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#cube resistor network

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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,

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

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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 =

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

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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

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

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