GATE 2025 EC Q.19: solution and answer

GATE 2025 · ECQ.19 · 1 mark · MCQ

Consider a part of an electrical network as shown below. Some node voltages, and the current flowing through the $3 Ω$ resistor are as indicated.

The voltage (in Volts) at node $X$ is ______.

GATE 2025 EC question 19: A horizontal chain of resistors running left to right. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
$20/3$
(B)
$32/3$
(C)
$22/3$
(D)
$2/3$

Answer and solution

ANSWER

$V_{X} = \frac{20}{3} V$ , option (A).

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

Call the junction where the $1 Ω$ , the $3 Ω$ and the right-hand $2 Ω$ meet node $N$ . You do not need to solve the rest of the network (the delta and the dotted part), because the given current already fixes $V_{N}$ .
Ohm's law on the $3 Ω$ resistor. 1 A flows downward, from the 9 V node into $N$ , so the voltage falls by $I R$ in that direction: $V_{N} = 9 - (1) (3) = 6 V$
Node $X$ has only two branches: the $2 Ω$ to the 8 V node and the $1 Ω$ to $N$ . By KCL at $X$ , the current arriving through one branch must leave through the other: $\frac{8 - V _{X}}{2} = \frac{V _{X} - 6}{1}$
Solve: $8 - V_{X} = 2 V_{X} - 12 \Rightarrow 3 V_{X} = 20 \Rightarrow V_{X} = \frac{20}{3} V \approx 6.67 V$ .
Check with the voltage-divider view. The $2 Ω$ and $1 Ω$ are in series across $8 - 6 = 2 V$ , so the current through them is $\frac{2}{3} A$ . The $1 Ω$ drops $\frac{2}{3} V$ , so $V_{X} = 6 + \frac{2}{3} = \frac{20}{3} V$ . Going the other way, the $2 Ω$ drops $\frac{4}{3} V$ and $8 - \frac{4}{3} = \frac{20}{3} V$ . The value lies between 6 V and 8 V, as it must. This is option (A).

KEY IDEA

You only need the part of the network next to $X$ . Ohm's law on the $3 Ω$ resistor gives the junction voltage, $V_{N} = 9 - 1 \times 3 = 6 V$ . Node $X$ has just two branches, so the $2 Ω$ and the $1 Ω$ carry the same current and act as a voltage divider between 8 V and 6 V. The delta and the dotted part never enter the calculation.

WATCH OUT FOR

Each wrong option matches a specific slip. (B) $32/3$ : taking the 1 A as flowing up, out of $N$ , which gives $V_{N} = 9 + 3 = 12 V$ instead of 6 V. (C) $22/3$ : swapping the divider weights, i.e. putting the larger drop across the $1 Ω$ instead of the $2 Ω$ . (D) $2/3$ : giving the series current, $2/3 A$ , instead of the node voltage. Some students also waste time trying to reduce the delta, which the given 1 A makes unnecessary.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2025 EC Q.19: solution and answer

Answer and solution

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