In calculating the load for a residential general purpose circuit that will have lights on it, can I use the wattage of fluorescent energy saving bulbs instead of the incandescent ones, since that is what I will be using? It doesn't sound likely, but thought I would ask anyway.

Lighting calculations are based on the outside dimensions of a dwelling presented in square feet (Use watts per square foot). Dwelling units are 3 watts per sq ft.

Exclude: Open porches, garages, and unfinshed spaces in finding the outside dimensions.

Yes, any individual branch circuit can or should be calculated with the actual load that it will serve.

A 15 amps circuit can have a total of 1800 watts, if the load will run 3 hours or more the circuit derated by 80%, for a total of 1440 watts. The load for a 20 amp circuit is 2400 watts noncontinuous, 1920 watts for 3 hours or more.


Washington did not continue to explain how to use the information he provided, it is to calculate how many total minimum general lighting circuits a home would need. If a home has 200 sq ft, and general lighting is at 3/wsf, the total wattage would be 6000 watts divided by 120 volts, the home will have a calculated lighting load of 50 amps.

50 amps can be served by a minimum of 4 -15 amp circuits, or 3- 20 amp circuits.The total lighting load of 6000 watts is then used to be added to other loads to begin calculating the service rating.

Tk is correct.

Just to add:)

Although what I said was correct. I would like to add that it seems that you are speaking on inductive lighting loads. An inductive lighting load includes: lighting units that have ballasts, transformers or the like.

In your case, do not add up the watts on these fixtures. Instead, you will be using the total ampere rating of the units. Example: Lets say you have a fluorescent fixture with two 40W lamps. Although the two lamps add up to be 80W 40W(2)=80W<---this isn't the correct way to calculate inductive loads. Instead, you will check the ballast for the ampere rating. Lets say this fluorescent has an ampere rating of .9A. Calculate: E(I)=W so---120(.9)=108w. The number you would use=108w