Electricity · Lesson 7 of 9
Heating Effect of Electric Current
“Electric current can bring the heat, sometimes usefully and sometimes a little too enthusiastically.”
• Explain how electrical energy becomes heat in a resistive circuit. • Derive and apply Joule’s law of heating. • Select an appropriate heating formula from the known quantities. • Compare heating when current, resistance or time changes. • Explain the operation of heating devices, lamp filaments and electric fuses. • Select a suitable fuse rating from appliance current.
The connecting wire of an iron stays comparatively cool while its element becomes red-hot. Both carry current, yet their resistances and shapes are very different. When charge moves through a resistance, electrical energy is transferred to the material. Faster atomic motion appears as a rise in temperature: the heating effect of electric current.
The production of heat when electrical energy is dissipated in a conductor carrying current.
A source does work W = VQ in moving charge Q through a potential difference V. Since Q = It, the energy supplied in time t is W = VIt. In a purely resistive circuit this electrical energy is dissipated as heat, so H = VIt. Substituting V = IR produces Joule’s law.
The squared dependence on current is especially important. Doubling current produces four times as much heat in the same resistance and time. At fixed current, increasing resistance increases heating. At fixed voltage, however, H = V²t/R shows that increasing resistance reduces current enough to reduce the total heat rate. Always identify what is being held constant.
Practical Applications of Heating Effect of Electric Current
Electric irons, toasters, ovens, kettles and heaters use high-resistivity elements to produce heat. Alloys are preferred because they can provide substantial resistance in a compact coil, tolerate high temperatures and resist oxidation. The long coiled form also fits a useful resistance into a small space.
Electric Bulb Filament
A bulb filament is heated until it becomes incandescent and emits light. Tungsten is used because it is strong and has a very high melting point, about 3380°C. The filament is thin and thermally isolated so it can reach a high temperature. Chemically inactive nitrogen and argon slow evaporation and chemical damage. Most energy still becomes heat; only a smaller part is emitted as visible light.
Electric Fuse
A protective device connected in series that melts and breaks a circuit when current exceeds a safe value.
A fuse wire is chosen with an appropriate melting point and current rating. Metals and alloys based on materials such as aluminium, copper, iron or lead may be used for suitable fuse elements. The wire is commonly enclosed in a porcelain or similar heat-resistant cartridge with metal ends. Excess current produces rapid I²R heating, melts the element and opens the path. Common ratings include 1 A, 2 A, 3 A, 5 A and 10 A. The chosen rating should be just above the normal operating current so normal use does not melt it, but an unsafe current does.
The supply cord of a heater has low resistance and a comparatively large cross-sectional area, so it produces little heat. The element uses a much higher-resistance alloy and is designed to retain heat. Although the same series current may pass through both, H = I²Rt shows that the greater-resistance element releases far more heat in the same time.
Selecting a Heating Formula
| Known quantities | Convenient relationship | Reason |
|---|---|---|
| V, I and t | H = VIt | Direct substitution |
| I, R and t | H = I²Rt | Direct form of Joule’s law |
| V, R and t | H = V²t/R | Current need not be calculated separately |
| Power and t | H = Pt | Power is energy transferred per second |
Problem
A 20 Ω iron carries 5 A for 30 s. Find heat produced.
- 1.Given: R = 20 Ω, I = 5 A and t = 30 s.
- 2.Use H = I²Rt because I, R and t are known.
- 3.H = 5² × 20 × 30 = 25 × 600 = 15,000 J.
- 4.The heat produced is 1.5 × 10⁴ J.
Problem
A resistor produces 100 J each second and has resistance 4 Ω. Find its current and voltage.
- 1.In 1 s, H = 100 J, with R = 4 Ω and t = 1 s.
- 2.From H = I²Rt, I = √(H/Rt) = √(100/4) = √25 = 5 A.
- 3.Then V = IR = 5 × 4 = 20 V.
- 4.Check: VIt = 20 × 5 × 1 = 100 J.
Problem
An iron operates at 840 W on maximum and 360 W on minimum at 220 V. Find current and resistance in each setting.
- 1.Use P = VI, so I = P/V.
- 2.Maximum: I = 840/220 = 3.82 A. Then R = V/I = 220/3.82 ≈ 57.6 Ω.
- 3.Minimum: I = 360/220 = 1.64 A. Then R = 220/1.64 ≈ 134 Ω.
- 4.At fixed supply voltage, the lower-resistance setting draws more current and therefore has greater power.
- 5.A check with P = V²/R gives approximately 840 W and 360 W.
Fuse-Rating Example
Problem
An iron is rated 1.0 kW at 220 V. Choose from 3 A, 5 A and 10 A fuses.
- 1.Convert power: 1.0 kW = 1000 W.
- 2.Normal current I = P/V = 1000/220 = 4.55 A.
- 3.A 3 A fuse would melt during normal use. A 10 A fuse allows much more than the normal current.
- 4.The 5 A fuse is the suitable available rating because it is just above 4.55 A.
Never replace a fuse with thick wire, foil or a higher rating merely to stop it melting. A repeatedly melting fuse indicates a fault or overload that should be checked by a qualified person.
Quiz
If current doubles through the same resistor for the same time, how does heat change?
Which formula is most direct when V, R and t are known?
Why is tungsten suitable for a bulb filament?
How is a fuse connected?
An appliance normally draws 4.5 A. Which available fuse is most suitable?
Practice Problems
- Find heat produced by 2 A through 10 Ω for 60 s. Solution: H = I²Rt = 2² × 10 × 60 = 2400 J.
- A 50 V source transfers 96,000 C in one hour. Find energy transferred. Solution: W = VQ = 50 × 96,000 = 4.8 × 10⁶ J.
- A 12 Ω resistor is across 24 V for 5 min. Find heat. Solution: t = 300 s. H = V²t/R = 24² × 300/12 = 14,400 J.
- Compare heat from 3 A and 6 A through the same resistor for equal time. Solution: H₂/H₁ = (6/3)² = 4. The 6 A current produces four times the heat.
- A 660 W kettle works at 220 V. Find normal current and choose a 3 A or 5 A fuse. Solution: I = P/V = 660/220 = 3 A. A fuse must be just above normal current, so choose 5 A; a 3 A fuse may melt during normal operation.
Key Takeaways
• Electrical energy becomes heat when current passes through resistance. • Joule’s law is H = I²Rt. • Doubling current produces four times the heat at fixed R and t. • Formula choice depends on which electrical quantities are known. • Alloys suit heating elements; tungsten suits incandescent filaments. • A fuse is connected in series and melts during excessive current. • Fuse rating should be just above normal operating current. • Fixed-current and fixed-voltage comparisons can lead to different resistance trends.