
#1
Nov2112, 10:09 AM

P: 147

How does one find the electrical resistance of a homogenous sphere of uniform density?
By connecting two wires across the diameter of the sphere, and assuming Pouillet's Law [tex]R=ρ\frac{L}{A}[/tex] holds. 



#2
Nov2112, 01:15 PM

Mentor
P: 10,774

I don't think this is welldefined, if you try to calculate the resistance between two points on the sphere. If you have some finite area where you connect electrodes, you get some finite resistance value, but I think you will need a numerical simulation to calculate it.




#3
Nov2112, 02:04 PM

P: 997

You need to truncate the sphere by lopping off a small section at each end. That way the electrodes have some area. If a flat planar electrode pair is pressed against opposite sides of a true sphere, the resistance is infinite since the surface area between the sphere contact & electrode is literally zero. If the sphere ends are flattened, a finite area exists, & the resistance is finite.
Integration will compute the actual resistance. If a sphere is 10 cm in diameter, & the opposite ends of the sphere are lopped off resulting in 1.0 cm diameter contact surface area, you can compute resistance from those values. Did I help? Claude 



#4
Nov2112, 05:49 PM

Sci Advisor
PF Gold
P: 2,020

Electrical Resistance of a Sphere?
Claude, that is an approximation. Ut not exact because the material you excluded is in parallel and will result in lowering your R value. Conformal mapping might work for planar case (R of disk) followed by pi rotation. Numerical simulation will definitely work.




#5
Nov2112, 08:19 PM

P: 147

I tried numerical approximation but it leads to contradictory results. (See below) 



#6
Nov2112, 08:48 PM

P: 147

First up, I have a solid sphere of radius [tex]r_0[/tex]
I approximated by creating cylinders that "fill up" the sphere from inside, hence it is an underapproximation. Let's look at one hemisphere. I split up the hemisphere into q cylinders of the same height. [tex]q\inℤ^{+}[/tex] The height of each cylinder: [tex]h=\frac{r_0}{q}[/tex] The radius of each cylinder, r_{n} [tex](r_n)^{2}+(r_x)^{2}=(r_0)^{2}[/tex] [tex](r_n)^{2}+(nh)^{2}=(r_0)^{2}[/tex] [tex](r_n)^{2}=(r_0)^{2}(nh)^{2}[/tex] [tex](r_n)^{2}=(r_0)^{2}n^{2}\frac{(r_0)^{2}}{q^{2}}[/tex] [tex](r_n)^{2}=(r_0)^{2}\left[\frac{q^{2}n^{2}}{q^{2}}\right][/tex] The area of each cylinder [tex]A_n=π(r_0)^{2}\left[\frac{q^{2}n^{2}}{q^{2}}\right][/tex] Total Resistance (ignoring the last cylinder): [tex]R_T=ρ\frac{h}{A_1}+ρ\frac{h}{A_2}+ρ\frac{h}{A_3}+....+ρ\frac{h}{A_{q1}}[/tex] [tex]R_T=ρh×\sum_{n=1}^{q1} \frac{1}{A_n}[/tex] [tex]\frac{1}{A_n}=\frac{1}{π(r_0)^{2}}\left[\frac{q^{2}}{q^{2}n^{2}}\right][/tex] [tex]R_T=\frac{ρh}{π(r_0)^{2}}×\sum_{n=1}^{q1} \left[\frac{q^{2}}{q^{2}n^{2}}\right] =\frac{ρ\frac{r_0}{q}}{π(r_0)^{2}}×\sum_{n=1}^{q1} \left[\frac{q^{2}}{q^{2}n^{2}}\right] =\frac{ρ}{πr_0}×\frac{1}{q}\sum_{n=1}^{q1} \left[\frac{q^{2}}{q^{2}n^{2}}\right][/tex] [tex]R_T=\frac{ρ}{πr_0}×\frac{1}{q}\sum_{n=1}^{q1} \left[\frac{q^{2}}{q^{2}n^{2}}\right] , q→∞[/tex] That should be the resistance of a hemisphere. The problem is, [tex]\frac{1}{q}\sum_{n=1}^{q1} \left[\frac{q^{2}}{q^{2}n^{2}}\right] , q→∞[/tex] does not converge to a single value. It apparently keeps on increasing as q inccreases. Hence, it would seem that a sphere can have any amount of resistance. This is quite contradictory. 



#7
Nov2112, 10:54 PM

Sci Advisor
PF Gold
P: 2,020





#8
Nov2112, 11:00 PM

Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,154





#9
Nov2212, 03:53 AM

P: 147

Why would there be no conductivity at the surface? Also, if I ignore the above, I should still be able to get an exact Pouillet value? 



