Question

An AC source of angular frequency $\omega$ is fed across a resistor $r$ and a capacitor $C$ in series. The current registered is $I$. If now the frequency of source is changed to $\omega /3$ (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency $\omega$.

• A. $\sqrt{\frac{3}{5}}$
• B. $\sqrt{\frac{2}{5}}$
• C. $\sqrt{\frac{1}{5}}$
• D. $\sqrt{\frac{4}{5}}$

Answer 1

Answer: (A)

$\sqrt{\frac{3}{5}}$

At angular frequency $\omega$ , current through RC circuit is given by
$I_{rms}=\frac{V_{rms}}{\sqrt{R^2+X_C^2}}=\frac{V_{rms}}{\sqrt{R^2+{\left(1/\omega C\right)}^2}}$......(1)
When angular frequency is changed to $\frac{\omega }{3}$
then the current becomes,
$\frac{I_{rms}}{2}=\frac{V_{rms}}{\sqrt{R^2+{\left(\frac{1}{\left(\frac{\omega }{3}\right)C}\right)}^2}}=\frac{V_{rms}}{\sqrt{R^2+{\left(\frac{3}{\omega C}\right)}^2}}$...........(2)

Dividing (i) by (ii)
$2=\frac{\sqrt{R^2+{\left(3/\omega C\right)}^2}}{\sqrt{R^2+{\left(1/\omega C\right)}^2}}$

or $4[R^2+{\left(\frac{1}{\omega C}\right)}^2]=R^2+{\left(\frac{3}{\omega C}\right)}^2$
$\Rightarrow 3R^2=\frac{5}{{\omega }^2{C}^2}\Rightarrow \frac{1/\omega C}{R}=\sqrt{\frac{3}{5}}$
$\Rightarrow \frac{X_C}{R}=\sqrt{\frac{3}{5}}$