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Errors and analysis

  • Thread starter Quantumkid
  • Start date
  • #1
we write [tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex] where R is the equivalent resistance of [tex]R_1[/tex] and [tex]R_2[/tex].

Let there is an error in the measurements of [tex]R_1[/tex] and [tex]R_2[/tex] of [tex]\pm[/tex] [tex]\Delta R_1[/tex] and
[tex]\pm [/tex][tex]\Delta R_2[/tex] respectively.
Is it correct that
[tex]\frac{\Delta R}{R^2}=\frac{\Delta R_1}{{R_1}^2}+\frac{\Delta R_2}{{R_2}^2}[/tex] ???

EDIT: Corrected
 
Last edited:

Answers and Replies

  • #2
24
0
There is an error in the measurements of [tex]R_1[/tex] and [tex]R_2[/tex] of [tex]\Delta R_1[/tex] and [tex]\Delta R_2[/tex] respectively.
It is correct that [tex]\frac{\Delta R}{R^2}=\frac{\Delta R_1}{{R_1}^2}+\frac{\Delta R_2}{{R_2}^2}[/tex]
 
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  • #3
Bump!!:surprised
Nobody tried to solve it.
 
  • #4
66
0
I'm a bit new to this but I believe what you want is propagation of error, which in this case would be given by:

[tex]\Delta R = \sqrt{(\frac{\partial R}{\partial R_{1}} \Delta R_{1})^{2} + (\frac{\partial R}{\partial R_{2}} \Delta R_{2})^{2}}[/tex]

and the derivatives would be:
[tex]\frac{\partial R}{\partial R_{1}} = \frac{1}{R_{1}^{2}} (R_{1}^{-1} + R_{2}^{-1})^{-2}[/tex]

and similarly for R2.


The best online explanation I could find is here:
http://teacher.pas.rochester.edu/PHY_LABS/AppendixB/AppendixB.html
scroll down to almost the bottom where is has the title "Propagation of Errors".
 
  • #5
133
0
BTW, this
[tex]\frac{\partial R}{\partial R_{1}} = \frac{1}{R_{1}^{2}} (R_{1}^{-1} + R_{2}^{-1})^{-2}[/tex]
reduces to

[tex]\frac{\partial R}{\partial R_{1}} = \frac{R_{2}^{2}}{(R_{1} + R_{2})^{2}}[/tex]

Also, this reference
The best online explanation I could find is here:
http://teacher.pas.rochester.edu/PHY_LABS/AppendixB/AppendixB.html
scroll down to almost the bottom where is has the title "Propagation of Errors".
is really great. Thanks.
 

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