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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?
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What methods have you been using?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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
 5 years ago
Best ResponseYou've already chosen the best response.0You then multiply by R to get Delta R  your error.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have been using differentials to approximate error but i dont know how to differentiate an equation like this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in an example i have we used partial derivatives to R with respect to R3, but we didnt try to find the maximum error

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...i know what's going on. do you need an answer right this minute? i have to go for a while.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont need it right this minute but it would be best if i could get it tonight at some point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[dR=\frac{9}{1600} \Omega \rightarrow dR=\frac{9}{1600}\Omega\]but check that  I'm rushing.
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