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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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kropot72Best ResponseYou've already chosen the best response.0
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)
 one year ago

RomeoRajBest ResponseYou've already chosen the best response.0
sorry brother but the right answer is 125mA
 one year ago

kropot72Best ResponseYou've already chosen the best response.0
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.
 one year ago

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

kropot72Best ResponseYou've already chosen the best response.0
The current multiplying factor 50/R(p) is correct using the resistance values in the question.
 one year ago

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

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

kropot72Best ResponseYou've already chosen the best response.0
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 \[\frac{125\times 2.381}{50}=5.95mA\]
 one year ago
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