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Wheatstone bridge -- Why set all 4 resistances equal?

  1. Mar 15, 2015 #1
    Can someone explain me,why Wheatstone bridge is most sensitive when all four resistances say A,B,C and D are equal?as far as i know condition for Wheatstone Bridge is A/B=C/D.
     
  2. jcsd
  3. Mar 15, 2015 #2

    jim hardy

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    Wheatstone bridges used to employ a galvanometer as a null detector.

    Solve the bridge for current through the galvanometer .....
     
  4. Mar 15, 2015 #3
    bc914d21-7037-4633-9c3a-d2797649e12e.gif

    Recipe:

    From theoretical considerations of the bridge circuit, current through the ideal galvanometer can be understand as function of 4 variables:
    IAC=V⋅f(R1,R2,R3,R4)​
    Definition of the sensitivity is:
    S=dIAC/dRx
    where Rx can stand for any of R1,R2,R3 or R4.

    Say Rx=R4 and given the condition of the balance R4=R2⋅R3/R1, one searches for the stationary points of multivariable function S by solving the system:

    ∂S/∂R1=0
    ∂S/∂R2=0
    ∂S/∂R3=0
    ...​
     
  5. Mar 15, 2015 #4
    You can vary the thing and still get good results. i.e. the right leg has the same ratio as the left leg is good and sufficient.

    To see why the system works better with equal values, find di / dRtest (i in galvenometer) as a function of the values with one value, Rtest set. Then set di/dR to zero to find the maximum sensitivity...
     
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