21

12

22

1/R 1/R

VV VV

I, I []

1/R 1/R

RR

a

y−



−−

== =



−



(b) 1

1112 23

3

1/R 0

I V /R , I V /R [ ] 01/R

B

y

==∴=





(c)

12 2

112

1

232

12

12 2

221

2

232

32

1/R 1/R 1/R

VVV

I[]

1/R 1/R 1/R

RR

1/R 1/R 1/R

VVV

I,[][]

1/R 1/R 1/R

RR ab

y

yy

+−



−

=+ ∴=



−+



+−



−

=+ + =



−+



CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23.

11 2 2 1 1

22 12 1 1 1

21111

112122121

212

V 8I 0.1V V 10V 80I

1

I V /12 0.05V I (10V 80I ) 0.05V

12

5 1 20 53 20

I V I V I

6 20 3 60 3

60 20 400 60 4000 600

V III IV I I80I

53 3 53 53 53 53

7

240 600

V I I[]

53 53 z

=+ ∴= −

=+ ∴= −+



∴= + − = −







∴= + = + ∴= + −





∴=− + ∴=.547 1.1321 ()

4.528 11.321



Ω



−



CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24.

(a)

(b) 2

12

1

V 1.7429 3.050

I I 1 A 0.9607

V 2.723 2.266

+

== ∴ = =

+

12112

11 221 221

21 2

1 1

2 2

12

1122

12

2

I 0.02V 0.2V 0.5V 0.5V

I 0.7V 0.52V I 0.1V 0.125V 0.5V 0.5V

I 0.4V 0.625V

I 0.52 0.7 I

I 0.625 0.4 I

0.625I 0.52I

V 2.723I 2.266I , V

0.7 0.52 0.2295 0.2295

0.4 0.625

0.4I 0.7 I

V0.

=− + + −

∴= − = + + −

∴=− +

−

+

∴= = = + =

−

+

∴= 12

2.723 2.266

1.7429I 3.050I [ ] ( )

1.7429 3.050

2295 z

=+∴= Ω





CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25. 4 1.5

[ ] ( ), R 5 , R 2

10 3 SL

z

=Ω=Ω=Ω





(a) 11 22 122 2 2 121

V 4I 1.5I , V 10I 3I , V 2I 10I 3I G I / I 2= + = + =− = + ∴ = =−

(b) 11

21

1

10I 6I

GV/V 4

4I 3I

v

−

== =

−

(c) GGG8=− =

pVI

(d) 1

1111

1

V

V4I3II Z 1

I

in

=−=∴==Ω

(e) 11121 22 22 2

1108

V 5I 4I 1.5I I I V I 3I I Z 1.3333

666

out

=− = + ∴ =− ∴ =− + = ∴ = Ω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26. 1000 100

[] ( )

2000 400

z

=Ω



−



(a)

(b) 26

500

P 15.625 500 10 122.07 mW

−

=××=

(c)

1122 12 112 2

21221 111

26

1 2 1 200

V 1000I 100I , V 2000I 400I , 10 200I V ,V 500I

20 2000

500I 2000I 400I , I I ; 10 200I 1000I I

99

20

I 7.031 mA, I I 15.625 mA P 7.031 200 10 9.888 mW

9

−

=+ =−+ =+=−

∴− =− + = ∴ = + +

∴= ∴= = ∴ = × × =

1 2 200 500

P 10I 70.31 mW(gen) P P P P 70.31 9.89 122.07

S port S

== ∴ =−−= −− ∴

2

P61.65 mW

port −

∴

=−

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

85 4 4

11 1 12

54 4

112

44

221 21

444

212

4

1

44 4

2

1544

24 44

10 , I 10 V 5 10 V 10 (V V )

I (10 6 10 ) V 10 V

I 10 V 0.01V 10 (V V )

I (0.01 10 ) V (10 10 )V

I10

I10 10 (10

V10 6 10 10

10 10 10 10

−−−

−− −

−−

−−−

−

−− −

−−−

−− −−

==+×+ −

∴= + × −

=++ −

∴= − + +

−

++

∴= =

+× −

−+

jj

j

jj

j

jj

j

ω

44

11

12

6

12

54

1

24 21

2

2622

133.15 47.64

10 )I 10 I

1.0621 10 92.640 94.15 2.642

10 6 10 I

9416 86.78

10 10 I

V565.0 3.60

1.0621 10 92.64

−

−−

−+

−−

−

∴= ∠− °Ω

+

×∠ =∠−°Ω

+×

∴= ∠ °Ω

−

==∠−°Ω

×∠ °

z

jj

z

j

z

j

z

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28.

111 1 2 1 2

2121 22 2 22

222 2 2

20 2

[ ] ( ), V 100 0 V, R 5 , R 25

40 10

100 5I V , V 20I 2I 100 25I 2I

1 1 25 25

V 40I 10I I V I 100 V I 2I

40 4 40 4

517 817

100 V I V 160 I 160 6.8I

84 54

V 160 V, R 6.8

SSL

th th

z

=Ω=∠°=Ω=Ω





=+ = + ∴ = +

=+ ∴= − ∴= − +

∴= − ∴=+× =+

∴= =Ω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29. 92

[] 20 0.2 S

hΩ−



=



(a)

(b)

11 22 1 2111 1

111 11 1 21 2

V 9I 2V , I 20I 0.2V , V 1I V Eliminate V

10 2

VV I V I9I2V, V10I2V [] 20 0.2 S

new

h

′

=− = + =+

Ω−



′′ ′

∴= −∴−= − = − ∴ =





11 22 1 22 22

2222

11 222 1 2 2 2 1 2

21 2112 1 2

11 2

V 9I 2V, I 20I 0.2V, V 1I V

Eliminate V V V I

V 9I 2V 2I , I 20I 0.2V 0.2I 1.2I 20I 0.2V

I 16.667I 0.16667V V 9I 2V 2(16.667 I 0.1667 V )

V 42.38 I 1.6667V [ ]new

h

′

=− = + =+

′

∴= −

′′

=− + = + − ∴ = +

′′ ′

∴= + = − + +

′

∴= − ∴ 42.33 1.6667

16.667 0.16667 S

Ω−



=



CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30.

11 221 212 21

11 1 1

11121 2

2

100 0.01

R 100 , R 500 [ ] 20 1 mS

Z : V 100I 0.01V , I 20I 0.001V 20I 0.5I 1.5I 20I

20

V 100I 0.01( 500) I 33.33I Z 33.33

1.5

0.01

Z : V 100I 100I 0.01V I V

200

0.01

I202

Ω



=Ω =Ω =





= + =+ =− ∴=

∴= + − = ∴ = Ω

=− = + ∴ =−

=−

SL

in

out

h

22

V 0.001 I 0 Z

00



+=∴=∞



 out

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31.

(a)

(b)

1

12 12

20

V 1.2

From above, 9.6

I 0.125

I

zz

=

===Ω

(c)

1

12 1 2 0 2

10 1 4 2

22 22

112

V / V | Let V 1 V

I 0.1 A, I 0 I 0.2I

0.1 I 0.2 I 0.8I , I 0.125 A

V 0.3 4(0.2)(0.125) 1 1.2 V 1.2

I

h

=

Ω

==

∴↓= =∴ ←=

∴=− = =

∴= − += ∴=

1

12 1 2 2

0

21

112

I / V SC input Let V 1 V

1.3 1.3

I 0.1 0.425 A, I 0.2(0.425)

44

I 0.24 A 0.24 S

V

y

=

==

=+= = −

∴=− ∴ =

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32. 1000 1

[] 4 500 S

h

µ

Ω−



=



(a)

(b) 2

3

21

4000 68.97

V 25.86 10 68.97 V P 4.756 W

1.5 1000

K

−

=××= ∴= =

(c)

112 12

43 3

21 2 2 1 2

21 111

26

200

100 200 I 1000I V 1200I V

I 4I 5 10 V 10 V 4I 1.5 10 V

4000 4000

V I 100 1200I I I 25.86 mA

1.5 1.5

P 25.86 10 200 133.77 mW

−− −

−

=+−=−

=+× =− ∴=−×

∴=− ∴ = + ∴=

∴= ××=

3

2

P 100 25.86 10 2.586 W (gen)

P 2.586 0.1338 4.756 2.304W

−

=× × =

∴= − − −−

S

port

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

(a)

(b)

()

5

11212

44

22 12 12

24

2122

4

21 2 4

V 1000 I 10 V 1000I 0.01V

V 10 I 100V I 10 (100V V )

I 10 (1000I 0.01V ) 10 V

1000 0.01

I10I210V [] 10 2 10 S

h

−

−−

−

=+=+

=− ∴= +

∴= + +

Ω



∴= +× ∴ =



×



1 112

46

1221222 2

6

22

V 200I 1000I 0.01V

111

I V I 10I 2 10 V V V 116.67 10 V

12,000 12,000 5000

Z V / I 10 /116.67 8.571

out k

−−

=− = +

−−

∴= ∴= +× = + + ×

∴= = = Ω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34.

(a)

(b)

12

1

11 2

12 12

2

11 2

21

VV

I1/R 1/R

VIRV RR

[] 1/R 1/R

II VV

IRR

[ ] parameters are all

R1

VIRV [] 10

II

y

z

h

∴= − −

=+ 

=

−

=− 

=− +

∞

=+ 

∴=



−

=− 

121 1 2

1

12212

12

2

21

[ ] parameters are

RR

VVVRIRI [] RR

V

IIVRIRI

R

VV

01

V

II []11/R

R

y

z

h

∞



==+∴=





=− = +

=



=− + ∴ =

−



CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35.

12 1

12 46 3241 305

[] ,[] ,[] ,[]

34 15 2350 2 31

442

ybc d

−





−−

    

=== =

    

−− −−

    

−



(a) 1246 64

[][] 34 15 838

yb −−



==



−



(b) 4 6 1 2 22 16

[][] 1 5 3 4 14 22

by −

 

==

 

−

 

(c) 46 324 1 0 2646 4

[][] 15 235 0 131321 1

bc −−

  

==

  

−− −

  

(d)

12 1

3241305 3 29

[][] 2 3 5 0 2 3 1 3 19 22

442

cd

−





−−−

 



==

 



−−−−−

 



−



(e) 6 4 3 2 9 6 64 34

[][][][] 8 38 3 19 22 138 738 908

ybcd −−− − −

  

==

  

−− −−−

  

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36.

(a)

(b)

1 11 2 12 2 1 21 2 22 2

22 12 11

112 12

2211 22

122

11 2 1 2 2 2

1

VVI, IV I

V V 1.5V V 1.5V V

V 10I V 1.5V , I 20 25 10

0.19 1

I 0.19V 0.31V , V V I

0.31 0.31

V 0.6129V 3.226 I

Then, 10I V (V 1.5V ) 2.5(0.6129V 3.226I ) V

I 0.05

tt tt=− =−

−−−

=+− =+ +

∴= − = −

∴= −

=− − = − −

∴= 22

0.6129 3.226

323V 0.8065 I [ ] 0.05323S 0.8065

t

−

Ω



−∴=





12212211

12222

22 22

Let R 15

V 0.06129V 3.226I , I 0.05323V 0.8065I , V 15I

15I 15(0.05323V 0.8065 I ) 0.6129V 3.226I

1.4114V 15.324I Z V / I 10.857

S

out

−

=Ω

∴= − = − =−

∴− =− − = −

∴= ∴==Ω

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37.

11 12 112

12221 22

122222

122

V 5I 0.3V V 1.3V 5I V

I 0.1V V / 4 I I 0.35V I

1.3V 5(0.35V I ) V 2.75 V 5I

2.115 3.846

V 2.115 V 3.846I [ ] 0.35 S 1

t

+

=− + ∴ =+

=+−∴= −

∴= −+= −



Ω

∴= − ∴ =





CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38.

