is my connection correct ??

[gbynum]

and btw how you are sure of comparing AWG with mm2 ?

you can find a formula (can load into a spreadsheet or use a scientific calculator) and tables here, also.

https://www.rapidtables.com/calc/wire/awg-to-mm.html

[zolar]

thanks a lot gbynum for the link .

thanks a lot suneagle , really nice to hear that info.

so with the load of 400w after the inverter , a cable of 4mm squar more than enough since only 1.8amps will flow between inverter and load,,,, but
where is the rest of amps gone since it will flow 16.6amps between batteries and inverter ? is it lost ?

[SunEagle]

thanks a lot gbynum for the link .

thanks a lot suneagle , really nice to hear that info.

so with the load of 400w after the inverter , a cable of 4mm squar more than enough since only 1.8amps will flow between inverter and load,,,, but
where is the rest of amps gone since it will flow 16.6amps between batteries and inverter ? is it lost ?

No it is not lost.

The additional 16.6amps is the possible extra amount that could go through the wires if the load on the inverter is increased to 2kw.

When it comes to electrical circuit wiring you need to size everything at the highest load which makes it safe if you ever need that many watts. If you only sized the DC wires for a couple of amps and for some reason your load increases then you run the possibility of burning up the wires or blowing a fuse. Neither is desirable.

[zolar]

sorry I think I didn't askclearly or I misunderstand ,
if we forget about inverter size ( I mean it could be 2kw less or more ) and suppose I am sure that daily I will go only for that load ( 400w )
when I said 16.6amps and 1.8amps ( cuz I divide 400w /24v between battery and inverter so it will be 16.6 amps ) and after inverter ( I divide 400w / 220v so it will be 1.8 amps ) that's what I am asking about < where is extra amps went after it comes out from battery since I will use only 1.8 amps to generate the load.

sorry brother for not obvious question.

[SunEagle]

sorry I think I didn't askclearly or I misunderstand ,
if we forget about inverter size ( I mean it could be 2kw less or more ) and suppose I am sure that daily I will go only for that load ( 400w )
when I said 16.6amps and 1.8amps ( cuz I divide 400w /24v between battery and inverter so it will be 16.6 amps ) and after inverter ( I divide 400w / 220v so it will be 1.8 amps ) that's what I am asking about < where is extra amps went after it comes out from battery since I will use only 1.8 amps to generate the load.

sorry brother for not obvious question.

If your load is 400watts at 220v then it will draw 1.8 amps on the wires between the inverter and load. But that same load of 400watts goes through the inverter at 24V (battery voltage) which will draw 16.6 amps on the wires between the battery and inverter.

Because the voltage increases but the wattage stays the same the amps will drop. The formula is Watts = Amps x Voltage. It is just the physics of electricity.

[zolar]

yes brother, I know the formula that's why I gave those values,
but generally as I understand now , you mean ( in solar system ) what ever the load is , so there will be 2 values of amps flow ( 1 before the inverter and 1 after the inverter ) and that value stay there as long as the load working.
for example as in my load 400w ; while the load is working so there is a 16.6amps withdrawn from battery due to 24v and its existing between the battery and inverter even though the current will be less ( 1.8 ) after the inverter since the voltage became higher ( 220v )

thanks a lot brother for explanation , and sorry took some of your valuable time , but really I learned a lot from you and others in this helpful nice site.