Asked by Anonymous
A 3.1-µF capacitor has a voltage of 31 V between its plates. What must be the current in a 5.6-mH inductor so that the energy stored in the inductor equals the energy stored in the capacitor?
Answers
Answered by
Damon
(1/2) L i^2 = (1/2) C V^2
i^2 = (C/L) V^2
i^2 = (3.1*10^-6 / 5.6*10^-3) (961)
i^2 = 532 * 10^-3 = 53.2 * 10^-4
i = 7.3 * 10^-2 = .073 amps
i^2 = (C/L) V^2
i^2 = (3.1*10^-6 / 5.6*10^-3) (961)
i^2 = 532 * 10^-3 = 53.2 * 10^-4
i = 7.3 * 10^-2 = .073 amps
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