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. Equivalent emf: εₑq = 5 + 9 = 14 V . Total resistance: Rtₒtₐl = 1 + 2 + 7 = 10 Ω . Current: I = (εₑq/Rtₒtₐl) = (14/10) = 1.4 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