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Voltage drop and AWG wire gauge calculator

Voltage drop over a DC, single-phase or three-phase cable run, plus AWG wire size to diameter, mm², kcmil and resistance per km, and back.

Updated Checked against 6 worked examples

V
Try
Voltage drop
%
Voltage drop: 3.91 %
Shown to 2 decimal places, half-even
Voltage drop
4.68962V
Voltage at the load
115.31V
Power lost in the wire
70.3444W
Nearest AWG
12 AWG
Diameter
2.05253mm / 0.0808081in
Cross-section area
3.30877mm² / 6.52995kcmil
Resistance at 20 °C
5.211Ω/km / 1.588Ω/kft
Resistance is for solid copper at 20 °C; at a 75 °C conductor temperature it is about 22% higher.
Above the 3% branch-circuit guidance (NEC 210.19(A) note) but within 5% overall.

30 m of 12 AWG copper carrying 15 A drops 4.69 V (3.91% of 120 V), leaving 115.31 V at the load and wasting 70.34 W as heat.

Voltage drop against the usual limits

≤ 3% branch3–5%> 5%035123.91%
How it's calculated S
  1. Diameter from the gauge number

    d=0.005 in×92(36−(12))/39=0.0808081 in=2.05253 mmd = 0.005\,\text{in}\times 92^{(36 - (12))/39} = 0.0808081\ \mathrm{in} = 2.05253\ \mathrm{mm}
  2. Area and resistance

    A=πd24=3.30877 mm2,R′=ρA=1.7241×10−80.00000330877=5.21069 Ω/kmA = \frac{\pi d^2}{4} = 3.30877\ \mathrm{mm^2},\quad R' = \frac{\rho}{A} = \frac{1.7241 \times 10^{-8}}{0.00000330877} = 5.21069\ \mathrm{\Omega/km}
  3. Voltage drop

    ΔV=2 IR′L=2(15)(0.00521069)(30)=4.68962 V\Delta V = 2\,I R' L = 2(15)(0.00521069)(30) = 4.68962\ \mathrm{V}

    The current flows out and back, so the wire length counts twice; reactance is ignored.

  4. As a share of the supply

    4.68962120×100=3.908%\frac{4.68962}{120}\times100 = 3.908\%
  5. Power lost in the conductors

    P=2 I2R′L=70.3444 WP = 2\,I^2R'L = 70.3444\ \mathrm{W}

About the voltage drop and AWG wire gauge calculator

Voltage drop is the voltage lost along a cable because the conductor has resistance. The calculator takes the resistance per metre from the cross-section (R′ = ρ/A, with ρ at 20 °C) and multiplies it by the current and the one-way length: twice for DC and single-phase, because the current goes out and back, and by √3 for a balanced three-phase run. It also converts any AWG size to diameter, mm², kcmil and ohms per km with the ASTM B258 formula, or finds the nearest AWG for a measured diameter.

The default, 15 A over 30 m of 12 AWG copper at 120 V, drops 4.69 V, or 3.91%. That is above the 3% branch-circuit figure in the NEC informational note; 10 AWG brings the same run down to 2.46%.

The result is a resistive estimate at 20 °C. Copper's resistance is about 22% higher at a 75 °C operating temperature, and long AC runs of large cables add reactance, so leave a margin when the result is close to a limit.

Worked examples

12 AWG copper

Calculate
AWG → size
Wire gauge (AWG)
12 AWG
Conductor
Copper
Diameter
2.05253 mm
Cross-section area
3.30877 mm²
Resistance at 20 °C
5.211 Ω/km

Checked against: Python 3.8 math: 0.005 in × 92^(24/39) = 2.0525254 mm; ρ/A with ρ = 1.7241e-8 Ω·m (AWG table: 2.053 mm, 3.31 mm², 5.211 Ω/km)

4/0 AWG is exactly 0.46 in

Calculate
AWG → size
Wire gauge (AWG)
4/0 (0000)
Conductor
Copper
Diameter
0.46 in
Cross-section area
211.6 kcmil

Checked against: ASTM B258 definition: 4/0 = 0.4600 in = 211.6 kcmil

36 AWG is exactly 0.005 in

Calculate
AWG → size
Wire gauge (AWG)
36 AWG
Conductor
Copper
Diameter
0.127 mm
Cross-section area
0.025 kcmil

Checked against: ASTM B258 definition: 36 AWG = 0.0050 in

2.05 mm wire

Calculate
Diameter → AWG
Conductor diameter
2.05 mm
Conductor
Copper
Nearest AWG
12 AWG
Exact gauge number
12.01

Checked against: Python 3.8 math: n = 36 − 39·log₉₂(2.05/0.127) = 12.0106

Questions

What is the maximum allowed voltage drop?

