A galvanometer has a resistance of 5Ω. By using a shunt wire of resistance 0.05Ω, the galvanometer could be converted to an ammeter capable of reading 2Amp....

A galvanometer has a resistance of 5Ω. By using a shunt wire of resistance 0.05Ω, the galvanometer could be converted to an ammeter capable of reading 2Amp. What is the current through the galvanometer?

Answer Details

Shunt resistance is usually given as
R = igrg/(I - ig)
Where R = 0.05Ω
rg = 5Ω
I = 2M
∴ ig = IR/(rg + R)
= (2 x 0.05)/(5 + 0.05)
= 0.10/5.05
= 0.0198A
= 19.8mA
= 20mA