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.