# Birds sitting on electric wires: potential difference between the legs

We have seen birds sitting on uninsulated electric wires of high voltage transmission lines overhead without getting harmed, because sitting on only one wire doesn't complete any circuit.

But what about the potential difference between their legs? Is this not a small complete circuit? Because the wire has a potential gradient, there should be a potential difference between the bird's feet. Is this potential difference so very small that we can say the bird is sitting at a single point on the wire? If a bird of a sufficiently large size, with a wide gap between its feet, sits on a single wire, shouldn't the bird receive a shock if the potential difference is sufficient?

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Yakshesmash ! We see that birds do not die when both legs are on same wire and there is a no connections to ground. Therefore, it is a resonable to conclude that the potential difference between their legs is a zero or very small. Chenqui. –  Borat Sagdiyev Apr 21 '14 at 18:16
Comment to the question (v6): In such homework type of questions, OP is encouraged to perform a crude back of an envelope calculation as a sanity check, and possibly include it in the post. –  Qmechanic Apr 21 '14 at 20:57

The potential difference between two points on a wire carrying a current is given by Ohm's Law, $V = R\cdot I$.

Since wires used for long-distance power transmission have, by design, a very low resistance per unit length, and the distance between the two extremities of your hands is very small (~10cm), even for large currents the potential difference is not dangerous at all.

For example, if 10cm of gauge 4/0 aluminum wire (cross-sectional area 1.07 $\mathrm{cm^2}$) have a total resistance of $2.63714\cdot 10^{-5} \Omega$, so if a large current of 300A flow through (maximum rating for this gauge of cable, actually), then the potential difference between the ends will be $V=2.63714\cdot 10^{-5} \Omega \cdot 300\mathrm{A}=0.0079\mathrm{V}$ which is not noticeable by human skin.

The same goes for the feet of a bird, which are separated by an even smaller distance.

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Note that in AC lines (which includes some but not all "high tension" lines) the potential difference between two points that are "nearish" can be dominated by the oscillation of the signal rather than the ohmic attenuation. With 50 or 60 Hz signals and standard issue birds this is not an issue, but the bare statement was bothering me. –  dmckee Apr 20 '14 at 0:01
@dmckee For a worst-case 400kV @ 60Hz I get a peak difference of 71mV across 10cm due to oscillation (assuming perfect sine wave moving at c). –  eBusiness Apr 20 '14 at 22:09

Here is a circuit representing the system. $R_{wire}$ is the resistance of the section of wire between the bird's legs. $R_{bird}$ is the resistance of the bird (which you can measure by sticking the two probes of the multimeter to the bird's two feet - if the cable is insulated, you will have to add the resistance of the insulation as well).

# When the bird lands, do everyone's lights dim? (does the bird affect how much electricity goes through)

When the bird lands, the resistance between the two points (where its feet touch the wire) changes, so first we must determine whether the current coming from the transformer at the beginning of the power line changes. The resistance would go from $R_{wire}$ to:

$$R_T = \frac{1}{\frac{1}{R_{wire}}+\frac{1}{R_{bird}}}=\frac{R_{wire} \cdot R_{bird}}{R_{wire} + R_{bird}}$$

As evidenced by the fact that we use metal cables, and not birds, to transmit electricity, $R_{wire} << R_{bird}$:

$$R_{wire} + R_{bird} \approx R_{bird} \Rightarrow R_T \approx \frac{R_{wire} \cdot R_{bird}}{R_{bird}}=R_{wire}$$

Therefore, the resistance does not change much, and the current should also stay about the same because $I=V / R$. (Actually, the current will increase very slightly, because the bird's resistance will be in parallel with the wire's resistance, and this will decrease the overall resistance of the power line very slightly - thanks Nate Eldredge and Max)

# Does the bird experience extreme voltage?

The potential difference between two points is $V_0 = I \cdot R$. $I$ here is the total current passing through the wire, which we have already established does not differ much with the bird or without. So:

• Without the bird we have $V_0=I \cdot R_{wire}$.
• With the bird we have $V_{bird}=I \cdot R_T \approx I \cdot R_{wire}$ (see previous section).

Therefore the voltage experienced by the bird can be approximated with $I \cdot R_{wire}$. Once again, the wire is very conductive, so $R_{wire}$ will be small; $I$ may be large but not very large. $V_0$ will probably be a volt or less, likewise for $V_{bird}$.

Alternatively, we can observe that resistance is proportional to length, and therefore so is voltage: $$\frac{R_{wire}}{R_{line}} = \frac{L_{wire}}{L_{line}} = \frac{V_{wire}}{V_{line}}$$

Here:

• $R_{line}$ is the resistance between the two endpoints of the whole line
• $V_{line}$ is the potential between the two endpoints of the whole line (typically tens of kV)
• $L_{line}$ is the length of the entire power line (typically several kilometers)
• $L_{wire}$ is the length of wire spanned by the bird's legs (typically a few centimeters)

Therefore you can appreciate that the right side of the equation is a very small number, so likewise, $V_{wire}$ must be less than a volt - and the bird experiences approximately $V_{wire}$ potential difference as well.

# Does the bird experience extreme current?

Despite low voltage, high current may still be dangerous to animals. As pointed out before, the amount of current passing through the bird-wire block is $I_T=V/R_T \approx V/R_{wire}$.

