I have been thinking about this and I know that other people have answered this on here, but there's one part that still baffles me, and it has to do with parallel circuits.

If I connect a battery to a resistor, and connect another in parallel to it, and measure the current across both, there will be a current! That is, if I connect a $6\ \mathrm V$ battery to a $100\ \mathrm\Omega$ resistor ($R_1$) and connect a $200\ \mathrm\Omega$ resistor in parallel to it ($R_2$), I would still measure $6\ \mathrm V$ across both (voltage is preserved in parallel circuits, correct?) and my current (per Ohm's Law) is

$$I=\frac VR\Rightarrow I=\frac{6\ \mathrm V}{100\ \mathrm\Omega}=0.06\ \mathrm A $$

$$I=\frac VR\Rightarrow I=\frac{6\ \mathrm V}{200\ \mathrm\Omega}=0.03\ \mathrm A $$

So that means one resistor has $6o\ \mathrm{mA}$ and the other has $30\ \mathrm{mA}$. Well and good, but why doesn't this apply to a bird?

That is, a bird dropping its feet on a wire isn't completing a circuit between two different potentials but it is making a parallel circuit. This is what confuses a lot of people I think including me. If the usual laws for parallel circuits apply, why doesn't it apply to birds on a wire?

One explanation I hear is that birds aren't connecting two places of differing potential – but if that was the case then why does my parallel circuit work? One resistor should register no current (or very little) – and I know if I make the resistor large enough (the one in parallel, say, $R_2$) the current draw will be smaller. Is that what is happening? The resistance of the bird is large enough that the current drawn is small?

Let's say a bird has $1\ \mathrm{M\Omega}$ of resistance. A $600\ \mathrm V$ wire would still put $0.6\ \mathrm{mA}$ through the animal.

But that doesn't satisfy me because we are dealing with a $\mathrm{kV}$ scale wire a lot of the time. You'd need for the bird, which is effectively a bag of water and such, to have a lot of resistance for that to work, but maybe it does.

I am always reading that in order for the circuit to be complete the bird (or person) must be grounded, but that doesn't make sense to me because then no parallel circuit would work from batteries! Or even the house current, which is basically a lot of circuits in parallel.

I feel like I am missing something here, and if anyone can tell me what it is that would be greatly appreciated.

  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Mar 27 at 5:42
  • $\begingroup$ Note that the real-world item corresponding to the battery in your scenario "if I connect a battery to a resistor, and connect another in parallel to it, and measure the current across both, there will be a current" is the generator in the power plant. And yes, if you touch the two terminals of the generator, creating a parallel connection which bridges almost all of the resistance between the terminals like you do with your battery, you die in a spectacular fashion. $\endgroup$ – Peter - Reinstate Monica Mar 28 at 17:09
  • $\begingroup$ Does anyone have access to a bird, to measure the resistance between its feet ? My google-fu is weak and I can't find a real-world measurement, and I have no pet bird. $\endgroup$ – Criggie Mar 28 at 22:39
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    $\begingroup$ @Criggie I needed a while but came up with this. I incorporated it in my answer. They seem to measure ~1kΩ/leg which I found rather small, given the dry claws etc. Now that I think of it, they were measuring chickens in some clamp which may significantly differ from a perching bird. But the error will exaggerate the current. $\endgroup$ – Peter - Reinstate Monica Mar 29 at 7:28

11 Answers 11


A birds legs are pretty close together. An electrical transmission wire has very little resistance.

This means that the voltage as a function of distance barely changes. So the voltage difference between two birds feet is essentially 0, because the potential on each foot is practically the same. The potential difference between the wire and the ground might be large; but the bird isn't offering any pathway between the wire and anything at much lower voltage. It only offers a pathway between it's two legs, and so voltage difference remains small.

To add on to that, the bird has a lot more relative resistance than the wire, since the wire is supposed to minimize voltage drop across it. This means that most of the current will also flow through the wire, and relatively little current would flow through the bird.

The bird isn't really at risk unless it can connect the high voltage line to something of a significantly different potential, which the same line a few inches further down isn't.

