# Errors and analysis

1. Jul 4, 2010

### 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

Last edited: Jul 5, 2010
2. Jul 4, 2010

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

Last edited: Jul 4, 2010
3. Jul 13, 2010

### Quantumkid

Bump!!:surprised
Nobody tried to solve it.

4. Jul 13, 2010

### 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".

5. Jul 13, 2010

### dulrich

BTW, this
reduces to

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

Also, this reference
is really great. Thanks.