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Electrical Resistance of a Sphere?

by greswd
Tags: electrical, resistance, sphere
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marcusl
#19
Dec14-12, 02:33 PM
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The lines should look similar to those for the TE11 mode in a spherical cavity resonator....
mfb
#20
Dec14-12, 02:37 PM
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You underestimate the total resistance with your approach as you do not consider current flow in radial direction (which contributes to resistance, too). This is easier to see if you take extreme cases - attach two small disks of radius r (the electrodes) to a cylinder with radius R>>r and height h~r. If you increase R, your calculated resistance would drop with 1/R^2, but the real resistance will approach a finite value.

I did some two-dimensional numerical simulation, but the borders are messy to consider.
For b=0.96, I get ~2.3 ρ/a as resistance, where your formula gives 1.24. I think my simulation overestimates the resistance a bit, so the real value is somewhere in between.
For b=0.52, I get ~0.38 ρ/a as resistance, where your formula gives 0.367. Here, the radial resistance is not significant.

Fixing a=ρ=1 now, as they are just prefactors anyway:

For (b-1)<<1, it is possible to calculate an interesting lower bound of the resistance as following:
Replace the disk (electrode) by a half-sphere. This replaces conducting material by an ideal conductor and clearly reduces the resistance.
Replace the half-ball by a half-ball around the electrode and connect the second electrode to the outer boundary. Use this as half the full resistance. This adds conducting materials where the original problem has none and moves the original symmetry area towards the electrode, so it again underestimates the resistance.
This modified problem has a spherical symmetry, so we can solve it with a one-dimensional integral. Let r be the radius of the electrode, ##r^2+b^2=1##.

$$R > 2 \int_r^1 \frac{1}{2\pi r'^2} dr' = \frac{1}{\pi} (\frac{1}{r}-1) = \frac{1}{\pi}(\frac{1}{\sqrt{1-b^2}}-1)$$

Compare this with your formula at b=0.99:
R > 1.93
R with your formula: 1.68

It gets worse with larger b, as the radial flow gets more important:
b=0.9999:
R > 22.2
R with your formula: 3.15
marcusl
#21
Dec14-12, 02:45 PM
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Quote Quote by the_emi_guy View Post
http://www.academia.edu/1841457/The_...e_by_Soliverez

I will try to find time to check out his result with my simulator.
Thanks for finding this reference! I took a quick look and the approach looks right (note the Laplace equation in spherical coordinates in Eq. (29), whose solution is typically expressed in Legendre polynomials). His Fig. 3 shows believable streamlines and equipotentials, with EP's that are normal to the boundary as I said above. Another useful point is that the problem is simplified when the electrodes are taken to match equipotential surfaces, as at top and bottom of Fig. 3. As Soliverez points out, these approach hemispheres when they are small.

Here's the really interesting part: the resistance of the sphere ends up being just the sum of resistances of the top and bottom contacts. To see this, use the "spreading resistance" of a hemispherical contact pressed into a semi-infinite slab that was worked out ages ago (you see it in books on semiconductor contacts, for example), [tex]R_{contact}=\frac{1}{2\pi b \sigma}[/tex] where b is the radius of the hemispherical contact. Double it to account for top and bottom contacts in series and you get Soliverez's Eq. (43)! We should have realized from the very start that, for b<<a, the spherical conductor can be considered infinite in comparison to the size of the contacts, so all significant resistance is contributed by the small contact areas.
cabraham
#22
Dec14-12, 03:58 PM
P: 1,035
marcusl - I don't understand how you can dispute me when the link above to Carlos' paper gives the same answer I gave. In his math, he uses "z0" to denote the terminal location on the z axis, where I use "b" for the same. However, I use "c" as the radius of the circular cross sectional area intercepted by the terminal plane, where Carlos uses "b". His "b" is not the same as my "b'. Carlos' "b" is Claude's "c", & Carlos' "z0" is Claude's "b". Also his equation has a factor of 2 in the denominator where mine doesn't. Carlos' integral is over the whole sphere top to bottom, where Claude's (moi) is center to top with a factor of 2 multiplied in. That is the descrepency in the 2 equations. If you compute the R value using Carlos' equation & Claude's you should get the same answer.

