Q. An ac source of angular frequency $\omega$ is fed across a resistor $r$ and a capctior $C$ in series. The current registered is $I$. If the frequency of source is changed to $\omega/3$ (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) $\sqrt{\frac{4}{5}}$
Explanation
(a) At angular frequency $\omega$, the current in $RC$ circuit is given by
$i_{rms} = \frac{V_{rms}}{\sqrt{R^{2} + \left( \frac{1}{\omega C} \right)^{2}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Also $\frac{i_{rms}}{2} = \frac{V_{rms}}{\sqrt{R^{2} + \left( \frac{1}{\frac{\omega}{3}}C \right)^{2}}} = \frac{V_{rms}}{\sqrt{R^{2} + \frac{9}{\omega^{2}C^{2}}}}\ \ $ …(ii)
From equation (i) and (ii), we get
$3R^{2} = \frac{5}{\omega^{2}C^{2}} \Rightarrow \frac{\frac{1}{\omega C}}{R} = \sqrt{\frac{3}{5}} \Rightarrow \frac{X_{C}}{R} = \sqrt{\frac{3}{5}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.