Topic: Network Theory
For the circuit shown in the given figure, the voltage V_AB is
- A. 6 V
- B. 10 V
- C. 25 V
- D. 40 V
Answer: A. 6 V
Explanation: Applying KCL at node B with node A as reference (0V): (V_B - 10)/5 + V_B/5 + (V_B + 50)/10 = 0. Solving for V_B: 2(V_B - 10) + 2V_B + V_B + 50 = 0 => 5V_B + 30 = 0 => V_B = -6 V. Thus V_BA = -6V, so V_AB = 6V.