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
Chapter 20: Transformers  What For...
Transformers get their name from the kind of
work they dothey TRANSFORM ELECTRICAL POWER. Suppose you had to supply a galley stove
at 220 volts, the ship's lighting system at 110 volts, and a 6volt signalbell system.
Three different generators for three different voltages? Absolutely not! Use transformers,
and then one generator can supply all three loadseach at its proper voltage. Your generator
would have a standard output, say 110 volts. One transformer would STEP this UP to 220
volts. Another would STEP it DOWN to 6 volts.
This sounds like something for nothingbut it isn't. TRANSFORMERS DO NOT INCREASE
OR DECREASE POWER. Remember that power is voltage multiplied by current. And when a
transformer changes voltage it also changes current. If you INCREASE the voltage, the
current is DECREASED. One goes up and the other goes down power INPUT. The VOLTAGE and
CURRENT of a circuit ARE CHANGED BY TRANSFORMERS. But the TOTAL POWER remains the SAME.
How It Works
The transformer is a mutual induction circuit. And that should tell you what to expect

1. There is an iron core.
2. There are TWO electrical circuits.
3. Energy is transformed from one circuit to the other by a field of
flux.
Look at figure 218. This is a SIMPLE transformer. Modern transformers are not built
this way, but every important transformer principle can be illustrated by this diagram.
Figure 218.  Simple transformer.
Figure 219.  Transformer flux.
FIRST, notice the core  it forms a closed iron path for flux. This is the path of
energy transfer between the two circuits. The efficiency of a transformer depends on
this core and it is always designed to carry the maximum flux.
SECOND, notice the two circuits  they are formed in coils around the iron of the
core. The coil, or winding, that receives energy from the source is called the PRIMARY.
In figure 218, the winding on the left is connected to the line, or source  it is the
primary. The other winding, the one connected to the load, is called the SECONDARY. In
the figure, the winding on the right supplies the load  it is the secondary. Never let
the relative number of turns on the two windings confuse you. The source winding is always
the primary and the load winding is always the secondary. If the load and source connections
of figure 218 were exchangedthe names of the windings would also exchange.
THIRD, notice the flux set up by the transformer primary. Figure 219 shows this better
than figure 218. You know that this is a mutual induction circuit. And you know something
else about ityou know that flux lines must be cut for transfer of energy. Neither the
primary nor the secondary coils can move. So the flux field must move. And that means
just one thing  ONLY A.C. and PULSATING D.C. can be used in transformers  they produce
moving fields.
Imagine that 60 cycle a.c. is being fed into the primary of figure 219. A field blossoms
out around EVERY PRIMARY TURN. The iron preserves this field and carries it to the secondary.
At the secondary EVERY TURN is cut by the field. Voltage is induced in the secondary
and power has 'been transferred from primary to secondary.
In 60 cycle a.c., you know that current changes direction 120 times each second (two
alternations to the cycle). Which means that the secondary is cut 120 times a second.
Therefore, the frequency of the secondary is EXACTLY the same as the frequency of the
primary.
HOW MUCH?
The question, "How much?" always comes up. That's because you are using transformers
to change voltage. And what's the sense of using a device unless you know what you're
going to get out of it?
Figure 220 is a typical transformer setup. Notice the voltmeter readings. They tell
the story of transformer voltage. The voltage of the primary (E_{p}) is 110 volts.
The voltage of the secondary (E_{s}) is 220 volts. You'll notice that the secondary
is unloaded  an open circuit. Therefore, the secondary current is zero. But, how much
current is there in the primary? Looking at the circuit, you see that the primary is
connected directly across a 110 volt line. You'd expect a heavy current. But you're wrong
 the voltage of selfinduction (E_{si}) in the primary is very high. The tight
windings and the iron core cause every turn of the winding to be cut by almost the total
of the coil's flux. The result is that E_{si} is almost equal to E_{p}.
Not only equal  but OPPOSITE. And this nearly equal and opposite E_{si} chokes
the current down to an extremely small value. The small current that does flow is called
the MAGNETIZING CURRENT of a transformer.
Figure 220.  Stepup transformer.
Let's make figure 220 an example. The resistance of the coil is 10 ohms. And the voltage
of selfinduction is 109 volts. This gives you these values 
E_{p} = 1;1.0 v.
E_{si} = 109 v.
R = 10 Ω
I_{p} = (110  109)/10 = 1/10 amp. (by Ohm slaw)
The magnetizing current is only onetenth of an ampere.
Notice that E_{si} was subtracted from Ep. Remember that Ep and E_{si}
are OPPOSITE  they cancel each other, so one must be subtracted from the other to get
THE NET VOLTAGE ACTING ON THE CURRENT. It's like the counteremf in a dc motor.
