(PDF) Introductory Circuit Analysis Twelfth Edition

Online Instructor’s Manual to accompany Introductory Circuit Analysis Twelfth Edition Robert L. Boylestad Prentice Hall Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Contents CHAPTER 1 1 CHAPTER 2 10 CHAPTER 3 15 CHAPTER 4 25 CHAPTER 5 32 CHAPTER 6 42 CHAPTER 7 56 CHAPTER 8 68 CHAPTER 9 92 CHAPTER 10 111 CHAPTER 11 130 CHAPTER 12 149 CHAPTER 13 156 CHAPTER 14 164 CHAPTER 15 174 CHAPTER 16 196 CHAPTER 17 203 CHAPTER 18 222 CHAPTER 19 252 CHAPTER 20 265 CHAPTER 21 279 CHAPTER 22 311 CHAPTER 23 318 CHAPTER 24 333 CHAPTER 25 342 CHAPTER 25 352 iii ___________________________________________________________________________________________ Copyright ©2010 Pearson Education, Inc., publishing as Prentice Hall, 1 Lake Street, Upper Saddle River, New Jersey 07458. All rights reserved. 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Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. 10 9 8 7 6 5 4 3 2 1 ISBN-13: 978-0-13-504082-9 ISBN-10: 0-13-504082-5 Chapter 1 1.  2.  3.  4. = 5. 6. d 20,000 ft  1 mi   60 s   60 min  = = 1363.64 mph t 10 s  5,280 ft  1 min   1 h   1h  4 min   = 0.067 h  60 min  d 31 mi =  = 29.05 mph t 1.067 h  3 ft  100 yds   1 yd   1 mi   5,280 ft  = 0.0568 mi   60 mi  1 h  1 min  = 0.0167 mi/s h  60 min   60 s  t= 7. a. b. c. d   0.0568 mi = 3.40 s 0.0167 mi/s 95 mi  5,280 ft   1 h  1 min  = 139.33 ft/s h  mi   60 min   60 s  d 60 ft t=  = 0.431 s  139.33 ft/s d 60 ft  60 s   60 min   1 mi  =  = 40.91 mph t 1 s 1 min   1 h   5,280 ft  8.  9.  10.  11. MKS, CGS, C = 12.  0.7378 ft - lb  1000 J   = 737.8 ft-lbs 1J   5 5 5 (F  32)  (68  32)  (36) = 20 9 9 9 SI: K = 273.15 + C = 273.15 + 20 = 293.15 Chapter 1 1 13.  3 ft  12 in.   2.54 cm  0.5 yd     = 45.72 cm 1 yd   1 ft   1 in.  14. a. °F = 2(°C) + 30° = 40° + 30° = 70° b. °F = c. very close d. 30°C  90°F vs. 86° 5°C  40°F vs 41° a. 14.6 b. c. 1,046.1 d. e. 3.14159 = 3.1 a. 14.60 b. c. 1,046.06 d. e. 3.14159 = 3.14 a. 14.603 b. c. 1,046.060 d. e. 3.14159 = 3.142 a. 104 15. 16. 17. 18. 9 9 (C)  32  (20)  32 = 68° 5 5 e. 100 b. 106 56.0 1 = 0.0625 = 0.1 16 56.04 1 = 0.0625 = 0.06 16 56.042 1 = 0.0625 = 0.063 16 c. 103 d. 103 c. 2.4  106 d. 60  103 f. 101 5  103 a. 15  103 e. 4.02  104 f. 2  1010 20. a. b. c. d. 4.2  103 + 48.0  103 = 52.2  103 = 5.22  104 90  103 + 360  103 = 450  103 = 4.50  105 50  105  6  105 = 44  105 = 4.4  104 1.2  103 + 0.05  103  0.4  103 = 0.85  103 = 850 21. a. b. c. d. e. f. (102)(103) = 105 = 100  103 (102)(103) = 101 = 10 (103)(106) = 1  109 (102)(105) = 1  103 (106)(10  106) = 10 (104)(108)(1028) = 1  1024 19. 2 b. Chapter 1 22. a. b. c. d. (50  103)(2  103) = 100  100 = 100 (2.2  103)(2  103) = 4.4  100 = 4.4 (82  106)(1.2  106) = 98.4 (30  104)(4  103)(7  108) = 840  101 = 8.40  103 23. a. b. c. d. e. f. 102/104 = 102 = 10  103 102/103 = 105 = 10  106 104/103 = 107 = 10  106 107/102 = 1.0  109 1038/104 = 1.0  1042 100 / 10 2 = 101/102 = 1  103 24. a. b. c. d. (2  103)/(8  105) = 0.25  108 = 2.50  107 (4  103)/(4  106) = 4/4  109 = 1  109 (22  105)/(5  105) = 22/5  100 = 4.4 (78  1018)/(4  106) = 1.95  1025 25. a. c. (102)3 = 1.0  106 (104)8 = 100.0  1030 26. a. b. c. d. 27. a. (2  102)2 = 4  104 (5  103)3 = 125  109 (4  103)(3  103)2 = (4  103)(9  104) = 36  101 = 360 ((2  103)(0.8  104)(0.003  105))3 = (4.8  103)3 = (4.8)3  (103)3 = 110.6  109 = 1.11  1011 (103)2 = 1.0  106 (10 2 )(10 4 ) = 102/103 = 1.0  105 3 10 3 2 (10 ) (10 2 ) (10 6 )(10 2 ) 10 4   4 = 1.0  108 4 4 10 10 10 3 4 (10 )(10 ) = 107/104 = 1.0  1011 4 10 b. c. d. e. (10 )(10 ) (10 ) 10  2 2 28. a. b. c. Chapter 1 (104)1/2 = 10.0  103 (107)9 = 1.0  1063 (1  104)3(102)/106 = (1012)(102)/106 = 1010/106 = 1.0  1016 2 f. b. d. 2 3 3  1 1 = 1.0  101  3 (10 )(10 ) 10 4 (3  10 2 ) 2 (10 2 ) = (9  104)(102)/(3  104) = (9  106)/(3  104) = 3  102 = 300 4 3  10 (4  10 4 ) 2 16  10 8 = 2  105  3 3 (20) 8  10 (6  104 ) 2 36  108 = 9.0  1012  2 2 4 (2  10 ) 4  10 3 d. (27  10 6 )1 / 3 3  10 2 = 1.5  107 = 150.0  109  5 5 2  10 2  10 e. (4  10 3 ) 2 (3  10 2 ) (16  10 6 )(3  10 2 ) 48  10 8 = 24.0  1012   2  10  4 2  10  4 2  10  4 f. (16  106)1/2(105)5(2  102) = (4  103)(1025)(2  102) = 8  1020 = 800.0  1018 g.  (3  103 )3  1.60  102  (2  102 )(8  104 )       5 2 (7  10 ) 2 1/ 2 (27  109 )(2.56  104 )(16  102 )1/ 2 49  1010 (69.12  105 )(4  101 ) 276.48  106   49  1010 49  1010 4 3 = 5.64  10 = 56.4  10  29. Scientific: Engineering: 30. Scientific Engineering: 4 a. b. 2.05 × 101 5.04 × 104 c. 6.74 × 104 d. 4.60 × 102 a. b. 20.46 × 100 50.42 × 103 c. 674.00 × 106 d. 46.00 × 103 a. b. 5.0 × 102 4.5 × 101 c. d. 1/32 = 0.03125 = 3.125 × 102 3.14159 = 3.142 × 100 a. b. 50.0 × 103 0.045 × 103 c. d. 31.25 × 103 3.142 × 100 Chapter 1 +2 31. a. 6  104 = 0.06  10+6 = 0.06 × 10+6 3 3 b. 0.4  103 = 400  106 = 400 × 106 +3 2 c. +3 50  105 = 5000  103 = 5  106 = 0.005  109 = 0.005 × 109 3 +2 3 3 +4 d. +3 3 12  107 = 0.0012  103 = 1.2  106 = 1200  109 = 1200 × 109 4 +3 +3 3 32. a. 0.05  100 s = 50  103 s = 50 ms +3 +3 b. 2000  106 s = 2  103 s = 2 ms 3 3 c. 0.04  103 s = 40  106 s = 40 s +3 +6 d. 8400  1012 s  0.0084  106 s = 0.0084 s 6 Chapter 1 5 +3 increase by 3 100 e. 100  103  103 m = 0.1  103 m = 0.1 km 3 33. 34. 35. 6 a.  60 s  1.5 min   = 90 s 1 min  b.  60 min   60 s  2 × 102 h   = 72 s   1 h  1 min  c.  1 s  0.05 s   6  = 0.05  106 s = 50  103 s 10 s  d.  1 mm  0.16 m  3  = 0.16  103 mm = 160 mm 10 m  e.  1 ns  1.2  107 s  9  = 1.2  102 ns = 120 ns 10 s  f. 1 min   1 h  1 day  4  108 s   = 4629.63 days    60 s   60 min   24 h  a. 100 cm  = 8000  103 cm = 8 cm 80  103 m    1m  b.  1 m   1 km  60 cm  = 60  105 km    100 cm  1000 m  c.  1 m  12 × 103 m  6  = 12  103 × 10+6 m = 12 × 103 m 10 m  d.  1m  1m  4 2 60 cm2    = 60  10 m 100 cm   100 cm  a.  1m  100 in.   = 2.54 m  39.37 in.  b. 12 in.  4 ft    1 ft   1m   39.37 in.  = 1.22 m   Chapter 1 c. d. 36. 37. 38. 39. 40. 41.  4.45 N  6 lb   = 26.7 N  1 lb   1 N   1 lb  60  103 dynes  5   = 0.13 lb 10 dynes   4.45 N  e.  1 in.   1 ft  150,000 cm    = 4921.26 ft  2.54 cm  12 in.  f.  5280 ft  12 in.   1 m  0.002 mi     = 3.22 m  1 mi   1 ft   39.37 in.  1 yd  5280 ft   = 1760 yds  3 ft   12 in.   1 m  5280 ft    = 1609.35 m, 1.61 km  1 ft   39.37 in.  5280 ft, 60 mi 5280 ft h 1 mi 12 in. 1 ft 1m 39.37 in. 1h 60 min 1000 m   39.37 in.   1 ft   1 mi  10 km       1 km   1 m  12 in.   5280 ft  d 6.214 mi 1 mi = = 40.39 min ,t=  1 mi  6.5 min 6.5 min 1 min 60 s = 26.82 m/s = 6.214 mi  3 ft  12 in.  100 yds    = 3600 in  3600 quarters 1 yd   1 ft  500 mi = 8.33 h = 8 h:19.8 min  60 mi/h d 500 mi = 6.67 h = 6h:40.2 min 75 mph: t =   75 mi/h difference = 1h:28.6 minutes 60 mph: t= d  cm   0.016 h   60 min   60 s   1 m  = 345.6 m d = t = 600  s    1 h  1 min  100 cm  Chapter 1 7 42.  14 ft  d = 86 stories    story  d d 1605 steps  1 minute  t    802.5 seconds  = 13.38 minutes 2 steps t   60 seconds  second = 43. 44.  14 ft   1 mile  d = (86 stories)    1204 ft   = 0.228 miles  5,280 ft   story  min 10.22 min  = 44.82 min/mile mile 0.228 miles 5 min 1 mile  5,280 ft  1056 ft  ,  mile 5 min  1mile  minute a.  1 Btu  5 J = 4.74  103 Btu  1054.35 J  b.  1 m3  1 gallon   4 3 24 ounces    = 7.1  10 m  128 ounces   264.172 gallons  c.  86,400 s  5 1.4 days   = 1.21  10 s 1 day   d.  264.172 gallons  1 m3   1 m3   46. 6(4 × 2 + 8) = 96 47. (42 + 6/5)/3 = 14.4 48. 2 5  3  8 pints   = 2113.38 pints  1 gallon  2 2 = 5.044 49. MODE = DEGREES: cos 21.87 = 0.928 50. MODE = DEGREES: tan1(3/4) = 36.87 51. 8  14 ft  distance = 86 stories   = 1204 ft  story  d d 1204 ft = 1.14 minutes t  t  1056 ft min = 45.   1 step   9  = 1605 steps  ft   12  400 /(6 2   10 / 5) = 7.071 Chapter 1 52. 205  106 53. 1.20  1012 54. 6.667  106 + 0.5  106 = 7.17  106 Chapter 1 9 Chapter 2 1.  2. a. F= k Q1Q2 (9  109 )(1 C)(2 C) = 18  109 N  r2 (1 m) 2 b. F=k Q1Q2 (9  109 )(1 C)2 C  = = 2  109 N (3 m) 2 r2 c. F= k Q1Q2 (9  109 )(1 C)(2 C) = 0.18  109 N  2 2 r (10 m) d. Exponentially, a. r = 1 mi: 3. r3 10 m F 18  109 N = 10 while 1  = 100  r1 1 m F2 0.18  109 N  5280 ft  12 in.   1 m  1 mi   = 1609.35 m    1 mi   1 ft   39.37 in.  kQ1Q2 (9  109 )(8  106 C)(40  10-6 C) 2880  103 F=   r2 (1609.35 m)2 2.59  106 = 1.11 N b. r =10 ft: 12 in.   1 m  10 ft    = 3.05 m  1 ft   39.37 in.  kQ1Q2 2880  103 2880  103 = 0.31 N F=   9.30 r2 (3.05 m) 2 c. 1 in.  1 m  = 1.59 mm 16  39.37 in.  kQ1Q2 2880  103 2880  103    1138.34  103 N r2 (1.59  103 m) 2 2.53  106 = 1138.34 kN F= 4.  5. Q1 = Q2 = Q; F1  6. F= 10 kQ1Q2 r r2 F1r12 kQ 2 kQ 2 k 2 ; Q F     2 2 k r12 r22 r2 kQ1Q2  F  F1r12  r12 and F1 F    2 r22  k  (9  109 )(20  106 ) 2 = 10 mm 3.6  104 Chapter 2 7. F= a. b. kQ1Q2 kQ Q  1.8  1 22  kQ1Q2 = 4(1.8) = 7.2 2 r (2 m) kQ1Q2 7.2 = 72 mN  2 r (10) 2 Q1/Q2 = 1/2  Q2 = 2Q1 7.2 = kQ1Q2 = (9  109)(Q1)(2Q1) = 9  109 2Q12 F=   7 .2 7 .2  Q12  Q1  = 20 C 9 18  10 18  109 Q2 = 2Q1 = 2(2  105 C) = 40 C W 1.2 J  = 60 kV Q 20 C 8. V= 9. W = VQ = (60 V)(8 mC) = 0.48 J 10. Q=  6.242  1018 electrons  W 120  J 15  = 6 mC   = 37.45 × 10 electrons   V 20 mV 1C 11. Q= W 72 J =8C  V 9V 12. a. b. W = QV = (1 × 1012 electrons)(40 V) = 40 × 1012 eV 1C   40 × 1012 eV  18  = 6.41 μJ 6.242 10 electrons   13. I= Q 12 mC  = 4.29 mA t 2.8 s 14. I = Q 312 C = 2.60 A  t (2)(60 s) 15. Q = It = (40 mA)(0.8)(60 s) = 1.92 C 16. Q = It = (250 mA)(1.2)(60 s) = 18.0 C 17. t= 18. Q 6 mC  =3s I 2 mA 1C   21.847  1018 electrons  18  = 3.5 C  6.242  10 electrons  Q 3.5 C I=  = 0.29 A t 12 s Chapter 2 11 19. Q = It = (4 mA)(90 s) = 360 mC  6.242  1018 electrons  18 360 mC   = 2.25  10 electrons 1 C   20. I= 21. 22. Q 86 C = 1.194 A > 1 A (yes)  (1.2)(60 s) t 1C   0.84  1016 electrons  18  = 1.346 mC 6.242  10 electrons   Q 1.346 mC I=  = 22.43 mA t 60 ms a. Q = It = (2 mA)(0.01 s) = 2  1011 C  6.242  1018 electrons   1 ¢  2  1011 C    1C    electron  = 1.25  108 ¢ = $1.25  106 = 1.25 million Q = It = (100 A)(1.5 ns) = 1.5  1013 C  6.242  1018 electrons   $1  1.5  1013 C    = 0.94 million 1C    electron  (a) > (b) b. 23. 24. Q = It = (200  103 A)(30 s) = 6 C W 40 J V= = 6.67 V  Q 6C  420 C  Q = It =   (0.5 min) = 210 C  min  W 742 J = 3.53 V V=  Q 210 C W 0.4 J  = 0.0167 C V 24 V Q 0.0167 C I=  = 3.34 A t 5  103 s 25. Q= 26. I= 27. Ah = (0.8 A)(75 h) = 60.0 Ah 28. t(hours) = 12 Ah rating 200 Ah =5A  40 h t (hours) Ah rating 32 Ah = 25 h  I 1.28 A Chapter 2 29.  60 min   60 s  6 40 Ah(for 1 h): W1 = VQ = VIt = (12 V)(40 A)(1 h)   1 min  = 1.728  10 J 1 h    60 min 60 s    6 60 Ah(for 1 h): W2 = (12 V)(60 A)(1 h)   1 min  = 2.592  10 J 1 h    Ratio W2/W1 = 1.5 or 50% more energy available with 60 Ah rating. 1 min   1 h  = I(16.67  103 h) For 60 s discharge: 40 Ah = It = I 60 s      60 s   60 min  40 Ah = 2400 A and I = 16.67  10-3 h 1 min   1 h  60 Ah = It = I 60 s  = I(16.67  103 h)     60 s   60 min  60 Ah = 3600 A and I = 16.67  10-3 h I2/I1 = 1.5 or 50 % more starting current available at 60 Ah 30. 0.75(18 Ah) = 13.5 Ah   250 mA 31. (18 Ah  15.5 Ah)/18 Ah × 100% = 13.89% 32. At 100 mA, discharge time  120 H; At 25 mA, discharge time  425 h;  300 h more at 25 mA 33. I= 3 Ah = 500 mA 6.0 h  60 min   60 s  Q = It = (500 mA)(6 h)   = 10.80 kC   1 h  1 min  W = QV = (10.8 kC)(12 V)  129.6 kJ 34.  35.  36.  37. a. b. c. Chapter 2  2.54 cm  = 1.27 cm 0.5 in   1 in   30 kV  = 38.1 kV 1.27 cm   cm   270 kV  1.27 cm  = 342.9 kV  cm  342.9 kV:38.1 kV = 9:1 13 38.  39.  40.  41.  14 Chapter 2 Chapter 3 1. a. 0.5 in. = 500 mils b. 1000 mils  0.02 in.   = 20 mils  1 in.  c. 1 1000 mils  in. = 0.25 in.   = 250 mils 4  1 in.  d. e. 2. 3.  39.37 in  1000 mils  10 mm = 10 × 103 m  = 393.7 mils  1 m   1 in  3 12 in.  10 mils  0.01 ft   = 120 mils   1 ft   1 in.  f.  1 in.  1000 mils  0.1 cm   = 39.37 mils   2.54 cm   1 in.  a. ACM = (30 mils)2 = 900 CM b. 0.016 in. = 16 mils, ACM = (16 mils)2 = 256 CM c. 1" = 0.125" = 125 mils, ACM = (125 mils)2 = 15.63  103 CM 8 d.  1 in.  1 cm    2.54 cm  e. 12 in.  1000 mils  2 3 0.02 ft    1 in.  = 240 mils, ACM = (240 mils) = 57.60  10 CM 1 ft    f.  39.37 in.  1000 mils  = 157.48 mils, ACM = (157.48 mils)2 = 24.8 × 103 CM 4 × 103 m      1 m   1 in  1000 mils  2 3  1 in.  = 393.7 mils, ACM = (393.7 mils) = 155  10 CM   ACM = (dmils )2  dmils = ACM a. d = 1600 CM = 40 mils = 0.04 in. b. d= 820 CM = 28.64 mils = 0.029 in. c. d= 40,000 CM = 200 mils = 0.2 in. Chapter 3 15 d. d= 625 CM = 25 mils = 0.025 in. e. d= 6.25 CM = 2.5 mils = 0.0025 in. f. d= 3  103 CM = 54.77 mils = 0.055 in. 4. 0.02 in. = 20 mils, ACM = (20 mils)2 = 400 CM (200 ) l = 5.19  R =  = (10.37) 400 CM A 5. a. A=   80  l  = 544 CM  17 R  2 .5   b. d= ACM  544 CM = 23.32 mils = 23.3  103 in. 6. 7. 1 " = 0.03125" = 31.25 mils, ACM = (31.25 mils)2 = 976.56 CM 32 l RA (2.2 )(976.56 CM) = 3.58 ft R= l=  R  600 a. ACM =  d= larger c. smaller  = 9. a. b. c. 16 942.73 CM = 30.70 mils = 30.7  103 in. b. 8. l (10.37)(300) = = 942.73 CM 3.3  A RA (500 )(94 CM) = 47  nickel  l 1000 1/32 = 0.03125 = 31.25 mils, ACM = (31.25 mils)2 = 976.56 CM l RA (3.12 )(976.56 CM) R= l   = 293.82 ft A  10.37 293.82 1000 (5)(293.82)  x = 1.47 lbs x 5 lb 1000 9 9 C  32  (40)  32 = 40° 5 5 9 9 105° C: F = C  32  (105)  32 = 221° 5 5 F° = 40°  221° 40° C: F = Chapter 3 10. a. 3   0.375 = 375 mils 8 4.8 = 4800 mils  4 /  CM  6 A = (375 mils)(4800 mils) = 1.8 × 106 sq. mils   = 2.29 × 10 CM 1 sq mil   b. 11. a. 1  = 0.083 in. = 83 mils 12 ACM = = (83 mils)2 = 6.89 × 103 CM 2.29  106 CM (#12) = 332.37 wires 6.89  103 CM 3" = 3000 mils, 1/2" = 0.5 in. = 500 mils Area = (3  103 mils)(5  102 mils) = 15  105 sq. mils  4 /  CM  5 15  105 sq mils   = 19.108  10 CM  1 sq mil  R=  b. l (10.37)(4) = 21.71  = A 19.108  105 CM l (17)(4) = = 35.59  A 19.108  105 CM Aluminum bus-bar has almost 64% higher resistance. R=  12. l2 = 2l1, A2 = A1/4, 2 = 1  2l2 R2 2l A l A A  2  22 1  1 1 =8 1l1 R1 1l1 A2 l1 A1 / 4 A1 and R2 = 8R1 = 8(0.2 ) = 1.6  R = 1.6   0.2  = 1.4  13. A= d 2 4 d= 4A   4(0.04 in.2 )  = 0.2257 in. dmils = 225.7 mils ACM = (225.7 mils)2 = 50,940.49 CM l 1 1 R1 lA lA A1   11 2  1 2 (1 = 2) R2  l2  2l2 A1 l2 A1 2 A2 R l A (800 m)(300 ft)(40,000 CM) = 942.28 m and R2 = 1 2 1  (200 ft)(50,940.49 CM) l1 A2 Chapter 3 17 14. a. b. 15. #12 = 6,529.9 CM, #14 = 4,106.8 CM 6,529.9 CM  4,106.8 CM  100% = 59% larger 4,106.8 CM #12 20 A #12 6,529.9 CM  = 1.33,  = 1.59 #14 4,106.8 CM #14 15 A Imax ratio = 1.33 vs Area ratio = 1.59 1.59  1.33  100% = 19.55% higher ratio for area 1.33 a. #9 13,094 CM  = 2 yes #12 6,529.9 CM b. #0 105,530 CM  = 16.16 yes #12 6,529.9 CM #0 150 A  = 7.5 #12 20 A 16. a. b. 17. 18. 18 #10 10,381 CM   10.16  10 yes # 20 1,021.5 CM # 20 1,021.5 CM  = 103.28 # 40 9.89 CM yes  100 l (10.37)(30) 311.1 CM = 51,850 CM  #3   R 6 m 6  103 but 110 A  #2 a. A=  b. A=  a. A/CM = 230 A/211,600 CM = 1.09 mA/CM b.   1.09 mA  1 CM  1000 mils  1000 mils  = 1.39 kA/in.2      CM  sq mils   1 in.   1 in.  4  c.  1 in.2  2 5 kA   = 3.6 in. 1.39 kA  l (10.37)(30) 311.1 CM = 103,700 CM  #0   R 3 m 3  10 3 Chapter 3 19. 20. 234.5  10 234.5  80 ,  2 R2 R2 = (314.5)(2 ) = 2.57  244.5 236  0 236  100  R2 0.02  (0.02 )(336) = 0.028  R2 = 236 5 5 (F  32)  (32  32) = 0 (=32F) 9 9 5 C = (68  32)  20 (=68F) 9 234.5  20 234.5  0  4 R2 (234.5)(4 ) = 3.69  R2 = 254.5 21. C= 22. a. b. c. d. Chapter 3 5 5 F  32  (70  32) = 21.11° 9 9 5 °C = (60  32) = 15.56° 9 234.5  21.11 234.5  15.56  0.025  R2 (250.06)(0.025 ) = 24.46 mΩ R2 = 255.61 °C = 5 (50  32) = 10° 9 234.5  21.11 234.5  10  0.025  R2 (244.5)(0.025 ) = 23.91 mΩ R2 = 255.61 °C = Part a: 25 mΩ  24.46 mΩ = 0.54 mΩ Part b: 24.45 mΩ  23.91 mΩ = 0.55 mΩ Linear 40°F  23.91 mΩ  0.55 mΩ = 23.36 mΩ 5 (30  32) = 34.44° 9 234.5  21.11 234.5  34.44  25 m R2 (25 m)(200.06) R2 = = 19.57 mΩ 255.61 Yes, 25 mΩ  19.57 mΩ = 5.43 mΩ °C = 19 e. 23. a. b. 24. 5 (120  32) = 48.89° 9 234.5  21.11 234.5  48.89  25 m R2 (25 m)(283.39) = 27.72 mΩ R2 = 255.61 Yes, 2.72 mΩ °C = 234.5  4 234.5  t2 ,  1 1 .1  234.5  4 234.5  t2 ,  1 0.1  t2 = 27.85C t2 = 210.65C 68F  20C a. 234.5+20 234.5  T2  1 2 2(254.5)  234.5  T2 1 T2  274.5C b. #10 = 0.9989 /1000 c. d mils  ACM  10,381 CM  101.89 mils din = 0.102 in  25. 26. 20 1  10 1 1 1 = 0.003929  0.00393   Ti  20C 234.5  20 254.5 a. 20 = b. R = R20[1 + 20(t  20C)] 1  = 0.8 [1 + 0.00393(t  20)] 1.25 = 1 + 0.00393t  0.0786 1.25  0.9214 = 0.00393t 0.3286 = 0.00393t 0.3286 = 83.61C t= 0.00393 R = R20[1 + 20(t  20C)] = 0.4 [1 + 0.00393(16  20)] = 0.4 [1  0.01572] = 0.39  Chapter 3 27. Table: 1000 of #12 copper wire = 1.588  @ 20C 5 5 C = (F  32) = (115  32) = 46.11C 9 9 R = R20[1 + 20(t  20C)] = 1.588 [1 + 0.00393(46.11  20)] = 1.75  28. R = 29. R = 30. a. 31. 10 k  3.5 k = 6.5 k 32. 6.25 k and 18.75 k 33.  34. a. b. c. d. e. 820   5%, 820   41 , 779   861  220   10%, 220   22 , 198   242  91 k  20%, 91 k  18.2 k, 77.8 k  109.2 k 9.1 k  5%, 9100   455 , 8,645   9,555 k 3.9 MΩ  20%, 3.9 MΩ  0.78 MΩ, 3.12 MΩ  4.68 M 35. a. b. c. d. 68  = Blue, Gray, Black, Silver 0.33  = Orange, Orange, Silver, Silver 22 k = Red, Red, Orange, Silver 5.6 M = Green, Blue, Green, Silver 36. a. 10   20% 8 15 Ω ± 20% 12 10 Ω ± 10% 9 15 Ω ± 10% 13.5 Rnominal 22  (PPM)( T) = (200)(65  20) = 0.198  6 10 106 R = Rnominal + R = 22.198  Rnominal 100  (PPM)( T) = (100)(50  20) = 0.30  6 10 106 R = Rnominal + R = 100  + 0.30  = 100.30  b. 37. 38. 2 times larger b. 4 times larger no overlap, continuance 12 18 no overlap 11 16.5 470 Ω ± 10% = 470 Ω ± 47 Ω = 423 Ω  517 Ω Yes No change Chapter 3 21 39. a. b. c. d. 621 = 62  101  = 620  = 0.62 k 333 = 33  103  = 33 k Q2 = 3.9  102  = 390  C6 = 1.2  106  = 1.2 M 40. a. G= 1 1 = 8.33 mS  R 120  b. G= 1 = 0.25 mS 4 k c. G= a. Table 3.2, /1000 = 1.588  1 1 G=  = 629.72 mS R 1.588  A 6529.9 CM (Table 3.2) = 629.69 mS (Cu) or G =  l (10.37)(1000) b. G= 6529.9 CM = 384.11 mS (Al) (17)(1000) a. G1 = 1 1 1 = 100 mS, G2 = = 50 mS, G3 = = 10 mS 10  20  100  b. G2:G1 = 50 mS: 100 mS = 1:2 whereas R2:R1 = 20 Ω:10 Ω = 2:1. The rate of change is the same although one is increasing and the other decreasing. c. inverse  linear 41. 42. 43. 1 = 0.46 S 2.2 M Ga > Gb > Gc vs. Rc > Rb > Ra l 2 5  2 A1 = A1, l2 = 1   l1  1 , 2 = 1 3 3 3  3 A  l1  1 1  3  A1 1 G1 l1 l A   22 1    A2 1l1 A2 5  5 G2  2 l1  A1  l2 3  A2 = 1 G2 = 5G1 = 5(100 S) = 500 S 44.  45.  46.  22 Chapter 3 47.  48. 1  2.54 cm  in. = 0.083 in.   = 0.21 cm 12  1 in.  d2 (3.14)(0.21 cm) 2 = 0.035 cm2 4 4 RA (2 )(0.035 cm 2 )  = 40,603 cm = 406.03 m l=  1.724  106 A= 49. a.  1 "  2.54 cm   2.54 cm  = 1.27 cm, 3 in.     = 7.62 cm 2  1"   1 in .  12 in.   2.54 cm  4 ft    = 121.92 cm  1 ft   1 in.  l (1.724 x 10 -6)(121.92 cm) R=  = 21.71 Ω A (1.27 cm)(7.62 cm)  (2.825  106 )(121.92 cm) = 35.59 Ω  A (1.27 cm)(7.62 cm) b. R=  c. increases d. decreases  = 100  d =  = 250  106 = 2.5 cm 100 50. Rs = 51. R l (150 )(1/ 2 in.) R = Rs l  w = s = = 0.15 in. R 500  w 52. a. d = 1 in. = 1000 mils ACM = (103 mils)2 = 106 CM RA (1 m)(106 CM) 1 =  = 1 CM-/ft l 103 ft b. 1 in. = 2.54 cm  d 2  (2.54 cm)2 A=  = 5.067 cm2 4 4 12 in.   2.54 cm  l = 1000 ft    = 30,480 cm  1 ft   1 in.  d 2 = Chapter 3 100 RA (1 m)(5.067 cm 2 )  = 1.66  107 -cm l 30,480 cm 23 c. k=  2 1.66  107 -cm = 1.66  107  1 CM- / ft 1 53.  54.  55.  56.  57.  58. a. 50C specific resistance  105 -cm 50C specific resistance  500 -cm 200C specific resistance  7 -cm b. negative c. No d. = a. Log scale: 10 fc  3 k 100 fc  0.4 k b. negative c. no—log scales imply linearity d. 1 k   30 fc 10 k   2 fc R 10 k   1 k  = = 321.43 /fc fc 30 fc  2 fc 59.   cm 300  30 270   cm  3.6 -cm/C   125  50 75 C T and 60. 24 a. @ 0.5 mA, V  195 V @ 1 mA, V  200 V @ 5 mA, V  215 V b. Vtotal = 215 V  195 V = 20 V c. 5 mA:0.5 mA = 10:1 compared to 215 V: 200 V = 1.08:1 R = 321.43 /fc  fc Chapter 3 Chapter 4 1. V = IR = (5.6 mA)(220 ) = 1.23 V 2. I= V 24 V  = 3.53 A R 6.8  3. R= V 24 V = 16 k = I 1.5 mA 4. I= 12 V V  = 300 A R 40 103  5. V = IR = (3.6 A)(0.02 M) = 0.072 V = 72 mV 6. I= V 120 V  = 2.4 mA R 50 k 7. R= V 120 V = 54.55  = I 2.2 A 8. I= V 120 V  = 15 mA R 8 k 9. R= V 120 V = = 28.57  I 4.2 A 10. R= V 4.5 V = 36  = I 125 m A 11. R= V 24 mV = 1.2 k = I 20  A 12. V = IR = (12 A)(0.5 ) = 6 V 13. a. R= V 120 V = 12.63  = I 9.5 A b.  60 min   60 s  t = 2 h   = 7200 s  1 h   1 min  W = Pt = VIt = (120 V)(9.5 A)(7200 s) = 8.21  106 J 14. V = IR = (2.4 A)(3.3 M) = 7.92 V 15.  Chapter 4 25 16. b. 17.  18.  19.  20. P= 21. t= 22. a. b. (0.13 mA)(500 h) = 65 mAh W  t 540 J 540 J = 2.25 W  60 s 240 s 4 min 1 min W 640 J  = 16 s P 40 J/s  60 min   60 s  8 h   = 28,800 s  1 h   1 min  W = Pt = (2 W)(28,000 s) = 57.6 kJ kWh = (2 W)(8 h) = 16  103 kWh 1000 23. P = VI = (3 V)(1.4 A) = 4.20 W W 12 J  = 2.86 s t= P 4.2 W 24. P = EI = (6 V)(750 mA) = 4.5 W 25. P = I2R = (7.2 mA)2 4 k = 207.36 mW 26. P = I2R  I = 27. I= 28. I= 29. E= 26 P  R 240 mW = 10.44 mA 2.2 k P 2W  = 129.10 mA R 120  V = IR = (129.10 mA)(120 ) = 15.49 V E 22 V  = 1.31 mA R 16.8 k P = I2R = (1.31 mA)2 16.8 k = 28.83 mW   60 min   60 s   W = Pt = (28.83 mW) 1 h     = 103.79 J   1 h  1 min   P 324 W  = 120 V I 2.7 A Chapter 4 30. I= P  R 1W = 461.27 A 4.7 M no PR  (42 mW)(2.2 k)  92.40 = 9.61 V 31. V= 32. P = VI, I = 33. V= 34. a. P = EI and I = b. Ah rating = (0.13 mA)(500 h) = 66.5 mAh P 100 W  = 0.833 A V 120 V V 120 V R=  = 144.06  I 0.833 A P 450 W  = 120 V I 3.75 A V 120 V R=  = 32  I 3.75 A P 0.4  103 W = 0.13 mA  E 3V P 100 W   5  103 = 70.71 mA R 20 k V = PR  (100 W)(20 k) = 1.42 kV 35. I= 36. P = EI = (220 V)(30 A) = 6.6 kW 6.6 kW = 8.85 HP PHP = 746 W/HP 37. a.  V 2   12 V  2 W = Pt =  t    60 s = 86.4 J  R   10   b. Energy doubles, power the same 39. 40.  1  4 weeks  12 h  3   [5 months] = 260 h week  1 month    (230 W)(260 h) kWh = = 59.80 kWh 1000 a. Chapter 4  60 min   60 s  W = Pt = (60 W)(10 h)  = 2.16 × 106 Ws  1 h   1 min  27 41. 42. b. 1 Ws = 1 J  2.16 × 106 J c. W = Pt = (60 W)(10 h) = 600 Wh d. 600 Wh = 0.6 kWh 1000 W/1 kWh e. Cost = (0.6 kWh)(11 ¢/kWh) = 6.6 ¢ a. kWh = b. I= c. Plost = Pi  Po = Pi  Pi = Pi(1  ) = 120 kW(1  0.82) = 21.6 kW Pt (21.6 kW)(10 h)  = 216 kWh kWhlost = 1000 1000 44. 28 P 120  103 W  = 576.92 A E 208 V $1.00 = 9.09 11¢ Pt (kWh)(1000) (9.09)(1000) kWh =  = 36.36 h t 1000 P 250 W #kWh = t= 43. Pt (1000)(kWh) (1000)(1200 kWh) P  = 120 kW 1000 P 10 h (kWh)(1000) (11.11)(1000)  = 2.32 h P 4800 a. $74 = $2.39/day 31 days b. $2.39 / day = 16¢/h 15 h/day c. 16¢ = 1.45 kWh 11¢/kWh d. 1.45 kWh = 24.16  24 bulbs 60 W e. no $1.00 = 9.09 kWh 11¢/kWh 9.09 kWh = 48.61 h 187 W Chapter 4 45. 46. 47. t = 5 h/day(365 days) = 1825 h P  t (339 W)(1825 h)  = 618.68 kWh kWh = 1000 1000 Cost = (618.68 kWh)(11¢/kWh) = $68.05 P  t (213 W)(1825 h) = 388.73 kWh  kWh = 1000 1000 Cost = (388.73 kWh)(11¢/kWh) = $42.76 Cost Savings = $68.05  $42.76 = $25.29 P  t (78 W)(4 h/day)(31 days) = 9.67 kWh  1000 1000 Cost = (11¢/kWh)(9.67 kWh) = $1.06 kWh = a. P = EI = (120 V)(100 A) = 12 kW b.  746 W  PT = 5 hp   + 3000 W + 2400 W + 1000 W  hp  = 10,130 < 12,000 W (Yes) c. W = Pt = (10.13 kW)(2 h) = 20.26 kWh (1600 W)(6 h)  (1200 W)(1/4 h)  (4800 W) 48. kWh = 1h 1 h  (900 W) 10 min 2 60 min)  (200 W)(2 h)  (50 W)(3.5 h) 1000 9600 Wh +300 Wh  2400 Wh + 150 Wh  400 Wh + 175 Wh = = 13.025 kWh 1000 (13.025 kWh)(11¢/kWh) = 1.43¢ (200 W)(4 h)  (6)(60 W)(6 h) + (1200 W) 20 min 49. kWh = 1h 60 min 1 6  (175 W)(3.5 h) + (250 W) 2 h  (30 W)(8 h) 1000 800 Wh  2160 Wh + 400 Wh + 612.5 Wh + 541.67 Wh + 240 Wh = = 4.754 kWh 1000 (4.754 kWh)(11¢/kWh) = 52.29¢ 50. = 51. = Po  100%  Pi 746 W hp 373  100%   100% = 90.98% 410 W 410 (0.5 hp) P (1.8 hp)(746 W/hp) Po , Pi = o  = 1960.29 W Pi  0.685 P 1960.29 W Pi = EI, I = i  = 16.34 A E 120 V Chapter 4 29 52. = Po (0.8 hp)(746 W/hp) 596.8  100%   100%   100% = 67.82% Pi (4 A)(220 V) 880 53. a. Pi = EI = (120 V)(1.8 A) = 216 W Pi = Po + Plost, Plost = Pi  Po = 216 W  50 W = 166 W b. % = 54. Pi = EI = 55. a. b. c. Po  Po 50 W  100% =  100% = 23.15% 216 W Pi I= Po (3.6 hp)(746 W/hp) = 16.06 A = E (0.76)(220 V) (2 hp)(746 W/hp) Pi = P o = = 1657.78 W  0.9 Pi = EI = 1657.78 W (110 V)I = 1657.78 W 1657.78 W = 15.07 A I= 110 V Po (2 hp)(746 W/hp) = 2131.43 W  0.7 Pi = EI = 2131.43 W (110 V)I = 2131.43 W 2131.43 W = 19.38 A I= 110 V Pi = Po = (15 hp)(746 W/hp) = 12,433.33 W  0.9 12, 433.33 W I = Pi = = 56.52 A E 220 V 56. Pi = 57. T = 1  2 = 0.75 = 0.85  2 2 = 0.88 58. T = 1  2 = (0.87)(0.75) = 0.6525  65.25% 59. T = 1  2 = 0.78 = 0.92 2 = 30 0.78 = 0.867  86.7% 0.9 Chapter 4 60. 61. a. T = 1  2  3 = (0.93)(0.87)(0.21) = 0.170  17% b. T = 1  2  3 = (0.93)(0.87)(0.80) = 0.647  64.7% 64.7%  17%  100% = 280.59% 17% T = Po = 1  2 = 1  21 = 212 Pi 12  Po  1 = 2 Pi Po 128 W  = 0.4 2 Pi 2(400 W) 2 = 21 = 2(0.4) = 0.8  2 = 40%, 2 = 80% Chapter 4 31 Chapter 5 1. a. b. c. d. e. f. E and R1 R1 and R2 E1, E2, and R1 E1 and R1; E2, R3 and R4 R3, R4 and R5; E and R1 R2 and R3 2. a. b. c. d. RT = 0.1 k + 0.39 k + 1.2 k + 6.8 kΩ = 8.49 k RT = 1.2  + 2.7  + 8.2  = 12.1  RT = 8.2 k + 10 k + 9.1 k + 1.8 k + 2.7 k = 31.8 k RT = 47  + 820  + 91  + 1.2 k = 2158.0  3. a. b. RT = 1.2 k + 1 k + 2.2 k + 3.3 k = 7.7 k RT = 1 k + 2 k + 3 k + 4.7 k + 6.8 k = 17.5 k 4. a. b. c. 1 M 100 , 1 k RT = 100  + 1 k + 1 M + 200 k = 1.2011 M vs. 1.2 M for part b. 5. a. b. RT = 10  + 33  + 56 Ω, Reading = 99  RT = 2.2 k + 0.82 kΩ + 1.2 k + 3.3 k, Reading = 7.52 k 6. a. b. RT = 129 k = R + 56 k + 22 k + 33 k, Reading = 18 k RT = 103 k = 24 k + R1 + 43 k + 2R1 = 67 k + 3R1, R1 = 12 k R2 = 24 k 7. a. b. c. 1.2 k 0  8. a. RT = 10  + 12  + 18  = 40  E 72 V Is = = 1.8 A  RT 40  V1 = I1R1 = (1.8 A)(10 ) = 18 V, V2 = I2R2 = (1.8 A)(12 ) = 21.6 V, V3 = I3R3 = (1.8 A)(18 ) = 32.4 V Ps = EIs = (72 V)(1.8 A) = 129.6 W P18 = V3I3 = (32.4 V)(1.8 A) = 58.32 W b. c. d. e. 9. a. b. c. 32 the most: R3, the least: R1 R3, RT = 1.2 k + 6.8 k + 82 k = 90 k 45 V E Is = = 0.5 mA  RT 90 k V1 = I1R1 = (0.5 mA)(1.2 k) = 0.6 V, V2 = I2R2 = (0.5 mA)(6.8 k) = 3.4 V, V3 = I3R3 = (0.5 mA)(82 k) = 41 V, results agree with part (a) Chapter 5 10. d. e. Ps = EIs = (72 V)(1.8 A) = 129.6 W P18Ω = V3I3 = (32.4 V)(1.8 A) = 58.32 W a. RT = 12 k + 4 k + 6 k = 22 k E = IRT = (4 mA)(22 k) = 88 V RT = 12  + 22  + 82  + 10  = 126  E = IRT = (500 mA)(126 ) = 63 V b. 11. a. a. b. c. d. b. a. b. c. d. 12. 13. V 5.2 V =4A  R 1.3  E = IRT = (4 A)(9 ) = 36 V RT = 9  = 4.7  + 1.3  + R, V4.7  = (4 A)(4.7 ) = 18.8 V V1.3  = (4 A)(1.3 ) = 5.2 V V3  = (4 A)(3 ) = 12 V I= R=3 V 6.6 V  = 3 mA R 2.2 k V3.3 k = (3 mA)(3.3 k) = 9.9 V I= E = 6.6 V + 9 V + 9.9 V = 25.5 V V 9V  = 3 k R= I 3 mA V2.2 k = 6.6 V, V3 k = 9 V, V3.3 k = 9.9 V 1 1 36 V E = 8.18 mA, Vm = E  (36 V) = 18 V  2 2 RT 4.4 k a. Im = b. RT = 1 k + 2.4 k + 5.6 k = 9 k 22.5 V E Im = = 2.5 mA, Vm = 2.5 mA(2.4 k + 5.6 k) = 20 V  9 k RT c. V3.3kΩ = a. RT = 3 k + 1 k + 2 k = 6 k E 120 V = 20 mA Is =  6 k RT VR1 = (20 mA)(3 k) = 60 V 3.3 k(12 V) = 8.8 V 4.5 k Vm = 12 V  8.8 V = 3.2 V 12 V Im = = 2.67 mA 4.5 k VR2 = (20 mA)(1 k) = 20 V VR3 = (20 mA)(2 k) = 40 V Chapter 5 33 b. PR1 = I12 R1 = (20 mA)2  3 k = 1.2 W PR2 = I 22 R2 = (20 mA)2  1 k = 0.4 W PR3 = I 32 R3 = (20 mA)2  2 k = 0.8 W c. PT = PR1  PR2  PR3 = 1.2 W + 0.4 W + 0.8 W = 2.4 W d. PT = EIs = (120 V)(20 mA) = 2.4 W e. the same f. R1  the largest g. dissipated h. R1: 2 W, R2 : 1/2 W, R3: 1 W 14. P = 21 W = (1 A)2R, R = 21  V1 = I1R1 = (1 A)(2 ) = 2 V, V2 = I2R2 = (1 A)(1 ) = 1 V V3 = I3R3 = (1 A)(21 ) = 21 V E = V1 + V2 + V3 = 2 V + 1 V + 21 V = 24 V 15 P = 8 W = I21 , I = 8 = 2.828 A P = 16 W = I2R1 = (2 .828 A)2R1, R1 = 2  RT = 32  = 2  + R2 + 1  = 3  + R2, R2 = 29  E = IRT = (2.828 A)(32 ) = 90.5 V 16. a. 34  1  RT = NR1 = 8 28   = 225   8  120 V E = 0.53 A I=  RT 225  2 b. 8  P = I 2R =  A   15  c.  8  225  V = IR =  A    = 15 V   15  8 d. All go out!  1   64  225   28      =8W   8   225  8  Chapter 5 17. Ps = PR1  PR2  PR3 EI = I2R1 + I2R2 + 24 (R1 + R2)I2  EI + 24 = 0 6I2  24 I + 24 = 0 I2  4 I + 4 = 0  (4)  (4) 2  4(1)(4) 4  16  16 4   =2A 2(1) 2 2 24  =6 P = 24 W = (2 A)2R, R = 4 I= 18. a. b. c. Vab + 4 V + 24 V  12 V = 0, Vab = 28 V + 12 V = 16 V Vab + 4 V + 8 V  16 V = 0, Vab = 16 V  12 V = 4 V Vab + 12 V  18 V + 6 V  12 V = 0, Vab = 30 V  18 V = 12 V 19. a. ET = 8 V  16 V + 20 V = 12 V, I = b. 20. a. 12 V = 1.17 A 10.3  2V = 173.91 mA ET = 4 V + 18 V  12 V = 2 V, I  11.5  P = 8 mW = I2R, R = I= 21. 22. 8 mW 8 mW = 2 k  2 (2 mA)2 I E 20 V  E = 2 mA (CW),  RT 3 k  2 k 16 V V 12 V = 8 mA, R =  = 1.5 k 2 k I 8 mA E E  4 V  10 V E  14 V = 8 mA (CCW) I=   RT 2 k   1 .5 k  3.5 k E = 42 V b. I= a. +10 V + 4 V  12 V  V = 0 V = 14 V  12 V = 2 V b. +30 V + 20 V  8 V  V = 0 V = 50 V  8 V = 42 V c. 6 V  22 V  V1 + 36 V = 0 V = 36 V  28 V = 8 V a. I= b. V2 = IR = (1.5 A)(2 Ω) = 3 V c. Chapter 5 E = 10 V 12 V = 1.5 A 8 60 V  12 V  V1  3 V = 0 V1 = 60 V  15 V = 45 V 35 23. a. b. 24. a. b. 25. 26. a. b. d. + 10 V  V1 + 6 V  2 V  3 V = 0, V1 = 11 V +10 V  V2  3 V = 0, V2 = 7 V 10 k V3: V2 = 10 k:1 k = 10:1 V3: V1 = 10 k:100  = 100:1 RE (10 k)(60 V) V3 = 3  = 54.05 V 0.1 k  1 k  10 k RT ( R  R3 ) E (1 k  10 k)(60 V) V = 2 = 59.46 V  11.1 k RT a. V= 40 (30 V) = 20 V 40   20  b. V= (2 k  3 k)(40 V) (5 k)(40 V) = 20 V  4 k  1 k  2 k  3 k 10 k c. 36 V1.8Ω = IR = (3 A)(1.8 Ω) = 5.4 V 24 V  V1  10 V  5.4 V = 0, V1 = 24 V  15.4 V = 8.6 V V2.7Ω = IR = (3 A)(2.7 Ω) = 8.1 V 10 V  8.1 V  V2 = 0 V2 = 10 V  8.1 V = 1.9 V (50 V)(2 ) 1 V 50 V , R2 = = 100   1V 2 R2 (100 V)(2 ) 1 V 100 V , R3 = = 200   2 1V R3 c. 27. +10 V  V2 = 0 V2 = 10 V +10 V  6 V  V1 = 0 V1 = 4 V +24 V  10 V  V1 = 0 V1 = 14 V +10 V  V2 + 8 V = 0 V2 = 18 V (1.5   0.6   0.9 )(0.72 V) (3 )(0.72 V) = 0.36 V  (2.5   1.5   0.6   0.9   0.5 ) 6 k Chapter 5 28. a. V1 20 V (1.2 )(20 V) = 12 V , V1 =  1.2  2  2 V2 20 V (6.8 )(20 V) , V2 = = 68 V  6.8  2  2 E = V1 + 20 V + V2 = 12 V + 20 V + 68 V = 100 V b. c. d. 29. 120 V  V1  80 V = 0, V1 = 40 V 80 V  10 V  V3 = 0, V3 = 70 V V 1000 V 68 (1000 V) = 680 V  2 , V2 = 100  68  100  1000 V V1 2 (1000 V) = 20 V , V1 =  100  2  100  E = V1 + V2 + 1000 V = 20 V + 680 V + 1000 V = 1700 V V1 = 0 V 10 k(50 V  30 V) V2 = 10 k  3.3 k  4.7 k 10 k(20 V) = 11.11 V = 18 k Vx = E1  V3.3kΩ 3.3 k(20 V) V3.3kΩ = 18 k = 3.67 V Vx = 50 V  3.67 V = 46.33 V 2V V 2 k(2 V)  2 , V2 = =4V 1 k 2 k 1 k 2V V 3 k(2 V)  4 , V4 = =6V 1 k 3 k 1 k 2V = 2 mA 1 k E = 2 V + 4 V + 12 V + 6 V = 24 V I= Chapter 5 37 30. a. b. 31. a. b. 32. R (20 V) 2.2 k  1.8 k  R 4(4 kΩ + R) = 20R 16 kΩ + 4R = 20R 16R = 16 kΩ 16 R= k = 1 kΩ 16 4V= (6 M  R )(140 V) 6 M  R  3 M 100(9 MΩ + R) = 840 MΩ + 140R 900 MΩ + 100R = 840 MΩ + 140R 40R = 60 MΩ 60 R= M = 1.5 MΩ 40 100 V = 8V = 160  50 mA R (12 V) 160 (12 V) , Rx = 80  in series with the bulb Vbulb = 8 V = bulb  160   Rx Rbulb  Rx Rbulb = VR = 12 V  8 V = 4 V, P = V 2 (4 V) 2 = 0.2 W, 1/4 W okay  R 80  VR1  VR2 = 72 V 1 VR  VR2 = 72 V 5 2 72 V  1 = 60 V VR2 1   = 72 V, VR2  1.2  5 VR VR 60 V 72 V  60 V 12 V R2 = 2  = 15 k, R1 = 1  = 3 k  I R2 I R1 4 mA 4 mA 4 mA 33. RT = R1 + R2 + R3 = 2R3 + 7R3 + R3 = 10R3 R (60 V) VR3 = 3 = 6 V, VR1 = 2VR3 = 2 (6 V) = 12 V, VR2  7VR3 = 7(6 V) = 42 V 10 R3 34. a. VR3  4VR2 = 4(3VR1 )  12VR1 E = VR1  3VR1  12VR1 RT = R1 + 3R1 + 12R1 = 16R1 = R1 = 38 64 V = 6.4 k 10 mA 6.4 k = 400 , R2 = 3R1 = 1.2 k, R3 = 12R1 = 4.8 k 16 Chapter 5 35. 36. 6.4 M 64 V = 400 k, R2 = 1.2 M, R3 = 4.8 M = 6.4 M, R1 = 10 A 16 I1 10 mA R  400 k = 103 also  = 103 and 1  I  10  A R1 400  b. RT = a. Va = 12 V + 5 V = 17 V Vb = 5 V + 16 V = 21 V Vab = 17 V  21 V = 4 V b. Va = 14 V Vb = 14 V + 6 V + 10 V = 30 V Vab = 14 V  30 V = 16 V c. Va = 10 V + 3 V = 13 V Vb = 8 V Vab = 13 V  (8 V) = 21 V a. 60 V + 20 V 80 V  = 0.8 A 18   82  100  Va = 60 V  I(18 Ω) = 60 V  (0.8 A)(18 Ω) = 60 V  14.4 V = 45.6 V I =  I = 100 V + 60 V 160 V  = 20 mA 2 k  2 k  2 k  2 k 8 k Va = 10 V  I(2 kΩ) = 100 V  (20 mA)(2 kΩ) = 100 V  40 V = 60 V 37. 38. 47 V  20 V 27 V = = 3 mA (CCW) 2k +3k + 4k 9k V2k = 6 V, V3k = 9 V, V4k = 12 V I= a. Va = 20 V, Vb = 20 V + 6 V = 26 V, Vc = 20 V + 6 V + 9 V = 35 V Vd = 12 V, Ve = 0 V b. Vab = 6 V, Vdc = 47 V, Vcb = 9 V c. Vac = 15 V, Vdb = 47 V + 9 V = 38 V VR 4 V + 4 V 8V 12 V  4 V 8 V  = 1 A, R1 = 1 =  = 8 Ω, 8 8 I 1A 1A VR 8V 4V 4V =4Ω R3 = 3 =  I 1A 1A I R2 = Chapter 5 39 39. VR2 = 48 V  12 V = 36 V VR3 = VR3 12 V = 0.75 k = I 16 mA = 20 V R3 = VR4 VR2 36 V = 2.25 k I 16 mA = 12 V  0 V = 12 V R2 = VR4 20 V = 1.25 k I 16 mA VR1  E  VR2  VR3  VR4 R4 = = = 100 V  36 V  12 V  20 V = 32 V VR 32 V = 2 k R1 = 1 = I 16 mA 40. 41. 42. 40 a. Va = 8 V + 14 V = +6 V, Vb = 14 V Vc = +I(10 Ω)  6 V with 14 V + 6 V 20 V  =1A I= 10   10  20  Therefore, Vc = (1 A)(10 Ω)  6 V = 10 V  6 V = 4 V Vd = 0 V b. Vab = Va  Vb = 6 V  14 V = 8 V Vcb= Vc  Vb = 4 V  14 V = 10 V Vcd = Vc  Vd = 4 V  0 V = 4 V c. Vad = Va  Vd = 6 V  0 V = 6 V Vca = Vc  Va = 4 V  6 V = 2 V V0 = 0 V, V4 = (2 kΩ)(6 mA) + 3 V = 12 V + 3 V = 15 V, V7 = 4 V V10 = V1  V0 = 12 V  0 V = 12 V, V23 = V2  V3 = 4 V  (8 V) = 4 V + 8 V = 12 V V30 = V3  V0 = 8 V  0 V = 8 V, V67 = V6  V7 = 4 V  4 V = 0 V 4 V + 8 V 12 V  =3A V56 = V5  V6 = 3 V  4 V = 1 V, I = 4 4 V0 = 0 V, V03 = V0  V3 = 0 V  0 V = 0 V, V2 = (3 mA)(3.3 kΩ) = 9.9 V V23 = V2  V3 = 9.9 V  0 V = 9.9 V, V12 = V1  V2 = 20 V  9.9 V = 10.1 V  Ii   Io Ii = 4 mA + 3 mA + 10 mA = 17 mA Chapter 5 43. 44. 45. a. VL = ILRL = (2 A)(28 ) = 56 V Vint = 60 V  56 V = 4 V V 4V =2 Rint = int  I 2A b. VR = 60 V  56 V VNL  VFL  100% = 7.14%  100% = VFL 56 V a. VL = 3.3 (12 V) 39.6 V = 11.85 V  3.3   43 m 3.343  b. VR = 12 V  11.85 V VNL  VFL  100% = 1.27%  100% = 11.85 V VFL c. Is = IL = a. I= E 12 V 12 V = 1.36 mA   RT 2 k  6.8 k 8.8 k b. I= E 12 V 12 V = 1.33 mA   RT 8.8 k  0.25 k 9.05 k c. not for most applications. Chapter 5 11.85 V = 3.59 A 3.3  Ps = EIs = (12 V)(3.59 A) = 43.08 W Pint = I2Rint = (3.59 A)2 43  = 0.554 W 41 Chapter 6 1. a. b. c. d. e. f. g. R2 and R3 E and R3 R2 and R3 R2 and R3 E, R1, R2, R3, and R4 E, R1, R2, and R3 E2, R2 and R3 2. a. b. R3 and R4, R5 and R6 E and R1, R6 and R7 3. a. RT = b. RT = c. RT = d. e. 4. 