# Help with Lab Report Ohm's law

I have some problems with my lab report.
The teacher is dealing with us as expert in physics but I don't understand many things.
She told us to find the uncertainty and percentage error , even though she didn't explain anything and I'm completely lost.
I tried my best but I'm not even sure of what I did. This is my first lab report.

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CWatters
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I had a quick look but I don't understand experiment 3...

The circuit appears to show two resistors in series but where they calculate "R value from mathematical formula" it assumes they are in parallel?

Then later it says..

From the graph

R=(X1-X2)/(Y1-y2) = (4.5-3.5)/(0.45-0.35) = 1/0.1 = 10 Ohms
That doesn't seem to come from the previous graph.

Is there a missing page?

What was the full scale and the accuracy class of the instruments used?

I had a quick look but I don't understand experiment 3...

The circuit appears to show two resistors in series but where they calculate "R value from mathematical formula" it assumes they are in parallel?

Then later it says..

That doesn't seem to come from the previous graph.

Is there a missing page?
Sorry about it. I put the wrong circuit but it should be fine now.

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What was the full scale and the accuracy class of the instruments used?
I don't really know what do you mean by accuracy but what I remember that the smallest scale is 0.02 in the ammeter and the highest is 1.

Every measuring instrument has certain accuracy, which is usually given as the class of accuracy. It should have been labeled on the instruments somewhere. If not given, then usually half the interval between two adjacent marks on the scale is considered the accuracy. That should give you some ± on each value measured.

Every measuring instrument has certain accuracy, which is usually given as the class of accuracy. It should have been labeled on the instruments somewhere. If not given, then usually half the interval between two adjacent marks on the scale is considered the accuracy. That should give you some ± on each value measured.
So, If the scale is like: 0.02 - 0.04 - 0.06
Then the accuracy = + or - ( 0.01)?

That would mean ± 0.01.

That would mean ± 0.01.
I get it now. The voltmeter was: 0.2 - 0.4 - 0.6
then it would be: ± 0.1 but how do I use it ?

CWatters
Homework Helper
Gold Member
There is some information we are missing.

The circuit diagrams for experiment 2 and 3 are the same. Yet the data is different. Is one circuit meant to have the resistors in series and the other in parallel?

The question asks you to calculate the "percentage error" but between what and what?

There is some information we are missing.

The circuit diagrams for experiment 2 and 3 are the same. Yet the data is different. Is one circuit meant to have the resistors in series and the other in parallel?

The question asks you to calculate the "percentage error" but between what and what?
Yes, experiment 2 and 3 are different. the resistors in experiment 2 are in series and in 3 in parallel.

The question says: comment on your results and calculate the percentage error (ΔR).

Because you compute R as V/I, ## \Delta R = \frac {V \Delta I + I \Delta V} {I^2 + \Delta I^2} ##.

For the first experiment, let's consider the measurement at the bottom of the table: 0.1 A and 1 V. That should be I = 0.1 ± 0.01 A and V = 1 ± 0.01 V. Using the formula above, ## \Delta R = \frac {1 \cdot 0.01 + 0.1 \cdot 0.01} {0.1^2 + 0.01^2}
\approx 1.09
##, which means 1.09/10*100% = 10.9% error. The measurement at the top of the table: I = 0.5 ± 0.01 A, V = 5 ± 0.01. The same formula yields 0.22, or 2.2% error, which is much better.

Because you compute R as V/I, ## \Delta R = \frac {V \Delta I + I \Delta V} {I^2 + \Delta I^2} ##.

For the first experiment, let's consider the measurement at the bottom of the table: 0.1 A and 1 V. That should be I = 0.1 ± 0.01 A and V = 1 ± 0.01 V. Using the formula above, ## \Delta R = \frac {1 \cdot 0.01 + 0.1 \cdot 0.01} {0.1^2 + 0.01^2}
\approx 1.09
##, which means 1.09/10*100% = 10.9% error. The measurement at the top of the table: I = 0.5 ± 0.01 A, V = 5 ± 0.01. The same formula yields 0.22, or 2.2% error, which is much better.
Sorry, Could you explain it in more details ? I didn't understand, right from the formula.

That formula gives the error for division. Note this is not the only possible formula. There are other formulas, which give slightly different results.

What that really means is that both V and I have some error (= uncertainty) to them. You do not measure R directly, you compute it based on the measured values of V and I. The computation "propagates" their respective errors, and they become the error in R. This formula estimates it.

CWatters
Homework Helper
Gold Member
Voko - I agree with your calculations so far but does the OP need to take into account the fact that multiple data points were taken and the resistance calculated from the slope of the graph? It's 30 years since I did any statistics that covered this. Is it a matter of fitting error bars to the graph and working out the min and max slope?

Voko - I agree with your calculations so far but does the OP need to take into account the fact that multiple data points were taken and the resistance calculated from the slope of the graph? It's 30 years since I did any statistics that covered this. Is it a matter of fitting error bars to the graph and working out the min and max slope?
I know as much as you do. The results in the report are such that no fitting seems required. They are just exact integer multiples.That usually triggers some skepticism by the examiner. I am not sure what equipment and technique were used, but resistance is known to be quite tricky to measure. Especially for someone who does it the first time! Which is compounded by the fact, that adequate training did not happen. Personally, I would expect errors on the order of 10%.

I'll try to do the same experiment again today and rewrite the report. So, would you please tell me what to do this time , what to expect and how to deal with it? What is uncertainty ? How did we get the formula for ΔR for the percentage error ?

I think that the second graph is not exact. Is this OK?

Thanks a lot for your help and sorry for the inconvenience but I don't the teacher will explain it to us.

I = 0.1 ± 0.01 A and V = 1 ± 0.01 V.
Sorry for all the trouble but could you tell me how do I use this in the first formula? And why did we use the uncertainty of the ammeter with the Voltage ( If it is even the case)?

Forgive me but I'm struggling with my lab teacher.

Last edited:
I do not understand your questions. The uncertainties of the voltage and current measurement are taken from your description of the instruments.

Yes, but the uncertainty of the voltmeter was ± 0.1

Then I made a mistake. Sorry.

Thanks a lot.