# Homework Help: Electric Circuit Question

1. Jan 27, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

http://img26.imageshack.us/img26/7307/captureghl.png [Broken]

I have a square that is 5 cm by 5 cm. I have filled it in with graphite from a pencil. The top left corner I have labeled A. The top right corner B. The bottom right hand corner C. The bottom left hand corner D. I take a DMM probe and set it to measure resistance, I put the red probe at A, the black probe at C. I then take a second DMM and set it to measure DC voltage. I place the black probe at C.

At this point I take the red probe and place it around the corners of the squares within the square that are each a one centimeter by one centimeter and measure the voltage. I collect 25 data points.

I observed the following data through measurement. The units are Volts. Please note I labeled the corners in my picture.

http://img577.imageshack.us/img577/382/captureyfu.png [Broken]

My measured data is obviously not completely accurate, but according to my measured data, I would say that the further away you get from corner C the less the voltage is. This however isn't true along the lines DC and CB. The voltages seem to increase along these lines.

I'm asked to explain my collected data for a lab report.

2. Relevant equations

R = $\frac{ρl}{A}$

Where:
R = resistance
ρ = resistivity (in the case of graphite ρ≈1*$10^{-5} Ωm$
l = length
A = cross sectional area

A = wh

Where:
w = width
h = height (thickness of the graphite layer)

V = IR

Where:
V = voltage
I = current
R = resistance

3. The attempt at a solution

I started off by putting in the formula for resistance into Ohm's law in the hopes that I would be able to explain the observed data.

V = I$\frac{ρl}{A}$

Now, I guess I just don't understand this. Sense the square is completely filled in, isn't the surface area and length remain constant. I'm not looking at a wire but a completely filled in square. So no matter where I put the red probe within the square to measure the voltage between the red probe and black probe the same resistance exists.

I guess I just don't understand why I'm getting different measurements.

Last edited by a moderator: May 6, 2017
2. Jan 27, 2013

### rude man

In the first place I think it is likely that your graphite layer is (1) not uniform in thickness, and (2) uneven in texture and therefore conductivity. Also, your measurement setup looks bad. For example, the voltage in the extreme bottom right-hand corner should be zero regardless of the graphite layer error sources.

3. Jan 27, 2013

### GreenPrint

I agree. The collected data is probably not accurate. There is no way that the graphite layer is of uniform thickness or that I actually measured the voltages at the correct points. Yes, it's supposed to be zero. But, what exactly was my data supposed to show, and why? this is where I'm confused.

Thanks for any help =)

4. Jan 27, 2013

### Staff: Mentor

Careful, many multimeters in resistance mode will place the positive measurement voltage on the black lead, contrary to expectations. Did you measure the voltage at the probes with the other DMM?

Last edited: Jan 27, 2013
5. Jan 27, 2013

### GreenPrint

So you mean I was supposed to get negative voltages?

So the red probe at A and the black probe at C should be measuring resistance. This is done by giving of a voltage and finding the difference.

The black probe with the second DMM set to measure voltage at point C as well. I used the red probe from second DMM to measure the voltages

So no I did not measure the voltages at point A (the red probe set to measure resistance).

I did measure the voltage at the two other black probes at C. I should have gotten zero but I got close to it.

Not sure though if it helps.

6. Jan 27, 2013

### Staff: Mentor

I'm just saying, you can't always trust that the red lead will be positive and the black lead negative for a multimeter in resistance mode. It might have been handy to know for sure the size and polarity of the voltage driving your experiment.
A multimeter measures resistance by placing a small test voltage across the resistance and measuring the resulting current. R = E/I.
I'm not sure what the values you found are due to. I'm a bit mystified by the isolated 0.485V reading nearby point C.

7. Jan 27, 2013

### GreenPrint

Ok so then in that case my data is really screwed up. I'm however not going to lie about the data I received. It's probably because the thickness of the graphite layer wasn't uniform and I didn't actually measure the data at the right spots.

In an ideal case though of uniform thickness and correct measuring instruments, what should I have seen? Shouldn't the resistance been the same throughout the whole entire square and there the voltage readings as well?

8. Jan 27, 2013

### Staff: Mentor

Just from intuition I'd guess that there would be a general gradient of potential from the + lead down to zero at the - lead of the source along the diagonal connecting the source leads, and some symmetrical distribution of values surrounding the diagonal. Offhand I don't know a quick way to compute the distribution for a continuous conducting slab of material.

9. Jan 27, 2013

### GreenPrint

So this is because why? So between A and C, the shortest path would be the diagonal of the square. So voltage should be the highest on this line. The voltage should than decrease uni formally as you move away from the line.

So basically current flows through the square from point A to point C as if it where one giant wire and as you move away from the shortest path (the diagonal) the current has to go through more graphite. Surface area increases so resistance decreases so voltage does as well? The whole square experiences the same current just how one end of a node must experience the same current at the other node.

Is this the logic your suppose to use?

