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Chapter 10, Problem 1. Determine i in the circuit of Fig. 10.50. Figure 10.50 For Prob. 10.1. Chapter 10, Solution 1. We first determine the input impedance. 1H ⎯⎯ → jω L = j1x10 = j10 1F ⎯⎯ → 1 jω C = 1 = − j 0.1 j10 x1 −1 ⎛ 1 1 1⎞ Zin = 1+ ⎜ + + ⎟ = 1.0101− j0.1 = 1.015 < −5.653o ⎝ j10 − j 0.1 1⎠ I= 2 < 0o = 1.9704 < 5.653o 1.015 < −5.653o i(t) = 1.9704 cos(10t + 5.653o ) A = 1.9704cos(10t+5.65˚) 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 10, Problem 2. Solve for V o in Fig. 10.51, using nodal analysis. Figure 10.51 For Prob. 10.2. Chapter 10, Solution 2. Consider the circuit shown below. 2 Vo + 4∠0o V- _ –j5 j4 At the main node, 4 − Vo V V = o + o ⎯⎯ → 40 = Vo (10 + j ) 2 − j5 j 4 40 Vo = = 3.98 < 5.71o A 10 − j 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 10, Problem 3. Determine v o in the circuit of Fig. 10.52. Figure 10.52 For Prob. 10.3. Chapter 10, Solution 3. ω= 4 2 cos(4t ) ⎯ ⎯→ 2∠0° 16 sin(4 t ) ⎯ ⎯→ 16∠ - 90° = -j16 2H ⎯ ⎯→ jωL = j8 1 1 1 12 F ⎯ ⎯→ = = - j3 jωC j (4)(1 12) The circuit is shown below. 4Ω -j16 V -j3 Ω + − Vo 1Ω j8 Ω 6Ω 2∠0° A Applying nodal analysis, - j16 − Vo Vo Vo +2= + 4 − j3 1 6 + j8 ⎛ - j16 1 1 ⎞ ⎟V + 2 = ⎜1 + + 4 − j3 ⎝ 4 − j3 6 + j8 ⎠ o Vo = Therefore, 3.92 − j2.56 4.682∠ - 33.15° = = 3.835∠ - 35.02° 1.22 + j0.04 1.2207 ∠1.88° v o ( t ) = 3.835 cos(4t – 35.02°) 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 10, Problem 4. Determine i1 in the circuit of Fig. 10.53. Figure 10.53 For Prob. 10.4. Chapter 10, Solution 4. ⎯⎯ → 0.5H jω L = j 0.5 x103 = j 500 1 = − j 500 jω C j10 x2 x10 −6 Consider the circuit as shown below. 2µ F 1 ⎯⎯ → I1 50∠0o V = 3 2000 V1 + _ -j500 j500 + – 30I1 At node 1, 50 − V1 30I1 − V1 V + = 1 − j500 2000 j 500 50 − V1 But I1 = 2000 50 − V1 + j 4 x30( I1 = 50 − V1 =0 2000 50 − V1 ) + j 4V1 − j 4V1 = 0 2000 → V1 = 50 i1(t) = 0 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 10, Problem 5. Find io in the circuit of Fig. 10.54. Figure 10.54 For Prob. 10.5. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 10, Solution 5. 0.25H 2µ F jω L = j 0.25 x4 x103 = j1000 ⎯⎯ → ⎯⎯ → 1 jω C = 1 = − j125 j 4 x10 x2 x10 −6 3 Consider the circuit as shown below. Io 2000 Vo 25∠0o V + _ -j125 j1000 + – 10Io At node Vo, Vo − 25 Vo − 0 Vo − 10I o =0 + + 2000 j1000 − j125 Vo − 25 − j2Vo + j16Vo − j160I o = 0 (1 + j14)Vo − j160I o = 25 But Io = (25–Vo)/2000 (1 + j14)Vo − j2 + j0.08Vo = 25 Vo = 25 + j2 25.08∠4.57° 1.7768∠ − 81.37° = 1 + j14.08 14.115∠58.94° Now to solve for io, 25 − Vo 25 − 0.2666 + j1.7567 = = 12.367 + j0.8784 mA 2000 2000 = 12.398∠4.06° Io = io = 12.398cos(4x103t + 4.06˚) 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 10, Problem 6. Determine V x in Fig. 10.55. Figure 10.55 For Prob. 10.6. Chapter 10, Solution 6. Let Vo be the voltage across the current source. Using nodal analysis we get: Vo Vo − 4Vx 20 −3+ = 0 where Vx = Vo 20 20 + j10 20 + j10 Combining these we get: Vo Vo 4Vo − −3+ = 0 → (1 + j0.5 − 3)Vo = 60 + j30 20 20 + j10 20 + j10 Vo = 60 + j30 20(3) or Vx = = 29.11∠–166˚ V. − 2 + j0.5 − 2 + j0.5 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 10, Problem 7. Use nodal analysis to find V in the circuit of Fig. 10.56. Figure 10.56 For Prob. 10.7. Chapter 10, Solution 7. At the main node, 120∠ − 15 o − V V V = 6∠30 o + + 40 + j20 − j30 50 ⎯ ⎯→ 115.91 − j31.058 − 5.196 − j3 = 40 + j20 ⎛ 1 j 1⎞ V⎜⎜ + + ⎟⎟ ⎝ 40 + j20 30 50 ⎠ V= − 3.1885 − j4.7805 = 124.08∠ − 154 o V 0.04 + j0.0233 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 10, Problem 8. Use nodal analysis to find current io in the circuit of Fig. 10.57. Let i s = 6 cos(200t + 15°) A. Figure 10.57 For Prob. 10.8. Chapter 10, Solution 8. ω = 200, 100mH 50µF ⎯ ⎯→ ⎯ ⎯→ jωL = j200x 0.1 = j20 1 1 = = − j100 jωC j200x 50x10 − 6 The frequency-domain version of the circuit is shown below. 0.1 Vo 40 Ω V1 6∠15 o 20 Ω + Vo - Io V2 -j100 Ω j20 Ω 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. At node 1, or V V1 V − V2 6∠15 o + 0.1V1 = 1 + + 1 20 − j100 40 5.7955 + j1.5529 = (−0.025 + j 0.01)V1 − 0.025V2 (1) At node 2, V1 − V2 V = 0.1V1 + 2 40 j20 From (1) and (2), ⎯⎯→ 0 = 3V1 + (1 − j2)V2 ⎡(−0.025 + j0.01) − 0.025⎤⎛ V1 ⎞ ⎛ (5.7955 + j1.5529) ⎞ ⎜ ⎟=⎜ ⎟⎟ ⎢ 3 (1 − j2) ⎥⎦⎜⎝ V2 ⎟⎠ ⎜⎝ 0 ⎠ ⎣ or (2) AV = B Using MATLAB, V = inv(A)*B leads to V1 = −70.63 − j127.23, V2 = −110.3 + j161.09 V − V2 Io = 1 = 7.276∠ − 82.17 o 40 Thus, i o ( t ) = 7.276 cos(200t − 82.17 o ) 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 10, Problem 9. Use nodal analysis to find v o in the circuit of Fig. 10.58. Figure 10.58 For Prob. 10.9. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 10, Solution 9. 10 cos(10 3 t ) ⎯ ⎯→ 10 ∠0°, ω = 10 3 10 mH ⎯ ⎯→ 50 µF ⎯ ⎯→ jωL = j10 1 1 = = - j20 3 jωC j (10 )(50 × 10 -6 ) Consider the circuit shown below. 20 Ω V1 -j20 Ω V2 j10 Ω Io 10∠0° V + − 20 Ω + 4 Io 30 Ω Vo − At node 1, At node 2, 10 − V1 V1 V1 − V2 = + 20 20 - j20 10 = (2 + j) V1 − jV2 (1) V1 − V2 V V2 V1 , where I o = has been substituted. = (4) 1 + 20 - j20 20 30 + j10 (-4 + j) V1 = (0.6 + j0.8) V2 0.6 + j0.8 V1 = V2 (2) -4+ j Substituting (2) into (1) (2 + j)(0.6 + j0.8) 10 = V2 − jV2 -4+ j 170 or V2 = 0.6 − j26.2 Vo = Therefore, 30 3 170 V2 = ⋅ = 6.154∠70.26° 30 + j10 3 + j 0.6 − j26.2 v o ( t ) = 6.154 cos(103 t + 70.26°) 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 10, Problem 10. Use nodal analysis to find v o in the circuit of Fig. 10.59. Let ω = 2 krad/s. Figure 10.59 For Prob. 10.10. Chapter 10, Solution 10. ⎯ ⎯→ 50 mH 2µF ⎯ ⎯→ jωL = j2000x50 x10 − 3 = j100, 1 1 = = − j250 jωC j2000x 2x10 − 6 ω = 2000 Consider the frequency-domain equivalent circuit below. V1 36<0o 2k Ω -j250 j100 V2 0.1V1 4k Ω At node 1, 36 = V1 V V − V2 + 1 + 1 2000 j100 − j250 ⎯ ⎯→ 36 = (0.0005 − j0.006)V1 − j0.004V2 (1) At node 2, V1 − V2 V = 0.1V1 + 2 − j250 4000 ⎯⎯→ 0 = (0.1 − j0.004)V1 + (0.00025 + j0.004)V2 (2) Solving (1) and (2) gives Vo = V2 = −535.6 + j893.5 = 8951.1∠93.43o vo (t) = 8.951 sin(2000t +93.43o) kV 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 10, Problem 11. Apply nodal analysis to the circuit in Fig. 10.60 and determine I o . Figure 10.60 For Prob. 10.11. Chapter 10, Solution 11. Consider the circuit as shown below. –j5 Ω Io 2Ω 2Ω V1 o 4∠0 V + _ V2 j8 Ω 2Io 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. At node 1, V1 − 4 V − V2 =0 − 2I o + 1 2 2 V1 − 0.5V2 − 2I o = 2 But, Io = (4–V2)/(–j5) = –j0.2V2 + j0.8 Now the first node equation becomes, V1 – 0.5V2 + j0.4V2 – j1.6 = 2 or V1 + (–0.5+j0.4)V2 = 2 + j1.6 At node 2, V2 − V1 V2 − 4 V2 − 0 + + =0 − j5 2 j8 –0.5V1 + (0.5 + j0.075)V2 = j0.8 Using MATLAB to solve this, we get, >> Y=[1,(-0.5+0.4i);-0.5,(0.5+0.075i)] Y= 1.0000 -0.5000 -0.5000 + 0.4000i 0.5000 + 0.0750i >> I=[(2+1.6i);0.8i] I= 2.0000 + 1.6000i 0 + 0.8000i >> V=inv(Y)*I V= 4.8597 + 0.0543i 4.9955 + 0.9050i Io = –j0.2V2 + j0.8 = –j0.9992 + 0.01086 + j0.8 = 0.01086 – j0.1992 = 199.5∠86.89˚ 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 10, Problem 12. By nodal analysis, find io in the circuit of Fig. 10.61. Figure 10.61 For Prob. 10.12. Chapter 10, Solution 12. 20 sin(1000t ) ⎯ ⎯→ 20 ∠0°, ω = 1000 ⎯→ 10 mH ⎯ 50 µF ⎯ ⎯→ jωL = j10 1 1 = = - j20 3 jωC j (10 )(50 × 10 -6 ) The frequency-domain equivalent circuit is shown below. 2 Io V1 10 Ω V2 Io 20∠0° A 20 Ω -j20 Ω j10 Ω 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. At node 1, 20 = 2 I o + V1 V1 − V2 + , 20 10 where V2 j10 2V2 V1 V1 − V2 20 = + + j10 20 10 400 = 3V1 − (2 + j4) V2 (1) Io = At node 2, or V V 2V2 V1 − V2 + = 2 + 2 j10 10 - j20 j10 j2 V1 = (-3 + j2) V2 V1 = (1 + j1.5) V2 (2) Substituting (2) into (1), 400 = (3 + j4.5) V2 − (2 + j4) V2 = (1 + j0.5) V2 Therefore, V2 = 400 1 + j0.5 Io = V2 40 = = 35.74 ∠ - 116.6° j10 j (1 + j0.5) i o ( t ) = 35.74 sin(1000t – 116.6°) 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 10, Problem 13. Determine V x in the circuit of Fig. 10.62 using any method of your choice. Figure 10.62 For Prob. 10.13. Chapter 10, Solution 13. Nodal analysis is the best approach to use on this problem. We can make our work easier by doing a source transformation on the right hand side of the circuit. –j2 Ω 18 Ω j6 Ω + 40∠30º V + − Vx 3Ω 50∠0º V + − − Vx − 40∠30° Vx Vx − 50 + + =0 − j2 3 18 + j6 which leads to Vx = 29.36∠62.88˚ 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 10, Problem 14. Calculate the voltage at nodes 1 and 2 in the circuit of Fig. 10.63 using nodal analysis. Figure 10.63 For Prob. 10.14. Chapter 10, Solution 14. At node 1, 0 − V1 0 − V1 V2 − V1 + + = 20∠30° - j2 10 j4 - (1 + j2.5) V1 − j2.5 V2 = 173.2 + j100 At node 2, V2 V2 V2 − V1 + + = 20∠30° j2 - j5 j4 - j5.5 V2 + j2.5 V1 = 173.2 + j100 (1) (2) Equations (1) and (2) can be cast into matrix form as ⎡1 + j2.5 j2.5 ⎤⎡ V1 ⎤ ⎡ - 200 ∠30°⎤ = ⎢ j2.5 - j5.5⎥⎦⎢⎣ V2 ⎥⎦ ⎢⎣ 200 ∠30° ⎥⎦ ⎣ ∆= 1 + j2.5 j2.5 j2.5 - j5.5 ∆1 = = 20 − j5.5 = 20.74∠ - 15.38° - 200 ∠30° j2.5 = j3 (200∠30°) = 600∠120° 200 ∠30° - j5.5 1 + j2.5 - 200∠30° = (200 ∠30°)(1 + j5) = 1020∠108.7° j2.5 200∠30° ∆1 V1 = = 28.93∠135.38° ∆ ∆2 V2 = = 49.18∠124.08° ∆ ∆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 10, Problem 15. Solve for the current I in the circuit of Fig. 10.64 using nodal analysis. Figure 10.64 For Prob. 10.15. 