Thats a very good explanation. I could not have been bothered to type all that out .
Cheers M8Thats a very good explanation. I could not have been bothered to type all that out .
No prob M8!Thanks Special Location. That would've taken me a couple of days to type! That was exactly what I wanted to know - the "why" not the "how". Cheers!
Many a true word spoken in jest Slip'sI do remember the phrase 'that is outside the scope of this course' being used often when i sat the 2381. It must be either the same guy or the get out clause you they taught at lecturers college ; \
Morning chaps (& chapesses!).Electric shock currents...
1 - 2ma barely perceptible
5 - 10ma throw off, painful
10 - 15ma muscular contraction can't let go
20 - 30ma impaired breathing
50ma & above ventricular fibrillation & death!!!
Correct me if I'm wrong - but the max current on the cable to the garage due to a house fault on a TT system with a rod Ze of 100 ohms is 2.4A.You are permitted in the wiring regs to join the TT rod & MET together.. BUT if you do.. "All of the CPC's must be capable of carrying the max PFC for any part of the installation" because for example.. a fault in the shower on the house.. could make a path down to the earth rod in the garage!
Hi Tesla[The resistors would not be in parallel though would they!
It would be 1 resistor ie the the lighting cpc having the fault current flowing down it. Then you have to consider the said 6a flowing through Ze resistance eg 0.1a which gives V=6ax0.1; v=0.6V. So the voltage across the earth rod to earth will not be 250v at all, no-where near, but just 0.6v in this case. Now the calculation for the earth rod will be I=v/R ; Amps=0.6/100 ; =0.6ma
YOUR ALL MISSING THE POINT. THE PROBLEM IS THAT IT IS A PME SYSTEM AND THE PROBLEMS OF BROKEN NEUTRALS.
Hi Teslathe 6a is the MCB rating not the fault current.
In a short circuit fault at a light fitting the CPC & phase conductors are in direct contact, (no load in between them)
so the full supply voltage is now on the CPC.
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resistance eg 0.1a which gives V=6ax0.1; v=0.6V.
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Resistance is ohms not 'a' V=IxR so I=V/R
The voltage is fixed by the supply company so taking your figures
230v/0.1ohm = 2300amp (2.3kA) which is your PFC (which in this case is the bit that is going to blow the fuse!).
If earth and phase are combined at the instant of the fault 230v will be on the CPC. (cuz its touching the 230v live!)
If the CPC has two paths (no matter what resistance) they are in parallel!
(a TNS & a ROD) and the 230v will be across both of those earth paths.
i.e. go and connect as many resistors, (various values), as you like across a 12v battery.
The potential of the battery will be across all of the resistors i.e. 12v.
but the current flowing through them is different...
230v across 101ohm = 2.3V droped across CPC and 227.7 droped across 100ohm rod!
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Nice polite reply their SP, but your still completely wrong.
230v will not actually by magic appear at the cpc at the fault, i know it may seem to, but it doesn't....We come back to ohms law again, V=ixr
The cpc has a resistance itself which is added to the Ze eg 0.2r and Ze 0.1r is a resistance of 0.3r ( i put r in place of the ohms symbol)
Now when the current begins to flow in the cpc a voltage appears across this resistance growing as quick as the current grows, which is in fragments of a second and a graph needs to be observed to show this.
But what also happens is current is also being drawn along the live conductor and again using ohms law and a resistance to the fault of 0.2r( if the conductors are of the same csa) both conductors live and cpc and Ze will have a voltage across them also. If we forget about Ze for a minute the max voltage at the fault on the cpc will eventually become half of 230v, which is why lights go dim when a fault occurs, but a fuse or mcb will have blown long
before that.
So no 230v does not suddenly appear at the cpc
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My next point is how on earth does 230v appear at the earth rod from a fault on a lighting cct in the house.
Using your battery senario, it would be like having a number of resistors side by side with one end connected to the + and the othe end connected to the -. But these are not the cpc resistances but would represent the LOAD resistances in parralel.
Each cpc would be represented by a seperate resistor for each load joined together at the - end but o/c near to + end next to each load.
Being that none of the cpcs are joined together at the + end even if one does in your scenario it will not effect the other cpcs ( only the Ze can create a rise in voltage when a large current is drawn)
So to sum up, the only voltage that will appear across the earth rod during this fault, will be V=IxR, The current drawn from the fault (B type =5x rating and C type = 7 Times rating) X Ze
6x5=30a X 0.1r= 3v
Therefore current going down the earth rod will be I=V/r, 3v/100r =30mA
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Answer to the other bit their SP, Im a sparks but have also got a HND in electronics......looking forward to see your reply
Ooh ooh ooh, if Im not too late, can I have one please?I believe that was the reply OP was looking for Albert. Try not to get worked up Albert - Watch your blood pressure. OP was unsure - so he asked - That's what we are here for Sir... To help each other out. That where everyones experience comes in.
I am on the C&G course - although we had used swa, we haven't been told of the earthing requirements on long runs - So I can sympethise with the OP. I have picked up a heck of a lot of knowledge through buying books and studying, going to college, Prcticing in my workshop etc and more so from Guys like YOU, whom I respect.
I don't like forums that give carp replies, hence my reason for starting this one, Sir.
Let's keep it insult free (ish) & word it better. And that goes for us all - not just you Sir.
Now...Who's for a free Promo Sticker. :^O
I am off for a minute to order them.
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