Why watts and amps are not interchangeable
Watts measure how fast energy is delivered. Amps measure how much charge flows per second. You can only get from one to the other if you also know the voltage pushing that charge — and, on AC, how well the current lines up in time with the voltage. That last qualifier is power factor, and it is the reason a 5 kW motor and a 5 kW heater on the same 240 V supply draw different currents.
The conversion matters because the two quantities size different things. Watts size your energy bill: kilowatt-hours are watts multiplied by time, which is what the appliance energy cost calculator turns into dollars. Amps size your hardware: conductors, breakers, contactors, plugs and the copper in a transformer are all chosen from current, because current is what produces heat in a conductor. Get the current wrong and you either buy more copper than you need or you overload a circuit.
Three relationships cover every case. On DC, power is simply voltage times current. On single-phase AC, you insert power factor. On balanced three-phase AC, you insert power factor and multiply by √3, using the line-to-line voltage. Everything else on this page is a consequence of those three.
The formula, and where the √3 comes from
For DC, P = V × I, so I = P / V. There is no phase angle on DC, so power factor is 1 by definition and apparent power equals real power. A 60 W LED bar on a 12 V vehicle system draws 60 ÷ 12 = 5 A.
For single-phase AC, P = V × I × PF, so I = P / (V × PF). Voltage and current are both sinusoids. If they peak together — a purely resistive load such as a heating element — the product averages to V·I and power factor is 1. If the load is inductive, as a motor or transformer winding is, current lags voltage and the average of the product falls. Power factor is exactly that ratio: real watts divided by volt-amperes. A power factor of 0.8 means the conductors carry 25% more current than the real power alone would suggest, because 1 ÷ 0.8 = 1.25.
For balanced three-phase, P = √3 × VLL × Iline × PF. The √3 is not a fudge factor. Each of the three phases delivers VLN × I × PF, so total power is 3 × VLN × I × PF. In a wye system the line-to-line voltage is √3 times the line-to-neutral voltage, so VLN = VLL / √3, and 3 ÷ √3 = √3. Substituting gives the familiar form. The practical consequence: always use the line-to-line voltage — 208, 240, 400, 480 or 600 V — never the 120 or 277 V you measure to neutral. The three-phase power calculator works the same relationship in every direction.
Apparent power follows from the same expression without the power factor: S = √3 × VLL × I, expressed in volt-amperes. Transformers, generators and UPS units are rated in kVA rather than kW precisely because their limit is heating from current, not the useful work the load extracts. Size them with the transformer kVA sizing calculator.
Worked example: a 1,500 W heater, then a 100 kW three-phase load
Case 1 — single-phase. A 1,500 W portable heater on a 120 V receptacle. The element is resistive, so power factor is 1.
- Current. I = 1,500 ÷ (120 × 1.0) = 12.5 A.
- Apparent power. S = 120 × 12.5 = 1,500 VA = 1.5 kVA. At unity power factor, kVA and kW are the same number.
- Continuous sizing. A heater runs more than three hours, so the NEC treats it as a continuous load: 12.5 × 1.25 = 15.625 A.
- Breaker. The first standard rating at or above 15.625 A in NEC 240.6(A) is 20 A. A 15 A circuit will not legally carry this heater continuously, which is exactly why 1,500 W heaters trip 15 A bedroom circuits when anything else is plugged in.
Case 2 — three-phase. A 100 kW resistive process heater on a 480 V, three-phase supply, power factor 1.
- Denominator. √3 × 480 = 1.7320508 × 480 = 831.38.
- Current. I = 100,000 ÷ 831.38 = 120.28 A in each of the three line conductors.
- Continuous sizing. 120.28 × 1.25 = 150.35 A, so the first standard rating above it is 175 A.
Now make it a motor instead, with a power factor of 0.85 at full load. I = 100,000 ÷ (831.38 × 0.85) = 100,000 ÷ 706.68 = 141.51 A. The same real power, the same voltage, 17.6% more current — and because conductor heating goes as I²R, the copper now dissipates 1.176² = 1.384, or 38% more heat, for exactly the same useful output. That gap is what power factor correction buys back. Note that for an actual motor branch circuit the NEC requires you to size from the table values in Article 430 rather than from the nameplate current; use the motor full load amps calculator for that.
How to read the result
The current figure is the number you carry into every downstream decision, but it is not by itself a wire size. Conductor selection starts from ampacity tables in NEC Article 310, then applies correction factors for ambient temperature and for the number of current-carrying conductors bundled together, then checks the terminal temperature rating of the equipment — a 75 °C column value is not usable on a breaker listed for 60 °C terminations. Feed the current into the wire size and ampacity calculator rather than reading a table by eye.
On long runs, the binding constraint is usually voltage drop rather than ampacity. The NEC recommends, in informational notes rather than as a requirement, keeping branch-circuit drop to about 3% and the combined feeder-plus-branch drop to about 5%. A conductor perfectly adequate on ampacity can still deliver 108 V to a motor 200 feet away; the voltage drop calculator settles it.
Read the kVA figure whenever a generator, transformer or inverter is involved. Those machines are limited by current and voltage, not by the useful work the load extracts, so a 25 kVA alternator delivers 25 × 0.8 = 20 kW into a 0.8-power-factor load and only 25 × 0.6 = 15 kW into a 0.6-power-factor one. The kVA limit binds first in both cases, which is why sizing a standby set from kW alone leaves you with an undersized alternator whenever the load is reactive; the generator sizing calculator handles starting current as well.
