Physics · Current Electricity

JEE Main 2024 — 5 April, Shift 2 — Question 34

A galvanometer of resistance $100 \Omega$ when connected in series with $400 \Omega$ measures a voltage of upto 10 V . The value of resistance required to convert the galvanometer into ammeter to read upto 10 A is $\mathrm{x} \times 10^{-2} \Omega$ . The value of x is :

Option A:
2
Option B:
800
Option C:
20
Correct
Option D:
200

Answer: C

Step-by-step solution

$\quad \mathrm{i}_{\mathrm{g}}=\frac{10}{400+100}=20 \times 10^{-3} \mathrm{~A}$

For ammeter

Let shunt resistance $=S$

$\mathrm{i}_{\mathrm{g}} \mathrm{R}=\left(\mathrm{i}-\mathrm{i}_{\mathrm{g}}\right) \mathrm{S}$

$20 \times 10^{-3} \times 100=10 \mathrm{~S}$

$\mathrm{S}=20 \times 10^{-2} \Omega$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Physics
Chapter: Current Electricity
Topic: Electrical Measuring Instruments

Answer: C

Step-by-step solution

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