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Linear fit for data (resistance of a wire)

  • Thread starter Cade
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  • #1
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Homework Statement



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.

Homework Equations



Equation of a line is:
y = mx + c

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.
 

Answers and Replies

  • #2
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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
 
  • #3
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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:

VWoHo.png


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:
  • #4
sophiecentaur
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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.
 
  • #5
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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.
 
  • #6
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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))
 
  • #7
1,506
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I used your numbers and got something very similar! Where did you get the data, was it from measurements you made?
 
  • #8
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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?
 
  • #9
1,506
17
What was post #6? I am not certain how to find it
 
  • #11
1,506
17
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?
 
  • #12
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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:
 
  • #13
1,506
17
you have got it
Cheers
 
  • #14
sophiecentaur
Science Advisor
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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.
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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