ElecAS
Electrical Unit Converter: Amps, kW, kVA and Motor HP
Free, browser-based electrical unit converter for Australian and New Zealand design work: convert in any direction between line current in amps, real power in kW, apparent power in kVA and motor shaft power in horsepower. Choose single-phase or three-phase, set the nominal voltage (defaulting to 230 V single-phase and 400 V three-phase), and enter a power factor and motor efficiency where the conversion needs them: the √3 factor, the power factor and the efficiency are applied automatically, and the formula plus the substituted numbers are shown alongside every result so the working can be checked or copied into a calculation sheet. Power factor is only requested for the conversions that actually depend on it (amps to kVA and kW to HP do not), horsepower is treated as mechanical shaft output at 0.746 kW per HP with the electrical input derived through the motor efficiency, and three-phase results assume a balanced load.
Key facts
- Three-phase line current from real power is I = (kW × 1000) ÷ (√3 × V × PF). At 400 V and 0.8 power factor, 1 kW draws about 1.80 A.
- Apparent power converts to line current without a power factor: I = (kVA × 1000) ÷ (√3 × V) for three-phase, so 100 kVA at 400 V is 144 A.
- Australia and New Zealand use 230 V single-phase and 400 V three-phase at 50 Hz, and the converter defaults to those two voltages.
- 1 HP is 0.746 kW of mechanical shaft output; the electrical input a motor draws is HP × 0.746 ÷ motor efficiency.
Who this page is for
Electrical engineers, designers, electricians, estimators and apprentices needing a quick amps / kW / kVA / HP conversion while sizing circuits, checking motor full-load current, reading equipment nameplates, converting a transformer or generator kVA rating to line current, or sanity-checking a load schedule on Australian and New Zealand 230 V and 400 V systems.
Relevant standards
- AS 60038 (Standard Voltages: the 230 V single-phase and 400 V three-phase nominal defaults the converter starts from)
What this tool helps with
- Converts in any direction between amps, kW, kVA and motor HP: twelve conversion pairs, with the source and target units chosen independently.
- Handles single-phase and three-phase systems at any nominal voltage, defaulting to 230 V single-phase and 400 V three-phase and applying the √3 factor automatically for three-phase.
- Asks for a power factor only where the conversion depends on it (amps to kW, kW to kVA, kVA to HP) and hides the field where it does not (amps to kVA, kW to HP).
- Treats horsepower as mechanical shaft output at 0.746 kW per HP and converts through the motor efficiency, so HP to kW returns the electrical input power rather than the shaft power.
- Shows the formula and the substituted values behind every result, so the working can be checked or transcribed straight into a calculation sheet.
- Suits motor full-load current checks, nameplate readings, transformer and generator kVA to line-current conversions, and quick load-schedule sanity checks.
- A conversion utility only: it performs no AS/NZS compliance check; use Cable Selection, Voltage Drop and Maximum Demand for the standards-referenced design calculations.
How to use the ElecAS electrical unit converter
- Enter the value and choose the unit to convert from: Type the figure into the large input field and pick its unit: amps (A), power (kW), apparent power (kVA) or motor horsepower (HP).
- Choose the unit to convert to: Select the target unit from the four buttons. The unit already in use as the source is disabled, so the twelve valid conversion pairs are the only options offered.
- Set the system parameters: Choose single-phase or three-phase, which sets the voltage to 230 V or 400 V by default, and override the voltage if the system differs. Enter a power factor and a motor efficiency where those fields appear: they are shown only for the conversions that use them.
- Read the result and the working: The converted value is shown with its unit, alongside the formula applied and the substituted numbers, so the calculation can be checked or copied into a calculation sheet.
