15-amp Lighting Circuits

[jeff48356]

I know that the old rule for 15-amp lighting circuits is that you need one for every 500 square feet of living space, so a typical 1960's ranch with 1200 square feet and a basement would require 3 such circuits, and that's exactly what you'd find in these houses.

But that includes all the wall outlets in the bedrooms and living room, even the bathrooms, which were also part of these circuits. Nowadays, suppose a new house of the same size were built and wired (or an old one re-wired), and the lights were wired separately from the outlets. The bathroom outlet is on a separate 20-amp circuit, and the rest of the general-purpose outlets are also on 20-amp circuits (about 8 outlets per circuit). The 15-amp circuits are only used to power the lighting fixtures (including recessed cans, ceiling fans, and bath fan). Would three 15-amp lighting circuits still need to be installed under those circumstances? Or could you get by with only two, since there would never be anywhere near 15 amps drawn on any one circuit, even with all the lights and fans on at the same time?

 

[Buck Parrish]

Theoretically maybe, but It depends on how many recessed lights, fans etc... you have. We just put 22 cans in a kitchen. Then if you use LED's so they only pull 12 watts each thats 264 watts. If it is a standard 65 watt bulb it would pull 1430 watts.

 

[roger3829]

It's one circuit for every 600 sq ft, so you would need only two GP circuits. That means that all the lights and general purpose receptacles could possibly be on those two circuits.

 

[Dennis Alwon]

This is an argument that many have fought. Some say that the 3 watts per sq. ft is for calculation purposes only and that there is no limit in residential.

 

[infinity]

Article 220 tells you haw to calculate the minimum number of general purpose lighting and receptacle circuits. They can be 15 or 20 amp. With 20 amp circuits you'll be required to use less circuits.

 

[augie47]

I may learn something here, but if the circuit is for lighting only and not GP, I have not sized it on a watts/ft basis, but looked at the actual (or possible) load based on the fixture wattage.
If the fixtures are unknown a conservative WAG is in order.
As Buck pointed out, one houe may have 22 can lights in a kitchen, the next a single fluorescent.

 

[renosteinke]

I agree that you can use the watts/sq ft math as a general check, but ..... it's trumped in the specific application by the actual load on the circuit. ZYour circuit is powering lights, not square feet of flooring.

That is, you add up the load, using the biggest light bulbs allowed for the fixtures. So, for example, your usual keyless you would calculate at 100 watts (as marked on the fixture as the maximum bulb size), even though you know the new "energy" codes will result in 13 watt CFL's being used.

If the usual fluorescent fixtures are used, you go by the amperage listed on the ballast, and ignore the watt markings on the bulb. A fixture with 4-40watt bulbs and a .93A ballast is calculated at .93 amps. The markings on fluorescent bulbs are 'nominal' but the actual load is determined by the ballast.

Lighting is usually assumed to be a 'continuous' load, so a 15A circuit would be limited to 12 amps (1440 watts) of calculated load.

 

[Dennis Alwon]

I think if the OP means a lighting circuit without recep. then the 3 watts/ sq. foot is irrelevant. The wattage marked on the fixtures should be used to calculated the number of cir.

 

[JDB3]

A house that I recently wired had 6 recess cans with 65 watt bulbs & 1 chandelier with 8 60 watt bulbs, in the master bedroom. The dinning room had 4 recess with 65 watt bulbs & 1 chandelier with a max of 3 150 watt bulbs. Another house I put the dinning fixture & rope lighting around the ceiling, on its own circuit , these maxed out the circuit. Hard to anticipate what the owners will end up choosing ! :?

 

[ActionDave]

Lighting is usually assumed to be a 'continuous' load, so a 15A circuit would be limited to 12 amps (1440 watts) of calculated load.

Not in my house. My kids are constantly going behind me flipping off switches and nagging me, "Daaaad, you need to turn off lights when you leave the room, teacher says."

 

[resistance]

If we are talking dwell. Then take the sq footage and multiply it by 3va (per NEC T220.12). Example: 1200sq ft dwell(3va)= 3600va
Then take the ampacity size of the lighting breaker (which is 15 or 20 amps) and multiply it by the voltage (120v). Example: 15A(120v)= 1800va. Then divide: 3600va/1800va= 2 circuits. Then Note NEC 220.42?where it tells you that a demand factor from T220.42 shall not apply for general illumination.

 

[resistance]

Lighting is usually assumed to be a 'continuous' load, so a 15A circuit would be limited to 12 amps (1440 watts) of calculated load.

Dwell lighting (and motel room lighting) is not considered continuos.

 

[iwire]

Dwell lighting (and motel room lighting) is not considered continuos.

I agree.