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): $$\begin{array}{c}\frac{1}{R_A}=\frac{1}{5\text{ Ω}}+\frac{1}{\text{5 Ω}}+\frac{1}{\text{5 Ω}}=\frac{3}{\text{5 Ω}}\\ \Rightarrow R_A=\frac{\text{5 Ω}}{3}=1.67\text{ Ω}\end{array}$$ Circuit B (series): $$R_B=5\text{ Ω}+10\text{ Ω}+15\text{ Ω}=30\text{ Ω}$$ Circuit C (series): $$R_C=5\text{ Ω}+5\text{ Ω}+5\text{ Ω}=15\text{ Ω}$$ Circuit D (parallel): $$\begin{array}{c}\frac{1}{R_D}=\frac{1}{5\text{ Ω}}+\frac{1}{10\text{ Ω}}+\frac{1}{15\text{ Ω}}\\ =\frac{6}{\text{30 Ω}}+\frac{3}{30\text{ Ω}}+\frac{2}{\text{30 Ω}}=\frac{11}{\text{30 Ω}}\\ \Rightarrow R_D=\frac{\text{30 Ω}}{11}=2.7\text{ Ω}\end{array}$$ The lowest of these four values is 1.67 Ω, the equivalent resistance of Circuit A.

Similarly, putting any resistor in parallel with the first will reduce the equivalent resistance of the circuit. Therefore, Circuit A and Circuit D both have an equivalent resistance of less than 5 Ω. Together, all of this means that we only have to choose between Circuit B and Circuit C to find the highest equivalent resistance.

Now, the equivalent resistance of a series circuit is simply the sum of the individual ohm values. Circuit B has an equivalent resistance of 30 Ω, while Circuit C has an equivalent resistance of 15 Ω. Circuit B therefore has the highest equivalent resistance of the four circuits under consideration.

Since we determined in Problem 3 that Circuit A has the lowest equivalent resistance, it will provide the greatest amount of heat.

1. Answers will vary, but they should include major appliances such as washer, dryer, dishwasher, and refrigerator.
2. Answers will vary, but the current should be estimated from the power as I = P/V.
3. Answers will vary, but in many households it is the electric clothes dryer that consumes 10%–20% of the total household power.