Physics · Current Electricity

JEE Main 2024 — 29 January, Shift 1 — Question 40

A galvanometer having coil resistance $10 \Omega$ shows a full scale deflection for a current of 3 mA . For it to measure a current of 8 A , the value of the shunt should be:

Option A:
$3 \times 10^{-3} \Omega$
Option B:
$4.85 \times 10^{-3} \Omega$
Option C:
$3.75 \times 10^{-3} \Omega$
Correct
Option D:
$2.75 \times 10^{-3} \Omega$

Answer: C

Step-by-step solution

Given $\mathrm{G}=10 \Omega$

$\mathrm{I}_{\mathrm{g}}=3 \mathrm{~mA}$

$I=8 \mathrm{~A}$

In case of conversion of galvanometer into ammeter.

We have $I_{g} G=\left(I-I_{g}\right) S$

$S=\frac{I_{g} G}{I-I_{g}}$

$\mathrm{S}=\frac{\left(3 \times 10^{-3}\right) 10}{8-0.003}=3.75 \times 10^{-3} \Omega$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Current Electricity
Topic: Electrical Measuring Instruments

Answer: C

Step-by-step solution

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