What is the resistance of a hollow aluminum cylinder with given dimensions?

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To determine the resistance of a hollow aluminum cylinder, the relevant formula is R = (resistivity * length) / area. The inner and outer surfaces of the cylinder must be treated as equipotential surfaces, but the area used in the resistance calculation should be the cross-sectional area through which the current flows, which is the difference between the outer and inner areas. The resistivity of aluminum is provided, along with the cylinder's dimensions, allowing for the calculation of the resistance. Understanding the relationship between length and cross-sectional area is crucial for accurately determining resistance. The discussion emphasizes the importance of selecting the correct area for the calculation.
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Homework Statement



A hollow aluminum cylinder is 2.5 m 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=[(resistivity of hollow cylinder)*(length of hollow cylinder)]/Area

V=IR

The Attempt at a Solution



I was given the resisitivity of the aluminum cylinder and the length, but I am at a loss: the area inner part of the cylinder is less than the area of the outer part of the cylinder. My question is this: I don't know what area to use: do I use the area of the outer cylinder or the inner cylinder. Thanks.
 
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A is the cross-sectional area of the current flow.
Think about why both the length and A are in the equation for resistance. Imagine their effects on a microscopic level.

Can you do it now?
 

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