42 (36 )(18 ) = 12  36   18  1 1  3 1 1 1 1  10 S  0.5  103 S  33.33  106 S   1 k 2 k 30 k 1 = = 0.652 k 1.533  103 S 1 1 1   6 3 3 1 1 1 833.33  10 S  8.33  10 S  83.33  10 S 92.49  103 S   1.2  120 k 12 k = 10.81  18 k = 6 k 3 (6 k)(6 k) = 3 k RT = 6 k  6 k RT = 22  10  = 5.5 , RT  = =5 4 2 (5.5 )(5 ) = 2.62  RT = 5.5   5  RT = 1 1  3 1 1 1 1000  10 S  1  10 3 S  0.001  103 S   1  1 k 1 M 1 = = 0.99  1001.001  10 3 S f. RT = a. RT = 1 1   3 1 1 1 1  10 S  0.833  103 S  3.333  10 3 S   1 k  1 .2 k  0 .3 k  1 = = 193.57  5.166  10 3 S Chapter 6 5. 1  1 1  10 S  0.833  10 S  0.455  10 3 S  1  103 S b. RT = a. RT = 3   6  = 2  (2 )( R ) , R=8 RT = 1.6  = 2R 6 k = 2 k RT = 3 (2 k)( R) , R = 18 k RT = 1.8 k = 2 k  R (20 k)( R ) , R = 6.8 k RT = 5.08 k = 20 k  R 1 1  RT = 1.02  = 1 1 1 1 416.67  10 6 S +  147.06  10 6 S   2.4 k R 6.8 k R 1 1.02 kΩ = 1 563.73  106  R 1.020 k  =1 575 × 103 + R 1.020 k R= = 2.4 kΩ 425  103 b. c. d. e. 1 1 1 1    1 k 1.2 k 2.2 k 1 k 1 = 304.14  = 3.288  103 S RT = 6 kΩ = 3 3 R1 4 R1 = 24 kΩ 6. a. b. 1.2 k about 1 k c. RT = d. e. Chapter 6 1 1 1 1 1    1.2 k 22 k 220 k 2.2 M 1 = 6 6 833.333  10 S  45.455  10 S  4.545  10 6 S  0.455  10 6 S 1 = 1.131 k = 883.788  10 6 S (1.2 k)(22 k) 220 k, 2.2 M: RT = = 1.138 k 1.2 k  22 k RT reduced. 43 7. 1 1 1   = 1.18  1 1 1 0.25 S + 0.50 S  0.10 S 0.85 S   4  2  10  a. RT = b.  c. RT = 3 Ω  6 Ω = 2  8. 24   24  = 12  1 1 1 1    RT R1 12  120  1 + 0.08333 S + 0.00833 S 0.1 S = R1 1 + 0.09167 S 0.1 S = R1 1 = 0.1 S  0.09167 S = 0.00833 S R1 1 = 120  R1 = 0.00833 S 9. a. RT = b. VR1  VR2 = 36 V c. Is = d. Is = I1 + I2 6 A = 4.5 A + 1.5 A = 6 A (checks) a. I1 = b. RT = c. Is = 10. 44 (8 )(24 ) =6 8   24  E 36 V =6A  6 RT VR 36 V = 4.5 A I1  1  8 R1 VR 36 V = 1.5 A I2  2  R2 24  VR1 R1  VR 18 V VR 18 V 18 V = 0.5 A = 6 A, I2 = 2  = 2 A, I3 = 3  3 9 R3 36  R2 1 1  1 1 1  0 . 333 S 0 . 111 S  0.028 S   3  9  36  1 = 2.12  = 472  103 S E 18 V = 8.5 A  RT 2.12  Chapter 6 11. d. Is = I1 + I2 + I3 = 6 A + 2 A + 0.5 A = 8.5 A e. they match a. I R1  I R3  12. VR1 R1 VR3 R3  VR 24 V 24 V = 2.4 mA, I R2  2  = 20 mA, 10 k R 2 1.2 k  24 V = 3.53 mA 6.8 k 1 1  6  1 1 1 100  10 S  833.333  106 S  147.06  106 S   10 k 1.2 k 6.8 k 1 = 925.93  = 1.08  103 S b. RT = c. Is = d. Is = I1 + I2 + I3 = 2.4 mA + 20 mA + 3.53 mA = 25.93 mA e. they match a. RT  900  b. RT = c. I3 the most, I4 the least d. I R1  24 V E = 25.92 mA  RT 925.93  1 1 1 1 1    20 k 10 k 1 k 91 k 1 = 6 6 50  10 S  100  10 S  1  103 S  10.99  106 S 1 = = 862.07 , very close 1.16  103 S I R3 VR1 VR 60 V 60 V = 3.0 mA, I R2  2  = 6 mA R1 20 k R2 10 k VR VR 60 V 60 V = 60.0 mA, I R4  4  = 0.659 mA  3  R3 1 k R4 91 k  60 V E = 69.6 mA  RT 862.07 k Is = 3 mA + 6 mA + 60 mA + 0.659 mA = 69.66 mA (checks) e. Is = f. always greater Chapter 6 45 13. 14. RT = 6 Ω = b. P = 81 W = a. P= b. R2 = c. I1 = d. Is = I1 + I2 + I3 = 2 A + 2 A + e. Ps = EIs = (20 V)(9 A) = 180 W f. g. 15. (18 )( R2 ) 18   R2 108 Ω + 6R2 = 18R2 12R2 = 108 Ω 108  =9Ω R2 = 12 a. V 2 E2 E2   R R 9 and E2 = (9)(81) or E = 729 = 27 V V 2 E2  and E = R R PR  (100 W)(4 )  400 = 20 V E 20 V = 10 Ω  I2 2A V1 E 20 V =2A   R1 R1 10  20 V =4A+5A=9A 4 E 2 (20 V) 2 400 W E 2 (20 V) 2 = 40 W, PR2  = 40 W,   PR1   10 R2  20 V  R1 10    2A  Ps = P1 + P2 + P3 180 W = 40 W + 40 W + 100 W = 180 W (checks) (20 )(10.8 A) =9A 20   4  E = VR3  I 3 R3  (9 A)(4 ) = 36 V I3 = I R1 = 12.3 A  10.8 A = 1.5 A R1 = 16. 46 VR1 I R1  36 V = 24  1.5 A a. V = 48 V b. I2 = 48 V = 2.67 mA 18 k Chapter 6 17. c. Is = 48 V 48 V   I 2 = 16 mA + 4 mA + 2.67 mA = 22.67 mA 3 k 12 k d. P= V 2 E 2 (48 V)2   = 192 mW R R 12 k a. I R2  = 4 A  1 A = 3 A, R2 = b. R3 = c. I1  18.  19. a. VR3 I2  E 12 V =4Ω  I2 3 A E 12 V = 12 Ω  I3 1 A 12 V = 6 A, Is = I1 + 4 A = 6 A + 4 A = 10 A 2 1 1   6 1 1 1 1000  10 S  212.77  106 S  100  106 S   1 k 4.7 k 10 k 1 = = 761.61  1.313  103 S VR 60 V VR 60 V = 60 mA, I R2  2  = 12.77 mA I R1  1  R1 1 k R2 4.7 k RT = I R3  b. I3  VR2 VR3 R3  60 V = 6 mA 10 k PR1  VR1  I R1 = (60 V)(60 mA) = 3.6 W PR2  VR2  I R2 = (60 V)(12.77 mA) = 766.2 mW PR3  VR3  I R3 = (60 V)(6 mA) = 360 W c. d. e. 20. a. b. c. Chapter 6 E 60 V = 78.78 mA  RT 761.61  Ps = EsIs = (60 V)(78.78 mA) = 4.73 W Ps = 4.73 W = 3.6 W + 766.2 mW + 360 mW = 4.73 W (checks) R1 = the smallest parallel resistor Is = 120 V E = 66.667 mA  Rbulb 1.8 k R 1.8 k RT = = 225   N 8 E 120 V = 0.533 A Is =  RT 225  Ibulb = 47 21. V 2 (120 V)2 =8W  R 1.8 k d. P= e. Ps = 8(8 W) = 64 W f. none, Is drops by 66.667 mA Network redrawn: RT = 3.33 Ω  7.5 Ω = 2.31 Ω E 2 (60 V)2 = 1.56 kW Ps =  RT 2.31  22. 48 a. 5 × 60 W = 300 W 300 W = 2.5 A Ibulbs = 120 V 1200 W Imicro = = 10 A 120 V 320 W ITV = = 2.67 A 120 V 25 W = 208.33 mA IDVD = 120 V b. Is =  I = 2.5 A + 10 A + 2.67 A + 208.33 mA = 15.38 A No c. RT = d. Ps = E Is = (120 V)(15.38 A) = 1,845.60 W e. Ps = 1845.60 W = 300 W + 1200 W + 320 W + 25 W = 1845 W (checks) E 120 V = 7.8 Ω  I s 15.38 A Chapter 6 23. a. b. c. 8   12  = 4.8 , 4.8   4  = 2.182  24 V  8 V = 14.67 A I1 = 2.182  V 2 (24 V  8 V) 2 P4 = = 256 W  R 4 I2 = I1 = 14.67 A 24. Is = 8 mA + 6 mA = 14 mA I2 = 6 mA  2 mA = 4 mA 25. a.  Ii   Io 2A+3A+9A=6A+I 14 A = 6 A + I I = 14 A  6 A = 8 A b.  Ii   Io 8 mA = 2 mA + I1 I1 = 8 mA  2 m A = 6 mA  Ii   Io I1 + 9 mA = I2 I2 = 6 mA + 9 mA = 15 mA  Ii   Io I2 = 10 mA + I3 I3 = 15 mA  10 mA = 5 mA a.  Ii   Io 8 A = 3 A + I2 I2 = 8 A  3 A = 5 A, I3 = 3 A  Ii   Io I2 + I3 = I4 I4 = 5 A + 3 A = 8 A b.  Ii   Io Is = 36 mA + 4 mA = 40 mA  Ii   Io 36 mA = I3 + 20 mA I3 = 36 mA  20 mA = 16 mA  Ii   Io 4 mA + 20 mA = I4 I4 = 24 mA I5 = Is = 40 mA 26. Chapter 6 49 27. I R2 = 5 mA  2 mA = 3 mA E = VR2 = (3 mA)(4 k) = 12 V R1 = R3 = 28. VR1 I R1 VR3 I R3  12 V 12 V = 3 k  (9 mA  5 mA) 4 mA  12 V = 6 k 2 mA RT = 12 V E = 1.33 k  I T 9 mA a. R1 = b. E = I1R1 = (2 A)(6 ) = 12 V E 12 V = 1.33 A I2 =  9 R2 E 10 V =5  I1 2 A I2 = I  I1 = 3 A  2 A = 1 A E 10 V = 10  R=  1A I2 P 12 W =1A  V 12 V E 12 V = 12  R3 =  1A I3 I = I1 + I2 + I3 = 2 A + 1.33A + 1 A = 4.33 A I3 = 29. 50 a. 64 V = 64 mA 1 k 64 V I3 = = 16 mA 4 k Is = I1 + I2 + I3 I2 = Is  I1  I3 = 100 mA  64 mA  16 mA = 20 mA 64 V E R= = 3.2 k  I 2 20 mA I = I2 + I3 = 20 mA + 16 mA = 36 mA I1 = Chapter 6 b. V12  V1  PR1  (30 W)(30 ) = 30 V R1 E = V1 = 30 V E 30 V =1A I1 =  R1 30  Because R3 = R2, I3 = I2 , and Is = I1 + I2 + I3 = I1 + 2I2 2 A = 1 A + 2I2 1 I2 = (1 A) = 0.5 A 2 I3 = 0.5 A E 30 V = 60  R2 = R3 =  I 2 0.5 A P= PR2  I 22 R2 = (0.5 A)2  60  = 15 W 30. 31. 32. 6 1 1 I1  I1 = (9 A) = 4.5 A 12  2 2 6 I3 = I1  3I1 = 3(9 A) = 27 A 2 6 1 1 I4 = I1  I1 = (9 A) = 3 A 18  3 3 IT = I1 + I2 + I3 + I4 = 9 A + 4.5 A + 27 A + 3 A = 43.5 A I2 = 8 k(20 mA ) = 16 mA 2 k  8 k I2 = 20 mA  16 mA = 4 mA a. I1 = b. I2.4kΩ = 2.5 A = a. 1 k( IT ) 1 k( IT )  1 k  2.4 k 3.4 k 3.4 k(2.5 A) and IT = = 8.5 A 1 k I1 = IT  2.5 A = 8.5 A  2.5 A = 6 A 1 1 =  3 1 1 1 250  10 S  125  103 S  83.333  10 3 S   4  8  12  1 = = 2.18  458.333  103 R 2.18  (6 A) = 3.27 A I1 = Ix = T I, 4 Rx RT = 2.18  (6 A) = 1.64 A 8 2.18  I3 = (6 A) = 1.09 A 12  I4 = 6 A I2 = Chapter 6 51 b. 33. 9 (10 A) = 9 A 10 a. I1  b. I1/I2 = 10 /1  = 10, c. I1/I3 = 1 k/1  = 1000, I3 = I1/1000 = 9 A/1000  9 mA d. I1/I4 = 100 k/1  = 100,000, I4 = I1/100,000 = 9 A/100,000  90 A e. very little effect, 1/100,000 1 RT = 1 1 1 1    1  10  1 k 100 k 1 = 1 S  0.1 S  1  10 3 S  10  10 6 S 1 = = 0.91  1.10 S R 0.91  (10 A) = 9.1 A excellent (9 A) Ix = T I , I1 = 1 Rx f. I2 = I1 9 A   0.9 A 10 10 g. I2 = 0.91  (10 A) = 0.91 A excellent (0.9 A) 10  h. I3 = 0.91  (10 A) = 9.1 mA excellent (9 mA) 1 k i. 34. 4 Ω  4 Ω = 2 Ω 20 (8 A) 20 (8 A)  = 5.33 A I2 = 20   2   8  30  I 5.33 A I1 = 2  = 2.67 A 2 2 I3 = 8 A  I2 = 8 A  5.33 A = 2.67 A I4 = 8 A a. 0.91  (10 A) = 91 A excellent (90 A) 100 k 3 I (39 )(1 A) CDR: I36 = = 1 A, I = = 13 A = I2 3   36  3 I4 = I1 = I  1 A = 13 A  1 A = 12 A b. 52 I3 = I = 24 mA, V12kΩ = IR = (4 mA)(12 kΩ) = 48 V V 48 V = 12 mA I2 =  R 4 k I1 = I 4 mA  I2 = 24 mA  4 mA  12 mA = 8 mA Chapter 6 35. a. b. 36. 84 mA = I1 + I2 + I3 = I1 + 2I1 + 2I2 = I1 + 2I1 + 2(2I1) 84 mA = I1 + 2I1 + 4I1 = 7I1 84 mA and I1 = = 12 mA 7 I2 = 2I1 = 2(12 mA) = 24 mA I3 = 2I2 = 2(24 mA) = 48 mA VR 24 V R1 = 1  = 2 k I1 12 mA R2 = R3 = 37. 38. 39. R = 3(2 k) = 6 k 6 k(32 mA) = 24 mA I1 = 6 k  2 k I 24 mA = 8 mA I2 = 1  3 3 VR2 I2 VR3 I3  24 V = 1 k 24 mA  24 V = 0.5 k 48 mA b. c. PL = VLIL 72 W = 12 V  IL 72 W =6A IL = 12 V 6A I I1 = I2 = L  =3A 2 2 Psource = EI = (12 V)(3 A) = 36 W Ps1  Ps2 = 36 W + 36 W = 72 W (the same) d. Idrain = 6 A (twice as much) a. RT = 8   56  = 7  E 12 V I2 = I3 = = 1.71 A  7 RT 1 1 I1 = I 2  (1.71 A) = 0.86 A 2 2 16 V = 2 A, I = 5 A  2 A = 3 A 8 16 V V =2 IR = 5 A + 3 A = 8 A, R = R  8A IR I8  = Chapter 6 53 40. a. b. c. 41. a. b. c. 42. 43. 4.7 k(9 V) 42.3V  = 6.13 V 4.7 k  2.2 k 6.9 VL = E = 9 V VL = E = 9 V VL = b. c. 20 V = 5 A, I2 = 0 A 4 V1 = 0 V, V2 = 20 V Is = I1 = 5 A a. V2 = b. RT = 11 M  22 k = 21.956 k 21.956 k(20 V) = 16.47 V (very close to ideal) V2 = 21.956 k  4.7 k c. Rm = 20 V[20,000 /V] = 400 k RT = 400 k  22 k = 20.853 k 20.853 k(20 V) = 16.32 V (still very close to ideal) V2 = 20.853 k  4.7 k d: a. V2 = b. RT = 200 k  11 M = 196.429 k (196.429 k)(20 V) = 13.25 V (very close to ideal) V2 = 196.429 k  100 k c. Rm = 400 k RT = 400 k  200 k = 133.333 k (133.333 k)(20 V) = 11.43 V (a 1.824 V drop from Rint = 11 M level) V2 = 133.333 k  100 k a. e. 54 E 12 V 12 V = 1.188 mA   RT 0.1 k  10 k 10.1 k VL = IsRL = (1.19 mA)(10 k) = 11.90 V 12 V = 120 mA Is = 100  VL = E = 12 V Is = I1 = 22 k(20 V) = 16.48 V 22 k  4.7 k 200 k(20 V) = 13.33 V 200 k  100 k DMM level of 11 M not a problem for most situations VOM level of 400 k can be a problem for some situations. Chapter 6 44. a. Vab = 20 V b. Vab = c. Rm = 200 V[20,000 /V] = 4 M 4 M(20 V) = 16.0 V (significant drop from ideal) Vab = 4 M  1 M Rm = 20 V[20,000 /V] = 400 k 400 k(20 V) = 5.71 V (significant error) Vab = 400 k  1 M 11 M(20 V) = 18.33 V 11 M  1 M 45. not operating properly, 6 k not connected at both ends 6V = 1.71 k RT = 3.5 mA RT = 3 k  4 k = 1.71 k 46. Vab = E + I4 k  R4 k 12 V  4 V 8V  = 1.6 mA I4 k = 1 k  4 k 5 k Vab = 4 V + (1.6 mA)(4 k) = 4 V + 6.4 V = 10.4 V 4 V supply connected in reverse so that 12 V  4 V 16 V  = 3.2 mA I= 1 k  4 k 5 k and Vab = 12 V  (3.2 mA)(1 k) = 12 V  3.2 V = 8.8 V obtained Chapter 6 55 Chapter 7 1. a. b. c. d. e. f. 2. a. b. c. d. R1, R2,. and E are in series; R3, R4 and R5 are in parallel E and R1 are in series; R2, R3 and R4 are in parallel. E and R1 are in series; R2, R3 and R4 are in parallel. E1 and R1 are in series; E2 and R4 in parallel. E and R1 are in series, R2 and R3 are in parallel. E, R1, R4 and R6 are in parallel; R2 and R5 are in parallel. RT = 4  + 10   (4  + 4 ) + 4  = 4 Ω + 10   8  + 4  = 4  + 4.44  + 4  = 12.44  10  = 10  + 5  = 15  RT = 10  + 2 4 + 10  = 2  + 10  = 12  RT = 2 RT = 10  3. 2.2 k  10 k  = 1.8 k RT = 2 × 1.8 kΩ = 3.6 kΩ RT = 1 Ω  (1 Ω + 1 Ω + RT) = 1 Ω  (2 Ω + RT) 2   RT 2   RT =  1   2   RT 3   RT RT(3 Ω + RT) = 2 Ω + RT 3RT + RT2 = 2 Ω + RT 4. RT2 + 2RT  2 Ω = 0 2  (2) 2  4(1)(2) 2 2  4  8 2  12 2  3.464 =   2 2 2 RT = 1  1.732 = 0.732 Ω or 2.732 Ω Since RT < 1 Ω and positive choose RT = 0.732 Ω RT = 56 Chapter 7 5. R   RT = 7.2 kΩ = R1   R1  1  = R1  1.5R1  2 ( R1 )(1.5R1 ) 1.5 R12 1.5 R1   2.5 R1 2.5 R1  1.5 R1 2.5(7.2 k) 1.2 kΩ and R1 = 1.5 so that 7.2 kΩ = 6. a. b. c. d. e. f. g. 7. a. b. c. 8. yes I2 = Is  I1 = 10 A  4 A = 6 A yes V3 = E  V2 = 14 V  8 V = 6 V RT = 4   2  = 1.33  , RT = 4   6  = 2.4  RT = RT  RT = 1.33  + 2.4  = 3.73  20  = 10 , RT = RT  RT = 10  + 10  = 20  RT  RT = 2 20 V E =1A Is =  RT 20  Ps = EIs = Pabsorbed = (20 V)(1 A) = 20 W RT = R1  R2 = 10   15  = 6  RT = RT  (R3 + R4) = 6   (10  + 2 ) = 6   12  = 4  E E 36 V 36 V = =6A = 9 A, I1 = Is =  RT 4 6 RT 36 V 36 V E =3A I2 =   R3  R4 10   2  12  I1 = Is  I2 = 6 A  3 A = 3 A Va = I2R4 = (3 A)(2 ) = 6 V Redrawn: a. b. Va = 32 V 8 Ω  24 Ω = 6 Ω 6 (32 V) = 10.67 V Vb = 6   12  32 V 32 V  = 1.78 A 12 +6  18  RT = 72 Ω  18 Ω  18 Ω = 8.12 Ω I1 = 9Ω Chapter 7 57 Is = 9. a. b. 32 V E = 3.94 A  RT 8.12  Va = 36 V, Vb = 60 V Vc = 5 k(60 V) = 20 V 5 k  10 k  60 V  36 V = 24 mA, I1  1 k 60 V 60 V I8kΩ = 8 k = 7.5 mA, I10kΩ = = 4 mA 15 k 24 mA   I  24 mA + 7.5 mA = 31.5 mA  I 2  31.5 mA + 4 mA = 35.5 mA 10. 11. a. RT = 1.2 k + 6.8 k = 8 k, RT = 2 k  RT = 2 k  8 k = 1.6 k RT  = RT + 2.4 k = 1.6 k + 2.4 k = 4 k RT = 1 k  RT  = 1 k  4 k = 0.8 k b. Is = 48 V E = 60 mA  RT 0.8 k c. V= (1.6 k)(48 V) RTE = 19.2 V  RT  2.4 k  1.6 k  2.4 k RT = 2 R  2 R (R + R) = 2 R  2R  2 R = 2R 3 E 120 V  = 15 Ω I 8A 2R 3 15 Ω = and R = (15 ) = 22.5 Ω 3 2 2 R = 45 Ω RT = 12. 58 a. RT = (R1  R2  R3)  (R6 + R4  R5) = (12 k  12 k  3 k)  (10.4 k + 9 k  6 k) = (6 k  3 k)  (10.4 k + 3.6 k) = 2 k  14 k = 1.75 k 28 V 28 V E E = 16 mA, I2 = = 2.33 mA Is =   RT 1.75 k R2 12 k R = R1  R2  R3 = 2 k R = R6 + R4  R5 = 14 k Chapter 7 I6 = Chapter 7 2 k(16 mA) R( I s ) = 2 mA  R  R 2 k  14 k 59 b. V1 = E = 28 V R = R4  R5 = 6 k  9 k = 3.6 k V5 = I6 R = (2 mA)(3.6 k) = 7.2 V VR23 c. P= 13. a. I1  14. I1 = 15. a. 60 (28 V)2 = 261.33 mW 3 k 24 V = 6 A; VR2  24 V  8 V = 16 V, I 2  VR2 / R2 = 16 V/2 Ω = 8 A 4 8V I 3  = 0.8 A, I = I1 + I2 = 6 A + 8 A = 14 A 10  20 V = 425.5 mA 47  14 V 14 V I2 =  = 139.35 mA 160   270  100.47  b. 16. R3  a. R = R4 + R5 = 14 Ω + 6 Ω = 20 Ω R = R2  R = 20 Ω  20 Ω = 10 Ω R = R + R1 = 10 Ω + 10 Ω = 20 Ω RT = R3  R = 5 Ω  20 Ω = 4 Ω E 20 V =5A = Is = RT 4  20 V 20 V 20 V =1A I1 = = = R1  R 10  + 10  20  20 V =4A I3 = 5 I 1A I4 = 1 = (since R = R2) = = 0.5 A 2 2 Va = I3R3  I4R5 = (4 A)(5 Ω)  (0.5 A)(6 Ω) = 20 V  3 V = 17 V I  Vbc =  1  R2 = (0.5 A)(20 Ω) = 10 V 2 20 V E  R1  R4  ( R2  R3  R5 ) 3   3   (3   6   6 ) 20 V 20 V 20 V = = = 3  + 3   (3  + 3 ) 3  + 3   6  3  + 2  =4A I1 = Chapter 7 17. R4 ( I1 ) 3 (4 A)  R4  R2  R3  R5 3   3   6   6  12 A = 1.33 A = 6+3 I I3 = 2 = 0.67 A 2 b. CDR: I2 = c. I4 = I1  I2 = 4 A  1.33 A = 2.67 A Va = I4R4 = (2.67 A)(3 Ω) = 8 V Vb = I3R3 = (0.67 A)(6 Ω) = 4 V a. IE = b. IB = c. VB = VBE + VE = 2.7 V VC = VCC  ICRC = 8 V  (2 mA)(2.2 kΩ) = 8 V  4.4 V = 3.6 V d. VCE = VC  VE = 3.6 V  2 V = 1.6 V VE 2V = 2 mA  RE 1 k IC = IE = 2 mA VCC  (VBE  VE ) 8 V  (0.7 V + 2 V) = RB RB 220 k  8 V  2.7 V 5.3 V = = = 24 μA 220 k 220 k VRB = VBC = VB  VC = 2.7 V  3.6 V = 0.9 V 18. 19. 22 V 22 V  =1A 4   18  22  a. I= b. 22 V + Vi  22 V = 0, V1 = 44 V a. All resistors in parallel (between terminals a & b) RT = 16 Ω  16 Ω  8 Ω  4 Ω  32 Ω 8 Ω  8 Ω  4 Ω  32 Ω 4 Ω  4 Ω  32 Ω 2 Ω  32 Ω = 1.88  Chapter 7 61 b. All in parallel. Therefore, V1 = V4 = E = 32 V c. I3 = V3/R3 = 32 V/4 Ω = 8 A  d. 12 V = 1.2 mA 10 k V ab = Va  Vb = 12 V  (18 V) = 30 V 20. I= 21. a. b. 22. Is = I1 + I2 + I3 + I4 + I5 32 V 32 V 32 V 32 V 32 V + + + + = 16  8  4  32  16  =2A+4A+8A+1A+2A = 17 A E 32 V RT = = 1.88 Ω as above = I s 17 A a. Va = 6 V, Vb = 20 V Vab = Va  Vb = (6 V)  (20 V) = 6 V + 20 V = +14 V 20 V =4A 5 V 14 V I 2   ab  =7A 2 2 6V I 3  =2A 3 I3Ω = I2Ω + I 6V , I6V = I3Ω  I2Ω = 2 A  7 A = 5 A I + I6V = I5Ω, I = I5Ω  I6V = 4 A  (5A) = 9 A I 5   Applying Kirchoff's voltage law in the CCW direction in the upper "window": +18 V + 20 V  V8Ω = 0 V8Ω = 38 V 38 V = 4.75 A 8 18 V 18 V =2A I3Ω = = 3 + 6 9 I8Ω = KCL: I18V = 4.75 A + 2 A = 6.75 A b. 62 V = (I3Ω)(6 Ω) + 20 V = (2 A)(6 Ω) + 20 V = 12 V + 20 V = 32 V Chapter 7 23. I2R2 = I3R3 and I2 = I1 = I2 + I3 = I 3 R3 2 R3 R3 (since the voltage across parallel elements is the same)   20 10 R2 R3 +2 10 R  KVL: 120 = I112 + I3R3 =  3  2  12 + 2R3  10  and 120 = 1.2R3 + 24 + 2R3 3.2R3 = 96  96  = 30  R3 = 3.2 24. Assuming Is = 1 A, the current Is will divide as determined by the load appearing in each branch. Since balanced Is will split equally between all three branches. 10 1  V1 =  A  (10 ) = V 3 3  10 1  V2 =  A  (10 ) = V 6 6  10 1  V3 =  A  (10 ) = V 3 3  10 10 10 E = V1 + V2 + V3 = V + V + V = 8.33 V 3 6 3 E 8.33 V = 8.33  RT = = I 1A 25. 36 kΩ  6 kΩ  12 kΩ = 3.6 kΩ 3.6 k (45 V) V= = 16.88 V  27 V. Therefore, not operating properly! 3.6 k  + 6 k  6 kΩ resistor "open" R (45V) 9 k(45 V) = 27 V R = 12 k  36 k = 9 k, V =  R  6 k 9 k  6 k Chapter 7 63 26. a. RT = R5  (R6 + R7) = 6   3  = 2  RT = R3  (R4 + RT) = 4   (2  + 2 ) = 2  RT = R1 + R2 + RT = 3  + 5  + 2  = 10  240 V = 24 A I= 10  b. I4 = c. d. 27. 28. a. 4 ( I ) 4 (24 A) = 12 A  44 8 6 (12 A) 72 A =8A I7 =  6  3 9 V3 = I3R3 = (I  I4)R3 = (24 A  12 A)4 Ω = 48 V V5 = I5R5 = (I4  I7)R5 = (4 A)6 Ω = 24 V V7 = I7R7 = (8 A)2 Ω = 16 V P = I 72 R7 = (8 A)22 Ω = 128 W P = EI = (240 V)(24 A) = 5760 W RT = R4  (R6 + R7 + R8) = 2 Ω  7 Ω = 1.56 Ω RT = R2  (R3 + R5 + RT) = 2 Ω  (4 Ω + 1 Ω + 1.56 Ω) = 1.53 Ω RT = R1 + RT = 4 Ω + 1.53 Ω = 5.53 Ω b. I = 40 V/5.53 Ω = 7.23 A c. I3 = 2 ( I ) 2 (7.23 A)  = 1.69 A 2   6.56  2   6.56  2 (1.69 A) = 0.375 mA I7 = 2+7 PR7  I 2 R = (0.375 A)2 2 Ω = 0.281 W Network redrawn: 24 V =3A 8 P6Ω = I2R = (3 A)2  6 Ω = 54 W I 8Ω = I 6Ω = 64 Chapter 7 29. a. b. 30. R10 + R11  R12 = 1 Ω + 2 Ω  2 Ω = 2 Ω R4  (R5 + R6) = 10 Ω  10 Ω = 5 Ω R1 + R2  (R3 + 5 Ω) = 3 Ω + 6 Ω  6 Ω = 6 Ω RT = 2 Ω  3 Ω  6 Ω = 2 Ω  2 Ω = 1 Ω I = 12 V/1 Ω = 12 A I1 = 12 V/6 Ω = 2 A 6 (2 A) =1A I3 = 6+6 1A I4 = = 0.5 A 2 c. I6 = I4 = 0.5 A 12 A =6A 2 d. I10 = a. E = (40 mA)(1.6 k) = 64 V b. RL2 = c. I R1 = 72 mA  40 mA = 32 mA 48 V = 4 k 12 mA 24 V RL3 = = 3 k 8 mA I R2 = 32 mA  12 mA = 20 mA I R3 = 20 mA  8 mA = 12 mA R1 = R2 = R3 = 31. VR1 I R1 VR2 I R2 VR3 I R3 = 64 V  48 V 16 V = = 0.5 k 32 mA 32 mA = 48 V  24 V 24 V = = 1.2 k 20 mA 20 mA = 24 V = 2 k 12 mA I R1 = 40 mA I R2 = 40 mA  10 mA = 30 mA I R3 = 30 mA  20 mA = 10 mA I R5 = 40 mA I R4 = 40 mA  4 mA = 36 mA R1 = R2 = Chapter 7 VR1 I R1 VR2 I R2 = 120 V  100 V 20 V = 0.5 k = 40 mA 40 mA = 100 V  40 V 60 V = = 2 k 30 mA 30 mA 65 R3 = R4 = R5 = VR3 I R3 VR4 I R4 VR5 I R5 = 40 V = 4 k 10 mA = 36 V = 1 k 36 mA = 60 V  36 V 24 V = 0.6 k = 40 mA 40 mA P1 = I12 R1 = (40 mA)20.5 k = 0.8 W (1 watt resistor) P2 = I 22 R2 = (30 mA)22 k = 1.8 W (2 watt resistor) P3 = I 32 R3 = (10 mA)24 k = 0.4 W (1/2 watt or 1 watt resistor) P4 = I 42 R4 = (36 mA)21 k = 1.3 W (2 watt resistor) P5 = I 52 R5 = (40 mA)20.6 k = 0.96 W (1 watt resistor) All power levels less than 2 W. Four less than 1 W. 32. 80 V = 400 Ω  390 Ω 200 mA 40 V R2 = = 266.67 Ω  270 Ω 150 mA R1 = 33. a. yes, RL  Rmax (potentiometer) b. VDR: VR2 = 3 V = c. VR1 = E  VL = 12 V  3 V = 9 V (Chose VR1 rather than VR2  RL since numerator of VDR R2 (12 V) R2 (12 V) = R1  R2 1k 3 V(1 k ) = 0.25 k = 250  R2 = 12 V R1 = 1 k  0.25 k = 0.75 k = 750  equation "cleaner") R1 (12 V) R1  ( R2  RL ) 9R1 + 9(R2  RL) = 12R1 R1  3( R2  RL )   2 eq. 2 unk( RL = 10 k) R1  R2  1 k   3R2 10 k 3R2 RL  R1 = R2  RL R2  10 k and R1(R2 + 10 k) = 30 k R2 VR1 = 9 V = 66 Chapter 7 R1R2 + 10 k R1 = 30 k R2 R1 + R2 = 1 k: (1 k  R2)R2 + 10 k (1 k  R2) = 30 k R2 R22 + 39 k R2  10 k2 = 0 R2 = 0.255 k, 39.255 k R2 = 255  R1 = 1 k  R2 = 745  34. a. b. 80   1 k = 74.07  20   10 k = 19.96  74.07 (40 V) = 31.51 V Vab = 74.07  + 19.96  Vbc = 40 V  31.51 V = 8.49 V c. P= d. 35. 36. 37. 38. 80 (40 V) = 32 V 100  Vbc = 40 V  32 V = 8 V Vab = (31.51 V ) 2 (8.49 V ) 2 + = 12.411 W + 3.604 W = 16.02 W 80  20  (32 V ) 2 (8 V ) 2 + = 12.8 W + 3.2 W = 16 W 80  20  The applied loads dissipate less than 20 mW of power. P= a. ICS = 1 mA b. Rshunt = Rm I CS (100 )(1 mA) 0.1  = 20 A  1 mA 20 I max  I CS Ω = 5 mΩ (1 k )(50  A) 2Ω 25 mA  0.05 mA (1 k )(50  A) =1Ω 50 mA: Rshunt = 50 mA  0.05 mA 100 mA: Rshunt  0.5 Ω 25 mA: Rshunt = Vmax  VVS 15 V  (50  A)(1 k) = 300 kΩ = I CS 50  A a. Rs = b. Ω/V = 1/ICS = 1/50 µA = 20,000 5 V  (1 mA)(1000 ) = 4 kΩ 1 mA 50 V  1 V 50 V: Rs = = 49 kΩ 1 mA 500 V  1 V = 499 kΩ 500 V: Rs = 1 mA 5 V: Rs = Chapter 7 67 39. 40. 10 MΩ = (0.5 V)(Ω/V)  Ω/V = 20  106 1 = 0.05 μA ICS = 1/(Ω/V) = 20  106 a. Rs = b. xIm = 2 k zero adjust 3V E  1 kΩ  = 28 kΩ  Rm  = 2 100  A 2 Im Runk = E Rseries + Rm + zero adjust + Runk 2 zero adjust  E    Rseries  Rm +  2 xI m   3V 30  103  30  103  30 kΩ  x100  A x 3 1 1 x = , Runk = 10 kΩ; x = , Runk = 30 kΩ; x = , Runk = 90 kΩ 4 2 4 = 41.  40. a. Carefully redrawing the network will reveal that all three resistors are in parallel R 12  and RT =  =4Ω N 3 b. Again, all three resistors are in parallel and RT = a. Network redrawn: 42. R 18   =6Ω N 3 Rohmmeter = 1.2 kΩ  (3.1 kΩ + 1.2 kΩ + 1.65 kΩ) = 1.2 kΩ  5.95 kΩ = 1 kΩ b. All three resistors are in parallel Rohmmeter = 68 R 18   =6Ω N 3 Chapter 7 Chapter 8 1. 2. 3. 8 (6 A) = 4.8 A 8 2  I2 = 6 A  I1 = 6 A  4.8 A = 1.2 A a. I1 = b. Vs = I1R1 = (4.8 A)(2 Ω) = 9.6 V a. I1 = I2 = 20 mA b. V2 = I2R2 = (20 mA)(3.3 kΩ) = 66 V Vs = IRT = (20 mA)((1.2 kΩ + 3.3 kΩ) = 20 mA(4.5 kΩ) = 90 V E + VR1  Vs = 0, VR1 = (8 mA)(2.7 kΩ) = 21.6 V Vs = E + VR1 = 10 V + 21.6 V = 31.6 V  4. a. b. c. Vs = E = 24 V 24 V 24 V E I2 = =6A   4 R1  R2 1   3  I + Is = I2, Is = I2  I = 6 A  2 A = 4 A 5. V1 = V2 = Vs = IRT = 0.6 A[6   24   24 ] = 0.6 A[6   12 ] = 2.4 V 2.4 V V = 0.1 A I2 = 2  R2 24  16 (2.4 V ) R3Vs = 1.6 V V3 =  24  R3  R4 6. a. E 24 V 24 V 24  E = 12 A, I R2 = =3A = = = R1 2  R2  R3 6  + 2  8  KCL: I + Is  I1  I R2 = 0 I1 = I s = I1 + I R2  I = 12 A + 3 A  4 A = 11 A b. Vs = E = 24 V VDR: V3 = 7. 68 R3 E 2 (24 V) 48 V =6V = = R2  R3 6  + 2  8  a. I= E 22 V = 4.68 A, Rp = Rs = 4.7 Ω = Rs 4.7  b. I= 9V E = 4.09 mA, Rp = Rs = 2.2 kΩ = Rs 2.2 k CHAPTER 8 8. 9. a. E = IRs = (6 A)(12 Ω) = 72 V, Rs = 12 Ω b. E = IRs = (18 mA)(5.6 kΩ) = 100.8 V, Rs = 5.6 kΩ a. CDR: IL = Es = IR = (20 A)(100 Ω) = 2 kV Rs = 100 Ω Es 2 kV = 18.18 A I= = Rs + RL 100  + 10  b. 10. 11. Rs ( I ) 100 (20 A) = 18.18 A, IL  I = Rs  RL 100  + 10  a. E = IR2 = (2 A)(5.6 Ω) = 11.2 V, R = 5.6 Ω b. ET = 12 V + 11.2 V = 23.2 V, RT = 10 Ω + 5.6 Ω = 15.6 Ω c. I3 = a. IT = 6.2 A  1.2 A  0.8 A = 4.2 A b. Vs = IT R = (4.2 A)(4 ) = 16.8 V ET 23.2 V = 217.64 mA  RT  91  15.6   91  12. IT = 7 A  3 A = 4 A R ( I ) 6 (4 A) = 2.4 A CDR: I1 = 2 T = R1  R2 4  + 6  V2 = I1R1 = (2.4 A)(4 ) = 9.6 V 13. a. b. Conversions: I1 = E1/R1 = 9 V/3 Ω = 3 A, R1 = 3 Ω I2 = E2/R2 = 20 V/2 Ω = 10 A, R2 = 2 Ω IT = 10 A  3A = 7 A, RT = 3 Ω  6 Ω  2 Ω = 2 Ω  2 Ω =1Ω V ab = IT RT = (7 A)(1 Ω) = 7 V  14. 7V = 1.17 A 6 c. I 3 = a. I= b. IT = 8 mA + 5.45 mA  3 mA = 10.45 mA R = 6.8 k  2.2 k = 1.66 k V1 = ITR = (10.45 mA)(1.66 k) = 17.35 V CHAPTER 8 E 12 V = 5.45 mA, Rp = 2.2 kΩ = R2 2.2 k  69 15. c. V1 = V2 + 12 V  V2 = V1  12 V = 17.35 V  12 V = 5.35 V d. I2 = V2 5.35 V = 2.43 mA = R2 2.2 k  a. 1 7 4  4I1  8I3 = 0 6  2I2  8I3 = 0 I1 + I2 = I3 ──────────── 5 4 A, I 3  A 7 7 1 5 4  I1   A, I R2  I 2  A, I R3  I3  A 7 7 7 I1 =  A , I 2  I R1 b. 16. 4  Va = I3R3 =  A  (8 ) = 4.57 V 7  10 + 12  3I3  4I1 = 0 12  3I3  12I2 = 0 I1 + I2 = I3 ─────────────── a. I1 = 3.06 A I2 = 0.19 A I3 = 3.25 A I R1  I1 = 3.06 A, I R3  I 2 = 0.19 A = I12Ω I R2  I 3 = 3.25 A b. I12Ω = c. 17. 1.714 )(1.5 A) = 0.19 A 1.714   12  the same 10  I1 5.6 kΩ  I3 2.2 kΩ + 20 = 0 20 + I3 2.2 kΩ + I2 3.3 kΩ  30 = 0 I1 + I2 = I3 ───────────────────────── I1 = I R1 = 1.45 mA, I2 = I R2 = 8.51 mA, I3 = I R3 = 9.96 mA 70 CHAPTER 8 18. 1.2 kΩ I1 + 9  8.2 kΩ I3 = 0 10.2 kΩ I2 + 8.2 kΩ I3 + 6 = 0 I2 + I3 = I1 ────────────────────── a. I1 = 2.03 mA, I2 = 1.23 mA, I3 = 0.8 mA I R1 = I1 = 2.03 mA I R2 = I3 = 0.8 mA I R3  I R4 = I2 = 1.23 mA = I9.1kΩ b. V4 = I2R4 = (1.23 mA)(1.1 kΩ) = 1.35 V Va = 6 V  V4 = 6 V  1.35 V = 4.65 V 19. I1 = I R1 (CW), I2 = I R2 (down), I3 = I R3 (CW), I4 = I R4 (down) I5 = I R5 (CW) a. E1  I1R1  I2R2 = 0 I2R2  I3R3  I4R4 = 0 I4R4  I5R5  E2 = 0 I1 = I2 + I3 I3 = I4 + I5 ──────────────── c. I2(R1 + R2) + I3R1 + 0 = E1 I2(R2)  I3(R3 + R4) + I5R4 =0  I5(R4 + R5) = E2 0 + I3R4 ─────────────────────────── 3I2 + 2I3 + 0 = 10 1I2  9I3 + 5I5 = 0 0 + 5I3  8I5 = 6 ───────────── d. I3 = I R 3 = 63.69 mA (CW) CHAPTER 8 b. E1  I2(R1 + R2)  I3R1 = 0 I2R2  I3(R3 + R4) + I5R4 = 0 I3R4  I5(R4 + R5)  E2 = 0 ─────────────────── 71 20. a. b. 21 a. 4  4I1  8(I1  I2) = 0 8(I2  I1)  2I2  6 = 0 ─────────────── 5 1 I1 =  A , I2 =   A 7 7 1 I R1 = I1 =  A 7 5 I R2 = I2 =  A 7 4  1   5  I R3 = I1  I2 =   A     A  = A (dir. of I1 ) 7  7   7  4  Va = I R3 R3   A  (8 ) = 4.57 V 7  10  4I1  3(I1  I2)  12 = 0 12  3(I2  I1)  12I2 = 0 ───────────────── I1 = 3.06 A, I2 = 0.19 A I E1 = 3.06 A (CCW) I E2 = 3.06 A + 0.19 A = 3.25 A (up) I R2 = I1  I2 = (3.06 A)  (0.19 A) = 3.25 A b. PE2  I E2 E2 = (3.25 A)(12 V) = 39 W PR3  I R23 R3 = (0.19 A)2 12 Ω = 433.2 mW 22. a. 10  I1(5.6 kΩ)  2.2 kΩ(I1  I2) + 20 = 0 20  2.2 kΩ(I2  I1)  I2 3.3 kΩ  30 = 0 ──────────────────────────── I1 = 1.45 mA, I2 = 8.51 mA I R1 = I1 = 1.45 mA, I R2 = I2 = 8.51 mA I R3 = I2  I1 = 7.06 mA (direction of I2) b. 23. a. V3.3kΩ = I2R2 = (8.51 mA)(3.3 kΩ) = 28.1 V I1(1.2 kΩ) + 9  8.2 kΩ(I1  I2) = 0 I2(1.1 kΩ) + 6  I2 (9.1 kΩ)  8.2 kΩ(I2  I1) = 0 ────────────────────────────────── I1 = 2.03 mA, I2 = 1.23 mA I R1 = I1 = 2.03 mA, I R3  I R4  I 2 = 1.23 mA I R2  I1  I 2 = 2.03 mA  1.23 mA = 0.80 mA (direction of I1) b. 72 Va = 6 V  I2(1.1 kΩ) = 6 V  (1.23 mA)(1.1 kΩ) = 6 V  1.35 V = 4.65 V CHAPTER 8 24. 10  I12  1(I1  I2) = 0 1(I2  I1)  I2 4  5(I2  I3) = 0 5(I3  I2)  I3 3  6 = 0 ────────────────────── 3I1  1I 2  0  10 a. 1I1  10 I 2  5 I 3  0 0  5I 2  8I 3   6 I1 = 3.31 A, I2 = 63.69 mA, I3 = 789.8 mA b, c. Ignore d. I10V  = I1 = 3.31 A I 6V  = I3 = (789.8 mA) = 789.8 mA 25. a. with or I1 2.2 kΩ  (I1  I2)9.1 kΩ + 18 V = 0 18 V  (I2  I1)9.1 kΩ  7.5 kΩ I2  (I2  I3)6.8 kΩ = 0 6.8 kΩ(I3  I2)  3 V  3.3 kΩ I3 = 0 ────────────────────── 11.3 kΩ I1  9.1 kΩI2 = 18 V 23.4 kΩ I2  9.1 kΩI1  6.8 kΩ I3 = 18 V 10.1 kΩ I3  6.8 kΩ I2 = 3 V ────────────────────── 11.3 kΩ I1  9.1 kΩ I2 = 18 V 9.1 kΩ I2 + 23.4 kΩ I2  6.8 kΩ I3 = 18 V 6.8 kΩ I2 + 10.1 kΩ I3 = 3 V ───────────────────────────── b. I1 = 1.21 mA, I2 = 0.48 mA, I3 = 0.62 mA c. I E1  = I1  I2 = 1.21 mA  (0.48 mA) = 1.69 mA I E2  = I3 = (0.62 mA) = 0.62 mA 26. a. 16  4I1  3(I1  I2)  12  4(I1  I3) = 0 12  3(I2  I1)  10 I2  15  4(I2  I3) = 0 16  4(I3  I1)  4(I3  I2)  7I3 = 0 ───────────────────────────── b. I1 = 0.24 A, I2 = 0.52 A, I3 = 1.28 A c. I R5 = I1 = 0.24 A CHAPTER 8 73 27. 28. d. a. 6.8 k I1  4.7 k(I1  I2) + 6  2.2 k(I1  I4) = 0 6  4.7 k(I2  I1)  2.7 k I2  8.2 k (I2  I3) = 0 1.1 k I3  22 k(I3  I4)  8.2 k(I3  I2)  9 = 0 5  1.2 k I4  2.2 k(I4  I1)  22 k(I4  I3) = 0 ──────────────────────────────────── b. I1 = 0.03 mA, I2 = 0.88 mA, I3 = 0.97 mA, I4 = 0.64 mA c. I6V = I1  I2 = 0.03 mA  (0.88 mA) = 0.91 mA, P6V = E I6V = (6 V)(0.91 mA) = 5.46 mW a. Network redrawn: b. 2I1  6  4I1 + 4I2 = 0 4I2 + 4I1  1I2 + 1I3  6 = 0 1I3 + 1I2 + 6  8I3 = 0 c. I1 = 3.8 A, I2 = 4.20 A, I3 = 0.20 A PE2  E2 I 3 = (6 V)(0.2 A) = 1.2 W PE1  E1 I 2 = (6 V)(4.2 A) = 25.2 W PT  PE1  PE2 = 1.2 W + 25.2 W = 26.4 W 29. 74 a. 20 V  IB(270 kΩ)  0.7 V  IE(0.51 kΩ) = 0 IE(0.51 kΩ) + 8 V + IC(2.2 kΩ)  20 V = 0 IE = IB + IC ────────────────────────────── IB = 63.02 μA, IC = 4.42 mA, IE = 4.48 mA b. VB = 20 V  IB(270 k) = 20 V  (63.02 A)(270 k) = 20 V  17.02 V = 2.98 V VE = IERE = (4.48 mA)(510 ) = 2.28 V VC = 20 V  IC(2.2 k) = 20 V  (4.42 mA)(2.2 k) = 20 V 9.72 V = 10.28 V c.   IC/IB = 4.42 mA/63.02 A = 70.14 CHAPTER 8 30. 24 V  6I1  4I2  10I1 + 12 V = 0 and 16I1 + 4I2 = 36 I1  I2 = 6 A ─────────────────── I1 = I2 + 6 A 16[I2 + 6 A] + 4I2 = 36 16I2 + 96 + 4I2 = 36 20I2 = 60 I2 = 3 A I1 = I2 + 6 A = 3 A + 6 A = 3 A I24V = I6 = I10 = I12V = 3 A (CW) I4 = 3 A (CCW) 31. 20 V  4I1  6(I1  I2)  8(I3  I2)  1I3 = 0 10I1  14I2 + 9I3 = 20 I3  I1 = 3 A I2 = 8 A ──────────────────────── 10I1  14(8 A) + 9[I1 + 3 A] = 20 19I1 = 105 I1 = 5.526 A I3 = I1 + 3 A = 5.526 A + 3 A = 8.526 A I2 = 8 A I20V = I4Ω = 5.53 A (dir. of I1) I6 = I2  I1 = 2.47 A (dir. of I2) I8 = I3  I2 = 0.53 A (dir. of I3) I1 = 8.53 A (dir. of I3) CHAPTER 8 75 32. a. b. 33. a. b. 34. (4 + 8)I1  8I2 = 4 (8 + 2)I2  8I1 = 6 ─────────────  1   5  4 I8  I1  I 2    A     A   A  7   7  7 (4 + 3)I1  3I2 = 10  12 (3 + 12)I2  3I1 = 12 ───────────────── I 3   I 2  I1  0.19 A  (3.06 A) = 3.25 A a. a. I1(5.6 k + 2.2 k)  2.2 k (I2) = 10 + 20 I2(2.2 k + 3.3 k)  2.2 k (I1) = 20  30 ────────────────────────────── b. I E1 = I1 = 1.45 mA, I E2  = 8.51 mA, I E3  = I1  I2 = (1.45 mA)  (8.5 mA) = 9.96 mA 35. 36. a. b. I1(2 + 1)  1I2 = 10 I2(1 + 4 + 5)  1I1  5I3 = 0 I3(5 + 3)  5I2 = 6 ─────────────────── I1 = 3.31 A, I2 = 63.69 mA, I3 = 789.8 mA c. I R2  I1  I 2 = (3.31 A)  (63.69 mA) = 3.37 A a. b. (2.2 k + 9.1 k)I1  9.1 kI2 = 18 (9.1 k + 7.5 k + 6.8 k)I2  9.1 k I1  6.8 kI3 = 18 (6.8 k + 3.3 k)I3  6.8 kI2 = 3 ─────────────────────────── I1 = 1.21 mA, I2 = 0.48 mA, I3 = 0.62 mA c. I E1  = I1  I2 = 1.21 mA  (0.48 mA) = 1.69 mA I E2  = I3 = (0.62 mA) = 0.62 mA 76 CHAPTER 8 37. (3 Ω + 6 Ω)I1  6 Ω I2 = 9 V (6 Ω + 2 Ω)I2  6 Ω I1 = 20 V ──────────────────── a. 9 Ω I1  6 Ω I2 = 9 V 6 Ω I1 + 8 Ω I2 = 20 V ──────────────────── I1 = 5.33 A, I2 = 6.5 A b. 38. 39. Vab = 2 Ω (I2)  20 V = 2 Ω = 2 Ω(6.5 A)  20 V = 13 V  20 V = 7 V a. I1(6.8 k + 4.7 k + 2.2 k)  4.7 k I2  2.2 k I4 = 6 I2(2.7 k + 8.2 k + 4.7 k)  4.7 k I1  8.2 k I3 = 6 I3(8.2 k + 1.1 k + 22 k)  22 k I4  8.2 k I2 = 9 I4(2.2 k + 22 k + 1.2 k)  2.2 k I1  22 k I3 = 5 ─────────────────────────── b. I1 = 0.03 mA, I2 = 0.88 mA, I3 = 0.97 mA, I4 = 0.64 mA c. I 22k  = I4  I3 = (0.64 mA)  (0.97 mA) = 0.33 mA V22kΩ = I22kΩ  22 kΩ = (0.33 mA)(22 kΩ) = 7.26 V a. (1 Ω + 2 Ω + 4 Ω)I1  2 ΩI2  4 ΩI3 = 12 V (2 Ω + 2 Ω + 10 Ω)I2  2 ΩI1  10 ΩI3 = 20 V (4 Ω + 10 Ω + 8 Ω)I3  10 ΩI2  4 ΩI1 = 20 V ─────────────────────────── 7I1  2I2  4I3 = 12 2I1  14I2 + 10I3 = 20 4I1 + 10I2  22I3 = 20 ──────────────── b. I1 = 2.38 A, I2 = 0.195 A, I3 = 1.25 A c. Va = (I1  I3)4 Ω = (2.38 A  1.25 A)4 Ω = 4.5 V Vb = I38 Ω = (1.25 A)(8 Ω) = 10 V d. Vab = Va  Vb = 4.5 V  10 V = 5.5 V CHAPTER 8 77 40. a. At V1:  I i   I o 0 V1 V V  5A  1 2 2 8  Ii   Io At V2: V1  V2 V 3A 2 8 4 and 41. 1 1 1 V1     V2    5  2 8 8 1 1 1  V1    V2     3 8 8 4  ─────────────────────────── b. V1 = 10.27 V, V2 = 11.36 V c. V8Ω = V1  V2 = 10.27 V  (11.36 V) = 1.09 V d. I 2  = a. V1 10.27 V  = 5.14 A 2 2 V 11.36 V I 4  = 2  = 2.84 A 4 4 At V1:  I i   I o V1  12 A  I 6  and V1  I 6  54 V  V2  0 8 V  V  54 V V1 V or I = 1 2   2 9 A 6 6 6 V1 V1 V2  12 A    9A so that 0 = 8 6 6 1   1  1  or V1    V2  = 12 A + 9 A = 3 A  8  6    6   0 At V2:  I i   I o V2 V  2 20  5  V1 V2 V V  9 A  2  2 6 6 20  5  I or 1 1   1  1  and V2     V1  = 9 A   6   6  20  5   78 CHAPTER 8 42. b. 1   1  1  resulting in V1    V2  = 3 A   6  8  6  1 1   1   1 = 9 A V1   V2      6   6  20  5   ────────────────────────────── V1 = 29.29 V, V2 = 33.34 V c. I 20    a. V2 33.34 V  = 1.67 A 20  20  At V1:  I i   I o V1 V1  V2  2A 2 4 4A Ar V2:  I i   I o 2A or 43. V1  V2 V V  2  2 4 20  5  1 1 1 V1     V2    2 2 4  4 1 1 1 1 V1    V2    2 4  4 20 5  b. V1 = 4.8 V, V2 = 6.4 V c. I1: P = V1I1 = (4.8 V)(4 A) = 19.2 W I2: P = (V1  V2 ) I 2  (4.8 V  6.4 V)(2 A) = 3.2 W a. At V1:  I i   I o 0  6A  V1 V1  V2 V1  V2   5 3 2 At V2:  I i   I o 7A  V1  V2 V1  V2 V2 V    2 3 2 4 8 1 1  1   1  1 so that V1  = 6 A    V2     3  2   5  3  2  1 1 1  1   1  1 =7A V2      V1     3  2   4  8  3  2  ──────────────────────────────────── or 1.03V1  0.833V2 = 6 0.833V1 + 1.21V2 = 7 ─────────────────── CHAPTER 8 79 44. b. V1 = 2.59 V, V2 = 4 V c. V2Ω = V3Ω = V2  V1 = 4 V  (2.59 V) = 6.59 V V5Ω = V1 = 2.59 V V4Ω = V8Ω = V2 = 4 V a. Source conversion: I3 = 12 V = 3 A, Rp = R3 = 4 Ω 4 At V1:  I i   I o 0 V1 V V V  1 5 A 1 2 + 3 A 3 6 4 At V2:  I i   I o 3A  V1  V2 V2  4A 4 8 Rewritten: b. c. 45. 1 1  V2  1 V1     = 5 A  3 A  3  6  4   4  1   1   1 V1   V2   = 4 A + 3A   4  8  4  V1 = 14.86 V, V2 = 12.57 V 14.86 V I 6   = 2.48 A 6 a. Source Conversion: I2 = At V1:  I i   I o 5A 15 V = 5 A, Rp = R1 = 3 Ω 3 V1 V1  V2 V1  V3   3 6 6 At V2:  I i   I o V  V3 V1  V2 V2  3 A 2 6 4 5 At V3:  I i   I o V1  V3 V2  V3 V3   6 5 7 80 CHAPTER 8 Rewritten: 1 1  1 1  1    V1  V2  V3 = 5 A       3 6 6 6 6   1 1  1 1  1 V2     V1  V3 = 3 A       6 4 5 6 5   1 1  1 1  1 V3  V2  V1 = 0          6 5 7 5 6   ──────────────────────────────────── 46. b. V1 = 7.24 V, V2 = 2.45 V, V3 = 1.41 V c. V5 = V3  V2 = 1.41 V  (2.45 V) = 3.86 V  + a. Source Conversion: I2 = 16 V = 4 A, Rp = R2 = 4 Ω 4 At V1:  I i   I o 02 A+ V1 V1  V3 V1  V2   9  20  20  At V2:  I i   I o V1  V2 V2  V3 V   2 20  20  18  At V3:  I i   I o V1  V3 V2  V3 V   4A  3 20  20  4 Rewritten: 1 1  1 1  1 V1     V2  V3 = 2 A  20   9  20  20   20  1 1  1 1  1 V2     V1  V3 = 0  20   20  20  18   20  1 1  1 1  1 V3  V2  V1 = 4A     20   20  20  4   20  ──────────────────────────────────── CHAPTER 8 b. V1 = 6.64 V, V2 = 1.29 V, V3 = 10.66 V c. VR6 = V3  V1 = 10.66 V  (6.64 V) = 17.30 V  + 81 47. a. At V1:  I i   I o 05 A V1 V1  V2  2 2 At V2:  I i   I o V  V3 V1  V2 V2 V   2  2 2 9 7 2 At V3:  I i   I o V2  V3 V V 5 A  3  3 2 2 4 Rewritten: 1  1  1  2   2   V1  2  V2  0 = 5 A 1 1 1  1 1  1  2   9   7   2   V2  2  V1  2  V3 = 0 1 1  1  1 V3    V3  V2 = 5A  2 2  2  4  ──────────────────────────────────── 48. b. V1 = 5.31 V, V2 = 0.62 V, V3 = 3.75 V c. I9  V2 0.62 V  = 68.9 mA 9 9 a. At V1:  I i   I o 05 A+ V1 V1  V3  2 6 At V2:  I i   I o 5A2 A V2 4 At V3:  I i   I o V1  V3 V  2A  3 6 5 82 CHAPTER 8 Rewritten: 1  1  1 V1    V3 = 5 A     2 6 6    1  V2  =5A2A  4   1  1  1 V3  V1 = 2A       6 5 6   ─────────────────────── 49. b. V1 = 6.92 V, V2 = 12 V, V3 = 2.3 V c. I 4Ω = V2 12 V  =3A 4 4 a.  Ii   Io Node V1: 2A= V1 V1  V2  6  10  Supernode V2, V3: V V V V 0= 2 1  2  3 10  4  12  Independent source: V2  V3 = 24 V or V3 = V2  24 V 2 eq. 2 unknowns: V1 V1  V2  =2A 6  10  V2  V1 V2 V2  24 V   =0 10  4 12  ───────────────────── 0.267V1  0.1V2 = 2 +0.1V1  0.433V2 = 2 ──────────────── V1 = 10.08 V, V2 = 6.94 V V3 = V2  24 V = 17.06 V CHAPTER 8 83 50.  