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)

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

Answer 2

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 $3R^2 = \frac{ 3}{ \omega^2 C^2 }$ $\Rightarrow \frac{ 1}{ \frac{\omega C}{R}} = \sqrt{ \frac{ 3}{ 5 }}$ $\Rightarrow \frac{ X_C}{ R} = \sqrt{ \frac{ 3}{ 5 }}$