For an example of the numbers, Solomon Slow estimated in the comments:

Suppose the wire is equivalent to 000-gauge copper, 0.0618 ohms per 1000 ft. Suppose it's carrying close to its rated capacity: 300A. Suppose a bird, maybe the size of a dove, with legs that grip the wire about 1 inch apart. According to my calculation, the potential difference between points 1 inch apart along the length of that wire will be about 1.6 millivolts.

Emphasis mine. It should be pretty easy to check that estimation for yourself, but it really illustrates the problem.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Mar 28 at 17:17
  • $\begingroup$ Comments got moved to chat so your link no longer works. Also, how big a bird would we need for it to cook itself? $\endgroup$ – Ian Kemp Mar 28 at 20:36
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    $\begingroup$ can you detail the calculation? (distance between the legs, resistance of the bird) and add the expected current going through the bird for those 1.6μV? $\endgroup$ – njzk2 Mar 28 at 23:19
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    $\begingroup$ @njzk2 It's just $V=IR$, with $I$ assumed as 300 $A$ as per the comment, and with 0.0618 ohms per 1000 ft resistance, and 1 inch grip, you can calculated the resistance between the two points. That gets you 0.001545 $V$. I was writing it out in more detail but it got deleted, you should be able to check the math yourself. $\endgroup$ – JMac Mar 29 at 0:09

Let me say in advance: You are perfectly right! The wire is a resistor, the bird is a resistor, and a bird standing on a wire with both feet down is indeed a parallel resistor to the wire. This means indeed that current flows through the bird. It is just not much, because the wire is — by design! — a very bad resistor, and birdie is by comparison (and that is what counts here) a good enough resistor that not much current flows through it.

Here is a take with real numbers.

An overhead power line which can carry 1000 ampere has, according to this catalogue, p. 136, a specific resistance of $0.022\ \mathrm{\Omega/km}$, or $2.2\times10^{-2}\ \mathrm{\Omega/km}$. A ten centimeter stretch — assuming that the bird straddles this convenient distance — has therefore a resistance of $1/10\,000$ of that, or $2.2\times10^{-6}\ \mathrm\Omega$.

To push a current of 1000 ampere through the resistance of the straddled $10\ \mathrm{cm}$ stretch, $2.2\times10^{-6}\ \mathrm\Omega$, a potential difference of $$V=I\cdot R=1000\ \mathrm A\times2.2\times10^{-6}\ \mathrm\Omega=2.2\times10^{-3}\ \mathrm V$$ is needed. That is the potential difference between birdie's legs, a bit more than 2 millivolt. We could stop right here because we know that we can bridge a thousand times that potential difference without feeling anything, but let's for the fun continue. Last not least I have drawn a picture which I want to show.

According to research done with the goal of being nice to our dinner, a chicken's legs have a resistance of $1400\ \mathrm\Omega$ each. Because our bird is a little smaller, we assume only $2000\ \mathrm\Omega$ for both, also ignoring the resistance of the body between the legs, if only out of modesty.

This implies that the current flowing through birdie is $$I=V/R=2.2\times10^{-3}\ \mathrm V/2000\ \mathrm\Omega=1.1\times10^{-6}\ \mathrm A.$$ (Draining a AA battery with that current would take a hundred years or more.) The quoted chicken research paper also mentions that a stunning current of $81\ \mathrm{mA}$ is unreliable; that is $80\,000$ times the current flowing through birdie on the wire, so there is room for error, acid rain or voltage spikes.

Here is a circuit diagram depicting your parallel, unroasted bird. As you can see, I have drawn the wire as a sequence of adjacent resistors, each with a length of $10\ \mathrm{cm}$. Normal circuit diagrams simply ignore the micro-ohms and draw a straight wire. Shame on them! Guys, that confuses people! Every wire is a resistor! True, each $10\ \mathrm{cm}$ segment by itself is a very weak resistor; but a million of them are annoying enough that the power plant must up the ante to many kilovolt.

