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

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$\sqrt{\frac{5}{2}}$

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(D) The power factor of an $LR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $X_L = 2R$,the old power factor is $\cos \phi_1 = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.When a capacitor with $X_C = R$ is added in series,the new impedance $Z_{new} = \sqrt{R^2 + (X_L - X_C)^2}$.Substituting the values,$Z_{new} = \sqrt{R^2 + (2R - R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.The new power factor is $\cos \phi_2 = \frac{R}{Z_{new}} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$.The ratio of the new power factor to the old one is $\frac{\cos \phi_2}{\cos \phi_1} = \frac{1/\sqrt{2}}{1/\sqrt{5}} = \sqrt{\frac{5}{2}}$.

Column-$I$	Column-$II$
$(A)$ Capacitive reactance	$(i)$ Will continuously increase
$(B)$ Inductive reactance	(ii) Will remain constant
$(C)$ Resistance	(iii) Will first decrease and then increase
$(D)$ Total impedance	(iv) Will continuously decrease

$A$ series $LR$ circuit is connected to an ac source of frequency $\omega$ and the inductive reactance is equal to $2R$. $A$ capacitance of capacitive reactance equal to $R$ is added in series with $L$ and $R$. The ratio of the new power factor to the old one is

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