This note covers the full syllabus content for 9.2 Potential difference & power (as context) and 9.3 Resistance, resistivity & temperature dependence. It focuses on NTC thermistors (negative‑temperature‑coefficient) while also recognising PTC devices (positive‑temperature‑coefficient).
Example: A 10 Ω heater carries 2 A.
\(P = I^{2}R = (2\ \text{A})^{2}\times10\ \Omega = 40\ \text{W}\). The heater converts 40 J of electrical energy into heat each second – a direct illustration of why resistance matters in heating elements.
For a uniform conductor of length L (m) and cross‑sectional area A (m²):
R = ρ · L ⁄ A
Unit reminder: use metres for L and square metres for A so that R is in ohms.
| Material | ρ | Typical use |
|---|---|---|
| Copper | 1.68 × 10⁻⁸ | Wiring |
| Aluminium | 2.82 × 10⁻⁸ | Power lines |
| Silicon (intrinsic) | 6.4 × 10³ | Semiconductor devices |
| Nickel‑chrome alloy (thermistor) | ≈ 10⁻² – 10⁰ | Temperature sensing |
Resistivity of most metals varies approximately linearly with temperature:
ρ = ρ₀ [1 + α (T − T₀)]
Substituting into R = ρL/A gives the familiar resistance expression:
R = R₀ [1 + α (T − T₀)]
| Material | α (K⁻¹) |
|---|---|
| Copper | 3.9 × 10⁻³ |
| Aluminium | 4.3 × 10⁻³ |
| Nickel‑chrome (metallic alloy) | ≈ 1 × 10⁻³ – 2 × 10⁻³ |
In a semiconductor the number of charge carriers rises sharply with temperature, so resistivity – and therefore resistance – falls. The syllabus adopts the empirical β‑model:
R = R₀ · e^{ β (1/T − 1/T₀) }
Because \(1/T\) decreases as \(T\) increases, the exponent becomes more negative and \(R\) drops exponentially.
Given \(R₀ = 10\ \text{kΩ}\) at \(T₀ = 298\ \text{K}\), \(\beta = 3500\ \text{K}\). Find \(R\) at \(T = 350\ \text{K}\).
R = R₀·e^{β(1/T – 1/T₀)}
= 10 000 Ω·e^{3500(1/350 – 1/298)}
1/350 – 1/298 = 0.002857 – 0.003356 = –0.000499 K⁻¹
Exponent = 3500 × (–0.000499) = –1.7465
R = 10 000 Ω·e^{–1.7465} ≈ 10 000 Ω·0.174 ≈ 1.74 kΩ
The resistance has fallen to about 17 % of its 25 °C value, illustrating the strong NTC effect.
A PTC thermistor has a positive temperature coefficient: its resistance increases as temperature rises. In most exam questions it is sufficient to state the sign and give one everyday example (e.g., self‑resetting heater protection).
\(V{\text{out}} = V{\text{supply}}\dfrac{R{\text{thermistor}}}{R{\text{fixed}}+R_{\text{thermistor}}}\). The output voltage varies with temperature.
Remember to keep units consistent (L in m, A in m², ρ in Ω·m, T in K) and to quote the reference temperature when using any temperature‑coefficient formula.
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