Question

A series combination of resistor $(R)$ and capacitor $(C)$ is connected to an $AC$ 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 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)

The impedance of the circuit, $Z=\sqrt{R^{2}+X_{C}^{2}}$
When angular frequency of source is reduced to $\frac{1}{3}$,
the capacitor reactance is increased by $3$ times.
$\therefore Z'=\sqrt{R^{2}+\left(3 X_{C}\right)^{2}}=\sqrt{R^{2}+9 X_{C}^{2}}$
As current becomes half
$\therefore Z'=2 Z$
$\therefore 2 Z=\sqrt{R^{2}+9 X_{C}^{2}} $
$\Rightarrow 2 \sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+9 X_{C}^{2}} $
$ \Rightarrow 4\left(R^{2}+X_{C}^{2}\right)=R^{2}+9 X_{C}^{2} $
$\Rightarrow 3 R^{2}=5 X_{C}^{2} $
or $ \frac{X_{C}}{R}=\sqrt{\frac{3}{5}}=\sqrt{0.6}$