Non-roasted bird on a wire

I should perhaps add that we could reach the same conclusion easier. The voltage calculation is quite unnecessary if we assume that the bird does not change the overall current through the wire, before and after. Then the current (whatever voltage is driving it) simply splits according to the ratio of the resistances, which is about $10^9$, so that $1/10^9$th of the $1000\ \mathrm A$ going through the wire is going through the bird, which is $10^{-6}\ \mathrm A.$

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    $\begingroup$ The mix of standard and scientific notation is a little confusing, IMO. Wouldn't it be better to be consistent, e.g. to write 1000A as 1 * 10^3? $\endgroup$ – Ian Kemp Mar 28 at 20:34
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    $\begingroup$ @IanKemp Well, 1000A and 2000Ω (or rather, 1436Ω or whatever per leg) were given in the literature I quoted in this format. I thought thousands are easy enough to grasp. The others I kept scientific for easier computation. Oh, and in the image/circuit diagram I found that 0.000001 A was illustrating the point very nicely. Who cares about an exponent more or less (did I get them right?). $\endgroup$ – Peter - Reinstate Monica Mar 28 at 23:54
  • $\begingroup$ Good answer, particularly the last part regarding ignoring the voltage. It's true that the additional resistance of the bird will not increase the load on the circuit in a measurable way. $\endgroup$ – David Waterworth Mar 29 at 0:01
  • $\begingroup$ I've deleted a number of obsolete comments and/or responses to them. $\endgroup$ – David Z Mar 29 at 11:41

Electric potential is a difference between two points, and considering that the wire the bird is standing on has little resistance, the potential difference would be negligible. That means when the bird is standing with both feet on the wire, the potential difference between its two feet is minuscule and with its own high resistance, will most definitely not hurt it.

The point you have made, concerning that 600V is across the bird has assumed that the bird places one claw on the live wire (600V) and the other one on circuit ground (0V). In contrary to the actual case the bird is standing on two points, where they have very similar potential difference, and by Ohm's Law the current is basically zero. To aid understanding, you can assume the bird is an extremely large resistance resistor, and it sits parallel to a conducting wire.

enter image description here

And for the same reason, I believe why others say that this is not a parallel circuit is due to the fact that the resistor has negligible influence on the circuit.

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    $\begingroup$ Actually it is a parallel circuit. It's just that the wire has a much lower resistance than the bird, so the bird is getting very little current (not zero but probably below all sensory thresholds). $\endgroup$ – toolforger Mar 27 at 11:43
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    $\begingroup$ To respond to OP's confusion of parallel circuits, you could include 62 milliohm resistance in parallel with the megaohm one and observe all the current flowing through the wire and almost nothing through the bird. $\endgroup$ – Džuris Mar 27 at 12:27
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    $\begingroup$ The diagram is fundamentally flawed. There should be an additional resistor shown on the bottom wire with a value such as 2 ohms to limit the current flowing through the circuit. Otherwise, the wires would melt. $\endgroup$ – Itsme2003 Mar 27 at 18:29
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    $\begingroup$ The circuit diagram makes no sense. The diagram simultaneously shows that the two ends of the battery have the same potential and that there is a 600V potential difference. That cannot be. $\endgroup$ – user2705196 Mar 27 at 21:05
  • $\begingroup$ @Džuris That would be the correct depiction (although the resistance of a high-current overhead cable is more like 1E-6 Ohm/10cm). I created a diagram like that in my answer. $\endgroup$ – Peter - Reinstate Monica Mar 28 at 11:22

Let me first note that the voltage itself does not have any physical effect - the harm comes from the electric current. This happens in two ways:

  • via the Joule heat when high current runs through the body (this requires currents of several Amperes, that we rarely encounter in the everyday life, but it is relevant for the bird)
  • via triggering the heart fibrillation when the frequency of the alternate current is in resonance with the heart frequency, that is 50-60 Hz (colloquially this is known as electrocution)

