# Homework Help: Resistance Variable Cross Section

1. Jul 21, 2014

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

The question asks for resistance R of a disk with radius r and fixed width w, whose cross sectional area is variable. Unlike in the picture below, the resistor is not connected to the circuit on the flat ends, but on the cylindrical sides.

2. Relevant equations

3. The attempt at a solution

With $$R = \rho \frac{L}{A}$$L = 2r, ρ is given, but I'm having trouble setting up an expression for the cross sectional area. Normally, for cross sectional area of this type of problem it's just the area of a circle, and some integration will be required with the bounds being the radius at each end.

But here my initial thought is to set up an expression for a one dimensional line along the top of the disk, then multiple that by the width of the disk, w. The area will change with the radius though, so I'm not sure if integration is required.

From the equation for the circle of radius r, I get

$$y = \sqrt{r^2-x^2}$$

which then needs to be multiplied by 2, and then by w to get a sort of cross sectional area:

$$R = \rho \frac{2r}{2w*\sqrt{r^2-x^2}}$$

Am I on the right track?

2. Jul 21, 2014

### rude man

How are the connections made to the resistor? Is the curved area divided into two with conductor deposited along the entire curved surface except for two thin cracks, or ??? A picture would be a good thing to have here.

3. Jul 21, 2014

Attached is a picture. The disk has radius 2r, I'm assuming the wires connect somewhere on the cylindrical sides.

#### Attached Files:

• ###### disk.png
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4. Jul 21, 2014

### rude man

The picture still does not show exactly how the metalization is deposited. For example, were the flat ends connected instead, the metalization would be across the entire flat surfaces.

So let's assume two diametrically opposed and arbitrarily thin stripes running the length of the cylinder. Then you can integrate dR from o to r and multiply by two. You definitely need to integrate.