ENGR 43 Solutions

ENGR 43 Solutions - IRWIANS-A1-A10-hr 12/27/04 11:41 AM...

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CHAPTER 1 Answers to Selected Problems 1.1 1.7 1.11 (a) (b) (c) (d) 1.14 (a) (b) 1.18 (a) (b) 1.21 absorbed P 2 = 48 W P 2 = 48 W P 1 =- 72 W P 2 = 24 W P 1 = 12 W V 1 12 V I 2 A 5 V , + top 5 V , + bottom 5 V , + bottom 5 V , + top W = 1440 J I = 2 A 1.24 (a) absorbed absorbed supplied (b) absorbed absorbed supplied supplied 1.28 absorbed absorbed absorbed supplied P D.S. = 8 W P 36 V = 144 W P 3 = 56 W P 2 = 48 W P 1 = 48 W P 24 V = 48 W P 3 = 16 W P 2 = 24 W P 1 = 40 W P 12 V = 24 W P 2 = 8 W P 1 = 16 W CHAPTER 2 2.2 2.6 2.11 2.15 2.19 2.25 2.30 2.39 2.45 2.49 2.53 R AB = 2 k V R AB = 12 k V I L = 2.4 mA V o = 36 V V S 2 V V x = 10 V V be = 9 V V da 9 V I x 3 mA I 2 = 3 mA I 1 = 6 mA G x = 200 m S R x = 5 k V 2.57 2.64 2.68 2.72 2.77 2.82 2.87 2.92 2.98 2.103 2.109 2.113 I o = 0.67 A I o = 2 A g = 4 I o 4 mA V S = 30 V V S = 41 V V o = 3 V V o 12 V V S 92.4 V I o = 0.33 mA I 1 3 mA P min = 175.1 mW I min = 17.51 mA P max = 182.2 mW I max = 18.22 mA CHAPTER 3 3.1 3.5 3.9 V o 3.6 V I o = 0.6 mA I o = 1 mA 3.14 3.20 V B 0.33 V V A = 2.5 V I o 1 mA IRWIANS-A1-A10-hr 12/27/04 11:41 AM Page A1

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3.27 3.32 3.36 3.41 3.46 3.49 V o = 15 V V o = 1.33 V V o = 8 V V 4 = 6.5 V V 3 = 12 V V 2 = 12 V V 1 = 11.5 V V o = 5 V V o = 2 V 3.54 3.59 3.63 3.69 3.73 3.79 3.84 3.89 3.95 I o = 2.88 mA V o =- 3.33 V V o = 6 V V o = 7.57 V I o = 1.64 mA I o = 2.67 mA V o = 6.67 V I o = 2.88 mA V o 5 V A2 ANSWERS TO SELECTED PROBLEMS CHAPTER 4 4.1 4.9 I o = 606 m A V o V in = 7.06 1.5 0.5 0.75 V –2.25 V v o (V) t (s) 0.5 –0.5 –1.5 –2.5 –2 –1 1 0 12 4.12 4.17 (a) (b) (c) 4.21 4.25 4.30 4.35 4.38 v o = v 1 c 1 + R 3 R 2 + R 3 R 4 da - R 2 R 1 b V o 11.43 V V o 11.2 V V o = 0 V i o v 1 1 R I 4 V # V 2 # 28 V V o = 2 V V o = 4V 1 - V 2 I 3 = 1 mA I 2 = 0 A I 1 = 1 mA CHAPTER 5 5.1 5.5 5.9 5.14 5.19 5.22 5.28 5.33 5.37 5.44 I o = 1.6 mA V o = 65.79 V I o = 5 mA V o = 4.8 V I o = 1.6 mA I o = 0.909 mA V o = 14.4 V I o 2 mA I o 3.2 mA I o = 1.14 mA 5.50 5.55 5.59 5.65 5.70 5.76 5.81 5.85 5.94 5.98 R L = 99.0 V I o = 1.6 mA I o 0.5 mA V o = 2 V P 2 mA = 12 mW V o 5 V V o 0.316 V R Th = 400 V I o = 0.909 mA I o = 2.57 mA IRWIANS-A1-A10-hr 12/27/04 11:41 AM Page A2
ANSWERS TO SELECTED PROBLEMS A3 CHAPTER 6 6.3 6.6 6.9 6.13 6.17 6.22 (a) (b) 6.27 v ( t ) (V) 2.5 0 –2.5 21 0 8 6 4 t (ms) w ( t ) = 0.2 sin 2 377t J v ( t ) = 75.4 cos 377t V v ( t ) = c 0 2500t 2 V 4 V t 6 0 6 t 6 t 7 0 t 1 t 1 i ( t ) = e 0 - 12 0 8 0 mA mA t 6 0 6 t 6 t 1 6 t 6 t 2 6 t 6 t 7 0 t 1 t 2 t 3 t 3 v ( t ) = c 0 100 0.2 V t V V t 6 0 6 t 6 t 7 0 t 1 t 1 i ( t ) = 9.23 cos 377t A v = 40 V 6.33 6.38 (a) (b) (c) 6.43 6.49 6.54 6.60 6.67 6.73 v o ( t ) = 3.81 cos ( 377t ) V L AB = 6 mH C eq = 3.18 m F C T = 2 m F V 1 = 4 V V 2 = 8 V w C = 144 J P R 2 = 21.33 W L = 68 m H ; 10 % ±61.2 m H # L # 74.8 m H L = 2 nH ; 5 % ±1.9 nH # L # 2.1 nH L = 10 mH ; 10 % 9 mH # L # 11 mH i ( t ) = d 0 A t y 2 A ( 3 * 10 - 3 - t ) A 0 A t 6 0 # t # 2 ms # t # t \$ 0 2 ms 3 ms 3 ms CHAPTER 7 7.3 7.8 v o (V) t (s) 0 10 20 30

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ENGR 43 Solutions - IRWIANS-A1-A10-hr 12/27/04 11:41 AM...

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