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An AC source of angular frequency $\omega$ is fed across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit found to be $I$. Now the frequency of the source is changed to $\frac{\omega}{3}$, (maintaining the same voltage) the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?
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The correct answer is:
$\sqrt{\frac{3}{5}}$
At angular frequency $\omega$, current through resistance R and capacitance C is given by
$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+X_C^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}$...(i)
When angular frequency is changed to $\frac{\omega}{3}$, then the current becomes
$$
\frac{I_{\mathrm{rms}}}{2}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\frac{\omega}{3} C}\right)^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{3}{\omega C}\right)^2}}...(ii)
$$
Dividing Eq. (i) by Eq. (ii), we get
$\begin{array}{rlrl} & & 2=\frac{\sqrt{R^2+\left(\frac{3}{\omega C}\right)^2}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}} \\ & \Rightarrow & 4\left[R^2+\left(\frac{1}{\omega C}\right)^2\right] & =R^2+\left(\frac{3}{\omega C}\right)^2 \\ & \Rightarrow & 3 R^2 & =\frac{5}{\omega^2 C^2} \\ & \Rightarrow & \frac{1}{\omega_C} & =\sqrt{\frac{3}{5}} \\ \Rightarrow & \frac{X_C}{R} & =\sqrt{\frac{3}{5}}\end{array}$
$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+X_C^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}$...(i)
When angular frequency is changed to $\frac{\omega}{3}$, then the current becomes
$$
\frac{I_{\mathrm{rms}}}{2}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{1}{\frac{\omega}{3} C}\right)^2}}=\frac{V_{\mathrm{rms}}}{\sqrt{R^2+\left(\frac{3}{\omega C}\right)^2}}...(ii)
$$
Dividing Eq. (i) by Eq. (ii), we get
$\begin{array}{rlrl} & & 2=\frac{\sqrt{R^2+\left(\frac{3}{\omega C}\right)^2}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}} \\ & \Rightarrow & 4\left[R^2+\left(\frac{1}{\omega C}\right)^2\right] & =R^2+\left(\frac{3}{\omega C}\right)^2 \\ & \Rightarrow & 3 R^2 & =\frac{5}{\omega^2 C^2} \\ & \Rightarrow & \frac{1}{\omega_C} & =\sqrt{\frac{3}{5}} \\ \Rightarrow & \frac{X_C}{R} & =\sqrt{\frac{3}{5}}\end{array}$
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