Chapter 3, Problem 1. Determine Ix in the circuit shown in Fig. 3.50 using nodal analysis. 1 kΩ Ix 9V + _ 2 kΩ + _ 6V 4 kΩ

Figure 3.50 For Prob. 3.1.

Chapter 3, Solution 1 Let Vx be the voltage at the node between 1-kΩ and 4-kΩ resistors.

9 − Vx 6 − Vx Vk + = 1k 4k 2k Vx Ix = = 3 mA 2k

⎯⎯ Vx = 6 →

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 3, Problem 2. For the circuit in Fig. 3.51, obtain v1 and v2.

Figure 3.51

Chapter 3, Solution 2 At node 1,

− v1 v1 v − v2 − = 6+ 1 10 5 2
At node 2,

60 = - 8v1 + 5v2

(1)

v2 v − v2 = 3+ 6+ 1 4 2
Solving (1) and (2), v1 = 0 V, v2 = 12 V

36 = - 2v1 + 3v2

(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.

Chapter 3, Problem 3. Find the currents i1 through i4 and the voltage vo in the circuit in Fig. 3.52.

Figure 3.52

Chapter 3, Solution 3 Applying KCL to the upper node, 10 =

v0 vo vo v + + +2+ 0 10 20 30 60

v0 = 40 V

i1 =

v0 v v v = 4 A , i2 = 0 = 2 A, i3 = 0 = 1.3333 A, i4 = 0 = 666.7 mA 10 20 30 60

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 3, Problem 4. Given the circuit in Fig. 3.53, calculate the currents i1 through i4.

Figure 3.53

Chapter 3, Solution 4 v1 i1 4A
5Ω

2A i2
10 Ω 10 Ω

v2 i3
5Ω

i4 5A

At node 1, 4 + 2 = v1/(5) + v1/(10) At node 2, 5 - 2 = v2/(10) + v2/(5) v2 = 10 v1 = 20

i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A

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Chapter 3, Problem 5. Obtain v0 in the circuit of Fig. 3.54.

Figure 3.54 Chapter 3, Solution 5 Apply KCL to the top node.

30 − v 0 20 − v 0 v 0 + = 2k 5k 4k
Chapter 3, Problem 6.

v0 = 20 V

Use nodal analysis to obtain v0 in the circuit in Fig. 3.55.

Figure 3.55 Chapter 3, Solution 6 i1 + i2 + i3 = 0

v 2 − 12 v 0 v 0 − 10 + + =0 4 6 2

or v0 = 8.727 V
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Chapter 3, Problem 7. Apply nodal analysis to solve for Vx in the circuit in Fig. 3.56.

+ 2A 10 Ω Vx _ 20 Ω 0.2 Vx

Figure 3.56 For Prob. 3.7.

Chapter 3, Solution 7

V − 0 Vx − 0 −2+ x + + 0.2Vx = 0 10 20
0.35Vx = 2 or Vx = 5.714 V. Substituting into the original equation for a check we get, 0.5714 + 0.2857 + 1.1428 = 1.9999 checks!

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 3, Problem 8. Using nodal analysis, find v0 in the circuit in Fig. 3.57.

Figure 3.57

Chapter 3, Solution 8
3Ω

i1

v1 i2

i3

5Ω

+
V0

3V 2Ω

+

–

–
1Ω

+ 4V0 –

But

v1 v1 − 3 v1 − 4 v 0 + + =0 5 1 5 2 8 v 0 = v1 so that v1 + 5v1 - 15 + v1 - v1 = 0 5 5 or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V i1 + i2 + i3 = 0

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Chapter 3, Problem 9. Determine Ib in the circuit in Fig. 3.58 using nodal analysis. 60 Ib Ib 250 Ω + –

24 V

+ _

50 Ω

150 Ω

Figure 3.58 For Prob. 3.9.

Chapter 3, Solution 9 Let V1 be the unknown node voltage to the right of the 250-Ω resistor. Let the ground reference be placed at the bottom of the 50-Ω resistor. This leads to the following nodal equation:

V1 − 24 V1 − 0 V1 − 60I b − 0 =0 + + 250 50 150 simplifying we get 3V1 − 72 + 15V1 + 5V1 − 300I b = 0 But I b =

24 − V1 . Substituting this into the nodal equation leads to 250

24.2V1 − 100.8 = 0 or V1 = 4.165 V.
Thus, Ib = (24 – 4.165)/250 = 79.34 mA.

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Chapter 3, Problem 10.

Find i0 in the circuit in Fig. 3.59.

Figure 3.59
Chapter 3, Solution 10
3Ω

i1

v1 i2

i3

6Ω

+ v0 –
12V +

+ v1 8Ω

–

–

+ –

2v0

At the non-reference node,

12 − v1 v1 v1 − 2v 0 = + 3 8 6
But -12 + v0 + v1 = 0 Substituting (2) into (1), v0 = 12 - v1

(1)

(2)

12 − v1 v1 3v1 − 24 = + 3 8 6

v0 = 3.652 V

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Chapter 3, Problem 11.

Find Vo and the power dissipated in all the resistors in the circuit of Fig. 3.60. 1Ω Vo 4Ω

36 V

+ _

2Ω

– +

12 V

Figure 3.60 For Prob. 3.11.

Chapter 3, Solution 11

At the top node, KVL gives

Vo − 36 Vo − 0 Vo − (−12) + + =0 1 2 4
1.75Vo = 33 or Vo = 18.857V

P1Ω = (36–18.857)2/1 = 293.9 W P2Ω = (Vo)2/2 = (18.857)2/2 = 177.79 W P4Ω = (18.857+12)2/4 = 238 W.

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Chapter 3, Problem 12.

Using nodal analysis, determine Vo in the circuit in Fig. 3.61. 10 Ω Ix 30 V + _ 2Ω 5Ω 4 Ix 1Ω

+ Vo _

Figure 3.61 For Prob. 3.12.

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Chapter 3, Solution 12

There are two unknown nodes, as shown in the circuit below.

10 Ω

V1

1Ω

Vo

30 V

+ _

2Ω

4 Ix

5Ω

At node 1,

V1 − 30 V1 − 0 V1 − Vo =0 + + 10 2 1 16V1 − 10Vo = 30

(1)

At node o,

Vo − V1 V −0 =0 − 4I x + o 1 5 − 5V1 + 6Vo − 20I x = 0 But Ix = V1/2. Substituting this in (2) leads to
–15V1 + 6Vo = 0 or V1 = 0.4Vo

(2)

(3)

Substituting (3) into 1, 16(0.4Vo) – 10Vo = 30 or Vo = –8.333 V.

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Chapter 3, Problem 13.

Calculate v1 and v2 in the circuit of Fig. 3.62 using nodal analysis.

Figure 3.62

Chapter 3, Solution 13

At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts

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Chapter 3, Problem 14.

Using nodal analysis, find vo in the circuit of Fig. 3.63.

Figure 3.63

Chapter 3, Solution 14

5A

v1
1Ω 2Ω

v0

8Ω

4Ω 40 V +

20 V

+

–

–

At node 1,

v1 − v 0 40 − v 0 +5= 2 1 v1 − v 0 v v + 20 +5= 0 + 0 2 4 8

v1 + v0 = 70

(1)

At node 0,

4v1 - 7v0 = -20

(2)

Solving (1) and (2), v0 = 27.27 V

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Chapter 3, Problem 15.

Apply nodal analysis to find io and the power dissipated in each resistor in the circuit of Fig. 3.64.

Figure 3.64

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Chapter 3, Solution 15
5A

v1
1Ω 2Ω

v0

8Ω

4Ω 40 V +

20 V

+

–

–

Nodes 1 and 2 form a supernode so that v1 = v2 + 10 At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2) At node 3, 2 + 4 = 3 (v3 - v2) Substituting (1) and (3) into (2), 2 + 6v2 + 60 + 8v2 = 3v2 + 6 v1 = v2 + 10 =
54 11

(1) 2 + 6v1 + 8v2 = 3v3 (2) (3)

v3 = v2 + 2

v2 =

− 56 11

i0 = 6vi = 29.45 A
2 v1 ⎛ 54 ⎞ 2 P65 = = v1 G = ⎜ ⎟ 6 = 144.6 W R ⎝ 11 ⎠ 2

⎛ − 56 ⎞ P55 = v G = ⎜ ⎟ 5 = 129.6 W ⎝ 11 ⎠
2 2

2

P35 = (v L − v 3 ) G = (2) 2 3 = 12 W
2

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Chapter 3, Problem 16.

Determine voltages v1 through v3 in the circuit of Fig. 3.65 using nodal analysis.

Figure 3.65
Chapter 3, Solution 16
2S

v1 i0
2A 1S

v2

8S

v3
13 V +

+ v0 4S

–

–

At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3 But v1 = v2 + 2v0 and v0 = v2. Hence v1 = 3v2 v3 = 13V Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V
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(1)

(2) (3)

Chapter 3, Problem 17.

Using nodal analysis, find current io in the circuit of Fig. 3.66.

Figure 3.66

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Chapter 3, Solution 17 v1 i0
4Ω 10 Ω 2Ω

60 V
60 V +

v2 3i0

8Ω

–

At node 1,

60 − v1 v1 v1 − v 2 = + 4 8 2 60 − v 2 v1 − v 2 + =0 At node 2, 3i0 + 10 2 60 − v1 . 4

120 = 7v1 - 4v2

(1)

But i0 = Hence

3(60 − v1 ) 60 − v 2 v1 − v 2 + + =0 4 10 2
Solving (1) and (2) gives v1 = 53.08 V. Hence i0 =

1020 = 5v1 + 12v2

(2)

60 − v1 = 1.73 A 4

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Chapter 3, Problem 18.

Determine the node voltages in the circuit in Fig. 3.67 using nodal analysis.

Figure 3.67
Chapter 3, Solution 18

–+ v1 2Ω 4Ω

v2 5A
2Ω 8Ω

v3 + v1 10 V

+ v3 –

–

(a)

(b)

At node 2, in Fig. (a), 5 =

v 2 − v1 v 2 − v3 + 2 2

10 = - v1 + 2v2 - v3

(1)

At the supernode,

v 2 − v1 v 2 − v 3 v1 v 3 + = + 2 2 4 8 v3 = v1 + 10

40 = 2v1 + v3

(2) (3)

From Fig. (b), - v1 - 10 + v3 = 0

Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3

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Chapter 3, Problem 19. Use nodal analysis to find v1, v2, and v3 in the circuit in Fig. 3.68.

