How this instrument works
Efficiency is defined as the ratio of useful output to total input: η = P_out ⁄ P_in. Rearranged for the number you usually don't have, that becomes P_in = P_out ⁄ η — output divided by a fraction less than one, which is why input always comes out larger than output. The shape of the formula is the whole story: dividing by a proper fraction inflates the result, and the amount of inflation tells you exactly how much of what you pay for never reaches the shaft.
The mechanical output is what a motor's nameplate actually specifies — the torque-times-speed figure the manufacturer guarantees at the shaft, stamped in watts or horsepower. It is not the number your electricity meter sees. Between the wall and the shaft sit resistive losses in the copper windings (I²R heating), magnetic losses in the iron core, and mechanical losses in bearings and windage, all of which show up as the difference P_in − P_out rather than as motion. A typical general-purpose induction motor loses 5 to 15 percent of its input this way, worse if it is undersized for its load or running hot.
The formula has a sharp edge at η = 0: input power goes to infinity, which is the correct description of a stalled or locked-rotor motor drawing current and producing zero shaft work — every watt in becomes heat with nowhere else to go. At the opposite extreme, η = 1 gives P_in = P_out, the frictionless, lossless ideal no physical machine reaches. Real efficiencies for line-operated AC motors sit between roughly 0.7 and 0.97, tighter for larger frames, which is why the field here refuses a value of zero or below.
- Enter the motor's nameplate rating into Mechanical output power — switch the unit menu to hp if that is how it's stamped on the plate.
- Enter Motor efficiency as a decimal, not a percentage: an 85% rated motor is typed as 0.85.
- Read Required electrical input power — this is the real power draw the supply, wiring, or breaker actually has to carry.
- Switch that field's unit to kW for larger industrial motors where watts becomes an unwieldy number of digits.
Worked example — sizing the supply for a 1 hp motor
A motor rated at 750 W of mechanical output — the 1 hp figure stamped on its plate — runs at a typical 85% efficiency. Enter mechOutput as 750 and efficiency as 0.85: P_in = 750 ⁄ 0.85 = 882.352941176 W, which the instrument rounds for display to about 882.4 W. That is the real electrical power the motor pulls from the line to deliver its rated 750 W of shaft work.
The missing 132 W does not disappear; it becomes waste heat in the windings and bearings, which is exactly why a nameplate's mechanical rating always understates what the motor actually costs to run. Left on continuously, that 882 W draw is roughly 7,730 kWh a year — the efficiency figure alone, not the shaft rating, is what a utility bill or a generator sizing calculation needs to start from.
Questions
Why is electrical input power always higher than mechanical output power?
No real motor converts electricity to shaft work without loss. Resistive heating in the windings, magnetic losses in the iron core, and friction in the bearings all consume part of the input before it becomes motion, so dividing output by an efficiency below 1 always returns a larger input figure. Only a hypothetical η = 1 motor would draw exactly its rated output.
I typed my motor's efficiency as 85 instead of 0.85 — why is the answer wrong?
The Motor efficiency field expects a decimal fraction, not a percentage. Typing 85 divides 750 W by 85 and returns a nonsensical 8.8 W, since the formula treats 85 as 8,500% efficient. Convert the nameplate percentage first: an 85% rating goes in as 0.85.
Can I use the result to size a circuit breaker or conductor?
Treat it as a real-power sanity check, not the final word. Breaker and wire sizing for motor circuits follows standardized full-load current tables, such as NEC Table 430.250, which already fold in typical power factor and code-mandated safety margins rather than a raw watts-to-amps conversion from nameplate efficiency alone.
Why won't the calculator accept an efficiency of zero?
At zero efficiency the formula divides by zero, which describes a stalled or locked-rotor motor: current flows in, but no shaft work comes out, so every watt becomes heat with nowhere else to go. That condition is real but not a steady operating point, so the field requires a value greater than zero.
Does this figure include power factor for AC motors?
No. P_in here is real power in watts, the same quantity a wattmeter or utility meter reports. Apparent power in volt-amperes, which is what determines current draw on an AC circuit, also depends on power factor and needs voltage and current entered separately to compute.