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Electricity  Basic Navy Training Courses NAVPERS
10622  Chapter 8

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 8 THE SERIES CIRCUIT
WHY KNOW CIRCUITS?
You have a good chance to outrank the bull in the China shop  just try fooling
around with circuits before you know what you're doing! After burning up a few instruments
at one hundred dollars per copy, you'll probably wind up electrocuting yourself
or your shipmates!
Electrical circuits are the power carriers of the electrical system. When they
are connected properly, they are efficient and trouble free. BUT, foul them up by
improper connections and they'll put your whole system out of commission.
The few circuits you have been made acquainted with are SIMPLE circuits  only
a source, its load, and the connecting wires. Few practical circuits can be so simple.
For example, it would be a terrific waste if EVERY electrical load had its OWN generator
and its OWN feeders. Imagine the confusion of cables, lines, and wires if every
light, every motor, every telephone, every heater, had its own separate feeder lines
directly from the dynamo room. For reasons of ECONOMY and EFFICIENCY, therefore,
most circuits are, actually VERY COMPLEX. They are designed so that ONE generator
can feed MANY electrical loads. But no matter how complex any particular circuit
becomes, it is one of three general types  the SERIES, the PARALLEL, or the SERIESPARALLEL.
Every electrical load is designed to contain a specific resistance and operate
at a certain rated voltage. The resistance of the road controls the amount of current
at the rated voltage. The proper type of circuit connection insures the load of
its rated voltage and current. Imagine the fireworks if a searchlight got 200 volts
instead of 95 volts. Or  you'd wait a long time for chow from a 220 volt galley
stove connected on a 110 volt line. Both these things could happen by using incorrect
connections.
VOLTAGE IN SERIES CIRCUITS
The first typethe series circuit is a ONEPATH circuit. You can always recognize
a series connection by two factsit will NEVER HAVE MORE THAN ONE CONDUCTOR CONNECTED
TO ANY TERMINAL, and you will find only ONE PATH from source to load (or loads)
and back to source.
Figure 37 shows a simple series circuit. In A, a voltmeter is connected across
the total resistance. It reads the TOTAL voltage drop of the TOTAL resistance 
in this case, 6 volts. NOTICE that the voltage drop which occurs in the wires is
ignored. This is standard practice for short wires because the resistance is very
small. But for longer wires, connecting load and source, the resistance is large
and the voltage drop is appreciable. It must be considered in the circuit. In B,
the voltmeter is connected across only HALF the resistance. In this case, the voltmeter
reads onehalf of 6 volts or 3 volts, indicating that HALF voltage is used for HALF
the resistance. Now connect two voltmeterseach one across half the resistance 
as in C. Each voltmeter reads 3 volts. This gives you the law for voltage drops
in a series circuit. THE TOTAL VOLTAGE IN ANY SERIES CIRCUIT IS THE SUM OF ALL THE
VOLTAGE DROPS. Mathematically the law says 
E_{t }= E_{1 }+ E_{2 }+ E_{3},
etc.
when 
E_{t }= the total voltage;
E_{1} = the voltage drop of the first load;
E_{2 }= the voltage drop of the second load;
E_{3} =the voltage drop of the third load.
In figure 37, E_{3} is 0; hence,
E_{t }= E_{1 }+ E_{2} + E_{3},
etc. 6 v. = 3 v. + 3 v. + 0
Figure 37.  Voltage in a series circuit.
Two more examples of series circuits are shown in figure 38. In A, three equal
resistors, of 4 ohms each, are connected in series. The voltmeters indicate that
it requires 2 volts to force the current through each resistor. Notice that the
total voltage is equal to the sum of all the voltage drops 
E_{t }= E_{1} + E_{2}
+ E_{3}, etc. E_{t} = 2 + 2 +
2 = 6 v.
Figure 38.  Voltages across separate loads.
In B of figure 38, two UNEQUAL resistances are connected in series. In this
case, the voltage drops are not equal  THEY ARE PROPORTIONAL TO THE RESISTANCE.
Again 
E_{t }= E_{1} + E_{2}
+ E_{3}, etc. 12 v.= 3 v. + 9v.
The searchlight circuit is a good example of this principle of a series circuit.
The ship's voltage is usually around 120 volts. But the standard voltage for the
Navy's largest searchlight is only 80 volts. In order to reduce the 120 volts to
the operating standard of 80 volts, a resistor is placed in series with the light.
This resistor uses up about 40 volts, leaving a drop of 80 volts for the light.
(120 v. = 40 v. + 80 v.).