Figure 3.68
Chapter 3, Solution 19 At node 1,

V1 − V3 V1 − V2 V1 + + 2 8 4 At node 2,
5 = 3+

⎯ ⎯→

16 = 7V1 − V2 − 4V3

(1)

V1 − V2 V2 V2 − V3 = + 8 2 4 At node 3,
3+ 12 − V3 +

⎯ ⎯→

0 = −V1 + 7V2 − 2V3

(2)

V1 − V3 V2 − V3 + =0 8 2 4 From (1) to (3),

⎯ ⎯→

− 36 = 4V1 + 2V2 − 7V3 (3)

⎛ 7 − 1 − 4 ⎞⎛ V1 ⎞ ⎛ 16 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ − 1 7 − 2 ⎟⎜V2 ⎟ = ⎜ 0 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ 4 2 − 7 ⎠⎝ V3 ⎠ ⎝ − 36 ⎠
Using MATLAB, ⎡ 10 ⎤ −1 V = A B = ⎢ 4.933 ⎥ ⎥ ⎢ ⎥ ⎢ ⎣12.267⎦

⎯ ⎯→

AV = B

⎯ ⎯→

V1 = 10 V, V2 = 4.933 V, V3 = 12.267 V

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Chapter 3, Problem 20. For the circuit in Fig. 3.69, find v1, v2, and v3 using nodal analysis.

Figure 3.69
Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence V1 V2 V3 + + =0 ⎯ ⎯→ V1 + 4V2 + V3 = 0 (1) 4 1 4

. V1
.

V2

2Ω

V3

4Ω

1Ω

4Ω

Between nodes 1 and 3, − V1 + 12 + V3 = 0 ⎯ ⎯→

V3 = V1 − 12

(2)

Similarly, between nodes 1 and 2, V1 = V2 + 2i But i = V3 / 4 . Combining this with (2) and (3) gives

(3)

V2 = 6 + V1 / 2
Solving (1), (2), and (4) leads to V1 = −3V, V2 = 4.5V, V3 = −15V

(4)

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Chapter 3, Problem 21.

For the circuit in Fig. 3.70, find v1 and v2 using nodal analysis.

Figure 3.70

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Chapter 3, Solution 21
4 kΩ

v1

2 kΩ

v3 +

3v0

3v0 v2 +

+
3 mA v0 +
1 kΩ v3

+ v2 –

–

–

(a)

(b)

Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source. At node 1, v − v 2 v1 − v 3 3x10 −3 = 1 + 12 = 3v1 - v2 - 2v3 (1) 4000 2000 At node 2, v1 − v 2 v1 − v 3 v 2 + = 4 2 1

3v1 - 5v2 - 2v3 = 0

(2)

Note that v0 = v2. We now apply KVL in Fig. (b) - v3 - 3v2 + v2 = 0 From (1) to (3), v1 = 1 V, v2 = 3 V v3 = - 2v2 (3)

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Chapter 3, Problem 22.

Determine v1 and v2 in the circuit in Fig. 3.71.

Figure 3.71

Chapter 3, Solution 22

At node 1,

v − v0 12 − v 0 v1 = +3+ 1 2 4 8 v 1 − v 2 v 2 + 5v 2 = 8 1

24 = 7v1 - v2

(1)

At node 2, 3 + But, v1 = 12 - v1

Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V 456 = 41v1 - 9v2 Solving (1) and (2), v1 = - 10.91 V, v2 = - 100.36 V (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.

Chapter 3, Problem 23.

Use nodal analysis to find Vo in the circuit of Fig. 3.72. 2 Vo 4Ω 1Ω + + _ + Vo _ –

30 V

2Ω

16 Ω

3A

Figure 3.72 For Prob. 3.23. Chapter 3, Solution 23

We apply nodal analysis to the circuit shown below. 1Ω Vo + Vo _ 4Ω 2 Vo + – V1

30 V

+ _

2Ω

16 Ω

3A

At node o,

Vo − 30 Vo − 0 Vo − (2Vo + V1 ) = 0 → 1.25Vo − 0.25V1 = 30 + + 1 2 4 (1)
At node 1,

(2Vo + V1 ) − Vo V1 − 0 + − 3 = 0 → 5V1 + 4Vo = 48 4 16
From (1), V1 = 5Vo – 120. Substituting this into (2) yields 29Vo = 648 or Vo = 22.34 V.

(2)

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Chapter 3, Problem 24.

Use nodal analysis and MATLAB to find Vo in the circuit in Fig. 3.73.

8Ω + Vo _ 4A 4Ω 2A

1Ω

2Ω

2Ω

1Ω

Figure 3.73 For Prob. 3.24.

Chapter 3, Solution 24

Consider the circuit below.

8Ω + Vo _ 4A 2A
V3 V4

V1

V2

4Ω

1Ω

2Ω

2Ω

1Ω

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V1 − 0 V − V4 −4+ 1 = 0 → 1.125V1 − 0.125V4 = 4 1 8 V − 0 V2 − V3 +4+ 2 + = 0 → 0.75V2 − 0.25V3 = −4 2 4 V3 − V2 V3 − 0 + + 2 = 0 → −0.25V2 + 0.75V3 = −2 4 2 V − V1 V4 − 0 −2+ 4 + = 0 → −0.125V1 + 1.125V4 = 2 8 1
0 0 − 0.125⎤ ⎡4⎤ ⎡ 1.125 ⎢ − 4⎥ ⎥ ⎢ 0 0.75 − 0.25 0 ⎥ ⎢ V=⎢ ⎥ ⎢ − 2⎥ ⎢ 0 − 0.25 0.75 0 ⎥ ⎢ ⎥ ⎥ ⎢ 0 0 1.125 ⎦ ⎣2⎦ ⎣− 0.125 Now we can use MATLAB to solve for the unknown node voltages.

(1) (2) (3) (4)

>> Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0;-0.125,0,0,1.125] Y= 1.1250 0 0 -0.1250 0 0.7500 -0.2500 0 0 -0.2500 0.7500 0 -0.1250 0 0 1.1250 >> I=[4,-4,-2,2]' I= 4 -4 -2 2 >> V=inv(Y)*I V= 3.8000 -7.0000 -5.0000 2.2000 Vo = V1 – V4 = 3.8 – 2.2 = 1.6 V.
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Chapter 3, Problem 25.

Use nodal analysis along with MATLAB to determine the node voltages in Fig. 3.74. 20 Ω 4

1Ω 1

2

10 Ω

3

10 Ω 30 Ω

4A

8Ω

20 Ω

Figure 3.74 For Prob. 3.25. Chapter 3, Solution 25

Consider the circuit shown below. 20 4 10 1 1 2 10 3 30 4 8 20

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At node 1. V −V V −V 4= 1 2 + 1 4 1 20 At node 2, V1 − V2 V2 V2 − V3 = + 1 8 10 At node 3, V2 − V3 V3 V3 − V4 = + 10 20 10 At node 4, V1 − V4 V3 − V4 V4 + = 20 10 30

V ⎯⎯ 80 = 21 1 − 20V2 − V4 →

(1)

⎯⎯ →

0 = −80V1 + 98V2 − 8V3

(2)

⎯⎯ 0 = −2V2 + 5V3 − 2V4 →

(3)

V ⎯⎯ 0 = 3V1 + 6V3 − 11 4 →

(4)

Putting (1) to (4) in matrix form gives:
⎡ 80 ⎤ ⎡ 21 −20 0 −1 ⎤ ⎡ V1 ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ 0 ⎥ = ⎢ −80 98 −8 0 ⎥ ⎢V2 ⎥ ⎢0⎥ ⎢ 0 −2 5 −2 ⎥ ⎢V3 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ 0 6 −11⎦ ⎢V4 ⎥ ⎣0⎦ ⎣ 3 ⎣ ⎦

B =A V

V = A-1 B

Using MATLAB leads to V1 = 25.52 V, V2 = 22.05 V, V3 = 14.842 V, V4 = 15.055 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 3, Problem 26.

Calculate the node voltages v1, v2, and v3 in the circuit of Fig. 3.75.

Figure 3.75

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Chapter 3, Solution 26

At node 1,

V − V3 V1 − V2 15 − V1 = 3+ 1 + 20 10 5

⎯ ⎯→

− 45 = 7V1 − 4V2 − 2V3

(1)

At node 2, V1 − V2 4 I o − V2 V2 − V3 + = 5 5 5 V1 − V3 But I o = . Hence, (2) becomes 10 0 = 7V1 − 15V2 + 3V3 At node 3, V − V3 − 10 − V3 V2 − V3 3+ 1 + + =0 10 15 5

(2)

(3)

⎯ ⎯→

70 = −3V1 − 6V2 + 11V3

(4)

Putting (1), (3), and (4) in matrix form produces
⎛ 7 − 4 − 2 ⎞⎛ V1 ⎞ ⎛ − 45 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ 7 − 15 3 ⎟⎜ V2 ⎟ = ⎜ 0 ⎟ ⎜ − 3 − 6 11 ⎟⎜ V ⎟ ⎜ 70 ⎟ ⎝ ⎠⎝ 3 ⎠ ⎝ ⎠

⎯ ⎯→

AV = B

Using MATLAB leads to ⎛ − 7.19 ⎞ ⎜ ⎟ V = A −1B = ⎜ − 2.78 ⎟ ⎜ 2.89 ⎟ ⎝ ⎠ Thus, V1 = –7.19V; V2 = –2.78V; V3 = 2.89V.

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Chapter 3, Problem 27.

Use nodal analysis to determine voltages v1, v2, and v3 in the circuit in Fig. 3.76.

Figure 3.76

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Chapter 3, Solution 27

At node 1, 2 = 2v1 + v1 – v2 + (v1 – v3)4 + 3i0, i0 = 4v2. Hence, At node 2, 2 = 7v1 + 11v2 – 4v3 0 = – v1 + 6v2 – v3 (1) (2)

v1 – v2 = 4v2 + v2 – v3 At node 3,

2v3 = 4 + v2 – v3 + 12v2 + 4(v1 – v3) or In matrix form,
⎡7 11 − 4⎤ ⎡ v 1 ⎤ ⎡ 2 ⎤ ⎢1 − 6 1 ⎥ ⎢ v ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2 ⎥ ⎢ ⎥ ⎢4 13 − 7⎥ ⎢ v 3 ⎥ ⎢ − 4⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

– 4 = 4v1 + 13v2 – 7v3

(3)

7 11 − 4 2 Δ = 1 − 6 1 = 176, Δ 1 = 0 4 13 −7 −4

11 − 4 − 6 1 = 110 13 −7

7 2 −4 Δ2 = 1 0 1 = 66, 4 −4 −7

7 11 2 Δ 3 = 1 − 6 0 = 286 4 13 − 4

v1 =

Δ 1 110 Δ 66 = = 0.625V, v2 = 2 = = 0.375V Δ 176 Δ 176 v3 =

Δ3 286 = = 1.625V. 176 Δ

v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V.
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Chapter 3, Problem 28.

Use MATLAB to find the voltages at nodes a, b, c, and d in the circuit of Fig. 3.77.

