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Homework Help: Linear fit for data (resistance of a wire)

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data

    The resistance for a wire with cross-sectional radius r is given by R = k/(r^2). Convert this equation to a straight line for which k can be determined from the slope.

    2. Relevant equations

    Equation of a line is:
    y = mx + c

    3. The attempt at a solution
    R = k/(r^2)
    ln(R) = ln(k/(r^2))
    ln(R) = ln(k) - ln(r^2)
    ln(R) = ln(k) - 2*ln(r)

    I'm not sure how to linearize this equation, this is where I get stuck. ln(R) would represent y on the graph. k ought to equal the slope, but I can't see how to do that.
  2. jcsd
  3. Nov 16, 2011 #2
    What you have done is a perfectly good analysis but k is given by the intercept (lnk) on the graph.
    The slope of the ln-ln graph is -2. This is the value of ln-ln graphs.....enables you to determine the power law of a set of data.
    Since R =k/r^2 a graph of R against 1/(r^2) will be a straight line with gradient k
  4. Nov 16, 2011 #3
    The problem is that if I plot given values of R on the y-axis = R and given values of r on the x-axis = (1/(r^2)), my graph looks like this:


    Edit: This is my dataset
    R r
    0.81 0.0001
    0.73 0.00015
    0.60 0.0002
    0.38 0.00025
    0.1 0.0003
    Last edited: Nov 16, 2011
  5. Nov 16, 2011 #4


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    Why not plot a graph of R against 1/r2?
    That should give a straight line y=kx which should pass through the origin as when r = 0, so should R. The c would be zero.

    It is handy to get used to the idea of choosing what to plot against what in order to get a straight line graph. I remember needing quite a few examples before it became 'obvious' what to do.
  6. Nov 16, 2011 #5
    That graph I showed is R against 1/(r^2). My problem is that it is very curved, so I don't have a best-fit line.
  7. Nov 16, 2011 #6
    If I use some known value of k to produce my own table of r and R from R = k/(r^2), I do get a linear fit, so I guess I'm missing something involving the dataset.

    I have a similar small question, how would I convert this to a straight line such that a and b can be found? My known values are x and y.
    y = a*(1 + b*(x^2))

    y = a + a*b*(x^2)
    ln(y) = ln(a) + ln(a*b*(x^2))
    ln(y) = ln(a) + ln(a) + ln(b*(x^2)
    ln(y) = 2ln(a) + ln(b*(x^2))
    I can see that the intercept is 2ln(a), but I'm not sure how I would read a slope from ln(b*(x^2))
  8. Nov 16, 2011 #7
    I used your numbers and got something very similar! Where did you get the data, was it from measurements you made?
  9. Nov 16, 2011 #8
    The data was given to me as part of the question by my instructor.

    Could also you please also take a look at my second question in post #6?
  10. Nov 16, 2011 #9
    What was post #6? I am not certain how to find it
  11. Nov 16, 2011 #10
  12. Nov 16, 2011 #11
    Got it (a bit on the slow side!!)
    You can't do this:
    y = a + a*b*(x^2)
    ln(y) = ln(a) + ln(a*b*(x^2))
    You cannot add the ln of 2 numbers added together.When you add ln it means you are dealing with multiplication. You did this right in your post #1
    For this equation I would leave it as:
    y = a + abx^2
    Try plotting y against x^2..... what will be the gradient and what will be the intercept?
  13. Nov 16, 2011 #12
    When I added, I was thinking of ln(x) + ln(x) = ln(x*x) = 2*ln(x)

    D'oh... if the intercept is a and gradient is ab, then I can find b as gradient/a, thanks. :smile:
  14. Nov 16, 2011 #13
    you have got it
  15. Nov 16, 2011 #14


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    Sorry. I was being dumb. You are right. The graph looks so wrong that I assumed you'd not actually plotted it right. But the values certainly give that odd curve.

    I think the data must be wrong - unless it relates to something that can get hot for thin samples and not following Ohms Law. You'll have to ask your tutor, I think.
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