B Why don't birds get shocked on power lines?

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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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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.
 
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
 
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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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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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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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.

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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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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.
 
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?
 
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?
 
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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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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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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
 
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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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.....



, 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
 
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?
 
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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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?
 
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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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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)
 
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.
 
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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.
 
Now it makes sense to me. Thank you for all the help.
 

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