Converting between amps, kW, kVA and horsepower on Australian systems
Amps, kW and kVA: the relationships the converter uses
Apparent power is what the supply has to deliver and real power is what the load consumes; the power factor links them, so kW = kVA × PF and kVA = kW ÷ PF. Line current follows from apparent power alone: for three-phase I = (kVA × 1000) ÷ (√3 × VLL), and for single-phase I = (kVA × 1000) ÷ V. Because the power factor is already inside the kVA figure, converting kVA to amps never asks for one: which is why the converter hides the power factor field for that pair.
Going from real power to current does need the power factor: I = (kW × 1000) ÷ (√3 × VLL × PF) for three-phase and I = (kW × 1000) ÷ (V × PF) for single-phase. A 100 kW load at 400 V three-phase and 0.85 power factor draws 100,000 ÷ (1.732 × 400 × 0.85) = 169.8 A, and the same load expressed as apparent power is 100 ÷ 0.85 = 117.6 kVA: which converts back to the same 169.8 A. On single-phase, 5 kW at 230 V and 0.9 power factor is 5000 ÷ (230 × 0.9) = 24.2 A.
Motor horsepower, efficiency and full-load current
Horsepower is a mechanical rating: it describes the power delivered at the motor shaft, not the electrical power drawn from the switchboard. One mechanical horsepower is 0.746 kW at the shaft. The electrical input is always larger, because the motor loses some of what it draws to heat, windage and friction: so input kW = HP × 0.746 ÷ efficiency, and in the other direction HP = kW × efficiency ÷ 0.746. The converter follows that convention in both directions, which is why its HP to kW result is the electrical input power rather than 0.746 kW per HP.
Chaining the two relationships gives motor full-load current directly: I = (HP × 746) ÷ (√3 × VLL × PF × efficiency) for three-phase. A 10 HP motor at 400 V, 85% efficiency and 0.8 power factor draws 7.46 kW at the shaft, 8.78 kW electrical, 10.97 kVA and about 15.8 A per phase. Treat that as a design estimate: the nameplate full-load current governs for real motor circuits, because the actual efficiency and power factor vary with load, motor design class and speed, and both fall away sharply on a lightly loaded motor.
Choosing the voltage and power factor to enter
Selecting single-phase or three-phase sets the nominal voltage to 230 V or 400 V respectively, which are the AS 60038 standard voltages used across Australia and New Zealand. Both can be overridden with any value: 415 V or 240 V for an older installation, 11,000 V for a high-voltage rating, 110 V or 480 V for imported equipment. The three-phase voltage is the line-to-line value and the current returned is the line current, on the assumption that the three phases are balanced.
The power factor defaults to 0.8, a conservative figure for mixed industrial load. Resistive load such as heating and incandescent lighting sits at or close to 1.0; modern LED and electronic loads are typically 0.9 or better; induction motors run around 0.8 to 0.85 at full load and much lower when lightly loaded. Motor efficiency defaults to 0.85, which is reasonable for a small to mid-size induction motor: use the nameplate or IE-class figure where it is known. Both fields only appear for the conversions that use them.
What sits outside this converter
This is a conversion utility, not a compliance tool. It applies no diversity or maximum-demand assessment to the figures entered, sizes no cable or protective device, and checks nothing against AS/NZS 3000 or AS/NZS 3008.1.1: the current it returns is the load current at the conditions entered, and it is the starting point for those calculations rather than a substitute for them.
It also stays out of reactive-power territory: it does not calculate kVAr or size capacitor banks (the Power Factor Correction calculator does that), does not convert between line and phase quantities inside a star or delta winding, and does not convert cable sizes between mm² and AWG. Direct current, unbalanced three-phase and harmonic-distorted load are all outside its scope, as is any starting or inrush current: the result is a steady-state, balanced, sinusoidal figure.
Key terms
Line current
- The current flowing in one active conductor of the supply, in amps. It is what the converter returns for every conversion into amps, and on three-phase it is per phase with the load assumed balanced across all three.
Real power (kW)
- The power actually converted into work or heat by the load, in kilowatts. It is what an energy meter bills and what a motor nameplate quotes as its output in metric markets. Real power equals apparent power multiplied by the power factor.
