Solution manual fundamentals of electric circuits 3rd edition chapter04

Chapter 4, Problem 1.
Calculate the current io in the circuit of Fig. 4.69. What does this current become when
the input voltage is raised to 10 V?

Figure 4.69

Chapter 4, Solution 1.

+
−

8 (5 + 3) = 4Ω , i =
io =

1
1
=
1+ 4 5

1
1
i=
= 0.1A
2
10

Since the resistance remains the same we get i = 10/5 = 2A which leads to
io = (1/2)i = (1/2)2 = 1A.

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 4, Problem 2.
Find vo in the circuit of Fig. 4.70. If the source current is reduced to 1 μA, what is vo?

Figure 4.70

Chapter 4, Solution 2.

6 (4 + 2) = 3Ω, i1 = i 2 =
io =

1
A
2

1
1
i1 = , v o = 2i o = 0.5V
2
4

If is = 1μA, then vo = 0.5μV

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 4, Problem 3.
(a) In the circuit in Fig. 4.71, calculate vo and Io when vs = 1 V.
(b) Find vo and io when vs = 10 V.
(c) What are vo and Io when each of the 1-Ω resistors is replaced by a 10-Ω resistor
and vs = 10 V?

Figure 4.71
Chapter 4, Solution 3.

+
−

+

+
−

vo
(a) We transform the Y sub-circuit to the equivalent Δ .
3R 2 3
3
3
3
= R, R + R = R
R 3R =
4R

4
4
4
2
vs
vo =
independent of R
2
io = vo/(R)
When vs = 1V, vo = 0.5V, io = 0.5A
(b) When vs = 10V, vo = 5V, io = 5A
(c)

When vs = 10V and R = 10Ω,
vo = 5V, io = 10/(10) = 500mA

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 4, Problem 4.
Use linearity to determine io in the circuit in Fig. 4.72.

Figure 4.72

Chapter 4, Solution 4.
If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω

resistor is 2A.

+
v1
3 6 = 2Ω , vo = 3(4) = 12V, i1 =

vo
= 3A.
4

Hence Is = 3 + 3 = 6A
If

Is = 6A
Is = 9A

Io = 1
Io = 9/6 = 1.5A

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 4, Problem 5.
For the circuit in Fig. 4.73, assume vo = 1 V, and use linearity to find the actual value
of vo.

Figure 4.73
Chapter 4, Solution 5.

+
−

If vo = 1V,

If vs =

10
3

Then vs = 15

⎛1⎞
V1 = ⎜ ⎟ + 1 = 2V
⎝3⎠
10
⎛2⎞
Vs = 2⎜ ⎟ + v1 =
3
⎝3⎠

vo = 1
vo =

3
x15 = 4.5V
10

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 4, Problem 6.
For the linear circuit shown in Fig. 4.74, use linearity to complete the following table.
Experiment
1
2
3
4

Vs

Vo

12 V
-1V
--

4V
16 V
--2V

+
Vs

Linear
Circuit

+
_

Vo
–

Figure 4.74

For Prob. 4.6.

Chapter 4, Solution 6.
Due to linearity, from the first experiment,
1
Vo = Vs
3
Applying this to other experiments, we obtain:
Experiment
2
3
4

Vs

Vo

48

1V
-6 V

16 V
0.333 V
-2V

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 4, Problem 7.
Use linearity and the assumption that Vo = 1V to find the actual value of Vo in Fig. 4.75.
.
1Ω
4Ω

+
+
_

4V

3Ω

Figure 4.75

2Ω

Vo
_

For Prob. 4.7.

Chapter 4, Solution 7.
If Vo = 1V, then the current through the 2-Ω and 4-Ω resistors is ½ = 0.5. The voltage
across the 3-Ω resistor is ½ (4 + 2) = 3 V. The total current through the 1-Ω resistor is
0.5 +3/3 = 1.5 A. Hence the source voltage
vs = 1x1.5 + 3 = 4.5 V
If vs = 4.5
Then vs = 4

⎯⎯
→ 1V

⎯⎯
→

1
x4 = 0.8889 V = 888.9 mV.
4.5

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 4, Problem 8.
Using superposition, find Vo in the circuit of Fig. 4.76.
4Ω

1Ω

Vo

3Ω

+
_

5Ω
+
_

Figure 4.76

3V

9V

For Prob. 4.8.

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 4, Solution 8.
Let Vo = V1 + V2, where V1 and V2 are due to 9-V and 3-V sources respectively. To find
V1, consider the circuit below.

V1

3Ω
9Ω

1Ω
+
_

9 − V1 V1 V1
= +
3
9 1

9V

⎯⎯
→ V1 = 27 /13 = 2.0769

To find V2, consider the circuit below.
V1

9Ω

V2 V2 3 − V2
+
=
9
3
1

3Ω

+
_

3V

⎯⎯
→ V2 = 27 /13 = 2.0769
Vo = V1 + V2 = 4.1538 V

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 4, Problem 9.
Use superposition to find vo in the circuit of Fig. 4.77.
2Ω

4Ω

2Ω

6A
+
vo

1Ω

+
_

18 V

_
Figure 4.77

For Prob. 4.9.

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 4, Solution 9.
Let vo = v1 + v2, where v1 and v2 are due to 6-A and 20-V sources respectively. We find

v1 using the circuit below.

2Ω

2Ω
4Ω

6A
+
v1

1Ω

_

2//2 = 1 Ω,

v1 = 1x

4
(6 A) = 4 V
4+2

We find v2 using the circuit below.

2Ω

2Ω

+

v2

1Ω

4Ω

+
_

18 V

_

v2 =

1
(18) = 3 V
1+ 1+ 4
vo = v1 + v2 = 4 + 3 = 7 V

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 4, Problem 10.
For the circuit in Fig. 4.78, find the terminal voltage Vab using superposition.