(a)

(b)

1121 22

221122

112 1 22

2211 22

1121 22

211

V 2I V I 0.2V I

I 0.2V I V 1.4V 2I

1.4 2

1

V3IV I VI [] 0.2 S 1

6

1.5 3

1

I V I V 1.5V 3I [ ] 1

6 S1

6

11/ 7 4

1

V4IV I VI[] 1/7 S 1

7

111

I VIV

77

A

B

C

R

t

=+ ∴= −

=−∴=−

Ω



=+ ∴= −∴ =





Ω





=−∴=−∴=





Ω



=+ ∴= − =





=− =

22

V4I−

1.4 2 1.5 3 11/ 7 4 2.433 6.2 11/ 7 4

[] [][][] 0.2 1 1/ 6 1 1/ 7 1 0.4667 1.6 1/ 7 1

4.710 15.933

[] 0.9619 S 3.467

ABC

tttt

t

   

== =

   

   

Ω



∴=





CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

(a)

(b)

112 2222

12

V2IV 2IVV2I [] 01

II

A

t

=+=−+=− ∴ =





=−

5

1212 14 1414 18

,

0101 01 0101 01

1812 110 12 110

Also, 10

0101 0 1 01 0 1



==





Ω 

   

== →



   

   



CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40.

(a)

(b)

12

12 2

10

VV [] 1/R 1

IV/RI

a

t

=∴=





=−

12 2

12

1R

VVRI [] 01

II

b

t

=− ∴ =





=−

1 2 1 0 0.25 0 1 20 1 0

[] 0 1 0.1 1 0 4 0 1 0.02 1

1.2 2 0.25 5 1 0 0.3 14 1 0 0.58 14

[] 0.1 1 0 4 0.02 1 0.025 4.5 0.02 1 0.115 S 4.5

t

    

=    

    

Ω

  

∴= = =

  

  

12

1/ 0

VV/ [] 0

II

c

a

at a

a



=∴=





=−

CHAPTER SEVENTEEN (TWO-PORT) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

(a)

(b) 1 0 10/3 400/ 3 1 0 10 133.33

[]

0.05 1 1/ 6 55/6 0.05 1 0.625 S 9.167

new

tΩ

   

∴= =

   

   

1112

221

12 12

1122 1221

21212

I 0.1V , 0.1V 0.02(V V ) 0.2(V V V ) 0

I 0.08V 0.2(V V V )

11 5

0.32V 0.22V 0.2 V V V V

16 8

11 1 11 5

I V V Also, I 0.28 V V 0.2V 0.2 V

160 16 16 8

31 10

I V VV V

400 40 3

xx x x

xx

=−+ −+−−=

=+−+

∴=−∴=−



∴= − = − + −





∴=− + ∴ = − 2

122222

3.333 133.33

400 I[] 0.16667S 9.17

3

3.333 133.33

11 10 400 1 1 55

I V I V VI[]

0.16667S 9.167

160 3 3 16 6 6

t

Ω



=



Ω





∴= − − = − ∴ =

 

 

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1.

( ) 3 3cos(100 40 ) 4sin(200 10 ) 2.5cos300 Vvt t t t

ππ π

=− − °+ − °+

(a) V 3 0 0 0 3.000 V

av =−++=

(b) 2222

1

V 3 (3 4 2.5 ) 4.962 V

2

eff =+ ++ =

(c) 22

T 0.02

100

o

s

ππ

ωπ

== =

(d) (18 ) 3 3cos( 33.52 ) 4sin(2.960 ) 2.5cos(19.440 ) 2.459 Vvms=− − °+ °+ °=−

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2. (a)

t v t v

0 2 0.55 -0.844

0.05 2.96 0.6 0.094

0.1 3.33 0.65 0.536

0.15 2.89 0.7 0.440

0.2 1.676 0.75 0

0.25 0 0.8 -0.440

0.3 -1.676 0.85 -0.536

0.35 -2.89 0.9 -0.094

0.4 -3.33 0.95 0.844

0.45 -2.96 1 2

0.5 -2

(b)

(c) min 3.330v=

22

2

max

4 sin 2 7.2 cos4 0

4sin 2 7.2(cos 2 sin 2 )

4 16 414.72

4sin 2 7.2(1 2sin 2 ) 0.5817, 0.8595 sin 2

28.8

0.09881,0.83539 3.330(0.5593 for smaller max)

vtt

ttt

ttx t

tv

ππ ππ

πππ

ππ π

′=− + =

∴= −

−± +

∴=− ∴= =−=

∴= ∴ =

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3.

(a) T 10 , F 0.1(2 4 2 2) 1.200

av o

sa===×+×=

(b)

(c)

22

00

2

223

00

0

1

F (4 ) 0.2 (16 8 )

5

18

0.2 16 4 0.2 32 16 1.9322

33

=−= −+





=−+=−+=











∫∫

eff tdt ttdt

tt t

222

3

000

22

20

0

2

22

2 (4 )cos3 0.4 4cos0.6 0.4 cos0.6

10 10

11

1.6 sin 0.6 0.4 cos 0.6 sin 0.6

0.6 0.36 0.6

810 4

sin1.2 (cos1.2 1) sin 1.2 0.04581

39 3

=×−×= −



=− +





=− −− =−

∫∫∫

t

atdt tdtttdt

t

ttt

πππ

ππ π

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4.

(a) T = 8 − 2 = 6 s

(b) 1Hz

6

o

f=

(c) 2 rad/s

3

oo

f

π

ωπ

==

(d) 1(10 1 5 1) 2.5

6

o

a=×+×=

(e)

34

2

23

4

3

23

2

22 2

10sin 5sin

63 3

1 30 2 15 2

cos cos

32 3 2 3

1 15 4 7.5 8 1 15 7.5

cos2 cos cos cos2 (1.5) ( 1.5) 1.1937

3333



=+







=− −







 

∴= − − − − = − − − =−





 



∫∫

tt

bdtdt

tt

b

ππ

ππ ππ

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5.

4

3

34

3

2

23

4

3

34

3

23

2 6 6 1 10 5

10cos 5cos sin sin

66 63

10 1 1

sin3 sin 2 sin 4 sin 3 0

322

1 1 10 5

10sin 5sin cos cos

33

10 1

cos3 cos2 cos

32



=+=−







=−+−=







=+=−−





=− − +

∫∫

tt

adtdttt

b tdt t dt t t

ππ ππ

ππ

ππ π π

π

ππ ππ

ππ

π

22

33

110

4 cos3 ( 1) 1.0610

23

1.0610



−=−−=





+=ab

ππ

π

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6.

(a) 22

3.8cos 80 1.9 1.9cos160 , T 12.5 ms, ave value 1.9

160

π

π= + π = = =

π

tt

(b)

(c) 2

3.8cos70 3.8sin80 ; , , T 2 ; ave value 0

π

π− π ω =π ω=π = = =

π

oo

tttt s

3

3.8cos 80 (3.8cos80 )(0.5 0.5cos 160 )

1.9cos80 0.95cos240 0.95cos80 2.85cos80 0.95cos 240

2

T 25ms, ave value 0

80

π= π + π

=π+ π+ π= π+ π

π

== =

π

tt t

ttttt

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7. T = 2 s

(a)

(b) 4

1

4

=π

b

()

1

4

0

41

11

242 1

sin cos4

22 4

11 cos4

4

max when 4 , 0.125

2

×π

==−π

π

∴= − π

ππ

π= =

∫t

tt

bdtt

bt

tt s

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

( ) 5 8cos10 5cos15 3cos20 8sin10 4sin15 2sin 20

=+ − + − − +

gt t t t t t t

(a) 2

5 T 1.2566

5

π

ω= ∴ = =

os

(b) 510

4 3.183 Hz

2

=β= ==

ππ

oo

ff

(c) G 5=−

av

(d) 2 222222

1

G ( 5) (8 5 3 8 4 2 ) 116 10.770

2

=−+ +++++ = =

eff

(e)

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9.

[]

0.1 0.1

0.1

T 0.2, ( ) V cos5 , 0.1 0.1

2V cos5 cos10 5V cos(5 10 ) cos(10 5 )

0.2

11

5V sin(10 5 ) sin(10 5 )

10 5 10 5

V2 2

sin(10 5 )0.1 sin(10

21 21

−−

−

==π−<<

=ππ=π+π+π−π



=π+π+π−π



π+ π π− π



=π+π+

π+ −

∫∫

m

nm m

m

ft t t

a t n t dt n t n t dt

nt nt

nn

n

nn

22

5 )0.1

V2 2

sin( 0.5 ) sin( 0.5 )

21 21

V2V22 11

cos ( cos ) cos

21 21 2121

2V 4V

2121 cos

cos 41 41

11

V cos5 5V sin sin

0.2 5 2 2



π− π







=π+π+π−π



π+ −





=π+−π=π−





π+ − π + −



−− − π

=π =−

π−π−

ππ



=π=−−



π

m

mm

om m

n

nn

nnn

nn nn

nn n

nnn

atdt

0.1

2V

2V 4V 4V 4V 4V

( ) cos10 cos20 cos30 cos 40 ...

3153563

−



=





π





∴= + π− π+ π− π+

ππ π π π

∫m

mm m m m

vt tttt

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10.

(a) 1

even, wave

2−

(b) 0 for all ; 0; 0===

neveno

bnaa

(c)

123 2

2

1

13

0, 0

810620

5cos sin sin sin

12 6 3 6 3 6

20 20 20

sin sin 2.330, sin sin 2.122

36 3 23

=== =

ππππ



===−



ππ



ππ π

 

∴= −= = π−=−=−

 

πππ

 

∫

n

bbb a

nt nt n n

adt

nn

aa

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

(a)

(b) 222

Y 0.5(0.2 0.6 0.4 ) 0.5(0.56) 0.5292=++= =

eff

(c) (2ms) 0.2sin 0.2 0.6sin 0.4 0.4sin 0.6 1.0686= π+π+π=y

0

( ) 0.2sin1000 0.6sin 2000 0.4sin 3000

==

∴ = π+ π+ π

on

aa

yt t t t

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12.

(a) (b)

(c) (d)

(e)

3

55

2

3

55

2

55

4 2 5 32 6 5 3.2 15 10

[ ] 0, 8cos sin sin sin 0.8821

666103 33

4 2 5 32 6 15 10 3.2

[ ] 0, 8sin cos cos ( 0.5) 0.5093

6661033

825641215

[ ] 0, 8cos sin

12 12 12 10

ππππ



== = = − =



ππ



π−ππ

 

== = − =−−=

 

ππ

 

π

== = π

∫

tt

ab a dt

t

ba b dt

t

cb a dt

3

2

3

55

2

10

sin 3.801

66

81064121510

[ ] 0, 8sin cos cos 1.0186

12 12 12 10 6 6

ππ



−=





πππ

 

== =− − =

 

π

 

∫

∫t

da b dt

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13.

T = 4 ms

(a)

(b)

(c) 4 0 : 8sin125

−<< π

tt

(d)

0.004

0

1000 250 8

8sin125 cos125

4 125

16 16

cos 1 5.093

2

×

=π=π

−π

π



=− − = =



ππ



∫

o

a t dt t

0.004

1

0

0.004 0.004

1

00

0.004

0

1

2

4000 sin125 cos 0.004

4000 sin125 cos500 2000 (sin 625 sin 375 )

cos625 cos375 3.2 5.333

2000 (1 cos 2.5 ) (1 cos1.5 ) 0.6791

625 375

4000 sin125

π

=π

∴= π π = π− π

ππ



=− + =−π− −π=−



πππ π



=π

∫

∫∫

t

atdt

attdt ttdt

tt

b

0.004 0.004

00

sin500 2000 (cos375 cos625 )

11 11

2000 (sin1.5 ) (sin 2.5 ) 2000 2.716

375 625 375 625

π= π− π

−



=π−π=−=−





ππ ππ



∫∫

t t dt t t dt

[]

()

0.004

11

0

0.004

1

0

4000

0, 8sin125 cos250

8

cos375 cos125

2000 sin375 sin125 2000 375 125

5.333 16

1 cos1.5 cos 1 3.395

2

+

== π π

ππ



∴= π− π = − +



ππ



π



=−π+ −=−



ππ



∫

ba t tdt

tt

attdt

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14.

