Resistance of infinite sheet

When you find such a sheet jusr stick a meter on it!

I can't speak for the correctness of your model, but that integral is infinite, which presumably isn't much use.

And that makes perfect sense to me! If ground is always an infinite distance away, the resistance should be infinite. (Unless the resistivity is zero.)

-Paul

I would look in the charming little book: RANDOM WALKS AND ELECTRIC NETWORKS

Assume you are at the centre, take a pie-shaped segment, angle Theta, Radius R, Resistivity Rho

Total resistance of segment = Rho * Int (Theta*r, r=0..R) Theta * R^2 / 2, this does not converge as R -> Infinity

Total resistance from the centre of the infinite plane,

sum of all segments: = Rho * Int (Theta* R^2/2, Theta=0..2*Pi).

For an infinitely large plane, you have an infinite resistance to ground.

Niall

"Joel" writes in article dated Thu, 09 Jun 2005

14:03:57 GMT:

It might be helpful to start with an easier question. Let's say you had a sheet that was grounded at both ends across the width and had a terminal across the entire center width. The resistance from terminal to ground would be Rho*(L/W)/4.

Can we expand on that answer to some kind of discrete model which approaches the continuous one?

--Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer.

Second the recommendation. It is freely available (under GPL) on the Web:

[arXiv:math - Random Walks and Electric Networks] In brief, for dimensions higher than two the answer is finite.

regards, chip

This is similar to the integral which occurs in the characteristic impedance of coaxial cables, which all electrical engineers know, which finishes up as something like Zo.log(R2/R1) as I remember. Zo is the characteristic impedance of the dielectric, R2 is the radius of the outer and R1 is the radius of the innner. So for a circular sheet of radius R2 and a central contact of radius R1 the impedance will have this form. Resistive sheets are used to terminate radio frequency coaxial cables as dummy loads.

rusty.

I don't dispute your conclusion, but the segments wouldn't be summed- they'd be in parallel, so would combine as 1/(1/seg1+1/seg2...etc.) Also, as R is increased in a single segment, the contribution of each increase in radius to the total resistance of the segment would be less because of broader parallel path.

-- john

This is why the obvious answer (resistance is infinite) didn't match my intuition. As r increases, the additional resistance added to the total for each unit increase in radius drops, so you have a sum of an infinite sequence of values, each of which get ever closer to zero. Seems like I dimly remember from college calculus a lot of Taylor/Maclaurin series like this that are bounded and converge to some constant, but I could be mistaken.

Resistance to ground is infinite - the path being infinitely long.

Brian Whatcott

Your result is fine but: You have added a complication. your inner radius is 0. The effective cross section at this point is 0 so the cross section of the material is also 0 so (rho) l/area is infinite. In other words the integral from 0 to a finite r also blows up to infinity. Why not look at an inner radius rmin and outer radius rmax and get an expression for this case, then see what happens as rmin goes to 0 or rmax goes to infinity. EE's are not necessarily bailing for the next post- I certainly hope that they are not.

in article h%Xpe.26070$ snipped-for-privacy@newssvr14.news.prodigy.com, Joel at snipped-for-privacy@removethis.pacbell.net wrote on 6/9/05 7:03 AM:

If you are truly talking about the ground connection made at infinity, the resistance is infinite. It is also infinite if the connection is made as a point contact. The cylindrical symmetry leads to a logarithmic potential as a funtion of radius.

To turn your problem into a "real" problem it needs to be modified.

The problem is relatated to that of calculating the d-c inductance of an infinite long wire of finite diameter. In essence, such two dimenisonal problems are dealt with by using analytic functions and conformal transformation. Look up the Schwartz transformation.

Finite resistance per unit length will be found between two wires that can be of different diameters. The field configurations for such problems have been widely published. They look like a tansmission impedance chart in rectangular coordinates. Such a chart is the equivalent of a Smith chart transformed from the Smith chart's unit circle to an infinite plane.

It is also possible to solve for the resistance of a finite wire to the edges of a rectangle in which the perimiter is grounded. A Schwartz transformation for such a problem has been published decades ago. Be prepared, however, that it requires knowledge of elliptic functions to understand what is going on.

Bill

in article hf3qe.26497$ snipped-for-privacy@newssvr14.news.prodigy.com, Joel at snipped-for-privacy@removethis.pacbell.net wrote on 6/9/05 3:18 PM:

True and false. Just because the terms get smaller does not mean that they converge. The classic example is that of the harmonic series:

1 + 1/2 +1/3 + 1/4 + 1/5 + ...

Bill

Oops - using the same approach but this time getting conductance and resistance correct....

For the pie-slice, inner radius r1, outer radius r2, angle theta,

R slice = Rho * Area = Rho * Int(Theta*r, r=r1..r2), for simplicity assume rho = 1

R slice = Theta * (r2^2-r1^2)/2

C slice = 2 / (Theta * ( r2^2 - r1^2))

Adding the conductances of the slices..

C plane = 2/ (r2^2 - r1^2) * Int (1/ Theta, Theta=0..2*Pi) = 2 / (r2^2 - r1^2) * (log (2*pi) - 1))

R plane = (r2^2 - r1^2) / constant

as r2 -> Infinity and assuming r1 -> 0 [ignoring the messier physical connection realities], we see that R plane -> Infinity

Niall

Oops - using the same approach but this time getting conductance and resistance correct....

For the pie-slice, inner radius r1, outer radius r2, angle theta,

R slice = Rho * Area = Rho * Int(Theta*r, r=r1..r2), for simplicity assume rho = 1

R slice = Theta * (r2^2-r1^2)/2

C slice = 2 / (Theta * ( r2^2 - r1^2))

Adding the conductances of the slices..

C plane = 2/ (r2^2 - r1^2) * Int (1/ Theta, Theta=0..2*Pi) = 2 / (r2^2 - r1^2) * (log (2*pi) - 1))

R plane = (r2^2 - r1^2) / constant

as r2 -> Infinity and assuming r1 -> 0 [ignoring the messier physical connection realities], we see that R plane -> Infinity

Niall

I think this is the nugget: If you set the integral in my original post up as an infinite summation of concentric rings whose radial extent is constant (and neglect the innermost one), assume that the resistance of each is rho*dr*circumference, then factor out all the constant terms, you get:

R=Sum over r=rmin..infinity of: [(rho*deltar)/(2*pi*r)] = ((rho*deltar)/(2*pi)) * Sum over r=rmin..infinity of [1/r] -- exactly the harmonic series!

Seems conclusive that the answer is infinite, even without a point contact in the center.

J

...which increases without limit, in contrast to 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... which limits at 2

Brian W

I agree with this conclusion =>

That the Resistance is Infinite no matter where you are measuring from on the sheet, since reference is to Another Surface (or Ground) this would envelope the entire form of the infinite sheet into the equation.

• Exception I * From 1 point on the sheet too point 2 on the sheet. this a different animal, since distance and R per unit length may vary.

• Exception II * unless: a single unit of the sheet also has Infinite R. then it wouldn't make any difference either, only with reference to an R value per unit length as with Exception I. }:-)

Roy ~ E.E.Technician Mayor: Electronics Strong, Physics: Minor