Electric Circuit Question

gneill

There's no one current value you can plug in because the current is spread out over a 3D volume and we don't know off-hand how its distributed.

GreenPrint

Alright well I don't need to know how to calculate the voltage. However it would be interesting to figure out how to do this.

I have to explain my collected data. I however have no idea how to do this. I thought if I could figure out how to calculate this (assuming ideal conditions) what values would I expect through mathematical calculation.

I'm not sure.

gneill

One way is to use numerical methods. That's where you use software to integrate the differential equations that describe the system, or run a simulation that seeks a steady-state solution.

Here's a combined surface/contour plot for a crude integration based on a 21 x 21 grid of points covering a square. One corner has its simulated "potential" fixed at 1.00 (top corner in the red contour). The diagonally opposing corner has its value fixed at 0.00 (bottom point in the deep blue contour). The potentials of the rest of points in the field sort themselves out according to the simple rule of taking on the average of surrounding values over many iterations. Simple but effective for a homogenous and isotropic playground like a slab of uniform resistivity.



Note that the "unpowered" diagonal (BD in your original diagram) has taken on a fixed value of half the supply potential.

I wrote the simulation and produced the plot using Mathcad, but I imagine that similar results cold be obtained using Matlab.

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GreenPrint

rest of points in the field sort themselves out according to the simple rule of taking on the average of surrounding values over many iterations

How do you find the rest of the points if you don't know the surrounding values and only that one corner should be zero and the other the voltage produced by the DMM to measure resistance?

Also, what exactly did you integrate to get those values?

gneill

They are calculated with an algorithm. A loop goes through every entry in the array (except the two fixed corner points) and sets its new value to the average of the current values in all the entries immediately adjacent. The process is repeated until the values converge (stop changing).

GreenPrint

Could you please post the algorithm which you used?

gneill

Let V be an n x n array representing the square slab. n should be chosen large enough to make a reaslistic simulation.

1. Initialize the array:
- set V(0,0) to 1 {this is the "+" powered corner}
- set V(n,n) to 0 {this is the "-" (or grounded) powered corner}
- set all other entries to 1/2

2. For each array element V(i,j) except for the fixed corner values (so not V(0,0) or V(n,n)):
- set that array entry equal to the average of the elements that surround it

3. Repeat 2 until values stop changing (convergence).

rude man

The potential distribution is the solution of LaPlace's equation, namely del²V = 0. Here this is a 2nd order partial differential equation in two independent variables so it needs 4 boundary values to compute. We know the lower right-hand corner potential is 0 and the upper left-hand one is V₀. By symmetry we also know the potential at the middle of the square is V₀/2. Unfortunately right now I can't think of a fourth boundary value.

Again by symmetry, along the main (top left to bottom right) diagonal the potential drop a distance s from the top left corner = the drop from the same distance s fro the bottom right-hand corner. Also by symmetry, the potential at the opposite corners must be the same.

Somehow finite element software knows how to extract the additional needed information. Looking at gneill's graph it looks like the potential is constant along the entire opposite diagonal. This of course would give us a fourth boundary value. I might pursue that a bit further (or not ).

GreenPrint

So this interesting way to think of it... all geometry and symmetry.

I haven't take differential equations or multivariable calculus in a while.

del^2 V(x,y) = 0

So because you could fold the square in half diagonally and voltage decreases as you get away from the top left hand corner, this diagonal should be at all the same potential?

What exactly should V_0 be? This is also an interesting question. Isn't it different for every DMM?

rude man

Well, when you hooked up your second DMM in voltage mode, did you not record the voltage at the top left-hand corner?

As for an equipotential being along the off-diagonal, I noticed that from gneill's graph. If you look at his graph and the way the equipotentials bend near the V_0 and ground corners, I think it can be argued persuasively that the off-diagonal potential has to be a straight line. So the potential at either off-corner would be V_0/2, same as the middle where I know it's V_0/2.

GreenPrint

I still don't see how you are supposed to create a Voltage color map. Could you please help me with this? I used to know how to use MATLAB very well. It has been a year however since I have touched it. Are there other programs I could use?

gneill

Excel will produce contour plots (under the category "surface charts").