0.001

00

13579

1

odd and wave 0, 0, 0

2

T 10 0.01

81

10sin 200 8000 cos 200

0.01 200

40 40

(cos0.2 1) (1 cos0.2 )

2.432, 5.556, 5.093, 2.381, 0.27

−∴== =

==



−



=π= π





π





∴=− π−= − π

ππ

∴=====

∫

o n even

odd

aab

ms s

bntdt nt

n

bn n

nn

bbbbb02

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

()

/4

0

0.001

0

2

60.001

3

222 0

22

18

odd and wave, T 8 ( )sin

2T

2250 1000 1000 sin 250

T

1

Now, sin sin cos , 250

10

( ) 10 sin 250 250 cos 250

250

16 sin 0

4

−=∴= ω

π

ω= = π∴ = π

=− =π

=∴= π−π π

π

∴= −

π

∫

T

no

on

x

n

ms b f t n t dt

b t nt dt

x ax dx a ax ax a n

a

ft t b nt nt nt

n

bn12

35

22

16

cos 0 sin cos 0.2460

44 444

16 3 3 3 16 5 5 5

sin cos 0.4275 ; sin cos 0.13421

9 4 4 4 25 4 4 4

0

−

ππ πππ



−+∴=−=



π



ππ π ππ π

 

=−== −=

 

ππ

 

=

even

nn b

bb

b

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16.

(a) odd, T = 4

(b) even, T = 4:

(c) 1

odd, wave: T 8

2−=

(d) 1

even, wave, T 8:

2−=

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

(a)

(b)

2

1,

2

1,

2

1,

1 12.5

Ae , , (0) 0, (0) 1.25 1 6.25

21 2

(0) 1.25 1.25 tanh 0.2 0.55388

0.16 4 0.4

1 12.5 5

A 0.55388, 0.55388 1.25 cos5 sin5

1 6.25

∞

−

∞

−



==+==+ −



+π



π

∴=− =− π=

π+ π×



∴=− =− + + − +



+ππ



∑

t

nfnf

odd

f

odd

t

odd

iiiiii n

in

i e nt nt

nn

1,

22

20 1 2 20 20

5sin sin5,V(1)

0.4

V20 5

Z452410,I Z (4 10 ) 1 2.5

5 1 2.5 12.5 5

I1 6.25 (1 6.25 )

12.5 1 5 1

cos5 sin5

1 6.25 1 6.25

∞π

=+ ∴ = = −

πππ π

−

=+ =+ = = =−

π+ +

−+

∴=− =−

π+ π+

∴=− +

π+ π+

∑

ssnsn

odd

sn

nfn

n

fn

nt

v v nt j

nnn

jj

jn j n njn jn

jjn j

nnnn

intnt

nnn

2

1,

1 12.5 5

1.25 cos5 sin5

1 6.25

∞

∴= + − +



+ππ



∑

f

odd

intnt

nn

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18.

(a) 20.4

0 0.2 : 2.5(1 ) (0.2 ) 2.5(1 ) 1.78848 A

−−π

<< π = − ∴ π= − =

t

ti ei e

(b) 2( 0.2 )

0.2 0.4 : 1.78848 (0.4 ) 0.50902 A

−−π

π< < π = ∴ π =

t

ti e i

(c) 2( 0.4 )

0.4 0.6 : 2.5 (2.5 0.50902) , (0.6 ) 1.9335

−−π −

π< < π = − − π =

t

ti ei

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19.

(a)

(b) /4

Ae−

=t

n

v

(c)

()

1,

22

2

20 1

5sin5

20 sin5

20

V

1 1 20 / 1 20/ 1 20

Z2 2 V

5 2 10 2 1/ 10 10 1 20 1 20

20 1 20 20 1

V, 20cos5sin5

1 400 1 400

20 1 1

5sin5

1400

∞

=+

π

=π

=− π−π −π−

=+ =+ ∴ = × = ×

++−

−−

∴= × = − +

+ππ+

∴=+

π+

∑

s

odd

sn

ncn

cn cn

cf

vnt

n

vnt

n

jn

jn jn jn

jn jn jn jn jn jn

nj v n nt nt

nn n n

vnn

1,

20cos5

∞

−





∑

odd

nt nt

222

1, 1,

22

1,

/4

2

1,

20 20 1 1

(0) A 5 A 5

1 400 (1/ 20)

1tanh 5 tanh 1.23117

(1/ 20) 4(1/ 20) 20 2 40

1

A 0 5 1.23117 4.60811

20 1 1

( ) 4.60811 5 sin5 20cos5

1400

∞∞

∞

−

=++ =+−

π+ π +

ππ π

==π=

+×

∴=−+× =−

π



∴=− ++ −



π+



∑∑

∑

c

odd odd

odd

t

c

odd

vnn

n

vt e nt nt

nn

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20.

()

3

0.001 0.005

33 2 / 6 10 100

3

0 0.003

0.001 0.005

51000 1000

0 0.003

53

33

3

10 100 100

6

10 1 1

6 1000 1000

100 100

1 (1 1 1 1) 10.610

66

10.610; 10.610

2

−

−×π × − π

−π −π

−π − π − π

−



=−







−

=+



ππ





=++−=+−+=−

ππ

∴= =

=

∫∫

jt jt

jjj

ce e

ee

jj

eee j

jj

cj c

a

()

0.001 0.005

3

00.003

5

22

333 33 33

10 100cos100 100cos1000

6

210 1 sin 0 sin 5 sin3 0

6 1000

11

( ) 21.22 and 21.22

22



×π− π





×

=π−−π+π=

π

=−=− ∴= +=

∫∫

tdt tdt

cajbjbb ab

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

0.001 0.002

5 400 400

0 0.001

0.001 0.002

400 400

0 0.001

0.00

400 0.001 400

0

22

0.001

1

T 5 10 100

0.005

20,000 1000

1

20,000 ( 400 1)

160 400

jnt jnt

m

jnt jnt

n

jnt jnt

n

msc te dt e dt

ctedtedt

e

c j nt e

njn

−π −π



== +







∴= +





∴= π+ +

π−π

∫∫

()

2

33

0.4 0.8 0.4

122

2

1

(50 10 100 10 ) 0.15 200 30

0.005

111

20,000 (1 0.4 )

160 160 400

125 (1 72 )(1.60597 51.488 ) 12.66515 15.91548 90 (1 144 1 72 )

12.665(1 72

oo

jjj

ca

cej ee

j

−−

−π −π −π







∴= = × + × = × =



=+π−−−



πππ



= ∠− ° ∠ ° − + ∠ ° ∠− °− ∠− °

π

=∠−

2

)(1 1.2566) 12.665 15.915(1 144 1 72 )

20.339 20.513 12.665 18.709 108 24.93 88.61

3.16625 144 (1 2.5133) 3.16625 7.9575(1 288 1 144 )

8.5645 75.697 3.16625 15.1361 144 13.309 1

jj

cjj

° + − + ∠− °− ∠− °

= ∠− °− + ∠− °= ∠− °

=∠−°+ −+ ∠−°−∠−°

=∠−°− + ∠°=∠77.43°

(a)

(b)

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22.

Fig. 17-8a: V 8 V, 0.2 , 6000

oo

s f pps

τ

µ

== =

(a) 11

T , 6000, 0.2 5 MHz

6000 o

fsf

τµ τ

===∴==

(b) 6000 Hz

o

f=

(c)

(d) 66 6

333

3 6

2 10 8 0.2 10 sin(1/ 2 333 12,000 0.2 10

333.3 7.270 mV

6 10 1/ 6000 1/ 2 333 12,000 0.2 10

c

π

−−

−

×××××××

=∴= =

×××××

(e) 1/ 5 MHz

β

τ

==

(f)

(g)

66

3

8 0.2 10 sin(1/ 2 3 12,000 0.2 10

6000 3 18,000 (closest) 1/ 6000 0.0036

9.5998 mV

c

π

−−

×× ×× ××

×= ∴ =

∴=

2000 2200

2 2.2 Mrad/s kHz or 318.3 350.1 kHz

22

6 kHz 6 53 318; 324,330,336,342,348kHz 5

o

ff

ff n

ωππ

<< ∴ << <<

=∴=×= ∴=

66

227

8 0.2 10 sin(1/ 2 227 12,000 0.2 10 8.470 mV

1/ 6000 ( )

227 6 1362 kHz

c

f

π

−−

×× × × ××

==

′′

=×=

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23.

12 3

T 5 ; 1, 0.2 0.2, 0.5 0.25, 1 2, 0, 4===−=+ =−−=≥

o n

ms c c j c j c j c n

(a)

(b)

11 1 2 2 3 3

2 1, 0.4 0.4 0.4, 1 0.5, 2

( ) 1 0.4cos400 cos800 2cos1200 0.4sin 400 0.5sin800 4sin1200

nnnoo

ajbcacajbjbjajbjajb

vt t t t t t t

ππ π π π π

=−= ∴== −=−=− −=+ −=−

∴=+ + − + − +

(1 ) 1 0.4cos72 cos144 2cos 216 0.4sin 72 0.5sin144 4sin 216vms= + °+ °− °+ °− °+ °

(1 ) 0.332Vvms∴=−

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24.

(a)

(b) 4

1(sin172.8 sin115.2 ) 0.06203

4

c

π

=°−°=−

(c) 66

6

0.2 10 0.2 10 0.08

510

oo

ca −−

−

×+×

== =

×

(d) max

a little testing shows is max 0.08

o

cc

∴=

(e)

(f) 6

740 10

740 148 MHz

5

×

== =

o

f

β

()

6

0.6 10

6

0.4 10

6

5

10

T5 2 1cos2

5510

510

4 10 sin 43.2 sin 28.8

2

1sin 43.2 sin 28.8

n

t

sc n dt

cnn

n

cnn

n

µπ

π

−

×

−

×

−

=∴=× ×

×

∴=× °− °

∴= °− °

∫

()

33

1

0.01 0.08 0.8 10 sin 43.2 sin 28.8 0.8 10

125 sin 43.2 sin 28.8 1

ok for 740

−−

×=×∴ °− °≤×

∴°−°≤

>

nn

n

nn

n

π

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25.

T1/16, 32

o

ωπ

==

(a)

(b)

1/96

96 96

3

00

3

16 40

16 40 96

20 40

( 1) 4.244 V

33

jt j

j

c e dt e

j

cj e j j

ππ

π

ππ

−−

−

×

=−

−

∴= −=− =−

∫

33 3 3 3

2

33

64

2

Near harmonics are 2 32 Hz, 3 48 Hz

Only 32 and 48 Hz pass filter 2

2 8.488 0, 8.488 V

8.488 1

I 1.4536 31.10 A; P 1.4536 5 5.283 W

5 0.01 96 2

1 640

40

1/16 6

oo

nn n

jt

ff

ajb c

ajb c j a b

j

cedt

j

π

−

==

−=

−==− ∴= =

==∠−°=××=

+×

==

−

1/96

64 /96

0

22 2

2

( 1) 2.7566 4.7746 V

4

2 5.5132 9.5492 11.026 60

11.026 60

I 2.046 65.39 A

5 0.01 64

1

P 2.046 5 10.465 W P 15.748 W

2

j

tot

ej

ab c j

j

π

−−= −

−= = − = ∠−°

∠− °

∴= = ∠− °

+×

∴=× ×= ∴ =

∫

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26. ( ) 5[ ( 3) ( 2) ( 2) ( 3)]f t ut ut ut ut= +++−−−−

(a)

(b)

22 3

32 2

23 22 3 2

33 22 22

F( ) ( )

F( ) 5 10 5

510 5

F()( )( )( )

5510

()()()

55

(2)sin3 (

∞−ω

−∞

−−ω −ω −ω

−−

ωω −ωω −ω−ω

ω−ω ω−ω ω−ω

ω=

∴ω= + +

∴ω= − + − + −

−ω −ω −ω

=−++ −+−+

−ω −ω −ω

=− ω+

−ω −ω

∫

∫∫ ∫

jt

jt jt jt

jj jj j j

jj jj jj

jftedt

j e dt e dt e dt

jee ee ee

jj j

ee ee ee

jjj

j

jj

10

2)sin 2 ( 2)sin 2

10 10 20 10

F( ) sin 3 sin 2 sin 2 (sin3 sin 2 )

ω+ − ω

−ω

∴ω= ω− ω+ ω= ω+ ω

ωωω ω

jj

j

-3 -2 -1 0 1 2 3

f(t)

t

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

(a)

(b)

(c)

0

()

0

() (), 0 F( ) ()

11

F( )

∞∞

−−−−

−∞

∞

−+

=>∴= =

−

∴= =

++

∫∫

at j t at j t

aj t

f t e u t a j f t e dt e e dt

je

aj aj

ωω

ω

ωωω

6()

()

() ( ), 0 F( )

111

F( )

∞

−−+

∞

−+ −

−+

=−>∴=

−−



∴= = − =



+++

∫

o

oooo

o

at at

at a j t

o

t

at at a j t j t

aj t

t

ft e e ut t a j e e dt

je e e e e

aj aj aj

ω

ωω

ω

ωωωω

[]

()

0

()

222

0

() (), 0 F( )

11

F( ) ( ) 1 0 [ 1]

() () ()

∞

−−+

−+ ∞

=>∴=

∴= −+−=− −=

+++

∫

at a j t

aj t

ft te ut a j te dt

e

jajt

aj aj aj

ω

ωω

ωωω

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28.

