GATE 2026 IN Q.40: solution and answer

GATE 2026 · INQ.40 · 2 marks · MCQ

In the circuit shown below, the value of the output voltage $V_{0}$ is ______ V.

GATE 2026 IN question 40: A ladder network between a top rail and a bottom (reference) rail, drawn left to right. — Figure from the GATE 2026 IN paper. It is part of the question.

(A)
2
(B)
3
(C)
4
(D)
8

Answer and solution

ANSWER

$V_{0} = 3 V$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2026 IN (IIT Guwahati)

Take the bottom rail as reference. Name the top-rail nodes: A, the node that joins the tops of both left-hand 2 Ω resistors; B, the node where the first 4 V branch connects; and C, the output node, so $V_{C} = V_{0}$ . Nothing is connected across the output terminals, so $V_{0}$ is simply the node voltage at C.
Combine the resistors at A in parallel. Both left-hand 2 Ω resistors run from A to the reference, so together they are $2 ∥ 2 = 1 Ω$ .
Add the series resistor. With the 1 Ω from A to B added, the left part seen from node B is $1 + 1 = 2 Ω$ to the reference, and it contains no source.
Take the Thevenin equivalent at B. The branch at B is a 4 V source behind 2 Ω, in parallel with that passive 2 Ω. By Millman's theorem $V_{t h} = \frac{\frac{4}{2} + \frac{0}{2}}{\frac{1}{2} + \frac{1}{2}} = 2 V, R_{t h} = 2 ∥ 2 = 1 Ω$
Add the series resistor again. With the 1 Ω from B to C added, the left part seen from C becomes a 2 V source behind $1 + 1 = 2 Ω$ .
Apply Millman's theorem at C. The 2 V / 2 Ω equivalent is in parallel with the 4 V / 2 Ω branch at C, so $V_{0} = \frac{\frac{2}{2} + \frac{4}{2}}{\frac{1}{2} + \frac{1}{2}} = \frac{3}{1} = 3 V$
Check with KCL on the original circuit, using $V_{0} = 3 V$ . At C: $\frac{3 - V _{B}}{1} + \frac{3 - 4}{2} = 0$ , so $V_{B} = 2.5 V$ . At B: $\frac{2.5 - V _{A}}{1} + \frac{2.5 - 4}{2} + \frac{2.5 - 3}{1} = 0$ , so $V_{A} = 1.25 V$ . At A: $\frac{2.5 - 1.25}{1} = 1.25 A$ flows in from B, and $\frac{1.25}{2} + \frac{1.25}{2} = 1.25 A$ flows out through the two 2 Ω resistors, so every node balances. $V_{0} = 3 V$ , which is option (B).

KEY IDEA

Reduce the ladder step by step, starting at the passive left end and working toward the output. Parallel branches collapse by Millman's theorem (a single source transformation does the same job), and series resistors simply add. The output is open-circuited, so $V_{0}$ is the node voltage at C. The passive left section draws current from the sources, which pulls the output below 4 V.

WATCH OUT FOR

Seeing two equal 4 V sources and deciding that no current flows, which gives $V_{0} = 4 V$ (option C). The two left-hand 2 Ω resistors are a load of $1 Ω$ at A, and that load draws current through the 1 Ω resistors, so the output settles at 3 V. A second trap is adding the two sources to get 8 V (option D), but they are in parallel branches, not in series. Misreading the two left-hand 2 Ω resistors as being in series gives about 3.48 V, which matches none of the options. Their tops are joined by a plain wire, so they are in parallel.

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

GATE 2026 IN Q.40: solution and answer

Answer and solution

More Laws & node/mesh questions