Here is the "Electrician's Mate 3 - Navy Training Courses" (NAVPERS 10548)
in its entirety (or will be eventually). It should provide one of the Internet's best
resources for people seeking a basic electricity course - complete with examples worked
- U.S. Government Printing Office; 1949
ELECTRICAL CURRENTS AND CIRCUITS - OHM'S LAW ELECTRICITY
Most of you have had some experience with electricity. Possibly your experience
was limited to replacing a blown fuse or burned-out light bulb. If you were a top-notch
amateur, you may have done some wiring about your home; but now you are about to
become an Electrician's Mate, and it is necessary for you to learn the Navy's way
of doing things. This manual will start you off on the right foot. H you must change
your method of making electrical connections, installing switches, or operating
motors, don't be upset - the men of the Navy have learned by experience the best
way of doing the job.
Electricity is known by what it does (producing light, running motors, and operating
telephones) rather than by what it is. While there may still be some doubt as to
what electricity is, the laws governing what electricity does are well known and
Electricity can be your slave, or your master. If you learn the laws of electricity
well, and respect its abilities, it will work for you in a thousand and one ways.
But if you are careless, and do not learn what electricity is capable of doing,
it may destroy you.
THIS COURSE AND YOU
This course is not intended to make a designing engineer out of you. Instead,
it is to acquaint you with the basic laws of electricity, and to teach you how to
apply them in operating and maintaining the electrical equipment on your ship.
This book starts where the course on Basic Electricity leaves off. You will find
the contents much easier to master if you will first review the course on Basic
Electricity (Navpers 10622).
As an Electrician's Mate you are principally concerned with electrical circuits.
To start you off right, this chapter will give you a quick review of the basic ideas
about electrical circuits that you have already learned from your study of elementary
ELECTRICAL CURRENT IS LIKE WATER IN A PIPE
An electric current is a flow of electrons through a conductor. Therefore, current
electricity can be compared to water flowing through a pipe.
Figure 1. - A source of electrical potential.
Before an electrical current can flow, there must be a source of electrical pressure,
just as you must have a pump to build up water pressure. This electrical pressure
is called an ELECTROMOTIVE FORCE and is produced by a battery or a generator. The
electrical pressure between any two points in a circuit is called the POTENTIAL.
Look at figure 1. The cell plays the same part in causing a current to flow as the
pump does in forcing ~water through the pipe.
Figure 2. - Conductors.
Figure 3. - Electrical circuits - increased electrical
pressure - increased current flow.
Next there must be an "electrical pipe" to carry the flows In the case of electricity,
the pipes usually are solid piece of metal and are called CONDUCTORS. The pipe in
figure 2 plays the same part for water as the conductor does for electricity.
If you increase the pressure on the electrons in the conductor, a greater current
will flow, just as an increased pressure on the water in a pipe will increase its
flow (fig. 3).
Figure 4. - Electrical circuits. Reduce the size of
conductor - increase the resistance to current flow.
Figure 5. - Electrical circuits - the longer the conductor,
the greater the resistance to current flow.
If the conductor is made smaller, figure 4, the electrical resistance is increased,
just as a smaller pipe increases the resistance to the flow of water.
When a conductor is lengthened, the resistance is increased, just as a longer
pipe increases the resistance to the flow of water (fig. 5). Each time you add another
length of conductor in series, the resistance increases; hence it takes a greater
electrical pressure to force the same amount of current through a long conductor
than through a shorter one. You can see this comparison in figure 6. You must remember
that the type of metal used in the conductor affects the resistance also. Since
copper has a low resistance, it is used most frequently for ship and shore electrical
Figure 6. - Electrical circuits - adding resistances
in series increases the total resistance to current flow.
