That was already established and accepted.
A true Brexiter will listen to the truth and amend accordingly.
A labour supporter remainer will resort to ridicule rather than answer a simple question.[1]
[1] Assuming they knew the answer of course.
But BS1362 13A fuses do not work like that. They will pass 20A most of the time without ever tripping and 26 A for just long enough to melt the flex on an extension reel.
As for warnings. Who reads them?
That is a significant misunderstanding. Without joking, do your apprentices have a learning disability?
not if it's still coiled in a reel, The current carrying capacity is reduiced by 75%. ie to 12A, so: use a 10A fuse in case.
Does the phrase "hoist by one's own petard" mean anything to you?
That seemed to be a response after testing lots of cheap chinesium extension leads and finding that under high loads they would fail catastrophically. The 10A fuse will still likely carry 13A indefinitely, but will limit the scope for overloading as far as a 13A fuse would permit.
For a fully coiled lead you could probably melt it even with a 5A fuse. They normally state a max load not exceeding 1kW, and that is only a bit over 4A.
Garden shredders, and the bigger mowers... Hot air paint stripper Pressure washer
Do you mean reduced by 75%, so the current capability is nearer 4A?
Short term use even at 13A would be fine, however a 10m reel will dissipate 24W for a 10A load. I can assure you a well insulated reel will eventually get above 70C internally, the limit for PVC.
Yes, I read Shakespeare too.
Thanks for not quoting the 'deliberate' mistake. :-)
But there a EU ISO standard for it.
So 25m[1] of 1.25mm^2 flex, will have about 24.20 mOhms/m round trip (i.e. L+N).
So that's about 0.61 ohms for the whole lead. A 2.2kW tool suggests about 9.6A of load, so that would give a dissipation of about 56W... not massive, but enough to get it reasonably hot. At a full 13A load that gets to 100W.
[1] I have some 50m extension leads as well, in an enclosed spool, so 200W of dissipation on that might be more of a problem if used coiled.
Doesn't say much about the training does it? I expect the poor lad was nervous of the reaction to any mistake so didn't absorb the instructions properly. Treat them like monkeys and monkeys are what you get.
So solve 2r = (pi)r^2 and you will know what wire has a diameter numerically equal to its CSA. I can't be bothered. But I think trial and error iteration to two decimal places would be quicker than remembering the formula for quadratic equations.
It has a trivial solution:
r = 2/pi
In fact I would say assembling the equation was the more difficult part.
So 1.27mm dia has a CSA of 1.27mm^2
Or for oldies, very close to 50''' (thou)
or you could just say it is 4/pi = 1.27mm
For the plug or how to wire it?
Well the little bastard has an answer for everything.
In this case his reply was "I don't know"
70 is not the limit for 70C PVC cable.
NT
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