Finally, sanity-check the magnitude. On a 120 V circuit, 1,000 W is a little over 8 A. On 240 V it is a little over 4 A. On 480 V three-phase it is 1.2 A. If your answer is an order of magnitude away from those anchors, you have almost certainly mixed watts with kilowatts or used a line-to-neutral voltage in a three-phase formula.
Line current for common loads and supply voltages
| Real power | 120 V, 1-phase | 240 V, 1-phase | 208 V, 3-phase | 480 V, 3-phase |
|---|---|---|---|---|
| 500 W | 4.17 A | 2.08 A | 1.39 A | 0.60 A |
| 1,000 W | 8.33 A | 4.17 A | 2.78 A | 1.20 A |
| 1,500 W | 12.50 A | 6.25 A | 4.16 A | 1.80 A |
| 2,000 W | 16.67 A | 8.33 A | 5.55 A | 2.41 A |
| 3,000 W | 25.00 A | 12.50 A | 8.33 A | 3.61 A |
| 5,000 W | 41.67 A | 20.83 A | 13.88 A | 6.01 A |
| 10 kW | 83.33 A | 41.67 A | 27.76 A | 12.03 A |
| 25 kW | 208.33 A | 104.17 A | 69.39 A | 30.07 A |
| 50 kW | 416.67 A | 208.33 A | 138.79 A | 60.14 A |
√3 × 208 = 360.27 and √3 × 480 = 831.38; every three-phase figure above is the power divided by one of those two numbers.
The 125% rule, and what this calculator does not decide
NFPA 70, the National Electrical Code (2023 edition), defines a continuous load as one whose maximum current is expected to continue for three hours or more. Article 210 requires the branch-circuit conductor and the overcurrent device to have a rating not less than 125% of that continuous load, and Article 240.6(A) fixes the standard ampere ratings this calculator selects from: 15, 20, 25, 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 110, 125, 150 A and upward to 6,000 A. That same section adds 1, 3, 6, 10 and 601 A as standard ratings for fuses only, so a small control or DC circuit can legitimately be protected below 15 A even though the breaker list this calculator searches starts there. Motor circuits follow Article 430 instead, where the branch-circuit protective device is sized from table full-load currents and may be set well above the conductor ampacity to allow starting current. Nothing on this page substitutes for the code book or for a licensed electrician's judgement about your specific installation.
Mistakes that produce a wrong current
- Using line-to-neutral voltage in the three-phase formula. On a 208Y/120 V system the line-to-line voltage is 208 V. Entering 120 V overstates the current by a factor of 1.73.
- Applying √3 to a single-phase load fed from a three-phase panel. A 120 V circuit taken from one phase of a three-phase panel is a single-phase circuit. The √3 belongs only to balanced three-phase loads.
- Confusing kVA with kW. They are equal only at unity power factor. A load quoted in kVA is already apparent power: multiply it by 1,000 and divide by √3 × VLL to get amps directly, with no power-factor term at all. A 25 kVA three-phase load at 480 V is 25,000 ÷ 831.38 = 30.07 A.
- Ignoring efficiency on motors. A motor's nameplate horsepower is output shaft power. Electrical input is output ÷ efficiency, so a 10 hp (7,460 W) motor at 90% efficiency draws current corresponding to about 8,290 W of input.
- Treating the calculated current as the breaker size. Continuous loads need 125%, motor circuits follow Article 430, and conductors need their own ampacity check with derating.
- Using nameplate power factor at part load. An induction motor's power factor falls sharply below about half load — often to 0.5 or lower — so a lightly loaded motor draws far more current per watt than its nameplate suggests.
Key terms
- Real power (W)
- The average rate at which the load converts electrical energy into heat, light or mechanical work. This is what a kilowatt-hour meter records.
- Apparent power (VA, kVA)
- The product of RMS voltage and RMS current without regard to phase. It sets the thermal limit of transformers, generators and conductors.
- Power factor
- Real power divided by apparent power, between 0 and 1. It combines displacement (phase lag from inductance) and distortion (harmonics from non-linear loads).
- Line current
- Current in one of the supply conductors. In a balanced three-phase system all three line currents are equal, which is why one figure describes the circuit.
- Continuous load
- A load whose maximum current is expected to continue for three hours or more, per the NEC. It is sized at 125%.
Limits of this conversion, and when to use something else
This calculator assumes a balanced, sinusoidal supply and a steady load. Three real-world effects break those assumptions. First, harmonics: variable-frequency drives, LED drivers and switch-mode supplies draw non-sinusoidal current, so true power factor is lower than displacement power factor and the neutral of a three-phase, four-wire system can carry more current than any line conductor. Second, unbalance: if the three phases carry different currents the √3 relationship no longer describes the system, and you must treat each phase separately. Third, inrush: a motor draws several times its running current for the first seconds after start, and a transformer draws a large magnetising surge on energisation, neither of which appears in a steady-state conversion.
For motors specifically, the code does not want you to use nameplate amps for conductor and overcurrent sizing at all — Article 430 sends you to the full-load current tables. For battery and off-grid DC systems, the current side of the calculation drives cable size heavily because the voltages are low and the currents correspondingly large; a 3,000 W inverter on a 12 V bank draws 250 A before losses, which is why 24 V and 48 V systems dominate above about 1 kW. The inverter sizing calculator and battery bank sizing calculator pick that up.
If what you actually want is the energy cost rather than the hardware size, stop at the watts. Current tells you nothing about your bill; only watts multiplied by hours does.