Solution manual fundamentals of electric circuits 3rd edition chapter16

Chapter 16, Problem 1.
Determine i(t) in the circuit of Fig. 16.35 by means of the Laplace transform.

Figure 16.35
For Prob. 16.1.

Chapter 16, Solution 1.
Consider the s-domain form of the circuit which is shown below.
1

1/s

I(s)

+
−

1/s

s
I(s) =

i( t ) =

1s
1
1
= 2
=
1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2

⎛ 3 ⎞
e - t 2 sin ⎜⎜
t ⎟⎟
2
3
⎝
⎠

2

i( t ) = 1.155 e -0.5t sin (0.866t ) A

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Chapter 16, Problem 2.

Find v x in the circuit shown in Fig. 16.36 given v s .= 4u(t)V.

Figure 16.36
For Prob. 16.2.

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Chapter 16, Solution 2.

8/s

s
+

4
s

+
−

Vx

2

4

−
4
s + Vx − 0 + Vx − 0 = 0
8
s
2
4+
s

Vx −

Vx (4s + 8) −

(16s + 32)
+ ( 2s 2 + 4s) Vx + s 2 Vx = 0
s

Vx (3s 2 + 8s + 8) =

16s + 32
s

⎛
⎞
⎜
⎟
− 0.125 ⎟
− 0.125
s+2
0.25
⎜
+
= −16
+
Vx = −16
⎜ s
8⎟
4

8
4
s(3s 2 + 8s + 8)
s+ − j
s+ + j
⎜
⎟
3 ⎠
3
3
3
⎝
v x = ( −4 + 2e − (1.3333 + j0.9428) t + 2e − (1.3333 − j0.9428) t )u ( t ) V

⎛2 2
vx = 4u ( t ) − e − 4 t / 3 cos⎜⎜
⎝ 3

⎞ 6 − 4t / 3 ⎛ 2 2
t ⎟⎟ −
e
sin ⎜⎜
2
⎠
⎝ 3

⎞
t ⎟⎟ V
⎠

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Chapter 16, Problem 3.

Find i(t) for t > 0 for the circuit in Fig. 16.37. Assume i s = 4u(t) + 2 δ (t)mA. (Hint: Can
we use superposition to help solve this problem?)

Figure 16.37
For Prob. 16.3.

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Chapter 16, Solution 3.

In the s-domain, the circuit becomes that shown below.
1
I
4
+2
s

2

0.2s

We transform the current source to a voltage source and obtain the circuit shown below.
2
1
I
8
+4
s

+
_

0.2s

8
+4
20 s + 40 A
B
s
I=
=
= +
3 + 0.2 s s ( s + 15) s s + 15

40 8
−15 x 20 + 40 52

= ,
B=
=
15 3
−15
3
8 / 3 52 / 3
I=
+
s
s + 15
8
52
⎡
⎤
i (t ) = ⎢ + e −15t ⎥ u (t )
⎣3 3
⎦
A=

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Chapter 16, Problem 4.

The capacitor in the circuit of Fig. 16.38 is initially uncharged. Find v 0 (t) for t > 0.

Figure 16.38
For Prob. 16.4.

Chapter 16, Solution 4.

The circuit in the s-domain is shown below.
I

2

4I

+
5

+
_

Vo

1/s

1

–

Vo
⎯⎯
→ 5 I = sVo

1/ s
5 − Vo
But I =
2
I + 4I =

⎛ 5 − Vo ⎞
5⎜
⎟ = sVo
⎝ 2 ⎠

⎯⎯
→ Vo =

12.5
s +5/ 2

vo (t ) = 12.5e −2.5t V
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Chapter 16, Problem 5.

If i s (t) = e −2t u(t) A in the circuit shown in Fig. 16.39, find the value of i 0 (t).

