1000W inverter wire size at 24V
A 1000 W inverter running flat out on 24 V draws about 49.1 A from the battery at 85% efficiency, or 46.3 A at 90%:
I = P / (V x efficiency) = 1000 / (24 x 0.85) = 49.1 A, rounded up. Over 5 ft one way at 3% drop that needs 8 AWG copper.
Size the cable for the inverter's continuous rating, and check the cable and fuse sizes its manual gives.
Use 8 AWG (8.37 mm²) copper or larger.
Ampacity sets the size, not voltage drop. 10 AWG would keep the drop under 3%, but 49.1 A needs a conductor rated for at least that current: 8 AWG is rated 50 A (NEC 310.16, 75 C).
Check: 8 AWG (8.37 mm²)
| Load current | 49.1 A |
| Wire in the circuit (out and back) | 2 x 5 ft = 10 ft |
| Resistance of 8 AWG | 0.6282 ohm per 1000 ft; 0.00628 ohm for this circuit |
| Voltage drop | 0.308 V = 1.29% within 3% |
| Voltage at the load | 23.69 V |
| Power lost in the wire | 15.1 W |
| Ampacity of 8 AWG | 50 A (NEC 310.16, 75 C) ok |
Vd = I x 2L x R / 1000 = 49.1 x 2 x 5 x 0.6282 / 1000 = 0.308 V, and 0.308 / 24 = 1.29%. Length L in feet, R in ohms per 1000 ft.
| AWG | mm² | Drop (V) | Drop (%) | At load (V) | Rated (A) | Result |
|---|---|---|---|---|---|---|
| 18 | 0.82 | 3.135 | 13.06 | 20.86 | 10* | drop and current |
| 16 | 1.31 | 1.972 | 8.22 | 22.03 | 13* | drop and current |
| 14 | 2.08 | 1.24 | 5.17 | 22.76 | 20 | drop and current |
| 12 | 3.31 | 0.78 | 3.25 | 23.22 | 25 | drop and current |
| 10 | 5.26 | 0.49 | 2.04 | 23.51 | 35 | current too high |
| 8 | 8.37 | 0.308 | 1.29 | 23.69 | 50 | smallest that passes |
| 6 | 13.3 | 0.194 | 0.81 | 23.81 | 65 | passes |
| 4 | 21.2 | 0.122 | 0.51 | 23.88 | 85 | passes |
| 3 | 26.7 | 0.097 | 0.4 | 23.9 | 100 | passes |
| 2 | 33.6 | 0.077 | 0.32 | 23.92 | 115 | passes |
| 1 | 42.4 | 0.061 | 0.25 | 23.94 | 130 | passes |
| 1/0 | 53.5 | 0.048 | 0.2 | 23.95 | 150 | passes |
| 2/0 | 67.4 | 0.038 | 0.16 | 23.96 | 175 | passes |
| 3/0 | 85 | 0.03 | 0.13 | 23.97 | 200 | passes |
| 4/0 | 107 | 0.024 | 0.1 | 23.98 | 230 | passes |
* chassis-wiring guidance; 14 AWG and up: NEC Table 310.16, copper, 75 C column.
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1000W inverter cable size by length at 24V
| One-way length | 85%: 49.1 A, 3% | 90%: 46.3 A, 3% | 85%: 49.1 A, 10% |
|---|---|---|---|
| 3 ft (0.9 m) | 8* | 8* | 8* |
| 5 ft (1.5 m) | 8* | 8* | 8* |
| 6 ft (1.8 m) | 8* | 8* | 8* |
| 8 ft (2.4 m) | 8 | 8 | 8* |
| 10 ft (3 m) | 8 | 8 | 8* |
| 12 ft (3.7 m) | 6 | 8 | 8* |
| 15 ft (4.6 m) | 6 | 6 | 8* |
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.
Worked example
49.1 A at 24 V, 5 ft one way, 3% drop allowed. The current goes out and comes back,
so the circuit is 2 x 5 = 10 ft of wire, and the allowed drop is 24 x 3 / 100 =
0.72 V. Voltage drop is Vd = I x 2L x R / 1000, with R in ohms per 1000 ft:
- 10 AWG:
49.1 x 10 x 0.9989 / 1000 = 0.49 V, which is 2.04% of 24 V; drop is fine, but it is rated only 35 A. - 8 AWG:
49.1 x 10 x 0.6282 / 1000 = 0.308 V, which is 1.29% of 24 V; passes both checks.
Answer: 8 AWG (8.37 mm²), rated 50 A, the smallest size that passes both. Here the current rating, not the drop, decides it: 10 AWG would be enough for the drop alone.
When the battery sags
Current rises as battery voltage falls. The same 1000 W at 22 V and 85% draws 53.5 A, so if you run the inverter hard on a low battery, check that current against the cable's rating too. A fuse or breaker on the positive cable, close to the battery, protects the cable itself.
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
How many amps does a 1000W inverter draw at 24V?
About 49.1 A at full load and 85% efficiency (1000 / (24 x 0.85)), or 46.3 A at 90%. At light load it draws far less.
What size cable for a 1000W inverter at 24V over 10 feet?
8 AWG copper at 3% drop, sized for 49.1 A.