Watts to Amps Calculator

Instantly convert power (watts or kilowatts) to current (amps) for DC, single‑phase AC, and three‑phase AC systems. Designed for engineers, electricians, students, and informed DIYers who need accurate, standards‑aligned results—fast.

Calculator

System type
Enter real power. If you only have apparent power (VA), multiply by power factor to get watts.
For three‑phase, enter line‑to‑line voltage (VLL).

Results

Outputs update as you type and use a min-height to avoid layout shifts.

Current (I)
0 A
Formula used
I = P / V
LaTeX: DC: I = \dfrac{P}{V}; 1φ AC: I = \dfrac{P}{V\cdot PF}; 3φ AC: I = \dfrac{P}{\sqrt{3}\cdot V \cdot PF}

Displayed in LaTeX for clarity. See the full explanation below.

Data Source and Methodology

Authoritative source: IEEE Std 1459-2010 — Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions. IEEE, 2010. View on IEEE Xplore.

Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

DC:
I = \dfrac{P}{V}

Single-phase AC (real power):
I = \dfrac{P}{V \cdot PF}

Three-phase AC (balanced, using line-to-line voltage V):
I = \dfrac{P}{\sqrt{3}\cdot V \cdot PF}

Glossary of Variables

Worked Example

How It Works: A Step‑by‑Step Example

Suppose an AC single‑phase load consumes P = 1500 W at V = 120 V with PF = 0.90. Using the single‑phase formula:

I = \dfrac{P}{V \cdot PF} = \dfrac{1500}{120 \cdot 0.90} = \dfrac{1500}{108} = 13.888\ldots \text{ A} \approx 13.89 \text{ A}

Therefore, the running current is approximately 13.89 A.

Frequently Asked Questions (FAQ)

Do I need power factor for DC?

No. DC uses I = P / V and PF is effectively 1. The calculator hides PF for DC selections.

Why does three‑phase use √3?

In a balanced three‑phase system, real power P relates to line quantities by P = √3 · VLL · IL · PF, hence I = P / (√3 · V · PF).

What if I only know kW and line‑to‑line voltage?

Choose kW in the unit selector; the tool converts kW to W internally and applies the proper formula for the selected system.

Is apparent power (kVA) supported?

Enter real power by multiplying kVA by PF: P = S × PF. For example, 5 kVA at PF 0.8 → P = 4 kW.

Are results suitable for conductor or breaker sizing?

Use results as a starting point only. Apply local electrical codes (e.g., NEC/IEC), consider continuous load multipliers, ambient conditions, inrush, and equipment ratings.

Does this calculator handle unbalanced three‑phase loads?

No. It assumes a balanced system using IEEE definitions. For unbalanced systems, analyze each phase separately or use specialized tools.

Strumento sviluppato da Ugo Candido.
Contenuti verificati da CalcDomain Engineering Editorial Team.
Last reviewed for accuracy on: 14 September 2025.