Let me put this right up front: I’m not the brightest lamp on the truss. At times, I can make Jessica Simpson look like Marilyn vos Savant, who is listed in the Guinness Book of World Records for five years under “Highest IQ” for both childhood and adult scores. I’m not particularly proud or embarrassed about that, but admitting it has served me well. An empty head can be like being an empty vessel looking for understanding with which to be filled.
Last month I was in the Washington D.C. area leading a seminar on electricity for the entertainment electrician. There were a lot of very experienced and very good electricians in attendance. When we started talking about how to calculate the feeder cable current in a three-phase system, the Jessica Simpson in me made her appearance.
The classic formula for power in a three-phase system is: watts = volts × amps × power factor × 1.73. By manipulating that formula we can come up with the formula for three-phase current, which is: I = W ÷ (V × PF × 1.73).
One of the electricians in the class said that he uses a different formula for figuring out the current in a three-phase system: I = (W × 2) ÷ V. Even I could figure out that these two formulas are nothing alike.
At this point, I could have invoked the teacher/student clause, which is, “I am the teacher, you are the student; now shut up and listen to me.” But I realized that dismissing his approach would be tantamount to dismissing his 30 years of experience. Admitting you don’t know something, particularly in front of your colleagues, is kind of like when someone is waving to a person behind you, but you think they’re waving at you. So you wave back and then you realize…and you feel…so…stupid.
After fumbling for an answer to this conundrum, I punted. I called a break to think about it and when the break was over, I not-so-gracefully moved on, promising myself to address it once and for all in the next day’s class.
That night, I couldn’t sleep. I went to bed and for two solid hours, I stared at the ceiling and contemplated the question of why the two approaches yielded different results. My mind went blank. Eventually, the words of Lao-tsu came to my mind: “The usefulness of what is depends on what is not.” In other words, the usefulness of your mind depends not on what you already know, but what you might gain by not knowing, contemplating, and then understanding. The answer suddenly popped into my head. I fell asleep.
The next day, I went to the class and here’s what I tried to convey. Suppose we have, for example, a single 1500-watt 208V automated light connected across phase A and phase B of a three-phase wye-connected transformer. With nothing else is connected, the line current will be 1500W ÷ 208V = 7.2 amps (assuming unity power factor). In this case we can use the formula for a single-phase system because we only have one single-phase load connected.
If we then connect another automated light across phase B and phase C, then we have a more complex situation. Phase B is now feeding current to both loads. How do we figure out the resulting line current?
To simplify it and make it easier to understand, we can redraw it as a delta-configured secondary with the same two loads, as shown in the right half of the illustration in Fig. 1. If we pay close attention to the two drawings we’ll see that they are exactly the same even though they are illustrated differently. Now we can see that the line current in B is being drawn from two different places; phase A and phase B. Does that mean that we can simply add the magnitude of the two currents, I1 and I2, to get the resulting current?
Not exactly.
Because these two currents are 120 degrees out of phase with each other, the resulting current is something less than twice the current in one of the phases. The mathematical solution to the addition of these two out of phase sinusoidal currents is a little bit complex, but the simple answer is that they add up to 1.73 times the magnitude of any of the phase currents.
So as long as the loads are equal in value, then the line current is not twice the single-phase current, but 1.73 times the single phase current.
In this case, we have 7.2 amps in phase A and 7.2 amps in phase B. So the resulting line current is 7.2 × 1.73 amps = 12.5 amps.
If we added another automated light across phases A and C, then all three legs will have 12.5 amps going through them. We can confirm this by using the formula for three phase power, which is W = V × I × PF × 1.73. In this case, with all three lights hooked up the wattage is 4500, so we have 4500 = 208 × I × PF (we’ll assume it’s 1) × 1.73, or I = 4500 ÷ (208 x 1.73) = 12.5 amps three-phase.
It turns out that our friend, the 30-year electrician who uses the formula I = (W × 2) ÷ V, is close, but technically not correct. However, every good electrician always builds in a de-rating factor, usually about 20%. If we take that magic number, 1.73, and give it a 20% overhead, we end up with – you guessed it – 2, or something very close to it (1.73 × 1.2 = 2.076).
Now I can sleep at night knowing that the classic formula for a three-phase load does indeed produce the correct answer. But I might never have confirmed it had I not had a satisfactory philosophy of ignorance and welcomed the doubt and discussion that drives both science and art.
