4.2.4 Resistance
Learning Objectives
- Define electrical resistance and write the basic formula
R = V / I. - State the quantitative relationship
R = ρ L / A for a uniform metallic conductor. - State the qualitative relationships:
- Resistance increases with length.
- Resistance decreases with cross‑sectional area.
- Explain how the material (resistivity ρ) influences resistance.
- Design a simple voltmeter‑ammeter experiment to determine the resistance of a wire.
- Apply the relationships to solve typical IGCSE‑style problems.
1. Definition of Resistance
Resistance is the opposition a conductor offers to the flow of electric charge. It is quantified by the ratio of the potential difference across the conductor to the current flowing through it:
2. Quantitative Relationship for a Uniform Metallic Conductor
For a straight piece of metal of uniform cross‑section the resistance depends on three factors:
- Material – characterised by its resistivity ρ (Ω·m).
- Length L (m).
- Cross‑sectional area A (m²).
The law is
3. Qualitative Relationships (AO1)
| Variable kept constant | What happens to R? | Mathematical expression |
|---|
| Length L (ρ, A constant) | Resistance increases proportionally | R ∝ L |
| Cross‑sectional area A (ρ, L constant) | Resistance decreases inversely | R ∝ 1/A |
| Resistivity ρ (L, A constant) | Higher ρ → higher R | R = ρ L / A |
4. Material Dependence (Resistivity)
The resistivity ρ is an intrinsic property of the material. For the same geometry, different metals give different resistances. Typical values (at 20 °C) are:
| Material | Resistivity ρ (Ω·m) |
|---|
| Copper | 1.68 × 10⁻⁸ |
| Aluminium | 2.82 × 10⁻⁸ |
| Nichrome | 1.10 × 10⁻⁶ |
Thus, for identical length and area, a copper wire will have a lower resistance than an aluminium or nichrome wire.
5. Derivation of the Proportionalities
- Length: With ρ and A fixed,
R = ρ L / A shows R varies linearly with L. Doubling L doubles R. - Cross‑sectional area: With ρ and L fixed,
R = ρ L / A shows R varies inversely with A. Doubling A halves R. - Material: For given L and A, a larger ρ gives a proportionally larger R.
6. Practical Implications
- Long, thin wires → high resistance → useful for heating elements.
- Short, thick wires → low resistance → essential for power transmission.
- Choosing a material with low ρ (e.g., copper) reduces R for a given size.
- High‑voltage transmission lines use aluminium because its low density allows very large cross‑sections, offsetting its higher ρ compared with copper.
7. Simple Experiment: Determining the Resistance of a Wire
- Connect the unknown wire in series with an ammeter. Connect a voltmeter across the wire (parallel).
- Use a variable DC source (or a set of batteries) to apply several different voltages (e.g., 1 V, 2 V, 3 V).
- For each voltage, record the current reading.
- Calculate the resistance for each trial using
R = V / I and take the average.
Typical sources of error
- Internal resistance of the voltmeter (reduces the measured V).
- Contact resistance at the terminals.
- Temperature rise of the wire during the test (ρ increases with temperature).
8. Safety Reminder
- Never touch exposed leads while the circuit is powered.
- Use insulated probes and keep the work area dry.
- Switch off the power source before making or breaking connections.
9. Sample Calculations
Example 1 – Effect of Length
Copper wire: ρ = 1.68 × 10⁻⁸ Ω·m, L = 2.0 m, A = 1.0 × 10⁻⁶ m².
R = ρ L / A = (1.68 × 10⁻⁸) · 2.0 / 1.0 × 10⁻⁶ = 0.0336 Ω
If L is increased to 4.0 m (A unchanged):
R' = (1.68 × 10⁻⁸) · 4.0 / 1.0 × 10⁻⁶ = 0.0672 Ω
Resistance doubles – confirming R ∝ L.
Example 2 – Effect of Cross‑sectional Area
Same wire, double the area to A = 2.0 × 10⁻⁶ m² (L = 2.0 m).
R'' = (1.68 × 10⁻⁸) · 2.0 / 2.0 × 10⁻⁶ = 0.0168 Ω
Resistance is halved – confirming R ∝ 1/A.
10. Common Misconceptions
- “Longer wires always give a larger voltage drop.” The drop depends on both resistance and the current (
V = IR). - “All metals have the same resistance for a given size.” Resistivity differs; see the table in section 4.
- “Cross‑sectional area is the same as the diameter.” For a circular wire,
A = πr² = π(d/2)²; a small change in diameter produces a large change in area.
11. Summary Table
| Variable (constant) | Effect on R | Mathematical form |
|---|
| Length L (A, ρ constant) | R ↑ proportionally | R ∝ L |
| Area A (L, ρ constant) | R ↓ inversely | R ∝ 1/A |
| Resistivity ρ (L, A constant) | R ↑ with ρ | R = ρ L / A |
12. Practice Questions
- A 3.0 m length of aluminium wire (ρ = 2.82 × 10⁻⁸ Ω·m) has a cross‑sectional area of 0.5 mm². Calculate its resistance.
- The same aluminium wire is stretched so that its length becomes 4.5 m and its area reduces to 0.35 mm². What is the new resistance? Compare with the original value and comment on the effect of stretching.
- Two wires of the same material each have a resistance of 0.10 Ω. Wire A is 1.0 m long, Wire B is 2.0 m long. Which wire has the larger cross‑sectional area and by what factor?
- Explain why high‑voltage power‑transmission lines use aluminium conductors that are many square centimetres in cross‑section, even though aluminium’s resistivity is higher than copper’s.
- Design a simple experiment (using a voltmeter, ammeter and a DC source) to determine the resistance of a piece of nichrome wire. List the steps and at least two possible sources of error.
13. Suggested Diagram
Three wires of the same material drawn side‑by‑side:
- (i) Short & thick – low R
- (ii) Long & thin – high R
- (iii) Same length as (i) but thinner – intermediate R
Arrows indicating the direction of the proportionalities help visual learners.