DC:AC - inverter amps
Battery LiFePo4
If a single 3.2v cell used...
If the inverter is rated at, say, 85% efficiency,
Serial connection: 3.2v+3.2v+3.2v+3.2v = 4*3.2v = 12.8v
If 280AH cell used, Total power = 4*280=1120AH
By paralleling in 2 lines: 1120*2 = 2240AH
By paralleling in 3 lines: 1120*3 = 3360AH
Example with powering the laptop computer
Power consumption: 50w. (AC side)
If inverter has 85% efficiency:
50W / 0,85 = 58,8W.
58,8W / 12V = 4,9A.
4,9A + 0,5A = 5,4A will be consumed on 12v side.
Example with powering the 0.5kW (AC side)
500W divided by 0.85 = 588W.
588W divided by 12V = 49A.
49A + 0,5A = 49,5A
Example with 2kW induction cooker (AC side)
2000W divided by 0.85 = 2352W.
2352W divided by 12V = 196A.
196A + 0,5A = 196.5A
In 16 minutes the battery will be flat.
Example with 1kW induction cooker (AC side)
1000W divided by 0.85 = 1776W.
1776W divided by 12V = 148A.
148A + 0,5A = 148.5A.
In 22 minutes the battery will be flat.
Note: a good induction cooker allow you to change the power consumption down to 0.5kW.
In this case:
Example with induction cooker set to 0.5kW (AC side)
500W divided by 0.85 = 588W.
588W divided by 12V = 50A.
50A + 0,5A = 50.5A.
In 66 minutes the battery will be flat.