Converting amperes to watts
Amperes measure the flow of electric current, while watts measure the rate at which electrical energy is actually converted into useful work or heat. To move between them you also need the voltage — and, for alternating current, the power factor. This calculator applies the correct real-power formula for whichever circuit you select, giving the answer in both watts and kilowatts.
The underlying relationship is Joule's law, P = I × V, extended to alternating-current systems where voltage and current can be out of step. Because a single amp figure can mean very different wattages on a 12 V battery versus a 400 V industrial supply, always pair the current with the right voltage and circuit type.
How the calculation works
The formula changes with the circuit topology:
DC: P = I × V
Single-phase AC: P = PF × I × V
Three-phase AC (line-to-line): P = √3 × PF × I × V_LL
Three-phase AC (line-to-neutral): P = 3 × PF × I × V_LN
| Circuit type | Formula | Power factor | Typical use |
|---|---|---|---|
| DC | P = I × V | Always 1 | Batteries, solar panels, DC motors |
| Single-phase AC | P = PF × I × V | 0 – 1 | Household outlets, single appliances |
| Three-phase AC (L-L) | P = √3 × PF × I × V_LL | 0 – 1 | Enter line-to-line voltage (400 / 415 V) |
| Three-phase AC (L-N) | P = 3 × PF × I × V_LN | 0 – 1 | Enter phase voltage (230 / 240 V) |
Real power, apparent power and power factor
On an AC circuit, multiplying current by voltage gives the apparent power in volt-amperes (VA) — the load the supply and wiring must physically carry. The realpower in watts is the portion that does useful work, and it equals the apparent power scaled by the power factor (the cosine of the phase angle between voltage and current). The remainder is reactive power, which sloshes back and forth without net energy transfer.
The table below shows typical power factor values for common load types. If you cannot measure the actual PF with a clamp meter, use these figures as starting estimates:
| Device / load type | Typical PF | Notes |
|---|---|---|
| Resistive heater / incandescent bulb | 1.00 | Purely resistive — watts equal volt-amperes |
| LED driver (with PFC) | 0.90 – 0.99 | Active power-factor correction built in |
| Induction motor — full load | 0.80 – 0.90 | Standard default for most motor loads |
| Fluorescent lamp with ballast (PFC) | 0.90 – 0.95 | Corrected electronic ballast |
| Fluorescent lamp (uncorrected) | 0.45 – 0.60 | Magnetic ballast, no correction |
| PC power supply (80 Plus certified) | 0.90 – 0.99 | Active PFC required for 80 Plus Silver+ |
| UPS (uninterruptible power supply) | 0.70 – 0.95 | Varies widely by model and load level |
| Induction motor — no load | 0.20 – 0.35 | Highly reactive at light load |
If you are watching your electricity spend, converting the power you calculate here into running cost is the natural next step — an LED savings calculator does exactly that for lighting, and a quick percentage calculator helps you compare the wattage of two appliances at a glance.
Quick reference: amps to watts at common voltages
The table below shows watts for common current values at four standard supply voltages, assuming a resistive load (power factor = 1). For inductive loads, multiply the tabled value by the actual power factor.
| Current | 12 V DC (Automotive, battery) | 120 V AC (North American outlet) | 230 V AC (European outlet) | 240 V AC (UK outlet / US 240 V) |
|---|---|---|---|---|
| 1 A | 12 W | 120 W | 230 W | 240 W |
| 5 A | 60 W | 600 W | 1,150 W | 1,200 W |
| 10 A | 120 W | 1,200 W | 2,300 W | 2,400 W |
| 15 A | 180 W | 1,800 W | 3,450 W | 3,600 W |
| 20 A | 240 W | 2,400 W | 4,600 W | 4,800 W |
When you know the resistance rather than the voltage, the Ohm's law identity P = I² × R gives the same result (since V = I × R, substituting yields P = I × (I × R) = I² × R). Both routes are equivalent for DC and resistive AC loads.
Worked examples
These figures come straight from the same engine that powers the calculator, so they cannot drift from the tool:
| Scenario | Calculation | Power (W) | Power (kW) |
|---|---|---|---|
| DC — 12 V battery | 5 A × 12 V | 60 | 0.06 |
| Single-phase AC — 230 V | 0.8 × 10 A × 230 V | 1,840 | 1.84 |
| Three-phase AC — 400 V L-L | √3 × 0.9 × 10 A × 400 V | 6,235.38 | 6.235 |
Assumptions and limitations
- Three-phase results assume a balanced load — equal current and voltage on all three phases. Unbalanced systems need per-phase summation.
- All AC voltages and currents are taken as RMS values in sinusoidal steady state.
- The result is real (active) power in watts, not apparent power (VA) or reactive power (VAR).
- Power factor must be measured or known; an incorrect PF scales the answer in direct proportion.
- This is an educational estimate — it is not a substitute for a qualified electrician when sizing conductors, breakers, or power supplies.
Frequently asked questions
How do I convert amps to watts for a DC circuit?+
For direct current, multiply the amperage by the voltage: P (watts) = I (amps) × V (volts). For example, a 5 A device on a 12 V battery draws 5 × 12 = 60 W. DC has no reactive component, so no power factor is needed.
Why do I need a power factor for AC circuits?+
In an AC circuit, voltage and current can be out of phase due to inductors or capacitors in the load. Power factor (cos φ) measures how much of the apparent power (volt-amperes) actually does useful work. A motor with PF = 0.8 converts only 80% of the apparent power into real work; the rest is reactive and oscillates back and forth in the supply. For DC, PF is always 1.
What is the formula for single-phase AC amps to watts?+
Single-phase real power = PF × I × V. Enter the RMS current (amps), the RMS supply voltage (volts), and the power factor (0–1). A 10 A appliance on 230 V with PF 0.8 draws 0.8 × 10 × 230 = 1,840 W.
How does the three-phase amps-to-watts formula work?+
For a balanced three-phase system using line-to-line voltage: P = √3 × PF × I × V_LL. The √3 factor (approximately 1.732) arises from the 120-degree phase displacement between the three phases. Alternatively, using line-to-neutral (phase) voltage: P = 3 × PF × I × V_LN. Both formulas give the same answer because V_LL = √3 × V_LN.
What is the difference between line-to-line and line-to-neutral voltage in three-phase systems?+
Line-to-line voltage (V_LL) is measured between any two of the three live conductors — for example, 415 V in a UK industrial supply. Line-to-neutral voltage (V_LN) is measured between one live conductor and the neutral (star point) — for the same supply, V_LN = 415 / √3 ≈ 240 V. Using the wrong voltage in the formula introduces a √3 error (about 73%), so always confirm which reading your meter is giving.
How do I convert watts to kilowatts?+
Divide the watt figure by 1,000. For example, 6,235 W = 6.235 kW. Kilowatts are the standard unit on electricity bills and on most motor and appliance nameplates.
What power factor should I use if I do not know the actual value?+
A power factor of 0.8 is a common engineering default for inductive loads such as motors and fluorescent lighting. Resistive loads (incandescent bulbs, electric heaters) have PF approximately 1. Modern switched-mode power supplies often have PF correction to 0.95–0.99. If exact energy cost matters, read the actual PF from a clamp meter or the equipment datasheet.
Can this calculator be used to size a circuit breaker or cable?+
This calculator gives real (active) power in watts, which is a useful starting point. However, cable and breaker sizing must account for the full apparent power in volt-amperes (VA = I × V, regardless of power factor), duty cycle, ambient temperature, grouping derating, and local electrical regulations. Always consult a qualified electrician for installation work.
Why does my appliance label show VA instead of watts?+
Appliances with reactive loads — motors, transformers, UPS units — draw more current than a pure-watt figure implies. The VA (volt-ampere) rating reflects the apparent power the supply must handle. Real power (watts) is what the device actually converts to work or heat; the gap between VA and W is the reactive power that the supply wiring must still carry.
What happens if I enter zero for current or voltage?+
The result is zero watts, which is mathematically correct: no current means no charge flow, so no energy transfer; no voltage means no potential difference, so no work is done. The calculator handles this cleanly without errors.
Is this formula valid for all frequencies (50 Hz vs 60 Hz)?+
Yes. The real-power formula P = PF × I × V (and its three-phase variants) holds at any supply frequency because it is derived from the time-averaged product of RMS voltage and RMS current. The frequency determines the impedance of reactive components (which affects PF), but once you have the actual PF value, the watts calculation is frequency-independent.
How do I find the amps if I know the watts and voltage?+
Rearrange the formula: I = P / (PF × V) for single-phase AC, or I = P / (√3 × PF × V_LL) for three-phase. For DC, I = P / V. A watt-to-amps converter applies the same identities in reverse.
Disclaimer
Sources
- HyperPhysics — Electric Power (Georgia State University)
- University of Louisville — ECE252 Three-Phase Power (Dept. of Electrical & Computer Engineering)
- Wikipedia — AC Power
- Wikipedia — Power Factor
- Wikipedia — Three-Phase Electric Power
Formula and data last reviewed by the TheCalculatorVault team on 3 July 2026. Figures are for general information, not professional advice.
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