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 10, Solution 15. We apply nodal analysis to the circuit shown below. 5A 2Ω jΩ V1 V2 I -j20 V + − -j2 Ω 2I 4Ω At node 1, V V − V2 - j20 − V1 = 5+ 1 + 1 2 - j2 j - 5 − j10 = (0.5 − j0.5) V1 + j V2 (1) At node 2, V1 − V2 V2 , = j 4 V where I = 1 - j2 5 V2 = V1 0.25 − j 5 + 2I + (2) Substituting (2) into (1), j5 = 0.5 (1 − j) V1 0.25 − j j40 (1 − j) V1 = -10 − j20 − 1 − j4 160 j40 ( 2 ∠ - 45°) V1 = -10 − j20 + − 17 17 V1 = 15.81∠313.5° - 5 − j10 − V1 = (0.5∠90°)(15.81∠313.5°) - j2 I = 7.906∠43.49° A 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 10, Problem 16. Use nodal analysis to find V x in the circuit shown in Fig. 10.65. Figure 10.65 For Prob. 10.16. Chapter 10, Solution 16. Consider the circuit as shown in the figure below. j4 Ω V1 V2 + Vx – 2∠0o A 5Ω –j3 Ω 3∠45o 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. At node 1, V − 0 V1 − V2 −2+ 1 + =0 5 j4 (0.2 − j0.25)V1 + j0.25V2 = 2 (1) At node 2, V2 − V1 V2 − 0 + − 3∠45° = 0 j4 − j3 j0.25V1 + j0.08333V2 = 2.121 + j2.121 In matrix form, (1) and (2) become ⎡0.2 − j0.25 ⎢ j0.25 ⎣ (2) j0.25 ⎤ ⎡ V1 ⎤ ⎡ 2 ⎤ =⎢ ⎢ ⎥ ⎥ j0.08333⎦ ⎣V2 ⎦ ⎣2.121 + j2.121⎥⎦ Solving this using MATLAB, we get, >> Y=[(0.2-0.25i),0.25i;0.25i,0.08333i] Y= 0.2000 - 0.2500i 0 + 0.2500i 0 + 0.2500i 0 + 0.0833i >> I=[2;(2.121+2.121i)] I= 2.0000 2.1210 + 2.1210i >> V=inv(Y)*I V= 5.2793 - 5.4190i 9.6145 - 9.1955i Vs = V1 – V2 = –4.335 + j3.776 = 5.749∠138.94˚ 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 10, Problem 17. By nodal analysis, obtain current I o in the circuit of Fig. 10.66. Figure 10.66 For Prob. 10.17. Chapter 10, Solution 17. Consider the circuit below. j4 Ω 100∠20° V + − 1Ω Io V1 3Ω 2Ω V2 -j2 Ω 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. At node 1, 100∠20° − V1 V1 V1 − V2 = + j4 3 2 V1 100 ∠20° = (3 + j10) − j2 V2 3 (1) At node 2, 100∠20° − V2 V1 − V2 V2 + = 1 2 - j2 100 ∠20° = -0.5 V1 + (1.5 + j0.5) V2 (2) From (1) and (2), ⎡100∠20°⎤ ⎡ - 0.5 0.5 (3 + j) ⎤⎡ V1 ⎤ ⎢100∠20°⎥ = ⎢1 + j10 3 - j2 ⎥⎦⎢⎣ V2 ⎥⎦ ⎣ ⎦ ⎣ ∆= - 0.5 1.5 + j0.5 = 0.1667 − j4.5 1 + j10 3 - j2 ∆1 = ∆2 = 100∠20° 1.5 + j0.5 100∠20° - j2 = -55.45 − j286.2 - 0.5 100∠20° = -26.95 − j364.5 1 + j10 3 100∠20° ∆1 = 64.74 ∠ - 13.08° ∆ ∆2 V2 = = 81.17 ∠ - 6.35° ∆ V1 − V2 ∆ 1 − ∆ 2 - 28.5 + j78.31 = = Io = 2 2∆ 0.3333 − j 9 I o = 9.25∠-162.12° A V1 = 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 10, Problem 18. Use nodal analysis to obtain V o in the circuit of Fig. 10.67 below. Figure 10.67 For Prob. 10.18. Chapter 10, Solution 18. Consider the circuit shown below. 8Ω V1 j6 Ω 4Ω V2 j5 Ω + 4∠45° A 2Ω Vx − + 2 Vx -j Ω -j2 Ω Vo − 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. At node 1, V1 V1 − V2 + 2 8 + j6 200 ∠45° = (29 − j3) V1 − (4 − j3) V2 (1) 4∠45° = At node 2, V1 − V2 V V2 , + 2Vx = 2 + 8 + j6 - j 4 + j5 − j2 (104 − j3) V1 = (12 + j41) V2 12 + j41 V1 = V 104 − j3 2 (2) where Vx = V1 Substituting (2) into (1), (12 + j41) V − (4 − j3) V2 104 − j3 2 200 ∠45° = (14.21∠89.17°) V2 200∠45° V2 = 14.21∠89.17° 200∠45° = (29 − j3) - j2 - j2 - 6 − j8 V2 = V2 = V2 4 + j5 − j2 4 + j3 25 10∠233.13° 200∠45° Vo = ⋅ 25 14.21∠89.17° Vo = 5.63∠189° V Vo = 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 10, Problem 19. Obtain V o in Fig. 10.68 using nodal analysis. Figure 10.68 For Prob. 10.19. 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 10, Solution 19. We have a supernode as shown in the circuit below. j2 Ω V1 V2 4Ω V3 + 2Ω Vo -j4 Ω 0.2 Vo − Notice that Vo = V1 . At the supernode, V3 − V2 V2 V1 V1 − V3 = + + 4 - j4 2 j2 0 = (2 − j2) V1 + (1 + j) V2 + (-1 + j2) V3 At node 3, V1 − V3 V3 − V2 0.2V1 + = j2 4 (0.8 − j2) V1 + V2 + (-1 + j2) V3 = 0 (1) (2) Subtracting (2) from (1), 0 = 1.2V1 + j V2 (3) But at the supernode, V1 = 12 ∠0° + V2 V2 = V1 − 12 or (4) Substituting (4) into (3), 0 = 1.2V1 + j (V1 − 12) j12 V1 = = Vo 1.2 + j 12∠90° 1.562∠39.81° Vo = 7.682∠50.19° V Vo = 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 10, Problem 20. Refer to Fig. 10.69. If v s (t ) = Vm sin ωt and vo (t ) = A sin (ωt + φ ) derive the expressions for A and φ Figure 10.69 For Prob. 10.20. Chapter 10, Solution 20. The circuit is converted to its frequency-domain equivalent circuit as shown below. R + Vm∠0° + − jωL Vo 1 jωC − Let 1 = Z = jωL || jωC L C = jωL 1 − ω2 LC 1 jωC jωL jωL Z 1 − ω2 LC V Vm = Vm = Vo = jωL R (1 − ω2 LC) + jωL m R+Z R+ 1 − ω2 LC ⎛ ⎞ ωL Vm ωL ⎜90° − tan -1 ⎟ ∠ Vo = R (1 − ω2 LC) ⎠ R 2 (1 − ω2 LC) 2 + ω2 L2 ⎝ jωL + If Vo = A∠φ , then ωL Vm A= R 2 (1 − ω 2 LC) 2 + ω 2 L2 and φ = 90° − tan -1 ωL R (1 − ω 2 LC) 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 10, Problem 21. For each of the circuits in Fig. 10.70, find V o /V i for ω = 0, ω → ∞, and ω 2 = 1 / LC . Figure 10.70 For Prob. 10.21. Chapter 10, Solution 21. (a) Vo = Vi 1 jωC R + jωL + 1 jωC As ω → ∞ , (b) Vo = Vi 1 LC Vo = Vi , jωL R + jωL + As ω → ∞ , 1 LC 1 jωC 1 jRC ⋅ 1 = -j L R C LC − ω2 LC = 1 − ω2 LC + jωRC Vo = 0 Vi Vo 1 = = 1 Vi 1 At ω = 0 , At ω = 1 1 − ω LC + jωRC 2 Vo 1 = = 1 Vi 1 Vo = 0 Vi At ω = 0 , At ω = = , Vo = Vi −1 jRC ⋅ 1 = j L R C LC 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 10, Problem 22. For the circuit in Fig. 10.71, determine V o /V s . Figure 10.71 For Prob. 10.22. Chapter 10, Solution 22. Consider the circuit in the frequency domain as shown below. R1 R2 Vs + − 1 jωC jωL Let Z = (R 2 + jωL) || + Vo − 1 jωC 1 (R + jωL) R 2 + jωL jωC 2 Z= = 1 1 + jωR 2 − ω2 LC R 2 + jωL + jωC R 2 + jωL Vo 1 − ω2 LC + jωR 2 C Z = = R 2 + jωL Vs Z + R 1 R1 + 1 − ω2 LC + jωR 2 C Vo R 2 + jωL = 2 Vs R 1 + R 2 − ω LCR 1 + jω (L + R 1 R 2 C) 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 10, Problem 23. Using nodal analysis obtain V in the circuit of Fig. 10.72. Figure 10.72 For Prob. 10.23. Chapter 10, Solution 23. V − Vs V + + jωCV = 0 1 R jωL + jω C V+ jωRCV − ω2LC + 1 + jωRCV = Vs ⎛ 1 − ω2LC + jωRC + jωRC − jω3RLC2 ⎞ ⎜ ⎟ V = Vs 2 ⎟ ⎜ 1 − ω LC ⎝ ⎠ V= (1 − ω2 LC)Vs 1 − ω2LC + jωRC(2 − ω2LC) 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 10, Problem 24. Use mesh analysis to find V o in the circuit of Prob. 10.2. Chapter 10, Solution 24. Consider the circuit as shown below. 2Ω + o 4∠0 V I1 + _ j4 Ω –j5 Ω Vo I2 – For mesh 1, 4 = (2 − j 5) I1 + j 5 I1 For mesh 2, (1) 0 = j 5I 1 + ( j 4 − j 5) I 2 ⎯⎯ → I1 = 1 I2 5 (2) Substituting (2) into (1), 1 4 = (2 − j 5) I 2 + j 5I 2 5 Vo = j 4 I 2 = ⎯⎯ → I2 = 1 0.1 + j j4 = 3.98 < 5.71o V 0.1 + j 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 10, Problem 25. Solve for io in Fig. 10.73 using mesh analysis. Figure 10.73 For Prob. 10.25. 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 10, Solution 25. ω= 2 10 cos(2t ) ⎯ ⎯→ 10∠0° 6 sin(2t ) ⎯ ⎯→ 6 ∠ - 90° = -j6 2H ⎯ ⎯→ jωL = j4 1 1 0.25 F ⎯ ⎯→ = = - j2 jωC j (2)(1 4) The circuit is shown below. 4Ω j4 Ω Io 10∠0° V + − I1 -j2 Ω I2 + − 6∠-90° V For loop 1, - 10 + (4 − j2) I 1 + j2 I 2 = 0 5 = (2 − j) I 1 + j I 2 (1) For loop 2, j2 I 1 + ( j4 − j2) I 2 + (- j6) = 0 I1 + I 2 = 3 (2) In matrix form (1) and (2) become ⎡ 2 − j j ⎤ ⎡ I 1 ⎤ ⎡ 5⎤ ⎢ 1 1 ⎥ ⎢ I ⎥ = ⎢ 3⎥ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦ ∆ = 2 (1 − j) , ∆ 1 = 5 − j3 , ∆ 2 = 1 − j3 ∆1 − ∆ 2 4 = = 1 + j = 1.414 ∠45° 2 (1 − j) ∆ i o ( t ) = 1.4142 cos(2t + 45°) A I o = I1 − I 2 = Therefore, 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 10, Problem 26. Use mesh analysis to find current io in the circuit of Fig. 10.74. Figure 10.74 For Prob. 10.26. Chapter 10, Solution 26. 0.4H ⎯⎯ → 1µ F ⎯⎯ → jω L = j103 x0.4 = j400 1 jωC = 1 = − j1000 j10 x10−6 3 20sin103 t = 20 cos(103 t − 90o ) ⎯⎯ → 20 < −90 = − j 20 The circuit becomes that shown below. 2 kΩ –j1000 Io 10∠0o + _ + _ I1 j400 –j20 I2 For loop 1, −10 + (12000 + j 400) I1 − j 400 I 2 = 0 ⎯⎯ → 1 = (200 + j 40) I1 − j 40 I 2 (1) For loop 2, − j 20 + ( j 400 − j1000) I 2 − j 400 I1 = 0 ⎯⎯ → −12 = 40 I1 + 60 I 2 (2) In matrix form, (1) and (2) become ⎡ 1 ⎤ ⎡ 200 + j 40 − j 40 ⎤ ⎡ I1 ⎤ ⎢ −12 ⎥ = ⎢ 40 60 ⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎣ ⎦ ⎣ Solving this leads to I1 =0.0025-j0.0075, I2 = -0.035+j0.005 I o = I1 − I 2 = 0.0375 − j 0.0125 = 39.5 < −18.43 mA io = 39.5cos(103 t − 18.43o ) 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 10, Problem 27. Using mesh analysis, find I 1 and I 2 in the circuit of Fig. 10.75. Figure 10.75 For Prob. 10.27. Chapter 10, Solution 27. For mesh 1, - 40 ∠30° + ( j10 − j20) I 1 + j20 I 2 = 0 4 ∠30° = - j I 1 + j2 I 2 (1) 50 ∠0° + (40 − j20) I 2 + j20 I 1 = 0 5 = - j2 I 1 − (4 − j2) I 2 (2) For mesh 2, From (1) and (2), ⎡ 4∠30°⎤ ⎡ - j j2 ⎤⎡ I 1 ⎤ ⎢ 5 ⎥ = ⎢ - j2 - (4 − j2) ⎥⎢ I ⎥ ⎦⎣ 2 ⎦ ⎣ ⎦ ⎣ ∆ = -2 + 4 j = 4.472∠116.56° ∆ 1 = -(4 ∠30°)(4 − j2) − j10 = 21.01∠211.8° ∆ 2 = - j5 + 8∠120° = 4.44 ∠154.27° I1 = ∆1 = 4.698∠95.24° A ∆ I2 = ∆2 = 0.9928∠37.71° 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 10, Problem 28. In the circuit of Fig. 10.76, determine the mesh currents i1 and i2 . Let v1 = 10 cos 4t V and v 2 = 20 cos(4t − 30°) V. Figure 10.76 For Prob. 10.28. 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 10, Solution 28. 1 1 = = − j0.25 jωC j1x 4 The frequency-domain version of the circuit is shown below, where 1H ⎯ ⎯→ V1 = 10∠0 o , jωL = j4, ⎯ ⎯→ 1F V2 = 20∠ − 30 o . 