Question

From Michael on Skeptics Stackexchange:

How about a wire that's grounded? Safe to touch, right? WRONG.

________________ 30 amps -> ________________
| |
+ |
220V Load
- |
|______(YOU ARE HERE)______<- 30 amps________|
|
Ground

The wire you touched was not only at 0 volts, but also grounded, and yet, you are feeling pretty shitty in this diagram. You have ceased to be as a human, and you are now a part of a circuit, functioning as part of a return leg (pictured above) or as a "parallel path to ground" (not pictured above.)

I don't get how this can work. If the wire is at 0 volts and you are at 0 volts, then there is no potential difference and hence I'd expect no current. Is this physics correct?

Answer 1

The issue is that by touching a wire, you're augmenting the circuit with yourself as a resistor. (At first I wrote "inserting yourself" but as mmc's comment pointed out, that is a misleading phrase to use.) And whenever you change the layout of an electrical circuit, all the potentials and currents are subject to change. So the wire that is at ground potential before you touch it won't necessarily still be at the same potential after you touch it.

In this specific case, I'd guess that the 30 A value is the current that flows with only the battery and the load.

circuit without person

You can calculate that the resistance of the load has to be 7.3 ?.

If you could actually insert yourself into the circuit as shown in the diagram, you would be adding to the resistance of the circuit, which reduces the current. The resistance of the human body varies greatly depending on several factors, but a "typical" value might be on the order of 10 k? which reduces the current to a small fraction of an Ampere.

circuit with person in series

Even with that small current, though, you're going to experience a large voltage drop because your resistance is so large compared to the rest of the circuit. In the diagram above, there would be a voltage drop of 219.8 V across your body. By inserting yourself in the circuit, you've prevented the wire coming out of the load from being grounded.

Note that if you're trying to determine whether this is a dangerous situation to be in, the value you should be looking at is the current of 0.022 A, not the 219.8 V voltage drop. That's what it means to say that it's the current that kills you, not the voltage.

If you instead grab on to the wire while standing on the ground (which seems like a more realistic situation), you're not actually inserting yourself in the circuit. Instead you wind up with a setup more like this,

circuit with person in parallel

In this case the wire coming out of the load resistor is still at potential zero because it's connected to ground via a zero-resistance path. (Keep in mind that this is an ideal model; real wires do have some small nonzero resistance so in reality the wire wouldn't quite be at zero volts.) So the voltage difference across your body is going to be basically zero.

Besides, any current that flows between the circuit and the ground can do so by one of two paths, either through you or through the open wire. Since your bodily resistance is much higher than that of the wire, essentially all the current will go through the wire, not through your body.