#parallel resistors

Potentiometer, Conductivity and Special Cases

A circuit has a \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 5 \, \Omega

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. Parallel resistance: (1/R_p) = (1/5) + (1/10) = (2 + 1/10) = (3/10) ⇒ R_p = (10/3) ≈ 3.33 Ω . Total resistance: Rtₒtₐl = 1 + 3.33 = 4.33 Ω . Current: I = (ε/Rtₒtₐl) = (10/4.33) ≈ 2.31 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

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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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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and three resistors \( 2 \, \Ome

**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) = (1/2) + (1/4) + (1/8) = (4 + 2 + 1/8) = (7/8) ⇒ R_p = (8/7) ≈ 1.14 Ω . Total resistance: Rtₒtₐl = 1 + 1.14 = 2.14 Ω . Total current: I = (ε/Rtₒtₐl) = (10/2.14) ≈ 4.67 A . Voltage across parallel: V = I R_p = 4.67 × 1.14 ≈ 5.33 V

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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 circuit has a \( 12 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and two resistors \( 6 \, \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) = (1/6) + (1/3) = (1 + 2/6) = (3/6) = 0.5 ⇒ R_p = 2 Ω . Total resistance: Rtₒtₐl = 3 + 2 = 5 Ω . Current: I = (ε/Rtₒtₐl) = (12/5) = 2.4 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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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 9 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 5 \, \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) = (1/5) + (1/3) = (3 + 5/15) = (8/15) ⇒ R_p = (15/8) = 1.875 Ω . Total resistance: Rtₒtₐl = 1 + 1.875 = 2.875 Ω . Current: I = (ε/Rtₒtₐl) = (9/2.875) ≈ 3.13 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

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Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 14 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance and two resistors \( 8 \, \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) = (1/8) + (1/4) = (1 + 2/8) = (3/8) ⇒ R_p = (8/3) ≈ 2.67 Ω . Total resistance: Rtₒtₐl = 2 + 2.67 = 4.67 Ω . Current: I = (ε/Rtₒtₐl) = (14/4.67) ≈ 3.0 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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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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EMF, Internal Resistance and Cells Combination

A circuit has a \( 28 \, \text{V} \) battery with \( 4 \, \Omega \) internal resistance and three resistors \( 4 \, \Ome

**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. Parallel resistance: (1/R_p) = (1/4) + (1/8) + (1/16) = (4 + 2 + 1/16) = (7/16) ⇒ R_p = (16/7) ≈ 2.29 Ω . Total resistance: Rtₒtₐl = 4 + 2.29 = 6.29 Ω . Total current: I = (ε/Rtₒtₐl) = (28/6.29) ≈ 4.45 A . Voltage across parallel: V = I R_p = 4.45 × 2.29 ≈ 10.19 V . Current through

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