-->

Current Electricity

Question
CBSEENPH12039869

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

  • 1/499 G

  • 499/500 G

  • 1/500 G

  • 500/499 G

Solution

A.

1/499 G

For ammeter,

0.0002 I x G = 0.998 I x rg

rgfraction numerator 0.002 over denominator 0.998 end fraction G
That is,
straight r subscript straight g space equals space 0.002004 space straight G space equals space 1 over 499 straight x space straight G
Resistance of ammeter is approximately equal to resistance of shunt rB.