Q 12-07-129JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
Two circuits are shown in figure (a) and (b). At a frequency of ______ rad s$^{-1}$ the average power dissipated in one cycle will be the same in both the circuits.
Numerical value type. Enter your answer.
Answer: 500
Circuit (a): $P_a = \dfrac{V_{rms}^2}{R}$. Circuit (b): $P_b = \dfrac{V_{rms}^2R}{Z^2}$.
$P_a = P_b$ requires $Z = R$, i.e. resonance:
$$\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1\times40\times10^{-6}}} = \frac{1}{2\times10^{-3}} = 500\ \text{rad s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics