What is the heating effect of current?
A cell keeps pushing charges through a resistor. The source does work on them; in a resistor this work does not make anything move, so all of it becomes heat. Moving electrons keep colliding with the metal's atoms, passing on energy and making them vibrate faster, which we feel as a rise in temperature. This is the heating effect of electric current.
Derivation of Joule's law of heating
Let a current I flow through a resistor R for time t, with p.d. V across it. Charge moved: Q = It. Work done by the source: W = VQ = VIt. This work appears as heat: H = VIt. Using Ohm's law V = IR:
H = I²Rt (also H = VIt = V²t/R)
Joule's law of heating: heat produced in a resistor is (i) directly proportional to the square of the current, (ii) directly proportional to the resistance for a given current, and (iii) directly proportional to the time for which current flows. H is in joules when I is in A, R in Ω and t in s.
Applications of the heating effect
- Heating appliances: iron, toaster, oven, kettle, room heater and geyser use nichrome coils (high resistivity, high melting point, does not oxidise easily).
- Electric bulb: the filament is tungsten, which melts only at about 3380 °C, so it can glow white-hot. The bulb is filled with inert nitrogen and argon so the filament does not burn. Most of the energy still becomes heat; only a small part becomes light.
- Electric fuse: a short wire of a metal or alloy with a low melting point (such as a tin–lead alloy), joined in series with the appliance. If the current goes above the fuse's rating (1 A, 2 A, 5 A, 10 A, 15 A…), the wire heats up, melts and breaks the circuit, protecting wiring and appliances. Fuse rating is chosen slightly above the normal current, e.g. a 1 kW iron at 220 V draws about 4.5 A, so a 5 A fuse is used.
When is heating unwanted?
In wires, motors, computers and chargers the heat is wasted energy and can damage parts. That is why connecting wires are thick copper, electronic devices have fans or vents, and power is sent over long distances at very high voltage (so that the current, and the I²R loss, stays small).
Key formulas and definitions
- H = I² R t (Joule's law of heating)
- H = V I t = V² t / R
- Q = I t, W = V Q
- Units: H in joules (J), I in A, R in Ω, t in s
Worked examples
1. A current of 2 A flows through a 10 Ω resistor for 30 s. Find the heat produced.
H = I²Rt = 2² × 10 × 30 = 1200 J.
2. An iron of resistance 50 Ω draws 4 A. How much heat does it produce in 1 minute?
H = 4² × 50 × 60 = 48 000 J = 48 kJ.
3. A 20 Ω heater is connected to 220 V for 10 s. Find the heat produced.
H = V²t/R = 220² × 10 / 20 = 24 200 J.
4. If the current through a heater is made three times, how many times does the heat produced in the same time become?
H ∝ I², so heat becomes 3² = 9 times.
5. A kettle produces 60 000 J in 2 minutes with a current of 5 A. Find its resistance.
t = 120 s. R = H / (I²t) = 60 000 / (25 × 120) = 20 Ω.
6. A 1100 W heater works on 220 V. Which fuse would you use: 2 A, 5 A or 10 A?
Normal current I = P/V = 1100/220 = 5 A. A fuse must be a little above this, so choose the 10 A fuse; a 5 A fuse would blow in normal use.
Common mistakes
- Thinking heat is proportional to current. It is proportional to I², so doubling I gives 4 times the heat.
- Using minutes for t in H = I²Rt. Always convert to seconds.
- Connecting a fuse in parallel. A fuse must be in series so that the full current passes through it.
- Saying copper is good for heater coils. Copper has low resistivity and oxidises; nichrome is used.