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#equivalent emf

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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,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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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,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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Kirchhoff's Laws and Combination of Resistors

Two cells in parallel have emf \( 6 \, \text{V} \) and \( 4 \, \text{V} \) with internal resistances \( 2 \, \Omega \) a

**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. For parallel: εₑq = (ε₁ r₂ + ε₂ r₁/r₁ + r₂) . Substitute: εₑq = (6 × 1 + 4 × 2/2 + 1) = (6 + 8/3) = (14/3) ≈ 4.67 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 4.67 V,

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

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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

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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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,

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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EMF, Internal Resistance and Cells Combination

Two cells of emf \( 2.5 \, \text{V} \) and \( 3.5 \, \text{V} \) with internal resistances \( 0.5 \, \Omega \) and \( 1

**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. For series: εₑq = ε₁ + ε₂ = 2.5 + 3.5 = 6.0 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 6.0 V,

Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination

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