Energy efficient light bulbs

$${ 720W\over 230V}=3.1 Amps$$ (note1). Now using lower Watts per unit, 20 lights of 12Watts is 240W and far lower than the equations above and also 240W/230V=1Amp.

Why might you be able to have more lights because of low energy efficiency?

You will not get sold this option as easily as you could be since a light socket will take any bulb meaning you could fit 100W light bulbs. When for instance the electrician has left you will fit 4000W/230V in your light circuit commanding 17.4Amps if you decided on 40 points. Most of the light bulbs are probably not rated to operate, nor is the cable size or the fuse with such a high power rating. You would not do this as you want lots of low energy lights but it is not written in a big sign if you sell the house or forget. Having calculated 40 fittings x 12W/230V =2 Amps just 10% of these with higher power ratings of 100W would increase the current to 832 Watts (3.6Amps).

Now If you use 3x 100W lights and 17x 12W low energy bulbs this makes 504W. Electricians can use what is called diversity (note a) where 66.66% of the figure of full load can be used to calculate how many light fitting you can have. This is based on 100W per fitting despite a lot of certainlty a person is fully aware they can run an energy saving light circuit. Purchases can be made from most supermarkets, DIY and hardware stores. It is very easy.

By diversity of 66.66% on 20x 100 Watt lights you get 13x 100 Watt lights (2000Watt x66.66% = 1333W). 13 lights on at 1300W require a fuse rated above 5.7 Amps

$$ {1300Watt \over 230Volts}= 5.7 Amps $$ This is as high as you can get before a 6 Amp fuse for your circuit.

How can you have more low energy lights without a new lighting circuit

No you cannot just fit a higher fuse for more lights No

It is a matter of safety. Diversity allows you to establish that a circuit will only run on 66.66% of it's full quota but if you did run 2000W and 8.7amps the fuse rating now is above capacity and will have blown. This is also too high for the cable if it runs through more than 500mm of insulation (note 5). It is recommended to half the rating. If you fitted a higher fuse then you would need to have larger cable as a matter of safety to run the higher current without overheating. If diversity was used to run all 20 lamps at 2000W in these conditions a new fuse now needs a rating of 8.7 Amps (2x8.7=17.4Amps) and the cable is derated to need to cope with 17.4 Amps. It would be dangerous to have a larger fuse than for 1300W (5.4Amp) without new calculations for cable size so that you could use your 20 high power lamps.

Can you have more low energy lights with 2.5mm cable

The high current described 20x100 Watt units looking at 2.5mm cable shows if 2.5mm cable was used to run 8.7 Amps lighting it has a maximum length of 44m(note 2). Applying diversity it would increase to 67metres but I found it a low figure for cable not used in lighting so show the non diverisfied calculation to emphasise the gain is not significant.

$$ {18Mv/m x 8.7 Amps\over 1000 x L} = 6.9 Volts$$ (permissible-voltage-drop). L = 44metres.

It is not difficult to envisage requirement of 44m for 20 lights. The length would be even shorter for 1.5mm cable or 1mm. I would not do the maths for 1.5 and 1mm cable because if these cables run through insulation their rating changes and it is most likely the circuit is not allowed at above 6 Amps. I will do it here however so you do not get the idea I want to run 2.5mm cable. What I am demonstrating is that the current for the calculations is high unless using low energy lighting. I have had to use diverstiy ( note 5) here for the smaller cables to run the lighting circuit. Otherwise I would be using 8.7 Amps and the lengths would respectively fall to 27metres for 1.5mm and 18metres for 1mm

$$ {29Mv/m x 5.7 Amps(diversity)\over 1000 x L}$$ =6.9V (permissible). L=41m for 1.5mm- cable.

$$ {44Mv/m x 5.7 Amps(diversity)\over 1000 x L}$$ =6.9V (permissible). L=27m for 1mm- cable

With a high current then 2.5mm cable seems a solution. But why do maths for a high current when the circuit you are designing is low current. I see only one rule for this which is the 100Watt per fitting one (note 5). It is not dangerous when I considered a higher fuse should of course be fitted to a suitable circuit to increase capacity, but it is not necessary to want a higher fuse. The design is for a lower current which will not need diversity to stay nearly 3 times less than the fuse rating and not need a larger cable at all on that basis

