In a galvanometer there is a deflection of 10V divisions per mA. The internal resistance of a galvanometer is 50ohm. If a shunt of 2.5ohm is connected to the galvanometer, calculate the maximum current which the galvanometer can read.

Question

kropot72 · Answer

Let resistance of the parallel combination of the galvanometer and the shunt = R(p)
$$R(p)=\frac{2.5\times 50}{2.5+50}$$
The maximum current the galvanometer can read with the shunt connected is:
(maximum current without shunt) * 50/R(p)

anonymous · Answer

sorry brother but the right answer is 125mA

kropot72 · Answer

I did not give an answer. This part of the question does not make sense "there is a deflection of 10V divisions per mA". You will note I did not give a value for the maximum current without shunt.

anonymous · Answer

ya i think you are right but this question was given by my sir in exam and he made no correction to it...

kropot72 · Answer

The current multiplying factor 50/R(p) is correct using the resistance values in the question.

vincent-lyon.fr · Answer

This should read:
"In a galvanometer there is a deflection of 10 divisions per mA"
What is the maximum reading? 100 divisions?

anonymous · Answer

I am afraid that without the max deflection in mA (or Volts), there is no way to solve it

kropot72 · Answer

According to the questioner the maximum current to be read with the shunt connected is 125 mA. Therefore the maximum current to be read without the shunt would be $$\frac{125\times 2.381}{50}=5.95mA$$