Calculating Resistance of Coaxial Cable with Isolator

[bolzano95]

Homework Statement

Coaxial cable has radius a of copper core and radius b of copper shield. Between there is an isolator with specific resistance ζ. What is the resistance of this cable with length L between the core and the shield?

Homework Equations

First, I tried to solve this like this:
$$R= ζ \cdot \frac{l}{S}$$
In our case the length is dr, and therefore I suppose that the area of this ring is 2πrdr:
$$dR= ζ \cdot \frac{dr}{2πrdr}$$

The Attempt at a Solution

The solution sheet says: $$dR= ζ \cdot \frac{dr}{2πrL}$$
I know that something is wrong with my equation, because dr goes away and then I cannot integrate from a to b. But why is in the solution L instead of dr? As I understand problem instruction L= b-a. And therefore L=dr which doesn't make sense to me.

 

[kuruman]

The current flows in the radial direction. A volume element is $dV=rdrd\theta dz.$ The resistance of this is $dR=\frac{\zeta dr}{rd\theta dz}$ I would integrate over $r$ first to get the resistance of a radial sliver, $dR_{sliver}$. I can do that because the stacked elements $dV$ in the radial direction are in series and their resistances add. Then note that all such slivers are in parallel which is how they should be added.

 

[rude man]

Between there is an isolator with specific resistance ζ.

should be ".. insulator with specific resistivity ... "

 

[bolzano95]

True. Insulator with specific resistivity ζ. Thanks.

 

[bolzano95]

I got an answer from fellow student that the area of the element is 2πrL and the thickness of the element is dr. But I wouldn't say so. I would suppose that the area of the element is 2πrdr. What am I missing?

 

[kuruman]

I would suppose that the area of the element is 2πrdr. What am I missing?

Please read and understand post #2 to see what you are missing.