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Homework Help: R through a wire: What's the max error in diameter if your R has 1% uncertainty?

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a wire of some resistance, resistivity, and length.

    What is the maximum error in measuring the diameter that you can have if the resistance is to have 1% uncertainty?

    2. Relevant equations

    R = [itex]\frac{ρ*l}{\pi*r^{2}}[/itex] for which I substituted the area as [itex]\pi[/itex](.5 D) ^{2}.

    3. The attempt at a solution

    I picked an arbitrary R and found the corresponding D. Then, I took 1% of that arbitrary R and found the corresponding D. The issue is that D's were basically the same number. I'm not sure if I could just go out to more decimal places or if I'm just wrong....
    Last edited: Jan 24, 2013
  2. jcsd
  3. Jan 24, 2013 #2


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    You meant R = [itex]\frac{ρl}{\pi*r^{2}}[/itex] , didn't you?

    Solve the problem symbolically. If the resistance can differ from R by one percent it can be 1.01 R or 0.99 R. What are the corresponding diameters and how are they related to the original D? By what percent are they different from it?

  4. Jan 24, 2013 #3
    Typo, thank you!

    Big help!!! Turns out I was off by a decimal point. My answer makes a lot more sense now. Thank you very, very much.
  5. Jan 24, 2013 #4
    Another clarification-- the D in the denominator is squared. Does that mean I need to take the square root of my result?
  6. Jan 24, 2013 #5


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    What is your result at all?

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