What voltage drop is and why it costs you money
Voltage drop is the voltage consumed by the conductors themselves before any of it reaches the load. Every conductor has resistance, current through resistance produces a voltage difference, and that difference is subtracted from what the load sees. A 240 V circuit that drops 8 V delivers 232 V.
The consequences are practical. Motors are the worst affected: torque falls with the square of the applied voltage, so a 5% drop costs roughly a tenth of the starting torque and makes the motor draw more current to do the same work, which heats the windings. Resistive heaters and incandescent lamps lose output roughly as the square of voltage too. Electronic power supplies mostly compensate, but they compensate by drawing more current, which makes the drop worse. And the energy lost in the conductors is dissipated as heat inside the raceway, where it does nothing useful and adds to the thermal load the ampacity tables are trying to manage.
Voltage drop is not usually an NEC requirement. The 3% and 5% figures most electricians quote appear as informational notes to 210.19 for branch circuits and 215.2 for feeders, and informational notes are explanatory rather than enforceable. A handful of articles do make a drop limit mandatory — Article 695 sets one for fire pump feeders during starting, for example — and many engineers write the 3% figure into a specification, at which point it becomes contractually binding even though the code does not enforce it.
The circular-mil formula, term by term
The field formula is Ohm's law rearranged so you never have to look up a resistance.
K, the resistivity constant. K is the resistance of a conductor one foot long with an area of one circular mil. For copper it is taken as 12.9 and for aluminium as 21.2, both at about 75 °C. You can verify it against NEC Chapter 9, Table 8: 12 AWG stranded uncoated copper is listed at 1.98 Ω per 1,000 ft, and 1.98 × 6,530 cmil ÷ 1,000 = 12.93. Aluminium checks out the same way: Table 8 lists 1/0 aluminium at 0.201 Ω per 1,000 ft, and 0.201 × 105,600 ÷ 1,000 = 21.2. That is the whole justification for the two numbers, and it is why K rises when the conductor runs hot.
CM, the circular-mil area. A circular mil is the area of a circle one mil (0.001 in.) across, chosen so that the area of a round conductor is simply the square of its diameter in mils. Chapter 9, Table 8 lists the value for every size; 12 AWG is 6,530 cmil and 4/0 is 211,600 cmil. Because the formula divides by area, doubling the area halves the drop, and that is exactly what a second parallel set does.
The multiplier. On a single-phase or DC circuit the current travels out along one conductor and back along the other, so it passes through two conductor lengths: the multiplier is 2. On a balanced three-phase circuit there is no neutral current and each conductor carries its share of a shared return path; working the phasor arithmetic through gives a line-to-line drop of √3 times the drop in one conductor, so the multiplier is 1.732.
L, the one-way length. Enter the distance from the source to the load, measured along the route the conductors actually take, including the vertical drops. The multiplier handles the return trip. Doubling the length yourself and then applying the multiplier is the single most common error in this calculation and produces exactly twice the true answer.
Worked example: a 20 A circuit 100 ft out on 12 AWG
A 120 V, 20 A single-phase circuit runs 100 ft to a load on 12 AWG copper. Is that acceptable at a 3% target?
- Pick the constants. Copper, so K = 12.9. Single-phase, so the multiplier is 2. From Chapter 9 Table 8, 12 AWG is 6,530 cmil.
- Work the numerator. 2 × 12.9 = 25.8. 25.8 × 20 A = 516. 516 × 100 ft = 51,600.
- Divide by the area. 51,600 ÷ 6,530 = 7.90 V dropped in the conductors.
- Convert to a percentage. 7.90 ÷ 120 × 100 = 6.59%. The load sees 120 − 7.90 = 112.10 V.
- Compare with the target. 6.59% is more than double the 3% guideline, so 12 AWG fails on drop even though it is perfectly adequate on ampacity for a 20 A circuit.
