NEC Voltage Drop Calculation for HVAC Circuits (With Worked Example)

Get this straight before an exam or an inspector conversation: the NEC does not have a general mandatory voltage drop limit. Informational Note No. 4 to 210.19(A) recommends sizing branch-circuit conductors so the drop doesn't exceed 3% at the farthest outlet, with the combined drop on feeder plus branch circuit held to 5% total. Section 215.2(A) carries the mirror-image note for feeders. Under NEC 90.5(C), informational notes are explanatory only — they are not enforceable code text.

Three caveats. First, some jurisdictions adopt the 3%/5% figures as a mandatory local amendment, and plenty of engineered jobs write them into the spec — then they're binding on you either way. Second, a few articles do impose hard voltage drop limits, notably fire pump circuits (Article 695) and sensitive electronic equipment (Article 647) — read those sections when they apply. Third, physics doesn't care about enforceability: a compressor starting on a soft circuit pulls locked-rotor current longer, chatters contactors, and trips overloads on the hot afternoons when utility voltage is already sagging.

On a 240 V single-phase condenser circuit, 3% is 7.2 V — meaning at least 232.8 V at the disconnect under load. The 5% total budget is 12 V.

The standard exam and field method:

VD = (2 × K × I × D) / CM

Term	What it is	Unit
VD	Voltage drop	volts
2	Round trip — current flows out and back	—
K	Resistivity constant: 12.9 copper, 21.2 aluminum	ohm-cmil/ft
I	Load current	amps
D	One-way circuit length	feet
CM	Conductor area	circular mils

K is derived from the DC resistance values in NEC Chapter 9, Table 8 at a 75°C conductor temperature. The most common mistake: D is the one-way distance — the 2 in the formula already handles the return path. Double both and your answer is twice reality.

For three-phase circuits, swap the 2 for 1.732 (√3): VD = (1.732 × K × I × D) / CM.

Percent drop = VD ÷ system voltage × 100. Two rearrangements worth memorizing for exams:

• Minimum conductor size: CM = (2 × K × I × D) / VD_max
• Maximum run for a given wire: D_max = (VD_max × CM) / (2 × K × I)

The K method is an approximation — Chapter 9, Table 9 AC impedance matters on large conductors and steel raceway — but for branch circuits through 4/0 AWG it lands within a couple percent, and it's the method licensing exams expect.

These are the CM values the formula needs. Source: NEC Chapter 9, Table 8, Conductor Properties.

Size (AWG)	Circular mils
14	4,110
12	6,530
10	10,380
8	16,510
6	26,240
4	41,740
3	52,620
2	66,360
1	83,690
1/0	105,600
2/0	133,100
3/0	167,800
4/0	211,600

For HVAC work, 12 through 6 AWG covers most residential condensers and air handlers — worth committing those four to memory. Each AWG size step is roughly a 26% change in circular mils, which is why one size up buys you meaningful drop reduction.

The job: a condenser with a nameplate minimum circuit ampacity (MCA) of 28 A, fed at 240 V single-phase, with a 120 ft one-way run from the panel to the disconnect. Planned conductor: 10 AWG copper THWN-2.

Step 1 — Ampacity first. Voltage drop never overrides ampacity. NEC Table 310.16 gives 10 AWG copper at the 75°C column an ampacity of 35 A. 35 A ≥ 28 A MCA, so 10 AWG passes ampacity. Now check drop.

Step 2 — Gather inputs. K = 12.9 (copper), I = 28 A, D = 120 ft, CM = 10,380 (10 AWG, Chapter 9 Table 8).

Step 3 — Numerator. 2 × 12.9 × 28 × 120 = 86,688.

Step 4 — Divide by CM. 86,688 ÷ 10,380 = 8.35 V of drop.

Step 5 — Percent. 8.35 ÷ 240 = 3.48%. Over the 3% recommendation — the condenser would see about 231.6 V at design load.

Step 6 — Upsize and recompute. Try 8 AWG (16,510 CM): 86,688 ÷ 16,510 = 5.25 V, which is 2.19%. Inside the 3% target, with 234.75 V at the unit.

So the run that a load calc alone would put on 10 AWG lands on 8 AWG copper once distance enters the math. That crossover is the whole reason this calculation exists — ampacity tables know nothing about how far the condenser sits from the panel.

Upsize when the computed drop at design load exceeds 3% on the branch circuit (or blows the 5% total budget once the feeder's share is counted). Using nameplate MCA as I is slightly conservative — actual running load is typically RLA plus the condenser fan — which is fine; the margin covers hot-day amp draw.

A useful pre-check is the max-distance rearrangement. For this 28 A load on 10 AWG copper at 240 V: D_max = (7.2 × 10,380) / (2 × 12.9 × 28) = 103 ft. Past about 103 ft one-way, 10 AWG is out of headroom for this load and the calculation is really just confirming what you already know.

Three things ride along with an upsize:

• The equipment grounding conductor scales too. NEC 250.122(B): when ungrounded conductors are increased in size for voltage drop, the EGC must be increased proportionately by circular-mil area.
• The breaker does not change. Max overcurrent protection comes off the nameplate (MOCP). Bigger wire for voltage drop never means a bigger breaker.
• Check terminations. Confirm the disconnect and contactor lugs accept the larger conductor, and that you're still using the 75°C ampacity column only if the terminations are rated for it.

For aluminum, K = 21.2 — about 1.64× copper's resistivity — so matching copper's drop takes roughly two AWG sizes up. Run the same formula with 21.2 and aluminum's CM before assuming.