Ii   Io Supernode: 3A+4A=3A+ V1 V  2 20  40  V1 V   2 4 A  2 eq. 2 unk.  20  40  V2  V1  16 V  Subt. V2 = 16 V + V1 V (16 V  V1 ) 4A= 1  20  40  and V1 = 48 V V2 = 16 V + V1 = 64 V ────────────────── 51. a. 1 1 1 V1     V2   = 5  2 8 8 1 1 1  V1    V2    = 3 8 8 4  ──────────────────── V1 = 10.27 V, V2 = 11.36 V b. 52. VI1 = V1 = 10.27 V, VI 2 = V2 = 11.36 V a. 1 1  1 V1     V2   = 12 A + 9 A = 3 A 8 6   6  1 1 1 1 V2      V1   = 9 A 20 5 6   6 ────────────────────────────── V1 = 29.29 V, V2 = 33.34 V 84 CHAPTER 8 b. 53. V1  V6Ω  54 V  V2 = 0 V6 = V1  V2  54 V = 29.29 V  (33.34 V)  54 V = 49.95 V +  a. 1 1  1 1  1 V1     V2  V3 = 2 A  20   9  20  20   20  1 1  1 1  1 V2     V1  V3 = 0  20   20  20  18   20  1 1  1 1  1 V3  V2  V1 = 4A     20   20  20  4   20  V1 = 6.64 V, V2 = 1.29 V, V3 = 10.66 V b. 54. Original Network: V4Ω = 16 V  V3 = 16 V  10.66 V = 5.34 V 5.34 V I 4Ω = = 1.34 A 4 a. 1  1  1  2   2   V1  2  V2  0 = 5 A 1 1 1  1 1  1  2   9   7   2   V2  2  V1  2  V3 = 0 1 1  1  1  2   2   4   V3  2  V2 = 5A ──────────────────────────────────── CHAPTER 8 b. V1 = 5.31 V, V2 = 0.62 V, V3 = 3.75 V c. I9  V2 0.62 V  = 68.89 mA 9 9 85 55. 56. 1  1  1 V1    V3 = 5 A   2  6  6   1  V2  =5A2A  4   1  1  1 V3  V1 = 2A    6  5  6  ─────────────────────── a. b. V1 = 6.92 V, V2 = 12 V, V3 = 2.3 V c. I 2Ω = V2 6.92 V  = 3.46 A 2 2 a. 1 1  1 1  1 V1     V2  V3 = 12 A  2 1  2  2   2  1 1  1 1  1 V2     V1  V3 = 2 A  10   2  4  10   2  1 1  1 1  1 V3     V1  V2 = 2A  10   2  10  8   2  ──────────────────────────────────── and 2V1  0.5V2  0.5V3 = 12 0.5V1 + 0.85V2  0.1V3 = 2 0.5V1  0.1V2 + 0.725V3 = 2 ─────────────────────── V1 = 9.63 V, V2 = 4.49 V, V3 = 10.02 V b. 57. Vab = V2  V3 = 4.49 V  10.02 V = 5.53 V a. I1(6 Ω + 2 Ω + 10 Ω)  2 ΩI2  10 ΩI3 = 12 V I2(2 Ω + 5 Ω + 5 Ω)  2 ΩI1  5 ΩI3 = 0 I3(5 Ω + 20 Ω + 10 Ω)  10 ΩI1  5 ΩI2 = 0 or 18I1  2I2  10I3 = 12 2I1 + 12I2  5I3 = 0 10I1  5I2 + 35I3 = 0 ─────────────── I1 = 850.99 mA, I2 = 258.53 mA, I3 = 280.07 mA 86 CHAPTER 8 58. b. I R5 = I3  I2 = 280.07 mA  258.53 mA = 21.54 mA c. no d. no 2 Ω/10 Ω = 1 1  5 Ω/20 Ω = 5 4 a. 1 1  1 1  1 V1     V2  V3 = 2 A  5 6  2  5  2  1 1  1 1  1 V2  V1  V3 = 0     5  2  5  10   2  1 1  1 1  1 V3  V2  V1 = 0     5  5  5  20   5  ──────────────────────────────────── or 0.867V1  0.5V2  0.2V3 = 12 0.5V1 + 0.8V2  0.2V3 = 0 0.2V1  0.2V2 + 0.45V3 = 0 ─────────────────────── V2 = 5.7 V, V3 = 5.6 V 59. b. V5Ω = V2  V3 = 5.7 V  5.6 V = 0.1 V c. no d. no 2 Ω/10 Ω = 1 1  5 Ω/20 Ω = 5 4 a. Source conversion: E = IR = (12 m)(2 kΩ) = 24 V Rs = 2 kΩ I1(2 kΩ + 33 kΩ + 3.3 kΩ)  33 kΩI2  3.3 kΩI3 = 24 V I2(33 kΩ + 56 kΩ + 36 kΩ)  33 kΩI1  36 kΩI3 = 0 I3(36 kΩ + 3.3 kΩ + 5.6 kΩ)  3.3 kΩI1  36 kΩI2 = 0 ──────────────────────────────────── I1 = 0.97 mA, I2 = I3 = 0.36 mA b. I5 = I2  I3 = 0.36 mA  0.36 mA = 0 mA c, d. yes CHAPTER 8 87 60. a. 1 1 1 1   1 V1   V2  V3 = 12 mA    56 k  2 k 33 k 56 k 33 k 1 1 1 1   1 V2   V1  V3 = 0    36 k  33 k 3.3 k 36 k 33 k 1 1 1 1   1  V3  V2  V1 = 0    56 k  56 k 36 k 5.6 k  36 k ──────────────────────────────────── Rewritten: 548.16V1  30.3V2  17.86V3 = 12 × 103 30.3V1 + 361.11V1  27.78V3 = 0 17.86V1  27.78V2 + 224.21V3 = 0 ──────────────────────────── V2 = 2.01 V, V3 = 2.01 V b. VR5 = V2  V3 = 2.01 V  2.01 V = 0 V c, d. yes 61. Mesh Analysis (1 k + 2 k + 2 k)I1  2 k I2  2 k I3 = 10 (2 k + 2 k + 2 k)I2  2 k I1  2 k I3 = 0 (2 k + 2 k + 2 k)I3  2 k I1  2 k I2 = 0 ───────────────────────────────── I1 = I10V = 3.33 mA Nodal Analysis: Source conversion: I = 10 V/1 kΩ = 10 mA, R = 1 kΩ 1 1  1 1  1  V1  V2  V3 = 10 mA + +  2 k  1 k 2 k 2 k  2 k 1 1  1 1  1 V2  V1  V3 = 0  + +  2 k  2 k 2 k 2 k  2 k 1 1  1 1  1 V3  V2  V1 = 0 + +   2 k 2 k 2 k 2 k 2 k       ────────────────────────────────────── V1 = 6.67 V = E  IRs = 10 V  I(1 k) 10  6.67 V = 3.33 mA I= 1k 88 CHAPTER 8 62. Mesh Analysis Source conversion: E = 20 V, R = 10  (10 + 10 + 20)I1  10I2  20I3 = 20 (10 + 20 + 20)I2  10I1  20I3 = 0 (20 + 20 + 10)I3  20I1  20I2 = 0 ────────────────────── I1 = I20V = 0.83 A V  20 V  8.3 V = 11.7 V Is = V 11.70 V = 1.17A  10  Rs Nodal Analysis: 1 1 1 1 1 V1        V2    V3  2  10 10 20   20  10  1 1 1 1 1 V2        V1    V3  0  20 20 10   20   20  1 1 1 1 1 V3        V1    V2  0 10 20 20  10   20  ─────────────────────── V I Rs = 1 = 1.17 A Rs 63. I= = 20 V 4 2  2   +   + 3     + 4  5 5  5  20 V 4  + (3.14 )  (4.4 ) 5 = 7.36 A CHAPTER 8 89 64. RT = 2.27 k + [4.7 k + 2.27 k]  [1.1 k + 2.27 k] = 2.27 k + [6.97 k]  [3.37 k] = 2.27 k + 2.27 k = 4.54 k 8V = 1.76 mA I= 4.54 k (Y-Δ conversion) 400 V 400 V  I= 12 k  12 k  6 k 3 k = 133.33 mA 65. 66. a. I= 42 V 42 V  (18   18 )   (18   18 )  (18   18 ) 9   9   9   = 7 A (YΔ conversion) b.   Y conversion I s1 = 10 V 5V 15 V = 0.83 mA + = 18 k  18 k  18 k  67. 90 CHAPTER 8 68. a. b. 69. R = R1 + 1 k = 3 k R = R2 + 1 k = 3 k 3k = 1.5 k RT = 2 RT = 1 k + 1.5 k + 1 k = 3.5 k E 20 V = 5.71 mA Is = = RT 3.5 k  Using two   Y conversions: c  g: 27   9   27  = 5.4  a  h: 27   9   27  = 5.4  RT = 5.4   (13.5  + 5.4 ) = 5.4   18.9 = 4.2  CHAPTER 8 91 Chapter 9 1. a. RT (from source) = 4 Ω + 2 Ω  12 Ω = 4 Ω + 1.71 Ω = 5.71 Ω E1 16 V Is    2.8 A RT 5.71 V 2 (2.8 A)  0.4 A I12   2   12  E1: RT (from source) = 2 Ω + 4 Ω  12 Ω =2Ω+3Ω =5Ω E2 10 V Is   2A RT 5 4 (2 A)  0.5 A I12   4   12  E2: I12Ω = 0.5 A  0.4 A = 0.1 A b. I12  c. 2. 1.333 (1A) = 0.1 A 1.333   12  the same a. 24 (3 A) = 2.25 A 24   8  V   I R = (2.25)(4.7 Ω) = 10.575 V I  V   4.7 (12 V) = 1.763 V 4.7   3.3   24  V   10.575 V  1.763 V = 8.81 V  92 CHAPTER 9 b. c. 3. V  2 (10.575 V)2  = 23.79 W R 4.7  V  2 (1.763 V) 2 P= = 0.661 W  R 4.7  P= V 2 (8.81 V) 2  = 16.51 W R 4.7  d. P= e. 23.79 W + 0.661 W  16.51 W 24.45 W  16.51 W E: RT = 12   24   56   28.8  24 V E Is = = 0.833 A  RT 28.8  24 (0.833 A)  I 56 = 0.25 A 24   56  I: 24   56   16.8  12 (8 A) I   = 3.33 A 12   16.8  24 (3.33 A)    I 56 =1A 24   56  I 56   I   I   0.25 A+1 A = 1.25 A 4. E1: 42 V = 1.944 A 18  + 3.6  9 ( IT ) 9 (1.944 A) I1 =  96 15  = 1.17 A IT = E2: IT = E2 24 V =2A  RT 12  I24V = IT + I1 = 2 A + 1.17 A = 3.17 A (dir. of I1) CHAPTER 9 93 5. E: V2 = 6.8 k(36 V) = 13.02 V 6.8 k  12 k I: I2 = 12 k(9 mA) = 5.75 mA 12 k  6.8 k V2  I 2 R2 = (5.75 mA)(6.8 k) = 39.10 V V2 = V2  V2 = 13.02 V + 39.10 V = 52.12 V 6. 1.2 k  4.7 k  0.956 k 3.3 kΩ + 0.956 kΩ = 4.256 kΩ 4.256 k(5 mA) I  2.2 k  4.256 k  3.3mA I: E: I  I   I  = 3.3 mA + 0.986 mA = 4.286 mA 7. E1: I1 = 94 2.2 kΩ + 3.3 kΩ = 5.5 k Ω 5.5 k  4.7 k  2.53 k RT = 2.54 kΩ + 1.2 kΩ = 3.73 kΩ 8V Is  = 2.14 mA 3.73 k 4.7 k(21.4 mA) I   = 0.986 mA 4.7 k  5.5 k E1 12 V = 1.03 A  RT 6   5.88  CHAPTER 9 30 ( I1 ) 30 (1.03 A)  30   7  37   835.14 mA Vs = I(4 ) = (835.14 mA)(4 ) = 3.34 V I  I: 8 (6 A) =4A 8  4 Vs = I(4 ) = 4 A(4 ) = 16 V I = E2: RT = 12   (4  + 5 ) = 12   9  = 5.14  E 8V I = 2   0.875 A RT 4   5.14  12 ( I ) 12 (0.875 A) I =   0.5 A 12   9  21  Vs I (4 ) = 0.5 A(4 ) = 2 V Vs = Vs Vs  Vs = 16 V  3.34 V  2 V = 10.66 V 8. a. RTh = R3 + R1  R2 = 4 Ω + 6 Ω  3 Ω = 4 Ω + 2 Ω = 6 Ω R2 E 3 (18 V) ETh =  6V R2  R1 3   6  CHAPTER 9 95 9. b. I1 = a. RTh: ETh 6V = 0.75 A = RTh  R 6  + 2  6V = 166.67 mA I2 = 6  + 30  6V I3 = = 56.60 mA 6  + 100   RTh  3.3 k  1.2 k  2.4 k  3.3 k  0.8 k  4.1 k ETh: ETh  (120 mA)(2.4 k  1.2 k)  96 V RTh = 4.1 kΩ b. 96 V = 15.74 mA 6.1 k P = I2R = (15.74 mA)2 2 kΩ = 0.495 W R = 100 kΩ: 96 V = 0.922 mA I= 104.1 k P = I2R = (0.922 mA)2 100 kΩ = 85 mW I= 10. a. RTh:  RTh = 5 Ω + 5 Ω  5 Ω = 7.5 Ω ETh: ETh = 96 20 V = 10 V 2 CHAPTER 9 2 b.  ETh   10 V  R = 2 Ω: P =  R =  2  = 2.22 W     R R 7.5 2    Th  2 2 10 V   R = 100 Ω: P =   100  = 0.87 W    7.5 100   11. RTh = 3 Ω  8 Ω = 2.18 Ω 18 V + 12 V = 2.73 A 3 8  V3Ω = IR = (2.73 A)(3 Ω) = 8.19 V ETh = 18 V  8.19 V = 9.81 V I= 12. RTh: RTh = 5.6 kΩ  2.2 kΩ = 1.58 kΩ ETh: Superposition: I: ETh = IRT = 8 mA(5.6 k  2.2 k) = 8 mA(1.579 k) = 12.64 V E: 5.6 k(16 V) 5.6 k  2.2 k = 11.49 V ETh = + ETh = 11.49 V  12.64 V = 1.15 V  CHAPTER 9 97 13. RTh: RTh = 4   (2  + 6 Ω  3 ) = 2  ETh: 72 V =9A 6   3   (2   4 ) 3 ( IT ) 3 (9 A) =3A I2 =  3  6 9 ETh = V6 + V2 = (IT)(6 ) + I2(2 ) = (9 A)(6 ) + (3 A)(2 ) = 60 V IT = 14. a. RTh: RTh = 2.7 k  (4.7 k + 3.9 k) = 2.7 k  8.6 k = 2.06 k ETh: I = 3.9 k(18 mA) = 6.21 mA 3.9 k  7.4 k ETh = I(2.7 k) = (6.21 mA)(2.7 k) = 16.77 V b. 16.77 V 16.77 V  2.06 k  1.2 k 3.26 k = 5.14 mA I= 98 CHAPTER 9 15. a. RTh: RTh = 2  + 8  = 10  ETh: ETh = V16 20 V = 825.08 mA 20   4.24  5 ( IT ) 5 (825.08 mA) = 125.01 mA I =  5   28  33  ETh = V16Ω = (I)(16 Ω) = (125.01 mA)(16 Ω) = 2 V IT = b. 16. a. 20 Ω: ETh 2V 2V = 66.67 mA   RTh  R 10   20  30  2V 2V  = 33.33 mA 50 Ω: I = 10   50  60  2V 2V 100 Ω: I =  = 18.18 mA 10   100  110  I = RTh: RTh = 3. 3 k + 2.2 k  1.1 k = 3.3 k + 0.73 k = 4.03 k ETh: Superposition: E1: ETh = V2.2kΩ = 2.2 k(12 V) 2.2 k  1.1 k =8V CHAPTER 9 99 ETh = E2 = 4 V ETh = ETh + ETh = 8 V + 4 V = 12 V b. V= 17. 1.2 k(12 V) = 2.75 V 1.2 k  4.03 k RTh: RTh = 2.2 k  5.6 k = 1.58 k R = 1.58 k + 3.3 k = 4.88 k R = 4.88 k  6.8 k = 2.84 k RTh = 1.2 k + R = 1.2 k + 2.84 k = 4.04 k ETh: Source conversions: 22 V = 10 mA, Rs = 2.2 k I1 = 2.2 k 12 V I2 = = 2.14 mA, Rs = 5.6 k 5.6 k Combining parallel current sources: IT = I1  I2 = 10 mA  2.14 mA = 7.86 mA 2.2 k  5.6 k = 1.58 k 100 CHAPTER 9 Source conversion: E = (7.86 mA)(1.58 k) = 12.42 V R = Rs + 3.3 k = 1.58 k + 3.3 k = 4.88 k I= 12.42 V  6 V 6.42 V = 549.66 A  4.88 k  6.8 k 11.68 k V6.8kΩ = I(6.8 k) = (549.66 A)(6.8 k) = 3.74 V ETh = 6 V + V6.8kΩ = 6 V + 3.74 V = 9.74 V 18. a. RTh: RTh = 51 k  10 k = 8.36 k ETh: ETh = b. 10 k(20 V) = 3.28 V 10 k  51 k IERE + VCE + ICRC = 20 V but IC = IE and IE(RC + RE) + VCE = 20 V 20 V  VCE 20 V  8 V 12 V = 4.44 mA or IE =   2.2 k  0.5 k 2.7 k RC  RE c. ETh  IBRTh  VBE  VE = 0 E  VBE  VE 3.28 V  0.7 V  (4.44 mA)(0.5 k) and IB = Th  8.36 k RTh 2.58 V  2.22 V 0.36 V = 43.06 μA = = 8.36 k 8.36 k CHAPTER 9 101 19. d. VC = 20 V  ICRC = 20 V  (4.44 mA)(2.2 k) = 20 V  9.77 V = 10.23 V a. ETh = 20 V I = 1.6 mA = ETh 20 V 20 V = 12.5 k  , RTh  1.6 mA RTh RTh b. ETh= 60 mV, RTh = 2.72 k c. ETh = 16 V, RTh = 2.2 k 20. RTh = 4   (2   2 )  4 =2 2 2 (6 V) 12 V  = 1.5 V 2 4 2  8  V2Ω = V4Ω = 1.5 V ETh = V4Ω + V2Ω = 1.5 V + 1.5 V = 3 V V4Ω = 21. a. From Problem 8, RN = RTh = 6 Ω RT  6   3   4   6   1.714   7.714  E 18 V Is = = 2.333 A  RT 7.714  3 (2.333 A) =1A IN = 3 4  102 b. RTh = 6 Ω, ETh = INRN = (1 A)(6 Ω) = 6 V c. same results CHAPTER 9 22. a. From Problem 9, RN = RTh = 4.1 kΩ 2.4 k(120 mA) 2.4 k  (1.2 k  3.3 k)  87.80 mA I  1.2 k(87.80 mA) 1.2 k  3.3 k  23.41 mA RTh = 4.1 kΩ, ETh = INRN = (23.41 mA)(4.1 kΩ) = 96 V same results. IN  b. c. 23. From Problem 11, RN = RTh = 2.18 Ω IN = 6 A  1.5 A = 4.5 A 24. From Problem 12, RN = RTh = 1.58 kΩ IN = 8 mA  7.27 mA = 0.73 mA 25. From Problem 13, RN = RTh = 2 Ω 72 V = 18 A 4 72 V I 3Ω = = 16 A 3 6 2  6 (16 A) I 2Ω = = 12 A 6 2  IN = I4Ω + I2Ω = 18 A + 12 A = 30 A I 4Ω = CHAPTER 9 103 26. From Problem 15, RN = RTh = 10 Ω RT = 20 Ω + 5 Ω  (12 Ω + 1.778 Ω) = 23.67 Ω E 20 V Is = = 844.95 mA  RT 23.67  5 (844.95 mA) =224. 98 mA I12Ω = 5   (12   1.778 ) 16 (224.98 mA) IN = = 200 mA 16   2  27. 104 From Problem 17, RN = RTh = 4.04 kΩ CHAPTER 9 4.88 k(3.427 mA) 4.88 k  1.02 k  2.83 mA I  IN = 28. 6.8 k(2.83 mA) = 2.41 mA 6.8 k   k From Problem 20, RN = RTh = 2 Ω IN  29. a. 6V = 1.5 A 4 RN: RN = 4   12  = 3  E = 12 V: IN = CHAPTER 9 12 V =3A 4 105 I = 2 A: IN = 2 A IN = IN + IN = 3 A + 2 A = 5 A b. I: 3 (5A) 3      145.63 mA I  V   IR  (145.63 mA)(100 )  14.56 V E: 100 (72 V) 100   3   69.9 V V100   V   V  V    69.9 V  14.56 V  55.34 V 30. 31. 106 a. R = RTh = 6 Ω from Problem 8 b. ETh = 6 V from Problem 8 E2 (6 V)2 = 1.5 W Pmax = Th  4 RTh 4(6 ) a. R = RTh = 2.18 Ω from Problem 11 b. ETh = 9.81 V from Problem 11 2 ETh (9.81 V) 2 = 11.06 W Pmax =  4 RTh 4(2.18 ) CHAPTER 9 32. 33. 34. a. R = RTh = 2 Ω from Problem 13 b. ETh = 60 V from Problem 13 E2 (60 V) 2 = 450 W Pmax = Th  4 RTh 4(2 ) a. R = RTh = 4.04 kΩ from Problem 17 b. ETh = 9.74 V from Problem 17 E2 (9.74 V)2 = 5.87 mW Pmax = Th  4 RTh 4.04 k) a. R = RN = RTh = 2.18 Ω b. Pmax = I N2 RN (13.33A)2 2.18   = 96.84 W 4 4 2 35. 36.  ETh  Pmax =   R4  RTh  R4  with R1 = 0 Ω, ETh is a maximum and RTh is a minimum.  R1 = 0 Ω a. V, and therefore V4 wll be its largest value when R2 is as large as possible. Therefore, choose R2 = open-circuit ( Ω) and V2 P4 = 4 will be a maximum. R4 b. 37. No, examine each individually. The voltage VL will be a maximum when R = 500 Ω because the full voltage, E, will appear across RL. Pmax = CHAPTER 9 VL2 E 2 (12 V) 2 = 1.44 W   RL RL 500  107 IT = 4 A + 7 A = 11 A RT = 10   6   3  = 1.67  VL = ITRT = (11 A)(1.67 ) = 18.37 V V 18.37 V IL = L  = 6.12 A RL 3 38. 39. 5 V / 2.2 k  20 V / 8.2 k = 0.2879 V 1/ 2.2 k  1/ 8.2 k 1 Req = = 1.7346 k 1/ 2.2 k  1/ 8.2 k Eeq 0.2879 V  IL = = 39.3 μA Req  RL 1.7346 k  5.6 k Eeq = VL = ILRL = (39.3 µA)(5.6 k) = 220 mV 40. 41. 42. IT = 5 A  0.4 A  0.2 A = 4.40 A RT = 200   80   50 Ω  50  = 17.39  VL = ITRT = (4.40 A)(17.39 ) = 75.52 V V 76.52 V = 0.38 A IL = L  RL 200  (4 A)(4.7 )  (1.6 A)(3.3 ) 18.8 V + 5.28 V = 3.01 A  4.7   3.3  8 Req = 4.7  + 3.3  = 8  Req ( I eq ) 8 (3.01 A) = = 2.25 A IL = Req + RL 8  + 2.7  VL = ILRL = (2.25 A)(2.7 Ω) = 6.08 V Ieq =  (4 mA)(8.2 k)  (8 mA)(4.7 k)  (10 mA)(2 k) I eq = 8.2 k  4.7 k  2 k 32.8 V + 37.6 V  20 V = 3.38 mA = 14.9 k Req = 8.2 k + 4.7 k + 2 k = 14.9 k Req I eq (14.9 k)(3.38 mA) IL = = 2.32 mA = Req + RL 14.9 k  6.8 k VL = ILRL = (2.32 mA)(6.8 k) = 15.78 V 108 CHAPTER 9 43. 15 k  (8 k + 7 k) = 15 k  15 k = 7.5 k 7.5 k(60 V) = 45 V Vab = 7.5 k  2.5 k 45 V Iab = = 3 mA 15 k 44. 10 V  8 V 2 k  0.51 k  1.5 k = 498.75 A V0.51kΩ = (498.75 A)(0.51 k) = 0.25 V Vab = 10 V  0.25 V = 9.75 V Iba =  45. Vab = 0 V (short) Iab = 0 A (open) R2 any resistive value  R2 = short-circuit, open-circuit, any value 46. a. Is = b. c. CHAPTER 9 I 24 V = 1.5 mA, I = s = 0.5 mA 24 k 3 8 k  3 24 V = 0.83 mA 24 k  8 k  12 k 12 k( I s ) I= = 0.5 mA 12 k  8 k Is = yes 109 47. (a) 10 V 4 k  8 k  4 k  4 k 10 V = 2.67 k  2 k 10 V = = 2.14 mA 4.67 k IT = 8 ( I T ) = 1.43 mA, I2 = IT/2 = 1.07 mA 8  4 I = I1  I2 = 1.43 mA  1.07 mA = 0.36 mA I1 = (8 k  4 k)(10 V) 8 k  4 k  4 k  4 k = 5.72 V V I1 = 1 = 0.71 mA 8 k V2 = E  V1 = 10 V  5.72 V = 4.28 V V2 = 1.07 mA I2 = 4 k I = I2  I1 = 1.07 mA  0.71 mA = 0.36 mA (b) 48. a. b. 110 V1 = R1 ( I ) 3 (6 A) =2A  R1  R2  R3 3   2   4  V = I R2 R2 = (2 A)(2 ) = 4 V I R2  R2 ( I ) 2 (6 A) = 1.33 A  R1  R2  R3 3   2   4  V = I R1 R1 = (1.33 A)(3 ) = 4 V I R1  CHAPTER 9 Chapter 10 1. Q1 (9  109 )(4 C) = 36  103 N/C = 2 2 r (1 m) (a) E= k (b) E = k Q1 (9  109 )(4  C) = 36  109 N/C r2 (1 mm) 2 E (1 mm): E (2 m) = 36 × 109: 36 × 103 = 1 × 106  kQ kQ (9  109 )(2  C) r = =  = 15.81 m E 72 N/C r2 2. E = 3. C= 4. Q = CV = (0.15 F)(45 V) = 6.75 μC 5. a. b. 6. 7. 8. Q 1200 C  = 50 μF V 24 V  1m  1"   25.4 mm  39.37"  V 500 mV E = = = 19.69 V/m d 25.4 mm 25.4 mm  0.254 mm 100 V 500 mV E = = = 1.97 kV/m d 0.254 mm Q 160 C  = 23.53 V C 6.8  F V 23.53 V E = = = 4.71 kV/m d 5 mm V=  1m  0.1"   2.54 mm  39.37"  (0.1 m 2 ) A = 348.43 pF C = 8.85  1012εr = 8.85  1012(1) 2.54 mm d (0.1 m 2 ) A C = 8.85  1012 εr = 8.85  1012(2.5) = 871.06 pF 2.54 mm d 9. C = 8.85  1012εr 10. C = εrCo  εr = CHAPTER 10 A 8.85  1012 (4)(0.15 m 2 ) d= = 2.66 m d 2 F C 6.8 nF = 5 (mica) = Co 1360 pF 111 11. 12. C = 8.85  10-12(7) b. E = c. Q = CV = (24.78 nF)(200 V) = 4.96 C a. b. c. d. 13. 14. 15. (0.08 m 2 ) = 24.78 nF 0.2 mm a. d= V 200 V  = 106 V/m d 0.2 mm 1 (4.7 F) = 2.35 F 2 C = 2(4.7 F) = 9.4 F C = 20(4.7 F) = 94 F 1 (4)   3  (4.7 F) = 25.1 F C= 1    4 C= 8.85  1012  r A (8.85  1012 )(5)(0.02 m 2 ) = 130.15 µm  6800 pF C 106 m   39.37 in.  1000 mils  d = 130.15 µm     = 5.12 mils  1  m   1 m   1 in.   5000 V  5.12 mils   = 25.6 kV  mil  1200 V  mil  mica:  1200 V   = 0.24 mils 5000 V  5000 V  mil 1   1m   0.24 mils   = 6.10 m  1000 mils   39.37 in.  200 (22 F)/C = 4400 pF/C 1  106 4400 pF 4400 pF [80C] = 0.35 F [T]  C C 16. J = 5%, Size  40 pF  2 pF, 38 pF  42 pF 17. F = 1%, Size  47 × 101 = 470 F ± 4.7 F, 465.3 F  474.7 F 18. K = 10%, Size  18 × 102 pF = 1800 pF  180 pF, 1620 pF  1980 pF 19. a. τ b. C = E(1  et/τ) = 20 V(1  et/0.56 s) 112 = RC = (105 Ω)(5.6 μF) = 0.56 s CHAPTER 10 c. 1τ = 0.632(20 V) = 12.64 V, 3τ = 0.95(20 V) = 19 V 5τ = 0.993(20 V) = 19.87 V d. iC = 20 V t/τ e = 0.2 mAet/0.56 s 100 k R = Eet/τ = 20 Vet/0.56 s e. 20. 21. = RC = (106 )(5.6 F) = 5.6 s b. υC = E(1  et/τ) = 20 V(1  et/5.6s) d. iC = a. τ c. 1τ = 12.64 V, 3τ = 19 V, 5τ = 19.87 V e. Same as problem 21 with 5τ = 28 s and Im = 20 A a. τ b. C = E(1  et/τ) = 100 V(1  et/5.5 ms) c. 1τ = 63.21 V, 3τ = 95.02 V, 5τ = 99.33 V d. iC = 20 V t/τ e = 20 A et/5.6s 1M υR = Eet/τ = 20V et/5.6s = RC = (2.2 k + 3.3 k)1 µF = (5.5 k)(1 F) = 5.5 ms VR2  R2 E t/τ 100 V t/τ e = 18.18 mAet/5.5 ms e = RT 5.5 k 3.3 k(100 V) = 60 V  3.3 k  2.2 k = 60 Vet/5.5 ms e. CHAPTER 10 113 22. a. R = 68 kΩ + 22 kΩ = 90 kΩ τ = RC = (90 k)(18 µF) = 1.62 s b. C = E(1  et/τ) = (20 V + 40 V)(1  et/τ) C = 60 V(1  et/1.62s) c. iC = 60 V t/τ E t/τ e = e = 0.67 mAet/1.62s 90 k  R d. 23. a. 100 μs b. C = 12 V(1  e50µs/100µs) = 12 V(1  e0.5) = 12 V(1  0.607) c. C = 12 V(1  e1ms/100µs) = 12 V(1  e10) = 12 V(1  45.4  106) a. c.  = 20 ms, 5 = 5(20 ms) = 100 ms  20 ms  = RC, R =  = 2 k C 10  F C (20 ms) = 40 mV(1  e20ms/20ms) = 40 mV(1  e1) d. e. f. C = 40 mV(1  e10) = 40 mV(1  45  106)  40 mV Q = CV = (10 F)(40 mV) = 0.4 C  = RC = (1000  106 )(10 F) = 10  103 s = 12 V(.393) = 4.72 V 24. b.  12 V(999.95 × 103)  12 V = 40 mV(1  .368) = 40 mV(0.632) = 25.28 mV 1 min   1 h  5 = 50  103 s   = 13.89 h   60 s   60 min  25. a.  = RC = (4.7 k)(56 F) = 263.2 ms b. C = E(1 et/) = 22 V(1  et/263.2ms) iC = c. d. C(1 s) = 22 V(1  e1s/263.2ms) = 22 V(1  e3.8) = 22 V(1  22.37  103) = 21.51 V iC (1 s) = 4.68 mAe1s/263.2ms = 4.68 mA(22.37  103) = 0.105 mA C = 21.51 Vet/263.2ms iC = 114 22 V  t / 263.2ms E  t / e e  = 4.68 mAet/263.2ms R 4.7 k 21.51 V  t / 263.2ms = 4.58 mAet/263.2ms e 4.7 k CHAPTER 10 e. 26. a.  = RC = (3 k + 2 k)(2 F) = 10 ms C = 30 V(1  et/10ms) iC =  R1 30 V  t /10ms = 6 mAt/10ms e 5 k = iC R1 = (6 mA)(3 k)et/10ms = 18 Vet/10ms b. 100ms: e10 = 45.4  106 C = 30 V(1  45.4  106) = 30 V iC = 6 mA(45.4  106) = 0.27 A  R1 = 18 V(45.4  106) = 0.82 mV c. 200 ms:  = R2C = (2 k)(2 F) = 4 ms C = 30 Vet/4ms 30 V  t / 4ms e = 15 mAet/4ms 2 k At t = 0:  R2  iC R2  (6 mA)(2 k)e  t /10 ms iC =  = 12 Vet/10 ms At t = 200 ms:  R2  (15 mA)(2 k)e t / 4 ms = 30 Vet/4 ms d. CHAPTER 10 115 27. a.  = RC = (220 k)(22 pF) = 4.84 s C = 60 V 1  e  t/4.84  s  iC = b.  60 V  t /  = 272.73  Ae t/4.84  s e 220 k  = RC = (220 k + 470 kΩ)(22 pF) = (690 kΩ)(22 pF) = 15.18 s C = 60 V(1  e5) = 60 V(1  6.73 × 103) = 59.6 V C = 59.6 Vet/15.18μs iC = 272.73 µAe5 = 272.73 µA(6.73 × 103) = 1.84 μA 60 V iC(max) = = 86.96 µA 690 k iC = 86.96 μAet/15.18μS c. 116 CHAPTER 10 d. R = (470 kΩ)(86.96 µA)et/15.18ms = 40.87Vet/15.18μs 28. 29. a.  = RC = (2 m)(1000 F) = 2 s 5 = 10 s b. Im = c. yes a. C = Vf + (Vi  Vf)et/  = RC = (4.7 k)(4.7 F) = 22.1 ms, Vf = 40 V, Vi = 6 V C = 40 V + (6 V  40 V)et/22.1ms C = 40 V  34 Vet/22.1ms b. Initially VR = E + C = 40 V  6 V = 34 V V 34 V  t / 22.1ms e = 7.23 mA et/22.1ms iC = R e  t /   R 4.7 k V 12 V  = 6 kA R 2 m c. CHAPTER 10 117 30.  = RC = (2.2 k)(2000 F) = 4.4 s C = VCet/ = 40 Vet/4.4 s VC t /  40 V  t / 4.4 s e e  = 18.18 mAet/4.4 s R 2.2 k R = C = 40 Vet/4.4s IC = 31. C = Vf + (Vi Vf)et/  = RC = (820 )(3300 pF) = 2.71 s, Vf = 20 V, Vi = 10 V C = 20 V + (10 V  (20 V))et/2.71s C = 20 V + 10 Vet/2.71s (20 V  10 V) 10 V  = 12.2 mA 820  820  iC = iR = 12.2 mAet/2.71s Im = 32. 118 a. R = 10 kΩ + 8.2 kΩ = 18.2 kΩ τ = RC = (18.2 kΩ)(6.8 µF) = 123.76 ms C = Vf + (Vi  Vf)et/τ Vf = 20 V + 40 V = 60 V Vi = 8 V C = 60 V + (8 V  60 V)et/123.76 ms C = 60 V  68 Vet/123.76 ms 8 V + 20 V + 40 V = 3.74 mA Im = 18.2 k iC = 3.74 mAet/123.76 ms CHAPTER 10 b. 33. a. C = 140 mV(1  e1ms/2 ms) = 140 mV(1  e0.5) = 140 mV(1  0.6065) = 140 mV(0.3935) = 55.59 mV b. C = 140 mV(1  e10) = 140 mV(1  45.4  106) c. 100 mV = 140 mV(1  et/2 ms) 0.714 = 1  et/2 ms 0.286 = et/2 ms loge 0.286 = loge et/2 ms 1.252 = t/2 ms t = 1.252 (2 ms) = 2.5 ms d. C = 138 mV = 140 mV(1  et/2 ms)  139.99 mV 0.986 = 1  et/2 ms 14 × 103 = et/2 ms loge 14 × 103 = t/2 ms 4.268 = t/2 ms t = (4.268)(2 ms) = 8.54 s CHAPTER 10 119 34. τ = RC = (33 kΩ)(20 µF) = 0.66 s C = 12 V(1  et0.66 s) 8 V = 12 V(1  et0.66 s) 8 V = 12 V  12 Ve(1  et/0.66 s) 4 V = 12 Vet0.66 s 0.333 = et0.66 s loge 0.333 = t/0.66 s 1.0996 = t/0.66 s t = 1.0996(0.66 s) = 0.73 s    t =  loge 1  C  E    12 V  10 s =  loge 1    20 V  35. .4 916.29  103 10 s = 10.92 s 0.916  10.92 s  = RC  R =  = 54.60 k C 200  F = 36. a. τ = RC = (12 kΩ + 8.2 kΩ)(6.8 µF) = 137.36 ms C  60 V(1  e  t /  ) 48 V  60 V(1  e  t /  ) 0.8  1  e  t /  0.2  e  t /  log e 0.2  log e e  t /  b. c. 1.61  t /  t  (1.61)  (1.61)(137.36 ms) = 221.15 ms 60 V  t /  E iC  e  t /   e 20.2 k R  2.97 mAe  t /137.36 ms iC(221.15 ms) = 2.97 mAe221.15 ms/137.36 ms = 2.97 mAe1.61 = 2.97 mA (199.89 × 103) = 0.594 mA t = 2τ iC = 2.97 mAe2τ/τ = 2.97 mAe2 = 0.4 mA 0.135 P = EI = (60 V)(0.4 mA) = 24 mW 120 CHAPTER 10 37. a. m = R = Eet/ = 60 Ve1/ = 60 Ve1 = 60 V(0.3679) = 22.07 V b. c. 38. a. E t /  60 V  2 /  e  e = 6 Ae2 R 10 M = 6 A(0.1353) = 0.81 A iC = C = E(1  et/) t/2 s 50 V = 60 V(1  e ) 0.8333 = 1  et/2 s loge 0.1667 = t/2 s t = (2 s)(1.792) = 3.58 s  = RC = (10 M)(0.2 F) = 2 s Thevenin’s theorem: RTh: ETh: RTh = 8 k  24 k = 6 k ETh =  24 k(20 V) = 15 V 24 k  8 k  = RC = (10 k)(15 F) = 0.15 s C = E(1  et/) = 15 V(1  et/0.15 s) iC = CHAPTER 10 15 V  t / 0.15 E  t / e e  = 1.5 mAet/0.15 s R 10 k 121 b. 39. a. Source conversion and combining series resistors: E = (4 mA)(6.8 kΩ) = 27.2 V RT = 6.8 kΩ + 1.5 kΩ = 8.3 kΩ Vf = 27.2 V, Vi = 10 V  = RC = (8.3 k)(2.2 F) = 18.26 ms C = Vf + (Vi  Vf)et/ = 27.2 V + (10 V  (27.2 V))et/18.26 ms C = 27.2 V + 37.2 Vet/18.26 ms R(0+) = 27.2 V  (27.2 V))e-t/18.26 ms = 37.2 V 37.2 V  t /18.26ms e iC =  8.3 k iC = 4.48 mAet/18.26 ms 122 CHAPTER 10 b. 40. a. RTh = 3.9 k + 0   1.8 k = 3.9 k ETh = 36 V  = RC = (3.9 k)(20 F) = 78 ms C = Vf + (Vi  Vf)et/ = 36 V + (12 V  36 V)et/78 ms C = 36 V  24 Vet/78 ms R(0+) = 24 V  12 V = 24 V 24 V  t/78 ms e iC = 3.9 k iC = 6.15 mAet/78 ms b. CHAPTER 10 123 41. Source conversion: E = IR1 = (5 mA)(0.56 k) = 2.8 V R = R1 + R2 = 0.56 k + 3.9 k = 4.46 k RTh = 4.46 k  6.8 k = 2.69 k 4 V  2 .8 V 1.2 V  = 0.107 mA I= 6.8 k  4.46 k 11.26 k ETh = 4 V  (0.107 mA)(6.8 k) = 4 V  0.727 V = 3.27 V C = 3.27 V(1  et/)  = RC = (2.69 k)(20 F) = 53.80 ms C = 3.27 V(1  et/53.80 ms) 3.27 V  t /  e 2.69 k = 1.22 mA et/53.80 ms iC = 42. a.  = RC = (6.8 k)(39 F) = 265.2 ms C = Vf + (Vi  Vf)et/ = 20 V + (8 V  (20 V))et/265.2 ms C = 20 V + 12 Vet/265.2 ms R(0 +) = +8 V  20 V = 12 V 12 V  t / 265.2 ms e iC =  6.8 k iC = 1.76 mAet/265.2 ms 124 CHAPTER 10 b. 43. a.  = RThC = (1.67 M)(1 F) = 1.67 s RTh = 2 M  10 M = 1.67 M 10 M(24 V) ETh = = 20 V 10 M  2 M C = ETh(1  et/) = 20 V(1  e4/) = 20 V(1  e4) = 20 V(1  0.0183) = 19.63 V E t /  e R 20 V  t /1.67s 3 A = e 1.67 M 0.25 = et/1.67s loge 0.25 = t/1.67 s t = (1.67 s)(1.39) = 2.32 s iC = c. meter = C C = ETh(1  et/) 10 V = 20 V(1  et/1.67s) 0.5 = 1  et/1.67s 0.5 = et/1.67s loge 0.5 = t/1.67 s t = (1.67 s)(0.69) = 1.15 s CHAPTER 10 125 44. iC ao  C C t (40 V) = 80 mA 1 ms (0 V) 1  2 ms: iC = 2  106 = 0 mA 1 ms (20 V) 2  3 ms: iC = 2  106 = 40 mA 1 ms (10 V) = 6.67 mA 3  6 ms: iC = 2  106 3 ms (0 V) 6 9 ms: iC = 2  106 = 0 mA 3 ms (10 V) 9 12 ms: iC = 2  106 = 6.67 mA 3 ms 0  1 ms: iC = 2  106 45. iC ao  C C t (5 V ) = 1.18 A 20  s (15 V ) 20  30 s: iC = 4.7 F = 7.05 A 10  s (15 V ) = 7.05 A 30  40 s: iC = 4.7 F 10  s 0  20 s: iC = 4.7 F (0 V ) =0A 10  s (5 V ) 50  55 s: iC = 4.7 F = 4.7 A 5 s 40  50 s: iC = 4.7 F 55 s  60 s: iC = 4.7 F 126 (5 V ) = 4.7 A 5 s CHAPTER 10 60 s  70 s: iC = 4.7 F (0 V ) =0A 10  s 70 s  80 s: iC = 4.7 F (10 V ) = 4.7 A 10  s 80 s  100 s: iC = 4.7 F 46. (5 V ) = 1.175 A 20  s t C  C  (iC ) t C 0  4 ms: iC = 0 mA C = 0 V (2 ms) (40 mA) = 4 V 4  6 ms: iC = 40 mA C = 20 F (10 ms) 6  16 ms: iC = +40 mA C = (40 mA) = +20 V 20 F (4 ms) 16  20 ms: iC = 120 mA C = (120 mA) = 24 V 20  F 20  25 ms: iC = 0 mA C = 0 V iC = C CHAPTER 10 127 47. 48. 49. 50. 51. 6 F + 4 F = 10 µF, 8 µF + 12 µF = 20 µF 10 µF  20 F = 6.67 F CT = 6 F  12 F = 4 F CT = CT + 12 F = 4 F + 12 F = 16 F 6 F  CT (6 F)(16 F) CT = 6 F  CT = = 4.36 F  6 F  CT 6 F  16 F V1 = 10 V, Q1 = V1C1 = (10 V)(6 F) = 60 C CT = 6 F  12 F = 4 F, QT = CTE = (4 F)(10 V) = 40 C Q2 = Q3 = 40 C Q 40  C = 6.67 V V2 = 2  C2 6 F Q 40  C V3 = 3  = 3.33 V C3 12  F 360 µF + 200 µF = 560 µF 470 µF  560 µF = 255.53 µF QT = Q3 = CTE = (255.53µF)(56 V) = 14.5 mC Q 14.5 mC = 30.4 V V3 = 3  C3 470  F V1 = V2 = E  V3 = 56 V  30.4 V = 25.6 V Q1 = V1C1 = (25.6 V)(360 µF) = 9.2 mC Q2 = V2C2 = (25.6 V)(200 µF) = 5.1 mC steady  state  ignore 10 kΩ resistor 330 µF + 120 µF = 450 µF CT = 220 µF  450 µF = 147.76 µF QT = Q1 = CTE = (147.76 µF)(20 V) = 2.96 mC Q 2.96 mC = 13.45 V V1 = 1  C1 220  F V3 = V2 = E  V1 = 20 V  13.45 V = 6.55 V Q2 = C2V2 = (330 µF)(6.55 V) = 2.16 mC Q3 = C3V3 = (120 µF)(6.55 V) = 0.786 mC 4 k(48 V) = 32 V = V0.08F 4 k  2 k Q0.08F = (0.08 F)(32 V) = 2.56 C V0.04F = 48 V Q0.04F = (0.04 F)(48 V) = 1.92 C 52. V4k = 53. WC = 54. W= 128 1 1 CV 2  (120 pF)(12 V)2 = 8,640 pJ 2 2 Q2 Q= 2C 2CW  2(6 F)(1200 J) = 0.12 C CHAPTER 10 55. a. 56. a. (220 k  3.3 k)(12 V) = 9.85 V 2.2 k  3.3 k  1.2 k (3.3 k)(12 V) = 5.91 V V100F = 2.2 k  3.3 k  1.2 k 1 W200F = (200  F)(9.85 V) 2 = 970 mJ 2 1 W100F = (100  F)(5.91 V)2 = 1.75 mJ 2 1 1 WC = CV 2  (1000 F)(100 V)2 = 5 pJ 2 2 V200F = b. Q = CV = (1000 F)(100 V) = 0.1 C c. I = Q/t = 0.1 C/(1/2000) = 200 A d. P = VavIav = W/t = 5 J(1/2000 s) = 10,000 W e. t = Q/I = 0.1 C/10 mA = 10 s CHAPTER 10 129 Chapter 11 1. a. b. c. d. 2. 3.  4  104 Wb  = 4  102 Wb/m2 = 0.04 Wb/m2 2 A 0.01 m 0.04 T F = NI = (40 t)(2.2 A) = 88 At 104 gauss  3 0.04 T   = 0.4  10 gauss 1 T   B=  2.54 cm   1 m  = 5.08 mm 0.2   1  100 cm   2.54 cm   1 m  = 25.4 mm 1   1  100 cm   d 2  (5.08 mm) 2  = 20.27  106 m2 A= 4 4 N 2  A (200 t) 2 (4  10 7 )(20.27  10 6 m 2 ) L=  = 40.1 H  25.4 mm L= b. increase = change in µ r Lnew = µ rLo 4. L = N2 5. L= a. b. c. d. 6. 130 N 2  r o A (200 t ) 2 (500)(4  10 7 )(20.27  10 6 m 2 )  = 20.06 mH  25.4 mm a. r o   (200 t) 2 (1000)(4  107 )(1.5  104 m 2 ) = 50.27 mH 0.15 m N 2  r o A  L = (3)2Lo = 9Lo = 9(4.7 mH) = 42.3 mH 1 1 L = Lo = (4.7 mH) = 1.57 mH 3 3 (2)(2) 2 L = Lo = 16 (4.7 mH) = 75.2 mH 1 2 2 1 1   (1500) Lo 2 2 L =   = 375(4.7 mH) = 1.76 mH 1 2 a. 39  102 H  10%  3900 H  10%  3.9 mH ± 10% b. 68 × 100 H  5% = 68 F ± 5% CHAPTER 11 c. 47 μH ± 10% d. 15 × 102 µH ± 10% = 1500 µH ± 10% = 15 mH ± 10% 7. e= N d = (50 t)(120 mWb/s) = 6.0 V dt 8. e= N d d e 20 V    = 100 mWb/s dt dt N 200 t 9.    1  d 1   = 14 turns  N = e e= N   42 mV   dt  3 m Wb/s   d   dt  10. a. 11. b. diL = (22 mH)(1 A/s) = 22 mV dt di e = L L = (22 mH)(1 mA/ms) = 22 mV dt 2 mA diL e= L = (22 mH)( = 4.4 V 10  s dt a. = b. iL = c. L = Eet/ = 20 Vet/15 s R = iRR = iLR = E(1  et/) = 20 V(1  et/15 s) d. iL: 1 = 0.632 mA, , 3 = 0.951 mA, 5 = 0.993 mA L: 1 = 7.36 V, 3 = 0.98 V, 5 = 140 mV e= L L 300 mH  = 15 s R 20 k E 20 mV (1  et/) (1  e  t/ )  R 20 k = 1 mA(1  et/15 μs) e. CHAPTER 11 131 12. a. = L 4.7 mH  = 2.14 s R 2.2 k b. iL = E 12 V (1  et/) = 5.45 mA(1  et/2.14 s) (1  e t/ )  R 2.2 k c. L = Eet/ = 12 Vet/2.14 s R = iRR = iLR = E(1  et/) = 12 V(1  et/2.14 s) d. iL: 1 = 3.45 mA, , 3 = 5.18 mA, 5 = 5.41 mA L: 1 = 4.42 V, 3 = 0.60 V, 5 = 0.08 V e. 18 V (1  e  t /  ) RT 18V = 1.2 kΩ RT = 15 mA R (15  s) 1.2 k(15  s) L 5  15   = 15 µs: L   R 5 5  3.6 mH 13. iL = 14. a. iL = If + (Ii  If)et/ E 36 V L 120 mH   = 9.23 mA,  = = 30.77 s Ii = 8 mA, If = R 3.9 k R 3.9 k iL = 9.23 mA + (8 mA  9.23 mA)et/30.77 s iL = 9.23 mA  1.23 mAet/30.77 s +E   L   R = 0 and  L  E   R  R = iRR = iLR = (8 mA)(3.9 k) = 31.2 V  L = E   R = 36 V  31.2 V = 4.8 V L = 4.8 Vet/30.77 s b. 132 CHAPTER 11 15. a. Ii = 8 mA, If = 9.23 mA,  = L 120 mH  = 30.77 s R 3.9 k iL = If + (Ii  If)et/ = 9.23 mA + (8 mA  9.23 mA)et/30.77 s iL = 9.23 mA  17.23 mA et/30.77 s +E   L   R = 0 (at t = 0) but,  R = iRR = iLR = (8 mA)(3.9 k) = 31.2 V  L = E   R = 36 V  (31.2 V) = 67.2 V L = 67.2 V et/30.77 s b. 16. c. Final levels are the same. Transition period defined by 5 is also the same. a. Source conversion: L 2H  = 588.2 s R 3.4 k iL = If + (Ii  If)et/ 6V = 1.76 mA If = 3.4 k iL = 1.76 mA + (4 mA  1.76 mA)et/588.2s iL = 1.76 mA + 2.24 mA et/588.2s = R(0 +) = 4 mA(3.4 k) = 13.6 V KVL: +6 V  13.6 V  L(0+) = 0 L(0+) = 7.6 V t/588.2s L = 7.6 Ve b. CHAPTER 11 133 17. a. 20.8 V = 2 mA 10.4 k L 200 mH τ= = = 19.23 s R 10.4 k iL = If + (Ii  If )et/τ = 2 mA + (6 mA  2 mA)et/19.23 µs iL = 2 mA + 4 mAet/19.23 μs If = KVL: 20.8 V  62.4 V  υL(0+) = 0 υL(0+) = 41.60 V υL = 41.6 Ve-t/19.23 μs b. 18. a. = L = 8 Vet/0.278s, iL = b. 134 10 mH L = 0.278 s = 36 k R E (1  e t/ ) = 0.222 mA(1  et/0.278s) R 5  steady state L 10 mH  =  = 0.208 s R 12 k  36 k iL = Imet/ = 0.222 mAet/0.208s L = (0.222 mA)(48 k)et/ = 10 .66Vet/0.308s CHAPTER 11 19. a. b. L 1 mH  = 0.5 s R 2 k E 12 V iL = (1  e  t /  )  (1  e  t /  ) = 6 mA(1  et/0.5s) R 2 k L = Eet/ = 12 V et/0.5s = iL = 6 mA(1  et/0.5s) = 6 mA(1  e1s/0.5s) = 6 mA(1  e2) = 5.19 mA L 1 mH = 83.3 ns  =  iL = Imet/ R 12 k iL = 5.19 mAet/83.3ns t = 1 s: L = 12 Vet/0.5s = 12 Ve2 = 12 V(0.1353) = 1.62 V VL = (5.19 mA)(12 k) = 62.28 V L = 62.28 Vet/83.3ns c. CHAPTER 11 135 20. a. RTh = 6.8 k ETh = 6 V = L 5 mH  = 0.74 s R 6.8 k E 6V (1  e t/ )  (1  e t/ ) = 0.88 mA(1  et/0.74s) R 6.8 k L = Eet/ = 6 Vet/0.74s iL = b. Assume steady state and IL = 0.88 mA  = L 5 mH  = 0.33 s R 15 k iL = Imet/ = 0.88 mA et/0.33s L = Vmet/ Vm = ImR = (0.88 mA)(15 k) = 13.23 V L = 13.23 Vet/0.33s c. 136 CHAPTER 11 d. 21. 22. a. VR2 max = ImR2 = (0.88 mA)(8.2 k) = 7.22 V RTh = 2 k + 2.2 k + 6.2 k  3 kΩ = 6.22 k 6.2 k(12 V) = 8.09 V ETh = 6.2 k  3 k 8.09 V L 47 mH If = = 7.56 s = 1.3 mA ,  =  6.22 k R 6.22 k iL = 1.3 mA(1  et/7.56s) L = 8.09 Vet/7.56s b. 0.632(1.3 mA) = 0.822 mA 0.368(8.09 V) = 2.98 V a. Source conversion: E = IR = (4 mA)(12 k) = 48 V, ENet = 48 V  20 V = 28 V = iL = L 2 mH  = 55.56 ns R 36 k E 28 V (1  e  t/ )  (1  e  t/ ) = 0.778 mA(1  et/55.56ns) R 36 k L = Eet/ = 28 Vet/55.56ns b. t = 100 ns: iL = 0.778 mA(1  e100ns/55.56ns) = 0.778 mA(1  e1.8) = 0.65 mA 0.165 L = 28 Ve1.8 = 4.62 V CHAPTER 11 137 23. RTh = 2.2 k  4.7 k = 1.50 k 4.7 k(10 V) = 6.81 V ETh =  4.7 k  2.2 k L 10 mH =  = 6.67 s R 1.50 k a. E 6.81 V (1  e  t/ )   (1  e  t/ ) = 4.54 mA(1  et/6.67s) R 1.5 k L = Eet/ = EThet/τ = 6.81 Vet /6.67s iL =  b. t = 10 s: iL = 4.54 mA(1  e10s/6.67s) = 4.54 mA(1  e1.5) = 3.53 mA L = 6.81 V(0.223) = 1.52 V c. 0.223 L 10 mH  = 2.13 s R 4.7 k iL = 3.53 mAet/2.13s At t = 10 s VL = (3.53 mA)(4.7 k) = 16.59 V L = 16.59 Vet/2.13s  = d. 24. a. L = Eet/ = 0 .6 H 0 .6 H L = 5 ms   R1  R3 100   20  120  L = 36 Vet/5 ms L = 36 Ve25 ms/5 ms = 36 Ve5 = 36 V(0.00674) = 0.24 V b. 138 L = 36 Ve1 ms/5 ms = 36 Ve0.2 = 36 V(0.819) = 29.47 V CHAPTER 11 c.   E (1  e t/ )  R1  R1  R3   R1  iR1 R1  iL R1    36 V  (1  e t/5ms )  100  =  120   t/5 ms ))100  = (300 mA(1  e 5 ms/5 ms ) = 30 V(1  e1) = 30 V(1  e = 30 V(1  0.368) = 18.96 V d. e. 25. a. iL = 300 mA(1  et/5 ms) 100 mA = 300 mA(1  et/5 ms) 0.333 = 1  et/5 ms 0.667 = et/5 ms loge 0.667 = t/5 ms 0.405 = t/5 ms t = 0.405(5 ms) = 2.03 ms None-In parallel with supply. 16 V = 2 mA 4.7 k  3.3 k t = 0 s: Thevenin: RTh = 3.3 k + 1 k  4.7 k = 3.3 k + 0.82 k = 4.12 k 1 k(16 V) = 2.81 V ETh = 1 k  4.7 k iL = If + (Ii  If)et/ Ii = 2.81 V L 2H = 0.49 ms = 0.68 mA,  =  4.12 k R 4.12 k iL = 0.68 mA + (2 mA  0.68 mA)et/0.49 ms iL = 0.68 mA + 1.32 mAet/0.49 ms R(0+) = 2 mA(4.12 k) = 8.24 V KVL(0+): 2.81 V  8.24 V  L = 0 L = 5.43 V L = 5.43 Vet/0.49 ms If = b. CHAPTER 11 139 26. a. 8V = 5.33 mA, VL = 0 V 1.5 k RTh = (3 k  12 k)  4 k  1.5 k 2.4 k(20 V) = 7.5 V ETh = 2.4 k  4 k Steady-state: I L  RTh   1.5 k  1.5 k  0.75 k ETh   8 V  7.5 V  15.5 V  L 3 mH   4 s R 0.75 k 15.5 V = 20.67 mA Ii = 5.33 mA 0.75 k iL = If + (Ii  If)et/τ = 20.67 mA + (5.33 mA  20.67 mA)et/4 µs iL = 20.67 mA  15.34 mAet/4μs υL = 15.5 Vet/4μs If = b. iL (2 )  20.67 mA  15.34 mA  e2 0.135  18.6 mA  L (2 )  15.5Ve2  15.5V(0.135) = 2.09 V c. Ii = 18.6 mA υL + υR  8 V = 0 υL = 8 V  υR = 8 V  (18.6 mA)(1.5 kΩ) = 19.9 V L 3 mH   = 2 µs R 1.5 k 8V = 5.33 mA 1.5 k iL = If + (Ii  If)et/τ = 5.33 mA + (18.6 mA  5.33 mA)et/2µs = 5.33 mA + 13.27 mAet/2µs υL = 19.9 Vet/2µs Ii = 18.6 mA 140 If = CHAPTER 11 27. a. RTh = 2 M  10 M = 1.67 M 10 M(24 V) = 20 V ETh = 10 M  2 M ETh 20 V = 12 μA = RTh 1.67 M  L 5H   5 s   Rmeter 10 M iL = 12 Aet/5 s 10 A = 12 Aet/5 s 0.833 = et/5 s loge 0.833 = t/5 s 0.183 = t/5 s t = 0.183(5 s) = 0.92 s I L (0 ) = 28. b. L (0+) = iL(0+)Rm = (12 A)(10 M) = 120 V L = 120 Vet/5s = 120 Ve10s/5s = 120 Ve2 = 120 V(0.135) = 16.2 V c. L = 120 Ve5/ = 120 Ve5 = 120 V(6.74  103) = 0.81 V a. Closed Switch: RTh = 1.2 k  2.2 k = 0.776 kΩ 1.2 k(24 V) = 8.47 V ETh = 1.2 k  2.2 k CHAPTER 11 141 Open Switch: RTh   6.9 k  1.2 k  1.02 k 1.2 k(24 V)   ETh = 3.56 V 8.1 k 3.56 V + υR  υL = 0 υL = 3.56 V + (1.09 mA)(1.02 kΩ) = 7.57 V υL = 7.57Vet/1.18 ms L 1.24  = 1.18 ms R 1.02 k 3.56 V Iss = = 3.49 mA = If 1.02 k iL  I f  ( I i  I f )e t /    3.49 mA+ (  10.91 mA  (3.49 mA)  e t /1.18 ms iL = 3.49 mA  7.42 mAet/1.18 ms b. 29. 