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Help with a current, resistance, and resistivity problem

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data

    In the figure below, current is set up through a truncated right circular cone of resistivity 898 Ω·m, left radius a = 2.00 mm, right radius b = 2.30 mm, and length L = 1.64 cm. Assume that the current density is uniform across any cross section taken perpendicular to the length. What is the resistance of the cone?

    26-30.gif

    HINT: Consider a cross-sectional slice of thickness dx. How is the resistance of that slice related to the resistivity of the material? How do you relate the resistance of the slice to the distance x from the left end and the full distance L?

    2. Relevant equations

    R = ρ * (L/A)

    3. The attempt at a solution

    Okay so I haven't gotten very far with this one. I know you have to set up an integral with respect to L, so I tried an integral from 0 to 0.0164 m and I related the radius as (b-a / L) * dx + 22) So I had an integral of [itex]\int\frac{ρx dx}{\pi* ((\frac{b-a}{L} * x) + 2)^{2}}[/itex] but I had a really hard time trying to work this out. Can someone just set me off in the right direction?
     
  2. jcsd
  3. Oct 3, 2011 #2

    cepheid

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    Welcome to PF iiiiaann,

    I don't think your integrand is quite right. The infinitesimal resistance dR of an infinitesimal cross-sectional slice of thickness dx is just given by:

    dR = (ρdx)/A(x)

    where A is the area of the slice, and I have explicitly indicated that this area is a function of x. Since the cross sections are circular,

    A(x) = πr2(x)

    where r(x) is the radius of the slice. So, the question is, how does the radius of the slices vary with x? This is the part I think you didn't do quite right. Hint: the radius increases from a at x = 0 to b at x = L, and since it's a cone, you can assume that it rises linearly from this starting value to this ending value. Does that help you figure out the form of r(x)?
     
  4. Oct 3, 2011 #3
    would that make r with respect to x = (b-a) * x/L + a or in this problem that would be (2.3 mm - 2 mm) * x/1.64 cm + 2mm?
     
    Last edited: Oct 3, 2011
  5. Oct 3, 2011 #4

    cepheid

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    I agree with (b-a)x/L + a.
     
  6. Oct 3, 2011 #5
    so the integral would then be [itex]\int\frac{ρ dx}{\pi [((b-a) * \frac{x}{L}) + 2]^{2}}[/itex] from 0 cm to L cm

    or with all the numbers plugged in from this problem:

    [itex]\int\frac{898\Omega * m dx}{\pi [((0.0023 m - 0.002 m) * \frac{x}{0.0164 m}) + 0.002 m]^{2}}[/itex] from 0 m to 0.0164 m

    simplified:

    [itex]\int\frac{898\Omega * m dx}{\pi [0.01829x m + 0.002 m]^{2}}[/itex] from 0m to 0.0164 m

    without the units:

    [itex]\int\frac{898 dx}{\pi [0.01829x + 0.002]^{2}}[/itex] from 0 to 0.0164

    So assuming my above integral was right, you can pull ρ and ∏ out of the integral because they are constant and you get [itex]\frac{\rho}{\pi}\int\frac{dx}{(0.01829x + 0.02)^{2}}[/itex] from 0 to 0.0164 and that integral comes out to [itex]\frac{\rho}{\pi}*\left[(1/0.01829) * ln(0.01829x + 0.02)\right|^{0.0164}_{0}[/itex] or



    [itex]\frac{\rho}{\pi}*[\frac{ln((0.01829)(0.0164) + 0.02)}{0.01829} - \frac{ln((0.01829)(0) + 0.02)}{0.01829}][/itex]

    or [itex]\frac{\rho}{\pi * 0.01829}*[\ln((0.01829)(0.0164) + 0.02) - ln((0.01829)(0) + 0.02)][/itex]

    simplifying to [itex]\frac{898}{\pi * 0.01829}*[\ln(0.0203) - ln(0.02)][/itex]

    [itex]15628.33668 * ln(\frac{0.0203}{0.02})[/itex]

    15628.33668 * 0.014889

    = 232.684Ω

    I'm not sure on this answer, but also not sure where I may have gone wrong


    EDIT: realized i was using 0.02 in the ln's instead of 0.002, i re-did it and came out with 2183.95, does that seem more reasonable?

    EDIT 2: Still wrong, any help with finding my mistake?
     
    Last edited: Oct 3, 2011
  7. Oct 3, 2011 #6
    fml im an idiot, integrating wrong. I figured it out. Thanks for the help
     
  8. Oct 3, 2011 #7

    gneill

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    :smile: Well, you certainly went about the integration like a Trojan! I think it might have been easier to leave it in symbolic form for integration rather than carrying around all those numbers.

    The form of the integral is [itex] \large \int \frac{1}{u^2} du = \frac{-1}{u} [/itex].
     
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