Practice question

**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (12 × 5 × 10⁻⁶/6) = 10 × 10⁻⁶ = 1.0 × 10⁻⁵ Ω m . 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 1.0 ×