Let us now look at the parallel circuit formed by the bird and the piece of the wire between its legs. The voltage, i.e. the potential difference between the bird's legs is definitely not the same as the potential difference between the wire and the ground (which is known to be a few kV). The current in the wire (a few Amperes) is partitioned between the bird and the piece of wire between its legs: $$i = i_{bird} + i_{wire}.$$ The potential difference between the bird's legs is $$V = i_{bird}R_{bird} = i_{wire}R_{wire}.$$ Solving these three equations we obtain: \begin{array} ii_{bird} = \frac{iR_{wire}}{R_{wire} + R_{bird}},\\ i_{wire} = \frac{iR_{bird}}{R_{wire} + R_{bird}},\\ V = \frac{iR_{wire}R_{bird}}{R_{wire} + R_{bird}}. \end{array} The resistance of a human body ranges from 1000 to 100000 Ohms, depending on whether it is wet or not - this could be a good estimate for the bird. The resistance of a copper wire is a few Ohms per thousand feet (depending on the wire diameter). That is the piece between the bird's legs has resistance of a few mOhms. Thus, $$\frac{i_{bird}}{i} = \frac{R_{wire}}{R_{wire} + R_{bird}} \approx \frac{R_{wire}}{R_{bird}} \ll 10^{-6},$$ i.e., the current flowing through the bird is a millionth part of the current in the wire or even smaller! It is too minuscule to cause any real damage. In other words: the piece of the wire between the bird's legs short-circuits the bird.

I would like to return here to some aspects that are often overlooked when discussing electric circuits:

  • It is possible (and very common) to have high potential difference (voltage) without any current flowing. Capacitors routinely accumulate voltages up to kilo- and mega-volts without a current flowing. Human skin has breakdown voltage of about 500V, that is a constant potential of a few hundred volts will not cause any current flow (and any harmful effects) at all! This is equally relevant for a bird. AC current presents greater danger, because ac impedance of a human body is much lower at frequencies of 50-60Hz.
  • That voltage may exist without current is important to remember when applying the Joule's heat formula: $P = i^2 R$ and $P = V^2/R$ seem to tell the same thing, but both are applicable only when there is an actual current flow, not whenever a voltage is applied.
  • There are no perfect sources of potential difference - connecting anything to a circuit changes the potentials and the currents in this circuit. In particular, one distinguishes voltage bias and current bias when talking respectively about the circuits designed to maintain the same level of bias or the same level of current. Fuses are used to detect exceedingly high current and prevent it from damaging the circuit (but interrupting the current flow). The bird in question finds itself as a part of a circuit where the current rather than the voltage is maintained.
  • $\begingroup$ Basically even if a bird sits on a feet of wire between two terminals it will not die due to electric shock (also it could be fried by melting wire :) ) $\endgroup$ – Alexei Levenkov Mar 27 at 4:16
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    $\begingroup$ The wire carries the same current, as if there were no bird on it. So it is not melting. $\endgroup$ – Vadim Mar 27 at 5:53
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    $\begingroup$ As my father used to tell me, It's the amps that kill. $\endgroup$ – Will Crawford Mar 28 at 1:37
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    $\begingroup$ Also note that electric lines are designed to carry potential, not current. Current has the nasty property of heating the wire, which is just plain loss. Potential can be transmitted with virtually no losses. That's why we put many, many kilovolts on those electric lines, and why those electric lines do not operate anywhere near their actual current limit (the lines would be glowing in the night if that were the case!). The bird only feels the effect of the current, which the electricity corporations aim to optimize away. $\endgroup$ – cmaster - reinstate monica Mar 28 at 6:29
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    $\begingroup$ The danger of a current flowing through the heart is not defibrillation -- that would be the remedy, administered by a device aptly named "defibrillator" :-). The danger is the fibrillation. Paradoxically, defibrillation is also attempted by letting current flow through the heart; the poison is its own cure. Similia similibus curentur... $\endgroup$ – Peter - Reinstate Monica Mar 28 at 11:16

Qualitatively: the low electrical resistance of power lines means that there is negligible difference in electrical potential between two closely spaced points on the line. As a result, barely any current will flow through a (resistive) bird on the line, and the vast majority of current will run through the (conductive) line as normal.