It is obvious that no more water can flow out of a pipe than flows into it. All
evidence points to the fact that the same is true of an electric current; no more
current can leave a conductor than enters it. In practice this means that if you
have a battery, generator, or any other source of electromotive force, just as much
current returns to the source as flows away from it. This is known as KIRCHHOFF'S
Apply this law to a SERIES CIRCUIT, as illustrated in figure 7, and you see that
THE CURRENT FLOWING IN ALL PARTS OF THE CIRCUIT IS IDENTICAL. If this law is applied
to a PARALLEL CIRCUIT, as shown in figure 21, you see that THE SUM OF ALL CURRENTS
FLOWING TO A JUNCTION EQUALS THE SUM OF CURRENTS FLOWING AWAY FROM IT.
Figure 7. - Current in a series circuit is everywhere
Figure 8. - Electrical pressure is greatest at its
Remember this law. You will use it hundreds of times in solving electrical problems.
It is simple, but also very important.
Ever try to draw water from a tap near the end of a water main? Not much pressure?
The pressure is always greatest at the pump and decreases with each step as you
In electrical circuits, the pressure is the greatest at the source, or as in
figure 8, at the positive terminal of the battery; the pressure decreases throughout
the length of the circuit to zero pressure at the negative terminal. The difference
in pressure between any two points in the circuit is the pressure drop, or potential
difference. THE SUM OF ALL THE PRESSURE (VOLTAGE) DROPS IN A CIRCUIT IS EQUAL TO
THE APPLIED ELECTROMOTIVE FORCE. This principle is known as KIRCHHOFF'S SECOND LA
W. You will find many problems aboard ship concerning electrical pressure drops,
so keep this law in mind.
THE THREE FACTORS - VOLT, OHM, AMPERE
All electrical circuits have the two factors of PRESSURE and RESISTANCE determining
the third, CURRENT. Think of these factors as illustrated in figure 9; pressure
tends to move the electrons, and the resistance of the conductors tends to stop
Figure 9. - The three factors in an electrical circuit.
Figure 10. - Volts, ohms, and amperes in a circuit.
The term used to express the unit of electrical pressure is the VOLT. SO whenever
you see the terms VOLT or VOLTAGE used, remember it refers to the force or pressure
tending to keep the electrons moving. Voltage is sometimes called ELECTROMOTIVE
FORCE (emf). In most electrical formulas the symbol "E" is used to represent voltage.
The unit of resistance is the OHM, and like the volt it has an exact value. Briefly,
it is the opposition to the flow of current offered by a column of mercury 106.300
centimeters long and weighing 14.4521 grams, when at 0° centigrade. In electrical
equations the letter "R" is the symbol used to designate resistance.
The AMPERE expresses the number of electrons flowing past a point each second.
Since the electron is extremely small, a bundle of 6.3 billion-billion electrons
is used as unit and called a COULOMB. When a coulomb moves past a point in a second,
the amount of current flowing in the circuit is said to be one ampere.
Put the three factors - voltage, resistance, and current together as illustrated
in figure 10, and you have the basis for all electrical computations. The current
is determined by the applied voltage and the resistance offered by the circuit.
Many times you will see this relationship stated as in figure 11, where a current
of 1 ampere will flow when the electrical pressure is 1 volt and the resistance
of 1 ohm. It is just another way of stating OHM'S LAW.
Figure 11. - One form of Ohm's law.
MORE ABOUT OHM'S LAW
You learned in the preceding paragraphs that an increase in voltage means an
increase in current, but an increase in resistance means a decrease in current.
In other words the current flowing in a circuit is directly proportional to the
applied voltage, and inversely proportional to the resistance. That is the most
common way of stating Ohm's law. When you put this word statement into a mathematical
relationship you get -
Current = Pressure/Resistance
I = E/R
so if you know two of the values in Ohm's law, you can always find the third.
To aid in solving problems, Ohm's law can be written in two other forms. The
E = I x R
enables you to find the voltage if you know the current and resistance.
The second form enables you to find the resistance of the circuit if you know
the current and voltage. It is written -
R = E/I
Here are some practice problems.
