Question

The steady state output of the circuit shown in the figure is given by

$y\left(t\right)=A\left(\omega \right)\sin (\omega t+\varphi (\omega \left)\right).$ If the amplitude $|A\left(\omega \right)|=0.25,$ then the frequency $\omega$ is

• A. $\frac{1}{\sqrt{3}RC}$
• B. $\frac{2}{\sqrt{3}RC}$
• C. $\frac{1}{RC}$
• D. $\frac{2}{RC}$

Answer 1

The correct option is B

$\frac{2}{\sqrt{3}RC}$

Applying KCL at node A, we get

$\frac{V_A-\sin \omega t}{R}+\frac{V_A}{\frac{1}{j\omega C}}+\frac{V_A}{\frac{2}{j\omega C}}=0$ ... (i)

$V_A[\frac{1}{R}+j\omega C+\frac{j\omega C}{2}]$=$\frac{\sin \omega t}{R}$=$\frac{1\angle 0{}^{\circ }}{R}$

$V_A=\frac{2}{2+3\left(RC\right)\left(j\omega \right)}$ ... (ii)

Also $Y=\frac{V_A}{2}=\frac{1}{2+3j\omega RC}$ ... (iii)

$\because |A\left(\omega \right)|=\frac{1}{4}=0.25$

$\therefore \frac{1}{4}=\frac{1}{\sqrt{4+9{\omega }^2{\left(RC\right)}^2}}$

$=\frac{2}{\sqrt{3}RC}$