Question

Consider the opamp circuit as shown below has opamp which is ideal in nature.

The voltage (in volt) at node A, connected to the negative input of the opamp is

• A. 0.864 V
• B. 0.001 V
• C. 0.499 V
• D. 0.032 V

Answer 1

The correct option is C 0.499 V

By the virtual ground,

$V_0=-\frac{R_2}{R_1}{V}_{in}=\frac{-31k\Omega }{1k\Omega }\times 1V=-31V$

$V_0=-31V<-15V$ (not possible)



$\therefore$ Output voltage, $V_0=-15V$

By KCL at node-A,

$\frac{V_A-(-15)}{31k\Omega }+\frac{V_A-1}{1k\Omega }=0$

$\frac{V_A}{31k\Omega }+\frac{V_A}{1k\Omega }=\frac{-15}{31k\Omega }+\frac{1}{1k\Omega }$

$V_A[\frac{1}{31}+1]=\frac{-15}{31}+1$

$V_A\left(1.0322\right)=0.516$

$V_A=0.499V$