Volts dropped, percent of the supply and volts at the load, from the values printed in NEC 2026 Chapter 9, Table 8.
The calculators on /calc make you type every table value in, on purpose, because finding a factor in your own book is the skill the exam is testing. This page is the other job. The Table 8 values are already here, so a run you are pricing on the truck takes about ten seconds.
The method underneath is identical and it is written out below, cited to the section it comes from. If you are studying rather than working, the study page on voltage drop covers what the code does and does not require.
Table 8 prints direct current resistance at 75 degrees C in ohms per 1000 feet, and for the small sizes it prints each size twice. 14 AWG through 8 AWG each get a solid row and a seven strand row, and the stranded row is the larger number because the same circular mil area spun into strands is a slightly longer path.
6 AWG and everything above it appears in Table 8 as stranded only, so the construction control on this page stops mattering at that point and says so. Table 8 also carries a coated copper column for tin coated conductor, which is not what you are pulling in ordinary work, so this page uses the uncoated column throughout.
Circular mils come out of the same table and they do not change with construction. 12 AWG is 6530 circular mils whether it is solid or stranded, which is why the K factor method returns one answer per size and the resistance method returns two.
The resistance method uses the number Table 8 measured for that size: VD = 2 x L x R x I / 1000 for DC and single-phase, and VD = 1.732 x L x R x I / 1000 for three-phase, with L the one-way length in feet and R in ohms per 1000 feet.
The K factor method taught for the exam replaces the measured resistance with one constant, K = 12.9 for copper and K = 21.2 for aluminum, over the circular mil area: VD = 2 x K x I x L / CM, and 1.732 in place of the 2 on three-phase. K is a rounded figure for copper at 75 degrees C rather than a per size measurement, so the two methods separate slightly. On 12 AWG solid copper carrying 20 amperes 100 feet, the resistance method gives 7.72 volts and the K method gives 7.90 volts.
Neither is wrong. The item on the paper usually decides for you by what it hands you: a resistance in ohms per 1000 feet points at the table method, a circular mil figure points at K.
Pick the multiplier from the system first, because it is the piece people substitute last and get backwards. Then read the resistance in the right metal and the right construction row, divide it by the number of parallel sets, and only then multiply by length and current and divide by 1000.
Parallel sets divide the resistance, they do not multiply it. Two sets of 4 AWG carry the same current on twice the copper, so the resistance of the run halves and the drop halves with it. On the K side the same thing happens by multiplying the circular mils by the number of sets.
The percent is always taken against the source voltage, not against the voltage left at the load. Dividing by the arrived voltage produces a number that is close enough to look right and is wrong every time, and on a long run it is wrong by enough to change the answer you pick.
The 3 percent branch circuit figure and the 5 percent combined figure come from Informational Notes at 210.19 and 215.2, and 90.5(C) puts explanatory material of that kind outside the enforceable text. This page prints the comparison because the exam asks for it and because a specification usually repeats it, not because the code compels it.
Two articles do compel a number. 647.4(D) holds sensitive electronic equipment to 1.5 percent on a branch circuit and 2.5 percent combined, and 695.7 sets limits for fire pumps. The study page works through what each of those actually asks for.
A 120 volt single-phase branch circuit carries 20 amperes to a load 100 feet away on 12 AWG solid uncoated copper. Table 8 gives that conductor 1.93 ohms per 1000 feet, and one set is being pulled.
Substitute: 2 x 100 feet x 1.93 x 20, divided by 1000. That is 7.72 volts. Against 120 volts it is 6.43 percent, and 112.28 volts arrives at the load.
Run the same circuit through the K factor method and 12 AWG is 6530 circular mils, so 2 x 12.9 x 20 x 100 divided by 6530 is 7.90 volts. The 0.18 volt gap between the two answers is the rounding inside K, not a mistake in either one.
A three-phase case for contrast: 50 amperes on 4 AWG copper, 200 feet, 480 volts. Table 8 gives 0.308 ohms per 1000 feet, so 1.732 x 200 x 0.308 x 50 divided by 1000 is 5.33 volts. That is 1.11 percent of 480 volts, and 474.67 volts arrives.
The value used for your inputs is read from NEC 2026 Chapter 9 Table 8, 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.
Resistance columns are ohms per 1000 feet. The K factor method on this page uses K = 12.9 for copper and K = 21.2 for aluminum against the circular mil column above. Coated copper is a separate column in Table 8 and is not reproduced here.
Multiply the multiplier for the system by the one-way length in feet, by the conductor resistance in ohms per 1000 feet, by the load current, then divide by 1000. The multiplier is 2 for DC and single-phase and 1.732 for three-phase. On 12 AWG solid uncoated copper at 1.93 ohms per 1000 feet, 20 amperes over 100 feet of a 120 volt single-phase circuit gives 7.72 volts, which is 6.43 percent.
On ordinary branch circuits and feeders it is not. The 3 percent branch circuit figure sits in an Informational Note at 210.19 and the 5 percent combined figure sits in an Informational Note at 215.2, and 90.5(C) places informational notes outside the enforceable text. Two articles do print an enforceable number: 647.4(D) at 1.5 percent on a branch circuit and 2.5 percent combined for sensitive electronic equipment, and 695.7 for fire pumps.
VD = 1.732 x L x R x I / 1000, with L the one-way length in feet and R the Table 8 resistance in ohms per 1000 feet. The square root of three replaces the 2 that single-phase uses. On 4 AWG copper at 0.308 ohms per 1000 feet, 50 amperes over 200 feet of a 480 volt three-phase circuit gives 5.33 volts, or 1.11 percent.
The same way as single-phase, because a two-wire DC circuit also sends the current out on one conductor and back on the other, so the multiplier is 2. Table 8 prints direct current resistance to begin with, so nothing has to be adjusted for it. Set the system to DC on this page and the multiplier stays at 2.
K is one rounded constant applied to every size, 12.9 for copper and 21.2 for aluminum, while Table 8 prints a resistance measured for each size and construction. On the 12 AWG example the K method returns 7.90 volts against 7.72 volts from the resistance method. Both are accepted arithmetic, and the question usually tells you which one it wants by handing you either a resistance or a circular mil figure.
It depends on the current, the one-way length, the system voltage and the metal, so there is no one size to memorize. Switch this page to size the conductor, enter the percent you are allowed, and it walks Table 8 from 14 AWG upward and returns the first size that meets it. It also prints the resistance it used so you can check that row against your own book.
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