\u03a9 38 Z 1 2 k \u03a9 Z 2 5 k \u03a9 I 2 I 3 I 5 V I 5 Z 2 I 2 I 3 Z 2 E oc

# Ω 38 z 1 2 k ω z 2 5 k ω i 2 i 3 i 5 v i 5 z 2 i 2

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Ω 0 °
CHAPTER 18 247 38. Z 1 = 2 k Ω 0 ° Z 2 = 5 k Ω 0 ° I 2 = I 3 + I 5 V = I 5 Z 2 = ( I 2 I 3 ) Z 2 E oc = E Th = 21 V = 21( I 2 I 3 ) Z 2 = 2 2 1 21 oc E I Z Z 2 1 1 + 21 oc Z E Z = 21 Z 2 I 2 E oc = 2 2 2 1 21 21(5 k 0 )(2 mA 0 ) = 5 k 0 1 + 21 1 + 21 2 k 0 Ω ∠ ° ∠ ° Ω ∠ ° Ω ∠ ° Z I Z Z E Th = E oc = 3.925 V 0 ° 20 V V V = 0 and I N = I sc = I 2 = 2 mA 0 ° Z N = 3.925 V 0 = 2 mA 0 oc sc ∠ ° ∠ ° E I = 1.96 k Ω 39. a. Z 1 = 3 Ω + j 4 Ω , Z 2 = j 6 Ω Z Th = Z 1 || Z 2 = 5 Ω 53.13 ° || 6 Ω ∠− 90 ° = 8.32 Ω ∠− 3.18 ° Z L = 8.32 Ω 3.18 ° = 8.31 Ω j 0.46 Ω
248 CHAPTER 18 E Th = 2 2 1 + Z E Z Z = (6 90 )(120 V 0 ) 3.61 33.69 Ω ∠ − ° ∠ ° Ω ∠ − ° = 199.45 V ∠− 56.31 ° P max = 2 2 (3.124 V) 4 4(8.31 ) Th Th E = R Ω = 1198.2 W b. Z 1 = 3 Ω + j 4 Ω = 5 Ω 53.13 ° Z 2 = 2 0 ° Z N = Z Th = Z 1 || Z 2 = 5 Ω 53.13 ° || 2 Ω 0 ° = 10 53.13 2 + 3 4 + j Ω ∠ ° = 10 53.13 5 + 4 j ° = 10 53.13 6.403 38.66 Ω ∠ ° ° = 1.56 Ω 14.47 ° Z Th = 1.56 Ω 14.47 ° = 1.51 Ω + j 0.39 Ω Z L = 1.51 Ω j 0.39 Ω E Th = I ( Z 1 || Z 2 ) = (2 A 30 ° )(1.562 Ω 14.47 ° ) = 3.12 V 44.47 ° P max = 2 2 (3.12 V) 4 4(1.51 ) Th Th E = R Ω = 1.61 W 40. a. Z Th : Z 1 = 4 Ω 90 ° , Z 2 = 10 Ω 0 ° Z 3 = 5 Ω ∠− 90 ° , Z 4 = 6 Ω ∠− 90 ° E = 60 V 60 ° Z Th = Z 4 + Z 3 || ( Z 1 + Z 2 ) = j 6 Ω + (5 Ω ∠− 90 ° ) || (10 Ω + j 4 Ω ) = 2.475 Ω j 4.754 Ω = 11.04 Ω ∠− 77.03 ° Z L = 11.04 Ω 77.03 °
CHAPTER 18 249 E Th : E Th = 3 3 1 2 ( ) + + Z E Z Z Z = 5 90 )(60 V 60 ) 5 + 4 + 10 ( j j Ω ∠ − ° ° Ω Ω Ω = 29.85 V ∠− 24.29 ° P max = 2 4 Th Th / E R = (29.85 V) 2 /4(2.475 Ω ) = 90 W b. Z 1 = 3 Ω + j 4 Ω = 5 Ω 53.13 ° Z 2 = j 8 Ω Z 3 = 12 Ω + j 9 Ω Z Th = Z 2 + Z 1 || Z 3 = j 8 Ω + (5 Ω 53.13 ° ) || (15 