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 $\omega /3$, the current becomes half of the original current. Then the ratio of the capacitive reactance and resistance at the former frequency is:

• A. $\sqrt{0.6}$
• B. $\sqrt{3}$
• C. $\sqrt{2}$
• D. $\sqrt{6}$

Answer 1

Answer: (A)

$\sqrt{0.6}$

Let r.m.s. value of supply voltage be $v$.

Impedance of RC circuit is given by:

$Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}$

Then, r.m.s. current in the circuit is:

$i=\frac{v}{Z}=\frac{v}{\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}}$

$⟹\frac{i_2}{i_1}=\frac{\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}}{\sqrt{R^2+\frac{1}{(\omega /3{)}^2{C}^2}}}$

Given : $\frac{i_2}{i_1}=\frac{1}{2}$ and ${\omega }_2=\omega /3$

On substituting values and simplifying, we get:

$\frac{1}{4}=\frac{1+{\omega }^2{R}^2{C}^2}{9+{\omega }^2{R}^2{C}^2}$

$⟹\omega RC=\sqrt{5/3}$

Ratio at capacitive reactance and resistance in original frequency is:

$\frac{X_C}{R}=\frac{1}{\omega RC}=\sqrt{3/5}=\sqrt{0.6}$