question_answer

In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is Q the resistance of ammeter will be:

A) \[\frac{1}{499}G\]

B) \[\frac{499}{500}G\]

C) \[\frac{1}{500}G\]

D) \[\frac{500}{499}G\]

Correct Answer: C

Solution :

[c] As 0.2% of main current passes through the galvanometer hence\[\frac{998}{1000}I\]current through the shunt.                \[\left( \frac{2I}{1000} \right)G=\left( \frac{998I}{1000} \right)S\] \[\Rightarrow \,\,\,S=\frac{G}{499}\]              Total resistance of Ammeter \[R=\frac{SG}{S+G}=\frac{\left( \frac{G}{499} \right)G}{\left( \frac{G}{499} \right)+G}=\frac{G}{500}\]