- Solve for the size you need. A 3% drop on 120 V is 3.60 V. Rearranged, CM = 2 × 12.9 × 20 × 100 ÷ 3.60 = 14,333 cmil. The first standard size at or above that is 8 AWG at 16,510 cmil, so you pull 8 AWG.
Notice how much conductor the drop rule costs you: ampacity alone would let you use 12 AWG, and distance forces you two sizes up. Check the raceway afterwards with the conduit fill calculator, because 8 AWG is nearly three times the area of 12 AWG.
How to read the percentage
Compare the number against the target you actually owe, not a habit. The widely used design rule is 3% on the branch circuit, 5% on the combined feeder and branch circuit. If your feeder already drops 2%, the branch circuit gets 3% and no more.
Below about 2% you are comfortable and further upsizing rarely pays for itself. Between 2% and your target you are compliant with the usual design rule but should sanity-check the load: an air conditioner or a well pump that starts against a stiff load may be drawing several times its running current at the moment of start, and the momentary drop scales with it. Above the target, upsize the conductor rather than accepting it, because the drop is permanent and the conductor is the cheapest thing in the assembly to change before the walls close.
Two situations deserve special care. Low-voltage DC is unforgiving: at 12 V a 3% target is 0.36 V, which a 20 ft run of 10 AWG at 10 A already exceeds. Solar, RV and marine work therefore uses much larger conductors than the current alone suggests, which is why the solar charge controller sizing calculator and battery work always come back to this page. Long motor feeders deserve a check at locked-rotor current as well as running current, because a motor that cannot develop starting torque will sit and heat until the overload opens; size the running current first with the motor full-load amps calculator.
Maximum one-way run length for a 3% drop, copper
| Size | cmil | 120 V, 15 A | 120 V, 20 A | 240 V, 30 A | 240 V, 50 A | 480 V 3∅, 100 A |
|---|---|---|---|---|---|---|
| 14 AWG | 4,110 | 38 ft | 28 ft | 38 ft | 22 ft | — |
| 12 AWG | 6,530 | 60 ft | 45 ft | 60 ft | 36 ft | — |
| 10 AWG | 10,380 | 96 ft | 72 ft | 96 ft | 57 ft | — |
| 8 AWG | 16,510 | 153 ft | 115 ft | 153 ft | 92 ft | — |
| 6 AWG | 26,240 | 244 ft | 183 ft | 244 ft | 146 ft | — |
| 4 AWG | 41,740 | 388 ft | 291 ft | 388 ft | 232 ft | 269 ft |
| 2 AWG | 66,360 | 617 ft | 462 ft | 617 ft | 370 ft | 427 ft |
| 1/0 AWG | 105,600 | 982 ft | 736 ft | 982 ft | 589 ft | 680 ft |
| 4/0 AWG | 211,600 | 1,968 ft | 1,476 ft | 1,968 ft | 1,181 ft | 1,363 ft |
Rounded down to the nearest foot. Cells are left blank where the conductor would not normally be used at that current. For aluminium, multiply every length by 12.9 ÷ 21.2 = 0.608.
The 3% figure is guidance, not a code requirement
In NFPA 70, National Electrical Code, 2023 edition, the familiar 3% branch-circuit and 5% total figures appear in informational notes to 210.19 and 215.2. Section 90.5(C) states that informational notes are explanatory and are not enforceable as requirements. Specific articles can and do make a limit mandatory in their own scope, and a project specification can make it contractual. Always check what your specification and your authority having jurisdiction require before deciding that a 4% drop is acceptable.
Mistakes that produce a wrong drop figure
- Entering the round-trip length. The formula's multiplier already doubles it. Entering 200 ft for a 100 ft run gives exactly twice the true drop.
- Using the breaker size as the current. Drop depends on the current that actually flows. Sizing at the breaker rating is conservative and often wasteful; sizing at running current is the normal design basis.
- Applying √3 to a single-phase load fed from a three-phase panel. A 120 V or 277 V line-to-neutral load uses the factor 2, whatever the panel is.
