NEC calculators

Voltage drop calculator

Voltage drop in volts and percent, the voltage at the load, the longest run within your limit and the smallest conductor that meets it, by the NEC, IEC 60364 or BS 7671.

Calculated with electrician.wiki · https://electrician.wiki/calculators/voltage-drop/

Result · NEC 2023

5.90 V · 2.46 %

40 A over 150 ft of 6 AWG copper at 240 V single-phase drops 5.90 V (2.46 %), within the 3 % limit; 234.1 V reaches the load (NEC 2023, K-factor).

Voltage drop: 5.90 V
Percent (limit 3 %): 2.46 %
Voltage at the load: 234.1 V
Longest run within 3 %: 183 ft
Smallest size for 3 % (drop only): 6 AWG
Impedance used: 0.4916 Ω/kft (K ÷ CM)

Show the work

1. Where K comes from: Chapter 9, Table 8 resistance × area for 6 AWG copper
$K = \frac{R_{75} \cdot CM}{1,000 ft} = \frac{0.491 Ω/kft \cdot 26,240 cmil}{1,000 ft} = 12.88 Ω·cmil/ft$
Every size gives nearly the same product, so the method uses one constant: K = 12.9 for copper at 75 °C.
2. Voltage drop (single-phase, out and back)
$V_{D} = \frac{2 \cdot K \cdot I \cdot L}{CM} = \frac{2 \cdot 12.9 \cdot 40 A \cdot 150 ft}{26,240 cmil} = 5.90 V$
3. Percent of the supply voltage
$%VD = \frac{5.90 V}{240 V} = 2.46 %$
Within the 3 % limit.
4. Voltage at the load
$V_{load} = 240 V - 5.90 V = 234.1 V$
5. Longest run within 3 % (the drop rises in step with length)
$L_{\max} = 150 ft \cdot \frac{3 %}{2.46 %} = 183 ft$
6. Compared with the NEC’s recommended limits (informational notes, not requirements)
Branch circuit, to the farthest outlet: 3 % ✓ · Feeder: 3 % ✓ · Feeder and branch circuit together: 5 % ✓
Branch-circuit note: 210.19 Informational Note; feeder note: 215.2(A)(2) Informational Note No. 2 (NEC 2023).

References: NEC Chapter 9, Table 8 (2023): source of K · NEC 210.19 Informational Note (2023): branch circuits: 3 % · NEC 215.2(A)(2) Informational Note No. 2 (2023): feeders: 3 %, total 5 %

Conductor resistance at 75 °C, as the Chapter 9 tables give it. A lightly loaded conductor runs cooler and drops a little less.
K-factor method: resistance only, no reactance; fine for conductors smaller than about 1/0 AWG, optimistic for large ac feeders.
Two-wire circuit: the current flows out and back, so the length counts twice. For a balanced 120/240 V three-wire circuit, enter 240 V.

New to this? Voltage drop explained: causes, limits and fixes

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How to use the voltage drop calculator

Pick the code you work to first. The four NEC editions use the Chapter 9 conductor tables and US wire sizes; IEC 60364 uses the Annex G formula and metric cable sizes; BS 7671 uses the millivolt-per-amp-per-metre figures in Appendix 4. Then choose the system (dc, single-phase or three-phase), enter the supply voltage, the load current and the one-way length of the run, and the conductor. For three-phase circuits the voltage is the line-to-line value: 208, 400 or 480 V.

The result gives the drop in volts and as a percentage of the supply, the voltage left at the load, the longest run that stays within your limit, and the smallest conductor that would meet the limit at this length. The chart under the inputs plots the drop against length for your size and its neighbours, so you can see how much room you have before the line crosses the limit. If a long run needs more than one conductor per phase, set the number of parallel sets.