Figure 3.77

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Chapter 3, Solution 28

At node c, Vd − Vc Vc − Vb Vc = + ⎯ ⎯→ 0 = −5Vb + 11Vc − 2Vd (1) 10 4 5 At node b, Va + 45 − Vb Vc − Vb Vb + = ⎯ ⎯→ − 45 = Va − 4Vb + 2Vc (2) 8 4 8 At node a, Va − 30 − Vd Va Va + 45 − Vb + + =0 ⎯ ⎯→ 30 = 7Va − 2Vb − 4Vd (3) 4 16 8 At node d, Va − 30 − Vd Vd Vd − Vc = + ⎯ ⎯→ 150 = 5Va + 2Vc − 7Vd (4) 4 20 10 In matrix form, (1) to (4) become ⎛ 0 − 5 11 − 2 ⎞⎛ Va ⎞ ⎛ 0 ⎞ ⎟ ⎟⎜ ⎟ ⎜ ⎜ ⎜ 1 − 4 2 0 ⎟⎜ Vb ⎟ ⎜ − 45 ⎟ ⎜ 7 − 2 0 − 4 ⎟⎜ V ⎟ = ⎜ 30 ⎟ ⎟ ⎟⎜ c ⎟ ⎜ ⎜ ⎜ 5 0 2 − 7 ⎟⎜V ⎟ ⎜ 150 ⎟ ⎠ ⎠⎝ d ⎠ ⎝ ⎝

⎯ ⎯→

AV = B

We use MATLAB to invert A and obtain ⎛ − 10.14 ⎞ ⎟ ⎜ ⎜ 7.847 ⎟ −1 V = A B=⎜ − 1.736 ⎟ ⎟ ⎜ ⎜ − 29.17 ⎟ ⎠ ⎝ Thus, Va = −10.14 V, Vb = 7.847 V, Vc = −1.736 V, Vd = −29.17 V

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Chapter 3, Problem 29.

Use MATLAB to solve for the node voltages in the circuit of Fig. 3.78.

Figure 3.78
Chapter 3, Solution 29

At node 1, 5 + V1 − V4 + 2V1 + V1 − V2 = 0 ⎯ ⎯→ − 5 = 4V1 − V2 − V4 At node 2, V1 − V2 = 2V2 + 4(V2 − V3 ) = 0 ⎯ ⎯→ 0 = −V1 + 7V2 − 4V3 At node 3, 6 + 4(V2 − V3 ) = V3 − V4 ⎯ ⎯→ 6 = −4V2 + 5V3 − V4 At node 4, 2 + V3 − V4 + V1 − V4 = 3V4 ⎯ ⎯→ 2 = −V1 − V3 + 5V4 In matrix form, (1) to (4) become ⎛ 4 − 1 0 − 1⎞⎛ V1 ⎞ ⎛ − 5 ⎞ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎜ − 1 7 − 4 0 ⎟⎜V2 ⎟ ⎜ 0 ⎟ AV = B ⎯ ⎯→ ⎜ 0 − 4 5 − 1⎟⎜ V ⎟ = ⎜ 6 ⎟ 3 ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎜ − 1 0 − 1 5 ⎟⎜V ⎟ ⎜ 2 ⎟ ⎠⎝ 4 ⎠ ⎝ ⎠ ⎝ Using MATLAB, ⎛ − 0.7708 ⎞ ⎟ ⎜ ⎜ 1.209 ⎟ −1 V = A B=⎜ 2.309 ⎟ ⎟ ⎜ ⎜ 0.7076 ⎟ ⎠ ⎝ i.e.

(1) (2) (3) (4)

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V1 = −0.7708 V, V2 = 1.209 V, V3 = 2.309 V, V4 = 0.7076 V
Chapter 3, Problem 30.

Using nodal analysis, find vo and io in the circuit of Fig. 3.79.

Figure 3.79

Chapter 3, Solution 30 v2 I0 v1
10 Ω 100 V + 20 Ω 40 Ω

–+
120 V

v0 2

1 4v0

–

+ –

2I0

80 Ω

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At node 1,

v 1 − v 2 100 − v 1 4 v o − v 1 = + 40 10 20
But, vo = 120 + v2 v2 = vo – 120. Hence (1) becomes 7v1 – 9vo = 280 At node 2, Io + 2Io =

(1)

(2)

vo − 0 80

⎛ v + 120 − v o ⎞ v o 3⎜ 1 ⎟= 40 ⎝ ⎠ 80 or 6v1 – 7vo = -720 (3)

from (2) and (3),

⎡7 − 9⎤ ⎡ v 1 ⎤ ⎡ 280 ⎤ ⎢6 − 7⎥ ⎢ v ⎥ = ⎢ − 720⎥ ⎦ ⎣ ⎦⎣ o ⎦ ⎣
7 −9 = −49 + 54 = 5 6 −7

Δ=

Δ1 =

280 − 9 = −8440 , − 720 − 7

Δ2 =

7 280 = −6720 6 − 720

v1 =

Δ Δ1 − 8440 − 6720 = = −1688, vo = 2 = − 1344V Δ 5 Δ 5
Io = –5.6 A

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Chapter 3, Problem 31.

Find the node voltages for the circuit in Fig. 3.80.

Figure 3.80

Chapter 3, Solution 31

1Ω

v1

+ v0 – v2

2v0

v3 i0

2Ω

1A

4Ω

1Ω

4Ω

10 V

+

–

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At the supernode, 1 + 2v0 =

v1 v 2 v1 − v 3 + + 4 1 1

(1)

But vo = v1 – v3. Hence (1) becomes, At node 3, 4 = -3v1 + 4v2 +4v3 2vo + or (2)

v3 10 − v 3 = v1 − v 3 + 4 2
(3)

20 = 4v1 + 0v2 – v3

At the supernode, v2 = v1 + 4io. But io = v2 = v1 + v3 Solving (2) to (4) leads to,

v3 . Hence, 4
(4)

v1 = 4.97V, v2 = 4.85V, v3 = –0.12V.

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Chapter 3, Problem 32.

Obtain the node voltages v1, v2, and v3 in the circuit of Fig. 3.81. Figure 3.81

Chapter 3, Solution 32

5 kΩ

v3
10 V 20 V +–

–+ v1 v2
10 kΩ 4 mA

+ v1 loop 1
12 V +

+ loop 2 v3 –

–

–

(a)

(b)

We have a supernode as shown in figure (a). It is evident that v2 = 12 V, Applying KVL to loops 1and 2 in figure (b), we obtain, -v1 – 10 + 12 = 0 or v1 = 2 and -12 + 20 + v3 = 0 or v3 = -8 V Thus, v1 = 2 V, v2 = 12 V, v3 = -8V.

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Chapter 3, Problem 33.

Which of the circuits in Fig. 3.82 is planar? For the planar circuit, redraw the circuits with no crossing branches.

Figure 3.82

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Chapter 3, Solution 33 (a) This is a planar circuit. It can be redrawn as shown below.

5Ω 3Ω 1Ω

2Ω 6Ω

4Ω

2A

(b) This is a planar circuit. It can be redrawn as shown below.

4Ω

3Ω 5Ω 12 V + 2Ω

–

1Ω

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Chapter 3, Problem 34.

Determine which of the circuits in Fig. 3.83 is planar and redraw it with no crossing branches.

Figure 3.83

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Chapter 3, Solution 34

(a)

This is a planar circuit because it can be redrawn as shown below,
7Ω 1Ω 6Ω 10 V + 5Ω 4Ω 2Ω 3Ω

–

(b)

This is a non-planar circuit.

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Chapter 3, Problem 35.

Rework Prob. 3.5 using mesh analysis. Chapter 3, Problem 5 Obtain v0 in the circuit of Fig. 3.54.

Figure 3.54
Chapter 3, Solution 35

30 V

+

–

20 V

+

–

i1
2 kΩ 5 kΩ

i2

+ v0 –

4 kΩ

Assume that i1 and i2 are in mA. We apply mesh analysis. For mesh 1, -30 + 20 + 7i1 – 5i2 = 0 or 7i1 – 5i2 = 10 For mesh 2, -20 + 9i2 – 5i1 = 0 or -5i1 + 9i2 = 20 Solving (1) and (2), we obtain, i2 = 5. v0 = 4i2 = 20 volts.
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(1)

(2)

Chapter 3, Problem 36.

Rework Prob. 3.6 using mesh analysis. Chapter 3, Problem 6 Use nodal analysis to obtain v0 in the circuit in Fig. 3.55.

Figure 3.55

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Chapter 3, Solution 36
4Ω i1 i2

10 V +– i3

12 V

+

I1

–

6Ω

I2

2Ω

Applying mesh analysis gives, 12 = 10I1 – 6I2 -10 = -6I1 + 8I2

or

⎡ 6 ⎤ ⎡ 5 − 3⎤ ⎡ I 1 ⎤ ⎢− 5⎥ = ⎢− 3 4 ⎥ ⎢I ⎥ ⎦⎣ 2 ⎦ ⎣ ⎦ ⎣
Δ= 5 −3 6 −3 5 6 = 11, Δ1 = = 9, Δ 2 = = −7 −3 4 −5 4 −3 −5
I1 = Δ1 9 I = Δ2 = − 7 = , 2 11 Δ Δ 11

i1 = -I1 = -9/11 = -0.8181 A, i2 = I1 – I2 = 10/11 = 1.4545 A. vo = 6i2 = 6x1.4545 = 8.727 V.

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Chapter 3, Problem 37.

Rework Prob. 3.8 using mesh analysis. Chapter 3, Problem 8 Using nodal analysis, find v0 in the circuit in Fig. 3.57.

Figure 3.57
Chapter 3, Solution 37
3Ω 5Ω

+ v0 3V 2Ω

+

i1
1Ω

–

i2

4v0

–

+ –

Applying mesh analysis to loops 1 and 2, we get, 6i1 – 1i2 + 3 = 0 which leads to i2 = 6i1 + 3 -1i1 + 6i2 – 3 + 4v0 = 0 But, v0 = -2i1 Using (1), (2), and (3) we get i1 = -5/9. Therefore, we get v0 = -2i1 = -2(-5/9) = 1.1111 volts
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(1) (2) (3)

Chapter 3, Problem 38. Apply mesh analysis to the circuit in Fig. 3.84 and obtain Io.

4Ω

3Ω

24 V

+ _ 2Ω Io 1Ω

4A 2Ω

1Ω

1Ω

+ _

9V

4Ω 2A Figure 3.84 For Prob. 3.38.
Chapter 3, Solution 38 Consider the circuit below with the mesh currents.

