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In a ammeter 0.2% of main current passes...

In a ammeter `0.2%` of main current passes through the galvanometer. If resistance of galvanometer is `G`, the resistance of ammeter will be

A

`1/(499)G`

B

`(499)/(500)G`

C

`1/(500)G`

D

`500/(499)G`

Text Solution

Verified by Experts

The correct Answer is:
C
As 0.2% of main current passes through galvanometer hence `998/(1000) I` current through the shunt .
`((2I)/(1000))G=((998I)/(1000))SimpliesS=G/499`
Total resistance of Ammeter
`R = (SG)/(S+G)=((G/499)G)/((G/(499))+G)=G/500`
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