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#physics basics

2 public questions tagged with this topic.

Kirchhoff's Laws and Combination of Resistors

Two cells of emf \( 3 \, \text{V} \) and \( 7 \, \text{V} \) with internal resistances \( 0.5 \, \Omega \) and \( 1.5 \,

**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 emf: εₑq = 3 + 7 = 10 V . Total resistance: Rtₒtₐl = 0.5 + 1.5 + 10 = 12 Ω . Current: I = (εₑq/Rtₒtₐl) = (10/12) ≈ 0.83 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 0.83 A,

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

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Electric Current, Drift Velocity and Mobility

Why does a conductor exhibit zero net current in the absence of an electric field?

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Without an electric field, electrons move randomly due to thermal energy, with no preferred direction. The average velocity of electrons cancels out, resulting in zero net current. 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

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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