4.2.5 Electrical energy and electrical power
- Syllabus
- 0625–2026–2027
- Topic
- 4.2.5
- Level
- —
An electric circuit transfers energy from a source, through electrical working, to circuit components and then into the surroundings. Energy is transferred and conserved; it is not used up or destroyed.
| Stage | Example | Energy change |
|---|---|---|
| source | cell or battery | energy from its chemical store is transferred electrically |
| component | lamp | electrical transfer leads to light and heating |
| component | motor | electrical transfer leads to kinetic energy and heating |
| surroundings | air and nearby objects | transferred energy eventually spreads mainly by heating |
A mains supply is also an electrical energy source for the circuit. The appliance does not store all the received energy permanently: its useful output and any heating are eventually transferred to the surroundings.
Current is the movement of charge, not a flow of energy that disappears inside a component. Describe both the source and the receiving component when tracing an energy pathway.
Electrical power is the rate at which a component transfers electrical energy. One watt means one joule transferred each second.
P = IV
| Quantity | Symbol | Unit |
|---|---|---|
| electrical power | P | watt, W |
| current through the component | I | ampere, A |
| p.d. across the component | V | volt, V |
A 12 V supply delivers 0.35 A to a circuit: P=0.35×12=4.2 W. Rearrange the same equation as I=P/V or V=P/I when current or p.d. is required.
Use the current through and p.d. across the same component. Convert prefixes first: 1.0 kW = 1000 W, 0.10 mA = 0.00010 A and 400 kV = 400 000 V.
The electrical energy transferred depends on the power and on how long the transfer continues. Since P=IV, energy transferred is E=Pt=IVt.
E = IVt
| Quantity | Symbol | Unit for a joule answer |
|---|---|---|
| energy transferred | E | joule, J |
| current | I | ampere, A |
| potential difference | V | volt, V |
| time | t | second, s |
A 3.0 V torch lamp carries 20 mA for 5.0 minutes. Convert first: I=0.020 A and t=300 s. Then E=3.0×0.020×300=18 J.
For an answer in joules, use volts, amperes and seconds. Minutes are not seconds, and milliamperes are not amperes. The symbol E here is energy, so use the wording and unit to distinguish it from e.m.f.
One kilowatt-hour (kWh) is the energy transferred when a power of 1 kW operates for 1 hour. It is a unit of energy, not power.
1\text{ kWh}=1000\text{ W}\times3600\text{ s}=3.6\times10^6\text{ J}
A 2200 W heater runs for 48 minutes at 0.25perkWh.P=2.2kWandt=0.80h,soE=2.2\times0.80=1.76kWh.Cost=1.76\times0.25=$0.44$.
A utility 'unit' means 1 kWh. Do not multiply watts directly by hours and label the result kWh; divide watts by 1000 first. A tariff is a price per kWh, so cost is not found from power alone.