GATE 2026 EC Q.30: solution and answer

GATE 2026 · ECQ.30 · 1 mark · NAT

The output voltage $V_{o}$ (in Volt) for the network given in the Figure is ____. (rounded off to two decimal places)

GATE 2026 EC question 30: Four nodes: top rail T, bottom rail G, and two middle nodes A (left) and B (right). — Figure from the GATE 2026 EC paper. It is part of the question.

Answer and solution

ANSWER

$V_{o} = 1.00 V$ .

Official answer: 0.95 to 1.05Source: Official answer key, GATE 2026 EC (IIT Guwahati)

Label the nodes. Take the bottom rail G as the reference and call the top-rail voltage $V_{o}$ . The middle nodes are A (left) and B (right). The battery's long plate faces A, so it fixes $V_{A} - V_{B} = 6 V$ .
KCL at the top node T, working in mA and kΩ. The source pushes 1 mA into T, and that current leaves through the two upper resistors: $1 = \frac{V _{o} - V _{A}}{1} + \frac{V _{o} - V _{B}}{1} \Rightarrow 2 V_{o} - (V_{A} + V_{B}) = 1$
Draw a supernode around A, B and the battery. The current entering it through the two upper resistors must equal the current leaving it through the two lower resistors: $(V_{o} - V_{A}) + (V_{o} - V_{B}) = V_{A} + V_{B}$ The left side is 1 mA by the KCL at T, so $V_{A} + V_{B} = 1 V$ .
Substitute back into the KCL at T: $2 V_{o} - 1 = 1 \Rightarrow V_{o} = 1 V$ .
Check by superposition. With the battery acting alone (current source opened), T is joined to A and B by equal 1 kΩ resistors, and so is G. Each rail therefore sits at $\frac{V _{A} + V _{B}}{2}$ , so the battery contributes nothing to $V_{o}$ . With the current source acting alone (battery shorted), A and B merge into one node. The resistance seen by the source is then $(1 ∥ 1) + (1 ∥ 1) = 0.5 + 0.5 = 1 k Ω$ , so $V_{o} = 1 mA \times 1 k Ω = 1 V$ .
Consistency check. From $V_{A} - V_{B} = 6$ and $V_{A} + V_{B} = 1$ : $V_{A} = 3.5 V$ and $V_{B} = - 2.5 V$ . The upper resistors carry $(1 - 3.5) + (1 + 2.5) = - 2.5 + 3.5 = 1 mA$ out of T. The lower resistors carry $3.5 + (- 2.5) = 1 mA$ into G. Both match the 1 mA source, so $V_{o} = 1.00 V$ .

KEY IDEA

The four equal resistors form a balanced bridge, with the battery across the middle arm. By symmetry the battery shifts $V_{A}$ and $V_{B}$ equally in opposite directions, so it adds nothing to the rail-to-rail voltage. With the battery shorted, A and B become one node and the source sees $(1 ∥ 1) + (1 ∥ 1) = 1 k Ω$ . That gives $V_{o} = 1 mA \times 1 k Ω = 1 V$ . The supernode equations confirm it directly: $V_{A} + V_{B} = 1$ and $2 V_{o} - (V_{A} + V_{B}) = 1$ .

WATCH OUT FOR

Assuming the 6 V battery must appear in $V_{o}$ and adding or subtracting it (for example, answering 4 V or 7 V). In a balanced bridge the battery only splits the middle nodes to $V_{A} = 3.5 V$ and $V_{B} = - 2.5 V$ , and its effect cancels between the two rails. Another slip is to write a KCL at A or at B alone and then try to include the battery current, which is unknown. Enclose the battery in a supernode instead.

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2026 EC Q.30: solution and answer

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