Figure 16.39

For Prob. 16.5.
Chapter 16, Solution 5.
Io
s

1
s+2

2

2
s

⎞
⎛
⎟
⎛
⎞
1 ⎜⎜
1
2s
2s
⎟= 1 ⎜
⎟⎟ =
V=
⎜
s + 2 ⎜ 1 1 s ⎟ s + 2 ⎝ s 2 + s + 2 ⎠ (s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229)
⎜ + + ⎟
⎝s 2 2⎠

Io =

Vs
s2
=
2
(s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229)

(−0.5 − j1.3229) 2
(−0.5 + j1.3229) 2
1
(1.5 − j1.3229)(− j2.646) (1.5 + j1.3229)(+ j2.646)
=
+
+
s+2
s + 0.5 + j1.3229
s + 0.5 − j1.3229

(

)

i o ( t ) = e − 2 t + 0.3779e − 90° e − t / 2 e − j1.3229 t + 0.3779e 90° e − t / 2 e j1.3229 t u ( t ) A
or

(

)

= e − 2 t − 0.7559 sin 1.3229 t u ( t ) A

⎛
⎛ 7 ⎞⎞
2
sin ⎜⎜
t ⎟⎟ ⎟u ( t )A
or io(t) = ⎜ e − 2 t −
⎜
⎟
2
7
⎝
⎠⎠
⎝
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Chapter 16, Problem 6.

Find v(t), t > 0 in the circuit of Fig. 16.40. Let v s = 20 V.

Figure 16.40
For Prob. 16.6.

Chapter 16, Solution 6.

For t<0, v(0) = vs = 20 V
For t>0, the circuit in the s-domain is as shown below.

I
+

10
s

20
s

100mF = 0.1F

+
_

10 Ω

v
_

⎯⎯
→

1 10
=
sC s

20

s = 2
s +1
10 + 10
s
20
V = 10 I =
s +1
I=

v(t ) = 20e − t u (t )
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Chapter 16, Problem 7.

Find v 0 (t), for all t > 0, in the circuit of Fig. 16.41.

Figure 16.41
For Prob. 16.7.

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Chapter 16, Solution 7.
The circuit in the s-domain is shown below. Please note, iL(0) = 0 and vo(0) = o because
both sources were equal to zero for all t<0.

1

1
Vo

2/s

+
_

s

2/s
1/s

At node 1
2 / s − V1 V1 V1 − Vo
2
= +
⎯⎯
→
= V1 (2 + 1/ s ) − Vo
1

1
s
s
At node O,
V1 − Vo 1 Vo
s
+ =
= Vo
⎯⎯
→ V1 = (1 + s / 2)Vo − 1/ s
1
s 2/ s 2
Substituting (2) into (1) gives
1
1
2 / s = (2 + 1/ s )(1 + s / 2)Vo − (2 + ) − Vo
s
s
(4s + 1)
A
Bs + C
Vo =
= + 2
2
s ( s + 1.5s + 1) s s + 1.5s + 1

(1)

(2)

4s + 1 = A( s 2 + 1.5s + 1) + Bs 2 + Cs
We equate coefficients.
0 = A+ B or B = - A
s2 :
s:
4=1.5A + C
constant:
1 = A, B=-1, C = 4-1.5A = 2.5
3.25
7
x
4
7
− s + 2.5
s + 3/ 4
1
1
4
Vo = + 2
= −
+
2
2
s s + 1.5s + 1 s
⎛ 7⎞
⎛ 7⎞
2
2
( s + 3 / 4) + ⎜
⎟ ( s + 3 / 4) + ⎜

⎟
⎝ 4 ⎠
⎝ 4 ⎠
v(t ) = u (t ) − e −3t / 4 cos

7
7
t + 4.9135e−3t / 4 sin
t
4
4

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Chapter 16, Problem 8.

If v 0 (0) = -1V,obtain v 0 (t) in the circuit of Fig. 16.42.

Figure 16.42
For Prob. 16.8.

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Chapter 16, Solution 8.

