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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 :

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$1 : \sqrt{2}$

$1 : 3$

$\sqrt{2} : 1$

$1 : 2$

answer is B.

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Detailed Solution

$P = \frac{R}{Z} \Rightarrow P_{1} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}} = \frac{R}{R \sqrt{2}} (a s X_{L} = R)$

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

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

$P_{1} = 1$

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