142 a. iL = 100 mA(1  e1ms/20ms) = 100 mA(1  e1/20) = 100 mA(1  e0.05) = 100 mA(1  951.23 103) = 100 mA(48.77  103) = 4.88 mA b. iL = 100 mA(1  e100ms/20ms) = 100 mA(1  e5) = 99.33 mA c. 50 mA = 100 mA(1  et/) 0.5 = 1  et/ 0.5 = et/ 0.5 = et/ loge 0.5 = t/ t = ()(loge 0.5) = (20 ms)(loge 0.5) = (20 ms)(693.15  103) = 13.86 ms CHAPTER 11 30. d. 99 mA = 100 mA(1  et/20 ms) 0.99 = 1  et/20ms 0.01 = et/20ms 0.01 = et/20ms loge 0.01 = t/20 ms t = (20 ms)(loge 0.01) = (20 ms)(4.605) = 92.1ms a. IL (1τ) = 0.632Imax = 126.4 µA 126.4 = 200 μA Imax = 0.632 iL  I m (1  e  t /  ) b. 64.4  s   160  A  200  A 1  e     0.8  1  e 0.2  e 64.4  s  64.4  s  log e 0.2  1.61   31. 64.4  s  64.4  s  40  s 1.61 L L  40 s = 500  R L  20 mH c.  d. Im  a. L  open circuit equivalent 10 M(16 V) = 13.33 V VL = 10 M  2 M E  E  (200  A)(500 )  100 mV R b. RTh = 2 M  10 M = 1.67 M 10 M(16 V) = 13.33 V ETh = 10 M  2 M I Lfinal  CHAPTER 11 ETh 13.33 V = 7.98 A  RTh 1.67 M 143 iL = 7.98 A(1  et/3 s) c. 10 A = 7.98 A(1  et/3 s) 1.253 = 1  et/3 s 0.253 = et/3 s loge(0.253) = t/3s 1.374 = t/3s t = 1.374(3 s) = 4.12 s d. = L 5H  = 3 s R 1.67 M L = 13.33 V et/3 s = 13.33 V e12 s/3 s = 13.33 V e4 = 13.33 V(0.0183) = 0.244 V 32. eL = L i : t 15 mA  0  2 ms, eL =  (200 mH)  = 1.5 V  2 ms   45 mA  = 0.75 V 2  14 ms, eL = (200 mH)   12 ms   15 ms  = 3 V 14  15 ms, eL = (200 mH)   1 ms  15  19 ms, eL = 0 V  15 mA  19  22 ms, eL = (200 mH)  = 1 V  3 ms  22  24 ms, eL = 0 V 33. L = L i L t  15 mA  = 37.5 mV 0  2 ms: L = (5 mH)   2 ms   30 mA  = 37.5 mV 2  6 ms: L = (5 mH)   4 ms   15 mA  6  9 ms: L = (5 mH)  = 25 mV  3 ms  9  13 ms: L = 0 V  5 mA  13  14 ms: L = (5 mH)  = 25 mV  1 ms  14  17 ms: L = 0 V 144 CHAPTER 11  5 mA  = 12.5 mV 17  19 ms: L = (5 mH)   2 ms  34. L = 10 mH, 4 mA at t = 0 s i t L = L  i   L t L 0  5 s: L = 0 V, iL = 0 mA and iL = 4 mA 5 s (10 V) = 5 mA 5  10 s: iL = 10 mH 2 s 10  12 s: iL = (+60 V) = +12 mA 10 mH 12  16 s: L = 0 V, iL = 0 mA and iL = 11 mA 8 s 10 V = 8 mA 16  24 s: iL = 10 mH 35. a. L3  L4 = 3.5 mH  5.6 mH = 1.953 mH L2 + L3  L4 = 3.3 mH + 1.953 mH = 5.253 mH L = L1  (L2 + L3  L4) = 2.4 mH  5.253 mH = 1.647 mH LT = L4 + L = 9.1 mH + 1.647 mH LT = 10.75 mH 36. L2  L3 = 10 mH  30 mH = 7.5 mH L = L1 + L2  L3 = 47 mH + 7.5 mH = 54.5 mH L  L4 = 54.5 mH  22 mH = 15.67 mH CHAPTER 11 145 37. 33 mH + 1.8 mH = 5.1 mH 4.7 mH  5.1 mH = 2.45 mH 38. LT = 6.2 mH + 12 mH  36 mH + 24 mH = 39.2 mH CT = 9.1  F + 10 F  91 F = 9.1 F + 9.01 F = 18.11 F 39.2 mH in series with 18.11 F 39. 7 µF  42 µF = 6 µF 12 µF + 6 µF = 18 μF 5 mH + 20 mH = 25 mH Series combination of 2.2 kΩ resistor, 25 mH coil, 18 µF capacitor 40. a. RT = 2 k  8.2 k = 1.61 k, LT = 3 mH  2 mH = 1.2 mH L  1.2 mH = 745.3 µs = T  RT 1.61 k E iL = (1  e  t /  ) RT 36 V (1  e  t / 745.3 s ) = 22.36 mA(1  et/745.3µs) 1.61 k L = Eet/ = 36 Ve t/745.3µ s = b. 146 CHAPTER 11 41. a. Source conversion: E = 16 V, Rs = 2 k RTh = 2 k + 2 k  8.2 k = 2 k + 1.61 k = 3.61 k 8.2 k(16 V) = 12.86 V ETh = 8.2 k  2 k E 12.86 V L 30 mH = 8.31 s Im = Th   3.56 mA,  =  RTh 3.61 k R 3.61 k iL = 3.56 mA(1  et/8.31s)  L1   L2 = 12.86 V initially (t = 0+) L  10 mH 1 of total = (12.86 V) = 4.29 V 10 mH +20 mH 3  L = 4.29 Vet/8.31μs b. 42. a. RTh = 10 k  20 k = 6.67 k 20 k(20 V) ETh = = 13.33 V 20 k  10 k LT = 3 H + 4.7 H  10 H = 3 H + 3.197 H = 6.197 H L 6.197 H = 0.93 ms τ= T = R 6.67 k  υL = 13.33Vet/0.93 ms 13.33 V (1  et/τ) = 2 mA(1  et/0.93 ms) iL = 6.67 k  CHAPTER 11 147 b. c. 3.197 H  0.52  L 3 H + 3.197 H = (0.52)(13.33et/0.93 ms) = 6.93 Vet/0.93 ms  L3 =  L3 43. 44. 45. E 20 V =5A  4 R1 20 V 20 V E I2 = I R2  =2A   R2  R3 6   4  10  I1 = I R1 + I2 = 5 A + 2 A = 7 A I R1  I1 = I2 = 0 A V1 = V2 = E = 60 V 12 V = 3 A, I2 = 0 A 4 V1 = 12 V, V2 = 0 V I1 = 46. 6  V) = 10.34 V 6   20   3  (3   6 )(50 V) V1 = = 15.52 V 29  50 V I1 = = 1.72 A 20   3   6  I2 = 0 A V2 = 148 CHAPTER 11 Chapter 12 Φ: 1. CGS: 5  104 Maxwells, English: 5  104 lines B: CGS: 8 Gauss, English: 51.62 lines/in.2 2. Φ: 3. a. B=  4  104 Wb = 0.04 T = A 0.01 m 2 4. a. R= 0.06 m 300 l = = 4 2  A  2  10 m m b. R= l 0.0762 m 152.4 = = 4 2  A  5  10 m m c. R= 0.1 m 1000 l = = 4 2  A 1  10 m m SI 6  104 Wb, English 60,000 lines B: SI 0.465 T, CGS 4.65  103 Gauss, English 30,000 lines/in.2 from the above R (c) > R (a) > R (b) 5. R= F 400 At = = 952.4  103 At/Wb  4.2  104 Wb 6. R= F 120 gilberts = 1.67  103 rels (CGS) =  72,000 maxwells 7.  1m  = 0.1524 m 6 in.   39.37 in .  F 400 At = 2624.67 At/m H= = l 0.1524 m 8. µ= 9. B= 2 B 2(1200  104 T) = 4  104 Wb/Am = H 600 At/m  10  104 Wb = 0.33 T = A 3  103 m 2 Fig. 12.7: H  800 At/m NI = Hl  I = Hl/N = (800 At/m)(0.2 m)/75 t = 2.13 A CHAPTER 12 149 10. 11. 12.  3  104 Wb = = 0.6 T A 5  104 m 2 Fig. 12.7, Hiron = 2500 At/m Fig. 12.8, Hsteel = 70 At/m NI = Hl(iron) + Hl(steel) (100 t)I = (Hiron + Hsteel)l (100 t)I = (2500 At/m + 70 At/m)0.3 m 771 A = 7.71 A I= 100 B= a. N1I1 + N2I2 = Hl  12  104 Wb B= = =1T A 12  104 m 2 Fig. 12.7: H  750 At/m N1(2 A) + 30 At = (750 At/m)(0.2 m) N1 = 60 t b. μ= a. B 1T = 13.34  104 Wb/Am = H 750 At/m  1 Wb  80,000 lines  8 = 8  104  108 Wb = 8  104 Wb  10 lines   1m  = 0.14 m l(cast steel) = 5.5 in.   39.37 in .   1m  l(sheet steel) = 0.5 in.  = 0.013 m  39.37 in .   1m   1m  = 6.45  104 m2 Area = 1 in.2     39.37 in . 39.37 in.     8  104 Wb = 1.24 T = A 6.45  104 m 2 Fig 12.8: Hsheet steel  460 At/m, Fig. 12.7: Hcast steel  1275 At/m NI = Hl(sheet steel) + Hl(cast iron) = (460 At/m)(0.013 m) + (1275 At/m)(0.14 m) = 5.98 At + 178.50 At NI = 184.48 At B= b. 150 B 1.24 T = 9.73  104 Wb/Am = H 1275 At/m B 1.24 T Sheet steel: μ = = 26.96  104 Wb/Am = H 460 At/m Cast steel: μ = CHAPTER 12 13. N1I + N2 = + Hl Hl   cast steel cast iron (20 t)I + (30 t)I = " (50 t)I = " B=   1m   1m  with 0.25 in.2  = 1.6  104 m2    A  39.37 in .   39.37 in.  0.8  104 Wb = 0.5 T 1.6  104 m 2 Fig. 12.8: Hcast steel  280 At/m Fig. 12.7: Hcast iron  1500 At/m  1m  lcast steel = 5.5 in.  = 0.14 m  39.37 in .  B=  1m  = 0.064 m lcast iron = 2.5 in.   39.37 in .  (50 t)I = (280 At/m)(0.14 m) + (1500 At/m)(0.064 m) 50I = 39.20 + 96.00 = 135.20 I = 2.70 A 14. 15. a. lab = lef = 0.05 m, laf = 0.02 m, lbc = lde = 0.0085 m NI = 2Hablab + 2Hbclbc + Hfalfa + Hglg  2.4  104 Wb = 1.2 T  H  360 At/m (Fig. 12.8) B= = A 2  104 m 2 100I = 2(360 At/m)(0.05 m) + 2(360 At/m)(0.0085 m) + (360 At/m)(0.02 m) + 7.97  105(1.2 T)(0.003 m) = 36 At + 6.12 At + 7.2 At + 2869 At 100I = 2918.32 At I  29.18 A b. air gap: metal = 2869 At:49.72 At = 58.17:1 B 1.2 T = 3.33  103 Wb/Am = μsheet steel = H 360 At/m μair = 4π  107 Wb/Am μsheet steel: μair = 3.33  103 Wb/Am:4  107  2627:1  1m  4 cm  = 0.04 m 100 cm  (8  104 Wb  0.5  104 Wb) 36(7.5  104 ) 1 d 1 = (80 t)(0.9 A) = f = NI 1 0.02 2 dx 2 (0.04 m) 2 = 1.35 N CHAPTER 12 151 16. C = 2πr = (6.28)(0.3 m) = 1.88 m  2  104 Wb B= = = 1.54 T A 1.3  104 m 2 Fig. 12.7: Hsheet steel  2100 At/m Hg = 7.97  105Bg = (7.97  105)(1.54 T) = 1.23  106 At/m N1I1 + N2I2 = Hglg + Hl(sheet steel) (200 t)I1 + (40 t)(0.3 A) = (1.23  106 At/m)(2 mm) + (2100 At/m)(1.88 m) I1 = 31.98 A 17. a.  1m  = 2  103 m 0.2 cm  100 cm   d 2 (3.14)(0.01 m) 2 = 0.79  104 m2 A= = 4 4 NI = Hglg, Hg = 7.96  105 Bg   0.2  104 Wb   2  103 m (200 t)I = (7.96  105 )  4 2    0.79  10 m   I = 2.02 A b. 18.  2  104 Wb = 0.25 T = A 0.79  104 m 2 2 1 Bg A 1 (0.25 T ) 2(0.79  104 m 2 ) F = 2 o 2 4  107  2N Bg = Table: Section Φ(Wb) ab, gh 4 bc, fg 2  10 cd, ef 2  104 152 2  10 4 H l(m) 5  10 0.2 5  10 4 0.1 5  104 5  10 Hl 0.099 4 0.2 4 0.2 4 0.002 2  10 bg B(T) 4 5  10 ah de A(m2) CHAPTER 12  2  104 Wb = 0.4 T = A 5  104 m 2 Air gap: Hg = 7.97  105(0.4 T) = 3.19  105 At/m Hglg = (3.19  105 At/m)(2 mm) = 638 At Fig 12.8: Hbc = Hcd = Hef = Hfg = 55 At/m Hbclbc = Hfglfg = (55 At/m)(0.1 m) = 5.5 At Hcdlcd = Heflef = (55 At/m)(0.099 m) = 5.45 At Bbc = Bcd = Bg = Bef = Bfg = For loop 2:  F = 0 Hbclbc + Hcdlcd + Hglg + Heflef + Hfglfg  Hgblgb = 0 5.5 At + 5.45 At + 638 At + 5.45 At + 5.50 At  Hgblgb = 0 Hgblgb = 659.90 At 659.90 At = 3300 At/m and Hgb = 0.2 m Fig 12.7: Bgb  1.55 T with Φ2 = BgbA = (1.55 T)(2  104 m2) = 3.1  104 Wb ΦT = Φ1 + Φ2 = 2  104 Wb + 3.1  104 Wb = 5.1  104 Wb = Φab = Φha = Φgh  5.1  104 Wb = 1.02 T Bab = Bha = Bgh = T = A 5  104 m 2 BH curve: (Fig 12.8): Hab = Hha = Hgh  180 At/m Hablab = (180 At/m)(0.2 m) = 36 At Hhalha = (180 At/m)(0.2 m) = 36 At Hghlgh = (180 At/m)(0.2 m) = 36 At which completes the table! Loop #1:  F = 0 NI = Hablab + Hbglbg + Hghlgh + Hahlah (200 t)I = 36 At + 659.49 At + 36 At + 36 At (200 t)I = 767.49 At I  3.84 A 19. NI = Hl l = 2πr = (6.28)(0.08 m) = 0.50 m (100 t)(2 A) = H(0.50 m) H = 400 At/m Fig. 12.8: B  0.68 T Φ = BA = (0.68 T)(0.009 m2) Φ = 6.12 mWb CHAPTER 12 153 20. NI = Hab(lab + lbc + lde + lef + lfa) + Hglg 300 At = Hab(0.8 m) + 7.97  105 Bg(0.8 mm) 300 At = Hab(0.8 m) + 637.6 Bg Assuming 637.6 Bg  Hab(0.8 m) then 300 At = 637.6 Bg and Bg = 0.47 T Φ = BA = (0.47 T)(2  104 m2) = 0.94  104 Wb Bab = Bg = 0.47 T  H  270 At/m (Fig. 12.8) 300 At = (270 At/m)(0.8 m) + 637.6(0.47 T) 300 At  515.67 At  Poor approximation! 300 At  100%  58% 515.67 At Reduce Φ to 58% 0.58(0.94  104 Wb) = 0.55  104 Wb  0.55  104 Wb = 0.28 T  H  190 At/m (Fig. 12.8) B= = A 2  104 m 2 300 At = (190 At/m)(0.8 m) + 637.6(0.28 T) 300 At  330.53 At Reduce Φ another 10% = 0.55  104 Wb  0.1(0.55  104 Wb) = 0.495  104 Wb  0.495  104 Wb B= = = 0.25 T  H  175 At/m (Fig. 12.7) A 2  104 m 2 300 At = (175 At/m)(0.8) + 637.6(0.28 T) 300 At  318.53 At but within 5%  OK Φ  0.55  104 Wb 21. a. 1τ = 0.632 Tmax Tmax  1.5 T for cast steel 0.632(1.5 T) = 0.945 T At 0.945 T, H  700 At/m (Fig. 12.7)  B = 1.5 T(1  eH/700 At/m) b. H = 900 At/m: B = 1.5 T 1  e  900 At/m 700 At/m Graph:  1.1 T H = 1800 At/m:    = 1.09 T 1800 At/m  B = 1.5 T 1  e 700 At/m = 1.39 T Graph:  1.38 T H = 2700 At/m: B = 1.5 1  e  2700 At/m 700 At/m  = 1.47 T Graph:  1.47 T Excellent comparison! 154 CHAPTER 12 c. B = 1.5 T(1  eH/700 At/m) = 1.5 T  1.5 TeH/700 At/m B  1.5 T = 1.5 TeH/700 At/m 1.5  B = 1.5 TeH/700 At/m 1.5 T  B = eH/700 At/m 1.5 T B  H  loge 1  =  1.5 T  700 At/m B   and H = 700 loge 1    1.5 T  d. B = 1 T: 1T   H = 700 loge 1   = 769.03 At/m  1.5 T  Graph:  750 At/m B = 1.4 T:  1.4 T  H = 700 loge 1   = 1895.64 At/m  1.5 T  Graph:  1920 At/m e. B   H = 700 loge 1    1.5 T   0.2 T  = 700 loge 1    1.5 T  = 100.2 At/m Hl (100.2 At/m)(0.16 m) = 40.1 mA I= = N 400 t vs 44 mA for Ex. 12.1 CHAPTER 12 155 Chapter 13 1. a. b. c. d. e. 10 V 15 ms: 10 V, 20 ms: 0 V 20 V 20 ms 2 cycles 2. a. b. c. d. e. 200 μA 1 s: 200 μA, 7 s: 200 µA 400 μA 4 s 2.5 cycles 3. a. b. c. d. e. 40 mV 1.5 ms: 40 mV, 5:1 ms: 40 mV 80 mV 2 ms 3.5 cycles 4. a. T= b. c. d. 5. a. b. c. d. 1 f 1 T= f 1 T= f 1 T= f 1 = 5 ms 200 Hz 1  = 25 ns 40 MHz 1  = 50 s 20 kHz 1 =1s  1 Hz  1 1  = 1 Hz T 1s 1 1 f=  = 16 Hz 1 T s 16 1 1 f=  = 25 Hz T 40 ms 1 1 = 40 kHz f=  T 25 s f= 6. T= 1 = 1 ms, 5(1 ms) = 5 ms 1 kHz 7. T= 24 ms = 0.3 ms 80 cycles 156 CHAPTER 13 42 cycles = 7 Hz 6s 8. f= 9. a. Vpeak = (2.5 div.)(50 mV/div) = 125 mV b. T = (3.2 div.)(10 s/div.) = 32 s c. f= a.    Radians =  40 = 0.22 π rad  180  10. b. c. 11.    Radians =  170 = 0.94 rad  180  a.  180   = 60 Degrees =     3 b.  180  Degrees =  1.2 = 216    d. a. b. c. d. 13.     Radians =  60 = rad 3  180     Radians =  135 = 0.75 rad  180  d. c. 12. 1 1 = 31.25 kHz  T 32  s a. b. c. d.  180  1 Degrees =    = 18    10  180  Degrees =   0.6  = 108    2 2  = 3.49 rad/s T 1.8 s 2 = = 20.94  103 rad/s 3 0.3  10 s 2 = 785.4  103 rad/s = 8  106 s 2 = 1.57 × 106 rad/s = 6 4  10 s =  = 2 f = 2 (100 Hz) = 628.32 rad/s  = 2 f = 2 (0.25 kHz) = 1.57 × 103 rad/s  = 2 f = 2 (2 kHz) = 12.56  103 rad/s  = 2 f = 2 (0.004 MHz) = 25.13  103 rad/s CHAPTER 13 157 14. 15. a. 2  f= T 2 2 1 T=   f  754 rad/s  = 120 Hz, T = 8.33 ms f= 2 2  = 2 f = b. f=  12 rad/s  = 1.91 Hz, T = 523.6 ms 2 2 c. f=  6000 rad/s  = 954.93 Hz, T = 1.05 ms 2 2 d. f=  0.16 rad/s  = 25.46  103 Hz, T = 39.28 ms 2 2     (60)  radians   180  3 t=   / 3 rad  / 3 rad 1 1     = 2.78 ms 2 f 2 (60 Hz) (6)(60) 360  16.   /6     = 104.7 rad/s (30)    ,  = t   =  t 5  10 3 s  180  6 17. a. Amplitude = 20, f = b. Amplitude = 12, f = 120 Hz  10,000 rad/s  = 1591.55 Hz Amplitude = 106, f = 2 2  10,058 rad/s  = 1.6 kHz Amplitude = 8, f = 2 2 c. d.  377 rad/s  = 60 Hz 2 2 18.  19.  20. T= 21. i = 0.5 sin 72 = 0.5(0.9511) = 0.48 A 22. 158 2   2 1 = 40 ms, cycle = 20 ms 157 2  180  1.2   = 216     = 20 sin 216 = 20(0.588) = 11.76 V CHAPTER 13 23. 24. 6  103 = 30  103 sin  0.2 = sin   = sin1 0.2 = 11.54 and 180  11.54 = 168.46  = Vm sin  30 1 ms  360 T  360  T = 1 ms   = 12 ms  30  1 1 = 83.33 Hz f=  T 12  10 3 s  = 2 f = (2)(83.33 Hz) = 523.58 rad/s 40 = Vm sin 30 = Vm (0.5) 40 = 80 V Vm = 0 .5 and  = 80 sin 523.58t 25.  26.  27. a.  = 6 × 103 sin (2π 2000t + 30) b. i = 20  103 sin(2π 60t  60) 28. a.  = 120  106 sin(2π 1000t  80) 29.  = 12  103 sin(2π 2000t + 135°) 30.  = 8  103 sin(2π 500t +π/6) 31.  leads i by 90 32. i leads  by 40 33.  = 2 sin (t  30 + 90) i = 5 sin(t + 60) +60 in phase 34.  = 4 sin(t + 90 + 90 + 180 = 4 sint i = sin(t + 10 + 180) = sin(t + 190) 35. T= 1 1 = 1 ms  f 1000 Hz t1 = 120  T  2  1 ms  1     = ms 180  2  3  2  3 CHAPTER 13 i leads  by 190 159 36.   2 f  T 2   2 T 2  125.66  s 50,000 rad/s 40 40 (T )  (125.66  s) 360 360  13.96  s t1  37. T = 1 ms tpeak @ 30° 30 1 tpeak = (T )  ms 360 12 38. a. T = ( 8 div.)(1 ms/div.) = 8 ms (both waveforms) b. f= c. Peak = (2.5 div)(0.5 V/div.) = 1.25 V Vrms = 0.707(1.25 V) = 0.884 V d. Phase shift = 4.6 div., T = 8 div. 4.6 div. =  360 = 207 i leads e 8 div. or e leads i by 153 39. 40. 1 1 = 125 Hz (both)  T 8 ms 0  (6 V)(5 ms)  (3 V)(10 ms)  (3 V)(10 ms) 30 ms 30 V +30 V  30 V   1V 30 G 1 1   2 (4 ms)(20 mA)   (2 ms)(8 mA)  2 (2 ms)(8 mA) G 8 ms 40 mA  16 mA  8 mA 16 mA   8 8 = 2 mA (35 V)(5 ms)  41. G 1 2 (20 V)(20 ms)  (20 V)(15 ms)  1 (20 V)(7.5 ms)  2 75 ms 1 2 (20 V)(7.5 ms)  1 2 (20 V)(15 ms) + 0 175 V  200 V 300 V  75 V  75 V 150 V 75 400 V  575 V   2.33 V 75  160 CHAPTER 13 42. 43. 1 1 0  (30 mA)(3 ms)  (20mA)(2 ms) 2 2 G 7 ms 45 mA  20 mA   3.57 mA 7 a. b. c. 44. 45. 1 1 (4 V)(5 ms)  (8 V)(5 ms)  (8 V)(5 ms)  (4 V)(5 ms)  (8 V)(5 ms)  (8 V)(5 ms) 2 2 G 25 ms 20 V  20 V  40 V  20 V  20 V  40 V  25 = 0V The same 1 1 ( r 2 )  ( 202 )  628.32 2 2 628.32 628.32   15.71  15.71 mA Area = d 40 (15.71mA)( )  (5 mA)( ) G 2  5.36 mA Area = c. T = ( 2 div.)(0.2 ms/div) = 0.4 ms 1 1 = 2.5 kHz f=  T 0.4 ms Average = (2.5 div.)(10 mV/div.) = 25 mV a. T = (4 div.)(10 s/div.) = 40 s b. f= c. G= a. b. 46. 0V 1 1 = 25 kHz  T 40 s ( 2.5 div.)(1.5 div.)  (1 div.)(0.5 div.)  (1 div.)(0.6 div.)  (2.5 div.)(0.4 div.)(1 div.)(1 div.) 4 div. 3.75 div.  0.5 div.  0.6 div.  1 div.  1 div. 4 6.85 div. = = 1.713 div. 4 1.713 div.(10 mV/div.) = 17.13 mV = 47. a. b. c. Vrms = 0.7071(120 V) = 84.85 V Irms = 0.7071(6 mA) = 4.24 mA Vrms = 0.7071(8 V) = 5.66 V CHAPTER 13 161 48. a. b. c.  = 6.79 sin 377t i = 70.7  103 sin 377t  = 2.83  103 sin 377t 49. Vrms = 50. Vrms = = 51. G= a. (3 V) 2 (2 s)  (2 V)2 (2 s)  0  (1 V) 2 (2 s)  (  3 V) 2 (2 s)  (  2 V)2 (2 s) 12 s 54 2 V  4.5 V 2 = 2.12 V 12 (8 V)(4 ms)  (8 V)(4 ms) 0  =0V 8 ms 8 ms Vrms = 52. 1  ( 2 V)2 (4 s)  (2 V)2 (1 s)  (3 V)2  s  2  = 1.43 V 12 s (8 V) 2 (4 ms)  (  8 V) 2 (4 ms) =8V 8 ms T = (4 div.)(10 s/div.) = 40 s 1 1 = 25 kHz f=  T 40 s Av. = (1 div.)(20 mV/div.) = 20 mV Peak = (2 div.)(20 mV/div.) = 40 mV rms = b. 162 2 (40 mV)2 Vmax  (20 mV)2  = 34.64 mV 2 2 T = (2 div.)(50 s) = 100 s 1 1 f=  = 10 kHz T 100 s Av. = (1.5 div.)(0.2 V/div.) = 0.3 V Peak = (1.5 div.)(0.2 V/div.) = 0.3 mV rms = 53. V02  V02  2 (.3 V)2 Vmax  (.3 V)2  = 367.42 mV 2 2 a. CHAPTER 13 1 1 (2)(16)  (2)(16)  (2)(48)  96 2 2 Area = 96 + (4)(64) + (2)(4) = 96 + 256 + 8 = 360 b. c. d. 54. A1 = rms = 360  30 = 5.48 12 1 (4)(8)  4(8)  2(2) 16  32  4  = 3.67 G= 2 12 12 e. rms  1.5 (average value) a. Vdc = IR = (4 mA)(2 k) = 8 V Meter indication = 2.22(8 V) = 17.76 V b. Vrms = 0.707(16 V) = 11.31 V CHAPTER 13 163 Chapter 14 1.  2.  3. a. (377)(10) cos 377t = 3770 cos 377t b. (200)(0.6) cos(754t + 20) = 120 cos(754t + 20) c. ( 2 20)(157) cos(157t  20) = 4440.63 cos(157t  20) d. (200)(1) cos(t + 180) = 200 cos(t + 180) = 200 cos t a. Im = Vm/R = 150 V/3 Ω = 50 A, i = 50 sin 200t b. Im = Vm/R = 30 V/3 Ω = 10 A, i = 10 sin(377t + 20) c. Im = Vm/R = 6 V/3 Ω = 2 A, i = 2 sin(ωt + 100) d. Im = Vm/R = 12 V/3 Ω = 4 A, i = 4 sin(ωt + 220) a. Vm = ImR = (0.1 A)(7  103 Ω) = 700 V υ = 700 sin 1000t b. Vm = ImR = (2  103 A)((7  103 Ω) = 14.8 V υ = 14.8 sin(400t  120) a. 0 b. XL = 12.56f = 12.56(60 Hz) = 753.6  c. XL = 12.56f = 12.56(4 kHz) = 50.24 k d. XL = 12.56f = 12.56(1.2 MHz) = 15.07 M a. L= XL 2 k = = 22 mH 2 f 2 (14.47 kHz) b. L= XL 40 k = 1.2 H = 2 f 2 (5.3 kHz) a. XL = 2πfL  f = 4. 5. 6. 7. 8. f= 164 XL XL XL = = 2 L (6.28)(1 mH) 6.28  10 3 H 10  = 1.59 kHz 6.28  103 H CHAPTER 14 9. 10. b. f= 4 k XL = 636.94 kHz  3 6.28  10 H 6.28  10 3 H c. f= 12 k XL = 1.91 MHz  3 6.28  10 H 6.28  10 3 H a. Vm = ImXL = (5 A)(20 ) = 100 V υ = 100 sin(ωt + 90) b. Vm = ImXL = (40  103 A)(20 ) = 0.8 V υ = 0.8 sin(ωt + 150) c. i = 6 sin(ωt + 150), Vm = ImXL = (6 A)(20 ) = 120 V υ = 120 sin(ωt + 240) = 120 sin(ωt  120) a. b. 11. 12. XL = ωL = (400 rad/s)(0.1 H) = 40  Vm = ImXL = (5  106 A)(40 ) = 200 µV υ = 200  106 sin(400t + 110) a. Im = Vm 120 V = 2.4 A, i = 2.4 sin(ωt  90) = X L 50  b. Im = Vm 30 V = 0.6 A, i = 0.6 sin(ωt  70) = X L 50  a. b. 13. XL = ωL = (100 rad/s)(0.1 H) = 10  Vm = ImXL = (10 A)(10 ) = 100 V υ = 100 sin(100t + 90) XL = ωL = (60 rad/s)(0.2 H) = 12  Im = Vm/XL = 1.5 V/12  = 0.125 A i = 0.125 sin(60t  90) XL = ωL = (10 rad/s)(0.2 H) = 2  Im = Vm/XL = 16 mV/2  = 8 mA i = 8  103 sin(t + 2  90) = 8  103 sin(t  88) a. XC = 1 1 = = 2 fC 2 (0 Hz)(5  10 6 F) b. XC = 1 1 = 530.79  = 2 fC 2 (60 Hz)(5  106 F) c. XC = 1 1 = 15.92  = 2 fC 2 (2 kHz)(5  106 F) CHAPTER 14 165 d. XC  14. 15. 16. 17. XC = 1 1 C  2 fC 2 fX C 1 2 (265 Hz)(60 )  10  F a. C b. C a. f= 1 1 = 4.08 kHz = 2 CX C 2 (3.9  10 6 F)(10 ) b. f= 1 1 = 34 Hz = 2 CX C 2 (3.9  10 6 F)(1.2 k) c. f= 1 1 = 408.1 kHz = 2 CX C 2π(3.9  10 6 F)(0.1 ) d. f= 1 1 = 20.40 Hz = 6 2 CX C 2 (3.9  10 F)(2000 ) 1 2 (34 kHz)(1.2 k)  3900 pF Im = Vm/XC = 120 V/2.5  = 48 A i = 48 sin(ωt + 90) a. b. Im = Vm/XC = 4 × 103 V/2.5  = 0.16 A i = 1.6 × 103 sin(ωt + 130) a. υ = 30 sin 200t, XC = Im = b. 1 1 = 5 k = ωC (200)(1  106 F) Vm 30 V = 6 mA, i = 6  103 sin(200t + 90) = X C 5 k υ = 60  103 sin 377t, XC = Im = 166 1 1 = 62.83  = 6 2 fC 2 (2  10 Hz)(5  106 F) 1 1 = 2.65 k = ωC (377)(1  106 ) 3 V m 60  10 V = = 22.64 A, i = 22.64  106 sin(377t + 90)  2,650 XC CHAPTER 14 18. Vm = ImXC = (50  103 A)(10 ) = 0.5 V υ = 0.5 sin(ωt  90) a. 19. b. Vm = ImXC = (2  106)(10 ) = 20 V υ = 20  106 sin(ωt  30) a. i = 0.2 sin 300t, XC = 1 1 = 5.952 k  ωC (300)(0.56  106 F) Vm = ImXC = (0.2 A)(5.952 kΩ) = 1190.48 V, υ = 1190.48 sin(300t  90) b. i = 8  103 sin (377t 30°), XC = 1 1 = 4.737 k  ωC (377)(0.56  106 F) Vm = ImXC = (8  103 A)(4.737 k) = 37.81 V υ = 37.81 sin(377t  120) 20. 21. a. υ leads i by 90  L, XL = Vm/Im = 550 V/11 A = 50  X 50  = 132.63 mH L= L= ω 377 rad/s b. υ leads i by 90  L, XL = Vm/Im = 36 V/4 A = 9  1 1 L= = 147.36 μH = ωX L (754 rad/s)(9 ) c. υ and i are in phase  R V 10.5 V =7 R= m= I m 1.5 A a. b. c. 22. i = 5 sin(ωt + 90)   i leads  by 90  C  = 2000 sin ωt  Vm 2000 V = 400  XC = = 5A Im i = 2 sin(157t + 60)    leads i by 90  L  = 80 sin(157t + 150)  V 80 V 40  = 40 Ω, L = X L = = 254.78 mH XL = m = ω 157 rad/s Im 2 A  = 35 sin(ωt  20)   in phase  R i = 7 sin(ωt  20)  Vm 35 V =5 R= = Im 7 A  CHAPTER 14 167 23. 24.  1 1 1 1 =R  f = = = 3 6 2 fC 2 RC 2 (2  10 )(1  10 F) 12.56  103  79.62 Hz XC = 25. XL = 2πfL = R R 10,000  = 318.47 mH L= = 2 f 2 (5  103 Hz) 26. XC = XL 1  2 fL 2 fC 1 f2 = 4 2 LC 1 1 = 1.59 kHz = and f =  3 2 LC 2 (10  10 H)(1  106 F) 27. XC = XL 1 1 1 = 2 fL  C = 2 2 = = 5.07 nF 2 fC 4 f L 4(9.86)(2500  106 )(2  103 ) 28. a. P= (60 V)(15 A) cos 30 = 389.7 W, Fp = 0.866 2 b. P= (50 V)(2 A) cos 0 = 50 W, Fp = 1.0 2 c. P= (50 V)(3 A) cos 10 = 73.86 W, Fp = 0.985 2 d. P= (75 V)(0.08 A) cos 40 = 2.30 W, Fp = 0.766 2 2 29. 168 8A V m 48 V = 6 , P = I2R =  =  6  = 192 W Im 8A  2 V I (48 V)(8 A) P = m m cos  = cos 0 = 192 W 2 2  48 V  8 A  P = VI cos θ =    cos 0 = 192 W  2  2  All the same! R= CHAPTER 14 30. P = 100 W: Fp = cos θ = P/VI = 100 W/(150 V)(2 A) = 0.333 P = 0 W: Fp = cos θ = 0 300 =1 P = 300 W: Fp = 300 31. P = V m I m cos  2 (50 V) I m (0.5)  Im = 40 A 500 W = 2 i = 40 sin(ωt  50) 32. a. Im = Em/R = 34 V/6.8  = 3.53 A, i = 3.53 sin(2π60t + 20) b. P = I 2R = c. T = a. Im = b. L= c. L0W a. Em = ImXC = (30  103 A)(2.4 k) = 72 V e = 72 sin(2π500t  20  90) = 72 sin(2π500t  110) b. C= c. P=0W 33. 34. 35. a. 2 3.53 A 2 2 6.8 Ω = 42.38 W 6.28 = 16.67 ms ω 2 60 rad/s 6(16.67 ms) = 100.02 ms  0.1 s = Vm 128 V = 4.27 A, i = 4.27 sin(1000t  30) = X L 30  XL ω = 30  = 30 mH, standard = 30 mH 1000 rad/s 1 1 = 0.133 μF standatd = 0.13 μF = 3 ωX C (3.14  10 rad/s)(2.4 k) 1 1 1 = 50  = = 4 2 f C1 ωC1 (10 rad/s)(2  F) 1 1 = 10  X C2 = = 4 ωC 2 (10 )(10  F ) 84.85 V 60 E = 1.697 A 150 = E = 84.85 V 60 I1 = 50    90 Z C1 X C1 = I2 = E ZC 2 = 84.85 V 60 = 8.485 A 150 10    90 4 i1 = 2.4 sin(10 t + 150) i2 = 12 sin(104t + 150) CHAPTER 14 169 b. CT  2  F  10  F  12  F 1 1  XC  4  C (10 rad/s)(12  F)  8.33  E 84.8560 Is =  XCT 8.33   90 = 10.19 A150 is = 14.4 sin (104t + 150°) 36. a. L1  L2 = 60 mH  120 mH = 40 mH 3 X LT = 2πfLT = 2π(10 Hz)(40 mH) = 251.33  Vm = I m X LT =  24 A  (251.33 ) = 6.03 kV and υs = 6.03 kV sin(103t + 30 + 90) or υs = 6.03  103 sin(103t + 120) b. I m1 = Vm , X L1 = 2fL1 = 2(103 Hz)(60 mH) = 376.99  X L1 I m1 = 6.03  103 V = 16 A 376.99  and i1 = 16 sin(103t + 30) 3 X L2 = 2πfL2 = 2π(10 Hz)(120 mH) = 753.98 Ω I m2 = 6.03  103 V =8A 753.98  and i2 = 8 A sin(103t + 30) 37. a. c. e. 5.0 36.87 12.65 7.57 4123.11 104.04 b. d. f. 2.83 45 1001.25 2.86 0.894 116.57 38. a. c. e. 17.89 116.57 20.22 × 103 8.53  200 0° b. d. f. 8.94 26.57 8.49 × 103 135 1000 178.85 39. a. c. e. 4.6 + j3.86 j2000 47.97 + j1.68 b. d. f. 6.0 + j10.39 6 × 103  j2.2 × 103 4.7 × 104 j1.71 × 104 40. a. c. e. 42 + j0.11 3 × 103  j5.20 × 103 15 b. d. f. 1 × 103  j1.73 × 103 6.13 × 103 + j5.14 × 103 2.09 × 103  j1.20 170 CHAPTER 14 41. 42. 43. 44. 45. 46. 47. 48. a. 11.8 + j7.0 b. 151.90 + j49.90 c. 4.72 × 106 + j71 a. 5.20 + j1.60 b. 209.30 + j311.0 c. 21.20 + j12.0 a. 12.17 54.70° b. 98.37 13.38° c. 28.07 115.91° a. 12.0 + j34.0 b. 86.80 + j312.40 c. 283.90  j637.65 a. 8.00 20° b. 49.68 64.0° c. 40 × 10340° a. 6.0 50° b. 200 × 106 60° c. 109 170° a. 4 b. 4.15  j4.23 c. 6.69  j6.46 a. 10  j 5 = 10.0  j5.0 1  j0 b. 8 60 8 60 = 19.38  103 15.69  102  j 400 412.80 75.69 c. (6 20)(120   40)(8.54 69.44) 6.15  103 49.44 = 3.07  103 79.44  2   30 2   30 CHAPTER 14 171 49 a. (0.16 120)(300 40) 48 160  = 5.06 88.44 9.487 71.565 9.487 71.565 b.  1 1   8     2  4  4  10 20   j ( j )   36  j 30   8  1        46.861 39.81 j      2500   20   (2500 20)(8j)(0.0213 39.81) = 426 109.81 50. 51. a. x + j4 + 3x + jy  j7 = 16 (x + 3x) + j(4 + y  7) = 16 + j0 x + 3x = 16 4+y7=0 4x = 16 y = +7  4 x=4 y=3 b. (10 20)(x 60) = 30.64  j25.72 10x 40 = 40 40 10 x = 40 x=4 5x + j10 2  jy ────── 10x + j20  j5xy  j210y = 90  j70 (10x + 10y) + j(20  5xy) = 90  j70 10x + 10y = 90 x+y=9 x=9y a. 20  5xy = 70 20  5(9  y)y = 70 5y(9  y) = 90 y2  9y + 18 = 0 (9)  (9) 2  4(1)(18) 2 93 y= = 6, 3 2 y= For y = 6, x = 3 y = 3, x = 6 (x = 3, y = 6) or (x = 6, y = 3) b. 52. 172 80 0 = 4 θ = 3.464  j2 = 4 30 40  θ = 30 a. 160.0 30 b. 25  103 40 c. 70.71 90 CHAPTER 14 53. 54. 55. 56. a. 14.14 180 b. 4.24  106 90 c. 2.55 × 10670 a. 56.57 sin(377t + 20) b. 169.68 sin (377t + 10) c. 11.31  103 sin(377t  110) d. 6000 sin(377t  180) (Using peak values) ein = υa + υb  υa = ein  υb = 60 V 45  20 V 45 = 63.25 V  63.43 and ein = 63.25 sin (377t + 63.43) is = i1 + i2  i1 = is  i2 (Using peak values) = (20  106 A 60)  (6  106 A 30) = 20.88  106 A  76.70 i1 = 20.88  106 sin (t + 76.70) 57. ein = υa + υb + υc υa = ein  υb  υc = 120 V 30°  30 V 60°  40 V 120° = 108.92 V 0.33° ein = 108.92 sin(377t  0.33°) 58. Is = I1 + I2 + I3 I1 = Is  I2  I3 = 12.73 A 180°  5.66 A 180°  2[5.66 A 180°] = 12.73 A 180°  5.66 A 180°  11.32 A 180° = 4.25 A 0° i1 = 6.01 sin 377t CHAPTER 14 173 Chapter 15 1. 2. a. R  0 = 6.8   0 = 6.8  b. XL = L = (377 rads/s)(1.2 H) = 452.4  XL  90 = 452.4   90 = +j452.4  c. XL = 2fL = (6.28)(50 Hz)(47 mH) = 1.48  XL  90 = 1.48   90 = +j1.48  d. XC = e. XC = f. R  0 = 220   0 = 220  a. V = 10.61 V  10, I = 1  1 C (100 rad/s)(10  106 F) XC  90 = 1 k  90 = j1 k 1 1 = 33.86   3 2 fC 2 (10  10 Hz)(0.47  F) XC  90 = 33.86   90 = j33.86  V  10.61 V 10 = 3.54 A  10  R0 3  0 i = 5 sin (t + 10) b. = 1 k V = 4.24 V  10, I = V  4.24 V 10 = 4.24 A  80  X L 90 1  90 i = 6 sin (t  80) c. 3. 1 1 = 15.924   2 fC 2 (5 kHz)(2  F) 84.84 V 0 V  = 5.328 A 90 I=  X C   90 15.924    90 i = 7.534 sin (t + 90) V = 84.84 V  0, XC = a. I = (0.707)(4 mA  0) = 2.828 mA  0 V = (I  0)(R  0) = 2.828 mA  0)(22   0) = 62.216 mV  0  = 88  103 sin 1000t b. I = (0.707)(1.5 A  60) = 1.061 A  60 XL = 2πfL = 2π(200 Hz)(12 mH) = 15.08 Ω V = (I  )(XL  90) = (1.061 A  60)(15.08   90) = 16 V  150  = 22.62 sin(2π200t + 150) 174 CHAPTER 15 c. I = (0.707)(2 mA  40) = 1.414 mA  40 1 1  = 135.52 k XC =  C (157rad/s)(0.047  F) V = (I  )(XC  90) = (1.414 mA  40)(135.52 k  90) = 191.63 V  50 Vp = 2(191.63 V) = 270.96 V and  = 270.96 sin (157t  50) 4. 5. 6. 7. a. ZT = 6.8  + j8.2  = 10.65   50.33 b. ZT = 2   j6  + 10  = 12   j6  = 13.42   26.57 c. ZT = 1 k + j3.2 k + 5.6 k + j6.8 k = 6.6 k + j10 k = 11.98 k  56.58 a. ZT = 3  + j4   j5  = 3   j1  = 3.16   18.43 b. ZT = 1 k + j8 k  j4 k = 1 k + j4 k = 4.12 k  75.96 c. LT = 247 mH XL = L = 2fL = 2(103 Hz)(247  103 H) = 1.55 k 1 1 XC = = 1.59 k  3 2 fC 2 (10 Hz)(0.1  106 F) = 470  + j1.55 k  j1.59 k = 470   j40  = 471.70   4.86 E 120 V 0 = 2   70 = 0.684   j1.879  = R  jXC  I 60 A 70 a. ZT = b. ZT = c. ZT = a. ZT = 8  + j6  = 10   36.87 c. I = E/ZT = 100 V  0/10   36.87 = 10 A  36.87 VR = (I  )(R  0) = (10 A  36.87)(8   0) = 80 V  36.87 VL = (I  )(XL  90) = (10 A  36.87)(6   90) = 60 V  53.13 f. P = I2R = (10 A)2 8  = 800 W g. Fp = cos θT = R/ZT = 8 /10  = 0.8 lagging h. R = 113.12 sin(t  36.87) L = 84.84 sin(t + 53.13) i = 14.14 sin (t  36.87) CHAPTER 15 E 80 V 320 = 4 k  280 = 4 k  80 = 0.695 k  j3.939   I 20 mA 40 = R  jXC 8 kV 0 E = 40 k  60 = 20 k + j34.64 k = R + jXL  I 0.2 A   60 175 8. a. ZT = 18   j29.15  = 34.26  58.30 1 1  = 29.15 Ω XC = 2 fC 2 (60 Hz)(91  F) c. I= 120 V  20 E = 3.50 A 78.30 = ZT 34.26    58.30 VR = (I θ)(R 0) = (3.50 A 78.30)(18  0) = 63.0 V 78.30 VC = (I θ)(XC 90) = (3.50 A 78.30)(29.15  90) = 102.03 V 11.70 9. f. P = I2R = (3.50 A)2 18  = 220.5 W g. Fp = R/ZT = 18 /34.26 Ω = 0.525 leading h. i = 4.95 sin(377t + 78.30) υR = 89.1 sin(377t + 78.30) υC = 144.27 sin(377t  11.70) a. ZT = 4  + j6   j10  = 4   j4  = 5.66  45 c. XL = ωL  L = d. 176 6 = 16 mH  377 rad/s 1 1 1 = 265 μF  C= XC = = C  X C (377 rad/s)(10 ) XL = E 50 V 0 = 8.83 A 45 = ZT 5.66    45 VR = (I θ)(R 0) = (8.83 A 45)(4  0) = 35.32 V 45 VL = (I θ)(XL 90) = (8.83 A 45)(6  90) = 52.98 V 135 VC = (I θ)(XC 90) = (8.83 A 45)(10  90) = 88.30 V 45 I= f. E = VR + VL + VC 50 V 0 = 35.32 V 45 + 52.98 V 135 + 88.30 V 45 50 V 0 = 49.95 V 0  50 V  0 g. P = I2R = (8.83 A)2 4  = 311.88 W h. Fp = cos θT = i. i = 12.49 sin(377t + 45) e = 70.7 sin 377t υR = 49.94 sin(377t + 45) υL = 74.91 sin(377t + 135) υC = 124.86 sin(377t  45) R ZT = 4 Ω/5.66 Ω = 0.707 leading CHAPTER 15 10. 11. a. XL = ωL = (20 × 103 rad/s)(0.1H) = 2 kΩ 1 1 = 6.1 kΩ XC =  3  C (20  10 rad/s)(8200 pF) ZT = 1.2 kΩ + j2 kΩ  j6.1 kΩ = 1.2 kΩ  j4.1 kΩ = 4.27 kΩ 73.69° b.  c.  d. I= e.  f. E = VR + VL + VC 4.24 V 60° = 1.19 V 133.69° + 1.99 V 223.69° + 6.06 V 43.69° = (0.822 V + j0.80 V) + (1.44 V  j1.37 V) + (4.38 V + j 4.19 V) = 2.12 V+ j3.62 V 4.24 V 60°  4.20 V 59.65° E 4.24 V60 = 0.993 mA 133.69°  ZT 4.27 k  73.69 VR = IR = (0.993 mA 133.69°)(1.2 kΩ 0°) = 1.19 V 133.69° VL = IXL = (0.993 mA 133.69°)(2 kΩ 90°) = 1.99 V 223.69° VC = IXC = (0.993 mA 133.69°)(6.1 kΩ 90°) = 6.06 V 43.69° g. P = I2R = (0.993 mA)2(1.2 kΩ) = 1.18 mW h. Fp = i. i = 1.4 × 103 sin (20,000t + 133.69°) υR = 1.68 sin (20,000t + 133.69°) υL = 2.81 sin (20,000t + 223.69°) υC = 8.57 sin (20,000t + 43.69°) R 1.2 k = 0.281 leading  ZT 4.27 k 20 V (rms)  28.28 V (peak) 43.20 V(p  p)  21.60 V (peak) 22 (28.28 V) 22   R 475.20 + 21.60R = 622.16 146.96  = 6.8 Ω R= 21.60 Vscope = 21.60 V = CHAPTER 15 177 12. a.  22.8 V  = 8.06 V VL (rms) = 0.7071   2  V (rms) 8.06 V XL = L = 6.2 kΩ  I (rms) 1.3 mA XL = L = (1000rad/s)L = 6.2 kΩ  L = b. 6.2 k = 6.2 H 1000 rad/s E 2  VR2  VL2 (22 V)2 = VR2  (8.06 V) 2 484  VR2  64.96 VR2  419.04 VR  419.04  20.47 V R c. 13. a. VR (rms) 20.47 V   15.75 k 1.3 mA I (rms) 6.2 H  8.27 V  = 2.924 V VR(rms) = 0.7071   2  V (rms) 2.924 V = 292.4 μA I(rms) = R  10 k R2 E 2  VR2  VC2 b. (12 V) 2  (2.924 V) 2  VC2 144  8.55  VC2 VC2  135.35 VC  135.45  11.64 V VC (rms) 11.64 V  = 39.81 kΩ I (rms) 292.4  A 1 1 1 = 100 pF XC = C   2 fC 2 fX C 2 (40 kHz)(39.81 k) XC = 14. a. b. 178 (2 k  0)(120 V 60) 240 V 60 = 29.09 V 15.96 = 2 k  + j8 k  8.25 75.96 (8 k  90)(120 V 60) = 116.36 V 74.04 V2 = 8.25 k  75.96 V1 = ( 40  90)(60 V 5) 2400 V 95 = 48.69 V 40.75 = 6.8  + j 40  + 22  28.8 + j 40 (22  0)(60 V 5) 1.32 kV 5 = 26.78 V 49.25 V2 = = 49.29  54.25 49.29  54.25 V1 = CHAPTER 15 15. a. b. 16. 17. a. (20  90)(20 V 70) = 14.14 V 155 20  + j 20   j 40  (40    90)(20 V 70) = 28.29 V 25 V2 = 28.28    45 V1 = ZT = 4.7 k + j30 k + 3.3 k  j10 k = 8 k + j20 k = 21.541 k 68.199 ZT = 3.3 k + j30 k  j10 k = 3.3 k + j20 k = 20.27 k 80.631 Z E (20.27 k 80.631)(120 V 0) = 112.92 V 12.432 V1 = T = ZT 21.541 k 68.199 Z E ZT = 3.3 k  j10 k = 10.53 k 71.737 V2 = T ZT (10.53 k   71.737)(120 V 0) = 58.66 V 139.94 = 21.541 k 68.199 XL = ωL = (1000 rad/s)(20 mH) = 20  1 1 = = 25.64  XC = ωC (1000 rad/s)(39  F) ZT = 30  + j20   j25.64  = 30   j5.64  = 30.53  10.65 E 20 V  40 I= = 655.1 mA 50.65 = ZT 30.53    10.65 VR = (I  θ)(R  0) = (655.1 mA 50.65)(30  0) = 19.65 V 50.65 VC = (655.1 mA 50.65)(25.64  90) = 16.80 V 39.35 R 30  = 0.983 leading = ZT 30.53  b. cos θT = c. P = I2R = (655.1 mA)2 30  = 12.87 W f. VR = g. ZT = 30   j5.64  = R  jXC (30  0)(20 V  40) 600 V 40 = 19.66 V 50.65 = ZT 30.53    10.65 (25.64    90)(20 V 40) = 16.80 V 39.35 VC = 30.53    10.65 P = VI cos θ  8000 W = (200 V)(I)(0.8) 8000 A = 50 A I= 160 0.8 = cos θ θ = 36.87 V = 200 V 0, I = 50 A 36.87 V 200 V 0 = 4  36.87 = 3.2  + j2.4  = ZT = I 50 A   36.87 CHAPTER 15 179 18. P = VI cos   300 W = (120 V)(3 A) cos θ cos θ = 0.833  θ = 33.59 V = 120 V 0, I = 3 A 33.59 V 120 V 0 = 40  33.59 = 33.34  + j22.10  = ZT = I 3 A   33.59 RT = 33.34  = 2  + R  R = 31.34  19. a. b. ZT = R 2 + X L2 tan1XL/R f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz ZT 1.0 k 1.008 k 1.181 k 1.606 k 2.134 k 2.705 k XLE ZT f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz VL 0.0 V 0.623 V 2.66 V 3.888 V 4.416 V 4.646 V VL = c. f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz 180 θT 0.0 7.16 32.14 51.49 62.05 68.3 θL = 90  tan1 XL/R 90.0 82.84 57.85 38.5 27.96 21.7 CHAPTER 15 d. f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz 20. a. ZT = │ZT│ = 1 2 2 R  X C tan XC/R R 2  X C2 , θT = tan1XC/R f 0 kHz 1 kHz 3 kHz 5 kHz 7 kHz 10 kHz CHAPTER 15 VR = RE/ZT 5.0 V 4.96 V 4.23 V 3.11 V 2.34 V 1.848 V │ZT│  353.1  150.80  120.78  111.09  105.58  θT 90.0 73.55 48.46° 34.11 25.82 18.71 181 b. VC = XC E ( X C   90)(E 0) 90 + tan1XC/R = 2 2 R  jX C R  XC │VC│ = XC E R 2  X C2 │VC│ 10.0 V 9.59 V 7.49 V 5.61 V 4.36 V 3.21 V f 0 Hz 1 kHz 3 kHz 5 kHz 7 kHz 10 kHz c. θC = 90 + tan1 XC/R θC 0.0 16.45 41.54 55.89 64.18 71.29 f 0 Hz 1 kHz 3 kHz 5 kHz 7 kHz 10 kHz d. │VR│ = RE R  X C2 f 0 Hz 1 kHz 3 kHz 5 kHz 7 kHz 10 kHz 182 2 │VR│ 0.0 V 2.83 V 6.63 V 8.28 V 9.00 V 9.47 V CHAPTER 15 21. a. R 2  ( X L  X C ) 2  tan 1 ( X L  X C ) / R ZT = f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz b. ZT  19.31 × 103  3.40 × 103  1.21 × 103  1.16 × 103  1.84 × 103  │VC│ = XC E ZT f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz c. CHAPTER 15 θT 90.0 87.03 72.91 34.33 +30.75 +56.99 E ZT f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz │VC│ 120 V 120.62 V 136.94 V 192.4 V 133.45 V 63.29 V I = I 0.0 mA 6.21 mA 35.29 mA 99.17 mA 103.45 mA 65.22 mA 183 22. 1 1 1 = 1.54 kHz =Rf=  2 fC 2 RC 2 (220 )(0.47  F) a. XC = b. Low frequency: XC very large resulting in large ZT High frequency: XC approaches zero ohms and ZT approaches R c. f = 100 Hz: XC = 1 1 = 3.39 k  2 fC 2 (100 Hz)(0.47  F) ZT  XC f = 10 kHz: XC = 1 1 = 33.86   2 fC 2 (10 kHz)(0.47  F) ZT  R d. e. 23. a. b. 24 184  1 1 = 8.47 k  2 fC 2 (40 kHz)(0.47  F) X 8.47  = 2.2  = tan1 C   tan 1 R 220  f = 40 kHz: XC = 1 1  0  0 = 0.147 S 0° R 6.8  1 1 YT =    90 = 5 mS 90° X L 90 200  YT = 1 1 1   90 = 0.5 mS 90° ZT X C   90 2 k c. YT = a. ZT = b. 22   2.2  = 2  (2 0)(6    90) 12    90 = 1.90  18.43 ZT =  2   j6  6.32    71.57 1 1 = 0.53 S 18.43° = 0.5 S + j0.17 S = G + jBC YT =  ZT 1.90   18.43 c. YT = (10 0)(60  90) = 9.86  9.46 10   j 60  1 1 YT = = 0.10 S 9.46° = 0.1 S  0.02 S = G  jBL  ZT 9.86 9.46 1 1 1 + + 3 k  0 6 k  90 9 k    90 = 0.333  103 0 + 0.167  103 90 + 0.111  103 90 = 0.333  103 S  j0.056  103 S = 0.34 mS 9.55 = G  jBL CHAPTER 15 25. a. ZT = 4.7  + j8  = 9.28   59.57, YT = 0.108 S 59.57 YT = 54.7 mS  j93.12 mS = G  jBL b. ZT = 33  + 20   j70  = 53   j70  = 87.80  52.87 YT = 11.39 mS 52.87 = 6.88 mS + j9.08 mS = G + jBC c. ZT = 200  + j500   j600  = 200   j100  = 223.61  26.57 YT = 4.47 mS 26.57 = 4 mS + j2 mS = G + jBC 26. a. YT = b. YT = I 60 A 70 = 0.5 S 70 = 0.171 S + j0.470 S = G + jBC = E 120 V 0 1 1 = 5.85 , XC = = 2.13  R= G BC I 20 mA 40 = 0.25 mS 280 = 0.25 mS 80 = E 80 V 320 = 0.043 mS + j0.246 mS = G + jBC 1 1 R= = 23.26 k, XC = = 4.07 k G BC 27. I 0.2 A   60 = 0.25 mS 60 = 0.0125 mS  j0.02165 mS = G  jBL = E 8 kV 0 1 1 R= = 80 k, XL = = 46.19 k G BL c. YT = a. YT = c. E = Is/YT = 2 A 0/111.8 mS 26.57= 17.89 V 26.57 E  = 17.89 V 26.57/10  0 = 1.79 A 26.57 IR = R 0 E  IL = = 17.89 V 26.57/20  90 = 0.89 A 63.43 X L 90 f. P = I2R = (1.79 A)2 10  = 32.04 W g. Fp = h. e = 25.30 sin(377t + 26.57) iR = 2.53 sin(377t + 26.57) iL = 1.26 sin(377t  63.43) is = 2.83 sin 377t CHAPTER 15 1 1 = 0.1 S  j0.05 S = 111.8 mS 26.57 + 10  0 20  90 G YT = 0.1 S = 0.894 lagging 111.8 mS 185 28. a. XC = b. YT = c. 29. 