Question

In an ammeter there are 20 divisions between the 0 mark and 0.5 mark.

Find least count of ammeter ?​

Answer 1

Answer:

Given:-

INITIAL READING = 0v

FINAL READING = 0.5v

Potential Difference = 0.5-0 = 0.5v

No. of divisions = 20

$least \: count = \frac{potential \: difference}{no. \: of \: divisions}$

$= \frac{0.5}{20} = 0.025 \: volt$