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(A) At angular frequency $\omega$,the current $I$ in the $RC$ series circuit is given by:$I = \frac{V}{\sqrt{R^2 + X_C^2}} = \frac{V}{\sqrt{R^2 + (1/\

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$\sqrt{3/5}$

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(A) At angular frequency $\omega$,the current $I$ in the $RC$ series circuit is given by:$I = \frac{V}{\sqrt{R^2 + X_C^2}} = \frac{V}{\sqrt{R^2 + (1/\omega C)^2}}$ ......$(i)$When the frequency is changed to $\omega' = \omega/3$,the new capacitive reactance becomes $X_C' = \frac{1}{(\omega/3)C} = 3X_C$. The new current $I'$ is given as $I/2$:$I/2 = \frac{V}{\sqrt{R^2 + (3X_C)^2}}$ ......$(ii)$Dividing equation $(i)$ by equation $(ii)$:$2 = \frac{\sqrt{R^2 + 9X_C^2}}{\sqrt{R^2 + X_C^2}}$Squaring both sides:$4 = \frac{R^2 + 9X_C^2}{R^2 + X_C^2}$$4R^2 + 4X_C^2 = R^2 + 9X_C^2$$3R^2 = 5X_C^2$$\frac{X_C^2}{R^2} = \frac{3}{5}$$\frac{X_C}{R} = \sqrt{\frac{3}{5}}$

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