1 1  = 20.4 kΩ 2 fC 2 (60 Hz)(0.13  F) 1 1 + = 0.1 mS 0 + 0.049 mS 90 10 k  0 20.4 k    90 = 0.111 mS 26.10 2 mA 20 Is = 18.02 V 6.1 = 0.111 mS 26.10 YT E 18.02 V   6.1 = = 1.80 mA 6.1 IR = 10 k  0 ZR 18.02 V   6.1 E IC = = = 0.883 mA 83.90 20.4 k    90 ZC E= e. Is = IR + IC 2 mA 20 = 1.80 mA 6.1 + 0.883 mA 83.90 = (1.79 mA  j0.191 mA) + (0.094 mA + j0.878 mA) = 1.88 mA + j0.687 mA 2 mA 20°  2 mA 20.07° f. P = I2R = (1.80 mA)2 10 k = 32.4 mW g. Fp = h. ω = 2f = 377 rad/s a. c. G YT = 0.1 mS = 0.9 leading 0.111 mS is = 2.83  103 sin(ωt + 20) iR = 2.55  103 sin(ωt  6.57) iC = 1.25  103 sin(ωt + 83.44) e = 25.48 sin(ωt  6.57) 1 1 1   1.2  0 2  90 5    90 = 0.833 S 0 + 0.5 S 90 + 0.2 S 90 = 0.833 S  j0.3 S = 0.89 S 19.81 ZT = 1.12   19.81 YT = XC = 1 1 1  C= = 531 μF = C  X C (377 rad/s)(5 ) XL = ωL  L = X L = ω 186 2 = 5.31 mH 377 rad/s CHAPTER 15 d. g. Is = IR + IL + IC 2.121 A 60 = 2.00 A 79.81 + 1.20 A 10.19 + 0.48 A 169.81  2.121 A 60 = 2.13 A 60.01 2 P = I R = (2.00 A)2 1.2  = 4.8 W h. Fp = i. e = 3.39 sin(377t + 79.81) iR = 2.83 sin(377t + 79.81) iL = 1.70 sin(377t  10.19) iC = 0.68 sin(377t + 169.81) f. 30.  (0.707)(3 A) 60 2.121 A 60 = E = Is = = 2.40 V 79.81 0.885 S   19.81 0.885 S   19.81 YT E  2.397 V 79.81 IR = = 2.00 A 79.81 = R 0 1.2  0 E  2.397 V 79.81 = IL = = 1.20 A 10.19 2  90 X L 90 2.397 V 79.81 E  IC = = = 0.48 A 169.81 5    90 X C   90 a. G YT = 0.833 S = 0.941 lagging 0.885 S XL = L = (1000 rad/s)(3.9 H) = 3.9 kΩ, 1 1  = 8.33 kΩ XC =  C (1000 rad/s)(0.12  F) 1 1 1 YT = + + 3 k  0 3.9 k  90 8.33 k    90 = 0.333 mS 0 + 0.256 mS 90 + 0.120 mS 90 = 0.333 mS  j0.136 mS = 0.36 mS 22.22 d. E = I/YT = 3.54 mA 20/0.36 mS 22.22 = 9.83 V 2.22 E  IR = = 9.83 V2.22/3 k 0 = 3.28 mA 2.22 R 0 E  IL = = 9.83 V2.22/3.9 k 90 = 2.52 mA 87.78 X L 90 E  = 9.83 V2.22/8.33 k 90 = 1.18 mA 92.22 IC = X C   90 g. P = I2R = (3.28 mA)23 k = 32.28 mW h. Fp = G/YT = 0.333 mS/0.36 mS = 0.925 leading i. e = 13.9 sin(1000t + 2.22) iR  4.64  103 sin(1000t + 2.22) iL  3.56  103 sin(1000t  87.78) iC = 1.67  103 sin(1000t + 92.22) CHAPTER 15 187 31. a. b. (60  90)(20 A 40) 1200 A 130 = 18.78 60.14°  22   j 60  63.91 69.86 (22  0)(20 A 40) 440 A 40 I2 =  = 6.88 29.86° 63.91  69.86 63.91 69.86 I1 = (12   j 6 )(6 A 30) (13.42   26.57)(6 A 30)  12   j 6   j 4  12  j 2 80.52 A 3.43 = 6.62 A 12.89° = 12.17   9.46 I1 = I2 = c. 32. a. (4  90)(6 A 30) 24 A 120 = 1.97 A 129.46°  12.17   9.46 12.17   9.46 ( j10   j 40 )(4 A 0) (30  90)(4 A 0)   j 20   j10   j 40  50  90 = 2.4 A 0° I1 = I2 = (20  90)(4A 0) = 1.6 A 0° 50  90 ZT = (R 0 )( X C   90 ) = RX C 90 + tan1XC/R 2 2 R  jX C R  XC │ZT│ = f 0 Hz 1 kHz 2 kHz 3 kHz 4 kHz 5 kHz 10 kHz 20 kHz 188 RX C R  2 X C2 │ZT│ 40.0  35.74  28.22  22.11  17.82  14.79  7.81  3.959  θT = 90 + tan1XC/R θT 0.0 26.67 45.14 56.44 63.55 68.30 78.75 89.86 CHAPTER 15 b. IRX C │VC│ = R 2 + X C2 │VC│ 2.0 V 1.787 V 1.411 V 1.105 V 0.891 V 0.740 V 0.391 V 0.198 V f 0 kHz 1 kHz 2 kHz 3 kHz 4 kHz 5 kHz 10 kHz 20 kHz c. │IR│ = VC R │IR│ 50.0 mA 44.7 mA 35.3 mA 27.64 mA 22.28 mA 18.50 mA 9.78 mA 4.95 mA f 0 kHz 1 kHz 2 kHz 3 kHz 4 kHz 5 kHz 10 kHz 20 kHz 33. a. ( R 0)( X L 90) ZR ZL = = ZR  ZL R  jX L ZT = │ZT│ = f 0 Hz 1 kHz 5 kHz 7 kHz 10 kHz CHAPTER 15 = I[ZT(f)] RX L R  2 X L2 RX L R  2 X L2 90  tan1XL/R θT = 90  tan1XL/R | ZT | 0.0 k 1.22 k 3.91 k 4.35 k 4.65 k θT 90.0 75.86 38.53 29.6 21.69 189 │IL│ = b. 34. f | IL | 0 Hz 1 kHz 5 kHz 7 kHz 10 kHz  31.75 mA 6.37 mA 4.55 mA 3.18 mA IR = c. YT = E XL E 40 V = 8 mA (constant) = R 5k R 2  X C2 RX C 90  tan1XC/R f │YT│ 0 Hz 25.0 mS 1 kHz 27.98 mS 2 kHz 35.44 mS 3 kHz 45.23 mS 4 kHz 56.12 mS 5 kHz 67.61 mS 10 kHz 128.04 mS 20 kHz 252.59 mS 35. YT = 1 (use data of Prob. 36),  TY =  TZ ZT f 0 Hz 1 kHz 5 kHz 7 kHz 10 kHz 190 θT 0.0 26.67 45.14 56.44 63.55 68.30 78.75 89.86 YT  0.82 mS 0.256 mS 0.23 mS 0.215 mS θT 90.0 75.86 38.53 29.6 21.69 CHAPTER 15 36. a. YT = G 0 + BL 90 + BC 90 B  BL = G 2  ( BC  BL ) 2 tan1 C G f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz f 0 Hz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz b. ZT = 1 YT f 0 kHz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz CHAPTER 15 │YT│ XL  0 , ZT = 0 , YT =   1.86 mS 1.02 mS 1.00 mS 1.02 mS 1.04 mS │θT│ 90.0 57.51 12.63 +1.66 +9.98 +16.54 , T Z =   T Y ZT 0.0  537.63  980.39  1 k 980.39  961.54  θT 90.0 57.52 12.63 1.66 9.98 16.54 191 c. VC(f) = I[ZT(f)] │VC│ 0.0 V 5.38 V 9.80 V 10 V 9.80 V 9.62 V f 0 kHz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz d. IL = VL (f) VC ( f )  XL XL f 0 kHz 1 kHz 5 kHz 10 kHz 15 kHz 20 kHz 37. a. b. IL 10.0 mA 8.56 mA 3.12 mA 1.59 mA 1.04 mA 0.765 mA Rp = Rs2  X s2 (20 ) 2  (40 ) 2 = 100  (R)  20  Rs Xp = Rs2  X s2 2000  = 50  (C)  40 Xs Rp = Rs2  X s2 (2 k ) 2 + (8 k ) 2 = 34 k (R) = Rs 2 k Rs2  X s2 (2 k ) 2 + (8 k ) 2 = 8.5 k (L) Xp = = Xs 8k 38. a. R p X p2 (8.2 k )(20 k ) 2 = 7.02 k = Rs = 2 X p  R p2 (20 k ) 2 + (8.2 k ) 2 Xs = R p2 X p X p2  R p2 = (8.2 k ) 2(20 k ) = 2.88 k 467.24 k  ZT = 7.02 k  j2.88 k b. Rs = Xs = R p X p2 X p2  R p2 R p2 X p X p2  R p2 = (68 )(40 ) 2 = 17.48  (40 ) 2 + (68 ) 2 = (68 ) 2(40 ) = 29.72  6224  2 ZT = 17.48  + j29.72  192 CHAPTER 15 39. a. CT = 2 μF 1 1 = 79.62  XC = = 3 C 2 (10 Hz)(2  F) XL = ωL = 2π(103 Hz)(10 mH) = 62.80  YT = 1 1 1   220  0 79.62    90 62.8  90 = 4.55 mS 0 + 12.56 mS 90 + 15.92 mS 90 = 4.55 mS  j3.36 mS = 5.66 mS 36.44 E = I/YT = 1 A 0/5.66 mS 36.44 = 176.68 V 36.44 E  = 176.68 V 36.44/220  0 = 0.803 A 36.44 IR = R 0 E  = 176.68 V 36.44/62.80 90 = 2.813 A 53.56 IL = X L 90 40. b. Fp = G/YT = 4.55 mS/5.66 mS = 0.804 lagging c. P = I2R = (0.803 A)2 220  = 141.86 W f. Is = IR + 2IC + IL I  I  IL and IC = s R 2 1 A 0  0.803 A 36.44  2.813 A   53.56 = 2 1  (0.646  j 0.477)  (1.671  j 2.263) 1.317  j1.786 =  2 2 IC = 0.657 + j0.893 = 1.11 A 126.43 g. ZT = 1 1 = 176.7  36.44 = YT 5.66 mS   36.44 = 142.15  + j104.96  = R + jXL P = VI cos θ = 3000 W 3000 W 3000 W 3000 = 0.75 (lagging) = = cos θ = VI (100 V)(40 A) 4000 θ = cos1 0.75 = 41.41 I 40 A   41.41 = 0.4 S 41.41 = 0.3 S  j0.265 S = GT  jBL = YT = E 100 V 0 1 1 1 GT = 0.3 S = + = 0.05 S + 20  R R 1 and R = =4 0.25 S CHAPTER 15 193 XL = 41. 1 BL  1 = 3.74  0.265 S a. b. c. 42. 194 a. e and  R2 b. e and is CHAPTER 15 43. c. iL and iC (I): (a) θdiv. = 0.8 div., θT = 4 div. 0.8 div.  360 = 72 θ= 4 div. 1 leads 2 by 72 (b) 1: peak-to-peak = (5 div.)(0.5 V/div.) = 2.5 V  2.5 V  V1(rms) = 0.7071   = 0.88 V  2  2: peak-to-peak = (2.4 div.)(0.5 V/div.) = 1.2 V  1.2 V  V2(rms) = 0.7071   = 0.42 V  2  (II): (c) T = (4 div.)(0.2 ms/div.) = 0.8 ms 1 1 = = 1.25 kHz (both) f= T 0.8 ms (a) θdiv. = 2.2 div., θT = 6 div. 2.2 div. θ=  360 = 132 6 div. 1 leads 2 by 132 (b) 1: peak-to-peak = (2.8 div.)(2 V/div.) = 5.6 V  5.6 V  V1(rms) = 0.7071   = 1.98 V  2  2: peak-to-peak = (4 div.)(2 V/div.) = 8 V 8V V2(rms) = 0.7071   = 2.83 V  2  (c) CHAPTER 15 T = (6 div.)(10 s/div.) = 60 μs 1 1 = = 16.67 kHz f= T 60  s 195 Chapter 16 1. 2. ZT = j4  + b. Is = c. I1 = 3.5 A 22.65 d. I2 = e. VL = Is XL = (3.5 A 22.65)(4  90) = 14 V 112.65 a. ZT = 3  + j6  + 2  0  8  90 = 3  + j6  + 1.94  14.04 = 3  + j6  + 1.88   j0.47  = 4.88  + j5.53  = 7.38  48.57 b. Is = c. IC = = d. 3. 196 (8    90)(12   0) = 3.69   j1.54  4 22.65  j8  12 a. a. 14 V 0 E = 3.5 A 22.65 = 4    22.65 ZT (8    90)(3.5 A 22.65) = 1.94 A 33.66 12   j8  30 V 0 E = 4.07 A 48.57 = ZT 7.38  48.57 Z R2 I s Z R2 + Z C = (2  0)(4.07 A   48.57) 2   j8  8.14 A   48.57 = 0.987 A 27.39 8.25   75.96 Z L E (6  90)(30 V 0) 180 V 90 = = ZT 7.38  48.57 7.38  48.57 = 24.39 V 41.43 VL = ZT = 12  90  (9.1   j12 Ω) = 12  90  15.06  52.826 180.72 37.17 = 9.100 = 19.86  37.17 E 60 V 0 = 3.02 A 37.17 19.86  37.17 b. Is = c. (CDR) I2 = ZT = (12 Ω 90°)(3.02 A   37.17) 36.24 A 52.83 = j12  + 9.1   j12  9.1 0 = 3.98 A 52.83 CHAPTER 16 4. 5. (12    90)(60 V 0) 720 V   90 = 9.1   j12  15.06   52.826 = 47.81 V 37.17 d. (VDR) VC = e. P = EI cos θ = (60 V)(3.02 A)cos(37.17) = 181.20(0.797) = 144.42 W a.  6.8 k  0   8 k 90  (4 k   90)(6 k 90)  2  ZT = 2 k 0 +  6.8 k  j 4 k  j 6 k  j 8 k 2 24 k 0 27.2 k 90 = 2 k +  2 90 3.4 k  j8 k  8.69 k 66.97 = 2 k + j12 k 3.13 k  23.03    2.88 k  j1.22 k = 2 k + 7.88 kΩ  j12 k + j1.22 k ZT = 4.88 k  j10.78 k = 11.83 k 65.64 b. V2 = IZT' = (4 mA 0)(3.13 k 23.03) = 12.52 V 23.03 V 12.52 V 23.03) = 1.57 A 66.97 IL = 2  X L2 8 k 90 c. Fp = a. 400 Ω 90  400  90 = b. VC = c. P = EI cos θ = (100 V)(0.25 A) cos 36.86 = (25)(0.8) = 20 W 4.88 k R = 0.413 (leading)  ZT 11.83 k 400    90 = 200  90 2 Z = 100   j200  = 223.61  63.43 Z = j200  + j600  = +j400  = 400  90 (223.61    63.43)(400  90) 89444.00 26.57 ZT = Z  Z =  (100   j 200 )  j 400  223.61 63.43 = 400  36.86 100 V 0 E = 0.25 A 36.86 I= = 400    36.86 ZT CHAPTER 16 (200    90)(100 V 0) 20, 000 V   90 = 89.44 V 26.57 = 100   j 200  223.61   63.43 197 6. a. Z1 = 3  + j4  = 5  53.13 E 120 V 0 I1 = = 24 A 53.13 = Z1 5  53.13 b. VC = c. VR1 = (I1 θ)R 0 = (24 A 53.13)(3  0) = 72 V 53.13 (13    90)(120 V 0) 1560 V   90 = 260 V 0 =  j13  + j 7  6   90 Vab + VR1  VC = 0 +- Vab = VC  VR1 = 260 V 0  72 V 53.13 +- = 260 V  (43.20 V + j57.60 V) = 216.80 V + j57.60 V = 224.32 V 14.88 7. Z1 = 10  0 Z2 = 80  90  20  0 1600  90 1600  90  = 20  j80 82.462 75.964 = 19.403  14.036 Z3 = 60  90 a. ZT = (Z1 + Z2)  Z3 = (10  + 18.824  + j4.706 )  60  90 1752.36    80.727 = 29.206  9.273  6  90 = 28.824  j 4.706  j 60 1752.36    80.727 = = 28.103  18.259 62.356   62.468 E 40 V 0 = 1.42 A 18.26 I1 = = ZT 28.103    18.259 198 Z 2E (19.403  14.036)(40 V 0) 776.12 V 14.036 = = 29.206  9.273 29.206 9.273 Z 2 + Z1 = 26.57 V 4.76 b. V1 = c. P = EI cos θ = (40 V)(1.423 A)cos 18.259 = 54.07 W CHAPTER 16 8. a. Z1 = 2  + j1  = 2.236  26.565, Z2 = 3  0 Z3 = 16  + j15   j7  = 16  + j8  = 17.889  26.565 1 1 1 1 1 1 YT = + + = + + Z1 Z 2 Z 3 2.236  26.565 3  0 17.889  26.565 = 0.447 S 26.565 + 0.333 S 0 + 0.056 S 26.565 = (0.4 S  j0.2 S) + (0.333 S) + (0.05 S  j0.025 S) = 0.783 S  j0.225 S = 0.82 S 16.03 1 1 = 1.23  16.03 ZT = = YT 0.82 S   16.03 b I1 = E 60 V 0 = 26.83 A 26.57 = Z1 2.236  26.565 E 60 V 0 = 20 A 0 = Z 2 3  0 60 V 0 E I3 = = 3.35 A 26.57 = Z3 17.889  26.565 I2 = c. Is = 60 V 0 E = 48.9 A 16.03 = ZT 1.227  16.032 ? Is = I1 + I2 + I3 ? 48.9 A 16.03 = 26.83 A 26.57 + 20 A 0 + 3.35 A 26.57 = (24 A  j12 A) + (20 A) + (3 A  j1.5 A)  = 47 A + j13.5 A = 48.9 A 16.03 (checks) 9. d. Fp = a. X L1 G 0.783 S = 0.955 (lagging) = YT 0.820 S = ωL1 = 2(103 Hz)(0.1 H) = 628  X L2 = ωL2 = 2(103 Hz)(0.2 H) = 1.256 k 1 1 = 0.159 k = 3 C 2 (10 Hz)(1  F) ZT = R 0 + X L1 90 + XC 90  X L2 90 XC = = 300  + j628  + 0.159 k 90  1.256 k 90 = 300  + j628   j182  = 300  + j446  = 537.51  56.07 b. Is = CHAPTER 16 E 50 V 0 = 93 mA 56.07 = ZT 537.51  56.07 199 c. (CDR): I1 = Z L2 I s Z L2 + ZC = (1.256 k  90)(93 mA   56.07) + j1.256 k   j 0.159 k  116.81 mA 33.93 = 106.48 mA 56.07 1.097 90 ZC I s (0.159 k    90)(93 mA   56.07) I2 = = Z L2 + Z C 1.097 k  90 = 14.79 mA   146.07 = 13.48 mA 236.07 1.097 90 = 13.48 mA 123.93 = d. V1 = (I2 θ)( X L2 90) = (13.48 mA 123.92)(1.256 k 90) = 16.93 V 213.93 Vab = E  (Is θ)(R 0) = 50 V 0  (93 mA 56.07)(300  0) +- = 50 V  27.9 V 56.07 = 50 V  (15.573 V  j23.149 V) = 34.43 V + j23.149 V = 41.49 V 33.92 10. e. P = I s2 R = (93 mA)2300  = 2.595 W f. Fp = a. ZT = 1.2 k + R 300 Ω = 0.558 (lagging) = ZT 537.51 Ω = 1.2 k + (1.2 k 0)(1.8 k   90) 2.4 k 90  1.2 k  j1.8 k 2 2.16 k  90  1.2 k 90 2.16  56.31 = 1.2 k + 1 k 33.69 + j1.2 k = 1.2 k + 832.05   j554.70  + j1.2 k = 2.03 k + j645.30  = 2.13 k 17.63 200 b. V1 = IR1 = (20 mA 0)(1.2 k 0) = 24 V0 c. I1 = d.  (2.4 k)  V2 = I(X L1  X L2 ) = (20 mA 0)  90  = 24 V 90   2 e. Vs = IZT = (20 mA 0)(2.13 k 17.63) = 42.60 V 17.63 (1.2 k 0)(20 mA 0) 2.4 A 0  = 11.11 mA 56.31 1.2 k  j1.8 k 2.16  103   56.31 CHAPTER 16 11. Z1 = 2   j2  = 2.828  45 Z2 = 3   j9  + j6  = 3   j3  = 4.243  45 Z3 = 10  0 a. 1 1 1 1 1 1 + + = + + Z1 Z 2 Z 3 2.828    45 4.243    45 10  0 = 0.354 S 45 + 0.236 S 45 + 0.1 S 0 = 0.59 S 45 + 0.1 S 0 = 0.417 S + j0.417 S + 0.1 S YT = 0.517 S + j 0.417 S = 0.66 S 38.89 1 1 = 1.52  38.89 ZT = = YT 0.66 S 38.89 YT = 12. b. V1 = c. I1 = d. Is = (2  0)(60 V0) 120 V 0 = 42.43 V 45  2   j  2.828   45 60 V 0 60 V 0 60 V 0 E    Z  3   j 9   j 6  3   j 3  4.243   45 = 14.14 A 45 60 V 0 E = 39.47 A 38.89  ZT 1.52    38.89 Z = 12   j20  = 23.32  59.04 R4 0  Z = 20  0  23.32  59.04 = 12.36  27.03 Z = R3 0 + R4 0  Z = 12  + 12.36  27.03 = 12  + (11.01   j5.62 ) = 23.01   j5.62  = 23.69  13.73 R2 0  Z = 20  0  23.69   13.73 = 10.92  6.29 ZT = R1 0 + R2 0  Z = 12  + 10.92 6.29 = 12  + (10.85   j1.25 ) = 22.85   j1.2  = 22.88  3.01 E 100 V 0 Is = = 4.37 A  +3.01 = ZT 22.88    3.01 I R1 = I I R3 = R2 0 I s (20  0)(4.37 A 3.01) 87.40 A 3.01 = = 43.38   7.44 R2 0  Z " 20   23.01   j 5.62  = 2.01 A 10.45 R4 0I R3 (20  0)(2.01 A 10.45) 40.20 A 10.45 40.20 A 10.45  = 20  + 12   j 20  32   j 20  37.74   32.01 R4 0 + Z = 1.07 A 42.46 I4 = CHAPTER 16 = 201 13. R3 + R4 = 2.7 k + 4.3 k = 7 k R = 3 k  7 k = 2.1 k Z = 2.1 k  j10  (CDR) (CDR) 14. (40 k  0)(20 mA 0) 40 k  + 2.1 k   j10  = 19 mA +0.014 as expected since R1  Z I (of 10  cap.) = (3 k  0)(19 mA 0.014) 57 mA 0.014 = 3k + 7 k 10 = 5.7 mA 0.014 P = I2R = (5.7 mA)2 4.3 k = 139.71 mW I4 = Z = X C2   90  R1 0 = 2  90  1  0 2    90 2    90  1  j2 2.236   63.435 = 0.894  26.565 Z = X L2 90 + Z = +j8  + 0.894  26.565 = = +j8  + (0.8   j4 ) = 0.8  + j4 = 4.079  78.69 IXL = 2 X C1   90I X C1   90  Z  (2   90)(0.5 A 0) 1 A   90  0.8  j 2  j 2   (0.8   j 4 ) 1 A   90 = 0.464 A  158.99 2.154 68.199 X C2   90 I X C (2    90)(0.464 A   158.99) 0.928 A   248.99 2 = = I1 = X C2   90 + R1  j2  + 1  2.236   63.435 = = 0.42 A 174.45 202 CHAPTER 16 Chapter 17 1.  2. Z = −j5  + 2  0  5  90 = −j5  + 1.72  + j0.69  = 4.64  −68.24 E 60 V 30 = 12.93 A 98.24 I= = Z 4.64    68.24 3. Z = 10  0  6  90 = 5.15  59.04 E = IZ = (2 A 120)(5.15  59.04) = 10.30 V 179.04 4. a. I= b. V = (hI)(R) = (50 I)(50 kΩ) = 2.5  106 I Z = 50 k 0 5. V 16 V = 4  103 V R 4  103 Z = 4 k 0 = Clockwise mesh currents: E  I1Z1  I1Z2 + I2Z2 = 0 I2Z2 + I1Z2  I2Z3  E2 = 0 ──────────────────── [Z1 + Z2]I1  Z2I2 = E1 Z2I1 + [Z2 + Z3]I2 = E2 ────────────────── Z 2 E1  E2  Z 2 + Z3  I R1 = I1 =  Z1 + Z 2  Z 2 Z 2 6. Z1 = R1 0 = 4  0 Z2 = XL 90 = 6  90 Z3 = XC 90 = 8  90 E1 = 10 V 0, E2 = 40 V 60 =  Z 2 + Z3  E1  Z 2 E2 Z1Z 2 + Z1Z3 + Z 2 Z 3 = 5.15 A 24.5  Z 2 + Z3  By interchanging the right two branches, the general configuration of Problem 5 will result and I50Ω = I1 = Z2  Z3  E1  Z 2 E 2 Z1Z 2 + Z1Z3 + Z 2 Z 3 = 0.44 A 143.48 CHAPTER 17 Z1 = R1 = 50  0 Z2 = XC 90 = 60  90 Z3 = XL 90 = 20  90 E1 = 5 V 30, E2 = 20 V 0 203 7. a. Z1 = 12  + j12  = 16.971  45 Z2 = 3  0 Z3 = j1  E1 = 20 V 50 E2 = 60 V 70 E3 = 40 V 0 I1[Z1 + Z2]  Z2I2 = E1  E2 I2[Z2 + Z3]  Z2I1 = E2  E3 ─────────────────── (Z1 + Z2)I1  Z2I2 = E1  E2 Z2I1 + (Z2 + Z3)I2 = E2  E3 ───────────────────── Using determinants: (E  E2 )(Z 2 + Z 3 ) + Z 2 (E 2  E3 ) = 2.55 A 132.72 I R1 = I1 = 1 Z1Z 2 + Z1Z3 + Z 2 Z 3 8. Clockwise mesh currents: E1  I1Z1  I1Z2 + I2Z2 = 0 I2Z2 + I1Z2  I2Z3  I2Z4 + I3Z4 = 0 I3Z4 + I2Z4  I3Z5  E2 = 0 ─────────────────────────  Z2 I 2 [Z1 + Z 2 ]I1 Z2 Z1 = 4  + j3 , Z2 = j1  Z3 = +j6 , Z4 = j2  Z5 = 8  E1 = 60 V 0, E2 = 120 V 120 +0 = E1  Z4 I3 = 0 I1 + [Z 2 + Z 3 + Z 4] I 2  Z 4 I 2 + [ Z 4 + Z 5]I 3 =  E 2 0 ─────────────────────────────────── I R1 = I 3 =  Z 2Z 4  E1 +  Z 22   Z1 + Z 2  Z 2 + Z3 + Z 4 E2  Z1 + Z 2  Z 2 + Z3 + Z 4  Z 4 + Z5   Z1 + Z 2 Z 24   Z 4 + Z5 Z 22 = 13.07 A   33.71° 204 CHAPTER 17 9. Z1 = 15  0, Z2 = 15  0 Z3 = j10  = 10  90 Z4 = 3  + j4  = 5  53.13 E1 = 220 V 0 E2 = 100 V 90 I1(Z1 + Z3)  I2Z3  I3Z1 = E1 I2(Z2 + Z3)  I1Z3  I3Z2 = E2 I3(Z1 + Z2 + Z4)  I1Z1  I2Z2 = 0 ───────────────────────  I3Z1 = E1 I1(Z1 + Z3)  I2Z3 I1Z3 + I2(Z2 + Z3)  I3Z2 = E2 I1Z1  I2Z2 + I3(Z1 + Z2 + Z4) = 0 ─────────────────────────────────── Applying determinants: I3 =  ( Z1 + Z 3)( Z 2 )E 2  Z1Z 3E 2 + E1  Z 2 Z 3 + Z1( Z 2 + Z 3)   ( Z1 + Z 3) ( Z 2 + Z 3)( Z1 + Z 2 + Z 4 )  Z 22 + Z 3  Z 3( Z1 + Z 2 + Z 4 )  Z1Z 2   Z1   Z 2 Z 3  Z1( Z 2 + Z 3)  = 48.33 A 77.57  or I3 = E1 E 2 if one carefully examines the network! Z4 10. Z1 = 5  0, Z2 = 5  90 Z3 = 4  0, Z4 = 6  90 Z5 = 4  0, Z6 = 6  + j8  E1 = 20 V 0, E2 = 40 V 60 I1(Z1 + Z2 + Z4)  I2Z2  I3Z4 = E1 I2(Z2 + Z3 + Z5)  I1Z2  I3Z5 = E2 I3(Z4 + Z5 + Z6)  I1Z4  I2Z5 = 0 ─────────────────────────  Z2I2  Z4I3 = E1 (Z1 + Z2 + Z4) I1  Z5I3 = E2 Z2I1 + (Z2 + Z3 + Z5)I2  Z5I2 + (Z4 + Z5 + Z6)I3 = 0 Z4I1 ───────────────────────────────────────── Using Z = Z1 + Z2 + Z4, Z = Z2 + Z3 + Z5, Z = Z4 + Z5 + Z6 and determinants: E1 (Z  Z   Z 52 )  E 2 (Z 2 Z  + Z 4 Z5 ) I R1 = I1 = Z  (Z  Z   Z52 )  Z 2 (Z 2 Z  + Z 4 Z5 )  Z 4 (Z 2 Z5 + Z 4 Z  ) = 3.04 A 169.12° CHAPTER 17 205 11. Z1 = 10  + j20  Z3 = 80  0 Z5 = 15  90 Z7 = 5  0 E1 = 25 V 0 Z2 = j20  Z4 = 6  0 Z6 = 10  0 Z8 = 5 Ω  j20  E2 = 75 V 20 I1(Z4 + Z6 + Z7)  I2Z4  I4Z6 = E1 I2(Z1 + Z2 + Z4)  I1Z4  I3Z2 = 0 I3(Z2 + Z3 + Z5)  I2Z2  I4Z5 = E2 I4(Z5 + Z6 + Z8)  I1Z6  I3Z5 = 0 ─────────────────────────  Z4 I2 +0  Z6I4 = E1 (Z4 + Z6 + Z7) I1 Z4I1 + (Z1 + Z2 + Z4)I2  Z2I3 +0 =0  Z5I4 = E2 0  Z2 I2 + (Z2 + Z3 + Z5)I3 +0  Z5I3 + (Z5 + Z6 + Z7)I4 = 0 Z6I1 ──────────────────────────────────────────────────── Applying determinants: I R1 = I 80 = 0.68 A 162.9 12. Z1 = 5 k 0 Z2 = 10 k 0 Z3 = 1 k + j4 k = 4.123 k 75.96 I1(Z1 + Z2)  Z2I2 = 28 V I2(Z2 + Z3)  Z2I1 = 0 ────────────────── (Z1 + Z2)I1  Z2I2 = 28 V Z2I1 + (Z2 + Z3)I2 = 0 ─────────────────── Z 2 28 V IL = I2 = = 3.17  103 V 137.29 Z1Z 2 + Z1Z 3 + Z 2 Z3 206 CHAPTER 17 Source Conversion: E = (I θ)(Rp 0) = (50 I)(40 k 0) = 2  106 I 0 Z1 = Rs = Rp = 40 k 0 Z2 = j0.2 k Z3 = 8 k 0 Z4 = 4 k 90 13. I1(Z1 + Z2 + Z3)  Z3I2 = E I2(Z3 + Z4)  Z3I1 = 0 ──────────────────── (Z1 + Z2 + Z3)I1  Z3I2 = E Z3I1 + (Z3 + Z4)I2 = 0 ──────────────────── Z3 E IL = I2 = = 42.91 I 149.31 (Z1 + Z 2 + Z3 )(Z3 + Z 4 )  Z 23 14. 6Vx  I1 1 k  10 V 0 = 0 10 V0  I2 4 k  I2 2 k = 0 ─────────────────────── Vx = I2 2 k I1 1 k + I2 12 k = 10 V 0 I2 6 k = 10 V 0 ────────────────────── 10 V 0 I2 = I 2k = = 1.67 mA 0 = I2kΩ 6 k I1 1 k + (1.667 mA 0)(12 k) = 10 V 0 I1 1 k + 20 V 0 = 10 V 0 I1 1 k = 10 V 0 10 V 0 = 10 mA 0 I1 = I1k = 1k CHAPTER 17 207 E1 = 5 V 0 E2 = 20 V 0 Z1 = 2.2 k 0 Z2 = 5 k 90 Z3 = 10 k 0 I = 4 mA 0 15. E1  I1Z1  Z2(I1  I2) = 0 Z2(I2  I1) + E2  I3Z3 = 0 ─────────────────── I3  I2 = I Substituting, we obtain: I1(Z1 + Z2)  I2Z2 = E1 I1Z2  I2(Z2 + Z3) = IZ3  E2 ──────────────────── Determinants: I1 = 1.39 mA 126.48, I2 = 1.341 mA 10.56, I3 = 2.693 mA 174.8 I10kΩ = I3 = 2.69 mA 174.8 Z1 = 1 k 0 Z2 = 4 k + j6 k E = 10 V 0 16. Z1(I2  I1) + E  I3Z3 = 0 I1 = 6 mA 0, 0.1 Vs = I3  I2, Vs = (I1  I2)Z1 ───────────────────────────────── Substituting: (1 k)I2 + (4 k + j6 k)I3 = 16 V 0 (99 )I2 + I3 = 0.6 V 0 ──────────────────────────── Determinants: I3 = I6 kΩ = 1.38 mA 56.31 208 CHAPTER 17 17. Z1 = 4 Ω 0 Z2 = 5 Ω 90 Z3 = 2 Ω 90 I1 = 3 A 0 I2 = 5 A 30 I1 = I3 + I4 1  1 1  V1 V1  V 2  V 1  +   V 2   = I1 + Z1 Z2  Z1 Z 2   Z2  or V1[Y1 + Y2]  V2[Y2] = I1 I1 = I4 = I5 + I2 1 1  1 V1  V 2 V 2 = + I 2  V 2  +   V1   =  I 2 Z2 Z3  Z2   Z 2 Z3  or V2[Y2 + Y3]  V1[Y2] = I2  Y2V2 = I1 [Y1 + Y2]V1 Y2V1 + [Y2 + Y3]V2 = I2 ────────────────────── [Y2 + Y3 ] I1  Y2 I 2 V1 = = 14.68 V 68.89 Y1 Y2 + Y1 Y3 + Y2 Y3 [Y1 + Y2 ] I 2 + Y2 I1 = 12.97 V 155.88 V2 = Y1 Y2 + Y1 Y3 + Y2 Y3 18. Z1 = 3  + j4  = 5 53.13 Z2 = 2  0 Z3 = 6  0  8  90 = 4.8  36.87 I1 = 0.6 A 20 I2 = 4 A 80 0 = I1 + I3 + I4 + I2  0 = I1 + V1 + V1 V 2 + I 2 Z1 Z2 1 1  1  V1  +   V 2   =  I1  I 2  Z1 Z 2   Z2  or V1[Y1 + Y2]  V2[Y2] = I1  I2 ────────────────────── I2 + I4 = I5 V  V2 V2 = I2 + 1 Z2 Z3 CHAPTER 17 209  1  1  1  V2  +   V1   = + I 2  Z2   Z 2 Z3  or V2[Y2 + Y3]  V1[Y2] = I2 ────────────────── and [Y1 + Y2]V1  Y2V2 = I1  I2 Y2V1 + [Y2 + Y3]V2 = I2 Applying determinants: [ Y 2 + Y 3][I1 + I 2] + Y 2I 2 = 5.12 V 79.36 Y1Y 2 + Y1Y 3 + Y 2 Y 3 Y1I 2  I1Y 2 V2 = = 2.71 V 39.96 Y1Y 2 + Y1Y 3 + Y 2 Y 3 V1 = 19. Z1 = 5  0 Z2 = 6  90 Z3 = 4  90 Z4 = 2  0 E = 30 V 50 I = 0.04 A 90 I1 = I2 + I3 1 1 1 V E1  V1 = V1 + (V1  V 2)  +   2 = E1 V1  + Z1 Z2 Z3 Z1  Z1 Z 2 Z3  Z 3 or V1[Y1 + Y2 + Y3]  Y3V2 = E1Y1 I3 + I = I4 V1  V 2 Z3 +I= 1 1  V V2  V2  +   1 = I Z4  Z3 Z4  Z3 or V2[Y3 + Y4]  V1Y3 = I resulting in V1[Y1 + Y2 + Y3]  V2Y3 = E1Y1 V1[Y3] + V2[Y3 + Y4] = +I ────────────────────── Using determinants: V1 = 19.86 V 43.8 and V2 = 8.94 V 106.9 210 CHAPTER 17 20. Z1 = 10  0 Z2 = 10  0 Z3 = 4  90 Z4 = 2  0 Z5 = 8  90 E = 50 V 120 I = 0.8 A 70 I1 = I2 + I5 1 1 1 1 1 1 E E  V1 (V1  V 2) V V  V2 = 1 + + 1  V1  + + +   V2  + = Z1 Z2 Z5 Z3 Z 5  Z1  Z1 Z 2 Z 3 Z 5   Z3 or V1[Y1 + Y2 + Y3 + Y5]  V2[Y3 + Y5] = E1Y1 I3 + I5 = I4 + I 1 1 1 1 1 V1  V 2 V1  V 2 V 2 + = + I  V2  + +   V1  +  = I Z3 Z5 Z4  Z3 Z 4 Z5   Z3 Z5  or V2[Y3 + Y4 + Y5]  V1[Y3 + Y5] = I resulting in V1[Y1 + Y2 + Y3 + Y5]  V2[Y3 + Y5] = E1Y1 V1[Y3 + Y5] + V2[Y3 + Y4 + Y5] = I ─────────────────────────────── Applying determinants: V1 = 19.78 V 132.48 and V2 = 13.37 V 98.78 21. Z1 = 15  0 Z2 = 10  90 Z3 = 15  0 Z4 = 3  + j4  1 1 1 1 1 +  V2  + V1  V3 = 0 Z3  Z1 Z 2 Z 3  Z1  1 1 1  1 1  + +  220 V 0  100 V 90 = 0 V2   15  15   j10  15   15  3 3 V 2 133.34  10  j100  10  14.67  j 6.67 V2  16.05 V 24.55 = 96.30 V 12.32 166.67  103 36.37 V1 = E1 = 220 V 0, V3 = E2 = 100 V 90 CHAPTER 17 211 22. E1 = 25 V 0 E2 = 75 V 20 Z1 = 10  + j20  Z2 = 6  0 Z3 = 5  0 Z4 = 20  90 Z5 = 10  0 Z6 = 80  0 Z7 = 15  90 Z8 = 5   j20 Ω    V1: V1 V 2 + V1 V 4 + V1 E1 = 0 Z1 Z2 Z3     V2: V 2 V1 + V 2 V 4 + V 2 E 2 V 3 = 0 Z1 Z4 Z6 +   V 3 E2 V 2 V 3 V 4 V 3 + + =0 V3: Z6 Z7 Z8    V4: V 4 V1 + V 4 V 2 + V 4 V 3 + V 4 = 0 Z2 Z4 Z7 Z5 ─────────────────────── Rearranging:  1 1 1  V V E V1  + +   2 4 = 1 Z1 Z 2 Z3  Z1 Z 2 Z 3   1 1 1  V1 V 4 V 3   = E2 V2  + +   Z6 Z1 Z 4 Z 6  Z1 Z 4 Z 6   1 1 1  V V E V3  + +   2 4 =  2 Z6 Z 7 Z6  Z 6 Z 7 Z8   1 1 1 1  V V V V4  + + +   1 2 3 =0  Z 2 Z 4 Z7 Z5  Z 2 Z 4 Z 7 Setting up and then using determinants: V1 = 14.62 V 5.86, V2 = 35.03 V 37.69 V3 = 32.4 V 73.34, V4 = 5.67 V 23.53 212 CHAPTER 17 23. 1 4 Ω 0° = 0.25 S 0 1 Y2 = 1 Ω 90 = 1 S 90 1 Y3 = 5  0 = 0.2 S 0 1 Y4 = 4    90 = 0.25 S 90 1 Y5 = 8 Ω 90° = 0.125 S 90 I1 = 2 A 30 I2 = 3 A 150 Y1 = V1[Y1 + Y2]  Y2V2 = I1 V2[Y2 + Y3 + Y4]  Y2V1  Y4V3 = I2 V3[Y4 + Y5]  Y4V2 = I2 ───────────────────────────  Y2 V2 + 0 = I1 [Y1 + Y2]V1 Y2V1 + [Y2 + Y3 + Y4]V2  Y4 V3 = I2 0  Y4 V2 + [Y4 + Y5]V3 = I2 ──────────────────────────────────── 2 I1  (Y 2 + Y 3 + Y 4)(Y 4 + Y 5)  Y 4   I 2[Y 2 Y5] V1 =  Y1 + Y 2 (Y 2 + Y3 + Y 4)(Y 4 + Y 5)  Y 24   Y 22(Y 4 + Y5) = Y  = 5.74 V 122.76 ( + ) ( + ) V2 = I1Y 2 Y 4 Y 5 I 2 Y 5 Y1 Y 2 = 4.04 V 145.03 Y 2 I 2  (Y1 + Y 2)(Y 3 + Y 4)  Y 2   Y 2 Y 4I1 V3 =  = 25.94 V 78.07 Y CHAPTER 17 213 24. 1 4 Ω 0° = 0.25 S 0 1 Y2 = 6  0 = 0.167 S 0 1 Y3 = 8  0 = 0.125 S 0 1 Y4 = 2    90 V1[Y1 + Y2 + Y3]  Y2V2  Y3V3 = I1 = 0.5 S 90 V2[Y2 + Y4 + Y5]  Y2V1  Y4V3 = 0 1 Y5 = V3[Y3 + Y4 + Y6]  Y3V1  Y4V2 = I2 5  90 ─────────────────────────── = 0.2 S 90  Y2V2  Y3V3 = I1 [Y1 + Y2 + Y3]V1 1 Y2V1 + [Y2 + Y4 + Y5]V2  Y4V3 = 0 Y6 = 4  90 Y3V1  Y4V2 + [Y3 + Y4 + Y6]V3 = I2 = 0.25 S 90 ────────────────────────────────────────── I1 = 4 A 0 I2 = 6 A 90 I  ( Y + Y + Y )( Y + Y + Y )  Y   I  Y Y + Y ( Y + Y + Y )  V1 = Y = ( Y  Y + Y )  ( Y + Y + Y )( Y  Y + Y )  Y   Y  Y ( Y + Y + Y )+ Y Y   Y  Y Y + Y ( Y + Y + Y )  Y1 = 2 1 2 4 5 3 4 6 4 2 2 4 3 3 4 5 2  1 2 3 2 4 5 3 4 6 4 2 2 3 4 6 3 4 3 2 4 3 2 4 5 = 15.13 V 1.29 I1 (Y 2)(Y 3 + Y 4 + Y 6)+ Y 3Y 4  +I 2  Y 4(Y1 + Y 2 + Y 3)  Y 2 Y 3 V2 = = 17.24 V 3.73 Y 2 I1 (Y3)(Y 2 + Y 4 + Y 5) + Y 2 Y 4  + I 2  Y 2 - (Y1 + Y 2 + Y 3)(Y 2 + Y 4 + Y5)  V3 = Y = 10.59 V 0.11 25. Left node: V1  Ii   Io 4Ix = Ix + 5 mA 0 + V1  V 2 2 k Right node: V2  Ii   I o 8 mA 0 = V2 V  V1 + 2 + 4I x 1k  2 k V1 4 k    90 Rearrange, reduce and 2 equations with 2 unknowns result: V1[1.803 123.69] + V2 = 10 V1[2.236 116.57] + 3 V2 = 16 ────────────────────── Determinants: V1 = 4.37 V 128.66 V2 = V1k = 2.25 V 17.63 Insert Ix = 214 CHAPTER 17 Z1 = 1 kΩ 0 Z2 = 2 kΩ 90 Z3 = 3 kΩ 90 I1 = 12 mA 0 I2 = 4 mA 0 E = 10 V 0 26.  Ii   Io V1 V 2 + + I2 Z1 Z 3 and V1 + V 2 = I1  I2 Z1 Z 3 with V2  V1 = E 0 = I1 + Substituting and rearranging: 1 1 E V1  +  = I1  I2  Z3  Z1 Z 3  and solving for V1: V1 = 15.4 V  178.2 with V2 = VC = 5.41 V  174.87 27. Left node: V1  Ii   Io 2 mA 0 = 12 mA 0 + and 1.5 V1  V2 = 10 Right node: V2  Ii   I o V1 V  V2 + 1 2 k 1k V 2  V1 V 2  6 V x  1k 3.3 k  and 2.7 V1  3.7 V2 = 6.6 0 = 2 mA 0 + Using determinants: CHAPTER 17 V1 = V2k = 10.67 V 0 = 10.67 V 180 V2 = 6 V 0 = 6 V 180 215 Z1 = 2 k 0 Z2 = 1 k0 Z3 = 1 k 0 I = 5 mA 0 28. V1 V + 3I1 + 2 Z1 Z3 with I1 = V1 V 2 Z2 and V2  V1 = 2Vx = 2V1 or V2 = 3V1 V1: I = Substituting will result in: or and with 29. 1 1 3 3 V1  +  + 3 V1    = I  Z1 Z 2   Z3 Z 2  1 6 3 +  =I V1    Z1 Z 2 Z 3  V1 = Vx = 2 V 0 V2 = 6 V 0 Ei  = 1  103 Ei R1 0 1 = 0.02 mS 0 Y1 = 50 k 1 = 1 mS 0 Y2 = 1k Y3 = 0.02 mS 0 I2 = (V1  V2)Y2 I1 = V1(Y1 + Y2)  Y2V2 = 50I1 V2(Y2 + Y3)  Y2V1 = 50I2 = 50(V1  V2)Y2 = 50Y2V1  50Y2V2 ──────────────────────────────────────────── (Y1 + Y2)V1  Y2V2 = 50I1 51Y2V1 + (51Y2 + Y3)V2 = 0 ─────────────────────── (50)(51)Y2 I1 = 2451.92 Ei VL = V2 = (Y1 + Y2 )(51Y2 + Y3 )  51Y22 216 CHAPTER 17 30. a. b. yes Z1 = Z 2 Z3 Z4 3 8  103 0 5  10 0 = 2.5  103 90 4  103 90 2 90 = 2 90 (balanced) Z1 = 5 k 0, Z2 = 8 k 0 Z3 = 2.5 k 90, Z4 = 4 k 90 Z5 = 5 k 90, Z6 = 1 k 0 I1[Z1 + Z3 + Z6]  Z1I2  Z3I3 = E I2[Z1 + Z2 + Z5]  Z1I1  Z5I3 = 0 I3[Z3 + Z4 + Z5]  Z3I1  Z5I2 = 0 ─────────────────────── [Z1 + Z3 + Z6]I1  Z1I2  Z3I3 = E  Z5I3 = 0 Z1I1 + [Z1 + Z2 + Z5]I2 Z3I1  Z5I2 + [Z3 + Z4 + Z5]I3 = 0 ─────────────────────────────────────── I2 = I3 = E Z1( Z 3 + Z 4 + Z 5 ) + Z 3 Z 5  Z  = ( Z1 + Z 3 + Z 6 )[( Z1 + Z 2 + Z 5 )( Z 3 + Z 4 + Z 5 )  Z 5 ]  Z 1[ Z 1( Z 3 + Z 4 + Z 5 )  Z 3 Z 5 ]  Z 3[ Z 1 Z 5 + Z 3 ( Z1 + Z 2 + Z 5 )] 2 E Z1Z5 + Z 3 (Z1 + Z 2 + Z5 )  I Z 5 = I2  I3 = CHAPTER 17 Z E Z1Z 4  Z 3 Z 2  Z = E  20  106 90  20  106 90  Z =0A 217 c. V1[Y1 + Y2 + Y6]  Y1V2  Y2V3 = I V2[Y1 + Y3 + Y5]  Y1V1  Y5V3 = 0 V3[Y2 + Y4 + Y5]  Y2V1  Y5V2 = 0 ─────────────────────────  Y1V2  Y2V3 = I [Y1 + Y2 + Y6]V1 Y1V1 + [Y1 + Y3 + Y5]V2  Y5V3 = 0  Y5V2 + [Y2 + Y4 + Y5]V3 = 0 Y2V1 ───────────────────────────────────────── V2 = V3 = I Y1( Y 2 + Y 4 + Y 5 ) + Y 2 Y 5  10 V 0 I = Es = 1 k  0 Rs = 10 mA 0 1 Y1 = 5 k   0 = 0.2 mS 0 1 Y2 = 8 k  0 = 0.125 mS 0 1 Y3 = 2.5 k  90 = 0.4 mS 90 1 Y4 = 4 k  90 = 0.25 mS 90 1 Y5 = 5 k    90 = 0.2 mS 90 1 Y6 = 1 k  0 V2 = 1 mS 0 Y  = ( Y1 + Y 2  Y 6 )[( Y 1 + Y 3 + Y 5)( Y 2 + Y 4 + Y 5)  Y 5 ]  Y1[ Y1( Y 2 + Y 4 + Y 5 )  Y 2 Y 5]  Y 2[ Y 1Y 5 + Y 2( Y1+ Y 3 + Y 5 )] 2 I Y1Y5 + Y2 (Y1 +Y3 +Y5 ) Y V Z 5 = V2  V3 = I Y1Y4  Y4 Y3  Y = I  0.05  103   90  0.05  103   90  Y =0V 31. a. Z1 = Z 2 Z3 Z4 3 4  10 0 4  103 0 ? 3 4  10 90 4  103   90 1 90  1 90 (not balanced) b. The solution to 26(b) resulted in E(Z1Z 5  Z 3 (Z1  Z 2  Z5 ) I3 = I X C = Z where and and 218 ZΔ = (Z1 + Z3 + Z6)[(Z1 + Z2 + Z5)(Z3 + Z4 + Z5)  Z 52 ]  Z1[Z1(Z3 + Z4 + Z5)  Z3Z5]  Z3[Z1Z5 + Z3(Z1 + Z2 + Z5)] Z1 = 5 k 0, Z2 = 8 k 0, Z3 = 2.5 k 90 Z4 = 4 k 90, Z5 = 5 k 90, Z6 = 1 k 0 I X C = 1.76 mA 71.54 CHAPTER 17 c. The solution to 26(c) resulted in I Y1Y5 + Y2 (Y1 +Y3 +Y5 ) V3 = VX C = Y YΔ = (Y1 + Y2 + Y6)[(Y1 + Y3 + Y5)(Y2 + Y4 + Y5)  Y52 ]  Y1 [Y1(Y2 + Y4 + Y5) + Y2Y5]  Y2[Y1Y5 + Y2(Y1 + Y3 + Y5)] Y1 = 0.2 mS 0, Y2 = 0.125 mS 0, Y3 = 0.4 mS 90 Y4 = 0.25 mS 90, Y5 = 0.2 mS 90 where with Y6 = 1 mS 0, I = 10 mA 0 V3 = VX C = 7.03 V 18.46 Source conversion: and 32. Z1Z4 = Z3Z2   (R1  jXC) Rx + jX Lx = R3R2  XC =  1 1 = 1 k = 3 C (10 rad/s)(1  F) (1 k  j1 k) Rx + jX Lx = (0.1 k)(0.1 k) = 10 k and Rx + jX Lx =  Rx = 5 , Lx = 33. 10  103  10  103 = 5 Ω + j5 Ω = 1  103  j1  103 1.414  103   45 X Lx  = 5 = 5 mH 10 rad/s 3 1 1 1 = k = C (1000 rad/s)(3  F) 3 1 Z1 = R1  X C1 90 = (2 k 0)   k 90 = 328.8  80.54 3 Z2 = R2 0 = 0.5 k 0, Z3 = R3 0 = 4 k 0 Z4 = Rx + j X Lx = 1 k + j6 k X C1 = Z1 = Z 2 Z3 Z4 328.8    80.54 0.5 k 0 ? 4 k 0 6.083  80.54 82.2  10380.54  82.2  103 80.54 (balanced) 34. Apply Eq. 17.6. CHAPTER 17 219 35. For balance: R1(Rx + j X Lx ) = R2(R3 + j X L3 ) R1Rx + jR1 X Lx = R2R3 + jR2 X L3 R2 R3 R1 and R1ωLx = R2ωL3  R1Rx = R2R3 and Rx = R1 X Lx = R2 X L3 so that Lx = 36. R2 L3 R1 Z1 = 8  90 = j8  Z2 = 4  90 = +j4  Z3 = 8  90 = +j8  Z4 = 6  90 = j6  Z5 = 5  0 a. Z1Z 2 = 5  38.66 + Z1 Z 2 + Z5 Z1Z 5 = 6.25  51.34 Z7 = Z1 + Z 2 + Z5 Z 2 Z5 Z8 = = 3.125  128.66 Z1 + Z 2 + Z5 Z = Z7 + Z3 = 3.9  + j3.12  = 4.99  38.66 Z = Z8 + Z4 = 1.95   j3.56  = 4.06  118.71 Z  Z = 10.13  67.33= 3.90   j9.35  ZT = Z6 + Z  Z = 7.80   j6.23  = 9.98  38.61 120 V 0 E = = 12.02 A 38.61 I= ZT 9.98    38.61 Z6 = 37. 220 12   j 9  = 4   j3  ZY = Z  = 3 3 CHAPTER 17 ZT = 2  + 4  + j3  + [4   j3  + j3 ]  [4   j3  + j3 ] = 6   j3  + 2  = 8   j3  = 8.544  20.56 E 60 V 0 = 7.02 A 20.56 I= = 8.544    20.56 ZT ZΔ = 3ZY = 3(3  90) = 9  90 Z = 9  90  (12   j16 ) = 9  90  20  53.13 = 12.96  67.13 38. 2Z 2 2 2 = Z = [12.96  67.13] = 8.64  67.13 3 Z + 2Z 3 100 V 0 E I= = 11.57 A 67.13 = ZT 8.64  67.13 ZT = Z  2Z = 39. ZΔ = 3ZY = 3(5 ) = 15  Z1 = 15  0  5  90 = 4.74  71.57 Z2 = 15  0  6  90 = 5.57  68.2 = 2.07  + j5.17  Z3 = Z1 = 4.74  71.57 = 1.5   j4.5  ZT = Z1 (Z2 + Z3) = (4.74  71.57)  (2.07  + j5.17  + 1.5   j4.5 ) = (4.74  7.57)  (3.63  10.63) = 2.71  23.87 100 V 0 E = = 36.9 A 23.87 I= ZT 2.71    23.87 CHAPTER 17 221 Chapter 18 1. Z1 = 3  0, Z2 = 8  90, Z3 = 6  90 Z2  Z3 = 8  90  6  90 = 24  90 30 V 30 E1 = = 1.24 A 112.875 Z1 + Z 2  Z 3 3   j 24  Z3 I (6    90)(1.24 A 112.875) = 3.72 A 67.125 I = = 2  90 Z 2 + Z3 I= Z1 Z2 = 3  0  8  90 = 2.809  20.556 60 V 10 E2 = I= Z3 + Z1  Z 2  j 6  + 2.630  + j 0.986  = 10.597 A 72.322 I = Z1 I (3  0)(10.597 A 72.322) = 3.721 A 2.878 = Z1 + Z 2 3  + j8  I L1 = I + I = 3.72 A 67.125 + 3.721 A 2.878 = 1.446 A  j3.427 A + 3.716 A + j0.187 A = 5.162 A  j3.24 A = 6.09 A 32.12 2. 