Installation design
Design current from electrical input power
- where
- design current (A)
- power (W)
- supply RMS voltage (V)
- power factor
Explain this equation
Gives the design current, I_b, that a single-phase load draws in normal service. It is the first step in choosing protection and cable size.
P is the electrical input power in watts. U is the RMS supply voltage. cos φ is the power factor as a decimal. I_b is the current in amperes.
Single-phase active power is U × I × cos φ, so dividing P by U × cos φ gives the current. A lower power factor means more current for the same power.
2.3 kW at 230 V and unity power factor gives 10 A; at 0.8 it rises to 12.5 A. As a rough guide, each kW at 230 V and unity is about 4.35 A.
Watch out: enter watts, not kW. If you only have a motor's shaft output, divide by its efficiency first to get electrical input. Current alone does not select a cable.
Find the current that the load is expected to draw before selecting protection and conductors. Divide the electrical input power by voltage and power factor.
For a motor nameplate giving mechanical output, first divide output power by efficiency to obtain electrical input. Do not apply efficiency twice.
Enter your values to see the result.
Common mistake
Putting kW into a formula expecting W, or using mechanical output as electrical input.
Limits
Sinusoidal single-phase model. Non-linear loads require true power factor and manufacturer data. Current alone does not select a cable.
Sources
The tools calculate the displayed equation. They do not automatically choose cable sizes, approve installations, supply current regulatory tables, certify equipment or replace inspection and testing.