As circuits become more complex, you will have resistors (or other components) that are in a combination of series and parallel configurations. Reducing such a circuit to a total resistance is not something you will often need to do in the field. However, it is common to have circuits that combine both series and parallel sections. It is often important to follow where circuits branch to follow dividing current paths. The key is to follow where the current paths split. When two components are connected, but there is one current path, the components are in series. If there is a split to two or more current paths, the components are in parallel.

In the first circuit above, the positive terminal of the battery is connected to R1 and nowhere else. There is only one current path at that point. On the other side of R1 the circuit splits into two current paths, one through R2 and the other through R3. Since there are now two current paths we have a parallel circuit; R2 and R3. They are in parallel with each other, but, together, they are in series with R1. The second circuit above is electrically the same as the first. A circuit is not a series or parallel circuit because of the way the schematic is drawn but because of the way the components are connected. In the second circuit, R3 is still connected in parallel with R2. In either circuit, the current path splits where R1 meets R2 and R3, then recombines where the opposite ends of R2 and R3 are connected together.

In the next circuit you can see how the rules for series and parallel circuits combine in a series-parallel circuit.

First, look at the current. Like a series circuit, the 3 amps flow around the entire circuit. However, where the path splits, 2 of the 3 amps go through the 15 ohm resistor and the remaining 1 amp goes through the 30 ohm resistor (greater current through the lesser resistor). The current recombines where these two resistors are connected together for the original total of 3 amps. This satisfies Kirchhoff's Current Law. Next, look at the voltages.

The battery supplies 60 volts. Half of this is lost across the 10 ohm resistor (10 ohms x 3 amps = 30 volts). The other 30 volts are lost across the combination of the 15 ohm and 30 ohm resistors in parallel. You can demonstrate this through each individual resistor (15 ohms x 2 amps = 30 volts or 30 ohms x 1 amp = 30 volts)

You can also demonstrate this through both resistors as a whole (30 ohms in parallel with 15 ohms = 10 ohms, 10 ohms multiplied by 3 amps = 30 volts).

Be careful, don't forget your parallel circuit rules. Do not add the 30 volts across the 15 ohm resistor to the 30 volts across the 30 ohm resistor to get 60 volts.

It is the same 30 volts across both resistors in combination. 30 volts is lost across the 10 ohm resistor then another 30 volts is lost across the combination of the 15 ohm and 30 ohm resistors in parallel. This equals the 60 volts supplied by the battery and satisfies Kirchhoff's Voltage Law.

Also, don't make the common mistake of placing the battery voltage of 60 volts across the 10 ohm resistor. They are in series with each other and the voltages will be different.

Be sure you don't apply the voltage at one part of a series circuit to another part of the circuit (series circuit rule: the voltages are different).