How to Convert Watts in Amps (DC, AC Single-Phase, and AC Three-Phase)
Converting watts to amps is a one-line calculation, but the voltage you plug into it changes with the situation: a phone charger sees 5 V DC, a US wall outlet delivers 120 V AC, a kitchen range runs on 240 V AC, and industrial motors sit on three-phase circuits. Use the wrong voltage — or skip the power factor for AC — and the current you size a wire or breaker for will be wrong by a factor of two or more.
The rule of thumb that gets it right most of the time is Amps = Watts ÷ Volts. That formula is exact for DC circuits and for AC single-phase circuits driving resistive loads (heaters, incandescent bulbs, electric ovens, resistive water heaters). Once you add a motor, an LED driver, a fluorescent ballast, or any other inductive load, you also need the power factor (PF) — typically 0.8 to 0.95 for household motors and 0.85 to 0.9 for industrial induction motors at full load. The calculation for a three-phase system adds another factor of √3 (≈1.732) on top of PF and line-to-line voltage.
This guide walks through each of those cases with the exact formulas, a 120 V reference table, and worked examples you can copy. By the end you will know which formula to use, which number to plug into it, and which service panels, breakers, and wire gauges those numbers map onto in the real world.
What you need before starting
Three numbers, in this order:
- Watts (W) — the rated power of the device, printed on the label, in the spec sheet, or calculated as Volts × Amps if you already know the other two.
- Volts (V) — the supply voltage the device will actually see. For US wall outlets that is 120 V for most receptacles and 240 V for electric dryers, ovens, water heaters, and most EV chargers. For DC circuits (batteries, solar, automotive) the common values are 5 V, 12 V, 24 V, and 48 V.
- Power factor (PF) — only required for AC circuits with reactive loads. Use PF = 1 for resistive loads, PF ≈ 0.95 for fluorescent lamps, PF ≈ 0.85 for an induction motor at full load, and PF ≈ 0.9 for a synchronous motor. If the device nameplate lists a PF, use that number instead of the table values.
Once you have all three numbers, pick the formula that matches your circuit type below.
The basic formulas
| Circuit type | Formula | When to use |
|---|---|---|
| DC | $I = \dfrac{P}{V}$ | Batteries, solar panels, USB, automotive, any DC source |
| AC single-phase | $I = \dfrac{P}{V \times PF}$ | US/Canadian wall outlets, residential appliances |
| AC three-phase (line-to-line) | $I = \dfrac{P}{\sqrt{3} \times V \times PF}$ | Industrial motors, commercial HVAC, three-phase service panels |
| AC three-phase (line-to-neutral) | $I = \dfrac{P}{3 \times V \times PF}$ | Rare; legacy three-phase systems with neutral |
P is power in watts, V is RMS voltage, PF is power factor (1.0 for purely resistive loads), and √3 ≈ 1.732.
Why √3 shows up in three-phase
A three-phase system delivers power across three windings offset by 120°. The total instantaneous power is constant — it does not pulse the way single-phase power does — which is why three-phase is more efficient for motors. The math works out that the per-line current is lower by a factor of √3 compared to a single-phase circuit delivering the same power, which is why a 10-horsepower three-phase motor draws roughly 28 A while the equivalent single-phase motor pulls around 70 A. That same √3 factor is the reason three-phase service from the utility is more efficient to transmit over distance.
120V reference table (US residential single-phase)
These numbers assume a standard 120 V single-phase US wall outlet and a resistive load (PF = 1.0). For motor-driven appliances, divide the current by the appropriate PF.
| Watts | Current at 120 V (A) |
|---|---|
| 10 W | 0.083 A |
| 50 W | 0.417 A |
| 100 W | 0.833 A |
| 200 W | 1.667 A |
| 500 W | 4.167 A |
| 1000 W | 8.333 A |
| 1200 W | 10.000 A |
| 1500 W | 12.500 A |
| 1800 W | 15.000 A |
| 2400 W | 20.000 A |
| 3000 W | 25.000 A |
| 5000 W | 41.667 A |
A 1500 W space heater on a 120 V outlet draws exactly 12.5 A — which is why most portable heaters ship with a warning to plug them into a 15 A or 20 A receptacle and to avoid extension cords. Push that same 1500 W through a 120 V / 15 A circuit that already has a lamp and a laptop charger running and you trip the breaker.
Step-by-step: convert watts in amps for the four real situations
Step 1 — DC circuit (battery, solar panel, USB device)
Use $I = P / V$. Power in watts divided by the battery or bus voltage gives the current in amps directly.
Worked example: a 100 W solar panel charging a 12 V battery bank. $I = 100 / 12 = 8.33$ A. The charge controller and the wiring from the panel to the battery need to be sized for at least 8.33 A continuous; the NEC 80% rule means a 10 A fused circuit is the minimum practical size, and you would normally round up to a 15 A fuse to leave margin for irradiance spikes above the 100 W rating.
