DC voltage drop formula
Voltage drop in a DC circuit is the current times the resistance of the wire. With resistance in ohms per 1000 ft and the
one-way length L in feet: Vd = I x 2L x R / 1000. Divide by the system voltage and multiply by 100 for the percentage.
Use 8 AWG (8.37 mm²) copper or larger.
Voltage drop sets the size. 10 AWG would drop 3.33%, over your 3% limit. Ampacity alone would allow 14 AWG.
Check: 8 AWG (8.37 mm²)
| Load current | 20 A |
| Wire in the circuit (out and back) | 2 x 10 ft = 20 ft |
| Resistance of 8 AWG | 0.6282 ohm per 1000 ft; 0.01256 ohm for this circuit |
| Voltage drop | 0.251 V = 2.09% within 3% |
| Voltage at the load | 11.75 V |
| Power lost in the wire | 5.03 W |
| Ampacity of 8 AWG | 50 A (NEC 310.16, 75 C) ok |
Vd = I x 2L x R / 1000 = 20 x 2 x 10 x 0.6282 / 1000 = 0.251 V, and 0.251 / 12 = 2.09%. Length L in feet, R in ohms per 1000 ft.
| AWG | mm² | Drop (V) | Drop (%) | At load (V) | Rated (A) | Result |
|---|---|---|---|---|---|---|
| 18 | 0.82 | 2.554 | 21.28 | 9.45 | 10* | drop and current |
| 16 | 1.31 | 1.606 | 13.39 | 10.39 | 13* | drop and current |
| 14 | 2.08 | 1.01 | 8.42 | 10.99 | 20 | too much drop |
| 12 | 3.31 | 0.635 | 5.29 | 11.36 | 25 | too much drop |
| 10 | 5.26 | 0.4 | 3.33 | 11.6 | 35 | too much drop |
| 8 | 8.37 | 0.251 | 2.09 | 11.75 | 50 | smallest that passes |
| 6 | 13.3 | 0.158 | 1.32 | 11.84 | 65 | passes |
| 4 | 21.2 | 0.099 | 0.83 | 11.9 | 85 | passes |
| 3 | 26.7 | 0.079 | 0.66 | 11.92 | 100 | passes |
| 2 | 33.6 | 0.063 | 0.52 | 11.94 | 115 | passes |
| 1 | 42.4 | 0.05 | 0.41 | 11.95 | 130 | passes |
| 1/0 | 53.5 | 0.039 | 0.33 | 11.96 | 150 | passes |
| 2/0 | 67.4 | 0.031 | 0.26 | 11.97 | 175 | passes |
| 3/0 | 85 | 0.025 | 0.21 | 11.98 | 200 | passes |
| 4/0 | 107 | 0.02 | 0.16 | 11.98 | 230 | passes |
* chassis-wiring guidance; 14 AWG and up: NEC Table 310.16, copper, 75 C column.
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Why the length is doubled
Current leaves the battery on the positive conductor and returns on the negative. Both have resistance, so the circuit length is twice the distance to the load. A 10 ft run is 20 ft of wire. Calculators that ask for the total length expect you to have doubled it already; this one asks for one-way length and doubles it for you.
Worked example
20 A at 12 V, 10 ft one way, 3% drop allowed. The current goes out and comes back,
so the circuit is 2 x 10 = 20 ft of wire, and the allowed drop is 12 x 3 / 100 =
0.36 V. Voltage drop is Vd = I x 2L x R / 1000, with R in ohms per 1000 ft:
- 10 AWG:
20 x 20 x 0.9989 / 1000 = 0.4 V, which is 3.33% of 12 V; over the 3% limit. - 8 AWG:
20 x 20 x 0.6282 / 1000 = 0.251 V, which is 2.09% of 12 V; passes both checks.
Answer: 8 AWG (8.37 mm²), rated 50 A, the smallest size that passes both.
Power lost in the wire
The power turned into heat in the wire is P = I x Vd. At 20 A and a 0.4 V drop that is
8 W, all of it warming the cable. Halving the drop halves that loss.
Metric version
With length in metres and resistance in ohms per km: Vd = I x 2L x R / 1000, the same shape.
