How Do We Calculate Equivalent Resistance with Measurement Errors?

[Quantumkid]

we write \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} where R is the equivalent resistance of R_1 and R_2.

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

EDIT: Corrected

 

[mo_0820]

There is an error in the measurements of R_1 and R_2 of \Delta R_1 and \Delta R_2 respectively.
It is correct that \frac{\Delta R}{R^2}=\frac{\Delta R_1}{{R_1}^2}+\frac{\Delta R_2}{{R_2}^2}

 

[Quantumkid]

Bump!
Nobody tried to solve it.

 

[Gear.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:

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

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

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".

 

[dulrich]

BTW, this

\frac{\partial R}{\partial R_{1}} = \frac{1}{R_{1}^{2}} (R_{1}^{-1} + R_{2}^{-1})^{-2}

reduces to

\frac{\partial R}{\partial R_{1}} = \frac{R_{2}^{2}}{(R_{1} + R_{2})^{2}}

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.