JEE Main 2022 — Alternating Current Question with Solution

At steady state charge stored on the capacitor,

$$q_{\max }=CV$$

$$=500\times 10^{-6}\times 100$$

$$=5\times 10^{-2}C$$

Energy stored in the capacitor,

$$U_{\max }=\frac{q_{\max }^2}{2C}$$

Now, when electrostatic energy of capacitor converted to magnetic field energy then all energy of capacitor is transferrd to the inductor.

$$\therefore$$ Maximum energy stored in the inductor

$$U_{L\max }=\frac{1}{2}L{I}_{\max }^2$$

$$\therefore$$ $$\frac{1}{2}L{I}_{\max }^2=\frac{q_{\max }^2}{2C}$$

$$\Rightarrow I_{\max }=\frac{q_{\max }}{\sqrt{LC}}$$

$$=\frac{5\times {10}^{-2}}{\sqrt{50\times {10}^{-3}\times 500\times {10}^{-6}}}$$

$$=\frac{5\times {10}^{-2}}{5\times {10}^{-3}}$$

$$=10A$$