Work, power and energy consumption

A lamp is labelled 10 W, but the energy meter reads kWh. Why can these numbers not be compared directly? Combine the familiar ideas of energy per coulomb and coulombs per second.

At voltage U, the field does U joules of work per coulomb transferred. For charge q, work is A = Uq, in joules. Doubling transferred charge at unchanged voltage doubles the energy transferred; with zero transfer during an interval this work is zero. A device converts energy to heating, motion, light or chemical energy. Charge is conserved: the device receives energy, rather than using electrons as fuel.

To find energy over time t, use constant current: q = It. Substituting into A = Uq gives A = UIt for constant U. Doubling duration doubles energy. Halving current over the same time transfers half the charge and half the energy. Units check the meaning: В·А·с = (Дж/Кл)·(Кл/с)·с = Дж. If the operating conditions change, one product of constant values cannot describe the whole interval.

Divide the energy transferred by the duration. For constant voltage and current, power is P = A/t = UI. One watt is one joule per second: 1 Вт = 1 Дж/с. Increasing U at the same I gives more energy per coulomb; increasing I at the same U gives more coulombs per second. Both increase power. With zero current, this section's power is zero even if voltage is present.

Equal power does not mean equal daily energy use: A = Pt also depends on operating time. A higher-power device working for less time may consume the same energy. If a heater is on for half an hour and off for the other half, do not multiply its on-state power by the whole hour. Add the energies of the stages, or use average power over the full interval. The nameplate value refers to a specified mode that need not continue all the time.

An electricity meter adds up energy transferred to loads. Its unit therefore combines power and duration. A 1 kW device operating for 1 h uses 1 kWh; 2 kW for half an hour gives the same result. The kilowatt-hour is an energy unit like the joule, but more convenient for household quantities. Power alone, without operating time, cannot tell us how much the meter reading will increase.

Convert to SI: 1 кВт·ч = 1000 Вт·3600 с = 3 600 000 Дж = 3,6 МДж. Half a kilowatt-hour is therefore half this energy. In A = Pt, use kW and hours to get kWh, or W and seconds to get J. Writing kW/h divides power by time and denotes a different quantity; it cannot replace the product kWh.

So far U and I have been constant. In an AC circuit their values, and sometimes directions, change. Average power therefore depends on how those changes line up in time. RMS voltage times RMS current gives a resistive heater's average power, but not that of every motor or electronic device. We will study AC later; for now, do not apply a constant-operation rule to any pair of readings without checking conditions.

A heater draws a constant average power of 1.5 kW for 20 min. Find the energy in kWh and J.

Power is constant on average over the whole interval. Convert time: 20 мин = 1/3 ч. Then A = Pt = 1,5·1/3 = 0,5 кВт·ч.

In joules this is 0,5·3 600 000 = 1 800 000 Дж, or 1.8 MJ.

Check with another unit set: 1500 Вт·1200 с = 1 800 000 Дж. In one third of an hour, the device receives one third of the energy it would use in an hour.

0,5 кВт·ч = 1 800 000 Дж = 1,8 МДж. The meter reading would increase by half a kilowatt-hour for this operation.

For constant operation, P = UI combines energy per charge with charge transferred per second, while A = UIt gives energy over time.

Kilowatt-hours and joules measure energy; watts measure its transfer rate.

Consumption depends on actual operating time and mode, not just the power printed on the case.

Some electrical energy heats the wires. Next we will study how heating depends on current and why protection is needed.