Voltage Drop Limits in NEC 210.19: Recommendation vs. Requirement and When It Controls Wire Size
The inspector marks up your permit drawings: “Voltage drop exceeds 3% — upsize conductors.” You look up NEC 210.19 and find the language is in an Informational Note, not a mandatory code section. The inspector’s red mark is correct. The NEC is also correct. Both things are true simultaneously, and understanding why determines whether you can push back on an upsize requirement or whether you need to reroute the run.
What NEC 210.19 Actually Says
Related: service entrance conductor sizing for 200A service covers the similar calculation for service-level conductors where 310.12 changes the math.
NEC 210.19(A) sets the minimum ampacity and size for branch circuit conductors. The voltage drop language appears in Informational Note No. 4 to that section:
Source: NFPA 70 (National Electrical Code), 2023 edition, Section 210.19(A) Informational Note No. 4.
“Conductors for branch circuits as defined in Article 100, sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating, and lighting loads, or combinations of such loads, and where the maximum total voltage drop on both feeders and branch circuits to the farthest outlet does not exceed 5 percent, provide reasonable efficiency of operation.”
Informational Notes are not requirements. NFPA 70 (the NEC) defines Informational Notes in Section 90.5: “Informational notes are informational only and are not enforceable as requirements.” So the 3% and 5% values are recommendations for “reasonable efficiency of operation,” not mandatory limits in the NEC itself.
The same structure applies to NEC 215.2(A)(1) Informational Note No. 2, which has parallel language for feeders: 3% voltage drop recommended for feeders, 5% total (feeder + branch circuit combined).
Some AHJs adopt the Informational Note recommendations as mandatory local amendments to the NEC. Before a voltage drop dispute with an inspector, check whether your jurisdiction has adopted the 3% / 5% limits as a mandatory code amendment rather than an informational note. Several large cities have done exactly this. The state building code amendments are the definitive source, not the base NEC text.
When Voltage Drop Controls the Wire Size
Two factors make voltage drop the controlling calculation rather than ampacity: long runs and light loads. A 20-amp circuit serving a receptacle outlet 200 feet from the panel may be adequately sized from an ampacity standpoint with 12 AWG (20A at 75°C per NEC Table 310.16) but fail the 3% voltage drop recommendation when fully loaded.
The voltage drop formula for a single-phase circuit:
$$VD = \frac{2 \times K \times I \times L}{CM}$$where:
- K = resistivity constant: 12.9 for copper, 21.2 for aluminum
- I = current in amperes
- L = one-way length of run in feet
- CM = conductor cross-sectional area in circular mils
The “2 ×” in the formula accounts for the round-trip path (current travels out through the hot conductor and returns through the neutral). One-way length is the distance from the panel to the farthest outlet, not the total wire length. This is the most common calculation error — using total wire length doubles the computed voltage drop.
K = 12.9 (Cu) or 21.2 (Al); I in amps; L = one-way distance in feet; CM = circular mils from NEC Chapter 9, Table 8
Worked Example — 200-Foot Run to a 20A Receptacle
Circuit: 120V, 20A branch circuit, 200 ft one-way run, copper conductors, 12 AWG.
From NEC Chapter 9, Table 8: 12 AWG solid copper = 6,530 CM; 12 AWG stranded = 6,530 CM (same for this calculation).
$$VD = \frac{2 \times 12.9 \times 20 \times 200}{6530} = \frac{103,200}{6530} = 15.8 \text{ V}$$Percentage voltage drop on a 120V circuit: 15.8 / 120 × 100 = 13.2%. That’s well above the 3% recommendation of 3.6V on a 120V circuit. The 12 AWG conductor is sized correctly for 20A ampacity but fails the efficiency recommendation significantly at full load over this run length.
Rearrange the formula to solve for required CM:
$$CM_{required} = \frac{2 \times K \times I \times L}{VD_{max}} = \frac{2 \times 12.9 \times 20 \times 200}{3.6} = \frac{103,200}{3.6} = 28,667 \text{ CM}$$From NEC Table 8: 10 AWG copper = 10,380 CM (not enough), 8 AWG = 16,510 CM (not enough), 6 AWG = 26,240 CM (close but still short), 4 AWG = 41,740 CM (adequate). 4 AWG copper for a 20A receptacle circuit that’s 200 feet from the panel to maintain ≤3% voltage drop at full load.
Four gauge wire for a 20A receptacle is a significant cost premium. This is where the engineering decision matters: is this circuit expected to operate at or near 20A continuously? If the actual connected load is an intermittent 10A, the voltage drop at 10A with 12 AWG is 6.6V (5.5%) — still outside the 3% branch circuit recommendation, but the 5% combined limit would allow this if the feeder voltage drop is minimal. At 5A (25% load), 12 AWG gives 3.3V drop (2.75%) — within 3%.
The voltage drop calculator runs this calculation across the full load range so you can evaluate the actual operating point against the recommendation rather than assuming worst-case loading for every circuit.
The 3% Branch / 5% Combined Rule in Practice
The 5% combined limit (feeder + branch circuit) gives you flexibility when the feeder voltage drop is minimal. If the panel serving the 200-ft circuit is a subpanel with a very short feeder from the main panel, the feeder voltage drop might be 0.5%. That leaves 4.5% of the 5% combined budget available for the branch circuit — and a 10 AWG conductor at that 200-ft run produces roughly 7.9V drop (6.6%), still over 5% at full load. At typical loads (10–15A for a general-purpose receptacle in commercial use), the numbers improve.
The practical guidance most AHJs follow even where the 3%/5% is advisory:
- Lighting circuits: the 3% limit is important because voltage affects light output; most lighting specs call it out
- Motor circuits: voltage drop affects motor torque and starting current; NEC 430.14(A) has specific requirements about conductor sizing for motors
- General-purpose receptacle circuits: 3% is a guideline; actual loading rarely reaches 100% of the OCPD rating on a continuous basis
Three-Phase Voltage Drop Calculation
For three-phase circuits, the formula changes — the factor 1.732 (√3) replaces the factor of 2:
$$VD_{3\phi} = \frac{1.732 \times K \times I \times L}{CM}$$Three-phase circuits have an inherent voltage drop advantage over single-phase: the three-phase factor produces about 13% lower voltage drop for the same conductor size and length compared to single-phase at the same current. This is one of the reasons three-phase distribution is used for long-run feeders in commercial and industrial settings.
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