AC/DC
ELECTRICAL
SYSTEMS
LEARNING
ACTIVITY
PACKET
COMBINATION CIRCUITS
BB227-BC05UEN
LEARNING ACTIVITY PACKET 5
COMBINATION CIRCUITS
INTRODUCTION
This LAP will continue to build on series circuits and parallel circuits discussed in
previous LAPs by combining them into combination circuits.
Combination circuits combine the desirable characteristics of series and parallel
circuits. In fact, many of the circuits found in industry, home, and commercial electrical
systems are actually combination circuits.
ITEMS NEEDED
Amatrol Supplied
1
T7017 AC/DC Electrical Learning System
FIRST EDITION, LAP 5, REV. A
Amatrol, AMNET, CIMSOFT, MCL, MINI-CIM, IST, ITC, VEST and Technovate are trademarks or registered trademarks of Amatrol,
Inc. All other brand and product names are trademarks or registered trademarks of their respective companies.
Copyright © 2012 by AMATROL, INC.
All rights Reserved. No part of this publication may be reproduced, translated, or transmitted in any form or by any means, electronic,
optical, mechanical, or magnetic, including but not limited to photographing, photocopying, recording or any information storage and
retrieval system, without written permission of the copyright owner.
Amatrol,Inc., 2400 Centennial Blvd., Jeffersonville, IN 47130 USA, Ph 812-288-8285, FAX 812-283-1584 www.amatrol.com
BB227-BC05UEN COMBINATION CIRCUITS
Copyright © 2012 Amatrol, Inc.
2
TABLE OF CONTENTS
SEGMENT
1 CHARACTERISTICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
OBJECTIVE
OBJECTIVE
SKILL
OBJECTIVE
SKILL
1
2
1
3
2
SEGMENT
2 LIGHTING CIRCUITS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
OBJECTIVE
SKILL
SKILL
OBJECTIVE
4
3
4
5
SEGMENT
3 VOLTAGE DIVIDERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
OBJECTIVE
OBJECTIVE
SKILL
SKILL
6
7
6
7
SEGMENT
4 TROUBLESHOOTING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Define a series-parallel circuit
Describe a method for identifying the series and parallel sections of a circuit
Trace the current path in a combination circuit
List the seven steps for solving a combination circuit
Solve a combination circuit
Describe how switches are used in combination circuits and give an application
Connect and operate a basic lighting circuit
Connect and operate a ceiling fan circuit
Describe the function of a variable resistor and give an application
Activity 1 Rheostat operation
SKILL 5 Connect and operate a rheostat as a light dimmer
OBJECTIVE 8
OBJECTIVE 9
SKILL 8
OBJECTIVE 10
SKILL 9
Describe the function of a voltage divider and give an application
Describe the operation of three types of voltage dividers
Design a voltage divider network
Connect and operate a voltage divider network
Explain the effect of a short circuit
Describe the four steps for troubleshooting a short circuit
Locate a short circuit
Describe the three basic steps for troubleshooting an open circuit
Locate an open circuit
BB227-BC05UEN COMBINATION CIRCUITS
Copyright © 2012 Amatrol, Inc.
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SEGMENT 1
CHARACTERISTICS
OBJECTIVE 1
DEFINE A SERIES-PARALLEL CIRCUIT
A series-parallel or combination circuit, as shown in figure 1, contains
elements of both series and parallel circuits. Components that are connected in
series display the characteristics of a series circuit, while components connected in
parallel display the characteristics of a parallel circuit.
R1
+
R2
R3
Figure 1. A Series-Parallel Combination Circuit
Two important points to remember about the characteristics of series-parallel
circuits are:
• Components that are connected in series have the same current flowing
through them.
• Components that are connected in parallel have the same voltage across
them.
These two points will be very helpful in understanding the operation of a
series-parallel or combination circuit.
BB227-BC05UEN COMBINATION CIRCUITS
Copyright © 2012 Amatrol, Inc.
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OBJECTIVE 2
DESCRIBE A METHOD FOR IDENTIFYING THE SERIES AND
PARALLEL SECTIONS OF A CIRCUIT
The series and parallel sections of a combination circuit can be identified by
tracing the current path to find nodes that split the current and nodes that recombine the current, as shown in figure 2.
R1
NODE
I1
+
R2
I1
I2
I3
R3
NODE
Figure 2. Tracing the Current Path
In figure 2, there are no nodes between the power supply and R1. Therefore,
R1 is in series with the power supply and all of the current from the power supply
flows through R1. When the current reaches the node where R2 and R3 are located,
it splits between them. This means that R2 and R3 are in parallel. The current in a
parallel section must recombine at some point, so you should trace the two lines
to find the node where they recombine. In this case, it is easy to see where the
currents recombine.
You can also represent that components are in parallel as R2 || R3. The two
vertical lines (||) mean these two resistors are in parallel with each other.
Being able to identify which components are in series and which are in parallel
is very important when designing or troubleshooting a combination circuit. It
should always be your first step in analyzing a combination circuit.
BB227-BC05UEN COMBINATION CIRCUITS
Copyright © 2012 Amatrol, Inc.
5
SKILL 1
TRACE THE CURRENT PATH IN A COMBINATION CIRCUIT
Procedure Overview
In this procedure, you will determine which components are in series
and which are in parallel in a combination circuit by tracing the current path
through the circuit.

