Continuous Load and the 125 Percent Rule

Why this matters

A circuit sized to the load instead of to 125 percent of a continuous load works on the day of the inspection and fails on the third hour of the coldest night of the year, when the baseboard breaker trips with the house at its coldest or, worse, does not trip and cooks its own lug. The rule is short, but where it applies, what it multiplies, and how it combines with derating are where electricians lose points with inspectors. This card owns the rule; every sibling card that sizes a continuous load cites it.

The rule in one sentence, and the two sections behind it

A continuous load is one where the maximum current is expected to continue for three hours or more (the Article 100 definition, as adopted). For a branch circuit serving one, NEC 2023, as adopted by your jurisdiction, sizes two things off the same arithmetic:

• The conductor (210.19(A)(1)). In substance, the conductor must pass two separate checks: its ampacity, before any adjustment or correction, must be at least the noncontinuous load plus 125 percent of the continuous load; and its ampacity after adjustment and correction must be at least the actual load. The conductor has to pass both. They are not multiplied together.
• The overcurrent device (210.20(A)). The breaker or fuse rating must be at least the noncontinuous load plus 125 percent of the continuous load, unless the assembly, including the overcurrent device, is listed for operation at 100 percent of its rating. Residential panelboards and the breakers in them essentially never carry that listing, so in a house you apply the 125 percent every time.

Feeders carry the same structure in 215.2(A)(1) and 215.3, as adopted, which is how the rule follows a heating load back to a subpanel.

Why 125 and not some other number

Standard molded-case breakers are not built to carry their full rating continuously inside a loaded enclosure. The heat from a breaker running at 100 percent for hours, added to the heat from its neighbors and from the conductor terminations, pushes the trip element and the lug past what the listing tests. Sizing at 125 percent of the continuous load is the same thing as loading the breaker to no more than 80 percent of its rating (1 divided by 1.25 is 0.80). Both statements are the rule; use whichever one answers the question in front of you:

• "What breaker does this load need?" Multiply the continuous amps by 1.25.
• "How much continuous load can this circuit carry?" Multiply the breaker rating by 0.80. A 20 A circuit carries 16 A continuous; a 30 A circuit carries 24 A; a 50 A circuit carries 40 A.

Which loads are continuous

Some loads are continuous by code, whatever the customer says about how they use them. Others are continuous by judgment, and the three-hour test decides.

Motors are a separate case. 430.22 sizes a single motor's conductors at 125 percent of its full-load current for its own reasons, and a pool pump that runs eight hours is sized by Article 430, not by this card. See related 'How to Wire a Pool Pump Motor Circuit'.

Worked example 1: a 4500 W storage water heater

A 50 gal electric tank, nameplate 4500 W at 240 V, two elements, non-simultaneous (the nameplate wattage is what one element draws, and the thermostats never let both run together).

LOAD             4500 W / 240 V          = 18.75 A
CONTINUOUS       422.13: yes
MIN OCPD         18.75 A x 1.25          = 23.44 A
                 next standard size up   = 25 A
MIN CONDUCTOR    23.44 A, before derating
  12 AWG NM      20 A (60 C column)      -> fails
  10 AWG NM      30 A (60 C column)      -> passes
OCPD CHOSEN      30 A, 2-pole
  check 1        30 A >= 23.44 A         -> passes
  check 2        30 A <= 30 A cap on 10  -> passes
  check 3        unmarked 422.11 limit:
                 18.75 A x 1.50 = 28.1 A,
                 next size up            = 30 A -> passes

Three checks ride on that 30 A. The 125 percent sets the floor at 23.44 A. The 240.4(D) small-conductor cap holds 10 AWG at 30 A. And 422.11, as adopted, caps the overcurrent device for a single non-motor appliance at the nameplate's marked rating, or, where the nameplate carries no marking and the appliance is rated over 13.3 A, at 150 percent of its rated current with the next standard size permitted (in substance; check the current edition). If the nameplate marks a maximum breaker, that marking governs. A 25 A breaker on 10 AWG would also pass every check; 30 A is the usual choice because the installation instructions commonly call for it and 25 A 2-pole breakers are not always on the truck.

The mistake that comes back from inspection: 12 AWG on a 25 A breaker because "18.75 A fits on 12 gauge." The load fits; the 125 percent does not, and 240.4(D) would hold 12 AWG at 20 A regardless.

If the nameplate says the elements may operate simultaneously, the load is the sum of both elements and the arithmetic restarts from that sum. The circuit wiring itself is a HowTo: see related 'How to Wire an Electric Water Heater Circuit'.

Worked example 2: baseboard heat on one 240 V circuit

A finished basement gets three baseboard heaters: two at 2000 W and one at 1000 W, all 240 V, on one line-voltage thermostat per heater.

