Understanding 3 phase 480V lighting circuit

Question

I understand that you would use 480V light fixtures as opposed to 277V because the 480V have a lower current draw, resulting in a more efficient system. What I don't understand is how this circuit works. In the following paragraphs, I'm using the term "supply" as current flowing to the load and "return" as current flowing back to the breaker panel.

In a simple single-phase lighting circuit, you would have two wires: say, 120V hot wire and a neutral wire. Current flows from the hot to the light fixture and back through the neutral, assuming conventional direction of current flow. This completes the circuit and you can calculate the wattage by measuring the current pulled and the voltage supplied.

However, I don't see a 3 phase lighting circuit with 480V light fixtures work like that. Consider a 3 ph system A, B and C. If the A and B phase are connected to the fixture in a "480V single phase" manner, which phase supplies and which one returns current? If A supplies and B returns, then what about a load that's connected across B and C? That would mean C supplies and B is the one carrying the return current at all times, or else you will have a conductor with two currents flowing in opposite directions.

OR is the fixture drawing current from both phases? If so, then that's not single phase. Also, wouldn't you use 3 single-pole breakers as opposed to 1 3-pole breaker to supply the 3 phases. Please advise.

Answer 1

I understand that you would use 480V light fixtures as opposed to 277V because the 480V have a lower current draw, resulting in a more efficient system.

The lamps aren't necessarily more efficient. It's just that there are less losses in power transmission.

In a simple single-phase lighting circuit, you would have two wires: say, 120V hot wire and a neutral wire. Current flows from the hot to the light fixture and back through the neutral, assuming conventional direction of current flow.

Correct.

This completes the circuit and you can calculate the wattage by measuring the current pulled and the voltage supplied.

You have to remember that they can be somewhat out of phase giving a power-factor < 1 which needs to be taken into consideration for power (watts) calculations but you are on the right track.

However, I don't see a 3 phase lighting circuit with 480 V light fixtures work like that. Consider a 3 ph system A, B and C. If the A and B phase are connected to the fixture in a "480V single phase" manner, which phase supplies and which one returns current?

Figure 1. The voltage difference between two phases is a sinewave.

Since there is no neutral connection you can look at it either way. Current will alternate between A and B.

If A supplies and B returns, then what about a load that's connected across B and C?

Current will also alternate between B and C.

That would mean C supplies and B is the one carrying the return current at all times, or else you will have a conductor with two currents flowing in opposite directions.

Figure 2. Two currents flowing in opposite directions cancel out as indicated at point (1) when the B phase is at zero volts (assuming resistive load). At other times B has to carry the sum of the return currents with a peak value when B is at maximum.

OR is the fixture drawing current from both phases? If so, then that's not single phase.

A load connected between two phases has only two wires and sees a sinusoidal voltage applied to it. As far as the load is concerned it has a single-phase supply. (It knows nothing about neutral.)

Also, wouldn't you use 3 single-pole breakers as opposed to 1 3-pole breaker to supply the 3 phases.

Single-phase breakers would leave the other phase connected and the potentially faulty circuit still live. Good practice would be to use a 2-pole breaker on each load or a 3-phase breaker to isolate everything at once.

Answer 2

Do you know any 3 Phase powered Luminaires? I don't. Neither do any other answers show it. They are only single phase lights, but there may be 3 phase distribution lighting to share load on 3 phases.

The main reason would be cost reduction on power distribution cables in a stadium.

If you had 480W 1A stadium lights on 480V delta 20A breakers each lamp is single delta phase and AB , AC or BC then one might choose to derate the breaker to 80% then,

20 x .8 = 16 amps per line.
16 / 1.732 = 9.2 amps per phase.
9 fixtures per phase x 3 = 27 fixtures.

The margin allows surge power start.

Any questions?

Answer 3

You're forgetting this is AC. Even for a single phase system, half the time one line is the supply and the other the return, then it swaps round a fraction of a second later.

The same goes for 3-phase. If you connect a load between phases A and B, half the time A will have a higher voltage than B, and the other half B will have a higher voltage than A.