Why do birds on power lines not act as resistors?

In summary, birds are able to sit safely on power lines because their resistance is much higher compared to the resistance of the wire they are sitting on. This causes the majority of the current to flow through the wire instead of the bird. However, large birds such as eagles and owls can bridge the gap between wires, resulting in a bright flash and potential electrocution. In low power lines, birds do not complete the circuit and therefore do not get electrocuted, but in high power lines, they can potentially complete the circuit and suffer electrocution.
  • #1
steve0606
11
0
Imagine a circuit with a cell and no resistors. If I connected a resistor to two points in the circuit, would a current flow through it? Does a bird on a power line act like this? Why does it not get electrocuted?
Sorry if this is a stupid question - I'm terrible with electronics! :blushing:
 
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  • #2
They do.

The resistance of the bird is very high compared to the resistance of the bit of cable between their feet.
 
  • #3
Simon Bridge said:
They do.

The resistance of the bird is very high compared to the resistance of the bit of cable between their feet.

Thanks for your reply. If that's true, why do most people say its because "It doesn't form a complete circuit".
 
  • #4
steve0606 said:
Thanks for your reply. If that's true, why do most people say its because "It doesn't form a complete circuit".

It's a reasonable simplification of the answer.

If I grab a live wire but I am otherwise insulated, I won't get a shock. It would be overly nitpicky to talk about the closed circuits formed by each finger.
 
  • #5
Try explaining electric circuits to most people and you'll find out why - besides it's pretty much the same thing... basically no current flows through them like sea-level doesn't change when you take a bucket of water out of the ocean. The effect is a bit like lightning choosing to strike the lightning rod and not some to the rod and some to the building.

otoh: I heard of squirrels getting fried when they hop from the wire to a transformer case. Presumably birds dump their surplus charge when they fly.
 
  • #6
So what you're saying is that a bird would have a current flowing through it, but because its resistance is much higher than that of the cable, it is negligible. So we may say that it isn't really a proper circuit.
 
  • #7
steve0606 said:
Thanks for your reply. If that's true, why do most people say its because "It doesn't form a complete circuit".

Because most people are right.

Birds are generally smaller than the space between wires on a power distribution system. So, they don't get a high voltage across them and they don't get electrocuted.

However large birds, like eagles and owls can easily bridge the gap between the wires, especially if two birds are fighting while on different lines.

Then you get a bright flash, burning feathers and a smell of roast chicken.


(Welcome Simon Bridge)
 
  • #8
That would be the oxidization of collagen
 
  • #9
when a bird lands on a wire he will be at the same voltage as that wire. so only enough current flows into bird to raise him to that voltage.
so long as that current is less than he can feel in his feet, he'll be unaware.

if you notice, birds typically avoid wires more than about 40kv. it tickles their feet.

we had a 500 kv line through the swamp. every evening about dusk there'd be faults, which were traced to big wading swamp birds roosting on the steel crosarms above the wire. if you've ever seen one of those birds 'poop', well it's a stream rather like a garden hose. should the bird be right over the wire when he goes, he gets vaporized with a big flash.
the transmission folks welded uncomfortable plates to crossarms right over the wires so the birds would roost a few feet away.

old jim
 
  • #10
Electricity is lazy. It has to work to go thru the bird. Or it can flow easily thru a low resistance wire. There is a small trickle current going thru the bird, but not enough to effect it.

Picture this...two resistors in parallel. One really low resistance...one really high resistance. One resistor (the wire) has a really low resistance...say .005 Ohm per inch. Say the bird has a resistance of 2,000 ohms. Again, the current is lazy and chooses the wire for 99.9999% of it's current.

Now if the bird could reach out and grab phase B while standing on phase A...the bird would practically vaporize in a millisecond. Just sayin:)
 
  • #11
well , flamigoes do sleep standing in shallow water on one foot.
i've always wondered if this habit evolved to protect them from nearby lightning strokes, which could cause significant voltage to appear between their feet were both in the water.
 
  • #12
Critters commonly get electrocuted at pole transformers. People who hunt with falcons or hawks are weary of this. I've also seen a squirrel get toasted as he stepped from the wire to the transformer.
 
