anonymous
  • anonymous
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.
Physics
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SOLVED
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chestercat
  • chestercat
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kropot72
  • kropot72
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
  • anonymous
sorry brother but the right answer is 125mA
kropot72
  • kropot72
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.

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anonymous
  • anonymous
ya i think you are right but this question was given by my sir in exam and he made no correction to it...
kropot72
  • kropot72
The current multiplying factor 50/R(p) is correct using the resistance values in the question.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
This should read: "In a galvanometer there is a deflection of 10 divisions per mA" What is the maximum reading? 100 divisions?
anonymous
  • anonymous
I am afraid that without the max deflection in mA (or Volts), there is no way to solve it
kropot72
  • kropot72
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 \[\frac{125\times 2.381}{50}=5.95mA\]

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