An a.c. voltage source V=V_0 sin omega t is c | vedclass

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(B) Given: $R = X_C = \frac{1}{\omega C}$.Initial impedance $Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + R^2} = \sqrt{2} R$.Initial peak current $I_0 = \frac

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$\frac{I_0}{\sqrt{2}}$

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(B) Given: $R = X_C = \frac{1}{\omega C}$.Initial impedance $Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + R^2} = \sqrt{2} R$.Initial peak current $I_0 = \frac{V_0}{Z} = \frac{V_0}{\sqrt{2} R}$,which implies $\frac{V_0}{R} = \sqrt{2} I_0$.When angular frequency $\omega' = \frac{\omega}{\sqrt{3}}$,the new capacitive reactance $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega / \sqrt{3}) C} = \sqrt{3} \left(\frac{1}{\omega C}\right) = \sqrt{3} R$.New impedance $Z' = \sqrt{R^2 + (X_C')^2} = \sqrt{R^2 + (\sqrt{3} R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.New peak current $I_0' = \frac{V_0}{Z'} = \frac{V_0}{2R} = \frac{1}{2} \left(\frac{V_0}{R}\right) = \frac{1}{2} (\sqrt{2} I_0) = \frac{I_0}{\sqrt{2}}$.

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