Electrical Resistance between 2 points in homogeneous resistive plane?

[memoguy]

Hi there!

Just say I have large square piece of some homogeneous resistive material like graphite. How would I go about determining the resistance between any two given points?

Further, just say I supply a voltage across two arbitrary points, can I determine the voltage difference between any other two arbitrary points? I don't even know where to begin with this.

To give you some context, what I am thinking about here is whether I can take a large sheet of resistive material and place a voltage across two known point P and Q. Then, if I know the distance between two further points A and B, and I measure the voltage across them, I want to be able to take this voltage reading and hence work out the position of the points A and B relative to points P and Q. I am just trying to work out if this is an at all feasible idea.

Cheers!

[AndyC_772]

This sounds like something that would make an interesting exam question, albeit one which my coffee-deficient brain wouldn't be up to solving right now.

Suppose you define your co-ordinate system such that your two known points are at (-1,0) and (1,0). By symmetry, any point on the y-axis will be at a potential halfway between the two, and the entire solution will of course be symmetrical about the x-axis.

Beyond that, I'm sure it's possible to come up with an equation giving the potential at any arbitrary point, in terms of the potentials at surrounding, nearby points. Let the definition of "nearby" tend towards zero, and you have an equation that might be solvable using calculus.

Personally I'd start with a sheet of copper clad board, a power supply, an accurate voltmeter and a marker pen.

[tszaboo]

You know the infinite resistor grid puzzle?
http://www.mathpages.com/home/kmath668/kmath668.htm
Well, you just have to apply a lim R-> 0 and m-> 1/R to it, where R is the resistance and m is the distance between the two points. You should assume n=m also, as it doesn't make a difference.
Should be simple if you know algebra.

[wiss]

Is your plane thick or thin?

[electr_peter]

All is too easy with homogeneous resistances all around. For real cases with different resistances, multiple voltage/current sources, other constraints just build a simulation of a plane and calculate from here.

[memoguy]

Thank you all for your helpful responses. I must just indicate that I am quite the beginner at all this, I am a first year engineering student, only really been learning this stuff for a couple of weeks. Just to be clear, this is absolutely unrelated to assessment, just something interesting I have been thinking about and have been unable to get a straight answer from any of my tutors.

I am interested where you are going with that idea NANDBlog. I am not quite I understand exactly what you are saying however Because this grid is finite, so I am not certain of the applicability of the infinite grid idea? The maths seems to be quite insane for that problem!

[memoguy]

(and lets just assume the plane is thin, just because I think it will make the math easier  )

[JacquesBBB]

The following is from a rapid thinking.

The problem is that the resistance in a point is not well defined.

Using the surface resistivity ro, we have for the resistance  res = ro * L / a .
where L is the length of the conductor and a  its width.

If you  measure the resistance from a point at the center of a circle of radius R2 to its circumference,
(which is  an idealized case simpler),
You will find

res = ro/(2 pi) int_0^R2  dr / r  = ro/(2 pi) [ Log r]_0^R2  which is diverging.
the resistance becomes infinite at the origin.

This is why resistance measures are defined between two circles of radius R1 and R2 : then you have

res = ro/(2 pi) Log (R2/R1)

See for example
http://www.trekinc.com/pdf/1005_Resistivity_Resistance.pdf

Of course, in the real world, things maybe somewhat different, as  you never have infinitely small point, and
the physics in the vicinity of the point is much more complicated than what is described in the
resistivity measure.

[robrenz]

You will not be able to determine position of the voltage probes in 2 dimensions using this setup. I have done just what you are talking about using a Micro Ohmmeter and 4 wire connections. It is very educational because you can see the pattern of current flow from the current source points but you can get the same readings from a symmetric placement of the A B probes or from a different location with a different probe spacing. Also if the line thru your A B probes is perpendicular to the line thru your P Q connections and A and B are symmetric about the P Q line you will get no voltage reading.

[memoguy]

Thanks robrenz. Yeah, I have just been thinking about it a bit and it seems like it is not possible to determine position, only distance. There would be infinite positions in the grid where the same voltage would be read, all on a circle.

So say if i limit my question to just determining distance, I wonder how one would do that?

[GK]

It's rare that I come a cross one of these physics questions that routinely pop up on these forums that isn't covered in depth in either one or more of my ancient analog computing texts. This is from Analog Simulation, solution of field problems, Walter J. Karplus, 1958, McGraw Hill. I don't have the time to scan the whole chapter . I've been derided for being "stuck in the past" for collecting these old volumes. Apparently the fundamentals change with time; go figure.
 
 

[T3sl4co1l]

http://pveducation.org/pvcdrom/characterisation/four-point-probe-resistivity-measurements

Note that the resistance between a given pair of probes might be extremely high, because you must account for contact resistance.  The small contact point, and also the spherical zones near said points, contribute a lot of resistance -- in fact for an infinitesimal point, infinite resistance!  The four probe method is required to perform this measurement.

