Q 12-07-056JEE MainJEE Main 2024 (27 Jan, Shift 2)Easy
A series LCR circuit with $L = \dfrac{100}{\pi}\ \text{mH}$, $C = \dfrac{10^{-3}}{\pi}\ \text{F}$ and $R = 10\ \Omega$ is connected across an AC source of $220\ \text{V}$, $50\ \text{Hz}$ supply. The power factor of the circuit would be ______.
Numerical value type. Enter your answer.
Answer: 1
$$X_L = 2\pi fL = 2\pi\times50\times\frac{0.1}{\pi} = 10\ \Omega$$
$$X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi\times50\times\frac{10^{-3}}{\pi}} = 10\ \Omega$$
$X_L = X_C$, so the circuit is at resonance, $Z = R$ and the power factor $\cos\phi = R/Z = 1$.
Solution by Sreeraj P, M.Sc Physics