Linear fit for data (resistance of a wire)

In summary, the equation of a line is y = mx + c, where y is the line's y-value, m is the slope of the line, and c is the y-intercept. 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 kThe 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:
  • #1
Cade
92
0

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.
 
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  • #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
 
  • #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:

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
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
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
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
I used your numbers and got something very similar! Where did you get the data, was it from measurements you made?
 
  • #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?
 
  • #9
What was post #6? I am not certain how to find it
 
  • #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?
 
  • #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:
 
  • #13
you have got it
Cheers
 
  • #14
Cade said:
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.
 

What is a linear fit for data and how is it used in determining the resistance of a wire?

A linear fit for data is a mathematical model that describes the relationship between two variables by fitting a straight line through the data points. In the context of determining the resistance of a wire, a linear fit can be used to plot the relationship between voltage (V) and current (I) in a wire, as described by Ohm's law (V = IR). By fitting a straight line to this data, the slope of the line (represented by the resistance, R) can be calculated, providing a measure of the wire's resistance.

What is the significance of the slope of the linear fit in determining the resistance of a wire?

The slope of the linear fit in a voltage-current plot represents the resistance of the wire. This is because Ohm's law states that the resistance of a wire is equal to the ratio of voltage to current. Therefore, the slope of the linear fit (in units of ohms) gives the value of the wire's resistance.

What factors can affect the accuracy of a linear fit for determining the resistance of a wire?

There are several factors that can affect the accuracy of a linear fit for determining the resistance of a wire. These include variations in the wire's temperature, imperfections in the wire's material, and errors in the measurement of voltage and current. Additionally, the length and thickness of the wire can also affect the resistance and thus the accuracy of the linear fit.

What is the difference between a linear fit and a best-fit line?

A linear fit is a mathematical model that fits a straight line through a set of data points, while a best-fit line is a line that is drawn through data points to represent the overall trend of the data. In the context of determining resistance of a wire, the linear fit is used to calculate the exact value of the resistance, while the best-fit line is used to visualize the relationship between voltage and current.

How can a linear fit be used to predict the resistance of a wire at different values of voltage and current?

Once a linear fit has been calculated for a wire's voltage-current plot, it can be used to predict the resistance at any given value of voltage or current. This is because the linear fit represents the overall relationship between these two variables, and can be used to interpolate or extrapolate values within the range of the data points.

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