What Breaker and Wire for a 3.8 kW Inverter at 113 °F With 3 Conductors?
Short answer: a 20 A breaker, 12 AWG copper conductors and a 12 AWG equipment grounding conductor. A 3.8 kW inverter puts out 15.8 A continuous at 240 V, and continuous duty means everything is sized at 125% of that — 19.8 A.
Sizing the overcurrent device
- Continuous output — 3.8 kW ÷ 240 V = 15.8 A.
- 125% continuous — 15.8 × 1.25 = 19.8 A.
- Standard OCPD — rounded up per 240.6(A) to 20 A.
- EGC — Table 250.122 for a 20 A device: 12 AWG copper.
The inverter nameplate is the authority on continuous output current. Where the nameplate differs from 15.8 A — and it often does, because kW ratings are rounded — size from the nameplate, not from the marketing number.
Every conductor, at 113 °F with 3 in the raceway
Temperature correction 0.87 × fill adjustment 1 — the conductor has to carry 19.8 A after both.
| Conductor | 90 °C base | After derating | Terminal (75 °C) | Result |
|---|---|---|---|---|
| 12 AWG | 30 A | 26.1 A | 25 A | passes |
| 10 AWG | 40 A | 34.8 A | 35 A | passes |
| 8 AWG | 55 A | 47.9 A | 50 A | passes |
| 6 AWG | 75 A | 65.3 A | 65 A | passes |
| 4 AWG | 95 A | 82.7 A | 85 A | passes |
| 3 AWG | 110 A | 95.7 A | 100 A | passes |
| 2 AWG | 130 A | 113.1 A | 115 A | passes |
| 1 AWG | 150 A | 130.5 A | 130 A | passes |
| 1/0 | 170 A | 147.9 A | 150 A | passes |
12 AWG clears with 6.3 A to spare, so there is room if the conditions turn out slightly worse than assumed.
The 110.14(C) trap
The conductor also has to satisfy its terminal rating, which is almost always the 75 °C column even when the wire itself is rated 90 °C. You derate from the 90 °C ampacity, but you compare against the 75 °C terminal — using the 90 °C column at the breaker lug is one of the most common sizing errors, and it fails inspection rather than failing safely.
Related
Computed with OneLine Studio's NEC engine — the same deterministic code tables that size a real permit set, not an estimate. Conductor ampacity from NEC 310.16, correction factors from 310.15(B)(1), fill adjustment from 310.15(C)(1). Always confirm the enforced NEC edition and any local amendments with your AHJ.
FAQ
What size breaker for a 3.8 kW inverter?
20 A at 240 V. The inverter's 15.8 A continuous output is multiplied by 125% per the continuous-duty rule, giving 19.8 A, rounded up to the next standard size in 240.6(A). Always confirm against the nameplate continuous output current.
What wire size for a 3.8 kW inverter at 113 °F?
12 AWG copper with 3 current-carrying conductors in the raceway. After a 0.87 temperature correction and a 1 fill adjustment it delivers 26.1 A against the 19.8 A required.
What size EGC?
12 AWG copper, from Table 250.122 for a 20 A overcurrent device. If you upsize the ungrounded conductors for voltage drop, 250.122(B) requires the EGC to be increased proportionally too.
Educational reference, reviewed 2026-07. A design aid, not a substitute for a licensed electrician or PE. Confirm the enforced NEC edition and local amendments with your AHJ.
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