1. Examine the circuit in figure 3.
R1
+
R2
R3
R4
Figure 3. Combination Circuit

2. Trace the current path through the circuit and note whether each component
is in series or parallel.
R1 _____________________________________________ (Series/Parallel)
R2 _____________________________________________ (Series/Parallel)
R3 _____________________________________________ (Series/Parallel)
R4 _____________________________________________ (Series/Parallel)

R1 and R4 are in series. R2 || R3.
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
3. Examine the circuit in figure 4 and determine which loads are in series and
which are in parallel with other loads.
R1 _____________________________________________ (Series/Parallel)
R2 _____________________________________________ (Series/Parallel)
R3 _____________________________________________ (Series/Parallel)
R4 _____________________________________________ (Series/Parallel)
R5 _____________________________________________ (Series/Parallel)
R1 and R5 are in series. R2 || R3 || R4.
R1= 10
+
R2=
20
R3=
R4=
50
25
R5= 10
Figure 4. Combination Circuit

4. Examine the circuit in figure 5 and determine which components are in series
and which are in parallel.
R1 _____________________________________________ (Series/Parallel)
R2 _____________________________________________ (Series/Parallel)
R3 _____________________________________________ (Series/Parallel)
R1 || R2. R3 is in series.
+
R1 =
R2 =
25
25
R3 = 10
Figure 5. Combination Circuit
BB227-BC05UEN COMBINATION CIRCUITS
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OBJECTIVE 3
LIST THE SEVEN STEPS FOR SOLVING A COMBINATION CIRCUIT
Solving a circuit involves determining unknown values of resistance, current,
or voltage in a combination circuit. The seven general steps for solving a circuit
are:
• Simplify the circuit to a series circuit by finding the effective equivalent
resistance (REQ) of each parallel section of the circuit.
• Use R to calculate the total resistance (R ) of the circuit by adding them to
EQ
T
the series resistances.
• Calculate total current (I ) using R (Ohms Law).
T
T
• Calculate the voltage drop across any series resistance or equivalent resistance you need to know using Ohms Law.
• Calculate the branch currents in a parallel section using the voltage drop
across REQ and Ohms Law.
• Use the branch currents and resistance values to calculate the voltage of the
parallel resistances.
• Make a summary of the voltage drops and currents for each resistance to
make sure they add up.
You may have to alter the order of the steps sometimes, depending on the
information that you are given for the circuit. But, in most cases, these seven steps
should help you find the information you need to know about the circuit.
BB227-BC05UEN COMBINATION CIRCUITS
Copyright © 2012 Amatrol, Inc.
8
SKILL 2
SOLVE A COMBINATION CIRCUIT
Procedure Overview
In this procedure, you will solve combination circuits using the seven-step
process. You will calculate voltage and current values for each resistance in the
circuit and redraw the circuit with all the values listed. In step 1, you will be
led through the process. In steps 2 and 3, you will do it yourself.

1. Perform the following substeps to simplify the combination circuit shown in
figure 6.

The total resistance in a combination circuit can be found by combining the
equations used to find total resistance in a series circuit and a parallel circuit.
For example, the total resistance of the combination circuit in figure 6 is
found by first calculating the equivalent resistance of the two resistors that
are in parallel, R2 and R3. This resistance is then added to the resistances that
are in series with it, R1 and R4.

