An AC source of angular frequency omega is co | vedclass

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(C) 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 + (\f

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

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(C) 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 reactance becomes $X_C' = \frac{1}{(\omega/3)C} = 3X_C$. The new current is $I' = \frac{I}{2}$. Thus,$\frac{I}{2} = \frac{V}{\sqrt{R^2 + (3X_C)^2}}$ ... (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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