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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` |
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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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