Question 424: AC fundamentals

Question

A series combination of resistor 'R' and capacitor '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

Accepted Answer

Initial current $$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{z}}$$, Final current $$\mathrm{I}^{\prime}=\frac{\mathrm{V}}{\mathrm{z}}$$
$$\begin{aligned}
& \frac{1}{2}=\frac{\mathrm{I}^{\prime}}{\mathrm{I}}=\frac{\mathrm{z}}{\mathrm{z}^{\prime}} \quad 	herefore \frac{\mathrm{z}}{\mathrm{z}^{\prime}}=\frac{1}{2} \
& 	herefore \frac{\mathrm{z}^2}{\mathrm{z}^{\prime 2}}=\frac{1}{4} \quad 	herefore \frac{\mathrm{R}^2+\mathrm{X}_{\mathrm{c}}^2}{\mathrm{R}^2+\mathrm{X}_{\mathrm{c}}^{\prime 2}}=\frac{1}{4} \
& 	herefore 4 \mathrm{R}^2+4 \mathrm{X}_{\mathrm{c}}^2=\mathrm{R}^2+\mathrm{X}_{\mathrm{c}}^{\prime 2} \
& 	herefore 3 \mathrm{R}^2=\mathrm{X}_{\mathrm{c}}^{\prime 2}-4 \mathrm{X}_{\mathrm{c}}^2 \quad 	ext{......(1)}
\end{aligned}$$
$$\mathrm{X}_{\mathrm{c}}=\frac{1}{\omega C}, \mathrm{X}^{\prime}=\frac{3}{\omega C} \quad 	herefore \mathrm{X}^{\prime}=3 \mathrm{X}_{\mathrm{c}}$$
Putting this value of $$X^{\prime}$$ in (1) we get
$$3 R^2=9 X_c^2-4 X_c^2=5 X_c^2$$
$$\begin{aligned}
& 	herefore \frac{\mathrm{X}_{\mathrm{c}}^2}{\mathrm{R}^2}=\frac{3}{5}=0.6 \
& 	herefore \frac{\mathrm{X}_{\mathrm{c}}}{\mathrm{R}}=\sqrt{0.6}
\end{aligned}$$

Answer

$$\sqrt{0.6}$$

Answer

$$\sqrt{6}$$

Answer

$$\sqrt{3}$$

Answer

$$\sqrt{2}$$

AC fundamentals

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