Why don't birds get shocked on power lines?

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  • #1
John3509
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I know the answer, because they are not grounded and do not complete any circut, but I have a different analysis that I would like to explore.

First, why do you need a closed circuit? Because if you have a battery pushing current along a wire that ends, electrons will accumulate at the end of that wire and will eventually produce an EMF that negates the battery's EMF, is this correct?

Second, current that flows to homes is alternating, so imagine you have an alternating current source and wires that extend in both directions and do not connect to anything, once the current reaches the point where the EMF from the displaced electrons cancels out the source EMF the current reverses, basically the alternating current is pushing electrons to alternating ends of the wires, likes a half of a capacitor being charged and discharged repeatability.

Third, if you put a light bulb in this circuit, it should light light up, right? So I would imagine the bird would get shocked, because even tho there is no where for the electrons to flow out of the bird once they enter as required by Kirchhoff's law, since the current is alternating,they would go out the same path and the bird would be pumped full of electrons then drained of them.

But, the birds don't get shocked, so whats wrong with my model?
 

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  • #2
Dale
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Your model incorrectly identifies what it means to “get shocked”. You indicate that merely having excess electrons is sufficient to be shocked. However, if you consider the experience of rubbing your feet on the carpet and getting shocked you will realize that you build up the charge without being shocked, and it is the process of rapidly losing that charge which shocks you.

Being shocked requires a high current density, not merely excess charge. Birds on a wire have no high current density so they are not shocked.
 
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  • #3
John3509
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However, if you consider the experience of rubbing your feet on the carpet and getting shocked you will realize that you build up the charge without being shocked, and it is the process of rapidly losing that charge which shocks you.

Yes, but once the polarity reverses then the electrons are sucked out
 
  • #4
b.shahvir
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Electron current flow is only requirement for shock phenomenon. Don't see birds on extra high voltage lines because of high density leakage and capacitive currents. EHV leakage currents would be unbearable for the birds.
 
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  • #5
Dale
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Yes, but once the polarity reverses then the electrons are sucked out
With the same very low current density as they were provided. Hence no shock.

Again, getting shocked requires high current density
 
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  • #6
davenn
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Because if you have a battery pushing current along a wire that ends, electrons will accumulate at the end of that wire and will eventually produce an EMF that negates the battery's EMF, is this correct?

not it isn't .... the electrons don't accumulate, current only flows along the wire when there is a circuit

Second, current that flows to homes is alternating, so imagine you have an alternating current source and wires that extend in both directions and do not connect to anything, once the current reaches the point where the EMF from the displaced electrons cancels out the source EMF the current reverses, basically the alternating current is pushing electrons to alternating ends of the wires, likes a half of a capacitor being charged and discharged repeatability.


no it doesn't, because in an AC circuit, the electrons don't move any significant distance .... that is, they oscillate back and forwards a tiny distance about a point in the wire
 
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  • #7
spareine
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Birds seem to avoid transmission lines above 50 kV. The ground wires, which do not carry current, except in the event of a lightning strike, need to be distinguished from the power transmission lines. The ground wires are the ones that are not separated from the pylon by a stack of insulating discs. On this photo of a 150 kV line, the birds are sitting exclusively on the ground wires, and evidently avoid the power lines.

242581


A bird landing on a power line resembles a helicopter with a lineman approaching the power line. When the helicopter approaches the line, a continuous spark develops. (youtube) The helicopter has a capacitance to earth and to the line. The capacitance to earth for a somewhat spherical object with a diameter of 10 m is C1 = 2πεD = 0.5 nF. If the voltage is 500 kV and 50 Hz, the estimated AC current is 0.1 A. That is not impossible, when looking at the spark on the photo. The bird is 50 times smaller, its capacitance is 10 pF, and the current from a 150 kV power line is 0.5 mA. That is small, but maybe the birds are able to feel it.

helikopter1.png
 
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  • #8
John3509
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not it isn't .... the electrons don't accumulate

They have to. That's how capacitor circuits work, the difference here is that the capacitor plates are the ends of the wires and instead of being close together they are spread out, so the capacitance is small but there will still be some accumulation at the ends, If you have a circuit with flowing current and you cut the circuit the current does not stop instantaneously, the battery still produces and EMF along the wire, current only stops because an electric field builds up in the opposite direction from the accumulation. Atleast that is how my text book (Pearson) explains it.
 
