I = P / VNo power factor needed for DC circuits.
Find how many amps an inverter draws from the battery bank for a given AC load. Inverter efficiency (typically 85-92%) increases the DC input current beyond a simple watts/volts calculation. Use this to size DC fuses, cables, and battery capacity.
Power factor varies by load. Use the equipment value when available.
I = P / VCustom voltage is always supported.
I = P / VNo power factor needed for DC circuits.
I = P / (V × PF)Single-phase AC — multiply voltage by power factor.
I = P / (√3 × V × PF)Three-phase AC — divide by √3 ≈ 1.732.
At 90% efficiency: DC amps = 2000 / (12 x 0.90) = 185.2 A. This is why 12V inverters above 2000W require very heavy cables (2/0 AWG or larger).
Inverters convert DC to AC with some power loss. A 90% efficient inverter delivering 2000W AC actually draws 2000 / 0.90 = 2222W from the battery. Use the correct DC current for cable and fuse sizing.