4Ω

3Ω

24 V

+ _

I3 4A 2Ω Io 1Ω

I4

1Ω

2Ω

1Ω

I1

I2

+ _

9V

4Ω
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2A

I1 =-2 A 1(I2–I1) + 2(I2–I4) + 9 + 4I2 = 0 7I2 – I4 = –11

(1) (2)

–24 + 4I3 + 3I4 + 1I4 + 2(I4–I2) + 2(I3 – I1) = 0 (super mesh) –2I2 + 6 I3 + 6I4 = +24 – 4 = 20 (3) But, we need one more equation, so we use the constraint equation –I3 + I4 = 4. This now gives us three equations with three unknowns. 0 − 1⎤ ⎡I 2 ⎤ ⎡− 11⎤ ⎡7 ⎢− 2 6 6 ⎥ ⎢ I ⎥ = ⎢ 20 ⎥ ⎢ ⎥⎢ 3 ⎥ ⎢ ⎥ ⎢ 0 − 1 1 ⎥ ⎢I 4 ⎥ ⎢ 4 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ We can now use MATLAB to solve the problem. >> Z=[7,0,-1;-2,6,6;0,-1,0] Z= 7 0 -1 -2 6 6 0 -1 0 >> V=[-11,20,4]' V= -11 20 4 >> I=inv(Z)*V I= -0.5500 -4.0000 7.1500 Io = I1 – I2 = –2 – 4 = –6 A. Check using the super mesh (equation (3)): 1.1 – 24 + 42.9 = 20!
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Chapter 3, Problem 39.

Determine the mesh currents i1 and i2 in the circuit shown in Fig. 3.85.

Figure 3.85

Chapter 3, Solution 39

For mesh 1, − 10 − 2 I x + 10 I 1 − 6 I 2 = 0 But I x = I 1 − I 2 . Hence,
10 = −2I1 + 2I 2 + 10I1 − 6I 2 ⎯ ⎯→ 5 = 4I1 − 2I 2 For mesh 2, 12 + 8I 2 − 6 I 1 = 0 ⎯ ⎯→ 6 = 3I 1 − 4 I 2 Solving (1) and (2) leads to I 1 = 0.8 A, I 2 = -0.9A

(1) (2)

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Chapter 3, Problem 40.

For the bridge network in Fig. 3.86, find Io using mesh analysis.

Figure 3.86
Chapter 3, Solution 40
2 kΩ 6 kΩ 30V +
2 kΩ i2

6 kΩ

–

i1
4 kΩ i3 4 kΩ

Assume all currents are in mA and apply mesh analysis for mesh 1. 30 = 12i1 – 6i2 – 4i3 for mesh 2, 0 = - 6i1 + 14i2 – 2i3 for mesh 2, 0 = -4i1 – 2i2 + 10i3 Solving (1), (2), and (3), we obtain, io = i1 = 4.286 mA.
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15 = 6i1 – 3i2 – 2i3

(1)

0 = -3i1 + 7i2 – i3

(2)

0 = -2i1 – i2 + 5i3

(3)

Chapter 3, Problem 41.

Apply mesh analysis to find io in Fig. 3.87.

Figure 3.87

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Chapter 3, Solution 41

10 Ω

i1
2Ω 6V +– 1Ω

i2
4Ω 8V +

i3

5Ω

–

i i2 i3 0

For loop 1, For loop 2, For loop 3, 6 = 12i1 – 2i2

3 = 6i1 – i2

(1) (2)

-8 = – 2i1 +7i2 – i3 -8 + 6 + 6i3 – i2 = 0 2 = – i2 + 6i3

(3)

We put (1), (2), and (3) in matrix form, ⎡6 − 1 0⎤ ⎡ i1 ⎤ ⎡ 3⎤ ⎢ 2 − 7 1 ⎥ ⎢i ⎥ = ⎢ 8 ⎥ ⎢ ⎥⎢ 2 ⎥ ⎢ ⎥ ⎢ 0 − 1 6⎥ ⎢ i 3 ⎥ ⎢ 2 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

6 −1 0 6 3 0 Δ = 2 − 7 1 = −234, Δ 2 = 2 8 1 = 240 0 −1 6 0 2 6
6 −1 3 Δ 3 = 2 − 7 8 = −38 0 −1 2 At node 0, i + i2 = i3 or i = i3 – i2 =
Δ3 − Δ2 − 38 − 240 = 1.188 A = − 234 Δ

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Chapter 3, Problem 42.

Determine the mesh currents in the circuit of Fig. 3.88.

Figure 3.88

Chapter 3, Solution 42

For mesh 1, (1) − 12 + 50 I 1 − 30 I 2 = 0 ⎯ ⎯→ 12 = 50 I 1 − 30 I 2 For mesh 2, − 8 + 100 I 2 − 30 I 1 − 40 I 3 = 0 ⎯ ⎯→ 8 = −30 I 1 + 100 I 2 − 40 I 3 For mesh 3, (3) − 6 + 50 I 3 − 40 I 2 = 0 ⎯ ⎯→ 6 = −40 I 2 + 50 I 3 Putting eqs. (1) to (3) in matrix form, we get 0 ⎞⎛ I 1 ⎞ ⎛12 ⎞ ⎛ 50 − 30 ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ − 30 100 − 40 ⎟⎜ I 2 ⎟ = ⎜ 8 ⎟ ⎜ − 40 50 ⎟⎜ I 3 ⎟ ⎜ 6 ⎟ ⎝ 0 ⎠⎝ ⎠ ⎝ ⎠ Using Matlab,

(2)

⎯ ⎯→

AI = B

⎛ 0.48 ⎞ ⎜ ⎟ I = A B = ⎜ 0.40 ⎟ ⎜ 0.44 ⎟ ⎝ ⎠ i.e. I1 = 0.48 A, I2 = 0.4 A, I3 = 0.44 A
−1

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Chapter 3, Problem 43.

Use mesh analysis to find vab and io in the circuit in Fig. 3.89.

Figure 3.89
Chapter 3, Solution 43
20 Ω a 80 V +

–

i1

30 Ω

+
30 Ω Vab

i3
20 Ω 80 V + 30 Ω b

–

i2
20 Ω

–

For loop 1, For loop 2, For loop 3, 80 = 70i1 – 20i2 – 30i3 80 = 70i2 – 20i1 – 30i3 0 = -30i1 – 30i2 + 90i3 Solving (1) to (3), we obtain i3 = 16/9 Io = i3 = 16/9 = 1.7778 A Vab = 30i3 = 53.33 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.

8 = 7i1 – 2i2 – 3i3 8 = -2i1 + 7i2 – 3i3 0 = i1 + i2 – 3i3

(1) (2) (3)

Chapter 3, Problem 44.

Use mesh analysis to obtain io in the circuit of Fig. 3.90.

Figure 3.90
Chapter 3, Solution 44
6V +

2Ω

i3
1Ω

4Ω

i2

6V 5Ω

+

–

i1
3A i1 i2

Loop 1 and 2 form a supermesh. For the supermesh, 6i1 + 4i2 - 5i3 + 12 = 0 For loop 3, Also, -i1 – 4i2 + 7i3 + 6 = 0 i2 = 3 + i1

(1) (2) (3)

Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 – i3 = -1.7333 A
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Chapter 3, Problem 45.

Find current i in the circuit in Fig. 3.91.

Figure 3.91
Chapter 3, Solution 45
4Ω 8Ω

i3

i4

2Ω 30V +

6Ω

–

i1

3Ω

i2

1Ω

For loop 1, For loop 2, For the supermesh, But

30 = 5i1 – 3i2 – 2i3 10i2 - 3i1 – 6i4 = 0 6i3 + 14i4 – 2i1 – 6i2 = 0

(1) (2) (3) (4)

i4 – i3 = 4 which leads to i4 = i3 + 4

Solving (1) to (4) by elimination gives i = i1 = 8.561 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 3, Problem 46.

Calculate the mesh currents i1 and i2 in Fig. 3.92.

Figure 3.92

Chapter 3, Solution 46

For loop 1, − 12 + 11i1 − 8i2 = 0 For loop 2, − 8i1 + 14i2 + 2vo = 0 But vo = 3i1 , − 8i1 + 14i2 + 6i1 = 0

⎯ ⎯→

11i1 − 8i2 = 12

(1)

⎯ ⎯→

i1 = 7i2

(2)

Substituting (2) into (1), 77i2 − 8i2 = 12 ⎯ ⎯→

i 2 = 0.1739 A and i1 = 7i2 = 1.217 A

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Chapter 3, Problem 47.

Rework Prob. 3.19 using mesh analysis. Chapter 3, Problem 3.19 Use nodal analysis to find V1, V2, and V3 in the circuit in Fig. 3.68.

Figure 3.68

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Chapter 3, Solution 47

First, transform the current sources as shown below. - 6V + 2Ω I3

V1 4Ω + 20V -

8Ω

V2

4Ω

V3

I1

2Ω

8Ω I2 + 12V -

For mesh 1, − 20 + 14 I 1 − 2 I 2 − 8 I 3 = 0 ⎯ ⎯→ 10 = 7 I 1 − I 2 − 4 I 3 For mesh 2, 12 + 14 I 2 − 2 I 1 − 4 I 3 = 0 ⎯ ⎯→ − 6 = − I1 + 7 I 2 − 2I 3 For mesh 3, − 6 + 14 I 3 − 4 I 2 − 8 I 1 = 0 ⎯ ⎯→ 3 = −4 I 1 − 2 I 2 + 7 I 3 Putting (1) to (3) in matrix form, we obtain ⎛ 7 − 1 − 4 ⎞⎛ I 1 ⎞ ⎛ 10 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎯ ⎯→ AI = B ⎜ − 1 7 − 2 ⎟⎜ I 2 ⎟ = ⎜ − 6 ⎟ ⎜ − 4 − 2 7 ⎟⎜ I ⎟ ⎜ 3 ⎟ ⎝ ⎠⎝ 3 ⎠ ⎝ ⎠ Using MATLAB, ⎡ 2 ⎤ −1 I = A B = ⎢0.0333⎥ ⎯ ⎯→ I 1 = 2.5, I 2 = 0.0333, I 3 = 1.8667 ⎢ ⎥ ⎢1.8667 ⎥ ⎣ ⎦ But
20 − V ⎯ ⎯→ V1 = 20 − 4 I 1 = 10 V 4 V2 = 2( I 1 − I 2 ) = 4.933 V

(1) (2) (3)

I1 =

Also,

I2 =

V3 − 12 8

⎯ ⎯→

V3 = 12 + 8I 2 = 12.267V

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Chapter 3, Problem 48.

Determine the current through the 10-kΩ resistor in the circuit in Fig. 3.93 using mesh analysis.

