Question

An ac voltage source $V=V_0\sin \left(\omega t\right)$ 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 $\frac{\omega }{\sqrt{3}}$ 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 $\frac{I_0}{\sqrt{2}}$

The peak value of the current is:

$I_0=\frac{V_0}{\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}}=\frac{V_0}{\sqrt{2}R}$

When the angular frequency is changed to $\frac{\omega }{\sqrt{3}}$, the new peak value is:

$I_0^{\prime }=\frac{V_0}{\sqrt{R^2+\frac{3}{{\omega }^2{C}^2}}}=\frac{V_0}{\sqrt{4R^2}}=\frac{V_0}{2R}$

$\therefore I_0^{\prime }=\frac{I_0}{\sqrt{2}}$