Full-load amps, and why there are two of them
Full-load current is the line current an induction motor draws when it is delivering its rated shaft horsepower at rated voltage, rated frequency and rated speed. It is the anchor number for everything downstream: conductor size, conduit fill, disconnect rating, contactor size, overload relay setting, transformer capacity and voltage drop all trace back to it.
The confusion that costs electricians re-inspections is that two different full-load currents exist for the same motor, and they are not interchangeable. Nameplate full-load amps (FLA) is the manufacturer’s measured current for that specific machine. It reflects the real efficiency and real power factor of that design, so a NEMA Premium motor and a 1970s standard-efficiency motor of the same horsepower have noticeably different nameplate FLAs. Full-load current (FLC) is the generic value published in NEC Table 430.248 for single-phase motors and Table 430.250 for three-phase motors. It is deliberately conservative and deliberately generic, so that a conductor sized today still works when the motor is replaced with a different brand in fifteen years.
NEC 430.6(A)(1) settles which one governs. Where the Code says to use a motor’s current rating for conductor ampacity, ampacity correction, disconnect ratings and branch-circuit short-circuit and ground-fault protection, you use the table value, not the nameplate. The single documented exception runs the other way: overload protection under 430.32 is sized on the nameplate FLA, because an overload relay is protecting that particular motor’s windings.
This calculator gives you both, side by side, so you never have to guess which one a plan reviewer is going to check.
Where the horsepower-to-amps formula comes from
The formula is three ideas stacked in order. Start at the shaft and work backwards to the supply.
First, convert mechanical output to watts. One horsepower is 745.7 watts, universally rounded to 746 in motor work. A 10 HP motor delivering rated load is putting 7,460 W of mechanical power into the coupling.
Second, add the losses. A motor is not free. Stator copper loss, rotor loss, core loss, friction and windage all mean the electrical input exceeds the shaft output. Divide by efficiency to get input watts: at 91.7% efficiency, 7,460 W of output needs 7,460 ÷ 0.917 = 8,135 W of input. Note the direction — you divide, never multiply, because input is always the larger number.
Third, convert watts to amps. In a balanced three-phase circuit, real power is P = √3 × VL × IL × PF. The √3 is not a fudge factor: it appears because line-to-line voltage in a wye system is √3 times the phase voltage, while line current equals phase current, and the three phases sum with 120° between them. Power factor enters because only the component of current in phase with the voltage does work; the magnetising current an induction motor needs to build its rotating field contributes amperes without contributing watts. Rearranging for current gives I = Pin ÷ (√3 × V × PF).
Put together, I = HP × 746 ÷ (√3 × V × η × PF). For a single-phase motor, delete the √3 and nothing else changes. If you want to work in the other direction — from measured amps back to power — use the three-phase power calculator, which handles the kW, kVA and kVAR triangle directly.
Two properties of the formula are worth internalising. Current is inversely proportional to voltage, so the same motor on 460 V draws almost exactly half what it draws on 230 V. And current is inversely proportional to the product η × PF, so a poor power factor raises amperes without raising the work done — which is exactly the problem a power factor correction capacitor is installed to fix.
Worked example: 10 HP, 460 V, three-phase, 91.7% efficient, 0.85 PF
Take a 10 HP NEMA Premium four-pole motor on a 480 V nominal system (rated 460 V), with 91.7% nameplate efficiency and 0.85 nameplate power factor.
- Shaft output in watts. 10 × 746 = 7,460 W.
- Electrical input. 7,460 ÷ 0.917 = 8,135.2 W, or 8.135 kW.
- Build the denominator. √3 = 1.73205, so 1.73205 × 460 = 796.74 V. Multiply by the power factor: 796.74 × 0.85 = 677.23.
- Divide. 8,135.2 ÷ 677.23 = 12.01 A calculated full-load current.
- Apparent power. 8.135 kW ÷ 0.85 = 9.57 kVA. Cross-check: √3 × 460 × 12.01 ÷ 1000 = 9.57 kVA.
