.... Checked garage wires E-N still 1500Ω, E-N & L-N still O/C. .... At that point I've proved an unexpected low resistance E-N and L is clear, (>200MΩ) on the garage circuit.
I was in the house CU where tripping proplems were occuring and I proved a fault on the only garage circuit.
It also depends on what one considers to be low resistance compared to ∞, the IR test thinks it is.
Assuming the N-E link is immediately before the RCD I'll place a figure of 1mΩ, assuming the faulty circuit is my garage example of say 25m of 2.5mm² so in the order of ½Ω it doesn't take too much calculation to establish that in order to get 30mA in the fault will rewuire 500x30mA or 15A so yes it will trip under a modest current
Yes, I should have acknowledged that a trip could occur 'with a modest load current' in a TN-C-S installation if the N-E fault was 'relatively close' to the RCD.
However, perhaps we can agree that an N-E fault of a given impedance at a point on the circuit which has a given R1+R2 is much less likely in a TN-C-S installation than in a non-TN-C-S one ?
Yes, if the only load in the installation is on the circuit with the N-E fault, then essentially half of the neutral current will NOT go through the RCD - so, as you say,a load current of 60 mA should be enough to result in a trip
True, but I personally would not use the phrase "very low resistance" to refer to a situation in which all I knew was that the resistance was lower than an IR test could quantify.
Yes, I should have acknowledged that a trip could occur 'with a modest load current' in a TN-C-S installation if the N-E fault was 'relatively close' to the RCD.
However, perhaps we can agree that an N-E fault of a given impedance at a point on the circuit which has a given R1+R2 is much less likely in a TN-C-S installation than in a non-TN-C-S one ?
Yes, if the only load in the installation is on the circuit with the N-E fault, then essentially half of the neutral current will NOT go through the RCD - so, as you say,a load current of 60 mA should be enough to result in a trip
True, but I personally would not use the phrase "very low resistance" to refer to a situation in which all I knew was that the resistance was lower than an IR test could quantify.
Yes, I should have acknowledged that a trip could occur 'with a modest load current' in a TN-C-S installation if the N-E fault was 'relatively close' to the RCD.
However, perhaps we can agree that an N-E fault of a given impedance at a point on the circuit which has a given R1+R2 is much less likely in a TN-C-S installation than in a non-TN-C-S one ?
Yes, if the only load in the installation is on the circuit with the N-E fault, then essentially half of the neutral current will NOT go through the RCD - so, as you say,a load current of 60 mA should be enough to result in a trip
True, but I personally would not use the phrase "very low resistance" to refer to a situation in which all I knew was that the resistance was lower than an IR test could quantify.
However, perhaps we can agree that an N-E fault of a given impedance at a point on the circuit which has a given R1+R2 is much less likely in a TN-C-S installation than in a non-TN-C-S one ?
I think you may be getting yourself confused here? 60 mA through a person may well be lethal, but we are talking about the current through (the N side of) a load protected by the same RCD that would cause an N-E fault on that circuit to result in a trip of the RCD. For that to happen would presumably require someone to be in contact with both L and N (with a body resistance of about 3.8 kΩ) and for there simultaneously to be a N-E fault - which sounds like an almost unbelievable combination of circumstances, doesn't it?
The only thing I was trying to 'get at' was that, had it not been for subsequent events/revelations, you might have wasted a fair bit of time searching for the cause of a N-E fault in the wring garage!
Yes, you have subsequently clarified - but at the time when I had only seen your original reference to "a very low resistance", that choice of words resulted my in interpreting them (I would say 'reasonably'!) as meaning something which was not necessarily the case!
I think you may be getting yourself confused here? 60 mA through a person may well be lethal, but we are talking about the current through (the N side of) a load protected by the same RCD that would cause an N-E fault on that circuit to result in a trip of the RCD. For that to happen would presumably require someone to be in contact with both L and N (with a body resistance of about 3.8 kΩ) and for there simultaneously to be a N-E fault - which sounds like an almost unbelievable combination of circumstances, doesn't it?
Not getting confused just juggling figures around. However it could be a case of a fault not tripping the RCD and someone trying to faultfind with it live.
The only thing I was trying to 'get at' was that, had it not been for subsequent events/revelations, you might have wasted a fair bit of time searching for the cause of a N-E fault in the wring garage!
It is a possibility but why would I look for a fault in a space which has been isolated but the power is on?
I stand by my statement: "I proved the fault onto the garage circuit"
Yes, you have subsequently clarified - but at the time when I had only seen your original reference to "a very low resistance", that choice of words resulted my in interpreting them (I would say 'reasonably'!) as meaning something which was not necessarily the case!
I must confess that I hadn't noticed that the last line has changed - but, as you can see, what happened when you tried to 'edit' the original was that a second copy got posted, with just that last line changed!
Not getting confused just juggling figures around. However it could be a case of a fault not tripping the RCD and someone trying to faultfind with it live.
Fault-finding live things is obviously always going to be hazardous, but I still don't understand how the 60 ma (N current from loads) you calculated has got anything to do this. Am I missing something?
You presumably wouldn't - but is it not quite possible that you would have started your 'looking for a fault' (at least, say, to the extent of unscrewing faceplates etc.) without yet realising that the power was still 'on' in the garage you were looking in?
I don't think that there is any point in our going on about this. I can but repeat that your mention of "a very low resistance" resulted in my thinking something rather different from what the situation actually was!
I must confess that I hadn't noticed that the last line has changed - but, as you can see, what happened when you tried to 'edit' the original was that a second copy got posted, with just that last line changed!
Fault-finding live things is obviously always going to be hazardous, but I still don't understand how the 60 ma (N current from loads) you calculated has got anything to do this. Am I missing something?
You presumably wouldn't - but is it not quite possible that you would have started your 'looking for a fault' (at least, say, to the extent of unscrewing faceplates etc.) without yet realising that the power was still 'on' in the garage you were looking in?
I don't think that there is any point in our going on about this. I can but repeat that your mention of "a very low resistance" resulted in my thinking something rather different from what the situation actually was!
I suppose that, in some senses (particularly with the "at least...." caveat I added), I inevitably am - but in those sort of circumstances, it probably is inevitable ....
... in the case of sockets circuits, you might not be equipped to safely 'prove/test for dead' without first unscrewing a faceplate. In the case of circuits supplying loads which are hard-wired into FCUs, switches/isolators or 'outlet plates', you would have to unscrew something to expose terminals before you could "test for dead". Even with a lighting circuit, you would have to either unscrew the faceplate of a switch or remove a 'light bulb' (in both cases from a possibly live circuit) before you could "test for dead".
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