Question

An ac voltage source $V=V_0\sin \omega t$ is connected across resistance $R$ and capacitance $C$ as shown in figure. It is given that $R=\frac{1}{\omega C}$. The peak current is $I_0$. If the angular frequency of the voltage source is changed to $\frac{\omega }{\sqrt{3}}$ by keeping $R$ and $C$ fixed, then the new peak current in the circuit is-

• A. $\frac{I_0}{2}$
• B. $\frac{I_0}{\sqrt{2}}$
• C. $\frac{I_0}{\sqrt{3}}$
• D. $\frac{I_0}{3}$

Answer 1

The correct option is B $\frac{I_0}{\sqrt{2}}$

Given,

$R=\frac{1}{\omega C}\text{and}{X}_C=\frac{1}{\omega C}$

$Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+R^2}=R\sqrt{2}$

$X_C^{\prime }=\frac{1}{{\omega }^{\prime }C}=\frac{\sqrt{3}}{\omega C}=\sqrt{3}R$

$\therefore Z^{\prime }=\sqrt{R^2+(\sqrt{3}R{)}^2}=2R$

$\Rightarrow \frac{I^{\prime }}{I_0}=\frac{\frac{V_0}{Z^{{}^{\prime }}}}{\frac{V_0}{Z}}=\frac{Z}{Z^{{}^{\prime }}}$

$\Rightarrow \frac{I^{\prime }}{I_0}=\frac{R\sqrt{2}}{2R}=\frac{1}{\sqrt{2}}$

$\Rightarrow I^{\prime }=\frac{I_0}{\sqrt{2}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.