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. • U.S. Government Printing Office; 1945  618779
The study of parallel circuits typically
follows on the heels of series circuits because at least for resistance and inductance,
the math is easier. Capacitance in parallel, on the other hand, uses the equations
and methods of resistance and inductance in series. Unfortunately, though, for newcomers, series
capacitance uses the equations and methods of resistance and inductance in parallel.
Sure, most RF Cafe visitors covered all that stuff years ago, but as I've
mentioned before,
there are always new people coming into the electrical and electronics craft. These
NAVPER Basic Navy Training Courses were and still are heralded as being excellent
introductory material for students entering the realm. I don't know if it's still
so, but back in the post World War II era and up through the 1980s, civilian
employers assigned great regard to and preference for U.S. Navy (and, ahem, Air
Force) electronic technicians when hiring not only because of the classroom training
but for the intense work experience with troubleshooting, repair, and maintenance.
Chapter 9: Parallel Circuits: Multiple Circuits
Parallel circuits are multiple
circuits  they have more than one path between the two terminals of the
source.
Compare the two circuits in figure 47.
In both cases the current is forced by a potential to flow from  to +. In the
series circuit, the current travels along only one path. But in the parallel
circuit the current divides and, in this case, travels along three paths.
In the series circuit only one conductor is connected to each terminal.
But, in the parallel circuit, more than one conductor may be connected
to a terminal. Imagine that you are laying out a road system between town A and
town B. You have the choice of two systems. 1. One road directly between the two
towns. 2. A network of several roads running parallel to each other between the
two towns. The first type is like the series circuit and the second is like the
parallel circuit.
Voltage in Parallel Circuits
Figure 47.  Series and parallel compared.
Figure 48.  Voltage in a parallel circuit.
Carry the comparison a little farther  consider town A to be on top of a 1,000
foot hill and town B to be in the valley below.
In order to go from A to B you must drop 1,000 feet. In the series system, the
one road drops this 1,000 feet. In the parallel system every road
drops 1,000 feet. The same problem in a parallel electrical system is illustrated
in figure 48.
There is a potential difference of 1,000 volts across every path between A and
B. In the electrical system A and B are the terminals of the generator. No matter
which path (branch) you take from A to Bthe voltage force on the current is 1,000
volts. This gives you the law for voltage in a parallel circuit 
The voltage is the same across all branches of a parallel circuit.
Or mathematically 
E_{t} = E_{1} = E_{2}=
E_{3}, etc.
When 
E_{t} = total or source voltage;
E_{1} = voltage drop through first load;
E_{2} = voltage drop through second load;
E_{3} = voltage drop through third load.
Compare the 100 ohm load to the 200 ohm load in figure 48. Note that these loads
have different resistances but their voltages are the same. This is similar
to the problem of the two roads from town A to town B. One road may be narrow and
rough  its resistance is high. The other may be broad and smooth  its resistance
is low. But both roads, regardless of resistance, drop 1,000 feet.
Current in Parallel Circuits
Assuming equal quality of road construction, which system would carry the greatest
traffic from town A to town B, the oneroad series, or the networkparallel? The
parallel, of course, because it has more paths (roads). Say each road can carry
10 cars per minute. Then the oneroad series system carries just ten cars per minute.
But the threeroad parallel system carries 30 cars per minute.
Figure 49.  Current in a parallel circuit
Figure 49 shows a 3branch parallel with equal loads of 100 ohms resistance.
The generator voltage is 1,000 volts. The current in each branch can be calculated
by Ohm's law 
I_{1} = E_{1}/R_{1} =
1,000/100 = 10 amps.
Since the voltage and resistance are the same for the other branches, their current
is 10 amperes also. Now, since each load draws 10 amperes from the generator, the
total current is 
10 + 10 + 10 = 30 amps.
This gives you the law for current in a parallel circuit 
The total current in a parallel circuits is the sum of the currents
in all of the branches.
Or mathematically 
I_{t} = I_{1} + I_{2}
+ I_{3}, etc.
