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A 6 microfarad capacitor is fully charged by a 6 V battery. The battery is then disconnected and a pure (resistanceless) 0.2 mH inductor is connected across the capacitor. At the instant when one-third of the total energy resides in the magnetic field of the inductor, the current in the circuit is:
- 0.6 A
- 0.1 A
- 0.4 A
- 0.2 A
Correct answer: 0.6 A
Solution
Total energy U = (1/2)C V² = 0.5 * 6e-6 * 36 = 1.08e-4 J. One-third of this is in the inductor: (1/2)L i² = U/3 = 3.6e-5 J. So i² = 2(U/3)/L = 7.2e-5/2e-4 = 0.36, giving i = 0.6 A.
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