I've been googling for hours and went through over a hundred answers. Now, some say the bird doesn't form a closed loop, some say the current is so small that it doesn't kill the bird. From as much as I understand electricity, I'm sure the bird does make a parallel connection to the wire, so there must be some electrons moving through it's body.

Could someone please explain which is the right answer and why? And if it's the case that a parallel circuit is formed, how can I calculate the current going through the birds body?

  • $\begingroup$ Isn't the wire insulated? It looks insulated from the ground. $\endgroup$ – Paul Apr 6 '13 at 2:19
  • $\begingroup$ @Paul No, the wires are not insulated, if they were so the insulation would have melted down due to heat produced by high power transmission through wires. $\endgroup$ – ABC Apr 6 '13 at 2:38
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    $\begingroup$ @exploringnet I think you are wrong, that high tension wires cannot be insulated . Remember there exist underground high tension wires too. Also if the inherent wire resistance were so high to induce a melt in plastic from the heating the birds would fry from the heat. The wires are not insulated because it is cheaper not to be and just trust to the air in the distance between them . $\endgroup$ – anna v Apr 6 '13 at 3:13
  • $\begingroup$ Yeah, I've been reading up on it, I always thought they were insulated because they look insulated, but I guess that's just what they look like. Clearly the insulation wouldn't melt because they are held apart from one another by insulators. $\endgroup$ – Paul Apr 6 '13 at 18:41

The copper cable inside the telephone wire has some finite resistance per unit length, call it $\alpha$. If current $I_0$ is flowing through the wire, then the voltage drop across a length $L$ of cable is $V=I_0\alpha L$. Here $L$ is the distance between the bird's feet. The bird forms a parallel branch whose resistance $R$ is probably almost entirely due to the resistance of the insulator on the wire under the bird's feet, not the bird's body itself. The current though the bird is $I_B=V/R=I_0\alpha L/R$. Because $\alpha$ is small and $R$ is large, this current is extremely small compared to $I_0$.

The people who tell you $I_B$ is zero are basically right. They're applying approximations that are excellent in most situations and that are excellent in this situation. Specifically, we normally take $\alpha\approx0$ for a wire, so there is no voltage drop between any two points on the same wire.

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  • $\begingroup$ Yes ,we mostly analyse for $\alpha\approx 0$ , and say " birds feet are equi-potential and thus bird has no current passing through it. $\endgroup$ – ABC Apr 6 '13 at 2:45
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    $\begingroup$ It depends on the power lines but some aren't insulated. My initial thought was the current would be exactly zero but your point that the tiny resistance in the wire and spacing of the feet is insightful. Surely though the current through the bird is so small that it can't be measured. $\endgroup$ – Brandon Enright Apr 6 '13 at 2:53
  • $\begingroup$ Birds also sit on high tension wires with no obvious problems, but your estimate holds for that too. $\endgroup$ – anna v Apr 6 '13 at 3:18
  • $\begingroup$ Just as an estimate, Aluminum has resistivity of 2.8e-9. the common gauges are between 12 and 750 mm^2. Assuming the birds' feet are 2 cm apart, resistance is R = pL/A, so for 12mm^2 that's 7.3uOhm, for 750mm^2 that's 0.1 uOhm. Voltage drop is the same for the wire and the birds, IW/RW = IB/RB, IB = IW(RB/RW). That means the bird gets IW*1e-5 / Ohm or IW*1e-7 / Ohm. I'm not sure how much current is drawn on average from those lines, but presumably a wet bird could actually get a serious shock (across its surface, anyway) if the transmitted power is on the order of 1-10kA. $\endgroup$ – Paul Apr 6 '13 at 18:39

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