3
\$\begingroup\$

This question already has an answer here:

When I probe my city mains with one probe in the live and one in earth (which should be 0 volt) it shows a voltage of around 250v. But when I probe the neutral and the earth it shows no voltage. I know that current runs in one direction for 50 times in a second. So the neutral should act like live for 50 times in a second. Then neutral should show some voltage with the earth, which it doesn't. If you touch the neutral wire you won't get shocked but if you touch the live the wire you get shocked why and how?

\$\endgroup\$

marked as duplicate by Huisman, JRE, Leon Heller, Dmitry Grigoryev, Nick Alexeev Jul 22 at 14:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    \$\begingroup\$ i can get shocked by live even if there is no load due to parasitic capacitance , but why does nuetral not behave the same \$\endgroup\$ – Jyotir Jul 21 at 5:00
  • \$\begingroup\$ so that means if touch the nuetral(which i am not going to) it will still shock me.my main question here is why is nuetral 0 volt or ground \$\endgroup\$ – Jyotir Jul 21 at 5:02
  • 5
    \$\begingroup\$ You have a misunderstanding about voltage measurements. \$\endgroup\$ – DKNguyen Jul 21 at 5:03
  • 2
    \$\begingroup\$ please help me clear the misunderstanding \$\endgroup\$ – Jyotir Jul 21 at 5:05
  • \$\begingroup\$ In case someone thinks tldr; YOU CAN GET SHOCKED BY TOUCHING NEUTRAL \$\endgroup\$ – Huisman Jul 21 at 6:06
14
\$\begingroup\$

We force it to be that way

Mains power is wired as an isolated system, with an asterisk. The asterisk came about for some very good reasons. The "safeness" of neutral is a side-effect, and an optional one.

If mains power were an isolated system (And I've run it that way, and it works), and you are grounded presumably... then it wouldn't matter if you touched pole 1 or center (I won't call it "neutral"). No current would flow. The hot and center have no relationship with earth (except through you, and with only one "wire", it's an open circuit). The system "floats".

An isolated system is exactly what you expect.

However, we build mains power to be resilient when something goes wrong. Things can go wrong with isolated systems, and one of the scariest is a transformer leak. If transformer primary leaks (even a little) into the secondary, or if there is capacitive coupling, then it de-isolates the isolated system, and "pulls it up" to thousands of volts compared to ground. Now we have a problem. In that lathe motor, coffee maker or LED light, the insulation is not rated for thousands of volts.

The equipotential bond makes the neutral

To prevent the secondary ("isolated system") from floating at high voltages, we intentionally add an equipotential bond to force a relationship to earth. You might use a transformer for the equipotential bond, e.g. in 3-phase delta (non-wild-leg) to put earth in the middle. You could also use a car battery, giving the system a 12VDC bias from earth. But usually, you use the cheapest equipotential bond available: a piece of wire. You bond one of the conductors to ground, typically "center". **Because it is bonded to earth, you label it 'Neutral'.

It really doesn't matter which supply wire you bond to neutral. Ideally you want to minimize the voltage (to earth) of the hottest hot, so the best choice is in the electrical "center" ... however, 240V wild-leg delta is an example of not doing that.

So to answer your question, neutral is cold because we made it cold.

Neutral is not quiescent; it pulses at line frequency just like the hot. The effect of the equipotential bond is to dynamically change the bias of the whole transformer secondary, to keep neutral at earth potential and make hot move away from it.

Other useful reasons

A desired side-effect of the equipotential bond is that if there is a hot-earth fault, there is a high-current path via ground wire, conduit etc. back to the neutral-earth equipotential bond, and ultimately back to neutral. This completes the circuit, allows high current to flow, and causes a circuit breaker trip, which arrests the ground fault. Remember, current wants to return to source, not to ground. It doesn't care about ground, except that the equipotential bond makes it care.

For a variety of reasons, there needs to be exactly one equipotential bond. Another one would create redundant (paralleled) paths for normal neutral (return) current, and that causes all sorts of mischief.

\$\endgroup\$
  • \$\begingroup\$ +1 ad infinitum for "Remember, current wants to return to source, not to ground. It doesn't care about ground, except that the equipotential bond makes it care." \$\endgroup\$ – Shamtam Jul 22 at 12:35
19
\$\begingroup\$

When I probe my city mains with one probe in the live and one in earth (which should be 0 volt) it shows a voltage of around 250 V.

That's correct - if a little high.

But when I probe the neutral and the earth it shows no voltage.

That's good too. That line has been "neutralised" by a connection to earth at your supply transformer and, depending on your local regulations, at the supply entrance to your building.

I know that current runs in one direction for 50 times in a second. So the neutral should act like live for 50 times in a second of a second[?]. Then neutral should show some voltage with the earth, which it doesn't.

No, this thinking is not correct.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Voltage measurements during positive and negative mains peaks.

