What is the Resistance of a Cylinder Conductor Made of Aluminum?

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To calculate the resistance of a cylindrical aluminum conductor with a radius of 2 mm and a length of 2.33 m, the formula R = ρL/A is used, where ρ is the resistivity (2.75E-8 Ωm), L is the length, and A is the cross-sectional area. The area is calculated as π times the radius squared, which is critical for obtaining the correct resistance value. An initial miscalculation led to an incorrect resistance of approximately 0.51 Ω instead of the correct 0.0051 Ω. The error was identified as a misunderstanding of unit conversions, specifically the radius measurement. This highlights the importance of careful unit handling in electrical calculations.
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Homework Statement


A cylindrical conductor of radius 2 mm and length 2.33 m is made of aluminum (p=2.75E-8\Omegam). Calculate the Resistance of the conductor.


Homework Equations


Unknown (not given).


The Attempt at a Solution


I purely do not know how to do it. I'm thinking that we find the volume of the conductor then multiply it by the 2.75 but I don't think that works.
 
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Look up "resistivity".
 
Welcome to PF!

Hi JSGandora! Welcome to PF! :smile:

(have a rho: ρ and an omega: Ω :wink:)

Look at the units … Ω.m, not Ω.m3 :wink:

(actually, that m is really m2/m, ie area/length :biggrin:)
 
Thank you! :)

I got the following equation:
R=\frac{\rho L}{A}
where R is the resistance, p is the resistivity (by the way, what is the symbol for the "p"-like symbol called?), and A is the cross sectional area. Therefore, we have
R=\frac{2.75\Omega m\times10^{-8}(2.33m)}{(0.0002m)^2\pi}\approx0.51\Omega?

The answer key says the answer is 0.0051\Omega. It looks like I am wrong by a factor of 100. I can't see why though...hmm...
 
Last edited:
erm :redface: … too many 0s in your 2mm ! :wink:

(and of course, it's called "rho")
 
OHHH, failure on my part. Thanks for pointing that out, I thought there were 100 millimeters in one centimeter when I did that. ><
 
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