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  • #9
John3509
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With the same very low current density as they were provided. Hence no shock.

Again, getting shocked requires high current density

I'm not sure I understand. Lets say the bird does touch the neutral wire and forms a closed circuit. The current density is the amount of electrons flowing trough a certain cross sectional area, right? Lets say the bird has a current density "A". Im assuming it will be shocked.
The only difference in the situation I previously described is that the electrons enter and exit trough the same path, the cross sectional area of the bird is the same, the speed at which the electrons flow is dependent on the voltage of the wires, the birds resistance is not different, so why is the current density less?
 
  • #10
John3509
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Birds seem to avoid transmission ... That is small, but maybe the birds are able to feel it.

View attachment 242582

Very insightful, although a bit advanced for me, I'm not sure I understand it properly but from what I got is that I am somewhat correct that the bird, or helicopter, do kind of behave like a capacitor with electrons constantly being pumped into it and sucked out with the alternating current but that the birds are small so it wont kill them? Is that correct?
 
  • #11
russ_watters
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Yes, but once the polarity reverses then the electrons are sucked out
There just aren't very many of them.
They have to. That's how capacitor circuits work...
Birds (and people) make very poor capacitors.
 
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  • #12
spareine
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.. that I am somewhat correct that the bird, or helicopter, do kind of behave like a capacitor with electrons constantly being pumped into it and sucked out with the alternating current but that the birds are small so it wont kill them? Is that correct?

Yes, in my diagram the helicopter can be replaced by the bird just before touching the wire. The part of the bird closest to the power line, its feet, is one plate of capacitor C2, and the part that is furthest from the power line, say its head, is one plate of capacitor C2. When AC current is flowing through the bird's body, the excess charge is jumping between its head and feet. A little later, when the spark develops or when the bird is touching the wire, C2 is shorted out, and the excess charge is jumping between the bird's head and the "earth". The excess charge equals Q = CV = 150 kV · 10 pF = 1.5 μC = 10^13 electrons.

Yes, the size of the bird matters, smaller birds experience smaller currents, whereas larger birds and helicopters experience larger currents.
 
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  • #13
Dale
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The current density is the amount of electrons flowing trough a certain cross sectional area, right?
Yes. So estimate what the current is in the two cases. Let’s assume a 120 kV wire at 60 Hz, a bird resistance of 2 kOhm, and a bird self-capacitance of 10 pF. Do two circuits, one where the bird closes the circuit and one where the bird does not so it is only the self capacitance that is relevant.

At 60 Hz, a 10 pF capacitor has an impedance of 270 MOhm so 120 kV/270 MOhm is 0.4 mA for the bird on a wire. Conversely the completed circuit resistance of 2 kOhm leads to 120 kV/ 2 kOhm which is 60 A. That is 150000 times larger. Hence the bird is shocked only by completing the circuit and not merely by being on the wire
 
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  • #14
davenn
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They have to. That's how capacitor circuits work,


no it isn't how a capacitor works ... you should read up on capacitors

The capacitor is in a circuit so for every charge that enters one plate, one charge ( same polarity) leaves the other plate and heads back to the power supply

the capacitance is small but there will still be some accumulation at the ends,

again NO
The net charge in a capacitor is zero ... for every - charge on one plate there is a + charge on the other plate

If you have a circuit with flowing current and you cut the circuit the current does not stop instantaneously, the battery still produces and EMF along the wire, current only stops because an electric field builds up in the opposite direction from the accumulation. Atleast that is how my text book (Pearson) explains it.

no, it's that close to instantaneous, for all practical purposes, it is instantaneous
 
  • #15
John3509
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no it isn't how a capacitor works ... you should read up on capacitors

The capacitor is in a circuit so for every charge that enters one plate, one charge ( same polarity) leaves the other plate and heads back to the power supply

Yes it is, you must be misunderstanding me be because you say it isnt but then explain them to me by basically repeating what I wrote, :confused:

I know how capacitors work. the battery EMF pumps one plate full of electrons and sucks electrons from the other creating opposite charges until the EMF from the disbalencee in charges becomes to great for the battery EMF to push more charges, and the capacitor is charged and current stops.

