## anonymous 5 years ago 1/R=1/R1+1/R2+1/R3 If R1= 5ohms, R2=4ohms, and R3=10 ohms and these measurements are accurate to 0.05 ohms, how do i find the maximum possible error?

1. anonymous

What methods have you been using?

2. anonymous

It's been a long time since I've worked with errors in physics. I remember something along the lines of $\frac{\Delta R}{R}=\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}+\frac{\Delta R_3}{R_3}$

3. anonymous

You then multiply by R to get Delta R - your error.

4. anonymous

we have been using differentials to approximate error but i dont know how to differentiate an equation like this

5. anonymous

brb

6. anonymous

in an example i have we used partial derivatives to R with respect to R3, but we didnt try to find the maximum error

7. anonymous

ok...i know what's going on. do you need an answer right this minute? i have to go for a while.

8. anonymous

I dont need it right this minute but it would be best if i could get it tonight at some point

9. anonymous

ok

10. anonymous

If you have a function in several variables, say,$z=f(x,y)$the full differential in z is given by the chain rule. We may take the derivative of z to some dummy variable, t, then$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$so that,$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

11. anonymous

The maximum error is estimated as such:$dR=-\frac{1}{R_1^2}dR_1-\frac{1}{R_2^2}dR_2-\frac{1}{R_3^2}dR_3$from what was discussed.

12. anonymous

$dR=-\frac{9}{1600} \Omega \rightarrow |dR|=\frac{9}{1600}\Omega$but check that - I'm rushing.