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hw6sol

# hw6sol - Purdue University School of Electrical and...

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II (S P ’10) Homework #6 Solution (21–24) Problem 21 R s = 2 Ω , R L = 8 Ω , R 1 = 190 Ω , L = 20H v in ( t ) = 100 [V] for all time. When the switch has been in position A for a long time (left figure), the current through the inductor is v in / ( R s + R L ) = 10 [A] which is the initial condition for the inductor for t 0. So, i L ( 0 ) = 10 [A]. For t 0, the equivalent s-domain circuit accounting for the initial condition is shown on the right. I 1 = V in ( s )+ Li L ( 0 ) R s + R 1 + R L + Ls = 10 s + 5 s ( s + 10 ) V out ( s ) = R 1 I 1 = 190 10 s + 5 s ( s + 10 ) v out ( t ) = 95 ( 1 + 19 e 10 t ) u ( t ) [V] p R 1 ( t ) = i 1 ( t ) v out ( t ) = v 2 out / R 1 = 95 2 19 e 10 t + 1 2 [W] E diss = t 0 p R 1 ( τ ) d τ = 47 . 5 t 180 . 5 e 10 t 857 . 4 e 20 t + 1038 [J] 1/ 5

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo Problem 22–23 (a) v C 1 ( 0 ) and v C 2 ( 0 ) v C 1 ( 0 ) = 0 [V] , v C 2 ( 0 ) = 0 [V] (b) v C 1 ( t ) and v C 2 ( t ) for 0 t 1 V C 1 ( s ) = 20 s 1 / ( C 1 s ) R 1 + 1 / ( C 1 s ) = 100 s ( s + 5 ) v C 1 ( t ) = 20 ( 1 e 5 t ) u ( t ) [V] v C 2 ( t ) =
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hw6sol - Purdue University School of Electrical and...

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