Just like in a wire that does not make a full circuit the electrons will accumulate at one end of the wire until it stops the current, hence the comparison to a capacitor, the main difference that compactor plates are more efficient. You say there is no accumulation, here is text from my textbook, University Physics from Pearson that literally says there will be.

For a conductor to have a steady current, it must be part of a path that forms a closed loop or complete circuit. Here’s why. If you establish an electric field inside an isolated conductor with resistivity that is not part of a complete circuit, a current begins to flow with current density (Fig. 25.11a). As a result a net positive charge quickly accumulates at one end of the conductor and a net negative charge accumulates at the other end (Fig. 25.11b). These charges themselves produce an electric field in the direction opposite to causing the total electric field and hence the current to decrease. Within a very small fraction of a second, enough charge builds up on the conductor ends that the total electric field inside the conductor. Then as well, and the current stops altogether (Fig. 25.11c). So there can be no steady motion of charge in such an incomplete circuit.

The current causes charge to build up at the ends.
The charge buildup produces an opposing field E2, thus reducing the current.



not it isn't .... the electrons don't accumulate, current only flows along the wire when there is a circuit

I dont mean to be rude, but the issue here is why it doesn't harm the birds, not whether there will be any current or not, which after reading other peoples responses here as well as my textbook, i'm certain there will be.
 
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  • #16
davenn
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I know how capacitors work. the battery EMF pumps one plate full of electrons and sucks electrons from the other creating opposite charges until the EMF from the disbalencee in charges becomes to great for the battery EMF to push more charges, and the capacitor is charged and current stops.


Sorry, but no you don't
this is at best a very bad description, at worse just not correct 😕

there is no pushing or sucking ...

For a conductor to have a steady current, it must be part of a path that forms a closed loop or complete circuit. Here’s why. If you establish an electric field inside an isolated conductor with resistivity that is not part of a complete circuit, a current begins to flow with current density (Fig. 25.11a). As a result a net positive charge quickly accumulates at one end of the conductor and a net negative charge accumulates at the other end (Fig. 25.11b). These charges themselves produce an electric field in the direction opposite to causing the total electric field and hence the current to decrease. Within a very small fraction of a second, enough charge builds up on the conductor ends that the total electric field inside the conductor. Then as well, and the current stops altogether (Fig. 25.11c). So there can be no steady motion of charge in such an incomplete circuit.

The current causes charge to build up at the ends.
The charge buildup produces an opposing field E2, thus reducing the current.

I have to disagree with some of that as well, he shoots himself in the foot by stating there is an electric field IN the conductor. That statement is wrong, the field in the conductor is zero

have a read of this.....
https://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Fields-and-Conductors


, but the issue here is why it doesn't harm the birds,


It will if the voltage on the transmission line is high enough....
As was shown to you earlier post #7, birds keep away from EHT transmission lines as they can easily sense the
very high electric field around the lines.
If you or I or a bird could stand on an EHT cable (unprotected) with our feet spread apart,
there would be enough of a potential difference between our feet that we ( the bird) would be shocked
As our or the birds body forms a parallel resistance with the section if power line between the 2 feet

Dale in post #13 was hinting at that



Dave
 
  • #17
John3509
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Yes, in my diagram the helicopter can be replaced by the bird just before touching the wire. The part of the bird closest to the power line, its feet, is one plate of capacitor C2, and the part that is furthest from the power line, say its head, is one plate of capacitor C2. When AC current is flowing through the bird's body, the excess charge is jumping between its head and feet. A little later, when the spark develops or when the bird is touching the wire, C2 is shorted out, and the excess charge is jumping between the bird's head and the "earth". The excess charge equals Q = CV = 150 kV · 10 pF = 1.5 μC = 10^13 electrons.

Yes, the size of the bird matters, smaller birds experience smaller currents, whereas larger birds and helicopters experience larger currents.