Figure 4.78
Chapter 4, Solution 10.
Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources
respectively.

+−
+
−

+−

+

+

vab1

vab2

For vab1, consider Fig. (a). Applying KVL gives,
- vab1 – 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V
For vab2, consider Fig. (b). Applying KVL gives,
-

vab2 – 3vab2 + 10x2 = 0, which leads to vab2 = 5
vab = 1 + 5 = 6 V

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 4, Problem 11.
Use the superposition principle to find io and vo in the circuit of Fig. 4.79.
10 Ω

io

20 Ω

+ vo –

6A

4 io

40 Ω

Figure 4.79

–
+

30 V

For Prob. 4.11.

Chapter 4, Solution 11.

Let vo = v1 + v2, where v1 and v2 are due to the 6-A and 80-V sources respectively. To
find v1, consider the circuit below.
I1

va

10 Ω
+ V1 _

6A

40 Ω

20 Ω

vb

4 i1

At node a,

6=

va va − vb
+
40
10

⎯⎯
→ 240 = 5va − 4vb

(1)

At node b,
–I1 – 4I1 + (vb – 0)/20 = 0 or vb = 100I1

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.


But

i1 =

va − vb
10

which leads to 100(va–vb)10 = vb or vb = 0.9091va

(2)

Substituting (2) into (1),
5va – 3.636va = 240 or va = 175.95 and vb = 159.96
However,

v1 = va – vb = 15.99 V.

To find v2, consider the circuit below.
io

10 Ω
+ v2 _

40 Ω

f

vc

20 Ω
e

4 io

–
+

30 V

0 − vc
(−30 − vc )
+ 4io +
=0
50
20
(0 − vc )
But io =

50
5vc (30 + vc )
−
=0
⎯⎯
→
50
20
0 − vc 0 + 10 1
i2 =
=
=
50
50
5
−

vc = −10 V

v2 = 10i2 = 2 V
vo = v1 + v2 =15.99 + 2 = 17.99 V and io = vo/10= 1.799 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 4, Problem 12.

Determine vo in the circuit in Fig. 4.80 using the superposition principle.

Figure 4.80

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 4, Solution 12.
Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V
sources respectively. For vo1, consider the circuit below.

+ v

1

−

+ v

1

−

6||3 = 2 ohms, 4||12 = 3 ohms. Hence,
io = 2/2 = 1, vo1 = 5io = 5 V
For vo2, consider the circuit below.

+
−

+ v

2

−

+
−

+

+ v

2

−

3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5
vo2 = (5/8)v1 = (5/8)(16/5) = 2 V
For vo3, consider the circuit shown below.

+ v

3

−

+
−

+ v

3

−

+

+
−

7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975
v = (-5/7)v2 = -7.125
vo = 5 + 2 – 7.125 = -125 mV
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 4, Problem 13.
Use superposition to find vo in the circuit of Fig. 4.81.

4A

8Ω
+–
12 V
2A

5Ω

10 Ω

+
vo
_

Figure 4.81

For Prob. 4.13.

Chapter 4, Solution 13.
Let vo = v1 + v2 + v 3 , where v1, v2, and v3 are due to the independent sources. To
find v1, consider the circuit below.
8Ω

2A

10 Ω

5Ω

+
v1

_

v1 = 5 x

10
x2 = 4.3478
10 + 8 + 5

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.


To find v2, consider the circuit below.
4A

8Ω

+
5Ω

10 Ω

v2 = 5 x

v2
_

8
x4 = 6.9565
8 + 10 + 5

To find v3, consider the circuit below.
8Ω

12 V
+ –

10 Ω

5Ω

+
v3
_

5
⎛
⎞
v3 = −12 ⎜
⎟ = −2.6087
⎝ 5 + 10 + 8 ⎠
vo = v1 + v2 + v 3 = 8.6956 V =8.696V.

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 4, Problem 14.
Apply the superposition principle to find vo in the circuit of Fig. 4.82.

Figure 4.82

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 4, Solution 14.
Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A
sources respectively. For vo1, consider the circuit below.

+
−

+

6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V
For vo2, consider the circuit below.

−+

+

+

3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V
For vo3, consider the circuit below.

+

− v
6||(4 + 2) = 3, vo3 = (-1)3 = –3

3

+

vo = 10 + 1 – 3 = 8 V
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 4, Problem 15.
For the circuit in Fig. 4.83, use superposition to find i. Calculate the power delivered to
the 3-Ω resistor.

Figure 4.83

Chapter 4, Solution 15.

Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For
i1, consider the circuit below.

+
−

4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 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.


For i3, consider the circuit below.

+
−
+

vo’
2||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4
i3 = vo’/4 = -1
For i2, consider the circuit below.

2||4 = 4/3, 3 + 4/3 = 13/3
Using the current division principle.
i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375
i = 2.5 + 0.375 - 1 = 1.875 A
p = i2R = (1.875)23 = 10.55 watts

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 4, Problem 16.
Given the circuit in Fig. 4.84, use superposition to get io.

Figure 4.84

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 4, Solution 16.
Let io = io1 + io2 + io3, where io1, io2, and io3 are due to the 12-V, 4-A, and 2-A sources.
For io1, consider the circuit below.

+
−
10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A
For io2, consider the circuit below.

2 + 5 + 4||10 = 7 + 40/14 = 69/7

i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9
For io3, consider the circuit below.

3 + 2 + 4||10 = 5 + 20/7 = 55/7
i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9
io = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA

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 4, Problem 17.
Use superposition to obtain vx in the circuit of Fig. 4.85. Check your result using PSpice.

Figure 4.85

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.