A series combination of resistor ' $\mathrm{R}$ ' and capacitor ' $\mathrm{C}$ ' is connected to an a.c.…
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
$\sqrt{0.6}$
$\sqrt{6}$
$\sqrt{3}$
$\sqrt{2}$
Solution
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)
$3 R^2=9 X_c^2-4 X_c^2=5 X_c^2$
$\frac{\mathrm{X}_{\mathrm{c}}^2}{\mathrm{R}^2}=\frac{3}{5}=0.6$
$\frac{\mathrm{X}_{\mathrm{c}}}{\mathrm{R}}=\sqrt{0.6}$