Quantitatively: the resistive loss on high-voltage power lines is typically 0.5% per 100 miles on a 765 kV line, which means a bird sitting on the line with its feet 10 cm apart has a voltage of about

$$\frac{0.005*765,000\ \mathrm{V}}{100*1600\ \mathrm{m}} \times 0.10\ \mathrm{m} \approx 0.0025\ \mathrm{V}$$

applied across its body. A safe bet for a bird's electrical resistance seems to be $500\ \mathrm{\Omega}$, such that the resulting current through the bird will be

$$0.0025\ \mathrm{V} / 500\ \mathrm{\Omega} = 5 \times 10^{-6} \ \mathrm{A}$$

through the bird's body. The bird will be fine :)

  • $\begingroup$ THAT helps. Though I wasn't aware potential difference behaved that way because when you hook up circuits in the house, for instance, I know the wiring might be long relative to what's wired elsewhere - but I assume that all my outlets provide ~120V (in the US) -- so why is it that my short-ish connection in one part of the house is basically the same voltage as the other part that is 2x further away? $\endgroup$ – Jesse Mar 26 at 13:54
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    $\begingroup$ "Power line operating at 500kV with a length of 100km" does not mean that there's a 5V/m gradient along the length of the wire. "Power line" means a power transmission line. There's a power supply (e.g., a generating station) at one end, and there's a load at the other end. Most of the voltage is dropped across the load. I don't know what the actual voltage gradient along the length of the conductors in a long-hau l transmission line would be, but I'm guessing, a few millivolts per meter at most. $\endgroup$ – Solomon Slow Mar 26 at 14:04
  • $\begingroup$ @SolomonSlow that makes sense, and I did not consider it. So the actual voltage drop per meter of power line would be much lower still. $\endgroup$ – Jeff Mar 26 at 14:07
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    $\begingroup$ In addition to @SolomonSlow comment, the electricity network consists of power lines at various voltages connected by transformers which change the voltage up or down. At the power station the voltage might be transformed from 11kV to 500kV which then flows through the transmission network for many hundreds of km's, it is then stepped down to say 11kV (sub-transmission) for smaller distances then finally 400V (distribution) to the premisis. $\endgroup$ – David Waterworth Mar 26 at 23:48
  • $\begingroup$ @SolomonSlow power loss due to power line resistance is usually given as 5%. So if 95% of the energy are dissipated in the load, you have 95% of the voltage drop at the load and 5% on the wire. So you can divide the between-feet voltage by another factor of 20. (Just very, very ballpark figures.) $\endgroup$ – toolforger Mar 27 at 11:45

The size of the voltage in the wire is not what determines the current through the bird. It is the difference in the voltage between the bird's feet. If the wire is charged by a 600V source, the voltage between the birds legs is not 600V. The actual voltage difference is very small. The birds legs are touching two points very close to each other on the wire. The 600V drop is over the entire length of the wire from the power source back to the power source. The bird would have to be touch parts of the wire miles apart in order to feel any significant amount of voltage. The electrical wire is also a good conductor, which means its resistance is small and so the voltage drops only a small amount over any distance that a single animal could reach.

To be more concrete, say you hook up a 9V battery to a light bulb with two wires. The wire connected to the positive terminal will be at a voltage of 9V. The entire wire will have a voltage of 9V. This means that, if you connect a pair of voltmeter probes to the same wire, you will measure a potential difference of zero, because both probes will be measuring the same potential. It is only when the probes are in contact with different wires that you will measure the 9V difference.

Remember, it is differences in voltage that cause current, not just an amount of voltage. I can be perfectly safe standing in a 600-foot tower. I am only in danger if I fall from that tower and hit the ground 600 feet below. If the bird on the wire could somehow reach the ground or another wire at a different voltage at the same time, then it would feel the full 600V potential difference, because two parts of its body would be feeling different voltages, which would cause a current to flow.

Your parallel circuit works because the first resistor causes a large drop in electrical potential from one side to the other. So, when you connect a second resistor in parallel, there is a potential difference across the second resistor. The bird on a wire is more akin to this diagram:

Resistance diagram with bird

The pluses indicate the constant higher voltage on the top wire and the negatives indicate the constant lower voltage on the bottom wire. It is only when a resistor straddles a positive-negative difference that a current flows through it. The bird resistor at top, with its legs touching parts of the wire with identical voltages, will have no current flowing through it. All of the current will flow through the much lower resistance wire.

The bird will have a resistance of, say, 1 M$\Omega$, while the wire between the birds feet will have a resistance of, say, 0.1 $\Omega$. Calculate the amount of current that will flow in the 1M$\Omega$ resistor in a parallel configuration.