When 
I_{t} = the total current;
I_{1} = the current through the first
load;
I_{2}= the current through the second
load;
I_{3} = the current through the third
load.
Resistance in Parallel Circuits
Referring back to figure 49, notice that the current path from  to + is over
the three wires. View A of figure 50 shows only one of these conductors cut to show
a crosssection. Say that this conductor is one square inch in a crosssectional
area. The current flowing through this one wire is very much like water
flowing through ONE pipe. But if you combine the three wires, you would
have a cross section like B in figure 50. This combination has three square inches
instead of one square inch of crosssectional area. Now the current flowing through
the three wires, is very much like water flowing through three pipes. It
is just three times as easy to force the same current through the 3wire parallel
as it is through a onewire series (same size of wire). If it is three times as
easy  the resistance is onethird as much.
Figure 50.  Resistance in a parallel circuit
What was that? The resistance is less when there are more wires? Exactly  because
the more branches you add to a parallel circuit, the easier it
becomes to force current from  to +. This gives you a general idea of resistances
in parallel  the more loads added in parallel, the less the total resistance.
To get the actual law of resistances in parallel, you must derive it mathematically.
(1) You know that I_{t} = I_{1} + I_{2}
+ I_{3};
and you know that I_{t} =
E_{t}/R_{t}; I_{1} = E_{1}/R_{1}; I_{2}
= E_{2}/R_{2}; I_{3} = E_{3}/R_{3}
(2) If in the formula I_{t} = I_{1} + I_{2}
+ I_{3}, the E/R values are substituted for all values of I,
it becomes 
E_{t}/R_{t} = E_{1}/R_{1} + E_{2}/R_{2}
+ E_{3}/R_{3}
(3) Now, since the parallel circuits, E_{t} = E_{1}
= E_{2} = E_{3}, you can substitute E_{t} for all
values of E 
E_{t}/R_{t}
= E_{t}/R_{1} + E_{t}/R_{2} + E_{t}/R_{3}
(4) and dividing each member of the equation by E_{t} 
1/R_{t}
= 1/R_{1} + 1/R_{2} + 1/R_{3}, etc.
When 
R_{t} = the total resistance;
R_{1} = the resistance of the first load;
R_{2} = the resistance of the second load;
R_{3} = the resistance of the third load.
Or: stated in words 
In parallel circuits, the reciprocal of the total resistance equals the sum
on the reciprocals of the resistances of all branches.
Using the problem in figure 49, what is the total resistance of the circuit?
1/R_{t} = 1/R_{1} + 1/R_{2}
+ 1/R_{3}
1/R_{t} = 1/100 + 1/100 + 1/100
1/R_{t} = 3/100
R_{t }= 100/3 = 33.33 ohms.
This proves that three equal resistors, in parallel, have a total resistance
of only onethird the resistance of each resistor.
Ohm's law provides an easier method of calculating the total resistance of a
parallel circuit. Remember, you use the TOTAL values of the circuit when solving
for the total values.
R_{t} = E_{t}/I_{t} =
1000/30 = 33.33 ohms.
Problems in Parallel Circuits
The parallel circuit is used for independent loads  that is, loads
which are not controlled by other loads. Most electrical loads are of this
type. You certainly would want the ship's running lights to be independent of the
cook's galley. And it is definitely good sense to have the bilge pump motors independent
of the steering motors!
The circuits which follow are typical parallel circuits. When you study them,
remember the six rules in Chapter 8 for solving problems  these rules apply equally
well to parallel circuits.
Example 1 
Figure 51.  Example 1
Two lamps and a heater are connected in parallel across a 110volt line. Each
load is controlled by a separate switch and draws the current indicated in figure
51.
Calculate the resistance of each load and the total resistance.
For first load: R_{1} = E_{1}/I_{1} = 110/10
= 11 ohms.
For first load: R_{2} = E_{2}/I_{2} = 110/2
= 55 ohms.
