Question

For the circuit shown below:



The output voltage $V_o$ is______.

• A. 14 V
• B. - 1 V
• C. 15 V
• D. - 15 V

Answer 1

The correct option is B - 1 V

Applying KCL at node a,

$\frac{6-V_a}{4}=\frac{V_a-V_o}{10}$

But $V_a=V_b=4$, and so

$\frac{6-4}{4}=\frac{4-V_o}{10}$

$V_o=-1V$