Q 12-07-057JEE MainJEE Main 2024 (29 Jan, Shift 1)Easy
A capacitor of capacitance $100\ \mu\text{F}$ is charged to a potential of $12\ \text{V}$ and connected to a $6.4\ \text{mH}$ inductor to produce oscillations. The maximum current in the circuit would be:
Answer: (B) $1.5\ \text{A}$
Energy is conserved between the capacitor and the inductor:
$$\frac12CV^2 = \frac12LI_{max}^2 \;\Rightarrow\; I_{max} = V\sqrt{\frac CL} = 12\sqrt{\frac{100\times10^{-6}}{6.4\times10^{-3}}} = 12\times0.125 = 1.5\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics