Error Propagation: Calculating Uncertainty of a Ratio

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SUMMARY

This discussion focuses on error propagation in calculating the uncertainty of a ratio, specifically the formula for the ratio of two resistances, R1 and R2, with given uncertainties. The user references an example from the RIT website, which illustrates how to calculate uncertainty using relative errors. The key takeaway is that the uncertainty in the ratio is derived from the sum of the relative uncertainties of the individual measurements, leading to a final uncertainty value that is critical for accurate results. The user expresses confusion regarding the source of the uncertainty value used in the example, specifically the calculation of 0.3.

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Error propagation

Homework Statement


Calculate:

[tex]\frac{ - \frac{R_{2}}{R_{1}}}{1 + \frac{1}{A} + \frac{R_{2}}{A R_{1}}}[/tex]

Homework Equations



[tex]R_{1} = 10000 \pm 5 \%[/tex]
[tex]R_{2} = 10000 \pm 5 \%[/tex]
[tex]A = 1000[/tex]

The Attempt at a Solution



I try to follow the example of at the website http://www.rit.edu/~uphysics/uncertainties/Uncertaintiespart1.html and in there example
[tex]x = ( 2.0 \pm 0.2)[/tex]
[tex]y = (3.0 \pm 0.6)[/tex]
[tex]z = \frac{x}{y}[/tex]

This is what they do in their example:
[tex]z = \frac{2.0}{3.0} = 0.6667[/tex]
[tex]\Delta z = 0.3 (0.6667 ) = 0.2[/tex]
[tex]z = (0.7 \pm 0.2)[/tex]

Now what i don't really understand is where they get [tex]0.3[/tex] from?
It seems that they just divide the uncertainty [tex]\frac{0.2}{0.6} = .33[/tex].

But, if i do this in my example i get [tex]\frac{500}{500} = 1[/tex]. Then when i multiply this agianst [tex]\frac{10000}{10000} = 1[/tex] i get 100% error. Yikes!

I kind of feel embarrassed asking this because i should have learned this a long time ago in physics but it was one of those things i never really took the time to actually understand.
 
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Now what i don't really understand is where they get from?
It seems that they just divide the uncertainty
The 0.3 comes from the addition of the relative errors of x and y. Find equation 2a on that page.
 
Thank you sir
 

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