RCD tripping in Consumer Unit

Hence, for, say, 50m of cable that would be 5,000 pF (5 nF) - so, about 640 kΩ at 50 Hz, resulting in an L-E 'leakage' current of about 0.36mA at 230V. Since the N-E potential difference will usually be fairly negligible, the N-E leakage current (which will reduce the L-N imbalance) is also likely to be fairly negligible but, in any event, the maximum L-N imbalance would be about 0.36 mA for 50m of T+E.
If I understand correctly that is not correct; the N-E leakage will add to the total leakage.
 
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If I understand correctly that is not correct; the N-E leakage will add to the total leakage.
Yes, you're right - my apologies. The L-E leakage (through cable capacitance) will add to the L current (as seen by the RCD), whereas the N-E leakage (through cable capacitance) will subtract from the N current (as seen by RCD), so the L-N imbalance seen by the RCD will be the sum of those two leakages.

However, that makes very little difference to the bottom line of what I said - since, as I said (and as Mr Ohm would confirm), the N-E leakage (current) will be 'fairly negligible', since the N-E potential difference will also usually be 'fairly negligible' - so the L-N imbalance seen by the RCD will primarily be that due to the (much larger) L-E leakage.

Kind Regards, John
 
I obviously can't speak for others but, probably because all of my meters (including the 'fairly expensive' ones) are pretty (in some cases very!) old, I don't have any which can measure AC currents in the mA range.

Even my older ones (I am pretty sure) will cover that range.
 
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