Question

A series combination of resistor $(R)$ and capacitor $(C)$ is connected to an $AC$ source of angular frequency ${}^{\prime }{\omega }^{\prime }$. 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.
$\therefore Z^{\prime }=\sqrt{R^2+{\left(3X_C\right)}^2}=\sqrt{R^2+9X_C^2}$
As current becomes half
$\therefore Z^{\prime }=2Z$
$\therefore 2Z=\sqrt{R^2+9X_C^2}$
$\Rightarrow 2\sqrt{R^2+X_C^2}=\sqrt{R^2+9X_C^2}$
$\Rightarrow 4\left(R^2+X_C^2\right)=R^2+9X_C^2$
$\Rightarrow 3R^2=5X_C^2$
or $\frac{X_C}{R}=\sqrt{\frac{3}{5}}=\sqrt{0.6}$