Temperature Dependence of Resistance and Resistivity

Temperature Dependence of Resistance and Resistivity

Two cells in parallel have emf \( 15 \, \text{V} \) and \( 5 \, \text{V} \) with internal resistances \( 3 \, \Omega \)

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. For parallel: (1/rₑq) = (1/r₁) + (1/r₂) = (1/3) + (1/1) = (1 + 3/3) = (4/3) . rₑq = (3/4) = 0.75 Ω . 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 and P = I²R, evaluation yields 0.75 Ω,

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Temperature Dependence of Resistance and Resistivity

Two cells in parallel have emf \( 10 \, \text{V} \) and \( 4 \, \text{V} \) with internal resistances \( 5 \, \Omega \)

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. For parallel: εₑq = (ε₁ r₂ + ε₂ r₁/r₁ + r₂) . Substitute: εₑq = (10 × 2 + 4 × 5/5 + 2) = (20 + 20/7) = (40/7) ≈ 5.71 V . 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 and P = I²R, evaluation yields 5.71 V,

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Temperature Dependence of Resistance and Resistivity

Two cells of emf \( 4 \, \text{V} \) and \( 6 \, \text{V} \) with internal resistances \( 1 \, \Omega \) and \( 2 \, \Om

**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. For series: εₑq = ε₁ + ε₂ = 4 + 6 = 10 V . Internal resistance: rₑq = r₁ + r₂ = 1 + 2 = 3 Ω . 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 and P = I²R, evaluation yields 10 V, 3 Ω,

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Temperature Dependence of Resistance and Resistivity

In a circuit with resistors in parallel, why does the total resistance decrease compared to the smallest individual resi

**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. In parallel, the reciprocal of total resistance is the sum of reciprocals of individual resistances ( 1/Rtₒtₐl = 1/R₁ + 1/R₂ + ·s ). This adds more paths for current, reducing the effective resistance below the smallest individual value. 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

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Temperature Dependence of Resistance and Resistivity

Why does a conductor’s resistance remain finite even when its length approaches zero?

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. Resistance R = rho l / A . As l to 0 , R to 0 , but resistivity ( rho ) is a material property that remains finite, ensuring R is small but non-zero unless l = 0 exactly, which is impractical. 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 and P =

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Temperature Dependence of Resistance and Resistivity

In a circuit with two identical cells connected in series, if one cell’s polarity is reversed, what happens to the total

**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. For two identical cells with emf ε , in series, total emf = ε + ε = 2ε . If one is reversed, it becomes ε - ε = 0 , as the emfs cancel out. 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 and P = I²R, evaluation yields

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Temperature Dependence of Resistance and Resistivity

Two cells in parallel have emf \( 8 \, \text{V} \) and \( 4 \, \text{V} \) with internal resistances \( 2 \, \Omega \) a

**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. For parallel: (1/rₑq) = (1/r₁) + (1/r₂) = (1/2) + (1/1) = (1 + 2/2) = (3/2) . rₑq = (2/3) ≈ 0.67 Ω . 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 and P = I²R, evaluation yields 0.67 Ω,

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Temperature Dependence of Resistance and Resistivity

A copper wire carries \( 3.4 \, \text{A} \) with a drift speed of \( 1.0 \times 10^{-4} \, \text{m/s} \). If \( n = 8.5

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (3.4/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁴) . Calculate: A = (3.4/1.36 × 10⁵) ≈ 2.5 × 10⁻⁵ m² . 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 and P = I²R,

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Temperature Dependence of Resistance and Resistivity

Why does the terminal voltage of a battery become equal to its emf when no current flows?

**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. Terminal voltage V = ε - I r . When I = 0 (open circuit), the internal voltage drop I r = 0 , so V = ε , matching the emf. 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 and P = I²R, evaluation yields No internal voltage drop,

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Temperature Dependence of Resistance and Resistivity

Two cells of emf \( 3 \, \text{V} \) and \( 6 \, \text{V} \) with internal resistances \( 1 \, \Omega \) and \( 2 \, \Om

**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. For series: εₑq = ε₁ + ε₂ = 3 + 6 = 9 V . 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 and P = I²R, evaluation yields 9.0 V,

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Temperature Dependence of Resistance and Resistivity

In a conductor, if the electric field is suddenly doubled while keeping the conductor's properties unchanged, what happe

**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. Drift velocity ( v_d ) is given by v_d = (e E tau/m) , where E is the electric field, e is the electron charge, tau is the relaxation time, and m is the electron mass. If E is doubled, v_d becomes 2v_d , assuming tau and other properties remain constant. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0,

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Temperature Dependence of Resistance and Resistivity

In a circuit, a \( 20 \, \text{V} \) battery with negligible internal resistance is connected across a cubical network o

**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. Equivalent resistance of cube network: Rₑq = (5/6) R = (5/6) × 2 = (10/6) = (5/3) Ω . Total current: I = (V/Rₑq) = (20/(5/3)) = 20 × (3/5) = 12 A . 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 and P = I²R, evaluation yields 12 A,

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