Nor can I.But overall, if the OP would rather have/stock/use 1.5mm² instead of 1mm², I really cannot see why some people should get so agitated about it.
Kind Regards, John
Nor can I.But overall, if the OP would rather have/stock/use 1.5mm² instead of 1mm², I really cannot see why some people should get so agitated about it.
What did you expect to find?Sorry, couldn't find the answer elsewhere... thanks
I'd forgotten that we need to talk about downlights....
Does volt drop not depend on the current? How can you quote lengths without knowing how many lamps are used and the wattage?It depends on what you have in stock, and how long is the total run, in general it is easier to get 1mm² wires into ceiling roses than 1.5mm² and since the ceiling rose is also a junction box and rated at 6 amp in the main lighting circuits are only 6 amp so really no need for 1.5mm² and even with 1.5mm² line and neutral the earth is still 1mm² so very little gain with earth loop impedance, only gain is volt drop so around 39 meters with 1mm² and 44 meters with 1.5mm² so not that much in it.
Lastly, anybody lying in a bath to relax will not thank you for installing that type of lighting.
It does - and the current reducing after each light.Does volt drop not depend on the current?
I assumed (never assume) Eric was using 6A but it doesn't compute.How can you quote lengths without knowing how many lamps are used and the wattage?
Yes, I also would have imagined/suspected that he was working with the latter. But, as you go on to say ....I assumed (never assume) Eric was using 6A but it doesn't compute. Also, if lights evenly spaced, an average of 3A throughout the circuit ???
Quite. No matter what current he was assuming, eric's figures of 39m (with 1.0mm²) and 44m (with 1.5mm²) are far too close to be correct.Anyway, the difference should be at the ratio of 29:44, or 2:3, or (2x1)to(2x1.5)
Other than that I presume you meant 44 mV/A/m, I see nothing wrong with that. The implication would be that eric was assuming a current of about 4.0A (to get 39m) - could that be (for some reason) 2A at the end of the circuit and the other 4A evenly spaced throughout the rest of the circuit?Where am I going wrong? ... 6.9V (3% of 230). 1mm² is 44mA/V/m ... 6900mV ÷ 44 ÷ 3 = 52m. (52 ÷ 29 x 44 = 78) .... 6A would only be 26m.
You need to read, I quoted 6 amp.Does volt drop not depend on the current? How can you quote lengths without knowing how many lamps are used and the wattage?
a junction box and rated at 6 amp
As per exchange between EFLI and myself above, could you please explain how you arrived at the circuit lengths you reported?You need to read, I quoted 6 amp.
You appeared to be talking about the maximum cable lengths in terms of voltage drop (hence assuming that Zs was OK) and, since voltage drop is measured from the origin of the installation, Ze is surely irrelevant?Using the formula given in BS7671 to work out the correction and then going by the volt drop of 6.9 volts, so for a typical Ze of 0.35Ω ....
That would be the sensible/'proper' course, but it seems that an awful lot of people don't do it - since I imagine that it would be pretty rare (in any 'ordinary' house) for that calculation to indicate that 1.0mm² cable was not adequate, yet it quite often gets installed. As you go on to say ...You calculate the size of the cable required and use that size.
... and, as above, it seems that you are far from alone. When you say "as a default", I presume you mean that they just install 1.5mm² cable, without any consderation of whether or not 1.0mm² would have been adequate?FYI, a lot of us just fit 1.5 as a default.
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