Figure 3.93

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Chapter 3, Solution 48

We apply mesh analysis and let the mesh currents be in mA. 3k Ω I4 4k Ω 2k Ω Io I2 + 8V 10k Ω 5k Ω

1k Ω + 12 V I1

I3

6V +

For mesh 1, − 12 + 8 + 5I 1 − I 2 − 4 I 4 = 0 ⎯ ⎯→ 4 = 5I1 − I 2 − 4 I 4 (1) For mesh 2, − 8 + 13I 2 − I 1 − 10 I 3 − 2 I 4 = 0 ⎯ ⎯→ 8 = − I 1 + 13I 2 − 10 I 3 − 2 I 4 (2) For mesh 3, (3) − 6 + 15 I 3 − 10 I 2 − 5 I 4 = 0 ⎯ ⎯→ 6 = −10 I 2 + 15 I 3 − 5 I 4 For mesh 4, − 4 I 1 − 2 I 2 − 5I 3 + 14 I 4 = 0 (4) Putting (1) to (4) in matrix form gives −1 − 4 ⎞⎛ I 1 ⎞ ⎛ 4 ⎞ 0 ⎛ 5 ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎜ − 1 13 − 10 − 2 ⎟⎜ I 2 ⎟ ⎜ 8 ⎟ AI = B ⎯ ⎯→ ⎜ 0 − 10 15 − 5 ⎟⎜ I ⎟ = ⎜ 6 ⎟ 3 ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎜ − 4 − 2 − 5 14 ⎟⎜ I ⎟ ⎜ 0 ⎟ ⎠⎝ 4 ⎠ ⎝ ⎠ ⎝ Using MATLAB, ⎛ 7.217 ⎞ ⎟ ⎜ ⎜ 8.087 ⎟ I = A −1 B = ⎜ 7.791 ⎟ ⎟ ⎜ ⎜ 6 ⎟ ⎠ ⎝ The current through the 10k Ω resistor is Io= I2 – I3 = 0.2957 mA
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Chapter 3, Problem 49.

Find vo and io in the circuit of Fig. 3.94.

Figure 3.94

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Chapter 3, Solution 49

3Ω

i3
1Ω 2Ω

2Ω

i1
2i0 i1 0 (a) 1Ω i2

i2

16 V

+

–

2Ω

2Ω

+ i1 v0 or

+ v0 –

–

i2

16V +

–

(b)

For the supermesh in figure (a), 3i1 + 2i2 – 3i3 + 16 = 0 At node 0, For loop 3, i2 – i1 = 2i0 and i0 = -i1 which leads to i2 = -i1 -i1 –2i2 + 6i3 = 0 which leads to 6i3 = -i1 (1) (2) (3)

Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A i0 = -i1 = 10.667 A, from fig. (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V.

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Chapter 3, Problem 50.

Use mesh analysis to find the current io in the circuit in Fig. 3.95.

Figure 3.95
Chapter 3, Solution 50

4Ω

i1
10 Ω

2Ω

i3

8Ω 60 V +

–

i2
3i0 i2 i3

For loop 1,

16i1 – 10i2 – 2i3 = 0 which leads to 8i1 – 5i2 – i3 = 0

(1)

For the supermesh, -60 + 10i2 – 10i1 + 10i3 – 2i1 = 0 or -6i1 + 5i2 + 5i3 = 30 (2) (3)

Also, 3i0 = i3 – i2 and i0 = i1 which leads to 3i1 = i3 – i2 Solving (1), (2), and (3), we obtain i1 = 1.731 and i0 = i1 = 1.731 A

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Chapter 3, Problem 51.

Apply mesh analysis to find vo in the circuit in Fig. 3.96.

Figure 3.96
Chapter 3, Solution 51
5A

i1
8Ω 2Ω 1Ω

i3 i2
4Ω

+

40 V +

20V +

For loop 1, For loop 2, For loop 3,

i1 = 5A -40 + 7i2 – 2i1 – 4i3 = 0 which leads to 50 = 7i2 – 4i3 -20 + 12i3 – 4i2 = 0 which leads to 5 = - i2 + 3 i3 i2 = 10 A, i3 = 5 A

(1) (2) (3)

Solving with (2) and (3), And,

v0 = 4(i2 – i3) = 4(10 – 5) = 20 V.

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Chapter 3, Problem 52.

Use mesh analysis to find i1, i2, and i3 in the circuit of Fig. 3.97.

Figure 3.97

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Chapter 3, Solution 52

+ v0 2 Ω

i2 3A

–
VS +

8Ω

i2 i3

–

i1
4Ω

i3

+ –

2V0

For mesh 1, 2(i1 – i2) + 4(i1 – i3) – 12 = 0 which leads to 3i1 – i2 – 2i3 = 6 For the supermesh, 2(i2 – i1) + 8i2 + 2v0 + 4(i3 – i1) = 0 But v0 = 2(i1 – i2) which leads to -i1 + 3i2 + 2i3 = 0 For the independent current source, i3 = 3 + i2 Solving (1), (2), and (3), we obtain, i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A. (2) (3) (1)

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Chapter 3, Problem 53.

Find the mesh currents in the circuit of Fig. 3.98 using MATLAB. 2 kΩ

I5 6 kΩ 8 kΩ

I3 1 kΩ

8 kΩ

I4 4 kΩ

3 mA

12 V

+ _

I1

3 kΩ

I2

Figure 3.98 For Prob. 3.53.

Chapter 3, Solution 53

Applying mesh analysis leads to; –12 + 4kI1 – 3kI2 – 1kI3 = 0 –3kI1 + 7kI2 – 4kI4 = 0 –3kI1 + 7kI2 = –12 –1kI1 + 15kI3 – 8kI4 – 6kI5 = 0 –1kI1 + 15kI3 – 6k = –24 I4 = –3mA –6kI3 – 8kI4 + 16kI5 = 0 –6kI3 + 16kI5 = –24

(1) (2) (3) (4) (5)

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Putting these in matrix form (having substituted I4 = 3mA in the above), ⎡ 4 − 3 − 1 0 ⎤ ⎡ I1 ⎤ ⎡ 12 ⎤ ⎢− 3 7 0 0 ⎥ ⎢I 2 ⎥ ⎢ − 12 ⎥ ⎥ ⎥k ⎢ ⎥ = ⎢ ⎢ ⎢ − 1 0 15 − 6⎥ ⎢ I 3 ⎥ ⎢− 24⎥ ⎥ ⎥ ⎢ ⎥ ⎢ ⎢ 0 − 6 16 ⎦ ⎣ I 5 ⎦ ⎣− 24⎦ ⎣0 ZI = V Using MATLAB, >> Z = [4,-3,-1,0;-3,7,0,0;-1,0,15,-6;0,0,-6,16] Z= 4 -3 -1 0 -3 7 0 0 -1 0 15 -6 0 0 -6 16 >> V = [12,-12,-24,-24]' V= 12 -12 -24 -24 We obtain, >> I = inv(Z)*V I=
1.6196 mA –1.0202 mA –2.461 mA 3 mA –2.423 mA

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Chapter 3, Problem 54.

Find the mesh currents i1, i2, and i3 in the circuit in Fig. 3.99.

Figure 3.99

Chapter 3, Solution 54

Let the mesh currents be in mA. For mesh 1, − 12 + 10 + 2 I 1 − I 2 = 0 ⎯ ⎯→ 2 = 2 I 1 − I 2 For mesh 2, − 10 + 3I 2 − I 1 − I 3 = 0 For mesh 3, − 12 + 2 I 3 − I 2 = 0
⎯ ⎯→ ⎯ ⎯→

(1)

10 = − I 1 + 3I 2 − I 3

(2)

12 = − I 2 + 2 I 3

(3)

Putting (1) to (3) in matrix form leads to ⎛ 2 − 1 0 ⎞⎛ I 1 ⎞ ⎛ 2 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ − 1 3 − 1⎟⎜ I 2 ⎟ = ⎜10 ⎟ ⎜ 0 − 1 2 ⎟⎜ I ⎟ ⎜12 ⎟ ⎝ ⎠⎝ 3 ⎠ ⎝ ⎠ Using MATLAB,

⎯ ⎯→

AI = B

⎡ 5.25 ⎤ I = A B = ⎢ 8.5 ⎥ ⎢ ⎥ ⎢10.25⎥ ⎣ ⎦
−1

⎯ ⎯→ I 1 = 5.25 mA, I 2 = 8.5 mA, I 3 = 10.25 mA

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Chapter 3, Problem 55.

In the circuit of Fig. 3.100, solve for i1, i2, and i3.

Figure 3.100

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Chapter 3, Solution 55
10 V b

I2 i1 I2

+

c

1A 1A

4A

6Ω

I1

d

I3

2Ω

i2
4A
a 12 Ω

I4

i3
4Ω +–

I3

I4
8V

0

It is evident that I1 = 4 For mesh 4, 12(I4 – I1) + 4(I4 – I3) – 8 = 0 6(I2 – I1) + 10 + 2I3 + 4(I3 – I4) = 0 or -3I1 + 3I2 + 3I3 – 2I4 = -5

(1) (2) (3) (4)

For the supermesh At node c,

I2 = I 3 + 1

Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A At node b, At node a, At node 0, i1 = I2 – I1 = -1A i2 = 4 – I4 = 0A i3 = I4 – I3 = 2A

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Chapter 3, Problem 56.

Determine v1 and v2 in the circuit of Fig. 3.101.

Figure 3.101

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Chapter 3, Solution 56 + v1 –
2Ω 2Ω

i2
2Ω

2Ω 2Ω

+ v2 12 V

+

–

i1

i3

–

For loop 1, 12 = 4i1 – 2i2 – 2i3 which leads to 6 = 2i1 – i2 – i3 For loop 2, 0 = 6i2 –2i1 – 2 i3 which leads to 0 = -i1 + 3i2 – i3 For loop 3, 0 = 6i3 – 2i1 – 2i2 which leads to 0 = -i1 – i2 + 3i3 In matrix form (1), (2), and (3) become, ⎡ 2 − 1 − 1⎤ ⎡ i1 ⎤ ⎡6⎤ ⎢ − 1 3 − 1⎥ ⎢i ⎥ = ⎢0⎥ ⎢ ⎥⎢ 2 ⎥ ⎢ ⎥ ⎢ − 1 − 1 3 ⎥ ⎢i 3 ⎥ ⎢0⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

(1) (2) (3)

2 −1 −1 2 6 −1 Δ = − 1 3 − 1 = 8, Δ2 = − 1 3 − 1 = 24 −1 −1 3 −1 0 3
2 −1 6 Δ3 = − 1 3 0 = 24 , therefore i2 = i3 = 24/8 = 3A, −1 −1 0 v1 = 2i2 = 6 volts, v = 2i3 = 6 volts

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Chapter 3, Problem 57.

In the circuit in Fig. 3.102, find the values of R, V1, and V2 given that io = 18 mA.

Figure 3.102

Chapter 3, Solution 57

Assume R is in kilo-ohms. V2 = 4kΩx18mA = 72V , V1 = 100 − V2 = 100 − 72 = 28V Current through R is 3 3 iR = io , V1 = i R R ⎯ ⎯→ 28 = (18) R 3+ R 3+ R This leads to R = 84/26 = 3.23 k Ω

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Chapter 3, Problem 58.

Find i1, i2, and i3 the circuit in Fig. 3.103.