  • #13
I don't think bird don't get electrocuted because they have high resistance...

I mean we are still talking about 20kV or so. If they did somehow complete the circuit, current would pass through them.

But here is my view.

They are NOT in fact completing the circuit.

If a bird sits on a wire, there is no way that it can complete the circuit, it is just sitting on a bare wire. That wire might be hot from the passing current, but anyway this is the same as you jumping and holding the phase wire, or touching a live wire, with just one hand.

But this only goes for low power lines, like 20 kV. For 500 kV, there is a different story.

And no they do not act like a resistor. Do an experiment at home.

Take a 9 V battery, connect it to a resistor, let's say 500 ohms, and then short circuit that resistor, in parallel.

Current won't go through the resistor any more, as a poster before said, current is lazy and it wants to go through a path with less resistance, ergo your short circuit wire will get very hot.

Your bird is your resistor, and your line is your short circuit.
 
  • #14
vk6kro said:
However large birds, like eagles and owls can easily bridge the gap between the wires, especially if two birds are fighting while on different lines.

I have seen a bright flash on the verge of my field of view, looked that fast enough to see falling body, then heard the arch (I was about half a kilometer from the 15 kV line). There was a flock of startled Common Wood Pigeons flying in my direction, I know they often sit on these lines. So I think what have happened was that when they started one of them shorted the wires - for a pigeon this is relatively large bird.
 
  • #15
Bassalisk said:
I don't think bird don't get electrocuted because they have high resistance...

I mean we are still talking about 20kV or so. If they did somehow complete the circuit, current would pass through them.

But here is my view.

They are NOT in fact completing the circuit.

If a bird sits on a wire, there is no way that it can complete the circuit, it is just sitting on a bare wire. That wire might be hot from the passing current, but anyway this is the same as you jumping and holding the phase wire, or touching a live wire, with just one hand.

But this only goes for low power lines, like 20 kV. For 500 kV, there is a different story.

And no they do not act like a resistor. Do an experiment at home.

Take a 9 V battery, connect it to a resistor, let's say 500 ohms, and then short circuit that resistor, in parallel.

Current won't go through the resistor any more, as a poster before said, current is lazy and it wants to go through a path with less resistance, ergo your short circuit wire will get very hot.

Your bird is your resistor, and your line is your short circuit.

Well...a wire isn't actually a short circuit...it has resistance. It is just taught that way in school to make a difficult subject more understandable.

For example...when u take a 480 volt three phase phasor diagram...and you ADD the two 277 line to neutral vectors together, you get 277 volts. That doesn't make sense since we know that line to line voltage of 277 is 480. NOW...if you take phase A MINUS phase B...now you get a vector that equals 480. And when I say MINUS...that flips one of the vectors 180 degrees.

For voltage to show up on a meter, it has to be the difference between to points.

That being said you need a large voltage drop between the birds feet. Whats the length between a birds feet. 1.5 inches maybe. The wire it's sitting on is likely 1" in diameter.

How much voltage drop occurs between that 1.5 inches? Not a lot...

I don't have a code book handy, but let's say that the line has a resistance .005 ohms per 1.5 inches. I believe a one inch line in diameter can hold about 1,000 amps.

Using the old V=IR...005 X 1000 = 5 volts...less than a nine volt battery...bird doesn't feel a thing.

Hmmmm...notice that I have not even considered the voltage of the line. Amps are amps...ohms are ohms.

Think about that...even the vets on this site. I think Jim Hardy said they don't stand on lines higher than 40 KV. I think this may be more of the magnetic field ticklilng their feet then the voltage. The voltage matters when you are line to line.

I say that if you have a 1" diameter line with .005 ohms per 1.5 inch...with 1,000 amps thru it...the bird feels the same voltage whether the line is 2,400 volts...or 240,000 volts.

V=IR in either case.

Just throwing out ideas here...that make some logical sense. Feel free to comment.
 
  • #16
psparky said:
Well...a wire isn't actually a short circuit...it has resistance. It is just taught that way in school to make a difficult subject more understandable.

