SERIESPARALLEL DC CIRCUITS
In the preceding discussions, series and parallel dc circuits have been considered separately. The technician will encounter circuits consisting of both series and parallel elements. A circuit of this type is referred to as a COMBINATION CIRCUIT. Solving for the quantities and elements in a combination circuit is simply a matter of applying the laws and rules discussed up to this point.
SOLVING COMBINATIONCIRCUIT PROBLEMS The basic technique used for solving dc combinationcircuit problems is the use of equivalent circuits. To simplify a complex circuit to a simple circuit containing only one load, equivalent circuits are substituted (on paper) for the complex circuit they represent. To demonstrate the method used to solve combination circuit problems, the network shown in figure 354(A) will be used to calculate various circuit quantities, such as resistance, current, voltage, and power.
Figure 354.  Example combination circuit.
Examination of the circuit shows that the only quantity that can be computed with the given information is the equivalent resistance of R_{2} and R_{3}.
Given:
Solution:
Now that the equivalent resistance for R_{2} and R_{3} has been calculated, the circuit can be redrawn as a series circuit as shown in figure 354(B).
The equivalent resistance of this circuit (total resistance) can now be calculated.
Given:
Solution:
The original circuit can be redrawn with a single resistor that represents the equivalent resistance of the entire circuit as shown in figure 354(C).
To find total current in the circuit:
Given:
Solution:
To find total power in the circuit:
Given:
Solution:
To find the voltage dropped across R_{l}, R_{2}, and R_{3}, refer to figure 354(B). R_{eq1} represents the parallel network of R_{2 } and R_{3}. Since the voltage across each branch of a parallel circuit is equal, the voltage across R_{eq1} (E_{eq1}) will be equal to the voltage across R_{ 2} (E_{R2} ) and also equal to the voltage across R_{3} (E_{R3}).
Given:
Solution:
To find power used by R_{l}:
Given:
Solution:
To find the current through R_{2} and R_{3}, refer to the original circuit, figure 354(A). You know E_{R2} and E_{R3} from previous calculation.
Given:
Solution:
To find power used by R_{2} and R_{3}, using values from previous calculations:
Given:
Solution:
Now that you have solved for the unknown quantities in this circuit, you can apply what you have learned to any series, parallel, or combination circuit. It is important to remember to first look at the circuit and from observation make your determination of the type of circuit, what is known, and what you are looking for. A minute spent in this manner may save you many unnecessary calculations.
Having computed all the currents and voltages of figure 354, a complete description of the operation of the circuit can be made. The total current of 3 amps leaves the negative terminal of the battery and flows through the 8ohm resistor (R_{1}). In so doing, a voltage drop of 24 volts occurs across resistor R_{1}. At point A, this 3ampere current divides into two currents. Of the total current, 1.8 amps flows through the 20ohm resistor. The remaining current of 1.2 amps flows from point A, down through the 30ohm resistor to point B. This current produces a voltage drop of 36 volts across the 30ohm resistor. (Notice that the voltage drops across the 20 and 30ohm resistors are the same.) The two branch currents of 1.8 and 1.2 amps combine at junction B and the total current of 3 amps flows back to the source. The action of the circuit has been completely described with the exception of power consumed, which could be described using the values previously computed.
It should be pointed out that the combination circuit is not difficult
to solve. The key to its solution lies in knowing the order in which the
steps of the solution must be accomplished. 
