A circuit with no voltage difference, but current flowing 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?
 A: If you are in the circuit at that point then there is a 240V potential different across you!  One side of you is at 0v and there is no potential difference between you and the 240V line (except for the wire resistance) and so that side of you is at 140V
If you mean you are standing on the ground and touch the circuit at that point then, no the line is at approximately 0 and there is no potential difference.
Obviously in the real world don't do this anyway - just in case!
A: I am stunned -- STUNNED! -- that no one mentioned the difference in voltage between "neutral" and "safety ground".
Anyone with a multimeter can jam the probes into the "neutral" and "ground" slots of a standard 3-pin electrical outlet and measure for himself the difference in voltage between those two points.
That difference is not zero -- even when both lines are properly and solidly connected to "earth ground" at a single point.
This situation involves small resistances that normally can be neglected, but not in this case.
This is not a situation with "no voltage difference".


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


The wire you touched was not only at 0
  volts, but also grounded

No. One end of the wire connected to the ground post is grounded at 0 volts, but the other end that you are holding is not exactly 0 volts.
...
...

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?

The wire is not at 0 volts, and therefore there is current through your body.
Let's redraw this diagram, including a few resistances that are normally negligible:


   ________________ 30 amps -> ________________
  |                                            |
  +                                            |
 220V                                        Load: R1
  -                                            |
  |_r2__________r6_______r4____<- 30 amps______|
  |                         |
  |r3_____(YOU ARE HERE)_r5_|
  |
  |
Ground



When 2 physical wires touch each other, there is a contact resistance.
(Rule of thumb: it's typically around 0.1 Ohm). That's r2, r3, r4, r5.
When you have current flowing from one end of a long wire to the other end, the resistance (r6) is often very negligible, but not in this case.
A copper-conductor 16 AWG extension cord has a resistance of about 4 milli-Ohm per foot.
So the "neutral" conductor of a 20 foot long cord has about 0.08 Ohm of resistance from the copper (not including the contact resistance).
Even though most of the 30 A bypasses the human and flows in the copper wires,
a small amount of that 30 A goes through the human.
The voltage across the human is approximately 30 A * ( 0.1 Ohm + 0.08 Ohm + 0.1 Ohm) =~= 8 V.
Out of 220 VAC, a mere 8 VAC is often negligible -- but not in this case.
If some human has a resistance of 100 Ohms (it's typically an order of magnitude more),
The current through that human would be approximately 8 V / 100 Ohm =~= 80 mA.
This is more than the 60 mA AC which can cause death from ventricular fibrillation.
A: 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 $30\ \mathrm{A}$ value is the current that flows with only the battery and the load.

You can calculate that the resistance of the load has to be $7.3\ \Omega$.
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 $10\ \mathrm{k\Omega}$ which reduces the current to a small fraction of an Ampere.

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 $219.8\ \mathrm{V}$ 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 $0.022\ \mathrm{A}$, not the $219.8\ \mathrm{V}$ 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,

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.
A: There will  be voltage across you since you are part of the circuit. How much? Well, apply Kirchoff's law,
$220 = IR + v$
where $I$ is the current in the circuit and also through your body which is $30 amp$ in this case, $R$ is the resistance of the load. $v$ is the voltage across you.
$v = 220 - IR$
or $v = 220 - 30R$
This is the voltage at a point just right of (you are here). You see it is not at 0 volt.
The voltage at a point just left of (you are here) is 0 however.
Hence 30 amps current will flow through your body which is dangerous. 
Don't try this experiment.
