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Graph of Ln(R) against 1/T

  1. Mar 30, 2013 #1
    53lqme.jpg

    The resistance,R, of a thermistor varies with absolute temperature according to the following equation:

    R= R0 ek/T


    I was asked to find the constant k using the graph.

    I put the equation given in the form

    ln R= k/t + lnR0

    k would be the gradient. So I found the gradient from the graph and I got 2968.80

    then I was asked to find Ro

    so I used a point off the graph and the gradient

    point (2.88,4.3)

    4.3 = lnRo + (2968.80/2.88)

    => lnRo = -1.031 x 106

    then I took anti logs and got Ro = 0 . Which i think is wrong...

    Guidance appreciated.
     
  2. jcsd
  3. Mar 30, 2013 #2

    TSny

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    Should the 2.88 be multiplied by 10 to some power?

    Do you really want to divide here?
     
  4. Mar 30, 2013 #3
    Oh yeah the 2.88 should be 2.88 x 10^-3

    and yeah wouldn't I divide? Isn't it k/T?
     
  5. Mar 30, 2013 #4

    TSny

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    k/T says to divide by T. But you're graphing 1/T, not T.
     
  6. Mar 30, 2013 #5
    oh I see thank you.
     
  7. Mar 30, 2013 #6
    I got Ro to be 0.014
     
  8. Mar 30, 2013 #7

    TSny

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    That looks ok. But, I find the graph has a slope (gradient) closer to 2850. That would change the result for Ro somewhat.
     
  9. Mar 30, 2013 #8
    Oh what did you use for your values to find the gradient? I used (3.52*10^-3 , 6.2)

    (2.88 * 10^-3, 4.3)
     
  10. Mar 30, 2013 #9

    TSny

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    When finding the gradient, it's best to pick two points on the line rather than picking two data points. The data points generally don't lie directly on the best fit line. Also, it's best to pick two points fairly far apart.

    I picked the two points shown in the attached figure.
     

    Attached Files:

  11. Mar 30, 2013 #10
    ohh okay thank you
     
  12. Mar 30, 2013 #11

    TSny

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    Also, note that you cannot expect to get a gradient more precise than 3 significant figures for this graph.
     
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