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 $\frac{\omega }{4}$ (but maintaining the same voltage), the current in then 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. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

At angular frequency $\omega$, the current in RC circuit is given by

$i_{rms}=\frac{V_{rms}}{\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}}.......\left(i\right)$

Also $\frac{i_{rms}}{2}=\frac{V_{rms}}{\sqrt{R^2+{\left(\frac{1}{\frac{\omega }{4}c}\right)}^2}}=\frac{V_{rms}}{\sqrt{R^2+\frac{16}{{\omega }^2{C}^2}}}......\left(ii\right)$

From equation (i) and (ii) we get

$3R^2=\frac{12}{{\omega }^2{C}^2}\Rightarrow \frac{\frac{1}{\omega C}}{R}=\sqrt{\frac{3}{12}}\Rightarrow \frac{X_c}{R}=\sqrt{\frac{3}{12}}$