You modeled the feet and the head as separate capacitor plates, I was not expecting that, how I imagined it is since current would enter the bird at one point, pass trouigh its body, and exit somewhere when it is in a circuit, since it only has an entry when it sits on the power line the electrons would just be pumped and compressed in the body, like the whole bird would have an excess charge, not just the feet.

The helicopter and receives a larger current because it can hold more excess electrons before the current to it stops?
 
  • #18
John3509
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Sorry, but no you don't
this is at best a very bad description, at worse just not correct 😕

there is no pushing or sucking ...

OK, well you are not helping here. I do actually, and considering that you for some reason would not believe that charges will accumulate at the ends of the conductor, that you are the only one here who questions my understanding of capacitors, or no one else here has a problem with my semantics, (seriously, who has a problem with using the words pushing or sucking electrons...), and since I just qoted my textbook so you know I have it open right in front of me, I will trust my own understanding over your very limited assessment because its clear you nitpicking just to keep up the "you don't know what you are talking about" attitude. 😕

I have to disagree with some of that as well, he shoots himself in the foot by stating there is an electric field IN the conductor. That statement is wrong, the field in the conductor is zero

No, there were missing texts that did not copy and paste

that talk about the electric field, even without those it should still be clear what the author trying to say.
 
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  • #19
John3509
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Yes. So estimate what the current is in the two cases. Let’s assume a 120 kV wire at 60 Hz, a bird resistance of 2 kOhm, and a bird self-capacitance of 10 pF. Do two circuits, one where the bird closes the circuit and one where the bird does not so it is only the self capacitance that is relevant.

At 60 Hz, a 10 pF capacitor has an impedance of 270 MOhm so 120 kV/270 MOhm is 0.4 mA for the bird on a wire. Conversely the completed circuit resistance of 2 kOhm leads to 120 kV/ 2 kOhm which is 60 A. That is 150000 times larger. Hence the bird is shocked only by completing the circuit and not merely by being on the wire

Ok, so trying to understand this more mechanically than mathematically, since the current does not have an exit point on the bird when it sits on the wire, the buildup of charge in the bird basically acts as a massive resistor, would this be a good interpretation?
 
  • #20
Dale
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The part of the bird closest to the power line, its feet, is one plate of capacitor C2, and the part that is furthest from the power line, say its head, is one plate of capacitor C2.
You modeled the feet and the head as separate capacitor plates, I was not expecting that,
I also would not model the bird as two plates. To do that you would need to say that the rest of the bird is a dielectric, but birds conduct reasonably well (chicken conductivity is used reliably in slaughterhouses)

Two plates separated by a dielectric is a traditional capacitor, which has “mutual capacitance”. What we want here is “self capacitance” which treats the bird as a single (strangely shaped) plate.

Basically, you can think of it as a standard capacitor where the second plate is at infinity. This is what I described in my previous post.

the buildup of charge in the bird basically acts as a massive resistor
No, it acts as a small capacitor (connected to ground).
 
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  • #21
John3509
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I ... (connected to ground).

So I understand now that it has a low current, but the one thing I still don't understand,
Lets forgot the bird and simply, just an AC source, a wire and one capacitor plate at the end.
Current is amount of electrons passing trough a certain area per time.
Lets look at a place on the wire directly in front of the capacitor,(point A) why is the current lower with the capacitor present, what physically happens, does it decrease the speed at which the electrons pass that point, does it reduce the density, like 1 lanes vs 2 lanes (using the road analogy for current)
 
  • #22
Dale
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why is the current lower with the capacitor present,
Compared to what?
 
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  • #23
John3509
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Compared to what?

Oh right, to a closed circuit without the capacitor, I suppose you would have to place a resistor in its place? Im trying to model what you did in post 13.
 
  • #24
Dale
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why is the current lower with the capacitor present, what physically happens, ... [compared to] a closed circuit without the capacitor
The current is lower with the capacitor because as charge accumulates on the capacitor it produces a voltage which opposes the current flow. The smaller the capacitance the less charge is required to produce the voltage.
 
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  • #25
John3509
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Now it makes sense to me. Thank you for all the help.
 