The voltage drop formulas

Every method multiplies current by the impedance of the conductors that carry it. A two-wire circuit, dc or single-phase, sends the current out on one conductor and back on the other, so the one-way length counts twice. A balanced three-phase circuit has no return current in the neutral, and the drop between lines is √3 times the drop in one conductor. The NEC K-factor method writes that as:

VD = \frac{2 \cdot K \cdot I \cdot L}{CM}

VD = \frac{\sqrt{3} \cdot K \cdot I \cdot L}{CM}

L is the one-way length in feet, CM the conductor area in circular mils and K the resistance of a conductor one circular mil in area and one foot long: 12.9 Ω for copper and 21.2 Ω for aluminum at 75 °C. On the reference circuit, 40 A over 150 ft of 6 AWG copper at 240 V, that gives 5.90 V, or 2.46 % of the supply. The longest run within 3 % is 183 ft, and the next smaller size, 8 AWG, would drop 3.91 %.

For ac circuits the NEC also gives Chapter 9, Table 9: ac resistance R and inductive reactance X for each size in PVC, aluminum and steel conduit. Only the part of the drop in phase with the supply voltage reduces the voltage at the load, so the table combines them into an effective impedance at the load’s power factor:

Z_{e} = R \cdot PF + X \cdot \sin \cdot (arccos PF)

VD = \frac{2 \cdot I \cdot Z_{e} \cdot L}{1,000}

Table 9 prints Ze only at a power factor of 0.85, rounded to two figures, so the calculator computes it from R and X at whatever power factor you enter. For dc there is no reactance and no skin effect: the drop uses the Table 8 dc resistance, with solid or stranded construction for 18 to 8 AWG.

IEC 60364-5-52 Annex G uses the same physics with standard constants:

u = b \cdot (ρ_{1} \cdot \frac{L}{S} \cdot cos φ + λ \cdot L \cdot sin φ) \cdot I_{B}

b is 2 for single-phase and 1 for three-phase, ρ1 is the resistivity in service (0.0225 Ω·mm²/m for copper, 0.036 for aluminium), λ is a typical reactance of 0.08 mΩ/m, L is in metres and S in mm². The percentage is u divided by the line-to-neutral voltage. BS 7671 skips the formula: Appendix 4 tabulates the drop directly in mV per amp per metre for each cable type, so Vd = (mV/A/m) × Ib × L ÷ 1000.

Where K comes from

K is not a separate code value. It falls out of Chapter 9, Table 8, which lists the dc resistance of every size at 75 °C. For 1/0 copper the table gives 0.122 Ω per 1000 ft and an area of 105,600 cmil, so one foot of a one-circular-mil conductor would measure 12.88 Ω. For aluminum the same size gives 21.23 Ω. Repeat that for every size and the product barely moves:

K = R × area ÷ 1000 ft, from the 75 °C dc resistances of Chapter 9, Table 8 (stranded). The conventional constants are 12.9 for copper and 21.2 for aluminum.
Size	Area (cmil)	Cu R (Ω/kft)	Cu K	Al R (Ω/kft)	Al K
14 AWG	4,110	3.14	12.91	5.17	21.25
12 AWG	6,530	1.98	12.93	3.25	21.22
10 AWG	10,380	1.24	12.87	2.04	21.18
8 AWG	16,510	0.778	12.84	1.28	21.13
6 AWG	26,240	0.491	12.88	0.808	21.20
4 AWG	41,740	0.308	12.86	0.508	21.20
2 AWG	66,360	0.194	12.87	0.319	21.17
1/0 AWG	105,600	0.122	12.88	0.201	21.23
2/0 AWG	133,100	0.0967	12.87	0.159	21.16
4/0 AWG	211,600	0.0608	12.87	0.1	21.16
250 kcmil	250,000	0.0515	12.88	0.0847	21.18
500 kcmil	500,000	0.0258	12.90	0.0424	21.20
1000 kcmil	1,000,000	0.0129	12.90	0.0212	21.20

That is why one constant per metal works. The small spread comes from stranding and rounding in the table. If you need the last few tenths of a percent, use the Chapter 9 tables method, which reads the resistance of the actual size.

