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 the source is changed to $\frac{\omega }{3}$ (but maintaining the same voltage), then 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 $RC$ circuit is given by
$i_{r m s}=\frac{V_{r m s}}{\sqrt{R^{2} + \left(\frac{1}{\omega C}\right)^{2}}}$ ......(i)
$Also \, \frac{i_{r m s}}{2}=\frac{V_{r m s}}{\sqrt{R^{2} + \left(\frac{1}{\frac{\omega }{3} C}\right)^{2}}}=\frac{V_{r m s}}{\sqrt{R^{2} + \frac{9}{\left(\omega \right)^{2} C^{2}}}}$ ......(ii)
From equation (i) and (ii) we get
$3R^{2}=\frac{5}{\omega ^{2} C^{2}}\Rightarrow \frac{\frac{1}{\omega C}}{R}=\sqrt{\frac{3}{5} \, }\Rightarrow \frac{X_{C}}{R}=\sqrt{\frac{3}{5}}$