An a.c. source is applied to a series LR circ | vedclass

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(D) The power factor of an $LR$ circuit is given by $\cos \phi = R / Z$, where $Z = \sqrt{R^2 + X_L^2}$.For the initial $LR$ circuit, $X_L = 3R$, so $

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$1 : \sqrt{2}$

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(D) The power factor of an $LR$ circuit is given by $\cos \phi = R / Z$, where $Z = \sqrt{R^2 + X_L^2}$.For the initial $LR$ circuit, $X_L = 3R$, so $Z_1 = \sqrt{R^2 + (3R)^2} = \sqrt{10R^2} = R\sqrt{10}$.Thus, $X_1 = R / (R\sqrt{10}) = 1 / \sqrt{10}$.After adding the capacitor in series, the circuit becomes an $LCR$ circuit with $X_{net} = X_L - X_C = 3R - R = 2R$.The new impedance is $Z_2 = \sqrt{R^2 + (2R)^2} = \sqrt{5R^2} = R\sqrt{5}$.Thus, $X_2 = R / (R\sqrt{5}) = 1 / \sqrt{5}$.The ratio $X_1 / X_2 = (1 / \sqrt{10}) / (1 / \sqrt{5}) = \sqrt{5} / \sqrt{10} = 1 / \sqrt{2}$.

An $a.c.$ source is applied to a series $LR$ circuit with $X_L = 3R$ and power factor is $X_1$. Now a capacitor with $X_c = R$ is added in series to the $LR$ circuit and the power factor is $X_2$. The ratio $X_1$ to $X_2$ is

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