Resistivity experimentally determined

A group of students prepares a large batch of conductive dough, a soft substance that can conduct electricity, and then molds the dough into several cylinders with various cross-sectional areas (A) and lengths (L). Each student applies a potential difference (\Delta V) across the ends of a dose cylinder and determines the resistance of the cylinder. The results of their experiments are shown in the table below.

So, what they're doing is they're taking this material, this conductive dough, and they're making cylinders of different cross-sectional areas and different lengths. So they're varying the area, the cross-sectional area, and the length, but it's all being made out of this same material. Then, depending on how they vary that, you get different resistances.

The students want to determine the resistivity of the dose cylinders. Just as a reminder, resistivity is different than resistance. A particular shape of a given material will have a specific resistance based on its length and cross-sectional area. All of that will have a given resistance, but resistivity is a property of the material. The relationship between resistance and resistivity can be written as:

[
R = \rho \frac{L}{A}
]

And so, if you have a higher resistivity, all else equal, that would make a higher resistance. If you have a higher length, that also makes a higher resistance. However, if you have a higher cross-sectional area, then that lowers the resistance. If you were to solve for (\rho), you could write:

[
\rho = R \frac{A}{L}
]

Then it says to indicate below which quantities could be graphed to determine a value for the resistivity of the dose cylinders. You may use the remaining columns in the table above as needed to record any quantities, including units that are not already in the table. So pause this video and think about how you would do that.

Well, I can think of two ways to do that. You could make a scatter plot where on the horizontal axis you put ( \frac{L}{A} ) and on the vertical axis you put resistance. The resistivity would be the slope of a line that you try to fit to those points. So that's one way to do it. Another way to do it is to make your horizontal axis ( \frac{A}{L} ), and then you can make your vertical axis your (\rho) and try to fit a horizontal line to those points because your resistivity should not be dependent on ( \frac{A}{L} ).

I'm actually going to pick this second option because if I'm doing this primarily by hand, it's at least easier for me to fit a horizontal line than to fit a non-horizontal line as I would have to do there. So let's do that. My vertical axis I'm going to do resistivity, which will be in ohm meters. Then, in my horizontal axis, I am going to do my area divided by length, which is going to be in meters.

Now to be able to do this, I need to calculate these things. I want to calculate my area divided by length. (0.00049) divided by (0.03) is equal to (0.016). So we have (0.016), two significant digits here. I can then calculate (\rho) by multiplying my area divided by length times resistance. My measured resistance was (23.6). It's important to realize that when you do things experimentally, things are not going to be perfect. There's a lot of noise in the real world, so I'll round that to (0.39).

Now, I'll keep doing that calculation, and for the sake of time, I will just speed up the video. Now we can move to this second part. It says on the grid below to plot the appropriate quantities to determine the resistivity of the dose cylinders.

Clearly, scale and label all axes, including units as appropriate. On our horizontal axis, we want to put our cross-sectional area divided by the length of our resistors, which will be in meters. Then, in our vertical axis, we want to plot our resistivity, which will be in ohm meters.

Now we need to think about the scales. The scale of this vertical axis, if we go back over here, we see that all of our calculations for (\rho) are between around (30) hundredths and (41) hundredths. We could have reasonable scales here: (0.1) for (10) hundredths, (0.2) for (20) hundredths, (0.3) for (30) hundredths, and (0.4) for (40) hundredths.

For our area divided by length, if we view all of this in terms of ten-thousandths, we go between (38) ten-thousandths all the way to (160) ten-thousandths. So let's make this (50) ten-thousandths as (0.0050), (100) ten-thousandths as (0.0100), (150) ten-thousandths as (0.0150), (200) ten-thousandths as (0.0200), and (250) ten-thousandths.

I think I have my scales right, so now we just plot our points. The first one has a resistivity of (0.39) and our (A/L) is (160) ten-thousandths of a meter. So we would go look up (160) right over here which is (150), then we'd want to go one-fifth of the way between this and this.

That would place us at (0.39) ohm meters for the resistivity. Now if I go to my second one, (98) ten-thousandths, (0.31) would be right around there. Then I get to (66) ten-thousandths and (0.41), which would be somewhere there. Lastly, (38) ten-thousandths and (0.4) would fit around here.

So those are our four data points. Now they say to use the above graph to estimate a value for the resistivity of the dose cylinders. The resistivity should be constant; it shouldn't depend on our area divided by our length. The goal is to fit a horizontal line that gets as close as possible.

Ideally, we would have a computer, but we're going to hand-do it. We have three of these points that are pretty close to (40) hundredths. So, I would say a line that looks something like this, and this point right over here seems to be approximately (0.37).

Hence, a reasonable estimate would give us (0.370) ohm meters. We're done! Another way to think about this is you could have done it potentially even without the graph by just taking the mean of these values. However, they wanted us to use a graph, and one technique would be to fit a horizontal line; another would have been plotting ( \frac{L}{A} ) here and resistance on our vertical axis. Then, our estimate of the slope of the fitted line would give us our estimate of (\rho), but that would have been harder to do than what we did here.