A bird stands on a dc electric transmission line carrying 4100 A (Fig. 18–34). The line has $2.5 \times 10^{-5} \; \Omega$ resistance per meter, and the bird’s feet are 4.0 cm apart. What is the potential difference between the bird’s feet?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. This bird is standing on a powerline carrying 4,100 amps. The two feet of the bird are located four centimeters apart or 0.04 meters. The resistance of this powerline is 2.5 times 10 to the negative 5 ohms per meter. So we multiply that by the number of meters between the bird's feet and we find the resistance between these two positions. This is the potential difference across the bird. Meters cancel, meaning we have one microohm of resistance which is really small. So the potential difference between the two feet is current times resistance. That's 4,100 amps times one times 10 to the minus 6 ohms which is 4.1 millivolts. And so this is why birds don't get shocked when they stand on powerlines because the voltage between their feet is what matters and then the voltage is very small. The powerline is rated at probably in the one outside the house will be rated at a few thousand volts I guess. But that's with respect to the ground. And so if the bird was to put one foot on the powerline, the other foot on the ground, and it had to be like a massive albatross type bird or something, then that would be a problem and it would get shocked for sure. As it is, potential difference between these two feet that are both on the powerline is actually very small even if the line is not insulated.