Continuing with our examination of Example D3 in the NEC, we’re trying to determine how many circuits will carry the 44A lighting load.

In Part 3, we said using three 15A breakers is the wrong solution. Why is that?

First, this approach assumes you can evenly divide the 44A lighting load among three circuits. Is that feasible? You have a total of 9,000VA. This is a store, so it is probably using overhead fluorescent fixtures. Maybe these are rated 250VA each, meaning you have 36 of them. You could put a dozen on each circuit. In this case, it works out.

But suppose the fixture wattages don’t divide so evenly. Or you have higher wattage fixtures in the back and lower wattage fixtures in the front where the sun comes through the windows. You must examine the actual load, not just perform a simple calculation.

A second issue is the fact that lighting is a continuous load. Thus, the breaker must be rated for 125% of the load. Another way of looking at this math is that you can’t load a lighting breaker past 80%. Going with three 20A breakers would solve this problem, but going this route means that you’re still not looking at the actual load.

A third issue arises if these are magnetic ballasts. You’ll need to derate your circuits to avoid overheating the conductors. If you specify electronic ballasts, this problem goes away.

The most likely answer is that you need at least four 15A breakers.

« Part 3 | Part 5 » | Source: Mark Lamendola | Mindconnection