1. What current will flow in a circuit containing a heater element
with a resistance
of 10 ohms when the applied
voltage is 110?
Solution: I = E/R I =
110/10 I = 11 amperes
2. If you know a certain horn requires 6 amperes to operate it properly
resistance of the circuit is 2 ohms,
what is the required voltage?
Solution: E = IR E = 6
x 2 E = 12 volts.
3. What resistance may a lamp requiring 3 amperes have, if only
6 volts are
available to light it?
Solution: R = E/I R =
6/3 R = 2 ohms
In electrical circuits, resistance is indicated by a zigzag line. Turn back to
figures 9, 10, or 11 and you will see this illustrated. Many times this symbol will
indicate all forms of resistances whether they be coils, heating devices, or just
the resistance of a conductor.
RESISTANCES IN SERIES
When you apply Ohm's law to circuits containing a single resistance, finding
the current or voltage is a simple matter of substituting this single resistance
value in the equations of Ohm's law, and then solving for the answer, as you did
in the examples above. But if the circuit contains several resistances, you must
find the combined resistance before the correct values of current and applied voltage
can be found.
Resistances connected in a manner that provides only one path for the current
to follow through the circuit and back to the source, are said to be connected in
SERIES. The circuit in figure 12 is an example of a series connection. All current
that flows through R1 must flow through R2, R3
and R4 before it returns to the source. Thus the total opposition to
the flow of current will be the sum of all the individual resistances, or RT
= R1 + R2 + R3 + R4
Figure 12. - Resistance of a series circuit.
The devices connected in a series circuit are linked together by conductors and
connections. These also have resistance, and their resistances must be added to
the resistance of the devices connected in the circuit to get the total resistance
of a series circuit. Since the resistance of the conductor depends on its length,
size, and the material it is made of the total resistance of a series circuit is
determined by the factors illustrated in figure 13.
Figure 13. - Factors that determine the resistance
of a circuit.
It is important to remember about this extra resistance due to conductors and
connections, because in low resistance circuits (battery circuits, firing circuits,
etc.) the greatest part of the circuit resistance is due to the conductors and connections
VOLTAGE DROPS IN A SERIES CIRCUIT
Figure 14 is an illustration of three lamps in series, and if the lamps are considered
to be the only resistances in the circuit the total resistance of the circuit will
RT = R1 + R2
and the current will be -
IT = I1 =I2 =I3
Figure 14. - Voltage in a series circuit.
The combined voltage drop across the lamps will be the sum of the individual
voltage drops -
I1R1 + I2R2 + I3R3
But in a series circuit the current is everywhere equal, so II! I2!and 13 are
the same as IT! and the combined voltage drop across the lamps will be-
ITRT = ITR1
+ ITR2 + ITR3
ITRT = IT (R1 + R2 + R3)
Since in this circuit the lamps are considered to contain all the resistances,
the total voltage drop of the circuit must be equal to the total applied voltage
ET = IT (R1 +
R2 + R3)
Thus you may say; THE SUM OF THE INDIVIDUAL IR DROPS ABOUT A SERIES CIRCUIT IS
EQUAL TO THE APPLIED VOLTAGE.
Figure 15. - A series circuit problem.
Here are two sample problems of resistors in series. More may be found in the
problems for this chapter given in the QUIZ at the end of this chapter.
1. The generator in figure 15 produces an emf of 120 volts, and
the circuit contains
resistors of 3,4, and
5 ohms. What is the current?
First find the total resistance -
RT = R1 + R2 + R3
RT = 3 + 4 + 5
RT = 12 ohms
Now substitute the values of E and RT in the formula for Ohm's law
I = E/R
I = 120/12 I = 10 amperes
2. The series circuit illustrated in figure 16 is carrying
6 amperes of current and
3 lamps of 2,4, and 6 ohms. What is the applied voltage? What is the
IR drop across each lamp?