For example...when u take a 480 volt three phase phasor diagram...and you ADD the two 277 line to neutral vectors together, you get 277 volts. That doesn't make sense since we know that line to line voltage of 277 is 480. NOW...if you take phase A MINUS phase B...now you get a vector that equals 480. And when I say MINUS...that flips one of the vectors 180 degrees.

For voltage to show up on a meter, it has to be the difference between to points.

That being said you need a large voltage drop between the birds feet. Whats the length between a birds feet. 1.5 inches maybe. The wire it's sitting on is likely 1" in diameter.

How much voltage drop occurs between that 1.5 inches? Not a lot...

I don't have a code book handy, but let's say that the line has a resistance .005 ohms per 1.5 inches. I believe a one inch line in diameter can hold about 1,000 amps.

Using the old V=IR...005 X 1000 = 5 volts...less than a nine volt battery...bird doesn't feel a thing.

Hmmmm...notice that I have not even considered the voltage of the line. Amps are amps...ohms are ohms.

Think about that...even the vets on this site. I think Jim Hardy said they don't stand on lines higher than 40 KV. I think this may be more of the magnetic field ticklilng their feet then the voltage. The voltage matters when you are line to line

I say that if you have a 1" diameter line with .005 ohms per 1.5 inch...with 1,000 amps thru it...the bird feels the same voltage whether the line is 2,400 volts...or 240,000 volts.

V=IR in either case.

Just throwing out ideas here...that make some logical sense. Feel free to comment.

I am well aware of that. But bird is approximately 20 cm thick. A power line has internal resistance of roughly 0,3 ohms/km(only counting the active). So internal resistance of 20 cm is 0,00006 Ohms.

A human has resistance around 1-3kOhms. Let's say bird has half of that. For my engineering purposes, I call that short circuit.(the power line)

Of course I didn't mean a short circuit like in the high school. But what do you call a solo wire that is going in parallel with resistor?
 
  • #17
I hear ya...but we can't call it a short in this case. The resistance of the "short" is inducing a difference in potential (a voltage drop if you will) resulting in a trickle current thru the bird!
 
  • #18
psparky said:
I hear ya...but we can't call it a short in this case. The resistance of the "short" is inducing a difference in potential resulting in a trickle current thru the bird!

Agreed. Bird has to be at same potential as the power line right? So maybe 20 pAmps go through bird? Is that what you are saying?
 
  • #19
I think so...The voltage drop between the birds feet due to the resistance of the wire is the voltage the bird feels. That voltage drop then reacts to the birds resistance...then inducing the 20 pico amps.

To sum of my point of view.

V=IR :)
 
  • #20
psparky said:
I think so...The voltage drop between the birds feet due to the resistance of the wire is the voltage the bird feels. That voltage then reacts to the birds resistance...then inducing the 20 pico amps.

To sum of my point of view.

V=IR :)

I might even ask my professor about this. Is this current significantly larger when we are talking about 500 kV power lines? Thats why the get roasted?
 
  • #21
And to correct myself earlier, the voltage (above 40 K) shouldn't really effect the magetic field since the amps are 1,000 in either case.

But somehow it does...voltage does crazy stuff when it gets really high.

It's amazing how much you can learn just by typing on your keyboard and this forum. Really makes you think!
 
  • #22
Bassalisk said:
I might even ask my professor about this. Is this current significantly larger when we are talking about 500 kV power lines? Thats why they get roasted?

Absolutely not.

A 1" diameter wire will carry roughly 1,000 amps whether it is a 1 volt line or a...

1,000,000,000,000,000,000,000 volt line.

Wire size is the same for any voltage assuming the current is the same. The only thing that changes is the size of the insulator around the wire (when talking about commercial buildings and such...not open air conductors) I would assume the metal might have to be different to withstand super high voltage...but again...amps are amps.
 
  • #23
Keep in mind that the whole purpose of high voltage lines is to keep the amps as low as possible...while making the same power.

Lower amps means lower wire size...which means less money.

And money is pretty much the reason behind everything.
 
  • #24
psparky said:
Keep in mind that the whole purpose of high voltage lines is to keep the amps as low as possible...while making the same power.

