# Errors and analysis

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

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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 $$\frac{\Delta R}{R^2}=\frac{\Delta R_1}{{R_1}^2}+\frac{\Delta R_2}{{R_2}^2}$$

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Bump!!:surprised
Nobody tried to solve it.

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".

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.