GATE 2025 IN – Question 61
The galvanometer G in the shown circuit has internal resistance 100 Ω. Its current $I_G$ is ______ μA (rounded to the nearest integer).
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 9 to 9
Explanation
With the galvanometer removed, the left midpoint is $4(1010/2010)$ V and the right midpoint is 2 V.
Hence $V_{th}=0.00995025$ V, left positive.
With the supply shorted, $R_{th}=(1000\parallel1010)+(1000\parallel1000)=1002.4876\ \Omega$.
Therefore $I_G=V_{th}/(R_{th}+100)=9.0253\ldots\ \mu\mathrm A$.
In the shown arrow direction, the answer is **9 μA**.