Lower amps means lower wire size...which means less money.

And money is pretty much the reason behind everything.

Yes yes I know that. Joule losses. But why do they get roasted at 500k power line?
 
  • #25
Bassalisk said:
Yes yes I know that. Joule losses. But why do they get roasted at 500k power line?

Who says they do get roasted on a 500 KV line?

If they do...I'm not sure why. Even if the current was 5 times larger I don't think it would be enough make a big enough voltage across their feet.

Perhaps there is something we are missing, but my explanation above sounds pretty good.

Anyone else?
 
  • #26
jim hardy said:
when a bird lands on a wire he will be at the same voltage as that wire. so only enough current flows into bird to raise him to that voltage.
so long as that current is less than he can feel in his feet, he'll be unaware.

if you notice, birds typically avoid wires more than about 40kv. it tickles their feet.

we had a 500 kv line through the swamp. every evening about dusk there'd be faults, which were traced to big wading swamp birds roosting on the steel crosarms above the wire. if you've ever seen one of those birds 'poop', well it's a stream rather like a garden hose. should the bird be right over the wire when he goes, he gets vaporized with a big flash.
the transmission folks welded uncomfortable plates to crossarms right over the wires so the birds would roost a few feet away.

old jim

Is this the post you are reffereing to birds frying at 500 KV?

If so, it's because the bird is pooping to a lower wire...therefore inducing a line to line voltage.

Completing the circuit if you will.
 
  • #27
psparky said:
Is this the post you are reffereing to birds frying at 500 KV?

If so, it's because the bird is pooping to a lower wire...therefore inducing a line to line voltage.

Completing the circuit if you will.

Yes this is the post. Because of line voltage?
 
  • #28
Bassalisk said:
Yes this is the post. Because of line voltage?

That is factually correct. He says they **** a stream.

The bird might as well have it's mouth on one line and its wet *** on the other.

500KV across the bird.

What's resistance of a bird? 100 ohms maybe?

500,000 volts divided by 100 equals 500 amps thru it's body.

That'll leave a mark. Takes .1 amp to kill a human for reference.
 
  • #29
psparky said:
That is factually correct. He says they **** a stream.

The bird might as well have it's mouth on one line and its wet *** on the other.

500KV across the bird.

What's resistance of a bird? 100 ohms maybe?

500,000 volts divided by 100 equals 500 amps thru it's body.

That'll leave a mark. Takes .1 amp to kill a human for reference.


Oh so these power lines are actually close? so they can complete the circuit?
 
  • #30
Incidentally, the line to neutral voltage of the said 500 KV line is...289KV.

Think about it.
 
  • #31
There seems to be a lot of mis-information being allowed to go uncorrected here.

A bird that flies directly to a power line and grabs it makes no circuit. Therefore its resistance is immaterial (or futile, if you want a pun).

But what does happen is that it will become charged to the potential of the line. Assume bird is at 0 potential with respect to ground (just flown off of the ground) then it will need to receive enough charge to raise its potential to the line potential.

It will have a capacitance. Let's make a guess here, I'd say it's free-space capacitance is a bit like a sphere, let's call it 1pF. To charge to 500kV requires energy in the amount of 1/2.[5e5]^2.[1e-12] = ~0.1 J.

I reckon you'd not really feel a discharge much less than 0.5 to 1 J, but you'd likely feel a bit of a spark at that level. Maybe birds do feel it, but they get their kicks from landing on such wires.

Here resistance does play a part. That capacitance represents a total of Q=CV, or ~ a microCoulomb of charge. If that discharges in a microsecond then the bird would get an Amp of current. Again, it might not notice that anyhow, because the total energy in the pulse is small. If the bird's grip and its legs are a lower resistance, then the charge-up would be slower and the current lower.

To see how this charge-up process actually works, you should go looking for youtube videos of the guys that work on these wires - live - from helicopters. The helicopter comes up to the line and the guy on the back holds out a probe to faciliate charging up of the helicopter to the line potential. You see a spark of a few seconds duration flowing between the probe and the power line, then the guy hooks a wire onto the line, to maintain the potential, then he does his work. Those guys must be on serious danger money (because of the height - the voltage is immaterial so long as you respect it), some jobs they actually clip onto the lines and crawl off the helicopter onto the lines, and shimmy up it. A hat-tip to live helo power-line workers!
 
