Voltage Drop Practice Questions

A. No percentage at all, since the figures are in an informational note

The branch-circuit note describes a conductor sized so the drop at the farthest outlet stays inside 3 percent, with the total across feeder and branch circuit inside 5, and calls that reasonable efficiency of operation. That is a description of good practice rather than a rule. An item asking what the code requires has an answer, and any percentage sitting in the options is there to be left alone.

NEC 2026 210.19

B. As explanatory material that is not enforceable as a requirement

The code sorts its text into mandatory rules, permissive rules and explanatory material, and puts informational notes in the third bucket by name. Mandatory rules are marked by shall and shall not, permissive ones by shall be permitted and shall not be required. Once you have that sorting, the whole voltage drop question answers itself without any arithmetic.

NEC 2026 90.5(A), 90.5(B), 90.5(C)

C. At 215.4(A)(2)

The feeder minimum rating and size section carries the note under its ampacity adjustment and correction subdivision, and the branch-circuit note points across at it by number. Both notes carry the same figures for the same reason. Knowing where the pair lives is worth more than the numbers, because an item usually asks which text is enforceable rather than what the percentage is.

NEC 2026 215.4(A)(2)

D. 3 percent at the farthest outlet and 5 percent total across feeder and branch circuit

The 3 belongs to the branch circuit at its farthest outlet and the 5 is the total once the feeder is added in, and both notes state it the same way. The third option has the two numbers the right way round but assigned backwards, which is how an item collects somebody who remembers a pair of figures without remembering which is which.

NEC 2026 210.19, 215.4(A)(2)

A. Because the current travels out on one conductor and back on the other

The conductor the current actually sees is the round trip, so a load 150 feet away sits at the end of 300 feet of copper. Nothing about it is a safety factor, which matters because a safety factor would be something you could argue about and this is not. Picture the path before you write anything and the multiplier stops being something you memorize.

NEC 2026 210.19

B. 1.732

The square root of three takes the place of the 2 because the three phase currents sit 120 degrees apart and their drops do not simply add. The 0.866 option is that same square root divided by two, which is the ratio between the two answers rather than the multiplier itself, so it produces a number that looks reasonable and is wrong by a factor of two.

NEC 2026 210.19

C. The one way length in both versions

Use the one way distance in both, because the multiplier already carries the return path. Doubling the length and then multiplying by 2 as well doubles the answer, and using the one way length with no multiplier at all halves it. Those two errors and the wrong multiplier are one mistake wearing different clothes, and all three come from not picturing the path first.

NEC 2026 210.19, 215.4(A)(2)

D. Chapter 9 Table 8

Table 8 gives direct current resistance in ohms per 1000 feet, and it splits copper from aluminum, solid from stranded, and coated copper from uncoated. Chapter 9 Table 1 is raceway fill percentages and answers a different question entirely. Divide the per thousand figure by 1000 to get ohms per foot, or keep it in thousands and divide the length instead. Mixing the two is a factor of a thousand and you will notice.

NEC 2026 Chapter 9 Table 8

A. 7.13 volts

Two times 120 feet times 15 amperes is 3600, and 3600 times 0.00198 ohms per foot is 7.13 volts. Halving it to 3.56 is the one way length with no multiplier. The 6.95 answer uses the solid copper figure of 1.93 instead of the stranded one, which is the right arithmetic off the wrong row of the table.

NEC 2026 Chapter 9 Table 8

B. 5.94 percent

Divide 7.13 by 120 and you get 5.94 percent, which is well past the figure the notes describe. Nothing in the code stops that circuit being installed, and nothing at the end of it will thank you either. The 3 and 5 percent options are the figures from the notes rather than the answer to the arithmetic, which is a distinction items on this subject lean on hard.

NEC 2026 210.19

C. 8.50 volts

Multiply 1.732 by 250 feet by 40 amperes to get 17,320, then by 0.000491 ohms per foot for 8.50 volts. Using the single phase multiplier of 2 gives 9.82, which is close enough to look right and is the answer of somebody who read the current and skipped the phases. Doubling the length as well gives 17.0.

NEC 2026 Chapter 9 Table 8

D. 1.77 percent

Divide 8.50 by 480 and you get 1.77 percent, comfortably inside what the note describes. Higher voltage is what buys that: the same volts lost are a smaller share of a bigger number, which is the whole reason long runs get designed at 480 rather than 208. Divide by the system voltage, not by the drop, and not by the voltage at the load.

NEC 2026 215.4(A)(2)

A. 9.70 volts

Two times 150 times 40 is 12,000, and 12,000 times 0.000808 is 9.70 volts. The 5.89 answer is the same run worked on the copper row at 0.491, which is the mistake of reading the size correctly and the material wrongly. Aluminum of the same size has roughly two thirds the conductivity, so it drops proportionally more over the same distance.