1 j4 j4 1 -j0.25 + + V1 - I1 V1 = 10∠0 o , 1 I2 V2 - V2 = 20∠ − 30 o Applying mesh analysis, 10 = (2 + j3.75)I1 − (1 − j0.25)I 2 (1) − 20∠ − 30 o = −(1 − j0.25)I1 + (2 + j3.75)I 2 (2) From (1) and (2), we obtain 10 ⎛ ⎞ ⎛ 2 + j3.75 − 1 + j0.25 ⎞⎛ I1 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ − 17.32 + j10 ⎠ ⎝ − 1 + j0.25 2 + j3.75 ⎠⎝ I 2 ⎠ Solving this leads to I1 = 2.741∠ − 41.07 o , I 2 = 4.114∠92 o Hence, i1(t) = 2.741cos(4t–41.07˚)A, i2(t) = 4.114cos(4t+92˚)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 10, Problem 29. By using mesh analysis, find I 1 and I 2 in the circuit depicted in Fig. 10.77. Figure 10.77 For Prob. 10.29. Chapter 10, Solution 29. For mesh 1, (5 + j5) I 1 − (2 + j) I 2 − 30 ∠20° = 0 30 ∠20° = (5 + j5) I 1 − (2 + j) I 2 (1) For mesh 2, (5 + j3 − j6) I 2 − (2 + j) I 1 = 0 0 = - (2 + j) I 1 + (5 − j3) I 2 (2) From (1) and (2), ⎡30∠20°⎤ ⎡ 5 + j5 - (2 + j) ⎤⎡ I 1 ⎤ ⎢ 0 ⎥ = ⎢ - (2 + j) 5 - j3 ⎥⎢ I ⎥ ⎣ ⎦ ⎣ ⎦⎣ 2 ⎦ ∆ = 37 + j6 = 37.48∠9.21° ∆ 1 = (30 ∠20°)(5.831∠ - 30.96°) = 175∠ - 10.96° ∆ 2 = (30 ∠20°)(2.356 ∠26.56°) = 67.08∠46.56° I1 = ∆1 = 4.67∠–20.17° A ∆ I2 = ∆2 = 1.79∠37.35° 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 10, Problem 30. Use mesh analysis to find vo in the circuit of Fig. 10.78. Let v s1 = 120 cos(100t + 90°) V, v s 2 = 80 cos 100t V. Figure 10.78 For Prob. 10.30. Chapter 10, Solution 30. 300mH ⎯⎯ → jω L = j100 x300 x10 −3 = j30 200mH ⎯⎯ → jω L = j100 x200 x10 −3 = j20 400mH ⎯⎯ → jω L = j100 x400 x10 −3 = j40 1 = − j200 j100 x50 x10−6 The circuit becomes that shown below. 50 µ F ⎯⎯ → 1 jω C = j40 j20 20 Ω + 120∠90o + _ 10 Ω I1 j30 –j200 vo I2 - I3` + _ 80∠0o 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 mesh 1, −120 < 90o + (20 + j30) I1 − j 30 I 2 = 0 ⎯⎯ → j120 = (20 + j 30) I1 − j 30 I 2 For mesh 2, − j 30 I1 + ( j 30 + j 40 − j 200) I 2 + j 200 I 3 = 0 ⎯⎯ → 0 = −3I1 − 13I 2 + 20 I 3 For mesh 3, 80 + j200I 2 + (10 − j180)I 3 = 0 → −8 = j20I 2 + (1 − j18)I 3 (3) (1) (2) We put (1) to (3) in matrix form. 0 ⎤ ⎡ I1 ⎤ ⎡ j12⎤ ⎡2 + j3 − j3 ⎢ − 3 − 13 20 ⎥⎥ ⎢⎢I 2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ ⎢ ⎢⎣ 0 j20 1 − j18⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ − 8⎥⎦ This is an excellent candidate for MATLAB. >> Z=[(2+3i),-3i,0;-3,-13,20;0,20i,(1-18i)] Z= 2.0000 + 3.0000i 0 - 3.0000i 0 -3.0000 -13.0000 20.0000 0 0 +20.0000i 1.0000 -18.0000i >> V=[12i;0;-8] V= 0 +12.0000i 0 -8.0000 >> I=inv(Z)*V I= 2.0557 + 3.5651i 0.4324 + 2.1946i 0.5894 + 1.9612i Vo = –j200(I2 – I3) = –j200(–0.157+j0.2334) = 46.68 + j31.4 = 56.26∠33.93˚ vo = 56.26cos(100t + 33.93˚ 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 10, Problem 31. Use mesh analysis to determine current I o in the circuit of Fig. 10.79 below. Figure 10.79 For Prob. 10.31. 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 10, Solution 31. Consider the network shown below. 80 Ω 100∠120° V + − I1 Io -j40 Ω j60 Ω I2 -j40 Ω 20 Ω I3 + − 60∠-30° V For loop 1, - 100∠120° + (80 − j40) I1 + j40 I 2 = 0 10 ∠20° = 4 (2 − j) I 1 + j4 I 2 (1) j40 I 1 + ( j60 − j80) I 2 + j40 I 3 = 0 0 = 2 I1 − I 2 + 2 I 3 (2) 60∠ - 30° + (20 − j40) I 3 + j40 I 2 = 0 - 6∠ - 30° = j4 I 2 + 2 (1 − j2) I 3 (3) For loop 2, For loop 3, From (2), 2 I 3 = I 2 − 2 I1 Substituting this equation into (3), - 6 ∠ - 30° = -2 (1 − j2) I 1 + (1 + j2) I 2 (4) From (1) and (4), ⎡ 10∠120° ⎤ ⎡ 4 (2 − j) j4 ⎤⎡ I 1 ⎤ = ⎢ - 6∠ - 30°⎥ ⎢ - 2 (1 − j2) 1 + j2⎥⎢ I ⎥ ⎣ ⎦ ⎣ ⎦⎣ 2 ⎦ ∆= ∆2 = 8 − j4 - j4 = 32 + j20 = 37.74∠32° - 2 + j4 1 + j2 8 − j4 10∠120° = -4.928 + j82.11 = 82.25∠93.44° - 2 + j4 - 6∠ - 30° Io = I2 = ∆2 = 2.179∠61.44° 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 10, Problem 32. Determine V o and I o in the circuit of Fig. 10.80 using mesh analysis. Figure 10.80 For Prob. 10.32. Chapter 10, Solution 32. Consider the circuit below. j4 Ω Io + 4∠-30° V 2Ω Vo I1 + 3 Vo I2 -j2 Ω − For mesh 1, where (2 + j4) I 1 − 2 (4∠ - 30°) + 3 Vo = 0 Vo = 2 (4∠ - 30° − I 1 ) Hence, (2 + j4) I 1 − 8∠ - 30° + 6 (4 ∠ - 30° − I 1 ) = 0 4 ∠ - 30° = (1 − j) I 1 I 1 = 2 2 ∠15° or Io = 3 Vo 3 = (2)(4∠ - 30° − I 1 ) - j2 - j2 I o = j3 (4 ∠ - 30° − 2 2 ∠15°) I o = 8.485∠15° A Vo = - j2 I o = 5.657∠-75° V 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 10, Problem 33. Compute I in Prob. 10.15 using mesh analysis. Chapter 10, Solution 33. Consider the circuit shown below. 5A I4 2Ω jΩ I -j20 V + − I1 -j2 Ω I2 2I I3 4Ω For mesh 1, j20 + (2 − j2) I 1 + j2 I 2 = 0 (1 − j) I 1 + j I 2 = - j10 For the supermesh, ( j − j2) I 2 + j2 I 1 + 4 I 3 − j I 4 = 0 Also, I 3 − I 2 = 2 I = 2 (I 1 − I 2 ) I 3 = 2 I1 − I 2 (1) (2) (3) For mesh 4, I4 = 5 Substituting (3) and (4) into (2), (8 + j2) I 1 − (- 4 + j) I 2 = j5 (4) (5) Putting (1) and (5) in matrix form, ⎡ 1− j j ⎤⎡ I 1 ⎤ ⎡ - j10 ⎤ ⎢8 + j2 4 − j⎥⎢ I ⎥ = ⎢ j5 ⎥ ⎦ ⎣ ⎦⎣ 2 ⎦ ⎣ ∆ = -3 − j5 , ∆ 1 = -5 + j40 , ∆ 2 = -15 + j85 ∆ − ∆ 2 10 − j45 I = I1 − I 2 = 1 = = 7.906∠43.49° A ∆ - 3 − j5 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 10, Problem 34. Use mesh analysis to find I o in Fig. 10.28 (for Example 10.10). Chapter 10, Solution 34. The circuit is shown below. Io I2 5Ω 3A 20 Ω 8Ω 40∠90° V + − -j2 Ω I3 10 Ω I1 j15 Ω j4 Ω For mesh 1, - j40 + (18 + j2) I 1 − (8 − j2) I 2 − (10 + j4) I 3 = 0 For the supermesh, (13 − j2) I 2 + (30 + j19) I 3 − (18 + j2) I 1 = 0 (1) (2) Also, I2 = I3 − 3 (3) Adding (1) and (2) and incorporating (3), - j40 + 5 (I 3 − 3) + (20 + j15) I 3 = 0 3 + j8 I3 = = 1.465∠38.48° 5 + j3 I o = I 3 = 1.465∠38.48° 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 10, Problem 35. Calculate I o in Fig. 10.30 (for Practice Prob. 10.10) using mesh analysis. Chapter 10, Solution 35. Consider the circuit shown below. 4Ω j2 Ω I3 8Ω 1Ω -j3 Ω 10 Ω 20 V + − I1 -j4 A I2 -j5 Ω For the supermesh, - 20 + 8 I 1 + (11 − j8) I 2 − (9 − j3) I 3 = 0 (1) I 1 = I 2 + j4 (2) Also, For mesh 3, (13 − j) I 3 − 8 I 1 − (1 − j3) I 2 = 0 Substituting (2) into (1), (19 − j8) I 2 − (9 − j3) I 3 = 20 − j32 Substituting (2) into (3), - (9 − j3) I 2 + (13 − j) I 3 = j32 From (4) and (5), ⎡ 19 − j8 - (9 − j3) ⎤⎡ I 2 ⎤ ⎡ 20 − j32 ⎤ ⎢ - (9 − j3) 13 − j ⎥⎢ I ⎥ = ⎢ j32 ⎥ ⎦ ⎣ ⎦⎣ 3 ⎦ ⎣ ∆ = 167 − j69 , (3) (4) (5) ∆ 2 = 324 − j148 ∆ 2 324 − j148 356.2∠ - 24.55° = = ∆ 167 − j69 180.69∠ - 22.45° I 2 = 1.971∠-2.1° A I2 = 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 10, Problem 36. Compute V o in the circuit of Fig. 10.81 using mesh analysis. Figure 10.81 For Prob. 10.36. Chapter 10, Solution 36. Consider the circuit below. j4 Ω -j3 Ω + 4∠90° A I1 2Ω Vo I2 + − 12∠0° V − 2Ω 2Ω I3 2∠0° A Clearly, I 1 = 4 ∠90° = j4 and I 3 = -2 For mesh 2, (4 − j3) I 2 − 2 I 1 − 2 I 3 + 12 = 0 (4 − j3) I 2 − j8 + 4 + 12 = 0 - 16 + j8 I2 = = -3.52 − j0.64 4 − j3 Thus, Vo = 2 (I 1 − I 2 ) = (2)(3.52 + j4.64) = 7.04 + j9.28 Vo = 11.648∠52.82° 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 10, Problem 37. Use mesh analysis to find currents I 1 , I 2 , and I 3 in the circuit of Fig. 10.82. Figure 10.82 For Prob. 10.37. 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 10, Solution 37. I1 + 120∠ − 90 o V - 120∠ − 30 V + Ix Z Z=80-j35 Ω I2 Iz Iy o Z I3 For mesh x, ZI x − ZI z = − j120 (1) ZI y − ZI z = −120∠30 o = −103.92 + j60 (2) − ZI x − ZI y + 3ZI z = 0 (3) For mesh y, For mesh z, Putting (1) to (3) together leads to the following matrix equation: − j120 0 (−80 + j35) ⎞⎛ I x ⎞ ⎛ ⎛ (80 − j35) ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ 0 (80 − j35) (−80 + j35) ⎟⎜ I y ⎟ = ⎜ − 103.92 + j60 ⎟ ⎜ ⎜ (−80 + j35) (−80 + j35) (240 − j105) ⎟⎜ I ⎟ ⎜ ⎟ 0 ⎝ ⎠⎝ z ⎠ ⎝ ⎠ ⎯ ⎯→ AI = B Using MATLAB, we obtain ⎛ - 0.2641 − j2.366 ⎞ ⎜ ⎟ I = inv(A) * B = ⎜ - 2.181 - j0.954 ⎟ ⎜ - 0.815 − j1.1066 ⎟ ⎝ ⎠ I1 = I x = −0.2641 − j2.366 = 2.38∠ − 96.37 o A I 2 = I y − I x = −1.9167 + j1.4116 = 2.38∠143.63o A I 3 = − I y = 2.181 + j0.954 = 2.38∠23.63o 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 10, Problem 38. Using mesh analysis, obtain I o in the circuit shown in Fig. 10.83. Figure 10.83 For Prob. 10.38. 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 10, Solution 38. Consider the circuit below. Io I1 2∠0° A 2Ω j2 Ω 1Ω I2 + − -j4 Ω 4∠0° A I3 I4 10∠90° V 1Ω A Clearly, I1 = 2 (1) (2 − j4) I 2 − 2 I 1 + j4 I 4 + 10 ∠90° = 0 (2) For mesh 2, Substitute (1) into (2) to get (1 − j2) I 2 + j2 I 4 = 2 − j5 For the supermesh, (1 + j2) I 3 − j2 I 1 + (1 − j4) I 4 + j4 I 2 = 0 j4 I 2 + (1 + j2) I 3 + (1 − j4) I 4 = j4 At node A, I3 = I4 − 4 Substituting (4) into (3) gives j2 I 2 + (1 − j) I 4 = 2 (1 + j3) From (2) and (5), ⎡1 − j2 j2 ⎤⎡ I 2 ⎤ ⎡ 2 − j5⎤ ⎢ j2 1 − j⎥⎢ I ⎥ = ⎢ 2 + j6⎥ ⎦ ⎣ ⎦⎣ 4 ⎦ ⎣ ∆ = 3 − j3 , (3) (4) (5) ∆ 1 = 9 − j11 - ∆ 1 - (9 − j11) 1 = = (-10 + j) ∆ 3 − j3 3 I o = 3.35∠174.3° A Io = -I2 = 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 10, Problem 39. Find I 1 , I 2 , I 3 , and I x in the circuit of Fig. 10.84. Figure 10.84 For Prob. 10.39. Chapter 10, Solution 39. For mesh 1, (28 − j15)I1 − 8I 2 + j15I 3 = 12∠64 o (1) − 8I1 + (8 − j9)I 2 − j16I 3 = 0 (2) j15I1 − j16I 2 + (10 + j)I 3 = 0 (3) For mesh 2, For mesh 3, In matrix form, (1) to (3) can be cast as j15 ⎞⎛ I1 ⎞ ⎛⎜12∠64 o ⎞⎟ −8 ⎛ (28 − j15) ⎟⎜ ⎟ ⎜ (8 − j9) − j16 ⎟⎜ I 2 ⎟ = ⎜ 0 ⎟ ⎜ −8 ⎜ ⎟ ⎜ j15 − j16 (10 + j) ⎟⎠⎜⎝ I 3 ⎟⎠ ⎜ 0 ⎟ ⎝ ⎝ ⎠ Using MATLAB, or AI = B I = inv(A)*B I1 = −0.128 + j0.3593 = 0.3814∠109.6 o A I 2 = −0.1946 + j0.2841 = 0.3443∠124.4 o A I 3 = 0.0718 − j0.1265 = 0.1455∠ − 60.42 o A I x = I1 − I 2 = 0.0666 + j0.0752 = 0.1005∠48.5 o 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 10, Problem 40. Find io in the circuit shown in Fig. 10.85 using superposition. Figure 10.85 For Prob. 10.40. 