AWG # 10 wire on 30 amp circuit.
8 gauge will be sufficient with less than a half volt drop
What size breakers are needed for a 30kva transformer 208 volt feed 600 volt out put
A 'volt ampere' (not 'volt amp'!) is the unit for theapparent power of a load in an a.c. circuit. It is simply the product of the supply voltage and the load current.
Four size D torch cells are needed for a 6-volt torch. Each size D torch cell typically provides 1.5 volts, so combining four of them will give a total voltage of 6 volts.
At a minimum you would need to run 5 parallel sets of 1000 Kcmil/mcm 90c rated copper wire. This is not calculating for other factors such as ambient temp., volt drop, or conduit fill.
The size of conductor needed will be a #14. It should be copper and have an insulation factor of 90 degrees C. The fact that it is a three phase load does not enter into the calculation of the wire size.
8 gauge will be sufficient with less than a half volt drop
If a precision voltage is needed from 200 to 230 an auto transformer could be used. If the load is a 230 volt motor to be operated on 200 then add 15% to the nameplate full load amperage for the calculation for overload protection.
What size breakers are needed for a 30kva transformer 208 volt feed 600 volt out put
A 'volt ampere' (not 'volt amp'!) is the unit for theapparent power of a load in an a.c. circuit. It is simply the product of the supply voltage and the load current.
The generator should be about ¾ hp.
Four size D torch cells are needed for a 6-volt torch. Each size D torch cell typically provides 1.5 volts, so combining four of them will give a total voltage of 6 volts.
They are not compatible. You cannot connect any AC device to a DC battery without a inverter. The size of the inverter determines what it will run. And the wattage of the fan determines what size inverter to buy.
At a minimum you would need to run 5 parallel sets of 1000 Kcmil/mcm 90c rated copper wire. This is not calculating for other factors such as ambient temp., volt drop, or conduit fill.
The wire size depends on how much current it will conduct.
3000 divided by 240 approx 13 amps?
The formula you are looking for is Watts = Amps x Volts. Amps = Watts/Volts. This comes to 4 amps load. Minimum size fuse would be 5 amps.