Amps to kWh: Worked Examples at 230 V and 415 V

An amp reading is not an electricity unit

An electricity “unit” on many bills means one kilowatt-hour: one kilowatt of active electrical power sustained for one hour. An ampere is current at one instant or over a short measurement window. There is no fixed amps-to-units multiplier. You need voltage, the supply arrangement, AC power factor and time.

Open the amps-to-kWh calculator with the 415 V example. It reports both the intermediate kW and the final kWh so the units can be checked.

Example 1: 230 V single phase

Suppose a load draws 10 A at 230 V with a measured power factor of 0.90. It operates at approximately that condition for six hours.

P = V × I × PF = 230 × 10 × 0.90 = 2,070 W = 2.07 kW

E = P × t = 2.07 kW × 6 h = 12.42 kWh

The answer is 12.42 estimated units for that six-hour interval. If the current changes, the answer changes. A 10 A reading made just as the machine starts is not a valid stand-in for six hours of operation.

Example 2: 415 V balanced three phase

Suppose each line carries 20 A, the line-to-line voltage is 415 V, and the power factor is 0.80. At this operating condition for eight hours:

P = √3 × VLL × I × PF = 1.73205 × 415 × 20 × 0.80 = 11,500 W ≈ 11.50 kW

E = 11.50 kW × 8 h ≈ 92.0 kWh

The three-phase formula uses line-to-line voltage and line current. Do not enter 415 V and then multiply by three again. The amps-to-kW examples explain the power step in more detail.

The power factor is not optional data to guess away

For AC, V × I gives apparent power, not necessarily active power. At 415 V and 20 A, apparent power is about 14.38 kVA. Multiplying by the assumed 0.80 power factor produces about 11.50 kW. Entering PF = 1 without evidence would instead estimate 14.38 kW, materially changing the eight-hour energy figure.

For a DC load, the simple steady-state relationship is P = V × I; there is no AC power-factor field. For a nonlinear AC load, a suitable power analyser or energy meter is preferable to inferring active power from a basic clamp-meter reading.

What this estimate does not say

The calculation does not know how often the equipment starts, how long it idles, whether a motor is lightly loaded, or whether the supply voltage and PF vary. It also does not know your billing rate, demand charge, taxes or meter accuracy. To model changing operation, split the day into measured intervals as described in the changing-load worksheet. For actual billed energy, use the utility or submeter reading and its measurement period.