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Homework Help: Errors and analysis

  1. Jul 4, 2010 #1
    we write [tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex] where R is the equivalent resistance of [tex]R_1[/tex] and [tex]R_2[/tex].

    Let there is an error in the measurements of [tex]R_1[/tex] and [tex]R_2[/tex] of [tex]\pm[/tex] [tex]\Delta R_1[/tex] and
    [tex]\pm [/tex][tex]\Delta R_2[/tex] respectively.
    Is it correct that
    [tex]\frac{\Delta R}{R^2}=\frac{\Delta R_1}{{R_1}^2}+\frac{\Delta R_2}{{R_2}^2}[/tex] ???

    EDIT: Corrected
     
    Last edited: Jul 5, 2010
  2. jcsd
  3. Jul 4, 2010 #2
    There is an error in the measurements of [tex]R_1[/tex] and [tex]R_2[/tex] of [tex]\Delta R_1[/tex] and [tex]\Delta R_2[/tex] respectively.
    It is correct that [tex]\frac{\Delta R}{R^2}=\frac{\Delta R_1}{{R_1}^2}+\frac{\Delta R_2}{{R_2}^2}[/tex]
     
    Last edited: Jul 4, 2010
  4. Jul 13, 2010 #3
    Bump!!:surprised
    Nobody tried to solve it.
     
  5. Jul 13, 2010 #4
    I'm a bit new to this but I believe what you want is propagation of error, which in this case would be given by:

    [tex]\Delta R = \sqrt{(\frac{\partial R}{\partial R_{1}} \Delta R_{1})^{2} + (\frac{\partial R}{\partial R_{2}} \Delta R_{2})^{2}}[/tex]

    and the derivatives would be:
    [tex]\frac{\partial R}{\partial R_{1}} = \frac{1}{R_{1}^{2}} (R_{1}^{-1} + R_{2}^{-1})^{-2}[/tex]

    and similarly for R2.


    The best online explanation I could find is here:
    http://teacher.pas.rochester.edu/PHY_LABS/AppendixB/AppendixB.html
    scroll down to almost the bottom where is has the title "Propagation of Errors".
     
  6. Jul 13, 2010 #5
    BTW, this
    reduces to

    [tex]\frac{\partial R}{\partial R_{1}} = \frac{R_{2}^{2}}{(R_{1} + R_{2})^{2}}[/tex]

    Also, this reference
    is really great. Thanks.
     
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