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ch09_ism - Sinusoidal Steady State Analysis 9 Assessment...

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9 Sinusoidal Steady State Analysis Assessment Problems AP 9.1 [a] V = 170/ 40 V [b] 10 sin(1000 t + 20 ) = 10 cos(1000 t 70 ) · . . I = 10/ 70 A [c] I = 5/36 . 87 + 10/ 53 . 13 = 4 + j 3 + 6 j 8 = 10 j 5 = 11 . 18/ 26 . 57 A [d] sin(20 , 000 πt + 30 ) = cos(20 , 000 πt 60 ) Thus, V = 300/45 100/ 60 = 212 . 13 + j 212 . 13 (50 j 86 . 60) = 162 . 13 + j 298 . 73 = 339 . 90/61 . 51 mV AP 9.2 [a] v = 18 . 6 cos( ωt 54 ) V [b] I = 20/45 50/ 30 = 14 . 14 + j 14 . 14 43 . 3 + j 25 = 29 . 16 + j 39 . 14 = 48 . 81/126 . 68 Therefore i = 48 . 81 cos( ωt + 126 . 68 ) mA [c] V = 20 + j 80 30/15 = 20 + j 80 28 . 98 j 7 . 76 = 8 . 98 + j 72 . 24 = 72 . 79/97 . 08 v = 72 . 79 cos( ωt + 97 . 08 ) V AP 9.3 [a] ωL = (10 4 )(20 × 10 3 ) = 200 Ω [b] Z L = jωL = j 200 Ω 9–1
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9–2 CHAPTER 9. Sinusoidal Steady State Analysis [c] V L = I Z L = (10/30 )(200/90 ) × 10 3 = 2/120 V [d] v L = 2 cos(10 , 000 t + 120 ) V AP 9.4 [a] X C = 1 ωC = 1 4000(5 × 10 6 ) = 50 Ω [b] Z C = jX C = j 50 Ω [c] I = V Z C = 30/25 50/ 90 = 0 . 6/115 A [d] i = 0 . 6 cos(4000 t + 115 ) A AP 9.5 I 1 = 100/25 = 90 . 63 + j 42 . 26 I 2 = 100/145 = 81 . 92 + j 57 . 36 I 3 = 100/ 95 = 8 . 72 j 99 . 62 I 4 = ( I 1 + I 2 + I 3 ) = (0 + j 0) A , therefore i 4 = 0 A AP 9.6 [a] I = 125/ 60 | Z | / θ z = 125 | Z | /( 60 θ Z ) But 60 θ Z = 105 · . . θ Z = 45 Z = 90 + j 160 + jX C · . . X C = 70 Ω; X C = 1 ωC = 70 · . . C = 1 (70)(5000) = 2 . 86 µ F [b] I = V s Z = 125/ 60 (90 + j 90) = 0 . 982/ 105 A ; · . . | I | = 0 . 982 A AP 9.7 [a] ω = 2000 rad/s ωL = 10 Ω , 1 ωC = 20 Ω Z xy = 20 j 10 + 5 + j 20 = 20( j 10) (20 + j 10) + 5 j 20 = 4 + j 8 + 5 j 20 = (9 j 12) Ω
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Problems 9–3 [b] ωL = 40 Ω , 1 ωC = 5 Ω Z xy = 5 j 5 + 20 j 40 = 5 j 5 + (20)( j 40) 20 + j 40 = 5 j 5 + 16 + j 8 = (21 + j 3) Ω [c] Z xy = 20( jωL ) 20 + jωL + 5 j 10 6 25 ω = 20 ω 2 L 2 400 + ω 2 L 2 + j 400 ωL 400 + ω 2 L 2 + 5 j 10 6 25 ω The impedance will be purely resistive when the j terms cancel, i.e., 400 ωL 400 + ω 2 L 2 = 10 6 25 ω Solving for ω yields ω = 4000 rad/s . [d] Z xy = 20 ω 2 L 2 400 + ω 2 L 2 + 5 = 10 + 5 = 15 Ω AP 9.8 The frequency 4000 rad/s was found to give Z xy = 15 Ω in Assessment Problem 9.7. Thus, V = 150/0 , I s = V Z xy = 150/0 15 = 10/0 A Using current division, I L = 20 20 + j 20 (10) = 5 j 5 = 7 . 07/ 45 A i L = 7 . 07 cos(4000 t 45 ) A , I m = 7 . 07 A AP 9.9 After replacing the delta made up of the 50 Ω , 40 Ω , and 10 Ω resistors with its equivalent wye, the circuit becomes
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9–4 CHAPTER 9. Sinusoidal Steady State Analysis The circuit is further simplified by combining the parallel branches, (20 + j 40) (5 j 15) = (12 j 16) Ω Therefore I = 136/0 14 + 12 j 16 + 4 = 4/28 . 07 A AP 9.10 V 1 = 240/53 . 13 = 144 + j 192 V V 2 = 96/ 90 = j 96 V jωL = j (4000)(15 × 10 3 ) = j 60 Ω 1 jωC = j 6 × 10 6 (4000)(25) = j 60 Ω Perform source transformations: V 1 j 60 = 144 + j 192 j 60 = 3 . 2 j 2 . 4 A V 2 20 = j 96 20 = j 4 . 8 A Combine the parallel impedances: Y = 1 j 60 + 1 30 + 1 j 60 + 1 20 = j 5 j 60 = 1 12 Z = 1 Y = 12 Ω V o = 12(3 . 2 + j 2 . 4) = 38 . 4 + j 28 . 8 V = 48/36 . 87 V v o = 48 cos(4000 t + 36 . 87 ) V
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Problems 9–5 AP 9.11 Use the lower node as the reference node. Let V 1 = node voltage across the 20 Ω resistor and V Th = node voltage across the capacitor. Writing the node voltage equations gives us V 1 20 2/45 + V 1 10 I x j 10 = 0 and V Th = j 10 10 j 10 (10 I x ) We also have I x = V 1 20 Solving these equations for V Th gives V Th = 10/ 45 V.
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