Notice his natural log term. The argument has "a+z0" in the numerator, with "a-z0" in the denominator. Same as mine. Examine carefully & his answer is identical to mine. He uses conductivity "sigma", where I use resistivity "rho". Please review carefully. Carlos & I agree. BR.

Claude
cabraham
#23
Dec14-12, 04:09 PM
P: 1,035
Also, the equation:

R = ρl/A

is valid over the region of integration. Laplace's equation is certainly upheld here, but it is a long roundabout way of computing the resistance. Carlos is indeed accurate, but my approach gets the same answer with much less effort.

However, for the situation I described at the end of my sheet, where the terminals are not diametrically opposite, but oblique, then Carlos' approach is exactly how I would tackle the problem. Again, from looking his equation over, we seem to agree.

marcusl - Would you please recompute the examples you gave above while duly noting that "b" in Carlos' method is not my "b" To convert, remember that a "b" of 0.1 in Carlos' math, with a radius "a" of 1 for the whole sphere, then we must use Pythagoreus' theorem. My "b" would be sqrt(1 - (0.1)2) = 0.995, quite a difference. As Carlos' "b" becomes vanishingly small, my "b" approaches unity (if sphere radius "a" is unity).

Please make due note of this important distinction. Thanks to all for your feedback.

Claude
cabraham
#24
Dec14-12, 04:23 PM
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Quote Quote by marcusl View Post
Thanks for finding this reference! I took a quick look and the approach looks right (note the Laplace equation in spherical coordinates in Eq. (29), whose solution is typically expressed in Legendre polynomials). His Fig. 3 shows believable streamlines and equipotentials, with EP's that are normal to the boundary as I said above. Another useful point is that the problem is simplified when the electrodes are taken to match equipotential surfaces, as at top and bottom of Fig. 3. As Soliverez points out, these approach hemispheres when they are small.

Here's the really interesting part: the resistance of the sphere ends up being just the sum of resistances of the top and bottom contacts. To see this, use the "spreading resistance" of a hemispherical contact pressed into a semi-infinite slab that was worked out ages ago (you see it in books on semiconductor contacts, for example), [tex]R_{contact}=\frac{1}{2\pi b \sigma}[/tex] where b is the radius of the hemispherical contact. Double it to account for top and bottom contacts in series and you get Soliverez's Eq. (43)! We should have realized from the very start that, for b<<a, the spherical conductor can be considered infinite in comparison to the size of the contact so all significant resistance is contributed by the small contact areas.
My approach says the same. Just note the distinction that my "b' differs from that of Carlos. For a Carlos "b" of 0.10, the equivalent Claude "b" is 0.995, & the resistance is ρ/a times 1.9055, over 5 times greater than the 0.350 value when Carlos "b" is 0.8660 with Claude's "b" at 0.5 times a.

But if Carlos "b" decreases to 0.01 times "a", we get a Claude "b" of 0.99995. My resistance is then equal to ρ/a times 3.3730. As the radius of the termination decreases, the overall resistance increases due to smaller terminal contact area, just as the discussions above indicate. Anyway I just thought I would mention it. BR.

Claude
marcusl
#25
Dec14-12, 05:25 PM
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Claude, we're still not seeing eye to eye.

Your expression for resistance in the attachment to post #12 is [tex]R=\frac{\rho}{\pi a} \log \frac{a+z_0}{a-z_0}[/tex] where I've used z_0 in place of b just to be consistent with usage in my previous post. This can be rewritten as [tex]R=\frac{\rho}{\pi a} \log \frac{1+z_0/a}{1-z_0/a}=\frac{\rho}{\pi a} \log (1+z_0/a)- \log (1-z_0/a).[/tex] Since z_0/a is less than one, we can apply the expansion [tex] \log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...[/tex] to get
[tex]R=\frac{2\rho}{\pi a} \left[\frac{z_0}{a} + \frac{1}{3}\left(\frac{z_0}{a}\right)^3 + ...\right].[/tex] Substituting [tex]z_0=\sqrt{a^2-b^2}[/tex] for z0, where b is the contact radius, gives
[tex]R=\frac{2\rho}{\pi a} \left[ \sqrt{1 - \left(\frac{b}{a}\right) ^2} +
\sqrt[3]{1 - \left(\frac{b}{a}\right) ^2 } + ...\right].[/tex]
Comparison to Soliverez's expression
[tex]R=\frac{1}{\pi \sigma b} [/tex] shows a poor match--the a's and b's aren't even in the same places.