Now to get back to the why and wherefore of that 220 volts on the secondary. It's
like this  the magnetizing current's flux produced 109 volts of selfinduction on the
11 turns of the primary. That's just slightly under 10 volts per turn. Call it an even
10 volts. Is there any reason why this flux is not producing a like voltage  10 volts
per turn  on the secondary winding? No reason in the world  both coils are on the same
core, so whatever flux cuts  one, cuts the other. BUT  and it's a big "but"  the secondary
has 22 turns. And at 10 volts per turn  that's 220 volts.
This tells you that the ratio of voltages is the same as the ratio of turns. Mathematically
it says 
E_{p}/E_{s} = T_{p}/T_{s}
in which
E_{p} = voltage in primary;
E_{s} = voltage in secondary;
T_{p} = turns in primary coil;
T_{s} = turns in secondary coil.
Suppose a transformer has 600 volts and 2,400 turns on the primary. What will be the
voltage of a 400 turn secondary?
E_{p}/E_{s} = T_{p}/T_{s}
600/E_{s} = 2,400/400 = 100 volts.
Here's another way to work this problem. How many volts per turn in the primary? 600
volts divided by 2,400 turns  1/4 volt per turn. The volts per turn of the primary and
secondary are just about equal, so 400 turns multiplied by 1/4 volt is 100 volts on the
secondary. Notice that this transformer is a stepdown job. The voltage goes from 600
volts to 100 volts.
Transformers are highly efficient. As high as 98 percent. That's why you can ignore
the difference between E_{p} and E_{si} in calculating volts per turn.
Now put a 44ohm load on the transformer of figure 220. You have the circuit of figure
221. The current in the secondary is 
I_{s} = E_{s}/R_{s} = 220/44 = 5 amps.
Right here is a good place for you to get this fact straight. THE SECONDARY CURRENT
IS CONTROLLED BY THE LOAD. If the load had been 22 ohms instead of 44 ohms, the current
would have been 10 amperes instead of 5 amperes.
Figure 221.  Loaded transformer.
The 5 ampere secondary current is OPPOSITE in direction to the primary current. You
can prove this by the coil hand rule. Or  reason it out this way  the current in the
secondary is produced by the voltage induced by the primary field. You know that E_{si}
is opposite to E_{p} and you know that both E_{s} and E_{si}
are produced by the same primary field. Therefore, Es must be opposite to E_{p}.
This is the same general rule you got for mutual induction circuits.
Now, what is the EFFECT of opposite primary and secondary currents? Simply thisTHEIR
FIELDS TEND TO CANCEL EACH OTHER. The 5 ampere secondary sets up a field which destroys
some of the primary flux. Result  less E_{si} and more current in the primary.
The primary current will increase until the primary field strength balances the secondary's
field strength. Field strengths are determined by ampereturns (IT). When the two fields
are equal, their ampereturns are equal 
I_{p}T_{p} = I_{s}T_{s}
or I_{p}/Is = Ts/T_{p}
Which says that, the currents in the two windings are INVERSELY proportional to their
number of turns.
In the example 
I_{p}/I_{s} = T_{s}/T_{p}
I_{p}/5 = 22/11
and I_{p} = 10 amps.
Which means that the primary current increases to 10 amperes for a secondary load
of 5 amperes. Suppose another transformer has a primary of 70 turns, a secondary
of 350 turns, and a 30 ampere load. What is the primary current?
I_{p}/I_{s} = T_{s}/T_{p}
I_{p}/30 = 350/70
and Ip = 150 amps.
You can prove this is correct by using the fact that primary and secondary fields
are of equal strength.
I_{p}T_{p} = I_{s}T_{s}
150 x 70 = 350 x 30
10,500 = 10,500
This brings you to an important fact about transformers. The load current in the SECONDARY
controls the current in the PRIMARY. This control is centered in the action of the secondary
flux. Every change of secondary flux changes the E_{si} on the primary and consequently
the primary current. It is an automatic control. No load on the secondary reduces primary
current almost to zero (only magnetizing current flows). As the secondary is loaded,
its current. increases and the primary current increases with it.
InputOutput
Since transformers are so highly efficient, their efficiency is usually considered
to be 100 percent. This assumption introduces a small error but for all practical purposes
it is close enough. Considered 100 percent efficient, the transformer's input power and
output power must be equal. Therefore 
P_{p} = P
I_{p}E_{p} = l_{s}E_{s}
Figure 222.  Power in a transformer.
Let's check this against an example. Figure 222 is a good practical problem. And these
are the things you can solve for  Es, Is, Iv, Ps, and Pv.
For E_{s} 
E_{p}/E_{s} = T_{s}/T_{p}
50/E_{s} = 250/1000
E_{s} = 200 v.
For I_{s} 
I_{s} = E_{s}/R_{s} = 200/8 = 2.5 amps.
For I_{p} 
I_{p}/I_{s} = T_{s}/T_{p}
I_{p}/2.5 = 1000/250
I_{p} = 10 amps.
For P_{s} 
P_{s} = E_{s}I_{s}
P_{s} = 200 x 2.5 = 500 w.
