GATE 2025 IN Q.61: solution and answer

GATE 2025 · INQ.61 · 2 marks · NAT

In the circuit shown, the galvanometer (G) has an internal resistance of 100 Ω. The galvanometer current $I_{G}$ is ________ μA (rounded off to the nearest integer).

GATE 2025 IN question 61: A Wheatstone bridge drawn as a diamond. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

ANSWER

The galvanometer current is $I_{G} \approx 9.03 μ A$ , which rounds to 9 μA and flows from the left node to the right node, the same way as the arrow.

Official answer: 9Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Label the left mid-node $A$ (between the 1000 Ω and 1010 Ω resistors) and the right mid-node $B$ (between the two 1000 Ω resistors). Remove G and find the Thevenin equivalent seen across $A$ - $B$ .
Voltage divider on the left arm: $V_{A} = 4 \times \frac{1010}{1000 + 1010} = \frac{4040}{2010} \approx 2.00995 V$ .
Voltage divider on the right arm: $V_{B} = 4 \times \frac{1000}{1000 + 1000} = 2 V$ .
Open-circuit (Thevenin) voltage: $V_{t h} = V_{A} - V_{B} = \frac{4040 - 4020}{2010} = \frac{2}{201} V \approx 9.950 mV$ . $A$ is the positive terminal, so the current flows from $A$ to $B$ through G, the same direction as the arrow; $I_{G}$ is positive.
Thevenin resistance: replace the 4 V source by a short, so the top and bottom vertices are both at ground. Then from $A$ the two left resistors are in parallel, and from $B$ the two right resistors are in parallel: $R_{t h} = (1000 ∥ 1010) + (1000 ∥ 1000) = \frac{1000 \times 1010}{2010} + 500 = 502.49 + 500 = 1002.49 Ω$ (exactly $\frac{201500}{201} Ω$ ).
Reconnect G (100 Ω) across the Thevenin equivalent: $I_{G} = \frac{V _{t h}}{R _{t h} + R _{G}} = \frac{2/201}{\frac{201500}{201} + 100} = \frac{2/201}{\frac{221600}{201}} = \frac{2}{221600} A \approx 9.025 μ A$
Rounded off to the nearest integer: $I_{G} \approx 9 μ A$ .

KEY IDEA

A slightly unbalanced Wheatstone bridge with a galvanometer of finite resistance is solved with Thevenin's theorem. Remove G. The open-circuit voltage between the two mid-nodes is the difference between the two voltage dividers. With the source shorted, the Thevenin resistance is (upper-left ∥ lower-left) + (upper-right ∥ lower-right). Then $I_{G} = V_{t h} / (R_{t h} + R_{G})$ .

WATCH OUT FOR

Using $I_{G} = V_{t h} / R_{G}$ gives about 99.5 μA because it leaves out the bridge's own Thevenin resistance of about 1002.5 Ω. Another common error is to put the arm resistors in series (1000 + 1010) instead of in parallel when finding $R_{t h}$ : once the source is shorted, the top and bottom vertices are the same node. Rounding $V_{A}$ to 2.01 V too early gives $V_{t h}$ = 10 mV and about 9.07 μA. That still rounds to 9 here, but on a tighter question the lost precision would matter.

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 16 Theorems questions, with notes

GATE 2025 IN Q.61: solution and answer

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