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:

1. Why do birds sitting on electric wires not get shocked?

2. Birds on a wire (again) - how is it that birds feel no current? They are just making a parallel circuit, no?

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.