What is a watt? The difference between power and energy
The watt is a unit of power: it describes how quickly energy is transferred or converted. Its symbol is W, and 1 W = 1 joule per second (1 J/s). A device can convert electrical energy into heat, motion, light or other forms; wattage describes the rate of that conversion.
What do W, kW, Wh and kWh mean?
1 kW = 1,000 W, and 1 MW = 1,000 kW. A kilowatt is a unit of power. 1 Wh is the energy transferred at 1 W for one hour. A kilowatt-hour equals 1 kWh = 1,000 Wh = 3,600,000 J = 3.6 MJ.
kWh does not mean kilowatts divided by hours. It multiplies power by time. The difference between kW and kWh is fundamental: one describes power, the other energy over a period. Power is commonly represented by P, and energy by E.
How is energy calculated from power?
For constant power, use E = P × t. W × hours = Wh; kW × hours = kWh. Convert minutes to hours first: 15 minutes = 0.25 hours. For joules, multiply watts by seconds.
| Power | Duration | Energy (Wh) | Energy (kWh) |
|---|---|---|---|
| 20 W | 8 hours | 160 Wh | 0.16 kWh |
| 100 W | 10 hours | 1,000 Wh | 1 kWh |
| 1,000 W | 1 hour | 1,000 Wh | 1 kWh |
| 2,000 W | 15 minutes (0.25 hours) | 500 Wh | 0.5 kWh |
A 100 W device running for 10 hours and a 1,000 W device running for one hour use the same energy: 1 kWh. Their power demands are different. A higher-power device can use less total energy if it runs for less time. Socket and installation selection cannot be based on total kWh alone: socket load capacity and operating current also matter.
The relationship between watts, volts and amperes
Volts describe voltage, amperes describe current, and watts describe power. The volt–ampere–watt distinction is not simply different names for the same quantity. Under steady DC conditions, P = V × I: 24 V × 2 A = 48 W. Both voltage and amperage are needed; current alone does not determine wattage.
A watts-to-amperes calculation therefore requires supply voltage. Likewise, converting amperes to watts depends on the circuit conditions. The AC–DC distinction changes how the calculation is applied.
Does voltage × current always give watts in AC?
No. In a single-phase AC circuit, RMS voltage multiplied by RMS current gives apparent power in VA: S = VRMS × IRMS. Active power is P = S × PF, where PF is power factor. Under sinusoidal conditions PF = cos φ; with harmonics, cos φ alone may be insufficient.
For illustration, 230 V × 2 A = 460 VA. At a power factor of 0.80, active power is 368 W. Consequently, kVA and kW are not always equal. The VA–watt distinction matters when selecting supplies and backup equipment. Power factor is not the same as device efficiency.
For a balanced three-phase system, total active power can be calculated as P = √3 × Vline × Iline × PF. Unbalanced loads require individual phase assessment.
Does nameplate wattage mean continuous consumption?
Not necessarily. A label may specify input power, output power, a nominal value or maximum capacity. For example, an adapter capable of delivering 100 W does not have to draw 100 W continuously: input power depends on its load and losses.
An electric heater may cycle under thermostat control. Air conditioners and refrigerators can draw varying power. Use E = Paverage × t, or add the energy used during separate operating periods. Multiplying nameplate wattage by the entire day can mislead when the actual average is unknown.
Motor power versus electrical consumption
A motor's nameplate kW rating usually describes nominal mechanical shaft output. At the same operating point, electrical input is higher because of losses. Efficiency η = Pout / Pin. An illustrative 7.5 kW shaft output at 90% efficiency requires approximately 7.5 / 0.90 = 8.33 kW of electrical input; drive losses are additional.
Actual motor consumption varies with load. Rated output alone does not establish daily kWh. HP–kW conversion is also a conversion between power units, not an energy calculation.
How are consumption and meter readings assessed?
An electricity meter records accumulated energy. Monthly consumption calculations combine average power, daily operating time and the number of operating days. A constant 20 W device used for eight hours a day over 30 days consumes 0.020 × 8 × 30 = 4.8 kWh.
Long operating hours matter for LED lighting; small continuous loads matter for standby devices. A suitable smart plug or energy meter can record total Wh/kWh. Measuring voltage and current separately with a multimeter is not sufficient to determine AC active power under every condition.
The basic energy-charge calculation is kWh × the applicable unit price. When assessing the cost of 1 kWh, consider other bill items and tariff conditions separately. A fixed monetary figure requires the current applicable price.