Figure 16. - Another series circuit problem.
First find the total resistance -
RT = R1 + R2
RT = 2 + 4 + 6
RT = 12 ohms
Using the equation ET = IRT and substituting the correct
values of I and RT -
ET = 6 x 12
ET = 72 volts
Since each lamp is carrying 6 amperes, the individual IR drops will be -
For L1 - E1 = 6 x 2 = 12
For L2 - E2 = 6 x 4 = 24
For L3 - E3 = 6 x 6 = 36
RESISTANCES IN PARALLEL
In parallel circuits, the appliances are connected in such a manner that the
total circuit current is divided between them. Each appliance then provides a branch
path for the current to follow. Look at figure 17. The lamps form three branch paths
for the current. Naturally the total resistance is less than what it would be if
only one of the lamps were present.
Figure 17. - Parallel circuit.
You may think of the lamps in parallel as a water pipe having three outlets.
When the outlets are all the same size, the total resistance offered will be -
Total resistance = Resistance of one outlet / Number of outlets
or, if the three lamps in figure 17 are of the same resistance, the total resistance
will be -
RT = Resistance of one lamp /3
So if each lamp has a resistance of 12 ohms, the total resistance will be -
RT = 12/3 =4 ohms.
UNEQUAL RESISTANCES IN PARALLEL
Seldom will you find parallel circuits that contain resistances all of equal
value. The task of finding the total resistance is then a little more complicated.
In this case the simplest way to find the total resistance is to use the formula
1/RT = 1/R1 + 1/R2
Figure 18. - Resistance in parallel.
Now try a problem to see how this system works out. In figure 19 you have three
resistances (30, 50, and 75 ohms) in parallel, and you wish to know the total current
when the applied voltage is 102.
Figure 19. - A problem on resistances in parallel.
First find the total resistance using the formula
1/RT = 1/R1 + 1/R2
Substitute the values of resistance for R1, R2, and R3
1/RT = 1/30 + 1/50 + 1/75
Next find the least common denominator and add
1/RT = (5 + 3 + 2)/150
1/RT = 10/150
To find RT cross multiply and solve -
10 RT = 150
RT = 15 ohms.
Now you find the current from Ohm's law. Since E is 120 volts -
I = E/R
I = 120/15 I = 8 amperes
VOLTAGES IN PARALLEL CIRCUITS
In parallel circuits, the voltage across each branch of the circuit is equal
to the voltage across every other branch. Thus, in figure 20, the voltages across
L1, L2, and L3 are all the same and are equal to
the applied voltage; that is -
E = E1 = E2 = E3
Figure 20. - Voltages in parallel circuits.
This principle holds regardless of the number of branches, or the relative resistance
of each. The VOLTAGES ACROSS ALL LEGS OF A PARALLEL CIRCUIT ARE EQUAL.
CURRENT IN PARALLEL CIRCUITS
Since the voltages across all parallel branches of a circuit are equal, the most
current, like most water, will flow through the branch of lowest resistance. Turn
back to figure 17 again. The resistance of the 4-inch outlet is less than that of
the 2-inch, so more water will flow through it. The same thing is true in electrical
circuits; the most current flows through the lowest resistance.
In figure 19 you have three unequal resistances; 30, 50, and 75 ohms. Naturally
the most current will flow through the 30-ohm lamp, and the least through the 75-ohm
lamp, but how much through each? The solution to the problem is just one of applying
Ohm's law to each leg of the circuit. The voltage applied to the lamp in each leg
is 120, since the voltages across all parallel branches of a circuit are equal.
Here is the solution
The current through L1 is -
I1 = 120/30 = 4.0 amperes
The current through L2 is -
I2 = 120/50 = 2.4 amperes
The current through L3 is -
I3 = 120/75 = 1.6 amperes
How about the total current? You read in a preceding paragraph Kirchhoff's FIRST
LAW, which says that AS MUCH CURRENT FLOWS AWAY FROM A POINT AS FLOWS INTO IT.
