#total current

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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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 8 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two resistors \( 4 \, \Omega

**Total voltage drop** across series equals source voltage because loop rule Σ V = ε, with internal resistance r included V = ε - I r. For 4 Ω,8 Ω,16 Ω parallel, 1/R_p =1/4+1/8+1/16=7/16, R_p=16/7≈2.29 Ω, then total resistance with internal 3 Ω is 5.29 Ω, current I=18/5.29≈3.4 A. Parallel resistance: (1/R_p) = (1/4) + (1/4) = (2/4) = 0.5 ⇒ R_p = 2 Ω . Total resistance: Rtₒtₐl = 2 + 2 = 4 Ω . Current: I = (ε/Rtₒtₐl) = (8/4) = 2 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

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Kirchhoff's Laws and Combination of Resistors

A circuit with a \( 9 \, \text{V} \) battery and \( 1 \, \Omega \) internal resistance has three resistors: \( 2 \, \Ome

**Total voltage drop** across series equals source voltage because loop rule Σ V = ε, with internal resistance r included V = ε - I r. For 4 Ω,8 Ω,16 Ω parallel, 1/R_p =1/4+1/8+1/16=7/16, R_p=16/7≈2.29 Ω, then total resistance with internal 3 Ω is 5.29 Ω, current I=18/5.29≈3.4 A. Parallel resistance: (1/R_p) = (1/2) + (1/3) + (1/6) = (3 + 2 + 1/6) = 1 ⇒ R_p = 1 Ω . Total resistance: Rtₒtₐl = 1 + 1 = 2 Ω . Total current: I = (ε/Rtₒtₐl) = (9/2) = 4.5 A . Applying I = n e A v_d, R = ρ l/A,

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Kirchhoff's Laws and Combination of Resistors

A \( 40 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 4 = (20/6) ≈ 3.33 Ω . Total current: I = (V/Rₑq) = (40/(20/6)) = 40 × (6/20) = 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 12.0 A,

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 10 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two resistors \( 4 \, \Omega

**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. Parallel resistance: (1/R_p) = (1/4) + (1/6) = (3 + 2/12) = (5/12) ⇒ R_p = (12/5) = 2.4 Ω . Total resistance: Rtₒtₐl = 2 + 2.4 = 4.4 Ω . Current: I = (ε/Rtₒtₐl) = (10/4.4) ≈ 2.27 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

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 8 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 3 \, \Omega

**Kirchhoff's loop rule** (energy conservation) states algebraic sum of potential differences around closed loop zero, Σ ε - Σ I R =0, ensures total voltage drop across series resistors equals source emf. Junction rule (charge conservation) Σ I_in = Σ I_out ensures current continuity. Parallel resistance: (1/R_p) = (1/3) + (1/6) = (2 + 1/6) = (3/6) = 0.5 ⇒ R_p = 2 Ω . Total resistance: Rtₒtₐl = 1 + 2 = 3 Ω . Current: I = (ε/Rtₒtₐl) = (8/3) ≈ 2.67 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,

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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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EMF, Internal Resistance and Cells Combination

A \( 24 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 4 = (20/6) = (10/3) ≈ 3.33 Ω . Total current: I = (V/Rₑq) = (24/(10/3)) = 24 × (3/10) = 7.2 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

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EMF, Internal Resistance and Cells Combination

A \( 18 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 3 = 2.5 Ω . Total current: I = (V/Rₑq) = (18/2.5) = 7.2 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 7.2 A,

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EMF, Internal Resistance and Cells Combination

A circuit has a \( 12 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 2 \, \Omega

**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Parallel resistance: (1/R_p) = (1/2) + (1/6) = (3 + 1/6) = (4/6) = (2/3) ⇒ R_p = (3/2) = 1.5 Ω . Total resistance: Rtₒtₐl = 1 + 1.5 = 2.5 Ω . Current: I = (ε/Rtₒtₐl) = (12/2.5) = 4.8 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's

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EMF, Internal Resistance and Cells Combination

A \( 45 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**EMF ε** is work done by non-electrostatic forces per unit charge, terminal voltage V = ε - I r, r internal resistance (Ω), I current (A). When external R = r, total resistance 2r, current I = ε/2r, power in external R is I²R = ε²/4r, total power ε²/2r, so half power dissipated externally, half internally. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 5 = (25/6) ≈ 4.17 Ω . Total current: I = (V/Rₑq) = (45/(25/6)) = 45 × (6/25) = 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

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