How to ameliorate this Wheatstone bridge?

[JulienB]

Homework Statement

Hi everybody! I just ran an experiment testing the accuracy of the Wheatstone bridge, which indeed yielded very small error intervals. However, we then ran a test of reproducibility and found that the deviation between the results was very high! Any idea how that could be corrected? (it's for the discussion about the uncertainty of the experiment).

Homework Equations

I've attached a picture of the experiment setup. ##R_x## is the resistor we measure, ##R_N## is the variable resistor, ##P## is the potentiometer (with wire of length ##l=100##cm with a positional sensor ##B##) and ##I## indicates where we measured the voltage (which we set to zero by moving the positional sensor). Here are the values we got for one resistor when changing ##R_N## dramatically:

##R_N = 10 \Omega \implies R_x = (12.27 \pm 0.06) \Omega##
##R_N = 1 \Omega \implies R_x = (14.9 \pm 0.4) \Omega##
##R_N = 400 \Omega \implies R_x = (8.6 \pm 0.4) \Omega##

The Attempt at a Solution

The first value (also the most accurate) is the closest to the value of the resistor (it's ##12 \Omega## according to the manufacturer). When using a value for ##R_N## close to the real ##R_x##, the positional sensor of the potentiometer is more or less in the middle (##50##cm). When ##R_N=1 \Omega##, ##B## is at ##93.7##cm; When ##R_N=400 \Omega##, ##B## is at ##2.1##cm; we took those measurements purposefully on the edges of the wire of the potentiometer.

As you can see, the deviation between the values is huge. I assumed this came from the inhomogeneity of the wire of the potentiometer, but I am not sure. Can anyone suggest me ideas on how to improve this experiment?

Thank you in advance for your answers.Julien.

 

[Merlin3189]

The problem I was trying to express is that when you don't know the value of the resistor prior to the experiment, the Wheatstone bridge seems to be of very little help. The problem that you mention about the error interval for me clearly shows the limitation of the method: if I had used RN=12Ω instead of 10Ω, then 12Ω might have been included inside the error interval. However the value 12Ω is also not a proper reference value, as it is given by the manufacturer with its own error interval (±1%).

I think one would have a range of possibilities for Rn and after a preliminary measurement use ##R_N## close to that value in order to get a more accurate reading. In your case you had only 3 standard resistances and you did indeed pick the 10Ω as the one which would give the most accurate result. And you could obviously see that, even if you had not known the true answer: 10Ω is the closest to 12.3, 14.9 and 8.6Ω.

I still think you need to look at how you estimated your errors. I think you just told me a new source of error, the 1% tolerance of the 10Ω standard resistor. Since your result is proportional to this value, I can't see how you can claim a result better than 1%. So shouldn't your result be less accurate than 12.27 ± 0.12Ω ?
(I may have mistaken you here: perhaps you're saying, a 12Ω resistor would be only 1%, but the 10Ω resistor was 0.5% or better? I'm not sure what you are saying about the 12Ω and 1%. I don't think you should worry about using 10Ω rather than 12Ω. Think about why it is that we want them to be roughly equal like 12Ω and 10Ω, rather than very different like 12Ω and 100Ω?)

Ah no when I mentioned the 1%, it was about the error given by the manufacturer with the value of the resistor (so (12.00±0.12)Ω), which I used as a reference so to say.

But you're right about the error when using the Wheatstone bridge of course, the error is too small and that's what I am pointing out in my discussion. ...

The formula to retrieve Rx was: ##R_x = R_N \frac{x}{l-x}\ \ ## (where l is the length of the wire).

To calculate the error, we used the formula for the propagation of error of Gauß and included the error for RN (given by manufacturer of the resistors), x and l (reading error and manufacturer's error)

I also think that the point of the experiment in my case was that we think about the setup, discover that the inhomogeneity of the wire greatly influences the results (and the error intervals) and discuss how that problem could be solved.

Your formula for ##R_x## shows the uncertainty in ##R_N## propagates directly to ##R_x##.
Say your value for ##\frac x {l-x}## were 1.2, then ##R_x = 1.2 R_N##
If your ##R_N## is 100Ω with 1% tolerance, it could be from 99 to 101Ω
Then ##R_x ## becomes 1.2 x 100 = 120Ω with a tolerance from 118.8Ω to 121.2, which is still 1% uncertainty.

With the ##x## measurement on the wire, I just use the rule of thumb that % errors add when you multiply or divide two quantities which each have uncertainty.
So if your wire were 1m long and you measure ##x=54.6cm \ \ ## then ##(l-x)=45.4cm## and your ratio is ##\frac{54.6}{45.4} = 1.20264##

Say the error in this measurement were 0.1%, then the ratio could be from (0.1%less)/(0.1% more) to (0.1% more) /(0.1% less)
That is ##\frac{0.999}{1.001} \times{1.2026}\ \ to \frac{1.001}{0.999}\times{1.2026} = 0.998\times{1.20264}\ \ to \ 1.002\times{1.20264}##
## =1.20024 \ to \ 1.20505 \ \ ##
So now the error is about 1.0024 in 1.20264 ≅ 0.2%, which is the sum of 0.1% in ##x## and 0.1% in ##(l-x)##

The significance of having the arms matched lies in finding the null near the centre (not exactly at it necessarily.)
If x= 50 cm with an error of 1mm, then (l-x) = 50 cm with an error of 1mm, That's 0.002 or 0.2% for both, giving 0.4% for their ratio.
If x= 40 cm with an error of 1mm, then (l-x) = 60 cm with an error of 1mm. That's 0.0025 and 0.0017 or 0.25% and 0.17% giving 0.42% error for the ratio. Not a lot of difference.
But once you get to 10cm and 90cm, the 1mm error becomes 1% and 0.11%, making the ratio error 1.11%, nearly 3x as big as the mid position.

I think the uniformity of the wire is probably a minor distraction for now, until we get the other uncertainties nailed.

About improving the experiment, I meant of course the setup and for example, maybe shortening the wire would yield better results (I doubt that it would be the case though). In any experiment we perform it is usual that we discuss how to improve the results of the experiment, and this time I don't have many ideas I'm afraid. :).

So you look at the sources of error / uncertainty that you identify in your analysis to determine the accuracy of your result. Then think about what you might do to reduce them or their effect.

If you explain how you arrived at your ±0.06 and ±0.04, then we'll have a list of the sources of error.

 

[ehild]

Other source of error originates from the meter. What was its sensitivity? How much change of the position on the potentiometer caused 1 scale deflection on the meter?