Can we connect the 2 phases lines ( 110 V each ) to RCD breaker? | on ElectriciansForums

Discuss Can we connect the 2 phases lines ( 110 V each ) to RCD breaker? in the UK Electrical Forum area at ElectriciansForums.net

If I remember correctly, Saudi is based on the US style system of using 110v for low load circuits and then using +ph-ph to derive 208v for more energy hungry systems. As both phases would be 'stolen from' in the event of earth leakage then I would assume you'd need earth monitoring via the cpc rather than ph-ph (ph/N in the UK). So yes, you can connect both phases via a 2p RCD device feeding a 208v circuit, just don't expect it to work!
 
If the load is Simpley 2 phase then all will be good, however, as @Rockingit has suggested if there is a center tap N involved then any non balanced load will trip the breaker.
most 3 phase rcd's would be able to cope, just leave the L3 disconnected and ensure both phases and N pass through it.
 
If the load is Simpley 2 phase then all will be good, however, as @Rockingit has suggested if there is a center tap N involved then any non balanced load will trip the breaker.
most 3 phase rcd's would be able to cope, just leave the L3 disconnected and ensure both phases and N pass through it.
I’ve always believed that a 3ph RCD is able to work because by monitoring each of the phases it’s able to basically calculate by electronics what the neutral current should be whether it’s actually present or not and sense an imbalance based on the maths of that rather than as in a 1ph RCD just looking for hysteresis caused by imbalance. (Happy to be schooled on that!).

The challenge in the OP’s scenario is that it’s two phases 180deg apart so they are always going to be in balance because that’s how a pd of 208v is derived. If you get a short to fault, the voltage in that short will be the same sum-to-zero value of the two phases - so there’s only ever mathematical zero passing through the RCD which means that Ohms law basically fails and consequently there is (mathematically, to a single ph device measuring hysteresis) also zero fault current.
 
Meanwhile in reality if you have a bolted fault to E (with 3P or 2P+N supply) that theoretically gives you zero imbalance due to line voltages summing to zero, in practice you ain't going to see such exquisite balance that fault has not got 30mA or whatever spare to trip the RCD.

Assuming such a balanced fault is not got the necessary kA-size experience to take out the OCPD faster than the RCD could trip anyway!
 
Meanwhile in reality if you have a bolted fault to E (with 3P or 2P+N supply) that theoretically gives you zero imbalance due to line voltages summing to zero, in practice you ain't going to see such exquisite balance that fault has not got 30mA or whatever spare to trip the RCD.

Assuming such a balanced fault is not got the necessary kA-size experience to take out the OCPD faster than the RCD could trip anyway!
I think the issue here is that we’re not talking about 3P or 2P+N. We’re talking about effectively 1ph on two poles. There is no neutral, nor is there a conventional 3ph imbalance.
 
I think the issue here is that we’re not talking about 3P or 2P+N. We’re talking about effectively 1ph on two poles. There is no neutral, nor is there a conventional 3ph imbalance.
But those two poles would, ideally, sum to zero the same way 3P bolted fault would, or indeed 2P+N would.

My point is you will never see such perfect balance on any real-world fault to earth that would not have a fraction of an amp error to trip the RCD, should it be needed for protection (i.e. OCPD not able to act / fast enough).
 

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