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Percentage Errors

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data

    What are the percentage errors for lengths measured with meter rulers and potential differences measured with voltmeters?

    I have a question asking me which of the two measurements can be measured more accurately, based on which one's percentage error is lower.



    2. Relevant equations



    3. The attempt at a solution

    Is it done this way:

    For length, % error = 1/1000 *100

    For p.d., % error = 1/10 *100

    ???

    Please clarify, thanks.
     
  2. jcsd
  3. Oct 7, 2007 #2

    Kurdt

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    Percentage error is given by the formula:

    % error = (|measured value - actual value|/actual value)*100

    It would help if you posted the question in full since we can't guess what numbers you are supposed to be using.
     
  4. Oct 7, 2007 #3
    The measurements for the lengths are:
    0.05m, 0.15m, 0.30m, 0.45m, 0.60m, 0.75m and 0.90m

    The measurements for the p.d.'s are:
    0.05V, 0.10V, 0.20V, 0.30V, 0.40V, 0.50V and 0.60V

    I guess these are the measured values for the equation, but what would be the actual values then?
     
  5. Oct 7, 2007 #4

    Kurdt

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    Is that the question in full? If not could you post it. I would expect in a question like this that you would be given the true value and the value that people have measured unless I've misinterpreted.
     
  6. Oct 7, 2007 #5
    Sorry for being unclear, but this is an experiment with a potentiometer actually. The measurements were what I observed during the experiment; none were given. The question following the instructions asks "Which can you measure more accurately, L or V; ie, which has the smaller percentage error?"

    Hope this helps.
     
  7. Oct 7, 2007 #6

    Kurdt

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    In that case the actual value will be the theoretical or expected value.
     
  8. Oct 7, 2007 #7
    Is that the mean?
     
  9. Oct 7, 2007 #8

    Kurdt

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    If you took more than one measurement for each length (You're normally advised to take at least 3 in a lab), then there is another method of obtaining the percentage error in measurement.

    % error = (1/2)*((max value - min value)/ avg. value)

    Since I do not know what was expected of you in your lab I couldn't tell you which method to use. I'm beginning to think this one sounds the best at the minute.

    What information have you been given with your lab about errors?
     
  10. Oct 7, 2007 #9
    All the experiment tells me to do is to observe and record the p.d. reading and the distance the contact point is touching, and to use the whole range of the potentiometer wire to record a series of readings for the p.d. and the length of the wire. I did do the experiment two or three times to back-check the p.d readings and I got the exact same values each time.

    Honestly, I thought the question about accuracy in measurements was just about the scale divisions, and that the length would be measured more accurately than the voltage because of this.

    The lab mentions nothing else about errors. It just goes on to tell me to plot a graph of V against L.
     
    Last edited: Oct 7, 2007
  11. Oct 7, 2007 #10

    Kurdt

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    Ok chasing the problem down. In that case you will probably be correct in that it comes down to scale divisions.
     
  12. Oct 7, 2007 #11
    Ok, so if that is the correct way, how would this be done then? Is it anything close to what I suggested in my original post?
     
  13. Oct 8, 2007 #12
    Can anyone help here?
     
  14. Oct 8, 2007 #13

    Kurdt

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    The problem with this question is that percentage error could mean one of a number of things. I'm surprised your lab did not give you any guidance about how to report errors. If you wish I can mail a handbook about the treatment of errors in the lab to you, but I would suggest you ask an advisor as to what they expect when you report errors in your write up.
     
  15. Oct 8, 2007 #14
    Ok. Thanks for your help though.
     
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