Voltage Drop — Formulas, NEC Rules, and Worked Examples

Voltage drop is one of the few NEC topics where the code book gives you a recommendation, not a hard rule — and licensing exams test whether you know the formulas, the constants, and what to do when the math forces you to upsize.

What the NEC actually says

Voltage drop is addressed in two informational notes, not in mandatory text:

"Informational Note" means it's a recommendation, not a code requirement. But most exams treat the 3% / 5% values as the answer, and many state amendments and AHJs adopt them as enforceable. For sensitive equipment (motors, electronic loads), the manufacturer's tolerance is usually tighter than 5%.

One section does impose a hard limit: 647.4(D), sensitive electronic equipment on technical-power systems, requires drop ≤ 1.5% on branch circuits and ≤ 2.5% combined. And NEC 647.4(D) applies only to those technical systems.

The formulas

Two formulas cover almost everything you'll see on the journeyman exam:

Single-phase, two-wire: VD = (2 × K × I × D) / CM
Three-phase, balanced: VD = (1.732 × K × I × D) / CM

Where:

The "K = 12.9" assumes 75 °C operating temperature. At 60 °C, copper K drops to about 11; at 90 °C, it rises to about 14. Most exam problems use 12.9 unless told otherwise.

Circular mils — the lookup you actually need

From Chapter 9, Table 8 (DC resistance, but the CM column is what we use):

Size (AWG/kcmil)Circular mils
144,110
126,530
1010,380
816,510
626,240
441,740
266,360
1/0105,600
2/0133,100
4/0211,600
250 kcmil250,000
500 kcmil500,000

Worked example 1 — 12 AWG Cu, 20 A, 100 ft (single-phase 120 V)

Run a 20 A general-purpose receptacle circuit 100 feet to a detached garage on 12 AWG Cu THWN at 120 V single-phase. What's the voltage drop?

VD = (2 × 12.9 × 20 × 100) / 6,530 VD = 51,600 / 6,530 VD = 7.9 V

7.9 V on 120 V = 6.6%. That blows the 3% branch-circuit recommendation (3.6 V) and even the 5% combined limit. We need to upsize.

Back-solve for the CM we need at 3% drop (3.6 V):

CM ≥ (2 × 12.9 × 20 × 100) / 3.6 CM ≥ 14,333

10 AWG (10,380 CM) is too small. 8 AWG (16,510 CM) clears it. Use 8 AWG Cu if you want to stay under 3% drop on this run.

Worked example 2 — 4/0 Al, 100 A, 250 ft (three-phase 480 V feeder)

A feeder runs 250 ft from a service to a sub-panel at 480 V three-phase, carrying 100 A continuous. Conductor: 4/0 aluminum.

VD = (1.732 × 21.2 × 100 × 250) / 211,600 VD = 918,160 / 211,600 VD = 4.34 V

4.34 V on 480 V = 0.90%. Comfortably under the 3% feeder recommendation. 4/0 Al is fine on this feeder run.

If the same 100 A load were on a 100 ft branch circuit downstream — say, 4 AWG Al for the branch (41,740 CM):

VD_branch = (1.732 × 21.2 × 100 × 100) / 41,740 VD_branch = 367,264 / 41,740 VD_branch = 8.80 V → 1.83%

Combined 0.90% + 1.83% = 2.73% — under the 5% combined limit. Both runs comply with the NEC recommendation.

Worked example 3 — back-solving for wire size

You need to feed a 50 A motor at 240 V single-phase, 200 ft from the panel, on aluminum. What's the smallest conductor that keeps you under 3% drop (7.2 V)?

CM ≥ (2 × 21.2 × 50 × 200) / 7.2 CM ≥ 424,000 / 7.2 CM ≥ 58,888

Look up CM ≥ 58,888 in the table. 2 AWG Al is too small (66,360 CM is for copper — but in Table 8 for aluminum the same AWG sizes have the same circular-mil area; the K constant captures the resistance difference). 2 AWG (66,360 CM) clears it. Use 2 AWG aluminum for the 50 A motor feeder.

Common pitfalls

Quick-reference cheat sheet

MaterialK @ 75 °CUsed in
Copper12.9Most branch circuits, sensitive loads
Aluminum21.2Service entrance, large feeders, utility laterals
RecommendationBranch cktCombined
NEC IN (210.19, 215.2)≤ 3%≤ 5%
647.4 (sensitive equipment)≤ 1.5%≤ 2.5%
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