At one of the bird feet, the current will split into $I_{wire}$ (which goes through the wire) and $I_{bird}$ (which goes through the bird), and then combine at the other foot. Because $V_T = V_{bird} = V_{wire}$, we can conclude that $I_{bird} = V_T/R_{bird}$ and $I_{wire} = V_T/R_{wire}$, therefore current and resistance of either component is inversely proportional:

$$\frac{I_{bird}}{I_{wire}} = \frac{V_T/R_{bird}}{V_T/R_{wire}} = \frac{R_{wire}}{R_{bird}}$$

We previously established that $R_{wire} << R_{bird}$, so $I_{wire} >> I_{bird}$. Current must be conserved (otherwise the bird must be stealing electrons) so $I_{wire} + I_{bird} = I_T > I_{wire} >> I_{bird}$.

$I_T$ can be pretty large for the higher capacity lines, but it's not that large - it's on the order of hundreds of amperes. Even though even 0.1 A is considered lethal to humans, the bird will experience a current $I_{bird}$ which is much smaller than this.

Recall the inverse proportion between current and resistance: Typically, animal bodies have a resistance of a few $M \Omega$ or a few hundred $k \Omega$ (original research), while good metal wires a few centimeters long will have less (often much less) resistance than 1 $\Omega$. So the current passing through the bird will be a few $\mu A$ at most - harmless.

# Is it dangerous for the bird to open its legs wide?

A critical factor is the ratio of the resistance of the bird's body $R_{bird}$ to the resistance of the section of wire between its 2 legs $R_{wire}$. First let's consider the the effect of opening legs on total current.

With legs closed, we get total resistance of the power line $R_{closed} = R_{line} + R_1 + \frac{1}{\frac{1}{R_2} + \frac{1}{R_{bird}}}$. With legs open, $R_{open} = R_{line} + \frac{1}{\frac{1}{R_1 + R_2} + \frac{1}{R_{bird}}}$. $R_{closed} > R_{open}$ (intuitively, you are replacing more of the wire with a more conductive bird/wire composite module). Accordingly, the total current through the whole power line will be higher when legs are open $I_{closed} < I_{open}$.

Furthermore, as Ilmari Karonen pointed out, increasing $R_{wire}$ increases both the potential experienced by the bird and how much of the (now higher) total current "splits" off into the bird part of the circuit.

If the bird increases the distance between its legs hundredfold, the increase in total current on the line will be negligible. $V_{wire} = V_{bird}$ will go up hundredfold, and correspondingly, the bird will experience hundredfold-stronger current. However, for a normal bird, if we repeat our original analysis we will find that even 100 cm of cable still has negligible resistance compared to a bird, so I doubt real birds would notice a difference.

What if you stretched a bird's legs so much that they could span the whole power line? Besides looking ridiculous, the bird would now experience tremendous potential difference. But in stretching the bird, you would also make it very thin (which increases resistance) and make it very long (which also increases resistance). So $R_{bird}$ would also be much larger and the current would still be very small. The bird would probably experience some form of discomfort, but not due to electrical phenomena.

What if you had a giant bird that is so big, its two legs could span the whole power line, even without stretching? Resistance is proportional to length, but inversely proportional to thickness. So if the bird was well-proportioned, it would have the same resistance as a small bird. However, now the resistance of $R_{wire}$ is non-trivial - many kilometers of even very conductive wire can have significant resistance. As said earlier, if 100 A passes through the power line, the bird need only get 0.1% of that to be at risk of death, so if the bird is long enough to span enough kilometers of power line that the resistance of the line is at least a few $k\Omega$, it will experience a very dangerous shock. Although a bird that big would also have other problems, such as the square-cube law, or current going through its head to make lightning in the upper layers of the atmosphere.

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Regarding your first question, the resistance of the bird is in parallel with the resistance of the wire, so the effective resistance will actually decrease (slightly). We definitely should not expect our lights to dim when the bird lands; if anything they would get brighter. –  Nate Eldredge Apr 20 '14 at 1:25
Having the bird spread its legs will (very slightly) increase both the current and the voltage through the bird (as one would intuitively expect), not decrease them: increasing $R_{wire}$ increases both $I_{bird} = I_{wire} \dfrac{R_{wire}}{R_{bird}}$ and $V_{bird} = V_{wire}$. –  Ilmari Karonen Apr 20 '14 at 2:41
That bird has webbed feet! No way could it perch on a power line! –  Nick Johnson Apr 21 '14 at 7:35
@Superbest: you should edit your answer to account for what Nate Eldredge said (I also noticed that mistake) –  Max Apr 21 '14 at 9:50
+1 for The bird would probably experience some form of discomfort, but not due to electrical phenomena. –  Cruncher Apr 21 '14 at 14:09

I suppose you could view your situation as a circuit with two parallel resistors; those are a part of the cable and the bird. However the resistance of the cable is many orders of magnitude smaller than the bird's (the cable is effectively a short-circuit), so there won't be any appreciable current through the bird.

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Even if the bird just bathed in salt water and is dripping wet, the bird's resistance will still be considerably higher than the wire's. Both that segment of wire and the bird have the same voltage drop across them, but it's unlikely that the bird will notice an uncomfortable current through it. –  Phil Perry Apr 21 '14 at 17:35

Maybe the confusion here is between potential difference, Volts, and the field strength, Volts/meter. If the man is just hanging in air, the field strength is relatively low and his primary danger comes from birds pecking his eyes out.

But if he hangs close to the ground, he's still at the wire's potential, but now instead of maybe 50 meters, he's 10 cm from the ground, so the field strength is very great. Once the field strength exceeds the dielectric breakdown value for air (homework: look it up :-) ), current will flow through him. That will be bad.

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sorry the question was not clear I modified the question –  Angelika Apr 19 '14 at 19:50

## protected by Qmechanic♦Apr 21 '14 at 18:12

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