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Finding Resistvity

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data
    calculate the resistivity of the germanium sample
    setup below given the resistance and the dimensions
    http://img25.imagevenue.com/img.php?image=91205_lab5_122_1170lo.JPG


    2. Relevant equations
    Resistivity=(resistance*Length)/(Area)


    3. The attempt at a solution
    Ok so I know how to find the resistivity of the setup if the wires were not on the corners of the sample, but since it is on the corner i know that i have to use an integral, so i have to use an integral right?

    then dResistivity=(Resistance*dx)/A
    putting the sample on a coordinate system
    and after finding the value of y in terms of x
    A = Thickness*(sqrt(50)-2x)
    is this right so far??
    i tried to do this but i keep getting a resistivity that is off from the resistivity of germanium
     
    Last edited: Nov 14, 2008
  2. jcsd
  3. Nov 14, 2008 #2

    LowlyPion

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    And what is the set-up again?
     
  4. Nov 14, 2008 #3

    LowlyPion

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    What is the minimum contact cross section at the corner?

    As that approaches 0, the resistivity approaches infinity doesn't it?
     
  5. Nov 14, 2008 #4
    lol what do you mean?
     
  6. Nov 14, 2008 #5

    LowlyPion

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    When area approaches 0 ...
     
  7. Nov 15, 2008 #6
    ok so there is an infinite resistivity when the area approaches zero do you have any hints on what to do next and how to set up the integral...I'm still kind of lost...do i integrate this from the middle of the square to the corner because of symmetry?
     
  8. Nov 15, 2008 #7

    LowlyPion

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    Are you sure your equation for resistivity is correct is my point.

    http://en.wikipedia.org/wiki/Resistivity
     
  9. Nov 15, 2008 #8
    ooo whoops i wrote the equation down wrong
    i was suppose to do
    dR=(resistivity*dx)/A
    then integrate right? or do i use
    dResistivity=(Resistance*A)/dx?

    ok so i did it with the
    dR=(resistivity*dx)/A
    but i got ln(0) so it approaches -inf??
     
    Last edited: Nov 15, 2008
  10. Nov 15, 2008 #9

    LowlyPion

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    What is it you are trying to calculate?

    The resistivity?

    Or the resistance?

    Your original post says resistivity.
     
  11. Nov 15, 2008 #10
    im given the resistance of 2755Ohms at 300K
    and i'm trying to find the resistivity.
    so i'm guessing since i have the resistance i integrate all the resistances since they are in series so they just add?

    this is what ive done so far
    http://img226.imagevenue.com/img.php?image=91044_lab51_122_395lo.jpg
     
    Last edited: Nov 15, 2008
  12. Nov 15, 2008 #11

    LowlyPion

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    Where did this come from?

    Maybe if you gave a complete statement of the problem it would be a little easier?
     
  13. Nov 15, 2008 #12
    o i'm sorry i shouldve been more specific i actually measured this in a lab and this is one of the questions for my lab report
    so i measured the resistance of the germanium at different temperatures and now i'm just trying to find the resistivity given the setup i posted ealier
    the dimensions of the sample is also on my first post...the only problem now is that i'm not sure how to setup the integral for finding this resistivity
     
  14. Nov 17, 2008 #13
    ok for my end result i calculated a resistivity of .256(Ohm meters) using a resistance of 2755Ohms but the accepted value is .46 so am i doing this correct?
     
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