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In an ammeter 0.2 % of maon current pass...

In an ammeter 0.2 % of maon current passes through the galvnometer . 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
`S=(I_(G))/(I_(S)).G=(I_(G))/(I-I_(G)).G=((0.2)/(100))/(1-(0.2)/(100))G`
` = (0.2)/(100)xx(100)/(99.8)G=(1)/(499)G`
`therefore` Resistance of the ammeter
`=(SG)/(S+G)=((1)/(499)G.G)/((1)/(499)G+G)=((1)/(499)G)/((500)/(499))=(1)/(500)G`
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