Galvanometer Spring Weakness and Current Adjustment

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The discussion centers on calculating the new current required for full-scale deflection of a galvanometer after its restoring spring weakens by 26.4%, originally needing 36.8 µA. It is established that the deflection angle is directly proportional to the current, leading to the equation kI = k'I'. As the spring constant k diminishes, the current I must increase to maintain the same deflection. Participants clarify that simply multiplying the original current by the percentage of weakening does not yield the correct answer. The conversation emphasizes using algebra to derive the new current based on the weakened spring constant.
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Hi; Could someone please help me with the following question: If the restoring spring of a galvanometer weakens by 26.4% over the years, what current will give full-scale deflection if it originally required 36.8uA? Thanks for your help.
 
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Jodi said:
Hi; Could someone please help me with the following question: If the restoring spring of a galvanometer weakens by 26.4% over the years, what current will give full-scale deflection if it originally required 36.8uA? Thanks for your help.

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 here

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?
 
Are you saying to do 0.264 x 36.8E-6? Because that doesn't give me the right answer. Thanks.
 
Jodi said:
Are you saying to do 0.264 x 36.8E-6? Because that doesn't give me the right answer. Thanks.

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. Then

k'I' = kI

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