Physics · Current Electricity

JEE Main 2025 — 23 January, Evening Shift — Question 59

A galvanometer having a coil of resistance $30 \Omega$ need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\frac{30}{\mathrm{X}} \Omega$ , where X is

Option A:
447
Option B:
298
Option C:
149
Correct
Option D:
596

Answer: C

Step-by-step solution

$\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}}=\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right) \mathrm{r}_{\mathrm{s}}$

$20 \times 10^{-3} \times 30=(3-0.02) \times \mathrm{r}_{\mathrm{s}}$

$\mathrm{r}_{\mathrm{s}}=\left(\frac{0.6}{2.98}\right)=\frac{30}{\mathrm{x}}$

$\mathrm{x}=\left(\frac{2.98 \times 30}{0.6}\right)=149$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Current Electricity
Topic: Electrical Measuring Instruments