quiz5 - d dt v C t 5 v C t 5 v in t with v in t 4 t and v...

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EE 202/Quiz 5 Jan 27, 2010 Morning Section The circuit given below for appropriate R and C has differential equation d dt v C ( t ) 0.5 v C ( t ) 0.5 v in ( t ) with v in ( t ) 10 ( t ) and v C (0 ) = 5 V. Then find v C ( t ). Solution . ( s 0.5) V C ( s ) 0.5 V in ( s ) v C (0 ) implies V C ( s ) 0.5 ( s 0.5) V in ( s ) v C (0 ) ( s 0.5) 5 ( s 0.5) 5 ( s 0.5) implies v C ( t ) 10 e 0.5 t u ( t ) Afternoon Section The circuit given below for appropriate values of R and C has differential equation
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Unformatted text preview: d dt v C ( t ) 5 v C ( t ) 5 v in ( t ) with v in ( t ) 4 ( t ) and v C (0 ) 10 V. Then find v C ( t ) . Solution . ( s 5) V C ( s ) 5 V in ( s ) v C (0 ) implies V C ( s ) 5 ( s 5) V in ( s ) v C (0 ) ( s 5) 20 ( s 5) 10 ( s 5) implies v C ( t ) 30 e 5 t u ( t ) V...
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This document was uploaded on 02/03/2012.

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