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. Parallel resistance: (1/R_p) = (1/4) + (1/4) = (2/4) = 0.5 ⇒ R_p = 2 Ω . Total resistance: Rtₒtₐl = 2 + 2 = 4 Ω . Current: I = (ε/Rtₒtₐl) = (8/4) = 2 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