04

40

0

1

4

44

1

00

4 0: ( ) 2.5( 4); 0 4: ( ) 2.5(4 )

F( ) 2.5( 4) 2.5(4 5)

ln 1 , let I 2.5(4 ) ( )

I 2.5(4 ) F( ) 2.5 (4 )( )

F( ) 5 (4 )cos 20

−ω −ω

−

ωτ

ωτ ω − ω

−<< = + << = −

∴ω= + + −

=τ ∴ = −τ − τ

∴= −τ τ∴ ω= − +

∴ω=−ω=×

∫∫

∫

∫∫

jt jt

st j

jjtjt

tft t tft t

jtedt edt

ted

ed j te e dt

jttdt

4

44

00

0

4

0

2

222

2

1sin 5 cos

20 5

F( ) sin 4 (cos sin )

20 5 5 5

sin 4 (cos4 1) 4 sin 4 (1 cos4 )

25 sin2

or, F( ) sin 2 10

ω− ω

ω

∴ω= ω− ω+ω ω

ωω

=ω− ω−−ωω=−ω

ωω ω ω

×ω



ω= ω= 

ωω



∫∫

ttdt

jttt

j

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



29.

(1 ) (1 )

(1 ) (1 ) (1 ) (1 )

() 5sin F( ) 5sin

5

F( ) ( )

2

5[]

2

51 1

F( ) ( ) ( )

2(1 ) (1 )

2.5 2.5

()(

11

π−ω

−π

π−−ω

−π

π−ω − +ω

−π

π−ω −π−ω −π+ω π+ω

−πω πω

=−π<<π∴ω=

∴ω= −

=−



ω= − − −



−ω − +ω



−

=−+−

−ω +ω

∫

jt

jt jt j t

jt jt

jj jj

jj

ft t t j te dt

jeeedt

j

eedt

j

jee ee

jj j

ee

222

)

2.5 2.5 1 1

(2sin) (2sin)5sin

11 11

11 10sin 10sin

5sin ( 1) 11 1

−πω πω

−+

−



=πω−πω=πω−−



−ω +ω −ω +ω



+ω+ −ω πω πω



=πω− =− =



−ω −ω ω −



jj

ee

jjj

jj

j

,

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



30.

/2

/2 /2

/2

(1 ) (1 )

/2

( ) 8cos [ ( 0.5 ) ( 0.5 )]

F( ) 8cos 4 ( )

4

11

4(1 ) (1 )

1

4()

(1 )

π

−π

ππ

−ω − −ω

−π −π

π−ω − +ω

−π

π

−ω −−ω

−π

−πω

=+π−−π

∴ω= = +



=+







=−



−ω +ω





=−−

−ω

∫∫

∫

jt jt jt jt

jt jt

jt

jt jt j t

j

ft tut ut

jtedteeedt

eedt

ee ee

jj

je j e

j

/2 /2 /2

22

1

(1 )

11 11

4 2cos 2cos 8cos

1212211

2cos/2

8cos 16

21 1

πω − πω πω





−−−





+ω



πω πω πω





=× +× = +





−ω +ω −ω +ω





πω πω

==

−ω −ω

jjj

je je

j

(a) 0 F( 0) 16ω= ∴ =j

(b) 16cos72

0.8, F( 0.8) 13.734

0.36

°

ω= = =j

(c) 16cos(3.1 90 )

3.1, F( 3.1) 0.2907

1 3.12

×°

ω= = =−

−

j

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



31.

(a)

(b)

(c)

[]

()

2

22

1

F( ) 4 ( 2) ( 2) ( ) F( )

2

4212

() 2

24 5

( ) 2sin 2 sin 2 (0.8) sin1.6 1.5909

2

∞ω

−∞

ωω −

−−

ω = ω+ −ω ω− ∴ = ω ω

π

∴= ω= = −

ππ π

∴= = ∴ = =

ππ π

∫

jt

jt jt jt jt

rad

ju ft ejd

ft e d e e e

jt j t

ft j t t f

tt

22

0

(2 ) ( 2 )

0

2

4

F( ) 4 ( ) 2

22

()

21 1 21 1 24

(1 0) (0 1)

22 224

88

( ) (0.8) 0.5488

(4 ) 4.64

∞

−ω −ω+ω

−∞

∞

+ω −+ω

−∞

ω= ∴ = ω

π

∴= ω+ ω

ππ



=−+−=+=





π+ −+ π+ − π+



∴= ∴ = =

π+ π×

∫

∫∫

jt

jt j t

je ft e d

ft e d e d

jt jt jt jt t

ft f

t

[]

()

0.5 0.5

0.5

() (0.50.5)

0.5

0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5

F( ) 4cos ( 0.5) ( 0.5)

421

() cos

22

1

11 1

() ( )

ωπω−πωω

−−

π+ ω − π− ω

−

π+ − π− − π+ π−

ω= πω ω+ − ω−

∴= πω× ω= + ω

ππ



=+ ω



π

=−+−

ππ+ −π+

∫∫

∫

jt j j jt

jjt j jt

jjt jjt jjtjj

juu

ft ed e e ed

ee d

ee e e

jt j t

()

()()

0.5 0.5 0.5 0.5

22 22

11 1

() ( )

1 1 1 2cos0.5 1 1

2cos0.5 2cos0.5

24

2cos0.5 cos0.5 (0.8) 0.3992

−−









=++−−



ππ+ −π+





=−=−





π π+ −π+ π π+ −π+



−



==∴=



−π π −



t

jt jt jt jt

je je je je

jt j t

t

tt

tt tt

ttf

tt

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



32. 1.5

( ) 20 ( 2) V=−−

t

vt e u t

(a)

(b)

(c) B (2) 0.3891=−

v

(d) F ( 2) 0.3983=

vj

(e)

φ

v(j2) = 282.3o or -77.69o

2

1.5 1.5

2

(1.5 ) 3 2 3

F( ) 20 ( 2) 20

20 20 20

F ( 0) 0.6638

1.5 1.5 1.5

∞−

−ω −ω

−∞ −∞

−

−ω −+ ω −

−∞

ω= −− =

==∴==

−ω −ω

∫∫

tjt tjt

v

jt j v

jeutedtedt

eeje

jj

32

34

20

F( ) A ( ) B( ) 1.5

20

F ( 2) 0.39830 282.31 0.08494 0.38913

1.5 2

A (2) 0.08494

−ω

−

ω= ω+ ω= −ω

∴= = ∠°= −

−

∴=

j

vvv

j

v

jee

j

jee j

j

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



33.

[]

I( ) 3cos10 ( 0.05 ) ( 0.05 )ω= ωω+π−ω−πjuu

(a)

(b)

0.05

22

0.05

/20

12

W4 I() 9cos10

2

18 1 1 9 9 1

cos20 0.1 sin 20 0.9 J

22 20

∞π

−∞ − π

π

−π

=× ω ω= ωω

ππ



=+ωω=×π+ ω=



πππ



∫∫

∫

jd d

d

991

(1 cos20 ) 0.45 2 2sin 20

20

0.05 2 0.1sin 20 , 0.04159 rad/s

ω

−ω



+ωω==ω+× ω



ππ



∴π=ω+ ωω=

∫x

x

xx

xxx

d

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



34. 4

() 10 ()

−

=t

ft te ut

(a)

(b)

(c)

()

2

16

100

)( +

=

ω

jF

mJ/Hz 90.63 )( 2

0=

=

ω

jF , mJ/Hz 7.669 )( 2

4=

=

ω

jF

8

228 2

1

00 0

W ( ) 100 100 (64 16 2)

(512)

100 2 0.3906 J

512

∞

∞∞ −

−

Ω== =× ++

−

=×=

∫∫ t

te

f t dt t e dt t t

(4 )

4(4)

2

00

22

10

F( ) {10 ( )} 10 [ (4 ) 1

(4 )

10 10

F( )

(4 ) 16

∞

∞−+ω

−−+ω

ω= = = − +ω −

+ω

=∴ω=

+ω ω+

∫jt

tjt

e

jteuttedt jt

j

F

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



35. 2

() 8 V

−

=t

vt e

(a) 24

1

0

W ( ) 2 64 32 J

∞∞

−

Ω

−∞

==× =

∫∫

t

v t dt e dt

(b)

(c)

2

0

(2 ) (2 )

0

(2 ) (2 )

2

0

F( ) () 8

F( ) 8 8

88 8832

F( )

22 224

∞∞

−

−ω −ω

−∞ −∞

∞

−ω − +ω

−∞

∞

−ω − +ω

−∞

ω= =

∴ω= +

= − =+==ω

−ω +ω −ω +ω +ω

∫∫

t

jt jt

v

jt jt

v

jt jt v

j e v t dt e e dt

j e dt e dt

ee j

jj jj

1

22 11

22 2

1

11 11

22

11

1

11

1

2

1

132 32 1

0.9 32 tan

2(4) 28(4)16 2

16 1 2 2

0.9 2 tan

8( 4) 16 2 4 2

2

0.45 tan 2.7174 rad/s (by SOLVE)

42

ω

−

−ω

−



ωω

×= ω= +



πω+ πω+





ωω ωω

∴=× + = +



πω+ πω+



ωω

∴π= + ∴ω=

ω+

∫d

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



36.

(a)

(b)

(c)

(d) Prove: { ( )} F( ) Let 1 in (c) above−=−ω =ft j k

F

(e)

Prove: { ( )} { ( )} ( ) Let

{( )} () {()}

∞

−ω −ω

−∞

∞−ω −ω

−ωτ

−∞

−= = − −=τ

∴−=τ =

∫

o

oo

jt jt

ooo

jt jt

j

o

ft t e ft ft t e dt t t

f t t f e e dt e f t

FF

Prove: { ( )} { ( )} Let , ,

,{()}() ()

We assume ( ) 0 { ()} { ()}

∞−ω −ω −ω

−∞

∞

−ω −ω

−∞ −∞

=ω = = =−ω

==∴ = +ω

±∞ = ∴ = ω

∫

jt jt jt

jt jt

df

f t j f t e dt u e du j e

dt

dv df v f f t f t e j f t e dt

fftjft

FF

F

FF

/

1

Prove: { ( )} F ( ) Let , 0

11

{( )} () F

1

If 0, limits are interchanged and we get: F

1

{( )} F

∞−ω

−∞

∞−ωτ

−∞

ω



== τ=>





ω



∴=τ τ=





ω



<−





ω



∴=





∫

jt

jk

j

fkt fkte dt ktk

kk

j

fkt f e d

kkk

j

kkk

j

fkt kk

F

Prove: { ( )} F( ) Now, F( ) ( )

dF( ) ( )( ) { ( )} { ( )} ( )}

∞−ω

−∞

∞−ω

−∞

=ω ω=

ω

∴=− =− ∴ =ω

ω

∫

jt

d

tf t j j j f t e dt

d

jft jte dt j tft tf f j ft

d

F

FF F

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



37.

