# Bird on power line

Discussion in 'General Electronics Chat' started by Basher, Apr 26, 2013.

1. ### Basher Thread Starter New Member

Nov 19, 2012
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I was reading the online textbook provided by All About Circuits, and stumbled upon their safety chapter: http://www.allaboutcircuits.com/vol_1/chpt_3/3.html

To quote the third paragraph: "There is no such thing as voltage "on" or "at" a single point in the circuit, and so the bird contacting a single point in the above circuit has no voltage applied across its body to establish a current through it"

I was wondering, wouldn't the birds two legs (and body) act as a wire/resistor? Wouldn't there be a current generated because of this resistance?

2. ### Brownout Well-Known Member

Jan 10, 2012
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Personally, I don't agree with that statement. However, both of the bird's feet are at the same potential, so no current flows. Now, if the birds's feet were on different potential, he would feel the effects of current, perhaps briefly.

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3. ### DerStrom8 Well-Known Member

Feb 20, 2011
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Exactly right Brownout.

Basher, current is given by the voltage *difference* divided by the resistance. Let's say the wire has 10,000 volts running through it, with respect to ground. The voltage on the bird's right leg will be 10,000 volts, and the voltage on the bird's left leg will also be 10,000 volts. The voltage *difference* is 0, so divide 0 by the bird's resistance will still give you 0 amps through the bird.

Make sense?

Matt

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4. ### Basher Thread Starter New Member

Nov 19, 2012
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DerStrom8:

I suppose that makes sense. But isn't that assuming that the bird has no resistance?

Second question: What if a human were to hang off with two arms? Wouldn't the body act as a resistor, or will it be the same as the bird situation?

5. ### Brownout Well-Known Member

Jan 10, 2012
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Ask yourself this: if you took a resistor of say, 100 ohms, and connected +5v to one lead and the same +5v to the other lead, how much current flows through the resistor?

6. ### #12 Expert

Nov 30, 2010
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Assume any resistance you want. Zero volts divided by any resistance is zero amps.

7. ### DerStrom8 Well-Known Member

Feb 20, 2011
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No, you're missing the entire point.

Current is the voltage difference ACROSS the element divided by the resistance. If you have 10,000v on one side, and 10,000v on the other side, the voltage DIFFERENCE is zero volts. 0 volts divided by ANY resistance is still going to give you 0 amps. That goes back to first grade math!

Same deal with a person hanging from the wire.

8. ### Basher Thread Starter New Member

Nov 19, 2012
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I guess when you put it like that, it's zero amps. I thought there is a voltage drop over the resistor, but I guess I was wrong. Thanks for your input

9. ### DerStrom8 Well-Known Member

Feb 20, 2011
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I think you're making the same error a lot of other people make. Resistors do not actively drop a voltage. It's simply a way of saying how much voltage is ACROSS a resistor. If you connect one end to 5 volts and the other end to 0 volts, the voltage drop will be 5-0 = 5 volts. If you connect one end to 5 volts and the other end to 2 volts, your voltage drop will be 5-2 = 3 volts. In the case of the bird, the voltage on the left leg and the voltage on the right leg are the same, so the voltage drop (difference) is zero volts. Therefore, no current flows through the bird.

Matt

10. ### #12 Expert

Nov 30, 2010
16,705
7,358
There is only a voltage drop across a resistor when you apply a voltage to it, a different voltage on each end. When you apply the same voltage on each end, there is no current flow because there is no voltage difference from one end to the other. When you connect two arms or legs to the same wire, there is no voltage difference between the two contacting points. Resistors do not create voltage all by themselves.

11. ### Markd77 Senior Member

Sep 7, 2009
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Really it's assuming that the cable has no resistance. There is current flowing through it, so there will be voltage differences along the cable. It's a pretty good assumption though, because the voltage difference in even a few metres of cable would be hard to measure they are so small.

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12. ### Basher Thread Starter New Member

Nov 19, 2012
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I see! Thank you and #12 for the clarification. I was under the impression that voltage was different on each lead when a resistor is put into a circuit. But these explanations make perfect sense. Appreciate the input!

13. ### killivolt Active Member

Jan 10, 2010
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When I was at a friends house as a young man and watched his father dig up a 3 phase cable, he cut all three lines.

Standing on a piece of wood in the trench, only touching one cable at a time he braided them together.

I just barely learned AC theory, I couldn't believe my eyes.

Still to this day, I wouldn't even think about it.

Edit: He aslo had rubber boots.

14. ### bountyhunter Well-Known Member

Sep 7, 2009
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Birds killed on power lines usually have their wing tips touch across the two lines (which have a very high voltage difference between them) as they fly into perch on the top of the pole, which incinerates the bird.

If a bird only touches one line, he's safe as long as there is no path to ground or any other voltage.

15. ### atferrari AAC Fanatic!

Jan 6, 2004
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I've seen instead, faintly glowing in night time, fluorescent tubes or what looks like that (?), hanging from two points of one (said one) high (said high) voltage line.

Aren't they using the small voltage difference along the line?

Sure, the do glow.

16. ### aws505 Member

Mar 11, 2013
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Remember that they are sitting on A/C voltage lines, so the voltage on the line varies with time and also space. Try calculating the wavelength. If you put your multimeter leads across an A/C voltage line at a distance of λ/2, for example, then your multimeter will read the full "A/C voltage" on the line. If you put your a multimeter leads across an A/C line at a distance that is much smaller than λ, then what will your multimeter read?

Of course, don't try these things because you should never poke around live A/C lines. Simply do the calculations, and maybe try punching a few numbers in the caluculator. Try estimating how far apart the birds feet are and comparing that to the wavelength of the voltage on the A/C line.

17. ### ErnieM AAC Fanatic!

Apr 24, 2011
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No, not the small voltage difference along the line.

That glow is from the BIG voltage between the wire and the Earth itself (or literally the ground). The current is being conducted thru the air, something that normally does not happen but is possible when the air is very humid or there is some object where it should not be: years back we had a tree limb over the power line and it would glow and buzz on rainy days.

Or you may be seeing St. Elmo's fire. where the voltage source is the same as that which produces lightning.

18. ### takao21203 Distinguished Member

Apr 28, 2012
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I'd not recommend to have contact with a high voltage line at all.

Static electricity will flow at very high voltages. Even if you are isolated. It depends on the total area of the human body surface I think.

19. ### TheComet Member

Mar 11, 2013
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In theory, if the bird's body capacitance is large enough, it could be fried when sitting on a power line because it's AC.

It has to be one fat bird though. Polly want a cracker NOW.

I'm guessing that's most likely due to the electromagnetic fields being generated by the wires themselves. Have you ever held a fluorescent tube up near a plasma ball? They begin to glow too. I think what you are describing is the same effect.

That's highly unlikely. Even air with over 100% humidity doesn't conduct electricity enough to cause a significant amount of flowing current. See my quote above.

TheComet

20. ### MrChips Moderator

Oct 2, 2009
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You're missing something. There is current flowing in the wire.
For there to be current there must be a voltage difference.
The bird on the wire is like two resistances in parallel, the resistance across the bird and the resistance in the wire.
The resistance through the bird is like 100MΩ.
The resistance through the wire is like 0.000001Ω.
For any given current through the wire, calculate the potential across the wire
and then the current through the bird.