Practice question
Question
A wire has a resistance of \( 25 \, \Omega \) at \( 20^\circ \text{C} \) and \( 27.5 \, \Omega \) at \(
90^\circ \text{C} \). 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: 27.5 = 25 [1 + α (90 - 20)] . Solve: 27.5 = 25 + 1750α ⇒ 1750α = 2.5 ⇒ α = (2.5/1750) ≈ 1.43 × 10⁻³ °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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