(a)

(b) ( ) 4[sgn( 1) ( )] {4sgn( 1) ( )} { 4 ( )} 4=−δ∴ −δ=−δ=−ft t t t t

FF

(c)

(10 30 ) (10 30 )

30 10 30 10 / 6 / 6

/6 /6

4

( ) 4sin(10 30 ) {4sin(10 30 ) 2

{ 2 2 } 2 2 ( 10) 2 2 ( 10)

4 [ ( 10) ( 10)]

−° − −°

−° °− −π π

−π π





=−°∴ −°= − =







−+ =−πδω−+πδω+

=− π δω− − δω+

jt jt

jjt j jt j j

jj

ft t t e e

j

je e je e je je

je e

FF

F

( ) 4[sgn( ) ( 1)] {4[sgn( ) ( 1)] {4sgn(1) ( 1)} {4 ( 1)} 4 −ω

= δ− ∴ δ− = δ− = δ− = j

ft tt tt t t e

FF F

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



38.

(a)

(b)

(c)

( ) A cos( ) F( ) {A cos cos Asin sin }

Acos { [ ( ) ( )]} Asin [ ( ) ( )]

A{cos [ ( ) ( )] sin [ ( ) ( )]}

F( ) A[ ( ) ( )]

φ−φ

=ω+

φ

∴ω=

φ

ω−

φ

ω=



π

φ

πδω+ω +δω−ω −

φ

δ ω−ω −δ ω+ω =





π φ δ ω+ω +δ ω−ω + φ δ ω−ω −δ ω+ω

∴ω=π δω−ω+ δω+ω

ooo

oo oo

jj

oo

ft t j t t

j

je e

F

2

21

( ) 3sgn( 2) 2 ( ) ( 1) F( ) 3 2 ( )

61

F( ) 2 ( )

−ω −ω



=−−δ−−∴ω=××−−πδω+



ωω





∴ω=− −− πδω−



ωω



jj

ft t t ut j e e

jj

jje e j

22 22

1

() sinh () F( ) [ ] ()

2

11 11

F( ) 2 2 2( )

−



=∴ω=−





+ω+ −ω −

∴ω= − = =

−+ω +ω − −ω ω+

kt kt

ft ktut j e e ut

kj kj k

jkj kj k k

F

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



39.

(a)

(b)

so f(5) = 0.1039∠73.52o

(c)

1

3

33

1

F( ) 3( 3) 3( 1) () [3( 3) 3( 1)]

2

3313

() ( )

22 2

3

(5) (1 5 1 15 ) 0.10390 106.48

10

∞ω

−∞

ωω+−

−−

ω = ω+ − ω− ∴ = ω+ − ω− ω

π

∴= = = −

ππ π

∴=− ∠−∠− = ∠− °

π

∫

jt

jt jt jt jt

rad rad

ju u ft u u ed

ft e dt e e e

jt j t

fj

3

F( ) 3 ( 3 ) 3 ( 1)

F( ) 3 F ( )

3

( ) 3 ( ) ( ) (5) 0 0.10390 106.48

2

−

ω= −−ω+ ω− →

∴ω=− ω

=δ − − ∴ = − ∠− °

π

a

jt j t

ju u

jj

ft t e e f

jt

3

21

F( ) 2( ) 3(3 ) 3( 1) Now, {2( )} 2

13 1

( ) ( ) (5) 0.10390 106.48 0.3618 15.985

2

− +

ω= δω+ −−ω+ ω− δω = =

ππ



∴=+− − ∴=− ∠− °= ∠ °



ππ π



jt j t

juu

ft e e f

jt

F

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



40.

(a)

(b)

(c) 3

222

6(3 ) 6(3 )

F( ) ( ) 3 cos2 ( )

(3 ) 4 (3 ) 2

−

+ω +ω

ω= = ∴ =

+ω + +ω +

t

jj

jfttut

jj

33

F( ) 3 3 ( 1)

1

1.5

( ) 3 ( ) 1.5sgn( ) 3 ( )

−

ω= + + +δω−

+ω ω

∴= + +δ+

π

tjt

jjj

ft eut t t e

1sin8/2

F( ) 5sin 4 8 2.5

8/2

() 2.5[ ( 4) ( 4)]

ω

ω= ω= ×

ωω

∴= +−−

j

f t ut ut

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



41.

1

/2

1

/2

22

1

/2 /2

22 22

/2 /2

22

T 4, periodic; find exp l form

110

4

1

2.5 /2 /4

11 11

2.5 2/4 /2/4

14

2.5 ( ) (

/2

−π

−

−π

−

−π π

′

=

∴=





∴= −





−π −π









∴= + − +





−π π π π





=−−+

ππ

∫jn t

n

jn t

n

jn jn

n

jn jn

ctedt

t

ce jn n

ce e

jn n jn n

ee

jn n

/2 /2

22

/2

22

)

510

2cos 2sin

22

10 20

() cos sin

22

10 20

F( ) cos sin 2

222

−π π

∞π

−∞

∞

−∞



−





ππ



=× + −



ππ



ππ



∴= −



ππ



πππ



∴ω= − πδω−





ππ



∑

jn jn

jn t

ee

jn n

j

nn

jn n

ft j e

nn

jn n n

jj

nn

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



42.

1

/2 3

/2

0.002

500

0.002

0.001 0.001 0.002

500 500 500

0.002 0 0.001

T 4 , ( ) 10 ( ) 6 ( 0.001) 4 ( 0.003)

1210

( ) , 500

T4

250 ( )

250 4 10 4

250

5

−ω

−

−π

−

−−π −π −π

−

==−−−−

π

=ω==π

=



=++





=

∫

∫∫∫

o

Tjn t

no

T

jnt

jnt jnt jnt

ms f t ut ut ut

cftedt

fte dt

edt edt edt

j0.001 0.001 0.002

500 500 500

0.002 0 0.001

/2 /2 /2

4104

00

4( ) 10( 1) 4( )

2

4( ) 4( ) 10( 1)

2

42sin 42sin

22

−π −π −π

−

π−π −π −π−π

π−π π−π −π



++





π



=−+−+−



π



=−−−+−



π

=−

π

jnt jnt jnt

jn jn jn jn jn

eee

n

jee e ee

n

jee ee e

n

jn

jj

n

/2

/2 /4 /4 /4

/4 /4

/4

10( 1)

11

4sin 5( 1) 5 ( ) 4sin

22

11

5 2sin 4sin 10 sin 4sin

42 42

1

F( ) 2 10 sin 4

−π

−π −π −π π

−π −π

−π



π+ −





ππ

 

=− + −= − −

 

ππ

 



ππ ππ





=−−= −





ππ





π

∴ω=π −

π

jn

jn jn jn jn

jn jn

jn

ne

nn

je je e e

nn

nn nn

je j e

nn

n

je

n4sin ( 500 )

2

∞

−∞

π



δω− π





∑nn

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



43.

1

F( ) 20 ( 20 )

!1

11 1 1 1

20 ( ) ( 20) ( 20) ( 40) ( 40)

11 11 11 21 3

11

( 60) ( 60) ...

77

20 20

10 ( ) [ ( 20) ( 20)] [ ( 40) ( 40)]

23

20 20

[ ( 60) ( 60) [ (

725

∞

−∞

ω= δω−

+



= δω+ δω+ + δω− + δω+ + δω−

++ + +





+δω+ +δω− +



= δ ω + πδ ω+ + πδ ω− + πδ ω+ + πδ ω − +

ππ

πδ ω+ + πδ ω− + πδ

ππ

∑

jn

n

80) ( 80)] ...

10 20 20 20 20

( ) cos 20 cos 40 cos60 cos80 ...

22 3 7 25

20111 1

0.25 cos 20 cos40 cos60 cos80 ...

23725

20 1 1 1 1

(0.05) 0.25 cos1 cos2 cos3 cos4 ... 1.3858

23725

ω+ +πδ ω− +

∴=+ + + + +

ππ π π π



=++++ +



π



∴= + ++++=



π

rad

ft ttt t

ttt t

f

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



44. Input ( ) 5[ ( ) ( 1)] ( ) ( ) ( )

t

xt ut ut yt xz ht z dz

−∞

== −− = −

∫

(a) () 2 ()ht ut

= (b) () 2 ( 1)ht ut

=− (c)

() 2 ( 2)ht ut

=−

t < 0:

y(t) = 0 1: ( ) 0tyt

<= 2: ( ) 0tyt

<=

01:t<< 12:t<< 23:t<<

0

( ) 10 10==

∫

t

y t dz t ∫== t

tdzty

1

1) - 10( 10 )(

∫== t

tdzty

2

2) - 10( 10 )(

t > 1: t > 2: t > 3:

∫== t

t

dzty

1-

10 10 )( ∫== t

t

dzty

1-

10 10 )( ∫== t

t

dzty

1-

10 10 )(

5

h( z)

z

2

y(t)

t

10

1

x(t – z)

z

t-1 t

x(t – z)

z

t-1 t

h( z)

z

2

y(t)

t

10

3

2

x(t – z)

z

t-1 t

h( z)

z

2

y(t)

t

10

1 2

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



45. () 5[ () ( 2)]; () 2[ ( 1) ( 2)]xt ut ut ht ut ut=−− =−−−

or….

same answers as above

1

0

2

() ( ) ( )

1: ( ) 0

1 2: ( ) 10 10( 1)

23:()10

3 4: ( ) 10 10(2 2) 10(4 )

4: ( ) 0

( 0.4) 0; (0.4) 0; (1.4) 4

(2.4) 10; (3.4) 6; (4.4) 0

−∞

−

=−

<=

<< = = −

<< =

<< = = −+ = −

>=

∴− = = =

===

∫

t

yt xz ht z dz

tyt

tyt dzt

tyt

tyt dz t t

tyt

yyy

0

1

2

() ( ) ( )

1: ( ) 0

1 2 : ( ) 10 10( 1)

23:()10

3 4: ( ) 10 10(2 2) 10(4 )

4: ( ) 0

∞

−

=−

<=

<< = = −

<< =

<< = = −+ = −

>=

∫

t

yt xt zhzdz

tyt

tyt dzt

tyt

tyt dz t t

tyt

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



46.

2

() 2()

0

22

0

22

() 3[ ], () ()

() ( ) ( )

3[ ]

1

3[] 3 2

3 ( 1) 1.5 ( 1)

( ) 3(1 ) 1.5(1 ) 1.5 3 1.5 , 0

−−

−∞

−− − −

−−

−− −−

=− =

=−



=−





=−− −

∴=−− − =−+ >

∫

tt

t

tz tz

t

tzt t Z

tt t t

tttt

ht e e xt ut

yt xzht zdz

eedz

ee e e

yt e e e e t

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



47.

