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A series combination of resistor $(R)$, capacitor (C) is connected to an A.C. source of angular frequency $\omega$. Keeping the voltage same, if the frequency is changed to $\omega / 3$, the current becomes half of the original current. Then the ratio of the capacitive reactance and resistance at the former frequency is
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$\sqrt{0.6}$
The impedance of the circuit, $Z=\sqrt{R^{2}+X_{C}^{2}}$
When angular frequency of source is reduced to $\frac{1}{3}$, the capacitor reactance is increased by 3 times. $\therefore \quad Z^{\prime}=\sqrt{R^{2}+\left(3 X_{C}\right)^{2}}=\sqrt{R^{2}+9 X_{C}^{2}}$
As current becomes half $\therefore \quad Z^{\prime}=2 Z$
$\begin{array}{ll}\therefore & 2 Z=\sqrt{R^{2}+9 X_{C}^{2}} \\ \Rightarrow & 2 \sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+9 X_{C}^{2}} \\ \Rightarrow & 4\left(R^{2}+X_{C}^{2}\right)=R^{2}+9 X_{C}^{2} \\ \Rightarrow & 3 R^{2}=5 X_{C}^{2} \\ \text { or } & \frac{X_{C}}{R}=\sqrt{\frac{3}{5}}=\sqrt{0.6}\end{array}$
When angular frequency of source is reduced to $\frac{1}{3}$, the capacitor reactance is increased by 3 times. $\therefore \quad Z^{\prime}=\sqrt{R^{2}+\left(3 X_{C}\right)^{2}}=\sqrt{R^{2}+9 X_{C}^{2}}$
As current becomes half $\therefore \quad Z^{\prime}=2 Z$
$\begin{array}{ll}\therefore & 2 Z=\sqrt{R^{2}+9 X_{C}^{2}} \\ \Rightarrow & 2 \sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+9 X_{C}^{2}} \\ \Rightarrow & 4\left(R^{2}+X_{C}^{2}\right)=R^{2}+9 X_{C}^{2} \\ \Rightarrow & 3 R^{2}=5 X_{C}^{2} \\ \text { or } & \frac{X_{C}}{R}=\sqrt{\frac{3}{5}}=\sqrt{0.6}\end{array}$
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