The following substeps will lead you through this process.
R1= 25
(R2 R3= REQ)
+
24V
R2=
50
R3=
10
R4= 25
Figure 6. Calculating Total Resistance
BB227-BC05UEN COMBINATION CIRCUITS
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A. Calculate the equivalent resistance of R2 and R3 in the circuit in figure 6.
REQ = _____________________________________________ (Ohms)
REQ = _____________________________________________ (Ohms)
This is found as follows:
1
REQ
1
REQ
=
=
1
R2
1
50
+
+
1
R3
1
10
REQ = 8.33 ohms
The equivalent resistance is 8.33 ohms.
B. Calculate the total resistance of the circuit.
Since this equivalent resistance is in series with R1 and R4, you can now
find the total resistance by applying the formula for calculating total resistance in a series circuit.
RT = _______________________________________________ (Ohms)
The answer is found as follows:
RT = R1 + REQ + R4 (Where REQ is the equivalent resistance
of R2 and R3 in parallel.)
RT = R1 + REQ + R4
RT = 25 + 8.33 + 25
RT = 58.33
The circuit in figure 6 has the same resistance as a circuit with just one
load with the resistance of 58.33 ohms.
This completes steps 1 and 2 of the seven step process.
C. Calculate the total current in the circuit of figure 6.
Since the source voltage and the total resistance are known, they can be
used to calculate the total current (IT) using Ohms Law.
IT = _______________________________________________ (Amps)
The answer is found as follows:
IT =
VT
RT
IT =
24
58.33
IT = .41
Total current in the circuit is approximately 0.41 amps.
BB227-BC05UEN COMBINATION CIRCUITS
Copyright © 2012 Amatrol, Inc.
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D. Calculate the voltage drops across the resistors in the circuit in figure 6.
In figure 7, the circuit from figure 6 has been reduced to a series circuit
using the equivalent resistance calculated for the two parallel resistors.
Since the current is known, the voltage drop across each resistance can
now be calculated using Ohms Law.
R1 = 25
+
REQ = 8.33
24V
R4 = 25
Figure 7. Combination Circuit to a Series Circuit
VR1 = _______________________________________________ (VDC)
VREQ = ______________________________________________ (VDC)
VR4 = _______________________________________________ (VDC)
The answers are found as follows:
VR1 = IT × R1
VR1 = 0.41 × 25
VR1 = 10.25 volts
VREQ = IT × REQ
VREQ = 0.41 × 8.33
VREQ = 3.42 volts
VR4 = IT × R4
VR4 = 0.41 × 25
VR4 = 10.25
The voltage drop across R1 is 10.25 volts. The voltage drop across REQ is
3.42 volts. The voltage drop across R4 is also 10.25 volts.
NOTE
The total calculated voltage is not quite 24V because we rounded off our
calculations.
BB227-BC05UEN COMBINATION CIRCUITS
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Since the voltage across the equivalent resistance of R2 || R3 is 3.42 volts,
we can say that the voltage drop across each of the parallel resistors, R2
and R3 is 3.42 volts, as shown in figure 8.
VR1= 10.25V
+
24V
R2
R3
VR2/3 = 3.42V
VR4 = 10.25V
Figure 8. Voltage Drops in a Combination Circuit
E. Calculate the current in each branch of the circuit in figure 6.
Figure 9 shows the updated circuit.
VR1 = 10.25V
R1=25
IT
+
I R2
24V
R2=50
IR3
R3=10
VREQ=3.42V
IT
R4=25
VR4=10.25V
Figure 9. Current Flow in a Combination Circuit
BB227-BC05UEN COMBINATION CIRCUITS
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The current in each branch of the parallel section can now be calculated
by dividing the voltage drop of each load by its resistance.
IR2 = _______________________________________________ (Amps)
IR3 = _______________________________________________ (Amps)
The answers are found as follows:
IR2 =
VR 2
R2
IR2 =
3.42
50
I R 2 = .068 A
I R3 =
VR3
R3
I R3 =
3.42
10
I R3 = .342 A
The current in branch R2 (IR2) is 0.068 A or 68 mA. The current in branch
R3 (IR3) is 0.342 A or 342 mA.
The branch currents are shown in figure 10.
R1
+
24V
R2
I R2 =
68mA
I R3 =
342mA
R3
R4
Figure 10. Branch Currents in a Combination Circuit
If the branch currents are added, the sum should equal the total current
that flows through the circuit. This is according to Kirchhoffs Current
Law.
BB227-BC05UEN COMBINATION CIRCUITS
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F. Draw the circuit from figure 9 with all values shown.
Your circuit should look like figure 11.
R1= 25
VR1 = 10.25V
IT = 0.41mA
+
24V
VR3 = 3.42V
VR2 = 3.42V R3=10
R2 = 50
I R3 = 342mA
I R2 = 68mA
R4= 25
VR4 = 10.25V
Figure 11. Circuit Solved

