How to Apply the NEC 3% Voltage Drop Rule

What low voltage does to the load

Every foot of conductor has resistance, and current through resistance costs volts. The load at the far end sees the source voltage minus that drop. Resistive loads simply put out less: a heater at 108 V instead of 120 V delivers 81% of its rated heat, because power falls with the square of the voltage. Motors are worse off. An induction motor tries to deliver the same horsepower regardless of voltage, so it draws more current, runs hotter and loses starting torque; a compressor that starts fine on a short test lead can trip its overload at the end of a long, undersized run. LED drivers and electronic power supplies tolerate a range of input voltage, but they pull more current at the bottom of that range, and the lights flicker whenever a motor starts on the same feeder. None of this is theoretical: a well pump, a detached shop, a parking lot light pole or a long run to a remote panel is where voltage drop shows up.

3% and 5% are recommendations, with exceptions

The 2023 NEC does not set a general voltage drop limit. The informational notes to 210.19 (branch circuits) and 215.2 (feeders) say that a drop of no more than 3% at the farthest outlet, and no more than 5% for feeder and branch circuit combined, will provide reasonable efficiency of operation. Informational notes are not enforceable, so an inspector cannot fail a branch circuit on those notes alone. The Code does make limits mandatory in a few places. Article 647, for sensitive electronic equipment on 120 V line-to-line, 60 V to ground systems, limits branch-circuit drop to 1.5% and the feeder-plus-branch total to 2.5% for fixed equipment (647.4(D)), and tighter still for receptacle circuits. Article 695 requires that fire pump controller voltage not fall more than 15% below rated while the motor starts, or more than 5% at the load terminals with the motor running at 115% of full-load current (695.7). Some energy codes and many project specifications turn the 3% and 5% guidance into a hard requirement. Your adopted edition, local amendments and the AHJ govern; everywhere else, 3% is a design target, and a good one.

The formula and a worked example

For a single-phase circuit the resistance method is:

VD = 2 × K × I × L ÷ CM

K is the resistivity constant in ohm-circular-mils per foot: 12.9 for copper and 21.2 for aluminum, both at about 75°C. I is the load current in amps, L is the one-way length in feet (the 2 accounts for the return conductor), and CM is the conductor area in circular mils from NEC Chapter 9, Table 8: 6,530 for 12 AWG, 10,380 for 10 AWG, 16,510 for 8 AWG, 26,240 for 6 AWG and 41,740 for 4 AWG. Divide VD by the source voltage for the percentage.

Take a 20 A load 150 feet from a 120 V panel on 12 AWG copper: VD = 2 × 12.9 × 20 × 150 ÷ 6,530 = 11.85 V, or 9.9%. The outlet sees about 108 V. Going to 10 AWG gives 7.46 V (6.2%), 8 AWG gives 4.69 V (3.9%), and 6 AWG gives 2.95 V, or 2.5%, the first size under 3%. Turned around, 12 AWG at a full 20 A stays under 3% for only about 45 feet at 120 V, which is why a long 20 A circuit to a shed is so often run in 10 AWG. The voltage drop calculator does this arithmetic for any size and also reports the smallest conductor that meets your limit.

Three-phase circuits use 1.732 in place of 2, so VD = 1.732 × K × I × L ÷ CM, and the percentage is figured on the line-to-line voltage. Everything else is the same.

Four ways to fix it

1. Upsize the conductor. Going from 12 AWG to 10 AWG cuts the drop by 37%; from 12 AWG to 6 AWG cuts it by 75%. When you upsize for voltage drop, 250.122(B) requires the wire-type equipment grounding conductor to grow in proportion, and the raceway may need to grow with it (check the conduit fill calculator).
2. Shorten the run. Drop is proportional to length. A subpanel near a group of loads turns one long branch circuit into a short feeder, which can carry a 2% or 3% drop of its own, plus short branch circuits.
3. Raise the voltage. For the same power, 240 V draws half the current of 120 V, and the drop as a percentage of a doubled base is one quarter. A well pump or a shop wired at 240 V instead of 120 V is the classic fix; 480 V instead of 208 V does the same for a large motor.
4. Split the load. Two circuits each carrying half the current have half the drop, and two runs of smaller wire are sometimes cheaper than one big one.

When resistance is enough and when reactance matters

The K-factor method is a DC resistance calculation. It ignores the inductive reactance of the conductors and the power factor of the load, and it assumes a 75°C conductor. For branch circuits and smaller feeders, in sizes up to about 1/0 AWG, that is accurate and slightly conservative, since a lightly loaded conductor runs cooler and drops a little less than the formula says. Reactance grows with conductor size and conductor spacing, and it starts to matter when large conductors (roughly 1/0 AWG and up, and certainly 250 kcmil and above), a long AC feeder in steel raceway, and a low power factor load such as a large motor come together. For those, use the effective impedance in NEC Chapter 9, Table 9, which is tabulated at 0.85 power factor for each raceway type, or the manufacturer's data, in place of K. For DC and for resistive loads the resistance method is exact at any size.

One last habit: size for ampacity first with the wire size calculator, then check voltage drop, and install the larger of the two answers. Voltbox carries both calculators offline, and voltage drop is a dependable question on the electrician practice exam.