230V phase voltage and 400V phase-to-phase voltage - why such a condition

Please tell me how the electricity works "in the socket". If what it is I know what is alternating current. I am interested in the very operation of this. Perhaps it will be best to say it on an example I propose a welder.

I know that to connect the welder to 230V, we connect the ground and 0 (omitting the protective conductor), and to connect it to 400V, 2 phases are connected. And now I wonder what is the role of 0 in this second connection?

Maybe you know a link where the operation of single-phase and three-phase current is precisely explained?

In the case of the power you write about, there is no neutral wire ..
The welder is powered by phase-to-phase voltage (400 V).
Pathologically ...
Three-phase power supply, where there are three phases, conventionally referred to as R, S, T, the additional power supply cable is the so-called 0 (zero) ..
The voltages between each of the R, S, T phases and zero are 240 V.
The voltage between the phases RS, ST, RT is always 400 V ...

And why is it that 400V voltage is generated between RS, ST, RT, and there is no voltage between SS. However, I would be interested in a more extensive answer explaining the exact operation of the current "in the socket".

How do you want the voltage between S -S phase ??? There is only one S phase ...
And in more detail ... Wait, I'll try to scan the preview material ...

I know that it is not possible, but to understand this, I must first find out exactly what this S is and how it works
Well, I will be very grateful if you upload this material

The material is quite extensive, and I do not know what knowledge you have on the basics of physics and electrical engineering ...

Below you have a link to the page on three-phase current ...

http://wazniak.mimuw.edu.pl/index.php?title=PEE_Modu%C5%82_6

If that doesn't get you closer, then you have to start with the basics of physics ...

this material is understandable to me up to a point. He explained a few things to me :)

I still have a question, what is the difference between zero and phase?
And why when we connect three phases to some device, the voltage is the same as if we connected two phases?

The difference is fundamental ...
The voltage measured on the neutral conductor in relation to the protective conductor (e.g. ground) is zero Volt ....
The easiest way ... is the wire that comes out of the power plant (the nearest power transformer with no voltage .......

As for the second part of your question, I can see that you didn't understand the idea of three-phase current ... or you don't understand the concept of voltage ...
The voltage is the potential difference always measured between ONLY TWO points ....

yes I know that the voltage is the potential difference between two points, I just mean what happens when there are three phases, what does it change? and how is the device connected then?
As far as I know, in the new power sockets there are now 5 wires (i.e. now it is also zero) why has it been changed and what does it give us?

And I have a question, what is mass?

Added after 3 [minutes]:

and if there is no voltage on the 0 wire, how does it happen that when we connect the device, the current flows from one side to another and then from the other side (from the other wire)

Perhaps this text will explain some things to you at least a little

And how is it that there is a potential difference of 400 between each of the two phases? After all, if we assume that the potential S = 800V, the potential R = 400V, then we will not select such potential T so that the difference of ST and RT potentials is equal to 400V. So I am to say that the potential changes over time?

We have a three-phase system in the network. The generator produces three voltages of 240V (rms values) shifted by 120 ° in relation to each other
http://osilek.mimuw.edu.pl/index.php?title=PEE_Modu%C5%82_6
And it is because of this phase shift that the stresses do not add up algebraically but geometrically

You can also look here to read the introduction to this exercise, it should also make things a little clearer.
www.elektrotechnika.po.opole.pl/cwiczenia/uklady_3f/uklady_3f.html

You have a voltage waveform on one phase with respect to the neutral conductor N:


$$U(t)=230\cdot\sqrt{2}\cdot sin(2*\Pi\cdot 50Hz\cdot t)$$

230V is the RMS voltage (i.e. the equivalent value of the DC voltage which, when applied to the same resistance, will cause the release of the same amount of heat as the course of the alternation)

In an electrical socket, the RMS voltage can vary from -10% to + 5% (and will be up to + 10%, if not already) of the nominal value of 230V (i.e. from 207V to 241.5 or 253V).
These fluctuations result from various voltage drops on the line between the transformer substation and the consumer, which result from the current drawn from the line and its resistance (and this depends, among other things, on the length of the line). So those at the substation can have a voltage of 230V + 5% (or + 10%) and those at the end of a long line 230V (-10%) and these values may fluctuate during the day depending on the momentary load on the line by other consumers.


There is a shift of 120 degrees between the waveforms on the phase conductors, which causes the voltage between the phases to be:

$$Umff=Um\cdot sin(\omega\cdot t)-Um\cdot(\omega\cdot t+120)=Um\cdot (sin(\omega\cdot t)-(\omega\cdot t-120))$$

$$\omega = 2\cdot\Pi\cdot 50Hz$$

$$Um$$ - voltage amplitude

from the formula for the difference of sines we have:

$$sin \alpha - sin \beta=2\cdot cos \frac{\alpha+\beta}{2}\cdot sin \frac{\alpha -\beta}{2}$$

$$\alpha=\omega\cdot t$$

$$\beta=\omega\cdot t -120$$

$$Umff=Um\cdot (sin(\omega\cdot t)-sin(\omega\cdot t-120))=Um\cdot 2 \cdot cos (\frac{2 \cdot\omega t -120}{2})\cdot sin(\frac{\omega t - \omega t +120}{2})=Um\cdot 2 \cdot cos (\omega t -60)\cdot sin(60)=Um\cdot\sqrt{3}\cdot cos(\omega t-60)=Usk\cdot\sqrt{2}\cdot\sqrt{3}\cdot cos(\omega t-60)$$

So the rms value of the phase-to-phase voltage is the product of the rms value of the phase voltage and the root of three and is:

$$Uff=230V*\sqrt{3}=398,37V$$ of course, this value may vary between -10% and + 5% (+ 10%) of this value.