Note that it matters that your "sheet" is "thin", i.e., thickness << distance between probes.

Tim

[codeboy2k]

Analog Simulation, solution of field problems, Walter J. Karplus, 1958, McGraw Hill.

Nice old book! I'm reading it now.  The book appears to be in the public domain and can be read online here:

http://hdl.handle.net/2027/mdp.39015036279555

No need to scan anymore pages, GK

[free_electron]

this got the smell of a field solver all over it...

take a sheet of tracing paper every line is 1K

start with a 5x5 matrix. pick

[GK]

The book appears to be in the public domain and can be read online here:

Oh, that's cool.

[MrAl]

Hello,

I believe that the Laplace equation in two dimensions can help here.
With one or more voltage excitations, the position of a probe might be found.
If we look at the sheet as a plane, if we apply a voltage across two adjacent sides at x1 and x2, we get a voltage gradient along that same span.  If we remove that voltage and then apply a voltage across the other two adjacent sides at y1 and y2, we get a voltage gradient along that perpendicular span.  Two measurements at the same point for each of these conditions should tell us where the point is located.
There may be other solutions, such as applying two high frequency sine waves 90 degrees apart across opposite adjacent sides so that one amplitude and phase measurement might reveal the position, but there are other problems that come in when using high frequency waves that are probably too undesirable and maybe even the frequency would need to be too high.
Just in case the voltage swapping method doesnt work (i havent tested it fully yet) it may just need some alteration or perhaps the addition of more voltage sources, such as one at each corner or something like that.
The solution may not be simple (again i havent tried this yet) so you might have to resort to using numerical techniques to solve the Laplace equation with the more complicated boundary conditions.

[MrAl]

Hello again,

A quick test calculation of the linear voltage swapping method shows a problem with the basic idea.
The basic idea was that a voltage across a sheet would produce a linear gradient, and that would pinpoint a place along one axis, and then swapping voltages would show the place along the other axis, and thus we arrive at a particular x,y position.
Unfortunately, Laplaces equation shows that the gradient is anything but linear, and in fact could be a combination of sinusoidal factors which produce equipotential lines that are not straight at all but are curved.  This makes things a lot more difficult, because in order to get a linear gradient we'd have to use a more complicated function along the boarder(s).  One way would be to connect individual wires to the edge, and then energize each wire with a different voltage, where the voltage function along the edge is an inverse of the solution to the Laplace equation so that we get voltages that are the same along the horizontal.

A quick finite difference solution lead to the following grid pattern, where each number represents the relative voltage at that point:
0.00  0.00  0.00  0.00  0.00  0.00  0.00  0.00  0.00  0.00  0.00
0.00  0.01  0.02  0.03  0.03  0.04  0.03  0.03  0.02  0.01  0.00
0.00  0.02  0.04  0.06  0.07  0.07  0.07  0.06  0.04  0.02  0.00
0.00  0.04  0.07  0.10  0.11  0.12  0.11  0.10  0.07  0.04  0.00
0.00  0.06  0.11  0.14  0.17  0.18  0.17  0.14  0.11  0.06  0.00
0.00  0.08  0.15  0.21  0.24  0.25  0.24  0.21  0.15  0.08  0.00
0.00  0.12  0.22  0.29  0.33  0.35  0.33  0.29  0.22  0.12  0.00
0.00  0.17  0.31  0.40  0.45  0.47  0.45  0.40  0.31  0.17  0.00
0.00  0.27  0.46  0.56  0.61  0.62  0.61  0.56  0.46  0.27  0.00
0.00  0.49  0.68  0.76  0.79  0.80  0.79  0.76  0.68  0.49  0.00
1.00  1.00  1.00  1.00  1.00  1.00  1.00  1.00  1.00  1.00  1.00

As we can see if we follow any of the approximate equipotentials we dont get a straight line, we get a curve.  For example, follow all the "0.04" entries, and we see it curves up and then down again.
This doesnt void the method, but it does make it a lot harder to do, and maybe not practical.
If we could generate a wave of the inverse function (or similar) we could solve it perfectly, but i fear that would take a frequency way too high.

I included the 3d plot above so we can visualize what is happening.
The lower row is the excitation of 1 volt DC and the upper side is 0 volts.  The left and right sides are not connected to anything yet, and that would be the first test condition.

[wiss]

Just by looking at the solution,  the left and right sides are connected,  to 0...

[Roger44]

Hi

The resistance between two electrodes of radius r0  and 2S apart on an infinite plane of resistivity lambda is simply   

R = arcosh(s/r0)/pi/lambda

If you know the voltage between the electrodes you can deduce voltage at any point on the plane. Alternatively if you take sufficient measures to identify an equipotential circle (radius and position on the y axis), you must be able to work back to find the voltage between the electrodes (x axis) then deduce voltages anywhere else on the plane. Have you got an infinite plane