A galvanometer has a resistance of 5Ω. By using a shunt wire of resistance 0.05Ω,…

Physics JAMB 1999

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?

  • 2mA
  • 10mA
  • 20mA checkmark
  • 25mA

The correct answer is: C

Explanation

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
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