Single-Phase Current Calculation: kW and kVA Examples at 230 V

Read the power label before choosing the formula

At 230 V, a 5 kW electrical load and a 5 kVA load do not necessarily draw the same current. A 5 kW motor shaft rating introduces a third question because electrical input must also supply the motor’s losses.

Open the single-phase example. The system selector is already set to single phase. The voltage remains an input, not a universal assumption about every supply.

Electrical input: 5 kW at PF 0.9

For active electrical input power:

I = P / (V × PF)

I = 5000 / (230 × 0.9) = 24.15 A

Power factor is dimensionless. The 5 kW becomes 5000 W before division. If PF were one, current would be 5000 / 230 = 21.74 A. The higher current at PF 0.9 follows from needing more apparent power to deliver the same active power at the same voltage.

Do not insert efficiency here if 5 kW already describes electrical input. That would count losses a second time.

Apparent power: 5 kVA at 230 V

For apparent power:

I = S / V = 5000 / 230 = 21.74 A

No PF is needed to obtain current from kVA. At PF 0.9, this operating point corresponds to 4.5 kW active power, not 5 kW. The converter can display that separate active-power estimate, but changing PF must not change the kVA-derived current.

Shaft output: 2.2 kW with 85% efficiency

Suppose 2.2 kW is mechanical output, efficiency is 0.85 and operating PF is 0.8. These are example assumptions, not default properties of all motors.

Electrical input = 2200 / 0.85 = 2588.24 W

I = 2588.24 / (230 × 0.8) = 14.07 A

Use nameplate and manufacturer information for the actual operating point. Efficiency and PF can change with loading. A calculation based on rated output is not a prediction of starting current or a substitute for a measured operating current.

Where the voltage is measured matters

Use the voltage across the load terminals. A single-phase load can be connected line-to-neutral or, where designed for it, between two lines. Having a three-phase supply upstream does not turn each individual branch load into a balanced three-phase calculation.

This is why √3 does not belong in the examples above. For a complete balanced three-phase load, use the separate line-current formula and line-to-line voltage. The line/phase converter helps identify the quantities, but it does not choose equipment wiring.

Current is an input to further checks

After obtaining current, voltage drop still depends on the route and conductor information. Cable capacity depends on installation conditions and traceable data. Protection requires checks that are not contained in the power equation.

Retain the power basis, voltage, PF and any efficiency used alongside the result. When comparing with a clamp reading, ensure the load was in the operating state represented by your calculation.

Calculate another single-phase case, or use the kVA worked examples if the available rating is apparent power.