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 $R=\frac{1}{\omega C}$.Then peak current is $I_0$. If the angular frequency of the voltage source is changed to $\frac{\omega }{\sqrt{3}}$ keeping R and C fixed, then the new peak current in the circuit is

• A. $\frac{l_0}{2}$
• B. $\frac{l_0}{\sqrt{2}}$
• C. $\frac{l_0}{\sqrt{3}}$
• D. $\frac{l_0}{3}$

Answer 1

The correct option is B $\frac{l_0}{\sqrt{2}}$

The peak value of the current is
$l_o\frac{v_0}{\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}}=\frac{v_0}{R\sqrt{2}}$
When the angular frequency is changed to $\frac{\omega }{\sqrt{3}}$
The new peak value is :
$l_0=\frac{v_0}{\sqrt{R^2+-\frac{3}{{\omega }^2{c}^2}}}=\frac{v_0}{2R}$
$\therefore l_0=\frac{l_0}{\sqrt{2}}$