Figure 3.103
Chapter 3, Solution 58
30 Ω

i2
30 Ω 10 Ω 10 Ω 30 Ω

i1

+

i3
120 V

–

For loop 1, 120 + 40i1 – 10i2 = 0, which leads to -12 = 4i1 – i2 For loop 2, 50i2 – 10i1 – 10i3 = 0, which leads to -i1 + 5i2 – i3 = 0 For loop 3, -120 – 10i2 + 40i3 = 0, which leads to 12 = -i2 + 4i3 Solving (1), (2), and (3), we get, i1 = -3A, i2 = 0, and i3 = 3A

(1) (2) (3)

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Chapter 3, Problem 59.

Rework Prob. 3.30 using mesh analysis. Chapter 3, Problem 30. Using nodal analysis, find vo and io in the circuit of Fig. 3.79.

Figure 3.79

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Chapter 3, Solution 59
40 Ω

–+ I0
10 Ω 20 Ω

i2

120 V

+
100V +

i1

–

4v0

+ – 2I0 i2

i3

v0

80 Ω

–

i3

For loop 1, -100 + 30i1 – 20i2 + 4v0 = 0, where v0 = 80i3 or 5 = 1.5i1 – i2 + 16i3 For the supermesh, 60i2 – 20i1 – 120 + 80i3 – 4 v0 = 0, where v0 = 80i3 or 6 = -i1 + 3i2 – 12i3 Also, 2I0 = i3 – i2 and I0 = i2, hence, 3i2 = i3 ⎡ 3 − 2 32 ⎤ ⎢ − 1 3 − 12⎥ ⎢ ⎥ −1 ⎥ 3 ⎢0 ⎣ ⎦ ⎡ i1 ⎤ ⎡10⎤ ⎢i ⎥ = ⎢ 6 ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢i 3 ⎥ ⎢ 0 ⎥ ⎣ ⎦ ⎣ ⎦

(1)

(2) (3)

From (1), (2), and (3),

3 − 2 32 3 10 32 3 − 2 10 Δ = − 1 3 − 12 = 5, Δ2 = − 1 6 − 12 = −28, Δ3 = − 1 3 6 = −84 0 3 −1 0 0 −1 0 3 0
I0 = i2 = Δ2/Δ = -28/5 = -5.6 A v0 = 8i3 = (-84/5)80 = -1.344 kvolts

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Chapter 3, Problem 60.

Calculate the power dissipated in each resistor in the circuit in Fig. 3.104.

Figure 3.104

Chapter 3, Solution 60
0.5i0

4Ω

10 V

8Ω

v1
1Ω 10 V +

v2
2Ω

–

i0

At node 1, (v1/1) + (0.5v1/1) = (10 – v1)/4, which leads to v1 = 10/7 At node 2, (0.5v1/1) + ((10 – v2)/8) = v2/2 which leads to v2 = 22/7 P1Ω = (v1)2/1 = 2.041 watts, P2Ω = (v2)2/2 = 4.939 watts P4Ω = (10 – v1)2/4 = 18.38 watts, P8Ω = (10 – v2)2/8 = 5.88 watts
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Chapter 3, Problem 61.

Calculate the current gain io/is in the circuit of Fig. 3.105.

Figure 3.105

Chapter 3, Solution 61 v1 is
20 Ω

v2

10 Ω i0

+ v0 –

30 Ω

– + 5v0

40 Ω

At node 1, is = (v1/30) + ((v1 – v2)/20) which leads to 60is = 5v1 – 3v2 But v2 = -5v0 and v0 = v1 which leads to v2 = -5v1 Hence, 60is = 5v1 + 15v1 = 20v1 which leads to v1 = 3is, v2 = -15is i0 = v2/50 = -15is/50 which leads to i0/is = -15/50 = –0.3

(1)

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Chapter 3, Problem 62.

Find the mesh currents i1, i2, and i3 in the network of Fig. 3.106.

Figure 3.106

Chapter 3, Solution 62
4 kΩ A 8 kΩ B 2 kΩ

100V +

–

i1

i2

i3

+

–

40 V

We have a supermesh. Let all R be in kΩ, i in mA, and v in volts. For the supermesh, -100 +4i1 + 8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3 At node A, At node B, i1 + 4 = i2 i2 = 2i1 + i3 (1) (2) (3)

Solving (1), (2), and (3), we get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA.

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Chapter 3, Problem 63.

Find vx, and ix in the circuit shown in Fig. 3.107.

Figure 3.107
Chapter 3, Solution 63
10 Ω A

5Ω 50 V +

–

i1

i2 + –
4ix

For the supermesh, -50 + 10i1 + 5i2 + 4ix = 0, but ix = i1. Hence, 50 = 14i1 + 5i2 At node A, i1 + 3 + (vx/4) = i2, but vx = 2(i1 – i2), hence, i1 + 2 = i2 Solving (1) and (2) gives i1 = 2.105 A and i2 = 4.105 A vx = 2(i1 – i2) = –4 volts and ix = i2 – 2 = 2.105 amp
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(1) (2)

Chapter 3, Problem 64.

Find vo, and io in the circuit of Fig. 3.108.

Figure 3.108

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Chapter 3, Solution 64 i1
50 Ω A

i2 10 Ω + − i0
10 Ω

i2

i1
100V +

+ –

4i0

i3

40 Ω

–

0.2V0

2A B

i1

i3

For mesh 2,

20i2 – 10i1 + 4i0 = 0

(1) (2)

But at node A, io = i1 – i2 so that (1) becomes i1 = (16/6)i2 For the supermesh, -100 + 50i1 + 10(i1 – i2) – 4i0 + 40i3 = 0 or At node B, But, 50 = 28i1 – 3i2 + 20i3 i3 + 0.2v0 = 2 + i1 v0 = 10i2 so that (4) becomes i3 = 2 + (2/3)i2

(3) (4) (5)

Solving (1) to (5), i2 = 0.11764, v0 = 10i2 = 1.1764 volts, i0 = i1 - i2 = (5/3)i2 = 196.07 mA

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Chapter 3, Problem 65.

Use MATLAB to solve for the mesh currents in the circuit of Fig. 3.109.

Figure 3.109

Chapter 3, Solution 65

For mesh 1, –12 + 12I1 – 6I2 – I4 = 0 or 12 = 12I 1 − 6 I 2 − I 4 For mesh 2, For mesh 3, For mesh 4, For mesh 5, –6I1 + 16I2 – 8I3 – I4 – I5 = 0 –8I2 + 15I3 – I5 – 9 = 0 or 9 = –8I2 + 15I3 – I5 –I1 – I2 + 7I4 – 2I5 – 6 = 0 or 6 = –I1 – I2 + 7I4 – 2I5 –I2 – I3 – 2I4 + 8I5 – 10 = 0 or 10 = − I 2 − I 3 − 2 I 4 + 8I 5 (2) (3) (4) (5) (1)

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Casting (1) to (5) in matrix form gives 1 0 ⎞⎛ I1 ⎞ ⎛12 ⎞ ⎛ 12 − 6 0 ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎜ − 6 16 − 8 − 1 − 1 ⎟⎜ I 2 ⎟ ⎜ 0 ⎟ ⎜ 0 − 8 15 0 − 1 ⎟⎜ I ⎟ = ⎜ 9 ⎟ ⎟⎜ 3 ⎟ ⎜ ⎟ ⎜ 7 − 2 ⎟⎜ I 4 ⎟ ⎜ 6 ⎟ ⎜ −1 −1 0 ⎜ 0 − 1 − 1 − 2 8 ⎟⎜ I ⎟ ⎜10 ⎟ ⎠⎝ 5 ⎠ ⎝ ⎠ ⎝

⎯ ⎯→

AI = B

Using MATLAB we input: Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8] and V=[12;0;9;6;10] This leads to >> Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8] Z= 12 -6 0 -1 0 -6 0 -1 16 -8 -1 -8 15 0 -1 0 7 -1 -1 -2 0 -1 -1 -2 8

>> V=[12;0;9;6;10] V= 12 0 9 6 10 >> I=inv(Z)*V I= 2.1701 1.9912 1.8119 2.0942 2.2489 Thus,
I = [2.17, 1.9912, 1.8119, 2.094, 2.249] A.
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Chapter 3, Problem 66.

Write a set of mesh equations for the circuit in Fig. 3.110. Use MATLAB to determine the mesh currents. 10 Ω 10 Ω

8Ω 12 V + _ 6Ω

4Ω I1 + _ 2Ω I2 24 V + _ 6Ω

8Ω 40 V

2Ω I4 4Ω

8Ω 30 V + _

I3

8Ω 4Ω I5 + _ 32 V

Figure 3.110 For Prob. 3.66. Chapter 3, Solution 66

The mesh equations are obtained as follows. −12 + 24 + 30I1 − 4I2 − 6I3 − 2I4 = 0 or 30I1 – 4I2 – 6I3 – 2I4 = –12 −24 + 40 − 4I1 + 30I2 − 2I4 − 6I5 = 0 or –4I1 + 30I2 – 2I4 – 6I5 = –16 –6I1 + 18I3 – 4I4 = 30 –2I1 – 2I2 – 4I3 + 12I4 –4I5 = 0 –6I2 – 4I4 + 18I5 = –32 (1)

(2) (3) (4) (5)

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Putting (1) to (5) in matrix form ⎡ 30 − 4 − 6 − 2 0 ⎤ ⎡ − 12 ⎤ ⎢− 4 30 0 − 2 − 6⎥ ⎢ − 16 ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ − 6 0 18 − 4 0 ⎥ I = ⎢ 30 ⎥ ⎥ ⎥ ⎢ ⎢ ⎢− 2 − 2 − 4 12 − 4⎥ ⎢ 0 ⎥ ⎢ 0 − 6 0 − 4 18 ⎥ ⎢− 32⎥ ⎦ ⎦ ⎣ ⎣ ZI = V Using MATLAB, >> Z = [30,-4,-6,-2,0; -4,30,0,-2,-6; -6,0,18,-4,0; -2,-2,-4,12,-4; 0,-6,0,-4,18] Z= 30 -4 -6 -2 0 -4 30 0 -2 -6 -6 0 18 -4 0 -2 0 -2 -6 -4 0 12 -4 -4 18

>> V = [-12,-16,30,0,-32]' V= -12 -16 30 0 -32 >> I = inv(Z)*V I=
-0.2779 A -1.0488 A 1.4682 A -0.4761 A -2.2332 A
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Chapter 3, Problem 67.

Obtain the node-voltage equations for the circuit in Fig. 3.111 by inspection. Then solve for Vo. 2A

4Ω

2Ω + Vo _

3 Vo

10 Ω

5Ω

4A

Figure 3.111 For Prob. 3.67. Chapter 3, Solution 67

Consider the circuit below.

