Question

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`

Answer 1

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)`