For first load: R_{3} = E_{3}/I_{3} = 110/2
= 55 ohms.
For first load: R_{t} = E_{t}/I_{t} = 110/14=
7.86 ohms.
There is also the reciprocal method of calculating the TOTAL resistance 
1/R_{t} = 1/R_{1} + 1/R_{2}
+ 1/R_{3}
1/R_{t} = 1/11 + 1/55 + 1/55
1/R_{t} = 5/55 + 1/55 + 1/55
1/R_{t} = 7/55
R_{t} = 55/7 = 7.86 ohms.
You will notice that the reciprocal method is much tougher  use Ohm's Law
for total resistance whenever possible.
It was required that each load be separately controlled. In order to control
a load, the switch is in series with that particular load, but
the loads are in parallel with each other. Thus, opening the
heater switch opens the heater circuit, but does not
turn off the lights. The same is true of the light switches  they control only
the load with which they are in series.
Example 2 
Figure 52.  Example 2
Figure 53.  Connections for Example 2
A motor has four coils. The resistance of each coil is 60 ohms, and each coil
must be connected to receive 2 amperes of current. (1) How would you connect these
coils in a 120v. line? (2) What is the total current drawn from the line? (3) What
is the total resistance of your connections? (4) How much power is consumed in each
coil? In all the coils?
(1) You would first draw your circuit as in figure 52. Then calculate the voltage
required to force 2 amperes through 60 ohms of resistance.
E_{1} = I_{1}R_{1} = 2
x 60 = 120 v.
Since each coil needs 120 volts, they will have to be paralleled to get the full
line voltage. You would now complete your diagram by making this connection. Remember,
in order to be in parallel, each coil will have to be connected directly
to the line. Figure 53 shows two ways you might complete your diagram. Both are
correct. In this case, A, is the better method, because it uses less wire.
(2) I_{t} = I_{1} + I_{2} + I_{3}, etc.
I_{t} = 2 + 2 + 2 + 2 = 8 amps.
(3) R_{t} = E_{t}/I_{t} = 120/8 = 15 ohms.
(OR 1/R_{t} = 1/60 + 1/60 + 1/60 + 1/60
= 4/60
1/R_{t}
= 4/60
R_{t}
= 60/4  15 ohm.)
(4) P_{1} = E_{1}I_{1}
P_{1} = 120 x 2 = 240 watts  for each coil
P_{t} = E_{t}I_{t}
(RF Cafe note: original had P_{1} = E_{t}I_{t})
P_{t} = 120 x 8 = 960 watts  total.
(RF Cafe note: original had P_{1} = 120 x 8)
Example 3 
Figure 54.  Example 3
Figure 55.  Example 4
Figure 54 shows a resistance bank of four resistors. These resistors are connected
in parallel to permit the maximum current to pass. If each resistor has 100 ohms
resistance, what is the current passed from the 60volt battery? How much current
does each resistor carry?
1/R_{t} = 1/R_{1} + 1/R_{2}
+ 1/R_{3}, etc.
1/R_{t} = 1/100 + 1/100 + 1/100 + 1/100
1/R_{t} = 4/100
R_{t} = 100/4 = 25 ohms.
I_{t} = E_{t}/R_{t} =
60/25 = 2.4 amps, from the battery
I_{t} = E_{t}/R_{t} =
60/100 = 0.6 amps, from the battery
Because the resistance in each resistor is the same, the current is the same.
Thus 
I_{1} = 1_{2} = I_{3}
= I_{4} = 0.6 amp for each resistor.
Example 4 
To be sure that you understand the current paths in a parallel circuit, the example
in figure 55 has ammeters and arrows in all lines. The ammeters read the current
flowing, and the arrows show the direction. Trace this circuit through carefully.
If you understand each reading, you can account for all the current flowing. And
by using Ohm's law, you can prove that all the EMF of the generator is used up in
the circuit.
Chapter 9 Quiz
(click here)
Posted March 5, 2019
