An a.c. source of angular frequency ' \(\omega\) ' is fed across a resistor ' \(\mathrm{R}\) ' and a capacitor ' \(\mathrm{C}\) ' in series. The current registered is I. If now the frequency of source is changed to \(\frac{\omega}{3}\) (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency ' \(\omega\) ' will be

\(\begin{aligned} & \text { Initial current } I=\frac{V}{Z}, \text { Final current } I^{\prime}=\frac{V}{Z^{\prime}} \\ & \frac{1}{2}=\frac{I^{\prime}}{I}=\frac{Z}{Z^{\prime}} \quad \therefore \frac{Z}{Z^{\prime}}=\frac{1}{2} \\ & \therefore \frac{Z^2}{Z^{\prime 2}}=\frac{1}{4} \quad \therefore \frac{R^2+X_c^2}{R^2+X_c^{\prime 2}}=\frac{1}{4} \\ & \therefore 4 R^2+4 X_c^2=R^2+X_c^{\prime 2} \\ & \therefore 3 R^2=X_c^{\prime 2}-4 X_c^2 \quad \ldots .(1) \\ & X_c=\frac{1}{\omega C}, X^{\prime}=\frac{3}{\omega C} \quad \therefore X^{\prime}=3 X_c \\ & \text { Putting this value of } X^{\prime} \text { in }(1) \text { we get } \\ & 3 R^2=9 X_c^2-4 X_c^2=5 X_c^2 \\ & \therefore \frac{X_c^2}{R^2}=\frac{3}{5}=0.6 \\ & \therefore \frac{X_c}{R}=\sqrt{\frac{3}{5}}\end{aligned}\)