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    $\begingroup$ This helps a lot too, I didn't realize that if I connected in the configuration you are showing that a voltmeter would read zero (though it makes perfect sense) $\endgroup$ – Jesse Mar 26 at 13:59
  • $\begingroup$ Also your answer helps me get a MUCH better sense of this, and as I have to go over this with students I wanted to be sure I could do so coherently $\endgroup$ – Jesse Mar 26 at 13:59
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    $\begingroup$ I would also tend to think of the wire as a large number of really really small resistors in series. The bird resistor is then a high resistance connected in parallel to one of the tiny line resistors. $\endgroup$ – David Waterworth Mar 26 at 23:50
  • $\begingroup$ @DavidWaterworth That works, too. $\endgroup$ – Mark H Mar 27 at 0:50
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    $\begingroup$ @DavidWaterworth My thinking exactly -- see my answer :-). $\endgroup$ – Peter - Reinstate Monica Mar 28 at 7:38

What is true for birds equally applies to humans: you can touch a wire carrying a significant current, and you won't get shocked unless there is a voltage difference between you and the wire:

enter image description here

Note that voltage itself doesn't matter. In case of a ground fault, it may well happen that the "live" (high voltage) wire becomes safe to touch:

enter image description here

Pictures are taken from the article about shock current path

Birds don't actually have much higher resistance than humans and they also get shocked just as well when they manage to complete the circuit for the shock current to pass.

  • $\begingroup$ So technically a man touching a live wire at, say, his home exposed wiring, such as when installing a new ceiling lamp, usually shouldn't feel anything too? Because that is usually not the case. $\endgroup$ – Gnudiff Mar 28 at 7:02
  • $\begingroup$ @Gnudiff The problem is AC. Each contact will change between voltages with high frequency. So theoretically if you would touch a contact in exactly the right moment for a very short time you wouldn't get shocked. $\endgroup$ – Paul Mar 29 at 15:08
  • $\begingroup$ @Paul but the high voltage lines use AC too? $\endgroup$ – Gnudiff Mar 29 at 18:25
  • $\begingroup$ Especially yes. I'm not an expert on this, but as far as I know the main advantage of AC power is that you lose less power over long distances. $\endgroup$ – Paul Mar 30 at 8:57
  • $\begingroup$ @Gnudiff If you have DC in your house then you definitely won't feel a thing. The pictures I have posted have DC voltage sources in them, not AC. DC lines do exist, and they have lower power losses compared to AC. Touching a live AC wire is possible if you wear a Faraday suit (that's how critical AC lines are repaired under voltage). $\endgroup$ – Dmitry Grigoryev Mar 30 at 15:03

Let's say a bird has 1 MΩ of resistance. A 600 V wire would still put 0.6 mA through the animal.

This is the crux of your misunderstanding. There are two or more wires exiting the electrical generator. 600 volts is the difference in electric potential between them, not a property of either wire itself.

The bird makes a circuit not between the two wires, but rather between two sections of the same wire. While there is indeed a very small difference in electric potential between two sections of the wire due to the nonzero resistance of real conductors, the power utility will always use a heavy enough wire to keep this electric potential very small since it represents energy lost heating the wire rather than being delivered to customers.

You can do some simple experiments to illustrate this yourself with a battery and a voltmeter. You'll notice the voltmeter has two leads, again because voltage is an electric potential difference and it takes two points to measure a difference. Try connecting a light bulb, or an LED and resistor to the battery and measuring the voltage when the meter is placed in parallel to the battery versus two ends of the same wire.


Bird is sitting at the terminal of the source as long as bird is not connected to the ground. Resistance of the bird is so much high as compared to the resistance of the piece of wire between its two legs to whome the bird is parallel.


That is, a bird dropping its feet on a wire isn't completing a circuit between two different potentials

That's not correct. If a wire carries, say, 10kv over a kilometer, then it's carrying 10V over each meter, or 10mV over each mm.

It's not clear what you're asking here.

Is that what is happening? The resistance of the bird is large enough that the current drawn is small?

