# Resistance of a disk

1. Mar 10, 2010

### baronbob

Resistance=resisivity(of a certain material)Xlength/cross-sectional area
resisivity is expressed in ohm-cm
I have a graphite material that is 6ohm-cm
the material is .032" thick
I will cut the graphite material into a 3' disk
I will place an electrode in the center of the disk and have another electrode completely around the circumferance of the disk.
What will the resistance of the disk be?
How will the resistance between the two electrodes vary with disk diameter?
Will the resistance be proportinal to the thickness of the disk? (ie will the resistance of a .064" disk be twice that of a .032" given identical diameters?)
Later I will as to calculate resistance with a thickness that varies from the center to the circumfrence.

2. Mar 10, 2010

### Okefenokee

Is this homework? People around these parts frown upon being asked to do homework.

I'll give you a hint: The differential resistive element will be "$$\sigma\frac{dr}{r d \theta \tau}$$" where "(rho)" is the resistivity of the material, "$$d r$$" is the differential radius, "$$d \theta$$" is a differential element of the angle theta, "$$r$$" is the radius, and "$$\tau$$" is the thickness of the disc. Note that the element is in the form of "resistivity*length/cross_section" just like your formula. You'll need a double integral in "$$r$$" and "$$\theta$$" to calculate the total resistance.

That will be revealed by the integral

yes

3. Mar 10, 2010

### Staff: Mentor

Welcome to the PF. Oke took the words right out of my mouth (dang you!). What is the context of your question? That will help to guide our assistance in your questions.

Are you comfortable with integral calculus? If so, the answers to your questions are straighforward. If not, it may be harder for us to help you to calculate this. We can probably still help, but the help will be different. Again, that help will depend on the context of the question.

4. Mar 11, 2010

### Okefenokee

Ahh, this turns out to be more interesting that I thought. This integral doesn't converge. The reason is that at the center, r = 0, the cross section of the resistor becomes 0 so it's resistance becomes infinite. We need to examine the geometry of how you connect to the disc very carefully. Is it going to be touching the surface or embedded in the center? That would make a small difference.

Anyway, the solution to the integral is:

$${\frac{\sigma}{2\pi \tau} log(r)}\right]^{r=OuterRadius}_{r=InnerRadius}$$

As an estimate, we can use the width of the electrode and simply integrate from r = (width of electrode) to (outer radius). I suspect that it will be really close to the true value. For a detailed analysis, we need to consider the electric fields at the contact and trace the current density vectors. It might be better to simply measure the resistance for many diameters. We know that the result should be on a logarithmic scale. We could expect a nearly straight line on a log plot of the resistance vs. diameter. As the diameter of the disc gets really small (near the size of the electrode, we should expect the effect of current-bunching at the contact to become prominent. The resistance will be a little higher than you might expect if you extrapolate downward from larger diameters.

That's all overkill though. I'm sure the first integral is close enough.

baron, you asked me in a notification for a way to contact me. You can do it here if you like. Each message you place in this thread will be forwarded to my Email.

Last edited: Mar 11, 2010
5. Mar 11, 2010

### Studiot

Are you sure it's not inversely proportional?

6. Mar 11, 2010

### baronbob

Thanks for your response. I am new to this form, but have all ready managed to meet the police first off. We can use the diameter of the center connection point, call it .025". The application is one of heating. I have a piece of graphite I have purchased in an effort to design a disk shaped heater. The end material I will be using can vary in thickness. I too am concerned about the current passing through a smaller and smaller cross-section and therefore plan to increase the material thickness closer to the center. Examining a current vector from the center to the circumference, a decrease in thickness increases resistance and therefore wattage (ixi)xR. That is why I hinted towards an expression varied to material thickness at the end of my initial thread. AS far as my ability to deal with integral calculus, it has been a long time (Carnegie Tech 1967). If necessary I will return to any level of understanding necessary to fully understand this project. I hope I have or can convince you and or the police that my interest is not homework related. Thank you for your help thus far. If we ever meet I would enjoy talking further about this project and others our lab gets involved with.

7. Mar 11, 2010

### Okefenokee

Yeah, I'm pretty sure this is not a homework problem. Unless you go to one of the insanely hard schools that I've heard horror stories about.

We can find a "thickness" function to suit your needs by setting up some differential equations. First let's simplify this a bit. Consider a cylindrical shell of differential width. The resistance of the shell will be: (Note that "R" is resistance and "r" is radius)

$$dR = \sigma \times \frac{dr}{2 \pi \times r \times \tau(r)}$$

The thickness (tau) is now a function. The total current passing through the shell should be the same as what goes into the electrode. We'll call it "i". I'm thinking that you want the heat output to be the same at all points on the disc. In other words, a constant.

$$Power = C = i^2 \times \sigma \times \frac{dr}{2 \pi \times r \times \tau(r)}$$

Rearranging, we get:

$$\tau(r) = i^2 \times \sigma \times \frac{dr}{2 \pi \times r \times C}$$

There you go. You don't even have to do anything fancy. You need a thickness profile in the form of (1/r) to distribute the heat evenly. If you think about it, it makes perfect sense. A (1/r) profile should cancel out a logarithmic resistance. The profile will be really thick in the middle where the resistance would normally be highest for a simple disc.

You could do more here to control the heating profile if you wanted. You could make C into a funtion of r and do some differential equation solving. Also, it looks like you need something that will be roughly horn shaped.

Do you have access to Ansoft? I only used it once back in college but it seems like this would be a perfect use for it. You could draw up a design, simulate it, then tweak it to your heart's content. You wouldn't have to trust the math of us internet schmucks either.

Last edited: Mar 11, 2010
8. Mar 11, 2010

### Okefenokee

Oops,

I made a little mistake in that last post. We want the same power per unit arc length at any given radius. The power has to go up as the radius gets larger. That way, the power distributes out evenly around the circumference to the same value for any given radius. That means that we want this:

$$Power = C \times r = i^2 \times \sigma \times \frac{dr}{2 \pi \times r \times \tau(r)}$$

Rearranging, we get:

$$\tau(r) = i^2 \times \sigma \times \frac{dr}{2 \pi \times r^2 \times C}$$

You'll want a (1/r^2) profile. It may be kinda hard to build. It'll get thin really fast.

9. Mar 11, 2010

### Studiot

10. Mar 11, 2010

### Okefenokee

You were right, it's inversely proportional.

It will be 1/2 not 2 times as much.

11. Mar 11, 2010

### baronbob

Thanks folks!! I wil be puting all this to a test here in the next couple of weeks. I am testing a new Thermal program called win-therm. I may be able to combine that program with a modifacation to a solid scape type program that will electrically then thermally simulate objects with both electrical conductivity and thermal conductivity. How is that??

12. Mar 12, 2010

### Studiot

Look forward to some pictures, Bob.