For P_{p} 
P_{p} = E_{p}I_{p}
P_{p} = 50 x 10 = 500 w.
Note that P_{p} = P_{s}.
Summary of Thory
A summation of the preceding information gives you three important equations and two
important facts.
EQUATIONS 
1. E_{p}/E_{s} = T_{s}/T_{p}
2. I_{p}/I_{s} = T_{s}/T_{p}
3. I_{p}E_{p} = I_{s}E_{s}
FACTS 
1. The voltage of the secondary is always opposite in direction to the
voltage of the primary.
2. The current drawn by the primary is controlled by the secondary load
current.
Modern Construction
Perhaps you are wondering why transformers are so efficient. The reason is their lack
of moving parts. When moving parts are eliminated from a device, mechanical friction,
the biggest source of loss, is gone. Losses become very small.
Modern transformers are designed with just one idea  cut down the losses to as small
a value as possible. In general, the losses that are unavoidable are divided into two
classes  IRON losses and COPPER losses.
Iron losses occur in the core. Part of these losses is due to the resistance of the
molecular magnets. They must turn over every time the a.c. reverses. In a 60cycle transformer,
the molecules must shift around 120 times each second. Molecules resist this shifting;
and their desire to stand still is called HYSTERESIS. You might say that hysteresis is
really FRICTION. And you know that friction produces heat.
The iron cores themselves act as wires. They are cut by the flux of their winding's
fields, and carry small induced currents. These induced currents are called EDDY CURRENTS
because they flow entirely within the iron core. Eddy currents produce heat  they are
really small short circuits within the core.
Thus, IRON LOSSES are made up of two factors  HYSTERESIS and EDDY CURRENTS. Both
produce heat , and both represent losses which must be subtracted from the output power.
COPPER LOSSES occur within the windings. They are due to just one thing  the heat
generated by the current in the winding conductors. Now here is an important fact  copper
(and most other conductors) increases its resistance as it gets hotter. This means that
if the heat resulting from iron and copper losses is allowed to accumulate, the windings
will get hotter. And the hotter they get, the higher is their resistance. Which increases
the power loss due to resistance. It's a vicious circle. Losses produce heat and heat
produces even higher losses and so on.
Iron Losses  Down
Figure 223.  Transformer cores.
Hysteresis losses are reduced by using either a soft iron or a special transformer
steel containing silicon. These metals allow their molecules to shift easily  with a
minimum of friction. Eddy currents are broken up by constructing the cores out of thin
plates of iron instead of a solid piece. This LAMINATED structure breaks up the eddy
currents by insulating each plate from its neighbors. Eddy currents still exist  but
they are so small that the loss they cause can be neglected.
Figure 223 shows a number of transformer cores. Notice that they are all complete
magnetic circuits, all are laminated, and all use iron generously.
Copper Losses  Down
Reducing copper losses is a pipe. The windings are designed to be as short and as
heavy as possible. Both short turns and large wire tend to decrease resistance and reduce
heat.
Finally, the overall loss in the windings, due to accumulated heat, is cut down by
special cooling devices. Oil baths, radiators, and air blowers are all used to keep transformers
cool.
Core Shapes
There are three general types of core design. Each has its own special advantage.
But all put both primary and secondary windings on the same leg of iron.
Figure 224A.  Core type.
Figure 224B.  Shell type.
Figure 224C.  Modified shell type.
Figure 225.  Types of winding design.
Figure 224 shows all three types. A  the CORE TYPE, is best for high voltage use.
Its winding turns are short and IR drops are at a minimum. B  the SHELL type has longer
turns because the middle leg is twice the size of the outside legs. The IR drops are
larger but the flux path is shorter. Therefore, this type is best for heavy current loads.
C  the MODIFIED SHELL type is a combination of both core and shell. The modified type
has some of the advantages of both the simple types.
Winding Designs
There are two winding designs which can be used on ANY of the three types of cores.
In figure 225, you'll see crosssections of both types.
A  the CYLINDRICAL design consists of a primary cylinder and a secondary cylinder.
One is fitted over the top of the other. And then the whole winding assembly is slipped
over the iron core. B  the PANCAKE design separates each winding into sections or pancakes.
Then alternate pancakes of primary and secondary are slipped over the core. In the final
assembly, all pancakes of the primary are connected together in series, and all pancakes
of the secondary are. connected together in series.
The cylindrical winding is a little cheaper to construct, but it is harder to repair.
The pancake winding costs a little more but is easier to repair because it is broken
up into sections. You'll run into both designs of winding and probably all three types
of core. The best advice is to understand each, but leave them as is.
Summary
Transformers are built to do just one job  change voltage and current values. They
are designed to do it as efficiently as possible. Cores are laminated, winding turns
are shortened, and artificial cooling is used. All these make the transformer the most
troublefree and efficient device used by electricians.
Chapter 20 Quiz
(click
here)
