When current flows through a purely resistive circuit — resistors and a battery, nothing that moves or spins — the source energy is dissipated entirely as heat. That is the heating effect, and it is what an electric heater, iron, toaster and geyser are built on.
If the circuit does useful work as well, as a fan does, only the rest of the energy becomes heat. That is why a fan still gets warm after running a long time.
Joule's law of heating
For a steady current I through a resistance R for time t:
H = I²Rt
The heat produced is proportional to:
- the square of the current — double the current, four times the heat
- the resistance of the conductor
- the time for which the current flows
The derivation the board asks for is short. The work done moving charge Q through potential difference V is VQ, so the power the source supplies is P = VQ/t = VI, and the energy in time t is VIt. In a purely resistive circuit all of it becomes heat, so H = VIt, and substituting V = IR gives H = I²Rt.
Electric power
Electric power is the rate at which electrical energy is consumed — or the rate at which work is done. SI unit: the watt (W).
P = VI = I²R = V²/R
Power is 1 watt when a current of 1 ampere flows through a device across a potential difference of 1 volt. That sentence is a 2-mark answer on its own and is asked almost every year.
Which form to use is the whole skill:
- given V and R → use V²/R
- given I and R → use I²R
- given a rating like 100 W, 220 V → find R from R = V²/P first
Commercial unit. Energy is sold in kilowatt hours. 1 kW h = 3.6 × 10⁶ J, and "1 unit" on a bill means 1 kW h.
Where the marks are
Two situations come up again and again, and they pull in opposite directions.
Bulbs in parallel (as in a house): each gets the full 220 V, so each runs at its marked power. A 100 W bulb is brighter than a 60 W bulb.
Bulbs in series: the same current passes through both, so P = I²R decides it — the bulb with the higher resistance glows brighter. Since R = V²/P, the higher-resistance bulb is the lower-rated one. So in series the 60 W bulb is brighter than the 100 W. This reversal is the classic trap.
Practical applications
Filament lamp. The filament is tungsten — a very high melting point (3380 °C) and high resistivity, so it stays hot enough to glow white without melting. Bulbs are filled with chemically inert nitrogen or argon to stop the filament oxidising and burning away.
Heating elements in an iron, toaster or heater are alloys such as nichrome, chosen for high resistivity and for not oxidising readily at high temperature.
The electric fuse is the heating effect used as protection. A short length of wire of suitable melting point is placed in series with the circuit; if the current exceeds a safe value the wire melts and breaks the circuit. Its rating must be above the normal operating current of the appliance.
Working a fuse question is always the same two steps: find the current the appliance actually draws with I = P/V, then compare it to the fuse rating. A 2 kW heater on 200 V draws 10 A, so a 4 A fuse would melt immediately — it cannot be used.
What the board actually asks
This is the heaviest topic in Electricity — about 3.5 marks a paper across 2023–2026, and it appears as 5-mark long answers more often than anything else in the chapter.
- Define electric power and express it in terms of V and R, then a numerical.
- State Joule's law of heating and write it for an appliance of resistance R on a source of potential difference V.
- Fuse rating: can this fuse be used with this appliance? Compute I = P/V and say why.
- Heating elements in series or parallel (a heater with three elements and switches): the same elements give different heat depending on how they are connected.
- Energy cost: an oven of a given rating used for so many hours — convert to kW h.
Two things cost marks routinely. Substituting a rating into P = I²R without first finding the actual current in that circuit. And giving heat in joules when the question asked for kilowatt hours, or the reverse — check the unit the question wants before you start.