An a.c. source of angular frequency omega is | vedclass

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(C) The current in an $RC$ series circuit is given by $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_C^2}}$, where $X_C = \frac{1}{\omega C}$ is the capaci

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$\sqrt{\frac{7}{8}}$

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(C) The current in an $RC$ series circuit is given by $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_C^2}}$, where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.At frequency $\omega$, the current is $I = \frac{V}{\sqrt{R^2 + X_C^2}}$.When the frequency is changed to $\omega' = \omega/4$, the new capacitive reactance becomes $X_C' = \frac{1}{(\omega/4)C} = 4X_C$.The new current is $I' = \frac{V}{\sqrt{R^2 + (4X_C)^2}} = I/3$.Substituting the expression for $I$, we get $\frac{V}{\sqrt{R^2 + 16X_C^2}} = \frac{1}{3} \cdot \frac{V}{\sqrt{R^2 + X_C^2}}$.This simplifies to $\frac{\sqrt{R^2 + 16X_C^2}}{\sqrt{R^2 + X_C^2}} = 3$.Squaring both sides, we get $R^2 + 16X_C^2 = 9(R^2 + X_C^2)$.Expanding the equation: $R^2 + 16X_C^2 = 9R^2 + 9X_C^2$.Rearranging the terms: $7X_C^2 = 8R^2$.Therefore, the ratio of resistance to reactance is $\frac{R}{X_C} = \sqrt{\frac{7}{8}}$.

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