Watts to Amps Formula: DC & AC

Converting watts to amps requires the voltage of the circuit and, for AC loads that are not purely resistive, the power factor. The formula is simplest for DC circuits and gets a power factor term for single-phase AC and both a √3 and a power factor term for three-phase AC. This page covers all three variants with worked numbers for resistive, inductive, and three-phase loads.

Formulas

DC: I = P / V

AC 1-Phase: I = P / (V × PF)

AC 3-Phase: I = P / (V × √3 × PF)

To find current in amps, divide power in watts by voltage. For AC circuits, also divide by the power factor because reactive loads draw extra current beyond what the DC formula predicts. Three-phase circuits use the √3 factor (1.732) because power is distributed across three conductors.

Worked Examples

Example 1: 1500W space heater at 120V AC (resistive, PF = 1.0)

I = 1500 ÷ 120 = 12.5 amps. Resistive loads have a power factor of 1.0, so the DC formula gives the correct AC current for a heater, toaster, or incandescent bulb.

Example 2: 1500W motor at 120V AC (PF 0.85)

I = 1500 ÷ (120 × 0.85) = 1500 ÷ 102 = 14.71 amps. Reactive loads draw about 15-20% more current than the DC formula predicts at the same real power.

Example 3: 300W solar array at 12V DC

I = 300 ÷ 12 = 25 amps. True DC from a battery or solar panel, where power factor does not apply.

Example 4: 5000W three-phase load at 480V (PF 0.9)

I = 5000 ÷ (480 × √3 × 0.9) = 5000 ÷ 748.25 = 6.68 amps per line.

Common Mistakes

Forgetting power factor on AC circuits. The DC formula underestimates AC current draw by 15-20% for typical reactive loads.
Mixing up line-to-line (V_LL) and line-to-neutral (V_LN) voltages in the three-phase formula. Use I = P ÷ (√3 × V_LL × PF) with line-to-line voltage, or I = P ÷ (3 × V_LN × PF) with line-to-neutral voltage. The two forms give the same answer when V_LL = √3 × V_LN, but swapping them without adjusting the formula gives a result off by a factor of √3.
Assuming the current result directly tells you the breaker size. NEC requires 125% sizing for continuous loads and separate motor branch-circuit rules (Article 430).

Try the Calculator

Use the interactive Watts to Amps Calculator for instant results with any values. Every result page shows the formula applied with your specific numbers.

All Formulas

functionsOhm's Law Formula functionsEnergy Cost Formula functionsWire Sizing Formula functionsVoltage Drop Formula functionsThree-Phase Power functionsPower Factor functionsHP to Amps Formula functionsSeries & Parallel functionsAC vs DC functionsElectrical Units functionsNEC Reference

This is a formula reference page for educational use. Always consult a licensed electrician and verify compliance with the National Electrical Code (NEC) and local electrical codes before applying any of these formulas to real installations.

Standards & References

This page cites the following electrical codes and standards. Always consult the current edition of your local adopted standard for authoritative requirements.

NEC 210.19(A) Informational Note 4. Branch-circuit conductors sized to prevent a voltage drop exceeding 3% at the farthest outlet. Combined with feeders, total voltage drop should not exceed 5%.
National Electrical Code (NFPA 70), Article 210, Branch Circuits. Reference →
NEC 240.4(D). Small conductor rule: overcurrent protection shall not exceed 15A for 14 AWG, 20A for 12 AWG, and 30A for 10 AWG copper, regardless of ampacity table values.
National Electrical Code (NFPA 70), Article 240, Overcurrent Protection. Reference →
IEC 60038. IEC standard voltages. Defines 230/400V as the nominal low-voltage supply for 50Hz systems, which harmonized European residential supply in 1995.
International Electrotechnical Commission. Reference →

Disclaimer: The information on this page is provided for reference. Always consult a licensed electrician and the current edition of your local adopted electrical code before performing electrical work.