Module 1  Introduction to Matter, Energy, and Direct Current
Pages i  ix,
11 to 110,
111 to 120,
121 to 130,
141 to 150,
151 to 160,
161 to 165,
21 to 210,
211 to 220,
221 to 229,
31 to 310,
311 to 320,
321 to 330,
331 to 340,
341 to 350,
351 to 360,
361 to 370,
371 to 380,
381 to 390,
391 to 3100,
3101 to 110,
3111 to 3120,
3121 to 3126, Appendix
I,
II,
III,
IV,
V,
Index
To find the current through R2 and R3, refer to the original circuit, figure
354(A). You know ER2 and ER3 from previous calculation.
Given:
Solution:
To find power used by R2 and R3, using values from previous
calculations: Given:
Solution:
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Now that you have solved for the unknown quantities in this circuit, you can
apply what you have learned to any series, parallel, or combination circuit. It
is important to remember to first look at the circuit and from observation make
your determination the type circuit, what is known, and what you are looking for.
A minute spent in this manner may save you many unnecessary calculations.
Having computed all the currents and voltages figure 354, a complete description
the operation the circuit can be made. The total current 3 amps leaves the negative
terminal the battery and flows through the 8ohm resistor (R1). In so doing, a voltage
drop 24 volts occurs across resistor R1. At point A, this 3ampere current divides
into two currents. the total current, 1.8 amps flows through the 20ohm resistor.
The remaining current 1.2 amps flows from point A, down through the 30ohm resistor
to point B. This current produces a voltage drop 36 volts across the 30ohm resistor.
(Notice that the voltage drops across the 20 and 30ohm resistors are the same.)
The two branch currents 1.8 and 1.2 amps combine at junction B and the total current
3 amps flows back to the source. The action the circuit has been completely described
with the exception power consumed, which could be described using the values previously
computed.
It should be pointed out that the combination circuit is not difficult to solve.
The key to its solution lies in knowing the order in which the steps the solution
must be accomplished.
Practice Circuit Problem
Figure 355 is a typical combination circuit. To make sure you understand the
techniques solving for the unknown quantities, solve for ER1.
Figure 355.  Combination practice circuit.
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It is not necessary to solve for all the values in the circuit to compute the
voltage drop across resistor R1 (E R1). First look at the circuit and determine
that the values given do not provide enough information to solve for ER1 directly.
If the current through R1 (IR1) is known, then ER1 can be computed by applying
the formula:
The following steps will be used to solve the problem.
1. The total resistance (RT) is calculated by the use equivalent resistance.
Given:
Solution:
Redraw the circuit as shown in figure 355(B). Given:
Solution:
Solution:
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Redraw the circuit as shown in figure 355(C). Given:
Solution:
2. The total current (IT) is now computed. Given:
Solution:
3. Solve for the voltage dropped across Req2. This represents the voltage dropped
across the network R1, R2, and R3 in the original circuit.
Given:
Solution:
4. Solve for the current through Req1. (Req1 represents the network R1
and R2 in the original circuit.) Since the voltage across each branch a parallel
circuit is equal to the voltage across the equivalent resistor representing the
circuit:
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Given:
Solution:
5. Solve for the voltage dropped across R1 (the quantity you were asked
to find). Since Reqi represents the series network R1 and R2 and total
current flows through each resistor in a series circuit, IR1 must equal IReq1.
Given:
Solution:
Q42. Refer to figure 355(A). If the following resistors were replaced
with the values indicated: R1 = 900, R3 = 1k , what is the total
power in the circuit? What is ER2?
REDRAWING CIRCUITS FOR CLARITY
You will notice that the schematic diagrams you have been working with have shown
parallel circuits drawn as neat square figures, with each branch easily identified.
In actual practice the wired circuits and more complex schematics are rarely
laid out in this simple form. For this reason, it is important for you to recognize
that circuits can be drawn in a variety ways, and to learn some the techniques for
redrawing them into their simplified form. When a circuit is redrawn for clarity
or to its simplest form, the following steps are used.
i. Trace the current paths in the circuit.
2. Label the junctions in the circuit.
3. Recognize points which are at the same potential.
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4. Visualize a rearrangement, "stretching" or "shrinking," connecting
wires.
5. Redraw the circuit into simpler form (through stages if necessary).
To redraw any circuit, start at the source, and trace the path current flow through
the circuit. At points where the current divides, called JUNCTIONS, parallel branches
begin. These junctions are key points reference in any circuit and should be labeled
as you find them. The wires in circuit schematics are assumed to have NO RESISTANCE
and there is NO VOLTAGE drop along any wire. This means that any unbroken wire is
at the same voltage all along its length, until it is interrupted by a resistor,
battery, or some other circuit component. In redrawing a circuit, a wire can be
"stretched" or "shrunk" as much as you like without changing any electrical characteristic
the circuit.
Figure 356(A) is a schematic a circuit that is not drawn in the boxlike fashion
used in previous illustrations. To redraw this circuit, start at the voltage source
and trace the path for current to the junction marked (a). At this junction the
current divides into three paths. If you were to stretch the wire to show the
three current paths, the circuit would appear as shown in figure 356(B).
