Temperature variation of resistance

In summary: You have to be careful with the powers of 10. Looks like you might have dropped a 0 somewhere in entering the numbers. Sometimes these problems unfortunately become more about arithmetic than physics. Just be careful is the only advice I can give, but then I guess you already know that.In summary, a copper wire with a radius of .45 cm has a potential difference of 5.0 volts at 20 degrees Celsius. If the temperature of the wire is increased to 200 degrees Celsius, the potential difference required to produce a current of 5.0 amps is 241.24 volts.
  • #1
jsantor1
7
0
So the problem I am working on is.. A 100 cm long copper wire of radius 0.45 cm has a potential difference across it sufficient to produce a current of 5.0 A at 20°C. Find a) What is the potential difference. b) If the temperature of the wire is increased to 200°C, what potential difference is now required to produce a current of 5.0 A

I solved part a easily by finding the resistance of the wire and plugging that into V=IR with the solution being 1.34mV.

Now for part b i used the equation R=R0[1+alpha(T-T00)] and plugged back into V=IR but my solution keeps coming out wrong... please help me out here I am totally confused...
 
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  • #2
jsantor1 said:
So the problem I am working on is.. A 100 cm long copper wire of radius 0.45 cm has a potential difference across it sufficient to produce a current of 5.0 A at 20°C. Find a) What is the potential difference. b) If the temperature of the wire is increased to 200°C, what potential difference is now required to produce a current of 5.0 A

I solved part a easily by finding the resistance of the wire and plugging that into V=IR with the solution being 1.34mV.

Now for part b i used the equation R=R0[1+alpha(T-T00)] and plugged back into V=IR but my solution keeps coming out wrong... please help me out here I am totally confused...

Where did you get your first answer? What is the driving potential?
 
  • #3
The first part you use R= resistivity* L/A. This gives you the resistance and with the currecnt use V=IR to solve for potential difference...
 
  • #4
anyone?...
 
  • #5
jsantor1 said:
The first part you use R= resistivity* L/A. This gives you the resistance and with the currecnt use V=IR to solve for potential difference...

I get a different answer for the initial resistance then. I used 16.8 nΩm and I get 3.73 x 10-4Ω.

What value of α are you using?
 
  • #6
well my answer is correct according to the book...the value for the resistivity of copper I am using is 1.7E-8
 
  • #7
jsantor1 said:
well my answer is correct according to the book...the value for the resistivity of copper I am using is 1.7E-8

Yes. You're right. I made an error calculating the area. Sorry.

Now what are you using for your temperature α ?

I have 3.9 x 10-3 per degree C for copper.
 
  • #8
That's what I'm using as well, getting a new resistance of .04824 and then getting a potential difference of 241.24mV which is coming out wrong for some reason
 
  • #9
jsantor1 said:
That's what I'm using as well, getting a new resistance of .04824 and then getting a potential difference of 241.24mV which is coming out wrong for some reason

Using the calculator at
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/restmp.html

I get 4.56 x 10-4Ω (using the 2.68 x 10-4 Ω for 20 C)

That yields 2.28 mV.
 
  • #10
hmm that worked..do u know why my calculator is not computing these correctly?
 
  • #11
jsantor1 said:
hmm that worked..do u know why my calculator is not computing these correctly?

You have to be careful with the powers of 10. Looks like you might have dropped a 0 somewhere in entering the numbers. Sometimes these problems unfortunately become more about arithmetic than physics. Just be careful is the only advice I can give, but then I guess you already know that.
 
  • #12
thank you...
 

What is the relationship between temperature and resistance?

The resistance of a material typically increases with increasing temperature. This is because at higher temperatures, the atoms in the material vibrate more and create more collisions, making it more difficult for electrons to flow through the material.

How does temperature affect the conductivity of a material?

As temperature increases, the conductivity of a material typically decreases. This is because at higher temperatures, the atoms in the material vibrate more, disrupting the flow of electrons and decreasing the material's ability to conduct electricity.

Why does the resistance of metals increase with temperature?

Metals have a positive temperature coefficient of resistance, meaning that their resistance increases with increasing temperature. This is due to the increased atomic vibrations at higher temperatures, which cause more collisions and impede the flow of electrons.

How can the temperature coefficient of resistance be calculated?

The temperature coefficient of resistance can be calculated by dividing the change in resistance by the change in temperature. It is typically expressed in units of ohms per degree Celsius (Ω/°C) or ohms per degree Kelvin (Ω/K).

What are some practical applications of understanding the temperature variation of resistance?

Understanding the temperature variation of resistance is important in the design and operation of electronic devices, as it allows for accurate predictions of how a material's resistance will change with temperature. This information is also useful in industries such as energy and transportation, where temperature variations can affect the performance of materials such as wires and conductors.

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