222 Z1 = 8  90, Z2 = 5 Ω 90 I = 0.3 A 60, E = 10 V 0 Z2 I (8  90)(0.3 A 60) 2.4 A 150 I =  = Z1 + Z 2 + j8   j 5  3 90 = 0.8A 60 E 10 V 0 10 A 0 I =  = Z1 + Z 2 + j8   j 5  3 90 = 3.33 A 90 IC = I  I = 0.8 A 60  3.33 A 90 = (0.4 A + j0.69 A) + j3.33 A = 0.4 A + j4.02 A = 4.04 A 84.32 CHAPTER 18 3. E: Z1 = 3  90, Z2 = 7  90 E = 10 V 90 Z3 = 6  90, Z4 = 4  0 Z = Z1 (Z3 + Z4) = 3  90  (4   j6 ) = 3  90  7.21  56.31 = 4.33  70.56 ZE V1 = Z + Z 2 (4.33  70.56)(10 V 90) = (1.44  + j 4.08 )  j 7 43.3 V 160.56 = = 13.28 V 224.31 3.26   63.75 13.28 V 224.31 I = V1 = 3  90 Z1 = 4.43 A 134.31 I: CDR: Z = Z3 + Z1 Z2 = j6  + 3  90  7  90 = j6  + 5.25  90 = j6  + j5.25  = j0.75  = 0.75 90 Z4 I (4  0)(0.6 A 120) 2.4 A 120 = = I3 = 4   j 0.75  4.07   10.62 Z 4 + Z = 0.59 A 130.62 Z2I3 (7    90)(0.59 A 130.62) 4.13 A 40.62 = = I =  j7  + j3  4   90 Z 2 + Z1 = 1.03 A 130.62 IL = I  I (direction of I) = 4.43 A 134.31  1.03 A 130.62 = (3.09 A + j3.17 A)  (0.67 A + j0.78 A) = 2.42 A + j2.39 A = 3.40 A 135.36 CHAPTER 18 223 4. AC: 1 1 1   2 fC C (1000)(4.7  F) = 212.77  XL = 2fL = L = (1000)(47 mH) = 47  XC = Z1 = 212.77  90, Z2 = 47   0, Z3 = 22  + j47  = 51.89  64.92 Z2 Z3 = 29.23  30.66 ZT = Z1 + Z2 Z3 = j212.77  + 25.14  + j14.91  = 25.14   j197.86  = 199.45  82.76 Is = E 20 V 60 = 0.1 A 142.76  ZT 199.45    82.76 Z3I S (51.89  64.92)(0.1 A 142.76) 5.19 A 207.68   22   j 47   47  83.49 34.26 Z3  Z 2 I = 62.16 mA  173.42 3 and i = 62.16  10 sin (1000t + 173.42) I= DC: 5V 5V  22   47  69  = 72.46 mA I= i = 72.46 mA + 62.16  10 5. sin (1000t + 173.42) DC: AC: 224 3 (6  0)(I ) 6  + 3   j1  (6  0)(4 A 0) = 9   j1  24 A 0 = 9.055  6.34 = 2.65 A 6.34 IC = CHAPTER 18 VC = ICXC = (2.65 A 6.34)(1  90) = 2.65 V 83.66 = 12 V + 2.65 V 83.66 υC = 12 V + 3.75 sin(ωt  83.66) 6. E = 20 V 0 Z1 = 10 k 0 Z2 = 5 k  j5 k = 7.071 k 45 Z3 = 5 k 90 I = 5 mA 0 Z = Z1  Z2 = 10 k 0  7.071 k 45 = 4.472 k 26.57 (CDR) (4.472 k   26.57)(5 mA 0) 22.36 mA   26.57 ZI = = 4 k  j 2 k + j5 k 5 36.87 Z + Z 3 = 4.472 mA 63.44 I = Z = Z2  Z3 = 7.071 k 45  5 k 90 = 7.071 k 45 ZE (7.071 k 45)(20 V 0) 141.42 V 45 = =  (5 k + j 5 k) + (10 k) 15.81 18.435 Z + Z1 = 8.945 V 26.565 V  8.945 V 26.565 = = 1.789 mA 63.435 = 0.8 mA  j1.6 mA I = 5 k 90 Z3 I = I + I = (2 mA  j4 mA) + (0.8 mA  j1.6 mA) = 2.8 mA  j5.6 mA = 6.26 mA 63.43 (VDR) 7. V = Z1 = 20 k 0 Z2 = 10 k 90 I = 2 mA 0 E = 10 V 0 I = CHAPTER 18 Z1 (hI ) (20 k  0)(100)(2 mA 0) = 0.179 A 26.57 = Z1 + Z 2 20 k + j10 k 225 E 10 V 0 = 22.36 k  26.57 Z1 + Z 2 = 0.447 mA 26.57 IL = I  I (direction of I) = 179 mA 26.57  0.447 mA 26.57 = 178.55 mA 26.57 I = V: 8. Z1 = 5 k 0, Z2 = 1 kΩ 90 Z3 = 4 k 0 V = 2 V 0, μ = 20 VL =  Z 3(  V ) (4 k 0)(20)(2 V 0) = 17.67 V 6.34 = 5 k  j1 k + 4 k Z1 + Z 2 + Z 3 I: CDR: I = Z1I Z1 + Z 2 + Z 3 (5 kΩ 0)(2 mA 0) 9.056 kΩ   6.34 = 1.104 mA 6.34 = VL = IZ3 = (1.104 mA 6.34)(4 k 0) = 4.416 V 6.34 VL = VL + VL = 17.67 V 6.34  4.416 V 6.34 = 22.09 V 6.34 9. Z1 = 20 k 0 Z2 = 5 k + j5 k I = Z1 (hI ) (20 k 0)(100)(1 mA 0) = 78.45 mA 11.31 = Z1 + Z 2 20 k + 5 k + j 5 k V (20)(10 V 0) Z1 + Z 2 25.495 k 11.31 = 7.845 mA 11.31 I = = IL = I  I (direction of I) = 78.45 mA 11.31  7.845 mA 11.31 = 70.61 mA 11.31 226 CHAPTER 18 Z1 = 2 k 0, Z2 = 2 k 0 VL = ILZ2 IL = hI + I = (h + 1)I VL = (h + 1)IZ2 and by KVL: VL = IZ1 + E V E so that I = L Z1 10. V  E VL = (h + 1)IZ2 = (h + 1)  L  Z2  Z1  Subt. for Z1, Z2 VL = (h + 1)(VL  E) VL(2 + h) = E(h + 1) (h + 1) 51 VL = E= (20 V 53) = 19.62 V 53 (h + 2) 52 11. I1: I1 = 1 mA 0 Z1 = 2 k 0 Z2 = 5 k 0 KVL: V1  20 V  V = 0 I = Z V1 21 V  I = or V = 1 I Z1 Z1 21 V1 = 21 V V = I5Z2 = [I1  I]Z2 Z1 I = I1Z2  IZ2 21 Z  I 1 + Z 2  = I1Z 2 21   Z2 and I = [I1] = Z1 + Z2 21 CHAPTER 18 ( 5 k  0)(1 mA 0) = 0.981 mA 0  2 k  0    + 5 k  0 21   227 I2: V1 = 20 V + V = 21 V Z V 21 V  V = 1 I I = 1 = Z1 Z1 21 Z1 V I5 = = I Z 2 21 Z 2 I = I2  I5 = I2  Z1 I 21 Z 2  Z  I 1 + 1  = I2  21 Z 2  2 mA 0 I2 = 1.963 mA 0 I = = Z1 2 k 1+ 1+ 21(5 k) 21 Z 2 I = I + I = 0.981 mA 0 + 1.963 mA 0 = 2.94 mA 0 12. E1: 10 V 0  I 10   I 2   4 Vx = 0 with Vx = I 10  Solving for I: 10 V 0 = 192.31 mA 0 I= 52  Vs = 10 V 0  I(10 ) = 10 V  (192.31 mA 0)(10  0) = 8.08 V 0 228 CHAPTER 18 I:  Ii   Io Vx 5 Vx + = 10  2  5 A + 0.1 Vx + 2.5 Vx = 0 2.6 Vx = 5 A 5 Vx =  V = 1.923 V 2.6 Vs = Vx = (1.923 V) = 1.923 V 0 Vs = Vs  Vs = 8.08 V 0 + 1.923 V 0 = 10 V 0 5 A 0 + 13. ZTh: Z1 = 3  0, Z2 = 4  90 E = 100 V 0 ZTh = Z1  Z2 = (3  0  4  90) = 2.4  36.87 = 1.92  + j1.44  ETh: Z 2E (4  90)(100 V 0) = Z 2 + Z1 5  53.13 = 80 V 36.87 ETh = 14. ZTh: ZTh = Z3 + Z1  Z2 = +j6 k + (2 k 0  3 k 90) = +j6 k + 1.664 k 33.69 = +j6 k + 1.385 k j0.923 k = 1.385 k + j5.077 k = 5.26 k 74.74 ETh: ETh = = CHAPTER 18 Z 2E (3 k   90)(20 V 0) = Z 2 + Z1 2 k  j 3 k 60 V   90 = 16.64 V 33.69 3.606   56.31 229 15. From #31. ZTh = Z1  Z2 ZTh = ZN = 21.31  32.2 ETh = IZ = IZTh = (0.1 A 0)(21.31  32.12) = 2.13 V 32.2 16. From #31. ZTh = ZN = 6.81 Ω 54.23 = 3.98   j5.53  Z1 = 2  0, Z3 = 8  90 Z2 = 4  90, Z4 = 10  0 E = 50 V 0 ETh = V2 + V4 Z2 E V2 = Z 2 + Z1  (Z3 + Z 4 ) ( 4  90)(50 V 0) = + j 4  + 2  0  (10   j8 ) = 47.248 V 24.7 V1 = E  V2 = 50 V 0  47.248 V 24.7 = 20.972 V 70.285 Z 4 V1 (10  0)(20.972 V   70.285) = V4 = = 16.377 V 31.625 10   j8  Z 4 + Z3 ETh = V2 + V4 = 47.248 V 24.7 + 16.377 V 31.625 = (42.925 V + j19.743 V) + (13.945 V  j8.587 V) = 56.870 V + j11.156 V = 57.95 V 11.10 17. ZTh: Z1 = 10  0, Z2 = 8  90 Z3 = 8  90 ZTh = Z3 + Z1  Z2 = j8  + 10  0  8  90 = j8  + 6.247  51.34 = j8  + 3.902  + j4.878  = 3.902   j3.122  = 5.00  38.66 230 CHAPTER 18 ETh: Superposition: (E1) (8  90)(120 V 0) 10  + j8  960 V 90 = 12.806 38.66 = 74.965 V 51.34 ETh = (I) ETh = VZ 2 + VZ3 = IZ3 + I(Z1  Z2) = I(Z3 + Z1  Z2) = (0.5 A 60)(j8  + 10  0  8  90) = (0.5 A 60)(j8  + 3.902  + j4.878 ) = (0.5 A 60)(3.902 Ω  j3.122 Ω) = (0.5 A 60)(4.997 Ω 38.663) = 2.499 V 21.337 ETh = ETh + ETh = 74.965 V 51.34 + 2.449 V 21.337 = (46.83 V + j58.538 V) + (2.328 V + j0.909 V) = 49.158 V + j59.447 V = 77.14 V 50.41 18. ZTh: ZTh = Z = 10   j10  = 14.14  45 ETh: CHAPTER 18 ETh = E  VZ = 20 V 40  IZ = 20 V 40  (0.6 A 90)(14.14  45) = 20 V 40  8.484 V 45 = (15.321 V + j12.856 V)  (6 V + j6 V) = 9.321 V + j6.856 V = 11.57 V 36.34 231 19. a. AC: ETh: ETh = 1   90  212.77    90 C Z3 = 22  + L 90 = 22  + j47  = 51.89   64 Z1 = Z 3E (51.89  64.92)(20 V 60) = 6.21 V  207.36  22   j 47   j 212.77  Z3  Z1 ZTh: (212.77    90)(51.89  64.92)  j 212.77   22   j 47  = 66.04  57.36  = 35.62  + j55.61  ZTh = Z1  Z2 = DC: ETh: ETh = 5 V RTh: RTh = 22  232 CHAPTER 18 b. AC: ETh ZTh  R L 6.21 V  207.36 = 35.62   j 55.61   47  6.21 V  207.36 = 82.62   j 55.61  6.21 V 207.36 = 99.59  33.94 = 62.36 mA 173.42 I= DC: 5V 5V  22   47  69  = 72.46 mA I= i = 72.46 mA + 62.36  103 sin (1000t + 173.42) matching the results of Problem 4. 20. a. ZTh: ZTh = Z R1  Z R2 = 6  + 3  = 9  DC: ETh = 12 V AC: ETh = IZ R1 = (4 A  0)(6  0) = 24 V  0 ETh = 12 V + 24 V  0 (DC) (AC) CHAPTER 18 233 DC: VC = 12 V ZC E AC: VC = ZC + Z RTh b. (1    90)(24 V 0)  j1  + 9  24 V   90 = 9.055   6.34 VC = 2.65 V 83.66 = υC = 12 V+ 2.65 V 83.66 = 12 V + 3.75 sin(ωt  83.66) 21. a. ZTh: 1 Z1 = 10 k 0 5 Z2 = 5 k  j5 k = 7.071 k 45 ZTh = Z1  Z2 = (10 k 0)  (7.071 k 45) = 4.47 k 26.57 Source conversion: E1 = (Iθ)(R10) = (5 mA 0)(10 k 0) = 50 V 0 ETh = Z 2 (E + E1 ) Z 2 + Z1 (7.071 k    45)(20 V 0 + 50 V 0) (5 k   j 5 k ) + (10 k ) (7.071 k    45)(70 V 0) = (15 k   j 5 k ) 494.97 V   45 = 15.811   18.435 = 31.31 V 26.57 = b. I= ETh 31.31 V   26.565 = ZTh + Z L 4.472 k    26.565 + 5 k  90 31.31 V   26.565 31.31 V   26.565 = 4 k   j 2 k  + j5 k  4 k  + j3 k  31.31 V   26.565 = = 6.26 mA 63.44 5 k  36.87 = 234 CHAPTER 18 22. Z1 = 10 k  0 Z2 = 10 k  0 Z3 = 1 k  90 ZTh = Z3 + Z1  Z2 = 5 k  j1 k  5.1 k 11.31 ETh: (VDR) 23. ETh = Z 2 (20 V ) (10 k  0)(20 V ) = 10 V = Z 2 + Z1 10 k  + 10 k  ZTh: 0 Z1 = 40 k 0 .2 Z 2 = 0.2 k 90 Z 3 = 5 k 0 ZTh = Z3  (Z1 + Z2) = 5 k 0  (40 k  j0.2 k) = 4.44 k 0.03 I = Z1 (100 I ) Z1 + Z 2 + Z3 (40 k  0)(100 I ) 45 k    0.255 = 88.89 I 0.255 = ETh = IZ3 = (88.89 I 0.255)(5 k 0) = 444.45  103 I 0.26 24. ZTh:  ZTh = Z1 = 20 k 0 ETh: ETh = (hI)(Z1) = (100)(2 mA  0)(20 k  0) = 4 kV 0 CHAPTER 18 235 E: ETh = ETh + ETh = 4 kV 0 + 10 V 0 = 3990 V 0 25. ZTh: Z1 = 5 k 0 Z2 = j1 ZTh = Z1 + Z2 = 5 k  j1 k = 5.10 k 11.31 ETh: ETh     V + VZ1  =   V  IZ1 = (20)(2 V 0) − (2 mA 0)(5 k 0) = −50 V 0 26. ZTh: Z1 = 20 k  0 Z2 = 5 k  0 ZTh = Z1 + Z2 = 25 k  0 ETh: ETh = V  (hI)(Z1) = (20)(10 V  0)  (100)(1 mA 0)(20 k  0) = 1800 V 0 236 CHAPTER 18 27. ETh: (Eoc) hI = I Z1 = 2 k 0 I = 0 and hI = 0 with Eoc = ETh = E = 20 V 53 Isc: Isc = (h + 1)I = (h + 1)(10 mA 53) = 510 mA 53 ZTh = 28. Eoc 20 V 53 = 39.22  0 (negative resistance) = I sc 510 mA 53 ETh: Eoc = 21 V Z1 = 5 k 0 V = I1Z1 = (1 mA 0)(5 k 0) = 5 V 0 Eoc = ETh = 21(5 V 0) = 105 V 0 V = I2Z1 = (2 mA 0)(5 k 0) = 10 V 0 Eoc = ETh = V + 20 V = 21 V = 210 V 0 Isc: Isc = I1 20 V = V  V = 0 V and I = 0 A  Isc = I2 Isc = Isc + Isc = 3 mA 0 Eoc = Eoc + Eoc = 315 V 0 = ETh 315 V 0 = 105 k 0 ZTh = Eoc = 3 mA 0 I sc CHAPTER 18 237 29. Eoc: (ETh) KVL: 6 Ix(2 k)  Ix(1 k) + 8 V 0  Ix(3.3 k) = 0 8 V 0 = 0.491 mA 0 Ix = 16.3 k  Eoc = ETh = Ix(3.3 k) = 1.62 V 0 Isc: 8V = 2.667 mA 0 3k E 1.62 V 0 = 607.42  0 ZTh = oc = 2.667 mA 0 I sc Isc = 30. From Problem 13: ZN = ZTh = 1.92  + j1.44  = 2.4  36.87 I N: Z1 = 3  0, Z2 = 4  90 E 100 V 0 Isc = IN = = 3  0 Z1 = 33.33 A 0 238 CHAPTER 18 31. Z1 = 20  + j20  = 28.284  45 Z2 = 68  0 ZN = Z1  Z2 = (28.284  45)  (68  0) = 21.31  32.2 Isc = I = IN = 0.1 A 0 32. From Problem 17: ZN = ZTh = 5.00  38.66 I N: Superposition: (E1) ZT = Z1 + Z2  Z3 = 10  + 8  90  8  90 64  0 = 10  + 0 = very large impedance E Is = =0A ZT and V Z1 = 0 V with V Z2 = V Z3 = E1 = 120 V 0 120 V 0 so that Isc = E1 =    90 8 Z3 = 15 A 90 (I) Isc = I = 0.5 A 60 IN = Isc + Isc = + j15 A + 0.5 A 60 = + j15 A + 0.25 A + j0.433 A = 0.25 A + j15.433 A = 15.44 A 89.07 CHAPTER 18 239 33. a. ZN: E = 20 V 0, I2 = 0.4 A 20 Z1 = 6  + j8  = 10  53.13 Z2 =   j12  = 15  53.13 ZN = Z1  Z2 = (10  53.13)  (15  53.13) = 9.66  14.93 I N: (E) (I2) Isc = E/Z1 = 20 V 0/10  53.13 Isc = I2 = 0.4 A 20 = 2 A 53.13 IN = Isc + Isc = 2 A 53.13 + 0.4 A 20 = 2.15 A 42.87 34. ZN: E1 = 120 V 30, Z1 = 3  0 Z2 = 8   j8 , Z3 = 4  90 ZN = Z3 + Z1  Z2 = 4  90 + (3  0)  (8   j8 ) = 4.37  55.67 = 2.47  + j3.61  I N: I= E1 120 V 30 = ZT Z1 +Z 2  Z3 120 V 30 3  + (8   j8 )  4  90 120 V 30 = 6.65  46.22 = 18.05 A 16.22 = (I ) (8   j8 )(18.05 A   16.22) = 22.83 A 34.65 Isc = IN = Z 2 = 8   j8  + j 4  Z 2 + Z3 240 CHAPTER 18 35. a. Z1 = 212.77    90 Z3 = 22  + j47  = 51.89   64 AC: I N: E 20 V 60 = 94 mA  150  Z1 212.77    90 ZN = ZTh (problem 19) = 66.04 57.36 = 35.62  + j55.61  IN = DC: I N: IN = 5V = 227.27 mA 22  RN = RTh = (problem 19) = 22  b. AC: I N: Z N (I N ) (66.04  57.36)(94 mA 150)  35.62   j 55.61   47  Z N  47  6.21 A 207.36 = 62.68 mA 173.22 = 99.08 34.14 I= DC: I= 3 and i = 72.46 mA + 62.68  10 Same as Problem 4 and 19. CHAPTER 18 22 (227.27 mA) = 72.46 mA 22   47  sin (1000t + 173.22) 241 36. a. From #20 ZN = ZTh = 9  0 DC: IN = E RT = 12 V = 1.33 A 9 AC: IN = = R1 I (6  0)(4 A 0) = R1 + R2 9  0 24 V 0 = 2.67 A 0 9   0 IN = 1.33 A + 2.67 A 0 b. DC: VC = IR = (1.33 A)(9 Ω) = 12 V AC: Z = 9  0  1  90 = 0.994  83.66 VC = IZ = (2.667 A 0)(0.994  83.66) = 2.65 V 83.66 VC = 12 V + 2.65 V 83.66 37. a. Note Problem 21(a): ZN = ZTh = 4.47 k 26.57 Using the same source conversion: E1 = 50 V 0 Defining ET = E1 + E = 50 V 0 + 20 V 0 = 70 V 0 Z1 = 10 k 0 Z2 = 5 k  j5 k = 7.071 k 45 Isc = ET 70 V 0 = 7 mA 0 = Z1 10 k  0 IN = Isc = 7 mA 0 242 CHAPTER 18 b. ( ) (4.472 k    26.565)(7 mA 0) I = ZN IN = Z N + Z L 4.472 k    26.565 + 5 k  90 31.30 mA   26.565 31.30 mA   26.565 = = 4  j 2 + j5 4 + j3 31.30 mA   26.565 = = 6.26 mA 63.44 as obtained in Problem 21. 5 36.87 38. ZN: Z1 = 10 k 0, Z2 = 10 k 0 Z3 = j1 k ZN = Z3 + Z1  Z2 = 5 k  j1 k = 5.1 k 11.31 I N: V2 = (Z 2  Z 3 )20 V (Z 2  Z 3 ) + Z1 (0.995 k    84.29)(20 V) 0.1 k   j 0.99 k  + 10 k  V2 = 1.961 V 78.69 = IN = Isc = 39. ZN: V2 1.961 V   78.69 = 1.96  103 V 11.31 = Z3 1 k    90 Z1 = 40 k 0, Z2 = 0.2 k 90 Z3 = 5 k 0 ZN = Z3  (Z1 + Z2) = 5 k 0  (40 k  j0.2 k) = 4.44 k 0.03 I N: Z1 (100 I ) Z1 + Z 2 ( 40 k 0)(100 I ) = 40 k   0.286 = 100 I 0.29 IN = Isc = CHAPTER 18 243 40. ZN: Z1 = 5 k 0, Z2 = 1 k 90  ZN = Z1 + Z2 = 5 k  j1 k = 5.1 k 11.31 . IN: V (20)(2 V 0) Z1 + Z 2 5.1 k   11.31 = 7.843 mA 11.31 Isc = (I): = Z1 (I ) Z1 + Z 2 (5 k 0)(2 mA 0) = 5.1 k   11.31 = 1.96 mA 11.31 Isc = IN = Isc + Isc = 7.843 mA 11.31 + 1.96 mA 11.31 = 9.81 mA 11.31 41. ZN: Z1 = 20 k 0, Z2 = 5 k 0 V = 10 V 0, µ = 20, h = 100 I = 1 mA 0 ZN = Z1 + Z2 = 25 k 0 IN: (hI) Z1 (hI ) Z1 + Z 2 (20 k 0)(hI ) = 20 k 0 + 5 k0 = 80 mA 0 Isc = (µV) Isc = V = Z1 + Z 2 = 8 mA 0 (20)(10 V 0) 25 k IN (direction of Isc) = Isc  Isc = 80 mA 0  8 mA 0 = 72 mA 0 244 CHAPTER 18 42. Z1 = 2 k 0 Z2 = 5 k 0 I2 = I3 + I5 V = I5Z2 = (I2  I3)Z2 Eoc = ETh = 21 V = 21(I2  I3)Z2  E  = 21 I 2  oc  Z 2 Z1    Z2  Eoc 1 + 21  = 21 Z2I2 Z1   21 Z 2 I 2 21(5 k  0)(2 mA 0) = Eoc = Z  5 k 0  2 1 + 21 1 + 21  Z1  2 k 0  ETh = Eoc = 3.925 V 0 20 V  V  V = 0 and IN = Isc = I2 = 2 mA 0 3.925 V 0 ZN = Eoc = = 1.96 k 2 mA 0 I sc 43. Z1 = 1 k 0 Z2 = 3 k 0 Z3 = 4 k 0 Eoc 21 Eoc V = I = I1 + I2, I1 = Z1 21 Z1 V2 = 21 V = Eoc  V = CHAPTER 18 245 I2 = and  1 Eoc Eoc E 1  , I = I1 + I2 = + oc = Eoc  +  Z2 21 Z1 Z2  21 Z1 Z 2   + 21 Z1  I = Eoc  Z 2   21 Z1Z 2  21 Z1Z 2 I (21)(1 k  0)(3 k  0)(2 mA 0) = Z 2 + 21Z1 3 k  + 21(1 k  0) ETh = Eoc = 5.25 V 0 Eoc = Z V3 21 V  V = 3 Isc = Z3 Z3 21 V = I1Z1 I = I1 + I Isc = Isc =  Z + Z3  Z 2 I  I =  2  I sc Z 2 + Z3  Z2   Z3 Z + Z3  V  Z 2 + Z3  + 2 +  I sc =   I sc Z1  Z 2  Z2   21 Z1 I 2 mA 0 = 0.79 mA 0 Isc = = 4 k 7 k Z3 Z3 + Z 2 + + 21 k  3 k  Z2 21 Z1 I = I1 + I =  IN = 0.79 mA 0 E 5.25 V 0 ZN = oc = = 6.65 k 0 I sc 0.79 mA 0 44. Z1 = 3  + j4 , Z2 = j6   ZTh = Z1  Z2 = 5  53.13  6  90 = 8.32  3.18 ZL = 8.32  3.18 = 8.31   j0.46  246 CHAPTER 18 ETh = Z2 E Z 2 + Z1 (6    90)(120 V 0) 3.61    33.69 = 199.45 V 56.31 2 (3.124 V ) 2 = 1198.2 W Pmax = E Th = 4RTh 4(8.31 ) = Z1 = 3  + j4  = 5  53.13 Z2 = 2 Ω 0  ZN = ZTh = Z1  Z2 = 5  53.13  2  0 10  53.13 = 2 + 3 + j4 10 Ω 53.13 = 5 + j4 10  53.13 = 6.403 38.66 = 1.56  14.47 ZTh = 1.56  14.47 = 1.51  + j0.39  ZL = 1.51   j0.39  45. ETh = I(Z1 Z2) = (2 A 30)(1.562  14.47) = 3.12 V 44.47 2 (3.12 V ) 2 = 1.61 W Pmax = E Th = 4RTh 4(1.51 ) 46. ZTh: Z1 = 4  90, Z2 = 10  0 Z3 = 5  90, Z4 = 6  90 E = 60 V 60 ZTh = Z4 + Z3  (Z1 + Z2) = j6  + (5  90)  (10  + j4 ) = 2.475   j4.754  = 11.04  77.03 ZL = 11.04  77.03 CHAPTER 18 247 ETh: ETh = Z 3 (E ) Z3 + Z1 + Z 2 ( 5    90)(60 V 60)  j5  + j 4  + 10  = 29.85 V 24.29 = 2 / 4 RTh = (29.85 V)2/4(2.475 ) = 90 W Pmax = E Th 47. Z1 = 3  + j4  = 5  53.13 Z2 = j8  Z3 = 12  + j9  ZTh = Z2 + Z1  Z3 = j8  + (5  53.13)  (15  36.87) = 5.71  64.30 = 2.475   j5.143  ZL = 5.71  64.30 = 2.48 Ω + j5.15  ETh + V Z3  E2 = 0 ETh = E2  VZ3 Z3 (E 2  E1 ) Z3 + Z1 = 168.97 V 112.53 VZ3 = ETh = E2  VZ3 = 200 V 90  168.97 V 112.53 = 78.24 V 34.16 2 / 4 RTh = (78.24 V)2/4(2.475 ) = 618.33 W Pmax = E Th 48. 248 E  0 1 V  0 = 1 mA 0 = R10 1 k  0 ZTh = 40 k 0 ETh = (50 I)(40 k 0) = (50)(1 mA 0)(40 k 0) = 2000 V 0 2 ( 2 kV ) 2 = 25 W Pmax = E Th = 4 RTh 4(40 k ) I= CHAPTER 18 49. ETh: Z1 = 2 k 0 Z2 = = 3 k 90 Z3 = 6 k 90 Z2E (3 k 90)(20 V 0)  Z 2  Z1  j 3 k  2 k 60 V   90 = 16.62 V 33.69  3.61   56.31 ETh  ZTh:  ZTh  Z3  Z1  Z 2 (2 k 0)(3 k   90) 2 k  j 3 k   j 6 k  1.66 k   33.69 ZTh = +j6 k +   j 6 k  1.38 k  j 920.8   1.38 k  j 5.08 k  5.26 k   74.80 ZL = 5.36 k 74.80 = 1.38k  j5.08 k b. 50. Pmax = 2 ETh (16.62 V) 2 = 50.04 mW  4 RTh 4(1.38 k) From #20, ZTh = 9 , ETh = 12 V + 24 V 0 a.  ZL = 9  b. 2 (12 V ) 2 (24 V ) 2 Pmax = E Th = = 4 W + 16 W = 20 W + 4 RTh 4(9 ) 4(9 ) or ETh = V 02 + V12eff 2 = 26.833 V 2 (26.833 V ) and Pmax = E Th = = 20 W 4 RTh 4(9 ) CHAPTER 18 249 51. 52. a. Problem 21(a): ZTh = 4.47 k 26.57 = 4 k  j2 k ZL = 4 k + j2 k ETh = 31.31 V 26.57 b. 2 Pmax = E Th / 4 RTh = (31.31 V)2/4(4 k) = 61.27 mW a. ZTh = 2 k 0  2 k 90 = 1 k  j1 k R L = R Th +  X Th + X Load  2 2 = (1 k) 2 + (1 k + 2 k) 2 = (1 k) 2 + (1 k) 2 = 1.41 k 53. b. Rav = (RTh + RLoad)/2 = (1 k + 1.41 k)/2 = 1.21 k 2 (50 V ) 2 = 516.53 mW Pmax = E Th = 4 Rav 4(1.21 k) a. ZTh: 1 1 = 2 fC 2 (10 kHz)(4 nF)  3978.87  XL = 2fL = 2(10 kHz)(30 mH)  1884.96  Z1 = 1 k 0, Z2 = 1884.96  90 Z3 = 3978.87  90 ZTh = (Z1 + Z2)  Z3 = (1 k + j1884.96 )  3978.87  −90) = 2133.79  62.05  3978.87  −90) = 3658.65  36.52 XC =  ZL = 3658.65  36.52 = 2940.27  − j2177.27  1 1 = 7.31 nF C= = 2 fX C 2 (10 kHz)(2177.27 ) b. RL = RTh = 2940.27  c. ETh  Z3 (E) (3978.87    90)(2 V0) = = 3.43 V−25.53) Z3  Z1  Z 2 1 k + j1884.96   j 3978.87 2 / 4 RTh = (3.43 V) 2 /4(2940.27 Ω) = 1 mW Pmax = ETh 250 CHAPTER 18 (4 k 0)(4 mA 0) = 1.33 mA 0 4 k + 8 k Vab = (Iab)(8 k 0) = 10.67 V 0 54. Iab = 55. a. 4 k(E) 1 = (20 V 0) 4 k + 12 k 4 = 5 V 0 5 V 0 = 0.83 mA 0 I= 6 k V= b. 6 k ( E ) 1 = (20 V 0) 2 6 k + 6 k = 10 V 0 10 V 0 = 0.83 mA 0 I= 12 k V= 56. 100 V 0 = 50 mA 0 2 k   0 50 V 0 I2 = 4 k  90 = 12.5 mA 90 Z1 = 2 k 0 Z2 = 4 k 90 Z3 = 4 k 90 IT = I1  I2 = (50 mA 0  12.5 mA 90) = 50 mA + j12.5 mA = 51.54 mA 14.04 Z = Z1  Z2 = (2 k 0)  (4 k 90) = 1.79 k 26.57 ZIT (1.79 k  26.57)(51.54 mA 14.04) IC = = 1.6 k + j 0.8 k  j 4 k Z + Z 3 = 25.77 mA 104.04 I1 = CHAPTER 18 251 Chapter 19 1. a. PT = 60 W + 45 W + 25 W = 130 W b. QT = 0 VARS, ST = PT = 130 VA c. 130 VA S = 0.542 A ST = EIs, Is = T = E 240 V d. P 60 W = 204.2  (0.542 A ) 2 V = IsR = (0.542 A)(204.2 ) = 110.68 V V1 = V2 = E  V = 240 V  110.68 V = 129.32 V 2 2 (129.32 V ) 2 V P1 = V 1 , R1 = 1 = = 371.6  45 W R1 P1 P = I s2 R, R = 2 Is 2 = 2 2 (129.32 V ) V V2 , R2 = 2 = = 668.9  25 W R2 P2 V 129.32 V V 129.32 V = 0.193 A = 0.348 A, I2 = 2 = I1 = 1 = R1 371.6  R2 668.9  P2 = e. 2. a. ZT = 3   j5  + j9  = 3  + j4  = 5  53.13 50 V 0 E I= = 10 A 53.13 = ZT 5  53.13 R: L: C: P = I2R = (10 A)2 3  = 300 W P=0W P=0W b. R: C: L: Q = 0 VAR QC = I2XC = (10 A)2 5  = 500 VAR QL = I2XL = (10 A)2 9  = 900 VAR c. R: C: L: S = 300 VA S = 500 VA S = 900 VA d. PT = 300 W QT = QL  QC = 400 VAR(L) ST = 2 2 PT + QT = EI = (50 V)(10 A) = 500 VA 300 W = 0.6 lagging Fp = PT = S T 500 VA e. 252  CHAPTER 19 f. WR =  VI   VI  VI VI : WR = 2   = 2  = f1  f2   2 f1  f1 V = IR = (10 A)(3 ) = 30 V (30 V)(10 A) =5J WR = 60 Hz 3. g. VC = IXC = (10 A)(5 ) = 50 V VI (50 V)(10 A) WC = = 1.33 J = 1 (2 )(60 Hz) VL = IXL = (10 A)(9 ) = 90 V VI (90 V)(10 A) = 2.39 J WL = = 1 376.8 a. PT = 0 + 100 W + 300 W = 400 W QT = 200 VAR(L)  600 VAR(C) + 0 = 400 VAR(C) ST = Fp = b. 5. a. PT 400 W = 0.707 (leading) = ST 565.69 VA  PT = EIs cos θT 400 W = (100 V)Is(0.7071) 400 W = 5.66 A Is = 70.71 V Is = 5.66 A 135 c. 4. PT2 + QT2 = 565.69 VA PT = 600 W + 500 W + 100 W = 1200 W QT = 1200 VAR(L) + 600 VAR(L)  1800(C) = 0 VAR ST = PT = 1200 VA b. 1200 W Fp = P T = =1 S T 1200 VA c.  d. Is = a. PT = 200 W + 100 W + 0 + 50 W = 350 W QT = 50 VAR(L) + 100 VAR(L)  200 VAR(C)  400 VAR(C) = 450 VAR(C) ST = CHAPTER 19 S T 1200 VA = 6 A, 1  0 = E 200 V Is = 6 A 0 2 2 PT + QT = 570.09 VA 253 6. 350 W PT = 0.614 (leading) = S T 570.09 VA b. Fp = c.  d. PT = EIs cos θT 350 W = (50 V)Is(0.614) 350 W = 11.4 A Is = 30.7 V Is = 11.4 A 52.12 a. IR = b. c. 60 V 30 = 3 A 30 20  0 P = I2R = (3 A)2 20  = 180 W QR = 0 VAR S = P = 180 VA 60 V 30 = 6 A 60 10  90 PL = 0 W QL = I2XL = (6 A)2 10  = 360 VAR(L) S = Q = 360 VA IL = PT = 180 W + 400 W = 580 W QT = 600 VAR(L) + 360 VAR(L) = 960 VAR(L) ST = (580 W) 2 + (960 VAR) 2 = 1121.61 VA 580 W Fp = PT = = 0.517 (lagging) θ = 58.87 1121.61 VA ST d. ST = EIs 1121.61 VA S = 18.69 A Is = T = E 60 V  I s = 30  58.87 = 28.87 Is = 18.69 A 28.87 2 2 7. a. b. (20 V) R: P = E = = 200 W R 2 PL,C = 0 W R: Q = 0 VAR C: 2 (20 V)2 = 80 VAR(C) QC = E = 5 XC L: (20 V ) QL = E = = 100 VAR(L) 4 XL 2 254 2 CHAPTER 19 S = 200 VA S = 80 VA S = 100 VA c. R: C: L: d. PT = 200 W + 0 + 0 = 200 W QT = 0 + 80 VAR(C) + 100 VAR(L) = 20 VAR(L) ST = (200 W)2 + (20 VAR) 2 = 200 VA 200 W Fp = PT = = 0.995 (lagging)  5.73 S T 200.998 VA 8. e.  f. Is = a. S T 200.998 VA = = 10.05 A E 20 V Is = 10.05 A5.73 50 V 60 = 10 A 6.87 5  53.13 PR = I2R = (10 A)2 3  = 300 W PL = 0 W PC = 0 W R  L: I= b. QR = 0 VAR QL = I2XL = (10 A)2 4  = 400 VAR 50 V 60 = 5 A 150 IC = 10    90 QC = I2XC = (5 A)2 10 Ω = 250 VAR c. SR = P = 300 VA SL = QL = 400 VA SC = QC = 250 VA d. PT = PR = 300 W QT = 400 VAR(L)  250 VAR(C) = 150 VAR(L) ST = (300 W) 2 + (150 VAR) 2 = 335.41 VA 300 W Fp = PT = = 0.894 (lagging) S T 335.41 VA e. f.  S T 335.41 VA = = 6.71 A E 50 V 0.894  26.62 lagging θ = 60  26.62 = 33.38 Is = 6.71 A 33.38 Is = CHAPTER 19 255 9. ac. XL = ωL = (400 rad/s)(0.1 H) = 40  1 1 XC = = C (400 rad/s)(100  F) = 25  Z1 = 40  90, Z2 = 25  90 Z3 = 30  0 ZT = Z1 + Z2  Z3 = +j40  + (25  90)  (30  0) = +j40  + 19.21  50.19 = +j40  + 12.3   j14.76  = 12.3  + j25.24  = 28.08  64.02 Is = E 50 V 0 = 1.78 A 64.02 28.08  64.02 = ZT V2 = Is(Z2  Z3) = (1.78 A 64.02)(19.21  50.19) = 34.19 V 114.21 34.19 V   114.21 I2 = V 2 = = 1.37 A 24.21 25    90 Z2 34.19 V   114.21 = 1.14 A 114.21 I3 = V 2 = 30  0 Z3 d. Z1: P = 0 W, QL = I s2 X L = (1.78 A)2 40  = 126.74 VAR(L), S = 126.74 VA Z2: P = 0 W, QC = I 22 X C = (1.37 A)2 25  = 46.92 VAR(C), S = 46.92 VA Z3: P = I 32 R = (1.14 A)2 30  = 38.99 W, QR = 0 VAR, S = 38.99 VA PT = 0 + 0 + 38.99 W = 38.99 W QT = +126.74 VAR(L)  46.92 VAR(C) + 0 = 79.82 VAR(L) 2 2 PT + QT = 88.83 VA ST = 38.99 W Fp = PT = = 0.439 (lagging) S T 88.83 VA e.  f. WR = f1 = g. 1 400 rad/s = 63.69 Hz = 2 6.28 WL = WC = 256 VR I R V2 I 3 (34.19 V)(1.14 A) = 0.31 J = = 2f1 2 f1 2(63.69 Hz) VL I L = (Is X L )Is VC I C = V2 I 2 1 1 1 1 = = I s2 X L 1 = (1.78 A) 2 40  = 0.32 J 400 rad/s (34.19 V)(1.37 A) = 0.12 J 400 rad/s CHAPTER 19 10. a. b. 11. 12. a. S T 10,000 VA = = 50 A E 200 V 0.5  60 leading  Is leads E by 60 E 200 V 0 ZT = = 4 Ω 60 = 2   j3.464  = R  jXC = 50 A 60 Is Is = Fp = PT  PT = FpST = (0.5)(10,000 VA) = 5000 W ST S T 5000 VA = = 41.67 A E 120 V Fp = 0.8  36.87 (lagging) E = 120 V 0, I = 41.67 A 36.87 E 120 V 0 = 2.88  36.87 = 2.30  + j1.73  = R + jXL Z= = I 41.67 A   36.87 I= b. P = S cos θ = (5000 VA)(0.8) = 4000 W a. PT = 0 + 300 W = 300 W QT = 600 VAR(C) + 200(L) = 400 VAR(C) 2 2 PT + QT = 500 VA P 300 W Fp = T = = 0.6 (leading) ST 500 VA ST = b. ST 500 VA = = 16.67 A E 30 V Fp = 0.6  53.13 Is = 16.67 A 53.13 Is = c.  d. Load: 600 VAR(C), 0 W R = 0, L = 0, QC = I2XC  XC = QC 2 = 600 VAR = 2.159  (16.67 A)2 I Load: 200 VAR(L), 300 W C = 0, R = P/I2 = 300 W/(16.67 A)2 = 1.079  Q 200 VAR = 0.7197  XL = 2L = (16.67 A) 2 I ZT = j2.159  + 1.0796  + j0.7197  = 1.08   j1.44  CHAPTER 19 257 13. a. PT = 0 + 300 W + 600 W = 900 W QT = 500 VAR(C) + 0 + 500 VAR(L) = 0 VAR ST = PT = 900 VA Fp = PT = 1 ST b. Is = c.  S T 900 VA = = 9 A, Is = 9 A 0 E 100 V d. 2 Z1: Z3: 14. a. 4 100 V 0 = 5A 90 Z1 20    90 I2 = Is  I1 = 9 A  j5 A = 10.296 A 29.05 P 300 W 300 = 2.83  R = 2= = 2 106 (10.296 A) I XL,C = 0  P 600 W = 5.66  R= 2= I 2 (10.296 A)2 Q 500 XL = 2 = = 4.72 , XC = 0  I 2 (10.296 A)2 I1 = Z2: 2 QC = V  X C = V = 10 = 20  XC QC 500 E = PT = 200 W + 30 W + 0 = 230 W QT = 0 + 40 VAR(L) + 100 VAR(L) = 140 VAR(L) ST = 2 2 PT + QT = 269.26 VA 230 W = 0.854 (lagging)  31.35 Fp = PT = S T 269.26 VA b. 258 S T 269.26 VA = 2.6926 A = E 100 V Is = 2.69 A 31.35 Is = CHAPTER 19 c. 2 Z2: Z3: 15. a. 4 R = V = 10 = 50  P 200 XL,XC = 0  100 V 0 = 2 A 0 I1 = 50  0 I2 = Is  I1 = 2.6926 A 31.35  2 A 0 = 2.299 A  j1.40 A  2.0 A = 0.299 A  j1.40 A = 1.432 A 77.94 Z1: 30 W P Q 40 VAR = 14.63 , XL = 2 = = 19.50  = 2 2 I 2 (1.432 A) I 2 (1.432 A)2 XC = 0  R= XL = Q 100 VAR = 48.76 , R = 0 , XC = 0  = I 22 (1.432 A)2 PT = 100 W + 1000 W = 1100 W QT = 75 VAR(C) + 2291.26 VAR(C) = 2366.26 VAR(C) ST = 2 2 PT + QT = 2609.44 VA 1100 W = 0.422 (leading)  65.04 Fp = PT = S T 2609.44 VA b. 2609.44 VA ST = EI  E = S T = = 521.89 V I 5A E = 521.89 V 65.07 c. I Z1 = S S 125 VA = 0.2395 A = = V1 E 521.89 V I Z2 = S S 2500 VA = 4.79 A = = V2 E 521.89 V CHAPTER 19 259 Z1: R= 100 W P = 1743.38  = 2 I Z1 (0.2395)2 Q = I Z21 X C  X C = Z2: R= XC = 16. P I Z21 X C = Q I Z21 X C Q 75 VAR = 1307.53  = 2 I Z1 (0.2395 A)2 1000 W = 43.59  (4.790 A) 2 = 2291.26 VAR = 99.88  (4.790 A) 2 a. 0.7  45.573 P = S cos θ = (10 kVA)(0.7) = 7 kW Q = S sin θ = (10 kVA)(0.714) = 7.14 kVAR(L) b. QC = 7.14 kVAR = V2 XC (208 V) 2 V2 = 6.059  = Q C 7.14 kVAR 1 1 1  C= = 438 μF XC = = 2 fC 2 fX C (2π)(60 Hz)(6.059 ) XC = c. Uncompensated: 10,000 VA S = 48.08 A Is = T = E 208 V Compensated: 7,000 W S = 33.65 A Is = T = P T = E E 208 V d. 260 cos θ = 0.9 θ = cos10.9 = 25.842 x tan θ = 7 kW x = (7 kW)(tan 25.842) = (7 kW)(0.484) = 3.39 kVAR y = (7.14  3.39) kVAR = 3.75 kVAR CHAPTER 19 QC = 3.75 kVAR = XC = C= V2 XC (208 V)2 V2 = 11.537  = QC 3.75 kVAR 1 1 = 230 μF = 2 fX C (2π)(60 Hz)(11.537 ) Uncompensated: Is = 48.08 A Compensated: ST = (7 kW) 2 + (3.39 kVAR) 2 = 7.778 kVA S T 7.778 kVA = = 37.39 A E 208 V Is = 48.08 A  37.39 A = 10.69 A Is = 17. a. PT = 5 kW, QT = 6 kVAR(L) ST = 2 2 PT + QT = 7.81 kVA b. 5 kW = 0.640 (lagging) Fp = PT = S T 7.81 kVA c. Is = d. XC = S T 7,810 VA = 65.08 A = E 120 V 2 1 (120 V) 2 , QC = I2XC = E = 2 fC XC XC and e. (120 V)2 14, 400 = 2.4  = 6000 QC 1 1 = 1105 μF C= = 2 fX C (2 )(60 Hz)(2.4 ) XC = ST = EIs = PT 5000 W = 41.67 A  Is = P T = E 120 V CHAPTER 19 261 18. a. Load 1: Load 2: P = 20,000 W, Q = 0 VAR θ = cos10.7 = 45.573 x 10 kW x = (10 kW)tan 45.573 = (10 kW)(1.02) = 10,202 VAR(L) tan θ = Load 3: θ = cos10.85 = 31.788 x 5 kW x = (5 kW)tan 31.788 = (5 kW)(0.62) = 3098.7 VAR(L) tan θ = PT = 20,000 W + 10,000 W + 5,000 W = 35 kW QT = 0 + 10,202 VAR + 3098.7 VAR = 13,300.7 VAR(L) ST = b. QC = QL = 13,300.7 VAR 2 (103 V)2 XC = E = = 75.184  Q C 13,300.7 VAR C= c. 2 2 PT + QT = 37,442 VA = 37.442 kVA 1 1 = 35.28 μF = 2 fX C (2π)(60 Hz)(75.184 ) Uncompensated: 37.442 kVA S Is = T = = 37.44 A E 1 kV Compensated: ST = PT = 35 kW 35 kW S = 35 A Is = T = E 1 kV ⌬Is = 37.44 A  35 A = 2.44 A 19. 262 a. ZT = R1 + R2 + R3 + jXL  jXC = 2  + 3  + 1  + j3   j12  = 6   j9  = 10.82  56.31 50 V 0 E = = 4.62 A 56.31 I= ZT 10.82    56.31 CHAPTER 19 b. 20. 21. P = VI cos θ = (50 V)(4.62 A) cos 56.31 = 128.14 W a-b: P = I2R = (4.62 A)2 2  = 42.69 W b-c: P = I2R = (4.62 A)2 3  = 64.03 W a-c: 42.69 W + 64.03 W = 106.72 W a-d: 106.72 W c-d: 0 W d-e: 0 W f-e: P = I2R = (4.62 A)2 1  = 21.34 W a. ST = 660 VA = EIs 660 VA = 5.5 A Is = 120 V θ = cos10.6 = 53.13  E = 120 V 0, Is = 5.5 A 53.13 P = EI cos θ = (120 V)(5.5 A)(0.6) = 396 W Wattmeter = 396 W, Ammeter = 5.5 A, Voltmeter = 120 V b. ZT = a. R= E 120 V 0 = 21.82  53.13 = 13.09  + j17.46  = R + jXL = I 5.5 A   53.13 P I 2 b. R= c. R= XL 49.75  = 132.03 mH = 2 f (2π)(60 Hz) P I 2 P I 2 22. a. = 90 W = 10  (3 A) 2 = E 200 V 60 W = 100  = 15 , ZT = = 2 I 2A (2 A) ZT2  R 2 = (100 Ω) 2  (15 Ω)2 = 98.87  XL = L= E 200 V 80 W = 50  = 5 Ω, ZT = = 2 I 4A (4 A ) ZT2  R 2 = (50 ) 2  (5 Ω) 2 = 49.75  XL = L= = XL 98.87  = 262.39 mH = 2 f 376.8 XL = 2πfL = (6.28)(50 Hz)(0.08 H) = 25.12  R 2 + X L2 = (4 Ω) 2 + (25.12 )2 = 25.44  ZT = 60 V = 2.358 A Z T 25.44  P = I2R = (2.358 A)2 4  = 22.24 W I= CHAPTER 19 E = 263 b. P 30 W = 2.07 A = 7 R E 60 V = 28.99  ZT = = I 2.07 A I= XL = L= c. 28.13 Ω XL = = 89.54 mH 2 f (2π)(50 Hz) P = I2R = (1.7 A)2 10  = 28.9 W E 60 V = 35.29  ZT = = I 1.7 A XL = L= 264 (28.99 Ω) 2  (7 Ω) 2 = 28.13  (35.29 Ω)2  (10 Ω) 2 = 33.84  XL 38.84  = 107.77 mH = 2 f 314 CHAPTER 19 Chapter 20 1. a. ωs = s 250 rad/s = 39.79 Hz = 2 2 fs = b. ωs = 3. 1 = 3496.50 rad/s (0.51 H)(0.16  F) s 3496.50 rad/s = 556.49 Hz = 2 2 fs = 2. 1 1 = = 250 rad/s LC 1 H)(16  F) 1 = 22,173 rad/s (0.27 mH)(7.5  F)  22,173 rad/s = 3528.93 Hz fs = s = 2 2 c. ωs = a. XC = 30  d. VR = IR = (25 mA)(10 ) = 250 mV = E VL = IXL = (25 mA)(30 ) = 750 mV VC = IXC = (25 mA)(30 ) = 750 mV VL = VC b. ZTs = 2  c. e. 30  Qs = X L = = 15 (med Q) R 2 a. XL = 2 k b. I= c. VR = IR = (120 mA)(100 ) = 12 V = E VL = IXL = (120 mA)(2 k) = 240 V VC = IXC = (120 mA)(2 k) = 240 V VL = VC = 20 VR d. Qs = e. XL = 2πfL, L = f. I= E 50 mV = = 25 mA 2 ZTs P = I2R = (25 mA)2 2  = 1.25 mW E 12 V = 120 mA = ZTs 100  X L 20000  = = 20 (high Q) R 100  XL 2 k = 63.7 mH = 2 f 2 (5 kHz) 1 1 1 ,C= = = 15,920 pF XC = 2 fC 2 fX C 2 (5 kHz)(2 k) CHAPTER 20 265 f. g. 4. f s 5 kHz = 250 Hz = 20 Qs BW 0.25 kHz = 5 kHz + = 5.13 kHz 2 2 BW 0.25 kHz = 5 kHz  = 4.88 kHz f1 = fs  2 2 f2 = fs + 1  L= 1 1 = 3.91 mH = 2 (2 f s ) C (2 1.8 kHz ) 2 2  F a. fs = b. XL = 2πfL = 2π(1.8 kHz)(3.91 mH) = 44.2  1 1 = = 44.2  XC = 2 fC 2 (1.8 kHz)(2  F) XL = XC c. Erms = (0.707)(20 mV) = 14.14 mV E 14.14 mV = 3.01 mA Irms = rms = R 4.7  d. P = I2R = (3.01 mA)2 4.7  = 42.58 μW e. ST = PT = 42.58 μVA g. h. 266 BW = 2 LC f. Fp = 1 44.2  Qs = X L = = 9.4 R 4.7  f 1.8 kHz = 191.49 Hz BW = s = 9.4 Qs 2 1  R 1 R 4   +  +   2  2 L 2  L  LC    2  1  4.7  1  4.7   4   + + =   2  2(3.91 mH) 2  3.91 mH  (3.91 mH)(2  F)    1 601.02  11.324  103  =  2  = 1897.93 Hz 2 1  R 1 R 4    f1 = + +   2  2 L 2  L  LC    1  601.02  11.324  103  =  2  = 1.71 kHz 1 1 PHPF = Pmax = (42.58 µW) = 21.29 μW 2 2 f2 = CHAPTER 20 5. 6. a. BW = fs/Qs = 6000 Hz/15 = 400 Hz b. f2 = fs + c. Qs = d. PHPF = a. L= b. f2 = fs + BW/2 = 10,000 Hz + 250 Hz/2 = 10,125 Hz f1 = fs  BW/2 = 10,000 Hz  125 Hz = 9,875 Hz c. d. 7. BW = 6000 Hz + 200 Hz = 6200 Hz 2 BW f1 = fs  = 6000 Hz  200 Hz = 5800 Hz 2 XL  XL = QsR = (15)(3 ) = 45  = XC R 1 1 1 Pmax = (I2R) = (0.5 A)2 3 = 375 mW 2 2 2 200  XL = 3.185 mH = 2 f 2 (104 Hz) R 5 = BW =  250 Hz 2 L 2 (3.185 mH) 200  f 10,000 Hz = 40, BW = s = = 250 Hz or Qs = X L = X C = R R 5 40 Qs 200  = 40 Qs = X L = R 5 E 0 30 V 0 = 6 A 0, VL = (I 0)(XL 90) = R0 5  0 = (6 A 0)(200  90) = 1200 V 90 VC = (I 0)(XC 90) = 1200 V 90 I= e. P = I2R = (6 A)2 5  = 180 W a. BW = b. Qs = X L  XL = QsR = (10)(2 ) = 20  R c. L= fs  Qs = fs/BW = 2000 Hz/200 Hz = 10 Qs XL 20  = = 1.59 mH 2 f (6.28)(2 kHz) 1 1 = 3.98 μF C= = 2 fX C (6.28)(2 kHz)(20 ) CHAPTER 20 267 8. 9. d. f2 = fs + BW/2 = 2000 Hz + 100 Hz = 2100 Hz f1 = fs  BW/2 = 2000 Hz  100 Hz = 1900 Hz a. BW = 6000 Hz  5400 Hz = 600 Hz b. BW = fs/Qs  fs = QsBW = (9.5)(600 Hz) = 5700 Hz c. Qs = d. L= XL 19  = 0.53 mH = 2 f 2 (5700 Hz) C= 1 1 = 1.47 μF = 2 fX C 2 (5.7 kHz)(19 ) XL  XL = XC = QsR = (9.5)(2 ) = 19  R E E 5V = 10  R= = R 500 mA IM BW = fs/Qs  Qs = fs/BW = 8400 Hz/120 Hz = 70 X Qs = L  XL = QsR = (70)(10 Ω) = 700  R XC = XL = 700  XL 700  = 13.26 mH L= = 2 f (2π)(8.4 kHz) IM = C= 1 1 = 27.07 nF = 2 fX C (2π)(8.4 kHz)(0.7 k ) f2 = fs + BW/2 = 8400 Hz + 120 Hz/2 = 8.46 kHz f1 = fs  BW/2 = 8400 Hz  60 Hz = 8.34 kHz 10. Qs = X L  XL = QsR = 20(2 Ω) = 40  = XC R f BW = s  fs = QsBW = (20)(400 Hz) = 8 kHz Qs 40  = 795.77 H L= XL = 2 f 2 (8 kHz) 1 1 C= = 497.36 nF = 2 fX C 2 (8 kHz)(40 ) f2 = fs + BW/2 = 8000 Hz + 400 Hz/2 = 8200 Hz f1 = fs  BW/2 = 8000 Hz  200 Hz = 7800 Hz 11. a. b. 268 fs = s 2  106 rad/s = = 1 MHz 2 2 f 2  f1 = 0.16  BW = f2  f1 = 0.16 fs = 0.16(1 MHz) = 160 kHz fs CHAPTER 20 c. 12. 2 2 (120 V)2 = 720  P= VR R= VR= R P 20 W R R 720  BW = = 0.716 mH  L= = 2 L 2 BW (6.28)(160 kHz) 1 1 1 fs = C= = 35.38 pF = 2 6 2 2 4 f s L 4 (10 Hz ) 2(0.716 mH) 2 LC d. 2 f s L 2 (106 Hz)(0.716 mH) X Q = X L = 80  R = L = = 56.23  = 80 80 80 R a. Q = XL R X 2 fL 2 (1MHz)(100  H) R = L = = 50.27  = 12.5 Q Q f 2  f1 1 = = 0.2 fs Qs X 1 2 fL 2 (1 MHz)(100  H) 628.32  = = 5= L = = 0.2 R R R R 628.32  = 125.66 R= 5 R = Rd + R Qs = 125.66  = Rd + 50.27  and Rd = 125.66   50.27  = 75.39  13. c. XC = a. fp = 1 = XL 2 fC 1 1 C= = 253.3 pF = 2 fX C 2 (1 MHz)(628.32 ) 1 2 LC = 2 = 159.16 kHz 2 (0.1 mH)(10 nF) b. c. IL = 4V 4V VL = = 40 mA = X L 2 f p L 100  IC = 4V 4V VL = 40 mA = = X C 1/ 2 f p C 100  CHAPTER 20 269 14. 15. d. 2 k 2 k = Qp = R s = = 20 X Lp 2 f p L 100  a. fs = b. Q = c. Since Q  10, fp  fs = 13.4 kHz d. XL = 2πfpL = 2π(13.4 kHz)(4.7 mH) = 395.72  1 1 = 395.91  = XC = 2 f p C 2 (13.4 kHz)(30 nF) XL = XC e. ZTp = Q2 R = (49.46)2 8  = 19.57 k f. VC = IZ T p = (10 mA)(19.57 k) = 195.7 V g. Q  10, Qp = Q = 49.46 f p 13.4 kHz BW = = 270.9 Hz = 49.46 Qp h. IL = IC = Q IT = (49.46)(10 mA) = 494.6 mA a. fs = b. Q = 1 2 LC = 1 = 13.4 kHz 2 (4.7 mH)(30 nF) X L 2 fL 2 (13.4 kHz)(4.7 mH) = 49.46  10 (yes) = = 8 R R 1 2 LC = 1 = 1.027 MHz 2 (200  H)(120  F) XL X 2 (1.027 MHz)(200  H) = 86.04   R  L  15 R Q Zp= Q2 R  (15) 2 86.04  = 19.36 k c. 270 2 2 P = I R = (120 mA) (950.9 ) = 13.69 W CHAPTER 20 d. XL = 2 fL  2 (1.027 MHz)(200  H) = 1.291 k 86.04 0)(114.1 V) VR = = 7.587 V 86.04   j1.291 k P = VR2 / R  (7.587 V)2 86.04  = 669 mW 13.69 W: 669 mW  20:1 16. a. Q =  X L 100  = 5  10 = RL 20  R2 + X 2 (20 )2 + (100 ) 2 1 XL = 104   X = = = C 100  XL R2 + X L2 X C b. + 10,400  = 342.11  ZT = Rs  Rp = Rs  R  X L = 1000   20 R c. E = IZTp = (5 mA 0)(342.11  0) = 1.711 V 0 2 IC = 2 E 1.711 V 0 = 16.45 mA 90 = X C   90 104    90 ZL = 20  + j100  = 101.98  78.69 E 1.711 V 0 = 16.78 mA 78.69 IL = = Z L 101.98  78.69 d. e. 17. 100  = 795.77 H L= XL = 2 f 2 (20 kHz) 1 1 = 76.52 nF C= = 2 fX C 2 (20 kHz)(104 ) R 342.11  = 3.29 = XC 104  BW = fp/Qp = 20,000 Hz/3.29 = 6079.03 Hz Qp =  2  106  2  Hz  (1 mH)  2  X X 2000   = 57.14  Q  35  L  R  L  35 35 35 R Q  10 : Q p  CHAPTER 20 fp BW  2  106 / 2 Hz  20 100,000 Hz 271 R  Q2 R R  (35) 2 57.14   Q p  20  XL 2000 And 40,000 = R  70, 000 So R = 93.33 k  use R = 91 k (standard value) 1 1  Q p  10, X C  X L  2000   2 fC  2  106  2  Hz  C  2  C = 250 pF  use C = 240 pF (standard value) 18. a. fs = 1 2 LC = 1 = 102.73 kHz 2 (80  H)(0.03  F) 2 (1.5 )2 0.03  F C = 102.73 kHz(.99958) fp = fs 1  R  = 102.73 kHz 1  L 80  H = 102.69 kHz 1  2C  fm = fs 1   R   = 102.73 kHz(0.99989) = 102.72 kHz 4 L  Since fs  fp  fm  high Qp b. XL = 2πfpL = 2π(102.69 kHz)(80 μH) = 51.62  1 1 = = 51.66  XC = 2 f p C 2 (102.69 kHz)(0.03  F) XL  XC c. ZTp = Rs  Q2 R 51.62  = 34.41 Q = X L = R 1.5  ZTp = 10 k   (34.41) 21.5  = 10 k   1.776 k  = 1.51 k R s  Q  R  Z T p 1.51 k  = 29.25 = = XL X L 51.62  f 102.69 kHz BW = p = = 3.51 kHz 29.25 Qp 2 d. Qp = e. Converting the voltage source to a current source: E 100 V = 10 mA Is =  Rs 10 k And Rs = Rp = 10 k R s I s = 10 k (10 mA) = 8.49 mA Then IT = 2 R s + Q  R  10 k  + 1.78 k  IC = IL  Q IT = (34.41)(8.49 mA) = 292.14 mA 272 CHAPTER 20 19. f. VC = IZTp = (10 mA)(1.51 k) = 15.1 V a. fs = 1 2 LC = 1 = 7.12 kHz 2 (0.5 mH)(1  F) R2 C (8 Ω) 2 (1  F) = 7.12 kHz 1  = 7.12 kHz(0.9338) = 6.65 kHz 0.5 mH L fp = fs 1  1  (8 Ω) 2 (1  F)  1  R 2C  fm = fs 1     = 7.12 kHz 1    = 7.12 kHz (0.9839) 4  0.5 mH  4 L  = 7.01 kHz Low Qp b. c. d. XL = 2πfpL = 2π(6.647 kHz)(0.5 mH) = 20.88  1 1 = = 23.94  XC = 2 fC 2 (6.647 kHz)(1  F) XC > XL (low Q) 2 2 2 2 R + X L = 500   (8 ) + (20.88 ) = 500   62.5  ZTp = Rs  Rp = Rs   8 R = 55.56  ZT p Qp = X Lp BW = e. fp Qp = 55.56  = 2.32 23.94  = 6.647 kHz = 2.87 kHz 2.32 One method: VC = IZTp = (40 mA)(55.56 ) = 2.22 V IC = IL = 20. VC 2.22 V = 92.73 mA = X C 23.94  VC 2.22 V 2.22 V = 99.28 mA = = R  + jX L 8 + j 20.88 22.36  f. VC = 2.22 V a. ZT p = R2 + X L2 = 50 k R (50 )2 + X L2 = (50 k)(50 ) XL = CHAPTER 20 250  104  2.5  103 = 1580.3  273 b. Q= X L 1580.3 = 31.61  10 = 50 R  XC = XL = 1580.3  21. XL 1580.3 Ω = 15.72 kHz = 2 L 2π(16 mH) c. XL = 2πfpL  fp = d. XC = a. Q = 20 > 10  fp = fs = b. Q = 1 1 1  C= = 6.4 nF = 2 f p C 2 f s X C 2 (15.72 kHz)(1580.3 ) 1 2 LC = 1 = 3558.81 Hz 2 (200 mH)(10 nF) X L 2 fL 2 fL 2 (3558.81 Hz)(0.2 H)  R = = 223.61  = = R R 20 Q ZTp = Rs  R p = Rs  Q2 R = 40 k  (20)2 223.61  ZTp = 27.64 k Converting the voltage source to a current source: E 200 V = 5 mA Is   Rs 40 k Rp = Rs = 40 k VC = IZTp = (5 mA)(27.64 k) = 138.2 V c. P = I2R = (5 mA)227.64 k = 691 mW d. Qp = BW = 22. f p Qp = 3558.81 Hz = 575.86 Hz 6.18 a. Ratio of XC to R suggests high Q system.  XL = 400  = XC b. Q = c. Qp = BW = 274 27.64 k  R Rs  R p = 6.18 = = 2 (3558.81 Hz)(0.2 H) XL XL X L 400  = 50 = R 8 R XL fp Qp = Rs  R p XL = Rs  Q 2R XL = 20 k   (50)2 8  10 k  = 25 = 400  400   fp = QpBW = (25)(1000 Hz) = 25 kHz CHAPTER 20 23. d. V C max = IZTp = (0.1 mA)(10 k) = 1 V e. f2 = fp + BW/2 = 25 kHz + a. XC = 1 kHz = 25.5 kHz 2 1 kHz f1 = fp  BW/2 = 25 kHz  = 24.5 kHz 2 R2 + X L2  X L2  XLXC + R2 = 0 XL X L2  100 XL + 144 = 0 XL = (  100)  (100) 2  4(1)(144) 2 104  576 = 50   48.54  2 XL = 98.54  or 1.46  = 50   b. c. Q = Qp = X L 98.54  = 8.21 = R 12  Rs  R p X Lp R2 + X L2 (12 ) 2 + (98.54 ) 2 40 k  R 12  = 100  XC 40 k  = 40 k  821.18  804.66  = = 8.05 100  100  BW = fp/Qp  fp = QpBW = (8.05)(1 kHz) = 8.05 kHz = 24. d. VCmax = IZTp = (6 mA)(804.66 ) = 4.83 V e. f2 = fp + BW/2 = 8.05 kHz + a. 1 kHz = 8.55 kHz 2 1 kHz f1 = fp  BW/2 = 8.05 kHz  = 7.55 kHz 2 1 1 fs = = = 41.09 kHz 2 LC 2 (0.5 mH)(30 nF) fp = fs 1  R2 C (6 )2 30 nF = 41.09 kHz 1  = 41.09 kHz(0.9978) = 41 kHz 0.5 mH L 1  (6 ) 2(30 nF)  1  R 2C  fm = fs 1     = 41.09 kHz 1    = 41.09 kHz(0.0995) 4  0.5 mH  4 L  = 41.07 kHz High Qp CHAPTER 20 275 b. 80 V 0 = 4 mA 0, Rs = 20 k 20 k  0 X 2 fL 2 (41 kHz)(0.5 mH) Q = L = = 21.47 (high Q coil) = 6 R R I= 2 2 R + X L (6 ) 2 + (128.81 ) 2   20 k Rs  R p R = 6 = Qp = 2 2 (6 ) 2 + (128.81 ) 2 R + X L X Lp 128.81  XL 20 k   2.771 k  2.434 k  = = = 18.86 (high Qp) 129.09  129.09  Rs  c. ZTp = Rs  Rp = 20 k  2.771 k = 2.43 k d. VC = IZTp = (4 mA)(2.43 k) = 9.74 V e. BW = f. XC = fp Qp 1 1 = = 129.39  2 fC 2 (41 kHz)(30 nF) V 9.736 V = 75.25 mA IC = C = X C 129.39  IL = 25. 