K-factor or Table 9: when the difference matters

The K method ignores reactance. Below about 1/0 AWG that hardly matters, because resistance dominates. In large conductors resistance falls with area but reactance barely changes, so it takes over. The table compares the impedance each method uses:

Impedance per conductor that each method uses, copper, ohms per 1000 ft. Table 9 values are computed from its R and X at each power factor.
Size	K ÷ CM	Table 9, PF 1, PVC	PF 0.85, PVC	PF 0.85, steel	Steel PF 0.85 vs K
12 AWG	1.98	2	1.73	1.74	−12 %
6 AWG	0.492	0.49	0.443	0.45	−8 %
2 AWG	0.194	0.19	0.185	0.2	+3 %
1/0 AWG	0.122	0.12	0.125	0.131	+7 %
4/0 AWG	0.061	0.062	0.0743	0.0804	+32 %
250 kcmil	0.0516	0.052	0.0658	0.0733	+42 %
500 kcmil	0.0258	0.027	0.0435	0.0499	+94 %
1000 kcmil	0.0129	0.015	0.0322	0.0395	+206 %

Two things stand out. For small conductors at power factor 0.85, Table 9 gives a lower figure than K, because only 85 % of the resistive drop is in phase with the voltage. For a 500 kcmil copper feeder in steel conduit it is the other way round: 380 A over 250 ft at 480 V drops 4.25 V (0.88 %) by the K method but 8.22 V (1.71 %) from Table 9. The calculator warns you when the K method understates the drop like this. On the reference circuit the two methods agree closely: 5.90 V by K, 5.88 V from Table 9 at power factor 1, and 5.32 V at 0.85.

Worked examples

40 A over 150 ft of 6 AWG copper at 240 V (NEC, K-factor)

A 240 V branch circuit to a detached shop. Is 6 AWG enough for a 40 A load at 150 ft?

1. Where K comes from: Chapter 9, Table 8 resistance × area for 6 AWG copper
$K = \frac{R_{75} \cdot CM}{1,000 ft} = \frac{0.491 Ω/kft \cdot 26,240 cmil}{1,000 ft} = 12.88 Ω·cmil/ft$
2. Voltage drop (single-phase, out and back)
$V_{D} = \frac{2 \cdot K \cdot I \cdot L}{CM} = \frac{2 \cdot 12.9 \cdot 40 A \cdot 150 ft}{26,240 cmil} = 5.90 V$
3. Percent of the supply voltage
$%VD = \frac{5.90 V}{240 V} = 2.46 %$
4. Voltage at the load
$V_{load} = 240 V - 5.90 V = 234.1 V$
5. Longest run within 3 % (the drop rises in step with length)
$L_{\max} = 150 ft \cdot \frac{3 %}{2.46 %} = 183 ft$
6. Compared with the NEC’s recommended limits (informational notes, not requirements)
Branch circuit, to the farthest outlet: 3 % ✓ · Feeder: 3 % ✓ · Feeder and branch circuit together: 5 % ✓

40 A over 150 ft of 6 AWG copper at 240 V single-phase drops 5.90 V (2.46 %), within the 3 % limit; 234.1 V reaches the load (NEC 2023, K-factor).

A 480 V feeder in steel conduit (NEC, Table 9)

380 A on 500 kcmil copper, 250 ft, power factor 0.85. Here reactance makes up about half of the impedance.

1. Effective impedance at PF 0.85 (Chapter 9, Table 9, steel conduit, copper)
$Z_{e} = R \cdot PF + X \cdot \sin \cdot (arccos PF) = 0.029 Ω/kft \cdot 0.85 + 0.048 Ω/kft \cdot 0.5268 = 0.04994 Ω/kft$
2. Voltage drop (three-phase, line to line)
$V_{D} = \frac{\sqrt{3} \cdot I \cdot Z_{e} \cdot L}{1,000} = \frac{\sqrt{3} \cdot 380 A \cdot 0.04994 Ω/kft \cdot 250 ft}{1,000} = 8.22 V$
3. Percent of the supply voltage
$%VD = \frac{8.22 V}{480 V} = 1.71 %$
4. Voltage at the load
$V_{load} = 480 V - 8.22 V = 471.8 V$
5. Longest run within 3 % (the drop rises in step with length)
$L_{\max} = 250 ft \cdot \frac{3 %}{1.71 %} = 438 ft$
6. Compared with the NEC’s recommended limits (informational notes, not requirements)
Branch circuit, to the farthest outlet: 3 % ✓ · Feeder: 3 % ✓ · Feeder and branch circuit together: 5 % ✓

380 A over 250 ft of 500 kcmil copper at 480 V three-phase drops 8.22 V (1.71 %), within the 3 % limit; 471.8 V reaches the load (NEC 2023, Chapter 9 Table 9, PF 0.85).

