Question

In the circuit shown in the figure, the input signal is $v_i\left(t\right)=5+3cos\omega t$

The steady state output is expressed as $v_0\left(t\right)=P+Qcos(\omega t-\varphi ).$ If $\omega CR=2$, the values of P and Q are

• A. $P=0andQ=\frac{6}{\sqrt{5}}$
• B. $P=0andQ=\frac{3}{\sqrt{5}}$
• C. $P=5andQ=\frac{6}{\sqrt{5}}$
• D. $P=5andQ=3$

Answer 1

The correct option is A $P=0andQ=\frac{6}{\sqrt{5}}$

Input signal,

$v_i\left(t\right)=5+3cos\omega t$

$v_i\left(t\right)=V_1+V_2\left(t\right)$

By applying superposition theorem

Case-I

When $V_1$ is applied





$V_{01}\left(t\right)=V_1{e}^{-t/RC}$

$\therefore$ Steady state output $V_{01}\left(t\right)=0$

Case-II

When $V_2\left(t\right)$ is applied


$\because Z=\sqrt{R^2+\frac{1}{{\omega }^2{C}^2}}$

$\therefore V_{02}\left(t\right)=\frac{V_2\left(t\right)}{Z\angle \varphi }.R=\frac{3cos(\omega t-\varphi )}{\sqrt{1+R^2{\omega }^2{C}^2}}.R\omega C$

Since $\omega RC=2$

$\therefore V_{02}\left(t\right)=\frac{6cos(\omega t-\varphi )}{\sqrt{1+4}}=\frac{6}{\sqrt{5}}cos(\omega t-\varphi )$

$\therefore$ Steady state output $v_0\left(t\right)$

$=v_{01}\left(t\right)+v_{02}\left(t\right)$

$=0+\frac{6}{\sqrt{5}}cos(\omega t-\varphi )$

$\therefore P=0andQ=\frac{6}{\sqrt{5}}$