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 10, Solution 40. Let i O = i O1 + i O 2 , where i O1 is due to the dc source and i O 2 is due to the ac source. For i O1 , consider the circuit in Fig. (a). 4Ω 2Ω iO1 + − 8V (a) Clearly, i O1 = 8 2 = 4 A For i O 2 , consider the circuit in Fig. (b). 4Ω 2Ω IO2 10∠0° V + − j4 Ω (b) If we transform the voltage source, we have the circuit in Fig. (c), where 4 || 2 = 4 3 Ω . IO2 2.5∠0° A 4Ω 2Ω j4 Ω (c) By the current division principle, 43 I O2 = (2.5∠0°) 4 3 + j4 I O 2 = 0.25 − j0.75 = 0.79∠ - 71.56° Thus, i O 2 = 0.79 cos(4t − 71.56°) A Therefore, i O = i O1 + i O 2 = 4 + 0.79 cos(4t – 71.56°) 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 10, Problem 41. Find v o for the circuit in Fig. 10.86, assuming that v s = 6 cos 2t + 4 sin 4t V. Figure 10.86 For Prob. 10.41. 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 10, Solution 41. We apply superposition principle. We let vo = v1 + v2 where v1 and v2 are due to the sources 6cos2t and 4sin4t respectively. To find v1, consider the circuit below. -j2 + + _ 6∠0o 2Ω V1 – 1/ 4F ⎯⎯ → 1 jω C = 1 = − j2 j 2 x1/ 4 2 (6) = 3 + j 3 = 4.2426 < 45o 2 − j2 Thus, v1 = 4.2426cos(2t + 45o ) To get v2, consider the circuit below –j V1 = + 4∠0o + _ 2Ω V2 – 1/ 4F ⎯⎯ → 1 jω C = 1 = − j1 j 4 x1/ 4 2 (4) = 3.2 + j11.6 = 3.578 < 26.56o 2− j v2 = 3.578 sin(4t + 26.56o ) V2 = Hence, vo = 4.243cos(2t + 45˚) + 3.578sin(4t + 25.56˚) 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 10, Problem 42. Solve for I o in the circuit of Fig. 10.87. Figure 10.87 For Prob. 10.42. Chapter 10, Solution 42. Let I o = I1 + I 2 where I1 and I2 are due to 20<0o and 30<45o sources respectively. To get I1, we use the circuit below. I1 j10 Ω 60 Ω o 20∠0 V + _ 50 Ω –j40 Ω Let Z1 = -j40//60 = 18.4615 –j27.6927, Z2 = j10//50=1.9231 + j9.615 Transforming the voltage source to a current source leads to the circuit below. I1 Z2 Z1 –j2 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. Using current division, Z2 I1 = (− j 2) = 0.6217 + j 0.3626 Z1 + Z 2 To get I2, we use the circuit below. j10 Ω I2 50 Ω 60 Ω –j40 Ω + _ 30∠45o V After transforming the voltage source, we obtain the circuit below. I2 Z2 Z1 0.5∠45o Using current division, − Z1 I2 = (0.5 < 45o ) = −0.5275 − j 0.3077 Z1 + Z 2 Hence, I o = I1 + I 2 = 0.0942 + j 0.0509 = 0.109 < 30o 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 10, Problem 43. Using the superposition principle, find i x in the circuit of Fig. 10.88. Figure 10.88 For Prob. 10.43. 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 10, Solution 43. Let I x = I 1 + I 2 , where I 1 is due to the voltage source and I 2 is due to the current source. ω= 2 5 cos(2t + 10°) ⎯ ⎯→ 5∠10° 10 cos(2t − 60°) ⎯ ⎯→ 10 ∠ - 60° 4H ⎯ ⎯→ jωL = j8 1 1 1 F ⎯ ⎯→ = = -j4 8 jωC j (2)(1 / 8) For I 1 , consider the circuit in Fig. (a). -j4 Ω 3Ω I1 + − j8 Ω 10∠-60° V (a) I1 = 10∠ - 60° 10 ∠ - 60° = 3 + j8 − j4 3 + j4 For I 2 , consider the circuit in Fig. (b). -j4 Ω 5∠10° A 3Ω I2 j8 Ω (b) I2 = - j8 - j40 ∠10° (5∠10°) = 3 + j8 − j4 3 + j4 1 (10∠ - 60° − j40∠10°) 3 + j4 49.51∠ - 76.04° Ix = = 9.902∠ - 129.17° 5∠53.13° i x = 9.902 cos(2t – 129.17°) A I x = I1 + I 2 = Therefore, 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 10, Problem 44. Use the superposition principle to obtain v x in the circuit of Fig. 10.89. Let v s = 50 sin 2t V and i s = 12 cos(6t + 10°) A. Figure 10.89 For Prob. 10.44. 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 10, Solution 44. Let v x = v1 + v 2 , where v1 and v2 are due to the current source and voltage source respectively. For v1 , ω = 6 , 5 H ⎯ ⎯→ jωL = j30 The frequency-domain circuit is shown below. 20 Ω 16 Ω Is Let Z = 16 //(20 + j30) = + V1 - 16(20 + j30) = 11.8 + j3.497 = 12.31∠16.5 o 36 + j30 V1 = I s Z = (12∠10 o )(12.31∠16.5 o ) = 147.7∠26.5 o For v2 , ω = 2 , 5 H j30 ⎯ ⎯→ v1 = 147.7 cos(6 t + 26.5 o ) V jωL = j10 The frequency-domain circuit is shown below. 20 Ω 16 Ω ⎯ ⎯→ j10 + V2 + Vs - - Using voltage division, 16(50∠0 o ) 16 V2 = Vs = = 21.41∠ − 15.52 o 16 + 20 + j10 36 + j10 ⎯ ⎯→ v 2 = 21.41sin(2t − 15.52 o ) V Thus, v x = 147.7 cos(6 t + 26.5 o ) + 21.41sin( 2 t − 15.52 o ) 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 10, Problem 45. Use superposition to find i (t ) in the circuit of Fig. 10.90. Figure 10.90 For Prob. 10.45. 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 10, Solution 45. Let i = i1 + i2 , where i1 and i2 are due to 16cos(10t +30o) and 6sin4t sources respectively. To find i1 , consider the circuit below. I1 16<30o V 20 Ω + _ jX X = ω L = 10 x300 x10 −3 = 3 16 < 30o I1 = = 0.7911 20 + j 3 i1 = 0.7911cos(10t + 21.47o ) A To find i2 , consider the circuit below. I2 20 Ω + _ 6∠0o V jX X = ω L = 4 x300 x10 −3 = 1.2 6 < 0o I2 = − = 0.2995 < 176.6o 20 + j1.2 i1 = 0.2995 sin(4t + 176.6o ) A Thus, i = i1 + i2 = 0.7911cos(10t + 21.47o ) + 0.2995 sin(4t + 176.6o ) A = 791.1cos(10t+21.47˚)+299.5sin(4t+176.6˚) 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 10, Problem 46. Solve for vo (t ) in the circuit of Fig. 10.91 using the superposition principle. Figure 10.91 For Prob. 10.46. Chapter 10, Solution 46. Let v o = v1 + v 2 + v 3 , where v1 , v 2 , and v 3 are respectively due to the 10-V dc source, the ac current source, and the ac voltage source. For v1 consider the circuit in Fig. (a). 6Ω 2H + 1/12 F + − v1 10 V − (a) The capacitor is open to dc, while the inductor is a short circuit. Hence, v1 = 10 V For v 2 , consider the circuit in Fig. (b). ω= 2 2H ⎯ ⎯→ jωL = j4 1 1 1 F ⎯ ⎯→ = = - j6 12 jωC j (2)(1 / 12) + 6Ω -j6 Ω 4∠0° A V2 j4 Ω − (b) 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. Applying nodal analysis, V V V ⎛1 j j ⎞ 4 = 2 + 2 + 2 = ⎜ + − ⎟ V2 6 - j6 j4 ⎝ 6 6 4 ⎠ V2 = Hence, 24 = 21.45∠26.56° 1 − j0.5 v 2 = 21.45 sin( 2 t + 26.56°) V For v 3 , consider the circuit in Fig. (c). ω=3 2H ⎯ ⎯→ jωL = j6 1 1 1 F ⎯ ⎯→ = = - j4 12 jωC j (3)(1 / 12) 6Ω j6 Ω + 12∠0° V + − -j4 Ω V3 − (c) At the non-reference node, 12 − V3 V3 V3 = + 6 - j4 j6 12 V3 = = 10.73∠ - 26.56° 1 + j0.5 Hence, v 3 = 10.73 cos(3t − 26.56°) V Therefore, v o = 10 + 21.45 sin(2t + 26.56°) + 10.73 cos(3t – 26.56°) 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 10, Problem 47. Determine io in the circuit of Fig. 10.92, using the superposition principle. Figure 10.92 For Prob. 10.47. 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 10, Solution 47. Let i o = i1 + i 2 + i 3 , where i1 , i 2 , and i 3 are respectively due to the 24-V dc source, the ac voltage source, and the ac current source. For i1 , consider the circuit in Fig. (a). 1Ω 24 V 1/6 F 2H − + i1 2Ω 4Ω Since the capacitor is an open circuit to dc, 24 i1 = =4A 4+2 For i 2 , consider the circuit in Fig. (b). ω=1 2H ⎯ ⎯→ jωL = j2 1 1 F ⎯ ⎯→ = - j6 6 jωC 1Ω j2 Ω -j6 Ω I2 10∠-30° V + − I1 2Ω I2 4Ω (b) For mesh 1, - 10 ∠ - 30° + (3 − j6) I 1 − 2 I 2 = 0 10 ∠ - 30° = 3 (1 − 2 j) I 1 − 2 I 2 (1) 0 = -2 I 1 + (6 + j2) I 2 I 1 = (3 + j) I 2 (2) For mesh 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. Substituting (2) into (1) 10 ∠ - 30° = 13 − j15 I 2 I 2 = 0.504 ∠19.1° i 2 = 0.504 sin( t + 19.1°) A Hence, For i 3 , consider the circuit in Fig. (c). ω=3 2H ⎯ ⎯→ jωL = j6 1 1 1 F ⎯ ⎯→ = = - j2 6 jωC j (3)(1 / 6) 1Ω j6 Ω -j2 Ω I3 2Ω 2∠0° A 4Ω (c) 2 || (1 − j2) = 2 (1 − j2) 3 − j2 Using current division, 2 (1 − j2) ⋅ (2∠0°) 2 (1 − j2) 3 − j2 = I3 = 2 (1 − j2) 13 + j3 4 + j6 + 3 − j2 I 3 = 0.3352 ∠ - 76.43° Hence i 3 = 0.3352 cos(3t − 76.43°) A Therefore, i o = 4 + 0.504 sin(t + 19.1°) + 0.3352 cos(3t – 76.43°) 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 10, Problem 48. Find io in the circuit of Fig. 10.93 using superposition. Figure 10.93 For Prob. 10.48. Chapter 10, Solution 48. Let i O = i O1 + i O 2 + i O 3 , where i O1 is due to the ac voltage source, i O 2 is due to the dc voltage source, and i O3 is due to the ac current source. For i O1 , consider the circuit in Fig. (a). ω = 2000 50 cos(2000t ) ⎯ ⎯→ 50∠0° 40 mH ⎯ ⎯→ 20 µF ⎯ ⎯→ jωL = j (2000)(40 × 10 -3 ) = j80 1 1 = = - j25 jωC j (2000)(20 × 10 -6 ) 80 || (60 + 100) = 160 3 50 30 = I= 160 3 + j80 − j25 32 + j33 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. Using current division, - 80 I -1 10∠180° = I= 80 + 160 3 46∠45.9° I O1 = 0.217 ∠134.1° Hence, i O1 = 0.217 cos(2000 t + 134.1°) A For i O 2 , consider the circuit in Fig. (b). I O1 = i O2 = 24 = 0.1 A 80 + 60 + 100 For i O3 , consider the circuit in Fig. (c). ω = 4000 2 cos(4000t ) ⎯ ⎯→ 2∠0° 40 mH ⎯ ⎯→ jωL = j (4000)(40 × 10 -3 ) = j160 20 µF ⎯ ⎯→ 1 1 = = - j12.5 jωC j (4000)(20 × 10 -6 ) For mesh 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. I1 = 2 (1) For mesh 2, (80 + j160 − j12.5) I 2 − j160 I 1 − 80 I 3 = 0 Simplifying and substituting (1) into this equation yields (8 + j14.75) I 2 − 8 I 3 = j32 (2) For mesh 3, 240 I 3 − 60 I 1 − 80 I 2 = 0 Simplifying and substituting (1) into this equation yields I 2 = 3 I 3 − 1.5 (3) Substituting (3) into (2) yields (16 + j44.25) I 3 = 12 + j54.125 12 + j54.125 I3 = = 1.1782∠7.38° 16 + j44.25 Hence, I O 3 = - I 3 = -1.1782 ∠7.38° i O 3 = -1.1782 sin( 4000t + 7.38°) A Therefore, i O = 0.1 + 0.217 cos(2000t + 134.1°) – 1.1782 sin(4000t + 7.38°) A Chapter 10, Problem 49. 