A series combination of resistor ' \(\mathrm{R}\) ' and capacitor ' \(\mathrm{C}\) ' is connected to an a.c. source of angular frequency ' \(\omega\) '. Keeping the voltage same, if the frequency is changed to \(\frac{\omega}{3}\) the current becomes half of the original current. Then the ratio of capacitive reactance and resistance at the former frequency is

Initial current \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}}\) and Final current \(\mathrm{I}^{\prime}=\frac{\mathrm{V}}{\mathrm{Z}^{\prime}}\)
\(
\because \mathrm{I}^{\prime}=\frac{\mathrm{I}}{2}
\)
\(
Z^{\prime}=2 Z
\)
\(
\frac{\mathrm{R}^2+\mathrm{X}_{\mathrm{c}}^2}{\mathrm{R}^2+\mathrm{X}_{\mathrm{c}}^{\prime 2}}=\frac{1}{4}
\)
\(
4 R^2+4 X_c^2=R^2+X_c^2
\)
\(
\begin{aligned}
& 3 R^2=X_c^{\prime 2}-4 X_c^2 \\
& X_c=\frac{1}{\omega C}, \quad X_c^{\prime}=\frac{3}{\omega C} \\
& X_c{ }^{\prime}=3 X_c
\end{aligned}
\)
From (1)
\(
3 R^2=9 X_c^2-4 X_c^2=5 X_c^2
\)
\(
\frac{\mathrm{X}_{\mathrm{c}}^2}{\mathrm{R}^2}=\frac{3}{5}=0.6
\)
\(
\frac{\mathrm{X}_{\mathrm{c}}}{\mathrm{R}}=\sqrt{0.6}
\)