1
1 2
F
⎯⎯
→
=
2
sC s
We analyze the circuit in the s-domain as shown below. We apply nodal analysis.

1

3
s

1

Vo

2
s

+
_
+

_

3
1
4
− Vo − − Vo
− Vo
s
+ s
+s
=0
2
1
1
s
Vo =

⎯⎯
→ V0 =

+
_

4
s

−1
s

14 − s

s ( s + 4)

A
B
+
s s+4

14
18
= 7 / 2,
B=
= −9 / 2
4
−4
7/2 9/2
Vo =
−
s
s+4
A=

⎛7 9
⎞
vo (t ) = ⎜ − e −4t ⎟ u (t )
⎝2 2
⎠

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Chapter 16, Problem 9.

Find the input impedance Z in (s) of each of the circuits in Fig. 16.43.

Figure 16.43
For Prob. 16.9.
Chapter 16, Solution 9.

The s-domain form of the circuit is shown in Fig. (a).
2 (s + 1 s)
2 (s 2 + 1)
Z in = 2 || (s + 1 s) =
= 2
2 + s + 1 s s + 2s + 1

(a)

1
s
2
1/s

2

2/s
1

(a)

(b)

s

(b)

The s-domain equivalent circuit is shown in Fig. (b).
2 (1 + 2 s) 2 (s + 2)
2 || (1 + 2 s) =
=
3+ 2 s
3s + 2
5s + 6
1 + 2 || (1 + 2 s) =
3s + 2
⎛ 5s + 6 ⎞
⎟
s ⋅⎜
⎝ 3s + 2 ⎠
⎛ 5s + 6 ⎞
s (5s + 6)
⎟=
= 2
Z in = s || ⎜
⎝ 3s + 2 ⎠
⎛ 5s + 6 ⎞ 3s + 7s + 6
⎟

s +⎜
⎝ 3s + 2 ⎠

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Chapter 16, Problem 10.

Use Thevenin’s theorem to determine v 0 (t), t > 0 in the circuit of Fig. 16.44.

Figure 16.44
For Prob. 16.10.

Chapter 16, Solution 10.

⎯⎯
→ 1s and iL(0) = 0 (the sources is zero for all t<0).
1H
1
1 4
F
⎯⎯
→
= and vC(0) = 0 (again, there are no source
sC s
4

contributions for all t<0).

To find ZTh , consider the circuit below.
1

s

ZTh
2

ZTh = 1//( s + 2) =

s+2
s+3

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To find VTh, consider the circuit below.
1

s
+

10
s+2

+
_

VTh

2

-

s + 2 10
10
=

s+3 s+2 s+3
The Thevenin equivalent circuit is shown below
VTh =

ZTh
+

VTh

+
_

4/s

Vo
–

40
4
3
10
40
3
s

=
=
Vo =
VTh =
.
4
4 s + 2 s + 3 s 2 + 6s + 12 ( s + 3)3 + ( 3) 2
+ ZTh
+
s
s s+3
4
s

vo (t ) = 23.094e −3t sin 3t

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Chapter 16, Problem 11.

Solve for the mesh currents in the circuit of Fig. 16.45. You may leave your results in the
s-domain.

Figure 16.45
For Prob. 16.11.

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Chapter 16, Solution 11.
In the s-domain, the circuit is as shown below.

