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

Here, resistance of the galvanometer = G Current through the galvanometer, IG=0.2% of I=0.2100I=1500I∴ Current through the shunt, IS=I-IG=I-1500I=499500I As shunt and galvanometer are in parallel∴  IGG=ISS1500IG=499500S or S=G499 Resistance of the ammeter RA is 1RA=1G+1S=1G+1G499=500GRA=1500G