2. Perform the following substeps to solve the circuit in figure 12.
R1 = 25
R3 =
50
+
R2 =
50
120V
R5 = 100
R4 =
25
R6 = 25
Figure 12. Combination Circuit
A. Calculate the equivalent resistance of R2 || R3 (REQ1) and R5 || R6 (REQ2) for
the circuit in figure 12.
REQ1 = _____________________________________________ (Ohms)
REQ2 = _____________________________________________ (Ohms)
They should be REQ1 = 25 ohms and REQ2 = 20 ohms.
BB227-BC05UEN COMBINATION CIRCUITS
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B. Calculate the total resistance (RT) of the circuit.
RT = _______________________________________________ (Ohms)
RT should = 95 ohms.
C. Calculate the total current (IT) using the source voltage and RT.
IT = _______________________________________________ (Amps)
IT should = 1.263A.
D. Calculate the voltage drops across the series resistances, including the
equivalent resistances.
VR1 = _______________________________________________ (Volts)
VREQ1 = _____________________________________________ (Volts)
VR4 = _______________________________________________ (Volts)
VREQ2 = _____________________________________________ (Volts)
They should be VR1 = 31.58V, VREQ1 = 31.58V, VR4 = 31.58V, and
VREQ2 = 25.26V.
E. Determine the current in the branches of the parallel sections.
IR2 = _______________________________________________ (Amps)
IR3 = _______________________________________________ (Amps)
IR5 = _______________________________________________ (Amps)
IR6 = _______________________________________________ (Amps)
They should be IR2 and IR3 = .632A, IR5 = .253A, and IR6 = 1.01A.
F. Calculate the voltage of each parallel resistance using the current and
resistance values.
VR2 = _______________________________________________ (Volts)
VR3 = _______________________________________________ (Volts)
VR5 = _______________________________________________ (Volts)
VR6 = _______________________________________________ (Volts)
They should be VR2 and VR3 = 31.58V, and VR5 and VR6 = 25.26V. VR2 and
VR3 should be the same because they are in parallel, as should VR5 and VR6.
G. Redraw the circuit and list the voltage and current for each resistor along
with its resistance.
BB227-BC05UEN COMBINATION CIRCUITS
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
3. Now solve the circuit in figure 13. Show all your work.
R1 = 10
R2 = 10
+
12V
R3 =
25
R6 = 10
R4 =
100
R5 = 50
Figure 13. Combination Circuit
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SEGMENT 1
SELF REVIEW
1. A(n) ____________ circuit contains elements of both a series circuit
and a parallel circuit.
2. In order to identify the series and parallel sections of a combination
circuit, ______________ the current path.
3. The representation R1 | | R2 indicates that R1 and R2 are __________.
4. Identifying which components are in series and which are in parallel
should be the first step in _____________ a combination circuit.
5. In a combination circuit, components connected _____________ have
the same current flowing through them.
6. The last step in solving a circuit is to make a summary of the voltage
drops and currents for each ______________ to make sure they add up.
7. The current will be ________ between the branches of the parallel
section of a combination circuit.
8. There are seven general steps for ___________ a combination circuit.
BB227-BC05UEN COMBINATION CIRCUITS
Copyright © 2012 Amatrol, Inc.
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SEGMENT 2
LIGHTING CIRCUITS
OBJECTIVE 4
DESCRIBE HOW SWITCHES ARE USED IN COMBINATION CIRCUITS
AND GIVE AN APPLICATION
Loads are not the only devices that can be connected in series, parallel, or
combinations. Switches can also be connected in many different configurations to
achieve the desired control of a circuit.
This is especially true of circuits found in many residential and commercial
buildings. Switches can operate several devices at one time or can operate individual components within a circuit. Three examples are:
• Commercial building lighting system
• Electrical outlets controlled by a switch
• Ceiling fan with lights
Commercial Building Lighting System
Figure 14 shows how the lighting system of a typical commercial building is
connected. In this case, a switch is connected in series with the parallel lamps. This
configuration allows the switch to control when the voltage is supplied to all the
lamps. When the switch is closed (turned on), each of the lamps receives the source
voltage. When the switch is opened (turned off), the lamps turn off.
LAMPS
120 VAC
LIGHT
SWITCH
Figure 14. A Common Commercial Building Lighting Circuit
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Electrical Outlets Controlled By a Switch
Another example of a combination circuit can be found in many home power
outlet circuits, as shown in figure 15. In this case, the outlets are connected in
parallel with each other but in series with the switch. This allows the switch to turn
the outlet power on or off. Many homes have circuits like this at the front door for
control of a table lamp.
HOT
120
VAC
TYPICAL
OUTLET
GROUND
Figure 15. Outlets Connected to a Switch
A Ceiling Fan with Lights
Now, consider the ceiling fan with lights, as shown in figure 16. In this circuit,
the series combination of S2 and the fan motor is in parallel with the series combination of S3 and the lamps. A switch (S1) is in series with the parallel combination
of S2 and S3 to control when voltage is applied to the circuit.
WALL SWITCH
S1
S2
120
VAC
CEILING
FAN
MOTOR
S3
M
Figure 16. A Typical Ceiling Fan Circuit
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This configuration creates a variety of possibilities, as shown in figure 17. If
the main switch (S1) is on and S2 and S3 are also on, the fan and the lights will be
on. To run the fan with the lights off, you could turn off S3 and leave S2 on. To turn
on the lights and turn off the fan, you would turn on S3 and turn off S2. Finally,
you could turn the fan and the lights both off while leaving the main switch (S1)
on by turning off S2 and S3.
S1(ON)
S2(ON)
120
VAC FAN
MOTOR
(ON)
S3(ON)
S1(ON)
S2(ON)
120
VAC
M
FAN
MOTOR M
(ON)
LAMPS
(OFF)
LAMPS
(ON)
S1(ON)
S2(OFF)
S3(ON)
120
VAC
S3(OFF)
S1(ON)
S2(OFF)
S3(OFF)
120
VAC
FAN
MOTOR M
(OFF)
FAN
MOTOR M
(OFF)
LAMPS
(ON)
LAMPS
(OFF)
Figure 17. Operating Options for a Ceiling Fan with Lights
BB227-BC05UEN COMBINATION CIRCUITS
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SKILL 3
CONNECT AND OPERATE A BASIC LIGHTING CIRCUIT
Procedure Overview
In this procedure, you will connect a lighting circuit such as one you might
find in a commercial building. You will operate the circuit and observe the
status of the lamps. You will also measure the voltage across the lamps with
the switch on and off.


1. Connect the circuit shown in figure 18.
Make sure that the circuit is connected to the 24 volt (outer) terminals of the
power supply on the T7017.
24 VAC
Figure 18. Lighting Circuit

2. Perform the following substeps to operate the power supply.
A. Place the AC/DC switch in the AC position.
B. Turn on the power supply.

3. Close (turn on) the knife switch and observe the lamps.
Lamp 1 status __________________________________________(On/Off)
Lamp 2 status __________________________________________(On/Off)
Lamp 3 status __________________________________________(On/Off)

All lamps should be on because the current can flow through them.
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
4. Measure the voltage across each lamp with the DMM.
Voltage Lamp 1 = ________________________________________ (VAC)
Voltage Lamp 2 = ________________________________________ (VAC)
Voltage Lamp 3 = ________________________________________ (VAC)
Each lamp should have the source voltage across it, 24 VAC ± 2 VAC.

5. Now open (turn off) the knife switch and observe the lamps.
Lamp 1 status __________________________________________(On/Off)
Lamp 2 status __________________________________________(On/Off)
Lamp 3 status __________________________________________(On/Off)


All lamps should now be off because the current flow has been stopped.
6. Measure the voltage across each lamp with the DMM.
Voltage Lamp 1 = ________________________________________ (VAC)
Voltage Lamp 2 = ________________________________________ (VAC)
Voltage Lamp 3 = ________________________________________ (VAC)



All the lamps should have zero volts across them.
7. Turn off the power supply, disconnect the circuit, and store all components.
You have now successfully connected and operated a typical lighting circuit.
BB227-BC05UEN COMBINATION CIRCUITS
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SKILL 4
CONNECT AND OPERATE A CEILING FAN CIRCUIT
Procedure Overview
In this procedure, you will connect a circuit that resembles a typical ceiling
fan circuit. You will then control the operation of the fan motor and the lights
with the three switches of the circuit.

