Current Electricity

JEE Main level, single correct, medium

323 of 380

Question 323 of 380

A galvanometer of resistance 100Ω100 \Omega when connected in series with 400Ω400 \Omega measures a voltage of upto 10 V10 \mathrm{~V}. The value of resistance required to convert the galvanometer into ammeter to read upto 10 A10 \mathrm{~A} is x×10−2Ωx \times 10^{-2} \Omega. The value of xx is :

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Answer: (C) 2020

ig=10400+100=20×10−3 A\mathrm{i}_{\mathrm{g}}=\frac{10}{400+100}=20 \times 10^{-3} \mathrm{~A}
For ammeter
Let shunt resistance =S=\mathrm{S}
$\begin{aligned}
& \mathrm{i}_{\mathrm{g}} \mathrm{R}=\left(\mathrm{i}-\mathrm{i}_{\mathrm{g}}\right) \mathrm{S} \\
& 20 \times 10^{-3} \times 100=10 \mathrm{~S} \\
& \mathrm{~S}=20 \times 10^{-2} \Omega
\end{aligned}$

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