EE1002 Notes Part1

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Unformatted text preview: osed will be same as the voltage before switch was closed. vc (t ) Vi Ke st t 0 K Vi Final solution will be vc (t ) Vi e t RC . After time interval RC , the capacitor voltage will be at 0.368 times the original voltage. After five times this time interval, the remaining voltage on the capacitor is negligible. Charging capacitor form a DC source through a resistance t 0 vC 71 EE1002 Introduction to Circuits and Systems From KVL, starting with the negative polarity of the DC supply and summing the voltage drops, we get Vs vR vC 0 vR iR RC Vs RC RC dvc dt dvc vc 0 dt dvc vc Vs dt Assuming the capacitor was discharged before the switch was closed, and the fact that capacitor voltage cannot change instantaneously: vc (t ) vc (t ) 0 Trying the expression of K1 K 2 e st , where K1 , K 2 , s are constants to be determined, we get: (1 RCs ) K 2 e st K1 Vs As the coefficient of e st must be zero i.e. s Thus the solution becomes, vc (t ) Vs K 2 e 1 and K1 Vs . RC t RC This solution is valid at time t=0 i.e. vc (t 0) 0 Vs K 2 e0 K 2 Vs . With K1 ,...
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## This document was uploaded on 01/20/2014.

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