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- Thread starter Jodi
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OlderDan

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You need to make some assumption here about the construction of a galvanometer. I am going to assume you have an ideal case where the angular deflection of the needle is directly proportional to the current because the magnetic field is constructed to achieve that condition, as illustrated hereJodi said:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html

You can then write a direct variation equation that says

Deflection angle = kI

The "weakening of the spring" is saying that k is diminished by 26.4%. So what has to happen to I to achieve the former deflection?

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Are you saying to do 0.264 x 36.8E-6? Because that doesnt give me the right answer. Thanks.

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OlderDan

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No. To have the same deflection angle, the product of k times I must not change. If k is reduced by some fraction, the current is going to have to increase. Let k' and I' be the "new" values for the weakened spring and k and I be the "old" values that gave the same deflection. ThenJodi said:Are you saying to do 0.264 x 36.8E-6? Because that doesnt give me the right answer. Thanks.

k'I' = kI

A little bit of algebra from here will get you to the answer.

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