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Question

In a series $L R$ 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 $\frac{P_{1}}{P_{2}}$ is:

$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3}}{\sqrt{2}}$
2 : 1

Solution

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