Model the three circuits as one tied bus and label the box with that result. You tie the circuit buses together through a very small impedance in a second study scenario, let the software solve one combined fault and one clearing sequence, and put that number on the box. Adding the three separate labels has no calculation behind it. On a 480 V box fed by three 200 ft [61 m] 3/0 AWG feeders, the three labels imply 26.8 kA where the tied bus gives 14.2 kA, and the slowest breaker — not the fastest — sets how long the arc burns. IEEE 1584 starts from a full three-phase fault in the first cycle and gives no credit for spreading time, so the tied-bus result is the conservative number your label needs.
Can you add the incident energy of three circuits in one junction box?
No. The three separate studies answer a smaller question than the box poses — a fault on one circuit alone — and their results sum to a state the network never produces: high when the breakers match, low when one of them is slow.
Incident energy is the outcome of one solved network — bolted current, arcing current, arc duration — not a conserved quantity you can total. With identical fast breakers, adding lands near double the tied-bus result, because every single-circuit study spends the whole source impedance on its own fault. With one slow device in the can, adding lands about a quarter low, because each study ends at its own breaker while the real arc waits for the slowest one to open. The rules for multiple motor circuits in one junction box cover the fill and segregation side of sharing a can; the arc flash side is the half those rules never touch.
Does an arc on one set of terminals reach the other circuits?
Yes — test programs have watched faults that start on one terminal group involve the rest of the can, and no published propagation delay exists that you could subtract from the burn time.
What the tests have not produced is a usable number of cycles: a delay after which you could treat the second terminal group as still clean. Without that number you bound the event instead of predicting it, and the bound is the tied-bus scenario in the next section.
IEEE 1584 keeps you conservative while you wait for that number. Its calculation begins with a full three-phase arcing fault in the first cycle, so it gives no credit for spreading and takes no penalty either; it answers the everything-involved case. The 2018 edition also corrects the arcing current for enclosure size — the older equations were fitted on boxes around 20 in [508 mm] square — so give the study the real inside dimensions of your can rather than a cabinet default. Barriers between terminal groups and separate wiring ducts, the habits behind segregating DI and DO circuit groups in a junction box, slow a real arc even though no study takes the credit.
How do you model all three circuits contributing?
Tie the circuit buses through an impedance at least 100 times smaller than your smallest feeder — 0.1 mΩ against 17.2 mΩ in the example below — as a separate scenario, and keep the normal three-circuit model beside it.
The tie is arithmetic, not hardware: small enough to own no share of any loop, present enough that the solver sees one node. Everything that can feed a fault in the can now feeds it together, and the software recalculates one current and one clearing sequence from scratch. When your software fights parallel feeders, build the equivalent by hand: identical feeders divide their impedance by the count, mixed sizes solve as any parallel network, and you enter the single equivalent as one feeder. When each circuit can be fed two ways — one path through two breakers, one through three — run the tie twice, once per path.
| Scenario in the study file | How you build it | What it answers |
|---|---|---|
| Three circuits separate | the normal one-line, one feeder per circuit | a fault on one circuit alone |
| Tied bus, two-breaker path | 0.1 mΩ tie between the circuit buses, first path active | the combined case at its highest current |
| Tied bus, three-breaker path | the same tie, second path active | the combined case when a slow device sits in the path |
| The label | the higher tied-bus result at 18 in [455 mm] | the worst the box can present to anyone working on it energized |
How much fault current do three feeders make together?
For three identical 200 ft [61 m] 3/0 AWG copper feeders from a 480 V switchboard with 20 kA available: 8.9 kA at the box per circuit alone, 14.2 kA with all three tied — 1.6 times one circuit, not three times.
Work the feeder first. Conductor tables put 3/0 AWG copper in steel conduit at about 0.077 Ω resistance and 0.038 Ω reactance per 1000 ft, so 200 ft gives 15.4 + j7.6 mΩ, an impedance of 17.2 mΩ. A 480 V source with 20 kA available holds 480 ÷ (√3 × 20 000) = 13.9 mΩ behind the box.
- One circuit alone: 480 ÷ (√3 × 31.1 mΩ) = 8.9 kA
- All three tied: (17.2 ÷ 3) + 13.9 = 19.6 mΩ, so 480 ÷ (√3 × 19.6 mΩ) = 14.2 kA
- Three labels added: 3 × 8.9 = 26.8 kA, 1.9 times what the one shared source can deliver
The shared 13.9 mΩ source impedance is what keeps the combined fault from tripling. For scale, use the open-air Lee equation: 5.12 × 10⁵ × kV × kA × s ÷ mm² of working distance, the theoretical model IEEE 1584 still carries above 15 kV. At 455 mm, with every device clearing in 0.03 s, it gives 0.3 cal/cm² for one circuit, 0.5 for the tied bus and about 1 for three labels added. Your study will run the IEEE 1584 empirical model with its enclosure correction; those ratios are what carry over.
Why does the combined model change the breaker clearing time?
