Resistors - Combinations Question 2

QuestionYou are given three identical 6.0 Ω resistors. Find the total resistance of each different way of connecting all three, and state which arrangement draws the largest current from a 12 V supply of negligible internal resistance.Useful resultFor n identical resistors of resistance R: in series R_total = nR, and in parallel R_total = R/n.Arrangement 1: all three in seriesR_total = 6.0 + 6.0 + 6.0 = 18 ΩArrangement 2: all three in parallel1/R = 1/6.0 + 1/6.0 + 1/6.0 = 3/6.0, so R = 2.0 ΩArrangement 3: two in parallel, in series with the thirdThe parallel pair gives 6.0/2 = 3.0 Ω, soR_total = 3.0 + 6.0 = 9.0 ΩArrangement 4: two in series, in parallel with the thirdThe series pair gives 12 Ω, which is in parallel with 6.0 Ω:R_total = (12 × 6.0)/(12 + 6.0) = 72/18 = 4.0 ΩCurrents from 12 VUsing I = V/R: 18 Ω draws 0.67 A, 9.0 Ω draws 1.3 A, 4.0 Ω draws 3.0 A and 2.0 Ω draws 6.0 A.All in series gives the largest resistance and smallest current. All in parallel gives the smallest resistance and so draws the largest current, 6.0 A.Exam tip: check each answer against the limits. Every mixed arrangement must lie between R/n (all parallel) and nR (all series).