Research Paper, 41 pages (10000 words)

Solution of fundamental of electric circuits

Subjects: Business, Industries

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Published: September 17, 2022
Updated: September 17, 2022
University / College: Florida Atlantic University - Boca Raton
Language: English
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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. 482×1017 x [-1. 602×10-19 C] = -0. 10384 C (b) q = 1. 24×1018 x [-1. 602×10-19 C] = -0. 19865 C (c) q = 2. 46×1019 x [-1. 602×10-19 C] = -3. 941 C (d) q = 1. 628×1020 x [-1. 602×10-19 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 120? pC (e) q(t ) = 20e ? 4 t cos 50t ? C ( ) 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 = 1200? cos 120? t pA i = dq/dt = ? e ? 4t (80 cos 50 t + 1000 sin 50 t ) ? A PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.

If you are a student using this Manual, you are using it without permission. Chapter 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 + ? / 6) ? A, q(0) = 2 ? 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 q(t) = ? 10e -30t sin 40t + q(0) = (c) q(t) = ? 20 cos (10t + ? / 6 ) + q(0) = (2sin(10t + ? / 6) + 1) ? C (d) 10e -30t ( ? 0 sin 40 t – 40 cos t) 900 + 1600 = ? e – 30t (0. 16cos40 t + 0. 12 sin 40t) C 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 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. 10 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 = (b) At t = 6ms, i = dq 80 = = 40 A dt 2 q = 0A dt dq 80 = = –20 A dt 4 (c) At t = 10ms, i = 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 7. The charge flowing in a wire is plotted in Fig. 1. 24. Sketch the corresponding current. Figure 1. 4 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 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 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

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 55. In the circuit of Fig. 3. 100, solve for i1, i2, and i3. Figure 3. 100 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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 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 56. Determine v1 and v2 in the circuit of Fig. 3. 101. Figure 3. 101 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, Solution 56 + v1 – 2? 2? i2 2? 2? 2? + 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 ? = ? 1 3 ? 1 = 8, ? 2 = ? 1 3 ? 1 = 24 ? 1 ? 1 3 ? 1 0 3 2 ? 1 6 ? 3 = ? 1 3 0 = 24 , therefore i2 = i3 = 24/8 = 3A, ? 1 ? 1 0 v1 = 2i2 = 6 volts, v = 2i3 = 6 volts 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 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 = 4k? x18mA = 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 ? 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 58. Find i1, i2, and i3 the circuit in Fig. 3. 103. Figure 3. 103 Chapter 3, Solution 58 30 ? i2 30 ? 10 ? 10 ? 30 ? i1 + i3 20 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) 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 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, Solution 59 40 ? –+ I0 10 ? 20 ? i2 120 V + 100V + i1 – 4v0 + – 2I0 i2 i3 v0 80 ? – 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 ? 32 3 10 32 3 ? 2 10 ? = ? 1 3 ? 12 = 5, ? 2 = ? 1 6 ? 12 = ? 28, ? 3 = ? 1 3 6 = ? 84 0 3 ? 1 0 0 ? 1 0 3 0 I0 = i2 = ? 2/? = -28/5 = -5. 6 A v0 = 8i3 = (-84/5)80 = -1. 344 kvolts 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 60. Calculate the power dissipated in each resistor in the circuit in Fig. 3. 104. Figure 3. 104 Chapter 3, Solution 60 0. 5i0 4? 10 V 8? v1 1? 10 V + v2 2? – 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 P1? = (v1)2/1 = 2. 041 watts, P2? = (v2)2/2 = 4. 939 watts P4? = (10 – v1)2/4 = 18. 38 watts, P8? = (10 – v2)2/8 = 5. 88 watts PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

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) 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 62. Find the mesh currents i1, i2, and i3 in the network of Fig. 3. 106. Figure 3. 106 Chapter 3, Solution 62 4 k? A 8 k? B 2 k? 100V + – i1 i2 i3 + – 40 V We have a supermesh. Let all R be in k? , 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. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

