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.....\(A\)

Inductance \(\mathrm{L}=0.5\, \mathrm{mH}=0.5 \times 10^{-3}\, \mathrm{H}\)

Current \(I=?\)

Using energy conservation

\(\frac{1}{2} C V^{2}=\frac{1}{2} C V_{1}^{2}+\frac{1}{2} L I^{2}\)

\(\frac{1}{2} \times 0.2 \times 10^{-6} \times 10^{2}+0\)

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

\(\therefore \quad I=\sqrt{3} \times 10^{-1}\, \mathrm{A}=0.17\, \mathrm{A}\)