Method for determining error in an equation

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To estimate the output voltage (V_out) range for the equation V_out = 0.5(1 + R4/R3)(V1+V2) with R3 and R4 having a ±5% tolerance, one can use partial derivatives to calculate the uncertainty in V_out. Assuming V1 and V2 are known constants, the uncertainty can be determined by applying the formula dV_out = √(∑(∂V_out/∂R_i * dR_i)²) for R3 and R4. This method allows for the assessment of how variations in R3 and R4 affect V_out. A user calculated a 12.8% uncertainty but was advised to check for a factor of √2 in their calculations. Understanding this approach is essential for accurately determining the tolerance of V_out based on the tolerances of R3 and R4.
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Homework Statement


Ok, so suppose I have an equation like V_out = 0.5(1 + R4/R3)(V1+V2) and I know the R3 and R4 has a tolerance of +-5%. For such an equation and similar ones, how would I estimate the possible range of values V_out? For example, I'd like to find out something like V_out_theoretical * 96% < V_out < V_out_theoretical * 104% or rather, I want to figure out the "tolerance" (no, I don't think that's the proper word) for V_out based on the known tolerances of R3 and R4.

Sorry, I don't know the proper terminology for this sort of stuff.

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First are You assuming V1 and V2 are known exactly? Secondly do You know how to take derivatives?
 
You can assume that V1 and V2 are fixed. And yes, I can take derivatives.
 
the commonly accepted way to determine the uncertainty for V_out is to determine the derivatives of V_out with respect to R3 and R4 to determine the changes in V_out wrt R3 with R4 held fixed and wrt R4 with R3 held fixed. Using dR3 and dR4 as the uncertainty in this case the tolerances substitute into the following equation.

dV_{out} = \sqrt{\sum_{i=3}^{4} \left \lfloor \frac{\partial V_{out}}{\partial R_{i}}dR_{i}\right\rfloor^{2}}

the two terms in the square root represent the change in V_out due to a change in R3 and R4.

For a more detailed explanation see "Data Reduction and Error Analysis for the Physical Sciences" by Bevington and Robinson
https://books.google.com/books/abou...s_for_th.html?id=0poQAQAAIAAJ&source=kp_cover
 
So partial derivatives? Finally! I get to use Calculus III stuff!
 
So if R3 and R4 is 10000 Ohms, and V1 is 1.8 and V2 is 3.3, I get 12.8%. Have I done things correctly?
 
I think you are off by a factor of √2 can you find where you missed it?
 

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