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Resitivity of a wire

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Two wires both 2L. Wire 1 has a cross sectional area of A. Wire 2 cross sectional area starts at 2a but gradually gets thinner till it is A. Both have the same resitivity constant. Which one has more resistance?


    2. Relevant equations
    R=kL/A


    3. The attempt at a solution
    since constants the same I started with just l/a. Wire one has 2l/a resitivity, but for wire twos have no clue on how to solve for it.
     
  2. jcsd
  3. Feb 15, 2012 #2

    wukunlin

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  4. Feb 15, 2012 #3
    Thanks. However, is their any way to do this without calculus? I understand what it is saying but my class is mainly comprised of pre-calculus students witha few calculus students like me. Is their even a way to do this without calculus?
     
  5. Feb 15, 2012 #4

    BruceW

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    The question simply asks which has more resistance, not the exact answer. So you can use a reasonable argument, if you want to avoid doing the actual calculation.
     
  6. Feb 15, 2012 #5
    Yeah but I kinda want to know how to solve it
     
  7. Feb 15, 2012 #6

    BruceW

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    hmm, I can only think of using calculus. There is no other way I know to get the exact answer.

    Also, the question says "Wire 2 cross sectional area starts at 2a but gradually gets thinner till it is A." And wukunlin has taken this to mean that the wire is a truncated cone. This is the most simple, and most likely possibility. But we don't actually know what the exact shape is. So the question is too vaguely worded to be solved anyway. It is only when we make the assumption about the shape, then it can be solved.
     
  8. Feb 15, 2012 #7

    wukunlin

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    that pdf i linked actually did all the calculus for you, the result is

    [tex]R = \frac{ \rho l }{ \pi r_1 r_2 } [/tex]

    where as the typical cylindrical wire gives you

    [tex]R = \frac{ \rho l }{ \pi r^2 } [/tex]



    you can think of it this way, cut the resistor into tiny little sections, each section will have a portion of the original resistor. when all the section's resistances are added together you get the original resistance. If you cut the resistors into so many sections that the one with varying cross section effective turn into bits of cylinders with different widths, what will be their individual resistance be like comparing to the cylindrical resistor if you cut it into sections that are equally fine?
     
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