Finally I invite you to work out a numerical example like rho=sigma=1, a=1, contact radius b=c=0.01 and z0=0.99995.
Soliveres gets R=31.8 ohms, your expression gives 3.4 ohms. mfb was pointing out a similar discrepancy with his lower bound calculation.
cabraham
#26
Dec14-12, 06:38 PM
P: 1,035
Quote Quote by marcusl View Post
Claude, we're still not seeing eye to eye.

Your expression for resistance in the attachment to post #12 is [tex]R=\frac{\rho}{\pi a} \log \frac{a+z_0}{a-z_0}[/tex] where I've used z_0 in place of b just to be consistent with usage in my previous post. This can be rewritten as [tex]R=\frac{\rho}{\pi a} \log \frac{1+z_0/a}{1-z_0/a}=\frac{\rho}{\pi a} \log (1+z_0/a)- \log (1-z_0/a).[/tex] Since z_0/a is less than one, we can apply the expansion [tex] \log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...[/tex] to get
[tex]R=\frac{2\rho}{\pi a} \left[\frac{z_0}{a} + \frac{1}{3}\left(\frac{z_0}{a}\right)^3 + ...\right].[/tex] substituting [tex]z_0=\sqrt{a^2-b^2}[/tex] for z0, where b is the contact radius, gives
[tex]R=\frac{2\rho}{\pi a} \left[ \sqrt{1 - \left(\frac{b}{a}\right) ^2} +
\sqrt[3]{1 - \left(\frac{b}{a}\right) ^2 } + ...\right].[/tex]
Comparison to Soliverez's expression
[tex]R=\frac{1}{\pi \sigma b} [/tex] shows a poor match--the a's and b's aren't even in the same places.

Finally I invite you to work out a numerical example like rho=sigma=1, a=1, contact radius b=c=0.01 and z0=0.99995.
Soliveres gets R=31.8 ohms, your expression gives 3.4 ohms.
marcusl - Please recheck your log \frac{1+z_0/a}{1-z_0/a}= ---

<<your quote: Since z_0/a is less than one, we can apply the expansion [tex] \log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...[/tex] to get --- end quote>>

The quantity "z0/a is hardly less than 1. In fact it is almost equal to 1, for a contact radius of 0.1, z0 equals 0.9995, hardly less than 1 at all. Expanding the log function into its Taylor series requires many terms before convergence.

Your 2nd mistake is near the end. After replacing z0 with sqrt{a^2-b^2}, you get a quantity inside the brackets consisting of radicals containing 1 - (b/a)2. But this quantity is virtually 1 for b very small. Hence the radicals all have approx. unity value. But you put coefficients of unity before each radical. The coefficients should have a progression of:

1, 1/3, 1/5, 1/7, ---

The sum of said coefficients diverges. You erred at the top when you took my log expression, then did the Taylor series. I will re-examine this, byt frankly, my inscribed/circumscribed cylinders seem to support my final expression. I will examine later. Right now my next higher household authority is on me to put up the Christmas tree. I will get back to you later tonight. BR.

Claude
marcusl
#27
Dec14-12, 06:49 PM
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The typo in omitting the numerical factor 1/3 from the expression towards the end of my post doesn't change the conclusion. Try the numerical example I gave above, and see that results from the two methods differ by an order of magnitude.

If you use the spreading resistance for a disk instead of a hemisphere to better match your contact geometry
[tex]R\approxeq 2 R_{disk}=\frac{1}{2\sigma b}[/tex] then the discrepancy is even larger.
cabraham
#28
Dec14-12, 09:22 PM
P: 1,035
Quote Quote by marcusl View Post
The typo in omitting the numerical factor 1/3 from the expression towards the end of my post doesn't change the conclusion. Try the numerical example I gave above, and see that results from the two methods differ by an order of magnitude.

