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Temperature variation of resistance

  1. Oct 1, 2008 #1
    So the problem Im 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 im totally confused...
     
  2. jcsd
  3. Oct 1, 2008 #2

    LowlyPion

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    Where did you get your first answer? What is the driving potential?
     
  4. Oct 2, 2008 #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...
     
  5. Oct 2, 2008 #4
    anyone?...
     
  6. Oct 2, 2008 #5

    LowlyPion

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    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?
     
  7. Oct 2, 2008 #6
    well my answer is correct according to the book...the value for the resistivity of copper im using is 1.7E-8
     
  8. Oct 2, 2008 #7

    LowlyPion

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    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.
     
  9. Oct 2, 2008 #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
     
  10. Oct 2, 2008 #9

    LowlyPion

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    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.
     
  11. Oct 2, 2008 #10
    hmm that worked..do u know why my calculator is not computing these correctly?
     
  12. Oct 2, 2008 #11

    LowlyPion

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    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.
     
  13. Oct 2, 2008 #12
    thank you...
     
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