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different numerical and analytical resistance values

Posted 7 mai 2015, 08:48 UTC−4 Low-Frequency Electromagnetics Version 5.0 8 Replies

Aden
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Hello!

I have a reallllllly simple problem which I can't figure it out :(
here is the very veryyyyy simple version of my question.
In EC physics, I have a cylinder with constant voltage through it, I have calculated the constant current passing through the terminal and then according to the Ohm' law, I calculate the resistance: R=V/I

then, when I calculate the resitance analytically,
resistance= length/(electrical conductivity*area) , I always get a different value :(

does anyone have any idea , where the problem is ??

Regards
8 Replies Last Post 19 mai 2015, 04:54 UTC−4
Jeff Hiller COMSOL Employee

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Posted: 1 decade ago 7 mai 2015, 09:05 UTC−4
Hi Aden,
First method:
R=V/I=20/.628=31.8 Ohms
Second method:
R=L/(sigma*A)=L/(sigma*pi*r^2)=10/(.1*pi*1^2)=31.8 Ohms
Jeff
Hi Aden, First method: R=V/I=20/.628=31.8 Ohms Second method: R=L/(sigma*A)=L/(sigma*pi*r^2)=10/(.1*pi*1^2)=31.8 Ohms Jeff
Aden
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Posted: 1 decade ago 7 mai 2015, 09:15 UTC−4
Dear Jeff

I can't believe it :)
I got it!!! thanks for great help!!!! :)

Dear Jeff I can't believe it :) I got it!!! thanks for great help!!!! :)
Aden
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Posted: 1 decade ago 15 mai 2015, 07:42 UTC−4
Dear Jiff

Here I come with another problem :)
now I have 3 cylinders, which the size of the middle one is variable.

Cylinder 1: (R=1µm, L=1µm)
Cylinder 2: (r=0.1µm...1µm, l=0.1µm...1µm)
Cylinder 3: (R=1µm, L=1µm)

then with r=l= 0.1µm, the calculated R=V/I is twice the analytical solution (R=L/(sigma*A).
however for r=l=1µm , both solutions give almost the same answer.
do you have any idea where this difference come from??
(bcs of limits on file size, the attached file is not run yet).

Regards
Dear Jiff Here I come with another problem :) now I have 3 cylinders, which the size of the middle one is variable. Cylinder 1: (R=1µm, L=1µm) Cylinder 2: (r=0.1µm...1µm, l=0.1µm...1µm) Cylinder 3: (R=1µm, L=1µm) then with r=l= 0.1µm, the calculated R=V/I is twice the analytical solution (R=L/(sigma*A). however for r=l=1µm , both solutions give almost the same answer. do you have any idea where this difference come from?? (bcs of limits on file size, the attached file is not run yet). Regards

Attachments:

Jeff Hiller COMSOL Employee

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Posted: 1 decade ago 15 mai 2015, 08:07 UTC−4
The formula applies to a single cylinder, not to a structure of varying radius.
The formula applies to a single cylinder, not to a structure of varying radius.
Daniel Connelly
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Posted: 1 decade ago 15 mai 2015, 14:07 UTC−4
Yes -- to be clear, the formula is based on the assumption the current is flowing perpendicular to the area and is equally distributed across the area. However, upon entering or exiting the narrow cylinder, the current is squeezed into a tighter distribution. Since it can't spread out, the resistance will be greater.

Consider the current spreading out from a hemisphere of radius rs into infinity. You can calculate the resistance as the integral from rs to infinity dr / (2 pi sigma r^2) where sigma is the conductivity. The definite integral is 1 / (2 pi sigma rs). The same is for current flowing into a hemisphere of the same radius.

If I treat the cylindrical cap of radius rc as a hemisphere of the same area, I get:

2 pi rs^2 = pi rc^2 => rs = rc/sqrt(2)

So then the resistance is 1 / [ sqrt(2) pi sigma rc ]

Summing these components into and out of the narrow cylinder I get:

sqrt(2) / [ pi sigma rc ]

so in the limit of the inner cylinder being much narrower than the outer cylinders, and the inner cylinder being of length L, I should get a total resistance of approximately, modeling the end-cap of the cylinders as a hemisphere of the same area, of:

[ sqrt(2) / (pi rc) + lc / (pi rc^2) ] / sigma

Hopefully I did that right....

The key point here is there's a "spreading resistance" term to the total resistance.

In the limit where you had infinitely thin slices of cylinders of radius R1 and R2 such that R1 < R2 the current would never be able to spread to a radius wider than R1 and the total resistance would be equivalent to a cylinder of the same total length with radius R1.

Yes -- to be clear, the formula is based on the assumption the current is flowing perpendicular to the area and is equally distributed across the area. However, upon entering or exiting the narrow cylinder, the current is squeezed into a tighter distribution. Since it can't spread out, the resistance will be greater. Consider the current spreading out from a hemisphere of radius rs into infinity. You can calculate the resistance as the integral from rs to infinity dr / (2 pi sigma r^2) where sigma is the conductivity. The definite integral is 1 / (2 pi sigma rs). The same is for current flowing into a hemisphere of the same radius. If I treat the cylindrical cap of radius rc as a hemisphere of the same area, I get: 2 pi rs^2 = pi rc^2 => rs = rc/sqrt(2) So then the resistance is 1 / [ sqrt(2) pi sigma rc ] Summing these components into and out of the narrow cylinder I get: sqrt(2) / [ pi sigma rc ] so in the limit of the inner cylinder being much narrower than the outer cylinders, and the inner cylinder being of length L, I should get a total resistance of approximately, modeling the end-cap of the cylinders as a hemisphere of the same area, of: [ sqrt(2) / (pi rc) + lc / (pi rc^2) ] / sigma Hopefully I did that right.... The key point here is there's a "spreading resistance" term to the total resistance. In the limit where you had infinitely thin slices of cylinders of radius R1 and R2 such that R1 < R2 the current would never be able to spread to a radius wider than R1 and the total resistance would be equivalent to a cylinder of the same total length with radius R1.
Aden
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Posted: 1 decade ago 18 mai 2015, 03:33 UTC−4
Thanks Jeff...
Thanks Jeff...
Aden
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Posted: 1 decade ago 18 mai 2015, 03:34 UTC−4
Thanks Daniel!!!
it was very helpful and comprehensive!! :)
Thanks Daniel!!! it was very helpful and comprehensive!! :)
Aden
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Posted: 1 decade ago 19 mai 2015, 04:54 UTC−4
Dear Daniel
I tried the other way, and still I have almost the same value for R (analytical).
so, you guess which values are closer to the real Resistance values?
the one I get from my calculations or the one from the model?
if the results from my model are so far from the reality, what are your suggestions to improve it?
Dear Daniel I tried the other way, and still I have almost the same value for R (analytical). so, you guess which values are closer to the real Resistance values? the one I get from my calculations or the one from the model? if the results from my model are so far from the reality, what are your suggestions to improve it?

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