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Homework Help: Proof of Wheatstone bridge equation

  1. Dec 26, 2017 #1
    1. The problem statement, all variables and given/known data

    Prove the following equation:

    ## \Delta U=\frac {R_1R_4}{(R_1+R_4)^2}(\frac {\Delta R_1}{R_1}-\frac {\Delta R_2}{R_2}+\frac{\Delta R_3}{R_3}-\frac{\Delta R_4}{R_4})E##

    This is used in Wheatstone bridge

    whets.png

    2. Relevant equations

    U=RI

    3. The attempt at a solution
    This has been a real head-scratcher

    Two voltage dividers can be found for starters. Voltage's direction is assumed to be clockwise

    ##V_{in1}=I_2(R_2+R_3)##

    ##I_2=\frac{V_{in1}}{R_2+R_3}##

    ##V_{out1}=I_2R_3##

    ##V_{out1}=V_{in1}\frac{R_3}{R_2+R_3}##

    Similarly:

    ##V_{out2}=V_{in1}\frac{R_4}{R_1+R_4}##

    ##V_G## is voltage between A and B

    ##V_{out1}-V_{out2}=V_G##

    ##V_{in1}\frac{R_3}{R_2+R_3}-V_{in1}\frac{R_4}{R_1+R_4}=V_G##

    ##V_{in1}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})=V_G##

    ##V_{in1}=E##

    ##V_G=\Delta U## so then

    ##E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})=\Delta U##

    I have calculated voltages in different circuits and tried to think this problem in different ways, but the real problem is that how is ##\Delta R_i## inserted in to equations. Assumption goes that it is added by ##R_i+\Delta R_i##. Maybe that is incorrect?

    Help is very much appreciated!

    edit: Misspelling corrected

    Also particularizing that ##\Delta R_i## is a change in one resistance
     
    Last edited: Dec 26, 2017
  2. jcsd
  3. Dec 26, 2017 #2

    haruspex

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    Which suggests that the Δ in ΔU refers to the consequent change in U, not to the potential difference between A and B at a given set of R values.
     
  4. Dec 26, 2017 #3
    Yes that is true, ##\Delta U## is zero before the change of resistances.
     
  5. Dec 26, 2017 #4

    haruspex

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    So this equation:
    ##E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})=\Delta U##
    Should read
    ##E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})=U##
    and you need a different expression for ##\Delta U##.
     
  6. Dec 27, 2017 #5
    Right now I'm trying to figure out why this would not be possible:

    ##E(\frac{R_3+\Delta R_3}{R_2+\Delta R_2+R_3+\Delta R_3}-\frac{R_4+\Delta R_4}{R_1+\Delta R_1+R_4+\Delta R_4})=\Delta U##

    I can simplify it a bit, but is this the right way to go
     
  7. Dec 27, 2017 #6

    haruspex

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    I do not see how you get that. It looks wrong.
    You have an equation for U (second eqn in post #4). Write out the corresponding eqn for U+ΔU.
     
  8. Dec 29, 2017 #7
    Now that I thought about it, simply adding the change does not make so much sense.

    But then I got an idea to take partial derivates since it is about change. Adding those partial derivates together should give the overall change in voltage.

    ##U=E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})##

    Seems like marking ##U## as ##U_{BA}## is needed since I took potential difference with ##V_{out1}-V_{out2}=V_G##

    ##U_{BA}=(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E##

    This should be legal: ##\Delta U_{BA}=dU_{BA}##

    Thus ##dU_{BA}=\frac {\partial} {\partial R_1}U_{BA}\Delta R_1+\frac {\partial} {\partial R_2}U_{BA}\Delta R_2+\frac {\partial} {\partial R_3}U_{BA}\Delta R_3+\frac {\partial} {\partial R_4}U_{BA}\Delta R_4##

    Solving partial derivates each:

    ##\frac {\partial} {\partial R_1}U_{BA}\Delta R_1=\frac {\partial} {\partial R_1}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E\Delta R_1=-\frac {R_4}{(R_1+R_4)^2}E\Delta R_1##

    ##\frac {\partial} {\partial R_2}U_{BA}\Delta R_2=\frac {\partial} {\partial R_2}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E\Delta R_2=\frac{R_3}{(R_2+R_3)^2}E\Delta R_2##

    ##\frac {\partial} {\partial R_3}U_{BA}\Delta R_3=\frac {\partial} {\partial R_3}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E\Delta R_3=-\frac{R_2}{(R_2+R_3)^2}E\Delta R_3##

    ##\frac {\partial} {\partial R_4}U_{BA}\Delta R_4=\frac {\partial} {\partial R_4}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E\Delta R_4=\frac{R_1}{(R_1+R_4)^2}E\Delta R_4##

    ##dU_{BA}=-\frac {R_4}{(R_1+R_4)^2}E\Delta R_1+\frac{R_3}{(R_2+R_3)^2}E\Delta R_2-\frac{R_2}{(R_2+R_3)^2}E\Delta R_3+\frac{R_1}{(R_1+R_4)^2}E\Delta R_4##

    ##dU_{BA}=(-\frac{R_4}{R_1(1+\frac{R_4}{R_1})^2}\frac{\Delta R_1}{R_1}+\frac{R_3}{R_2(1+\frac{R_4}{R_1})^2}\frac{\Delta R_2}{R_2}-\frac{R_2}{R_3(1+\frac{R_1}{R_4})^2}\frac{\Delta R_3}{R_3}+\frac{R_1}{R_4(1+\frac{R_1}{R_4})^2}\frac{\Delta R_4}{R_4})E##

    I just kept playing with the identity ##\frac{R_2}{R_3}=\frac{R_1}{R_4}## and I got:

    ##dU_{BA}=\frac{R_1R_4}{(R_1+R_4)^2}(-\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}-\frac{\Delta R_3}{R_3}+\frac{\Delta R_4}{R_4})E##

    It has wrong signs because of ##V_{out1}-V_{out2}=V_G##

    So I think the equation is about ##U_{AB}##

    ##\Delta U_{AB}=dU_{AB}=-dU_{BA}=-\frac{R_1R_4}{(R_1+R_4)^2}(-\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}-\frac{\Delta R_3}{R_3}+\frac{\Delta R_4}{R_4})E##

    ##=\frac{R_1R_4}{(R_1+R_4)^2}(\frac{\Delta R_1}{R_1}-\frac{\Delta R_2}{R_2}+\frac{\Delta R_3}{R_3}-\frac{\Delta R_4}{R_4})E##
     
  9. Dec 29, 2017 #8

    haruspex

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    Looks good.
     
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