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 :
JEE MAIN 2024, Diffcult
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$\mathrm{i}_{\mathrm{g}}=\frac{10}{400+100}=20 \times 10^{-3} \mathrm{~A}$
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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