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. Using source transformation, find i in the circuit of Fig. 10.94. Figure 10.94 For Prob. 10.49. Chapter 10, Solution 49. 8 sin( 200t + 30°) ⎯ ⎯→ 8∠30°, ω = 200 5 mH ⎯ ⎯→ jωL = j (200)(5 × 10 -3 ) = j 1 mF ⎯ ⎯→ 1 1 = = - j5 jωC j (200)(1 × 10 -3 ) After transforming the current source, the circuit becomes that shown in the figure below. 5Ω 40∠30° V I= 3Ω I + − jΩ -j5 Ω 40 ∠30° 40 ∠30° = = 4.472∠56.56° 5 + 3 + j − j5 8 − j4 i = 4.472 sin(200t + 56.56°) A Chapter 10, Problem 50. 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. Use source transformation to find vo in the circuit of Fig. 10.95. Figure 10.95 For Prob. 10.50. Chapter 10, Solution 50. 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. 5 cos(10 5 t ) ⎯⎯→ 5∠0°, ω = 10 5 0.4 mH ⎯ ⎯→ 0.2 µF ⎯ ⎯→ jωL = j (10 5 )(0.4 × 10 -3 ) = j40 1 1 = = - j50 5 jωC j (10 )(0.2 × 10 -6 ) After transforming the voltage source, we get the circuit in Fig. (a). j40 Ω + 20 Ω 0.25∠0° -j50 Ω 80 Ω Vo − (a) Let Z = 20 || - j 50 = - j100 2 − j5 - j25 2 − j5 With these, the current source is transformed to obtain the circuit in Fig.(b). and Vs = (0.25∠0°) Z = j40 Ω Z + Vs + − 80 Ω Vo − (b) By voltage division, 80 80 - j25 Vs = ⋅ j 100 Z + 80 + j40 2 − j5 + 80 + j40 2 − j5 8 (- j25) Vo = = 3.615∠ - 40.6° 36 − j42 v o = 3.615 cos(105 t – 40.6°) V Vo = Therefore, Chapter 10, Problem 51. 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. Use source transformation to find I o in the circuit of Prob. 10.42. Chapter 10, Solution 51. Transforming the voltage sources into current sources, we have the circuit as shown below. –j2 j10 50 60 -j40 0.5∠45o j10 x50 = 1.9231 + j 9.615 50 + j10 V1 = − j 2 Z1 = 19.231 − j 3.846 − j 40 x60 Let Z 2 = − j 40 // 60 = = 18.4615 − j 27.6923 60 − j 40 V2 = Z 2 x0.5 < 45o = 16.315 − 3.263 Let Z1 = j10 // 50 = Transforming the current sources to voltage sources leads to the circuit below. Io Z2 Z1 V1 + _ + _ V2 Applying KVL to the loop gives −V1 + I o ( Z1 + Z 2 ) + V2 = 0 Io = ⎯⎯ → Io = V1 − V2 Z1 + Z 2 19.231 − j 3.846 − 16.316 + j 3.263 = 0.1093 < 30o A = 109.3∠30˚ mA 1.9231 + j 9.615 + 18.4615 − j 27.6923 Chapter 10, Problem 52. 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. Use the method of source transformation to find I x in the circuit of Fig. 10.96. Figure 10.96 For Prob. 10.52. Chapter 10, Solution 52. We transform the voltage source to a current source. 60∠0° Is = = 6 − j12 2 + j4 The new circuit is shown in Fig. (a). -j2 Ω Ix 2Ω Is = 6 – j12 A 6Ω j4 Ω 4Ω 5∠90° A -j3 Ω (a) Let 6 (2 + j4) = 2.4 + j1.8 8 + j4 Vs = I s Z s = (6 − j12)(2.4 + j1.8) = 36 − j18 = 18 (2 − j) Z s = 6 || (2 + j4) = 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. With these, we transform the current source on the left hand side of the circuit to a voltage source. We obtain the circuit in Fig. (b). Zs -j2 Ω Ix Vs 4Ω + − j5 A -j3 Ω (b) Let Z o = Z s − j2 = 2.4 − j0.2 = 0.2 (12 − j) Vs 18 (2 − j) = = 15.517 − j6.207 Io = Z o 0.2 (12 − j) With these, we transform the voltage source in Fig. (b) to a current source. We obtain the circuit in Fig. (c). Ix Io 4Ω Zo j5 A -j3 Ω (c) Using current division, Zo 2.4 − j0.2 (I o + j5) = (15.517 − j1.207) Ix = 6.4 − j3.2 Z o + 4 − j3 I x = 5 + j1.5625 = 5.238∠17.35° 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 10, Problem 53. Use the concept of source transformation to find V o in the circuit of Fig. 10.97. Figure 10.97 For Prob. 10.53. Chapter 10, Solution 53. We transform the voltage source to a current source to obtain the circuit in Fig. (a). -j3 Ω j4 Ω + 5∠0° A 4Ω j2 Ω 2Ω Vo -j2 Ω − (a) Let j8 = 0.8 + j1.6 4 + j2 Vs = (5∠0°) Z s = (5)(0.8 + j1.6) = 4 + j8 Z s = 4 || j2 = 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. With these, the current source is transformed so that the circuit becomes that shown in Fig. (b). -j3 Ω Zs j4 Ω + Vs + − 2Ω -j2 Ω Vo − (b) Z x = Z s − j3 = 0.8 − j1.4 V 4 + j8 = −3.0769 + j4.6154 Ix = s = Z s 0.8 − j1.4 With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c). Let j4 Ω + Ix 2Ω Zx -j2 Ω Vo − (c) Let 1.6 − j2.8 = 0.8571 − j0.5714 2.8 − j1.4 Vy = I x Z y = (−3.0769 + j4.6154) ⋅ (0.8571 − j0.5714) = j5.7143 Z y = 2 || Z x = With these, we transform the current source to obtain the circuit in Fig. (d). Using current division, j4 Ω Zy + Vy + − -j2 Ω Vo − (d) Vo = - j2 ( j5.7143) - j2 Vy = = (3.529 – j5.883) V Z y + j4 − j2 0.8571 − j0.5714 + j4 − j2 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 10, Problem 54. Rework Prob. 10.7 using source transformation. Chapter 10, Solution 54. 50 x(− j 30) = 13.24 − j 22.059 50 − j 30 We convert the current source to voltage source and obtain the circuit below. 50 //( − j 30) = 40 Ω + Vs =115.91 –j31.06V 13.24 – j22.059 Ω j20 Ω + - I 134.95-j74.912 V V - + - Applying KVL gives -115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0 or I = − 250.86 + j105.97 = −4.7817 + j1.8055 53.24 − j 2.059 But − Vs + (40 + j20)I + V = 0 ⎯ ⎯→ V = Vs − (40 + j20)I V = 115.91 − j31.05 − (40 + j20)(−4.7817 + j1.8055) = 124.06∠ − 154 o V which agrees with the result in Prob. 10.7. 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 10, Problem 55. Find the Thevenin and Norton equivalent circuits at terminals a-b for each of the circuits in Fig. 10.98. Figure 10.98 For Prob. 10.55. Chapter 10, Solution 55. (a) To find Z th , consider the circuit in Fig. (a). j20 Ω 10 Ω Zth -j10 Ω (a) ( j20)(- j10) j20 − j10 = 10 − j20 = 22.36∠-63.43° Ω To find Vth , consider the circuit in Fig. (b). Z N = Z th = 10 + j20 || (- j10) = 10 + j20 Ω 10 Ω + 50∠30° V + − -j10 Ω Vth − (b) 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. Vth = IN = (b) - j10 (50∠30°) = -50∠30° V j20 − j10 Vth - 50 ∠30° = = 2.236∠273.4° A Z th 22.36 ∠ - 63.43° To find Z th , consider the circuit in Fig. (c). -j5 Ω 8Ω Zth j10 Ω (c) Z N = Z th = j10 || (8 − j5) = ( j10)(8 − j5) = 10∠26° Ω j10 + 8 − j5 To obtain Vth , consider the circuit in Fig. (d). -j5 Ω Io 4∠0° A 8Ω j10 Ω + Vth − (d) By current division, 8 32 (4∠0°) = Io = 8 + j10 − j5 8 + j5 Vth = j10 I o = IN = j320 = 33.92∠58° V 8 + j5 Vth 33.92 ∠58° = = 3.392∠32° A 10 ∠26° Z th 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 10, Problem 56. For each of the circuits in Fig. 10.99, obtain Thevenin and Norton equivalent circuits at terminals a-b. Figure 10.99 For Prob. 10.56. Chapter 10, Solution 56. (a) To find Z th , consider the circuit in Fig. (a). j4 Ω 6Ω -j2 Ω Zth (a) ( j4)(- j2) Z N = Z th = 6 + j4 || (- j2) = 6 + = 6 − j4 j4 − j2 = 7.211∠-33.69° Ω By placing short circuit at terminals a-b, we obtain, I N = 2∠0° A Vth = Z th I th = (7.211∠ - 33.69°) (2∠0°) = 14.422∠-33.69° 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. (b) To find Z th , consider the circuit in Fig. (b). j10 Ω 30 Ω 60 Ω -j5 Ω Zth (b) 30 || 60 = 20 (- j5)(20 + j10) 20 + j5 = 5.423∠-77.47° Ω Z N = Z th = - j5 || (20 + j10) = To find Vth and I N , we transform the voltage source and combine the 30 Ω and 60 Ω resistors. The result is shown in Fig. (c). j10 Ω 4∠45° A 20 Ω a IN -j5 Ω (c) b 20 2 (4∠45°) = (2 − j)(4∠45°) 20 + j10 5 = 3.578∠18.43° A IN = Vth = Z th I N = (5.423∠ - 77.47°) (3.578∠18.43°) = 19.4∠-59° 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 10, Problem 57. Find the Thevenin and Norton equivalent circuits for the circuit shown in Fig. 10.100. Figure 10.100 For Prob. 10.57. 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 10, Solution 57. To find Z th , consider the circuit in Fig. (a). 5Ω -j10 Ω 2Ω Zth j20 Ω (a) ( j20)(5 − j10) 5 + j10 = 18 − j12 = 21.63∠-33.7° Ω Z N = Z th = 2 + j20 || (5 − j10) = 2 + To find Vth , consider the circuit in Fig. (b). 5Ω -j10 Ω 2Ω + 60∠120° V + − j20 Ω Vth − (b) j20 j4 (60 ∠120°) = (60∠120°) 5 − j10 + j20 1 + j2 = 107.3∠146.56° V Vth = IN = Vth 107.3∠146.56° = = 4.961∠-179.7° A Z th 21.633∠ - 33.7° 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 10, Problem 58. For the circuit depicted in Fig. 10.101, find the Thevenin equivalent circuit at terminals a-b. Figure 10.101 For Prob. 10.58. Chapter 10, Solution 58. Consider the circuit in Fig. (a) to find Z th . 8Ω Zth j10 Ω -j6 Ω (a) ( j10)(8 − j6) Z th = j10 || (8 − j6) = = 5 (2 + j) 8 + j4 = 11.18∠26.56° Ω Consider the circuit in Fig. (b) to find Vth . Io + 8Ω j10 Ω 5∠45° A Vth -j6 Ω (b) Io = 8 − j6 4 − j3 (5∠45°) = (5∠45°) 8 − j6 + j10 4 + j2 Vth = j10 I o = ( j10)(4 − j3)(5∠45°) = 55.9∠71.56° V (2)(2 + j) 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 10, Problem 59. Calculate the output impedance of the circuit shown in Fig. 10.102. Figure 10.102 For Prob. 10.59. Chapter 10, Solution 59. Insert a 1-A current source at the output as shown below. -j2 Ω 10 Ω V1 + – Vo j40 Ω 0.2 Vo + Vin 1A – v1 j 40 But vo = −1(− j 2) = j 2 0.2vo + 1 = j 2 x0.2 + 1 = V1 j 40 ⎯⎯ → V1 = −16 + j 40 Vin = V1 – Vo + 10 = –6 + j38 = 1xZin Zin = –6 + j38 Ω. 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 10, Problem 60. Find the Thevenin equivalent of the circuit in Fig. 10.103 as seen from: (b) terminals c-d (a) terminals a-b Figure 10.103 For Prob. 10.60. Chapter 10, Solution 60. (a) To find Z th , consider the circuit in Fig. (a). 10 Ω -j4 Ω a j5 Ω Zth 4Ω b (a) Z th = 4 || (- j4 + 10 || j5) = 4 || (- j4 + 2 + j4) Z th = 4 || 2 = 1.333 Ω To find Vth , consider the circuit in Fig. (b). 