Q 12-07-147JEE MainJEE Main 2018 (15 Apr, Shift 1)Medium
An ideal capacitor of capacitance $0.2\ \mu\text{F}$ is charged to a potential difference of $10\ \text{V}$. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance $0.5\ \text{mH}$. The current at a time when the potential difference across the capacitor is $5\ \text{V}$ is:
Answer: (B) $0.17\ \text{A}$
Energy is conserved in the ideal LC circuit:
$$\frac12CV_0^2 = \frac12CV^2 + \frac12LI^2$$
$$I = \sqrt{\frac{C(V_0^2 - V^2)}{L}} = \sqrt{\frac{0.2\times10^{-6}\times(100 - 25)}{0.5\times10^{-3}}} = \sqrt{0.03} \approx 0.17\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics