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I want to put 6 300w LED flood lights around a tennis court positioned centre and corners. The cable run is approx. 80 meters. I have a 1.5mm rubber external cable in free air mounted along the top of the mesh fence. 1800w at 240v should give me 7.5a which is within the cable rating but how do i account for voltage losses raising the current draw?
 
When you say rubber cable, do you mean H07RN-F??
 
OK, so normal logic would suggest that if you joined the dots A-B-C-D-E-F you end up with a radial circuit and whatever the calculated volt drop is (which I’ll happily look up for you after my bacon…) BUT as you’re doing a perimeter then you may as well rejoin A to F and make it a ring, in which case your losses are far less.
 
OK, so normal logic would suggest that if you joined the dots A-B-C-D-E-F you end up with a radial circuit and whatever the calculated volt drop is (which I’ll happily look up for you after my bacon…) BUT as you’re doing a perimeter then you may as well rejoin A to F and make it a ring, in which case your losses are far less.
I intend to join up A with F to close the loop. The lights are 300W 1.36A. Full loop is approx. 80m. Enjoy your bacon butty!
 
Actually, having just done the maths, you’ll HAVE to make it a ring anyway to keep within allowable tolerances.
 
I intend to join up A with F to close the loop. The lights are 300W 1.36A. Full loop is approx. 80m. Enjoy your bacon butty!

Actually, having just done the maths, you’ll HAVE to make it a ring anyway to keep within allowable tolerances.
OK will do - what will be the expected current load so I can select the size of fuse (I have a local fuse box as well as the breaker)?
 
Double check the manufacturers instructions regarding the type of protective device required, you may require a C type to allow for in rush currents of the LED driver. This will have a knock on effect regarding the impedance of the circuit.
 
One thing to be aware of is the simultaneous switch-on surge of 1800W at the same time which can run to several times the rated current. You should check the manufacturer's recommendation to see what they suggest, but you might need something like a 16A C-curve MCB (with RCD protection, of course, as it is outdoors).

Otherwise you might look at having two separately switched circuits for 3 lights at a time, etc.

EDIT: Just beaten to it by @Strima !
 
If we keep the maths simple - 1800W is basically 10A. 1.5mm H07 has a voltage drop of 32mV per Amp per Metre. Therefore, 0.032 x 10 x 80 = a voltage drop of 25.6v at the final light, which exceeds the minimum tolerance of 3%. Make a ring and that changes to 6.4v - which is just under the magic figure of 6.9v

As for sizing of breaker, the max allowance of 1.5mm H07 is only 17A anyway. Conductor resistance is 0.0133Ω /m therefore for a ring you'll have an R1R2 value of something in the region of 0.53Ω - a C16 RCBO will give you around 0.6 ohms worth 'to play with' on final figures..... so basically hope the origin of that circuit isn't too far from the origin of supply to start with!!
 

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