A 277-volt single-phase lighting circuit will carry 20 amperes 300 feet one way in aluminum, and the drop must be held to 3 percent of the supply voltage. Use 21.2 as the aluminum constant and the single-phase relationship in which drop equals 2 times the constant times the current times the one-way length divided by circular mils. Aluminum areas are 16,510 circular mils for 8 AWG, 26,240 for 6 AWG, 41,740 for 4 AWG and 66,360 for 2 AWG. What is the smallest of these conductors that holds the drop within the limit?
- A6 AWG aluminum
This runs the calculation with the copper constant of 12.9 on an aluminum conductor. Aluminum has roughly 1.6 times the resistivity of copper, so using copper's constant understates the area needed.
- 4 AWG aluminum
- C8 AWG aluminum
This leaves the factor of 2 out, which halves the required area to about 15,300 circular mils. The multiplier accounts for the return conductor and cannot be dropped.
- D2 AWG aluminum
This doubles the length to 600 feet and also keeps the factor of 2, counting the return path twice and roughly doubling the area the calculation appears to demand.
Why B is correct
Three percent of 277 volts is 8.31 volts. Rearranged for area, CM = (2 x K x I x L) / VD = (2 x 21.2 x 20 x 300) / 8.31 = 254,400 / 8.31 = 30,614 circular mils. That is more than the 26,240 of 6 AWG, so the next size up is required, and 4 AWG at 41,740 circular mils satisfies it with margin.
What this question is testing
Tests whether you pick the resistivity constant that matches the conductor material, whether you can rearrange the drop relationship to solve for circular mils, and whether you then round up to a real conductor size rather than reporting the raw calculated area.
On the job
Aluminum is common on long site-lighting and feeder runs because of cost, and it is exactly where drop bites hardest. The material saving evaporates if you have to go two sizes larger, so the calculation is really a cost comparison as much as a Code exercise. Crews who run the numbers before ordering avoid the situation where the conductor is on the truck and the drop calculation says it will not do.
Memory technique
Aluminum needs about two-thirds more copper's worth of metal for the same drop.
Exam tip
Solve for circular mils, then step up to the first listed size that meets or beats it. Never round down.
Where to look it up
Chapter 9, Table 8 lists circular mil areas for both copper and aluminum. The constants come from the exam, not the Code.