0

() ( 2) ( )

2

( ) (5 ), 2 5

3

∞

=−

=−<<

∫

yt xt hzdz

ht t t

(a)

(b)

55

22

220

() 10 (5 ) (5 )

33

Note: ( )is in window for 4 6

=×− = −

<<

∫∫

yt z dz z dz

hz t

5

2

20 1

() (5 )

32

10 (0 9) 30 at 5

3

yt z

t



=− −





=− − = =

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



48. (2)

0

( ) 5 ( 2), ( ) (4 16) [ ( 4) ( 7)], ( ) ( ) ( )

∞

−−

=−=−−−−=−

∫

t

xt e ut ht t ut ut yt xt z hz dz

(a) 6: ( ) 0 (5) 0tyt y

<=∴=

(b)

(c)

6(8 2)

4

66

44

6

6664

4

66 4 2 2 2

2

8: (8) 5 (4 16)

(8) 20 80

20 ( 1) 80 ( )

1

20 (5 3 ) 80 80 20 80 60

20(1 ) 22.71

−−−

−−

−−−−

−

== −

∴= −



=−−−





=−−+=+−

=+=

∫

∫∫

z

zz

z

ty e zdz

y e z e dz e e dz

e

ez eee

ee e e e e

e

7(10 2)

4

78

4

77

8887874

4

44

87 4 1 4 1 4

10: (10) 5 (4 16)

(10) 20 ( 4)

(10) 20 80 20 [ ( 1)] 80 ( )

20 (6 3 ) 80( ) 40 20 15.081

−−−

−

−−− −

−−−−−

== −

∴= −

∴= − = −− −

=−−−=+=

∫

∫∫

z

zzz

ty e zdz

yeezdz

y e ze dz e e dz e e z e e e

ee e ee e e

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



49.

0

00

0

( ) sin , 0 ; 0 elsewhere, Let ( ) ( )

() ( ) ( )

0: ( ) 0

0:()sin sin

1

() (sin cos )

2

1[ (sin cos ) 1]

2

1(sin cos )

2

−

∞

−+ −

−

=<<π =

=−

<=

<<π = × =



∴= −





=−+

∫

∫∫

t

tt

tz t z

t

tz

tt

t

ht t t xt e ut

yt xt z hzdz

tyt

tyt zedzeezdz

yt e e z z

ee t t

tte

(a) (1) 0.3345y+

=

(b) (2.5) 0.7409y=

(c)

0

:() sin

11

: ( ) (sin cos ) ( 1) 12.070

22

(4) 0.2211

π

−

π

−−π−

>π =



>π = − = + =





∴=

∫

tz

tz t t

yyteezdz

yyteezz ee e

y

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



50.

0

1

2

12

2

23

()0.8(1)[(1) (3)],

( ) 0.2 ( 2)[ ( 2) ( 3)]

() ( ) ( ) ,

3: ( ) 0

3 4: ( ) 0.8( 1)0.2( 2)

( ) 0.16 ( 2 2 2)

11

0.16 [ ( 1) 2 2 ] 0.16 ( 1)

32

∞

−

=− −−−

=−

<=

<< = −− −

∴= −−+−+

=−+++−=−++

∫

t

xt t ut ut

ht t ut ut

yt xt z hzdz

tyt

tyt tz zdz

yt tz t z z z dz

ztz tdz z t

1

12

2

32

32 2 2

32

(2 2 )

181 1

0.16 ( 1) ( 1)( 1) ( 1)4 (2 2 )( 1 2)

332 2

1181

( ) 0.16 ( 1)( 1) 2 2 2 6 2 6

3332

11 11

0.16 1 2 1 6 3 8

62 22

−

−

+−







=−−+++−−++−−−







∴= −+−+++ −−−−+−−+







 

= + −− +−−+ ++−

 



 



∫t

t

ztz

tttttt

yt t t t t t t t t t

tt t 32

1399

0.16 6222



=−+−





ttt

3

(4.8) 90.67 10y−

∴=×

3

332

2

(3.8) 13.653 10

11

4 5: ( ) 0.16( 1)( 2) 0.16 ( 1) (2 2 )

32

11

( ) 0.16 (27 8) ( 1)5 (2 2 )1

32

19 11

0.16 2.5 2.5 2 2 0.16 0.5

36

−

∴=×



<< = −− − = − + + + −







∴= − −+++−







=−+++−= −







∫

y

t y t t z z dz z t z t z

yt t t

ttt

(b)

(a)

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



51.

22

0

2( ) 2

0

22

0

2

() 10 (), () 10 ()

() ( ) ( )

() 10 10

100 100

() 100 ()

−−

∞

−− −

−−

−

==

=−

∴=

==×

∴=

∫

tt

t

tz z

t

tt

t

xt e ut ht e ut

yt xt zhzdz

yt e e dz

edz et

yt te ut

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



52. 4

() 5 ()

t

ht e ut

−

=

(a)

(b)

0.8

80.86.4

1

0.1

25

W 25 ( ) 1.3990 J

8

25

% 1.3990/ 100% 44.77%

8

−−−

Ω==−=



∴= × =





∫t

edt e e

2

21

120

0

1

5 1 25 25 1

H( ) W tan

41644

25 1 0.9224

W tan 0.9224 J % 100% 29.52%

42 25/8

−

Ω

−

Ω

ω

ω= ∴ = ω=

ω+ π ω + π

∴= = ∴= × =

π

∫

jd

j

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



53.

2

222

F( ) ( ) (2 2 ) ( )

(1 )(2 ) 1 2

−−

ω= = − ∴ = −

+ω +ω +ω +ω

tt

jfteeut

jj j j

(a) 234

1

0

4841

W(484) J

2343

∞−−−

Ω=−+ =−+=

∫tt t

eeedt

(b)

2

0.69315 2 0.69315

max

( ) 2 4 0, 2 4 0, 2, 0.69315

2( ) 0.5

tt t t

ft e e e e t

fe e

−− −

−−×

=− + = − + = = =

∴= − =

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



54.

(a)

(b)

(c)

(d)

23

1 1/6 1/2 1/3

F( ) (2 )(3 ) 2 3

11 1

( ) sgn( ) ( ) ( )

12 2 3

−−

ω= = − +

ω +ω +ω ω +ω +ω

∴= − +

tt

jjj jj j j

ft t e ut e ut

23

11/61/22/3

F( ) (2 )(3 ) 2 3

11 2

( ) sgn( ) ( ) ( )

12 2 3

−−

+ω

ω= = + −

ω +ω +ω ω +ω +ω

∴= + −

tt

j

jjj jj j j

ft t e ut e ut

2

23

(1 ) 1/ 6 1/ 2 4 / 3

F( ) (2 )(3 ) 2 3

11 4

( ) sgn( ) ( ) ( )

12 2 3

−−

+ω

ω= = − +

ω +ω +ω ω +ω +ω

∴= − +

tt

j

jjj jj j j

ft t e ut e ut

3

23

(1 ) 1/ 6 1 / 2 8 / 3

F( ) 1

(2 )(3 ) 2 3

11 8

() () sgn() () ()

12 2 3

−−

+ω

ω= = + + −

ω+ω +ω ω +ω +ω

∴=δ+ + −

tt

j

jjj j j j j

ft t t e ut e ut

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



55. ( ) 2 ( )

t

ht e ut

−

=

(a) 12

H( ) 2 11

ω= × =

+ω +ω

jjj

(b) V1111/

H( )

212V11/

ω

ω= = =

+ω + ω

o

i

j

jjj

(c) Gain = 2

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



56.

2

22

11

() 2

2

V( ) 11

() 2()2

12

( ) 2( ) 2 2( ) 2( )

V( ) 1

() 2()2 () 2()2

2248

Let V ( ) 1 ; 1 1

22 2

AB AB

V( ) 1 Let 0 0

11 11 1111

Let

ω+ ω+

ω

ω= = ω+ ω+

+ω+

ω

ω+ ω+− ω − ω

∴ω= =+

ω+ ω+ ω+ ω+

−± −

ω= ∴ = − = =− ±

++

∴=+ + = =∴+=

++ +− + −



o

jj

j

jjj

jj

jj j j

jjj jj

x

jx x x j

xx

xjxj j j

(1 1) (1 1)

AB B2 B

12AB2, AB2 0

11 1111

B B 22B B0 B 1 1 A 1 1

11 11 11 11

V( ) 1 , V( ) 1

11 11 ()11()11

() () (1 1) () (1 1) (

−− −+

+

=−∴ + =∴−= =+ ∴ + =

−+−

∴ − + + + + = ∴ =− − ∴ =− +

−+ −− − +

∴=+ + ω=− −

++ +− ω++ ω+−

∴=δ−− −+

oo

jt jt

o

j

xjj

jj j j

jj j j j

jj j j

xj

xjxj jjjj

vtt jeut jeu

45 45

)

() 2 () 2 ()

( ) 2 2 cos( 45 ) ( )

−°−− °+−

−

=δ − −

=δ − + °

j jtt j jtt

t

te uteut

tetut

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



57.

22

/6

5/ 10 /

V( ) 105/ 35 30( ) 1/ 7 6( )

10 10 / 6

V( ) 71

6( ) 7( ) 1 () ()

66

49 24 1 10 / 6 2 2

7/6 /2 , 1 V( )

36 36 6 ( 1/6)( 1) 1/6 1

( ) 2( ) ( )

−−

ωω

ω= =

ω+ + ω ω+ + ω

∴ω= =

ω+ ω+ ω+ ω+



∴ω=− ± − =− − ∴ ω= = −



 ω+ ω+ ω+ ω+



∴= −

c

tt

c

jj

jjjjj

jjj jj

jj

jjj j

vt e e ut

CHAPTER EIGHTEEN (Fourier Analysis) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



58. 23

() 5 (), () 4 ()

tt

ft e ut gt e ut

−−

==

(a)

(b)

0

22 3 2

00

2

23

()()

54 20

20 ( 1) V

()()

∞

−− −−

−

−−

∗= −

==

=− −

∴∗= −

∫

∫∫

tt

tz z t z

tt

fg ftzgzdz

ee edz e edz

ee

fg e e ut

23

54 20

F( ) , G( ) F( )G( )

23 (2)(3)

20 20

F( )G( ) 20( 2 ) ( )

23

−−

ω= ω= ∴ ω ω=

ω+ ω+ ω+ ω+

∴ω ω= − ∴∗= −

ω+ ω+

tt

jj jj

jj fge ut

jj

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



1.

123 1 3 1 2

2123

213 13

0

13 12

1232

12 3 1

Order , , : 2 6 5 2cos10 3 2 (1)

4 0.05 0.15 0.25 (2)

2 5 0.4 ( ) 8 (3)

(1) 2 6 3 2 5 2cos10 A

(2) 0.05 0.15 0.25 4 B

(3) 2 5 0.4 0

′′

−−=+ −+

′′

=−+

=− − + − +

′′

→− − =− + + + =

′′′

→−+==

′′ ′

→++= −

∫

t

iii i i t i i

iiii

iii iidt

ii ii t

iiii

ii i i 3

112 2 1 3

1

2

.4 C

30 20 50 20cos10 240 3.6 3.6

A0 6

B 0.15 0.25

C1 5 A( 1) B(6) C( 0.9) A 6B 0.9C 10A 60B 9C

20 6 2( 1) 6(0.35) 0.1

0.05 0.15 0.25

215

2A 6

0.05 B 0.25

2C5 10[ A( 0.25) B(2) C(

0.1

=

′

∴= + + + + + −

−

−− + − −− −

′

∴= = = = + +

−−

−−− −

−

−−

′==−−−+−−

−

i

iii t i ii

i

212 213

2123

3

312 21

0.2)] 2.5A 20B 2C

7.5 5 12.5 5cos10 80 0.8 0.8

6.7 85 0.8 12.5 5cos10

20A

0.05 0.15 B

21C

10[A(0.35) B( 2) C(0.3)] 3.5A 20B 3C

0.1

10.5 7 17.5 7cos10 80 1.2 1.

