Ohm's law, resistivity, and how resistance changes with temperature
Ohm's law: V = IR at steady temperature. R = ρl/A, where the resistivity ρ depends on the material and rises with temperature for metals.
Ohm's law, ohmic and non-ohmic conductors
Ohm's law: the current through a conductor is directly proportional to the potential difference across it, provided the temperature and other physical conditions stay the same. So V ∝ I, or V = IR. The constant R is the resistance: the property of a conductor by which it opposes the current. Its unit is the ohm (Ω): 1 Ω = 1 V A⁻¹. Most metals obey Ohm's law. Their I-V graph is a straight line through the origin, with slope 1/R, so they are ohmic. A semiconductor diode, a vacuum diode and a transistor do not obey it. Their graph is curved, so they are non-ohmic. Copper sulphate solution between copper plates also gives a straight I-V graph, so it too is ohmic.
Resistance and resistivity
Take wires of one material with the same cross-section but different lengths: a longer wire passes less current, so R ∝ l. Take wires of the same length but different thickness: a thicker wire passes more current, so R ∝ 1/A. Together, R = ρl/A. The constant ρ is the resistivity (specific resistance) of the material: the resistance of a wire 1 m long with a cross-section of 1 m². Its unit is the ohm metre (Ω m). Resistivity depends on the material and its temperature, not on the size of the wire. Its reciprocal is the conductivity σ = 1/ρ, in S m⁻¹, also written Ω⁻¹ m⁻¹ or mho m⁻¹.
How to draw the I-V graphs
- Draw V on the x-axis and I on the y-axis, crossing at the origin. Write the units: V, and A or mA.
- Ohmic conductor: a straight line through the origin. Label it 'slope = 1/R'.
- Semiconductor diode: almost flat along the V-axis for small forward voltages, then rising steeply. On the reverse side, a line very close to the axis.
- For two wires on one graph, the steeper straight line belongs to the wire of smaller resistance.
How resistance changes with temperature
For most metals, resistivity rises with temperature, almost in a straight line over a limited range: ρ = ρ₀[1 + α(T − T₀)]. Here ρ₀ is the resistivity at a reference temperature T₀ in °C, and α is the temperature coefficient of resistivity, in °C⁻¹. A wire's resistance follows the same law: R = R₀[1 + α(T − T₀)], so α = (R₂ − R₁)/[R₀(T₂ − T₁)]. In a hot metal the atoms vibrate more, so electrons collide more often. Alloys such as manganin and constantan have a high resistivity and a very small α (about 10⁻⁶ °C⁻¹ for manganin), so they make standard resistors. Nichrome also has a high resistivity, so it is used for heater coils. Your NIOS lesson puts nichrome in the very small α group too. Its α is in fact about 10⁻⁴ °C⁻¹: larger than manganin's, but still far below copper's 4 × 10⁻³ °C⁻¹. Metals with a steady, larger α are used in resistance thermometers. In semiconductors such as silicon and germanium, resistivity falls as temperature rises: α is negative. Some materials lose all resistance below a certain temperature: they are superconductors.
Worked example: resistivity and a stretched wire
A wire of length 2.0 m and area of cross-section 0.40 mm² has a resistance of 5.0 Ω. (a) Find the resistivity and conductivity of its material. (b) The wire is stretched uniformly to double its length. Find its new resistance.
- Given: l = 2.0 m, A = 0.40 mm² = 0.40 × 10⁻⁶ m², R = 5.0 Ω.
- (a) Formula: R = ρl/A, so ρ = RA/l.
- ρ = (5.0 Ω × 0.40 × 10⁻⁶ m²) ÷ 2.0 m = 1.0 × 10⁻⁶ Ω m.
- σ = 1/ρ = 1 ÷ (1.0 × 10⁻⁶ Ω m) = 1.0 × 10⁶ S m⁻¹.
- (b) Stretching keeps the volume Al fixed. The new length is 2l, so the new area is A/2.
- R' = ρ(2l)/(A/2) = 4ρl/A = 4R = 4 × 5.0 Ω = 20 Ω.
(a) ρ = 1.0 × 10⁻⁶ Ω m (close to nichrome), σ = 1.0 × 10⁶ S m⁻¹; (b) 20 Ω. The resistivity does not change on stretching.
Worked example: finding α from two readings
A metal wire used as a resistance thermometer has a resistance of 5.0 Ω in melting ice (0 °C) and 7.0 Ω in boiling water (100 °C). Find the temperature coefficient of resistance of the metal. What is its resistance at 50 °C?
- Formula: R = R₀[1 + α(T − T₀)], so α = (R − R₀)/[R₀(T − T₀)].
- α = (7.0 Ω − 5.0 Ω) ÷ (5.0 Ω × 100 °C) = 2.0 Ω ÷ 500 Ω °C = 4.0 × 10⁻³ °C⁻¹.
- At 50 °C: R = 5.0 Ω × (1 + 4.0 × 10⁻³ × 50) = 5.0 Ω × 1.2 = 6.0 Ω.
α = 4.0 × 10⁻³ °C⁻¹; R = 6.0 Ω at 50 °C
The tungsten filament of an old bulb has a much higher resistance when glowing than when cold. Electric irons and geysers use nichrome coils, because nichrome has a high resistivity and resists oxidation when red hot.
1 marks: “Name the SI unit of resistivity.” The ohm metre (Ω m).
2 marks: “A wire is folded in half, so that its length halves and its area of cross-section doubles. How does its resistance change?” R = ρl/A. The new R = ρ(l/2)/(2A) = R/4: the resistance becomes one quarter.
3 marks: “A wire has a resistance of 4.0 Ω at 20 °C and 4.4 Ω at 70 °C. Taking 20 °C as the reference temperature, find the temperature coefficient of resistance of its material.” α = (R₂ − R₁)/[R₀(T₂ − T₁)] = (4.4 − 4.0) ÷ (4.0 × 50) = 0.4 ÷ 200 = 2.0 × 10⁻³ °C⁻¹.
Long and thin means more resistance: R = ρl/A. Metals get worse conductors when hot; semiconductors get better.