Are the "bird sitting on a live wire" answers wrong?

Question

Long ago, my high school teacher wrote the popular question on board,

"Why doesn't a bird sitting on a live wire get electrocuted?"

He gave us four options (I don't remember all of them) among which was the obvious "since the bird's feet don't touch the ground" and naturally we chose this one.

He told us that this answer was actually not satisfactory (or rather incomplete) since the current in the wire is not a direct current but an alternating current.

The bird's body should be treated as a capacitor (since the resistance of the bird owing to its small longitudinal extent can be ignored as both feet are at almost the same voltage) which for small frequencies offers large impedance.

Because of this the current through the bird's body is negligible and it doesn't get shocked.

Now, the following answers and links therein:

suggest that its actually the no grounding that prevents the bird from getting fried. (Along with having both feet in effectively the same place)

Most people in the previous answers seemed to not have mentioned anything about the alternating nature of the current and impedances etc generated in the bird's body due to that.

Which explanation is more correct?

P.S: The explanation which had been given by my teacher seems more plausible to me.

My own answer to one of the quoted questions brings attention to what it actually means to be electrocuted - this is crucial to understanding why it doesn't happen. See also this answer in the biology community. — Roger V., Nov 19 '21 at 08:47
Alternating nature of the current is not anything special. Alternating with respect to what? Electrical potentials are meaningful only when referenced to something else, in this case the ground). The potential of the bird is alternating along with the wire with respect to the ground. — Durmus, Nov 19 '21 at 10:44
Isn’t the most correct – yet quite boring – answer that the electric lines are wrapped in thick insulation and so no contact between bird and circuit is ever formed? — Cory Klein, Nov 19 '21 at 14:07
@CoryKlein No. If you touch the electric line and there's a good enough path to ground, you absolutely will die (and catch fire in the process). The insulation is there to prevent arcing between the wires (and anything nearby), not to make the wires safe to touch! Do NOT try this at home. — Luaan, Nov 19 '21 at 14:32
Excellent point, exempting a semantic quibble that at home this actually IS safe to do. Out of curiosity, where on the gradient of “insulated wire in my home on 120V” to “high voltage transmission line” does the “it’s not safe to touch” rule begin to apply? — Cory Klein, Nov 19 '21 at 14:49
As @Roger Vadim points out, it seems the confusion is terminological. To sum up, the DC does not fry (By Joule's heating) and the AC does not electrocute because the current through the bird's body is very less due to the bird's high impedance, resistance and low longitudinal extent (small potential difference). Also, since my teacher mentioned electrocuted in the question he probably was looking for the answer that specifically menitons the AC effect whereas in reality both effects (AC and DC) could be taken into account to explain why the bird doesn't die... — Lost, Nov 19 '21 at 15:16
Thus, I conclude that the linked answers are correct and satisfactory Note: My usage of AC and DC for the transmisson lines is based on the fact that AC also has a DC component in it. — Lost, Nov 19 '21 at 15:17
@CoryKlein if the wires have enormous glass insulators as stand-offs, that indicates that these expensive parts are needed to prevent the wire from getting close to the support poles and each other. — JDługosz, Nov 19 '21 at 16:42
@CoryKlein Overhead, high voltage transmission lines are bare wire; this applies all the way down to the high side of the local distribution transformers. It's cheaper, requires less maintenance (cloth or rubber wire sheathing would be degraded by the weather and require periodic replacement) and the wires are adequately insulated from each other by the air, and from the poles by the insulating stand-offs. — zwol, Nov 19 '21 at 16:58
If you've seen what happens to a large bird or bat (or indeed, an arboreal mammal small enough to walk on a single wire but also large enough to touch two at once, a common sight where I live) it's quite clear that you don't have to touch the ground to die, contact with two such wires is sufficient. — Glen_b, Nov 20 '21 at 00:47
I understand birds don't perch on very high voltage lines (>500 kV), perhaps due to corona discharges from pointy parts generating uncomfortable sensations. — Neil_UK, Nov 20 '21 at 09:00
Self-capacitance is amazingly small. For the entire earth it's only 710uF. For this reason, the bird doesn't need to be considered a capacitor except at very high frequencies. — Matt Timmermans, Nov 20 '21 at 14:06
Has anyone suggested that the reason may be, because the wires are insulated? — nuggethead, Nov 20 '21 at 15:29
An answer that considers the capacitive impedance of the bird is more complete than one that does not, even if the conclusion is that it is insignificant. Your teacher missed at least one other issue, though: the inductive impedance of the wire between the bird's feet (similarly negligible at power-line frequencies). And while we are on the topic, a bird landing on a DC high-tension cable will experience a brief, tiny current as its miniscule capacitance charges up - and capacitive coupling means this would be true (at an even more negligible level) if, per @CoryKlein, the cable was insulated. — sdenham, Nov 21 '21 at 17:34
@nuggethead yes, in the third comment of the thread you're commenting in — coagmano, Nov 22 '21 at 00:54

