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Can I integrate to find resistance of an object of increasing radius?

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data
    I have to find the ratio of RB/RA where R is electrical resistance.
    Both objects A and B have circular cross section and length L, and are made of the same material.
    A has radius r0. B has radius r0/2 at the bottom and increases linearly to 2r0.


    2. Relevant equations

    R=pl/A

    3. The attempt at a solution
    RA is simple enough to determine, but I'm stuck with RB
    I have 2 potential theories as how to tackle this.
    One is that the thin end and the thick ends will cancel, due to symmetry, leaving me with a ratio of 1.
    The other involves R=pl/A becoming R=p∫0L1/A dx and somehow changing this to involve r0/2 and 2r0
     
  2. jcsd
  3. Nov 30, 2013 #2

    tiny-tim

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    Hi LiamG_G! :smile:

    B is a lot of discs in series. :wink:
     
  4. Dec 1, 2013 #3

    rude man

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    Pick that one!
    Your integral is OK, so what is A(x)?
     
  5. Dec 1, 2013 #4
    So A=∏r2 and r ranges from r0/2 to 2r0

    But with my integral, which I thought would be R=ρ∫0L (1/A)dx relates to the length of the cylinder, not it's radius.

    I think I need to find an equation for r at a specific point in the bar in terms of x (which ranges from 0 to L)
    Would this be somewhere along the right track?:
    radius of disk, rdisk=(3r0/2)(x/L) + r0/2
    This way, at x=0 (the bottom of the bar) the radius is equal to r0/2 and at x=L (the top of the bar) the radius is equal to 2r0
    So then I would also have to differentiate that I can replace dx with dr, but then I'm sure I would still have x in the integral, which leaves me with two variables.

    Am I at least on the right track? Where would I go next?
     
  6. Dec 1, 2013 #5

    tiny-tim

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    suppose you had n discs in series, of areas A1 … An (and all of the same thickness, ∆x)

    what would the formula be for the total resistance R?​

    (and i'm going out now o:))
     
  7. Dec 1, 2013 #6
    Very nice reasoning. This is exactly what you should do.
     
  8. Dec 1, 2013 #7

    rude man

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    You are very much on the right track.

    So now you have r(x), what is A(x)? A(x) contains r0 only, which is a constant. So you integrate with respect to x only. You never integrate w/r/t r.
     
  9. Dec 1, 2013 #8
    Actually, you could do it either way. For example, dx=((2L)/(3r0))dr
     
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