Resistors - Combinations Question 1

QuestionA 3.0 Ω resistor is connected in series with a parallel pair made of a 6.0 Ω resistor and a 12 Ω resistor. The combination is connected to a 14 V supply of negligible internal resistance. Calculate (a) the total resistance of the circuit and (b) the current drawn from the supply.StrategyReduce the network from the inside out. Replace each parallel group by a single equivalent resistor first, then add the resistors that are in series with it.(a) Total resistanceFor the parallel pair:1/R = 1/R₁ + 1/R₂ = 1/6.0 + 1/12 = 3/12R = 12/3 = 4.0 ΩCheck: the equivalent resistance of a parallel pair is always smaller than the smallest resistor in it (4.0 Ω is less than 6.0 Ω), because each extra branch gives the charge another path.This 4.0 Ω is in series with the 3.0 Ω resistor:R_total = 3.0 + 4.0 = 7.0 Ω(b) Supply currentI = V/R_total = 14 V / 7.0 Ω = 2.0 AShortcut: for exactly two resistors in parallel, use 'product over sum':R = R₁R₂/(R₁ + R₂) = (6.0 × 12)/(6.0 + 12) = 72/18 = 4.0 ΩIt only works for two resistors at a time. For three or more in parallel, add the reciprocals.