Or from the cross-section: Vd = I x 2L x 0.0172 / A, with A in mm² and 0.0172 ohm mm²/m the resistivity of copper at
20 C.
Copper resistance and rated amps by gauge
| AWG | mm² | ohm per 1000 ft | ohm per km | Rated amps | Basis |
|---|---|---|---|---|---|
| 18 | 0.82 | 6.385 | 20.9482 | 10 | chassis-wiring guidance |
| 16 | 1.31 | 4.016 | 13.1759 | 13 | chassis-wiring guidance |
| 14 | 2.08 | 2.525 | 8.2841 | 20 | NEC 310.16, 75 C |
| 12 | 3.31 | 1.588 | 5.21 | 25 | NEC 310.16, 75 C |
| 10 | 5.26 | 0.9989 | 3.2772 | 35 | NEC 310.16, 75 C |
| 8 | 8.37 | 0.6282 | 2.061 | 50 | NEC 310.16, 75 C |
| 6 | 13.3 | 0.3951 | 1.2963 | 65 | NEC 310.16, 75 C |
| 4 | 21.2 | 0.2485 | 0.8153 | 85 | NEC 310.16, 75 C |
| 3 | 26.7 | 0.197 | 0.6463 | 100 | NEC 310.16, 75 C |
| 2 | 33.6 | 0.1563 | 0.5128 | 115 | NEC 310.16, 75 C |
| 1 | 42.4 | 0.1239 | 0.4065 | 130 | NEC 310.16, 75 C |
| 1/0 | 53.5 | 0.09827 | 0.3224 | 150 | NEC 310.16, 75 C |
| 2/0 | 67.4 | 0.07793 | 0.2557 | 175 | NEC 310.16, 75 C |
| 3/0 | 85 | 0.0618 | 0.2028 | 200 | NEC 310.16, 75 C |
| 4/0 | 107 | 0.04901 | 0.1608 | 230 | NEC 310.16, 75 C |
Resistance: copper at 20 C. Rated amps for 14 AWG and larger: NEC (NFPA 70) Table 310.16, copper, 75 C column, which assumes 30 C ambient and no more than three current-carrying conductors together; hotter spaces and bundles lower it. 18 and 16 AWG are not in that table; their figures are common chassis-wiring guidance. NEC 240.4(D) also limits overcurrent protection on 14, 12 and 10 AWG to 15, 20 and 30 A.
12V quick chart at 10% drop
| Amps | 5 ft 1.5 m | 10 ft 3 m | 20 ft 6.1 m | 30 ft 9.1 m | 50 ft 15.2 m |
|---|---|---|---|---|---|
| 5 A | 18 | 18 | 16 | 14 | 12 |
| 10 A | 18 | 16 | 14 | 12 | 10 |
| 20 A | 14* | 14 | 10 | 10 | 6 |
| 30 A | 10* | 10* | 10 | 8 | 6 |
| 50 A | 8* | 8* | 6 | 6 | 3 |
| 100 A | 3* | 3* | 3* | 3 | 1/0 |
Copper, minimum AWG. * ampacity, not voltage drop, sets this size. "over 4/0" means no single copper conductor up to 4/0 passes: shorten the run, use a higher system voltage or run conductors in parallel. Length is one way; the math uses twice that, out and back.
Before you buy wire
- Ampacity figures for 14 AWG and larger are from NEC (NFPA 70) Table 310.16, copper, 75 C column, which assumes 30 C ambient and no more than three current-carrying conductors bundled together. Hot spaces, bundles and conduit lower them. 18 and 16 AWG figures are chassis-wiring guidance, not NEC ratings.
- On boats, follow ABYC E-11 for DC wiring. Engine rooms are hot, and conductors there are allowed less current than the same wire elsewhere.
- Use wire rated for the location: heat, oil, water, sunlight and abrasion all matter. Stranded copper is the norm in vehicles and boats, where wire vibrates.
- Fuse every positive run close to the source, sized to protect the wire. NEC 240.4(D) limits the overcurrent device on 14, 12 and 10 AWG to 15, 20 and 30 A.
- This calculator does arithmetic. It does not certify an installation; check your work against the standards and the equipment makers' instructions, or have it checked.
Questions
Is voltage drop the same at 12V and 24V for the same wire?
The volts lost are the same for the same current. As a percentage it halves at 24V, and the same power draws half the current, so the drop halves again.