Find Resistivity of Germanium Sample Setup

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The discussion centers on calculating the resistivity of a germanium sample based on its resistance and dimensions. The user is attempting to apply the formula for resistivity but struggles with the setup due to the placement of wires at the corners of the sample, which complicates the area calculation. There is confusion regarding the correct integral setup and whether to integrate from the center to the corner of the sample. The user measured the resistance at 2755 Ohms and calculated a resistivity of 0.256 Ohm-meters, which is lower than the accepted value of 0.46 Ohm-meters, indicating potential errors in their calculations. The conversation highlights the need for clarity in the problem statement and proper integration techniques to achieve accurate results.
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Homework Statement


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

Homework Equations


Resistivity=(resistance*Length)/(Area)

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
 
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And what is the set-up again?
 
What is the minimum contact cross section at the corner?

As that approaches 0, the resistivity approaches infinity doesn't it?
 
LowlyPion said:
What is the minimum contact cross section at the corner?

As that approaches 0, the resistivity approaches infinity doesn't it?

lol what do you mean?
 
atwong713 said:
lol what do you mean?

Resistivity=(resistance*Length)/(Area)

When area approaches 0 ...
 
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?
 
atwong713 said:
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?

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

http://en.wikipedia.org/wiki/Resistivity
 
LowlyPion said:
Are you sure your equation for resistivity is correct is my point.

http://en.wikipedia.org/wiki/Resistivity

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??
 
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What is it you are trying to calculate?

The resistivity?

Or the resistance?

Your original post says resistivity.
 
  • #10
LowlyPion said:
What is it you are trying to calculate?

The resistivity?

Or the resistance?

Your original post says resistivity.

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 I've done so far
http://img226.imagevenue.com/img.php?image=91044_lab51_122_395lo.jpg
 
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  • #11
atwong713 said:
im given the resistance of 2755Ohms at 300K
and I'm trying to find the resistivity.

Where did this come from?

Maybe if you gave a complete statement of the problem it would be a little easier?
 
  • #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
 
  • #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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