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Question

A conductor has a resistivity of \( 9 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times
10^{-3} \, ^\circ\text{C}^{-1} \) at \( 20^\circ \text{C} \). What is its resistivity at \( 60^\circ
\text{C} \)?

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Explanation

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 9 × 10⁻⁸ [1 + 4 × 10⁻³ (60 - 20)] . Calculate: rho_t = 9 × 10⁻⁸ [1 + 0.16] = 9 × 10⁻⁸ × 1.16 = 1.044 × 10⁻⁷ Ω m .

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