GATE 2021 EC – Question 29

Correct answer: 1

Explanation

The top wire and the horizontal wire above the middle resistor tie the left branch and the middle resistor to the same node A. Call the right-hand node B. The $6$ mA source pushes current from B to A. KCL at B: $\frac{V_A-V_B}{1\text{k}}=6\text{ mA}+2\text{ mA}$, so $V_A-V_B=8$ V. KCL at A: $6\text{ mA}=\frac{V_A}{1\text{k}}+\frac{V_A-4}{1\text{k}}+\frac{V_A-V_B}{1\text{k}}$, so $6=2V_A-4+8$ and $V_A=1$ V. Hence $v_0=V_A=1.0$ V.