Figure 356.  Redrawing a simple parallel circuit.
While these circuits may appear to be different, the two drawings actually
represent the same circuit. The drawing in figure 356(B) is the familiar boxlike
structure and may be easier to work with. Figure 357(A) is a schematic a circuit
shown in a boxlike structure, but may be misleading. This circuit in reality is
a seriesparallel circuit that may be redrawn as shown in figure 357(B). The drawing
in part (B) the figure is a simpler representation the original circuit and could
be reduced to just two resistors in parallel.
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Figure 357.  Redrawing a simple seriesparallel circuit.
Redrawing a Complex Circuit
Figure 358(A) shows a complex circuit that may be redrawn for clarification
in the following steps.
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Figure 358.  Redrawing a complex circuit.
NOTE: As you redraw the circuit, draw it in simple boxlike form. Each time
you reach a junction, a new branch is created by stretching or shrinking the wires.
Start at the negative terminal the voltage source. Current flows through R1
to a junction and divides into three paths; label this junction (a). Follow one
the paths current through R2 and R3 to a junction where the current
divides into two more paths. This junction is labeled (b).
The current through one branch this junction goes through R5 and
back to the source. (The most direct path.) Now that you have completed a path for
current to the source, return to the last junction, (b). Follow current through
the other branch from this junction. Current flows from junction (b) through R4
to the source. All the paths from junction (b) have been traced. Only one path from
junction (a) has been completed. You must now return to junction (a) to complete
the other two paths. From junction (a) the current flows through R7 back to
the source. (There are no additional branches on this path.) Return to junction
(a) to trace the third path from this junction. Current flows through R6
and R8 and comes to a junction. Label this junction (c). From junction (c)
one path for current is through R9 to the source. The other path for
current from junction (c) is through R10 to the source. All the junctions
in this circuit have
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now been labeled. The circuit and the junction can be redrawn as shown in figure
358(C). It is much easier to recognize the series and parallel paths in the redrawn
circuit.
Q43. What is the total resistance the circuit shown in figure 359? (Hint:
Redraw the circuit to simplify and then use equivalent resistances to compute for
RT.)
Figure 359.  Simplification circuit problem.
Q44. What is the total resistance the circuit shown in figure 360?
Figure 360.  Source resistance in a parallel circuit.
Q45. What effect does the internal resistance have on the rest the circuit
shown in figure 360?
EFFECTS OPEN AND SHORT CIRCUITS
Earlier in this chapter the terms open and short circuits were discussed. The
following discussion deals with the effects on a circuit when an open or a short
occurs.
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The major difference between an open in a parallel circuit and an open in a series
circuit is that in the parallel circuit the open would not necessarily disable the
circuit. If the open condition occurs in a series portion the circuit, there will
be no current because there is no complete path for current flow. If, on the other
hand, the open occurs in a parallel path, some current will still flow in the circuit.
The parallel branch where the open occurs will be effectively disabled, total resistance
the circuit will INCREASE, and total current will DECREASE.
To clarify these points, figure 361 illustrates a series parallel circuit. First
the effect an open in the series portion this circuit will be examined. Figure 361(A)
shows the normal circuit, RT = 40 ohms and IT = 3 amps. In figure 361(B)
an open is shown in the series portion the circuit, there is no complete path for
current and the resistance the circuit is considered to be infinite.
Figure 361.  Seriesparallel circuit with opens.
In figure 361(C) an open is shown in the parallel branch R3. There
is no path for current through R3. In the circuit, current flows through R1
and R2 only. Since there is only one path for current flow, R1 and R2
are effectively in series.
Under these conditions RT = 120 and IT = 1 amp. As you
can see, when an open occurs in a parallel branch, total circuit resistance increases
and total circuit current decreases.
A short circuit in a parallel network has an effect similar to a short in a series
circuit. In general, the short will cause an increase in current and the possibility
component damage regardless the type
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NEETS Table of Contents
 Introduction to Matter, Energy,
and Direct Current
 Introduction to Alternating Current and Transformers
 Introduction to Circuit Protection,
Control, and Measurement
 Introduction to Electrical Conductors, Wiring
Techniques, and Schematic Reading
 Introduction to Generators and Motors
 Introduction to Electronic Emission, Tubes,
and Power Supplies
 Introduction to SolidState Devices and
Power Supplies
 Introduction to Amplifiers
 Introduction to WaveGeneration and WaveShaping
Circuits
 Introduction to Wave Propagation, Transmission
Lines, and Antennas
 Microwave Principles
 Modulation Principles
 Introduction to Number Systems and Logic Circuits
 Introduction to Microelectronics
 Principles of Synchros, Servos, and Gyros
 Introduction to Test Equipment
 RadioFrequency Communications Principles
 Radar Principles
 The Technician's Handbook, Master Glossary
 Test Methods and Practices
 Introduction to Digital Computers
 Magnetic Recording
 Introduction to Fiber Optics
