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

Options

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

Answer

B

Explanation

`I_g G = (I - I_g) S`
`0.002 I G = (I - 0.002 I) S`
`S = (G)/(499)`
Resistance of ammeter
`R_A = (GS)/(G + S) = (G.(G)/(499))/(G +(G)/(499)) = (G)/(500)`.