The mains voltage peaks at \$ \sqrt 2 \$ times the RMS voltage. For your 250 V RMS supply that will be about 350 V peak. If you had a fast-acting DC meter with max / min peak hold function you would be able to take the readings indicated in Figure 1. Neutral stays at 0 V and the live wire polarity alternates.

If you touch the neutral wire you won't get shocked but if you touch the live the wire you get shocked why and how?

Because the neutral has been neutralised. There is no voltage on it with respect to ground.


A note of caution

Since all wires have resistance there is a low voltage on the mains neutral and this increases with the current. If, for example, the neutral from your socket has a resistance of 0.2 Ω back to the fuseboard then a current of 10 A will cause the neutral voltage to rise to \$ V = IR = 10 \times 0.2 = 2 \ \text V \$.

schematic

simulate this circuit

Figure 2. The disconnected neutral wire turns live.

Also be aware that if the neutral conductor breaks and anything is plugged in to the circuit then the neutral wire may go live. Never assume that a mains conductor is at 0 V. Isolate properly.

\$\endgroup\$
  • \$\begingroup\$ Is figure 2 an example of an incorrectly wired circuit? You aren't normally suppose to wire connections in a series like that right? Precisely due to the risk of the neutral going live. \$\endgroup\$ – Nelson Jul 22 at 1:35
  • \$\begingroup\$ @Nelson I'd say it's an example of a faulty circuit - i.e one where the neutral wire was broken, or indeed installed incorrectly. It wont work (light wouldn't turn on), untill the neutral is properly connected. What do you mean by "in series" here? This (with the neutral connected) is how it normally is - the lamp and the switch have to be in series, how else would it work? \$\endgroup\$ – JonasCz Jul 22 at 5:18
  • \$\begingroup\$ Oh, ok, that makes sense. The open circuit caused by a broken wire, not someone deliberately putting a junction box there. That makes sense. A broken neutral will definitely shock if someone is stupid enough to grab the two broken wires of the neutral and complete the circuit. \$\endgroup\$ – Nelson Jul 22 at 5:25
  • \$\begingroup\$ @Nelson: I've added some text to Figure 2 to clarify. Since the transformer secondary is earthed one could get a shock by touching 'NEUTRAL' and earth. The body's capacitance to earth may be enough to shock even without a direct connection. \$\endgroup\$ – Transistor Jul 22 at 6:11
  • 2
    \$\begingroup\$ @Nelson It's not so much "someone is stupid enough to grab the two broken wires", but an example of what could happen to someone poking around the lamp socket, believing themselves "safe" in the knowledge that "the neutral is at zero volts" ... a break in the neutral anywhere from the lamp back to the distribution panel could expose them to the full mains potential. \$\endgroup\$ – TripeHound Jul 22 at 10:54
10
\$\begingroup\$

The neutral does not act like the live because the neutral is tied to ground at a single point somwhere.

You are imagining the neutral and live both move above and below each other about ground. Relative to each other, the neutral and live do move above and below each other. This is if you think of them in complete isolation relative to each other and only each other.

In the real world there are more absolute voltage potentials like ground or earth. If left to float, the live and neutral move together above and below some common-mode voltage. But in the real world we tie the neutral to ground for safety purposes. We don't want the live and neutral floating to whatever common mode voltage they feel like. The equipment probably wouldn't care and would still function since it only cares about what live and neutral are relative to each other, but it's dangerous to you (who is at ground potential) if their common mode voltage floats up to 1kV relative to ground.

Since the neutral is tied to ground somewhere, then the neutral becomes more or less fixed to ground. In that case, live line ends up doing all the movement above and below the neutral.

That said, you can still get shocked by the neutral if there is a load. The neutral is tied to ground SOMEWHERE but that somewhere might be quite far from where you are looking at the neutral. That means there is an impedance between where the neutral connects to ground and where you are looking at the neutral.

So if there is no load current then nothing disturbs that impedance and the neutral sits at ground. But if there is a load current, then the voltage developed across that impedance can be enough to have the neutral voltage rise above ground to shockable levels when there is a load current.

\$\endgroup\$
0
\$\begingroup\$

As regards electricity, what matters is the potential difference (this is the reason why you need to touch two points to get shocked).

Usually, the neutral and ground are at the same potential and a simultaneous contact is harmless. On the opposite, touching the live and one of ground or neutral is to be avoided.

There can be faulty situations where there is a potential difference between ground and neutral (floating ground), even though the tension between neutral and live is normal.

\$\endgroup\$
0
\$\begingroup\$

Neutral stays close to the same potential as 'earth' because they are shorted at your local power source.

The AC of live is plus or minus compared to neutral.

Regarding shock, consider that one end of 'you' is at same potential as earth (e.g. your bare feet on the ground). Then only if the other end is at a big enough potential difference away, THEN you experience a shock.

\$\endgroup\$

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