Electricity · Lesson 5 of 9
Factors on Which the Resistance of a Conductor Depends
“Resistance depends on the conductor, proving that size, material, and temperature all matter.”
• Explain how length, cross-sectional area and material affect resistance. • Distinguish resistance from resistivity. • Apply R = ρl/A with consistent SI units. • Calculate cross-sectional area from wire diameter. • Predict resistance changes when wire dimensions change. • Relate material properties to electrical uses.
A long, thin wire and a short, thick wire made of the same metal do not oppose current equally. Charge carriers in the longer wire encounter a longer route, while the thicker wire provides more parallel pathways through its cross-section. Material structure also matters, which is why copper, tungsten and alloys serve different electrical purposes.
Experiments with wires show that resistance is directly proportional to length and inversely proportional to cross-sectional area when material and temperature are unchanged. Combining the relationships introduces a material constant called resistivity.
A material property defined by ρ = RA/l for a uniform conductor at a specified temperature.
Resistance belongs to a particular object and changes when its dimensions change. Resistivity characterises the material and does not depend on the chosen length or thickness, although it can change with temperature. Conductors have low resistivity, insulators very high resistivity, and alloys commonly have higher resistivity than their constituent metals.
Using Diameter Correctly
Because area depends on the square of diameter, doubling diameter makes area four times as large and reduces resistance to one quarter when length and material stay unchanged. Confusing diameter with area is one of the most frequent errors in wire problems.
Effect of Changing Dimensions
For the same material, compare two wires using R₂/R₁ = (l₂/l₁)(A₁/A₂). This ratio method often avoids calculating resistivity. If a wire is stretched without loss of material, its volume approximately remains constant. Increasing length then reduces area as well, so resistance rises for both reasons.
Choice of Materials
Copper and aluminium have low resistivity and are used for transmission wires. Tungsten combines a high melting point with suitable resistance, allowing a thin filament to become extremely hot. Alloys such as nichrome have relatively high resistivity, tolerate high temperatures and do not oxidise readily, making them suitable for heating elements.
| Material group | Electrical behaviour | Typical use |
|---|---|---|
| Low-resistivity metals | Carry current with less heating loss | Connecting and transmission wires |
| High-resistivity alloys | Produce useful heating in compact lengths | Toasters, irons and heaters |
| Very high-resistivity materials | Strongly restrict current | Electrical insulation |
Activity
Purpose: compare resistance indirectly through current. Use a low-voltage circuit containing a cell, key, ammeter and one test wire at a time. Compare equal-material wires of different lengths, wires of different thicknesses, and equal-sized wires of different materials. Keep the source voltage unchanged. Observation: the ammeter reading is smaller for greater resistance. Conclusion: length, area and material all affect resistance.
Problem-Solving Strategy
- Convert length to metres and diameter to metres.
- Calculate area with πd²/4 when diameter is given.
- Choose R = ρl/A or the ratio method for two wires.
- Track which quantities remain unchanged.
- Check that longer means larger R and thicker means smaller R.
Problem
A wire of resistivity 1.7 × 10⁻⁸ Ω m is 2 m long and has area 1.0 × 10⁻⁶ m². Find its resistance.
- 1.Given: ρ = 1.7 × 10⁻⁸ Ω m, l = 2 m and A = 1.0 × 10⁻⁶ m².
- 2.Use R = ρl/A.
- 3.R = (1.7 × 10⁻⁸ × 2)/(1.0 × 10⁻⁶) = 3.4 × 10⁻² Ω.
- 4.The small resistance is consistent with a low-resistivity metal and a relatively large area.
Problem
A 1 m wire has R = 26 Ω and diameter 0.30 mm. Find resistivity.
- 1.Convert d = 0.30 mm = 3.0 × 10⁻⁴ m.
- 2.A = πd²/4 = π(3.0 × 10⁻⁴)²/4 ≈ 7.07 × 10⁻⁸ m².
- 3.Rearrange R = ρl/A to ρ = RA/l.
- 4.ρ = 26 × 7.07 × 10⁻⁸ / 1 ≈ 1.84 × 10⁻⁶ Ω m.
- 5.This value is close to the tabulated resistivity of manganese at the stated temperature.
Problem
A wire has resistance 4 Ω. A second wire of the same material has half the length and twice the area. Find its resistance.
- 1.The material is the same, so ρ cancels in a ratio.
- 2.R₂/R₁ = (l₂/l₁)(A₁/A₂).
- 3.Here l₂/l₁ = 1/2 and A₁/A₂ = 1/2.
- 4.R₂/R₁ = 1/4, so R₂ = 4 Ω × 1/4 = 1 Ω.
- 5.Both changes reduce resistance, so a fourfold reduction is reasonable.
Problem
A bulb of resistance 1200 Ω and a heater of resistance 100 Ω are separately connected to 220 V. Compare their currents.
- 1.Both devices have the same potential difference, so use I = V/R for each.
- 2.Bulb current: I = 220/1200 = 0.183 A, approximately 0.18 A.
- 3.Heater current: I = 220/100 = 2.2 A.
- 4.The heater draws about twelve times the current because its resistance is much smaller.
- 5.This comparison shows why equal supply voltage does not imply equal appliance current.
Quiz
Which change increases resistance when material and temperature remain constant?
Which quantity primarily characterises the material rather than the particular wire?
If wire diameter doubles at fixed length, how does resistance change?
Why are alloys often used in heating elements?
Which unit belongs to resistivity?
Practice Problems
- A wire doubles in length with area unchanged. How does R change? Solution: Since R ∝ l, resistance doubles.
- A wire area becomes three times at unchanged length. How does R change? Solution: Since R ∝ 1/A, resistance becomes R/3.
- A wire has l = 5 m, A = 2 × 10⁻⁶ m² and ρ = 4 × 10⁻⁷ Ω m. Find R. Solution: R = ρl/A = (4 × 10⁻⁷ × 5)/(2 × 10⁻⁶) = 1 Ω.
- A 10 Ω wire is replaced by the same material and length but double diameter. Find new resistance. Solution: Double diameter gives four times area, so new R = 10/4 = 2.5 Ω.
- A wire is cut into five equal pieces. Find the resistance of each piece if the original resistance is R. Solution: Each piece has length l/5 with unchanged area and material, so each resistance is R/5.
Key Takeaways
• Resistance depends on length, area, material and temperature. • For a uniform wire, R = ρl/A. • Resistivity characterises a material at a specified temperature. • Doubling diameter makes area four times and resistance one quarter. • Ratio reasoning is efficient when comparing wires of the same material. • Low-resistivity metals suit transmission; high-resistivity alloys suit heating. • Units must be converted to metres and square metres before substitution.