A 12 V dc run (NEC, Table 8)

15 A for a 12 V pump 20 ft from the battery on 10 AWG copper. Low voltages turn small drops into large percentages.

1. DC resistance at 75 °C (Chapter 9, Table 8, 10 AWG copper, stranded)
$R_{75} = 1.24 Ω/kft$
2. Voltage drop (dc, out and back)
$V_{D} = \frac{2 \cdot I \cdot R_{75} \cdot L}{1,000} = \frac{2 \cdot 15 A \cdot 1.24 Ω/kft \cdot 20 ft}{1,000} = 0.74 V$
3. Percent of the supply voltage
$%VD = \frac{0.74 V}{12 V} = 6.20 %$
4. Voltage at the load
$V_{load} = 12 V - 0.74 V = 11.26 V$
5. Longest run within 3 % (the drop rises in step with length)
$L_{\max} = 20 ft \cdot \frac{3 %}{6.20 %} = 9.68 ft$
6. Compared with the NEC’s recommended limits (informational notes, not requirements)
Branch circuit, to the farthest outlet: 3 % ✗ · Feeder: 3 % ✗ · Feeder and branch circuit together: 5 % ✗

15 A over 20 ft of 10 AWG copper at 12 V dc drops 0.74 V (6.20 %), over the 3 % limit; 11.26 V reaches the load (NEC 2023, Chapter 9 Table 8).

32 A over 30 m of 6 mm² copper at 230 V (IEC 60364 Annex G)

A single-phase final circuit with cos φ = 0.8, the value Annex G uses when the load’s power factor is unknown.

1. Annex G constants
b = 2 (single-phase); ρ1 = 0.0225 Ω·mm²/m (copper, 1.25 × the 20 °C value); λ = 0.08 mΩ/m; cos φ = 0.8, sin φ = 0.6
2. Voltage drop, IEC 60364-5-52 Annex G
$u = b \cdot (ρ_{1} \cdot \frac{L}{S} \cdot cosφ + λ \cdot L \cdot sinφ) \cdot I_{B} = 2 \cdot (0.0225 \cdot \frac{30 m}{6 mm²} \cdot 0.8 + 0.00008 Ω/m \cdot 30 m \cdot 0.6) \cdot 32 A = 5.85 V$
3. Percent of the supply voltage
$Δu = 100 \cdot \frac{u}{U_{0}} = 100 \cdot \frac{5.85 V}{230 V} = 2.54 %$
4. Voltage at the load
$V_{load} = 230 V - 5.85 V = 224.1 V$
5. Longest run within 5 % (the drop rises in step with length)
$L_{\max} = 30 m \cdot \frac{5 %}{2.54 %} = 59 m$
6. Compared with IEC 60364-5-52 Table G.52.1
Lighting, public LV supply: 3 % ✓ · Other uses, public LV supply: 5 % ✓ · Lighting, private LV supply: 6 % ✓ · Other uses, private LV supply: 8 % ✓

32 A over 30 m of 6 mm² copper at 230 V single-phase drops 5.85 V (2.54 %), within the 5 % limit; 224.1 V reaches the load (IEC 60364-5-52, Annex G).

A 32 A cooker circuit in 6 mm² twin and earth (BS 7671)

25 m of 6242Y flat twin and earth, using the Table 4D5 figure of 7.3 mV/A/m.

1. BS 7671 Table 4D5: 6 mm², single-phase
7.3 mV/A/m
2. Voltage drop
$V_{d} = \frac{mV/A/m \cdot I_{b} \cdot L}{1,000} = \frac{7.3 \cdot 32 A \cdot 25 m}{1,000} = 5.84 V$
3. Percent of the supply voltage
$%VD = \frac{5.84 V}{230 V} = 2.54 %$
4. Voltage at the load
$V_{load} = 230 V - 5.84 V = 224.2 V$
5. Longest run within 5 % (the drop rises in step with length)
$L_{\max} = 25 m \cdot \frac{5 %}{2.54 %} = 49.2 m$
6. Compared with BS 7671 Appendix 4, section 6.4
Lighting, public LV supply: 3 % ✓ · Other uses, public LV supply: 5 % ✓ · Lighting, private LV supply: 6 % ✓ · Other uses, private LV supply: 8 % ✓

32 A over 25 m of 6 mm² twin and earth at 230 V single-phase drops 5.84 V (2.54 %), within the 5 % limit; 224.2 V reaches the load (BS 7671, Table 4D5, tabulated mV/A/m).