1. Connect the circuit shown in figure 19.
S1
(KNIFE SWITCH)
S3
(SELECTOR)
SWITCH
S2
(PUSHBUTTON)
+
12V
FAN
M
LAMP 1
LAMP 2
Figure 19. A Ceiling Fan Circuit

2. Perform the following substeps to operate the T7017 power supply.
A. Place the AC/DC switch in the DC position.
B. Turn on the power supply.

3. Close (turn on) the knife switch (S1) and observe the lamps and the fan.
Lamp 1 status __________________________________________(On/Off)
Lamp 2 status __________________________________________(On/Off)
Fan status _____________________________________________(On/Off)

They should all be off because S2 and S3 are off (open).
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
4. Now press (turn on) the pushbutton switch (S2) and observe the lamps and
the fan.
Lamp 1 status __________________________________________(On/Off)
Lamp 2 status __________________________________________(On/Off)
Fan status _____________________________________________(On/Off)


The fan should now be on because closing S2 allows current to flow through
that branch. The lamps do not come on because S3 is not on (open).
5. Release the pushbutton and rotate (turn on) the selector switch (S3). Observe
the lamps and the fan.
Lamp 1 status __________________________________________(On/Off)
Lamp 2 status __________________________________________(On/Off)
Fan status _____________________________________________(On/Off)


The fan should be off and the lights should be on because with S3 closed,
current can flow in the branch that contains the lamps.
6. Now energize (turn on) both S2 and S3 together and observe the lamps and
fan.
Lamp 1 status __________________________________________(On/Off)
Lamp 2 status __________________________________________(On/Off)
Fan status _____________________________________________(On/Off)


They should all be on because current can flow through both branches of the
circuit.
7. Leave S2 and S3 on and open (turn off) the knife switch (S1) and observe the
lamps and the fan.
Lamp 1 status __________________________________________(On/Off)
Lamp 2 status __________________________________________(On/Off)
Fan status _____________________________________________(On/Off)



They should all be off because there is no current flow to the branches because
S1 is open (off).
8. Turn off the power supply, disconnect the circuit, and store all components.
This procedure demonstrates how a typical ceiling fan with lights operates.
By using the three switches you can change the operation of the circuit.
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OBJECTIVE 5
DESCRIBE THE FUNCTION OF A VARIABLE RESISTOR AND GIVE AN
APPLICATION
Up to this point, the resistors you have used are fixed types. Another type of
resistor that is often used is a variable resistor. A variable resistor is one which can
have its resistance adjusted manually.
A variable resistor can have two or three terminals, usually three. The two
outside terminals act like a fixed resistor, the resistance between the two does not
change. The center terminal, however, is attached to a slider, which is controlled
by a knob, as shown in figure 20.
TERMINALS
RESISTOR
HOUSING
KNOB
Figure 20. Variable Resistor
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Figure 21 shows the schematic symbols for 2-terminal and 3-terminal variable
resistors.
2-TERMINAL
VARIABLE RESISTOR
3-TERMINAL
VARIABLE RESISTOR
CENTER
TERMINAL
(SLIDER)
Figure 21. Variable Resistor
The two common types of variable resistors used in most circuits are:
• Rheostat - A rheostat is a variable resistor that is used primarily in highpower circuits. A rheostat often only has two terminals, but can have three.
The rheostat supplied with the T7017 trainer uses three terminals.
• Potentiometer - A potentiometer is a variable resistor that is used primarily
in low-power (e.g. electronic) circuits. It usually has three terminals and is
typically smaller than a rheostat.
Variable resistors are used in many devices. Your radio uses variable resistors
to control the volume, tuning, balance, and so on. A television uses them to control
volume, color, tint, and brightness. Sometimes they are used to dim lighting in a
home or business. They can also be used to control the speed of a motor by controlling the current flow to the motor. Variable resistors are even used in some power
supply units to allow the output voltage to be adjusted to a desired level.
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Activity 1. Rheostat Operation
Procedure Overview
In this procedure, you will measure the resistance of a rheostat as it is
adjusted and note the effect the slider has on the resistance between it and
either outside terminal.
You should also note that the sum of the two resistances created by the
slider should always equal the resistance across the two outside terminals.

1. Prepare the DMM to measure resistance.

2. Locate the rheostat module and place it on the construction surface of the
T7017.

3. With the control knob on the rheostat facing you, turn the knob fully
counterclockwise.

4. Measure the resistance across the two outside terminals, as shown in figure
22.
Resistance = ___________________________________________ (Ohms)

It should be approximately 25 ohms.
30XR
DMM
NON
CONTACT
VOLTAGE
MIN MAX
V
200
HOLD
600 OFF 600
200
20
V
20
2
200m
2
200m
200
2m
20M
2M
20m
200k
200m
20k
2k
200
10 A
1.5V 9V
200
BATT
BATT 1.5V
10 A
200m
2m 20m
A
mA
10A
RHEOSTAT
MODULE
CAT
CAT
V
COM
A
600 V
300V
BATT 9V
200mA
MAX
FUSED
10A MAX
FUSED
MAX
600V
600V
Figure 22. Measurement of Rheostat Resistance

5. Turn the knob fully clockwise and measure the resistance across the two
outside terminals.
Resistance = ___________________________________________ (Ohms)

It should be the same as in step 4, approximately 25 ohms. The resistance
between the two outside terminals is always the maximum amount. The
slider does not affect this value.
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
6. Now, measure the resistance between the center terminal and one of the
outside terminals.
Resistance = ___________________________________________ (Ohms)

7. Measure the resistance between the center terminal and the other outside
terminal.
Resistance = ___________________________________________ (Ohms)

8. Add the resistances from step 6 and 7.
Total resistance = _______________________________________ (Ohms)

It should equal the resistance recorded in steps 3 and 4, approximately 25
ohms.

