Q 12-07-092JEE MainJEE Main 2022 (24 Jun, Shift 1)Medium
As shown in the figure, an inductor of inductance $200\ \text{mH}$ is connected to an AC source of emf $220\ \text{V}$ and frequency $50\ \text{Hz}$. The instantaneous voltage of the source is $0\ \text{V}$ when the peak value of current is $\dfrac{\sqrt a}{\pi}\ \text{A}$. The value of $a$ is ______.
Numerical value type. Enter your answer.
Answer: 242
In a purely inductive circuit the current lags the voltage by $90^\circ$, so the current is at its peak when the voltage is zero.
$$i_0 = \frac{V_0}{\omega L} = \frac{220\sqrt2}{2\pi\times50\times0.2} = \frac{220\sqrt2}{20\pi} = \frac{11\sqrt2}{\pi} = \frac{\sqrt{242}}{\pi}\ \text{A}$$
So $a = 242$.
Solution by Sreeraj P, M.Sc Physics