I'm piecing together a 2200W solar system for an off grid hunting cabin. Currently looking at a 2200W 12V pure sign wave inverter and I'm attempting to size the wire and fuse. Here is my logic:
- 2200 inverter watts / .85 inverter efficiency = 2588.24 DC watts / 12V DC = 215.69 max DC Amps
- Using 215.69 max DC amps, 12V DC, 12' conductor, variable load type, and terminated on a fuse as input into this calculator I get 1 AWG wire. Just for safety and availability I am proposing to up size the wire to 1/0 AWG.
- Using 1/0 AWG as input into this chart I get 300 DC Amps ANL Fuse.