ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2024
  • Hard
  • MCQ

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

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(C) \(20\)

Step-by-step Solution

Detailed explanation

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

For ammeter

Let shunt resistance \(=\mathrm{S}\)

\(i_g R=\left(i-i_g\right) S\)

\(20 \times 10^{-3} \times 100=10 S\)

\(S=20 \times 10^{-2} \Omega\)

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app