Let’s assume I use an ideal rotary phase converter to power a three-phase 10HP motor (100% efficient, PF = 1). Thus, I need S = 10 VA. On a 120V single-phase line, I would get three-phase line-line voltages of 240V. Here’s my question. I was under the assumption that when you analyze motors, you find the single-phase power with the product of V(line-neutral) X I (one-phase current on the line). For the 10 VA situation described above, you’d want the sum of the three VI products I just described to be 10 VA. I was under the impression that a balanced system would have 3 phases, with each phase carrying (sorry if this isn’t the right word, english isn’t my primary language) 3.33 VA. Question 1. Here’s my problem: In the situation I described, two of the line-neutral are 120V, but the other line-neutral voltage is 208V! (read on phase converters if you do not understand why this is so) However, I know rotary phase converters are considered to be perfectly balanced when the currents of the three phases are the same. This is where my brain melts. Assuming two line-neutral voltages of 120V and a line-neutral voltage of 208V, with three perfectly-balanced currents on the lines, does that mean two of the lines carry a certain power, while the 3rd line (the 208V) carries more power? Question 2. I also have the following question: Assuming I have an ideal phase converter, I can easily find the single-phase line-neutral voltage from either the three-phase line-line voltage, or from the line-neutral voltages of any of the three lines (assuming I know which line is the 208V “wild leg”). However, how can I find the single-phase current (the one entering the phase converter) if I know the currents in the three phases? Will it simply be a sum of all three currents? Thanks in advance!