haya.Learning made clear
Unit IV · Electricity
7 min readConcept sheetPart II (2 marks) and Part III (4 marks): the law, a numerical or the applications

Heating effect of current and Joule's law

Current heats a conductor. By Joule's law, H = I²Rt. Heaters use nichrome, fuses use a low-melting alloy and bulbs use tungsten.

Joule's lawH = I²Rt (heat in joule)
Heater elementNichrome (nickel + chromium): high resistivity, high melting point
Fuse wireAlloy of tin and lead: low melting point
Bulb filamentTungsten: very high melting point

Why does a wire get hot?

When current flows, the moving electrons keep colliding with the atoms of the wire. Each collision passes some energy to the atoms. The atoms vibrate faster, and the wire gets hot. This is the heating effect of current. Electrical energy turns into heat energy.

Joule's law and its derivation

A current I flows through a resistor R for time t, under a potential difference V.

  1. 1
    Work done: W = V × Q, and all of it appears as heat, so H = VQ.
  2. 2
    Charge Q = I × t, so H = V × I × t.
  3. 3
    By Ohm's law V = IR, so H = IR × I × t.
  4. 4★
    H = I²Rt, with H in joule.
  5. 5
    Joule's law: the heat produced in a conductor is directly proportional to (i) the square of the current, (ii) the resistance, (iii) the time of flow.

★ key step

Applications of the heating effect

DeviceMaterialWhy this material
Electric heater, iron box, water heater, toasterCoil of nichrome (alloy of nickel and chromium)High resistivity and high melting point; does not oxidise (react with oxygen) easily when red hot
Electric fuseThin wire of an alloy of tin and leadLow melting point: it melts and breaks the circuit when too much current flows
Filament bulbThin coil of tungstenVery high melting point, so it glows white-hot without melting. The bulb is filled with an unreactive gas such as argon or nitrogen so the filament lasts longer

Worked example: heat produced

Problem

A current of 2 A flows through a heater coil of resistance 50 Ω for 5 minutes. Find the heat produced. What happens to the heat if the current is doubled?

Working
  1. Given I = 2 A, R = 50 Ω, t = 5 minutes = 300 s.
  2. Formula: H = I²Rt.
  3. H = (2 A)² × 50 Ω × 300 s = 4 × 50 × 300 = 60,000 J.
  4. So H = 6 × 10⁴ J (60 kJ).
  5. If I is doubled, I² becomes 4 times, so H = 4 × 60,000 J = 2.4 × 10⁵ J.
Answer

6 × 10⁴ J; doubling the current makes it 4 times, 2.4 × 10⁵ J

In real life

The iron box, water heater and electric kettle at home all use a nichrome coil to turn electricity into heat. The same effect wastes energy in wires, so thick copper wires are used for heavy appliances.

Exam angle

2 marks: “State Joule's law of heating.” Heat produced in a conductor is directly proportional to the square of the current, the resistance and the time: H = I²Rt.

2 marks: “Why is nichrome used as the heating element of an electric iron?” It has high resistivity and a high melting point, and does not oxidise easily when red hot.

4 marks: “An electric iron of resistance 40 Ω draws a current of 5 A. Find the heat produced in 30 s.” H = I²Rt = 5² × 40 × 30 = 25 × 40 × 30 = 30,000 J = 3 × 10⁴ J.

Remember it

Heat loves current most: H = I²Rt, so double the current gives four times the heat.