Low energy circuit calculations with 100W units and diversity

No You would not want to change the fuse. It can probably only go up in size beyond the current design of the circuit that you wish to alter or add to. Sticking to the plan of more light points the main thing is that by using low energy bulbs you would not blow the 6 Amp or even a 3 Amp fuse using all 20 lights. (40x12W=2A & 20x12W=1A) Not getting ahead of reality (100W per outlet) gives 1 Amp for 20x12Watt or with some allowance for a few high energy bulbs 2.2Amp for 17x12W with 3x100W (the example of the first paragraph). Just a few high power bulbs doubles the circuit load and should not be ruled out of the design before applying the calculation. It falls way inside the permissible voltage drop, way beneath the fuse rating and nearly 3 times inside the need for applying diversity. I think this is important because the tables used (note 4) for the 100W per unit calculations must use lower current than 5.9Amps too. In fact they suggest the same, 500W=2.1A (see next paragraphs) To get maximums of 68metres for 1mm cable and 106metres for 1.5mm cable a figure of 2.1 Amps works. (see below)

$$ {29Mv/m x2.1A\over 1000 xL}$$=6.9(permissible-voltage-drop).L=113metres for 1.5mm- cable

$$ {44Mv/m x 2.1A\over 1000x L}$$ =6.9 (permissible). L=74metres for 1mm- cable

So why do we calculate lighting at 100Watt per outlet and how does this affect the circuit rules

If low watt lights begin to have different fittings then this can change (my reasoning). For now you are guided to use 100W per fitting design calculation and a low number of outlets as a result.

$$ {20 fittings x100Watts\over 230V}$$ = 8.7 Amp

$$ {29Mv/m xAmps x106metres\over 1000}$$=6.9V Load- is- 2.2Amps

It seems we are hanging on to 100W light fittings but not doing it for any reason that can be put in a calculation. This is a load of 506Watts with 106Metres run. There is a lot of scope for distance, alot of scope for power but not many lights as they have to calculated at 100W each. Now to return to the original premise. We have now a good cable length for whatever we want to do, we must protect it at 6Amp as stated which since we are aiming for 2.2Amp is good diversity and I had 504W from 3x 100Watt and 17x 12Watt making a stable circuit. We have 6 Amps for the protective device, can call that 230VX6Amps = 1380Watts but not add it to the distance because that is a distributed (note 4) result, but in the sums 1380 Watts is 13 lighting points from diversity 66.66% of 20 outlets. This is achieved with low energy taking the same number of allowed outlets but designing just over 1⁄3 (one third) of the current

So I hope you can see I do the maths myself and think about the job, best to stay simple aswell since before being qualified you can trip up on anything someone suggests you gamble upon. With lights I could make a 2 or 3 way switch arrangement for any points and organise the cable run and check for correct connections. I would like to buy testing equipment which is needed for every job professionally but want to get some basic jobs that are paid for in order to get the domestic electrician initial accreditation. As work it would have to be minor works and not a new circuit and I would propose being assessed for 2 or 3 jobs which count towards examination for becoming a domestic electrician on any work opportunity. That means I get to pass to an exam more quickly. I would be happy these kind of minor works certificated jobs. It is not for me to issue the actual minor works certificate yet with the backing of a scheme so if you got a paid check of the work or paid so I could pay for it and I could get feedback would be just as valuable personally. Ready to come and discuss minor works (note 8).

So to recap, 2.2Amps is the answer from a 106m cable run with 6Amp protection needing 100W minimum for each point, making 5 lights to a circuit. It is common knowledge that it is 12 or 13 points so I propose

1. to remain protected by a lighting circuit as stated at 6Amp though will accecpt finding of 3Amp cirucit and calculate for a shorter length
2. to have lighting ammount by parameters to 6Amp and 230V making 1380W at diversity being 2070W thus 20 light points
3. to have lighting ammount to by design no more than 2.2Amp for 13 points or 20 at a maximum if diversity can be applied
4. 20 lights points is 2.2amp demand for 17x 12Watt and 3x 100Watt by diversity
5. 2.2Amp demand is a figure found to be by distribution from 106Metre light circuit (ring 112m) or radial
6. if the last point in a ring circuit breaks it makes a shorter radial circuit
7. 13 points shall be measured for 2.2amps and not 20
8. the circuit shall be designed from 38Watt average for 13 points in order to reach the low energy maximum figure
9. 38Watt average protected by 6amp would allow for 35 lighting points so it not excessive
10. 12W low energy bulbs average is 3 times below the aim of 38Watts average and would be 13x 60Watt equivalent light bulbs