In a series LR circuit with X_L = R. Power fa | vedclass

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(B) In a series $LR$ circuit, the impedance is $Z_1 = \sqrt{R^2 + X_L^2}$.Given $X_L = R$, so $Z_1 = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.The p

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$1 : \sqrt{2}$

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(B) In a series $LR$ circuit, the impedance is $Z_1 = \sqrt{R^2 + X_L^2}$.Given $X_L = R$, so $Z_1 = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.The power factor $P_1 = \cos \phi_1 = \frac{R}{Z_1} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$.When a capacitor is added such that $X_c = X_L$, the circuit becomes an $LCR$ series circuit at resonance.In a resonant $LCR$ circuit, the net reactance $X = X_L - X_c = 0$.Thus, the impedance $Z_2 = R$.The power factor $P_2 = \cos \phi_2 = \frac{R}{Z_2} = \frac{R}{R} = 1$.The ratio $P_1 : P_2 = \frac{1}{\sqrt{2}} : 1 = 1 : \sqrt{2}$.

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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