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

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SUMMARY

The resistance of a hollow aluminum cylinder measuring 2.5 m in length, with an inner radius of 3.20 cm and an outer radius of 4.60 cm, can be calculated using the formula R = [(resistivity of aluminum) * (length)] / (cross-sectional area). The cross-sectional area for current flow must be determined by subtracting the inner area from the outer area, as the current flows through the hollow section. The correct area to use in the resistance calculation is the annular area between the inner and outer surfaces.

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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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