Calculate Resistance of Frustrum Wire

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    Resistance Wire
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Homework Help Overview

The discussion revolves around calculating the resistance of a wire shaped like a frustrum, which is defined as the solid remaining when a cone is truncated. The problem involves parameters such as the radii of the upper and lower planes, the vertical length of the frustrum, and the resistivity of the material.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the resistance formula involving resistivity, length, and cross-sectional area. There are inquiries about how to set up the integration process and what expression to use for the cross-sectional area. One participant suggests using elemental resistors in series.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to approach the integration and the expression for the cross-sectional area. There is a sense of progress as one participant expresses understanding of the solution process.

Contextual Notes

There is an emphasis on the need for integration and calculus in solving the problem, and participants are exploring the relationships between the dimensions of the frustrum and the resistance calculation.

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How do I calculate the resistance of a wire shaped like a frustrum (in layman's terms, the solid that remains when a cone is removed from the top of a bigger cone)?

Let the radius of the smaller upper plane of the frustrum be a, the radius of the larger lower plane of the frustrum be b, the vertical length of the frustrum be l, and the resistivity of the material be r.

I believe the equation
Resistance = Resistivity * Length / Cross-sectional Area
can be applied. Some integration and calculus will also be needed in solving this problem, but I do not know how to begin.

Should I take an elemental cross-sectional area of the wire and integrate?
 
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Yes. You are effectively adding all the elemental resistors in series.
 
How do I obtain the general expression to integrate?
What is the expression for the cross-sectional area I should use?

dR = r (dl) / (Expression for cross-sectional area?),
where d represents a small quantity
 
Use simple proportions for the radius of the cross section: [itex]R = b + \frac {a-b}{L} z[/itex] where z goes from 0 to L. The cross-sectional area is just [itex]\pi R^2[/itex].
 
All right, I finally understand how to solve the problem.

Thank you for your help!
 
You are most welcome!
 

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