Many "real life"circuits are a combination of elements in series and parallel. It is not possible to solve these circuits by direct application of these basic rules. However, adding some simple procedures for reducing the circuit to a simple series or parallel circuit will allow us to solve most circuits of interest.
The basic procedure is to use a step-by-step process of replacing combinations of elements that are in simple series or parallel with the equivalent single resistance value. Rules for addition of resistance in series or parallel can be used to do this. When a simple series or parallel circuit is obtained, this circuit is solved. Results can then be used to solve for values in the more complicated circuit, generally by simply reversing the process used to simplify the circuit.
The following example will demonstrate this method of solution.
Simplify the circuit in a step-by-step fashion by combining groups of resistors in series or parallel to an equivalent single resistor, thereby producing an equivalent circuit which can be more easily solved.
For the example shown, two combinations will be required.
Solve the simplified circuit by application of the basic rules for either a series or a parallel circuit.
For the example, applying the basic rules of series circuits and using Ohm's Law, we can solve for the current flow through and voltage drop across each element.
Applying Ohm's Law for the whole circuit,
Recalling the Rule for a simple series circuit from the Series-Resistance section
Then applying Ohm's Law to each element
Using the information from the equivalent circuit, work backwards in a step by step process towards the original circuit. For the example problem this will require two steps.
(3a) Knowing the voltage drop across R234 = 30 V, we see the voltage across the two parallel resistors R2 and R34 is 30 V. Therefore, we can solve for the current flow through each of the resistors.
(3b) Knowing the current flow through R34 = 1.5 amp, we now know the current flow through each of the two resistors (R3 and R4) in the series must be 1.5 amp. Therefore, we can solve for the voltage drop across each of the two resistors.
We have now successfully solved for the current flow through and the voltage drop across each element of the series-parallel combination circuit.