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)

At angular frequency $\omega$, the current in R-C circuit is given by
$I_{rms}=\frac{V_{rms}}{\sqrt{R^2+(\frac{1}{\omega C}{)}^2}}$ ..( i )
Also, $\frac{I_{rms}}{2}=\frac{V_{rms}}{\sqrt{R^2+[\frac{\frac{1}{\omega C}}{3}{]}^2}}=\frac{V_{rms}}{\sqrt{R^2+\frac{9}{{\omega }^2{C}^2}}}$ ...(ii)
From Eqs. (i) and (ii), we get
$3R^2=\frac{3}{{\omega }^2{C}^2}$
$\Rightarrow \frac{1}{\frac{\omega C}{R}}=\sqrt{\frac{3}{5}}$
$\Rightarrow \frac{X_C}{R}=\sqrt{\frac{3}{5}}$