Power invertor

[Rawze]

(08-11-2021 )Nilao Wrote:  The formula I use is Watts÷Output Volts×Input Volts=fuse size. So your typical 2500w inverter would be 2500/120×12=250 amps. Now that's if your using the correct wire size for the load. If your using say a wire that's .85 the proper gauge then you'd need a 300 amp fuse as the wire will need to carry more average to produce 2500w on the other end. So that 80A fuse will power a max output of

Just an observation here ...

I don't know where that formula came from but it is severely flawed. That formula seems to coincidentally get someone in a ball-park range for +12Volts system (by the sheer coincidence that 120 volts is 10x the range of 12 volts), but it is some damaged mathematics. Imagine that someone tries to use that formula on a 24-volt or 48-Volt system in a boat or Solar array. It would be completely wrong, the math goes the wrong way.

The chart that I presented above is not based on any formulas but rather real world numbers and ratings instead. It includes compensations for those pesky advertised ratings of inverters vs. what can actually be achieved safely and effectively. This is so that someone does not have to get all confused when considering all of those things.