#10
Nov2212, 09:18 AM

Sci Advisor
PF Gold
P: 2,020

Sorry, I misunderstood your cylinders to be concentric instead of disks.




#12
Dec1312, 05:38 PM

P: 997

Here it is computed, & the answer checks out. Not too hard. If you want to make it hard, offset the terminals at an oblique angle. That is tough. But this sheet computed resistance w/ the terminals diametrically opposite. That is too easy. A single integral solves it. Let me know if clarification is needed. Best regards.
Claude 



#13
Dec1412, 08:15 AM

Mentor
P: 10,774

You seem to assume that voltage is a function of the zcoordinate only. Can you prove this assumption? I doubt that it is true.




#14
Dec1412, 11:54 AM

P: 997

Since the disks are stacked in series, we must sum all disks along the zaxis. The integral is a single integral wrt "z", but the radis of wach disk & that of the sphere enter into the computation. The polar angle theta does not enter in since we have symmetry. If the sphere had a section removed in a way that the angle was less than 2∏ radians, we would have theta in the integral. Also, if we examine a section where b is half of a, we can compute the resistance of either cylinder with ease. I did that & demonstrated how the sphere section resistance falls in between the inscribed & circumscribed cylinders. That provides a sanity check but does not assure correctness. The resistances of the 2 cylinders are ρ/a multiplied by 0.318 and 0.424. The sphere section resistance must be in between these values. Although 0.350 lies in between 0.318 & 0.424, that itself is no guarantee. There are many values in between 0.318 & 0.424. However if my final answer was outside these values it would be wrong beyond a doubt. Since my answer lies in between these values, it might be correct. Either way, it cannot be far off. We know the right answer is between the limits I just described. If I erred, please point it out. Did I miss a sign, drop a factor, make an invalid assumption, etc.? Just let me know & I will correct it. BR. Claude 



#15
Dec1412, 12:01 PM

Sci Advisor
PF Gold
P: 2,020

I have to agree with mfb. The problem with summing the resistances of flat disks is that you are neglecting the lateral spread of current from a small disk at top to a larger disk beneath it. Equipotential surfaces are not normal to the z axis (assuming that electrodes are at z=±a) but rather curve up in the upper hemisphere or down in the lower so as to intersect the spherical surface at right angles. Proper solution of this problem requires solution of Laplace's equation in spherical coordinates, with two sets of BC's specifying a) the source and sink surfaces, and b) zero normal component of current at the spherical boundary. I expect solutions for equipotential surfaces and for streamlines to be expressed in terms of Legendre polynomials.
I don't think this is such a simple problem. 



#16
Dec1412, 01:27 PM

P: 997

As the cross section changes, the lateral spread of current will encounter a changing area resuslting in changing resistance. I've covered that issue well. Now for the equipotential surfaces not being normal to zaxis, I say they are. Draw a diagram please & attach it for us to review. If you don't, I'll draw one in the evening. Equipotential surfaces intersect the zaxis normally due to symmetry. If, however, the terminal planes were not diametrically opposite, but offset/oblique, you would be correct. That is indeed a much more complex problem. Please pay attention to my inscribed & circumscribed cylinder resistance values. They illustrate that I did indeed take the lateral current spread into account. Again, I am happy to accept correction, but please review my paper thoroughly. Skimming through it & making off the cuff rebuttals that are meritless gets us nowhere. I'm not out to "win" anything, an argument or otherwise, but I do notice that when someone, not necessarily me, produces info to a tough problem, there are always those who rebuke them while offering no evidence at all where the mistake is. I'm not accusing you of anything, just asking you to support your criticism. BR. Claude 



#17
Dec1412, 02:00 PM

Sci Advisor
PF Gold
P: 2,020

Claude, your equipotential surfaces are normal to the z axis because you forced them to be so by your choice of solution method. The resistance of a disk, likewise, is only [tex]R=\frac{\rho l}{A}[/tex] if the top and bottom surfaces are equipotentials, as though each were coated with a perfect metallic conductorconsistent with equipotentials normal to z. You can see easily that this cannot be a valid solution for the sphere. The BC (that current cannot flow across the sphere surface), together with Ohm's law [tex]\vec{J}=\sigma\vec{E}[/tex] means that [tex]E_{\perp}(r=a)=\left. \frac{\partial\phi}{\partial r}\right_{r=a}=0.[/tex] The equipotential surfaces *must* be normal to the spherical surface, not flat as you postulate. It is the curvature of these surfaces that permits current to spread.




#18
Dec1412, 02:32 PM

P: 580

http://www.academia.edu/1841457/The_...e_by_Soliverez
I will try to find time to check out his result with my simulator. 


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