Ω 36.87 ° ) = 5.71 Ω ∠− 64.30 ° = 2.475 Ω j 5.143 Ω Z L = 5.71 Ω 64.30 ° = 2.48 + j 5.15 Ω E Th + 3 Z V E 2 = 0 E Th = E 2 3 Z V 3 Z V = 3 2 1 3 1 ( ) + Z E E Z Z = 168.97 V 112.53 ° E Th = E 2 3 Z V = 200 V 90 ° 168.97 V 112.53 ° = 78.24 V 34.16 ° P max = 2 4 Th Th / E R = (78.24 V) 2 /4(2.475 Ω ) = 618.33 W 41. I = 1 0 1V 0 0 1 k 0 E = R ∠ ° ∠ ° ∠ ° Ω ∠ ° = 1 mA 0 ° Z Th = 40 k Ω 0 ° E Th = (50 I )(40 k Ω 0 ° ) = (50)(1 mA 0 ° )(40 k Ω 0 ° ) = 2000 V 0 ° P max = 2 2 2 kV) 4 4(40 k ) Th Th ( E = R Ω = 25 W
250 CHAPTER 18 42. a. Z Th = Z N = 8 0 ° (Problem 30(b)) Z L = 8 Ω 0 ° : Th N N = E I Z DC : (0.5 A)(8 ) 4 V Th N N = = Ω = E I Z AC : (1.25 A 0 )(8 0 ) 10 V Th N N = = ∠ ° Ω ∠ ° = E I Z b. P max = 2 2 2 (4 V) (10 V) 0.5 W + 3.13 W 4(8 ) 4(8 ) 4 Th Th E R = + = Ω Ω = 3.63 W 43. From #16, Z Th = 9 Ω , E Th = 12 V + 24 V 0 ° a. Z L = 9 Ω b. P max = 2 2 2 (12 V (24 V ) ) + 4 4(9 ) 4(9 ) Th Th E = R Ω Ω = 4 W + 16 W = 20 W or E Th = eff 2 2 0 1 + V V = 26.833 V and P max = 2 2 (26.833 V) 4 4(9 ) Th Th E = R Ω = 20 W 44. a. Problem 17(a): Z Th = 4.47 k Ω ∠− 26.57 ° = 4 k Ω j 2 k Ω Z L = 4 k Ω + j 2 k Ω E Th = 31.31 V ∠− 26.57 ° b. P max = 2 4 Th Th / E R = (31.31 V) 2 /4(4 k Ω ) = 61.27 mW 45. a. Z Th = 2 k Ω 0 ° || 2 k Ω ∠− 90 ° = 1 k Ω j 1 k Ω ( ) 2 2 Load 2 2 2 2 = (1 k + ( 1 k + 2 k ) ) = (1 k + (1 k ) ) = L Th Th = + + R R X X Ω Ω Ω Ω Ω 1.41 k Ω b. R av = ( R Th + R Load )/2 = (1 k Ω + 1.41 k Ω )/2 = 1.21 k Ω P max = 2 2 av (50 V) = 4 4(1.21 k ) Th E R Ω = 516.53 mW
CHAPTER 18 251 46. a. Z Th : X C = 1 1 2 2 (10 kHz)(4 nF) = fC π π 3978.87 Ω X L = 2 π fL = 2 π (10 kHz)(30 mH) 1884.96 Ω Z 1 = 1 k Ω 0 ° , Z 2 = 1884.96 Ω 90 ° Z 3 = 3978.87 Ω ∠− 90 ° Z Th = ( Z 1 + Z 2 ) || Z 3 = (1 k Ω + j 1884.96 Ω ) || 3978.87 Ω 90 ° ) = 2133.79 Ω 62.05 ° || 3978.87 Ω 90 ° ) = 3658.65 Ω 36.52 ° Z L = 3658.65 Ω ∠− 36.52 ° = 2940.27