NEC 430.22: Motor Branch-Circuit Conductor Sizing (FLC vs Nameplate)

The rule: 125% of the table full-load current

NEC 430.22(A) is short and blunt. Branch-circuit conductors supplying a single continuous-duty motor must carry an ampacity of at least 125% of the motor's full-load current. That extra 25% covers the motor running at full load continuously and the heating that comes with it, the same logic the Code applies to any continuous load.

The part that trips people is where the full-load current comes from. It comes from the NEC tables: Table 430.247 for DC, 430.248 for single-phase AC, 430.249 for two-phase, and 430.250 for three-phase AC. You read the value for the motor's horsepower and voltage. You do not read the motor nameplate. That instruction lives in 430.6(A)(1), and it governs both the conductor and the overcurrent device.

• Required conductor ampacity = table FLC x 1.25
• FLC comes from Tables 430.247 through 430.250 (by HP and voltage)
• Then read the conductor size off Table 310.16 at the correct temperature column

Why 430.6(A) splits the table FLC from the nameplate FLA

The Code deliberately separates two currents that live on the same motor. The table FLC is a standardized, conservative value tied to horsepower and voltage. The nameplate FLA is the specific current that one motor was tested to draw. They exist for different reasons.

Conductors and the short-circuit device protect the wiring, and that wiring should stay valid if a tradesman swaps in a different 10 HP motor next year. So those decisions ride on the standardized table value, which does not move. Overload protection guards that one physical motor against running too hot, so it has to track the real motor, which means the nameplate FLA. Once you can say which job you are doing, the current to use is automatic.

Worked example: single 10 HP, 230V, 3-phase motor (430.22)

A continuous-duty 10 HP, 230V, three-phase motor. The nameplate happens to read 26 A. Size the branch-circuit conductor.

• Step 1. Identify the job: this is conductor sizing, so 430.22 and the TABLE FLC apply. Ignore the nameplate 26 A.
• Step 2. Look up FLC in Table 430.250, 10 HP, 230V column: 28 A.
• Step 3. Apply 125% per 430.22: 28 x 1.25 = 35 A required ampacity.
• Step 4. Read Table 310.16, 75C copper column: 10 AWG = 35 A, 8 AWG = 50 A.
• Step 5. Smallest conductor that meets 35 A at 75C: 10 AWG copper (35 A is an exact match).
• Step 6. Answer: 10 AWG copper minimum.

The trap answer here is 32.5 A, which is 26 A nameplate x 1.25. It looks like a clean calculation, and it is wrong, because 430.6(A) told you to use the 28 A table value, not the nameplate.

Worked example: feeder for several motors (430.24)

When one feeder serves more than one motor, you move from 430.22 to 430.24. The rule changes shape: apply 125% to the largest motor's FLC only, then add every other motor's FLC at 100%. The 125% is a one-time adder on the single biggest motor, not a blanket multiplier on the whole group.

Feeder serving three continuous-duty 230V, three-phase motors: a 10 HP (FLC 28 A), a 5 HP (FLC 15.2 A), and a 3 HP (FLC 9.6 A), all from Table 430.250.

• Step 1. Find the largest motor by FLC: the 10 HP at 28 A.
• Step 2. Apply 125% to that one only: 28 x 1.25 = 35 A.
• Step 3. Add the other motors at 100% of their FLC: 15.2 A + 9.6 A = 24.8 A.
• Step 4. Total feeder demand: 35 + 24.8 = 59.8 A.
• Step 5. Read Table 310.16, 75C copper column: 6 AWG = 65 A.
• Step 6. Answer: 6 AWG copper minimum for the feeder.

The wrong answer the exam plants is the one where you take 125% of all three motors: (28 + 15.2 + 9.6) x 1.25 = 66.0 A. That pushes you to a larger conductor and a different lettered choice. Only the largest motor gets the 25% bump.

The traps the exam writes around

• Using the nameplate FLA when 430.22 calls for the table FLC. This is the single biggest motor-question miss, and the nameplate number is always offered as a choice.
• Applying 125% to every motor on a 430.24 feeder instead of only the largest one.
• Forgetting the 125% entirely and sizing the conductor straight off the FLC at 100%.
• Confusing the three sizing decisions: conductor (125% of table FLC, 430.22), overload (115-125% of nameplate FLA, 430.32), and short-circuit/ground-fault (Table 430.52 percentages of table FLC, 430.52).
• Reading the 460V column for a 230V motor, or pulling from single-phase Table 430.248 when the motor is three-phase (Table 430.250).
• Skipping the termination temperature rule: standard terminations cap you at the 75C column of Table 310.16 even when the wire is rated 90C.

How this shows up on the exam

Expect a direct single-motor conductor question: a horsepower and voltage, sometimes a nameplate current dangled as bait, and a request for minimum branch-circuit conductor ampacity or size. The work is table FLC, times 1.25, into Table 310.16. The whole question is testing whether you reached for the table or the nameplate.

Expect a multi-motor feeder question that hands you two or three motors and asks for feeder ampacity. The work is 125% of the largest FLC plus the sum of the rest at 100%. The discriminator is whether you applied the 25% once or to everyone.

And expect a discrimination question that gives you both numbers and asks for a specific sizing decision: conductor, overload, or short-circuit protection. There is no math trick here. It is purely whether you know which current and which percentage belong to which job. Keep the three straight and these are free points.

The key exception worth remembering

430.22(A) carries an exception for motors used in short-time, intermittent, periodic, or varying duty. For those, the conductor ampacity is based on a percentage of the nameplate current rating taken from Table 430.22(E), and that percentage depends on the duty classification and the motor's rated time. Continuous duty is still the default the exam assumes unless the question states otherwise, so unless you are told the motor is intermittent or varying duty, stay with the straight 125% of table FLC.