score 38 · Accepted Answer · answered Nov 18 '21 at 20:16

38

Which explanation is more correct?

The answer to the second question you cite is the best one.

In order to be "electrocuted" a non-trivial amount of current must flow through the body. The amount of current that flow is a function of the impedance of the bird and the voltage difference between the two contact points.

The second point is crucial here. The voltage difference the two contact points is essentially zero. A bird's feet are maybe a few centimeters apart and they touch THE SAME wire. The only voltage difference between the two feet comes from losses in the wire itself and these are minimal over such a short distance. Power line wires are specifically designed to have as little losses as is practical!

Large birds do indeed get electrocuted occasionally. That's simply because some part of their body touches (or gets too close) to something that's NOT the same wire their feet are on and that are at different potential. It doesn't need to be ground, any other phase wire will do the trick just as well (if not better).

The AC argument doesn't hold much water. The bird has indeed a capacitance but it's small and the AC line frequency is very low. If we assume a capacitance of maybe 50 pF for the bird (a human as about 100 pF) and 50 Hz, that comes out to an reactance of 16 $M\Omega$ as compared to a few $k\Omega$ for the resistive impedance of the bird. So the capacitance contributes only about 1/1000th of the total current.

answered Nov 18 '21 at 20:16

Hilmar

3,949

16

"if not better", are you rating based on how well-cooked the birds are after? – Jojo Nov 19 '21 at 12:31
1

Regarding the last sentence, it still seems considerable since any amount of current over 10 mA is capable of producing painful to severe shock, and currents between 100 and 200 mA are lethal to "human body". This means a total current of over 10 A should be enough to fry a bird. Or am I missing something? – polfosol Nov 19 '21 at 13:57
5

"that comes out to an reactance of 16 MΩ as compared to a few kΩ for the resistive impedance of the bird." – Aren't you comparing apples to oranges here? For the bird's self-capacitance, the relevant voltage is the voltage between the power line and the ground (or any other conductor at a constant potential), which is very high. On the other hand, for the bird's resistance, the relevant voltage is the voltage between the bird's feet, which is very low. So I think the bird's self-capacitance will generate the vast majority of the current. – Tanner Swett Nov 19 '21 at 14:18
@polfosol That's a casualty of excessive simplification. The human body can actually handle decent currents (and survive much higher currents, though you will get burned if the flow is sustained for long enough). The thing that usually kills you is current flowing through your heart specifically - it's easy enough to disrupt the complex electrical patterns controlling the heart, and stop it. In fact, that's what defibrillators do (no, they don't "start" hearts, like on most TV shows - they're used when the heart is in fibrillation, not when it's "stopped"). – Luaan Nov 19 '21 at 14:36
2

I'm confused: you've shown that the reactance due to the capacitance effect is 1000x larger. Doesn't that mean the capacitance effect is dominant in reducing the current through the bird? – Alwin Nov 19 '21 at 16:00
10

Downed power line safety training teaches you to move in "bunny hops" because the gradient in the ground for a downed HV power line can be large enough to kill at the spacing of your feet. In this case the potential difference is caused by short circuit current into the earth. – crasic Nov 19 '21 at 17:36
2

@alwin: the resistance is parallel to the reactance, not in series. – Hilmar Nov 20 '21 at 00:49

anna v · Answer 2 · 2021-11-20T04:37:11.613

10

Edit due to comment:

This is a partial answer, it addresses the title of the question:

Are the "bird sitting on a live wire" answers wrong?

which according to OP

suggest that its actually the no grounding that prevents the bird from getting fried. (Along with having both feet in effectively the same place)

Even if one treats the bird as a capacitor, from an answer in electronics.se

sitting on the wire it is like C2 in the image, and C2 is shorted, by-passed. The two points are at the same voltage, and the voltage on the conducting wire for each leg of the bird is the same, even if alternating.

