6
$\begingroup$

Voltage is potential difference, and current flows because of voltage. So if the voltage is zero, how can current flow through the neutral wire.

$\endgroup$
  • $\begingroup$ Is the circuit open? Please add details to your question to clarify all ambiguities $\endgroup$ – Jim Aug 11 '15 at 15:06
  • $\begingroup$ If you have ever put your hand on the 'neutral' wire (hint: do not do this, especially if you live in a 2xxV country) you will realise that the voltage on it is very often not even approximately zero. $\endgroup$ – tfb Jun 13 '16 at 14:20
  • $\begingroup$ It can't. If the wire is not a superconductor, then the voltage must be different at every point along the length of the wire while current is flowing. The voltage of the neutral conductor is only nominally 0 volts (i.e., zero volts in name only.) If the guage of the wire is heavy enough, then the voltage difference from one end to the other won't be enough to matter (as far as your electrician or your building inspector is concerned.) $\endgroup$ – Solomon Slow Jun 13 '16 at 15:17
6
$\begingroup$

You are taking a shortcut when you say, "The voltage is zero."

Voltage is always measured between two points. In electrical engineering, when we say the voltage at point X is V, we actually are measuring the voltage between point X and an implicit other point called "ground".

In the electric power grid, "neutral" is ground, by definition. So the voltage of the neutral wire is always zero... By definition.


The reality is a little different.

If you measure the voltage between any two points on a superconducting wire, you will measure zero volts no matter how much current is flowing, but the neutral wires in the power grid are not superconductors.

If you measure the voltage between two different points on a neutral wire that is carrying current, you will be able to measure a small difference.

Usually, that difference is small enough to be ignored for most purposes. If it's not small enough to be ignored, it means that the neutral wire is too small (diameter) for the amount of current that it's carrying.


current flows because of voltage

Well, yes, but no. Current will flow forever in the windings of a superconducting magnet with no applied voltage at all. It takes voltage to start the current flow, but once the current is flowing in the coil, no voltage is needed to keep it flowing.

Current only stops flowing in the coils of an ordinary electromagnet when the voltage is removed because of the wire's resistance.

It's like the difference between an airplane flying through the air, and a satellite orbiting the Earth far above the atmosphere. The plane won't fly for long if its engine quits because of the resistance it encounters as it moves through the air, but after its initial boost, the satellite will keep orbiting forever.

$\endgroup$
  • $\begingroup$ From a different vantage point - is this also why when I move a AC current sensing stick (non-contact type) along a "live" neutral wire, it doesn't alert? That's what I'm observing. (And since this is a specific case, it's a hot-side-only switch to lights, which are on, and energy returns via neutrals. The sensor things there isn't anything in the neutrals, but I know there is since the switch activated the lights). Thanks for your answer and anything you can add. $\endgroup$ – Scott Prive Feb 24 '19 at 23:49
  • $\begingroup$ @Crossfit_and_Beer, Sorry, I don't know how those voltage probe things work, but it would not make sense if they only worked when a load was drawing current from the line. The voltage probe is a safety device, and the live wire is dangerous whether current is flowing or not. My guess is, the capacitance between your hand and the device and, between the device and the line, allows a minute AC current that the device can detect. But, since your body normally should be close to the same potential as the neutral line, that minute current only flows when the probe comes near the "hot" line. $\endgroup$ – Solomon Slow Feb 25 '19 at 14:46
  • $\begingroup$ Thanks. The "can't be (due to) load" part makes perfect sense, since unpowered. I Googled a bit more and I see these devices can be magnetic/inductive or capacitive. I can't summarize and there's a bit left to understand, but this points me in the right direction :-) $\endgroup$ – Scott Prive Feb 28 '19 at 11:27
0
$\begingroup$

The key thing to remember here is that we are talking about an ideal wire. This means that there is (effectively) no resistance in the wire, and therefore no voltage drop along the wire. Now if there is a voltage difference between the two ends of the wire, that induces a constant electric field along the wire, causing current to flow.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.