An AC voltage source $V = V_{0} \sin ω t$ is connected across resistance $R$ and capacitance $C$ as shown in figure. It is given that $R = \frac{1}{ω C}$ . The peak current is $I_{0}$ . If the angular frequency of the voltage source is changed to $\frac{ω}{\sqrt{3}}$ , then the new peak current in the circuit is

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$\frac{I_{0}}{\sqrt{3}}$

$\frac{I_{0}}{\sqrt{2}}$

$\frac{I_{0}}{2}$

$\frac{I_{0}}{3}$

answer is B.

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Detailed Solution

$R = \frac{1}{ω C} = X_{C}$

$∴ Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{2} R (as X_{C} = R) \Rightarrow I_{0} = \frac{V_{0}}{Z} = \frac{V_{0}}{\sqrt{2} R} . . . . . . (i)$

When $ω$ becomes $\frac{1}{\sqrt{3}}$ times, $X_{C}$ will become $\sqrt{3}$ times or $\sqrt{3} R$ .

$\begin{array}{rcl} ∴ Z' & = & \sqrt{R^{2} + {(\sqrt{3} R)}^{2}} = 2 R \\ \Rightarrow & I_{0}' = \frac{V_{0}}{Z'} = \frac{V_{0}}{2 R} = \frac{I_{0}}{\sqrt{2}} \end{array}$

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