In a series LR circuit,X_L = 3R. Now,a capaci | vedclass

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(D) In the initial series $LR$ circuit,the impedance is $Z_1 = \sqrt{R^2 + X_L^2}$.Given $X_L = 3R$,so $Z_1 = \sqrt{R^2 + (3R)^2} = \sqrt{10R^2} = R\s

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$\sqrt{2}$

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(D) In the initial series $LR$ circuit,the impedance is $Z_1 = \sqrt{R^2 + X_L^2}$.Given $X_L = 3R$,so $Z_1 = \sqrt{R^2 + (3R)^2} = \sqrt{10R^2} = R\sqrt{10}$.The initial power factor is $\cos \phi_1 = \frac{R}{Z_1} = \frac{R}{R\sqrt{10}} = \frac{1}{\sqrt{10}}$.When a capacitor with $X_C = R$ is added in series,the circuit becomes an $LCR$ circuit.The new impedance is $Z_2 = \sqrt{R^2 + (X_L - X_C)^2}$.Substituting the values,$Z_2 = \sqrt{R^2 + (3R - R)^2} = \sqrt{R^2 + (2R)^2} = \sqrt{5R^2} = R\sqrt{5}$.The new power factor is $\cos \phi_2 = \frac{R}{Z_2} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}}$.The ratio of the new power factor to the old power factor is $\frac{\cos \phi_2}{\cos \phi_1} = \frac{1/\sqrt{5}}{1/\sqrt{10}} = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}$.

In a series $LR$ circuit,$X_L = 3R$. Now,a capacitor with $X_C = R$ is added in series. What is the ratio of the new power factor to the old power factor?

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