- Ignoring conductor temperature. K = 12.9 assumes about 75 °C. A conductor running near its 90 °C limit has noticeably higher resistance, and the drop rises with it.
- Forgetting reactance on large conductors. Above roughly 1/0 in a steel raceway, inductive reactance is a real part of the impedance and the resistance-only method under-reads.
What the K method assumes, and when to use Table 9 instead
This calculator models the conductor as a pure resistance at about 75 °C. That is an excellent model for DC and a good one for alternating current on conductors up to roughly 1/0 AWG in a non-metallic raceway at a high power factor. It is not exact for the cases it simplifies away.
Reactance. Alternating current in a conductor sets up a magnetic field, and that field opposes changes in current. The resulting inductive reactance adds to resistance vectorially. It grows with conductor spacing and is much larger inside a steel raceway than in PVC, and because the resistance of a large conductor is very small, reactance can dominate the impedance of a 500 kcmil feeder. NEC Chapter 9, Table 9 exists for exactly this case: it lists alternating-current resistance and reactance for each size in PVC, aluminium and steel conduit, plus an effective impedance column computed at 0.85 power factor.
Power factor. The K method implicitly assumes unity power factor, and the direction of the resulting error depends on the conductor. Table 9 computes an effective impedance as Ze = R·cosθ + X·sinθ, so dropping the power factor scales the resistive term down while scaling the reactive term up. On a small conductor, where R swamps X, the real drop at 0.85 power factor is lower than the resistance-only figure: 12 AWG copper in PVC is 2.0 Ω per 1,000 ft of resistance against 0.054 Ω of reactance, so Ze = 2.0 × 0.85 + 0.054 × 0.527 = 1.73, against the 1.98 the K method uses. On a 500 kcmil feeder in steel conduit the reactance is larger than the resistance and the same arithmetic runs the other way, pushing the effective impedance well above the resistance alone. Either way, correcting the power factor at the load cuts the current for the same real power and therefore cuts the drop, which is one of the arguments for the power factor correction capacitor calculator.
Temperature. Copper resistance follows R2 = R1(234.5 + T2) ÷ (234.5 + T1). Going from 75 °C to 90 °C is a factor of 324.5 ÷ 309.5 = 1.048, so a conductor running at its 90 °C limit has about 5% more resistance and drops about 5% more volts than this page reports. Using K = 12.9, which already corresponds to 75 °C rather than to 20 °C, builds most of that margin in; a cool conductor drops less, not more.
Where drop sits among the other conductor checks
Sizing a conductor is three independent tests and you take the largest answer. Ampacity asks whether the conductor can carry the current without overheating; that is NEC Table 310.16 with the corrections and adjustments, and it is handled by the wire size and ampacity calculator. Overcurrent protection asks whether the device protecting the conductor is small enough, which for 14, 12 and 10 AWG copper is capped by 240.4(D) regardless of the table ampacity. Voltage drop asks whether enough voltage survives the trip, and it is the test that most often wins on long runs.
The three interact. Upsizing for drop puts more copper in the raceway, so re-run the fill check. Upsizing for drop can also mean the equipment grounding conductor has to grow: 250.122(B) requires the equipment grounding conductor to be increased proportionally when the ungrounded conductors are increased in size for any reason, which the equipment grounding conductor calculator will work out for you. And on a three-phase feeder, checking the three-phase power calculation first will give you the current this page needs.
Key terms
- Circular mil
- The area of a circle 0.001 in. in diameter. A round conductor's area in circular mils is the square of its diameter in mils, which is why the unit exists at all.
- K
- The resistance in ohms of a conductor one foot long and one circular mil in area. Taken as 12.9 for copper and 21.2 for aluminium at about 75 °C.
- Effective impedance
- The combination of resistance and reactance that actually determines drop on an AC circuit. NEC Chapter 9, Table 9 tabulates it at 0.85 power factor for conductors in PVC, aluminium and steel conduit.