1Ω

10
s

+
_

I1

4

1
H
4

10
s
1
= (1 + ) I1 − sI 2
4
4
s
1
5
− sI1 + I 2 (4 + s ) = 0
4
4
In matrix form,

I2

1s

(1)
(2)

1 ⎤
⎡ s
− s ⎥

⎡10 ⎤ ⎢1 +
⎡ I1 ⎤
4
⎢ s ⎥=⎢ 4
⎥⎢ ⎥
⎢ ⎥ ⎢ 1
I
5
⎢⎣ 0 ⎥⎦ − s 4 + s ⎥ ⎣ 2 ⎦
4 ⎦⎥
⎣⎢ 4
1
9
∆ = s2 + s + 4
4
4
10
1
− s
40 50
4
s
∆1 =
=
+
5
s
4
0 4+ s
4

s 10
1+
5
4 s
∆2 =
=
1
2
− s 0
4
40 25
+
∆
50s + 160
s
2
I1 = 1 =
=
2
∆ 0.25s + 2.25s + 4 s(s 2 + 9s + 16)
I2 =

∆2
2.5
10
=
=
2
2
∆

0.25s + 2.25s + 4 s + 9s + 16

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Chapter 16, Problem 12.

Find v o (t) in the circuit of Fig. 16.46.

Figure 16.46
For Prob. 16.12.

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Chapter 16, Solution 12.
We apply nodal analysis to the s-domain form of the circuit below.
s

10/(s + 1)

+

−

Vo

1/(2s)

4

3/s

10
− Vo 3 V
s +1
o
+ =
+ 2sVo
s
s
4
10
10 + 15s + 15
+ 15 =
(1 + 0.25s + s 2 ) Vo =
s +1
s +1
Vo =

15s + 25
A
Bs + C

=
+ 2
2
(s + 1)(s + 0.25s + 1) s + 1 s + 0.25s + 1

A = (s + 1) Vo

s = -1

=

40
7

15s + 25 = A (s 2 + 0.25s + 1) + B (s 2 + s) + C (s + 1)
Equating coefficients :
0= A+B ⎯
⎯→ B = -A
s2 :
1
15 = 0.25A + B + C = -0.75A + C
s :
0
s :
25 = A + C
A = 40 7 ,

B = - 40 7 ,

C = 135 7

40
- 40 135
1
3
s+
s+
⎛
⎞
155
2
40
1
40
7
7
7
2
2
⎟
+⎜
⋅
=
−
Vo =
+
2
2
2
7 s +1 7 ⎛ 1⎞

s +1 ⎛ 1⎞
3 ⎠⎛ 1⎞
3 ⎝ 7
3
3
⎜s + ⎟ +
⎜s + ⎟ +
⎜s + ⎟ +
⎝ 2⎠
⎝ 2⎠
⎝ 2⎠
4
4
4

v o (t) =

⎛ 3 ⎞
⎛ 3 ⎞ (155)(2)
40 - t 40 - t 2
t ⎟⎟ +
e − e cos ⎜⎜
e - t 2 sin ⎜⎜
t ⎟⎟
7
7
⎝ 2 ⎠
⎝ 2 ⎠ (7)( 3 )

v o ( t ) = 5.714 e -t − 5.714 e -t 2 cos(0.866t ) + 25.57 e -t 2 sin( 0.866t ) V

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Chapter 16, Problem 13.

Determine i 0 (t) in the circuit of Fig. 16.47.

Figure 16.47
For Prob. 16.13.

Chapter 16, Solution 13.

Consider the following circuit.
1/s

2s

Vo

Io
2

1/(s + 2)

1

Applying KCL at node o,
Vo
Vo
1
s +1
=
+
=
V
s + 2 2s + 1 2 + 1 s 2s + 1 o
2s + 1
Vo =
(s + 1)(s + 2)
Io =

Vo
1
A
B
=
=
+
2s + 1 (s + 1)(s + 2) s + 1 s + 2

A = 1,
Io =

B = -1

1

1
−
s +1 s + 2

i o ( t ) = ( e -t − e -2t ) u(t ) A

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Chapter 16, Problem 14.

* Determine i 0 (t) in the network shown in Fig. 16.48.

Figure 16.48
For Prob. 16.14.

* An asterisk indicates a challenging problem.
Chapter 16, Solution 14.

