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How to ameliorate this Wheatstone bridge?

  1. Dec 13, 2016 #1
    1. The problem statement, all variables and given/known data

    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).

    2. Relevant 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##

    3. 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.
     

    Attached Files:

  2. jcsd
  3. Dec 13, 2016 #2

    Merlin3189

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    Gold Member

    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?)


    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.

    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.
     
  4. Dec 13, 2016 #3

    ehild

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    Homework Helper
    Gold Member

    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?
     
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