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  4. 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 P2. The ratio of P1 to P2 will be :

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Q. 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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Solution:

$ P=\frac{R}{Z} \Rightarrow P_1=\frac{R}{\sqrt{R^2+X_L^2}}=\frac{R}{R \sqrt{2}}\left(\text { as } X_L=R\right)$
$P_1=\frac{1}{\sqrt{2}} $
$ P_2=\frac{R}{\sqrt{R^2+\left(X_L-X_L\right)^2}}=P_2=1 $
$ \frac{P_1}{P_2}=\frac{1}{\sqrt{2}}$