Step 2 — AC single-phase, resistive load (heater, incandescent, oven)
Same formula: $I = P / V$, with PF = 1.0 implicitly. A resistive load converts every watt it draws into heat or light, so the current you calculate is the actual current drawn.
Worked example: a 240 V electric oven rated at 5000 W. $I = 5000 / 240 = 20.83$ A. The oven circuit in a US home is usually a 30 A or 40 A breaker on 10 AWG or 8 AWG copper wire — sized above the 20.83 A continuous draw to satisfy the NEC 80% rule for continuous loads.
Step 3 — AC single-phase, motor or inductive load
Now PF enters the calculation: $I = P / (V \times PF)$. For most fractional-horsepower motors (fans, small pumps, refrigerator compressors), PF sits between 0.6 and 0.85 at full load.
Worked example: a 1-horsepower (746 W) blower motor at 120 V with PF = 0.75. $I = 746 / (120 \times 0.75) = 8.29$ A. Compare that to the 6.22 A you would get assuming PF = 1.0 — using the wrong PF undersizes the wire by 25%.
Step 4 — AC three-phase motor (industrial, commercial, EV fast chargers)
Three-phase line-to-line: $I = P / (\sqrt{3} \times V \times PF)$, where √3 ≈ 1.732.
Worked example: a 10-horsepower (7460 W) three-phase induction motor at 480 V line-to-line with PF = 0.85 at full load. $I = 7460 / (1.732 \times 480 \times 0.85) = 10.6$ A. The same motor running on single-phase 240 V would draw roughly 36.5 A — which is why production equipment and commercial HVAC almost always use three-phase service when it is available.
Worked examples for common voltage scenarios
How many amps is 3000 watts at 120V?
Direct application of $I = P / V$: $I = 3000 / 120 = 25$ A. On a 120 V / 30 A breaker circuit, that leaves no room for anything else on the circuit, which is why a 3000 W EV charger is usually hardwired on a dedicated 30 A or 40 A branch circuit. Trying to share that circuit with anything else will trip the breaker under sustained load.
How many watts is 1 amp at 120V?
Rearranging the formula: $P = V \times I = 120 \times 1 = 120$ W. At 240 V, the same 1 A delivers 240 W — exactly double. This is one of the reasons 240 V circuits are popular for high-power loads: a 20 A circuit at 240 V delivers 4800 W, while the same 20 A at 120 V delivers only 2400 W.
Is 1200 watts equal to 10 amps at 120V?
Yes. $1200 / 120 = 10$ A. This is the textbook example used to introduce the relationship because the numbers round cleanly: a 1200 W hair dryer on a standard 120 V outlet draws exactly 10 A, which is right at the 80% continuous-load limit of a 15 A circuit.
What is 1500 watts in amps at 120V?
$1500 / 120 = 12.5$ A. This is the standard 1500 W space heater number printed on most product labels. A 15 A circuit can technically handle 12.5 A continuously, but NEC 240.4(D) limits continuous loads to 80% of the breaker rating — 12 A — making a 20 A circuit the safe minimum for a 1500 W portable heater.
What about 12V and 240V?
A 100 W device at 12 V draws $100 / 12 = 8.33$ A — almost ten times what the same device draws at 120 V (0.83 A). This is why low-voltage DC systems (12 V solar, automotive, marine) need much heavier wiring. At 240 V, the same 100 W device draws $100 / 240 = 0.417$ A — half what it draws at 120 V.
Typical power factor values
Use these as a starting point. The nameplate PF on the device is always more accurate.
| Load type | Typical PF |
|---|---|
| Resistive heater, incandescent bulb, resistive oven | 1.0 |
| Fluorescent lamp (with electronic ballast) | 0.95 |
| Incandescent lamp | 1.0 |
| Induction motor — full load | 0.85 |
| Induction motor — no load | 0.35 |
| Synchronous motor — rated load | 0.9 |
| LED driver (typical consumer) | 0.9 – 0.95 |
| Switch-mode power supply (laptop, TV) | 0.6 – 0.7 |
The “no load” PF of 0.35 for an induction motor is the trap: an unloaded motor still draws real current, but most of that current is reactive (it does no useful work), so the PF drops dramatically. Always size motor circuits using the full-load amps on the nameplate, not the no-load current you measure with a clamp meter.
Common mistakes to avoid
- Skipping the power factor on motor loads. A 1 HP motor at 120 V draws 8.29 A at PF = 0.75, not the 6.22 A you would get assuming PF = 1. Undersized wiring on a motor circuit is one of the most common causes of nuisance tripping.
