Q77 / 90 · JEE Main 2023 (30 Jan Shift 1) · Physics

In a series $L R$ 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 :

Solution

In case of $L R$ circuit:

Given: $X_{L} = R$

$P_{1} = \cos ϕ = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + {(X_{L})}^{2}}} = \frac{1}{\sqrt{2}}$

In case of $L C R$ circuit:

$P_{2} = \cos ϕ = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = 1$

Therefore,

$\frac{P_{1}}{P_{2}} = 1 : \sqrt{2}$

Hence, the correct answer is option: 'B'.