Russ I am going to give you the simple answer. It means you will have neutral current flowing of about 1.5 amps. If all three phases are perfectly balanced, you will not have any neutral current flowing in a 4 wire wye system.
Think of it like this: We have three horses bridled together to a common point pulling in 3 different directions. They are pulling 120 degrees adjacent to each other. Think a pie cut in thirds. If they apply the same force, they go no where. If one or two are pulling harder than one, the center of the circle travels at x degrees at Y speed. The math is very complicated and beyond the scope of this discussion.
But to get to what you want to know the Power adds to PT = P1 +P2 +P3 assuming there is no power factor involved with pure resistive load. Again the math is very complicated if we factor in Power factor which means angles again.
So in your questions lets assume the power factor = 1, or a purely resistive load like a resistor and then we can use Ohm's Law. The voltage is assumed to be 208 volts for a 208 delta system. So you have 1 amp in phase A and the power in Phase A = 208 volts x 1 amp = 208 watts. Phase B power = 208 volts x 2 amps = 416 watts. Phase C power = 208 volts x 3 amps = 624 watts Total power consumed = PA + PB + PC = 208 + 416 + 624 = 1248 watts.
So in a 3-phase 208/120 circuit to deliver 1248 watts the maximum current is 3 amps at this degree of unbalance. If balance so all 3 phase current are equal the current in each phase would be 2 amps in each phase. To do the same with a single phase circuit at 240 volts would require 5.2 amps, or 10.4 amps @ 120 volts.
Hopefully now you see why 3-phase and higher voltages are the way to go for higher power demand.
Here is an example: I supply a single phase 240/120 service rated at 200 amps at your house, and next door I deliver the same 200 amp service at the equivalent voltage 0f 208/120. How much power can each service deliver at a given moment in time.
Single phase can deliver 240 volts x 200 amps = 48,000 watts or 48 Kw
3-phase can deliver 208 volts x 200 amps x 3 phases = 124,800 watts or 124.8 Kw
Understand?
Think of it like this: We have three horses bridled together to a common point pulling in 3 different directions. They are pulling 120 degrees adjacent to each other. Think a pie cut in thirds. If they apply the same force, they go no where. If one or two are pulling harder than one, the center of the circle travels at x degrees at Y speed. The math is very complicated and beyond the scope of this discussion.
But to get to what you want to know the Power adds to PT = P1 +P2 +P3 assuming there is no power factor involved with pure resistive load. Again the math is very complicated if we factor in Power factor which means angles again.
So in your questions lets assume the power factor = 1, or a purely resistive load like a resistor and then we can use Ohm's Law. The voltage is assumed to be 208 volts for a 208 delta system. So you have 1 amp in phase A and the power in Phase A = 208 volts x 1 amp = 208 watts. Phase B power = 208 volts x 2 amps = 416 watts. Phase C power = 208 volts x 3 amps = 624 watts Total power consumed = PA + PB + PC = 208 + 416 + 624 = 1248 watts.
So in a 3-phase 208/120 circuit to deliver 1248 watts the maximum current is 3 amps at this degree of unbalance. If balance so all 3 phase current are equal the current in each phase would be 2 amps in each phase. To do the same with a single phase circuit at 240 volts would require 5.2 amps, or 10.4 amps @ 120 volts.
Hopefully now you see why 3-phase and higher voltages are the way to go for higher power demand.
Here is an example: I supply a single phase 240/120 service rated at 200 amps at your house, and next door I deliver the same 200 amp service at the equivalent voltage 0f 208/120. How much power can each service deliver at a given moment in time.
Single phase can deliver 240 volts x 200 amps = 48,000 watts or 48 Kw
3-phase can deliver 208 volts x 200 amps x 3 phases = 124,800 watts or 124.8 Kw
Understand?
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