Q 12-07-102JEE MainJEE Main 2022 (26 Jul, Shift 1)Easy
In a series LR circuit $X_L = R$ and the power factor of the circuit is $P_1$. When a capacitor with capacitance $C$ such that $X_L = X_C$ is put in series, the power factor becomes $P_2$. The ratio $\dfrac{P_1}{P_2}$ is
Answer: (B) $\dfrac{1}{\sqrt2}$
$P_1 = \dfrac{R}{\sqrt{R^2 + X_L^2}} = \dfrac{1}{\sqrt2}$. With $X_L = X_C$ the circuit is at resonance, so $P_2 = 1$.
$$\frac{P_1}{P_2} = \frac{1}{\sqrt2}$$
Solution by Sreeraj P, M.Sc Physics