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 + \bigg( \frac{ 1}{ \omega C}\bigg)^2 }} $ ..( i )
Also, $ \frac{ I_{ rms}}{ 2} = \frac{ V_{ rms}}{ \sqrt{ R^2 + \Bigg [ \frac{ \frac{1}{ \omega C}}{ 3}\Bigg]^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 }} $