Figure 23 shows the operation of a rheostat visually.

Depending on the sliders position, the resistance between it and either outside
terminal varies. Therefore, it effectively creates two different resistance
values.
RESISTANCE BETWEEN
AB=0 OHMS
BC=25 OHMS
AC= 25 OHMS
RESISTANCE BETWEEN
AB=12.5 OHMS
BC=12.5 OHMS
AC= 25 OHMS
RESISTANCE BETWEEN
AB=25 OHMS
BC=0 OHMS
AC= 25 OHMS
A
A
A
B
B
B
C
C
C
Figure 23. Operation of a Variable Resistor



9. Turn the knob counterclockwise about 1/4 turn and repeat steps 6 through 8.
10. Keep repeating step 8 until the knob is all the way counterclockwise.
Each time the two resistances added together should equal the total across the
two outside terminals. This is always true of a rheostat.
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SKILL 5
CONNECT AND OPERATE A RHEOSTAT AS A LIGHT DIMMER
Procedure Overview
In this procedure, you will connect and operate a rheostat in a circuit and
use it to dim a lamp by controlling the voltage across the lamp.

1. Connect the circuit shown in figure 24. Make sure the knob on the rheostat is
facing you before you connect it.
+
SOURCE SELECT
AC
24V
DC
LEFT
TERMINAL
RHEOSTAT
24V
12V
12V
LAMP
CENTER
TERMINAL
RHEOSTAT
MODULE
LAMP
MODULE
Figure 24. Light Dimmer Circuit

2. Place the AC/DC switch in the DC position.

3. Make sure the rheostat knob is turned fully clockwise before turning on the
power supply.

4. Turn on the power supply and observe the lamps status.
Lamp status ___________________________________________(On/Off)
Lamp status _______________________________________ (Bright/Dim)

The lamp should be on and bright.

5. Prepare the DMM to measure voltage.

6. Measure the voltage across the lamp.
Voltage = ______________________________________________ (Volts)

The voltage should be approximately 23.5 V.
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
7. Turn the knob about 1/4 turn counterclockwise and observe the lamp.
Lamp status __________________________________ (Brighter/Dimmer)


The lamp should start to dim somewhat.
8. Measure the voltage across the lamp.
Voltage = ______________________________________________ (Volts)


It should be lower than it was in step 6.
9. Turn the knob to about the half-way point and observe the lamp.
Lamp status __________________________________ (Brighter/Dimmer)


The lamp should be even dimmer.
10. Measure the voltage across the lamp.
Voltage = ______________________________________________ (Volts)


It should be lower than in step 8.
11. Turn the knob about 3/4 counterclockwise and observe the lamp.
Lamp status __________________________________ (Brighter/Dimmer)


The lamp should be even dimmer than before.
12. Measure the voltage across the lamp.
Voltage = ______________________________________________ (Volts)


It should be lower than in step 10.
13. Now turn the knob fully counterclockwise and observe the lamp.
Lamp status __________________________________ (Brighter/Dimmer)


The lamp should be very dim.
14. Measure the voltage across the lamp.
Voltage = ______________________________________________ (Volts)

It should be lower than in step 12.

