N and PE short-circuit on an unloaded socket with a differential

Question

I searched a lot of information on the Internet, including forum, I also asked a lot of people but I did not get a clear answer. If we have a single socket on one circuit, protected by a 30mA residual current circuit breaker and I short the N and PE wire together, will the switch switch off the...

basstec · Answer

Not necessarily. Differential works on the principle of the difference between the incoming and the returning current. If there is a differential device connected to only one outlet and this outlet is not loaded, it will not disconnect because there is no potential difference. If the receiver is connected and turned on, making a short circuit between pe and somehow part of the current flows to full and the difference between what influences through L and returns through N because some of it escapes in PE.

Anonymous · Answer

I know how the differential works and I also know that if the yellow-green wire short-circuits the blue wire, the differential will turn off if it is functional.

Łukasz-O · Answer

You probably don't quite know how the differential works, or you don't understand what your colleague basstec wrote. I will repeat again, the RCD will not work after N and PE short-circuited, if the given circuit is not loaded.

pavulon · Answer

The circuit does not have to be loaded after the RCD, it is enough that there is a different load in a given area, relatively close to the RCD, and then a short circuit N to PE will trip. After all, the N conductor is a work conductor and if there is a current, there is a voltage drop, if the voltage is sufficient to flow the RCD tripping current, it will release, if not, it will not.

krzysiek7 · Answer

I join the moderator. If the circuit is not loaded, no current flows, and if no current is flowing, there is no differential current, so the differential has no reason to switch off. My friend maryjandx , I suggest reading a bit.

niutat · Answer

And I will join my colleague Pavulon, the loaded N may have a voltage in relation to PE that is so high that their short circuit after the differential will cause it to work through the equalizing current in the N path of the differential. It has happened to me several times.
Of course, such a case is possible when the differential is powered by other circuits than the one on which we are experimenting.

krzysiek7 · Answer

My friend New Year's Eve maybe make up your mind. You write that you join your friend pavulon and you actually confirm the words moderator .

greg16 · Answer

Now I will add something from myself. The differential does not need to be biased to work. In the case of an N-PE short-circuit, a current flow through the N line may also occur behind an unloaded differential. Why ? because N may be at a potential other than PE because there are voltage drops across it. Of course, it also depends on where the point of division into PE and N is

pavulon · Answer

Well done, I can see that greg16 He also knows what's going on, I go to the workshop and quickly construct a small installation and capture the experience of triggering the rcd without load at shorted PE to N on a video and I will provide a link tomorrow morning.

tomek_tkkj · Answer

Colleague greg 16 is right as long as we are talking about a TNC-S network. There is no dividing point in TNS networks (and these are mainly built now) ... The socket in question is powered from the 1st phase, and this circuit is supplied from the 3-phase circuit, i.e. both have a common N conductor. is equal to 0 V, but has a certain potential (i.e. the N conductor is connected to a certain potential). If we close N and PE (with a potential equal to 0), we close the circuit and the flow of current (and only 30mA is enough to operate). If we had a symmetrically loaded (and this is almost never the case) 3-phase network, the differential would not work.

blaszczok · Answer

Hello. I just wanted to share some practical comments. The differential was without any load, the N wire was 5cm long and the PEN maybe 10cm and it was also triggered. So that you can see from this example the typical operation of the potential difference. Best regards, Michał.

greg16 · Answer

Of course I know I'm not saying that this will always happen, but it can. Many factors will influence it. One of them is the fact that the loaded N conductor may have a higher potential than the PE conductor because there is a voltage drop on it, but it is and it is enough for 30mA to flow. If someone does not believe it does not have to. And who has any knowledge of electrical engineering should know.

Łukasz-O · Answer

Naturally, it would really depend on the type of network and information where the distribution is, RCD and socket, etc. - my mistake. I considered the split circuit directly before the RCD.

Anonymous · Answer

The type of network does not matter here. not even a lot of choice. The RCD may operate, but does not have to operate. Buddy Greg 16! You hit 10, but you translate a bit vague, mixing up the potential unnecessarily. Because this is all about reducing the tension. Doubts arise because many electricians do not work with diagrams. Almost nobody. Even sometimes those in high places cannot draw a schematic drawing by drawing lines as lines. And not as resistors. In addition, their diagrams do not reflect reality. Buddy Pavulon - the RCD will work as it should, but your table layout must map the network with the correct resistances. If the layout on the table does not represent the network, you may draw wrong conclusions from the experimental layout. I have attached diagrams, where the layout a corresponds to what is in the topic. You can write if the commentary to these diagrams is correct in your opinion. Bronek

aklim · Answer

Just make an experience. Short PN and N with a milliammeter, the current through the milliammeter is the difference of the current through the differential.

niutat · Answer

I meant the unloaded circuit in which we are doing the experiment, not the entire installation.
I had to deal with such cases and the differential worked even though the circuit was turned off by a fuse.

Anonymous · Answer

We are talking about this case in the subject. Bronek

Wirnick · Answer

For the correct operation of the RCD, it must be turned on in accordance with the diagram on its housing or in the installation manual. The task is to measure the residual current in active conductors and switch off the current in active conductors when the nominal value of the residual current is exceeded. Switching off (tripping) is caused by errors or damage to the circuits. Installation errors were presented. In addition to the prophylactic differential current paths (PE conductor), there may be random paths (flame ion flow, foundation, ...).

N and PE short-circuit on an unloaded socket with a differential

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