EE411_HW3_Soln

# EE411_HW3_Soln - 10 7 2 36 28 5 100 50 01 1 − − − −...

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Ans:The upper half in a non-inverting amplifier and the lower is an inverting amplifier. So we have 5 . 5 2 5 . 3 30 60 20 20 50 = + = Ω Ω Ω Ω + Ω = i i i i o v v k k v k k k v v Ans: (a)We have 8 2 4 ) 100 50 ( 01 . 0 ) 0 ( 10 10 ) 0 ( = + = × = = + = B A B A dt dv C B A V v Solve the above equations, we can find A=28 and B=-18. (b)Energy stored is J Cv 5 . 0 10 01 . 0 2 1 ) 0 ( 2 1 2 2 = × × =

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(c)Capacitor current at t=1 sec is A e e Be Ae dt dv C 21 100 50 100 50
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Unformatted text preview: 10 7 . 2 ) 36 28 ( 5 . ) 100 50 ( 01 . ) 1 ( − − − − − × − = − × − = − − × = Ans:The circuit can be simplified as follows: So the equivalent capacitance between A&B is mF 13 . 20 35 10 20 1 10 10 1 5 10 15 1 1 10 = + + + + + + + +...
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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_HW3_Soln - 10 7 2 36 28 5 100 50 01 1 − − − −...

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