Figure 21. - Currents in a parallel circuit.
Thus, as shown in figure 21, at point A the total current must divide and flow
through the three lamps, and at point B the three currents from the lamps must all
unite to form the total current, which flows back to the battery. In other words,
THE TOTAL CURRENT MUST BE THE SUM OF THE INDIVIDUAL CURRENTS.
IT = I1 + I2
IT = 4.0 + 2.4 + 6
IT = 8 amperes.
SAMPLE PROBLEM - PARALLEL CIRCUITS
Figure 22 is a sample problem of resistances in parallel. You are~ asked to find
the resistance of L3.
Figure 22. - Sample problem on resistances in parallel.
First substitute the known values of resistance in -
1/RT = 1/R1 + 1/R2
1/4 = 1/6 + 1/20 + 1/R3
Then solve for R3 -
1/4 - 1/6 - 1/20 = 1/R3
(15 - 10 - 3)/60 = 1/R3
2R3 = 60
R3 = 30 ohms
In the preceding problem you used the same parallel resistance formula. The only
difference between this and other examples is the unknown you solved for.
You may have problems with 4, 5, 6 and even many more resistances in parallel.
In that case, the parallel resistance formula just gets longer, such as -
1/RT = 1/R1 + 1/R2
+ 1/R3 + 1/R4 + 1/R5 ...
for as many more as you wish to add.
More practice problems on parallel circuits can be found in the questions for
this chapter in the QUIZ at the end of this chapter. Try your luck.
SERIES - PARALLEL CIRCUITS
Many hook-ups of resistances in combinations of series and parallel circuits
will be found aboard ship. No new formulas need be used. You just break the complete
circuit into simple series and parallel circuits. Solve each part separately and
then combine the parts.
Here is a sample problem. In figure 23A five resistances are connected as indicated.
You are asked to find the total current.
The first step is to find the combined resistance of R1-R2
and R4-R5, by using the parallel resistance formula for each
For group R1-R2, the equivalent resistance, Rx,
1/Rx = 1/5 + 1/20
1/Rx = (4 + 1)/20
1/Rx = 5/20
Rx = 4 ohms.
For group R4-R5, the equivalent resistance, Ry,
1/Ry = 1/4 + 1/12
1/Ry = (3 + 1)/12
1/Ry = 4/12
Ry = 3 ohms
The combined resistance of R1-R2 is 4 ohms, and of R4-R5,
3 ohms. As far as the flow of current is concerned, a single resistance of 4 ohms
can replace R1-R12, and one of 3 ohms can replace R4-R5.
Thus for calculation it is possible to redraw the circuit as in figure 23B and get
a circuit with three resistances of 4, 6, and 3 ohms in series. The total resistance
will be -
RT = 4 + 3 + 6 ohms
and the current will be -
I = E/R
I = 130/13
I = 10 amperes
Figure 23B. - Equivalent circuit of figure 23A.
Figure 24A presents another example of a series-parallel problem. It has a few
more resistors than the preceding problem, but it is just as simple. You are told
that the current in the circuit, I, produces an m-drop across the 2-ohm resistor
of 24 volts. What voltage must the generator have?
Figure 24A. - A series-parallel network.
The first step is to find Rx, the combined resistance of the R1-R2-R3-R4
branch. Using the parallel resistance formula -
1/Rx = 1/5 + 1/5 + 1/20 + 1/20
1/Rx = 10/20
Rx = 2 ohms
Now for the R5-R6-R7 branch, Ry -
1/Ry = 1/40 + 1/20 + 1/20
1/Rx = 5/40
Rx = 8 ohms
Look at figure 24B. A single resistor Rx of 2 ohms is equivalent to
and replaces the R1-R2-R3-R4 parallel
circuit; while Ry of 8 ohms replaces R5-R6-R7.