Also do you know how to create a color map on a computer? I'm supposed to create a color map with this data and don't know of software to do it =(

10. Jan 27, 2013

### Staff: Mentor

I don't know what logic is expected to be used here; I just imagined the diagonal as a single thin resistor element which would have a graduated potential along it, and then added in parallel resistive paths of increasing end-to-end length surrounding it, bulking out the geometry of the slab. Dunno if this thought experiment model is realistic.

I'm reminded of heat transfer problems with heat sources and sinks, where differential equations with boundary values need to be solved over the slab to determine the temperature profile. Laplace and Fourier spring to mind... Not my cup of tea, really. Sorry I can't be more helpful.
I'd look for graphing software, maybe GNUPLOT, and look for keywords like "level plot", "surface plot", and "contour plot". Could be that matlab can do it... do a google search on "matlab plot gallery".

11. Jan 27, 2013

### GreenPrint

See that's the thing. I don't know what to think of this thing. It was easier I was just dealing with wires and components. This however confuses me. Ideally I'm suppose to have a uniform layer of graphite. I'm sure if I should think of it as one giant variable resistor perhaps where R = (rho*l)/A. But see... in that case it's not a variable resistor just one giant resistor with a defined A and l that dose not change no matter where put the red probe. But see like in this case every voltage should be the same. Like I don't know if current travels the shortest distance or it just travels through the whole square or what. I just don't know what to think of this square in terms of one resistor or many, and how placing your probe in a certain location has anything to do with it.

Like in your logic the current goes through the diagonal resistor first. But why not just through the whole square equally?

This is very confusing... especially when I was never told how to think of this... I'm just left to guess =(... I don't know what think of the square???

12. Jan 27, 2013

### GreenPrint

Like I don't even know what voltage difference I'm measure when I place the red probe in spots. Like in the picture below pretend the red circle is the red probe. Am I measure the voltage difference that's the red line despite the whole square being covered in graphite and the whole square is conductive of electricity with the red probe and not just the red line?

http://img594.imageshack.us/img594/377/captureldt.png [Broken]

Or something like this or perhaps the whole entire square sense it's all connected together.

http://img4.imageshack.us/img4/2931/capturevz.png [Broken]

Last edited by a moderator: May 6, 2017
13. Jan 27, 2013

### Staff: Mentor

No, the potential changes along the length of even standard resistors. In fact, that's how the usual variable resistor (potentiometer) works; it's essentially a long thin slab of resistive material with connections at either end and a "probe" that contacts it at moveable locations along its length and so picks off different potentials.
Current flows through all available paths, but the current distribution will depend upon the geometry of the object.
I didn't mean to imply that the current flows through the diagonal "first". It was just single well defined path that I could identify and build upon for a mental model.
Agreed. I'll ask around and see if I can find anyone who's seen this type of homework exercise before.

14. Jan 27, 2013

### Staff: Mentor

Voltage is a scalar value that tells you the electrical potential at some location with respect to some other location. The two locations can be anywhere in space, so the concept of potential difference between points on an object is valid. If you want, picture the graphite as a near infinitely dense network of resistors and you're measuring the potential between two nodes.

Last edited by a moderator: May 6, 2017
15. Jan 27, 2013

### GreenPrint

Interesting. So the .485 volts that makes no sense. This means that this point is .485 volts higher than point C. This is all this means. It's a scalar. Now what exactly is the length or cross sectional area that you are suppose to in vision for the resistance at this point, as opposed to other points?

I have no idea.

16. Jan 27, 2013

### Staff: Mentor

"How does the potential come about" is a different question than "What is the potential"; Measuring is easy, calculating is not so easy

If there's a trick to making the calculation simple for this geometry I haven't spotted it yet.

17. Jan 27, 2013

### GreenPrint

Perhaps I'm over complicating things. But lets say we take the red probe and place it as close as it can be to the black probe without touching it . Lets make corner C the origin of a coordinate system. To the left is positive. Upwards vertically in the same plane is positive. Vertically upwards out of the plane is positive. Assuming uniform thickness

So the probe would be at point (dx,dy)

I'll just call the width x and the length y.

So the voltage would be

$dV = I\frac{ρ dy}{dx*h}$

Now as you move away from the corner C... idk but I think maybe perhaps I'm on the right path???

18. Jan 28, 2013

### Staff: Mentor

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.

19. Jan 28, 2013

### 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.

20. Jan 28, 2013

### Staff: Mentor

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.

#### Attached Files:

• ###### Fig1.gif
File size:
11.5 KB
Views:
181
21. Jan 29, 2013

### 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?

Last edited: Jan 29, 2013
22. Jan 29, 2013

### Staff: Mentor

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).

23. Jan 29, 2013

### GreenPrint

Could you please post the algorithm which you used?

24. Jan 29, 2013

### Staff: Mentor

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).

25. Jan 29, 2013

### 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 ).

Last edited: Jan 29, 2013