How far each wire size can run

The drop rises in proportion to length, so every circuit has a longest run for a given limit. These figures come from the same engine as the calculator:

Longest one-way run in feet that stays within 3 %, copper, K-factor method, with the load at 80 % of the breaker rating (the most a continuous load may draw).
Wire	Breaker	Load	120 V single-phase	240 V single-phase	208 V three-phase	480 V three-phase
14 AWG	15 A	12 A	48	96	96	221
12 AWG	20 A	16 A	57	114	114	263
10 AWG	30 A	24 A	60	121	121	279
8 AWG	40 A	32 A	72	144	144	333
6 AWG	60 A	48 A	76	153	153	352
4 AWG	70 A	56 A	104	208	208	480
2 AWG	100 A	80 A	116	231	232	535
1/0 AWG	125 A	100 A	147	295	295	681
2/0 AWG	150 A	120 A	155	310	310	715
4/0 AWG	200 A	160 A	185	369	369	852

Doubling the voltage doubles the run for the same load, which is why 240 V circuits reach twice as far as 120 V circuits on the same wire. On the UK side, the same exercise with BS 7671 twin and earth:

Flat twin and earth (BS 7671 Table 4D5) at 230 V, loaded to the protective device rating: longest run within the Appendix 4 limit for the use.
Circuit	Cable	mV/A/m	Limit	Longest run
Lighting circuit, 6 A	1.5 mm²	29	3 %	39.7 m
Radial socket circuit, 20 A	2.5 mm²	18	5 %	31.9 m
Radial socket circuit, 32 A	4 mm²	11	5 %	32.7 m
Cooker circuit, 32 A	6 mm²	7.3	5 %	49.2 m
Electric shower, 45 A	10 mm²	4.4	5 %	58.1 m
Sub-main, 63 A	16 mm²	2.8	5 %	65.2 m

How much voltage drop is allowed?

The NEC recommends no more than 3 % on a branch circuit to the farthest outlet, 3 % on a feeder, and 5 % for feeder and branch circuit together. These are informational notes, so they are advice, not enforceable rules, and their location moved in 2023 and again in 2026:

Where the NEC’s voltage-drop recommendations sit in each edition
Edition	Branch circuits (3 %)	Feeders (3 %, 5 % total)
NEC 2017	210.19(A)(1) Informational Note No. 4	215.2(A)(1) Informational Note No. 2
NEC 2020	210.19(A)(1) Informational Note No. 3	215.2(A)(1) Informational Note No. 2
NEC 2023	210.19 Informational Note	215.2(A)(2) Informational Note No. 2
NEC 2026	210.19 Informational Note	215.4(A)(2) Informational Note No. 2

A few articles do make voltage drop mandatory. On a fire pump circuit the voltage at the controller’s line terminals may not fall more than 15 % while the motor starts, and the voltage at its load terminals may not fall more than 5 % below the motor rating while the motor runs at 115 % of full-load current (695.7). Sensitive electronic equipment systems under Article 647 are held to 1.5 % on a branch circuit and 2.5 % for feeder and branch circuit together (647.4(D)). Your engineer or the equipment maker may also set tighter limits; motors, LED drivers and EV chargers all have a voltage range on their data sheets.

IEC 60364-5-52 (Table G.52.1) and BS 7671 (Appendix 4, section 6.4) give the same figures for installations fed from a public low-voltage network: 3 % for lighting and 5 % for other uses. An installation with its own transformer may run to 6 % and 8 %. Where the main wiring is longer than 100 m, both allow an extra 0.005 % per metre beyond 100 m, up to 0.5 %. Motor starting and other short inrush events fall outside these limits, as long as the equipment still works within its own voltage range.

