An ideal capacitor of capacitance 0.2 μ F is charged to a potential difference of 10 V . The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inducta

Question

An ideal capacitor of capacitance 0.2 μ F is charged to a potential difference of 10 V . The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH . The current at a time when the potential difference across the capacitor is 5 V , is:

TestHub · Accepted Answer

Given: Capacitance, C = 0.2 μ F = 0.2 × 1 0 − 6 F Inductance L = 0.5 mH = 0.5 × 1 0 − 3 H Current I = ? Using energy conservation 2 1 ​ C V 2 = 2 1 ​ C V 1 2 ​ + 2 1 ​ L I 2 2 1 ​ × 0.2 × 1 0 − 6 × 1 0 2 + 0 = 2 1 ​ × 0.2 × 1 0 − 6 × 5 2 + 2 1 ​ × 0.5 × 1 0 − 3 I 2 ∴ I = 3 ​ × 1 0 − 1 A = 0.17 A

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