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I am an electrician seeking clarification on determining the ampacity for a 3-phase electrical panel with balanced single phase 2P loads. I have a specific scenario and would appreciate the expertise of the community to guide me in the right direction.

Scenario: I am connecting three equal balanced single-phase 208 V loads to a 3-phase 208 V wye panel. Each load is on a 100 A 2P breaker, and that is the minimum breaker size for the load. One breaker goes on L1+L2, one goes on L2+L3, and one goes on L1+L3. My question revolves around determining the correct ampacity required for the conductors feeding the 3-phase panel.

My understanding is that the square root of three (√3) should be applied to calculate the required ampacity (100 A × 208 V) × 3 / √3 / 208 V = 173.2 A). However, I don't understand why because each 2P breaker is essentially two 1P breakers, leading me to think that each pole should receive 200 A, thus requiring the conductors to have an ampacity of 200 A.

Could someone please clarify the correct approach for determining the ampacity of the feeder conductors in this scenario? I would greatly appreciate it if the explanation could be presented in plain terms suitable for an electrician.

Michael Frank
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  • Try our Electrical forum, here people know how to wire a switch or install a panel or run conduits, and can cite the CODE. – Ruskes Apr 29 '23 at 19:59
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    I’m voting to close this question because this site is for DIY home improvement only. – nobody Apr 29 '23 at 20:01
  • Are the loads between the 2 phases or between the phase and neutral? – Rohit Gupta Apr 29 '23 at 20:03
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    try this forum https://electronics.stackexchange.com – Ruskes Apr 29 '23 at 20:04
  • Rohit, yes these loads are between 2 phases each. Balanced so that one is L1 to L2, one is L2 to L3, and one is L1 to L3. Ruskes, I actually posted this in electronics.stackexchange.com first, but they told me to delete it and post it here... seems like this is a question with no home? – Michael Frank Apr 30 '23 at 00:50
  • VTC - Not Home Improvement/DIY. – gnicko May 02 '23 at 01:17

2 Answers2

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For a delta connected load which you have described. The incoming line current will be √3*phase current which for 100 amps would be 173.2 amps. It comes down to trigonometry. The line current is a combination of the currents from the two phases. Not sure if you want a more in depth math class. I've haven't thought of these proofs in a long time.

JD74
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  • It's wye... I appreciate your answer but it's the same result I gave in my question... point is, why? Not asking in terms of math, but in terms of physics. Physically they are three separate lines, and each line has two 100A single pole breakers on it - L1 has 100A from load 1 and 100A from load 2, so why is that not just simply 200A? Is it due to the interaction between the lines at the transformer? Or is it the interaction between the lines at the load? Without understanding why the result is 173A, I am just taking strangers word for it (no offense) which I just can't do at this level. – Michael Frank Apr 30 '23 at 00:45
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    @MichaelFrank The way you describe your loads as connected from L1 to L2, L2 to L3 and L3 to L1 that is a Delta connection which it sounds like you're connecting to a Wye source. The reason the line current on your source is 100 * √3 in simple terms it's the due to the interaction between the lines and the loads. Each line carries the current from two different loads. They are spaced 120 electrical degrees apart. That's where the trigonometry comes in. Using sines and cosines is how the formula is derived. – JD74 Apr 30 '23 at 02:09
  • This explained it! I never would have thought of the way the loads are connected to the wye source as a delta connection. It's just the way I have known to do it from experience without apparently understanding what it really is. Thank you! – Michael Frank Apr 30 '23 at 02:24
  • To expand my understanding I have another question. Does this mean that some of the power from one load is flowing into the other due to being 120 degrees apart? Or canceling the other out, or something? – Michael Frank Apr 30 '23 at 02:40
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    @MichaelFrank Power will be the voltage * current across the load. But yes the current and voltage are rising and falling across the two loads at different angles. It's hard to explain. I've included some Youtube links that can visualize it better. https://www.youtube.com/watch?v=qthuFLNSrlg&pp=ygUZMyBwaGFzZSBwb3dlciBjYWxjdWxhdGlvbg%3D%3D https://www.youtube.com/watch?v=FEydcr4wJw0&pp=ygUZMyBwaGFzZSBwb3dlciBjYWxjdWxhdGlvbg%3D%3D – JD74 Apr 30 '23 at 03:10
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    Current is flowing both directions between the two phases, yes, cancelling or reinforcing depending on where they are in their cycles. – keshlam Apr 30 '23 at 03:13
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Since the loads are symmetrical, just use VA.

So one load is a 100A / 208V breaker, or 20,800 VA.

All 3 loads are 62,400 VA.

Now what would happen if loads of the exact same VA were 120V single-phase loads that were connected "Wye"? They would be 20,800 VA each, or 173 amps each.

Since a balanced load has functionally zero current on neutral, switching the connection from wye to delta does not change the amount of power required for the same VA.

Harper - Reinstate Monica
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  • I'm following for the part where 100A * 208V = 20,800 VA. I don’t understand how all 3 loads together is 32,400 VA. Is that a typo? Wouldn’t you multiply the 20,800 * 3 = 62,400 VA? The final answer I am looking for is how many AMPS would be flowing through the bus bars of that main panel, and why? This is the practical answer I need because I have to size the panel and the conductors feeding it. And I need to understand why I am doing what I am doing. – Michael Frank Apr 30 '23 at 00:34
  • @MichaelFrank Yes, that was a typo. – Harper - Reinstate Monica Apr 30 '23 at 00:43