- Look up the code value. NEC Table 430.250, 10 HP row, 460 V column: 14 A. That is 16.5% above the calculated 12.01 A, which is normal — the table was built around older, less efficient designs.
- Size the conductors. NEC 430.22 requires 125% of the table FLC: 1.25 × 14 = 17.5 A minimum conductor ampacity. At 75°C that is satisfied by 12 AWG copper, subject to the terminal temperature limits and any derating.
Notice what would have happened if you had sized from the calculated 12.01 A: 1.25 × 12.01 = 15.0 A, which still lands on 12 AWG here. The gap between the two paths is not a fixed percentage across motor sizes — it depends on how far each particular motor's efficiency and power factor sit from the generic design the table was built around — so sooner or later it straddles an ampacity step and changes the wire size. Feed the 17.5 A figure into the wire size and ampacity calculator and then check the run length with the voltage drop calculator.
Which number to use where
Use the table FLC for anything that is about the circuit, and the nameplate FLA for anything that is about the motor.
Branch-circuit conductors (430.22). Minimum ampacity is 125% of the table FLC for a continuous-duty motor. Apply ambient and conduit-fill correction after that, not before.
Branch-circuit short-circuit and ground-fault protection (430.52). The maximum device rating is a multiple of the table FLC — 250% for an inverse-time breaker, 300% for a non-time-delay fuse, 175% for a dual-element time-delay fuse, with the next standard size up permitted when the calculated value does not correspond to one. The motor overload and breaker sizing calculator works this through with the article 430 exceptions applied.
Disconnecting means (430.110). Rated at least 115% of the table FLC.
Overload protection (430.32). This one uses the nameplate FLA. A motor with a service factor of 1.15 or greater, or a temperature rise of 40°C or less, gets an overload device set at no more than 125% of nameplate FLA; everything else gets 115%.
As a sanity check on the motor itself, compare your calculated current against the nameplate FLA printed on the same plate. They should agree within a few percent, because they are computed from the same efficiency and power factor. A large disagreement usually means the efficiency or power factor you entered is not the full-load value — nameplates sometimes print part-load figures, and power factor in particular collapses at light load. If you are measuring a running motor rather than sizing a new one, the motor efficiency calculator backs out actual efficiency and load percentage from measured volts, amps and power factor.
Finally, remember that full-load current is a steady-state number. A standard NEMA Design B motor draws several times its full-load current during an across-the-line start, for a few seconds — the exact locked-rotor current follows the code letter on the nameplate, whose locked-rotor kVA-per-horsepower bands are defined in NEMA MG 1. That inrush does not size conductors, but it does size the protective device, the contactor and any upstream transformer.
NEC Table 430.250 excerpt: three-phase full-load current in amperes
| Horsepower | 208 V | 230 V | 460 V | 575 V |
|---|---|---|---|---|
| 1 | 4.6 | 4.2 | 2.1 | 1.7 |
| 2 | 7.5 | 6.8 | 3.4 | 2.7 |
| 3 | 10.6 | 9.6 | 4.8 | 3.9 |
| 5 | 16.7 | 15.2 | 7.6 | 6.1 |
| 7.5 | 24.2 | 22 | 11 | 9 |
| 10 | 30.8 | 28 | 14 | 11 |
| 15 | 46.2 | 42 | 21 | 17 |
| 20 | 59.4 | 54 | 27 | 22 |
| 25 | 74.8 | 68 | 34 | 27 |
| 30 | 88 | 80 | 40 | 32 |
| 40 | 114 | 104 | 52 | 41 |
| 50 | 143 | 130 | 65 | 52 |
| 60 | 169 | 154 | 77 | 62 |
| 75 | 211 | 192 | 96 | 77 |
| 100 | 273 | 248 | 124 | 99 |
Reproduced for reference from NEC Table 430.250. Confirm against the edition your jurisdiction has adopted before stamping a design. Motors built for 200 V and 2300 V systems have their own columns in the full table.
