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Propagation of uncertainty problem

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the equivalent resistance viewed by A and B and its equivalent uncertainty:
    http://img707.imageshack.us/img707/5040/dadadadaz.png [Broken]

    R1 = 10ohm, 5% tolerance.
    R2 = 2ohm, 1% tolerance.
    R3 = 5ohm, 5% tolerance.
    R4 = 15ohm, 1% tolerance.

    3. The attempt at a solution

    The Req = (1/R1 + 1/R2)-1 + (1/R3 + 1/R4)-1 = 65/12ohm = 5,41ohm.

    Now, for the uncertainty, I separated the problem in two: first, find RA (10/6ohm), the equivalent resistance of the first two resistors and RB (15/4ohm), the equivalent resistance of the second two resistors, and then add them.

    I was given a formula for propagation of uncertainty for parallel resistors and for series resistors, based in the formula for a function f(x1, x2,..., xk): [tex]\mu[/tex]f = [tex]\sum[/tex]|(xk/f)*(f'xk)|*[tex]\mu[/tex]xk:

    For two parallel resistors: [tex]\mu[/tex] = (R2/(R1+R2))*[tex]\mu[/tex]R1 + (R1/(R1+R2))*[tex]\mu[/tex]R2

    For two series resistors: [tex]\mu[/tex] = (R1/(R1+R2))*[tex]\mu[/tex]R1 + (R2/(R1+R2))*[tex]\mu[/tex]R2

    If I do the calculus, I get that the total uncertainty is 3,28%: the uncertainty of A is 2/(10+2)*0.05 + 10/(10+2)*0.01 = 1/60, and the one of B is 15/(15+5)*0.05 + 5/(15+5)*0.01 = 1/25. Then the total uncertainty is (10/6)/(65/12)*1/60 + (15/4)/(65/12)*1/25 = 0.0328 = 3.28%.

    But the answer given by the problem says it has to be 8.7%.

    I'd like to know if I'm using the formulas right, or if I should be doing something with the uncertainties instead of that.

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 9, 2010 #2
    Anybody?
     
  4. May 9, 2010 #3

    ehild

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    Homework Helper
    Gold Member

    I got the same result as you.

    ehild
     
  5. May 9, 2010 #4
    Then the answer I was given is wrong? Or is there another method I could have used to reach that answer?

    Thanks.
     
    Last edited: May 9, 2010
  6. May 10, 2010 #5

    ehild

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    Homework Helper
    Gold Member

    Sometimes wrong answers are given. Your calculation was correct but a bit overcomplicated.

    Unless the function is a product or fractions of its variables, calculate with absolute uncertainties instead of relative ones.

    [tex]\Delta f = \Sigma (|\frac{\partial f }{\partial x_i}|\Delta x_i)[/tex]

    Divide the absolute error with the value of f if relative uncertainty is needed.




    ehild
     
    Last edited: May 10, 2010
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