Question:
The resistance of a galvanometer is $50 \mathrm{ohm}$ and the maximum current which can be passed through it is $0.002 \mathrm{~A}$. What resistance must be connected to it order to convert it into an ammeter of range $0-0.5 \mathrm{~A}$ ?
Correct Option: , 4
Solution:
(4) Using, $i_{g}=i \frac{S}{S+G}$
$0.002=0.5 \frac{S}{S+50}$
On solving, we get
$S=\frac{100}{498} \simeq 0.2 \Omega$
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