The connections to the socket wouldn't be in insulation. I'd guess that if you removed one strand of a seven strand conductor and then gradually wound the current up the switch or socket would get hot before the remaining six strands.
Bill
It helps a lot to use a pair of very small long nosed pliers to hold the last cable very near the end as you push it in.
Bill
maybe like the plumbers trench, you can never fit all the dirt back
AIUI, the suggestion is to cut off 1 strand of the last ~cm to go into the terminal - shouldn't need too much derating for that. The chap who managed to wire the downstairs ring in my last GF's house in '2.5' (equiv.) in stranded AND got it into the bell transformer must have had a shrink-ray.
I recall doing that myself on one occasion. It was the only real option and didn't give me any cause for concern with regards to the cable still vastly exceeding the maximum load for the circuit/connection.
Pete@
The effect of the insulation on the circuit as a whole would be that its cable reaches its maximum nominal conductor temperature (70 deg for PVC insulated T&E) at a lower current than it would if clipped direct to a surface or buried in plaster etc.
So your starting point would be a circuit with conductor temperatures already at their maximum. The reduction in CSA at the connection would then create a localised hot spot at that connection. If the situation is maintained, the likely outcome would be damage to the cable insulation close to the socket, and possibly damage to the socket itself.
(its mostly a moot point, since most general purpose circuits in insulation are likely to be more modern ones wired in single core cable anyway, also the nature of diversity means its unlikely that a general purpose circuit will run at full capacity for extended periods of time)
Its more a case of if there is de-rating required on the rest of the circuit for other reasons - then you use up the margin that would let you get away with what would look like (electrically) a slightly poorer connection at the socket.
Given you don't have to derate short lengths of cable through insulated areas, I would have thought a few mm of cable being a strand short could be treated the same way.
Where you would assume that heat conduction along the cable would mitigate a small local rise in temperature.
Just crimp them and extend a suitable wire to the terminal. Why does it need to be bodged at all? This a DIY group not a bodge it group.
There's an overlap. It's a grey area.
Bill
I should think that the contact area of the connector and/or the blades would be less than 6/7ths of the csa of the cable.
Bill
Yup, that is a better solution.
In article , snipped-for-privacy@mdfs.net writes
Strands of stranded cable will sometimes break off, either because of over-flexing during installation or because one or more strands were nicked when stripping the insulation.
Big difference, though, between that and deliberately cutting off strands to make several cables fit in a terminal.
Total CSA yes, but its the CSA of the area of the end of the conductor that (in conjunction with other factors) dictates its current carrying capacity.
In the grand scheme of things, its not a big deal - but it is poor practice to trim strands IMHO.
On a socket circuit you are unlikely to suffer ill effects in reality. However I would be very wary about doing it on a high current circuit like a shower feed with 10mm^2 T&E (where even well made connections can dissipate a few watts at 45A
If they dissipate a few watts they aren't well made. They should be in the milliohm range and dissipate in the milliwatt range.
I would describe 50 milli-ohms as "in the milli-ohm range", and that will dissapate (just less than) 2.5W which I would describe as "a few watts".
Never underestimate the effect of the square law when dealing with largish amperages. I've just tried 45 x 45 x 0.005 for a 5 milli ohm joint and get a value of 10.125W (50 milli ohms gives 101.25W!).
Although you were defending John's statement, you've somehow cocked up the calculation to the detriment of your argument. :-(
Are we sure? Lessee now, dissipation will be I x I x R. If I = 45A, and R = 50 / 1000, that gives us:
W = (45 x 45 x 50) / 1000 = 101.25
Seems quite high. I expect you want 1 milli-ohm or less.
6mm^2 cable has 3.08 mOhm resistance per metre, so even a 10m circuit can dissipate over 60W in just the cable.
What's the resistance at a connection point though?
What always struck me about houses in the US was that due to using cheese-paring 110 volts, they were quadrupling the heat generation at any given socket or connection.
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