Resistance of a cone section

  • Thread starter DrIxn
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  • #1
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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?
 

Answers and Replies

  • #2
gneill
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You sure that ##A = 2 \pi r## ?
 
  • #3
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Would it be something else? A bunch of little rings expanding out would make a circle yes?
 
  • #4
gneill
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2,866
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...
 
  • #5
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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?
 
  • #6
gneill
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20,925
2,866
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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