Question

Consider the op-amp circuit shown in the figure below:



The value of output voltage $V_0$ is equal to

• A. 65 V
• B. 55 V
• C. 60 V
• D. 50 V

Answer 1

The correct option is B 55 V

Applying KCL at node V${}^{+}$, we get,

$\frac{V^{+}-10}{60k\Omega }+\frac{V^{+}-15}{20k\Omega }+\frac{V^{+}}{30k\Omega }=0$



$\therefore \frac{V^{+}}{10k\Omega }=\frac{1}{6k\Omega }+\frac{3}{4k\Omega }$
$\therefore V^{+}=\frac{55}{6}$
$\therefore V_0=(1+\frac{R_2}{R_1})V^{+}=(1+5)\times \frac{55}{6}$
$V_0$ = 55 V