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A capacitor of capacitance 6 microfarad is fully charged to 6 V using a battery. The battery is disconnected and an ideal (resistanceless) inductor of inductance 0.2 mH is connected across the capacitor. What current flows through the inductor at the instant when one-third of the total energy is stored in the magnetic field of the inductor?

1. 0.1 A
2. 0.2 A
3. 0.4 A
4. 0.6 A

Correct answer: 0.6 A

Solution

Total energy E = (1/2)*C*V² = (1/2)*(6*10⁻⁶)*(6²) = 108 microjoules. Energy in inductor = E/3 = 36 microjoules. (1/2)*L*I² = 36*10⁻⁶. I² = 2*36*10⁻⁶ / (0.2*10⁻³) = 72*10⁻⁶ / (2*10⁻⁴) = 0.36. I = 0.6 A.

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