Whether there are higher order interactions with the electromagnetic fields that a bird might sense as a "tingle" or a sense of the tension in the wires next to it, is answered by the other questions.

edited Nov 20 '21 at 04:37

answered Nov 18 '21 at 19:22

anna v

233,453

3

Given that humans can and do sense high voltage a/c via capacitive coupling without touching a wire, so one plate is the wire and the other the human connected to earth, I'm not sure that circuit is the correct one, the two plates of the capacitor in question should be the bird and the nearest wire of a different phase. – Pete Kirkham Nov 19 '21 at 10:47
4

@PeteKirkham the question is about why birds can sit on a live wire without getting a shock. – anna v Nov 19 '21 at 11:42
2

the OP's teacher was saying that you need to show that the capacitive mechanism (which induces enough current to sense in a human) does not induce enough current to damage a smaller organism. This circuit is not showing that mechanism. It doesn't matter how many legs the bird has for that mechanism. – Pete Kirkham Nov 19 '21 at 19:36
1

@PeteKirkham I have edited to make clear to what my answer refers – anna v Nov 20 '21 at 04:37

score 6 · Answer 3 · answered Nov 19 '21 at 11:54

Nobody in the previous answers seemed to have mentioned anything about the alternating nature of the current and impedances etc generated in the bird's body due to that.

This is not correct, as my own answer to one of the cited questions explicitly mentions that ac and dc currents differ in their harmful effect. The danger of the ac current is in causing involuntary muscle contractions, and thus triggering heart fibrillation, which is the typical cause of death in everyday electrocution cases. This is possible with much lower potential differences than the harmful effects of the dc current, which are maily due to the Joule's heating (see references in this answer for more details). Yet, since the bird is in a parallel circuit with the wire, the currents generated are too weak to cause any harm.

Remark: note also that what is actually important is the current rathe rthan the potential difference: the latter may exist without any current flowing, and may drop significantly once the circuit is closed.

Thanks for pointing out your answer which I overlooked. I've made an edif in the question. Also, see the comment I've below my post. — Lost, Nov 19 '21 at 15:21

score 3 · Answer 4 · answered Nov 21 '21 at 23:29

All reasons are relevant. If only one condition were not met, the bird would be electrocuted:

If the bird were touching ground and wire, or two different phase wires, it's the only thing that stands between the electrons and a large voltage difference, and it will be well-done in no time.
If the bird were standing on a good resistor, comparable to its own resistance between its feet, you wouldn't need much current on the "wire" to kill the bird.
If the bird were touching a large enough capacitor, sufficient current would flow through the bird to kill it.
If the alternating frequency of the electric power source were high enough, the bird would get hot feet due to its own body acting as a capacitor. That wouldn't necessarily kill the bird, as such high frequencies don't interfere with nerve cells. That's the same trick as when a magician causes fluorescent light tubes to light up by touching them.

Fortunately for the birds, neither of these conditions are generally met:

Their two feet are touching the same wire, so there is no relevant voltage difference between their feet. And that wire is both very thick and made from a well-conducting metal, so it's resistance is totally dwarfed by the bird's resistance. Consequently, a negligible fraction of the wire's current flows through the bird. And that's the only thing the birds touch at all.
The body of the bird is small, and so is its capacitance. Also, the alternating frequency is low. The current that (un-)charges the bird's body capacitor is controlled by these factors. And they are so small that the current does not seem to cause the birds any discomfort.

Sidenote:
Many "simple" physics problems are like this. They look simple, they seem to have a simple answer. However, once you look closely, you see that they are actually not that simple. Quite frequently, it turns out that there is more than one effect at play, that somewhat (un-)important confounding factors have been ignored (like air drag in cannon-ball questions, etc.), and that the simple answer is only as good as these confounding factors are actually small. It is important to become aware of these ignored factors, so they can be taken into account when a change in experimental parameters causes them to become relevant.

Are the "bird sitting on a live wire" answers wrong?

4 Answers4

Linked