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  • #26
spareine
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You modeled the feet and the head as separate capacitor plates, I was not expecting that, how I imagined it is since current would enter the bird at one point, pass trouigh its body, and exit somewhere when it is in a circuit, since it only has an entry when it sits on the power line the electrons would just be pumped and compressed in the body, like the whole bird would have an excess charge, not just the feet.

I also would not model the bird as two plates. To do that you would need to say that the rest of the bird is a dielectric, but birds conduct reasonably well (chicken conductivity is used reliably in slaughterhouses)

Maybe this schematic is clearer. Capacitor C2 is the air gap between feet and power line before landing. After touchdown, it is just an ohmic contact. The only importance of C2 is that its air gap is the location of the electric arc. My thought was the bird may be more sensitive to a current in the form of an electric arc during landing, than to the same current spread out over a larger contact area after landing. There is no dielectric in the model, only air and conductors.

242635
 
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  • #27
spareine
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BTW, a convenient guinea pig model for the bird on the power line, with its capacitive current, is a person on insulating shoes, touching a neon screwdriver test light. The current that causes the neon lamp to light, is equal to the capacitive current flowing through the person's body. Without the person the current is zero. The self capacitance of a somewhat spherical person having a diameter of ~1 meter is C = 2πεD = 100 pF, corresponding at 50-60 Hz to an impedance of 30 MΩ. An experimental confirmation for this calculated impedance value is obtained by inserting a resistor between the finger and the screwdriver. A 10 MΩ resistor does not lower the brightness of the lamp, whereas a 100 MΩ resistor does.

242777
 
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  • #28
Mister T
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Third, if you put a light bulb in this circuit, it should light light up, right?

If the light bulb is sensitive enough. Likewise current flows through the bird.

If the bird is modeled as a point, no current flows through it. That is the usual intent of this very common question.
 
  • #29
SteveUSA
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I am no expert on electricity, but I have seen credible reports of large birds getting killed when their wings touch 2 wires on take-off. Hawks and ospreys love to perch and to build nests on tall electric towers. Mostly they are OK, but one occasionally gets fried.
 
  • #30
trainman2001
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However, big birds have another problem. I was at a stop light and some Canada Geese were taking off almost vertically up through a utility pole carrying probably 10kV AC distribution lines. One of the birds miscalculated and its large wing span caught two of the three conductors. It didn't end well. In this case we weren't talking about static charges, but full blown short circuit across a goose. To say, this "goose was cooked" was an understatement. It was shocking to look at (pun intended), and it did disturb me the rest of the day.
 
