An a.c. voltage source $\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}$. is connected across resistance ' R…
An a.c. voltage source $\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}$. is connected across resistance ' R ' and capacitance ' $C$ ' in series. It is given that $R=\frac{1}{\omega \mathrm{c}}$ and the peak current is $\mathrm{I}_0$. If the angular frequency of the voltage source is changed to $\left(\frac{\omega}{\sqrt{3}}\right)$, then the new peak current in the circuit is
$\frac{\mathrm{I}_0}{2}$
$\frac{\mathrm{I}_0}{\sqrt{2}}$
$\sqrt{2} \mathrm{I}_0$
$\sqrt{3} \mathrm{I}_0$
Solution
Given: $R=\frac{1}{\omega \mathrm{C}}=\mathrm{X}_{\mathrm{c}}$.
$\begin{aligned}
\therefore \quad Z & =\sqrt{R^2+X_C^2}=\sqrt{2} R \\
I_0 & =\frac{V_0}{Z}=\frac{V_0}{\sqrt{2} R} \\
& \Rightarrow \frac{I_0}{\sqrt{2}}=\frac{V_0}{2 R}...(i)
\end{aligned}$ When $\omega$ becomes $\frac{1}{\sqrt{3}}$ times, $X_C$ will become $\sqrt{3}$ times, i.e., $\sqrt{3} R$.
$\begin{array}{ll}
\therefore & Z^{\prime}=\sqrt{R^2+(\sqrt{3} R)^2}=2 R \\
\therefore & I_0^{\prime}=\frac{V_0}{Z^{\prime}}=\frac{V_0}{2 R}=\frac{I_0}{\sqrt{2}}...[From(i)]
\end{array}$