Short answer: Not at full power. 2,000 W pulls about 185 A at 12 V, but most 100 Ah LiFePO4 batteries allow about 100 A continuous (~1,100 W). Even if it could, it would last only about 30–35 minutes.
Math: 2,000 W ÷ (12 V × 0.9) ≈ 185 A. Battery energy: 1,280 Wh × 0.9 usable × 0.9 inverter ≈ 1,040 Wh ÷ 2,000 W ≈ 0.5 h. For a 2,000 W inverter, use at least two 100 Ah batteries in parallel (or one battery rated for 200 A continuous), and the inverter maker's specified cable and fuse, often 2/0 AWG and a 250 A fuse.
| Inverter | DC amps at 12V | Min. LiFePO4 (100A BMS each) |
|---|---|---|
| 1000 W | 93 A | 1 × 100 Ah |
| 2000 W | 185 A | 2 × 100 Ah |
| 3000 W | 278 A | 3 × 100 Ah |
Safety: These are general estimates, not a design. Follow your equipment manufacturer's instructions and local electrical code, and consult a qualified electrician for anything connected to shore power, a house, or high-current batteries.
Yes, about 93 A at full load and roughly 1 hour of runtime for LiFePO4, if the BMS allows ~100 A continuous.
No. 3,000 W needs about 280 A; use three or more 100 Ah batteries or a 24 V/48 V system.
As an Amazon Associate, RVSolarMath earns from qualifying purchases.