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  • #31
Sooty
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Into the lions den! Lordie, what a contentious issue! :nb)
So, why are birds not shocked ? Errr.. they can be... with very high voltages...
Well, at the risk of starting a war here, I will offer you another slant on this matter that if I am correct, wasn't even mentioned by anyone else! And that is, that although the various models discussing capacitors and the pushing and pulling of electrons may be an aspect of the situation, considering our birds/helicopters as capacitors may be relatively minor to the question of 'being shocked'.
Yes?:rolleyes:
Errr... yes. And what is the major factor?
Athmospheric Voltage Gradient... if the term doesn't upset somebody! :biggrin:
What's that all about? Well, when we are talking about massive voltages, it's not longer a case of 'in the conductor we have a voltage, and outside of the conductor, we don't'. With a few volts on our conductor, no big deal. However, when we have tens/hundreds of thousand of volts, we may have a voltage gradient created that may be of the order of 25kV per inch! That means that even if we come close to such a conductor without even touching it, we (and birds!) may not only be exposed to a high voltage gradient, but if the air is not especially dry (ie, damp, and so conductive!) such conducting air may be capable of providing enough -current- flow to be felt to a lesser or greater level!
"But what about -thousand of volts- of Static Electricity???" I can hear someone scream!
Well, that's the thing about 'static electricity'. It's 'harmless', not because it's 'a low voltage', but because it's being created in a scenario that is not capable of providing much -current-. After all, -power- (ie, that which can actually -do- something!) is the -multiple- of Voltage with Current - and voltage without current doesn't do much at all!
So, a power line is most definitely capable of providing a high current, (maybe that's why they cell 'em -Power- lines!:biggrin:) and when we/bird are in the middle of a high voltage gradient, a current may most definitely flow, and it's that current (not just voltage) that is a measure of what can be felt, and may harm us too!
As for -where- that flow may be -to-, the answer is, the environment/earth in the form of a Coronal Discharge.
Just to elaborate... (waffle some more?:cool:) it is interesting to see the pic in another post of a guy in a hoodie holding a sparking rod... but there is no explanatory caption... :confused:
So, I'll try to fill in a little. A hoodie...? Er, not really. A Faraday Suit! Eh? Like a Faraday Cage, except personal!
What is going on? Well, the guy is approaching a live high tension line, probably 'cos he has to work on it. Why hasn't it been shut down? Probably 'cos that would mean shutting down an entire area of the countryside. And anyway, with the right kit, he doesn't have to!
So, as he approaches the line, he reaches out with a conducting rod that is connected to his suit. This will bring -the suit's- voltage (and that of the guy inside!) up to that of the power line, (without any voltage gradients across any part of his body) and as we can see, (from the spark!) -massive- current is being passed! Not just because of the guy in the suit is acting as a capacitor, but because the power is passing into the suit, and then discharged into the environment -from the suit-, and -not- through/from the guy himself.
As for 'feeling it'... (ie, feeling the voltage gradient I mentioned above) I spoke to a friend who did this kind of work. And being an adventurous kind of guy, he said yes, he'd wondered about this too... and he decided to conduct a lill' experiment. Yes? He wondered what would happen if he tried -removing- his suit while connected up to a -very- high voltage line! 😁
And what happened? Well, he said that as he began to open up his 'hoodie', (no rush!) he began to 'feel the electricity in the air'! And after he got the hood of the suit less than halfway off his head, he could stand it no more, and promptly put it back on!
Shocking stuff!
 
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  • #32
Dale
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That voltage gradient is a result of the capacitance of the line, and the current is still the charging of a capacitor. You wrote as though you think what you described is a big controversial surprise, but it is what we have been discussing.
 
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  • #33
thetrellan
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Yes, in my diagram the helicopter can be replaced by the bird just before touching the wire. The part of the bird closest to the power line, its feet, is one plate of capacitor C2, and the part that is furthest from the power line, say its head, is one plate of capacitor C2. When AC current is flowing through the bird's body, the excess charge is jumping between its head and feet. A little later, when the spark develops or when the bird is touching the wire, C2 is shorted out, and the excess charge is jumping between the bird's head and the "earth". The excess charge equals Q = CV = 150 kV · 10 pF = 1.5 μC = 10^13 electrons.

Yes, the size of the bird matters, smaller birds experience smaller currents, whereas larger birds and helicopters experience larger currents.
I know birds are small, but it's my understand that electricity takes the shortest route, which in the bird's case won't take it through the head, but from foot to foot. That much power would probably just fry the whole body, but still. When I nearly electrocuted myself I had my hand in a socket, and power flowed between just the last two fingers. I know this because you can still see entrance and exit wounds marked by the skin graft that followed. The rest of me didn't like that at all, of course. I was just a dumb preschooler, rooted to the spot, muscles frozen. A cousin tackled me to safety. My entire body was affected, but had I put both hands in travel would have been between them, likely stopping my heart along the way, but still not passing through the head.
 
  • #34
Michaela SJ
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Don't see birds on extra high voltage lines because of high density leakage and capacitive currents. EHV leakage currents would be unbearable for the birds.
I once watched a beautiful Golden Eagle land with a wing crossed over a transmission line and its other wing on the spreader. The eagle immediately combusted and caused enough damage that PG&E had to look at the tower from one of its helicopters.

It was always my understanding that birds do not exhibit enough of a path through their two legs for a current to leak off the less resisted transmission line.

I still carry the scar from leaning against a high voltage distribution line while climbing a tree when I was 10 years old. Simply being in the tree gave enough of a circuit through my arm to create a nasty burn.
 
  • #35
b.shahvir
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Electricity works in mysterious ways.
 
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