CURRENT IN SERIES CIRCUITS
There is only one path for current in a series circuit. And the amount of current
passing any point in the circuit is the same as the amount of current passing any
other point in the circuit. This gives you the law for current in a series circuit.
THE CURRENT IN A SERIES CIRCUIT IS THE SAME IN ALL PARTS. Or mathematically 
I_{t }= I_{1} = I_{2}=
I_{3}, etc.
when 
I_{t} = total current;
I_{1} = current through the first load;
I_{2} = current through the second load;
I_{3} = current through the third load.
Suppose you connect a lamp, a switch, and a motor in series as in figure 39.
The current through each part of the circuit is the same.
Figure 39.  Control in a series circuit.
If the switch is opened the current through the switch becomes zero  open circuit.
Likewise, the current through both the motor and the lamp be come zero. The opened
switch "shutoff" the loads. Switches CONTROLLING electrical loads are
ALWAYS in series with the loads. In the same circuit, imagine that the lamp is broken.
A broken lamp is also an open circuit and would stop the current through the motor
and the switch. For this reason, two loads which are intended to be INDEPENDENT
of each other, are NEVER connected in series.
RESISTANCE IN SERIES CIRCUITS
Figure 40.  Resistance in series.
Two resistances are connected in series in figure 40. Tracing the circuit, note
that the current passes first through one resistor 'and then through the other.
This means that the current is opposed by the force of both resistances. In figure
40, each load has a resistance of 10 ohms, but since the current passes through
BOTH loads, the total opposition to current is 20 ohms. This gives you the law for
resistances connected in series  THE TOTAL RESISTANCE IN ANY SERIES CIRCUIT IS
EQUAL TO THE SUM OF ALL THE INDIVIDUAL RESISTANCES. Mathematically this says 
R_{t }= R_{1} + R_{2}
+ R_{3}, etc.
when 
R_{t} = total resistance,
R_{1} = resistance of the first load;
R_{2} = resistance of the second load;
R_{3} = resistance of the third load.
In the problem of figure 40 
R_{t}=R_{1}+R_{2}
20 Ω=10 Ω+10 Ω
This principle is employed in limiting the amount of current through a load
by inserting resistors in series with the load. For example, the rated amperage
through the lamp in figure 41 is 2 amperes. The lamp itself has 40 ohms resistance;
but on a 120 volt line, 40 ohms will permit 3 amperes to pass 
I = E/R = 120/40 = 3 amps.
Figure 41.  Controlling current by series resistance.
This means that, as long as the lamp is operated on 120 volts, it will pass
3 amperes. And 3 amperes in a 2 ampere lamp will melt the filament (burn it out)
. You can see that more. than the lamp's 40 ohms of resistance is required to reduce
the current to its safe value of 2 amperes. Using Ohm's law to calculate the total
resistance required to limit the current to 2 amperes 
R = E/I = 120/2 = 60 Ω
This means that a 20 ohm resistor would have to be added in series to the 40
ohms of the lamp. Then the total resistance is 60 ohms 
R_{t }= R_{1} + R_{2}
60 Ω = 40 Ω + 20 Ω
Now the current is limited to 2 amperes.
KEEP OFF THE ROCKS
Here are a few simple rules that will help keep you off the rocks 
1. Draw a schematic of your problem.
2. Label each value on the schematic that is known.
3. Determine WHAT it is that you want to know. Is it the CURRENT, VOLTAGE, RESISTANCE,
POWER, EFFICIENCY, INPUT OR OUTPUT?
4. Select the equation for your solution and use the EASIEST form of the equation.
If you are looking for CURRENT, use I = E/R. If you are looking for VOLTAGE, use
E = IR. If you are looking for RESISTANCE, use R = E/I.
5. (IMPORTANT) Apply the law to either a PART or the WHOLE circuit. DON'T MIX
UP YOUR VALUES. If you are looking for the resistance of a PART of a circuit, use
the values or current and voltage for that PART. If you are looking for the TOTAL
resistance, use the values of current and voltage for the TOTAL circuit.
6. Substitute the numerical values in your equation and solve.
You've been given a pretty big dose of formulas and equations all at once. They're
hard to take  so, here's a "crutch" to help you over the rough spots.
When you have a problem in current, voltage, or resistance sketch a "pie"
and divide it into three pieces. Label each piece as in figure 42. Now, using your
finger, cover up the quantity you want to know. What's left is the formula for solving
for that quantity.
For example, if you want to know current, cover I; E/R remains, and I = E/R.
