Practice question
Question
A platinum wire has a resistance of \( 6 \, \Omega \) at \( 0^\circ \text{C} \) and \( 6.48 \, \Omega
\) at \( 100^\circ \text{C} \). What is the temperature coefficient of resistivity?
Explanation
**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 6.48 = 6 [1 + α (100 - 0)] . Solve: 6.48 = 6 + 600α ⇒ 600α = 0.48 ⇒ α = (0.48/600) = 8 × 10⁻⁴ °C⁻¹ . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r
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