10 Ω V1 -j4 Ω V2 + 20∠0° V + − j5 Ω 4∠0° A 4Ω Vth − (b) 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. At node 1, 20 − V1 V1 V1 − V2 = + 10 j5 - j4 (1 + j0.5) V1 − j2.5 V2 = 20 (1) At node 2, V1 − V2 V2 = - j4 4 V1 = (1 − j) V2 + j16 4+ (2) Substituting (2) into (1) leads to 28 − j16 = (1.5 − j3) V2 28 − j16 V2 = = 8 + j5.333 1.5 − j3 Therefore, Vth = V2 = 9.615∠33.69° V (b) To find Z th , consider the circuit in Fig. (c). Zth c d 10 Ω -j4 Ω j5 Ω 4Ω (c) ⎛ j10 ⎞ ⎟ Z th = - j4 || (4 + 10 || j5) = - j4 || ⎜ 4 + 2 + j⎠ ⎝ - j4 Z th = - j4 || (6 + j4) = (6 + j4) = 2.667 – j4 Ω 6 To find Vth ,we will make use of the result in part (a). V2 = 8 + j5.333 = (8 3 ) (3 + j2) V1 = (1 − j) V2 + j16 = j16 + (8 3) (5 − j) Vth = V1 − V2 = 16 3 + j8 = 9.614∠56.31° 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 10, Problem 61. Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 10.104. Figure 10.104 For Prob. 10.61. 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 10, Solution 61. To find VTh, consider the circuit below 4Ω Vo a Ix + -j3 Ω o 2∠0 A 1.5Ix VTh – b 2 + 1.5Ix = Ix But Ix = –4 Vo = –j3Ix = j12 VTh = Vo + 6Ix = j12 − 24 V To find ZTh, consider the circuit shown below. 4Ω Vo Ix -j3 Ω 1+1.5 Ix = Ix 1A Ix = -2 −Vo + Ix(4 − j 3) = 0 ZTh = 1.5Ix ⎯⎯ → Vo = −8 + j6 Vo = −8 + j 6 Ω 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 10, Problem 62. Using Thevenin’s theorem, find vo in the circuit of Fig. 10.105. Figure 10.105 For Prob. 10.62. 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 10, Solution 62. First, we transform the circuit to the frequency domain. 12 cos( t ) ⎯ ⎯→ 12∠0°, ω = 1 2H ⎯ ⎯→ 1 F ⎯ ⎯→ 4 1 F ⎯ ⎯→ 8 jωL = j2 1 = - j4 jωC 1 = - j8 jωC To find Z th , consider the circuit in Fig. (a). 3 Io Io 4Ω Vx j2 Ω 1 Ix 2 -j4 Ω -j8 Ω + − 1V (a) At node 1, Vx Vx 1 − Vx + + 3Io = , 4 - j4 j2 Thus, where I o = - Vx 4 Vx 2 Vx 1 − Vx − = - j4 4 j2 Vx = 0.4 + j0.8 At node 2, I x + 3Io = 1 1 − Vx + - j8 j2 I x = (0.75 + j0.5) Vx − j 3 8 I x = -0.1 + j0.425 Z th = 1 = -0.5246 − j2.229 = 2.29∠ - 103.24° Ω Ix PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. To find Vth , consider the circuit in Fig. (b). 3 Io Io 4Ω j2 Ω V1 V2 1 12∠0° V + − 2 -j4 Ω -j8 Ω + Vth − (b) At node 1, V V − V2 12 − V1 = 3Io + 1 + 1 , 4 - j4 j2 24 = (2 + j) V1 − j2 V2 where I o = 12 − V1 4 (1) At node 2, V1 − V2 V + 3Io = 2 j2 - j8 72 = (6 + j4) V1 − j3 V2 (2) From (1) and (2), ⎡ 24⎤ ⎡ 2 + j - j2⎤ ⎡ V1 ⎤ ⎢ 72 ⎥ = ⎢ 6 + j4 - j3⎥ ⎢ V ⎥ ⎣ ⎦ ⎣ ⎦⎣ 2⎦ ∆ = -5 + j6 , Vth = V2 = Thus, ∆ 2 = - j24 ∆2 = 3.073∠ - 219.8° ∆ 2 (2)(3.073∠ - 219.8°) Vth = 2 + Z th 1.4754 − j2.229 6.146∠ - 219.8° Vo = = 2.3∠ - 163.3° 2.673∠ - 56.5° Vo = Therefore, v o = 2.3 cos(t – 163.3°) 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 10, Problem 63. Obtain the Norton equivalent of the circuit depicted in Fig. 10.106 at terminals a-b. Figure 10.106 For Prob. 10.63. 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 10, Solution 63. Transform the circuit to the frequency domain. 4 cos(200t + 30°) ⎯ ⎯→ 4∠30°, ω = 200 10 H ⎯ ⎯→ 5 µF ⎯ ⎯→ jωL = j (200)(10) = j2 kΩ 1 1 = = - j kΩ jωC j (200)(5 × 10 -6 ) Z N is found using the circuit in Fig. (a). -j kΩ j2 kΩ ZN 2 kΩ (a) Z N = - j + 2 || j2 = - j + 1 + j = 1 kΩ We find I N using the circuit in Fig. (b). -j kΩ 4∠30° A j2 kΩ 2 kΩ IN (b) j2 || 2 = 1 + j By the current division principle, 1+ j IN = (4 ∠30°) = 5.657 ∠75° 1+ j − j Therefore, i N = 5.657 cos(200t + 75°) A Z N = 1 kΩ 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 10, Problem 64. For the circuit shown in Fig. 10.107, find the Norton equivalent circuit at terminals a-b. Figure 10.107 For Prob. 10.64. 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 10, Solution 64. Z N is obtained from the circuit in Fig. (a). 60 Ω 40 Ω ZN -j30 Ω j80 Ω (a) Z N = (60 + 40) || ( j80 − j30) = 100 || j50 = (100)( j50) 100 + j50 Z N = 20 + j40 = 44.72∠63.43° Ω To find I N , consider the circuit in Fig. (b). 60 Ω 3∠60° A I1 40 Ω I2 -j30 Ω IN Is j80 Ω (b) I s = 3∠60° For mesh 1, 100 I 1 − 60 I s = 0 I 1 = 1.8∠60° For mesh 2, ( j80 − j30) I 2 − j80 I s = 0 I 2 = 4.8∠60° IN = I2 – I1 = 3∠60° 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 10, Problem 65. Compute io in Fig. 10.108 using Norton’s theorem. Figure 10.108 For Prob. 10.65. Chapter 10, Solution 65. 5 cos(2 t ) ⎯ ⎯→ 5∠0°, ω = 2 4H ⎯ ⎯→ 1 F ⎯ ⎯→ 4 1 F ⎯ ⎯→ 2 jωL = j (2)(4) = j8 1 1 = = - j2 jωC j (2)(1 / 4) 1 1 = = -j jωC j (2)(1 / 2) To find Z N , consider the circuit in Fig. (a). 2Ω ZN -j2 Ω -j Ω (a) Z N = - j || (2 − j2) = - j (2 − j2) 1 = (2 − j10) 2 − j3 13 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. To find I N , consider the circuit in Fig. (b). 5∠0° V 2Ω + − -j2 Ω -j Ω IN (b) IN = 5∠0° = j5 -j The Norton equivalent of the circuit is shown in Fig. (c). Io ZN IN j8 Ω (c) Using current division, Io = ZN (1 13)(2 − j10)( j5) 50 + j10 IN = = (1 13)(2 − j10) + j8 2 + j94 Z N + j8 I o = 0.1176 − j0.5294 = 0542∠ - 77.47° Therefore, i o = 542 cos(2t – 77.47°) 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 10, Problem 66. At terminals a-b, obtain Thevenin and Norton equivalent circuits for the network depicted in Fig. 10.109. Take ω = 10 rad/s. Figure 10.109 For Prob. 10.66. Chapter 10, Solution 66. ω = 10 0.5 H ⎯ ⎯→ jωL = j (10)(0.5) = j5 1 1 10 mF ⎯ ⎯→ = = - j10 jωC j (10)(10 × 10 -3 ) To find Z th , consider the circuit in Fig. (a). -j10 Ω Vx + 10 Ω Vo j5 Ω 2 Vo 1A − (a) Vx Vx + , j5 10 − j10 19 Vx V - 10 + j10 1+ = x ⎯ ⎯→ Vx = 10 − j10 j5 21 + j2 1 + 2 Vo = Z N = Z th = where Vo = 10Vx 10 − j10 Vx 14.142 ∠135° = = 0.67∠129.56° Ω 1 21.095∠5.44° PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. To find Vth and I N , consider the circuit in Fig. (b). 12∠0° V -j10 Ω − + + + 10 Ω -j2 A Vo j5 Ω I 2 Vo Vth − − (b) where Thus, (10 − j10 + j5) I − (10)(- j2) + j5 (2 Vo ) − 12 = 0 Vo = (10)(- j2 − I ) (10 − j105) I = -188 − j20 188 + j20 I= - 10 + j105 Vth = j5 (I + 2 Vo ) = j5 (−19I − j40) = − j95 I + 200 Vth = − j95 (188 + j20) + 200 = 29.73 + j1.8723 - 10 + j105 Vth = 29.79∠3.6° V IN = Vth 29.79∠3.6° = = 44.46∠–125.96° A Z th 0.67∠129.56° 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 10, Problem 67. Find the Thevenin and Norton equivalent circuits at terminals a-b in the circuit of Fig. 10.110. Figure 10.110 For Prob. 10.67. Chapter 10, Solution 67. Z N = Z Th = 10 //(13 − j5) + 12 //(8 + j6) = Va = 10 (60∠45 o ) = 13.78 + j21.44, 23 − j5 10(13 − j5) 12(8 + j6) + = 11.243 + j1.079Ω 23 − j5 20 + j6 Vb = (8 + j6) (60∠45 o ) = 12.069 + j26.08Ω 20 + j6 VTh = Va − Vb = 1.711 − j4.64 = 4.945∠ − 69.76 o V, V 4.945∠ − 69.76° I N = Th = = 0.4378∠ − 75.24 o A Z Th 11.295∠5.48° 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 10, Problem 68. Find the Thevenin equivalent at terminals a-b in the circuit of Fig. 10.111. Figure 10.111 For Prob. 10.68. Chapter 10, Solution 68. 1H ⎯ ⎯→ jωL = j10x1 = j10 1 1 1 F ⎯ ⎯→ = = − j2 1 20 jω C j10 x 20 We obtain VTh using the circuit below. Io 4Ω a + 6<0 - o + Vo/3 j10(− j2) = − j2.5 j10 − j2 Vo = 4I o x (− j2.5) = − j10I o 1 − 6 + 4I o + Vo = 0 3 + - -j2 j10 Vo 4Io b j10 //(− j2) = (1) (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. Combining (1) and (2) gives Io = 6 , 4 − j10 / 3 VTh = Vo = − j10I o = − j60 = 11.52∠ − 50.19 o 4 − j10 / 3 v Th = 11.52 sin(10 t − 50.19 o ) To find RTh, we insert a 1-A source at terminals a-b, as shown below. Io 4Ω a + + - Vo/3 -j2 j10 Vo 4Io - 1 4I o + Vo = 0 3 1 + 4I o = ⎯⎯→ 1<0o V Io = − o 12 Vo Vo + − j2 j10 Combining the two equations leads to Vo = 1 = 1.2293 − j1.4766 0.333 + j0.4 V Z Th = o = 1.2293 − 1.477Ω 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 10, Problem 69. For the differentiator shown in Fig. 10.112, obtain V o /V s . Find vo (t ) when v s (t) = V m sin ωt and ω = 1/RC. Figure 10.112 For Prob. 10.69. Chapter 10, Solution 69. This is an inverting op amp so that Vo - Z f -R = = = -jωRC Vs Zi 1 jωC When Vs = Vm and ω = 1 RC , 1 Vo = - j ⋅ ⋅ RC ⋅ Vm = - j Vm = Vm ∠ - 90° RC Therefore, v o ( t ) = Vm sin(ωt − 90°) = - Vm cos(ωt) 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 10, Problem 70. The circuit in Fig. 10.113 is an integrator with a feedback resistor. Calculate vo (t ) if v s = 2 cos 4 × 10 4 t V. Figure 10.113 For Prob. 10.70. Chapter 10, Solution 70. This may also be regarded as an inverting amplifier. 2 cos(4 × 10 4 t ) ⎯ ⎯→ 2 ∠0°, ω = 4 × 10 4 1 1 ⎯→ = = - j2.5 kΩ 10 nF ⎯ 4 jωC j (4 × 10 )(10 × 10 -9 ) Vo - Z f = Vs Zi where Z i = 50 kΩ and Z f = 100k || (- j2.5k ) = - j100 kΩ . 40 − j Vo j2 = Vs 40 − j Thus, If Vs = 2 ∠0° , Vo = Therefore, j4 4∠90° = = 0.1∠91.43° 40 − j 40.01∠ - 1.43° v o ( t ) = 0.1 cos(4x104 t + 91.43°) 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 10, Problem 71. Find vo in the op amp circuit of Fig. 10.114. Figure 10.114 For Prob. 10.71. Chapter 10, Solution 71. 8 cos(2t + 30 o ) 0. 5µF ⎯ ⎯→ ⎯⎯→ 8∠30 o 1 1 = = − j1MΩ jωC j2x 0.5x10 − 6 At the inverting terminal, Vo − 8∠30 o Vo − 8∠30 o 8∠30 o + = − j1000k 10k 2k ⎯ ⎯→ Vo (1 − j100) = 8∠30 + 800∠ − 60° + 4000 ∠ − 60° Vo = 6.928 + j4 + 2400 − j4157 4800∠ − 59.9° = = 48∠29.53o 1 − j100 100∠ − 89.43° vo(t) = 48cos(2t + 29.53o) 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 10, Problem 72. Compute io (t ) in the op amp circuit in Fig. 10.115 if v s = 4 cos10 4 t V. Figure 10.115 For Prob. 10.72. Chapter 10, Solution 72. 