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In an ammeter 0*2% of main current passe...

In an ammeter `0*2%` of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be

A

`(G)/(499)`

B

`(499)/(500)G`

C

`(G)/(500)`

D

`(500)/(499)G`

Text Solution

Verified by Experts

Fractional current passing through galvanometer,

`I_g/I=(S)/(G+S)=(0*2)/(100)=(1)/(500)` or `G+S=500S` …(i)
Resistance of ammeter `=(GS)/(G+S)=(GS)/(500S)=(G)/(500)`
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