- Using the line-to-neutral voltage on a three-phase motor instead of line-to-line. These two differ by √3. For a 480 V three-phase system, line-to-neutral is 277 V. Plugging 277 V into the three-phase line-to-line formula gives an answer that is off by a factor of 1.732.
- Confusing watts and watt-hours. A 1500 W heater running for 1 hour consumes 1.5 kWh of energy. The watt-hour rating on a battery tells you how long it can deliver a given number of watts — it does not change the watts-to-amps calculation.
- Mixing RMS and peak voltage. The formulas here use RMS voltage, which is what every standard voltmeter and multimeter displays. Peak voltage is √2 (≈1.414) times higher than RMS for a sine wave, but only matters when sizing insulation or designing peak-detecting circuits.
- Forgetting the NEC 80% rule. Continuous loads (running more than 3 hours) must be limited to 80% of the breaker rating. A 20 A breaker can therefore supply only 16 A continuous.
Frequently Asked Questions
How do you convert watts in amps?
Divide the wattage by the voltage. For DC and AC single-phase resistive loads, the formula is Amps = Watts ÷ Volts. For AC single-phase inductive loads, divide by Volts × Power Factor instead. For three-phase circuits, the formula is Amps = Watts ÷ (√3 × Volts × Power Factor).
How many amps is 1500 watts at 120 volts?
12.5 A. The math is 1500 W ÷ 120 V = 12.5 A. Because NEC limits continuous loads to 80% of the breaker rating, a 1500 W heater should sit on a 20 A circuit minimum, not a 15 A circuit.
Can I use the same formula for DC and AC?
For purely resistive AC loads and DC, yes. The moment a motor, transformer, or any reactive load is involved, you need a power factor term. The DC formula does not apply to AC motor circuits, even though the algebraic structure looks similar.
What voltage should I use?
The voltage the device actually sees at its terminals when it is running — not the nominal nameplate voltage. For a US wall outlet that is 120 V (or 240 V on a dedicated appliance circuit). For a 12 V battery that is 12 V (or 13.8 V when the alternator is running). Long wire runs can drop the voltage significantly; if the device is more than 50 ft (15 m) from the panel, measure at the load.
Why does three-phase have √3 in the formula?
A three-phase system delivers power across three windings offset by 120°. The math for total power uses √3 to capture the contribution of all three phases and the phase angle between them. Without √3, you would only account for one phase and understate the current by 1.732×. The same √3 explains why three-phase transmission lines carry more power per amp than single-phase lines of the same gauge.
How do I size a breaker or wire from the amps?
After you have the amps, the next step is breaker and wire sizing. For continuous loads, multiply the current by 1.25 to account for the NEC 80% rule. A 3000 W / 120 V heater draws 25 A; a 25 A continuous load needs a 30 A breaker and 10 AWG copper wire (rated for 30 A in most residential wiring methods). For motor circuits, follow the nameplate Full-Load Amps (FLA) and use the breaker sizing rules in NEC Article 430.
Key Takeaways
- Core formula: Amps = Watts ÷ Volts. This is exact for DC and for AC single-phase resistive loads.
- AC motors and reactive loads add a power factor term: Amps = Watts ÷ (Volts × PF). Typical PF values are 1.0 for resistive loads, 0.85 for induction motors at full load, and 0.95 for fluorescent ballasts.
- Three-phase adds √3: Amps = Watts ÷ (√3 × Volts × PF). Line-to-line and line-to-neutral three-phase formulas differ by another factor of √3.
- Worked examples worth memorizing: 1500 W at 120 V = 12.5 A; 100 W at 12 V = 8.33 A; 100 W at 240 V = 0.417 A.
- After converting watts in amps, multiply by 1.25 for continuous loads per NEC, then round up to the next standard breaker size (15 A, 20 A, 30 A, 40 A, 50 A).
Related Guides
- How to Calculate Amps, Volts, and Watts — the foundational watts / volts / amps triangle with worked kitchen equipment examples.
- Rapid Tables Watts to Amps Calculator — interactive calculator covering DC, AC single-phase, and AC three-phase with typical power factor table.
- Bosstab Amps & Watts Calculator — practical A = W / V walkthrough with appliance-specific examples.
- Watts to Amps Calculator — safety-oriented calculator from Electrical Safety First.
Sources
- Watts / amps / volts / power factor formulas and 120 V reference table values — Rapid Tables. RapidTables
- Worked examples and appliance-focused watts / volts / amps coverage — Webstaurant Store. WebstaurantStore Guide
- National Electrical Code (NEC) continuous-load rule of 80% of breaker rating — NFPA. NFPA 70 — National Electrical Code
- Power factor reference values for typical motor and lighting loads — Rapid Tables Power Factor Table. RapidTables PF Table