Yes. Not only does the bird have more resistance on a per length basis, but the path of the electricity through it is longer: the wire is connecting two points in nearly a straight line, but the electricity going through the bird has to go up its leg, across the body, and down the other leg.

A 600 V wire would still put 0.6 mA through the animal.

Voltage refers to change in electrical potential. Every voltage is a difference between electrical potential; there is no such thing as an absolute voltage. If an outlet is labeled as being 120V, that means the difference between the two positive and negative is 120V. Every voltage requires two points; otherwise it doesn't make sense to talk of a "difference". Saying "600 V wire" doesn't make sense unless it's taken as shorthand for something else, such as "a wire connecting two terminals with 600 V difference between them". It's the terminals, not the wire, that has the voltage difference. You can't look at any particular point in the wire and measure 600 V. If you were to take a voltmeter and hook both of its leads up to the same point, it would register 0V, regardless of how much voltage there is across the circuit as a whole.

So if you have 600V between the power plant and a house, and a bird is on the wire connecting them, the 600V isn't relevant because the bird isn't connecting the power plant and the house.

One analogy for electricity is water. Voltage is analogous to height. If you have a river that's 1000 km long, and during the course of the river, the altitude decrease 1000m, each kilogram of water has to release 1000m*9.8m/(s^2)*1kg of energy over the course of flowing down the river. But if you build a water mill and cut out a side channel and have some of the water flow through your water mill, you're not going to get that 1000m*9.8m/(s^2)*1kg of energy per kg of water, because that 1000m of altitude decrease is how much the water goes down over the course of the entire river, not how much it decreases in your side channel.

I am always reading that in order for the circuit to be complete the bird (or person) must be grounded

That would cause what's known as a "displacement current". Displacement current is where there isn't a circuit in the normal sense; normally in a circuit, charge moves around the circuit, but current in to any portion equal current out, so there's no net movement of charge. In displacement current, there is net movement of charge from one place to another. Since this requires an object to hold charge, how much displacement current is possible depends on the capacitance. There is displacement current when the bird lands on the wire, but since the capacitance of the bird isn't very large, the displacement current is small. The capacitance of the ground, however, is so massive that the current can continue pretty much indefinitely. So electrocution can happen either when you are connect two terminals with a voltage difference, or you connect the ground to a terminal with a voltage difference compared to the ground.


I may not be very entitled to answer the Physics part of the question however, the reason birds do not conduct electricity is rather naive- It's feet are made of a hard natural material which doesn't conduct electricity at all.

A bird's feet and toes are made up mostly of tough tendons and bones. The feet don't have very many nerves, blood vessels or muscles. This is what enables a bird to land on cold metal perches or walk on ice when temperatures drop....and also to sit on Electric poles and wires carrying a high voltage.

Your concept of Physics may or may not be correct, but the basic answer to your question is that feet of birds are very strong Insulators, therefore no current flows through them. Thus, it's useless to apply concepts of physics when the current just doesn't pass through....

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    $\begingroup$ Where does that quote even come from? It seems quite incorrect. I'm pretty darn sure if a bird put its two legs on two different wires at different voltage, current would flow through it, the resistance wouldn't stop the current completely. $\endgroup$ – JMac Mar 26 at 14:08
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    $\begingroup$ This is nonsense. The bird's feet do conduct electricity, and the material in quoteblocks either needs a very strong citation, or should not be in quotes. $\endgroup$ – Nij Mar 27 at 0:23
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    $\begingroup$ Perfect insulators are very rare, there's probably some truth in this though. A birds claws may have very high electrical resistance compared to say a human foot. The current flow through them would then be proportionally lower for the same voltage. Plus an electrical current hand-to-hand is worse than foot-to-foot due to the current path and the location of our heart. I agree though the quote is misleading $\endgroup$ – David Waterworth Mar 27 at 0:36
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    $\begingroup$ Anything conducts electricity if you throw enough volts at it. $\endgroup$ – Mark Mar 27 at 1:31
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    $\begingroup$ I found most of that quoted section at projectbeak.org/adaptations/feet.htm but the claim after the ellipsis seems to have been added by the answerer. $\endgroup$ – maxathousand Mar 27 at 15:04

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