NEC 2026 Chapter 9 Table 8

B. By solid and stranded construction, and by coated and uncoated copper

Four copper figures can apply to one size and picking the wrong one is a quiet error, because the answer stays in a plausible range. The table also prints circular mil area and overall diameter, and a note gives the equation for correcting resistance to a temperature other than 75 degrees. Insulation type does not appear, because the resistance belongs to the metal and not the jacket.

NEC 2026 Chapter 9 Table 8

C. When the item introduces power factor or reactance rather than treating the circuit as purely resistive

Table 9 gives alternating current resistance and reactance for conductors in a raceway, with columns for the raceway material and effective impedance figures at a stated power factor. Table 8 is direct current resistance and treats the run as a plain resistor. A stem that hands you a power factor is telling you which table it wants, and answering it out of Table 8 throws away the input it gave you.

NEC 2026 Chapter 9 Table 8, Table 9

D. Nothing but the wire

A 20 ampere circuit run in 10 AWG for distance is still a 20 ampere circuit, protected at 20 amperes and limited to what a 20 ampere circuit may supply. One thing does follow the wire: where ungrounded conductors are increased in size, the wire type equipment grounding conductor is increased proportionally, which is a rule in Article 250 rather than a voltage drop rule.

NEC 2026 210.19, 250.122

A. 4 AWG

Three percent of 208 is 6.24 volts. Divide that by 1.732 times 200 feet times 50 amperes, which is 17,320, to get 0.00036 ohms per foot, or 0.36 ohms per 1000 feet. In the uncoated copper column 6 AWG is 0.491 and misses, and 4 AWG is 0.308 and clears it. Going to 2 AWG works too but is larger than the calculation asks for.

NEC 2026 Chapter 9 Table 8

B. 2 AWG

Three percent of 240 is 7.2 volts. Divide by 2 times 175 feet times 60 amperes, which is 21,000, to get 0.000343 ohms per foot, or 0.343 per 1000 feet. In the aluminum column 4 AWG is 0.508 and misses while 2 AWG is 0.319 and clears. Running the same numbers on the copper column lands you two sizes small, which is why the material has to be read out of the stem first.

NEC 2026 Chapter 9 Table 8

C. Toward less resistance, which means toward the larger conductor

You solved for a maximum resistance, so any conductor at or below that figure works and anything above it fails the target you set. A conductor one size small is the option an item will put in the list on purpose, because it looks like the natural rounding and it misses. Ampacity is a separate check that the answer still has to pass on its own.

NEC 2026 Chapter 9 Table 8

D. Article 647, at 1.5 percent on a branch circuit and 2.5 percent combined

The tight numbers are the reason that article exists, and they sit in enforceable text rather than in a note. The article covers a separately derived system at 120 volts line to line and 60 volts to ground, used in commercial or industrial occupancies under close supervision by qualified personnel. A candidate carrying the slogan that voltage drop is only a recommendation gets this one wrong.

NEC 2026 647.1, 647.5(D)

A. 1 percent, with the load taken as 50 percent of the branch-circuit rating and a combined limit of 2.0 percent

The cord-connected case is tighter than the fixed equipment case, and it tells you what load to calculate with, which is unusual. The stated purpose is to hold the drop to 1.5 percent once a portable cord is added on the end of the branch circuit, so the extra half percent is the cord. Fixed equipment on Chapter 3 wiring methods stays at 1.5 and 2.5.

NEC 2026 647.5(D)

B. Not more than 5 percent below the motor voltage rating with the motor at 115 percent of its full load current rating

Notice the shape of it. The measurement is taken at 115 percent of full load rather than at full load, so the calculation you run is not the one the stem first appears to ask for. It is also referenced to the motor voltage rating and not to the controller rated voltage, which is the reference the starting limit uses. Two limits, two reference points, two conditions.

NEC 2026 695.7, 695.8

C. Mechanical starting on emergency run, or bypass operation of a variable speed pressure limiting controller, where the start is shown to work on the standby system

Two named conditions, and each one carries the same proof: somebody has to show the machine starting on the standby supply rather than assume it will. That demonstration is what the exception is buying, so an option that names the equipment and drops the demonstration has taken half the sentence. The running limit at the load terminals is a separate subsection and neither exception reaches it.

NEC 2026 695.8

D. 10.4 volts

Only the multiplier changed, so the new answer is the old one times 1.732 divided by 2, which is 12 times 0.866, or 10.4 volts. That gives you a check worth carrying: if your three phase answer comes out larger than your single phase answer on identical inputs, you have swapped the multipliers. The 13.9 option is the same arithmetic done upside down.

NEC 2026 215.4(A)(2)

A. The ampacity rules do not take voltage drop into consideration, and the notes at 210.19 and 215.4(A)(2) address it separately

The ampacity section says so in its own informational note and points at both voltage drop notes by number. That is why the two calculations are run separately and why the larger of the two answers is what you install. Derating is about the conductor heating up and voltage drop is about the load starving, and neither calculation can see the other one.

NEC 2026 310.14(A)