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(A) The current in an $RC$ series circuit is given by $I = \frac{V}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C}$.At initial frequency $\omega

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$\sqrt{3/5}$

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(A) The current in an $RC$ series circuit is given by $I = \frac{V}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C}$.At initial frequency $\omega$,$I = \frac{V}{\sqrt{R^2 + (1/\omega C)^2}}$ ... $(i)$At frequency $\omega' = \omega/3$,the new reactance is $X_C' = \frac{1}{(\omega/3)C} = \frac{3}{\omega C} = 3X_C$.The new current is $I' = I/2 = \frac{V}{\sqrt{R^2 + (3X_C)^2}}$ ... (ii)Dividing $(i)$ by (ii),we get $2 = \frac{\sqrt{R^2 + 9X_C^2}}{\sqrt{R^2 + X_C^2}}$.Squaring both sides: $4 = \frac{R^2 + 9X_C^2}{R^2 + X_C^2}$.$4R^2 + 4X_C^2 = R^2 + 9X_C^2$.$3R^2 = 5X_C^2$.$\frac{X_C^2}{R^2} = \frac{3}{5}$.Therefore,the ratio $\frac{X_C}{R} = \sqrt{\frac{3}{5}}$.

$A$ resistor and a capacitor are connected in series with an $AC$ source of angular frequency $\omega$. If the frequency is changed to $\omega / 3$ while keeping the voltage constant,the current becomes half. What is the ratio of the reactance to the resistance at the initial frequency?

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