Question

In 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 $P_2$ Calculate $P_1/P_2$

Answer 1

$P=VI\cos \varphi$

and $\cos \varphi =\frac{R}{\sqrt{R^2+X_L^2-X_C^2}}$

And, $I=\frac{V}{\sqrt{R^2+X_L^2-X_C^2}}$

$⟹P=\frac{V^{2}R}{R^2+X_L^2-X_C^2}$

$P_1=\frac{V^{2}R}{\sqrt{R^2+R^2}}=\frac{V^2}{\sqrt{2}R}$

$P_2=\frac{V^{2}R}{\sqrt{R^2+R^2-R^2}}=\frac{V^2}{R}$

$⟹\frac{P_1}{P_2}=\frac{1}{\sqrt{2}}=0.707$