41 kHz = 2.17 kHz 18.86 = 9.736 V 9.736 V VC = = 75.50 mA = R + jX L 6  + j128.81  128.95  2 f p L 2 f p L 2 (20 kHz)(2 mH) Q = X L =  = 3.14   R = 80 Q R R BW = fp/Qp  Qp = fp/BW = 20 kHz/1.8 kHz = 11.11 1 1 1 = 2 C= = 31.66 nF High Q fp  fs = 2 2 4 f p L 4 (20 kHz) 2 2 mH 2 LC Qp = R XC  R = QpXC = Qp 2 f p C = 11.11 = 2.79 k 2 (20 kHz)(31.66 nF) = (80) 3.14  = 20.1 k Rp R (20.1 k)(2.793 k) RsR p R = Rs  Rp =  Rs = = 3.24 k = R p  R 20.1 k  2.793 k Rs + R p Rp = 276 Q2 R 2 CHAPTER 20 26. VCmax 1.8 V = 9 k I 0.2 mA R p 9 k R Rs  R p R p Qp = = 300  = XC =  XL = = = XL XL XL 30 Qp VCmax  IZTp  ZTp = BW = fp Qp  fp = QpBW = (30)(500 Hz) = 15 kHz L= XL 300  = 3.18 mH = 2 f 2 (15 kHz) C= 1 1 = 35.37 nF = 2 fX C 2 (15 kHz)(300 ) Qp = Q (Rs=  ) = 27. a. = fs = 1 2 LC Q = = X 300 Ω XL  R = L = = 10  30 Qp R 1 2 (200  H)(2 nF) = 251.65 kHz X L 2 (251.65 kHz)(200  H) = 15.81  10 = R 20   fp = fs = 251.65 kHz b. ZTp = Rs  Q2 R = 40 kΩ  (15.81)2 20  = 4.44 k c. Qp = d. BW = e. 20 μH, 20 nF fs the same since product LC the same fs = 251.65 kHz X 2 (251.65 kHz)(20  H) = 1.581 Q = L = 20  R Low Q : Rs  Q 2 R  4.444 k = = 14.05 XL 316.23  fp Qp = 251.65 kHz = 17.91 kHz 14.05 fp = fs 1  R2 C (20 ) 2 (20 nF) = (251.65 kHz) 1  20  H L = (251.65 kHz)(0.775) = 194.93 kHz XL = 2πfpL = 2π(194.93 kHz)(20 µH) = 24.496 Ω R 2 + X L2 (20 )2 + (24.496 ) 2 Rp =  = 50  = 20  R ZTp = Rs  Rp = 40 k  50  = 49.94  CHAPTER 20 277 R 49.94  = 2.04 = X L 24.496  f p 194.93 kHz BW = = = 95.55 kHz 2.04 Qp Qp = f. 0.4 mH, 1 nF fs = 251.65 kHz since LC product the same 2 (251.65 kHz)(0.4 mH) Q = X L = = 31.62  10 20  R  fp = fs = 251.65 kHz ZTp = Rs  Q2 R = 40 k  (31.62)2 20  = 40 k  ( 20 k)  13.33 k Rs  Q2 R 13.33 k  = 21.08 = 632.47  XL f 251.65 kHz = 11.94 kHz BW = p = Qp 21.08 Qp = g. h. L 200  H = 100  103 = C 2 nF L 20  H part (e) = = 1  103 C 20 nF L 0.4 mH part (f) = = 400  103 C 1 nF Network L ratio increased BW decreased. C Also, Vp = IZTp and for a fixed I, ZTp and therefore Vp will increase with increase in the Yes, as L/C ratio. 278 CHAPTER 20 Chapter 21 1. 3  = 0.1875, d2 = 1 16 Value = 103  100.1875"/1" = 103  1.54 = 1.54 kHz 3 right: d1 =  = 0.75, d2 = 1 4 Value = 103  100.75"/1" = 103  5.623 = 5.62 kHz a. left: d1 = b. bottom: top: 5 15  = 0.3125, d2 =  = 0.9375 16 16 Value = 101  100.3125"/ 0.9375" = 101  100.333 = 101  2.153 = 0.22 V 11  = 0.6875, d2 = 0.9375 d1 = 16 Value = 101  100.6875"/ 0.9375" = 101  100.720 = 101  5.248 = 0.52 V d1 = a. 5 b. 4 c. 8 d. 6 e. 1.30 f. 3.94 g. 4.75 h. 0.498 a. 1000 b. 1012 c. 1.59 d. 1.1 e. 1010 f. 1513.56 g. 10.02 h. 1,258,925.41 4. a. 11.51 b. −9.21 c. 5. log10 48 = 1.68 log10 8 + log10 6 = 0.903 + 0.778 = 1.68 6. log10 0.2 = 0.699 log10 18  log10 90 = 1.255  1.954 = 0.699 7. log10 0.5 = 0.30 log10 2 = (0.301) = 0.30 8. log10 27 = 1.43 3 log10 3 = 3(0.4771) = 1.43 2. 3. CHAPTER 21 2.996 d. 9.07 279 9. 280 mW P2 = log10 = log10 70 = 1.85 4 mW P1 a. bels = log10 b. dB = 10 log10 P2 = 10(log10 70) = 10(1.845) = 18.45 P1 10. dB = 10 log10 P 2 P1 100 W 6 dB = 10 log10 P1 0.6 = log10 x 100 W x = 3.981 = P1 100 W P1 = = 25.12 W 3.981 11. dB = 10 log10 12. dBm = 10 log10 dBm = 10 log10 40 W P2 = 10 log10 = 10 log10 20 = 13.01 2W P1 P 1 mW 120 mW = 10 log10 120 = 20.79 1 mW 13. dBv = 20 log10 8.4 V V2 = 20 log10 = 20 log 10 84 = 38.49 0.1 V V1 14. dBυ = 20 log10 V2 V1 22 = 20 log10 Vo 20 mV 1.1 = log10 x Vo 20 mV Vo = 251.79 mV x = 12.589 = 15. 280 P 0.0002  bar 0.001  bar dBs = 20 log10 = 13.98 0.0002  bar 0.016  bar dBs = 20 log10 = 38.06 0.0002  bar Increase = 24.08 dBs dBs = 20 log10 CHAPTER 21 16. 60 dBs  90 dBs quiet loud 60 dBs = 20 log10 P1 = 20 log10x 0.002  bar 3 = log10x x = 1000 90 dBs = 20 log10 P2 = 20 log10y 0.002  bar 4.5 = log10y y = 31.623  103 P1 x 0.002  bar 103 = = P1 = y P2 P 2 31.623  103 0.002  bar and P2 = 31.62 P1 18. b. 19. V2 0.775 V V2 0.4 = log10 0.775 V a. a. 8 dB = 20 log10 V2 = 2.512 0.775 V V2 = (2.512)(0.775 V) = 1.947 V 2 (1.947 V)2 = 6.32 mW P= V = R 600  V2 5 dB = 20 log10 0.775 V V2 0.25 = log10 0.775 V V2 = 0.562 0.775 V V2 = (0.562)(0.775 V) = 0.436 V 2 (0.436 V) 2 V = 0.32 mW = P= 600  R Aυ = Vo = Vi XC R  2 X C2 90 + tan1 XC/R = 1 2  R    1  XC  tan1 R/XC 1 1 = 3617.16 Hz = 2 RC 2 (2.2 k)(0.02  F) V f = f c: Aυ = o = 0.707 Vi fc = CHAPTER 21 281 f = 0.1fc: At fc, XC = R = 2.2 k 1 1 1  1  XC = = =   = 10  2.2 k   = 22 k 2 fC 2 0.1 f c C 0.1  2 f c C  Aυ = f = 0.5fc = 1 fc : 2 XC = Aυ = f = 2fc: XC = Aυ = f = 10fc: XC = Aυ = b. θ = tan1 R/XC f = f c: f = 0.1fc: f = 0.5fc: f = 2fc: f = 10fc: 20. a. 1 2  R    1  XC  = 1 2  2.2 k     +1  22 k   = 1 (.1) 2 + 1 = 0.995  1  1 = 2  = 2  2.2 k  = 4.4 k 2 f cC   fc   2   C  2 1 1 = = 0.894 2 (0.5) 2 + 1  2.2 k     +1  4.4 k   1 = 2 fC 1 1 1  1 =   =  2.2 k   = 1.1 k 2 (2f c )C 2  2 f cC  2 1 2  2.2 k     +1  1.1 k   = 1 (2) 2 + 1 = 0.447 1 1  1  1 =   =  2.2 k   = 0.22 k 2 (10 f c )C 10  2 f cC  10 1 2  2.2 k     +1  0.22 k   = 1 (10) 2 + 1 = 0.0995 θ = tan1 = 45 1 = 5.71 10 1 θ = tan1 2.2 k/4.4 k = tan1 = 26.57 2 θ = tan1 2.2 k/1.1 k = tan1 2 = 63.43 θ = tan1 2.2 k/0.22 k = tan1 10 = 84.29 θ = tan1 2.2 k/22 k = tan1 1 1 = = 15.915 kHz 2 RC 2 ( 1 k )(0.01  F) f = 2fc = 31.83 kHz 1 1 XC = = = 500  2 fC 2 (31.83 kHz)(0.01  F) 500  V XC Aυ = o = = 0.4472 = 2 2 Vi (1 k )2 + (0.5 k ) 2 R X fc = C Vo = 0.4472Vi = 0.4472(10 mV) = 4.47 mV 282 CHAPTER 21 b. 1 1 f c = (15,915 kHz) = 1.5915 kHz 10 10 1 1 XC = = = 10 k 2 fC 2 (1.5915 kHz)(0.01  F) V 10 k  XC = = 0.995 Aυ = o = 2 2 Vi (1 k ) 2 + (10 k ) 2 R + XC f= Vo = 0.995Vi = 0.995(10 mV) = 9.95 mV c. 21. Yes, at f = fc, Vo = 7.07 mV 1 f c , Vo = 9.95 mV (much higher) at f = 10 at f = 2fc, Vo = 4.47 mV (much lower) fc = 500 Hz = 1 1 = 2 RC 2 (1.2 k )C 1 1 = 0.265 μF = 2 Rfc 2 (1.2 k )(500 Hz) V 1 Aυ = o = 2 Vi  R    1  XC  C= At f = 250 Hz, XC = 2402.33  and Aυ = 0.895 At f = 1000 Hz, XC = 600.58  and Aυ = 0.4475 θ = tan1R/XC 1 At f = 250 Hz = fc, θ = 26.54 2 At f = 1 kHz = 2fc, θ = 63.41 CHAPTER 21 283 22. 1 1 = = 67.73 kHz 2 RC 2 (4.7 k )(500 pF) a. fc = b. f = 0.1 fc = 0.1(67.726 kHz)  6.773 kHz 1 1 = = 46.997 k XC = 2 fC 2 (6.773 kHz)(500 pF) V 46.997 k  XC = Aυ = o = = 0.995  1 2 2 Vi R +X (4.7 k )2 + (46.997 k ) 2 C c. f = 10fc = 677.26 kHz 1 1 =  470  XC = 2 fC 2 (677.26 kHz)(500 pF) V 470  XC = = 0.0995  0.1 Aυ = o = 2 2 Vi R + XC (4.7 k ) 2 + (470 ) 2 d. Aυ = Vo = 0.01 = Vi XC 2 R + X C2 X C = 100 X 2 2 R +XC= C 0.01 R2 + X C2 = 104 X C2 R2 = 104 X C2  X C2 = 9,999 X C2 R 4.7 k   47  XC = = 9,999 99.995 1 1 1 XC = = f= = 6.77 MHz 2 fC 2 X CC 2 (47 )(500 pF) 23. a. Aυ = Vo R tan1 XC/R = = 2 2 Vi R  XC fc = f = f c: f = 2fc: X  1+  C   R  2 tan1 XC/R 1 1 = 3.62 kHz = 2 RC 2 (2.2 k)(0.02  F) Vo = 0.707 Vi At fc, XC = R = 2.2 k 1 1 1 1  1 XC = = =   =  2.2 k  = 1.1 k 2 fC 2 (2 f c )C 2  2 f cC  2 Aυ = Aυ = 284 1 1  1.1 k   1+    2.2 k   2 = 0.894 CHAPTER 21 f= 1 f c: 2 f = 10fc:  1  1 = 2  = 2  2.2 k  = 4.4 k f    2 f cC  2  c  C  2 1 Aυ = = 0.447 2  4.4 k   1+    2.2 k   1 1  1  2.2 k  =  = 0.22 k XC = = 2 (10 f c )C 10  2 f cC  10 XC = 1 = 0.995 2  0.22 k   1+    2.2 k    1  1 = 10  XC =  = 10  2.2 k   = 22 k  f   2 f cC  2  c  C  10  1 = 0.0995 Aυ = 2 22 k    1+   2.2 k    Aυ = f= b. f = f c, θ = 45 f = 2fc, θ = tan1 (XC/R) = tan1 1.1 k/2.2 k = tan1 1 f c, 2 θ = tan1 f = 10fc, θ = tan1 f= 4.4 k  = tan1 2 = 63.43 2.2 k  0.22 k  = 5.71 2.2 k  22 k  = 84.29 2.2 k  1 f c, 10 θ = tan1 a. f = fc: Aυ = Vo = 0.707 Vi b. fc = f= 24. 1 f c: 10 1 = 26.57 2 1 1 = = 15.915 kHz 2 RC 2 (10 k )(1000 pF) f = 4fc = 4(15.915 kHz) = 63.66 kHz 1 1 = = 2.5 k XC = 2 fC 2 (63.66 kHz)(1000 pF) V R 10 k  = Aυ = o = = 0.970 (significant rise) 2 2 Vi R  XC (10 k )2 + (2.5 k ) 2 CHAPTER 21 285 25. c. f = 100fc = 100(15.915 kHz) = 1591.5 kHz  1.592 MHz 1 1 = = 99.972  XC = 2 fC 2 (1.592 MHz)(1000 pF) 10 k  R = Aυ = = 0.99995  1 2 2 (10 k ) 2 + (99.972 ) 2 R + XC d. At f = fc, Vo= 0.707Vi = 0.707(10 mV) = 7.07 mV 2 (7.07 mV ) 2 Po = V o =  5 nW R 10 k  Aυ = fc = Vo = Vi 1 X  1+  C   R  2 tan1 XC/R 1 1 1  R= = 795.77  = 2 RC 2 f cC 2 ( 2 kHz)(0.1  F) R = 795.77   750   47   = 797   nominal values 1 = 1996.93 Hz using nominal values  fc = 2 (797 )(0.1  F) At 286 f = 1 kHz, Aυ = 0.458 f = 4 kHz, Aυ  0.9 θ = tan1 X C R f = 1 kHz, θ = 63.4 f = 4 kHz, θ = 26.53 CHAPTER 21 26. 1 1 = = 79.58 kHz 2 RC 2 (100 k )(20 pF) a. fc = b. f = 0.01fc = 0.01(79.577 kHz) = 0.7958 kHz  796 Hz 1 1 = = 9.997 M XC = 2 fC 2 (796 Hz)(20 pF) R 100 k  = Aυ = V o = = 0.01  0 2 2 Vi (100 k ) 2 + (9.997 M ) 2 R + XC c. f = 100fc = 100(79.577 kHz)  7.96 MHz 1 1 = = 999.72  XC = 2 fC 2 (7.96 MHz)(20 pF) V R 100 k  = = 0.99995  1 Aυ = o = 2 2 Vi (100 k ) 2 + (999.72 ) 2 R + XC d. Aυ = Vo R = 0.5 = Vi R 2 + X C2 R 2 + X C2 = 2R R2 + X C2 = 4R2 X C2 = 4R2  R2 = 3R2 3R 2 = 3R = 3 (100 k) = 173.2 k 1 1 1 f= XC = = 2 fC 2 X C C 2 (173.2 k )(20 pF) f = 45.95 kHz XC = 27. a. 1 1 = = 795.77 Hz 2 RC 2 (0.1 k )(2  F) 1 1 = 1.94 Hz f c2 = = 2 RC 2 (10 k )(8200 pF) low-pass section: f c1 = high-pass section: For the analysis to follow, it is assumed (R2 + jX C 2 )  R1  R1 for all frequencies of interest. At f c1 = 795.77 Hz: VR1 = 0.707 Vi X C2 = |Vo | = 1 = 24.39 k 2 fC2 24.39 k (VR1 ) (10 k ) 2 + (24.39 k ) 2 = 0.925 VRi Vo = (0.925)(0.707 Vi) = 0.654 Vi CHAPTER 21 287 At f c2 = 1.94 kHz: Vo = 0.707 VR1 1 = 41  2 fC1 R1Vi 100 (Vi ) = 0.925 Vi VR1 = = (100 ) 2 + (41 ) 2 R12 + X C21 X C1 = |Vo | = (0.707)(0.925 Vi) = 0.64 Vi (1.94 kHz  795.77 Hz) = 1.37 kHz 2 X C 1 = 58.1 , X C 2 = 14.17 k At f = 795.77 Hz + VR1 = Vo = 100 (Vi ) (100 ) 2 + (58.1 )2 = 0.864 Vi   14.17 k  VR1 (10 k ) 2 + (14.17 k ) 2 = 0.817 VR1 Vo = 0.817(0.864 Vi) = 0.706Vi V and Aυ = o = 0.706 ( maximum value) Vi After plotting the points it was determined that the gain should also be determined at f = 500 Hz and 4 kHz: f = 500 Hz: X C1 = 159.15 , X C2 = 38.82 k, VR1 = 0.532 Vi, Vo = 0.968 VR1 f = 4 kHz: Vo = 0.515 Vi X C1 = 19.89 , X C2 = 4.85 kΩ, VR1 = 0.981 Vi, Vo = 0.437 VR1 Vo = 0.429 Vi b. 288 Using 0.707(.706)  0.5 to define the bandwidth BW  3.4 kHz  0.48 kHz = 2.92 kHz and BW  2.9 kHz  2.9 kHz  with fcenter = 480 Hz +   = 1930 Hz 2   CHAPTER 21 28. 1 = 4 kHz 2 R1C1 Choose R1 = 1 k 1 1 C1 = = 39.8 nF  Use 39 nF = 2 f1R1 2 (4 kHz)(1 k ) f1 = 1 = 80 kHz 2 R2C2 Choose R2 = 20 k 1 1 C2 = = 99.47 pF  Use 100 pF = 2 f 2 R2 2 (80 kHz)(20 k ) f2 = 80 kHz  4 kHz = 42 kHz 2 At f = 42 kHz, X C1 = 97.16 , X C2 = 37.89 k Center frequency = 4 kHz + Assuming Z2  Z1 R1 (Vi ) = 0.995Vi |VR1 |= R12 + X C21 |Vo | = X C2 (VR1 ) R22 + X C21 = 0.884Vi Vo = 0.884 VR1 = 0.884(0.995Vi) = 0.88 Vi as f = f1: VR1 = 0.707Vi, X C2 = 221.05 k and Vo = 0.996 VR1 so that Vo = 0.996 VR1 = 0.996(0.707Vi) = 0.704Vi Although Aυ = 0.88 is less than the desired level of 1, f1 and f2 do define a band of frequencies for which Aυ  0.7 and the power to the load is significant. CHAPTER 21 289 29. a. b. c. fs = 1 2 LC = 1 = 98.1 kHz 2 (4.7 mH)(560 pF) XL 2 (98.1 kHz)(4.7 mH) = = 16.84 160  + 12  R + R f 98.1 kHz BW = s = = 5.83 kHz 16.84 Qs Qs = R 160 (1 V) V = 0.93 V and Aυ = o = 0.93 Vi= R  R 172  Vi 5.83 kHz BW = 95.19 kHz f1 = fs  = 98.1 kHz  Since Qs  10, 2 2 BW f2 = fs + = 101.02 kHz 2 At f = 95.19 kHz: XL = 2πfL = 2π(95.19 kHz)(4.7 mH) = 2.81 k 1 1 = 2.99 k XC = = 2 fC 2 (95.19 kHz)(560 pF) 160 (1 V 0) 160 V 0 Vo = = 172 + j 2.81 k   j 2.99 k  172  j180 160 V 0 = 0.643 V46.30 = 248.97   46.30 At f = fs: Vomax = At f = 101.02 kHz: XL = 2πfL = 2π(101.02 kHz)(4.7 mH) = 2.98 k 1 1 XC = = 2.81 k = 2 fC 2 (101.02 kHz)(560 pF) 160 (1 V 0) 160 V 0 Vo = = 172 + j 2.98 k   j 2.81 k  172 + j170 160 V 0 = 0.66 V44.66 = 241.83 44.66 d. f = fs: Vomax = 0.93 V f = f1 = 95.19 kHz, Vo = 0.707(0.93 V) = 0.66 V f = f2 = 101.02 kHz, Vo = 0.707(0.93 V) = 0.66 V 30. a. R2C  159.15 kHz L 2 LC 2 f p L 2 (159.15 kHz)(1 mH) X Q = L = = 62.5  10 = 16  R R fp = 1 1 ZTp = Q2 R = (62.5)2 16  = 62.5 k  4 k and Vo  Vi at resonance. 290 CHAPTER 21 However, R = 3.3 k affects the shape of the resonance curve and BW = fp/ Q cannot be applied. V For Aυ = o = 0.707, | X | = R for the following configuration Vi For frequencies near fp, XL  R and ZL = R + jXL  XL and X = XL  XC. For frequencies near fp but less than fp XC X L X= XC  X L and for Aυ = 0.707 XC X L =R XC  X L 1 and XL = 2πf1L 2 f1C the following equation can be derived: Substituting XC = f12 + 1 1 f1  2 =0 2 RC 4 LC For this situation: 1 1 = 48.23  103 = 2 RC 2 (3.3 k )(0.001  F) 1 1 = 2.53  1010 = 2 2 4 LC 4 (1 mH)(0.001  F) and solving the quadratic equation, f1 = 135.83 kHz BW = fp  f1 = 159.15 kHz  135.83 kHz = 22.32 kHz and 2 BW so that f2 = fp + = 159.15 kHz + 18.75 kHz = 177.9 kHz 2 b. fp 159.15 kHz = 3.57 BW 44.64 kHz BW = 2(18.75 kHz) = 37.5 kHz Qp = CHAPTER 21 = 291 31. b. a. Qs = 5000  5000  XL = 12.5 = = R + R 390  + 10  400  f s 5000 Hz = 400 Hz = 12.5 Qs 400 Hz = 4.8 kHz f1 = 5000 Hz  2 400 Hz = 5.20 kHz f2 = 5000 Hz + 2 BW = c. At resonance 10 (Vi ) Vo = 10  + 400  = 0.024 Vi 32. d. At resonance, a. Q = 10 Ω  2 k = 9.95 Ω 9.95 (Vi )  0.024 Vi as above! Vo = 9.95  + 400  X L 400  = 40 = R 10  2 2 Z T p = Q  R = (40) 20  = 32 k  1 k At resonance, Vo = 32 k  Vi = 0.97Vi 32 k  + 1 k  Vo = 0.97 Vi For the low cutoff frequency note solution to Problem 30: 1 1 f12 + f1  2 =0 2 RC 4 LC 1 1 C= = 19.9 nF = 2 fX C 2 (20 kHz)(400 ) and Aυ = L= XL 400  = = 3.18 mH 2 f 2 (20 kHz) Substituting into the above equation and solving f1 = 16.4 kHz BW = 20 kHz  16.4 kHz = 3.6 kHz with 2 and BW = 2(3.6 kHz) = 7.2 kHz f 20 kHz Qp = p = = 2.78 BW 7.2 kHz 292 CHAPTER 21 b.  c. At resonance Z T p = 32 k  100 k = 24.24 k 24.24 k  V i = 0.96Vi 24.24 k  + 1 k  V and Aυ = o = 0.96 vs 0.97 above Vi with Vo = At frequencies to the right and left of fp, the impedance Z T p will decrease and be affected less and less by the parallel 100 k load. The characteristics, therefore, are only slightly affected by the 100 k load. d. At resonance Z T p = 32 k  20 k = 12.31 k with Vo = 12.31 k  Vi = 0.925Vi vs 0.97 Vi above 12.31 k  + 1 k  At frequencies to the right and left of fp, the impedance of each frequency will actually be less due to the parallel 20 k load. The effect will be to narrow the resonance curve and decrease the bandwidth with an increase in Qp. 33. a. fp = 1 2 LC 1 = 726.44 kHz (band-stop) 2 (400  H)(120 pF) =   X Ls 90 + X L p 90  X C   90 = 0 jX Ls + jX Ls + X  90  X C   90  jX L p  jX C  X Lp X C j X Lp  X C jX Ls  j X Ls  Lp X Lp X C X Lp  XC X Lp X C X Lp  X C  =0  =0 =0 =0 X Ls X C  X Ls X L p + X L p X C = 0  Lp  Ls   Ls + =0 C C CHAPTER 21 293 1  Ls + L p  = 0 C Ls + L p = CLs L p LsLpω2  f= 34. a. Ls + L p 1 2 CLs L p 1 c. fc = f= f= 1 f c: 2 1 f c: 10 f = 10fc: f= 1 f c: 2 f = 2fc: 294  Ls = 2 XL 24.15 k  = 128.19 mH = 2 f 2 (30 kHz) 1 1 = 7.2 kHz = 2 RC 2 (0.47 k )(0.047  F) f = 2fc: d. 460  106 = 2.01 MHz (pass-band) 28.8  1019 1 2 1 1 = 12.68 mH = 2 2 4 f s C 4 (100 kHz ) 2(200 pF) 2 LC XL = 2πfL = 2π(30 kHz)(12.68 mH) = 2388.91  1 1 = = 26.54 k XC = 2 fC 2 (30 kHz)(200 pF) XC  XL = 26.54 k  2388.91  = 24.15 k(C) X Lp = X C(net) = 24.15 k fs = Lp = 35. a, b. = A dB = 20 log10 A dB = 20 log10 A dB = 20 log10 1 1+  f c /f  1 1 1  (10)2 Aυ = 1 + (0.1) 2 1 +  f c /f  1 1  (0.5) 2 2 = 1 1+ (2)2 = 7 dB = 20.04 dB 1 1 = 20 log10 = 0.969 dB 1  (0.5) 2 A dB = 20 log10 Aυ = 2 = 0.043 dB 1 1 + (2) 2 = 0.447 = 0.894 CHAPTER 21 e. 36. a. fc = 1 1 1 = 1.83 kHz = = 2 RC 2 (6.8 k   12 k )0.02  F 2 (4.34 k )(0.02  F) Vo 1 = Vi 1 + ( f /f ) 2 c   1 Vi and Vo =   1 + ( f /f ) 2  c   b. c. & d. CHAPTER 21 295 e. Remember the log scale! 1.5fc is not midway between fc and 2fc A dB = 20 log10 Aυ 1.5 = 20 log10 Aυ 0.075 = log10 Aυ V Aυ = o = 0.84 Vi f. 37. a, b. θ = tan1 fc/f Aυ = fc = c. Vo = A  = Vi 1 1 + (f/f c ) 2 tan1f/fc 1 1 = 13.26 kHz = 2 RC 2 (12 k )(1 nF) f = fc/2 = 6.63 kHz AdB  20 log10 1 1 + (0.5) 2 = 0.97 dB f = 2fc = 26.52 kHz AdB = 20 log10 1 1 + (2) 2 = 6.99 dB f = fc/10 = 1.326 kHz AdB = 20 log10 1 1 + (0.1) 2 = 0.04 dB f = 10fc = 132.6 kHz A dB = 20 log10 d. 296 f = fc/2: Aυ = f = 2fc: Aυ = 1 1 + (10) 2 1 1  (0.5) 2 1 1 + (2) 2 = 20.04 dB = 0.894 = 0.447 CHAPTER 21 38. e. θ = tan1 f/fc f = fc/2: θ = tan1 0.5 = 26.57 θ = tan1 1 = 45 f = f c: f = 2fc: θ = tan1 2 = 63.43 a. R2  XC = ( R2 )(  jX C ) R2 X C = j R2  jX C R2  jX C   jR2 X C    Vi R2  jX C  R2 X C Vi  Vo = = j R2 X C R1 ( R2  jX C )  jR2 X C R1  j R2  jX C -jR2 X C Vi  jR2 X C Vi = = R1R2  jR1 X C  jR2 X C R1R2  j ( R1  R2 ) X C R2 X C Vi R2 Vi = = jR1R2  ( R1  R2 ) X C j R1R2 + ( R  R ) 1 2 XC  R2    Vi R1  R2  R2 Vi  = = RR R1  R2  j 1 2 1 + j  R1R2  1 XC  R1  R2  X C R2 V R1  R2 and Aυ = o =  RR  Vi 1 + j  1 2  C  R1  R2  or Aυ = CHAPTER 21 R2 R1  R2   1   1+ j 2 f ( R1  R2 )C  297 defining fc = 1 2 ( R1  R2 )C   1   1+ j f/f c   1 R2   and Aυ =   tan 1 f/fc  R1  R2  1+ ( f/f c )2    R2  1  |Vi | with |Vo | = R1  R2  1 + (f/f c ) 2    R2 27 k  for f  fc, Vo = Vi = Vi = 0.852Vi R1  R2 4.7 k  + 27 k  Aυ = R2 R1  R2 at f = fc: Vo = 0.852[0.707]Vi = 0.602Vi 1 1 = 1.02 kHz fc =  2 ( R1  R2 )C 2 (4.7 k  27 k)0.039  F b. c. & d. 20 log10 298 4.7 k  + 27 k  R1  R2 = 20 log10 27 k  R2 = 20 log10 1.174 = 1.39 dB CHAPTER 21 e. AdB  1.39 dB  0.5 dB = 1.89 dB AdB = 20 log10 Aυ 1.89 = 20 log10 Aυ 0.0945 = log10 Aυ V Aυ = o = 0.80 Vi f. θ = tan1 f/fc 39. R2'  39 k  68 k = 24.79 k a. From Section 21.11, Aυ = j f/f1 Vo = Vi 1  jf/f c 1 1 = 642.01 Hz = 2 R2 C 2 (24.79 k )(0.01  F) 1 1 = fc = = 457.47 Hz 2  R1 + R2  C 2 (10 k  + 24.79 k )(0.01  F) f1 = 20log10 CHAPTER 21 f f  20log10 c f1 f1 457.5 Hz = 20 log10 642 Hz = −2.94 dB 299 b. f f = + tan1 1 f f1 θ = 45 θ = 54.52 θ = 90  tan1 f = f1: f = f c: 1 f = f1 = 321 Hz, θ = 63.44 2 1 f= f1 = 64.2 Hz, θ = 84.29 10 f = 2f1 = 1,284 Hz, θ = 26.57 f = 10f1 = 6420 Hz, θ = 5.71 40. a. VTh = 12 k  Vi = 0.682 Vi 12 k  + 5.6 k  RTh = 5.6 k  12 k = 3.82 k f =  Hz: (C  short circuit) 8.2 k  (0.682 Vi ) = 0.465 Vi Vo = 8.2 k  + 3.82 k  At f c : V0  0.707(0.465 Vi )  0.329 Vi 300 CHAPTER 21 R2 (0.682 Vi ) 0.682 R2 Vi = R1  R2  jX C R1  R2  jX C V 0.682 R2 j 2 f (0.682 R2 )C and Aυ = o = = Vi R1  R2  jX C 1 + j 2 f ( R1  R2 )C 1 1 j f/f1 so that Aυ = with f1 = = 1 + j f/f c 2 0.682R2 C 2 0.682(8.2 k )(0.1  F) = 284.59 Hz 1 1 and fc = = 2 ( R1  R2 )C 2 (3.82 k  + 8.2 k )(0.1  F) = 132.41 Hz voltage-divider rule: Vo = 132.41 Hz 284.59 Hz = 20 log10 0.465 = 6.65 dB 20 log10 f/f1 = 20 log10 b. θ = 90  tan1 f/fc = +tan1 fc/f = tan1 132.6 Hz/f or CHAPTER 21 301 f f1 Aυ = f 1 j fc 1 j 41. a. 1 1 = 19.41 kHz  2 R2C 2 (10 k)(820 pF) 1 1 fc =  2 ( R1  R2 )C 2 (10 k  91 k)(820 pF) = 1.92 kHz f1 = 20log10 R1  R2  20log10 10 =  20 dB R2 b. θ = tan1 f/f1  tan1 f/fc f = 10 kHz 10 kHz 10 kHz  tan1 = 27.25  79.13 = 51.88 θ = tan1 19.41 kHz 1.92 kHz f = fc: (f1 = 10 fc) fc f θ = tan1  tan 1 c = tan 1 0.1tan 1 1 = 5.71  45 = 39.29 10 f c fc 42. a. R1 no effect! Note Section 21.12. Vo 1 + j ( f/f1 ) = Vi 1 + j ( f/f c ) 1 f1 = = 2.84 kHz 2 (5.6 k )(0.01  F) 1 fc = = 904.3 Hz 2 (12 k  + 5.6 k )(0.01  F) Aυ = Note Fig. 21.65. Asymptote at 0 dB from 0  fc 6 dB/octave from fc to f1 12 k  + 5.6 k    9.95 dB from f1 on  20 log =  9.95 dB  5.6 k    302 CHAPTER 21 (b) Note Fig. 21.67. From 0 to 26.50 at fc and f1 θ = tan1 f/f1  tan1 f/fc At f = 1500 Hz (between fc and f1) θ = tan1 1500 Hz/2.84 kHz  tan1 1500 Hz/904.3 Hz = 27.83  58.92 = 31.09 43. a. V o 1  jf1/f  V i 1  jf c /f 1 1 f1 = = 945.66 Hz  2 R1C 2 (3.3 k)(0.051  F) 1 1 = 7.59 kHz fc = = 2 (R1  R2 )C 2 (3.3 k   0.47 k ) (0.051  F)   0.411 k  Aυ = 20 log10 R1  R2 3.3 k  + 0.47 k  =  20 log10 8.02 = 18.08 dB =  20 log10 0.47 k  R2 b. f f1 + tan1 c f f f f f  f1 :    tan 1 1  tan 1 c f1 f1 θ = tan1 7.59 kHz 945.66 Hz  45  82.89  37.89   tan 1 1  tan 1 945.66 Hz 7.59 kHz  tan 1 4 kHz 4 kHz  13.28  62.24  48.96 f  4 kHz:    tan 1 f  f c :    tan 1 f 945.66 Hz  tan 1 c 7.59 kHz fc   tan 1 0.125  tan 1 1  7.11  45  37.89 CHAPTER 21 303 44. a. Note Section 21.13. 1  j ( f1 /f ) Aυ = 1  j (f c /f ) f1 = fc = 1 1 = 964.58 Hz = 2 R1C 2 (3.3 k )(0.05  F) 1 1 = 7334.33 Hz = 2 ( R1  R2 )C 2 (3.3 k   0.5 k ) 0.05  F    0.434 k Note Fig. 21.72. 20 log10 3.3 k  + 0.5 k  R1  R2 = 17.62 dB = 20 log10 0.5 k  R2 Asymptote at 17.62 dB from 0  f1 +6 dB/octave from f1 to fc 0 dB from fc on b. θ = tan1 f1/f + tan1 fc/f Test at 3 kHz θ = tan1 964.58 Hz/3.0 kHz + tan1 7334.33 Hz/3.0 kHz = 17.82 + 67.75 = 49.93  50 Therefore rising above 45 at and near the peak 50 kHz vs 23 kHz  drop about 1 dB at 23 kHz due to 50 kHz break. Ignore effect of break frequency at 10 Hz. Assume 2 dB drop at 68 Hz due to break frequency at 45 Hz. Rough sketch suggests low cut-off frequency of 90 Hz. Checking: Ignoring upper terms 2 2  10 Hz   45 Hz   68 Hz  A dB  20 log10 1 +    20 log10 1 +    20 log10 1 +    f   f   f  = −0.0532 dB − 0.969 dB − 1.96 dB = −2.98 dB (excellent) 304 2 CHAPTER 21 High frequency cutoff: Try 20 kHz 2  f   f  A dB =  20log10 1 +    20 log10 1 +    23 kHz   50 kHz  = −2.445 dB − 0.6445 dB = −3.09 dB (excellent  BW = 20 kHz  90 Hz = 19,910 Hz 2  20 kHz f1 = 90 Hz, f2 = 20 kHz Testing: f = 100 Hz f f 10 Hz 45 Hz 68 Hz + tan 1 + tan 1  tan 1  tan 1 23 kHz 50 kHz f f f = tan1 0.1 + tan1 0.45 + tan1 0.68  tan1 0.00435  tan1 .002 = 5.71 + 24.23 + 34.22  0.249  0.115 = 63.8 vs about 65 on the plot θ = tan 1 45. a. 1 A = 100 Hz  130 Hz   f f   A max  1  j  1  j  1 + j 1 + j  20 kHz  50 kHz  f  f   Proximity of 100 Hz to 130 Hz will raise lower cutoff frequency above 130 Hz: Testing: f = 180 Hz: (with lower terms only) 2 A dB  100   130  = 20 log10 1     20 log10 1     f   f  2  100   130  = 20 log10 1     20 log10 1     180   180  = 1.17 dB  1.82 dB = 2.99 dB  3 dB CHAPTER 21 2 2 305 Proximity of 50 kHz to 20 kHz will lower high cutoff frequency below 20 kHz: Testing: f = 18 kHz: (with upper terms only) 2  f   f  A dB = 20 log10 1     20 log10 1     20 kHz   50 kHz  2  18 kHz   13 kHz  = 20 log10 1 +    20 log10 1 +    20 kHz   20 kHz  = 2.576 dB  0.529 dB = 3.105 dB 2 2 b. Testing: f = 1.8 kHz: 100 130 1.8 kHz 1.8 kHz + tan 1  tan 1  tan 1 1.8 kHz 1.8 kHz 20 kHz 50 kHz = 3.18 + 4.14  5.14  2.06 = 0.12  0 θ = tan1 47. flow = fhigh  BW = 36 kHz  35.8 kHz = 0.2 kHz = 200 Hz Aυ = 306 120  50  200   f    1  j  1  j  1  j 36 kHz  f  f   CHAPTER 21 0.05 1 1 +jf = = = 100 100 2000 +jf + 2000 0.05  j 1 j 1 j f 0.05 f f f +j 2000 and f = 2000 Hz = 1 f 1+ j 2000 48. Aυ = 49. Aυ = 50. Aυ = 200 1 1   200  j 0.1 f 1  j 0.1 f 1  j f 200 2000 1 f , = 1 and f = 2 kHz A dB = 20 log 20 2  f  2000 1+    2000  jf/ 1000 (1 + jf/ 1000)(1 + jf/10,000) CHAPTER 21 307 51. f  f   1  j 1  j  1000  2000  Aυ =  2 f    j 1   3000   2 2  f   f  AdB = 20 log10 1 +  1  + 20 log10 1 +  2  + 40 log10  1000   2000  52. 308 1  f  1+  3   3000  2 j 2 f f f j f = j = j = j , = j 1000 1000 1000 159.16 Hz 5000 795.78 Hz 2 CHAPTER 21 53. a. b. Woofer − 400 Hz: XL = 2πfL = 2π(400 Hz)(4.7 mH) = 11.81  1 1 = = 10.20  XC = 2 (400 Hz)(39  F) 2 fC R  XC = 8  0  10.2090 = 6.3  38.11 ( R  X C )(Vi ) (6.3    38.11)(Vi ) = Vo = ( R  X C )  jX L (6.3    38.11)  j 11.81  Vo = 0.673 96.11 Vi V and Aυ = o = 0.673 vs desired 0.707 (off by less than 5%) Vi Tweeter − 5 kHz: XL = 2πfL = 2π(5 kHz)(0.39 mH) = 12.25  1 1 = = 11.79  XC = 2 fC 2 ( 5 kHz)(2.7  F) R  XL = 8  0  12.25  90 = 6.7  33.15 (6.7  33.15)(Vi ) Vo = (6.7  33.15)  j 11.79  Vo = 0.678 88.54 Vi V and Aυ = o = 0.678 vs 0.707 (off by less than 5%) Vi Woofer − 3 kHz: XL = 2πfL = 2π(3 kHz)(4.7 mH) = 88.59  1 1 = = 1.36  XC = 2 fC 2 ( 3 kHz)(39  F) R  XC = 8  0  1.36  90 = 1.341  80.35 ( R  X C )(Vi ) (1.341    80.35)(Vi ) Vo = = ( R  X C )  jX L (1.341    80.35)  j 88.59  Vo = 0.015 170.2 Vi V and Aυ = o = 0.015 vs desired 0 (excellent) Vi Tweeter − 3 kHz: XL = 2πfL = 2π(3 kHz)(0.39 mH) = 7.35  1 1 = = 19.65  XC = 2 fC 2 (3 kHz)(2.7  F) R  XL = 8  0  7.35  90 = 5.42  47.42 ( R  X L )(Vi ) (5.42  47.42)( Vi ) = Vo = ( R  X L )  jX C (5.42  47.42)  j 19.65  Vo = 0.337 124.24 Vi V and Aυ = o = 0.337 (acceptable since relatively close to cut frequency for tweeter) Vi CHAPTER 21 309 c. Mid-range speaker − 3 kHz:  Z = 7.41   22.15    Z = 8.24   33.58    Z = 7.816   37.79 ZVi (7.816  37.79)Vi = = 1.11 8.83 Vi Z  jX C 7.816  37.79  j1.36  (7.41    22.15)Vi ZV1 = Vo = = 0.998 46.9 Vi Z + jX L 7.41    22.15  j 7.35  V Aυ = o = 0.998 (excellent) Vi V1 = and 310 CHAPTER 21 Chapter 22 1. a. 2 (40 mH) 2 = 50 mH M = k L p Ls  L s = M 2  (50 mH)(0.8) 2 L pk b. ep = N p es = kNs c. ep = L p es = M 2. a. d p dt d p dt di p dt di p dt = (20)(0.08 Wb/s) = 1.6 V = (0.8)(80 t)(0.08 Wb/s) = 5.12 V = (40 mH)(0.3  103 A/s) = 12 V = (80 mH)(0.03  103 A/s) = 24 V k=1 2 b. (a) 2 ( 40 mH ) = 32 mH Ls = M 2 = (50 mH)(1) 2 L pk (b) ep = 1.6 V, es = kNs (c) ep = 15 V, es = 12 V d p = (1)(80 t)(0.08 Wb/s) = 6.4 V dt k = 0.2 2 3. (a) 2 (40 mH ) Ls = M 2 = = 0.8 H (50 mH)(0.2) 2 L pk (b) ep = 1.6 V, es = kNs (c) ep = 15 V, es = 12 V d p dt = (0.2)(80 t)(0.08 Wb/s) = 1.28 V a. 2 ( 40 mH ) 2 Ls = M 2 = = 355.56 mH (50 mH)(0.3) 2 L pk b. ep = N p es = kNs c. d p dt d p dt = (300 t)(0.08 Wb/s) = 24 V = (0.9)(25 t)(0.08 Wb/s) = 1.8 V ep and es the same as problem 1: ep = 15 V, es = 12 V CHAPTER 22 311 4. 5. Ns 120 t Ep = (40 V) = 240 V Np 20 t a. Es = b. Φmax = a. Es = b. Φm(max) = 6. Ep = 7. f= 8. a. 40 V Ep = 7.51 mWb = 4.4 fN p 4.44(60 Hz)(20 t) 30 t Ns (40 V) = 5 V E p= 240 t Np 40 V Ep = = 625.63 μWb 4.44 fN p (4.44)(60 Hz)(240 t) 60 t Np Es = (240 V) = 20 V 720 t Ns 40 V Ep = 120 Hz = (4.44) N p  m (max) (4.44)(20 t)(3.75 mWb) 1 IL = aIp =   (2 A) = 0.4 A 5 2  VL = ILZL =  A  (2 ) = 0.8 V 5  2 b. 9. 10. 11. Zp = 1 Zin = a2ZL =   2  = 0.08  5 Vg Ip = 120 V = 30  4A 1 Vg = aVL =   (600 V) = 150 V 4 V g 150 V Ip = = = 37.5 A 4 Zi IL = Is = VL 240 V = 12 A = Z L 20  I s = a = N p  12 A = N p 0.05 A 50 Ip Ns 50(12) Np = = 12,000 turns 0.05 312 CHAPTER 22 12. a. a= 1 N p 400 t = = 1200 t 3 Ns 2 1 Zi = a ZL =   [12  + j12 ] = 1.333  + j1.333  = 1.885  45 3 Ip = Vg/Zi = 100 V/1.885 Ω = 53.05 A 2 13. IL = aIp = a. Zp = a2ZL  a = Zp = b. Zp ZL 10 V Vp = = 36  I p 20 V/ 72  36  =3 4 a= 1 1 1 Vs Ns 1 = =  V s = V p = (10 V) = 3 V 3 3 3 Vp Np 3 P= 14. 1 (53.05 A) = 17.68 A, VL = ILZL = (17.68 A)(16.97 Ω) = 300 V 3 b. Vs2 (3.33 V ) 2 = 2.78 W = Zs 4 a. Re = Rp + a2Rs = 4  + (4)2 1  = 20  b. Xe = Xp + a2Xs = 12  + (4)2 2  = 44  c. Vg d. Ip = e. aVL = Zp = 120 V 0 120 V 0 = = 0.554 A 11.73 20  + 192  + j 44  212  + j 44  2 or f.  g. VL = CHAPTER 22 a R LV g = Ipa2RL 2 ( R e  a R L)  j X e VL = aIpRL0 = (4)(0.554 A 11.73)(20  0) = 26.59 V 11.73 1 Ns Vg = (120 V) = 30 V 4 Np 313 15. 4t =4 Ns 1t Re = Rp + a2Rs = 4  + (4)2 1  = 20  Xe = Xp + a2Xs = 12  + (4)2 2  = 44  Zp = Z Re + Z X e + a 2 Z X L = 20  + j44  + j(4)2 20  = 20  + j44  + j320  = 20  + j364  = 364.55  86.86 Np a. a= b. Ip = c. VRe = (Iθ)(Re0) = (329.17 mA 86.86)(20  0) Vg Zp = = 120 V 0 = 329.17 mA 86.86 364.55  86.86 = 6.58 V 86.86 VX e = (Iθ)(Xe90) = (329.17 mA 86.86)(44  90) = 14.48 V 3.14 VX L = I(a2 Z X L ) = (329.17 mA 86.86)(320 Ω 90) = 105.33 V 3.14 16. a. a = Np/Ns = 4 t/1 t = 4, Re = Rp + a2Rs = 4  + (4)2 1  = 20  Xe = Xp + a2Xs = 12  + (4)2 2  = 44  Zp = Re + jXe  ja2XC = 20  + j44   j(4)2 20  = 20  + j44   j320  = 20   j276  = 276.72 Ω 85.86 b. Ip = c. V Re = (Ipθ)(Re0) = (0.43 A 85.86)(20  0) = 8.6 V 85.86 V X e = (Ipθ)(Xe90) = (0.43 A 85.86)(44  90) = 18.92 V 175.86 Vg Zp = 120 V 0 = 0.43 A 85.86 276.72    85.86 2 V X C = (Ipθ)(a XC90) = (0.43 A 85.86)(320  90) = 137.60 V 4.14 17.  18. Coil 1: L1  M12 Coil 2: L2  M12 LT = L1 + L2  2M12 = 4 H + 7 H  2(1 H) = 9 H 19. L T (  )  L1  L2  2M 12 M12 = k L1 L2 = (0.8) (200 mH)(600 mH) = 277 mH LT (+) = 200 mH + 600 mH + 2(277 mH) = 1.35 H 314 CHAPTER 22 20. M23 = k L2 L3  1 (1 H)(4 H) = 2 H L1 + M12  M13 = 2 H + 0.2 H  0.1 H = 2.1 H L2 + M12  M23 = 1 H + 0.2 H  2 H = 0.8 H L3  M23  M13 = 4 H  2 H  0.1 H = 1.9 H LT = 2.1 H  0.8 H + 1.9 H = 3.2 H Coil 1: Coil 2: Coil 3: 21. E1  I1[ Z R1 + Z L1 ]  I2[Zm] = 0 I2[ Z L2 + Z RL ] + I1[Zm] = 0 ────────────────────── I1( Z R1 + Z L1 ) + I2(Zm) = E1 I1(Zm) + I2( Z L2 + Z RL ) = 0 ─────────────────────── 22. Zi = Zp + Xm = ωM 90 ( M ) 2 ( M ) 2 = Rp + j X Lp + Zs + ZL Rs + jX Ls + RL Rp = 2 , X L p = ωLp = (103 rad/s)(8 H) = 8 k Rs = 1 , X Ls = ωLs = (103 rad/s)(2 H) = 2 k M = k L p L s = 0.05 (8 H)(2 H) = 0.2 H Zi = 2  + j8 k + (103 rad/s  0.2 H ) 2 1  + j 2 k  + 20  4  104  21  j 2  103 = 2  + j8 k + 0.21   j19.99  = 2.21  + j7980  Zi = 7980  89.98 = 2  + j8 k + 23. Np Vp 2400 V = 20 120 V a. a= b. 10,000 VA = VsIs  Is = c. Ip = d. a= Is = CHAPTER 22 Ns = Vs = 10,000 VA 10,000 VA = 83.33 A = 120 V Vs 10,000 VA 10,000 VA = = 4.17 A 2400 V Vp Vp Vs = 120 V 1 = 0.05 = 2400 V 20 10,000 VA = 4.17 A, Ip = 83.33 A 2400 V 315 24. Is = I1 = 2 A, Ep = VL = 40 V Es = Vs  VL = 200 V  40 V = 160 V 200 V (2 A) = 10 A VgI1 = VLIL  IL = Vg/VL  I1 = 40 V Ip + I1 = IL  Ip = IL  I1 = 10 A  2A = 8 A 25. a. Es = Ns Ep Np 25 t (100 V 0) = 25 V 0 = VL 100 t E 25 V 0 = 5 A 0 = IL Is = s = Z L 5  0 = 2 26. b. N   100 t  2 Zi = a ZL =  p  Z L    5  0 = (4) 5  0 = 80  0  25 t   Ns  c. 1 1 Z1/ 2 = Zi = (80  0) = 20  0 4 4 a. b. 2 2 15 t E2 = N 2 E1 = (60 V 0) = 10 V 0 90 t N1 45 t E3 = N 3 E1 = (60 V 0) = 30 V 0 90 t N1 10 V 0 I2 = E 2 = = 1.25 A 0 Z 2 8  0 30 V 0 I3 = E 3 = = 6 A 0 Z3 5  0 1 R1 = 1 1  2 ( N1 / N 2 ) R2 ( N1 / N 3 ) 2R3 1 1  2 (90 t /15 t ) 8  (90 t / 45 t ) 2 5  1 1 1 = 0.05347 S   R1 288  20  R1 = 18.70  = 27. a. N2  40 t  E1 =   (120 V 60) = 40 V 60 N1  120 t  E 40 V 60 = 3.33 A 60 I2 = 2 = Z 2 12  0 E2 = N3  30 t  E1 =   (120 V 60) = 30 V 60 N1  120 t  E 30 V 60 I3 = 3 = = 3 A 60 Z3 10  0 E3 = 316 CHAPTER 22 b. 1 R1 = 1 1  2 ( N1 / N 2 ) R2 ( N1 / N 3 ) 2R3 1 1  2 (120 t / 40 t ) 12  (120 t / 30 t ) 210  1 1 1  = = 0.0155 S 108  160  R1 1 = 64.52  R1 = 0.0155 S = 28. ZM = Z M12 = ωM12 90 E  I1Z1  I1 Z L1  I1(Zm)  I2(+Zm)  I1 Z L2 + I2 Z L2  I1(Zm) = 0 E  I1(Z1 + Z L1  Zm + Z L2  Zm)  I2(Zm  Z L2 ) = 0 or I1(Z1 + Z L1 + Z L2  2 Zm) + I2(Zm  Z L2 ) = E ────────────────────────────────────────────── I2Z2  Z L2 (I2  I1)  I1(+Zm) = 0 or I1(Zm  Z L2 ) + I2(Z2 + Z L2 ) = 0 ────────────────────────── E1  I1Z1  I1 Z L1  I2( Z M12 )  I3(+ Z M13 ) = 0 29. or E1  I1[Z1 + Z L1 ] + I2 Z M12  I3 Z M13 = 0 ──────────────────────────────── I2(Z2 + Z3 + Z L2 ) + I3Z2  I1( Z M12 ) = 0 or I2(Z2 + Z3 + Z L2 ) + I3Z2 + I1 Z M12 = 0 ──────────────────────────────── I3(Z2 + Z4 + Z L3 ) + I2Z2  I1(+ Z M13 ) = 0 or I3(Z2 + Z4 + Z L3 ) + I2Z2  I1 Z M13 = 0 ───────────────────────────── Z M12 I2 + [Z1 + Z L1 ]I1   Z M12 I1  [Z2 + Z3 + Z L2 ]I2 + Z M13 I3 = E1 Z2I3 = 0 Z M13 I1 Z2I2 + [Z2 + Z4 + Z L3 ]I3 = 0 ──────────────────────────────────────── CHAPTER 22 317 Chapter 23 1. 2. a. E = EL/ 3 = 208 V/1.732 = 120.1 V b. V = E = 120.1 V c. I = d. IL = I = 12.01 A a. E = EL/ 3 = 208 V/1.732 = 120.1 V b. V = E = 120.1 V c. Z = 12   j16 Ω = 20  53.13 d. IL = I = 6 A b. V = 120.1 V I = 3. 4. V  120.1 V  6A = 20  Z a. E = 120.1 V c. Z = (10  0  (10  90) = 7.071  45 120.1 V V I =  = = 16.98 A Z  7.071  d. IL = 16.98 A a. θ2 = 120, θ3 = 120 b. Van = 120 V 0, Vbn = 120 V 120, Vcn = 120 V 120 c. 5. V  120.1 V = 12.01 A = 10  R 120 V 0 Ian = V an = = 6 A 0 Z an 20  0 120 V   120 Ibn = V bn = = 6 A 120 20  0 Z bn 120 V 120 Icn = V cn = = 6 A 120 20  0 Z cn d. IL = I = 6A a. θ2 = 120, θ3 = +120 b. Van = 120 V 0, Vbn = 120 V 120, Vcn = 120 V 120 c. Z = 9  + j12  = 15  53.13 e. VL = 3 V = 3 (120 V) = 207.8 V 120 V 0 120 V   120 = 8 A 53.13, Ibn = = 8 A 173.13 15  53.13 15  53.13 120 V 120 = 8 A 66.87 Icn = 15  53.13 Ian = 318 CHAPTER 23 e. 6. a, b. c. IL = I = 8 A f. EL = 3 E = (1.732)(120 V) = 207.85 V The same as problem 4. Z = 6  0  8  90 = 4.8  36.87 120 V 0 = 25 A 36.87 Ian = V an = Z an 4.8    36.87 120 V   120 Ibn = V bn = = 25 A 83.13 Z bn 4.8    36.87 120 V 120 Icn = V cn = = 25 A 156.87 Z cn 4.8    36.87 d. 7. IL = I = 25 A e. VL VL = 3 V = 3 (120 V) = 207.84 V 220 V = 127.0 V 3 1.732 Z = 10   j10  = 14.42  45 V = Van = Vbn = Vcn = = 127 V V = = 8.98 A Z  14.142  IL = IAa = IBb = ICc = I = 8.98 A I = Ian = Ibn = Icn = 8. Z = 12  + j16  = 20 53.13 V  50 V = = 2.5 A Z  20  ZT  = 13  + j16  = 20.62  50.91 I = V = I Z T  = (2.5 A)(20.62 ) = 51.55 V VL = 9. a. 3 V =  3  (51.55 V) = 89.29 V 22 kV   30 = 12.7 kV 30 3 22 kV EBN =   150 = 12.7 kV 150 3 22 kV ECN = 90 = 12.7 kV 90 3 EAN = CHAPTER 23 319 b, c. IAa = Ian = E AN 12.7 kV   30 = (30  + j 40 ) + (0.4 k  + j1 k ) Z AN 12.7 kV   30 12.7 kV   30 = 430  + j1040  1125.39  67.54 = 11.29 A 97.54 12.7 kV   150 = 11.29 A 217.54 = 1125.39  67.54 12.7 kV 90 = = 11.29 A 22.46 1125.39  67.54 = IBb = Ibn = E BN Z BN E CN ICc = Icn = Z CN 10. 11. 12. 13. 320 d. Van = IanZan = (11.29 A 97.54)(400 + j1000) = (11.29 A 97.54)(1077.03  68.2) = 12.16 kV 29.34 Vbn = IbnZbn = (11.29 A 217.54)(1077.03 68.2) = 12.16 kV 149.34 Vcn = IcnZcn = (11.29 A 22.46)(1077.03 68.2) = 12.16 kV 90.66 a. E = EL/ 3 = 208 V/1.732 = 120.1 V c. I = a. E = EL/ 3 = 208 V/1.732 = 120.1 V c. Z = 6.8  + j14  = 15.564  64.09 208 V V = 13.36 A I =  = Z  15.564  d. IL = 3 I = (1.732)(13.36 A) = 23.14 A V  208 V = 10.4 A = Z  20  b. V = EL = 208 V d. IL = 3 I = (1.732)(10.4 A) = 18 A b. V = EL = 208 V b. V = 208 V Z = 18  0  18  90 = 12.728  45 a. E = VL/ 3 = 208 V/ 3 = 120.09 V c. I = d. IL = 3 I = (1.732)(16.34 A) = 28.30 A a. θ2 = 120, θ3 = +120 b. Vab = 208 V 0, Vbc = 208 V 120, Vca = 208 V 120 208 V V = = 16.34 A Z  12.728  CHAPTER 23 c. d. 14.  