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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

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kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0Let 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
 4 years ago
Best ResponseYou've already chosen the best response.0sorry brother but the right answer is 125mA

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0ya i think you are right but this question was given by my sir in exam and he made no correction to it...

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0The current multiplying factor 50/R(p) is correct using the resistance values in the question.

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.0This should read: "In a galvanometer there is a deflection of 10 divisions per mA" What is the maximum reading? 100 divisions?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am afraid that without the max deflection in mA (or Volts), there is no way to solve it

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0According 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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