GATE 2023 EE – Question 32
For the circuit shown in the figure, $V_1=8$ V, DC and $I_1=8$ A, DC. The voltage $V_{ab}$ in Volts is _________ (Round off to 1 decimal place).
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Show answer and explanation
Correct answer: 5.97 to 6.03
Explanation
Node $b$ is wired directly to the bottom-middle node $M$, so $V_b=V_M$. The voltage source fixes $V_T-V_M=8$ V for the top-left node $T$. The current source and the 2 Ω resistors only connect $T$/the bottom-left node with $M$ and do not affect this. Apply KCL at node $a$: current in through the 0.5 Ω from $T$ equals the current out through the two 3 Ω resistors (to $M$ and to $b=M$): $\dfrac{8-V_a}{0.5}=\dfrac{V_a}{3}+\dfrac{V_a}{3}$ (taking $V_M=0$). So $16-2V_a=\tfrac23V_a$, giving $V_a=6$ V and $V_{ab}=V_a-V_b=6.0$ V.