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I can't seem to find the answer I'm looking for, or perhaps I'm just really not understanding what I've read. I'm just confused.

So, I'm a college student in a Theatre Production program and an assignment we've been given in our lighting course asks the following:

"You will be staging a production in a small theatre, using only 120-volt conventional fixtures. Power is available from a disconnect dedicated exclusively to stage lighting. It provides 100 Amps, 120/208 Volts 3-Phase. What is the maximum amount of power at 120 Volts available for your lighting gear?"

Is this as simple as calculating the watts with the 100 Amps and listed 120 Volts? If so, is the 208 even really relevant here, or just something to make me overthink (with the whole 0.8PF and the square root of 3 and stuff)?

Many thanks in advance!

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  • \$\begingroup\$ Sounds like a "loaded" question to me... 120/208 is implying that a three-phase step-down transformer is used. Measuring from one of the three phases to neutral is 120v - measuring from phase-to-phase is 208v. So that is kind-of ancillary, but is helpful to know how the power works. So then, are there three 100A, 120vAC fixtures, or just one? \$\endgroup\$ – rdtsc Sep 19 '18 at 2:09
  • \$\begingroup\$ Awesome, thank you! And I believe what the question specifically is asking is how much power is available from a single, 100A disconnect that is 120/208V 3-Phase. Is this as simple as 100x120=12kW? \$\endgroup\$ – Jarrod Dunlop Sep 19 '18 at 2:14
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The 100 amp, 120/208 volts, 3-phase rating tells you that the 120 volts is available from each of three 120 volt line to-neutral circuits. Each of those three circuits provide 100 x 120 = 12,000 watts for 36,000 watts total.

You mentioned 0.8pf, but conventional incandescent fixtures would have 1.0 pf. If you have been told that the fixtures are LED fixtures with 0.8 pf, the power available would be 0.8 x 36,000 watts.

If 100 amp, 120/208 volts, 3-phase is available from a single disconnect it would be a 3-phase disconnect supplying the three circuits described.

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  • \$\begingroup\$ So three phase is important to then multiply it by, well, 3! Got it, thank you! And the 0.8 is something I found on the internet while trying to figure out how to answer this question, haha! Thank you very much for the quick answer! I appreciate it! \$\endgroup\$ – Jarrod Dunlop Sep 19 '18 at 2:17
  • \$\begingroup\$ @JarrodDunlop I think you made an assumption of \$\sqrt{3}\$ because you were thinking of current and voltage separately but if you multiply \$\sqrt{3}\times\sqrt{3},\$ you will get \$3\$. \$\endgroup\$ – KingDuken Sep 19 '18 at 3:02
  • \$\begingroup\$ Note that there is a trap with conventional fixtures used on conventional dimming, the third harmonic load current when dimming can be Significant and sums in the neutral. For phase controlled loads (Almost all conventional dimming) you can easily overload the neutral if it is not deliberately oversized (As the NEC requires for theatres), the harmonics in the load can also overheat delta-star transformers and sometimes special 'K' rated transformers are required. \$\endgroup\$ – Dan Mills Sep 19 '18 at 12:45

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