An ac source of angular frequency ω 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 ω/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 ω.

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

$\sqrt {\frac{2}{5}}$

$\sqrt {\frac{1}{5}}$

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

answer is A.

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Detailed Solution

At angular frequency ω, the current in RC circuit is given by

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

$also\,\,\frac{i_{rms}}{2}=\frac{V_{rms}}{\sqrt{R^2+\left ( \frac{1}{\frac{\omega}{3}C} \right )^2}}=\frac{V_{rms}}{\sqrt{R^2+\frac{9}{\omega^2{C}^2}}}$ .......... (2)

From equation (i) and (ii) we get

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