BME211HW10 - Homework#10 Solutions P14.6-4 vc 12i L 2 diL...

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Homework #10 Solutions P14.6-4 L c c L L 12 2 8 and d i d v v i i C dt dt - + + = - = - Taking the Laplace transform yields ( 29 ( 29 ( 29 ( 29 L L c L 8 12 2 0 V s I s sI s i s - + + - = - ( 29 ( 29 ( 29 c c L 0 I s C sV s v = - - c L (0) 0, (0) 0 v i = = Solving yields ( 29 c 2 4 6 2 C V s C s s s = + + (a) 1 F 18 C = ( 29 ( 29 ( 29 2 2 72 3 3 3 c c a b V s s s s s s = = + + + + + ( 29 ( 29 2 24 8 8 8, 8, and 24 3 3 c a b c V s s s s - - = = - = - = + + + + ( 29 3 3 8 8 24 V, 0 t t c v t e t e t - - = - - (b) 1 F 10 C = ( 29 ( 29 ( 29 40 1 5 1 5 c c a b V s s s s s s s = = + + + + + + ( 29 C 2 8 10 8, 10, and 2 1 5 a b c V s s s s - = = - = = + + + + ( 29 5 8 10 2 V, 0 t t c v t e e t - - = - +
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P14.6-5 ( 29 c L (0 ) 10 V, 0 0 A v i - - = = ( 29 6 c c 5 10 and 400 1 0 d v d i i i v dt dt - = + + = Taking Laplace transforms yields ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 6 c 2 2 5 2 c 1 400 5 10 10 10 40 400 2 10 400 0 0 200 400 I s sV s I s s s I s s I s V s s - - ° = - - = = + + + - + = + + so ( 29 ( 29 ( 29 200 1 sin 400 A 40 t i t e t u t - = - P14.6-7 KCL: 1 6 7 5 t v i e - + = KVL: 1 1 4 3 0 4 3 di di i v v i dt dt + - = = + Then 6 6 4 3 35 7 2 5 4 t t di i di dt i e i e dt - - + + = + = Taking the Laplace transform of the differential equation: 35 1 35 1 ( ) (0) 2 ( ) ( ) 4 6 4 ( 2)( 6) s I s i
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