Galvanometer Spring Weakness and Current Adjustment

In summary, the question is asking about the current needed to achieve full-scale deflection on a galvanometer with a weakened restoring spring. Assuming an ideal case where the deflection angle is directly proportional to the current, the weakened spring means that the constant of proportionality (k) is reduced by 26.4%. This means that the current (I) must increase to compensate and achieve the same deflection angle. Using the direct variation equation, k'I' = kI, we can solve for the new current needed.
  • #1
Jodi
23
0
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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  • #2
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?
 
  • #3
Are you saying to do 0.264 x 36.8E-6? Because that doesn't give me the right answer. Thanks.
 
  • #4
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.
 

1. What is a galvanometer?

A galvanometer is an instrument used to measure small electrical currents. It typically consists of a coil of wire suspended within a magnetic field, which causes the coil to move in response to the flow of electricity.

2. How does a galvanometer work?

A galvanometer works by utilizing the principles of electromagnetism. When an electric current flows through the coil of wire, it creates a magnetic field that interacts with the external magnetic field, causing the coil to rotate. The amount of rotation is proportional to the strength of the current.

3. What is the difference between a galvanometer and an ammeter?

A galvanometer is a sensitive instrument that is used to detect and measure small electrical currents, while an ammeter is used to measure larger currents. Galvanometers have a higher sensitivity and can measure currents in the microampere range, while ammeters typically measure currents in the milliampere range.

4. What are some common uses of galvanometers?

Galvanometers have a variety of uses in scientific and industrial settings. They are commonly used in electronic circuit testing, medical equipment, and in the manufacturing of electronic devices. They are also used in geophysical surveys to measure electrical conductivity in the ground.

5. What is the history of the galvanometer?

The galvanometer was invented by Johann Schweigger in 1820. It was originally used to measure current and voltage in electrical circuits, but its uses expanded over time. In the mid-19th century, Wilhelm Weber and Carl August von Steinheil developed the mirror galvanometer, which used a mirror and light beam to amplify the motion of the coil and provide more precise measurements. Today, galvanometers are still widely used in various fields of science and technology.

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