Galvanometer Spring Weakness and Current Adjustment

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SUMMARY

The discussion centers on calculating the new current required for full-scale deflection of a galvanometer when its restoring spring weakens by 26.4%. Originally, the galvanometer required 36.8 µA for full deflection. The relationship between the deflection angle and current is defined by the equation Deflection angle = kI, where k is the spring constant. As the spring constant k decreases due to weakening, the current I must increase to maintain the same deflection angle, leading to the conclusion that the new current can be derived using the equation k'I' = kI.

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  • Understanding of galvanometer operation and principles
  • Familiarity with direct variation equations
  • Basic algebra for manipulating equations
  • Knowledge of current measurement in microamperes (µA)
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Jodi
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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