GATE 2026 IN Q.32: solution and answer

GATE 2026 · INQ.32 · 1 mark · NAT

The potential difference between nodes $A$ and $B$ in a circuit is $v_{A} - v_{B} = 5 V$ . The work done in moving a charge of 1 coulomb from point $B$ to $A$ is ____ J. (answer in integer)

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Answer and solution

ANSWER

The work done in moving 1 C from $B$ to $A$ is $5 J$ .

Official answer: 5Source: Official answer key, GATE 2026 IN (IIT Guwahati)

Definition of potential difference: $v_{A} - v_{B}$ is the work done per unit charge (by an external agent, against the field) in moving a charge from $B$ to $A$ . So $W_{B \to A} = q (v_{A} - v_{B})$ .
Given: $q = 1 C$ and $v_{A} - v_{B} = 5 V$ .
$W_{B \to A} = q (v_{A} - v_{B}) = (1 C) (5 V) = 5 J$
Sign check: the charge goes from the lower-potential node $B$ up to the higher-potential node $A$ , so energy has to be supplied to it and the work done is positive, $+ 5 J$ . (Moving it the other way, from $A$ to $B$ , would give $- 5 J$ , meaning the field does the work.)

KEY IDEA

One volt is one joule per coulomb. The potential difference $v_{A} - v_{B}$ is the work needed per coulomb to move a charge from $B$ to $A$ , so $W_{B \to A} = q (v_{A} - v_{B})$ .

WATCH OUT FOR

Getting the direction backwards. Some students use $v_{B} - v_{A}$ and answer $- 5$ . Others think the charge goes "downhill" from $B$ to $A$ . $A$ is 5 V above $B$ , so moving a positive charge from $B$ to $A$ needs $+ 5 J$ of work.

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.32: solution and answer

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