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Help with conductors

  1. Mar 28, 2009 #1
    Help with conductors!!

    1. The problem statement, all variables and given/known data
    Determine the resistance per meter of a hollow cylindrical aluminum conductor with an outer diameter of 32mm and wall thickness 6mm.


    2. Relevant equations
    R = length / sigma * Area
    Sigma = conducitivity

    3. The attempt at a solution
    The answer is 53.4microohms.

    Well I know how the darn equation I supplied works. But how can this problem be solved without the length given it seems like that would need to be supplied? Does anyone know of any tips for this problem.

    Thanks so much
     
  2. jcsd
  3. Mar 28, 2009 #2

    mgb_phys

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    Re: Help with conductors!!

    It asks for ohms/m so the length is 1m !
     
  4. Mar 28, 2009 #3

    Doc Al

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    Re: Help with conductors!!

    You don't need the length. You are finding the resistance per meter, not the resistance.
     
  5. Mar 29, 2009 #4
    Re: Help with conductors!!

    I see the reason why the length is not needed and you can use 1. But in this case is the area... just half of 32 for the radius which is 16mm and since the wall thickness is 6m i would want to take the area from 0 to 10 mm???
     
  6. Mar 29, 2009 #5

    Doc Al

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    Re: Help with conductors!!

    You want the cross-sectional area of the aluminum-filled part of the cylinder. The part from r = 0 to r = 10mm is the empty part.
     
  7. Mar 29, 2009 #6
    Re: Help with conductors!!

    Okay so i understand what your are saying....So the set up should look like this.....

    1/(38.2*10^6*(pi*(6^2))) = 2.3146*10^-10 ???? Which does not even look remotely close to the answer. From what you stated it looks like the empty part is from 0 to 10 and then the radius in total is 16 so thats why i used the 6.
     
  8. Mar 29, 2009 #7

    Doc Al

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    Re: Help with conductors!!

    First find the cross-sectional area by itself. Hint: What's the area of the full cylinder with r = 16mm? What's the area of the hollow part? (Just set it up without crunching the numbers. Use standard units.)
     
  9. Mar 29, 2009 #8
    Re: Help with conductors!!

    Oh, I see ..... Now I understand. We take the total area and take away the hollow part leaving us with the filled area.... then plugging that into the equation gets us the result.....

    Thanks Doc you were such a good help..

    I hope you dont mind....I dont wanna bug you....but could you provide some guidance to my other topic "Help with resistance" I seem to get the right answer, however converting everything to meters I still get the wrong order.... I would really appreciate it

    Thanks a bunch
     
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