Question

The circuit shown below is driven by a sinusoidal input $v_i=V_{p}cos\left(t/RC\right)$. The steady output $V_0$ is

• A. $\left(V_p/3\right)cos\left(t/RC\right)$
• B. $\left(V_p/3\right)sin\left(t/RC\right)$
• C. $\left(V_p/2\right)cos\left(t/RC\right)$
• D. $\left(V_p/2\right)sin\left(t/RC\right)$

Answer 1

The correct option is A $\left(V_p/3\right)cos\left(t/RC\right)$

Redrawing the circuit s-domain
$V_i\left(s\right)=(R+\frac{1}{sC})I\left(s\right)+\frac{R.\frac{1}{sC}}{R+\frac{1}{sC}}I\left(s\right)$

$V_i\left(s\right)=\frac{1+sCR}{sC}I\left(s\right)+\frac{R}{1+sCR}I\left(s\right)$ ...(i)

$\because v_i=V_{p}cos\left(t/RC\right)$

So here,
$\omega =\frac{1}{RC}$

Now,
$V_i\left(s\right)=\frac{(1+j\omega CR)}{j\omega C}I\left(s\right)+\frac{R}{(1+j\omega CR)}I\left(s\right)$

$Put,\omega =\frac{1}{RC}$

So,
$V_i\left(s\right)=[\frac{(1+j)R}{j}+\frac{R}{1+j}]I\left(s\right)$

$\frac{V_i\left(s\right)}{I\left(s\right)}=\frac{3R}{(1+j)}$ ...(ii)

$I\left(s\right)=\frac{V_i\left(s\right)}{3R}\times (1+j)$

Now,
$V_0\left(s\right)=\left(R||\frac{1}{sC}\right)I\left(s\right)$

$\Rightarrow V_0\left(s\right)=\frac{R.\frac{1}{sC}}{R+\frac{1}{sC}}I\left(s\right)$

$\Rightarrow V_0\left(s\right)=\frac{R}{1+sCR}.\frac{V_i\left(s\right)}{3R}(1+j)$

$\Rightarrow V_0\left(s\right)=\frac{R}{1+j}.\frac{V_i\left(s\right)}{3R}(1+j)$

$\Rightarrow V_0\left(s\right)=\frac{V_i\left(s\right)}{3}$
In time domain,

$v_0\left(t\right)=\frac{1}{3}{v}_i\left(t\right)$

$\Rightarrow v_0\left(t\right)=\frac{V_p}{3}cos\left(\frac{t}{RC}\right)$