We first find the initial conditions from the circuit in Fig. (a).
1Ω

4Ω
+

5V

+
−

vc(0)

io

−
(a)

i o (0 − ) = 5 A , v c (0 − ) = 0 V
We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
1

4

Vo
Io

15/s

+
−

2s

5/s

4/s

(b)
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At node o,
Vo − 15 s Vo 5 Vo − 0
+
+ +
=0
1
2s s 4 + 4 s
15 5 ⎛
1
s ⎞
⎟V
− = ⎜1 + +
s s ⎝ 2s 4 (s + 1) ⎠ o
10 4s 2 + 4s + 2s + 2 + s 2
5s 2 + 6s + 2
=
Vo =
Vo
s
4s (s + 1)
4s (s + 1)

40 (s + 1)
Vo = 2
5s + 6s + 2
Vo 5
4 (s + 1)
5
+ =
+
2
2s s s (s + 1.2s + 0.4) s
5 A
Bs + C
Io = + + 2
s s s + 1.2s + 0.4

Io =

4 (s + 1) = A (s 2 + 1.2s + 0.4) + B s s + C s
Equating coefficients :
s0 :
4 = 0.4A ⎯
⎯→ A = 10
s1 :
2

s :

4 = 1.2A + C ⎯
⎯→ C = -1.2A + 4 = -8
0= A+B ⎯

⎯→ B = -A = -10

5 10
10s + 8
+ − 2
s s s + 1.2s + 0.4
10 (s + 0.6)
10 (0.2)
15
Io = −
2
2 −
s (s + 0.6) + 0.2
(s + 0.6) 2 + 0.2 2
Io =

i o ( t ) = [ 15 − 10 e -0.6t ( cos(0.2 t ) − sin( 0.2 t )) ] u(t ) A

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.


Chapter 16, Problem 15.

Find V x (s) in the circuit shown in Fig. 16.49.

Figure 16.49

For Prob. 16.15.

Chapter 16, Solution 15.
First we need to transform the circuit into the s-domain.
10
s/4
Vo
+
3Vx

Vx

−

5/s

+

+
−

5
s+2

5
Vo − 3Vx Vo − 0
s+2 =0
+
+
s/4

5/s
10
5s
5s
= 0 = (2s 2 + s + 40)Vo − 120Vx −
40Vo − 120Vx + 2s 2 Vo + sVo −
s+2
s+2
Vo −

But, Vx = Vo −

5
5
→ Vo = Vx +
s+2
s+2

We can now solve for Vx.
5 ⎞
5s
⎛
(2s 2 + s + 40)⎜ Vx +
=0
⎟ − 120Vx −
s + 2⎠
s+2
⎝
(s 2 + 20)
2(s + 0.5s − 40)Vx = −10

s+2
2

Vx = − 5

(s 2 + 20)
(s + 2)(s 2 + 0.5s − 40)

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.


Chapter 16, Problem 16.

* Find i 0 (t) for t > 0 in the circuit of Fig. 16.50.

Figure 16.50
For Prob. 16.16.

* An asterisk indicates a challenging problem.

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.

Chapter 16, Solution 16.

We first need to find the initial conditions. For t < 0 , the circuit is shown in Fig. (a).
2Ω

Vo

+

−

1Ω

1F

+
−

Vo/2

+

1H

3V

io

(a)

To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit.
Hence,
-3
i L (0) = i o =
= -1 A ,
v o = -1 V
3
⎛ - 1⎞
v c (0) = -(2)(-1) − ⎜ ⎟ = 2.5 V
⎝2⎠
We now incorporate the initial conditions for t > 0 as shown in Fig. (b).
2

+

Vo

−

1

1/s

s
5/(s + 2)

+
−

I1

2.5/s

Vo/2

+
−

I2
−
+

+

-1 V

Io
(b)

For mesh 1,
- 5 ⎛ 1⎞
1
2.5 Vo
+ ⎜ 2 + ⎟ I1 − I 2 +
+
=0
s+2 ⎝
s⎠
s
s

2
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.