Resistance Variable Cross Section

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Homework Help Overview

The discussion revolves around calculating the resistance of a cylindrical disk with a variable cross-sectional area, where the connections are made along the cylindrical sides rather than the flat ends. Participants are exploring the implications of this setup on the resistance calculation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set up an expression for resistance using the formula R = ρ(L/A), questioning how to define the cross-sectional area given the variable nature. Some participants inquire about the specifics of how connections are made to the resistor and suggest that a visual representation could clarify the setup.

Discussion Status

Participants are actively discussing the setup and implications of the problem, with some suggesting the need for integration to account for the variable cross-section. There is a recognition of the complexity involved in determining the cross-sectional area and how it affects resistance.

Contextual Notes

There is uncertainty regarding the exact configuration of the connections to the resistor and how the metalization is applied to the disk, which may affect the approach to solving the problem.

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Homework Statement



The question asks for resistance R of a disk with radius r and fixed width w, whose cross sectional area is variable. Unlike in the picture below, the resistor is not connected to the circuit on the flat ends, but on the cylindrical sides.


Homework Equations



2_3_4_1_eng.png


The Attempt at a Solution



With R = \rho \frac{L}{A}L = 2r, ρ is given, but I'm having trouble setting up an expression for the cross sectional area. Normally, for cross sectional area of this type of problem it's just the area of a circle, and some integration will be required with the bounds being the radius at each end.

But here my initial thought is to set up an expression for a one dimensional line along the top of the disk, then multiple that by the width of the disk, w. The area will change with the radius though, so I'm not sure if integration is required.

From the equation for the circle of radius r, I get

y = \sqrt{r^2-x^2}

which then needs to be multiplied by 2, and then by w to get a sort of cross sectional area:

R = \rho \frac{2r}{2w*\sqrt{r^2-x^2}}

Am I on the right track?
 
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How are the connections made to the resistor? Is the curved area divided into two with conductor deposited along the entire curved surface except for two thin cracks, or ? A picture would be a good thing to have here.
 
Attached is a picture. The disk has radius 2r, I'm assuming the wires connect somewhere on the cylindrical sides.
 

Attachments

  • disk.png
    disk.png
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The picture still does not show exactly how the metalization is deposited. For example, were the flat ends connected instead, the metalization would be across the entire flat surfaces.

So let's assume two diametrically opposed and arbitrarily thin stripes running the length of the cylinder. Then you can integrate dR from o to r and multiply by two. You definitely need to integrate.
 

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