If you use the spreading resistance for a disk instead of a hemisphere to better match your contact geometry
[tex]R\approxeq 2 R_{disk}=\frac{1}{2\sigma b}[/tex] then the discrepancy is even larger.
Sorry marcusl but I'm convinced it changes the conclusion by a lot, because omitting 1/3, 1/5, etc. results in convergence as opposed to divergence. Anyway I composed 2 Excel spread sheets using the summation of stacked disks. As we subdivide into finer increments we get better accuracy. I used 99 disks per hemisphere, then 989. As expected the 99 piece answer was 8% off, & the 989 piece answer was 1% off from my equation. Unfortunately the larger sheet file is too large to attach but I attached the smaller file.

I do not believe that you can replace my log term with a very short truncated Taylor series because the z0 quantity (my "b") is nearly unity. Please repeat your computations w/o such replacement, i.e. let a log function remain as writted w/o trying to simplify. Carlos did very well but his assumptions that this term is negligible & can be dropped might not be valid. Please examine my spread sheet. Thanks for your interest.

Claude
Attached Files
File Type: xls sphere section resistance99.xls (52.5 KB, 0 views)
marcusl
#29
Dec14-12, 09:38 PM
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You are missing the points (many of them). Your exact expression (not my expansion) is an order of magnitude too small compared both to mfb's lower bound and to Soliverez's expression.

I am done with this topic.
cabraham
#30
Dec15-12, 07:52 AM
P: 1,035
Quote Quote by marcusl View Post
You are missing the points (many of them). Your exact expression (not my expansion) is an order of magnitude too small compared both to mfb's lower bound and to Soliverez's expression.

I am done with this topic.
Is it possible that Soliverez' equation is an order of magnitude too high? Without proof you accept his, not mine. The Excel spread sheet confirms my answer. By manual summing each disk section we get answers much closer to mine than to CS. I guess it's pointless to argue. Nobody has been able to show where I went wrong other than to say my result is "too large" as if CS result is somehow immutable.

Claude
mfb
#31
Dec15-12, 08:01 AM
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Quote Quote by cabraham View Post
Is it possible that Soliverez' equation is an order of magnitude too high? Without proof you accept his, not mine.
We have two independent derivations (at the time I wrote my post, the link to Soliverez' was not there yet), both agree with each other.

On the other hand, we have your calculation which is based on an incorrect assumption.

The Excel spread sheet confirms my answer.
It uses the same, wrong assumption as your analytic approach.

Nobody has been able to show where I went wrong
We did so several times, and we ran out of different ways to explain it to you. You just ignore those explanations and examples.
cabraham
#32
Dec15-12, 08:11 AM
P: 1,035
Quote Quote by mfb View Post
We have two independent derivations (at the time I wrote my post, the link to Soliverez' was not there yet), both agree with each other.

On the other hand, we have your calculation which is based on an incorrect assumption.


It uses the same, wrong assumption as your analytic approach.


We did so several times, and we ran out of different ways to explain it to you. You just ignore those explanations and examples.
Just where did I go wrong? You haven't answered. I will review your posts carefully. What is wrong with summing the stacked disks in series. Each disk has a differing area so that R differs. Current laterally diffuses over differing areas, & ther disks are summed. It has to be right. Again, I will carefully review, but my 2 cylinder boundary limit affirms my answer. BR.

Claude
cabraham
#33
Dec15-12, 08:20 AM
P: 1,035
Quote Quote by mfb View Post
You underestimate the total resistance with your approach as you do not consider current flow in radial direction (which contributes to resistance, too). This is easier to see if you take extreme cases - attach two small disks of radius r (the electrodes) to a cylinder with radius R>>r and height h~r. If you increase R, your calculated resistance would drop with 1/R^2, but the real resistance will approach a finite value.

I did some two-dimensional numerical simulation, but the borders are messy to consider.
For b=0.96, I get ~2.3 ρ/a as resistance, where your formula gives 1.24. I think my simulation overestimates the resistance a bit, so the real value is somewhere in between.
For b=0.52, I get ~0.38 ρ/a as resistance, where your formula gives 0.367. Here, the radial resistance is not significant.

