Motor amps calculator
This computes a current from the horsepower, efficiency and power factor you supply — a nameplate calculation, not the tabulated full-load current your code may require for conductor and overload sizing. Use it to sanity-check, and size the circuit from your code book and the authority over the job, never from this figure alone.
Enter the motor's horsepower and voltage, read the efficiency and power factor off the nameplate, and the tool works out the running current. Because it uses the motor's own numbers rather than a lookup, it stays honest across motor types and code editions.
Ejemplo resuelto
A 5 hp single-phase motor at 240 V, 90% efficient with a 0.85 power factor, takes (5 × 746 ÷ 0.9) input watts = 4,144 W, then 4,144 ÷ (240 × 0.85) ≈ 20.3 A. Wire the same motor three-phase at 240 V and the √3 drops each line to about 11.7 A.
In kilowatts: a 4 kW three-phase motor on 400 V, 90% efficient with a 0.85 power factor, takes 4,000 ÷ 0.9 = 4,444 W in, then 4,444 ÷ (1.732 × 400 × 0.85) ≈ 7.55 A per line. The 5 hp motor above is 5 × 0.746 = 3.73 kW — switch to kilowatts and the tool shows it that way.
En la obra
The nameplate wins. If the plate is readable, the full-load amps on it are the number — the maker measured them. This page is for the motor with the plate painted over, or for a quick check that the plate says something sensible. A 5 hp single-phase at 240 V marked 28 A is a real motor; one marked 8 A is a misprint or a three-phase plate.
Why 120 V motors stop at a couple of horsepower. 3 hp on 120 V single-phase, 85% efficient, 0.8 power factor: 3 × 746 ÷ 0.85 = 2,633 W input, ÷ (120 × 0.8) = 27.4 A running. Starting current is several times that. On 240 V the same motor is 13.7 A, and on 480 V three-phase, 4.0 A.
A 10 hp three-phase at 480. 0.9 efficiency, 0.85 PF: 10 × 746 ÷ 0.9 = 8,289 W; ÷ (1.732 × 480 × 0.85) = 11.7 A. The √3 is what makes three-phase the cheap way to run big motors. What size conductor and overload that needs comes out of the code book, after this number goes in.
Pruebe un ejercicio
Quick referencefor motors
- Input watts
- hp × 746 ÷ efficiency
- Input watts from kW
- kW × 1,000 ÷ efficiency — a 4 kW plate at 0.9 is 4,444 W in
- Monofásico
- A = input W ÷ (V × PF)
- Trifásico
- A = input W ÷ (1.732 × V × PF)
- Tensión
- double the volts, halve the amps — 120 → 240 → 480
- Unidades
- 1 hp = 746 W · 1 kW = 1.34 hp
- What this page does not do
- the nameplate FLA beats it, and conductor, overload and breaker sizing are the code book's
Mistakes to catchbefore you trust the number
- Typing efficiency and power factor as percentages (90, 85) instead of decimals (0.9, 0.85) — the amps come out a hundred times too low.
- Using the calculated running amps as the motor's full-load current for sizing overloads or conductors; the code's FLC table figure is the one your book requires, and this page does not have it.
- Setting single phase for a three-phase nameplate: the current per line is different by √3.
What an inspector looks forthe electrical inspector
On this part of the job
- The nameplate. Conductor sizing, overload protection and disconnect sizing are worked from the motor's nameplate or from the code's own motor table, depending on the rule — not from a horsepower estimate.
- Overload protection sized and set for this motor, with the service factor considered.
- The disconnecting means: present, within sight where required, and rated for the job.
- Branch-circuit protection sized to the motor rules, which deliberately allow more than the conductor ampacity to let the motor start.
- Bonding of the motor frame and any controller enclosure.
Have this ready
- A photograph of the nameplate — it is unreadable once the motor is installed.
- Your code book's motor section.
- The starter or controller data sheet.
Every figure an inspector measures against comes from the code your jurisdiction has adopted, the stamped drawing, or the manufacturer's printed instructions — so none of them are printed here. Look the number up in your own book, type it into the tool, and the arithmetic is yours.
La fórmula
amps = (hp × 746 ÷ efficiency) ÷ (phase × V × PF)
in kilowatts: amps = (kW × 1000 ÷ efficiency) ÷ (phase × V × PF)
Horsepower becomes watts at 746 W each; dividing by efficiency gives the input power the supply must deliver. Dividing that by the voltage and power factor — and by √3 for three phase — turns power into current. Every term is a value you enter, so no code table is involved.
Preguntas
- How do I calculate motor amps from horsepower?
- Convert the horsepower to watts (1 hp is 746 watts), divide by the motor efficiency to get the input power, then divide that by the voltage — and by √3 as well for three phase — and by the power factor. The result is the running current the motor draws from the supply.
- Does this match the full-load current in the code?
- No, and that is the point. This is a calculated value from the motor’s own nameplate figures — its efficiency and power factor — not a figure read from a code table. The two are usually close, but where your rules require the tabulated full-load current for sizing, use that table, not this arithmetic.
- Where do I find efficiency and power factor?
- They are on the motor nameplate, often written as a percentage and a decimal respectively. Enter them as you read them — efficiency as a fraction like 0.9 and power factor as a decimal like 0.85. They vary by motor, so this tool asks for them rather than assuming a value.
- Why is my three-phase current lower than single phase?
- The √3 in the three-phase formula spreads the same power across three lines, so each line carries less current for the same horsepower. That is one of the practical reasons larger motors are wired three-phase — thinner conductors carry the load.
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