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

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(C) The power factor of an $LR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $X_L = 3R$,the power factor $X_1$ i

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$1: \sqrt{2}$

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(C) The power factor of an $LR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.Given $X_L = 3R$,the power factor $X_1$ is:$X_1 = \frac{R}{\sqrt{R^2 + (3R)^2}} = \frac{R}{\sqrt{R^2 + 9R^2}} = \frac{R}{\sqrt{10R^2}} = \frac{1}{\sqrt{10}}$.When a capacitor with $X_C = R$ is added in series,the circuit becomes an $LCR$ circuit.The impedance $Z'$ of the $LCR$ circuit is $Z' = \sqrt{R^2 + (X_L - X_C)^2}$.Given $X_L = 3R$ and $X_C = R$,we have $X_L - X_C = 3R - R = 2R$.Thus,$Z' = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5}$.The new power factor $X_2$ is:$X_2 = \frac{R}{Z'} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}}$.The ratio $X_1 : X_2$ is:$\frac{X_1}{X_2} = \frac{1/\sqrt{10}}{1/\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{10}} = \frac{1}{\sqrt{2}}$.Therefore,the ratio is $1: \sqrt{2}$.

An a.c. source is applied to a series $LR$ circuit with $X_L = 3R$ and the 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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