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hw6soln

# hw6soln - ECE 202 Linear Circuit Analysis II Purdue...

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ECE 202 - Linear Circuit Analysis II Purdue University, Spring 2009 Homework Set 3 Solutions Solution 21 When the switches are moved to position A, the circuit looks like the following:- + - 20/s 1/4s 1/16s 1/8s 1/16s V1(s) V2(s) V 1 = 20 s 1 4 s + 1 16 s - 1 1 4 s = 16 s v 1 ( t ) = 16 u ( t ) V 2 = 20 s 1 4 s + 1 16 s - 1 1 16 s = 4 s v 2 ( t ) = 4 u ( t ) v 1 (0 + ) = v 1 (2 - ) = 16 V v 2 (0 + ) = v 2 (2 - ) = 4 V At t=2, when the switch is moved to position B, we get the following circuit, Thus is the new time frame we have, + - 24/s 1/4s 1/8s 32 64 V2 24 s - V 2 4 s + 32 = V 2 (8 s ) + 64 1

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v 2 = 16 3 u ( t ) v 2 (2 + ) = 16 3 V v 1 (2 + ) = 24 - 16 3 = 56 3 V Solution 22 (a) Before t=0, the circuit reaches steady state so that capacitors can be assumed to be in open circuit. Thus we get, v 1 (0 - ) = v 2 (0 - ) = 16 2 R × R = 8 V (b) After S 2 is opened the circuit looks like the following, + - 16/s 2 1/s 8 V1 1/s 8 2 V2 1 2 16 s - V 1 + 8 = sV 1 V 1 = 8( s + 1) s ( s + 0.5) v 1 ( t ) = [16 - 8 e - 0.5 t ] u ( t ), 0 t < 1 sV 2 + 0.5 V 2 = 8 V 2 = = 8 s + 0.5 v 2 ( t ) = 8 e - 0.5 t u ( t ), 0 t < 1 v 1 (0 + ) = v 2 (0 + ) = 8 V Check:
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