In the US, an informational note to NEC 210.19 recommends at most 3% on a branch circuit and 5% in total for feeder plus branch circuit. The note is advice, not an enforceable rule, although some jurisdictions adopt it. IEC 60364-5-52 and BS 7671 give 3% for lighting and 5% for other loads when the installation is fed from a public low-voltage network, and 6% and 8% from a private supply.

How do you calculate voltage drop in a wire?

Multiply the current by the resistance of the wire that carries it. For DC or single-phase, ΔV = 2 × I × R′ × L, where R′ is the resistance per metre and L the one-way length; the 2 counts the outgoing and return conductors. For balanced three-phase, use √3 (1.732) instead of 2 and compare the result with the line-to-line voltage. 15 A over 30 m of 12 AWG copper (5.21 Ω/km) drops 4.69 V.

How does the AWG number relate to wire diameter?

ASTM B258 defines AWG by a formula: 36 AWG is 0.005 in and 4/0 (0000) is 0.46 in, with 39 equal ratio steps between them, so each gauge number changes the diameter by a factor of 1.1229. Three gauge numbers double or halve the cross-sectional area, and six double or halve the diameter. 12 AWG is 2.053 mm across with 3.31 mm² of metal.

Why is this resistance lower than in NEC Chapter 9 Table 8?

Table 8 lists DC resistance at 75 °C, while this calculator uses solid conductors at 20 °C with copper's resistivity from IEC 60028 (0.017241 Ω·mm²/m). For solid 12 AWG copper, Table 8 gives 1.93 Ω per 1000 ft against 1.59 here; the whole difference is the 55 °C of extra temperature. Stranded 12 AWG is listed slightly higher, at 1.98 Ω per 1000 ft.

How much bigger does an aluminium wire need to be than copper?

About 1.6 times the cross-section, or roughly two AWG sizes larger, for the same resistance and voltage drop. Electrical-grade aluminium conducts 61% as well as copper (IEC 60889), with a resistivity of 0.028264 Ω·mm²/m against copper's 0.017241. The final size must also meet the ampacity tables, and aluminium terminations need connectors rated for it.

How accurate is the voltage drop and AWG wire gauge calculator?

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 6 worked examples whose answers come from independent sources; for example, “12 AWG copper” is checked against Python 3.8 math: 0.005 in × 92^(24/39) = 2.0525254 mm; ρ/A with ρ = 1.7241e-8 Ω·m (AWG table: 2.053 mm, 3.31 mm², 5.211 Ω/km).

Where does the method come from?

ASTM B258 — Standard specification for standard nominal diameters and cross-sectional areas of AWG sizes; IEC 60028 — International standard of resistance for copper (0.017241 Ω·mm²/m at 20 °C); NFPA 70 (NEC) 210.19(A) Informational Note — 3% branch / 5% total voltage drop guidance; Wikipedia — American wire gauge (table of diameters, areas and copper resistance).

About this calculator

d=0.005 in×92(36−n)/39,R′=ρA,ΔV=k IR′L, k=2 (DC, 1-phase), 3 (3-phase)d = 0.005\,\text{in}\times 92^{(36-n)/39},\quad R' = \frac{\rho}{A},\quad \Delta V = k\,I R' L,\ k = 2\ (\text{DC, 1-phase}),\ \sqrt3\ (\text{3-phase})

Sources

  1. ASTM B258 — Standard specification for standard nominal diameters and cross-sectional areas of AWG sizes
  2. IEC 60028 — International standard of resistance for copper (0.017241 Ω·mm²/m at 20 °C)
  3. NFPA 70 (NEC) 210.19(A) Informational Note — 3% branch / 5% total voltage drop guidance
  4. Wikipedia — American wire gauge (table of diameters, areas and copper resistance)

Checked against references

6 worked examples with independently sourced answers ship with this calculator. They run in the test suite; you can run them here too.

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