Heating effect of electric current
Current flowing through a resistor turns electrical energy into heat, given by Joule's law of heating, H = I²Rt. Irons, heaters, bulbs and the fuse all use this.
Where does the heat come from?
A cell or battery is a source of electrical energy. In a circuit that has only a resistor, all this energy turns into heat. Suppose a current I flows through a resistor R for a time t, with a potential difference V across it. The charge that flows is Q = It, so the work done is W = VQ = VIt. Putting V = IR gives H = I²Rt. This is Joule's law of heating. The heat is directly proportional to the square of the current, to the resistance and to the time the current flows.
Uses of the heating effect
| Device | How it uses the heat | Material |
|---|---|---|
| Electric iron, toaster, oven, kettle, room heater | A coil of high resistance gets very hot | Nichrome, an alloy |
| Electric bulb | The filament gets white-hot and gives out light. The bulb is filled with nitrogen and argon, which are chemically inactive, so the filament lasts longer. | Tungsten (melting point 3380 °C) |
| Fuse | A thin wire melts and breaks the circuit when the current is too large | A metal or alloy of suitable melting point, for example of aluminium, copper, iron or lead |
How a fuse protects a circuit
- 1The fuse is joined in series with the appliance or circuit, so all the current passes through it.
- 2A fault or overload makes the current rise above the fuse rating.
- 3The heat produced, I²Rt, rises sharply, because it depends on the square of the current.
- 4The fuse wire melts quickly and breaks the circuit.
- 5The current stops, so the wiring and the appliance are saved from damage or fire.
Worked example: heat and fuse rating
(a) A heating coil of resistance 40 Ω carries a current of 5 A for 1 minute. How much heat is produced? (b) An electric iron is rated 880 W at 220 V. Find the current it draws, and choose a suitable fuse from 1 A, 2 A, 3 A, 5 A and 10 A.
- (a) Given: I = 5 A, R = 40 Ω, t = 1 min = 60 s.
- Formula: H = I²Rt.
- H = (5 A)² × 40 Ω × 60 s = 25 × 40 × 60 J = 60,000 J.
- H = 6 × 10⁴ J (60 kJ).
- (b) Power P = VI, so I = P/V = 880 W ÷ 220 V = 4 A.
- Choose the fuse just above the normal current: 5 A.
(a) 6 × 10⁴ J; (b) 4 A, so use a 5 A fuse
When heating is a problem
- Heating is often unwanted: it wastes energy in wires and can harm electronic parts in computers and phones.
- That is why long wires that carry large currents are made thick and of copper or aluminium.
- A warm plug or socket is a warning sign of a loose contact or too much current.
An immersion rod in a bucket of bath water and the coil of an electric kettle both use Joule heating. The same heat is a danger in a loose plug: a warm socket needs an electrician.
1 marks: “If the current through a resistor is doubled, by what factor does the heat produced change?” It becomes four times, because H ∝ I².
3 marks: “Why is tungsten used for bulb filaments? Why are bulbs filled with nitrogen and argon?” Tungsten has a very high melting point (3380 °C), so it glows white-hot without melting. Nitrogen and argon are chemically inactive, so the filament does not oxidise and lasts longer.
3 marks: “Find the heat produced in 30 s by a 20 Ω resistor carrying 2 A. What will the heat be if the current is 4 A?” H = I²Rt = 2² × 20 × 30 = 2400 J. At 4 A: 4² × 20 × 30 = 9600 J, four times as much.
Joule's heat is 'I squared R T': double the current, four times the heat.