An ideal 14-mF capacitor is connected in series with an ideal 6.0-mH inductor.The peak current in the wires between the capacitor and the inductor is 0.30A. What is the total electrical energy in this circuit? A)2.7 * 10^-4 J B)5.4 * 10^-4 J C)1.1 * 10^-3 J D)4.6 * 10^-3 J

Question

Quizplus · Accepted Answer

The total electrical energy in the circuit is stored in the capacitor and the inductor. The energy stored in the capacitor is given by 1/2 CV^2, where C is the capacitance and V is the voltage across the capacitor. The energy stored in the inductor is given by 1/2 LI^2, where L is the inductance and I is the current through the inductor. In this problem, the peak current in the wires is 0.30A, so the peak voltage across the capacitor is V = I/C = 0.30A / 14mF = 21V. The energy stored in the capacitor is therefore 1/2 CV^2 = 1/2 * 14mF * (21V)^2 = 2.7 * 10^-4 J. The energy stored in the inductor is 1/2 LI^2 = 1/2 * 6.0mH * (0.30A)^2 = 2.7 * 10^-4 J. The total electrical energy in the circuit is therefore 2.7 * 10^-4 J + 2.7 * 10^-4 J = 5.4 * 10^-4 J.

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