Question 321: AC fundamentals

Question

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 $R=\frac{1}{\omega c}$ 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

Accepted Answer

Given: $\mathrm{R}=\frac{\mathrm{l}}{\omega \mathrm{C}}=\mathrm{X}_{\mathrm{c}}$.
$$\begin{aligned}
	herefore \quad Z & =\sqrt{R^2+X_C^2}=\sqrt{2} R \
& I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{2} R} \
& \Rightarrow \frac{I_0}{\sqrt{2}}=\frac{V_0}{2 R}\quad	ext{.... (i)}
\end{aligned}$$
When $\omega$ becomes $\frac{1}{\sqrt{3}}$ times, $\mathrm{X}_{\mathrm{C}}$ will become $\sqrt{3} 	ext { times, i.e., } \sqrt{3} R.$
$$\begin{array}{ll} 
	herefore & \mathrm{Z}^{\prime}=\sqrt{\mathrm{R}^2+(\sqrt{3} R)^2}=2 R \
	herefore & \mathrm{I}_0^{\prime}=\frac{\mathrm{V}_0}{\mathrm{Z}^{\prime}}=\frac{\mathrm{V}_0}{2 \mathrm{R}}=\frac{\mathrm{I}_0}{\sqrt{2}} 	ext{...[From(i)]}
\end{array}$$

Answer

$\frac{\mathrm{I}_0}{2}$

Answer

$\frac{\mathrm{I}_0}{\sqrt{2}}$

Answer

$ \sqrt{2} \mathrm{I}_0$

Answer

$\sqrt{3} \mathrm{I}_0$

AC fundamentals

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