Ten loads on a clock. Each prompt names a current, a material and a temperature column, and you pick the smallest conductor that carries it. Every ampacity comes from NEC 2026 Table 310.16 and the 240.4(D) caps, printed in full further down.
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Levels one and two hand you a plain load. Level three mixes the two materials and marks some of the loads continuous, which puts the minimum conductor ampacity at 125 percent of the load under 210.19(A)(1). The 240.4(D) cap on 14, 12 and 10 AWG is applied at every level, so a load those sizes cannot be protected for steps you up a size.
The ampacity sprint asks you to read a cell. This one asks the question the exam actually asks, which is the same table pointed the other way: here is a load, what carries it. The work is a walk up the rows until a number is big enough, plus two things people forget under a clock.
Those two are the 125 percent on a continuous load and the 240.4(D) ceiling on the three small sizes. Run a level until both of them come to mind before you touch the table, then read continuous loads and size a real circuit in the wire size calculator.
Table 310.16 is written to answer one question: what does this conductor carry. Sizing is that question turned around. You settle on a number the conductor has to be good for, then walk down one column until you reach the first figure that covers it, and the row you stopped on is your size.
First settle the number. A load that is not continuous is just itself. Where a branch circuit supplies a continuous load or any mix of continuous and noncontinuous load, 210.19(A)(1) puts the minimum conductor size at an ampacity not less than the noncontinuous load plus 125 percent of the continuous load, and 215.4(A)(1) says the same thing for a feeder. Article 100 defines a continuous load as one where the maximum current is expected to continue for 3 hours or more.
Then settle the column. 110.14(C) requires the ampacity to be selected and coordinated so that it does not exceed the lowest temperature rating of any connected termination, conductor or device, and 110.14(C)(1)(a) holds equipment on circuits rated 100 amperes or less to the 60 degree ampacity unless it is listed and marked for more. On this page the prompt names the column, because the skill being drilled is the walk rather than the column choice.
Then walk. In the named column, take the first row whose ampere figure is at least the number you calculated. Stopping one row early is the mistake this game is built around, because a row that is 2 amperes short reads as close enough when the clock is running and there are 79 questions behind this one.
240.4(D) is the last gate and it touches only 14, 12 and 10 AWG. The overcurrent device on those sizes cannot exceed 15, 20 and 30 amperes on copper, or 15 and 25 amperes on aluminum, applied after any correction factors for ambient temperature and number of conductors. A load that needs a bigger device than that cannot ride on the size no matter what the 90 degree column prints, so it steps up. That is why a 25 ampere load on copper lands on 10 AWG in every one of the three columns.
The trap the exam sets is a continuous load that lands exactly on a row when you forget the multiplier. Forty amperes continuous on copper out of the 75 degree column needs 50 amperes of ampacity, which is exactly what 8 AWG prints, so 8 AWG is right either way and the mistake costs nothing. Move the load to 45 amperes and the required figure is 56.25, 8 AWG is short, and the answer is 6 AWG at 65. Same table, one multiplication between right and wrong.
The sprint shows 45 amperes, copper, the 75 degree column, and marks the load continuous.
Continuous puts 210.19(A)(1) in play, so the conductor has to be good for 125 percent of 45, which is 56.25 amperes. Walking down the copper side of the 75 degree column: 10 AWG gives 35 and 8 AWG gives 50, and both are short of 56.25. 6 AWG gives 65, which covers it, so 6 AWG is the answer.
8 AWG is on the option list on purpose, because 50 amperes covers 45 and that is the size you land on the moment you skip the 125 percent. A wrong answer nobody would pick teaches nothing, so the four options here are always the right row and three of its neighbours.
The value used for your inputs is read from NEC 2026 240.4(D) and Table 310.16, the way you would read it in the exam room, and the working above names the row it came from. The tables themselves belong to NFPA and are not printed on this page. Bring your own copy of the 2026 code, which is what the exam requires anyway, and check the row there.
On a narrow screen the table scrolls sideways to reach the aluminum columns. The aluminum column covers aluminum and copper-clad aluminum. The printed table also carries a 14 AWG aluminum line, footnoted so that it applies only to copper-clad aluminum and only for ampacity adjustment or correction. Those entries are left off this page and out of the game.
Work out what the conductor has to carry, settle which temperature column you are allowed to read, then walk down that column until you reach the first ampere figure that is at least the number you calculated. The row you stopped on is the size. On 14, 12 and 10 AWG check 240.4(D) before you write the answer down, because the overcurrent device on those three sizes has a ceiling of its own.
Where a branch circuit supplies a continuous load, or any mix of continuous and noncontinuous load, 210.19(A)(1) sets the minimum conductor size at an ampacity not less than the noncontinuous load plus 125 percent of the continuous load. 215.4(A)(1) puts the same rule on a feeder. Article 100 defines a continuous load as one where the maximum current is expected to continue for 3 hours or more, which is why lighting and similar loads keep turning up in these questions.
On ampacity alone 12 AWG copper looks like enough, since it prints 25 amperes in the 75 degree column and 30 in the 90 degree column. 240.4(D) holds the overcurrent device on 12 AWG copper to 20 amperes after any correction factors for ambient temperature and number of conductors have been applied, and a 25 ampere load needs more device than that. So the size steps up to 10 AWG, where the cap is 30.
110.14(C) requires the ampacity to be selected and coordinated so that it does not exceed the lowest temperature rating of any connected termination, conductor or device. 110.14(C)(1)(a) puts equipment on circuits rated 100 amperes or less on the 60 degree ampacity unless the equipment is listed and marked for something higher, and 110.14(C)(1)(b) is the rule for circuits above that. The 90 degree column stays available for correction and adjustment. This game names the column in the prompt so the drill stays on the walk up the rows.
No. It sizes out of one printed column and stops. 310.15(B) corrects the table figure for an ambient other than 30 degrees C, 310.15(C)(1) adjusts it when more than three current-carrying conductors share a raceway or a cable, and on a real circuit both of those land before the 240.4(D) check. The wire size calculator on this site runs the whole order with the correction and adjustment tables in it.
The right answer lands somewhere between 14 AWG and 500 kcmil, and the three wrong options are drawn from the rows either side of it. The aluminum rows start at 12 AWG here, because the 14 AWG aluminum line in the printed table is footnoted so that it applies only to copper-clad aluminum and only for ampacity adjustment or correction.
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