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Unformatted text preview: Homework #7 Solutions P8.315 At steadystate, immediately before t = 0: ( 29 10 12 0 0.1 A 10 40 16 4010 i = = + + After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be 2 2 20 1 so 0.3 A, 40 , s 40 2 Finally: ( ) (0.1 0.3) 0.3 0.3 0.2 A sc t t t t L I R R i t e e τ = = Ω = = = = + = P8.316 At steadystate, immediately before t = 0 After t = 0, we have: 6 so 12 V, 200 , (200)(20 10 ) 4 ms oc t t V R R C τ = = Ω = = = 0.004 Finally: ( ) (12 12) 12 12 V t v t e = + = P8.68 For t < 0, the circuit is: After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get: ( 29 ( 29 ( 29 ( 29 / 4000 0.00005 5 15 6 15 15 9 V t c t v t e e =  +    =  + P8.69 The input changes abruptly at time t = 0. The voltage v ( t ) may not be continuous at t = 0, but the capacitor voltage, v C ( t ) will be continuous. We will find v C ( t ) first and then use KVL to find v ( t ). The circuit will be at steady state before t = 0 so the capacitor will act like an open circuit. ( 29 ( 29 5 7 4.375 V 5 3 v v + = = = + After t = 0, we replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit....
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This note was uploaded on 03/07/2008 for the course BME 211 taught by Professor Cain during the Winter '07 term at University of Michigan.
 Winter '07
 Cain

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