Common mistakes

Entering the round-trip length. The formulas already count the return conductor. Enter the one-way distance from the panel to the load.
Using 120 V for a 240 V load. The percentage is of the voltage the load sees. A 240 V water heater on a 120/240 V service is a 240 V circuit.
Mixing line and phase voltages on three-phase. The √3 formulas give the line-to-line drop, so divide by the line-to-line voltage (480 V, not 277 V).
Trusting the K method on large ac feeders. It leaves out reactance, which matters from about 1/0 AWG and dominates in steel conduit at 500 kcmil and above.
Sizing for voltage drop alone. A size that meets 3 % may still be too small for the current. The calculator flags it; size for ampacity first with the wire size calculator, then check the drop.
Forgetting the grounding conductor. Upsizing the phase conductors for voltage drop makes the NEC equipment grounding conductor grow in proportion. The ground wire size calculator does that arithmetic.
Using the BS 7671 figure without the power factor. The tabulated (mV/A/m)z is a worst case. For cables of 25 mm² and up, the load power factor gives a closer answer.

Method and verification

The NEC methods use Chapter 9, Tables 8 and 9 as printed, unchanged from 2017 to 2026, at 75 °C. The K method uses the conventional 12.9 and 21.2. IEC results use the Annex G constants. BS 7671 results use Tables 4D5, 4D1B and 4E4B. Several BS 7671 resistive and reactive components for 25 mm² and larger come from a single published reproduction, while most of the tabulated z values are confirmed by a second manufacturer. The calculator names the single-source value whenever a result uses one. The Annex G constants carry the same kind of note until we confirm them against a second source.

The test suite checks hand-worked cases for each method: 5.90 V for the reference circuit, a Table 9 drop of 5.32 V at power factor 0.85, Annex G’s 5.85 V for 32 A over 30 m of 6 mm², and 5.84 V for the BS 7671 cooker circuit. It also checks that every printed Table 9 impedance is reproduced from R and X within the table’s rounding, and the properties every answer must have: longer runs drop more, larger conductors drop less, parallel sets divide the drop, and feet and metres give the same percentage. For the 52 A version of the reference circuit the smallest copper within 3 % is 4 AWG.

Updated 2026-10-14. Found a problem? Report it; every fix is logged publicly.

Questions people ask

How do you calculate voltage drop?

Multiply the current by the impedance of the conductors, counting both directions on a two-wire circuit. With the NEC K-factor method: VD = 2 × K × I × L ÷ CM for single-phase and √3 × K × I × L ÷ CM for three-phase, with K = 12.9 for copper. 40 A over 150 ft of 6 AWG copper drops 5.90 V, 2.46 % at 240 V.

What is the maximum voltage drop allowed by the NEC?

The NEC recommends 3 % for a branch circuit and 5 % for feeder and branch circuit combined, in informational notes that are not enforceable on their own. Fire pumps (695.7) and sensitive electronic equipment (647.4(D)) have mandatory limits, and local codes or specifications can make the 3 % figure a requirement.

Is the length one-way or round trip?

One-way. Enter the distance from the source to the load; the factor 2 (single-phase, dc) or √3 (three-phase) in the formula accounts for the return path.

What is K in the voltage drop formula?

The resistance of a copper or aluminum conductor one circular mil in area and one foot long at 75 °C: 12.9 Ω for copper and 21.2 Ω for aluminum. It comes from Chapter 9, Table 8: resistance per 1000 ft times area, divided by 1000 (1/0 copper gives 12.88).

Should I use the K-factor or Chapter 9 Table 9?

For conductors smaller than about 1/0 AWG either is fine. For larger ac feeders use Table 9 with the real power factor, because the K method leaves out reactance; for 500 kcmil copper in steel conduit it understates the drop by roughly half.

What voltage drop does BS 7671 allow?

Appendix 4 (section 6.4) recommends 3 % for lighting and 5 % for other uses when the installation is supplied from the public network, and 6 % and 8 % from a private supply. A 32 A cooker on 25 m of 6 mm² twin and earth drops 5.84 V, 2.54 %.

Does a bigger wire always fix voltage drop?

A bigger conductor lowers resistance in proportion to its area, but on large ac feeders the reactance stays almost the same, so upsizing gains less and less. Past that point, parallel sets, a higher voltage or a shorter route help more.