How Does Resistivity Vary in a Conical Material with Changing Dimensions?

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Homework Help Overview

The discussion revolves around the resistivity of a conically-shaped material in a circuit, where the resistivity varies with position along the cone. The problem involves calculating resistance using the relationship between resistivity, length, and cross-sectional area.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the varying radius and the cross-sectional area of the cone, questioning the initial assumption about the area formula. There is discussion about integrating with respect to the cone's dimensions and the implications of varying radius on the calculations.

Discussion Status

Some participants have provided clarifications regarding the correct formula for the cross-sectional area and suggested integrating with respect to the correct variable. There appears to be a shift towards a more accurate approach in calculating resistance, but no consensus has been reached on the final method.

Contextual Notes

Participants are working within the constraints of the problem as stated, including the specific resistivity function and the linear variation of the radius. There is an ongoing examination of the assumptions made about the geometry of the cone and its impact on the calculations.

DrIxn
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Homework Statement


A piece of conically-shaped material is placed in a circuit along the x-axis. The resistivity of this material varies as rho=(6*10^6)*x^4 (where x is measured in meters and rho is measured in ohm*meters), and its radius varies linearly as a function of x, ranging from r1=xinitial=3 cm to r2=xfinal=8.4 cm.


Homework Equations


R=rho*L/A


The Attempt at a Solution



Well current is flowing through the cone section so the cross sectional area is a circle, with the radius increasing proportional to x, in fact the are equal.

So dR=rho*dx/(2*pi*r) and dx=dr and x=r so plugging in for r

dR=(6*10^6)*r^4/(2*pi*r) dr = (3/pi * 10^6)*r^3 dr

which integrating from 0.03 to 0.084 m i got 11.69 Ohm, doesn't seem quite right did I mess up somewhere?
 
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You sure that ##A = 2 \pi r## ?
 
Would it be something else? A bunch of little rings expanding out would make a circle yes?
 
DrIxn said:
Would it be something else? A bunch of little rings expanding out would make a circle yes?

Shouldn't A be the cross-sectional area of the cone? It didn't appear that you were integrating in the radial direction...
 
Okay so the cross sectional area would be a circle, A=pi*r^2 , and since r varies linearly with the length of the cone A=pi*x^2

And integrating with respect to x..

(6/pi*10^6)*x^2 dx from xi to xf?
 
DrIxn said:
Okay so the cross sectional area would be a circle, A=pi*r^2 , and since r varies linearly with the length of the cone A=pi*x^2

And integrating with respect to x..

(6/pi*10^6)*x^2 dx from xi to xf?

Yup, that looks better.
 

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