Circuit A. There are two ways to approach this problem. The first is to reason using general properties of series and parallel circuits. The second is to calculate the equivalent resistance of each circuit, using the appropriate formula from the “Key equations” table above. Let’s first explore the problem by reasoning about parallel and series circuits. Start with a single 5 Ω resistor. Put any resistor in series with this first one, and you will increase the equivalent resistance. Circuits B and C are series circuits. Therefore, each of them has an equivalent resistance of more than 5 Ω. Similarly, putting any resistor in parallel with the first will reduce the equivalent resistance of the circuit. Circuits A and D are parallel circuits. Therefore, each has an equivalent resistance of less than 5 Ω. Together, all of this means that we only have to choose between Circuit A and Circuit D to find the lowest equivalent resistance. A given voltage can drive more current through a 5 Ω resistor than through a 10 Ω or 15 Ω one. Therefore, Circuit A will have a larger total current than Circuit D, and this means that Circuit A has the lowest equivalent resistance of the four circuits shown here. A second approach would be to substitute the individual resistances into the correct formula for equivalent resistance. Those formulas are$$\begin{array}{c}{R}_{eq}={R}_{1}+{R}_{2}+{R}_{3}\text{(series)}\\ \frac{1}{{R}_{eq}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\text{(parallel)}\end{array}$$Using the correct formula for each circuit, we get
Circuit A (parallel):$$\begin{array}{c}\frac{1}{{R}_{A}}=\frac{1}{5\text{\Omega}}+\frac{1}{\text{5\Omega}}+\frac{1}{\text{5\Omega}}=\frac{3}{\text{5\Omega}}\\ \Rightarrow {R}_{A}=\frac{\text{5\Omega}}{3}=1.67\text{\Omega}\end{array}$$Circuit B (series):$${R}_{B}=5\text{\Omega}+10\text{\Omega}+15\text{\Omega}=30\text{\Omega}$$Circuit C (series):$${R}_{C}=5\text{\Omega}+5\text{\Omega}+5\text{\Omega}=15\text{\Omega}$$Circuit D (parallel):$$\begin{array}{c}\frac{1}{{R}_{D}}=\frac{1}{5\text{\Omega}}+\frac{1}{10\text{\Omega}}+\frac{1}{15\text{\Omega}}\\ =\frac{6}{\text{30\Omega}}+\frac{3}{30\text{\Omega}}+\frac{2}{\text{30\Omega}}=\frac{11}{\text{30\Omega}}\\ \Rightarrow {R}_{D}=\frac{\text{30\Omega}}{11}=2.7\text{\Omega}\end{array}$$The lowest of these four values is 1.67 Ω, the equivalent resistance of Circuit A.
