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.
Best Answer
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.