15. Adjust the knob back and forth several more times and observe the lamp to
familiarize yourself with the operation.

16. Turn off the power supply, disconnect the circuit, and store all components.
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SEGMENT 2
SELF REVIEW
1. _____________ can be connected in series, parallel, or combinations
just like loads.
2. A(n) ________________ often has a switch to apply power to the
circuit, a switch to control the lights, and a switch to control the fan
motor.
3. A(n) _________ resistor can be manually adjusted.
4. A(n) _________ is a variable resistor used primarily in high-power
circuits.
5. A variable resistor most commonly used in low-power circuits is a(n)
____________.
6. Depending on the position of the _________ on a variable resistor, the
resistance between it and either outside terminal will vary.
7. A variable resistor can be used to _____ lighting in a home or business.
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SEGMENT 3
VOLTAGE DIVIDERS
OBJECTIVE 6
DESCRIBE THE FUNCTION OF A VOLTAGE DIVIDER AND GIVE AN
APPLICATION
A common application of Ohm’s Law and series circuits is a voltage divider
network. The function of a voltage divider network is to create a voltage that is
lower than the source voltage.
As shown in figure 25, the voltage divider consists of two or more resistors in
series with the source voltage. The voltage at point B with reference to ground is
some value between the source voltage and zero volts.
SOURCE
VOLTAGE
+12V
POINT B
R1
R2
VR1
VR2
Figure 25. A Voltage Divider Network
The main application of voltage divider networks is with electronic circuits,
which need different voltage levels such as +12 volts, +5 volts, and +3 volts to
operate. The voltage divider network provides these voltages from one power
supply instead of using three separate power supplies. This saves space and money.
Another application of a voltage divider is to provide a reduced voltage to drive a
load.
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OBJECTIVE 7 DESCRIBE THE OPERATION OF THREE TYPES OF VOLTAGE DIVIDERS
There are three points you must consider when designing or using a voltage
divider network:
• How constant the load voltage must be
• How much power the circuit will consume
• Whether load resistance is high or low
For these reasons, there are three types of voltage divider networks:
• Firm Voltage Divider
• Stiff Voltage Divider
• Loaded Voltage Divider
All three of these networks look like the network in figure 26. The difference
is in how the resistances for R1 and R2 are selected. Each of the three types of
networks is designed to deal with a certain type of application. The reason for this
is the load resistance affects the voltage level from the voltage divider.
Before you learn more about each type of network it is important to understand
why the voltage level supplied by any voltage divider changes if the load resistance
changes. To understand why, first examine the divider network shown in figure
26. As shown, a voltage divider network is actually a combination circuit because
the load is connected in parallel with one part of the circuit. In this case, it is R2.
This resistor is called the bleeder resistor. The voltage, VRL, supplied to the load RL
depends on the amount of the equivalent resistance of R2 and RL. If RL or R2 change
the voltage VRL will also change.
VT = 24 VOLTS
R1 = 100
VOLTAGE
DIVIDER
NETWORK
VRL= 6 VOLTS
+
24V
R2 = 10
RL = LOAD
RESISTANCE
Figure 26. A Combination of Circuit Serving as a Voltage Divider
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Firm and Stiff Voltage Dividers
To deal with the effects of changes in load, firm and stiff voltage dividers are
designed so that the bleeder resistor, in parallel with the load, is much less than the
load resistance. In this case, the voltage level is determined mostly by the bleeder
resistor, R2, not the load resistor, RL.
A firm voltage divider uses a bleeder resistor, R2, which is 1/10 the resistance
of the load. A stiff voltage divider uses a bleeder resistor which is 1/100 the resistance of the load, as shown in figure 27.
The firm voltage divider is used if the load voltage does not need to be very
constant. Its advantage is that it does not consume as much power as the stiff
voltage divider. The stiff voltage divider is used when the load voltage must be
more constant.
FIRM VOLTAGE DIVIDER
STIFF VOLTAGE DIVIDER
R1
R1
+
+
R2=
RL
10
R2=
RL
RL
100
RL
Figure 27. A Firm and a Stiff Voltage Divider
The relationship between the voltages and resistances for either a stiff or firm
voltage divider is shown in the following formula. This is actually an approximation because it assumes RL is infinite. This formula can be used to calculate the
load voltage if you know the resistance. It can also be used to size resistor R1 if you
know what the load voltage needs to be.
STIFF AND FIRM VOLTAGE DIVIDER FORMULA
VRL = VT ×
where
R2
R1 + R2
VRL = load voltage in volts
VT = total voltage drop in volts
R1 = resistance of R1 in Ohms
R2
= resistance of R2 in Ohms
The firm and stiff voltages dividers are commonly used to supply voltages
to devices which have relatively high resistances. An electronic amplifier is an
example.
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If the load resistance is relatively low, these two types of voltage dividers waste
a lot of power. For example, if you had a load which required 50 mA of current,
your divider would waste 500 mA of current using a firm divider (using the 10:1
ratio of the load resistor to the bleeder resistor) or 5 Amps using a stiff divider
(100:1 ratio). For these applications, a loaded voltage divider network should be
used.
Loaded Voltage Divider
A loaded voltage divider network looks exactly like a firm or stiff network
except the bleeder resistor is 10 times the resistance of the load resistor instead of
1/10 or 1/100, as shown in figure 28. With this approach, most of the power is used
by the load. The power wasted by the divider resistor is only 10%.
VT
R1
+
VRL
RL
R2 = 10 x RL
Figure 28. Loaded Voltage Divider
The relationship between the resistances and the voltages is shown in the
following formula. This is the same formula that is used for a stiff or firm divider
except that REQ (parallel resistance of RL and R2) is inserted for R2. This formula is
the exact calculation for the load voltage.
STIFF AND FIRM VOLTAGE DIVIDER FORMULA
VRL = VT ×
where
R1 + REQ
VRL = load voltage in volts
VT = total voltage drop in volts
R1 = resistance of R1 in Ohms
REQ
BB227-BC05UEN COMBINATION CIRCUITS
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REQ
= resistance of R2 and RL in Ohms
35
SKILL 6
DESIGN A VOLTAGE DIVIDER NETWORK
Procedure Overview
In this procedure, you will design circuits with all three types of voltage
dividers. You will be led through the first design of each type and will then
work through the others yourself.

1. Perform the following substeps to design a firm voltage divider network for
the circuit shown in figure 29.
R1
12V
VRL = 8V
R2
RL = 1M
Figure 29. Voltage Divider Circuit
A. Calculate the resistance of R2.
R2 = _______________________________________________ (Ohms)
The answer is as follows:
R2 =
RL
10
R2 =
1,000,000
10
R2 = 100,000
The resistance of R2 is 100,000 ohms.
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B. Calculate the resistance of R1.
R1 =(Ohms)
The basic formula is:
⎛ R2 ⎞
VRL = ⎜
× VT
⎝ R1 + R2 ⎟⎠
The answer is found by rearranging the formula as follows:
⎛ V × R2 ⎞
R1 = ⎜ T
− R2
⎝ VRL ⎟⎠
⎛ 12 × 100,000 ⎞
VRL = ⎜
⎟⎠ − 100,000
⎝
8
R1 = 150,000 − 100,000
The resistance of R1 is 50,000 ohms.