=− − −

′=−−− −−+

′

∴= − + − −

−

′==−−−+=−−−

−

′=−−− −−+

iii tiii

iiii t

i

iii tii

3

3123

2

9.3 87 1.2 17.5 7cos10

′

∴= − + − − ←

i

iiii t

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



2. 1, 2 2 1xyxy x y xy

′′ ′ ′

+=++ − = −−

(a)

(b)

(c)

Order , M by 2: 2 2 2 2 2; add: 3 4 1

411 122

and

333 333

xy x y x y x x y

xxy y xy

′′ ′

+=++ =++

′′

∴= + + =− + +

4112241224

3333339993

12 2

(3 4 1)

99 9

2 , or 2 0



′′ ′ ′ ′

= +−+ += −+ +=





′

−+ −−+

′′ ′ ′′ ′

∴= − − +=

xx xy xxy x

xxx

xxxxxx

4 1 1851 4

Let (0) 2 and (0) 5 (0) (2) ( 5)

33 33333

1 2 2 2 10 2 10

Also, (0) (2) ( 5)

33 33333

82

(0) 2 (0) (0) 2

33

24

(0) 2 (0) (0) 2 0

33

xyx

y

xxx

′

= =−∴ = + −+=−+=

′=− + − + =− − + =−

′′ ′

∴= −=−=



′′′ ′′ ′

=−=−=





CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



3.

1

11

2 3 ( ), 2 5 3, 2 0 Order , ,

2.5 1.5 (1)

25 1

2.5 1.5 2 3 0.5 (3)

36 3

2511 11 511

2(2)

3623 231246

xyzftxz zyx xyz

xz

zyzfzyz f

yzx y z fxy x y z f

′′ ′ ′′

−−= += − −=

′

∴=− +

′′

−+−−=∴=−−+−

′′ ′

∴=−=−−+− −∴=−−− +−

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



4. 1

234, 567,(0)2,(0)

3

xxyyxyx y

′′

=− − + = − + = =

(a)

(b) 56 (0)5(1)6(15)95yxyy

′′ ′ ′ ′′

=− ∴ =−− =−

(c) 5 6 (0) 5( 43) 6( 95) 215 570 355yxyy

′′′ ′′ ′′ ′′′

=− ∴ =−−−=−+=

11

2 3 , (0) 2(2) 3 4 1, (0) 5(2) 6 7 15

33

(0) 2( 1) 3(15) 2 45 43

xxyx y

x

 

′′ ′ ′ ′ ′

=− − =− − + =− = − + =

 

 

′′

∴=−−−=−=−

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



5.

65

0.2 60

300 5 500cos 120 (1)

5 10 2.4 3.4 6.8 10 (2)

Ls LC

LLC

CL L L C L

iv iv

iiv t

vi i i v i

π

−

′

∴=−−

∴=− − +

′′

×=+=∴=×

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



6.

(a)

(b)

(c)

()()

()

22 2

2 2

2

22 2 2

Order: , , 5 8 ( ) 0.2 1.6 ( ) (1)

1 1 19 1 10

3 8 () 2 8 () ()

461243

19 1 10 () (2)

36 12 9

1111

2 8 () () (3)

4288

Lx F L F L F

xFx x xxF

xx F

FL F x F L x F

ivv i v ut i v ut

vvvutvutvvvut

vvv ut

v i v v ut v i v v ut

′′

∴=−+ ∴=− +

′=− + − + + − = − +

′

∴= − +

′′

=− +− ∴ = − − +

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



7.

(a)

(b)

4

Order: ,

0.002 3 15 , 0.02

0.002 15 0.94 , 7500 470 (1)

2 10 0.02

5000 100 5000 (2)

LC

LC A LA C

LL CL LC

CLAsL Cs

CLCs

iv

iv i ii v

iivi iv

viiii vi

vivi

−

′=− − =

′′

∴=−+ =−+

′

× =−− +=−− +

∴=− − +

123

11 12

112

212 2

212

323323

Order: , ,

0.1 0.2 0.2 0.1 0.1

32(1)

0.2 0.1 0.1 0.04 0.04

0.5 0.7 0.2 (2)

0.5 0.04 0.04 , 0.08 0.08 (3)

s

vvv

vvvvv

vvvv

vvv v v

vvvv

vvvvvv

′=− + − +

′

∴=− ++

′=−− +

′

∴= − +

′′

=− =−

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



8.

2

0.02(12.5 12.5 12.5 )

1.25 0.25 0.02 0.25

0.2 0.016 0.2

0.002 15 3 12.5 12.5 12.5

0.002 27.5 15.5( 0.2 0.016 0.2 ) 12.5

0.002 24.4 0

AsAC

ALCs

AL Cs

LLACsAL

LLC LCss

LL

iiiiv

iivi

ii vv

iiiviii

iiv ivii

ii

=−−+

∴=−+ +

∴=− + +

′=− − + + − −

′

∴=−+−−+ ++

′

∴=−+.752 9.4

12,200 376 4700 (1)

Order: , Also, 0.0002 0.2 0.016 0.2

0.0002 0.8 0.016 0.8 4000 80 4000 (2)

Cs

LLCs

LC C s A L s L C s L

CL CsC LC s

vi

iivi

iv v i i i i i v i i

vi viv ivi

+

′

∴=− + +

′=−−=+ − − −

′′

∴=−−+∴=−−+

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9.

112 1 1 2

21 2 1

112 1 12

21 2 2 1 2

1212

112 1 1 2

21 2 1

0.1 10 10 (1)

0.2 5 (2)

0.5 , 2 2 (3)

0.05 2 2 20 40 40 20 (4)

Order: , , ,

0.1 10 10 (1)

0.2 5 (2)

0.

ss ss

vii v i i

viv i

ivv i vv

iv i iv i v i i v

vvii

vii v i i

viv i

′′

=− ∴ = −

′′

=− ∴ =−

′′

=− + ∴ =− +

′′

=+ − + ∴= − + +

′′

=− ∴ = −

′′

=− ∴ =−

1121 12

21 2 2 2

21 2 2 2 1 2

522(3)

0.05 2 2 , 3 2 2 0.4 0.2

0.05 0.8 0.4 2 20 24 12 (4)

sR R R s R s

ss s

ivvi vv

ivv i ii i iv i i v

ivv i v i i v i v

′′

=− + ∴ =− +

′=+− − =− − + ∴=− +

′′

∴=++−−∴=−+

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



10.

(a)

(b)

Order: , 0.2 20 20

100 5 100 (1)

0.1 10 (2)

LC L C L s

LLCs

CL C L

iv i v i i

iivi

vi v i

′

∴=−−+

′

∴=− − +

′′

=∴=

0.5 5 10 2 (1)

0.2 0.5 0.5

5 2.5 2.5 (2)

LC L L L C

CsCL

CL C s

iv i i i v

vvvi

vivv

′′

=− ∴=− +

′=−−

′

∴=−− +

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



11.

5202020

0.04 0.8 0.8

0.1 0.04 0.2 0.8

Order: , 0.4 2 8 (1)

0.2 20 20 20( 0.04 0.2 0.8 )

0.2 0.2 4 4 20 20 (2)

Rs R

Rs

s

ss

ivi ii

ivii

vii v i i

vi v v i i

ivi i v i i

iviiivii

∴=−+ −−

∴=− − +

′

∴=+=− ++

′

∴=− + +

′=− + + − − + +

′′

∴=−−+∴=−−+

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



12.

40.2

20

0.1 0.05 0.05 0.2

0.15 0.2 0.05

5 , 0.2 0.2 , 0.2 0.2

0.15 0.2 0.2 0.04 0.04 0.05

0.11 0.2 0.2 0.01 0.11 0.

xx

Cx C s x

Cx x s

xcsx c sx C s

CCsC ss

CCs s C

ii

vi v v i

vi i v

ivvi v vi v v

vvvvvv

vvvv v

=

′′′′

=− + −

′′′

∴=−+

′′′

=− + =− + =− +

′′′′

∴=−++−+

′′′

∴=−++ ∴=−2 sin 2 ( ) 0.1cos2 ( )

20 100 10

sin2 () cos2 () (1)

11 11 11

C

CC

vtut tut

vv tut tut

++

′

∴=− + +

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



13.

(a)

(b)

1

2

2112

2

212

12

123 2

,456,3 232(1)

789 1

4563 (2)

7891 (3)

L

LLLLC

C

LLL C

CLL C

it

qi a f t i i i v t

vt

iiivt

viiv t

−−−

    

     ′

==− =∴=−−−+

    

−− +

    

′=−++

′=−−++

06 1

,,

40 0

61, 4

0.25 C 0.25 F

111

L H

666

L

C

LCCL

CL

LC

i

qa f

v

ivvi

vi

iv

−

 

== =

 





′′

∴=− + =

′

∴=∴=

′=− + ∴ =

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



14.

123

1123

2123

2 123 3123

3123

0.1 5 5 5

50 50 50 10 (1)

0.2 5 15 15

25 75 75 5 (2) 0.5 5 15 40

10 30 80 2 (3)

50 50 50

25 75 75

10 3

LLLLs

LLLLs LLLLs

LLLLs

iiiiv

iiiiv iiiiv

iiiiv

a

′=− − − +

′

∴=− − − +

′=− − − +

′′

∴=− − − + =−− − +

′

∴=− − − +

−−−

=− − −

−−

10

5

080 2

s

v

fv

v





=



−



CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



15.

01

1

02

2

1

2

12 0 0

33 0 10 00 1 2

,210, 0; , ,

012 2

14 2 0 13 0 0

,,

12 0

00 1

012

13 0

C

R

L

R

v

vv

qv a f w b d

i

ii

w bq d q aq f w bq d w bq baq bf

ba

 

−

     

−

       

==− === =

     

−−

   

− 

     



′′′

=+∴=+ =+ = = +



−



=−





12

1

2

110 120 10

33 0 10

142 00 1 0

210 , 0

09 4 012 0

14 2 0

300 130 10

10

110 10

142 0

09 4 0

300 10

CC

C

L

bf

vv

wv

i



−

 



−− −

 



−= = =

 



−−

 

−



 



++







−

−− 



′

∴= + =





−







2

1

01 1 2 02 1 2 1 02 2 1

42

94

310

10, 4 2 , 9 4 , 3 10

CL

C

CC C C LR LR C

vi

v

vvv v v v ii v ii v





−+



−



+





′′ ′′

∴=++ =−− + = − = +

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



16.

1

2

3

4

3 1 2 cos2 1 2 0 3

,221,sin2, ,0111

130 0 2 1 13

10

210

1, / , , (0) 5

35

12 03

0111

2113



−

      



      



==−− = ==−

      



      

−−−



      







 

−

 

==+=+=

 

−

 

 





∴= −

−−



y

qt

y

qqa f ty b

y

qy

dqaqfqbydy

q

π

1

124

2

234

3

123 4

4

2 232 7

11(0)11

32 33 53

3 1 2 7 1 21 11 106 1

(0) 2 2 1 11 0 14 22 53

13053 0 733

 +++

  



  



+=−+++ ∴ =

 



 

−−+ +



  





−−+++





′=− − + = − − +



−−+



yyyy

yyyy q

yyyy y

y

q

97

17

26





=





CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



17.

1

4 , 2 3 3 , 18 2

99

21

666 66,4

6399

21

Also, 2 6 6 3 3 9 6

33

1

3

C

L

C

CLRLRRLsCRLRL

CR RLs R R L s L L s C L L

L LLs LLCs LCLss

L

v

qi

v

v iiiv i ivv i i i i

vv iii v i i i i i i v i i

iiiiiivvivivi

iv



=



′′ ′′

=− − − = − − + + = ∴ =

′′′

=− − − ∴ =− − − =− − − ∴ =− − −

′′ ′ ′

=− − − − − + + ∴ = − + −

′

∴= 111112

32(2)4

39273279

4212 111 1

4(1)

27 3 27 9 27 6 108 18

11

1/27 1/6 108 18

[] []

1/3 3 1 2

3

CL ss C CL C L s s

CCLssCCLss

ss

ivi v vi vi vi

vvivivvivi

vi

af

vi

′

−+ − =− −− + − +

′′

∴=− −− +∴=− −− +



−+



−−



∴= =



−





−





CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



18.

(a) 30 10 2 5 15ii i i

′′

=+ ∴=−+

(b)

(c) 55

3(1 ) A, 0.5 A

tt

zero state zero input

ieie

−−

=− =

(d)

555

0

555 5

(0) 0.5 A, 5 0.5 15

0.5 3( 1) 3 2.5 A, 0

−−

−− −

==−∴=+

∴= + − = − >

∫

t

ttZ

ttt t

iaieeedz

ieee e t

55 5

5

3, A 3 A , (0) 0.5 3 2.5 A

2.5 A, 3 A

tt t

fn

t

nf

iie i ei i e

iei

−− −

−

= = ∴= + = ∴= −

∴=− =

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



19.

