Practice question

**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. Kirchhoff’s junction rule states that the sum of currents entering a node equals the sum leaving it, ensuring charge conservation, as charge cannot accumulate at a point in a steady-state circuit. 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