Question

How do I solve for error calculation in a resistor?

Answer 1

In general, your percent error will be:
\[\eta=\left| \frac{\Omega_{accepted}-\Omega_{experimental}}{\Omega_{accepted}} \right|\]How exactly you implement that into your data analysis depends on what your data are.

Answer 2

I'm afraid I still don't understand. I'll be more specific. Let's say, I want to solve for the percent of error in resistance for these values:


Current I (in mA) | Voltage Delta V (in Volts)
0 | 0
24.4 | 2.43
48.2 | 4.97
72.5 | 7.23
96.4 | 9.95

Do I calculate for percentage error by obtaining the average value of all resistances, and then subtracting the average value from the obtained resistance value and proceeding to divide the subtracted value by average resistance value and multiply by 100?

And in doing so, the Delta V in my table throws me off. I know that for this scenario, I'll have to use \[R=\frac{ \Delta V}{ \Delta I} \] rather than \[R=\frac{ V }{ I }\]. Is that correct?

For example, is getting the change in V, done this way: "4.97-2.43" (for my values in my chart). I keep thinking that it's redundant, because the values per se, of 4.97 and 2.43 are already \[\Delta V\]....

Answer 3

So you do have some value for the resistance that isn't calculated from your data? For example, the resistance posted on the side of the resistor.

Answer 4

@Xishem: That's the problem. I did not see the resistor's rating, nor did my professor make it known. I suppose that the point was to solve for it theoretically, rather than simply checking the resistance rating on the device. Also, I've been trying to look for the kind of resistor used in the experiment, but to no avail. My professor noted it as being a "tunable resistor" (it was about a 3 inch by 3 inch chunky block, with switches containing numbers, on it). I've tried to replicate the experiment on my own, however I never managed to find that specific resistor used in the experiment with my professor. I have a feeling that the resistor was custom made.

Answer 5

In that case the original question isn't so clear. There are a few ways you can analyse your data.

One way is to plot current vs. voltage and the slope of the best-fit line will be your best value for the resistance of the resistor. The R^2 value will be related to the error of your measurements, but ONLY if you assume that the resistor follows exactly Ohm's law.

There are certainly good ways to perform error analysis, but it depends what kinds of results you are looking for.

Answer 6

Yeah, my professor sped-through the experiment. I was quite disappointed about it because personally, lab classes are my favourite.

I'm not too comfortable with asking him questions because when I asked him whether I can redo the experiment by myself, he interpreted my question as if I did not take notes when he did the experiment in class. That was not true at all. I only wanted to replicate the experiment because frankly, he didn't do a good job at it and I tried to take responsibility for my own learning. Even then he simply told me to stick with the measurements we derived during class...

*Sigh*. Anyway,

I actually plotted the graph and got y = 0.1025x - 0.0359
R² = 0.9993

I wasn't aware what the R^2 meant (I plotted it in Microsoft Excel). So the R^2 value is the error of my measurements? How did it manage to compute that (or rather, I want to know how I can do that manually)?

Answer 7

Also, I have some additional information from my professor: "assume up to 2% error for voltage, and up to 2% error on current measurement."

Answer 8

The R^2 value (or coefficient of determination) is a measure of how far the points are from the fit-line, on average. Determining a trendline and an R^2 value can be extremely tedious and the subject of error analysis is a whole semester or two on its own.

You can read more here: http://en.wikipedia.org/wiki/Linear_regression, and here: http://en.wikipedia.org/wiki/Coefficient_of_determination

Answer 9

One thing you can do is calculate the resistance for each set of datum and calculate the uncertainty in each value given the uncertainty of each of your measurements: 2% in voltage difference and 2% in current. Do you know how to approach that?

Answer 10

@Xishem: I've done Linear Regression awhile back... I can't remember it exactly, but I know I've done it before...

Regarding your last reply: "One thing you can do is calculate the resistance for each set of datum and calculate the uncertainty in each value given the uncertainty of each of your measurements: 2% in voltage difference and 2% in current. Do you know how to approach that?"

I actually do not. I've never done error calculation before, so I'm pretty clueless. You'll have to walk me through the steps, in this case...

Answer 11

I think you'll find it's simple enough. One way to find what your uncertainty in a value is is to figure out what the maximum possible value is and what the minimum possible value is.

Given that you know the minimum and maximum possible values for your current and voltage difference, can you calculate the minimum and maximum possible values for your resistance?

Answer 12

Ohhh definitely. So will I stick with using R=V/I instead of \[R=\Delta V/ \Delta I\]?

Answer 13

Be careful. I think you might be confused about what this means:\[\Delta V\]It doesn't mean change in V or uncertainty in V, it just means the difference in electrical potential between the two sides of the resistor.

Answer 14

To clarify, by Ohm's law:\[R=\frac{\Delta V}{I}\]

Answer 15

Similarly, the experimental values you measured are \[\Delta V\ and\ I\]

Answer 16

"It doesn't mean change in V or uncertainty in V, it just means the difference in electrical potential between the two sides of the resistor." Oh yes! That's right!!!!

So from there, I'll just do this?:

\[R=\frac{ (4.97-2.43)V }{(0.0482-0.0244)I}\]

and

\[R=\frac{ (9.95-7.23)V }{(0.0964-0.0725)A}\]

and what happens when I get their total values?

Answer 17

Ah. That's not quite what you want. You need to consider what the minimum and maximum ACTUAL values are for each datum pair.

Answer 18

You know that for any voltage measurement:\[\Delta V_{actual}=\Delta V_{experimental}\pm(0.02)\Delta V_{experimental}\]Does that equation make sense? The 0.02 comes from the 2% error.

Answer 19

Yes it does... So I will also apply the (0.02) to the current, is that correct?

And once I've done that, I will have new values for my chart which I can use to solve for the resistance, in the same manner as I would have, if there was no error?

Answer 20

Yes, it applies to current as well.

What you want to do is figure what the maximum actual resistance could have been given the uncertainties in your voltage and current measurements.

Basically what you do in this case is look at the minimum and maximum possible values for voltage and current:\[\Delta V_{\min}=\Delta V_{experimental}-(0.02)\Delta V_{experimental}\]\[\Delta V_{\max}=\Delta V_{experimental}+(0.02)\Delta V_{experimental}\]And so on with your currents. Then find the minimum possible resistance that could exist given the limits of your voltage and current measurements.

Answer 21

My \[\Delta V\] would be 2.43 and not 0, right?

I'm starting to see the picture now...

Answer 22

It's a bit more difficult to use the 0, 0 datum point because 2% error in 0 is... 0.

So, yeah, if you use the 24.4mA, 2.43V point, then you'll want to find, given the entire range of your uncertainty in the two values, what the minimum and maximum possible resistances are.

Answer 23

Once I have the maximum and minimum values for resistances, where do I go from there?

Answer 24

Well, from knowing your min and max values possible values for R, you can find the uncertainty as a percentage. You'll basically just do the same procedure you used for finding the min/max values for voltage, except backwards.

Answer 25

And from that you should be able to find some form:\[R_{actual}=R_{\exp}\pm(X\%) R_{\exp}\]Where R_exp and X should be constants.

Answer 26

I really appreciate the help! Thank You!