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I have multiple single-phase loads required to be powered from a 3 phase supply with neutral (208/120) V. I'm going to tap to (L1, L2) & (L2, L3) and (L1,L3) respectively. Assuming the load was 1400 W, rated input 100-240 V AC, how can I calculate the current for each phase?
Since later extra single phase L-N to be added into the system too. But first I have to understand how single phase (L-L) current distributed.

For starting example with 6 loads connected 2 units-(L1-L2), 2 units-(L1-L3), 2 units-(L1-L3)
1400 W X 2 units = 2800 W
I = P/(VLine X PF) considering PF= 1 just to ease understanding
I = P/(VLine)
I = 2800 W/208 V = 13.46 A
13.46 A is current calculate for L1 and 13.46 A L2 or 13.46 A is both L1&L2.

If 13.46 A is load for each 2 loads in parallel.
then 13.46 A + 13.46 A = 26.92 A for L1,
26.92 A for L2,
26.92 A for L3.

Is it correct?

I'm not really clear about this 13.46 A. Kindly advise.

SinglePhase2hot

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  • \$\begingroup\$ HI greybeard, thanks for the answer, do means like this. 1400W X 6 = 8400w total power then 8400W / 3 /208V = 13.46A > current for each phase? \$\endgroup\$
    – Epe
    Jan 26, 2023 at 7:12
  • \$\begingroup\$ No. 208 V is line-to-line voltage, I think of ("unqualified") line voltage as line-to-neutral - pick \$208/\sqrt 3\$ or 120. \$\endgroup\$
    – greybeard
    Jan 26, 2023 at 7:18
  • \$\begingroup\$ 120V is phase voltage (line to neutral) where from usual term. this what you meant? 8400W/3/120V = 23.33A it will be the same as 8400W/√3/208 = 23.33A. hmmm i think 23.33A will be the answer if load was connected phase to neutral. but if connected phase to phase as per diagram, is it 13.46A per phase ? or how. \$\endgroup\$
    – Epe
    Jan 26, 2023 at 7:50
  • \$\begingroup\$ According to en.wikipedia, you are right about the use of line voltage. (Phase voltage may differ.) Usage is different enough in my first language - that would be "triangle voltage" as oppose to "outer conductor", "star" (wye), "grid" or just unqualified "tension"/ voltage. Line-to-line has line current to mean something different from current in a load operated at line voltage - sheesh. \$\endgroup\$
    – greybeard
    Jan 26, 2023 at 15:03
  • \$\begingroup\$ All bets are off with reactive loads (without further specification). With a symmetrical resistive load (you assumed PF=1), just divide one third of total power by line-to-neutral voltage - at any given power, the current won't change whether the load is delta- or wye connected. \$\endgroup\$
    – greybeard
    Jan 26, 2023 at 15:05

2 Answers 2

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Three pairs of 1400 VA units may be either delta- or star-connected.

When delta-connected the input voltage to the units would be 208 V.

enter image description here

The current drawn by a pair at 208 V would be 2800 / 208 = 13.47 A and the resulting line current 13.47 x √3 = 23.33 A.

When star-connected the input voltage to the units would be 120 V.

enter image description here

The current drawn by a pair at 120 V would be 2800 / 120 = 23.33 A and the same would be the line current as well.

With the loads being balanced, no neutral connection would be required.

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  • \$\begingroup\$ Hi vu2nan. thanks a lot for this wonderful explanation and diagram. now it was clearer to me about equivalent delta. it was the way load connected making it delta. The resulting line current is 23.33A which obtain by 13.47A X √3 = 23.33A. At first i thought i can have the resultant line current by adding both L1+L2. That's why 26.92A. this resultant line current will be same? time √3 even it if i dont have balance load? ill popup with other question later since im planning to have combination of single phase load, 120V, 208V and 3 phase load. thanks again for you time and patience. \$\endgroup\$
    – Epe
    Jan 27, 2023 at 2:24
  • \$\begingroup\$ Anytime, Epe! With a balanced three-phase load and zero neutral current, the neutral need not be connected. With a load imbalance, not connecting the neutral would result in unequal voltages across the loads. \$\endgroup\$
    – vu2nan
    Jan 27, 2023 at 6:26
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That is not correct. What you are describing is equivalent to a delta-connected balanced 3 phase load.

26.92A would be your phase current. You are interested in the line current.

To calculate your line current, you must multiply your phase current by the square root of 3.

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  • \$\begingroup\$ Hi Pat, thanks for your answer! Sorry im not getting what do you means by "describing is equivalent to a delta-connected balanced 3 phase load." it was wye/star, where I can have 120V single phase and 208V 3phase and 208V single phase. From my understanding in star connection, I phase=I line. and for this single phase (L to L), does it work like single phase + neutral. Current at L = Current at N? \$\endgroup\$
    – Epe
    Jan 26, 2023 at 6:05
  • \$\begingroup\$ Hi Pat, That's not correct either. The phase current would be 13.47 A and the line current 13.47 x √3 = 23.33 A. \$\endgroup\$
    – vu2nan
    Jan 26, 2023 at 14:01
  • \$\begingroup\$ Vu2nan, the question is not consistent. Epe asks “If 13.46 A is load for each 2 loads in parallel…” which suggests a total phase current of 26.92A. They proceed to calculate this same result, but mistakenly represent it as a line current and ask if it is correct. I indicated that is not correct, and explained the relationship between line and phase current on a delta-connected balanced 3 phase load. \$\endgroup\$
    – Pat
    Jan 26, 2023 at 14:36
  • \$\begingroup\$ Hi Pat, In any case, your statement '26.92A would be your phase current.' is incorrect. \$\endgroup\$
    – vu2nan
    Jan 26, 2023 at 14:58
  • \$\begingroup\$ If you have 2 loads connected in parallel across each phase, with each load drawing 13.46A, then your phase current would be 26.92A per phase. In his first example with 1400W loads, then yes, the phase current would be 13.46A. \$\endgroup\$
    – Pat
    Jan 26, 2023 at 15:04

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