On Thu, May 30, 2019 at 8:12 AM Nicklas Karlsson < [email protected]> wrote:
> > Theoretically, yes, an inverter would then also have less I²R losses than > > at 48v. And there are 120v inverters ... > > Yes but there might be difference in voltage drop for transistors then > maximum current is lower. For an electric motor sum of currents is the same > so then wind for a higher voltage current need to make more turns. > > It is called "i squared r loss". Or ower lost to heat = i^2 x r Assume you need to runs a coffee maker in the morning that ses a 1000W heating element. Look at two cases 48 Volts and 480 Volts (1) at 48 volts we need 1000/48 amps or 20 amps to make coffee (2) at 480 volts we need 2 amps Power loss through abuot 0.1 ohm resistence for 20 amps is 20 x 20 x 0.01 about 40W or power Power loss through abuot 0.1 ohm resistence for 2 amps is 2 x 2 x 0.01 about 0.04W or 40 milliwates of power turned into heat You wates about 1000 times less power with the high volt system. That is why is is used. But it does not matter if the system is small or if you can find a productive use for the waste heat. Perhaps you bolt the inverter switch to the water heater? I know the the 0.1 number may be wrong but the ratio is the same, still 1000x. > > _______________________________________________ > Emc-users mailing list > [email protected] > https://lists.sourceforge.net/lists/listinfo/emc-users > -- Chris Albertson Redondo Beach, California _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
