GATE 2026 EE Q.24: solution and answer

GATE 2026 · EEQ.24 · 1 mark · MCQ

For the circuit shown, which one of the following options correctly identifies the Thevenin's equivalent parameters between nodes Y and Z?

GATE 2026 EE question 24: Four nodes: X (left), Y (top middle), Z (top right), W (bottom middle). — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
$V_{T H} = 100 V$ , $R_{T H} = 10 k Ω$
(B)
$V_{T H} = 140 V$ , $R_{T H} = 0 Ω$
(C)
$V_{T H} = 100 V$ , $R_{T H} = 0 Ω$
(D)
$V_{T H} = 140 V$ , $R_{T H} = 10 k Ω$

Answer and solution

ANSWER

$V_{T H} = 140 V$ and $R_{T H} = 0 Ω$ , which is option (B).

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

Take W as the reference node: $V_{W} = 0$ .
The 100 V source between X and W has its $+$ terminal toward X, so $V_{X} = V_{W} + 100 = 100 V$ .
The top 20 V source between X and Y has its $-$ terminal on the X side and its $+$ terminal on the Y side, so $V_{Y} = V_{X} + 20 = 120 V$ .
The right-hand 20 V source between Z and W has its $+$ terminal toward W and its $-$ terminal toward Z, so $V_{Z} = V_{W} - 20 = - 20 V$ .
Every node voltage is set by an ideal voltage source, so the $10 k Ω$ , $2 k Ω$ and $1 k Ω$ resistors only decide how much current each source supplies. None of them changes a node voltage. The figure marks $+$ at Y and $-$ at Z, so the open-circuit voltage is $V_{T H} = V_{Y} - V_{Z} = 120 - (- 20) = 140 V$ The same result comes from KVL along the source-only path Y $\to$ X $\to$ W $\to$ Z: $20 + 100 + 20 = 140 V$ .
To find $R_{T H}$ , deactivate the independent sources by replacing each voltage source with a short. This shorts X to Y (top 20 V source), X to W (100 V source) and W to Z (right-hand 20 V source), so Y, X, W and Z all become one node.
Y and Z are then joined by a zero-resistance path that sits in parallel with every resistor, so $R_{T H} = 0 Ω$ . Between Y and Z the network therefore behaves as an ideal 140 V source: option (B).

KEY IDEA

If a chain made only of ideal voltage sources connects the two terminals, the open-circuit voltage is just the KVL sum of those sources along the chain (here $20 + 100 + 20 = 140 V$ ), whatever the resistors are. Killing the sources turns that chain into a short across the terminals, so $R_{T H} = 0$ and the Thevenin equivalent is an ideal voltage source.

WATCH OUT FOR

There are two traps. (1) Reading the visible $10 k Ω$ between Y and Z as $R_{T H}$ (options A and D). Once the sources are shorted, that resistor is in parallel with a zero-resistance path through X and W, so it adds no resistance. (2) Misreading the polarity of the right-hand 20 V source, which has its $+$ toward W. If it is taken as opposing the other two sources, the sum becomes $20 + 100 - 20 = 100 V$ (options A and C) instead of the correct $140 V$ .

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 16 Theorems questions, with notes

GATE 2026 EE Q.24: solution and answer

Answer and solution

More Theorems questions