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Effective resistance of truncated conical cylinder

  1. Feb 18, 2005 #1
    Hi, I'm having trouble doing this problem:
    A truncated conical cylinder of graphite (bulk resistivity [tex] \rho = 1/\sigma [/tex]). The top of the cylinder has radius r = a, the bottom has r = b (b>a). Find the effective resistance between top and bottom of the cylinder. Show that the expression reduces to the usual one ([tex] R = \rho L / A [/tex]) when a = b.

    I know I need to do some kind of integration from a to b, but I really don't know how to set this up. Thanks.
  2. jcsd
  3. Feb 18, 2005 #2


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    Just treat all the tiny cross sections as a bunch of resistors in series. That is, get the resistance for each one in terms of p, r (which goes from a to b), and dy, and integrate over y.
    Last edited: Feb 18, 2005
  4. Feb 18, 2005 #3
    The professor said we shouldn't need to solve Laplace's equation for this problem. I'm thinking the effective resistance is the sum of resistive disks in series. And the integration would be integrating the height from 0 to L.

    But first of all, the potential at a point z in between 0 and L would be just V(z) = (Vo/L)z, and therefore E = -Vo/L in the z-direction.

    About the integration, do I have to both integrate the radius from a to b AND the height from 0 to L. Or should there be some kind of relationship between the radius and height (radius as a function of height)?
  5. Feb 18, 2005 #4
    Ok, by doing sum of resistive disks, I got to this point:
    [tex] dR = \frac{\rho dz}{\pi r^2} [/tex]
    where r is the radius of a disk at height dz. The problem now is how to relate r and z because I'm pretty sure there's a dependence between radius and height. thanks.
  6. Feb 18, 2005 #5


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    yea sorry about that, I realized there was a much easier way.

    Well r goes from a to b linearly as h goes from 0 to L, that's just the equation for a line. r = (b-a) h/L
  7. Feb 18, 2005 #6
    It's still not working out. r = b - (b-a)z/L, but I could not get the right resistance expression when a = b. There must be something wrong with this r expression.
  8. Feb 18, 2005 #7
    Nevermind. I made a mistake in the integration. It came out to be this:
    [tex] R = \frac{\rho L}{\pi ab} [/tex]

  9. Feb 18, 2005 #8
    What is z?

    If you measure h from top,

    [tex] r = a + \frac{h(b-a)}{L}[/tex]
  10. Feb 19, 2005 #9


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    His "z" is nearly the same as your "h", except he's measuring from the other end.
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