An AC source of angular frequency omega is co | vedclass

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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 + (\fr

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$\sqrt{\frac{3}{5}}$

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