In an ammeter 0.2% of main current passes through the galvonometer. If resistance of galvonometer is G, the resistance of ammeter will be _____

Options

A. `1/499G`
B. `499/500G`
C. `1/500G`
D. `500/499G`

Answer

C

Explanation

Here. G and S are in parallel,
`thereforeI_(G)R_(G)=I_(S)R_(S)`
Now `I_(G)=0.2%I=0.002I`
`thereforeI/I_(G)=1/0.002=500" "thereforen=500`
Now shunt `S=G/(n-1)=G/(500-1)=G/499`
Now resistance of Ammeter,
`1/R_(A)=1/G+1/S`
`therefore1/R_(A)=1/G+499/G`
`therefore1/R_(A)=500/G`
`thereforeR_(A)=G/500`