How Many Amps Is 1,080 Watts at 230V?
At 230V, 1,080 watts converts to 4.7 amps using the AC single-phase formula (Amps = Watts ÷ (V × PF)) at PF 1.0 for a resistive load. AC resistive at PF 1.0 and the DC baseline land on the same number at this voltage.
At 4.7A, the NEC 210.19(A) continuous-load sizing math (125% of the load, equivalently 80% of the breaker rating) points to a 15A breaker as the smallest standard size that covers this load continuously.
Use this citation when referencing this page.
Assumes an AC single-phase resistive load at PF 1.0. Typing a commercial L-L voltage (208/400/480V) re-routes the result to three-phase; 277V stays on single-phase because it's the L-N lighting leg of a 480Y/277V wye; 12/24V re-routes to DC.
Formulas
DC: Watts to Amps
I(A) = P(W) ÷ V(V)
AC Single Phase (PF = 0.85)
I(A) = P(W) ÷ (PF × V(V))
Circuit Sizing
Breaker Sizing
NEC 240.6(A) standard ampere ratings for branch-circuit and feeder breakers start at 15, 20, 25, 30, 35, 40, 45, and 50A and continue at 60A and above for feeder and large-appliance circuits. At 4.7A, the smallest standard breaker the raw current fits under is 15A. NEC 210.19(A) sizes conductor and OCP at 125% of any continuous load, equivalently 80% of breaker rating. Final selection still depends on the equipment nameplate, whether the load is continuous, conductor ampacity, and local code.
| Breaker Size | Max Continuous Load (80%) | Status for 4.7A |
|---|---|---|
| 15A | 12A | OK for continuous |
| 20A | 16A | OK for continuous |
| 25A | 20A | OK for continuous |
| 30A | 24A | OK for continuous |
| 35A | 28A | OK for continuous |
| 40A | 32A | OK for continuous |
| 45A | 36A | OK for continuous |
| 50A | 40A | OK for continuous |
Energy Cost
Running 1,080W costs approximately $0.18 per hour at the US average rate of $0.17/kWh (rates last reviewed April 2026). That is $1.47 for 8 hours or about $44.06 per month. See detailed cost breakdown.
AC Conversion Detail
The DC baseline for 1,080W at 230V is 4.7A. On an AC circuit with a power factor of 0.85, the current rises to 5.52A because reactive current flows alongside the real-power current.
| Circuit Type | Formula | Result |
|---|---|---|
| DC | 1,080 ÷ 230 | 4.7 A |
| AC Single Phase (PF 0.85) | 1,080 ÷ (230 × 0.85) | 5.52 A |
Power Factor Reference
Power factor is the main reason 1,080W draws more current on AC than DC. At PF 1.0 (pure resistive, like a heater), the load pulls 4.7A at 230V on the single-phase basis the rest of the page uses. At PF 0.80 (typical induction motor), the same 1,080W pulls 5.87A. That is an extra 1.17A just to overcome the reactive component. Use the typical values below as a starting point, not for precise engineering calculations.
| Load Type | Typical PF | 1,080W at 230V (single-phase) |
|---|---|---|
| Resistive (heaters, incandescent) | 1 | 4.7 A |
| Fluorescent lamps | 0.95 | 4.94 A |
| LED lighting | 0.9 | 5.22 A |
| Synchronous motors | 0.9 | 5.22 A |
| Typical mixed loads | 0.85 | 5.52 A |
| Induction motors (full load) | 0.8 | 5.87 A |
| Computers (without PFC) | 0.65 | 7.22 A |
| Induction motors (no load) | 0.35 | 13.42 A |
Same Wattage, Other Voltages
Related Calculations
Other Wattages at 230V
| Watts | AC 1Φ Amps PF 1.0 resistive | AC 1Φ Amps PF 0.85 motor |
|---|---|---|
| 300W | 1.3A | 1.53A |
| 350W | 1.52A | 1.79A |
| 400W | 1.74A | 2.05A |
| 450W | 1.96A | 2.3A |
| 500W | 2.17A | 2.56A |
| 600W | 2.61A | 3.07A |
| 700W | 3.04A | 3.58A |
| 750W | 3.26A | 3.84A |
| 800W | 3.48A | 4.09A |
| 900W | 3.91A | 4.6A |
| 1,000W | 4.35A | 5.12A |
| 1,100W | 4.78A | 5.63A |
| 1,200W | 5.22A | 6.14A |
| 1,300W | 5.65A | 6.65A |
| 1,400W | 6.09A | 7.16A |
| 1,500W | 6.52A | 7.67A |
| 1,600W | 6.96A | 8.18A |
| 1,700W | 7.39A | 8.7A |
| 1,800W | 7.83A | 9.21A |
| 1,900W | 8.26A | 9.72A |