In a series LR circuit $\mathrm{X}_{\mathrm{L}}=\mathrm{R}$ and power factor of the circuit is $\mathrm{P}_{1}$. When capacitor with capacitance C such that $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ is put in series, the power factor becomes $P_{2}$. The ratio $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}$ is

(C) $\frac{\sqrt{3}}{\sqrt{2}}$

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Alternating Current

In a series LR circuit

X_{L} = R

and power factor of the circuit is

P_{1}

. When capacitor with capacitance C such that

X_{L} = X_{C}

is put in series, the power factor becomes

P_{2}

. The ratio

\frac{P _{1}}{P _{2}}

Answer: (B)

For the LR circuit with

X_{L} = R

Z = R^{2} + X_{L}^{2} = 2 R

Power factor

P_{1} = cos ϕ = \frac{R}{Z} = \frac{1}{2}

When capacitor is added with

X_{L} = X_{C}

Z = R^{2} + (X_{L} - X_{C})^{2} = R

Power factor

P_{2} = \frac{R}{R} = 1

Therefore

\frac{P _{1}}{P _{2}} = \frac{1}{2}

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option