2A

V1

4Ω

V2

2Ω + Vo -

V3

3 Vo

10 Ω

5Ω

4A

0 ⎤ ⎡ 0.35 − 0.25 ⎡− 2 + 3Vo ⎤ ⎢− 0.25 0.95 − 0.5⎥ V = ⎢ ⎥ 0 ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎢ ⎥ − 0.5 0.5 ⎥ 6 ⎣ ⎦ ⎣ ⎦
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Since we actually have four unknowns and only three equations, we need a constraint equation. Vo = V2 – V3 Substituting this back into the matrix equation, the first equation becomes, 0.35V1 – 3.25V2 + 3V3 = –2 This now results in the following matrix equation, 3 ⎤ ⎡ 0.35 − 3.25 ⎡ − 2⎤ ⎢− 0.25 0.95 − 0.5⎥ V = ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎢6⎥ − 0.5 0.5 ⎥ ⎣ ⎦ ⎣ ⎦ Now we can use MATLAB to solve for V. >> Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5] Y= 0.3500 -3.2500 3.0000 -0.2500 0.9500 -0.5000 0 -0.5000 0.5000 >> I=[-2,0,6]' I= -2 0 6 >> V=inv(Y)*I V= -164.2105 -77.8947 -65.8947 Vo = V2 – V3 = –77.89 + 65.89 = –12 V. Let us now do a quick check at node 1. –3(–12) + 0.1(–164.21) + 0.25(–164.21+77.89) + 2 = +36 – 16.421 – 21.58 + 2 = –0.001; answer checks!
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Chapter 3, Problem 68.

Find the voltage Vo in the circuit of Fig. 3.112. 3A

10 Ω + 4A 40 Ω Vo _

25 Ω

20 Ω

+ _

24 V

Figure 3.112 For Prob. 3.68. Chapter 3, Solution 68

Consider the circuit below. There are two non-reference nodes. 3A

V1

10 Ω +

Vo

25 Ω

4A

40 Ω

Vo _

20 Ω

+ _

24 V

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⎡ +4+3 ⎤ ⎡ 7 ⎤ ⎡0.125 − 0.1⎤ ⎢ − 0.1 0.19 ⎥ V = ⎢− 3 + 24 / 25⎥ = ⎢− 2.04⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Using MATLAB, we get, >> Y=[0.125,-0.1;-0.1,0.19] Y= 0.1250 -0.1000 -0.1000 0.1900 >> I=[7,-2.04]' I= 7.0000 -2.0400 >> V=inv(Y)*I V= 81.8909 32.3636 Thus, Vo = 32.36 V. We can perform a simple check at node Vo, 3 + 0.1(32.36–81.89) + 0.05(32.36) + 0.04(32.36–24) = 3 – 4.953 + 1.618 + 0.3344 = – 0.0004; answer checks!

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Chapter 3, Problem 69.

For the circuit in Fig. 3.113, write the node voltage equations by inspection.

Figure 3.113

Chapter 3, Solution 69

Assume that all conductances are in mS, all currents are in mA, and all voltages are in volts. G11 = (1/2) + (1/4) + (1/1) = 1.75, G22 = (1/4) + (1/4) + (1/2) = 1, G33 = (1/1) + (1/4) = 1.25, G12 = -1/4 = -0.25, G13 = -1/1 = -1, G21 = -0.25, G23 = -1/4 = -0.25, G31 = -1, G32 = -0.25 i1 = 20, i2 = 5, and i3 = 10 – 5 = 5 The node-voltage equations are:

− 1 ⎤ ⎡ v 1 ⎤ ⎡20⎤ ⎡ 1.75 − 0.25 ⎢ − 0.25 1 − 0.25⎥ ⎢ v 2 ⎥ = ⎢ 5 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ − 0.25 1.25 ⎥ ⎢ v 3 ⎥ ⎢ 5 ⎥ ⎢ ⎣ −1 ⎦⎣ ⎦ ⎣ ⎦

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Chapter 3, Problem 70.

Write the node-voltage equations by inspection and then determine values of V1 and V2 in the circuit in Fig. 3.114.

V1 ix 4A 1S 2S

4ix

V2

5S

2A

Figure 3.114 For Prob. 3.70.

Chapter 3, Solution 70

⎡ 4I x + 4 ⎤ ⎡3 0⎤ ⎢0 5 ⎥ V = ⎢ − 4 I − 2 ⎥ x ⎣ ⎦ ⎣ ⎦
With two equations and three unknowns, we need a constraint equation, Ix = 2V1, thus the matrix equation becomes, ⎡ − 5 0⎤ ⎡4⎤ V=⎢ ⎥ ⎢ 8 5⎥ ⎣ ⎦ ⎣ − 2⎦ This results in V1 = 4/(–5) = –0.8V and V2 = [–8(–0.8) – 2]/5 = [6.4 – 2]/5 = 0.88 V.

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Chapter 3, Problem 71.

Write the mesh-current equations for the circuit in Fig. 3.115. Next, determine the values of I1, I2, and I3.

5Ω I1

I3

3Ω

10 V

+ _

1Ω 2Ω 4Ω I2 + _ 5V

Figure 3.115 For Prob. 3.71.

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Chapter 3, Solution 71

⎡ 9 − 4 − 5⎤ ⎡ 10 ⎤ ⎢− 4 7 − 1⎥ I = ⎢− 5⎥ ⎢ ⎥ ⎢ ⎥ ⎢− 5 − 1 9 ⎥ ⎢ 0 ⎥ ⎣ ⎦ ⎣ ⎦ We can now use MATLAB solve for our currents. >> R=[9,-4,-5;-4,7,-1;-5,-1,9] R= 9 -4 -5 -4 7 -1 -5 -1 9 >> V=[10,-5,0]' V= 10 -5 0 >> I=inv(R)*V I=
2.085 A 653.3 mA 1.2312 A

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Chapter 3, Problem 72.

By inspection, write the mesh-current equations for the circuit in Fig. 3.116.

Figure 3.116

Chapter 3, Solution 72

R11 = 5 + 2 = 7, R22 = 2 + 4 = 6, R33 = 1 + 4 = 5, R44 = 1 + 4 = 5, R12 = -2, R13 = 0 = R14, R21 = -2, R23 = -4, R24 = 0, R31 = 0, R32 = -4, R34 = -1, R41 = 0 = R42, R43 = -1, we note that Rij = Rji for all i not equal to j. v1 = 8, v2 = 4, v3 = -10, and v4 = -4 Hence the mesh-current equations are: 0 ⎤ ⎡ i1 ⎤ ⎡ 8 ⎤ ⎡ 7 −2 0 ⎢ − 2 6 − 4 0 ⎥ ⎢i ⎥ ⎢ 4 ⎥ ⎥ ⎢ ⎥⎢ 2 ⎥ = ⎢ ⎢ 0 − 4 5 − 1⎥ ⎢i 3 ⎥ ⎢ − 10⎥ ⎥ ⎢ ⎥⎢ ⎥ ⎢ 0 − 1 5 ⎦ ⎣i 4 ⎦ ⎣ − 4 ⎦ ⎣ 0

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Chapter 3, Problem 73.

Write the mesh-current equations for the circuit in Fig. 3.117.

Figure 3.117

Chapter 3, Solution 73

R11 = 2 + 3 +4 = 9, R22 = 3 + 5 = 8, R33 = 1+1 + 4 = 6, R44 = 1 + 1 = 2, R12 = -3, R13 = -4, R14 = 0, R23 = 0, R24 = 0, R34 = -1 v1 = 6, v2 = 4, v3 = 2, and v4 = -3 Hence,

⎡ 9 − 3 − 4 0 ⎤ ⎡ i1 ⎤ ⎡ 6 ⎤ ⎢− 3 8 0 0 ⎥ ⎢i 2 ⎥ ⎢ 4 ⎥ ⎥⎢ ⎥ = ⎢ ⎥ ⎢ ⎢− 4 0 6 − 1⎥ ⎢i3 ⎥ ⎢ 2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ 0 − 1 2 ⎦ ⎣i 4 ⎦ ⎣− 3⎦ ⎣0

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Chapter 3, Problem 74.

By inspection, obtain the mesh-current equations for the circuit in Fig. 3.11.

Figure 3.118

Chapter 3, Solution 74

R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 + R8, R44 = R3 + R5 + R8, R12 = -R4, R13 = -R6, R14 = 0, R23 = 0, R24 = -R5, R34 = -R8, again, we note that Rij = Rji for all i not equal to j.

⎡ V1 ⎤ ⎢− V ⎥ 2⎥ The input voltage vector is = ⎢ ⎢ V3 ⎥ ⎥ ⎢ ⎣ − V4 ⎦ ⎡R 1 + R 4 + R 6 ⎢ − R4 ⎢ − R6 ⎢ ⎢ 0 ⎣ − R4 R2 + R4 + R5 0 − R5 − R6 0 R6 + R7 + R8 − R8 0 ⎤ ⎡ i 1 ⎤ ⎡ V1 ⎤ ⎥ ⎢i ⎥ ⎢ − V ⎥ − R5 2⎥ ⎥⎢ 2 ⎥ = ⎢ − R8 ⎥ ⎢i 3 ⎥ ⎢ V3 ⎥ ⎥ ⎥⎢ ⎥ ⎢ R 3 + R 5 + R 8 ⎦ ⎣i 4 ⎦ ⎣ − V4 ⎦

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Chapter 3, Problem 75.

Use PSpice to solve Prob. 3.58. Chapter 3, Problem 58 Find i1, i2, and i3 the circuit in Fig. 3.103.

Figure 3.103

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Chapter 3, Solution 75
* Schematics Netlist * R_R4 R_R2 R_R1 R_R3 R_R5 V_V4 v_V3 v_V2 v_V1 $N_0002 $N_0001 30 $N_0001 $N_0003 10 $N_0005 $N_0004 30 $N_0003 $N_0004 10 $N_0006 $N_0004 30 $N_0003 0 120V $N_0005 $N_0001 0 0 $N_0006 0 0 $N_0002 0

i3

i1

i2

Clearly, i1 = –3 amps, i2 = 0 amps, and i3 = 3 amps, which agrees with the answers in Problem 3.44.

Chapter 3, Problem 76.
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Use PSpice to solve Prob. 3.27. Chapter 3, Problem 27 Use nodal analysis to determine voltages v1, v2, and v3 in the circuit in Fig. 3.76.

Figure 3.76

Chapter 3, Solution 76
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* Schematics Netlist * I_I2 R_R1 R_R3 R_R2 F_F1 VF_F1 R_R4 R_R6 I_I1 R_R5 0 $N_0001 DC 4A $N_0002 $N_0001 0.25 $N_0003 $N_0001 1 $N_0002 $N_0003 1 $N_0002 $N_0001 VF_F1 3 $N_0003 $N_0004 0V 0 $N_0002 0.5 0 $N_0001 0.5 0 $N_0002 DC 2A 0 $N_0004 0.25

Clearly, v1 = 625 mVolts, v2 = 375 mVolts, and v3 = 1.625 volts, which agrees with the solution obtained in Problem 3.27.

Chapter 3, Problem 77.

