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(B) The power factor of an $AC$ circuit is given by $\cos \phi = \frac{R}{Z}$.For the initial circuit ($RL$ circuit):$Z = \sqrt{R^2 + X_L^2} = \sqrt{R

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$1$

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(B) The power factor of an $AC$ circuit is given by $\cos \phi = \frac{R}{Z}$.For the initial circuit ($RL$ circuit):$Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (3R)^2} = \sqrt{R^2 + 9R^2} = \sqrt{10R^2} = R\sqrt{10}$.Old power factor,$\cos \phi = \frac{R}{R\sqrt{10}} = \frac{1}{\sqrt{10}}$.For the new circuit ($RLC$ circuit):$Z' = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (3R - 2R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$.New power factor,$\cos \phi' = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$.The ratio of the new power factor to the old power factor is:$\frac{\cos \phi'}{\cos \phi} = \frac{1/\sqrt{2}}{1/\sqrt{10}} = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5} = \frac{\sqrt{5}}{1}$.Comparing this with $\sqrt{5} : x$,we get $x = 1$.

An $AC$ circuit has an inductor and a resistor of resistance $R$ in series,such that $X_L = 3R$. Now,a capacitor is added in series such that $X_C = 2R$. The ratio of the new power factor to the old power factor of the circuit is $\sqrt{5} : x$. The value of $x$ is ...... .

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