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A wire with a resistance R is lengthened to 1.25

  1. Sep 6, 2003 #1
    My answer for this problem doesn't agree with the answer in the back of my textbook book:

    A wire with a resistance R is lengthened to 1.25 times its original length by being pulled through a small hole. Find the resistance of the wire after it has been stretched.

    R = pL/A

    p = resistivity
    L = length
    A = area of a cross section perpendicular to the length
    r = radius

    R1 = pL1/A1

    L2 = 1.25L1 so r1/r2 should equal 1.25 right.

    The decrease in radius should be proportional to the increase in length.

    So r2 = r1/1.25

    Therefore R2 = p(1.25L1)/[pi](r1/1.25)^2 = R1(1.25*1.56) = 1.95R1

    The answer in the back of my textbook is R2 = 1.56R1

    Where have I gone wrong?

    Thanks
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Sep 6, 2003 #2

    Hurkyl

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    Should it? How would you go about proving this?
     
  4. Sep 6, 2003 #3
    I'm assuming it's proportional. If it's not then I'm lost
     
  5. Sep 6, 2003 #4

    enigma

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    Re: Resistance

    Volume is constant, and volume is L*A, not L*r
     
  6. Sep 6, 2003 #5
    You're right. Stupid me!!

    The area would have to shrink proportionally: A2 = A1/1.25 so R2 = 1.25*1.25R1 = 1.56R1.

    Thank you very much for the help. :smile:
     
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