4 cos(10 4 t ) ⎯ ⎯→ 4 ∠0°, ω = 10 4 1 1 ⎯→ = = - j100 kΩ 1 nF ⎯ 4 jωC j (10 )(10 -9 ) Consider the circuit as shown below. 50 kΩ Vo + − 4∠0° V + − -j100 kΩ At the noninverting node, Vo 4 − Vo = 50 - j100 Io = Therefore, ⎯ ⎯→ Vo = Vo Io 100 kΩ 4 1 + j0.5 Vo 4 = mA = 35.78∠ - 26.56° µA 100k (100)(1 + j0.5) i o ( t ) = 35.78 cos(104 t – 26.56°) µA 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 10, Problem 73. If the input impedance is defined as Z in =V s /I s find the input impedance of the op amp circuit in Fig. 10.116 when R1 = 10 k Ω, R2 = 20 k Ω, C1 = 10 nF, and ω = 5000 rad/s. Figure 10.116 For Prob. 10.73. Chapter 10, Solution 73. As a voltage follower, V2 = Vo 1 1 = = -j20 kΩ 3 jωC1 j (5 × 10 )(10 × 10 -9 ) 1 1 ⎯→ = = -j10 kΩ C 2 = 20 nF ⎯ 3 jωC 2 j (5 × 10 )(20 × 10 -9 ) ⎯→ C1 = 10 nF ⎯ 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. Consider the circuit in the frequency domain as shown below. -j20 kΩ Is 10 kΩ 20 kΩ + − V1 VS + − V2 Io Vo -j10 kΩ Zin At node 1, Vs − V1 V1 − Vo V1 − Vo = + 10 - j20 20 2 Vs = (3 + j)V1 − (1 + j)Vo (1) At node 2, V1 − Vo Vo − 0 = 20 - j10 V1 = (1 + j2)Vo (2) Substituting (2) into (1) gives 2 Vs = j6Vo or 1 Vo = -j Vs 3 ⎛2 1⎞ V1 = (1 + j2)Vo = ⎜ − j ⎟ Vs ⎝3 3⎠ Vs − V1 (1 3)(1 + j) Vs = 10k 10k 1+ j = 30k Is = Is Vs Vs 30k = = 15 (1 − j) k Is 1 + j Z in = 21.21∠–45° kΩ Z in = 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 10, Problem 74. Evaluate the voltage gain A v = V o /V s in the op amp circuit of Fig. 10.117. Find A v at ω = 0, ω → ∞, ω = 1 / R1C1 , and ω = 1 / R2 C 2 . Figure 10.117 For Prob. 10.74. Chapter 10, Solution 74. Zi = R1 + 1 , jωC1 Zf = R 2 + 1 jωC 2 1 ⎛ C ⎞ ⎛ 1 + jω R 2 C 2 ⎞ V - Zf jωC 2 ⎟⎟ Av = o = =− = − ⎜⎜ 1 ⎟⎟ ⎜⎜ 1 Vs Zi C 1 j R C + ω 1 1 ⎠ ⎝ 2⎠⎝ R1 + jωC1 R2 + Av = – At ω = 0 , As ω → ∞ , Av = – C1 C2 R2 R1 At ω = 1 , R 1 C1 ⎛ C ⎞ ⎛ 1 + j R 2 C 2 R 1C1 ⎞ ⎟ A v = –⎜ 1 ⎟ ⎜ 1+ j ⎠ ⎝ C2 ⎠ ⎝ At ω = 1 , R 2C2 ⎛C ⎞⎛ ⎞ 1+ j ⎟ A v = –⎜ 1 ⎟ ⎜ ⎝ C 2 ⎠ ⎝ 1 + j R 1C1 R 2 C 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 10, Problem 75. In the op amp circuit of Fig. 10.118, find the closed-loop gain and phase shift of the output voltage with respect to the input voltage if C1 = C 2 = 1 nF, R1 = R2 = 100 k Ω , R3 = 20 k Ω , R4 = 40 k Ω , and ω = 2000 rad/s. Figure 10.118 For Prob. 10.75. 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 10, Solution 75. ω = 2 × 10 3 ⎯→ C1 = C 2 = 1 nF ⎯ 1 1 = = -j500 kΩ 3 jωC1 j (2 × 10 )(1 × 10 -9 ) Consider the circuit shown below. 100 kΩ -j500 kΩ -j500 kΩ V2 + − V1 40 kΩ VS + − 100 kΩ + Vo 20 kΩ − Let Vs = 10V. At node 1, [(V1–10)/(–j500k)] + [(V1–Vo)/105] + [(V1–V2)/(–j500k)] = 0 or (1+j0.4)V1 – j0.2V2 – Vo = j2 (1) [(V2–V1)/(–j5)] + (V2–0) = 0 or –j0.2V1 + (1+j0.2)V2 = 0 or V1 = (1–j5)V2 (2) At node 2, But V2 = V R3 Vo = o R3 + R4 3 From (2) and (3), V1 = (0.3333–j1.6667)Vo (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. Substituting (3) and (4) into (1), (1+j0.4)(0.3333–j1.6667)Vo – j0.06667Vo – Vo = j2 (1.077∠21.8˚)(1.6997∠–78.69˚) = 1.8306∠–56.89˚ = 1 – j1.5334 Thus, (1–j1.5334)Vo – j0.06667Vo – Vo = j2 and, Vo = j2/(–j1.6601) = –1.2499 = 1.2499∠180˚ V Since Vs = 10, Vo/Vs = 0.12499∠180˚. Checking with MATLAB. >> Y=[1+0.4i,-0.2i,-1;1,-1+5i,0;0,-3,1] Y= 1.0000 + 0.4000i 0 - 0.2000i -1.0000 1.0000 -1.0000 + 5.0000i 0 0 -3.0000 1.0000 >> I=[2i;0;0] I= 0 + 2.0000i 0 0 >> V=inv(Y)*I V= -0.4167 + 2.0833i -0.4167 -1.2500 + 0.0000i (this last term is vo) and, the 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 10, Problem 76. Determine V o and I o in the op amp circuit of Fig. 10.119. Figure 10.119 For Prob. 10.76. 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. Chapter 10, Solution 76. Let the voltage between the -jk Ω capacitor and the 10k Ω resistor be V1. 2∠30 o − V1 V1 − Vo V1 − Vo = + − j4k 10k 20k ⎯ ⎯→ (1) 2∠30 o = (1 − j0.6)V1 + j0.6Vo = 1.7321+j1 Also, V1 − Vo V = o 10k − j2k ⎯⎯→ V1 = (1 + j5)Vo (2) Solving (2) into (1) yields 2∠30° = (1 − j0.6)(1 + j5)Vo + j0.6Vo = (1 + 3 − j0.6 + j5 + j6)Vo = (4+j5)Vo 2∠30° Vo = = 0.3124∠ − 21.34 o V 6.403∠51.34° >> Y=[1-0.6i,0.6i;1,-1-0.5i] Y= 1.0000 - 0.6000i 0 + 0.6000i 1.0000 -1.0000 - 5.0000i >> I=[1.7321+1i;0] I= 1.7321 + 1.0000i 0 >> V=inv(Y)*I V= 0.8593 + 1.3410i 0.2909 - 0.1137i = Vo = 0.3123∠–21.35˚V. 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 10, Problem 77. Compute the closed-loop gain V o /V s for the op amp circuit of Fig. 10.120. Figure 10.120 For Prob. 10.77. 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 10, Solution 77. Consider the circuit below. R3 2 R1 1 + − VS At node 1, V1 V1 C2 R2 − + + Vo C1 − Vs − V1 = jωC V1 R1 Vs = (1 + jωR 1C1 ) V1 (1) At node 2, 0 − V1 V1 − Vo = + jωC 2 (V1 − Vo ) R3 R2 ⎛ R3 ⎞ V1 = (Vo − V1 ) ⎜ + jωC 2 R 3 ⎟ ⎝R2 ⎠ ⎛ ⎞ 1 ⎟ V1 Vo = ⎜1 + ⎝ (R 3 R 2 ) + jωC 2 R 3 ⎠ (2) From (1) and (2), Vo = ⎛ ⎞ Vs R2 ⎜1 + ⎟ 1 + jωR 1C1 ⎝ R 3 + jωC 2 R 2 R 3 ⎠ Vo R 2 + R 3 + jωC 2 R 2 R 3 = Vs (1 + jωR 1C 1 ) ( R 3 + jωC 2 R 2 R 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 10, Problem 78. Determine vo (t ) in the op amp circuit in Fig. 10.121 below. Figure 10.121 For Prob. 10.78. 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. Chapter 10, Solution 78. 2 sin(400t ) ⎯ ⎯→ 2∠0°, ω = 400 1 1 ⎯→ = = - j5 kΩ 0.5 µF ⎯ jωC j (400)(0.5 × 10 -6 ) 1 1 ⎯→ = = - j10 kΩ 0.25 µF ⎯ jωC j (400)(0.25 × 10 -6 ) Consider the circuit as shown below. 20 kΩ 10 kΩ V 1 2∠0° V + − -j5 kΩ V + − V 40 kΩ -j10 kΩ 10 kΩ 20 kΩ At node 1, V V − V2 V1 − Vo 2 − V1 = 1 + 1 + 10 - j10 - j5 20 4 = (3 + j6) V1 − j4 V2 − Vo (1) V1 − V2 V2 = 10 − j5 V1 = (1 − j0.5) V2 (2) At node 2, But 20 1 Vo = Vo 20 + 40 3 From (2) and (3), 1 V1 = ⋅ (1 − j0.5) Vo 3 Substituting (3) and (4) into (1) gives 1 4 1⎞ ⎛ 4 = (3 + j6) ⋅ ⋅ (1 − j0.5) Vo − j Vo − Vo = ⎜1 + j ⎟ Vo 3 3 6⎠ ⎝ 24 Vo = = 3.945∠ − 9.46° 6+ j Therefore, v o ( t ) = 3.945 sin(400t – 9.46°) V V2 = (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. Chapter 10, Problem 79. For the op amp circuit in Fig. 10.122, obtain vo (t ) . Figure 10.122 For Prob. 10.79. 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 10, Solution 79. 5 cos(1000 t ) ⎯ ⎯→ 5∠0°, ω = 1000 1 1 ⎯→ = = - j10 kΩ 0.1 µF ⎯ jωC j (1000)(0.1 × 10 -6 ) 1 1 ⎯→ = = - j5 kΩ 0.2 µF ⎯ jωC j (1000)(0.2 × 10 -6 ) Consider the circuit shown below. 20 kΩ -j10 kΩ 10 kΩ Vs = 5∠0° V − + + − 40 kΩ V1 − + -j5 kΩ Since each stage is an inverter, we apply Vo = + Vo − - Zf V to each stage. Zi i Vo = - 40 V1 - j5 (1) V1 = - 20 || (- j10) Vs 10 (2) and From (1) and (2), ⎛ - j8 ⎞⎛ - (20)(-j10) ⎞ ⎟ 5∠0° ⎟⎜ Vo = ⎜ ⎝ 10 ⎠⎝ 20 − j10 ⎠ Vo = 16 (2 + j) = 35.78∠26.56° Therefore, v o ( t ) = 35.78 cos(1000t + 26.56°) 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 10, Problem 80. Obtain vo (t ) for the op amp circuit in Fig. 10.123 if v s = 4 cos(1000t − 60°) V. Figure 10.123 For Prob. 10.80. Chapter 10, Solution 80. 4 cos(1000t − 60°) ⎯ ⎯→ 4∠ - 60°, ω = 1000 1 1 ⎯→ = = - j10 kΩ 0.1 µF ⎯ jωC j (1000)(0.1 × 10 -6 ) 1 1 ⎯→ = = - j5 kΩ 0.2 µF ⎯ jωC j (1000)(0.2 × 10 -6 ) The two stages are inverters so that ⎛ 20 20 ⎞⎛ - j5 ⎞ ⎟ Vo = ⎜ V ⎟⎜ ⋅ (4∠ - 60°) + 50 o ⎠⎝ 10 ⎠ ⎝ - j10 = -j 2 -j ⋅ ( j2) ⋅ (4∠ - 60°) + ⋅ Vo 2 2 5 (1 + j 5) Vo = 4∠ - 60° Vo = Therefore, 4∠ - 60° = 3.922 ∠ - 71.31° 1+ j 5 v o ( t ) = 3.922 cos(1000t – 71.31°) 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 10, Problem 81. Use PSpice to determine V o in the circuit of Fig. 10.124. Assume ω = 1 rad/s. Figure 10.124 For Prob. 10.81. 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. Chapter 10, Solution 81. We need to get the capacitance and inductance corresponding to –j2 Ω and j4 Ω. 1 1 − j2 ⎯⎯ → C= = = 0.5F ω X c 1x 2 X j4 ⎯⎯ → L = L = 4H ω The schematic is shown below. When the circuit is simulated, we obtain the following from the output file. FREQ VM(5) VP(5) 1.592E-01 1.127E+01 -1.281E+02 From this, we obtain Vo = 11.27∠128.1o 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 10, Problem 82. Solve Prob. 10.19 using PSpice. Chapter 10, Solution 82. The schematic is shown below. We insert PRINT to print Vo in the output file. For AC Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we print out the output file which includes: FREQ 1.592 E-01 which means that VM($N_0001) 7.684 E+00 VP($N_0001) 5.019 E+01 Vo = 7.684∠50.19o 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 10, Problem 83. Use PSpice to find vo (t ) in the circuit of Fig. 10.125. Let i s = 2 cos(10 3 t ) A. Figure 10.125 For Prob. 10.83. Chapter 10, Solution 83. The schematic is shown below. The frequency is f = ω / 2π = 1000 = 159.15 2π When the circuit is saved and simulated, we obtain from the output file FREQ 1.592E+02 Thus, VM(1) 6.611E+00 VP(1) -1.592E+02 vo = 6.611cos(1000t – 159.2o) 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 10, Problem 84. Obtain V o in the circuit of Fig. 10.126 using PSpice. Figure 10.126 For Prob. 10.84. Chapter 10, Solution 84. The schematic is shown below. We set PRINT to print Vo in the output file. In AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes: FREQ VM($N_0003) 1.592 E-01 1.664 E+00 VP($N_0003) -1.646 E+02 Namely, Vo = 1.664∠-146.4o 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 10, Problem 85. Use PSpice to find V o in the circuit of Fig. 10.127. Figure 10.127 For Prob. 10.85. Chapter 10, Solution 85. The schematic is shown below. We let ω = 1 rad/s so that L=1H and C=1F. 