The correct answer is 6 AWG. Before I cover the answer, I want to talk about the wording of the *Code* section.

Here is the wording from 210.19(A)(1):

(1) General. Branch-circuit conductors shall have an ampacity not less than the maximum load to be served. Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.

There is no question about the first sentence; it is very understandable. The problem is with the second sentence. This sentence is very confusing, especially the part that says "before the application of any adjustment or correction factors". It seems to be saying to multiply continuous loads by 125 percent and then apply the additional correction factors for ambient temperature and the adjustment factors for more than three current-carrying conductors. But this is not what it is saying. This is not just my opinion. david luchini was exactly right when he said "Example D3(a) in Annex D gives a good example of this." In this example, the calculations are separate calculations. Also, as an instructor teaching NFPA seminars, this is how we teach it. I spend quite a lot of time explaining this section, because what it sounds like is not what it is saying.

The continuous load calculation is separate. The continuous load aspect can be calculated with 110.14(C), especially when determining the minimum size conductor at the overcurrrent device termination. The correction factor and adjustment factor (sometimes referred to as the "conditions of use") is another calculation. Perform this calculation with the actual load, not with the extra 25 percent for continuous loads.

Besides the two separate calculations, we must also consider the size of the overcurrent protection. I thought it was great when david luchini (post #13) changed the number of conductors to make a point. I thought this was great because the change that david made is exactly one of my examples that will be in the September issue.

To satisfy 110.14(C) and 210.19(A)(1), multiply the load by 125 percent (39 ? 125% = 48.75 = 49). Select an 8 AWG conductor out of the 75?C column. Now see if that 8 AWG can carry 39 ampere after applying the correction and adjustment factors. Because the conductors are THHN, it is permitted to use the 90?C column (55 ? 0.91 ? 0.70 = 35). Since the load is 39 amperes, an 8 AWG is not permitted. Perform the same calculation using a 6 AWG conductor (75 ? 0.91 ? 0.70 = 47.775 = 48). Although the continuous load is 49 amperes, the conductors are only required to have a rating of the actual load of 39 amperes. Therefore, 6 AWG THHN copper conductors are permitted to supply this branch circuit