Q 12-07-072JEE MainJEE Main 2023 (30 Jan, Shift 1)Easy
In a series LR circuit with $X_L=R$, power factor is $P_1$. If a capacitor of capacitance $C$ with $X_C=X_L$ is added to the circuit the power factor becomes $P_2$. The ratio of $P_1$ to $P_2$ will be
Answer: (B) $1:\sqrt2$
$P_1=\dfrac{R}{\sqrt{R^2+X_L^2}}=\dfrac{1}{\sqrt2}$. With $X_C=X_L$ the circuit is at resonance, $P_2=1$.
$P_1:P_2=1:\sqrt2$.
Solution by Sreeraj P, M.Sc Physics