Watts to Amps Calculator

Example data table

Formula used

This tool converts real power to current using the appropriate electrical model.

• DC: I = P ÷ (V × η)
• AC single-phase: I = P ÷ (V × PF × η)
• AC three-phase: I = P ÷ (√3 × V × PF × η)

Where P is watts, V is volts, PF is power factor, and η is efficiency as a decimal. Design current applies: I_design = I × (1 + safety%) × (continuous? 1.25 : 1).

How to use this calculator

1. Select the supply type: DC, AC single-phase, or AC three-phase.
2. Enter the real power and choose W or kW.
3. Enter the operating voltage. For three-phase, use line-to-line voltage.
4. Set power factor and efficiency if the load is not purely resistive.
5. Add a safety factor if you want extra margin for variations.
6. Optionally apply the continuous-load factor to size protective devices.
7. Press Calculate to view results above the form.
8. Use the export buttons to save a CSV or PDF summary.

Field guide: watts to amps for jobsite planning

1) Why current matters on site

Amps drive conductor heating, breaker selection, and voltage drop. Converting watts to amps helps crews confirm that temporary power panels, extension feeds, and distribution boards are not overloaded. It also supports fast checks when multiple tools share the same circuit during peak activity.

2) Real power versus apparent power

Watts represent real power that performs work. In AC systems, apparent power (VA) includes reactive effects from motors, welders, and electronic drives. The calculator reports current from real watts, and it also estimates kVA so you can compare against generator or transformer capacity ratings.

3) Typical voltage and system choices

Common construction voltages include 120 V and 230 V for single-phase tools and lighting, plus 400–480 V for three-phase equipment. For three-phase inputs, use line-to-line voltage. If you are working from a battery plant or DC bus, select DC and enter the nominal system voltage.

4) Power factor as a practical input

Resistive heaters and incandescent loads are close to PF 1.00. Inductive equipment typically runs between 0.80 and 0.95, and lightly loaded motors can be lower. A realistic PF improves the current estimate and can explain why a nameplate current seems higher than a simple watts ÷ volts calculation.

5) Efficiency and drive losses

Efficiency accounts for losses in motors, inverters, and power supplies. For example, a 1.5 kW load at 90% efficiency draws roughly 1.67 kW from the source, increasing amps accordingly. Entering efficiency is especially useful when converting equipment output power to expected input current.

6) Planning margin and continuous loads

The safety factor adds a percentage margin for uncertainty, altitude, temperature, aging, or future load growth. The optional continuous-load factor multiplies the design current by 1.25, a common practice for long-duration loads. Use these features to size protective devices conservatively without guessing.

7) Breaker selection and coordination

The suggested breaker is the next standard rating above the design current. Final selection should consider conductor ampacity, ambient temperature derating, bundling, and equipment inrush. When multiple downstream circuits exist, confirm selectivity so the correct device trips first during faults or overloads.

8) Quick sanity checks with examples

A 1,800 W load at 120 V is about 15 A at PF 1.00. A 7,500 W three-phase load at 400 V, PF 0.90, and 92% efficiency is about 13 A. Use the example table to benchmark results before committing to cabling and protection.

FAQs

1) What if my equipment is rated in kVA, not watts?

kVA is apparent power. Convert to watts by multiplying kVA by 1000 and by power factor. If PF is unknown, use a conservative estimate such as 0.9 for many motor-driven loads.

2) Should I use line-to-neutral or line-to-line voltage?

For single-phase, use the voltage across the load, usually line-to-neutral. For three-phase, this calculator assumes line-to-line voltage, which matches most equipment nameplates and panel ratings.

3) Why do my measured amps exceed the calculation?

Common causes include lower power factor, lower efficiency, supply voltage sag, harmonics from electronic drives, or additional loads on the circuit. Startup or inrush current can also temporarily exceed steady-state values.

4) How do I choose a safety factor?

Use 5–15% for well-known loads and stable supplies. Use 15–30% when the load mix changes, cable runs are long, temperatures are high, or future expansion is likely.

5) Does this replace a full load calculation for compliance?

No. It is a planning tool. Final design should follow local electrical codes, verified conductor ampacity, protective device coordination, grounding, and voltage drop requirements for your installation.

6) What power factor should I enter for welders and compressors?

Many welders and compressors can run between 0.7 and 0.95 depending on model and loading. If you cannot confirm PF, choose a lower value to avoid underestimating current.

7) Can I use this for generator sizing?

Yes, as a starting point. Compare the kVA estimate to generator capacity, then account for motor starting kVA, transient response, altitude derating, and the generator’s allowable harmonic distortion limits.