Resistance of a hollow aluminum cylinder

AI Thread Summary
The discussion centers on calculating the resistance of a hollow aluminum cylinder with specified dimensions. The resistance is determined using the formula R=(ρ/2πL)ln(R/r), where R is the outer radius, r is the inner radius, and L is the length of the cylinder. There is confusion regarding whether the current flows radially through the cylinder and how the ohmmeter connects to the surfaces. The assumption is that the connection is equipotential across the surfaces, which influences the resistance calculation. Clarification on the approach and detailed calculations is needed for accurate results.
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Homework Statement



A hollow aluminum cylinder is 2.50m long and has an inner radius of 3.20 cm and an outer radius of 4.60 cm. Treat each surface (inner, outer, and the two end faces) as an equipotential surface.

At room temperature, what will an ohmmeter read if it is connected between the inner and outer surfaces?

Homework Equations



R=4.60cm
r=3.20cm

R=\rhoL/A

and possibly

R=(\rho/2piL)*ln(R/r)

The Attempt at a Solution



Well, the first part of the question (not asked here) was to find the resistance from one end of the hollow cylinder to the other (the faces). I found that: 2.00*10-5 \Omegas. The problem does not say that the current is flowing radially through the cylinder though, but is that to be assumed for doing this second part? Or not? I plugged everything into the second equation, but the answer is wrong. Should the resistance be the same as it is from end to end? Is it 0?
*confused*
 
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It is a curious question. Connected exactly how to the two surfaces? Connection at infinitely small points will lead to infinite resistance, so it seems we must assume the contact is spread, equipotentially, across each entire cylindrical surface.

That should mean your second equation is appropriate. See e.g. https://www.miniphysics.com/uy1-resistance-of-a-cylindrical-resistor.html.
Can't say any more without seeing your working and/or answer.
 

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