CH4 Solutions - Chapter 4: 4.1, 4.2, 4.4, 4.7, 4.10, 4.12,...

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1 Chapter 4: 4.1, 4.2, 4.4, 4.7, 4.10, 4.12, 4.14 4.29, 4.32, 4.33, 4.37, 4.38 4.43, 4.45, 4.47, 4.49, 4.51, 4.54, 4.55 4.68, 4.69, 4.72, 4.73 Q4.1. Solution Using the differential relationship for the inductor, we may obtain the voltage by differentiating the current: v L t () = L di L t dt = 0.5 di L t dt = 0.5 ×− 377 × 2sin 377 t + π 6 = 377sin 377 t + 6 = 377sin 377 t 5 6 V Q4.2. Solution. Using the defining differential relationship for the capacitor, (Eq. 4.4), we may obtain the current by differentiating the voltage: i C t = C dv C t dt = 100 × 10 6 dv C t dt = 10 4 dv C t ( ) dt a) i C t = 10 4 20 × 40sin 20 t 2 =− 0.08sin 20 t 2 = 0.08sin 20 t 2 + = 0.08sin 20 t + 2 A b) i C t ( ) = 10 4 100 × 20cos100 t [] = 0.2cos100 t A c) i C t = 10 4 80 × 60 cos 80 t + 6 0.48 cos 80 t + 6 = 0.48 cos 80 t + 6 = 0.48 cos 80 t 5 6 A d) i C t = 10 4 100 × 30sin 100 t + 4 0.3sin 100 t + 4 = 0.3sin 100 t + 4 = 0.3sin 100 t 3 4 A
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2 Q4.4, Solution. The magnetic energy stored in an inductor may be found from, (Eq. 4.16): () ( ) () () t i t i t Li t w L 2 2 2 2 2 1 2 1 = = = For 0 < < −∞ t , 0 = t w L For s t 0 1 0 < J 2 t t w L = For +∞ < t s 10 J 100 = t w L Q 4.7, Solution . The energy stored in a capacitor may be found from: ( )() () 2 2 2 05 . 0 1 . 0 2 1 2 1 t v t v t Cv t w C = = = For 0 < < −∞ t 0 = t w C For s t 10 0 < J 05 . 0 2 t t w C = For +∞ < t s 10 ( ) ( ) J 5 10 05 . 0 2 = = t w C Q4.10, Solution. Under steady-state conditions, all the currents are constant, no current can flow through the capacitors, and the voltage across any inductor is equal to zero. ( ) J V F V 76 . 11 43 . 3 2 2 1 2 1 43 . 3 8 4 2 6 2 2 2 2 2 4 4 4 4 4 2 = = = = + = = Ω Ω Ω Ω Ω F F F F v C w v v v v v v v 1 F = v 2 H = 0 w 1 F = 1 2 C 1 F v 1 F 2 = 1 2 1 F 0 2 = 0 i 2 H = v 4 Ω 8 = 0.43 A w 2 H = 1 2 L 2 H i 2 H 2 = 1 2 2 H 0.43 A 2 = 0.18 J v 3 F = v 4 Ω = 3.43 V w 3 F = 1 2 C 3 F v 3 F 2 = 1 2 3 F 3.43 V 2 = 17.65 J
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3 Q4.12, Solution.
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This homework help was uploaded on 04/19/2008 for the course ECE 3413 taught by Professor Bruce during the Spring '08 term at Mississippi State.

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CH4 Solutions - Chapter 4: 4.1, 4.2, 4.4, 4.7, 4.10, 4.12,...

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