Solve for V1 and V2 in the circuit of Fig. 3.119 using PSpice.
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2 ix

V1

5Ω

V2

5A

2Ω ix

1Ω

2A

Figure 3.119 For Prob. 3.77.

Chapter 3, Solution 77

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As a check we can write the nodal equations, ⎡ 1.7 − 0.2⎤ ⎡5⎤ V=⎢ ⎥ ⎢− 1.2 1.2 ⎥ ⎣ ⎦ ⎣ − 2⎦ Solving this leads to V1 = 3.111 V and V2 = 1.4444 V. The answer checks!

Chapter 3, Problem 78.
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Solve Prob. 3.20 using PSpice. Chapter 3, Problem 20 For the circuit in Fig. 3.69, find V1, V2, and V3 using nodal analysis.

Figure 3.69

Chapter 3, Solution 78
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The schematic is shown below. When the circuit is saved and simulated the node voltages are displaced on the pseudocomponents as shown. Thus,

V1 = −3V, V2 = 4.5V, V3 = −15V,

.

Chapter 3, Problem 79. Rework Prob. 3.28 using PSpice.
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Chapter 3, Problem 28 Use MATLAB to find the voltages at nodes a, b, c, and d in the circuit of Fig. 3.77.

Figure 3.77
Chapter 3, Solution 79 The schematic is shown below. When the circuit is saved and simulated, we obtain the node voltages as displaced. Thus, Va = −5.278 V, Vb = 10.28 V, Vc = 0.6944 V, Vd = −26.88 V

Chapter 3, Problem 80.
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Find the nodal voltage v1 through v4 in the circuit in Fig. 3.120 using PSpice.

Figure 3.120

Chapter 3, Solution 80
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* Schematics Netlist * H_H1 VH_H1 I_I1 V_V1 R_R4 R_R1 R_R2 R_R5 R_R3 $N_0002 $N_0003 VH_H1 6 0 $N_0001 0V $N_0004 $N_0005 DC 8A $N_0002 0 20V 0 $N_0003 4 $N_0005 $N_0003 10 $N_0003 $N_0002 12 0 $N_0004 1 $N_0004 $N_0001 2

Clearly, v1 = 84 volts, v2 = 4 volts, v3 = 20 volts, and v4 = -5.333 volts

Chapter 3, Problem 81.

Use PSpice to solve the problem in Example 3.4
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Example 3.4 Find the node voltages in the circuit of Fig. 3.12.

Figure 3.12

Chapter 3, Solution 81

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Clearly, v1 = 26.67 volts, v2 = 6.667 volts, v3 = 173.33 volts, and v4 = -46.67 volts which agrees with the results of Example 3.4. This is the netlist for this circuit. * Schematics Netlist * R_R1 R_R2 R_R3 R_R4 R_R5 I_I1 V_V1 E_E1 0 $N_0001 2 $N_0003 $N_0002 6 0 $N_0002 4 0 $N_0004 1 $N_0001 $N_0004 3 0 $N_0003 DC 10A $N_0001 $N_0003 20V $N_0002 $N_0004 $N_0001 $N_0004 3

Chapter 3, Problem 82.

If the Schematics Netlist for a network is as follows, draw the network.
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R_R1 R_R2 R_R3 R_R4 R_R5 V_VS I_IS F_F1 VF_F1 E_E1

1 2 2 3 1 4 0 1 5 3

2 0 0 4 3 0 1 3 0 2

2K 4K 8K 6K 3K DC DC VF_F1 0V 1

100 4 2 3 3

Chapter 3, Solution 82
2i0

+ v0 –
3 kΩ

1 4A

2 kΩ

2

+

3v0

3

6 kΩ

4

4 kΩ

8 kΩ

100V +

– 0

This network corresponds to the Netlist.

Chapter 3, Problem 83.

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The following program is the Schematics Netlist of a particular circuit. Draw the circuit and determine the voltage at node 2. R_R1 R_R2 R_R3 R_R4 V_VS I_IS 1 2 2 3 1 2 2 0 3 0 0 0 20 50 70 30 20V DC 2A

Chapter 3, Solution 83

The circuit is shown below.
1
20 Ω

2

70 Ω

3

20 V

+

–

50 Ω

2A

30 Ω

0

When the circuit is saved and simulated, we obtain v2 = –12.5 volts
Chapter 3, Problem 84.

Calculate vo and io in the circuit of Fig. 3.121.

Figure 3.121
Chapter 3, Solution 84
From the output loop, v0 = 50i0x20x103 = 106i0 (1) From the input loop, 3x10-3 + 4000i0 – v0/100 = 0 (2) From (1) and (2) we get, i0 = 0.5μA and v0 = 0.5 volt.

Chapter 3, Problem 85.
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An audio amplifier with resistance 9Ω supplies power to a speaker. In order that maximum power is delivered, what should be the resistance of the speaker?
Chapter 3, Solution 85

The amplifier acts as a source. Rs + Vs RL

For maximum power transfer, R L = R s = 9Ω
Chapter 3, Problem 86.

For the simplified transistor circuit of Fig. 3.122, calculate the voltage vo.

Figure 3.122
Chapter 3, Solution 86
Let v1 be the potential across the 2 k-ohm resistor with plus being on top. Then, [(0.03 – v1)/1k] + 400i = v1/2k (1) Assume that i is in mA. But, i = (0.03 – v1)/1 Combining (1) and (2) yields, v1 = 29.963 mVolts and i = 37.4 nA, therefore, v0 = -5000x400x37.4x10-9 = -74.8 mvolts (2)

Chapter 3, Problem 87.

For the circuit in Fig. 3.123, find the gain vo/vs.
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Figure 3.123
Chapter 3, Solution 87 v1 = 500(vs)/(500 + 2000) = vs/5 v0 = -400(60v1)/(400 + 2000) = -40v1 = -40(vs/5) = -8vs, Therefore, v0/vs = –8

Chapter 3, Problem 88.

Determine the gain vo/vs of the transistor amplifier circuit in Fig. 3.124.

Figure 3.124
Chapter 3, Solution 88
Let v1 be the potential at the top end of the 100-ohm resistor. (vs – v1)/200 = v1/100 + (v1 – 10-3v0)/2000 For the right loop, v0 = -40i0(10,000) = -40(v1 – 10-3)10,000/2000, or, v0 = -200v1 + 0.2v0 = -4x10-3v0 (2) (1)

Substituting (2) into (1) gives, (vs + 0.004v1)/2 = -0.004v0 + (-0.004v1 – 0.001v0)/20 This leads to 0.125v0 = 10vs or (v0/vs) = 10/0.125 = -80

Chapter 3, Problem 89.
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For the transistor circuit shown in Fig. 3.125, find IB and VCE. Let β = 100 and VBE = 0.7V. _ 3V + _ 1 kΩ 0.7 V + 100 kΩ
| |

15 V

Figure 3.125 For Prob. 3.89. Chapter 3, Solution 89

Consider the circuit below. _ 0.7 V C + 100 kΩ + IC VCE _ 3V + _ E For the left loop, applying KVL gives
VBE = 0.7 −3 − 0.7 + 100 x103 IB + VBE = 0 ⎯⎯⎯⎯ IB = 30 μ A → For the right loop, −VCE + 15 − Ic(1x10 3 ) = 0 But IC = β IB = 100 x30 μ A= 3 mA

|

|

15 V

1 kΩ

VCE = 15 − 3 x10 −3 x103 = 12 V

Chapter 3, Problem 90.
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Calculate vs for the transistor in Fig. 3.126, given that vo = 4 V, β = 150, VBE = 0.7V.

Figure 3.126

Chapter 3, Solution 90

1 kΩ

10 kΩ

IB
+ VBE

+
VCE –

i2
+

vs

+

–

-

i1

18V

500 Ω

+
V0

-

IE

–

For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0.7 + 10,000IB + 500(1 + β)IB which leads to vs + 0.7 = 10,000IB + 500(151)IB = 85,500IB But, v0 = 500IE = 500x151IB = 4 which leads to IB = 5.298x10-5 Therefore, vs = 0.7 + 85,500IB = 5.23 volts
Chapter 3, Problem 91.
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For the transistor circuit of Fig. 3.127, find IB, VCE, and vo. Take β = 200, VBE = 0.7V.

Figure 3.127
Chapter 3, Solution 91
We first determine the Thevenin equivalent for the input circuit. RTh = 6||2 = 6x2/8 = 1.5 kΩ and VTh = 2(3)/(2+6) = 0.75 volts
5 kΩ

IC
1.5 kΩ

IB
+ VBE

+
VCE –

i2
+

+ 0.75 V

–

-

i1

9V

400 Ω

+
V0

-

IE
B

–

For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500IB + 400(1 + β)IB
B B

IB = 0.05/81,900 = 0.61 μA
B

v0 = 400IE = 400(1 + β)IB = 49 mV
B

For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = βIB and IE = (1 + β)IB
B B

VCE = 9 – 5kβIB – 400(1 + β)IB = 9 – 0.659 = 8.641 volts
B B

Chapter 3, Problem 92.
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Find IB and VC for the circuit in Fig. 3.128. Let β = 100, VBE = 0.7V.

Figure 3.128
Chapter 3, Solution 92
10 kΩ

I1

5 kΩ

VC IC

IB

+
+ VBE
4 kΩ

VCE –

–

12V

+

+
V0

-

IE

–

I1 = IB + IC = (1 + β)IB and IE = IB + IC = I1 Applying KVL around the outer loop, 4kIE + VBE + 10kIB + 5kI1 = 12 12 – 0.7 = 5k(1 + β)IB + 10kIB + 4k(1 + β)IB = 919kIB IB = 11.3/919k = 12.296 μA Also, 12 = 5kI1 + VC which leads to VC = 12 – 5k(101)IB = 5.791 volts Chapter 3, Problem 93 Rework Example 3.11 with hand calculation.
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In the circuit in Fig. 3.34, determine the currents i1, i2, and i3.

Figure 3.34
Chapter 3, Solution 93
1Ω 4Ω

v1 i1
2Ω

i
2Ω

3v0

v2 i3 +
8Ω

2Ω

3v0

i2
4Ω

i + v0 + + v2 24V

+

+ v1 –

–

–

–

(a)

(b)

From (b), -v1 + 2i – 3v0 + v2 = 0 which leads to i = (v1 + 3v0 – v2)/2 At node 1 in (a), ((24 – v1)/4) = (v1/2) + ((v1 +3v0 – v2)/2) + ((v1 – v2)/1), where v0 = v2 or 24 = 9v1 which leads to v1 = 2.667 volts At node 2, ((v1 – v2)/1) + ((v1 + 3v0 – v2)/2) = (v2/8) + v2/4, v0 = v2 v2 = 4v1 = 10.66 volts Now we can solve for the currents, i1 = v1/2 = 1.333 A, i2 = 1.333 A, and i3 = 2.6667 A.

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