An ideal capacitor of capacitance 0.2 μF 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 :

Answers (1)

From energy conservation

\frac{1}{2}\times 0.2\times 10^{6}\times 10^{2}+0 = \frac{1}{2}\times 0.2\times 10^{-6}\times 5^{2}+\frac{1}{2}\times 0.5\times 10^{-3} 1^{2}

I=\frac{\sqrt{3}}{10}A = 0.17A

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