GATE 2026 EE Q.44: solution and answer

GATE 2026 · EEQ.44 · 2 marks · MCQ

The terminal voltage and current of a linear electrical network shown in Figure (a) are given in the table.

| Terminal voltage ( $v_{t}$ ) | Terminal current ( $i_{t}$ ) |
|---|---|
| 18 V | $- 0.5$ A |
| 30 V | 0.5 A |
| 36 V | 1.0 A |

The correct choice for the parameters ( $I_{N}$ , $R_{N}$ ) of the Norton equivalent circuit shown in Figure (b) is:

GATE 2026 EE question 44: Figure (a): a box labelled 'Electrical Network' with two terminals brought out to the right. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
$I_{N} = 3.0 A$ , $R_{N} = 24.0 Ω$
(B)
$I_{N} = 12.0 A$ , $R_{N} = 2.0 Ω$
(C)
$I_{N} = 2.0 A$ , $R_{N} = 12.0 Ω$
(D)
$I_{N} = 2.0 A$ , $R_{N} = 24.0 Ω$

Answer and solution

ANSWER

$I_{N} = 2.0 A$ , $R_{N} = 12.0 Ω$ — option (C).

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

Terminal equation of a linear one-port. Seen from X–Y, any linear network behaves as a Thevenin source $V_{t h}$ in series with $R_{t h}$ . Here $i_{t}$ flows from X INTO the network, i.e. into its $+$ terminal (passive sign convention), so $v_{t} = V_{t h} + R_{t h} i_{t}$ .
Use two rows of the table. $18 = V_{t h} - 0.5 R_{t h}$ and $30 = V_{t h} + 0.5 R_{t h}$ . Subtracting the first from the second gives $12 = R_{t h}$ , so $R_{t h} = 12 Ω$ .
Back-substitute. $V_{t h} = 30 - 0.5 (12) = 24 V$ .
Check with the third row. $v_{t} = 24 + 12 (1.0) = 36 V$ , which matches the table, so all three measurements lie on the same straight line and the parameters are consistent.
Thevenin to Norton. $R_{N} = R_{t h} = 12 Ω$ and $I_{N} = \frac{V _{t h}}{R _{t h}} = \frac{24}{12} = 2 A$
Check the polarity. In Figure (b) the source arrow points up toward X. On open circuit all of $I_{N}$ flows through $R_{N}$ , so $v_{X Y} = I_{N} R_{N} = 2 \times 12 = + 24 V$ , which agrees with $V_{t h}$ . (KCL at X with $i_{t}$ leaving the Norton circuit toward the network gives $v_{t} = R_{N} (I_{N} + i_{t}) = 12 (2 + i_{t})$ , which reproduces 18 V, 30 V and 36 V.) So $(I_{N}, R_{N}) = (2.0 A, 12.0 Ω)$ : option (C).

KEY IDEA

A linear one-port has a straight-line terminal characteristic $v_{t} = V_{t h} + R_{t h} i_{t}$ (with $i_{t}$ entering the + terminal). Two measured points fix it: the slope is $R_{t h} = R_{N}$ , the intercept at $i_{t} = 0$ is the open-circuit voltage $V_{t h}$ , and the Norton current is the short-circuit current $I_{N} = V_{t h} / R_{t h}$ . The third row serves as a check.

WATCH OUT FOR

Reading $i_{t}$ as leaving terminal X and writing $v_{t} = V_{t h} - R_{t h} i_{t}$ . That gives $R_{t h} = - 12 Ω$ , which is physically impossible and should tell you the sign is wrong. Two other slips: taking $V_{t h} = 24$ as the Norton resistance (that gives option D), or swapping the two values (that gives option B).

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.44: solution and answer

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