Register to reply 
Electrical Resistance of a Sphere? 
Share this thread: 
#19
Dec1412, 02:33 PM

Sci Advisor
PF Gold
P: 2,080

The lines should look similar to those for the TE11 mode in a spherical cavity resonator....



#20
Dec1412, 02:37 PM

Mentor
P: 11,890

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 twodimensional 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 (b1)<<1, it is possible to calculate an interesting lower bound of the resistance as following: Replace the disk (electrode) by a halfsphere. This replaces conducting material by an ideal conductor and clearly reduces the resistance. Replace the halfball by a halfball 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 onedimensional integral. Let r be the radius of the electrode, ##r^2+b^2=1##. $$R > 2 \int_r^1 \frac{1}{2\pi r'^2} dr' = \frac{1}{\pi} (\frac{1}{r}1) = \frac{1}{\pi}(\frac{1}{\sqrt{1b^2}}1)$$ 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 


#21
Dec1412, 02:45 PM

Sci Advisor
PF Gold
P: 2,080

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 semiinfinite 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. 


#22
Dec1412, 03:58 PM

P: 1,041

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 "z_{0}" 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' "z_{0}" 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+z_{0}" in the numerator, with "az_{0}" 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 


#23
Dec1412, 04:09 PM

P: 1,041

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 


#24
Dec1412, 04:23 PM

P: 1,041

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 


#25
Dec1412, 05:25 PM

Sci Advisor
PF Gold
P: 2,080

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}{az_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}{1z_0/a}=\frac{\rho}{\pi a} \log (1+z_0/a) \log (1z_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^2b^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 matchthe 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. 


#26
Dec1412, 06:38 PM

P: 1,041

<<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 "z_{0}/a is hardly less than 1. In fact it is almost equal to 1, for a contact radius of 0.1, z_{0} 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 z_{0} with sqrt{a^2b^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 reexamine 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 


#27
Dec1412, 06:49 PM

Sci Advisor
PF Gold
P: 2,080

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. 


#28
Dec1412, 09:22 PM

P: 1,041

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 


#29
Dec1412, 09:38 PM

Sci Advisor
PF Gold
P: 2,080

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. 


#30
Dec1512, 07:52 AM

P: 1,041

Claude 


#31
Dec1512, 08:01 AM

Mentor
P: 11,890

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


#32
Dec1512, 08:11 AM

P: 1,041

Claude 


#33
Dec1512, 08:20 AM

P: 1,041

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 


#34
Dec1512, 09:55 AM

Mentor
P: 11,890

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. 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. 


#35
Dec1512, 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.



#36
Dec1512, 07:14 PM

P: 1,041

I am reexamining 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 


Register to reply 
Related Discussions  
Electrical resistance  Introductory Physics Homework  2  
Electrical resistance  Classical Physics  2  
Resistance for a sphere in a half sphere containing liquid  Introductory Physics Homework  19  
Electrical Resistance  Introductory Physics Homework  2  
Electrical Resistance  Classical Physics  3 