Resistance, Resistivity and Ohm's Law

Resistance, Resistivity and Ohm's Law

A wire of length \( 4 \, \text{m} \) and resistance \( 10 \, \Omega \) is stretched to \( 8 \, \text{m} \). What is the

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 10 = 40 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 =

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Resistance, Resistivity and Ohm's Law

A wire of length \( 4 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**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. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (8 × 2 × 10⁻⁶/4) = 4 × 10⁻⁶ Ω m . 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 =

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Resistance, Resistivity and Ohm's Law

A wire of length \( 8 \, \text{m} \) and resistance \( 16 \, \Omega \) is stretched to \( 16 \, \text{m} \). What is the

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 16 = 64 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 64 Ω, consistent

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Resistance, Resistivity and Ohm's Law

In a circuit with two cells in series, why does the equivalent internal resistance increase compared to a single cell?

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. In series, internal resistances add ( rₑq = r₁ + r₂ ) because current flows through both resistances sequentially, increasing the total opposition to current flow. 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

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Resistance, Resistivity and Ohm's Law

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

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. For parallel: εₑq = (ε₁ r₂ + ε₂ r₁/r₁ + r₂) . Substitute: εₑq = (8 × 1 + 2 × 2/2 + 1) = (8 + 4/3) = (12/3) = 4 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

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Resistance, Resistivity and Ohm's Law

A \( 27 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**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. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 3 = (15/6) = 2.5 Ω . Total current: I = (V/Rₑq) = (27/2.5) = 10.8 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 =

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Resistance, Resistivity and Ohm's Law

A battery of emf \( 15 \, \text{V} \) and internal resistance \( 3 \, \Omega \) is connected to a resistor. If the termi

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. Voltage drop: I r = ε - V = 15 - 12 = 3 V . Current: I = (3/r) = (3/3) = 1 A . Resistance: R = (V/I) = (12/1) = 12 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 12 Ω,

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Resistance, Resistivity and Ohm's Law

A wire of length \( 12 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) has a resistance of

**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. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (24 × 3 × 10⁻⁶/12) = 6 × 10⁻⁶ Ω m . 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 =

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Resistance, Resistivity and Ohm's Law

When two identical resistors are connected in series and then in parallel, how does the equivalent resistance change?

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. For resistors R in series: Rₑq = R + R = 2R . In parallel: 1/Rₑq = 1/R + 1/R = 2/R ⇒ Rₑq = R/2 . The series resistance (2R) is four times the parallel resistance (R/2), as 2R / (R/2) = 4 . 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

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Resistance, Resistivity and Ohm's Law

A wire of length \( 3 \, \text{m} \) and resistance \( 6 \, \Omega \) is stretched to \( 6 \, \text{m} \). What is the n

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 6 = 24 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 24 Ω, consistent

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Resistance, Resistivity and Ohm's Law

In a circuit with two resistors in series, why does the voltage divide proportionally to their resistances?

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Voltage drop across each resistor is V = I R . With the same current in series, V ∝ R , so the total voltage splits in proportion to the resistances (e.g., V₁ / V₂ = R₁ / R₂ ). Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0,

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Resistance, Resistivity and Ohm's Law

Two cells in parallel have emf \( 6 \, \text{V} \) and \( 3 \, \text{V} \) with internal resistances \( 3 \, \Omega \) a

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. For parallel: εₑq = (ε₁ r₂ + ε₂ r₁/r₁ + r₂) . Substitute: εₑq = (6 × 1 + 3 × 3/3 + 1) = (6 + 9/4) = 3.75 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 3.75 V,

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