Combine Rs and Ry (4+8) and you have 12 ohms in parallel with
resistance R9, also of 12 ohms. So Rz, the effective resistance
of this parallel circuit, is -
1/Rz = 1/12 + 1/12
Rx = 6 ohms
Figure 24B. - Equivalent circuit for Figure 24A.
Figure 24C. - Most simplified form of circuit of figure
Now you have resolved your series-parallel circuit into an equivalent circuit
with three resistors in series, as shown in figure 24c. Thus the total resistance
of the circuit will be -
RT = 6 + 2 + 2
RT = 10 ohms
Look back at figure 24A again. A 24-volt IR drop is indicated across the 2-ohm
resistor. You know the resistance 2 ohms and the voltage 24, so by using Ohm's law
the current is found to be -
I = 24/2 = 12 amperes
Since in figure 24A, R10 is in series with the rest of the circuit,
12 amperes must be the total being delivered by the generator. In that case, the
IR drop across the 6-ohm resistor and the other 2-ohm resistor will be
E1 = 6 x 12
E1 = 72 volts.
E2 = 2 x 12
E2 = 24 volts.
The total IR drop across the whole circuit is -
E = 72 + 24 + 24
E = 120 volts.
And 120 volts is the emf being delivered by the generator.
POWER IN ELECTRICAL CIRCUITS
You learned in Basic Electricity that power is the RATE of doing work, and that
electrically, power is equal to the product of the current times the voltage, or
Like water flowing out of a pipe, the greater the volume and greater the pressure,
the larger will be the power.
Look back again at the problem you just completed. The current was 12 amperes,
with an applied E of 120 volts, so the power of the circuit is
P = 120 x 12
P= 1440 watts.
More problems dealing with power in series and parallel circuits are found in
the QUIZ at the end of the chapter. The more you practice, the better you will be
able to solve these problems.
HEATING IN ELECTRICAL CIRCUITS
You know from Ohm's law that E = IR, so you can substitute IR for E in the power
equation of the preceding paragraph and from P = E x I you get
P = IR x I
P = I2R
This means that the POWER, or the electrical energy used up each second, in forcing
electrons through a conductor depends on the RESISTANCE of the conductor and the
SQUARE OF the CURRENT. This power is the electrical energy used up, each second,
in forcing a current I through a conductor whose resistance is R. The electrical
energy thus "used up" does not disappear; it is merely transformed into heat which
warms the conductor. Therefore every electrical machine or piece of electrical equipment
that has conductors - a motor, generator, transformer, or even a cable - is heated
by the current going through it. The above formula tells us how much heating the
Consider an example: A motor which has an armature resistance of 1 ohm, when
operating at full load had an armature current of 5 amperes. Therefore the power
used up in heating the armature is -
P = I2R
P = 5 x 5 x 1 P = 25 watts
Now suppose the motor is overloaded so that the armature current goes up to 10
amperes. Then -
P = I2R
P = 10 x 10 x 1 P = 100 watts
This shows you what it means when you say "the power used up in heating a conductor
goes up as the square of the current in the conductor." It means that a small increase
in current produces a great increase in heating. Thus in the problem above, when
the current was only DOUBLED the heating increased FOURFOLD. For this reason it
is dangerous to overload electrical machines, because even a slight overloading
produces a great increase in heating. Often this increased heating will char or
break down the insulation around the conductor, and this leads to grounds, shorts,
etc., and soon the machine is ruined. In short, OVERLOADING of electrical machines
MEANS OVERHEATING, and overheating starts a destructive process which will rapidly
ruin the machine. So - when operating electrical machinery always be careful - Do
To help you tell when machines are being overloaded each electrical machine bears
a name plate which tells you how much current the machine takes at full load, and
how much it is heated up (what temperature rise it has) at this current. The temperature
rise is usually given as so many degrees rise "above ambient temperature". The AMBIENT
here means the temperature of the compartment in which the machine is located.
Chapter 1 Quiz