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Why can a bird survive on a high voltage transmission line?

Discussion in 'MCAT Study Question Q&A' started by manohman, 09.23.14.

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  1. manohman

    manohman 2+ Year Member

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    1. The problem statement, all variables and given/known data
    Birds can be seen perched on a high voltage power line yet seem unaffected by the high voltage. This is due to...

    A) Low resistance of the birds compared to the wire, minimizing current through the birds
    B) birds keeping only one foot on the wire, thus not completing a circuit
    C) low capacitance of the birds (?)
    D) Minimal potential difference across the birds (correct answer)


    Assuming current is constant across the transmission line (it is right?) since the distance between the birds feet is very small, and since resistance is determined by length, the resulting resistance across that portion of wire would be very small.

    Since current is constant across the wire, and the resistence is very small, then the resulting Voltage across those two points must be very small as well. The bird is effectively in parallel with the wire portion so it feels the same voltage,. Since the voltage is so small, it is essentially unaffected.

    So that is why D is the correct answer.

    What i dont understand is why a bird keeping only one foot on the wire wouldn't be a legitate reason either. There is no complete circuit so charge cant flow.

    I was thrown off by the "low capacitance of the birds" statement, and since q=CV , I knew that would be false. But does capacitance even apply here? I thought it only applied when you had two parallel plates?

    Finally, in this scenario, we know the voltage difference between the beginning and end of the wire (thats what we refer to when we say it's "high voltage" right?), current is constant throughout the wire right? I ask because it seemed odd to base my calculations of V=IR upon the constant current. Usually you use voltage to find the curent.

    Thanks guys.
     
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  3. type12

    type12 2+ Year Member

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    Lighting doesn't strike only camels.
     
  4. tdod

    tdod 2+ Year Member

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    "What i dont understand is why a bird keeping only one foot on the wire wouldn't be a legitate reason either. There is no complete circuit so charge cant flow." this is incorrect because it is false; birds put two feet on the power line, not one.

    Capacitance is incorrect because the current delivered to the first leg will be returned at the second leg; nothing about the bird indicates that it could function as a capacitor.
     
  5. manohman

    manohman 2+ Year Member

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    I see, because nothing about the bird allows it to store charge.

    are there any other substances than a parallel plate that could serve as a capacitor to store charge?
     
  6. Cawolf

    Cawolf 2+ Year Member

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    Capacitors will always consist of two conductors separated by a dielectric.

    This may be parallel plates, two wires, a coaxial wire, or many other things.

    In MCAT terms, I think that you can assume any capacitor you will encounter can be modeled by parallel plates.
     
  7. justadream

    justadream 5+ Year Member

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    @Cawolf

    This reminds me of our lively discussion about the exact same topic a few months ago.
     
  8. Cawolf

    Cawolf 2+ Year Member

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    Sure does!
     
  9. dynamicalan

    dynamicalan

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    Hello,

    In DC it does calculate the voltage. By the way that stands for direct current not Washington DC in AC, Pythagoras Therom will give you Z and this is calculated by the square root of x square plus r square. X is reactance and r is resistance. That's after you figure out capacitance reactance and inductive reactance as x and I forgot how to do that because it's been a few years. But, as far as I know this has nothing to do with it. I thought the two bird legs cancelled out the difference in potential between the two wires while if you or I touch a live wire it conducts current but if it's insulated it seems to me it wouldn't conduct current if you're grounded. In water: Of course that's it.
     
  10. Cawolf

    Cawolf 2+ Year Member

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    Is this real?
     
  11. dynamicalan

    dynamicalan

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    What do you mean real? You calculate the capacitive reactance by the receptical of the products of 2pi frequency and capatance while inductive reactance is found by the products of 2pi frequency and inductance. They work opposites to each other. You don't have to get complicated with complex formulas. But, I can't draw formulas in this post. For each frequencies this all changes. So to get impedance you add them together.

    This has nothing to do with DC or AC Impedence it rather has has to do with the difference of potential. If a person touches both wires at the same time then this closes the circuit with the person in the middle.

    By the way I'm not selling anything. In addition, at anytime I share information or teach. I never take money for any of this or have I ever.

    While I admit I don't deserve to post on here because I know the least of anyone on here but there's no reason to make degrading comments.
     
    Last edited: 10.04.14
  12. Cawolf

    Cawolf 2+ Year Member

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    I mean your post reminded me of something a spam account would type to generate posts and then start posting advertising links.
     
  13. dynamicalan

    dynamicalan

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    No! I was only responding to the question. Personally, I don't know that question would be on the MCAT but it seems to be and the post is already discussing electronic principles. I didn't start the thing. I have nothing to sell. I just thought because I studied electronics I may be able to help. All the discussion on reactance was stated previously by others.

    Actually, I sense snoobry and an attitude of critism and in my opinion you don't have a right to say anything at all inless someone starts spamming people on here. I wouldn't do what you did but then I'm not an aggressive personality. Maybe you have a right to but I don't know what that is.
     
    Last edited: 10.04.14
  14. neurodoc

    neurodoc Member 10+ Year Member

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    You are correct that B (the bird keeps only one foot on the conductor) would prevent electrocution. Electrocution occurs when there is a sufficient current flow through an organism to cause injury or death. Ohm's law describes the relationships between current, potential difference, and resistance. With one "foot" on the wire (and no other connection of the body to ground potential), there will be no current flow through the bird. The bird could still be "charged" like a capacitor, but without any way for that "charge" to flow through the bird's body to ground potential (regardless of the bird's "capacitance"), the bird will not be harmed.

    If the bird grasps the conductor with both "feet", the distance between it's feet will be a few inches at most, and there will be minimal potential difference between its points of contact and therefore minimal current flow through the bird. If it were a human and not a bird hanging from a high voltage line (and not in contact with anything else, the human should also escape electrocution.

    I agree that B should not have been included as an answer. Perhaps the folks who formulated this question assumed that birds don't "perch" on one "foot." In fact they usually don't do so, but that's irrelevant to the question.



    The quick answer is that
     

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