Calculate Resistance per Meter of a Hollow Cylindrical Aluminum Conductor

[FatoonsBaby71]

Help with conductors!

Homework Statement

Determine the resistance per meter of a hollow cylindrical aluminum conductor with an outer diameter of 32mm and wall thickness 6mm.

Homework Equations

R = length / sigma * Area
Sigma = conducitivity

The Attempt at a Solution

The answer is 53.4microohms.

Well I know how the darn equation I supplied works. But how can this problem be solved without the length given it seems like that would need to be supplied? Does anyone know of any tips for this problem.

Thanks so much

 

[mgb_phys]

It asks for ohms/m so the length is 1m !

 

[Doc Al]

Determine the resistance per meter

You don't need the length. You are finding the resistance per meter, not the resistance.

 

[FatoonsBaby71]

I see the reason why the length is not needed and you can use 1. But in this case is the area... just half of 32 for the radius which is 16mm and since the wall thickness is 6m i would want to take the area from 0 to 10 mm?

 

[Doc Al]

But in this case is the area... just half of 32 for the radius which is 16mm and since the wall thickness is 6m i would want to take the area from 0 to 10 mm?

You want the cross-sectional area of the aluminum-filled part of the cylinder. The part from r = 0 to r = 10mm is the empty part.

 

[FatoonsBaby71]

Okay so i understand what your are saying...So the set up should look like this...

1/(38.2*10^6*(pi*(6^2))) = 2.3146*10^-10 ? Which does not even look remotely close to the answer. From what you stated it looks like the empty part is from 0 to 10 and then the radius in total is 16 so that's why i used the 6.

 

[Doc Al]

First find the cross-sectional area by itself. Hint: What's the area of the full cylinder with r = 16mm? What's the area of the hollow part? (Just set it up without crunching the numbers. Use standard units.)

 

[FatoonsBaby71]

Oh, I see ... Now I understand. We take the total area and take away the hollow part leaving us with the filled area... then plugging that into the equation gets us the result...

Thanks Doc you were such a good help..

I hope you don't mind...I don't want to bug you...but could you provide some guidance to my other topic "Help with resistance" I seem to get the right answer, however converting everything to meters I still get the wrong order... I would really appreciate it

Thanks a bunch