EE411_HW5_Soln

# EE411_HW5_Soln - Ans All the current is flowing through the...

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Ans: All the current s i flowing through the parallel combination of the 2µF capacitor and the 50kΩ resistor will eventually show up at the negative input terminal of the OPAMP. So we have 0 1 . 0 10 10 10 ) ( ) ( 10 ) ( ) ( 0 > = × = × = × = t for V k k t i t v k t i t v s o s o µ

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Ans:At t=0 - , we have the following circuit by using delta-Y transformation V i V V v A V i L c L 20 2 ) 0 ( 10 4 20 ) 0 ( 5 4 20 ) 0 ( = × + = = = = At t=0 + , we have
20 ) 0 ( 0 10 ) 0 ( 2 1 20 20 0 10 ) 0 ( 2 1 4 ) 0 ( ) 0 ( = = = × + + + + + dt di dt di dt di i v L L L L c For this circuit, we have 4 1 , 4 2 0 = = = = LC L R ω α So t L e Bt A t i 4 ) ( ) ( + = But we know 0 20 ) 4 ( ) 0 ( 5 ) 0 ( = = × + = = = + B A B dt di A i L L So 0 5 ) ( 4 > = t for e t i t L

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Ans: (a)At t=0 - , we have V v 4 ) 0 ( = Since the voltage across a capacitor cannot change instantaneously, we have V v 4 ) 0 ( = + At t=0 + , switch is switched to the left,
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## This note was uploaded on 02/24/2011 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas.

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EE411_HW5_Soln - Ans All the current is flowing through the...

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