Wheatstone bridge -- Why set all 4 resistances equal?

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Discussion Overview

The discussion revolves around the sensitivity of the Wheatstone bridge circuit when all four resistances are equal. Participants explore the theoretical underpinnings of the bridge's operation, particularly in relation to the condition for balance and sensitivity in the context of using a galvanometer as a null detector.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants inquire why the Wheatstone bridge is most sensitive when all four resistances (A, B, C, and D) are equal, noting that the balance condition is A/B = C/D.
  • One participant suggests that the current through the galvanometer can be expressed as a function of the four resistances, indicating that sensitivity can be defined as the derivative of current with respect to one of the resistances.
  • Another participant argues that varying the resistances while maintaining the ratio on both sides of the bridge can still yield good results, implying that equal values may not be strictly necessary for effective operation.
  • There is a suggestion to analyze the sensitivity by finding the derivative of current through the galvanometer with respect to a test resistance, indicating a method to determine maximum sensitivity.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of equal resistances for maximum sensitivity, with some supporting the idea while others suggest that maintaining the ratio is sufficient. The discussion remains unresolved regarding the optimal configuration for sensitivity.

Contextual Notes

Participants reference theoretical considerations and mathematical formulations related to the Wheatstone bridge, but there are no explicit resolutions to the assumptions or dependencies involved in their arguments.

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

Solve the bridge for current through the galvanometer ...
 
arvindsharma said:
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.

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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
...​
 
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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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