In a series LR circuit,X_L=R,the power factor | vedclass

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(B) The power factor of an $LR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $X_L = R$,we have $P_1 = \frac{R}{\

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$1: \sqrt{2}$

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(B) The power factor of an $LR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $X_L = R$,we have $P_1 = \frac{R}{\sqrt{R^2 + R^2}} = \frac{R}{\sqrt{2R^2}} = \frac{1}{\sqrt{2}}$.When a capacitor $C$ is added such that $X_C = X_L$,the circuit becomes a series $LCR$ circuit at resonance.The impedance of the $LCR$ circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Since $X_L = X_C$,the impedance $Z = \sqrt{R^2 + 0} = R$.The new power factor is $P_2 = \frac{R}{Z} = \frac{R}{R} = 1$.Therefore,the ratio $P_1 : P_2 = \frac{1}{\sqrt{2}} : 1 = 1 : \sqrt{2}$.

In a series $LR$ circuit,$X_L=R$,the 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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