1.

In a ammeter `0.2%` of main current passes through the galvanometer. If resistance of galvanometer is `G`, the resistance of ammeter will beA. `(499)/(500) G`B. `(1)/(500) G`C. `(500)/(499) G`D. `(1)/(499) G`

Answer» Correct Answer - B
`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)`.


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