• Load. 2000 + 2000 + 1000 = 5000 W. At 240 V, 5000 divided by 240 is 20.83 A.
• 125 percent. 20.83 times 1.25 is 26.04 A. The next standard breaker is 30 A, and 10 AWG NM carries 30 A from the 60 C column. One 30 A circuit on 10 AWG serves all three.
• The alternative that does not work. One 20 A circuit on 12 AWG carries 16 A continuous, which at 240 V is 3840 W. 5000 W is over that by 1160 W.
• The alternative that does. Two 20 A circuits on 12 AWG: one feeding a 2000 W and the 1000 W heater (3000 W, 12.5 A, times 1.25 is 15.6 A, under 20), the other feeding the second 2000 W heater (8.33 A, times 1.25 is 10.4 A). The split costs a second home run and a second breaker but keeps 12 AWG at the thermostats and heater terminals, which matters where those terminals are not rated to take 10 AWG. Either way, check each thermostat's rating against the heater amps it switches.

The installation is its own HowTo: see related 'How to Install Electric Baseboard Heat and a Line-Voltage Thermostat'.

Worked example 3: a retail lighting circuit

A small shop is open 10 a.m. to 9 p.m., eleven hours, so its lighting is continuous. The fixtures are LED troffers whose driver label reads 42 VA input; that input figure is what the circuit carries, not the lamp wattage on the box.

• At 120 V on a 20 A circuit. Continuous capacity is 20 A times 0.80, or 16 A, which at 120 V is 1920 VA. 1920 divided by 42 is 45.7, so 45 fixtures (45 times 42 is 1890 VA). A 46th puts the circuit at 1932 VA, over the line.
• Add a noncontinuous load. The same circuit also feeds a display receptacle the owner uses for an hour at a time, counted at 360 VA, or 3 A. The rule adds noncontinuous load at 100 percent and continuous at 125: 3 A plus 1.25 times the lighting current must not exceed 20 A, so the lighting may draw at most 13.6 A (20 minus 3 is 17, divided by 1.25 is 13.6). At 120 V that is 1632 VA, or 38 fixtures (38 times 42 is 1596 VA; 39 would be 1638 VA).
• At 277 V on a 20 A circuit. 16 A times 277 V is 4432 VA, or 105 fixtures (105 times 42 is 4410 VA). This is why commercial lighting runs at 277 V when the building has it.

Now the derating check, because commercial circuits share raceways. Say six current-carrying 12 AWG THHN conductors share one EMT run. The adjustment factor for four to six conductors is 80 percent (Table 310.15(C)(1)), applied to the 90 C ampacity of 30 A, giving 24 A. The two 210.19(A)(1) checks are then: 125 percent of 16 A is 20 A, against the unadjusted 75 C ampacity of 25 A, which passes; and the actual 16 A load against the adjusted 24 A, which passes. The mistake is multiplying them, 16 A times 1.25 then divided by 0.80 for 25 A, and concluding 12 AWG is too small. The code does not stack the two factors.

What changes the answer

• A 100 percent rated assembly. Some commercial panelboards with 100 percent rated breakers lift the 125 percent on the overcurrent device, and only where the whole assembly is listed that way. The conductor sizing still applies under 210.19(A)(1).
• An energy management system. Where a listed system under Article 750 limits the load, the calculation is done on the managed maximum, not the connected total. That belongs to the EV and load-management cards, not here.
• A nameplate that specifies the circuit. Where the installation instructions name a minimum circuit or maximum breaker, 110.3(B) makes them part of the installation, and they often already include the 125 percent. Do not apply it twice: see related 'Reading an Appliance Nameplate for Circuit Sizing'.

How it fails in the field

The signature of a circuit sized to 100 percent of a continuous load is a breaker that trips after hours, not minutes: the baseboard circuit that trips around midnight in January, the shop lighting that drops late on a Saturday. A breaker that has run near its rating for years often shows discolored plastic around the load lug and a conductor whose insulation has gone brittle near the termination. When you find one, the fix is the circuit size, not a new breaker of the same rating; replacing the breaker resets the clock on the same failure.

References

• NFPA 70, National Electrical Code, 2023 edition as adopted by your jurisdiction: Article 100 (Continuous Load), 210.19(A)(1), 210.20(A), 215.2(A)(1), 215.3, 240.4(D), Table 310.15(C)(1), 422.11, 422.13, Article 424, 625.41
• See related: Branch Circuit Wire and Breaker Pairing Card
• See related: Reading an Appliance Nameplate for Circuit Sizing
• See related: How to Wire an Electric Water Heater Circuit; How to Install Electric Baseboard Heat and a Line-Voltage Thermostat