Because the tie puts the whole 14.2 kA on every breaker in the model, while each real breaker sees only its own share — about 4.7 kA here — so the software clears devices that have not opened.
Two of the example breakers are 200 A frames with instantaneous pickup at 10 × 200 A = 2000 A; their 4.7 kA share clears in about 0.03 s, under two cycles at 60 Hz. The third runs with instantaneous off for coordination, a 0.3 s short-time delay and pickup at 5 × 200 A = 1000 A, and 4.7 kA sits deep inside that band. The arc burns until this device opens, no matter how quickly the other two cleared.
A model that believes its own early trips clears all three in 0.03 s and labels the box 0.5 cal/cm². The defensible label holds the combined 14.2 kA for the longest time any contributing breaker takes: 14.2 kA × 0.3 s at 18 in [455 mm] is 5 cal/cm² on the same scale. It is an upper bound — once the two fast feeders clear, the current falls back toward the third feeder's own 8.9 kA — and a reviewer can follow it line by line. Adding fails here in the other direction: 0.3 + 0.3 + 3.2 = 3.8 cal/cm², about 25 % under. When the software auto-parallels the feeders and re-times every device, enter your hand-built 5.7 mΩ equivalent and set the duration from the device that actually holds the arc. The setting and wiring choices that shorten a burn are what arc flash compliant enclosure wiring walks through.
What number goes on the arc flash label?
The higher tied-bus result from the two supply paths, at your study's working distance — 5 cal/cm² at 18 in [455 mm] in this example — with the normal per-circuit scenarios kept in the file behind it.
NFPA 70E has the label carry the incident energy at its working distance, or the PPE category, and one box carries one label: the worst number that box can present. Keep the study file able to explain the number — which scenario gave the current, which supply path, which device's 0.3 s set the duration — so a reviewer never has to guess. At 5 cal/cm² your arc-rated clothing meets or beats 5, which in practice puts the job in an 8 cal/cm² kit; arc flash PPE for junction boxes walks that assessment. The label exists for energized work: when the job can be planned dead, lock the box out and test for absence of voltage before your hands go in, and the door number stays a formality.
When should you split the circuits into separate boxes?
When the combined label makes energized work impractical — most sites start pushing back around 8 cal/cm² and refuse it near 40 — or when one coordination setting drives the whole number and nobody will re-set it.
Separate cans take the aggregation question out of every future study: each box models on its own feeder, its own breaker, its own duration, and no arc in one can reaches the next. The trade is three boxes, three sets of glands, three labels and more wall, so let the numbers decide. At 5 cal/cm² on a box someone opens monthly, most sites live with the label and the PPE; at 25 cal/cm² set by one 0.5 s band nobody will touch, splitting is cheaper than the paperwork. When you split, size each can to the bending space NEC 312.6 sets at terminals by conductor size — how to size an electrical enclosure walks it — and run each box fresh, because the enclosure-size correction follows the can's dimensions. We build custom enclosures to your cutouts and material, delivered in 10 business days, with the clock starting at your approval of the drawing. The enclosure selector sorts the candidates when choosing is your next step.
Enclosures for this job
Catalog sizes that suit the job this article answers, picked from its subject; open one for its full specifications.
| Enclosure | Material | Outside H x W x D / display | Rating |
|---|---|---|---|
| 6 × 6 × 4 in | 304 stainless | 6 x 6 x 4 in | Rated NEMA 4X / IP66 by the manufacturer; not UL listed |
| 8 × 6 × 4 in | 304 stainless | 8 x 6 x 4 in | Rated NEMA 4X / IP66 by the manufacturer; not UL listed |
| 10 × 8 × 4 in | 304 stainless | 10 x 8 x 4 in | Rated NEMA 4X / IP66 by the manufacturer; not UL listed |
Questions
What impedance should the tie between the circuit buses have?
Keep it at least 100 times smaller than your smallest feeder so it owns no share of the loop: 0.1 mΩ or less against the 17.2 mΩ feeders in the worked example. Its only job is to make the circuit buses one node, so the result reflects the combined sources rather than the tie itself.
Should the working distance be shorter for a small wall box?
Pick the distance from the task, not from the can. At 18 in [455 mm] the example label reads 5 cal/cm²; hands working at 12 in [305 mm] raise it by (455 ÷ 305)² = 2.2, to about 11 cal/cm². State the distance on the label and compute at it.
Can fast breakers stop the arc reaching the other circuits?
You cannot count on it and you cannot model it: no measured propagation delay for a shared can exists to subtract, and IEEE 1584 gives no credit for spreading time in either direction. Clearing in 0.03 s cuts the energy hard, but the label is still built on full propagation, which is the tied-bus case.
How do you handle feeders of different sizes in the same box?
Solve them as a parallel network instead of dividing by the count: a 17.2 mΩ feeder beside a 40 mΩ feeder gives 1 ÷ (1/17.2 + 1/40) = 12.0 mΩ. The lower-impedance feeder carries the larger share of the fault, so its breaker settings decide the duration you put on the label.
