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 $\frac{\omega }{3}$ (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency $\omega$ will be.

• A. $\sqrt{\frac{3}{5}}$
• B. $\sqrt{\frac{5}{3}}$
• C. $\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A

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

According to given problem

$I=\frac{V}{Z}=\frac{V}{{[R^2+\left(\frac{1}{C^2{\omega }^2}\right)]}^{\frac{1}{2}}}\dots \dots \left(1\right)$

and $\frac{I}{2}=\frac{V}{{[R^2+\left(\frac{9}{C^2{\omega }^2}\right)]}^{\frac{1}{2}}}\dots \dots \left(2\right)$

Substituting the value of I from equation (1) in (2),

$4(R^2+\frac{1}{C^2{\omega }^2})=R^2+\frac{9}{C^2{\omega }^2},i.e.,\frac{1}{C^2{\omega }^2}=\frac{3}{5}{R}^2$

So then $\frac{X}{R}=\frac{\left(\frac{1}{C\omega }\right)}{R}=\frac{{\left[\left(\frac{3}{5}\right)R^2\right]}^{\frac{1}{2}}}{R}=\sqrt{\frac{3}{5}}$