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Chapter 1, Problem 1 How many coulombs are represented by these amounts of electrons: (a) 6.482 × 1017 (b) 1.24 × 1018 (c) 2.46 × 1019 (d) 1.628 × 10 20 Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1...

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Chapter 1, Problem 1 How many coulombs are represented by these amounts of electrons: (a) 6.482 × 1017 (b) 1.24 × 1018 (c) 2.46 × 1019 (d) 1.628 × 10 20

Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C

Chapter 1, Problem 2. Determine the current flowing through an element if the charge flow is given by (a) q(t ) = (3t + 8) mC (b) q(t ) = ( 8t 2 + 4t-2) C

(

)

(c) q (t ) = 3e -t − 5e −2 t nC (d) q(t ) = 10 sin 120π t pC (e) q(t ) = 20e −4 t cos 50t μC

Chapter 1, Solution 2 (a) (b) (c) (d) (e)

i = dq/dt = 3 mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200π cos 120π t pA i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) μ A

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Chapter 1, Problem 3. Find the charge q(t) flowing through a device if the current is: (a) i (t ) = 3A, q(0) = 1C (b) i ( t ) = ( 2t + 5) mA, q(0) = 0 (c) i ( t ) = 20 cos(10t + π / 6) μA, q(0) = 2 μ C (d) i (t ) = 10e −30t sin 40tA, q(0) = 0

Chapter 1, Solution 3 (a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C (b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC

(c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) μ C (d)

10e -30t ( −30 sin 40 t - 40 cos t) 900 + 1600 = − e - 30t (0.16cos40 t + 0.12 sin 40t) C

q(t) = ∫ 10e -30t sin 40t + q(0) =

Chapter 1, Problem 4. A current of 3.2 A flows through a conductor. Calculate how much charge passes through any cross-section of the conductor in 20 seconds.

Chapter 1, Solution 4 q = it = 3.2 x 20 = 64 C

Chapter 1, Problem 5. Determine the total charge transferred over the time interval of 0 ≤ t ≤ 10s when 1 i (t ) = t A. 2 Chapter 1, Solution 5 10

1 t 2 10 q = ∫ idt = ∫ tdt = = 25 C 2 4 0 0 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 1, Problem 6. The charge entering a certain element is shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 ms

Figure 1.23

Chapter 1, Solution 6 (a) At t = 1ms, i =

dq 80 = = 40 A dt 2

(b) At t = 6ms, i =

dq = 0A dt

(c) At t = 10ms, i =

dq 80 = = –20 A dt 4

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Chapter 1, Problem 7. The charge flowing in a wire is plotted in Fig. 1.24. Sketch the corresponding current.

Figure 1.24

Chapter 1, Solution 7

⎡ 25A, dq ⎢ i= = - 25A, dt ⎢ ⎣⎢ 25A,

0< t> 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

12 = − I 2 + 2 I 3

(2)

(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 ⎠ ⎝ ⎠

⎯ ⎯→

AI = B

Using MATLAB,

⎡ 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

I2

b

i1 4A

+

c

1A

I2

6Ω

1A

i2 4A

I3

d

I1

i3

I4

12 Ω

4Ω +–

a

2Ω

I4

I3

0

8V

It is evident that I1 = 4

(1)

For mesh 4,

(2)

12(I4 – I1) + 4(I4 – I3) – 8 = 0

For the supermesh At node c,

6(I2 – I1) + 10 + 2I3 + 4(I3 – I4) = 0 or -3I1 + 3I2 + 3I3 – 2I4 = -5

I2 = I 3 + 1

(3) (4)

Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A At node b,

i1 = I2 – I1 = -1A

At node a,

i2 = 4 – I4 = 0A

At node 0,

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 12 V

+

–

i1

2Ω

i3

+ v2

–

For loop 1, 12 = 4i1 – 2i2 – 2i3 which leads to 6 = 2i1 – i2 – i3

(1)

For loop 2, 0 = 6i2 –2i1 – 2 i3 which leads to 0 = -i1 + 3i2 – i3

(2)

For loop 3, 0 = 6i3 – 2i1 – 2i2 which leads to 0 = -i1 – i2 + 3i3

(3)

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⎥⎦

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

i1

10 Ω

30 Ω

i3

+

–

120 V

For loop 1, 120 + 40i1 – 10i2 = 0, which leads to -12 = 4i1 – i2

(1)

For loop 2, 50i2 – 10i1 – 10i3 = 0, which leads to -i1 + 5i2 – i3 = 0

(2)

For loop 3, -120 – 10i2 + 40i3 = 0, which leads to 12 = -i2 + 4i3

(3)

Solving (1), (2), and (3), we get, i1 = -3A, i2 = 0, and i3 = 3A 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 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

120 V

i2

10 Ω 20 Ω

+ 100V +

i1

–

4v0

i3

+ –

v0

80 Ω

– 2I0 i2

i3

For loop 1, -100 + 30i1 – 20i2 + 4v0 = 0, where v0 = 80i3 or 5 = 1.5i1 – i2 + 16i3

(1)

For the supermesh, 60i2 – 20i1 – 120 + 80i3 – 4 v0 = 0, where v0 = 80i3 or 6 = -i1 + 3i2 – 12i3

(2)

Also, 2I0 = i3 – i2 and I0 = i2, hence, 3i2 = i3

(3)

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

⎡ 3 − 2 32 ⎤ ⎢ − 1 3 − 12⎥ ⎢ ⎥ − 1 ⎥⎦ 3 ⎢⎣ 0

⎡ i1 ⎤ ⎡10⎤ ⎢i ⎥ = ⎢ 6 ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣i 3 ⎥⎦ ⎢⎣ 0 ⎥⎦

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

v2 1Ω

10 V

+

–

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 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 61.

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

Figure 3.105

Chapter 3, Solution 61 v1

20 Ω

v2

10 Ω i0

is

+ v0

–

30 Ω

– + 5v0

At node 1, is = (v1/30) + ((v1 – v2)/20) which leads to 60is = 5v1 – 3v2

40 Ω

(1)

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

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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

100V +

–

A

i1

8 kΩ

i2

B

2 kΩ

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

(1)

At node A,

i1 + 4 = i2

(2)

At node B,

i2 = 2i1 + i3

(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

(1) (2)

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 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 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

i2 10 Ω + −

A

i0 i1

10 Ω

i2

+ –

4i0

i3

40 Ω

100V +

–

2A

0.2V0

i1

For mesh 2,

B

20i2 – 10i1 + 4i0 = 0

i3

(1)

But at node A, io = i1 – i2 so that (1) becomes i1 = (16/6)i2

(2)

For the supermesh, -100 + 50i1 + 10(i1 – i2) – 4i0 + 40i3 = 0 or

50 = 28i1 – 3i2 + 20i3

(3)

At node B,

i3 + 0.2v0 = 2 + i1

(4)

But,

v0 = 10i2 so that (4) becomes i3 = 2 + (2/3)i2

(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

(1)

–6I1 + 16I2 – 8I3 – I4 – I5 = 0

(2)

–8I2...