Solution Manual - Fundamentals of Electric Circuits 3rd Edition Chapter 03 PDF

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Chapter 3, Problem 1. Determine Ix in the circuit shown in Fig. 3.50 using nodal analysis. 1 kΩ

4 kΩ Ix

9V

+ _

2 kΩ

+ _

6V

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 Ix =

⎯⎯ → Vx = 6

Vx = 3 mA 2k

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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,

v − v2 − v1 v 1 − = 6+ 1 2 10 5

60 = - 8v1 + 5v2

(1)

At node 2,

v − v2 v2 =3 +6 + 1 2 4

36 = - 2v1 + 3v2

(2)

Solving (1) and (2), v1 = 0 V, v2 = 12 V

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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 v v v + o + o +2+ 0 60 10 20 30

i1 =

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

v0 = 40 V

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

Figure 3.53

Chapter 3, Solution 4 2A

v1 i1 4A

5Ω

i2

v2 i3

10 Ω

10 Ω

i4 5Ω

5A

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

v1 = 20

At node 2, 5 - 2 = v2/(10) + v2/(5)

v2 = 10

i 1 = 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 − v0 v0 + = 2k 5k 4k

v0 = 20 V

Chapter 3, Problem 6. Use nodal analysis to obtain v0 in the circuit in Fig. 3.55.

Figure 3.55 Chapter 3, Solution 6 i 1 + i2 + i3 = 0

v 2 − 12 v0 v 0 − 10 + + =0 4 6 2

or v0 = 8.727 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 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!

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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

i3

5Ω

i2 + V0

3V 2Ω

+

–

+ 4V0 –

– 1Ω

v1 v1 − 3 v 1 − 4v 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 i 1 + i2 + i3 = 0

But

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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 V 1 = 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

v1

i1

+ v0 – 12V

+

–

6Ω

i3 i2

+ v1

8Ω

+ –

–

2v0

At the non-reference node,

12 − v1 v1 v1 − 2v 0 = + 6 3 8

(1)

But -12 + v0 + v1 = 0

v0 = 12 - v1

(2)

Substituting (2) into (1),

12 − v1 v1 3v 1 − 24 = + 6 3 8

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

36 V

Vo

+ _

4Ω

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 2 P4Ω = (18.857+12) /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

1Ω Ix

30 V

+ _

2Ω

5Ω 4 Ix

+ 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

30 V

+ _

1Ω

V1

Vo

2Ω

4 Ix

5Ω

At node 1,

V1 − 30 V1 − 0 V1 − V o =0 + + 10 2 1 16V1 − 10V o = 30

(1)

At node o,

Vo − V1 V −0 − 4Ix + o =0 1 5 − 5V1 + 6Vo − 20Ix = 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

v0

v1

8Ω

2Ω

1Ω

4Ω 40 V

20 V

–

+

+

–

At node 1,

40 − v0 v1 − v 0 +5= 2 1

At node 0,

v1 − v 0 v v + 20 + 5= 0 + 0 2 4 8

v1 + v0 = 70

4v1 - 7v0 = -20

(1)

(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

v0

v1 2Ω

1Ω

4Ω 40 V

8Ω

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)

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

v3 = v2 + 2

(2) (3)

Substituting (1) and (3) into (2), 2 + 6v2 + 60 + 8v2 = 3v2 + 6 v1 = v2 + 10 =

v2 =

− 56 11

54 11

i 0 = 6vi = 29.45 A 2

P65 =

v12 ⎛ 54 ⎞ = v12 G = ⎜ ⎟ 6 = 144.6 W R ⎝ 11 ⎠ 2

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

v2

v1 i0 2A

+

1S

v0

4S

8S

v3 13 V

+

–

–

At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3

(1)

But v1 = v2 + 2v0 and v0 = v2. Hence v1 = 3v2 v3 = 13V

(2) (3)

Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 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 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

2Ω 10 Ω

v2

60 V 60 V

8Ω

3i0

+

–

60 − v1 v 1 v1 − v 2 = + 4 8 2 60 − v 2 v1 − v2 At node 2, 3i0 + + =0 10 2

At node 1,

But i0 =

120 = 7v1 - 4v2

(1)

60 − v1 . 4

Hence

3(60 − v1 ) 60 − v 2 v 1 − v 2 =0 + + 4 10 2

1020 = 5v1 + 12v2

Solving (1) and (2) gives v1 = 53.08 V. Hence i0 =

60 − v1 = 1.73 A 4

(2)

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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

–+ v2

v1 2Ω

5A

v3 2Ω 8Ω

4Ω

10 V

+

+

v1

v3

–

–

(a)

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

At the supernode,

(b)

v 2 − v1 v2 − v3 + 2 2

10 = - v1 + 2v2 - v3

v 2 − v 1 v 2 − v 3 v1 v3 + = + 2 2 4 8

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

v3 = v1 + 10

40 = 2v1 + v3

(1)

(2) (3)

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

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

⎯ ⎯→

⎯ ⎯→

0 = −V1 + 7V2 − 2V3

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

3+

16 = 7V1 − V 2 − 4V3

+

⎯ ⎯→

(1)

(2)

− 36 = 4V 1 + 2V 2 − 7V 3 (3)

⎛ 7 − 1 − 4⎞⎛ V1 ⎞ ⎛ 16 ⎞ ⎟ ⎟⎜ ⎟ ⎜ ⎜ ⎜− 1 7 − 2⎟⎜ V2 ⎟ = ⎜ 0 ⎟ ⎜ 4 2 − 7⎟⎜ V ⎟ ⎜ − 36 ⎟ ⎠ ⎠⎝ 3 ⎠ ⎝ ⎝

⎯ ⎯→

Using MATLAB, ⎡ 10 ⎤ −1 V = A B = ⎢ 4.933 ⎥ ⎥ ⎢ ⎢⎣12.267⎥⎦

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

⎯ ⎯→

AV = B

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