Electrical power and energy

Power is the rate of energy transfer.

A watt (W) is one joule of energy transferred per second. One kilowatt (kW) is 1,000 W. Power tells you how quickly energy is being transferred at a particular time.

For steady DC, electrical power is P = V × I. A 24 V load taking 0.5 A receives 12 W.

Energy includes time.

A light using 20 W for five hours

20 W × 5 h = 100 Wh = 0.1 kWh. A kilowatt-hour is a quantity of energy, not a rate of use.

For constant power, energy = power × time. If power changes, add the energy from the different periods rather than assuming full power all day.

AC power: watts, VA and power factor.

With an AC supply, voltage and current change throughout each cycle. Their peaks need not occur together. Inductors and capacitors can temporarily store energy and return it, so current can flow without all of it contributing to net energy transfer.

For a single-phase load, three quantities help keep the picture clear. Here RMS (root mean square) means an effective voltage or current, based on its equivalent heating effect in a resistor.

Real power, P — watts (W): The average rate of net electrical energy transfer. It includes energy converted into useful output and energy lost as heat.
Apparent power, S — volt-amperes (VA): RMS voltage × RMS current. It describes the combination of voltage and current the supply and equipment must handle.
Power factor, PF — no unit: Real power ÷ apparent power. For a consuming load, P = S × PF.

One load, two different power figures

A single-phase load draws 2 A RMS at 230 V RMS, with a power factor of 0.80.

Apparent power: 230 × 2 = 460 VA.
Real electrical input: 460 × 0.80 = 368 W.

Power factor 0.80 does not mean the equipment is 80% efficient. Efficiency compares useful output with real input. Nor is 460 − 368 a calculation of wasted watts: VA and W describe different quantities.

For the same real power and voltage, a lower power factor means more RMS current. That can increase losses in the supply wiring. The impedance explanation connects this to resistance and energy storage; the three-phase lesson covers a different supply arrangement.

Optional: phase angle, distortion and reactive power

For sinusoidal voltage and current, PF = cos φ, where φ is their phase difference. In a consuming circuit, φ = 0 gives power factor 1. With distorted current, power factor also reflects waveform shape: a phase angle alone does not describe it.

Reactive power, Q, is expressed in vars. For the sinusoidal single-phase case, Q = V_RMSI_RMS sin φ and S² = P² + Q². It describes the alternating energy exchange associated with fields, rather than an extra amount of heat being consumed.

The general starting point for real power is the average of the instantaneous product v(t) × i(t). It is the product at matching instants, not the product of two separately averaged waveforms.

Input power and useful output differ.

Equipment does not convert all its input into the desired result. Motors also produce heat, for example. Efficiency compares useful output with input under specified conditions.

A motor’s nameplate commonly states rated mechanical output. Actual electrical input depends on the operating point and losses; it is not automatically equal to the marked kW.

A lift’s energy use changes through the day.

Travel, waiting, lighting and control equipment contribute differently. Some installations can also return energy to the supply during generating operation.

A meaningful comparison needs a defined period and operating conditions. A smaller motor rating alone does not establish that a lift will use less energy.

A closer look

For AC loads, multiplying RMS voltage by RMS current gives apparent power in volt-amperes for a single-phase circuit. Real power in watts also depends on power factor. Do not apply the simple DC calculation to every motor circuit.

Check your understanding

A constant 2 kW load runs for half an hour. How much energy does it use?

Reveal the explanation

1 kWh: 2 kW × 0.5 h. Using 2 kW for one hour would use 2 kWh.