5

500 500

0

500 500

0

500 500

2

0

500 500

22

5 10 0.04 ( ) 0.025

500 800 ( ) 500

800Z

800 Z

Z1

800 500 500

11

800 500 500 500

1.

1.6

−

′

∴× = −

′

∴=− + ∴=−

∴=

=





∴= −













∴= − +









∴= +

∫

CC

t

tZ

C

t

tZ

t

tZ

C

tt

C

vtut v

vvtuta

ve e dz

eedz

vee

t

vee

vt 500 3 6 500

2

61

( 1 ) 8 10 16 10 (1 ) A, 0

500 200

−−−−

−+ ∴ = = × − × − >

tt

C

eiv t et

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



20.

(a)

(b)

50 50 50 50 50 50

0

50 50

50

1000 500 100 20

0.3 15 100 50 ( ), 50 ( ), (0) A

3 3 15 3

20 20 1

500dZ 500 ( 1)

3350

20 10(1 ) 0

3

20 10

( ) 10 ( ) A

33

−− −−

−−

−

′′

=− + + =− + + = =

∴= × + = + −

∴= + − >



∴= −+ −





∫

LL LL L

t

ttZ ttt

L

tt

L

t

L

ii utii uti

ie e e e e e

ie et

iut eut

50 50

,,

50

20 A, 10(1 ) A

3

10

10 A, A

3

tt

L zero input L zero state

t

Lf Ln

iei e

iie

−−

−

==−

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



21.

2.5

2.5 2.5 2.5

2

00

2

100[ ( ) ( 0.5)]cos V

1

0.01 (0.8 ) 2.5 200[ ( ) ( 0.5)]cos

40

00

0 0.5 200 cos Z 200 (2.5cos Z sin Z)

6.25

200 (2.5cos si

6.25

−−

=−−

′′

=−∴=−+−−

<=



≤≤ = × = +



+



∴= +

+

∫

s

CsCCC

C

t

tZ

tZ t

C

vutut t

vvvvvutut t

tv

e

t v e e dz e

vt

π

ππππ

π

ππ

π

2.5

2

2.5

0.5

2.5

2

0

1.25

2.5 2.5

22

200

n ) 2.5

6.25

31.02cos 38.98sin 31.02

0.5 200 (2.5cos Z sin Z)

6.25

1

200 2.5 105.03 V

6.25 6.25

−

−−

−+

∴= + −



>= +



+





∴= − × =



++



t

C

Z

t

C

tt

C

te

vtte

e

tv e

e

ve e

ππ

πππ

π

ππ

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



22.

11

510()

40 4

vv ut

′

∴=−++

(a) 10 200 400 ( )vv ut

′

∴=− + +

(b)

(c)

10 10 10 10 10 10

0

10 10 10

020 V

0 20 600 20 60 ( 1)

20 60 60 60 40 V

−− −−

−− −

<∴=

>= + = + −

∴= + − ∴= −

∫

Z

ttZ ttt

tt t

tv

tve eedze ee

ve e v e

10

10 10

60V 40 V

60(1 ) V 20 V

t

forced nat

tt

zero state zero input

vve

veve

−

−−

==−

=− =

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



23.

22

0

22 2

0

0.5

220.52

0

2

15 30 3

2 4 [ ( ) ( 0.5)]

4 [ (Z) (Z 0.5)]

0 0.5 4 dZ 4 ( 1) 4 4 A

0.5 4 dZ 0 4 ( 1) 2.595 A

−

−−

−−−−

′=− +

′

∴=− + − −

∴= − −

≤≤ = = − = −

≥= += −=

∫

LLs

t

LL

t

tZZ

L

t

tZ tt t t

L

tZ t t

L

iv

iieutut

ieeeu u dz

tiee ee ee

tiee ee e

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



24.

85

[ ] , 0.01

10 10

at

−



==



−



(a)

(b)

(c) 1.0860 0.0547 0.9254 0.0457 1.0000 0

0.1095 1.1079 0.0915 0.9071 0 1.0000

ta ta

ee

−−



==



−



0.08 0.05 0.08 0.05 0.08 0.05

1

I ...

0.1 0.1 0.1 0.1 0.1 0.1

2

1 0 0.08 0.05 0.0114 0.0090 0.0114 0.009 0.08 0.05

11

0 1 0.1 0.1 0.0180 0.0150 0.018 0.015 0.1 0.1

26

ta

e−−−−



=− + −



−−−



−−−−

   

=− + −

   

−− − −

    ...

1.08 0.05 0.0057 0.0045 0.00181 0.00147 1.0860 0.0547

1

0.1 1.1 0.009 0.0075 0.00294 0.00240 0.1095 1.1079

6

+





−−− −

   

=+ − =

   

−− −−

   

1 0 0.08 0.05 0.0057 0.0045 0.0003 0.0002 0.9254 0.0457

0 1 0.1 0.1 0.009 0.0075 0.0005 0.0004 0.0915 0.9071

ta

e−−−

    

=+ + + =

    

−− −

    

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



25.

(a)

(b)

(c)

() 1 2 1

,cos,013

() 2 3 1

2 ( ), 3 cos , 2 3 ( )

xut

qyf ta

zut

x x y z ut y y z t z x y z ut

−

    

    

== =−

    

−−−−

    

′′′

∴=+ −+ =−+ + =− − −−

2121214

(0) 3 , /(0) 0 1 3 3 1 7 0.1

1231113

2 0.4 1.6

(0.1) (0) 0.1 (0) 3 0.7 2.3

1 0.3 1.3

qq t

qq q

−−

  

  

=− ∴ = − − + = ∆=

  

−−− −

  

−

  

  

′

=+ =−+ =−

  

  

2 4 1.8

0.05 (0.5) 3 0.05 7 2.65

1 3 1.15

1 2 1 1.8 1 3.65

(0.05) 0 1 3 2.65 cos0.05 7.0988

2 3 1 1.15 1 2.2

(0.1) (0.05) 0.05 (0.05)

tq

q

qq q

−

  

  

∆= ∴ =− + =−

  

  

−−





′=− −+ =



−−− −



′

=+ =

1.8 3.65 1.6175

2.65 0.05 7.0988 2.2951

1.15 2.2 1.26

−

 

 

−+ =−

 

 

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



26.

232

12 3 1 2 3

0 1 2 Now, I 0 1 2

31 1 3 1 1

det[ ] ( 1 )( 2 1 2) 3(4 3 3 ) 3 8 22 0

s

aass

s

ss s s s s s

−−−

  

  

=− −= −−

  

−−−

  

=−− ++−+++=−−++=

By trial and error, Solve, or high school algebra (Hormer’s method) we find s = -3.48361. Now

divide polynomial by s + 3.48361. Get quadratic, s2 – 0.48361s – 6.31592 = 0.

The remaining two roots are: s = -2.28282 and s = + 2.76641.

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



27.

12

2

12 11 12

25 25 25

11 11

225 2

32 3 2

det( I) det 7 12 2 0

14 1 4

Roots are 2, 5 Now, ,

11

2, 5 3, 33

22 5

, or

33 3

−− −− −−

−−− −

−−−

  

=−= =++−=

  

−−−

  

=− =− = + = +

∴=− =−∴−= = −

∴= + − =

st st

oo

tt tt tt

oo

ttt

oo

s

aas ss

s

s s e u us e u us

euueuuee uue e

ue e e u e 5

25

1

25

2525 25

25 252 5

2

3

52 032

11

33

I52 14

33

033

52 22

33 33

11 5244

33 3333

21

33

−

−−

−−−− −−

−− −−− −

−−

−



−



−





=+= + −







−







−







−−+ −



=



− −−+





+

∴=

tt

ta tt

o

tt

tttt tt

ta

tt tttt

t

ta

e

ee

eu ua e e

ee

eeee ee

e

ee eeee

ee

e

525

25 25

22

33

1112

3333

−−

−− −−



−



−+





ttt

tt t t

ee

eeee

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



28.

(a)

(b) 2

12

0 0.1

I , det( I) 10 9 1, 9

90 10

s

as as s s s s

s

−−



−= − =+ +∴=− =−



−−



(c)

(d)

(e)

10 36

11

90 9

0.1 3.6, 90 10

0 0.1 3.6

,

90 10 0

iv

i

qvvi v

ivviv

af

′=− −



= ′=−



′′

∴=− − = −

−−



∴= =



−



99

11 1

99 9

1

,9 8

11 11 91

,

88 88 88

−− −−

−− −−− −−

=− =− ∴− =

∴= − = − + = −

tt tt

oo

tt ttttt

o

euue uuee u

ue eue ee e e

99

1

9 9

999

99

1 0 0 0.1

11

I(9) ()

0 1 90 10

88

9 0 0 0.1 0.1

80 9 90 90 10 10

9 0.1 0.1

890 90 9

ta tt tt

o

tt t t

ta

tt t t t t

tt t t

ta

tt tt

eu ua ee ee

ee e e

eee e e e e

ee e e

eee ee

−− −−

−− − −

−− − − − −

−− − −

−− −−

−

  

=+= − + −

  

−

  

 

−−+

∴= +

 

−−−+

 



−−+

=−−+



99

91 1 1

8 8 80 80

90 90 1 9

88 88

tt t t

ta

tt tt

ee e e

e

ee ee

−− − −

−− −−





−−+



∴=



−−+





0

99

0

9

(0) 4

(0) (Z) (0) (0) 36

1

9 0.1 0.1

1

9

290 90 9

3.6

9 0.1 0.1

1

0

890 90 9

9

1

2

−

−− − −

−− −−

−−



=+ ==







−−+

=



−−+



−



−−+ 

+



−−+



−−

∴=

∫

t

ta ta Za

tt t t

tt tt

tZZ Z Z

ta

ZZ ZZ

tt

i

q e q e e f dz q v

ee e e

qee ee

ee e e

edz

ee ee

ee

q

99

99 9

0

0.9 0.9 9

0.45 dZ

90 90 9 81 90 90

−−

−−−−



+−

−



−−+ −



∫

t

tt ZZ

ta

tttt ZZ

ee ee

e

eeee ee

CHAPTER NINETEEN (WEB CHAPTER) SOLUTIONS

Engineering Circuit Analysis, 6th Edition Copyright 



9

99

9

999

11

99

4.05 0.05 0.45 99

40.5 4.5 90 10 90 10

180

9

4.05 0.05 9 0.1 0.1

0.45 99

8

40.5 4.5 90 90 9 90 1

tt

tt ta

tt tt

tt

tt tttt

t

ee

ee e

ee ee

ee

ee ee ee

qee ee ee e

−

−−

−− −− −−



−−+

−

=−



−



−−+





 

−−

−−+

=−

 

−−−+

 

−9

9

889889

080

4.05 0.05

4.05 4.5

180

81 80 9 9 8 9 1 8

0.45 99

8810 10 800 810 10 800 90 10 80 810 90 720

4.05

t

tt

ttt t ttt t

tt t t tt t t

e

ee

eee e eee e

e

−−

−− − −− −





−







−

=

−



 

−− − ++ −++ + −−

 

− 

−− − ++ −+++ −−

 

 

=9

9

99

9

18

71 72

0.05 0.45 99

7

40.5 4.5 640 720 80

4.05 0.05 4 4.05 0.05

40.5 4.5 36 40.5 4.5

4 8.1 0.1

36 81 9

tt

tt tt

tt

ee

e

ee ee

ee

qee

−−

−− −−

−−



+− +

−

−



−



−+





 

−−+

=−

 

−−+

 

−+ −

∴=−+ − (0)t

>





Chapter Two Exercise Solutions Engineering Circuit Analysis (6th Edition, 2001) Hayt Solution Manual

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