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Strain Gauge Formula Derivation

  1. Apr 1, 2015 #1
    Using R = ρL/A,
    I am trying to get this formula, which is the change in resistance due to change in length, area, and resistivity.

    ΔR = (∂R/∂L)ΔL + (∂R/∂A)ΔA + (∂R/∂ρ)Δρ

    I understand the above conceptually, but I am confused about why we are multiplying by terms ΔL, ΔA, and Δρ.
    Intuitively, I think it is just ΔR = (∂R/∂L) + (∂R/∂A) + (∂R/∂ρ), as it says "adding up the changes in resistance by changes in length, area, and resistivity equals total resistance change". And by unit analysis, I agree that multiplying by ΔL, ΔA, and Δρ is correct, but can anyone show me how to derive this mathematically?


    Just thought about it some more, and while it doesn't satisfy me, it makes a bit more sense.
    You can just take R = ρL/A and take the partial of it WRT L and get ∂R/∂L = ρ/A.
    Multiply denominator over to get ∂R = (ρ/A)∂L
    Rewrite as ΔR = (ρ/A)ΔL
    Realize that (ρ/A) is ∂R/∂L and substitute back in.
    ΔR = (∂R/∂L)ΔL, then do the same with A and ρ.
    However, this seems kind of backwards...
    Last edited: Apr 1, 2015
  2. jcsd
  3. Apr 1, 2015 #2


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    It's a math concept known as the total derivative:


    You are linearizing the change in resistance R in terms of the changes in length, area, and resistivity of the wire
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