Know that electrical energy is transferred to heat energy and other forms of energy in the resistor, or other circuit components, and then into the surroundings

4.2.4 Resistance

Objective

Understand that electrical energy is converted into heat energy (and sometimes other forms of energy) in a resistor or other circuit component, and that this energy is then transferred to the surroundings.

1. Energy conversion in a resistor

When a current \$I\$ flows through a resistor of resistance \$R\$, the electrical energy supplied by the source is dissipated in the resistor. The main form of this dissipated energy is heat, but in some components (e.g., a filament lamp) part of the energy is emitted as light.

2. Power dissipated in a resistor

The rate at which electrical energy is converted to heat is called the power \$P\$ of the resistor. The following equivalent expressions are used in IGCSE:

All three forms give the same result because \$V = IR\$ (Ohm’s law).

3. Energy transferred to the surroundings

The energy \$E\$ transferred in a time \$t\$ is:

\$E = Pt\$

Since \$P\$ is usually expressed in watts (J s⁻¹), \$E\$ will be in joules (J) when \$t\$ is in seconds.

4. Factors influencing the amount of heat produced

• Current (\$I\$): Heat varies with the square of the current (\$P \propto I^{2}\$). Doubling the current increases the heat fourfold.
• Resistance (\$R\$): For a given current, heat is directly proportional to resistance (\$P \propto R\$). For a given voltage, heat is inversely proportional to resistance (\$P \propto 1/R\$).
• Time (\$t\$): Longer operation means more total energy transferred.
• Material and dimensions: Resistivity, length, and cross‑sectional area determine \$R\$ (see Section 5).

5. Calculating resistance

Resistance of a uniform conductor is given by:

\$R = \rho \frac{L}{A}\$

where \$\rho\$ is the resistivity of the material (Ω m), \$L\$ is the length (m), and \$A\$ is the cross‑sectional area (m²).

6. Example calculation

1. A 10 Ω resistor has a current of 2 A flowing through it. Find the power dissipated and the energy released in 5 minutes.
2. Solution:

   • Power: \$P = I^{2}R = (2\ \text{A})^{2}\times10\ \Omega = 40\ \text{W}\$
   • Time: \$t = 5\ \text{min}=300\ \text{s}\$
   • Energy: \$E = Pt = 40\ \text{W}\times300\ \text{s}=12\,000\ \text{J}\$

7. Summary of key points

• Electrical energy is converted to heat (and sometimes light) in resistors.
• Power dissipated can be calculated using \$P=VI\$, \$P=I^{2}R\$, or \$P=V^{2}/R\$.
• The total energy transferred to the surroundings is \$E = Pt\$.
• Heat production increases with the square of the current and is directly proportional to resistance for a given current.
• Resistivity and the geometry of a conductor determine its resistance.

8. Practice questions

1. A 5 Ω heater is connected to a 12 V supply. Calculate:

   • The current through the heater.
   • The power it dissipates.
   • The energy released after 10 minutes.

2. A copper wire (resistivity \$1.68\times10^{-8}\ \Omega\!\cdot\!m\$) is 2 m long and has a cross‑sectional area of \$1.0\times10^{-6}\ \text{m}^{2}\$. Find its resistance and the power dissipated when a current of 3 A flows through it.
3. Explain why a filament lamp gets hotter than a metal heating element of the same resistance when both are connected to the same voltage source.