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Temperature coefficient of resistance

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data
    ok we did a temperature coefficient of resistance at class, we need to calculate the coefficient of resistance (a) ourselves.
    known variable:
    R = resistance
    ln R = log of resistance
    T = tempreature
    a = coefficient of resistance
    c = the intercept


    2. Relevant equations
    ln R = aT+c
    i.e. R= k exp(aT), where k=exp(c) is a constant.


    3. The attempt at a solution
    at first I thought I could calculate a as if it is gradient, and therefore worked out the intercept (c). But since a is the coefficient of resistance it didn't make sense to me to do it this way. So I'm lost
    Anyways this is my working:
    gradient = delta ln R/ delta T
    = (lnR_1-lnR_2/T_1-T_2) = -.912185/44
    = -0.02682897
    therefore intercept (c) = -.912185 - (-.02682897 * 44)
    = 10.8925618

    put it back to the equation
    lnR = a*T+10.893
    given T = 64, its correspond ln R = 4.957 (we measured these ourselves, assume correct)

    so:
    4.957 = a*64+10.893
    -5.936=a*64
    a=-5.936/64
    a=-0.0927

    put everything back to original equation:
    ln R = -0.0927 T + 10.893

    is this even correct??
    thanks
     
  2. jcsd
  3. May 19, 2010 #2
    if my above working was correct, here is the second part of my question.
    during the experiment we only measure the resistance every 2C drop (from 100C to 30C). I've plot a ln R vs T graph, how can a determine the a at temperature of 65C?
     
  4. May 19, 2010 #3
    arha! I was right in general, but had made a small mistake! I forgot to convert temperature into kelvin before I calculated the coefficient. when I convert it, everything looks more reasonable and correct!
     
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