Electricity • Topic 3 of 3

Heating Effect & Electric Power

Why does a wire get hot when current flows? As electrons drift through a resistor they collide with the atoms of the conductor, transferring energy that appears as heat. This is the heating effect of electric current, and it powers everyday devices like heaters, irons, toasters and the glowing filament of a bulb.

Joule's law of heating states that the heat produced in a resistor is proportional to (a) the square of the current, (b) the resistance, and (c) the time for which current flows. In symbols, $H=I^2Rt$, where $H$ is the heat in joules. Because the current is squared, even a small increase in current produces a large rise in heat — this is why overloaded wires can become dangerously hot.

Electric power is the rate at which electrical energy is consumed or converted: $P=\frac{W}{t}=VI$. Using Ohm's law ($V=IR$) we get three equivalent forms:

  • $P=VI$ — when voltage and current are known.
  • $P=I^2R$ — when current and resistance are known.
  • $P=\frac{V^2}{R}$ — when voltage and resistance are known.

The SI unit of power is the watt (W), where $1\ \text{W}=1\ \text{J/s}=1\ \text{V}\times1\ \text{A}$. A larger unit is the kilowatt ($1\ \text{kW}=1000\ \text{W}$).

Commercial unit of energy. Electricity bills are not charged in joules — that unit is far too small. Instead we use the kilowatt-hour (kWh), also called 1 unit. One kilowatt-hour is the energy used by a 1 kW appliance running for 1 hour: $1\ \text{kWh}=1000\ \text{W}\times3600\ \text{s}=3.6 \times 10^{6}\ \text{J}$.

Practical applications. A bulb filament is made of tungsten, which has a high melting point and high resistance so it glows white-hot. An electric fuse is a deliberate weak link: a thin wire of low melting point that melts and breaks the circuit when the current exceeds a safe value, protecting appliances from damage. Heating appliances use nichrome coils because of their high resistivity and resistance to oxidation at high temperature.

Heating effect: cell, fuse and glowing bulb circuitCellFuseBulb (filament)H = I squared x R x t (heat produced)P = VI = I squared R = V squared / R
Solved Examples
1
Worked Example
A current of 3 A flows through a $5\ \Omega$ resistor for 2 minutes. Calculate the heat produced.
Solution
  1. Step 1: Convert time: $t=2\ \text{min}=120\ \text{s}$.
  2. Step 2: Use Joule's law $H=I^2Rt$.
  3. Step 3: Substitute: $H=(3)^2 \times 5 \times 120=9 \times 5 \times 120$.
  4. Step 4: Compute: $H=5400\ \text{J}$.

Answer: $H=5400\ \text{J}$

2
Worked Example
An electric bulb is rated 60 W, 220 V. Find the current drawn and its resistance.
Solution
  1. Step 1: From $P=VI$, current $I=\frac{P}{V}=\frac{60}{220}$.
  2. Step 2: Compute: $I \approx 0.27\ \text{A}$.
  3. Step 3: Resistance $R=\frac{V^2}{P}=\frac{(220)^2}{60}=\frac{48400}{60}$.
  4. Step 4: Compute: $R \approx 806.7\ \Omega$.

Answer: $I \approx 0.27\ \text{A}$, $R \approx 806.7\ \Omega$

3
Worked Example
How much electrical energy in kWh is consumed by a 2 kW heater used for 3 hours?
Solution
  1. Step 1: Energy $=$ power $\times$ time (in kW and hours).
  2. Step 2: Substitute: $E=2\ \text{kW} \times 3\ \text{h}$.
  3. Step 3: Compute: $E=6\ \text{kWh}$ (i.e. 6 units).

Answer: $6\ \text{kWh}$ (6 units)

4
Worked Example
A device has a resistance of $40\ \Omega$ and operates at 200 V. Find the power consumed.
Solution
  1. Step 1: Use $P=\frac{V^2}{R}$.
  2. Step 2: Substitute: $P=\frac{(200)^2}{40}=\frac{40000}{40}$.
  3. Step 3: Compute: $P=1000\ \text{W}=1\ \text{kW}$.

Answer: $P=1000\ \text{W}=1\ \text{kW}$

5
Worked Example
Express 1 kWh in joules.
Solution
  1. Step 1: $1\ \text{kWh}=1000\ \text{W} \times 1\ \text{hour}$.
  2. Step 2: Convert the hour to seconds: $1\ \text{hour}=3600\ \text{s}$.
  3. Step 3: Multiply: $1\ \text{kWh}=1000 \times 3600=3.6 \times 10^{6}\ \text{J}$.

Answer: $1\ \text{kWh}=3.6 \times 10^{6}\ \text{J}$

6
Worked Example
Two bulbs rated 100 W and 60 W are connected to the same 220 V supply. Which bulb has greater resistance?
Solution
  1. Step 1: At fixed voltage, $R=\frac{V^2}{P}$, so $R$ is inversely proportional to $P$.
  2. Step 2: The lower-power bulb (60 W) therefore has the higher resistance.
  3. Step 3: Check: $R_{60}=\frac{(220)^2}{60} \approx 807\ \Omega$, while $R_{100}=\frac{(220)^2}{100} \approx 484\ \Omega$.

Answer: The 60 W bulb has greater resistance ($\approx 807\ \Omega$).

Key Points

  • Joule's law of heating: $H=I^2Rt$ — heat depends on the square of the current.
  • Electric power has three forms: $P=VI=I^2R=\frac{V^2}{R}$; SI unit watt (W).
  • Commercial unit of energy is the kilowatt-hour: $1\ \text{kWh}=3.6 \times 10^{6}\ \text{J}$ (1 unit).
  • A bulb filament (tungsten) glows because of high resistance and high melting point.
  • An electric fuse is a low-melting-point wire that breaks the circuit on excess current.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Joule's law of heating is given by:
Explanation: Heat produced is $H=I^2Rt$.
Q2.Which is NOT a correct formula for electric power?
Explanation: The correct forms are $P=VI=I^2R=\frac{V^2}{R}$; $\frac{R}{V^2}$ is wrong.
Q3.One kilowatt-hour equals:
Explanation: $1\ \text{kWh}=1000 \times 3600=3.6 \times 10^{6}\ \text{J}$.
Q4.An electric fuse is made of a wire with:
Explanation: A fuse uses a low-melting-point wire that melts and breaks the circuit on overload.
Practice more MCQs → 📝 Assignment · 30 marks