Fixing a=ρ=1 now, as they are just prefactors anyway:

For (b-1)<<1, it is possible to calculate an interesting lower bound of the resistance as following:
Replace the disk (electrode) by a half-sphere. This replaces conducting material by an ideal conductor and clearly reduces the resistance.
Replace the half-ball by a half-ball around the electrode and connect the second electrode to the outer boundary. Use this as half the full resistance. This adds conducting materials where the original problem has none and moves the original symmetry area towards the electrode, so it again underestimates the resistance.
This modified problem has a spherical symmetry, so we can solve it with a one-dimensional integral. Let r be the radius of the electrode, ##r^2+b^2=1##.

$$R > 2 \int_r^1 \frac{1}{2\pi r'^2} dr' = \frac{1}{\pi} (\frac{1}{r}-1) = \frac{1}{\pi}(\frac{1}{\sqrt{1-b^2}}-1)$$

Compare this with your formula at b=0.99:
R > 1.93
R with your formula: 1.68

It gets worse with larger b, as the radial flow gets more important:
b=0.9999:
R > 22.2
R with your formula: 3.15
I think I see the discrepency. Your "radial flow" would seem to originate from a point source contact terminal. I assumed, maybe incorrectly, that the contacts were flat planes w/ circular cross section, & that the electrodes from the power supply contacted the entire area of the sphere terminations. Here the current is primarily "vertical" w/ lateral diffusion (spreading) as charges transit downward.

But you mention "radial flow", & I am thinking that at the contact area, the current would not be normal to the equipotential contact surface. Right at the contact surfave the contact area is in fact an equipotential plane. Hence all current entering/exiting the sphere section must be normal to that contact plane as it is an equipotential surface. If there was a component of current oriented radially, then we would have current at an oblique angle to an equipotential surface, which I didn't think can happen.

Can you provide a quick hand sketch of current paths & equipotential surfaces detailing what you describe as "radial flow". Thanks.

Claude
mfb
#34
Dec15-12, 09:55 AM
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P: 11,580
What is wrong with summing the stacked disks in series.
It requires perfect radial flow of current.
It gives wrong predictions for the surface of the ball (equipotential areas have to be perpendicular to the surface, as shown above - they are not in your model).
It gives wrong predictions for the total resistance (see my lower bound derived above).
It gives wrong predictions for the limit of disks with radius R->infinity.

Again, I will carefully review, but my 2 cylinder boundary limit affirms my answer.
No, it just shows that your answer is between some lower and upper bounds - which are very weak for small contact areas.

Your "radial flow" would seem to originate from a point source contact terminal.
No, point sources are unphysical. I used small hemispheres.

If there was a component of current oriented radially, then we would have current at an oblique angle to an equipotential surface, which I didn't think can happen.
You need a radial component "somewhere" to have current flow everywhere in the sphere. In fact, you get radial flow everywhere, with the most significant part close to the edge of the electrodes.


I attached a sketch of a sphere, the vertical lines are current, the horizontal lines are equipotential surfaces. You can see how they are bent. The right/upper part is symmetric to the lower left part. The uppermost and lowermost horizontal line are the electrodes.
- equipotential surfaces have to be perpendicular to the surface of the ball
- current has to be perpendicular to the equipotential surfaces.
Attached Thumbnails
sphere.png  
Studiot
#35
Dec15-12, 10:08 AM
P: 5,462
I have always understood the analytical solution (current isolines) is the same as the solution for foundation pressure trajectories in the ground or stress trajectories from a point load in elasticity, first presented by Coulomb.
cabraham
#36
Dec15-12, 07:14 PM
P: 1,035
I am re-examining my stacked disk model, & it looks like I will have to deep six it. I presumed at first that current would laterally diffuse & distribute evenly throught each disk cross section. That is approximately true if we are near the center of the sphere & equipotential surfaces are almost flat. But near the poles, the curve is too pronounced for even distribution. I think the "UD" model (uniform density) results in too low an R value.

I will try to solve the Laplace equation as soon as I figure out how. It will require general curvilinear coordinates. I will post when I have the solution. BR.

Claude


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