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Why only 16A of 32A?

Jhenson29

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230 volts times 3 phases times 16 amps will give you 11 kW!.
Yes…but…only if they’re actually being used that way and referenced to the neutral. Basically, as three single phases, not 3-phase.

If not and they’re referenced line-line (as I suspect they are), then this is wrong (even if the answer comes out the same) and @B61 is right.
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WoKe

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whitex

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Line to Neutral in USA 2 phase system is 120V and Line to Line is 240V. In Europe however, since it's a 3 phase system, Line to Neutral is 220V (mainland) or 240V (UK), Line to Line is 380V (mainland) or 415V (UK). Line to line in a 3 phase system is not double because they are 120 degrees out of phase, rather than 180 degrees like in the USA (2 phase). The European 3-phase system is sometimes referred to as "harmonized 400V".

All that said, there are 3 phase systems in the USA too, L-N is 120V, L-L is 208V. I have personally encountered Tesla destination chargers installed on such systems, the the voltage on those is L-L, so 208V (which is a valid J1772 EVSE voltage btw). The only drawback of course that it takes longer to charge your car from such a charger. Then there are some high power application (think factories) which use a 480V 3 phase system (no step-down transformer) - probably no L2 chargers on those.
 
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Jhenson29

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All that said, there are 3 phase systems in the USA too, L-N is 120V, L-L is 208V.
Yeah, but the charging is still single phase in this case.

I understand the charging available in Europe to be actual 3-phase charging,
 

whitex

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Yeah, but the charging is still single phase in this case.

I understand the charging available in Europe to be actual 3-phase charging,
Yes, but in 3-phase loads, neutral is not used (just like in 2 phase in the USA), so the current is only between the 3 phases.
Porsche Taycan Why only 16A of 32A? 3-phase_flow


The above picture came from here:
https://en.wikipedia.org/wiki/Three-phase_electric_power

Btw, if you were to connect a balanced 3 phase loads between L and N, N would have no current (current from all 3 phases going in to the middle point in the above diagram always adds up to exactly 0). Same for USA 2 phase system, if you connect 2 identical loads, one between L1-N and the second one between L2-N, there would be no current flowing through N since the 2 phases are 180 degrees off, so their currents cancel out. This is also why when you run a 2 hot and 1 Neutral in the USA, the neutral does not have to be twice the gauge even though at first glance you might think that the Neutral could potentially return the sum of the L1 and L2 currents, but once you consider phase, you realize that Neutral will at most conduct the max of either L1 or L2. So if you L1 and L2 are sized for 50A each, then N can also be sized for 50A, not 100A as it would be in L1 and L2 were in phase (0 degrees apart).
 
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whitex

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Well, that’s what I assumed hence my comments about it really being 400V.
Yes, it is a 400V system, but the currents on each of the phases cannot just be added together numerically since they are not in phase.
 

Jhenson29

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Yes…but…only if they’re actually being used that way and referenced to the neutral. Basically, as three single phases, not 3-phase.

If not and they’re referenced line-line (as I suspect they are), then this is wrong (even if the answer comes out the same) and @B61 is right.
@Silverback I Thank linkade to an online calculator in another thread.

https://www.rapidtables.com/calc/electric/Amp_to_kW_Calculator.html

so there You can do all calculations ehatever way You want
 

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AC three phase amps to kilowatts calculation
Calculation with line to line voltage
The power P in kilowatts (kW) is equal to square root of 3 times the power factor PFtimes the phase current I in amps (A), times the line to line RMS voltage VL-L in volts (V) divided by 1000:

P(kW) = 3 × PF × I(A) × VL-L (V) / 1000

Calculation with line to neutral voltage
The power P in kilowatts (kW) is equal to 3 times the power factor PF times the phase current I in amps (A), times the line to neutral RMS voltage VL-N in volts (V) divided by 1000:

P(kW) = 3 × PF × I(A) × VL-N (V) / 1000
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