An AC source of angular frequency omega is fe | vedclass

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Question

(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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$\sqrt{3/5}$

Vedclass · Answer

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