Standard assumed: NEC (NFPA 70), Article 430
This calculator follows NFPA 70, the National Electrical Code, Article 430, as written in the 2023 edition. Table 430.248 covers single-phase alternating-current motors and Table 430.250 covers three-phase alternating-current motors. Section 430.6(A)(1) is the rule that directs you to those tables rather than the nameplate for conductor and device sizing. Adoption is by state and sometimes by municipality, so check which edition is enforced where the work is being done, and confirm the table values against that printing.
Mistakes that make a motor circuit fail inspection
- Sizing conductors from the nameplate. The most common one. On the 10 HP motor worked through above, the nameplate-derived 12.01 A gives a 15.0 A conductor requirement instead of the table's 17.5 A — 14% short of what NEC 430.22 asks for. Whether the two figures happen to land on the same AWG size varies with the motor; 430.6(A)(1) removes the choice either way.
- Reading the wrong voltage column. The table is indexed by motor rated voltage, not system nominal voltage. A motor on a 480 V system is a 460 V motor and belongs in the 460 V column.
- Interpolating between horsepower rows. The table has no rows between the listed ratings because motors are not built between them. If your nameplate says 12.5 HP, you have a metric-rated motor; convert its kW rating and use the nearest standard horsepower the manufacturer assigns.
- Using power factor as a percentage in a hand calculation. Write 0.85, not 85 — an 85 in the denominator makes the current a hundred times too small. This calculator caps the field at 1.0 so it cannot happen here, but it is a standard slip on a scratch pad.
- Forgetting that overloads use the nameplate. Setting an overload relay from the table value defeats the protection on an efficient motor whose real current is lower.
- Ignoring the drive. A motor fed from a variable-frequency drive draws current from the drive, and the drive draws a different current from the supply. Size the supply conductors from the drive’s rated input current per NEC 430.122, not from the motor table.
Where this sits among the other motor calculations
Full-load current is the first number in a chain, not the last. Once you have it, the branch circuit needs conductor ampacity, overcurrent protection, an overload setting, a disconnect and a check on voltage drop over the run. Torque and speed are a separate branch of the same nameplate: shaft torque follows from horsepower and speed, so a 10 HP motor at 1,760 RPM produces 10 × 5,252 ÷ 1,760 = 29.8 lb-ft of full-load torque — the motor torque calculator handles the conversions and the starting-torque multiples.
Upstream, several motors on one feeder are not simply added. NEC 430.24 sizes a feeder at 125% of the largest motor’s FLC plus 100% of the remaining motors’ FLCs, which is why a motor control centre schedule always lists every table value explicitly. That same sum drives transformer selection, and the inrush of the largest motor usually decides whether the transformer holds voltage during a start.
Outside North America, IEC practice replaces the horsepower table with a kW rating and a nameplate current, and there is no equivalent generic table — IEC 60034-1 specifies how ratings are declared, and the installation standard (IEC 60364 or a national derivative such as BS 7671) sizes cables from the declared current. If you are working from a European nameplate, enter the kW rating with the kW unit selected and read the calculated current; the NEC table columns will still appear, but they only apply if the installation is under the NEC.
Key terms
- FLA (full-load amps)
- The current a specific motor draws at rated output, as measured and printed by its manufacturer on the nameplate. Used for overload protection under NEC 430.32.
- FLC (full-load current)
- The generic current for a motor of a given horsepower and voltage as published in NEC Table 430.248 or 430.250. Used for conductors, disconnects and short-circuit protection under NEC 430.6(A)(1).
- Service factor
- A multiplier on rated horsepower that a motor can carry continuously without exceeding its insulation class rise. A 1.15 service factor motor is permitted a higher overload setting.
- Displacement power factor
- The cosine of the angle between the fundamental voltage and current waveforms. For an induction motor it is set by the magnetising current the rotating field requires, and it falls sharply at light load.
- NEMA Design B
- The general-purpose induction motor design covering most integral-horsepower industrial motors: normal starting torque, normal starting current and slip of 5% or less.