PractExam2 - 3 =-60*10 3 t 6 V t>= 100 µ s v(t =...

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Problem #3 The following current pulse is applied to the series capacitors i ( m A ) 1 0 50 100 150 t( µ s) -10 v(0 µ s) = 0v; Find v 1 (t) and v 2 (t) Solution Equivalent capacitance C= C 1 C 2 /(C 1 +C 2 ) = 0.125/0.75 = 1/6 µ F v(t) = 1/C i(t)dt + i(t 0 ) 0 <= t < 50 µ s : v(t) = 1/C*10*10 -3 t +0 = 6*10*10 3 t V 50 <= t < 100 µ s : v(t) = 1/C*(-10t) t 50 µ s + v(t=50 µ s) v(t=50 µ s) = 6*10*(50*10 -6 )*10 3 = 3V v(t) = 6*10 6 ( -10*10 -3 t + 10*10 -3 *50*10 -6
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Unformatted text preview: ) + 3 = -60*10 3 t + 6 V t >= 100 µ s : v(t) = v(t=100 µ s) = -60*10 3 *(100*10-6 ) + 6 = -6 + 6 = 0 V v1 = v*C 2 /(C 1 +C 2 ) = 1/3*60000t 0<= t < 50 µ s 1/3*-60000t + 6 50<= t <100 µ s 0 t>=100 µ s v2 = v*C 1 /(C 1 +C 2 ) = 2/3*60000t 0<= t < 50 µ s 2/3*-60000t + 6 50<= t <100 µ s 0 t >= 100 µ s + v _ C 1 =0.5 µ F C 2 =0.25 µ F i(t) + v 1 _ + v 2 _...
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This note was uploaded on 03/29/2012 for the course EE 215 taught by Professor Unknown during the Spring '05 term at University of Washington.

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PractExam2 - 3 =-60*10 3 t 6 V t>= 100 µ s v(t =...

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