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. When the circuit is saved and simulated, we obtain from the output file FREQ VM($N_0001) VP($N_0001) 1.592E-01 4.471E-01 1.437E+01 From this, we conclude that Vo = 447.1∠14.37˚ mV Checking using MATLAB and nodal analysis we get, >> Y=[1.5,-0.25,-0.25,0;0,1.25,-1.25,1i;-0.5,-1,1.5,0;0,1i,0,0.5-1i] Y= 1.5000 0 -0.5000 0 -0.2500 -0.2500 1.2500 -1.2500 -1.0000 1.5000 0 + 1.0000i 0 0 0 + 1.0000i 0 0.5000 - 1.0000i >> I=[0;0;2;-2] I= 0 0 2 -2 >> V=inv(Y)*I V= 0.4331 + 0.1110i = Vo = 0.4471∠14.38˚, answer checks. 0.6724 + 0.3775i 1.9260 + 0.2887i -0.1110 - 1.5669i 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 10, Problem 86. Use PSpice to find V 1 , V 2 , and V 3 in the network of Fig. 10.128. Figure 10.128 For Prob. 10.86. 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 10, Solution 86. The schematic is shown below. We insert three pseudocomponent PRINTs at nodes 1, 2, and 3 to print V1, V2, and V3, into the output file. Assume that w = 1, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit, we obtain the output file which includes: FREQ VM($N_0002) 1.592 E-01 6.000 E+01 FREQ VM($N_0003) 1.592 E-01 2.367 E+02 FREQ VM($N_0001) 1.592 E-01 1.082 E+02 VP($N_0002) 3.000 E+01 VP($N_0003) -8.483 E+01 VP($N_0001) 1.254 E+02 Therefore, V1 = 60∠30o V V2 = 236.7∠-84.83o V V3 = 108.2∠125.4o 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 10, Problem 87. Determine V 1 , V 2 , and V 3 in the circuit of Fig. 10.129 using PSpice. Figure 10.129 For Prob. 10.87. 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 10, Solution 87. The schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set Total Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After simulation, the output file includes: FREQ VM($N_0004) 1.592 E-01 1.591 E+01 FREQ VM($N_0001) 1.592 E-01 5.172 E+00 FREQ VM($N_0003) 1.592 E-01 2.270 E+00 VP($N_0004) 1.696 E+02 VP($N_0001) -1.386 E+02 VP($N_0003) -1.524 E+02 Therefore, V1 = 15.91∠169.6o V V2 = 5.172∠-138.6o V V3 = 2.27∠-152.4o 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 10, Problem 88. Use PSpice to find vo and io in the circuit of Fig. 10.130 below. Figure 10.130 For Prob. 10.88. 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 10, Solution 88. The schematic is shown below. We insert IPRINT and PRINT to print Io and Vo in the output file. Since w = 4, f = w/2π = 0.6366, we set Total Pts = 1, Start Freq = 0.6366, and End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes: FREQ VM($N_0002) 6.366 E-01 3.496 E+01 1.261 FREQ IM(V_PRINT2) IP 6.366 E-01 8.912 E-01 VP($N_0002) E+01 (V_PRINT2) -8.870 E+01 Therefore, Vo = 34.96∠12.6o V, Io = 0.8912∠-88.7o A vo = 34.96 cos(4t + 12.6o)V, io = 0.8912cos(4t - 88.7o )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 10, Problem 89. The op amp circuit in Fig. 10.131 is called an inductance simulator. Show that the input impedance is given by Ζin = Vin = jωLeq Ι in where Leq = R1R3 R4 C R2 Figure 10.131 For Prob. 10.89. 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. Chapter 10, Solution 89. Consider the circuit below. R1 R2 Vin 2 R3 1 Vin C 4 R4 3 − + Iin + − − + Vin At node 1, 0 − Vin Vin − V2 = R1 R2 R2 - Vin + V2 = V R 1 in (1) At node 3, V2 − Vin Vin − V4 = R3 1 jωC Vin − V2 - Vin + V4 = jωCR 3 (2) From (1) and (2), - Vin + V4 = - R2 V jωCR 3 R 1 in Thus, I in = Vin − V4 R2 = V R4 jωCR 3 R 1 R 4 in Z in = Vin jωCR 1R 3 R 4 = = jωL eq I in R2 where L eq = R 1R 3 R 4C R2 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 10, Problem 90. Figure 10.132 shows a Wien-bridge network. Show that the frequency at which the phase 1 shift between the input and output signals is zero is f = π RC , and that the necessary 2 gain is A v =V o /V i = 3 at that frequency. Figure 10.132 For Prob. 10.90. 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 10, Solution 90. 1 R = jωC 1 + jωRC 1 1 + jωRC Z3 = R + = jωC jωC Consider the circuit shown below. Let Z 4 = R || Z3 Vi + − R1 + Vo Z4 Vo = R2 R2 Z4 Vi − V R1 + R 2 i Z3 + Z 4 R Vo R2 1 + jωC − = R 1 + jωRC R 1 + R 2 Vi + 1 + jωC jωC = jωRC R2 − 2 jωRC + (1 + jωRC) R1 + R 2 Vo R2 jωRC = − 2 2 2 Vi 1 − ω R C + j3ωRC R 1 + R 2 For Vo and Vi to be in phase, Vo must be purely real. This happens when Vi 1 − ω2 R 2 C 2 = 0 1 ω= = 2πf RC or At this frequency, 1 2πRC Vo 1 R2 Av = = − Vi 3 R 1 + R 2 f= 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 10, Problem 91. Consider the oscillator in Fig. 10.133. (a) Determine the oscillation frequency. (b) Obtain the minimum value of R for which oscillation takes place. Figure 10.133 For Prob. 10.91. 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 10, Solution 91. (a) Let V2 = voltage at the noninverting terminal of the op amp Vo = output voltage of the op amp Z p = 10 kΩ = R o 1 jωC Z s = R + jωL + As in Section 10.9, Zp V2 = = Vo Z s + Z p Ro R + R o + jωL − j ωC ωCR o V2 = Vo ωC (R + R o ) + j (ω2 LC − 1) For this to be purely real, 1 ωo2 LC − 1 = 0 ⎯ ⎯→ ωo = fo = 1 2π LC = LC 1 2π (0.4 × 10 -3 )(2 × 10 -9 ) f o = 180 kHz (b) At oscillation, ωo CR o Ro V2 = = Vo ωo C (R + R o ) R + R o This must be compensated for by Vo 80 = 1+ =5 Av = V2 20 Ro 1 = R + Ro 5 ⎯ ⎯→ R = 4R o = 40 kΩ 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 10, Problem 92. The oscillator circuit in Fig. 10.134 uses an ideal op amp. (a) Calculate the minimum value of Ro that will cause oscillation to occur. (b) Find the frequency of oscillation. Figure 10.134 For Prob. 10.92. 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. Chapter 10, Solution 92. Let V2 = voltage at the noninverting terminal of the op amp Vo = output voltage of the op amp Zs = R o Z p = jωL || ωRL 1 1 = || R = 1 1 ωL + jR (ω2 LC − 1) jωC + jωC + jωL R As in Section 10.9, ωRL V2 ωL + jR (ω2 LC − 1) = = ωRL Vo Z s + Z p Ro + ωL + jR (ω2 LC − 1) V2 ωRL = Vo ωRL + ωR o L + jR o R (ω2 LC − 1) Zp For this to be purely real, ωo2 LC = 1 ⎯ ⎯→ f o = (a) 1 2π LC At ω = ωo , ωo RL V2 R = = Vo ωo RL + ωo R o L R + R o This must be compensated for by Vo Rf 1000k Av = = 1+ = 1+ = 11 V2 Ro 100k Hence, R 1 = ⎯ ⎯→ R o = 10R = 100 kΩ R + R o 11 (b) fo = 1 2π (10 × 10 -6 )(2 × 10 -9 ) f o = 1.125 MHz 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 10, Problem 93. Figure 10.135 shows a Colpitts oscillator. Show that the oscillation frequency is 1 fo = 2π LCT where CT = C1C 2 / (C1 + C 2 ) . Assume Ri >> X C 2 Figure 10.135 A Colpitts oscillator; for Prob. 10.93. (Hint: Set the imaginary part of the impedance in the feedback circuit equal to zero.) 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 10, Solution 93. As shown below, the impedance of the feedback is jωL 1 jωC2 ZT = 1 jωC1 ZT ⎛ 1 1 ⎞ ⎟ || ⎜ jωL + jωC1 ⎝ jωC 2 ⎠ -j ⎛ -j ⎞ 1 ⎜ jωL + ⎟ − ωLC 2 ωC1 ⎝ ωC 2 ⎠ ω = ZT = -j -j j (C1 + C 2 − ω2 LC1C 2 ) + jωL + ωC1 ωC 2 In order for Z T to be real, the imaginary term must be zero; i.e. C1 + C 2 − ωo2 LC1C 2 = 0 ωo2 = fo = C1 + C 2 1 = LC1C 2 LC T 1 2π LC T 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 10, Problem 94. Design a Colpitts oscillator that will operate at 50 kHz. Chapter 10, Solution 94. If we select C1 = C 2 = 20 nF C1 C 2 C1 CT = = = 10 nF C1 + C 2 2 Since f o = 1 2π LC T L= , 1 1 = = 10.13 mH 2 2 (2πf ) C T (4π )(2500 × 10 6 )(10 × 10 -9 ) Xc = 1 1 = = 159 Ω ωC 2 (2π )(50 × 10 3 )(20 × 10 -9 ) We may select R i = 20 kΩ and R f ≥ R i , say R f = 20 kΩ . Thus, C1 = C 2 = 20 nF, L = 10.13 mH R f = R i = 20 kΩ 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 10, Problem 95. Figure 10.136 shows a Hartley oscillator. Show that the frequency of oscillation is 1 fo = 2π C (L1 + L2 ) Figure 10.136 A Hartley oscillator; For Prob. 10.95. Chapter 10, Solution 95. First, we find the feedback impedance. C ZT L2 L1 ⎛ 1 ⎞ ⎟ Z T = jωL1 || ⎜ jωL 2 + jωC ⎠ ⎝ ⎛ j ⎞ ⎟ jωL1 ⎜ jωL 2 − ⎝ ω2 L1C (1 − ωL 2 ) ωC ⎠ ZT = = j j (ω2 C (L1 + L 2 ) − 1) jωL1 + jωL 2 − ωC In order for Z T to be real, the imaginary term must be zero; i.e. ωo2 C (L1 + L 2 ) − 1 = 0 1 ω o = 2π f o = C ( L1 + L 2 ) fo = 1 2π C (L 1 + L 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 10, Problem 96. Refer to the oscillator in Fig. 10.137. (a) Show that V2 1 = Vo 3 + j (ωL / R − R / ωL ) (b) Determine the oscillation frequency f o . (c) Obtain the relationship between R1 and R2 in order for oscillation to occur. Figure 10.137 For Prob. 10.96. 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 10, Solution 96. (a) Consider the feedback portion of the circuit, as shown below. jωL Vo V2 = + − V1 R V2 R jωL V R + jωL 1 ⎯ ⎯→ V1 = R + jωL V2 jωL (1) Applying KCL at node 1, Vo − V1 V1 V1 = + jωL R R + jωL ⎛1 ⎞ 1 ⎟ Vo − V1 = jωL V1 ⎜ + ⎝ R R + jωL ⎠ ⎛ j2ωRL − ω2 L2 ⎞ ⎟ Vo = V1 ⎜1 + R (R + jωL) ⎠ ⎝ (2) From (1) and (2), ⎛ R + jωL ⎞⎛ j2ωRL − ω2 L2 ⎞ ⎟V ⎟⎜1 + Vo = ⎜ R (R + jωL) ⎠ 2 ⎝ jωL ⎠⎝ Vo R 2 + jωRL + j2ωRL − ω2 L2 = V2 jωRL V2 = Vo 1 R − ω2 L2 3+ jωRL 2 V2 1 = Vo 3 + j (ωL R − R ωL ) 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. (b) Since the ratio V2 must be real, Vo ωo L R − =0 R ωo L R2 ωo L = ωo L ωo = 2πf o = fo = (c) R L R 2π L When ω = ωo V2 1 = Vo 3 This must be compensated for by A v = 3 . But R2 Av = 1+ =3 R1 R 2 = 2 R1 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.