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. The ratio of reactance to resistance at the original frequency $\omega$ is

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

Answer 1

Answer: (A)

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

$I=\frac{V}{\sqrt{R^2+(\frac{1}{wc}{)}^2}}$, for w

for $\frac{w}{3},\frac{I}{2}=\frac{V}{\sqrt{R^2+(\frac{3}{wc}{)}^2}}$$\frac{2}{\sqrt{R^2+\frac{9}{w^2{C}^2}}}=\frac{1}{\sqrt{R^2+\frac{1}{w^2{C}^2}}}$$4R^2+\frac{4}{w^2{c}^2}=R^2+\frac{9}{w^2{c}^2}$

$3R^2=\frac{5}{w^2{c}^2}$

$3R^2=5(x_c{)}^2$ [at w]

$\sqrt{\frac{3}{5}}=\frac{x_c}{R}$