Here is the "Electricity  Basic Navy Training Courses" (NAVPERS 10622) in its
entirety. It should provide one of the Internet's best resources for people
seeking a basic electricity course  complete with examples worked out. See
copyright. See
Table of Contents.
ΒΆ U.S. GOVERNMENT PRINTING OFFICE; 1945  618779
CHAPTER 10 SERIESPARALLEL CIRCUITS
COMBINATIONS
Many circuits are neither SIMPLE series nor SIMPLE parallel. They are COMBINATIONS of simple circuits.
Fortunately, it is easy to recognize the series or parallel connections within the combinations by a few simple
rules.
SERIES CONNECTIONS HAVE 
1. Only one path.
2. Only one conductor connected to a terminal.
3. All the current of one part passing through the
other part.
PARALLEL CONNECTIONS HAVE 
1. More than one path. 2. More than one conductor
connected to a terminal. 3. Divided currents.
The best way to understand the complicated
seriesparallel circuits is to analyze them part by part, or section by section. In this way, the correct circuit
law can be used to analyze each part. First, spot your series connections, and then locate the parallels.
The balance of this chapter consists of a number of circuits. They illustrate the important problems you will
meet in actual circuits. Go through each step carefullyhe sure you understand it. If you get stuck, you'll
probably find that you've forgotten one of the laws given in Chapters 6, 7, 8 or 9. To help you, all these laws
are brought together in a table at the end of this chapter. Turn back to the table if you need help. Remember, the
important thing is to KNOW. You don't really know that you KNOW unless you test yourself. When you have finished
going over the examples, try working them yourself without reference to this book. After you've finished each
example, check your answers and methods against the answers and methods given here.
EXAMPLE 1 During war,
ships must travel blacked out. No lights show except those used for communication and as formation guides. These
lights are screened to 'show only in one direction. Imagine a lighted compartment with a door or hatch opened
accidentallya beautiful target! To guard against any such accident, doors and hatches opening to exposed decks
are equipped with doorswitches. Doorswitches open the lighting circuit of the compartment so that the lights are
turned off every time the door is opened.
Draw the schematic diagram for wiring a compartment with two
overhead lights, separately controlled, and with a doorswitch.
Trace this circuit (figure 56) from  to
+. Following the arrows and assuming that all switches are closed, current leaves the negative terminal of the
source, flows along L1 (Line 1) to the first light terminal. Here the current divides (did you spot a parallel
connection?)  part goes to light No. 1 and part goes to light No.2. (You can determine how much current goes to
each light if you know the voltage and resistance  I = E/R). The current passes through the two lights and enters
L2 then through L2 and back to the positive terminal of the source.
Figure 56.  Compartment schematic diagram.
Try opening switch No. 1. What happens? The circuit through light No. 1 is opened, BUT the circuit through
light No. 2 is undisturbed. Try the same with switch No. 2. This switch affects only light No. 2. Switch No.1 is
IN SERIES WITH light No.1 and thereby CONTROLS light No. 1. But switch No. 1 is IN PARALLEL WITH light No. 2 and
CANNOT control light No. 2. Now open the doorswitch. This switch is in series with BOTH lights and CONTROLS both
lights. Ask yourself this question, "Does ALL the current of light No. 2 go through the doorswitch?" The answer
is "YES." And whenever the answer to this question is 'yes'  the two devices are in series.
EXAMPLE 2  Aboard ship you will have a ship's service generator. This generator furnishes the power for all
electrical circuits except propulsion. The generated power is fed through a main switchboard, then through
lighting panels, interior communication panels, etc., and to feeder boxes and branch feeder boxes. And finally, to
the power outlets  lights, heaters, and telephones.
Figure 57.  Water distribution.
These ship's distribution systems are a lot like the water distribution systems of small towns. Look at
figures 57 and 58 and compare the branching methods .. In figure 58 start at the power outlettrace backwards to
the power bus. (A BUS is simply a very large conductorusually a bar of copper.) Notice that the fuses PROTECTING
each line and box are in SERIES with the load. And of course, switches controlling each load would be in series
with that load.
If you start tracing at the bus, you will notice that the "FEEDINGOUT" or branching is
done by PARALLEL connections. Start tracing from the negative bus and go through the complete circuit of the
outlet. How many times is this circuit protected by fuses? The reason for this multiple protection is simple. Each
fuse has a capacity which just fits the circuit it protects. For instance, the branch box has 5ampere fuses
because the circuit from branch box to outlet is only large enough to handle 5 amperes. The feeder box is fused
for 25 amperes because the circuit from feeder box to branch box will stand only 25 amperes. Each fuse protects
its own circuit from overload or short circuit damage.
Figure 58.  Electrical distribution.
EXAMPLE 3  The electrical power to a Navy searchlight must do four things
1. Furnish an arc
between the carbon electrodes. 2. Run the feed motor. 3. Run the ventilating fan motor. 4.
Run the shutter motor.
Power is furnished to each of these loads by means of a fourbranch parallel
circuit. Figure 59 is a simplified diagram of the searchlight circuit. In addition to the loads, there is a
rheostat (an adjustable resistance) in series with the parallel group. This is necessary to reduce the ship's
voltage from about 120 volts to about 80 volts for searchlight operation. In figure 59 the current is labeled for
each load. (1) What is ,the voltage drop across the rheostat? (2) What is the voltage used across each branch of
the parallel? (3) What is the resistance of each branch of the parallel? (4) What is the total resistance?
Figure 59.  Simplified searchlight diagram.
(1) The rheostat is in series with the rest of the circuit, therefore, it carries all the current of the
circuit 
I_{t} = I_{1} + l_{2} + I_{3}, etc.
It = 150 +
2 + 2 + 3 = 157 amps.
The voltage drop of this rheostat is
E = IR = 157 X 0.25 = 39.25 volts.
(2) The parallel group is in series with the rheostat, therefore adding the voltages of the group and the rheostat
together gives the total voltage.
E_{t} = E_{1} + E_{2} + E_{3}, etc.
120 = 39.25 + E_{2}
E_{2} = 120  39.25 = 80.75 v.
Which means that the voltage drop across the rheostat is 39.25
volts and the drop across EACH branch of the parallel is 80.75 volts. You might look at it this way  you have 120
volts (ship's service) to use in forcing 157 amperes through the complete circuit. The rheostat used up 39.25
volts of this 120. Which leaves 80.75 volts for the balance of the circuit  the parallel group.
(3) The
resistance of each branch of the parallel appears to be 
(arc)
R_{1} = E_{1}/I_{1}
= 80.75/150 = 0.54 ohm.
(feed and shutter motors)
R_{2} =E_{2}/I_{2}
= 80.75/2 = 40.38 ohms each.
(ventilating fan motor)
R_{3} = E_{3}/I_{3} = 80.75/3 = 26.92 ohms.
Notice in these calculations,
that the DIFFERENT, RESISTANCES LIMIT the current to DIFFERENT VALUES for a particular load.
(4) The total
resistance can be calculated by two methods 
R_{t} = E_{t}/I_{t} = 120/157 =
0.77 ohm.
OR
the resistance for the parallel group is 
1/R_{t} = 1/R_{1}
+ 1/R_{2} + 1/R_{3} etc.
1/R_{t} = 1/0.54 + 1/40.38 + 1/40.38 + 1/26.92
R_{t}
= 0.52 ohm. (of parallel group only)
add this to the resistance of the rheostat which is in series 
R_{t} = R_{1} + R_{2}
+ R_{3}, etc.
R_{t} = 0.52 + 0.25 = 0.77 ohm.
It is important that you see how
much EASIER it is to find TOTAL RESISTANCE BY OHM'S LAW.
EXAMPLE 4  You know that searchlights are
brightbut how much power do they consume? In the searchlight of Example 3, the ARC ITSELF used 80.75 volts and
passes 150 amperes. Therefore, its power is 
P = EI
P = 80.75 x 150 = 12,112.5, say
12,110 watts.
And the total power consumed by the light and its apparatus is 
P = EI P = 120
x 157 = 18,840 watts.
This is approximately the same amount of power as consumed by a 20hp motor. SOME LIGHT!
EXAMPLE 5  The operating voltage of a submarine's motors is 120 volts at cruising speed. This voltage must
come from batteries of the leadacid type. But this type of storage battery produces only 2 volts per cell. How is
an emf of 120 volts produced by cells which themselves produce only 2 volts each? Look at the laws of voltage in
the series and the parallel circuits 
E_{t} = E_{1} + E_{2} + E_{3},
etc. (SERIES)
E_{t} = E_{1} = E_{2}
= E_{3}, etc. (PARALLEL)
It is evident that in the series connection, voltages will add. Two cells
of 2 volts in series would add, giving 4 volts. And ten cells in series would give 20 volts. To produce the 120
volts needed by the sub's motors requires 60 cells in series. A part of such a battery is shown in figure 60.
All batteries store only a certain amount of energy  the exact amount is known as their CAPACITY. Capacity is
measured in units of AMPEREHOURS  the number of amperes which can flow for a certain number of hours before the
battery is discharged. For example, a battery having a capacity of 100 amperehours will deliver a current of 10
amperes for 10 hours (10 x 10 = 100) or 5 amperes for 20 hours (5 x 20 = 100) or 50 amperes for 2 hours (50 x 2 =
100) before it is discharged.
Say that the sub's motors require a current of 200 amperes at 120 volts. This
would exhaust a 1,000 amperehour battery in 5 hours. But by using two batteries in PARALLEL the drain on each
battery is only 100 amperes 
I_{t} = I_{1} + I_{2}
Figure 60.  Cells in series.
Where I_{t} = 200 amperes, I_{1} and I_{2} are only 100 amperes each. Figure 61 shows
two batteries of 20 cells  the cells are in SERIES and the two batteries are in PARALLEL. Remember that CELLS IN
SERIES INCREASE THE VOLTAGE and CELLS IN PARALLEL INCREASE THE CURRENT.
EXAMPLE 6  Three vacuum tube
filaments rated at 0.3 ampere and 6.3 volts must be operated on a 110volt line. Obviously, the voltage is too
HIGH. Even connecting the tubes in series 
E_{t} = E_{1} + E_{2}
+ E_{3}
110 = 36.67 + 36.67 + 36.67
gives a voltage of 36.67 volts per unit whereas
each unit is designed for only 6.3 volts. This means applying about six times the rated voltage to each tube 
they would burn out in a split second. A series resistance will have to be added to the circuit to use up some of
the excess voltage. The question is  how MUCH resistance? You have a circuit involving 110 volts and you must
LIMIT the current by a resistor to 0.3 ampere.
Figure 61.  Cells in seriesparallel.
Therefore, you will need a total of 
R = E/I = 110/0.3 = 366.67 ohms of resistance.
You
already have 
R = E/I = 6.3/0.3 = 21 ohms of resistance in each tube.
If they are connected
in series, you have a total resistance of 
R_{t} = R_{1}
+ R_{2} + R_{3}
R_{t }= 21 + 21 + 21 = 63 ohms.
for the tubes.
If the tubes furnish 63 ohms out of a total requirement of 366.67 ohms, the balance, R2, is found to be 
R_{t}
= R_{1} + R_{2}
366.67 = 63 + R_{2}
R_{2} = 366.67  63 = 303.67 ohms.
This resistance would have to be furnished by the
resistor. The completed circuit would look like figure 62. Notice that the voltage is labeled at a number of
points in figure 62. This shows that the voltage drops as the current goes through each successive load. In other
words, some voltage is used up in pushing the current through each resistance. The voltage drop in each case, is
the difference in voltage between the two points.
Figure 62.  Example 6  Series.
You can prove this circuit. The voltage drop (often called IR drop) across the resistor is 
E =
IR = 0.3 x 303.67 = 91.1 volts
leaving E_{2}, which is found by the formula 
E_{t}
= E_{1}
+ E_{2}
E_{t} = 91.1 + E_{2} = 18.9 volts
This is the total for the
three tubes. They are in series so each uses onethird of this 18.9 volts, or 18.9/3 = 6.3 v. which is the rated
voltage.
PRACTICE
For practice, try to set up this circuit with the tubes in parallel, and a limiting resistor in series. Your
circuit should look like figure 63.
Figure 63.  Example 6  Parallel.
BEFORE OR AFTER
Perhaps you are wondering if it makes any difference whether a limiting resistor comes BEFORE or AFTER the
load. Think about a garden hose. Does it make any difference which valve you operate  the one at the meter, or
the one at the side of the house, or the one in the nozzle of the hose? Partially closing anyone of these valves
will limit the amount of water flowing through the hose. Likewise, placing a resistance any place in a circuit
will limit the current through every load that is in series with the resistance.
CIRCUIT LAWS
OHM'S LAW 
POWER EQUATION 
E = I/R I = E/R R = E/I 
P = EI E = P/I I = P/E 


SERIES CIRCUITS 
PARALLEL CIRCUITS 
E_{t} = E_{1} + E_{2} + E_{3}, etc. I_{t}
= I_{1} + I_{2} + I_{3}, etc. R_{t} = R_{1} + R_{2} + R_{3}, etc. 
E_{t} = E_{1} = E_{2} = E_{3}, etc. I_{t}
= I_{1} = I_{2} = I_{3}, etc. 1/R_{t} = 1/R_{1} = 1/R_{2} = 1/R_{3}, etc. 
Chapter 10 Quiz
(click
here)
