Resistance Variable Cross Section

[millifarads]

Homework Statement

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.

Homework Equations

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?

 

[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.

 

[millifarads]

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

 

[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.

 

[HalfLostAndSearching]

Yes, you are on the right track. Since the cross sectional area is variable, we need to take into account the changing radius as we move along the disk. Your approach of setting up an expression for a one dimensional line along the top of the disk and then multiplying by the width is a good start. However, you will need to integrate this expression over the entire cross sectional area to get the correct value for R.

To set up the integration, you can use the formula for the area of a circle, A = πr^2, and substitute in your expression for the radius. This will give you an expression for the cross sectional area in terms of x, the distance along the top of the disk. Then, you can integrate this expression from 0 to r (the radius of the disk) to get the total cross sectional area.

Once you have the cross sectional area, you can plug it into your original equation, R = ρL/A, to get the resistance in terms of the variable cross sectional area. Remember to use the correct value for L, which in this case would be the circumference of the disk, 2πr.

I hope this helps guide you towards a solution. Good luck!