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DC Circuits

# Series-Parallel DC Circuits

In the preceding discussions, series and parallel DC circuits have been considered separately.
However, there are 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.

The basic technique used for solving DC combination-circuit 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 below (view A) will be used to calculate various circuit quantities, such as resistance, current, voltage, and power.

Examination of the circuit shows that the quantity that can be easily computed with the given
information is the equivalent resistance of *R*_{2} and *R*_{3} (resistors in parallel).

Given: *R*_{2} = 20 Ω, *R*_{3} = 30 Ω

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 above (view B).

The equivalent resistance *R*_{eq} of this circuit (i.e. total resistance *R*_{t})
can now be calculated.

Given: *R*_{1} = 8 Ω, *R*_{eq1} = 12 Ω
(resistors in series)

Solution:

The original circuit can be redrawn with a single resistor that represents the equivalent resistance of the entire circuit as shown in figure above (view C).

To find total current in the circuit:

Given: *V*_{s} = 60 V, *R*_{t} = 20 Ω

Solution:

To find total power in the circuit:

Given: *V*_{s} = 60 V, *I*_{t} = 3 A;

Solution:

To find the voltage dropped across *R*_{1}, *R*_{2}, and *R*_{3},
refer to figure above (view 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} (*V*_{eq1}) will be equal to the
voltage across *R*_{2} (*V*_{R2}) and also equal to the voltage across
*R*_{3} (*V*_{R3}).

Given: *I*_{t} = 3 A, *R*_{1} = 8 Ω,
*R*_{eq1} = 12 Ω
(current through each part of a series circuit is equal to total current)

Solution:

To find power used by *R*_{1}:

Given: *V*_{R1} = 24 V, *I*_{t} = 3 A

Solution:

To find the current through *R*_{2} and *R*_{3}, refer to the original circuit,
figure above (view A). You know *V*_{R2} and *V*_{R3} from previous calculation.

Solution:

To find power used by *R*_{2} and *R*_{3}, using values from previous calculations:

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 above, a complete description of the
operation of the circuit can be made. The total current of 3 amps leaves the positive terminal of the
battery and flows through the 8-ohm resistor (*R*_{1}). In so doing, a voltage drop
of 24 V occurs across resistor *R*_{1}. At point A, this 3-ampere current divides into
two currents. Of the total current, 1.8 amps
flows through the 20-ohm resistor. The remaining current of 1.2 amps flows from point A, down through
the 30-ohm resistor to point B. This current produces a voltage drop of 36 volts across the 30-ohm
resistor. (Notice that the voltage drops across the 20- and 30-ohm 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.