Q 12-07-108JEE MainJEE Main 2022 (27 Jul, Shift 2)Medium
A series LCR circuit has $L = 0.01$ H, $R = 10\ \Omega$ and $C = 1\ \mu$F and it is connected to ac voltage of amplitude $(V_m)$ $50$ V. At frequency $60\%$ lower than resonant frequency, the amplitude of current will be approximately
Answer: (C) $238$ mA
$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-8}}} = 10^4\ \text{rad s}^{-1}$$
$60\%$ lower: $\omega = 4000\ \text{rad s}^{-1}$.
$X_L = \omega L = 40\ \Omega$, $\quad X_C = \dfrac{1}{\omega C} = 250\ \Omega$.
$$Z = \sqrt{10^2 + (250 - 40)^2} = \sqrt{44200} \approx 210.2\ \Omega$$
$$I_m = \frac{50}{210.2} \approx 0.238\ \text{A} = 238\ \text{mA}$$
Solution by Sreeraj P, M.Sc Physics