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 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. (1) (2) Chapter 3, Problem 64. Find vo, and io in the circuit of Fig. 3. 108. Figure 3. 108 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, Solution 64 i1 50 ? A i2 10 ? + ? i0 10 ? i2 i1 100V + + – 4i0 i3 40 ? – 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 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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) 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. 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 gt;> 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. 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 66. Write a set of mesh equations for the circuit in Fig. 3. 110. Use MATLAB to determine the mesh currents. 10 ? 10 ? 8? 12 V + _ 6? 4? I1 + _ 2? I2 24 V + _ 6? 8? 40 V 2? I4 4? 8? 30 V + _ I3 8? 4? 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) 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. Putting (1) to (5) in matrix form ? 30 ? 4 ? 6 ? 2 0 ? ? ? 12 ? ?? 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 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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. 105 -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! 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 68. Find the voltage Vo in the circuit of Fig. 3. 112. 3A 10 ? + 4A 40 ? Vo _ 25 ? 20 ? + _ 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 ? + Vo 25 ? 4A 40 ? Vo _ 20 ? + _ 24 V PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

If you are a student using this Manual, you are using it without permission. 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 ?? ? ? ? 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 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. 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 71. Write the mesh-current equations for the circuit in Fig. 3. 115. Next, determine the values of I1, I2, and I3. 5? I1 I3 3? 10 V + _ 1? 2? 4? I2 + _ 5V Figure 3. 15 For Prob. 3. 71. 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, 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 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 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 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 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 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 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 ? PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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 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. * 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. 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. 2 ix V1 5? V2 5A 2? ix 1? 2A Figure 3. 119 For Prob. 3. 77. Chapter 3, Solution 77 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc.

Chapter 3, Problem 78. 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. Solve Prob. 3. 20 using PSpice. Chapter 3, Problem 20 For the circuit in Fig. 3. 9, find V1, V2, and V3 using nodal analysis. Figure 3. 69 Chapter 3, Solution 78 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. 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. 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 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. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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 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.

Example 3. 4 Find the node voltages in the circuit of Fig. 3. 12. Figure 3. 12 Chapter 3, Solution 81 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.

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. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

This network corresponds to the Netlist. Chapter 3, Problem 83. 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. 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 ? 2 70 ? 3 20 V + – 50 ? 2A 30 ? 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, 3×10-3 + 4000i0 – v0/100 = 0 (2) From (1) and (2) we get, i0 = 0. 5?

A and v0 = 0. 5 volt. Chapter 3, Problem 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. An audio amplifier with resistance 9? 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 = 9? 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. nA, therefore, v0 = -5000x400x37. 4×10-9 = -74. 8 mvolts (2) Chapter 3, Problem 87. For the circuit in Fig. 3. 123, find the gain vo/vs. 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.

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 = -4×10-3v0 (2) (1) Substituting (2) into (1) gives, (vs + 0. 004v1)/2 = -0. 004v0 + (-0. 04v1 – 0. 001v0)/20 This leads to 0. 125v0 = 10vs or (v0/vs) = 10/0. 125 = -80 Chapter 3, Problem 89. 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 the transistor circuit shown in Fig. . 125, find IB and VCE. Let ? = 100 and VBE = 0. 7V. _ 3V + _ 1 k? 0. 7 V + 100 k? | | 15 V Figure 3. 125 For Prob. 3. 89. Chapter 3, Solution 89 Consider the circuit below. _ 0. 7 V C + 100 k? + IC VCE _ 3V + _ E For the left loop, applying KVL gives VBE = 0. 7 ? 3 ? 0. 7 + 100 x103 IB + VBE = 0 ???? IB = 30 ? A > For the right loop, ? VCE + 15 ? Ic(1×10 3 ) = 0 But IC = ? IB = 100 x30 ? A= 3 mA | | 15 V 1 k? VCE = 15 ? 3 x10 ? 3 x103 = 12 V Chapter 3, Problem 90. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

If you are a student using this Manual, you are using it without permission. For the transistor circuit of Fig. 3. 127, find IB, VCE, and vo. Take ? = 200, VBE = 0. 7V. Figure 3. 127 Chapter 3, Solution 91 We first determine the Thevenin equivalent for the input circuit. RTh = 6|| 2 = 6×2/8 = 1. 5 k? and VTh = 2(3)/(2+6) = 0. 75 volts 5 k? IC 1. 5 k? IB + VBE + VCE – i2 + + 0. 75 V – – i1 9V 400 ? + V0 – IE B – For loop 1, -0. 75 + 1. 5kIB + VBE + 400IE = 0 = -0. 75 + 0. 7 + 1500IB + 400(1 + ? )IB B B IB = 0. 05/81, 900 = 0. 61 ? A B v0 = 400IE = 400(1 + ? IB = 49 mV B For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = ? IB and IE = (1 + ? )IB B B VCE = 9 – 5k? IB – 400(1 + ? )IB = 9 – 0. 659 = 8. 641 volts B B Chapter 3, Problem 92. 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. Find IB and VC for the circuit in Fig. 3. 128. Let ? = 100, VBE = 0. 7V. Figure 3. 128 Chapter 3, Solution 92 10 k? I1 5 k? VC IC IB + + VBE 4 k? VCE – – 12V + + V0 – IE – I1 = IB + IC = (1 + ? )IB and IE = IB + IC = I1 Applying KVL around the outer loop, 4kIE + VBE + 10kIB + 5kI1 = 12 12 – 0. 7 = 5k(1 + ? )IB + 10kIB + 4k(1 + ? )IB = 919kIB IB = 11. 3/919k = 12. 296 ? A Also, 12 = 5kI1 + VC which leads to VC = 12 – 5k(101)IB = 5. 791 volts Chapter 3, Problem 93 Rework Example 3. 1 with hand calculation. 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. In the circuit in Fig. 3. 34, determine the currents i1, i2, and i3. Figure 3. 34 Chapter 3, Solution 93 ? 4? v1 i1 2? i 2? 3v0 v2 i3 + 8? 2? 3v0 i2 4? 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. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Find vo in the circuit of Fig. 4. 70. If the source current is reduced to 1 ? A, what is vo? Figure 4. 70 Chapter 4, Solution 2. 6 (4 + 2) = 3? , i1 = i 2 = io = 1 A 2 1 1 i1 = , v o = 2i o = 0. 5V 2 4 If is = 1? A, then vo = 0. 5? 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 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-? resistors is replaced by a 10-? resistor and vs = 10 V? Figure 4. 71 Chapter 4, Solution 3. + ? + vo + ? (a) We transform the Y sub-circuit to the equivalent ? . 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 = 10? 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 6? resistor is 6V so that the current through the 3? resistor is 2A. + v1 3 6 = 2? , vo = 3(4) = 12V, i1 = Hence Is = 3 + 3 = 6A If Is = 6A Is = 9A Io = 1 Io = 9/6 = 1. 5A vo = 3A. 4 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, ? 1? V1 = ? ? + 1 = 2V ? 3? 10 ? 2? Vs = 2? ? + v1 = 3 ? 3? If vs = 10 3 vo = 1 vo = 3 x15 = 4. 5V 10 Then vs = 15 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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 48 1V -6 V Vo 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. . 1? 4? + 4V + _ 3? 2? Vo _ Figure 4. 75 For Prob. 4. 7. Chapter 4, Solution 7. If Vo = 1V, then the current through the 2-? and 4-? resistors is ? = 0. 5. The voltage across the 3-? resistor is ? (4 + 2) = 3 V. The total current through the 1-? resistor is 0. 5 +3/3 = 1. 5 A. Hence the