A series combination of resistor R and capaci | vedclass

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(A) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$.The initial current is $I = \frac{V}{Z}$.When the frequency cha

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$\sqrt{0.6}$

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(A) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$.The initial current is $I = \frac{V}{Z}$.When the frequency changes to $\omega' = \frac{\omega}{3}$,the new capacitive reactance is $X_c' = \frac{1}{\omega' C} = \frac{1}{(\omega/3) C} = 3X_c$.The new impedance is $Z' = \sqrt{R^2 + (X_c')^2} = \sqrt{R^2 + (3X_c)^2}$.Given that the new current $I' = \frac{I}{2}$,we have $\frac{V}{Z'} = \frac{1}{2} \frac{V}{Z}$,which implies $Z' = 2Z$.Squaring both sides,$(Z')^2 = 4Z^2$,so $R^2 + 9X_c^2 = 4(R^2 + X_c^2)$.$R^2 + 9X_c^2 = 4R^2 + 4X_c^2$.$5X_c^2 = 3R^2$.$\frac{X_c^2}{R^2} = \frac{3}{5} = 0.6$.Therefore,the ratio $\frac{X_c}{R} = \sqrt{0.6}$.

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