Ee250_HW1_sol - SOLUTIONS TO HOMEWORK 1 1 i = 530 A v = 25 108 V 2 iC = 28.27 cos(45 t 18 mA iC = 28.2718 mA 3 vL = 82.87 sin(350 t 80 V = 82.87

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SOLUTIONS TO HOMEWORK 1 1. 53 0A 25 108 V i v =∠ ° =∠ − ° 2. 28.27cos(45 18 ) mA C it π =+ ° 28.27 18 mA C i ° 3. 82.87sin(350 80 ) V = 82.87cos(350 10 ) V 82.87 10 V L L vt t v ° ° ° 4. 5000 2 max max 50( )V 1 25mJ 2 t L di vL e dt WL I == 5. Energy dissipated in the resistor = initial stored energy in capacitor= 2 0 3.6mJ 2 Q C = Q 0 = 1.18 x 10 -4 C = 118 μC 6. I 1 = -0.09 A= -90.9 mA I 2 = -0.4545 A = -454.5 mA 7. [] () Let where ;then Let cos( ) sin( ) cos( ) sin( ) 2cos( ) whic jj Bj x j yx j y x j y x j y BB x j y j y x x Ar e r A A r e B x jy B x jy Ae re e re
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This note was uploaded on 12/20/2010 for the course E E 240 taught by Professor Shankar during the Spring '10 term at University of Alberta.

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Ee250_HW1_sol - SOLUTIONS TO HOMEWORK 1 1 i = 530 A v = 25 108 V 2 iC = 28.27 cos(45 t 18 mA iC = 28.2718 mA 3 vL = 82.87 sin(350 t 80 V = 82.87

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