Question

For the op-amp circuit shown below; $V_0$ is approximately equal to

• A. -6.24 V
• B. -4.52 v
• C. 0
• D. -3.32 V

Answer 1

The correct option is D -3.32 V

Break the given circuit in two parts

In the above figure, $V_1=V_3$

From figure (a) by voltage division rule,

$V_1=10\times \frac{20}{30}=6.66V$

$V_2=10\times \frac{20}{40}=5V$

$\text{Due to symmetry}{V}_0=A_V(V_2-V_1)$

$A_V=\frac{R_f}{R_1}=\frac{200}{100}=2$

$V_0=\left(2\right)(5-6.66)=-3.32V$