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 $\omega /\sqrt{3}$, then the new peak current in the circuit is $I_0/\sqrt{x}$. Find x.

Answer 1

Initial peak current,
$I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+(\frac{1}{\omega C}{)}^2}}$ ... (1)

It is given that,
$\frac{1}{\omega C}=R$ .... (2)

Put equation (2) in equation (1), we get
$\Rightarrow I_0=\frac{V_0}{\sqrt{2R^2}}=\frac{V_0}{\sqrt{2}R}$

Changed frequency given as,
${\omega }^{\prime }=\frac{\omega }{\sqrt{3}}$

Then new peak current,
$\therefore I_0^{\prime }=\frac{V_0}{\sqrt{R^2+(\frac{1}{{\omega }^{\prime }C}{)}^2}}=\frac{V_0}{\sqrt{R^2+\frac{3}{{\omega }^2{C}^2}}}$
$\therefore I_0^{\prime }=\frac{V_0}{\sqrt{R^2+3R^2}}=\frac{V_0}{\sqrt{4R^2}}$
$\Rightarrow I_0^{\prime }=\frac{V_0}{2R}$

Then the ratio,
$\frac{I_0^{\prime }}{I_0}=\left(\frac{V_0}{2R}\right)\times \left(\frac{\sqrt{R}}{V_0}\right)=\frac{1}{\sqrt{2}}$

Therefore,
$I_0^{\prime }=\frac{1}{\sqrt{2}}{I}_0$
Hence, the value of x is 2.