Unformatted text preview: can be replaced by open circuits. So, the current through the inductor is simply 2A. Thus the energy stored in the inductor is 1 2 × 2 2 × 2H = 4J. (b) The voltage on the the two capacitors in series is 2A × 3Ω = 6V. Thus by capacitor voltage divider v 1 F = 6 × 1 / 3 = 2V. Thus the energy stored in the 1F capacitor is 1 2 × 2 2 × 1F = 2J....
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This note was uploaded on 04/06/2010 for the course EE 102 taught by Professor Pingli during the Fall '08 term at Shanghai Jiao Tong University.
 Fall '08
 PingLi

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