Question

Consider the circuits shown in the figure below.

Where, $V_1\left(t\right)=10\sin \left(\omega t\right)and{V}_2\left(t\right)=-10\sin \left(\omega t\right)$ then the output voltage $V_o\left(t\right)$ is equal to

• A. 0
• B. $20\cos \left(\omega t\right)$
• C. $20\sin (\omega t+\pi )$
• D. $20\sin \omega t$

Answer 1

The correct option is C $20\sin (\omega t+\pi )$

Applying KCL at node $V_{a^{}}$ We get

$\frac{V_a-V_1\left(t\right)}{20k\Omega }+\frac{V_a-V_0\left(t\right)}{20k\Omega }=0$


$V_a=\frac{V_1\left(t\right)+V_0\left(t\right)}{2}$
Now ,
$V_b=\frac{V_2\left(t\right)}{2}$ (By voltage division)


Thus, due to virtual ground

$V_a=V_b$

i.e., $\frac{V_1\left(t\right)+V_0\left(t\right)}{2}=\frac{V_2\left(t\right)}{2}$
$V_0\left(t\right)=V_2\left(t\right)-V_1\left(t\right)$
= $-10\sin \left(\omega t\right)-10\sin \left(\omega t\right)$
=$-20\sin \left(\omega t\right)$
= $20\sin (\omega t+\pi )$