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 $\left|A\right(\omega \left)\right|=0.25$, 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)

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

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

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

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

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

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