Q. In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be :
A) $\frac{1}{499}G$
B) $\frac{499}{500}G$
C) $\frac{1}{500}G$
D) $\frac{500}{499}G$
Explanation
[Solution] $I_g = I/500$, so shunt S = G/499; ammeter resistance = $\frac{GS}{G + S} = \frac{G}{500}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.