Chapt08PP - 2009 CHAPTER 8 P.P.8.1 (a) At t = 0-, we have...

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© 2009 CHAPTER 8 P.P.8.1 (a) At t = 0-, we have the equivalent circuit shown in Figure (a). 50mF 10 Ω 2 Ω (a) + v 12V + 2 Ω (b) + v C 12V + v L + a i i(0-) = 12/(2 + 10) = 1 A , v(0-) = 2i(0-) = 2 V hence, v(0+) = v(0-) = 2V . (b) At t = 0+, the switch is closed. L(di/dt) = v L , leads to (di/dt) = v L /L But, v C (0+) + v L (0+) = 12 = 2 + v L (0+), or v L (0+) = 10 (di(0+)/dt) = 10/0.4 = 25 A/s C(dv/dt) = i C leading to (dv/dt) = i C /C But at node a, KCL gives i(0+) = i C (0+) + v(0+)/2 = 1 = i C (0+) + 2/2 or i C (0+) = 0, hence (dv(0+)/dt) = 0 V/s (c) As t approaches infinity, the capacitor is replaced by an open circuit and the inductor is replaced by a short circuit. v( ) = 12 V , and i( ) = 6 A . i
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P.P.8.2 (a) At t = 0-, we have the equivalent circuit shown in (a). i L (0-) = –6A, v L (0 ) = 0, v (0 ) = 0 R At t = 0 + , we have the equivalent circuit in Figure (b). At node b, i R (0 + ) = i L (0 + ) + 6, since i L (0 + ) = i L (0 ) = –6A, i (0 + ) = 0, R and v R (0+) = 5i R (0+) = 0. Thus, i L (0) = –6 A , v C (0) = 0 , and v R (0 + 0 ) = . (b) dv C (0 + )/dt = i C (0 + )/C = 4/0.2 = 20 V/s. To get (dv /dt), we apply KCL to node b, i = i R R L + 6, thus di R /dt = di L /dt. Since v = 5i R R , dv /dt = 5di /dt = 5di R R L /dt. But Ldi L /dt = v L , di L /dt = v L /L. + Hence, dv R (0 )/dt = 5v L (0 + )/L. Applying KVL to the middle mesh in Figure (b), –v C (0 + ) + v R (0 + ) + v L (0 + ) = 0 = 0 + 0 + v R (0 + ), or v (0 + ) = 0 R + Hence, dv R (0 )/dt = 0 = di L (0 + )/dt; di L (0 + )/dt = 0 , dv C (0 + )/dt = 20 V/s , dv R (0 + )/dt = 0 . 5 Ω (a) i L i R 6A 4A v L 2H 10 μ F 5 Ω (b) 6A + v C + v R b a
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(c) As t approaches infinity, we have the equivalent circuit shown below. 4 = 6 + i L ( ) leads to i L ( ) = –2A v ( ) = v ( ) = 4x5 = 20V C R –2 A Thus, i L ( ) = , v C ( ) = v R ( ) = 20 V P.P.8.3 2 10 x 2 x 5 1 LC 1 = (a) α = R/(2L) = 10/(2x5) = 1 10 , ω = = o 100 1 1 2 o 2 ± = ω α ± α –1 ± j9.95 s 1,2 = = . (b) Since α < ω , we clearly have an underdamped response. o P.P.8.4 For t < 0, the inductor is connected to the voltage source although it acts like a short circuit. i(0-) = 50/10 = 5 = i(0+) = i(0) The voltage across the capacitor is 0 = v(0-) = v(0+) = v(0). For t > 0, we have a source-free RLC circuit. = = = ω 9 1 x 1 1 LC 1 o 3 α = R/(2L) = 5/(2x1) = 2.5 4A 5 Ω i L 6A
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Since α < ω , we have an underdamped case. o 9 25 . 6 5 . 2 2 o 2 ± = ω α ± α s 1,2 = = -2.5 ± j1.6583 i(t) = e -2.5t [A cos1.6583t + A 1 2 sin1.6583t] We now determine A and A .
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This note was uploaded on 10/25/2009 for the course ECE 280 taught by Professor Staff during the Spring '08 term at George Mason.

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Chapt08PP - 2009 CHAPTER 8 P.P.8.1 (a) At t = 0-, we have...

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