Apparent power (kVA)
- The product of voltage and current without regard to phase angle, in kilovolt-amps. Transformers, generators and supply agreements are rated in kVA because it is the current-carrying duty that sizes them. Converting kVA to amps needs no power factor.
Power factor (PF)
- The ratio of real power to apparent power, between 0 and 1, equal to cosine of the phase angle between voltage and current. The converter defaults to 0.8 and asks for it only on the conversions that depend on it: amps to kW, kW to kVA and kVA to HP.
Horsepower (HP)
- A mechanical power rating equal to 0.746 kW measured at the motor shaft, common on imported and older motor nameplates. It is an output rating, so the electrical input a motor draws is always higher by the reciprocal of its efficiency.
Motor efficiency
- The proportion of electrical input power a motor converts to shaft output, defaulting to 0.85 in the converter and appearing only for conversions involving HP. It is what separates the mechanical HP rating from the electrical kW the circuit has to supply.
The √3 factor
- The multiplier 1.732 that appears in balanced three-phase power equations because the three phase currents are 120 degrees apart. The converter applies it automatically when the system type is set to three-phase and omits it for single-phase.
Balanced load
- The assumption that the load is shared equally across all three phases, so a single line current describes the whole circuit. Every three-phase result from the converter rests on it; a materially unbalanced installation has to be assessed phase by phase.
Frequently asked questions
How do I convert kW to amps in three-phase?
- For three-phase: I (A) = (kW × 1000) ÷ (√3 × VLL × power factor). At 400 V three-phase and unity power factor, 1 kW ≈ 1.44 A; at a power factor of 0.8 the same 1 kW draws about 1.80 A. For single-phase, drop the √3: I = (kW × 1000) ÷ (V × power factor), so 1 kW at 230 V and 0.8 power factor is about 5.43 A.
How do I convert kVA to kW?
- kW = kVA × power factor, and kVA = kW ÷ power factor. A 100 kVA load at 0.9 power factor delivers 90 kW of real power. Converting kVA straight to line current needs no power factor at all: I = (kVA × 1000) ÷ (√3 × VLL) for three-phase, so 100 kVA at 400 V is 144 A.
What is the difference between kW, kVA and HP?
- kW is the real electrical power the load consumes and kVA is the apparent power the supply has to deliver; the power factor links them (kW = kVA × PF). Horsepower is mechanical shaft output rather than electrical input: 1 HP = 0.746 kW at the shaft, so a motor delivering 10 HP (7.46 kW mechanical) at 85% efficiency draws about 8.78 kW electrical, which at 0.8 power factor is 10.97 kVA.
How do I convert kW to kVA?
- kVA = kW ÷ power factor, and kW = kVA × power factor. For a 100 kW load at 0.85 power factor, kVA = 100 ÷ 0.85 = 117.6 kVA. The three-phase line current then follows from the apparent power without needing the power factor again: I = kVA × 1000 ÷ (√3 × VLL), which for 117.6 kVA at 400 V is 169.8 A.
What voltage should I enter for Australian and New Zealand systems?
- Use 230 V for single-phase and 400 V line-to-line for three-phase: the AS 60038 standard voltages, which the converter selects automatically with the system type. Older installations and some tariff documents still quote 240 V / 415 V, and the actual voltage at the point of supply is typically within +10% / −6% of nominal, so any value can be entered where the site voltage is known.
How do I convert motor HP to amps?
- Convert the shaft rating to electrical input first, then to current: I = (HP × 746) ÷ (√3 × VLL × PF × efficiency) for three-phase, dropping the √3 for single-phase. A 10 HP motor at 400 V, 0.8 power factor and 85% efficiency draws 8.78 kW electrical and about 15.8 A per phase. Use the nameplate full-load current when sizing a real motor circuit, as actual efficiency and power factor vary with loading.