If the generator is 3-phase 120/208 v. it produces 104.21amps per phase using the formula 30,000 / (1.73 x 208 x .80). If it is 277/480 v. It gives 45.18 amps. A singlephase generator will give 125 amps at 240 volts and 62.5 amps at 480v.
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A 415 v three phase system has a line-to neutral voltage of 240 v on each of the three phase wires. Each wire supplies 250,000/3 or 83,333 VA so the current is 83,333 / 240 or 347 amps.
The current in each live wire is 347 amps. If a balanced load was delta-connected to it, the load current would be 200 amps at 415 volts.
Another Answer I suspect that you are really asking what the line current, rather than the phase current, would be.... however,
Assuming the machine is supplying a balanced load, to find out the line current (and, therefore, the phase current, in the case of a star-connected machine), you simply divide the kVA by (1.732 x line voltage):
current = (250 000) / (1.732 x 415) = 347 A.
This equation applies to both star and delta connections, so If the machine was delta connected, it would supply exactly the same line current, but the load current would be the line current divided by 1.732, or 200 A.
Here, power p=2000kva
v=550 v
Now P= 1.73*V*I*COS(PHY)
ASSUMING POWER FACTOR IS UNITY HENCE COS=1
HENCE I=(2000*1000)/(1.73*550*1)
=2101AMP
It depends upon the Generator system voltage. For 3 Phase, 600 Volt system, it will be 73 Amps For 3 Phase, 480 Volt system, it will be 90 Amps For 3 Phase, 208 Volt system, it will be 208 Amps
50 Amps Single Phase 20 Amps Three Phase
KVA means product of voltage and current. For 3 phase generator, its KVA = (1.732 X (Line Voltage) X Current)/1000.Put line voltage in this equation and get current.
depends on the number of cans and the specific gravity of the electrolyte
The formula you are looking for is I = W/E. Amps = Watts/Volts.