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.
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 \(=\mathrm{S}\)
\(i_g R=\left(i-i_g\right) S\)
\(20 \times 10^{-3} \times 100=10 S\)
\(S=20 \times 10^{-2} \Omega\)
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