Problem 2: A Galvanometer whose resistance
is 40 Ohms deflected full-scale for a potential difference of 100mV
across its terminals. How can it be converted into an ammeter of 5
ampere range?
Solution:
Step: 1 Overview
See the overview of problem 1.Step: 2 Calculation
Given:Rg=40 OhmsRequired:
I=5A
Vg=100mV=0.1V
Rs=?According to above equation:
Rs= (Ig/(I-Ig))Rg
But, here we do not the value of Ig,
So first calculating Ig
Ig=Vg/Rg
Ig=(0.1)/40=0.0025A
Now,
Rs= (0.0025/(5-0.0025))(40)
Rs= (0.0025/4.9975)(40)
Rs= 0.02 Ohms (Ans)
