Question

The wiper of the $20k\Omega$ potentiometer in figure is positioned half way then the voltage $V_0$ of the circuit is

• A. -1.5 V
• B. -1.0 V
• C. -0.75 V
• D. +2.5 V

Answer 1

The correct option is B -1.0 V


This is the modified circuit.
Due to virtual ground concept V_ = 0 V
Writing KCL at node A, we get

$\frac{V_A}{10}+\frac{V_A}{10}+\frac{V_A-2}{10}=0$

$3V_A-2=0$

$V_A=2/3Volts$

Now KCL at -ve terminal of OPAMP

$\frac{V\_-V_A}{10}+\frac{V\_-V_0}{15}=0$
$\frac{-2}{3\times 10}-\frac{V_0}{15}=0$

$V_0=\frac{-2}{3}\times \frac{15}{10}=-1Volt$