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- anonymous

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?

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- anonymous

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- anonymous

What methods have you been using?

- anonymous

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

- anonymous

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

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- anonymous

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

- anonymous

brb

- 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

- anonymous

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

- anonymous

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

- anonymous

ok

- 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,\[dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy\]

- 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.

- anonymous

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

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