GATE 2026 EE Q.57: solution and answer

GATE 2026 · EEQ.57 · 2 marks · NAT

The resistance values of the Wheatstone bridge shown are: $P = 2000 Ω$ , $Q = 500 Ω$ , $R = 3000 Ω$ . The battery voltage $E = 50 V$ . The battery has an internal resistance of $1 Ω$ and the Galvanometer (G) has a resistance of $50 Ω$ .

[Figure]

The value of the resistance S for balanced condition is ________ $Ω$ (Answer in integer)

GATE 2026 EE question 57: A Wheatstone bridge drawn as a diamond with four nodes: left, top, right and bottom. — Figure from the GATE 2026 EE paper. It is part of the question.

Answer and solution

ANSWER

The resistance S for balance is 750 Ω.

Official answer: 750Source: Official answer key, GATE 2026 EE (IIT Guwahati)

Read the topology from the figure. P (left node to top node) and R (left node to bottom node) meet at the left supply node. Q (top node to right node) and the variable arm S (bottom node to right node) meet at the right supply node. G is connected between the top and bottom nodes, and the battery E is connected across the left and right nodes. So P is opposite S, and Q is opposite R.
Balance condition. At balance G carries no current, so the same current $I_{1}$ flows through P and Q, and the same current $I_{2}$ flows through R and S. The top and bottom nodes are also at the same potential, so the drops from the left node are equal, $I_{1} P = I_{2} R$ , and so are the drops to the right node, $I_{1} Q = I_{2} S$ . Dividing the first equation by the second gives $\frac{P}{Q} = \frac{R}{S} ⟺ P S = QR$
Solve for the unknown arm: $S = \frac{Q R}{P} = \frac{500 \times 3000}{2000} Ω = 750 Ω$
The battery voltage ( $E = 50 V$ ), the battery's internal resistance ( $1 Ω$ ) and the galvanometer resistance ( $50 Ω$ ) do not enter the result. At balance no current flows through G, so its resistance has no effect. E and the source resistance only scale every current in the bridge by the same factor, so they cannot change the condition $P S = QR$ . These values matter only for the bridge's sensitivity or for the galvanometer current when the bridge is off balance.
Hence $S = 750 Ω$ .

KEY IDEA

A Wheatstone bridge is balanced when the products of opposite arms are equal. Here P is opposite S and Q is opposite R, so $P S = QR$ . This condition does not depend on the supply voltage, the source's internal resistance or the galvanometer's resistance.

WATCH OUT FOR

Pairing the wrong arms. Writing $P / Q = S / R$ gives $S = P R / Q = 12000 Ω$ . Always find which arms sit opposite each other in the figure: they are the arms that share neither a supply node nor a galvanometer node. A second mistake is trying to use the 1 Ω internal resistance, the 50 Ω galvanometer resistance or E = 50 V. They are distractors, because no current flows through G at balance.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 14 Bridges questions, with notes

GATE 2026 EE Q.57: solution and answer

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