208 V 0 Iab = V ab = = 9.46 A 0 Z ab 22  0 208 V 120 Ibc = V bc = = 9.46 A 120 22  0 Z bc 208 V 120 = 9.46 A 120 Ica = V ca = 22  0 Z ca e. IL = 3 I = (1.732)(9.46 A) = 16.38 A f. E = EL/ 3 = 208 V/1.732 = 120.1 V a. θ2 = 120, θ3 = +120 b. Vab = 208 V 0, Vbc = 208 V 120, Vca = 208 V 120 c.  d. Z = 100   j100  = 141.42 45 208 V 0 Iab = V ab = = 1.47 A 45 141.42    45 Z ab 208 V   120 Ibc = V bc = = 1.47 A 75 Z bc 141.42    45 208 V 120 Ica = V ca = = 1.47 A 165 Z ca 141.42    45 e. IL = 3 I = (1.732)(1.471 A) = 2.55 A f. E = EL/ 3 = 208 V/1.732 = 120.1 V 15. a, b. The same as problem 13. c.  d. Z = 3  0  4  90 = 2.4  36.87 208 V 0 Iab = V ab = = 86.67 A 36.87 Z ab 2.4  36.87 208 V   120 Ibc = V bc = = 86.67 A 156.87 Z bc 2.4  36.87 208 V 120 Ica = V ca = = 86.67 A 83.13 Z ca 2.4  36.87 CHAPTER 23 321 e. IL = 3 I = (1.732)(86.67 A) = 150.11 A 16. Vab = Vbc = Vca = 220 V Z = 10  + j10  = 14.142 45 220 V V = 15.56 A Iab = Ibc = Ica =  = 14.142  Z 17. a. f. E = 120.1 V 16 kV 0 16 kV 0 = Iab = V ab = Z ab 300  + j1000  1044.03  73.30 Iab = 15.33 A 73.30 16 kV   120 Ibc = V bc = = 15.33 A 193.30 Z bc 1044.03  73.30 16 kV 120 Ica = V ca = = 15.33 A 46.7 Z ca 1044.03  73.30 b. IAa  Iab + Ica = 0 IAa = Iab  Ica = 15.33 A 73.30  15.33 A 46.7 = (4.41 A  j14.68 A)  (10.51 A + j11.16 A) = 4.41 A  10.51 A  j(14.68 A + 11.16 A) = 6.11 A  j25.84 A = 26.55 A 103.30 IBb + Iab = Ibc IBb = Ibc  Iab = 15.33 A 193.30  15.33 A 73.30 = 26.55 A 136.70 ICc + Ibc = Ica ICc = Ica  Ibc = 15.33 A 46.7  15.33 A 193.30 = 26.55 A 16.70 c. EAB = IAa(10  + j20 ) + Vab  IBb(22.361  63.43) = (26.55 A 103.30)(22.361  63.43) + 16 kV 0  (26.55 A 136.70)(22.361  63.43) = (455.65 V  j380.58 V) + 16,000 V  (557.42 V  j204.32 V) = 17.01 kV  j176.26 V = 17.01 kV 0.59 EBC = IBb(22.361  63.43) + Vbc  ICc(22.361  63.53) = (26.55 A 136.70)(22.361  63.53) + 16 kV 120  (26.55 A 16.70)(22.361  63.53) = 17.01 kV 120.59 ECA = ICc(22.361  63.43) + Vca  IAa(22.361  63.43) = 17.01 kV 119.41 322 CHAPTER 23 18. 19. a. E = EL = 208 V c. I = a. E = EL = 208 V c. I = 20. a, b. c. V  120.09 V = = 7.08 A Z  16.971  208 V V = E L = = 120.1 V 3 1.732 d. IL = I   4 A b. V = EL 3 = 120.09 V d. IL = I = 7.08 A The same as problem 18. Z = 15  0  20  90 = 12  36.87 I = d. V  120.1 V = 4.00 A = 30  Z b. V  120.1 V =  10 A 12  Z IL = I  10 A 120 V 120 V = 69.28 V = 1.732 3 69.28 V Ian = Ibn = Icn = = 2.89 A 24  IAa = IBb = ICc = 2.89 A 21. Van = Vbn = Vcn = 22. Van = Vbn = Vcn = 120 V = 69.28 V 3 Z = 10  + j20  = 22.36 63.43 Ian = Ibn = Icn = V Z = 69.28 V = 3.10 A 22.36  IAa = IBb = ICc = I = 3.10 A 23. Van = Vbn = Vcn = 69.28 V Z = 20  0  15  90 = 12  53.13 69.28 V = 5.77 A 12  IAa = IBb = ICc = 5.77 A Ian = Ibn = Icn = 24. a. E = EL = 440 V c. I = CHAPTER 23 V  440 V =2A = Z  220  b. V = EL = E = 440 V d. IL = 3 I = (1.732)(2 A) = 3.46 A 323 25. a. E = EL = 440 V c. Z = 12   j9  = 15  36.87 26. a, b. c. The same as problem 24. Z = 22  0  22  90 = 15.56  45 440 V V = 28.28 A = Z  15.56  3 I = (1.732)(28.28 A) = 48.98 A d. IL = a. θ2 = 120, θ3 = +120 b. Vab = 100 V 0, Vbc = 100 V 120, Vca = 100 V 120 c.  d. 28. 3 I = (1.732)(29.33 A) = 50.8 A IL = I = 27. V = EL = 440 V V  440 V = 29.33 A = Z  15  I = d. b. 100 V 0 = 5 A 0 Iab = V ab = Z ab 20  0 100 V   120 Ibc = V bc = = 5 A 120 20  0 Z bc 100 V 120 Ica = V ca = = 5 A 120 20  0 Z ca 3 (5 A) = 8.66 A e. IAa = IBb = ICc = a. θ2 = 120, θ3 = +120 b. Vab = 100 V 0, Vbc = 100 V 120, Vca = 100 V 120 c.  d. Z = 12  + j16  = 20  53.13 100 V 0 V ab = = 5 A 53.13 Z ab 20  53.13 100 V   120 Ibc = V bc = = 5 A 173.13 Z bc 20  53.13 Iab = 324 CHAPTER 23 100 V 120 Ica = V ca = = 5 A 66.87 Z ca 20  53.13 29. 3 I = (1.732)(5 A) = 8.66 A e. IAa = IBb = ICc = a. θ2 = 120, θ3 = 120 b. Vab = 100 V 0, Vbc = 100 V 120, Vca = 100 V 120 c.  d. Z = 20  0  20  90 = 14.14  45 100 V 0 = 7.07 A 45 14.14    45 100 V   120 Ibc = = 7.07 A 75 14.14    45 100 V 120 = 7.07 A 165 Ica = 14.14    45 Iab = e. 30. IAa = IBb = ICc =  3  (7.07 A) = 12.25 A PT = 3I2 R = 3(6 A)2 12  = 1296 W QT = 3I2 X  = 3(6 A)2 16  = 1728 VAR(C) ST = 2 2 PT  QT = 2160 VA 1296 W = 0.6 (leading) Fp = PT = 2160 VA ST 31. V = 120 V, I = 120 V/20  = 6 A PT = 3I2 R = 3(6 A)2 20  = 2160 W QT = 0 VAR ST = PT = 2160 VA 2160 W Fp = PT = =1 S T 2160 VA 32. PT = 3I2 R = 3(8.98 A)2 10  = 2419.21 W QT = 3I2 X  = 3(8.98 A)2 10  = 2419.21 VAR(C) ST = Fp = PT2  Q T2 = 3421.28 VA PT 2419.21 W = 0.7071 (leading) = ST 3421.28 VA CHAPTER 23 325 33. V = 208 V V 2  (208 V ) 2 PT = 3    = 3  = 7210.67 W 18   R  V2  (208 V ) 2 QT = 3    = 3  = 7210.67 VAR(C) 18   X  2 2 PT  QT = 10,197.42 VA ST = 7210.67 W = 0.707 (leading) Fp = PT = 10,197.42 VA ST 34. PT = 3I2 R = 3(1.471 A)2 100  = 649.15 W QT = 3I2 X  = 3(1.471 A)2 100  = 649.15 VAR(C) PT2  QT2 = 918.04 VA ST = Fp = 35. PT 649.15 W = 0.7071 (leading) = ST 918.04 VA PT = 3I2 R = 3(15.56 A)2 10  = 7.26 kW QT = 3I2 X  = 3(15.56 A)2 10  = 7.26 kVAR PT2  Q T2 = 10.27 kVA ST = Fp = PT 7.263 kW = 0.7071 (lagging) = ST 10.272 kVA 2 36. PT = 3 2 V  3(120.1 V ) = = 2884.80 W 15  R 2 QT = 3 2 V  3(120.1 V ) = = 2163.60 VAR(C) 20  X 2 2 PT  QT = 3605.97 VA P 2884.80 W Fp = T = = 0.8 (leading) ST 3605.97 VA ST = 37. Z = 10  + j20  = 22.36  63.43 V L 120 V = = 69.28 V 3 1.732 69.28 V V = 3.098 A I =  = Z  22.36  V = PT = 3I2 R = 3(3.098 A)2 10  = 287.93 W 326 CHAPTER 23 QT = 3I2 X  = 3(3.098 A)2 20 Ω = 575.86 VAR PT2  QT2 = 643.83 VA ST = Fp = PT 287.93 W = 0.447 (lagging) = ST 643.83 VA 2 38. ST = Fp = 39. 2 V  3(440 V ) = = 26.4 kW 22  R QT = PT = 26.4 kVAR(L) PT = 3 PT2  QT2 = 37.34 kVA PT 26.4 kW = 0.707 (lagging) = ST 37.34 kVA Z = 12  + j16  = 20  53.13 I = V  100 V =5A = Z  20  PT = 3I2 R = 3(5 A)2 12  = 900 W QT = 3I2 X  = 3(5 A)2 16  = 1200 VAR(L) 2 2 PT  QT = 1500 VA P 900 W Fp = T = = 0.6 (lagging) ST 1500 VA ST = 40. PT = 3 ELIL cos θ 4800 W = (1.732)(200 V)IL (0.8) IL = 17.32 A I 17.32 A I = L = = 10 A 1.732 3 θ = cos1 0.8 = 36.87 200 V 0 V Z =  = = 20  36.87 = 16  + j12  I 10 A   36.87 41. PT = 3 ELIL cos θ 1200 W = 3 (208 V)IL(0.6)  IL = 5.55 A V 208 V V = L = = 120.1 V 3 1.732 θ = cos1 0.6 = 53.13 (leading) 120.1 V 0 V Z =  =   j 17.31  = 21.64  53.13 = 12.98       I 5.55 A 53.13 R XC CHAPTER 23 327 42. Δ: Z = 15  + j20  = 25 53.13 V I = Z = 125 V =5A 25  PT = 3I2 R = 3(5 A)2 15  = 1125 W QT = 3I2 X  = 3(5 A)2 20  = 1500 VAR(L) Y: V = VL/ 3 = 125 V/1.732 = 72.17 V Z = 3   j4  = 5 53.13 I = V  72.17 V = 14.43 A = 5 Z PT = 3I2 R = 3(14.43 A)2 3  = 1874.02 W QT = 3I2 X  = 3(14.43 A)2 4  = 2498.7 VAR PT = 1125 W + 1874.02 W = 2999.02 W QT = 1500 VAR(L)  2498.7 VAR(C) = 998.7 VAR(C) ST = Fp = 43. a. c. d. e. PT2  QT2 = 3161 VA PT 2999.02 W = 0.949 (leading) = 3161 VA ST E = 16 kV = 9,237.6 V 3 b. IL = I = 80 A 1200 kW = 400 kW 3 P4Ω = (80 A)24  = 25.6 kW PT = 3P = 3(25.6 kW + 400 kW) = 1276.8 kW P L = Fp = PT , ST = ST Fp = 1, 276.8 kW = 0.576 lagging 2,217.025 kVA 3 VLIL = 3 (16 kV)(80 A) = 2,217.025 kVA θL = cos1 0.576 = 54.83 (lagging) E AN0  80A   54.83° IAa =  Z T 54.83  given  for entire load 328 CHAPTER 23 44. f. Van = EAN  IAa(4  + j20 ) = 9237.6 V 0  (80 A 54.83)(20.396  78.69) = 9237.6 V 0  1631.68 V 23.86 = 9237.6 V  (1492.22 V + j660 V) = 7745.38 V  j660 V = 7773.45 V 4.87 g. Z = h. Fp(entire system) = 0.576 (lagging) Fp(load) = 0.643 (lagging) V an 7773.45 V   4.87 = = 97.168  49.95 80 A   54.83 I Aa = 62.52  + j 74.38        R XC i. 1276.8 kW  3(25.6 kW)  η = P o = P i P lost = = 0.9398  93.98% 1276.8 kW Pi Pi a.  b. V = 220 V = 127.02 V, Z = 10   j10  = 14.14 45 3 127.02 V V I =  = = 8.98 A Z  14.14  PT = 3I2 R = 3(8.98 A)2 10  = 2419.2 W Each wattmeter: 2419.2 W = 806.4 W 3 45. b. PT = 5899.64 W, Pmeter = 1966.55 W 46. a.  b. PT = P + Ph = 85 W + 200 W = 285 W c. 0.2  P  = 0.5 Ph 100 W Ph = P  = = 200 W 0.5 0.5 PT = Ph  P = 200 W  100 W = 100 W CHAPTER 23 329 48. a. 208 V 0 Iab = E AB = = 20.8 A 0 R0 10  0 208 V   120 208 V   120 E BC = 14.708 A 165 Ibc = =  R  jX L 10   j10  14.142   45 208 V 120 208 V 120 E CA Ica = = 14.708 A 165  = R  jX C 10   j10  14.142    45 b. IAa + Ica  Iab = 0 IAa = Iab  Ica = 20.8 A 0  14.708 A 165 = 20.8 A  (14.207A + j3.807 A) = 35.007 A  j3.807 A = 35.213 A 6.207 IBb + Iab  Ibc = 0 IBb = Ibc  Iab = 14.708 A 165  20.8 A 0 = (14.207 A  j3.807 A)  20.8 A = 35.007 A  j3.807 A = 35.213 A 173.79 ICc + Ibc  Ica = 0 ICc = Ica  Ibc = 14.708 A 165  14.708 A 165 = (14.207 A + j3.807 A)  (14.207 A  j3.807 A) = 7.614 A 90 c. P1 = VacIAa cos IVAaca Vca = Vca  θ  180 = 208 V 120  180 = 208 V 60 IAa = 35.213 A 6.207 P1 = (208 V)(35.213 A) cos 53.793 = 4.326 kW P2 = VbcIBb cos IVBbbc Vbc = 208 V 120 IBb = 35.213 A 173.79 P2 = (208 V)(35.213 A) cos 53.79 = 4.327 kW d. 330 PT = P1 + P2 = 4.326 kW + 4.327 kW = 8.653 kW CHAPTER 23 49. a. b. c. V = E = E L = 120.09 V 3 120.09 V Ian = V an = = 8.49 A Z an 14.142  120.09 V V = 7.08 A Ibn = bn = Z bn 16.971  120.09 V Icn = V cn = = 42.47 A Z cn 2.828  2 2 10  + I bn 12  + I cn2 2  PT = I an = (8.49 A)2 10  + (7.08 A)2 12  + (42.47 A)2 2  = 720.80 W + 601.52 W + 3.61 kW = 4.93 kW QT = PT = 4.93 kVAR(L) ST = Fp = 50. PT2  QT2 = 6.97 kVA PT = 0.707 (lagging) ST d. Ean = 120.09 V30, Ebn = 120.09 V150, Ecn = 120.09 V90 120.09 V   30 120.09 V   30 = 8.49 A 75 = Ian = E an = 10  + j10  14.142  45 Z an 120.09 V   150 120.09 V   150 Ibn = Ebn = = 7.08 A 195 = 12  + j12  16.971  45 Z bn 120.09 V 90 120.09 V 90 Icn = Ecn = = = 42.47 A 45 2  + j2  2.828  45 Z cn e. IN = Ian + Ibn + Icn = 8.49 A 75 + 7.08 A 195 + 42.47 A45 = (2.02 A  j8.20 A) + (6.84 A + j1.83 A) + (30.30 A + j30.30 A) = 25.66 A  j23.93 A = 35.09 A 43.00 Z1 = 12   j16  = 20  53.13, Z2 = 3  + j4  = 5  53.13 Z3 = 20  0 EAB = 200 V0, EBC = 200 V 120, ECA = 200 V 120 ZΔ = Z1Z2 + Z1Z3 + Z2Z3 = (20  53.13)(5  53.13) + (20  53.13)(20  0) + (5  53.13)(20  0) = 100  0 + 400  53.13 + 100  53.13 = 100  + (240   j320 ) + (60  + j80 ) = 400   j240  = 466.48  30.96 CHAPTER 23 331 Ian = E AB Z3  ECA Z 2 (200 V 0)(20  0)  (200 V 120)(5  53.13) = Z Z 4000 A 0  1000 A 173.13 = 10.71 A 29.59 466.48   30.96 E Z  E AB Z3 (200 V   120)(20    53.13)  (200 V 0)(20  0) Ibn = BC 1 = Z Z = 4000 A   173.13  4000 A 0 = 17.12 A 145.61 466.48   30.96 E Z  E BC Z1 (200 V 120)(5 53.13)  (200 V   120)(20   53.13) Icn = CA 2 = Z Z = 1000 A 173.13  4000 A   173.13 = 6.51 A 42.32 466.48   30.96 2 2 PT = I an 12  + I bn 4  + I cn2 20  = 1376.45 W + 1172.38 W + 847.60 W = 3396.43 W 2 2 QT = I an 16  + I bn 3  = 1835.27 VAR(C) + 879.28 VAR(L) = 955.99 VAR(C) = ST = Fp = 332 PT2  QT2 = 3508.40 VA PT 3396.43 W = 0.968 (leading) = ST 3508.40 VA CHAPTER 23 Chapter 24 1. a. positive-going d. Amplitude = 8 V  2 V = 6 V e. f. g. 2. Vb = 2 V tp = 0.2 ms V1 V 2  100% V 8V + 7.5 V = 7.75 V V= 2 8 V  7.5 V % tilt =  100% = 6.5% 7.75 V 1 1 1 prf = =  = 625 kHz T (2.0 ms  0.4 ms) 1.6 ms tp T  100%  0.2 ms  100% = 12.5% 1.6 ms negative-going d. 8 mV ( from base line level) b. +7 mV c. 3 μs 8 mV  7 mV 15 mV = = 7.5 mV 2 2  8 mV  (7 mV) % Tilt = V 1 V 2  100% =  100% V 7.5 mV 1 mV =  100% = 13.3% 7.5 mV V= f. T = 15 μs  7 μs = 8 μs 1 1 = 125 kHz prf = = T 8 s g. Duty cycle = a. positive-going d. Amplitude = (30  10)mV = 20 mV e. c. % tilt = a. e. 3. b. 3 s tp  100% =  100% = 37.5% T 8 s b. Vb = 10 mV c. 8 tp =   4 ms = 3.2 ms  10  V1 V 2  100% V 30 mV + 28 mV = 29 mV V= 2 30 mV  28 mV % tilt =  100%  6.9% 29 mV % tilt = CHAPTER 24 333 4. tr  (0.2 div.)(2 ms/div.) = 0.4 ms tf  (0.4 div.)(2 ms/div.) = 0.8 ms 5. tilt = V1  V2 V V = 0.1 with V = 1 2 V 2 Substituting V into top equation, V1  V2 0.95 V1 = 0.1 leading to V2 = or V2 = 0.905(15 mV) = 13.58 Mv V1  V2 1.05 2 6. 7. a. tr = 80% of straight line segment = 0.8(2 μs) = 1.6 μs b. tf = 80% of 4 μs interval = 0.8(4 μs) = 3.2 μs c. At 50% level (10 mV) tp = (8  1)μs = 7 μs d. prf = a. T = (4.8  2.4)div. 50  s/div. = 120 μs c. 8. 334 1 1 = = 50 kHz T 20  s b. f= 1 1  = 8.33 kHz T 120 s Maximum Amplitude: (2.2 div.)(0.2 V/div.) = 0.44 V = 440 mV Minimum Amplitude: (0.4 div.)(0.2 V/div.) = 0.08 V = 80 mV T = (3.6  2.0)ms = 1.6 ms 1 1 = 625 Hz prf = = T 1.6 ms tp 0.2 ms Duty cycle =  100% =  100% = 12.5% T 1.6 ms CHAPTER 24 9. T = (15  7)μs = 8 μs 1 1 = 125 kHz prf =  T 8 s tp (20  15)  s 5  100% =  100% = 62.5% Duty cycle =  100%  T 8 s 8 10. T = (3.6 div.)(2 ms/div.) = 7.2 ms 1 1 prf = = = 138.89 Hz T 7.2 ms tp 1.6 div. Duty cycle =  100% =  100% = 44.4% T 3.6 div. 11. a. T = (9  1)μs = 8 μs c. prf = d. Vav = (Duty cycle)(Peak value) + (1  Duty cycle)(Vb) tp 2 s  100% = 25% Duty cycle =  100%  T 8 s Vav = (0.25)(6 mV) + (1  0.25)(2 mV) = 1.5 mV  1.5 mV = 0 V or (2  s)(6 mV)  (2  s)(6 mV) Vav = =0V 8s e. Veff = 12. b. tp = (3  1)μs = 2 μs 1 1  = 125 kHz T 8 s (36  106 )(2 s)  (4 106 )(6 s) = 3.46 mV 8 s Eq. 24.5 cannot be applied due to tilt in the waveform. (Method of Section 13.6) Between 2 and 3.6 ms 1 (3.4 ms  2 ms)(2 V) + (3.6 ms  3.4 ms)(7.5 V) + (3.6 ms  3.4 ms)(0.5 V) 2 Vav = 3.6 ms  2 ms 1 (1.4 ms)(2 V) + (0.2 ms)(7.5 V) + (0.2 ms)(0.5 V) 2 = 1.6 ms 2.8 V + 1.5 V + 0.05 V = = 2.719 V 1.6 CHAPTER 24 335 13. Ignoring tilt and using 20 mV level to define tp tp = (2.8 div.  1.2 div.)(2 ms/div.) = 3.2 ms T = (at 10 mV level) = (4.6 div.  1 div.)(2 ms/div.) = 7.2 ms 3.2 ms t  100% = 44.4% Duty cycle = p  100%  T 7.2 ms Vav = (Duty cycle)(peak value) + (1  Duty cycle)(Vb) = (0.444)(30 mV) + (1  0.444)(10 mV) = 13.320 mV + 5.560 mV = 18.88 mV 14. Vav = (Duty cycle)(Peak value) + (1  Duty cycle)(Vb) tp Duty cycle = (decimal form) T (8  1)  s = 0.35 = 20  s Vav = (0.35)(20 mV) + (1  0.35)(0) = 7 mV + 0 = 7 mV 15. Using methods of Section 13.8: A1 = b1h1 = [(0.2 div.)(50 μs/div.)][(2 div.)(0.2 V/div.)] = 4 μsV A2 = b2h2 = [(0.2 div.)(50 μs/div.)][(2.2 div.)(0.2 V/div.)] = 4.4 µsV A3 = b3h3 = [(0.2 div.)(50 μs/div.)][(1.4 div.)(0.2 V/div.)] = 2.8 μsV A4 = b4h4 = [(0.2 div.)(50 μs/div.)][(1 div.)(0.2 V/div.)] = 2.0 μsV A5 = b5h5 = [(0.2 div.)(50 μs/div.)][(0.4 div.)(0.2 V/div.)] = 0.8 μsV Vav = 16. Using the defined polarity of Fig. 24.57 for υC, Vi = 5 V, Vf = +20 V and τ = RC = (10 k)(0.02 μF) = 0.2 ms a. 336 (4  4.4  2.8  2.0  0.8)  sV = 117 mV 120  s υC = Vi + (Vf  Vi)(1  et/τ) = 5 + (20  (5))(1  et/0.2 ms) = 5 + 25(1  et/0.2 ms) = 5 + 25  25et/0.2 ms υC = 20 V  25 Vet/0.2 ms CHAPTER 24 b. c. Ii = 0 iC = E  C 20 V   20 V  25 V e = R 10 k   t / 0.2 ms   = 2.5 mAet/0.2 ms d. 17. υC = Vi + (Vf  Vi)(1  et/RC) = 8 + (4  8)(1  et/20 ms) = 8  4(1  et/20 ms) = 8  4 + 4et/20 ms = 4 + 4et/20 ms υC = 4 V(1 + et/20 ms) 18. Vi = 10 V, Vf = 2 V, τ = RC = (1 k)(1000 μF) = 1 s υC = Vi + (Vf  Vi)(1  et/τ) = 10 V + (2 V  10 V)(1  et) = 10  8(1  et) = 10  8 + 8et υC = 2 V+ 8 Vet 19. Vi = 10 V, Ii = 0 A CHAPTER 24 τ = RC = (2 k)(10 μF) = 20 ms Using the defined direction of iC (10 V  2 V) t/τ e iC = 1k τ = RC = (1 k)(1000 μF) = 1 s 8 V t iC =  e 1k and iC = 8mAet 337 20. τ = RC = (5 k)(0.04 μF) = 0.2 ms (throughout) υC = E(1  et/τ) = 20 V(1  et/0.2 ms) (Starting at t = 0 for each plot) a. T= 1 1 = = 2 ms f 500 Hz T = 1 ms 2 5τ = 1 ms = b. T= T 2 1 1 = = 10 ms f 100 Hz T = 5 ms 2 5τ = 1 ms = c. T= 1T    5 2  1 1 = = 0.2 ms f 5 Hz T = 0.1 ms 2 T  5τ = 1 ms = 10   2 21. The mathematical expression for iC is the same for each frequency! τ = RC = (5 k)(0.04 μF) = 0.2 ms 20 V  t / 0.2 ms = 4 mAet/0.2 ms and iC = e 5 k a. T= 1 T = 2 ms, = 1 ms 500 Hz 2 5τ = 5(0.2 ms) = 1 ms = 338 b. T= c. T= T 2 1 T = 10 ms, = 5 ms 100 Hz 2 1T  5τ = 1 ms =   5 2  1 T = 0.2 ms, = 0.1 ms 5000 Hz 2 T  5τ = 1 ms = 10   2 CHAPTER 24 22. τ = 0.2 ms as above 1 = 2 ms T= 500 Hz T 5τ = 1 ms = 2 T 0  : υC = 20 V(1  et/0.2 ms) 2 T  T: Vi = 20 V, Vf = 20 V 2 υC = Vi + (Vf  Vi)(1  et/τ) = 20 + (20  20)(1  et/0.2 ms) = 20  40(1  et/0.2 ms) = 20  40 + 40et/0.2 ms υC = 20 V+ 40 Vet/0.2 ms T 23. 3 T : Vi = 20 V, Vf = +20 V 2 υC = Vi + (Vf  Vi)(1  et/τ) = 20 + (20  (20))(1  et/τ) = 20 + 40(1  et/τ) = 20 + 40  40et/τ υC = 20 V  40 Vet/0.2 ms υC = Vi + (Vf  Vi)(1  et/RC) Vi = 20 V, Vf = 20 V υC = 20 + (20  20)(1  et/RC) T = 20 V (for 0  ) 2 For T  T, υi = 0 V and υC = 20 Vet/τ 2 τ with = RC = 0.2 ms T T = 1 ms and 5τ = 2 2 3 T , υi = 20 V 2 υC = 20 V(1  et/τ) For T  For 3 T  2T, υi = 0 V 2 υC = 20 Vet/τ CHAPTER 24 339 24. 25. τ = RC = 0.2 ms T 5τ = 1 ms = 2 Vi = 10 V, Vf = +20 V T 0 : 2 υC = Vi + (Vf  Vi)(1  et/τ) = 10 + (20  (10))(1  et/τ) = 10 + 30(1  et/τ) = 10 + 30  30et/τ υC = +20 V  30 Vet/0.2 ms Zp: Zs: T  T: 2 Vi = 20 V, Vf = 0 V υC = 20 Vet/0.2 ms 1 1 = = 5.31 M 2 fC 2 (10 kHz)(3 pF) (9 M  0)(5.31 M    90) Zp = = 4.573 M 59.5 9 M   j 5.31 M  XC = CT = 18 pF + 9 pF = 27 pF 1 1 = 0.589 M XC = = 2 fC T 2 (10 kHz)(27 pF) Zs = Vscope = (1 M  0)(0.589 M    90) = 0.507 M 59.5 1 M   j 0.589 M  (0.507 M    59.5)(100 V 0) ZsVi = Z s  Z p (0.257 M   j 0.437 M ) + (2.324 M   j 3.939 M ) 50.7 106 V   59.5 1 (100 V 0) = 10 V 0 = 6 10 5.07  10   59.5 =  Z p = 59.5 =  Zs 26. Zp: XC = 1 1 = 3.333 M = 5 C (10 rad/s)(3 pF) (9 M  0)(3.333 M ) = 3.126 M 69.68 9 M   j 3.333 M  1 1 XC = = 0.370 M = 5 C (10 rad/s)(27 pF) Zp = Zs: (1 M  0)(0.370 M    90) = 0.347 M 69.68 1 M   j 0.370 M    Z p = Z s Zs = 340 CHAPTER 24 Vscope = (0.347 M    69.68)(100 V 0) ZsVi = Z s  Z p (0.121 M   j 0.325 M ) + (1.086 M   j 2.931 M ) 34.70  106 V   69.68 3.470  106   69.68 1  10 V 0 = (100 V 0) 10 = CHAPTER 24 341 Chapter 25 1. 2. I: a. no b. no c. yes d. no e. yes II: a. yes b. yes c. yes d. yes e. no III: a. yes b. yes c. no d. yes e. yes IV: a. no b. no c. yes d. yes e. yes b. i= c. 2Im  2 2 2   1 + cos(2ωt  90)  cos(4ωt  90) + cos(6ωt  90) +     3 15 35    1  4     2 2   2  i = 2 I m 1   cos(2ωt  90)  cos(4ωt  90)  cos(6ωt  90)   15 35   4 3  2Im  Im 2Im =  2 d. 2 2 2    cos (2ωt  90)  cos (4ωt  90) + cos (6ωt  90) +  i = 2 I m 1  + 15 35   4 3  3. 342 a. υ = 4 + 2 sin α b. υ = (sin α)2 CHAPTER 25 c. 4. i = 2  2 cos α a. b. CHAPTER 25 343 5. a. b. c. 344 CHAPTER 25 6. Vav = 100 V a. Veff = (50 V)2 + (25 V) 2 = 107.53 V 2 Iav = 3 A b. Ieff = 7. (100 V)2 + (3 A)2 + (2 A)2 + (0.8 A)2 = 3.36 A 2 a. Veff = (20 V) 2 + (15 V)2 + (10 V)2 = 19.04 V 2 b. Ieff = (6 A) 2 + (2 A) 2 + (1 A)2 = 4.53 A 2 8. PT = V0I0 + V1I1 cos θ1 +  + VnIn cos θn (50 V)(2 A) (25 V)(0.8 A) cos 53 + cos 70 = (100 V)(3 A) + 2 2 = 300 + (50)(0.6018) + (10)(0.3420) = 333.52 W 9. P= 10. a. DC: E = 18 V, Io = b. Ieff = (20 V)(6 A) (15 V)(2 A) (10 V)(1 A) cos 20 + cos 30 + cos 60 2 2 2 = 60(0.9397) + 15(0.866) + 5(0.5) = 71.87 W E 18 V = 1.5 A = R 12  ω = 400 rad/s: XL = ωL = (400 rad/s)(0.02 H) = 8  Z = 12  + j8  = 14.42  33.69 E 30 V/ 2 0 2.08 A 33.69 I= = = Z 14.42  33.69 2  2.08  i = 1.5 + 2   sin(400t  33.69)  2  i = 1.5 + 2.08 sin(400t  33.69) CHAPTER 25 (1.5 A)2 + (2.08 A) 2 = 2.10 A 2 345 c.  2.08 A    33.69  (12  0) DC: υR = E = 18 V, VR =  2   24.96 V =   33.69 2  24.96  υR = 18 + 2   sin(400t  33.69)  2  υR = 18 + 24.96 sin(400t  33.69) d. 2 V Reff = (18 V ) + e. DC: VL = 0 V (24.96 V ) 2 = 25.21 V 2  2.08 A  VL =    33.69  (8  90) 2   16.64 V = 56.31 2 υL = 0 + 16.64 sin(400t + 56.31) ω = 400 rad/s: 11. 346 (16.64 V)2 = 11.77 V 2 f. 2 V Leff = 0 + g. 2 R = (2.101 A)2 12  = 52.97 W P = I eff a. DC: IDC = b. Ieff = c. υR = iR = i(12 ) = 24 + 24.96 sin(400t  33.69) + 6 sin(800t  53.13) d. Veff = 24 V =2A 12  ω = 400 rad/s: Z = 12  + j(400 rad/s)(0.02 H) = 12  + j8  = 14.422  33.69 30 V 0 = 2.08 A 33.69 (peak values) I= 14.422  33.69 ω = 800 rad/s: Z = 12 Ω + j(800 rad/s)(0.02 H) = 12  + j16  = 20  53.13 10 V 0 = 0.5 A 53.13 (peak values) I= 20  53.13 i = 2 + 2.08 sin(400t  33.69) + 0.5 sin(800t  53.13) (2 A) 2 + (2.08 A)2 + (0.5 A 2 ) = 2.51 A 2 (24 V) 2 + (24.96 V) 2 + (6 V)2 = 30.09 V 2 CHAPTER 25 e. 12. DC: VL = 0 V ω = 400 rad/s: VL = (2.08 A 33.69)(8  90) = 16.64 V 56.31 ω = 800 rad/s: VL = (0.5 A 53.13)(16  90) = 8 V 36.87 υL = 0 + 16.64 sin(400t + 56.31) + 8 sin(800t + 36.87) (0) 2 + (16.64 V) 2 + (8 V) 2 = 13.06 V 2 f. Veff = g. 2 PT = I eff R = (2.508 A)2 12  = 75.48 W a. DC: I =  b. Ieff = c. DC: V = IR = (5 A)(12 ) = 60 V VR = (1.054 A 26.57)(12  0) = 12.648 V 26.57 ω = 600 rad/s: VR = (0.417 A 45)(12  0) = 5 V 45 υR = 60 + (1.414)(12.648)sin(300t  26.57)  (1.414)(5)sin(600t  45) υR = 60 + 17.88 sin(300t  26.57)  7.07 sin(600t  45) d. 2 V Reff = (60 V ) + 60 V = 5A 12  ω = 300 rad/s: XL = ωL = (300 rad/s)(0.02 H) = 6  Z = 12  + j6  = 13.42  26.57 E = (0.707)(20 V) 0 = 14.14 V 0 E 14.14 V 0 = 1.054 A 26.57 I= = Z 13.42  26.57 ω = 600 rad/s: XL = ωL = (600 rad/s)(0.02 H) = 12  Z = 12  + j12  = 16.97  45 E = (0.707)(10 V) 0 = 7.07 V 0 7.07 V 0 E = 0.417 A 45 I= = 16.97  45 Z i = 5 + (1.414)(1.054)sin(300t  26.57)  (1.414)(0.417)sin(600t  45) i = 5 + 1.49 sin(300t  26.57)  0.59 sin(600t  45) (5 A) 2 + (1.49 A) 2 + (0.59 A) 2 = 5.13 A 2 ω = 300 rad/s: CHAPTER 25 (17.88 V ) 2 + (7.07 V ) 2 = 61.52 V 2 347 e. DC: VL = 0 V ω = 300 rad/s: ω = 600 rad/s: VL = (1.054 A 26.57)(6  90) = 6.324 V 63.43 VL = (0.417 A 45)(12  90) = 5 V 45 υL = 0 + (1.414)(6.324)sin(300t + 63.43)  (1.414)(5)sin(600t + 45) υL = 8.94 sin(300t + 63.43)  7.07 sin(600t + 45) 13. f. (8.94 V ) 2 + (3.54 V ) 2 = 6.8 V V Leff = 2 g. 2 P = I eff R = (5.13 A)2 12  = 315.8 W a. DC: I = 0 A 1 1 = = 20  ωC (400 rad/s)(125  F) Z = 15   j20  = 25  53.13 E = (0.707)(30 V) 0 = 21.21 V 0 E 21.21 V 0 = 0.848 A 53.13 I= = Z 25    53.13 i = 0 + (1.414)(0.848)sin(400t + 53.13) i = 1.2 sin(400t + 53.13) ω = 400 rad/s: b. c. 348 XC = (1.2 A) 2 = 0.85 A as above 2 Ieff = DC: VR = 0 V ω = 400 rad/s: VR = (0.848 A 53.13)(15  0) = 12.72 V 53.13 υR = 0 + (1.414)(12.72)sin(400t + 53.13) υR = 18 sin(400t + 53.13) (18 V ) 2 = 12.73 V 2 d. V Reff = e. DC: VC = 18 V ω = 400 rad/s: VC = (0.848 A 53.13)(20  90) = 16.96 V 36.87 υC = 18 + (1.414)(16.96)sin(400t  36.87) υC = 18 + 23.98 sin(400t  36.87) f. 2 V C eff = (18 V ) + g. 2 R = (0.848 A)2 15  = 10.79 W P = I eff (23.98 V ) 2 = 24.73 V 2 CHAPTER 25 14. a. 400 400 cos 4ωt cos 2ωt   3 15  = 63.69 + 42.46 sin(2ωt + 90)  8.49 sin(4ωt + 90) ω = 377 rad/s: e = 63.69 + 42.46 sin(754t + 90)  8.49 sin(1508t + 90) e= 200  DC: XL = 0  VL = 0 V 1 ω = 754 rad/s: XC = 1 = = 1330  ωC (754 rad/s)(1  F) XL = ωL = (754 rad/s)(0.1 H) = 75.4  Z = (1 k 0)  75.4  90 = 75.19  85.69 E = (0.707)(42.46 V) 90 = 30.02 V 90 Z(E) (75.19  85.69)(30.02 V 90) Vo = = 1.799 V 94.57 = Z + ZC 75.19  85.69 + 1330    90 ω = 1508 rad/s: XC = 1 1 = = 6631.13  ωC (1508 rad/s)(1  F) XL = ωL = (1508 rad/s)(0.1 H) = 150.8  Z = (1 k 0)  150.8  90 = 149.12  81.42 E = (0.707)(8.49 V) 90 = 6 V 90 Z(E) (149.12  81.42)(6 V 90) Vo = =  Z + ZC 149.12  81.42 + 6631.13    90 = 1.73 V 101.1 υo = 0 + 1.414(1.799)sin(754t  94.57)  1.414(1.73)sin(1508t  101.1) υo = 2.54 sin(754t  94.57)  2.45 sin(1508t  101.1) 15. b. Voeff = c. P= (2.54 V ) 2 + (2.45 V ) 2 = 2.50 V 2 ( V eff ) 2 (2.50 V ) 2 = 6.25 mW = R 1k i = 0.318Im + 0.500 Im sin ωt  0.212Im cos 2ωt  0.0424Im cos 4ωt +  (Im = 10 mA) i = 3.18  103 + 5  103 sin ωt  2.12  103 sin(2ωt + 90)  0.424  103 sin(4ωt + 90) +  i  3.18  103 + 5  103 sin ωt  2.12  103 sin(2ωt + 90) DC: Io = 0 A, Vo = 0 V ω = 377 rad/s; XL = ωL = (377 rad/s)(1.2 mH) = 0.452  1 1 XC = = = 13.26  C  377 rad/s  (200  F) Z = 200   j13.26  = 200.44  3.79 I = (0.707)(5  103)A 0 = 3.54 mA 0 ZLI (0.452  90)(3.54 mA 0) = 7.98 μA 93.66 Io = =  ZL + Z j 0.452  + 200   j13.26  CHAPTER 25 349 Vo = (7.98 μA 93.66)(200  0) = 1.596 mV 93.66 XL = ωL = (754 rad/s)(1.2 mH) = 0.905  1 1 = = 6.63  XC = ωC (754 rad/s)(200  F) ω = 754 rad/s: Z = 200   j6.63  = 200.11  1.9 I = (0.707)(2.12 mA) 90 = 1.5 mA 90 Z LI (0.905  90)(1.5 mA 90) Io = = 6.8 μA 181.64 = Z L + Z j 0.905  + 200   j 6.63  Vo = (6.8 µA 181.64)(200  0) = 1.36 mA 181.64 υo = 0 + (1.414)(1.596  103)sin(377t + 93.66)  (1.414)(1.360  103)sin(754t + 181.64) 3 υo = 2.26  10 sin(377t + 93.66) + 1.92  103 sin(754t + 1.64) 16. a. b. 17. 350 60 + 70 sin ωt + 20 sin(2ωt + 90) + 10 sin(3ωt + 60) +20 + 30 sin ωt  20 sin(2ωt + 90) + 5 sin(3ωt + 90) DC: 60 + 20 = 80 ω: 70 + 30 = 100  100 sin ωt 2ω: 0 3ω: 10 60 + 590 = 5 + j8.66 + j5 = 5 + j13.66 = 14.55 69.9 Sum = 80 + 100 sin ωt + 14.55 sin(3ωt + 69.9) 20 + 60 sin α + 10 sin(2α  180) + 5 sin(3α + 180) 5 + 10 sin α + 0  4 sin(3α  30) DC: 20  5 = 15 α: 60 + 10 = 70  70 sin α 2α: 10 sin(2α  180) 3α: 5 180  4 30 = 5  [3.46  j2] = 8.46 + j2 = 8.69 166.7 Sum = 15 + 70 sin α + 10 sin(2α  180) + 8.69 sin(3α + 166.7) iT = i1 + i2 = 10 + 30 sin 20t  0.5 sin(40t+ 90) +20 + 4 sin(20t + 90) + 0.5 sin(40t+ 30) DC: 10 A + 20 A = 30 A ω = 20 rad/s: 30 A 0 + 4 A 90 = 30 A + j4 A = 30.27 A 7.59 ω = 40 rad/s: 0.5 A 90 + 0.5 A 30 = j0.5 A + 0.433 A + j0.25 A = 0.433 A  j0.25 A = 0.5 A 30 iT = 30 + 30.27 sin(20t + 7.59) + 0.5 sin(40t  30) CHAPTER 25 18. e = υ1 + υ2 = 20  200 sin 600t + 100 sin(1200t + 90) + 75 sin 1800t + 50 sin(1800t + 60) 10 + 150 sin(600t + 30) +0 DC: 20 V  10 V = 10 V ω: 600 rad/s: 200 V 0 + 150 V 30 = 102.66 V 133.07 ω = 1200 rad/s: 100 sin(1200t + 90) ω = 1800 rad/s: 75 V 0 + 50 V 60 = 108.97 V 23.41 e = 10 + 102.66 sin(600t + 133.07) + 100 sin(1200t + 90) + 108.97 sin(1800t + 23.41) CHAPTER 25 351 Chapter 26 Ei V 1.05  1.00 V 50 mV  ; Ii = R  = 1.064 mA Ii R 47  47  E 1.05 V = 986.84 Ω Zi = i  I i 1.064 mA 1. Zi = 2. Zi = 3. a. I i1  b. Z i2  c. Ei3  I i3 Z i3 = (1.5 mA)(4.6 kΩ) = 6.9 V Ei 120 V 0 = 19.35 Ω  10.8° = 19 Ω + j3.623 Ω  I i 6.2 A   10.8 XL 3.623  f = 60 Hz: R = 19 Ω, L =  = 9.61 mH 2 f 2 (60 Hz) Ei1  Z i1 Ei2 I i2 E g  Eo 20 mV = 10 μA 2 k  1.8 V = 4.5 kΩ 0.4 mA 4 V  3.8 V 0.2 V = 0.1 mA(p  p)  2 k 2 k Rs E 3.8 V(p  p ) = 38 kΩ Zo = o  I o 0.1 mA( p  p )  4. Io = 5. Eopeak  Eg peak  VRpeak = 2 V  0°  40 × 103 V  0° = 1.96 V  0° VRpeak 40 mV = 43.96 µA Rs 0.91 k E 1.96 V 0 Zo = o  = 44.59 kΩ 43.96  A IR Ipeak = 6.  Eopeak  2 0.6 V(rms) = 0.849 V   Eo( p p )  2 Eo( peak ) = 2(0.849 V) = 1.697 V E g  Eo 18 V  1.697 V = 51.5 µA(p  p) 2 k Rs E 1.697 V( p  p) Zo = o  = 32.95 kΩ I o 51.5  A(p  p ) Io = 352  CHAPTER 26 7. Zo = Eo p p  I o p p Eg p p  VRp p I o p p  0.8 V  0.4 V = 10 kΩ 40  A VR p p  2 div  0.2 V/div. = 0.4 V E g p p  4 div  0.2 V/div. = 0.8 V I o p p  8. VR p p 10 k  0.4 V = 40 μA 10 k Ei = IiZi = (10 µA  0°)(1.8 kΩ  0°) = 18 mV  0° E i(peak)  2 (18 mV) = 25.46 mV Ei( p p ) = 2(25.46 mV) = 50.92 mV A NL  9. Eo 4.05 V180 = 79.54  180° = 79.54  Ei 50.92 mV 0 Eo RL (5.6 k) = 392.98  A NL  (3200) Ei 5.6 k  40 k RL  Ro E E E  o  o i E g Ei E g a. A  b. AT with Ei = AT  10. Zi E g Zi  Rg and Ei Zi  E g Zi + Rg Eo Zi (2.2 k)   (392.98) = 320.21 2.2 k  0.5 k Ei Zi  Rg 1400 mV = 1200 1.2 mV 0 192 mV A  = 160 1.2 mV  A  Ro  RL  NL  1  A  A NL    1200   4.7 k   1  160   30.55 k 11. a. Eo RL  A NL RL  Ro Ei 2 k 160 = A NL = A NL (0.0667) 2 k  28 k A NL = 2398.8 A  CHAPTER 26 353 b. c. Eo = IoRL = (4 mA)(2 kΩ) = 8 V E A = o = 160 Ei Eo 8 V  = 50 mV Ei = 160 160 V Ii = Zi = 12. E g  Ei Rg  70 mV  50 mV = 50 µA 0.4 k Ei 50 mV = 1 kΩ  I i 50  A Ri RL  Ro (3200)(2.2 k) = 5.6 k  40 k = 154.39 a. Ai =  A NL b.  Rg  Z i  A iT   AT   RL   A Z    i  Zi  R g A iT   A   Rg  Z i      RL   Zi RL  Z i    A NL RL RL  Ro  RL    A NL Zi RL  Ro  (3200)(2.2 k) 5.6 k  40 k  154.39  13. c. Same result since Ii = Ig a. A G  A2 Ri RL  ( 392.98) 2 A = A NL 2.2 k 5.6 k  6.067 ×104 A G   A A i  (392.98)(154.39)  6.067 ×104 RL RL  Ro 5.6 k    (3200)   5.6 k  40 k   392.98 A i =  A NL Ri RL  Ro 2.2 k    (3200)    5.6 k  40 k   154.39 354 CHAPTER 26 b. AT  A Zi 2.2 k    (392.98)  = 320.21  2.2 k  0.5 k  Zi  Rg A iT   AT Rg  Z i RL  0.5 k  2.2 k   (320.21)   = 154.39  5.6 k  Rg  Ri   0.5 k  2.2 k   ( 320.21) 2  = 4.94 × 104 A GT   A2T      5.6 k  RL  A GT   AT A iT  (320.21)(154.39) = 494 × 104 14. a. Ai  Io Z   A i Ii RL (160)(0.75 k) 2 k  60  b. A GT  PL  A2T Pg AT  A  Rg  Ri   R  L Zi Z i  Rg (160)(0.75 k)  104.35 0.75 k  0.4 k  0.4 k  0.75 k  A GT  (104.35) 2    2 k = 6.261 × 103  15. a. AT  A1  A2 = (30)(50) = 1500 b. A iT  AT c. A i1   A1 ZiL A i2   A2 16. RL  1 k  = 187.5  (1500)   8 k   1 k    (30)  = 15  2 k  RL1 Zi1 Zi2 RL2  2 k   (50)  = 12.5  8 k  d. A iT  A i1  A i2 = (15)(12.5) = 187.5 as above a. AT  A1  A2  A3   6912 = (12) A 2 (32) A2 = 18 CHAPTER 26 355 b. A i1  4=  A1 Zi1 RL1   A1 Zi1 Zi2  (12)(1 k) Z i2 Z i2 = 3 kΩ c. A i3   A3 Zi3 RL3   (32)(2 k) 2.2 k = 29.09 A iT  A i1  A i2  A i3 = (4)(26)(29.09) = 3.025 × 103 17. a. z11 = z11 = E1 I1  Z1  (Z 2  Z3 ) I2 0 Z1Z 2  Z1Z 3 Z1 + Z 2  Z3 Z3I 2 Z1 + Z 2  Z3 (Z3I 2 )(Z1 ) E1 = I1Z1 = Z1 + Z 2  Z3 I= z12 = z21 = E2 I1 E2 I2  I1  0 Z1I 3 Z1 + Z 2  Z3 Mirror image of z12 I2 0  z 21  z22 = E1 I2 Z1Z 3 Z1 + Z 2  Z3 Mirror image of z11 I1  0  z 22  Z3  (Z1  Z 2 )  356 Z1Z3  Z 2 Z3 Z1 + Z 2  Z3 CHAPTER 26 b. 18.  z11  a. E1 I1 I2 0 z11  R4  R2  ( R1  R3 ) `  R4  z12  E1 I2 R2 ( R1  R3 ) R1  R2  R3 I1  0 R2 (I 2 ) ( R1  R2 )  R3 E1  I R2  I 2 R4 I   and z12 = R2 R3I 2  R4 I 2 R1  R2  R3 R2 R3 R R  R4 ( R1  R2  R3 ) E1   R4  2 3 R1  R2  R3 I 2 R1  R2  R3 E2 = I R3  I1 R4 R2 (I1 ) CDR: I  = ( R1  R3 )  R2 R2 R3I1 E2 = + I1R4 R1  R2  R3 and z21 = z 22  E2 I2 R2 R3 R R  R4 ( R1  R2  R3 ) E2   R4  2 3 R1  R2  R3 I1 R1  R2  R3 I1  0 Z22 = R4  R3  ( R1  R2 ) R ( R  R2 ) = R4  3 1 R3  ( R1  R2 ) CHAPTER 26 357 19. a. y11 = I1 E1 YT  Y1  (Y2  Y3 ) E2  0  Y1 (Y2  Y3 ) Y1  Y2  Y3  Y1Y2  Y1Y3 Y1  Y2  Y3 Nodal Analysis: V [Y1  Y2  Y3 ]  E2Y2 V  = I1/Y1 and  I1 [Y1  Y2  Y3 ]  E 2 Y2 Y1 y12 = y21 = I2 E1 I2 E2  E1  0  Y1Y2 Y1  Y2  Y3 Mirror image of y12 E2  0  y 21  y22 = I1 E2  Y1Y2 Y1  Y2  Y3 Mirror image of y11 E1  0 y 22  YT  Y2  (Y1  Y3 )  20. a. y11 = I1 E1 Y1Y2  Y1Y3 Y1  Y2  Y3 E2  0 y11  Y1  Y2  Y4  Y1   358 Y2 Y4 Y2  Y4 Y1 (Y2 + Y4 )  Y2 Y4 Y2  Y4 CHAPTER 26 y21 = E1 = I2 E1 (using the above diagram) E2  0 I  1 I2 I  1   (E  + E) =   2  2    I 2   Y1  Y2 Y4   Y2 Y4   Y  Y2  I YY and E1 = I 2  4 with y21 = 2   2 4  E1 Y2  Y4  Y4 Y2  y11 = E2 = E2  0 I  1 I1 I  1   (E  + E  ) =   1  1   I1   Y3  Y2 Y4   Y2 Y4  and y12 =  y22 = I2 E1 I2 E2 Y2 Y4 = y21 Y2  Y4 y22 = Y3 + Y2  Y4 = Y3 + E1  0 = Y2 Y4 Y2  Y4 Y3 (Y2 + Y4 ) + Y2 Y4 Y2  Y4 21. h11 = E1 I1 = ZT = Z1  Z 2  E2  0 Z1Z 2 Z1 + Z 2 Using the above figure: Z1 (I1 ) CDR: I2 = Z1 + Z 2 h11 = I2 I1 CHAPTER 26 = E2  0 Z1 Z1 + Z 2 359 h12 = E1 E2 I1  0 VDR: E1 = h12 = E1 E2 Z1E 2 Z1 + Z 2 = I1  0 Z1 Z1 + Z 2 Using above figure: h22 = I2 E2 : Z  = Z 3  (Z1  Z 2 )  I1  0 Z1Z3  Z 2 Z3 Z1 + Z 2  Z3 1 Z1  Z 2 + Z3 h22 =  Z  Z1Z3  Z 2 Z3 22. a. h11 = E1 I1 E2  0 = Z i  R1  (R 2  R 3  R 4 ) h12 = E1 E2 I1  0 E1 = I R2  I 2 R4 R3 (I 2 ) I  R1  R2  R3 R2 R3I 2  E1   I 2 R4 R1  R2  R3 E E2 and I2 = 2  Z  R4  R3  ( R1  R 2 )      R2 R3 E2  E1 =   R4    R1  R2  R3   R  R3 R1  R2 R3   4 R1  R2  R3  and h12 = 360 R2 R3  R4 ( R1  R2  R3 ) E1  E 2 R1 R3  R2 R3  R4 ( R1  R2  R3 ) CHAPTER 26 Z  R2  R3  R4 (Z  )(I1 ) I R1  Z  + R1 RI I  1 1 R1  Z  R4 I  R4  R1 (I1 )     R4  R3 R4  R3  R1  Z   R1 R4 I1  ( R3  R4 )( R1  Z ) I R3  I 2 =  I R1  I R3  h22 = I2 I1 E2  0  Z  R1 R4     Z  + R1 ( R3  R4 )( R1  Z )   h22 = I2 E2  Z I1 R1 R4 I1  Z  + R1 ( R3  R4 )( R1  Z )  I1  0 R1 R4  1  Z   R1  Z   R3  R4  1 ZT ZT  R4  R3  ( R1  R2 ) 1 h22 = R4  R3  ( R1  R2 ) A Y- conversion would have simplified the problem to on similar to Fig 26.70. 23. h11 = E1 I1 E2  0 Y   Y1  (Y2  Y3 ) Y1Y2  Y1Y3 Y1  Y2  Y3 1 Y1  Y2 + Y3 h11 =  Y  Y1Y2  Y1Y3 Y  h21 = I2 I1 CHAPTER 26 E2  0 361 From above figure:  Z3I1 I1/Y3 CDR: I2 =  Z3 + Z 2 1/Y3  1/Y2 and h21 = I2 I1  E2  0 1/Y3  Y2  1/Y3  1/Y2 Y2  Y3 h12 = E1 E2 I1  0 Z 3E2 1/Y3E 2  Z3 + Z 2 1/Y3  1/Y2 Y2 E 2 And E1 = Y2  Y3 VDR: E1 = with h12 = h22 = h22 = I2 E2 I2 E2 Y = I1  0 I1  0 : CDR  I2 = E2  0 = and h21 =  362 I1  0 Y2 Y2 + Y3 Y2 Y3 Y2 + Y3 = Y = h11 = I2 I1 = Y2  Y3 (from above figure) Y2 + Y3 24. h21 = E1 E2 E1 I1  E2  0 1 1  YT Y1  Y2  Y4  Z1 (I1 ) 1/ Y1 (I1 )  Z1 + Z 2 + Z 4 1/Y1  1/Y2  1/Y4 1/ Y1 (I1 ) Y2 Y4  Y1Y4  Y1Y2 Y1Y2 Y4 Y2 Y4 Y2 Y4  Y1Y4  Y1Y2 CHAPTER 26 h11 = E1 E2 I1  0 VDR: E1   I2 E2 1/Y1 (E 2 ) 1/Y1  1/Y2  1/Y4 Y2 Y4 Y2 Y4  Y1Y4  Y1Y2 and h12 = h22 = Z1 (E 2 ) Z1 + Z 2 + Z 4 = YT (using the above figure) I1  0 YT = Y3  Y1  Y2  Y4 Y1Y2 Y4 = Y3  Y1Y2  Y1Y4  Y2 Y4 25. a. b. Eq. 26.45: hf Ai =  1  ho Z L  a. = 47.62 Zi = h f ZL hi (1  ho Z L )  hr h f Z L 50(2 k)  99 1 k(1  0.05)  (4  104 )(50)(2 k) hr h f Z L E1  hi  I1 1  ho Z L  1 k  b.  1  1  (2 k)  40 k  Eq. 26.46: A  26. 50 Zo = ho  CHAPTER 26 (4  104 )(50)(2 k)  961.9   1   1  (2 k )  40 k  1 hr h f hi  Rs  1  200 k 1 (4  104 )(50)  40 k 1 k  0 363 27. Z11 = 1 kΩ  0°, z12 = 5 kΩ  90°, z21 = 10 kΩ  0°, z22 = 2 kΩ  j4 kΩ, ZL = 1 kΩ  0° z z E (5 k 90)(10 k) = 9,219.5 Ω  139.40°  z11  12 21  1 k  I z 22  Z L 2 k  j 4 k  1 k E z z (5 k 90)(10 k) = 29.07 kΩ  86.05° Zo = 2  z 22  12 21  2 k  j 4 k  I2 Rs  z11 1 k  1 k Zi = 28. 1/ y 22 Z L 1/ y 22 + Z L ZL  1  y 22 Z L 1/ y 22  Z L   ZL  E2 = y21E1   1  y 22 Z L    ZL  I1 = E1y11 + y12E2 = E1y11 + y12   y 21E1    1  y 22 Z L    I1 y y Z  y11  12 21 L 1  y 22 Z L E1 E 1 and Zi = 1  y y Z I1 y11  12 21 L 1  y 22 Z L 364 CHAPTER 26 1 Rs  y12 E 2  y12 E 2  y12 Rs E 2   E1 = 1 Y y11 Rs  1 y11  Rs Y   y11   y R E  I2 = y21E1 + y22E2 = y22  12 s 2  + y22E2  y11 Rs  1  I2 y y R   12 21 s + y22 E2 y11 Rs  1 and Zo = 29. 30. E2 I2 = E1  0 1 y y R y 22  12 21 s 1  y11 Rs z11z 22  z12 z 21 (4 k)(4 k)  (2 k)(3 k) = 2.5 kΩ  4 k z 22 z 2 k h12 = 12  = 0.5 z 22 4 k z 3 k = 0.75 h12 =  21  z 22 4 k 1 1 = 0.25 mS h22 =  z 22 4 k h11 = a. h = h11h22  h12h21 = (103)(20 × 106)  (2 × 104)(100) = 20 × 103  20 × 103 = 0  h 2  104 = 10 Ω z11 = h = 0 Ω, z12 = 12  Z 22 h12 20  106 S h 21 100 1 = 5 MΩ, z22 = z21 = = 50 Ω  6 h 22 h 22 20  10 S b. y11 = h12 2  104 1 1 = 2 × 107 S = 103 S, y12 =   3 h11 h11 10  103   h 100 y21 = 21  3 = 100 × 103 S, y22 = h = 0 S h11 h11 10  CHAPTER 26 365

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Introductory Circuit Analysis Twelfth Edition

Introductory Circuit Analysis Twelfth Edition

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