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RomeoRaj Group Title

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.

  • 2 years ago
  • 2 years ago

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  1. kropot72 Group Title
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    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)

    • 2 years ago
  2. RomeoRaj Group Title
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    sorry brother but the right answer is 125mA

    • 2 years ago
  3. kropot72 Group Title
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    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.

    • 2 years ago
  4. RomeoRaj Group Title
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    ya i think you are right but this question was given by my sir in exam and he made no correction to it...

    • 2 years ago
  5. kropot72 Group Title
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    The current multiplying factor 50/R(p) is correct using the resistance values in the question.

    • 2 years ago
  6. Vincent-Lyon.Fr Group Title
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    This should read: "In a galvanometer there is a deflection of 10 divisions per mA" What is the maximum reading? 100 divisions?

    • 2 years ago
  7. CarlosGP Group Title
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    I am afraid that without the max deflection in mA (or Volts), there is no way to solve it

    • 2 years ago
  8. kropot72 Group Title
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    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 \[\frac{125\times 2.381}{50}=5.95mA\]

    • 2 years ago
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