Q 12-07-156JEE MainJEE Main 2025 (4 Apr, Shift 2)Medium
An inductor of self inductance $1\ \text{H}$ is connected in series with a resistor of $100\pi\ \Omega$ and an ac supply of $100\pi\ \text{V}$, $50\ \text{Hz}$. The maximum current flowing in the circuit is ______ A.
Numerical value type. Enter your answer.
Answer: 1
$X_L = 2\pi fL = 2\pi\times50\times1 = 100\pi\ \Omega$, so
$$Z = \sqrt{(100\pi)^2 + (100\pi)^2} = 100\sqrt2\,\pi\ \Omega$$
Taking $100\pi\ \text{V}$ as the rms supply voltage:
$$I_{\text{rms}} = \frac{100\pi}{100\sqrt2\,\pi} = \frac1{\sqrt2}\ \text{A},\qquad I_{\max} = \sqrt2I_{\text{rms}} = 1\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics