Q144 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics
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
- Correct answerA$\sqrt{0.6}$
- B$\sqrt{6}$
- C$\sqrt{3}$
- D$\sqrt{2}$
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)
$$
3 R^2=9 X_c^2-4 X_c^2=5 X_c^2
$$
$$
\frac{\mathrm{X}_{\mathrm{c}}^2}{\mathrm{R}^2}=\frac{3}{5}=0.6
$$
$$
\frac{\mathrm{X}_{\mathrm{c}}}{\mathrm{R}}=\sqrt{0.6}
$$
Hence, the correct answer is option: 'A'.