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Side Flash Problem

  1. Oct 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Side Flash. Part of a bolt of lightning jumps through distance d in air and then travels through the person ( who has negligible resistance relative to that of air because of the highly conducting salty fluids within the body). The rest of the current travels through air alongside the tree, for a distance h. If d/h = .400 and the total current is I = 5000A, what is the current through the person?
    https://imagizer.imageshack.us/v2/1920x893q90/633/IF9c3g.png [Broken]

    2. Relevant equations
    R= rho* L/A

    3. The attempt at a solution

    W
    hile attempting this problem I could see that the person and the air along the tree form a circuit that is in parallel. So the voltage across the person and the air along the tree is the same. So the resistance of one side of the circuit multiplied by it's current will equal the other side of the circuit's resistance multiplied by it's current. I also know that the total current of 5000A will be equal to the amperage that flows through the person and along the tree.

    I became stuck here though. Knowing that the resistance equation rho*L/A was probably going to play a part. I found a solution manual that equated the current of one side multiplied by d equals the current of the other side multiplied by h.

    After reworking the equation you have that the larger current through the person is 5000/1.4 and the other current is simple to find from there.

    Can anyone give me some insight as to how you make the jump to knowing that i1*h = i2*d is something that is valid to do?

    Thank you in advance and I apologize for the ugly typing and equations, I was looking into latex but it just all seemed so overwhelming.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 1, 2014 #2

    BvU

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    The insight is based on realizing that there is one and only one voltage wrt ground at the point of branching.
    For that voltage, Ohms law says V = I1 * R1 and also V = I2 * R2.
    For this exercise, R is assumed to be proportional to d and h, respectively. With the same proportionality constant, which personally I find very far-fetched.
     
  4. Oct 1, 2014 #3

    gneill

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    Staff: Mentor

    I think the idea is that the resistivity of the man is so much smaller than that of air that the effective resistance of each path is that of the air carried portions alone. Then we can see how the resistances will be proportional to d and h.
     
  5. Oct 1, 2014 #4
    Right after reading this I wasn't sure what you meant. It nailed me when I was driving though! Thanks.
     
  6. Oct 1, 2014 #5
    Yes! That is what hit me. I didn't really realize that the resistivity of the man was supposed to be taken as that of an idea wire, and that now I just have like two resistors in parallel with equal resistivity just different lengths! So it clicked... Love that moment when that happens :D.
     
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