  • #32
Bassalisk said:
Oh so these power lines are actually close? so they can complete the circuit?

Wait a second here...
 
Last edited:
  • #33
cmb said:
There seems to be a lot of mis-information being allowed to go uncorrected here.

A bird that flies directly to a power line and grabs it makes no circuit. Therefore its resistance is immaterial (or futile, if you want a pun).

But what does happen is that it will become charged to the potential of the line. Assume bird is at 0 potential with respect to ground (just flown off of the ground) then it will need to receive enough charge to raise its potential to the line potential.

It will have a capacitance. Let's make a guess here, I'd say it's free-space capacitance is a bit like a sphere, let's call it 1pF. To charge to 500kV requires energy in the amount of 1/2.[5e5]^2.[1e-12] = ~0.1 J.

I reckon you'd not really feel a discharge much less than 0.5 to 1 J, but you'd likely feel a bit of a spark at that level. Maybe birds do feel it, but they get their kicks from landing on such wires.

Here resistance does play a part. That capacitance represents a total of Q=CV, or ~ a microCoulomb of charge. If that discharges in a microsecond then the bird would get an Amp of current. Again, it might not notice that anyhow, because the total energy in the pulse is small. If the bird's grip and its legs are a lower resistance, then the charge-up would be slower and the current lower.

To see how this charge-up process actually works, you should go looking for youtube videos of the guys that work on these wires - live - from helicopters. The helicopter comes up to the line and the guy on the back holds out a probe to faciliate charging up of the helicopter to the line potential. You see a spark of a few seconds duration flowing between the probe and the power line, then the guy hooks a wire onto the line, to maintain the potential, then he does his work. Those guys must be on serious danger money (because of the height - the voltage is immaterial so long as you respect it), some jobs they actually clip onto the lines and crawl off the helicopter onto the lines, and shimmy up it. A hat-tip to live helo power-line workers!

I always wondered why there was a huge spark between that probe and the line. Is it because the probe is initial at 0 potential with respect the to power line? And whole chopper is at 0 potential?
 
  • #34
Here's a clear show of the helo linesmen at work;

 
Last edited by a moderator:
  • #35
cmb said:
There seems to be a lot of mis-information being allowed to go uncorrected here.

A bird that flies directly to a power line and grabs it makes no circuit. Therefore its resistance is immaterial (or futile, if you want a pun).

But what does happen is that it will become charged to the potential of the line. Assume bird is at 0 potential with respect to ground (just flown off of the ground) then it will need to receive enough charge to raise its potential to the line potential.

It will have a capacitance. Let's make a guess here, I'd say it's free-space capacitance is a bit like a sphere, let's call it 1pF. To charge to 500kV requires energy in the amount of 1/2.[5e5]^2.[1e-12] = ~0.1 J.

I reckon you'd not really feel a discharge much less than 0.5 to 1 J, but you'd likely feel a bit of a spark at that level. Maybe birds do feel it, but they get their kicks from landing on such wires.

Here resistance does play a part. That capacitance represents a total of Q=CV, or ~ a microCoulomb of charge. If that discharges in a microsecond then the bird would get an Amp of current. Again, it might not notice that anyhow, because the total energy in the pulse is small. If the bird's grip and its legs are a lower resistance, then the charge-up would be slower and the current lower.

To see how this charge-up process actually works, you should go looking for youtube videos of the guys that work on these wires - live - from helicopters. The helicopter comes up to the line and the guy on the back holds out a probe to faciliate charging up of the helicopter to the line potential. You see a spark of a few seconds duration flowing between the probe and the power line, then the guy hooks a wire onto the line, to maintain the potential, then he does his work. Those guys must be on serious danger money (because of the height - the voltage is immaterial so long as you respect it), some jobs they actually clip onto the lines and crawl off the helicopter onto the lines, and shimmy up it. A hat-tip to live helo power-line workers!

Cool stuff.
 

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