If you want to know the voltage, cover E; IR remains, and E = IR. Likewise, if you
want to know resistance, cover R; E/I remains, and R = E/I. You can make the same
kind of "pie" for the power equation. It's shown in figure 43.
Figure 42.  Ohm's law.
Figure 43.  Power equation.
Simple, isn't it? Let's try it on a few problems.
PROBLEMS IN SERIES CIRCUITS
The BEST (not always the easiest) way to get a clear understanding of circuits
is to work circuit problems. You will learn a lot by going through each of the examples
which follow. Each example has been selected to illustrate a practical circuit problem.
EXAMPLE 1 
Three lamps are connected in series. Each has a resistance of 60 ohms. A generator
producing 120 v. of emf powers the circuit. What is the total current in the circuit,
and what is the current through each lamp?
Rule 1. Sketch the circuit  figure 44.
Figure 44.  Example 1.
Rule 2. Label values.
Rule 3. What is wanted? I_{t}, I_{1}, I_{2} and I_{3}.
Rule 4. Use I = E/R
Rule 5. Use I = E/R first, for the TOTAL current, and then, for the current
of EACH LAMP.
Rule 6. Use I_{t} = E_{t}/R_{t} = 120/180 = 2/3 amp.
(for I_{t})
(R_{t} = R_{1} + R_{2} + R_{3}) (for I_{1},
I_{2} and I_{3}) E_{1}/R_{1} = 40/60 = 2/3 amp.
(Voltage divides in a series circuit)
Is it strange that the answers are the same? No  I_{t} = I_{1}
= I_{2} = I_{3} in a series circuit. Therefore, the total current
is equal to the part currents. You will learn to make use of short cuts like this.
But until you are sure of what you're doing, it's. best to be complete AND CORRECT.
Suppose you had slipped up on Rule 5 (using a combination of a PART and the WHOLE).
Say you had used the TOTAL voltage and the resistance of a PART. See what happens

I = E/R = 120/40 = 3 amps. ABSOLUTELY WRONG.
EXAMPLE 2 
Figure 45.  Example 2.
Two lamps are connected in series on a 240v. line. One lamp has 40 ohms resistance,
the other has 80 ohms resistance (see figure 45). What is the total current and
power? What is the current and power of each lamp?
I_{t} = E_{t}/R_{t} = 240/120 = 2
amps. I_{t} = I_{1} = I_{2} therefore
I_{1} = I_{2} = 2 amps. P_{t} = E_{t}I_{t}
= 240 x 2 = 480 w.
The power of each of the two lamps can be found in two ways.
(1) Find the voltage drop across each lamp 
(RF Cafe note: original had E_{1}
and E_{2} reversed)
E_{1} =
I_{1} x R_{1} = 2 x 40 = 80 v.
E_{2} = I_{2} x R_{2} = 2 x 80 = 160 v.
Then  (RF Cafe note:
original had P_{1} and P_{2} reversed  thanks to Lou N.)
P_{1} = E_{1}
x I_{1} = 80 x 2 = 160 w.
(RF Cafe note: original had E_{2} x I_{2)}
P_{2} = E_{2} x I_{2} = 160 x 2 = 320 w.
OR 
(2) In the equation P = EI, substitute for E
the value of IR.
Then (RF Cafe note: original
had P_{1} and P_{2} reversed  thanks to Lou N.)
P = I x IR, or P = I^{2}R
P_{1} = I_{1}^{2}R_{1} = 4 x 40 = 160 w.
P_{2} = I_{2}^{2}R_{2} = 4 x 80 = 320 w.
Figure 46.  Example 3.
EXAMPLE 3 
Figure 46 shows three 1,000 watt heaters connected in series to a 240volt battery.
Each heater has a current of 12.5 amperes. (1) What is the total resistance? (2)
Resistance of each heater? (3) Total current? (4) Total power?
(1) R_{t} = E_{t}/I_{t}
= 240/12.5 = 19.2 Ω (2) R_{1} = P_{1}/I_{1}^{2}
= 1,000/(12.5 x 12.5) = 1,000/156.25 = 6.4 Ω
R_{1} = R_{2} = R_{3} (equal power  all are 1,000 w.)
R_{1} = R_{2} = R_{3} = 6.4 Ω
(3) I_{t} = I_{1} = I_{2} = I_{3} = 12.5 amps.
(4) P_{1} = E_{1}I_{1} = 240 x 12.5 = 3,000 w.
or P_{t} = P_{1 }+ P_{2} + P_{3} = 1,000 + 1,000
+ 1,000 = 3,000 w.
Chapter 8 Quiz
(click here)

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