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Homework Help: Error Expressions+Capacitance

  1. Feb 26, 2005 #1

    say I have something like Cp=C1+C2+C3 for the equivalent parallel combination capacitance.

    When doing the error expression, I figure we just use the addition rule in the case above.

    So, Delta(Cp)=Delta(C1)+Delta(C2)+Delta(C3).

    Apparently this is wrong and I need to use squares/roots somewhere but I don't see that.

    Help! :mad:
  2. jcsd
  3. Feb 26, 2005 #2
    Someone verify? *bump*
  4. Feb 26, 2005 #3
    I would appreciate any tips here lol, kind of urgent for my lab :yuck:
  5. Feb 27, 2005 #4
    Ok I looked at this thread and though 3 post someone must of answered the question. I guess not.

    What you did is not right you need to used Pythagorean theorem. The error does not add linearly like that. The Pythagorean theorem is not exactly correct, however it is very widely excepted. If we had:
    A\pm \delta A = B\pm \delta B +C \pm \delta C
    This produces:
    A \pm \delta(\sqrt{\delta{B}^2 +\delta{C}^2 })
    Three varibles can be derived by dA=dC+(dD+dE). With three variables we would have.
    \delta CP = \sqrt{\delta C_1^2 +\delta C_2^2+\delta C_3^2}

    This is usually in any basic undergrad statistics books.

    Does this make sense?
    Last edited: Feb 27, 2005
  6. Feb 27, 2005 #5
    Ok I understand that you used the pythagorean theorem, but my question is why? What makes this error expression for Cp so different from simple addition of the errors that we have to use the theorem? Thanks.
  7. Feb 27, 2005 #6
    Where have you seen direct addition of error before?

    It is used because when you have the value [itex]1+\pm 1[/itex] it could be any where from 0 to 2. When you have the value [itex]1+2 \pm 1 \pm 3[/itex] it could be any where between -1 and 7, however it is more likely to be in the middle then the edges. The extreme cases are less probable so to get the extreme case after the addition you must have two improbable things happen, which is much less likely then either one individually.

    event A happens 1/10 of the time
    event b happens 1/20 of the time

    So the event of A and B happening are:

    Does this make sense?
  8. Feb 27, 2005 #7
    Addition of errors is actually what we often use in the lab. For ex. z=x+y, dz=dx+dy. Well as for your explanation, I got up to your example, but failed to see connections from that point on. Anyway, what I was interested in was the expression and verification that addition of errors is wrong for this case.
  9. Feb 27, 2005 #8
    So you are satisfied with the expression?

    For the expression [itex]1+\pm 1[/itex] there is not an equal probability of finding the "real" answer anywhere between [itex]1+\pm 1[/itex]. In fact the "real" answer does not even have to be found in 0 to 2 for the above example. [itex]\pm 1[/itex] represents a certain confidence level that the measurement/answer is in that range. To my knowledge the most used standard is 90% confidence level. So the expression [itex]1+\pm 1[/itex] could be said as: “The method of measurement used allows for 90% confidence that that the “true” value is 1 plus or minus 1.”

    My example was just trying to point out how(Taking [itex]1+\pm 1[/itex] as an example) the real value equaling 2 or 0 and the extreme cases and are therefore less likely then .5 or 1.5.

    I was just trying to show with example that two extreme cases does not equal another extreme case. The relationship is not linear. The error where 90% confidence is obtained shrinks after adding two random errors together. Each error has a Gaussian probability distribution if two identical Gaussian distribution are multiplied together you get a more narrow Gaussian distribution.

    I just remembered with my last sentence that a requirement that this method is used is that the two errors be random and independent from each other.
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