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 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. $\sqrt{\frac{4}{5}}$

Answer 1

The correct option is A

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

At angular frequency w, the current in RC circuit is given by
$i_{max}=\frac{V_{max}}{\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}}$ ...(i)
Also $\frac{i_{max}}{2}=\frac{V_{max}}{\sqrt{R^2+{\left(\frac{1}{\frac{\omega }{3}C}\right)}^2}}=\frac{V_{max}}{\sqrt{R^2+\frac{9}{{\omega }^2{C}^2}}}$ ....(ii)
From equation (i) and (ii) we get
$3R^2=\frac{5}{{\omega }^2{C}^2}\Rightarrow \frac{\frac{1}{\omega C}}{R}=\sqrt{\frac{3}{5}}\Rightarrow \frac{X_c}{R}=\sqrt{\frac{3}{5}}$