2. Design a firm voltage divider network for the circuit shown in figure 29 for
an RL of 75 K.
R2 = _________________________________________________ (Ohms)
R1 = _________________________________________________ (Ohms)


R2 is 7,500 ohms and R1 is 3,750 ohms.
3. Perform the following substeps to design a stiff voltage divider for the circuit
shown in figure 30.
R1
24V
VRL = 16V
R2
RL = 50 k
Figure 30. Voltage Divider Circuit
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A. Calculate the resistance of R2.
R2 = _______________________________________________ (Ohms)
R2 =
RL
100
R2 =
50,000
100
The resistance of R2 is 500 ohms.
B. Calculate the resistance of R1.
R1 = _______________________________________________ (Ohms)
⎛ V × R2 ⎞
R1 = ⎜ T
− R2
⎝ VRL ⎟⎠
⎛ 24 × 500 ⎞
R1 = ⎜
⎟ − 500
⎝
16 ⎠
R1 = 750 − 500
The resistance of R1 is 250 ohms.

4. Design a stiff voltage divider network for the circuit shown in figure 30 for an
RL of 85K.
R2 = _________________________________________________ (Ohms)
R1 = _________________________________________________ (Ohms)


R2 is 850 ohms and R1 is 425 ohms.
5. Perform the following substeps to design a loaded voltage divider for the
circuit shown in figure 31.
R1
11.6V
VRL = 3.6V
R2
RL = 5
Figure 31. Loaded Voltage Divider
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A. Calculate the resistance of R2.
R2 = _______________________________________________ (Ohms)
Resistor R2 is chosen to be 10 times larger than RL as follows:
R2 = RL × 10
R2 = 5 × 10
The resistance of R2 is 50 ohms.
B. Calculate REQ (the resistance of R2 || RL).
NOTE
The design of the loaded voltage dividers considers the effect of the load
in parallel with the bleeder resistor.
REQ = ______________________________________________ (Ohms)
REQ =
1
1
1
+
R2 RL
REQ =
1
.02 + .2
REQ =
1
.22
The solution is as follows:
The resistance of REQ is 4.55 ohms.
C. Calculate the value of R1.
R1 = _______________________________________________ (ohms)
This calculation uses the same voltage divider formula used to design stiff
and firm dividers except the combined resistance of R2 and RL (REQ) is
used instead of just R2.
This can be rearranged as before:
⎛ REQ ⎞
VRL = ⎜
⎟ × VT
⎝ R1 + REQ ⎠
R1
⎛ VT × REQ ⎞
=⎜
− REQ
⎝ VRL ⎟⎠
R1
⎛ 11.6 × 4.55 ⎞
=⎜
⎟⎠ − 4.55
⎝
3.6
R1
= 14.66 − 4.55
The resistance of R1 is 10.1 ohms.
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
6. Design a loaded voltage divider for the circuit shown in figure 31. Use a VRL
of 10V and an RL of 10 ohms.
R2 = _________________________________________________ (Ohms)
R1 = _________________________________________________ (Ohms)


R2 is 100 ohms and R1 is 1.45 ohms.
7. Design a loaded voltage divider for the circuit shown in figure 31. Use a VRL
of 4V and an RL of 2 ohms.
R2 = _________________________________________________ (Ohms)
R1 = _________________________________________________ (Ohms)

SKILL 7
R2 is 20 ohms and R1 is 3.46 ohms.
CONNECT AND OPERATE A VOLTAGE DIVIDER NETWORK
Procedure Overview
In this procedure, you will connect and operate a loaded voltage divider
network. You will then take measurements to verify its operation.

1. Perform the following substeps to set the rheostat (RL) to the required load
resistance value.
NOTE
In the next step, the rheostat will be used in a voltage divider circuit to
simulate the load (RL). The two (2) 25-ohm resistors will be used in series to
simulate R2.
A. Set the DMM to measure resistance.
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B. With the control knob of the rheostat facing you, measure the resistance
between the left terminal and the center terminal as shown in figure 32.
30XR
NON
CONTACT
VOLTAGE
MIN MAX
V
200
600 OFF 600
V
200
20
20
2
200m
2
200m
200
2m
20M
2M
20m
200k
200m
20k
2k
200
1.5V 9V
200
BATT
10 A
200m
2m 20m
A
mA
CAT
CAT
V
COM
A
600 V
300V
BATT 9V
200mA
MAX
FUSED
10A MAX
FUSED
RHEOSTAT
MODULE
LEFT
TERMINAL
10 A
BATT 1.5V
10A
DMM
HOLD
CENTER
TERMINAL
MAX
600V
600V
Figure 32. Resistance Across teh Rheostat
C. Turn the knob until the resistance between the left and center terminals is
5.0 ohms.


2. Connect the circuit shown in figure 33 using the rheostat you just set in step
2 for RL and the two (2) 25 ohm resistors for R2.
The circuit is equivalent to the one that you calculated the values for in the
previous skill (step 5).
R1=10
+
12V
25
(
R2= 50
)
TWO 25
RESISTORS
IN SERIES
RL =5
RHEOSTAT
25
Figure 33. Circuit to Simulate Figure 31
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
3. Perform the following substeps to operate the circuit and make measurements.
A. Place the AC/DC switch in the DC position.
B. Turn on the power supply on the T7017.
C. Set the DMM to measure DC Volts.
D. Now measure the voltage across RL.
VRL = _______________________________________________ (Volts)
It should be approximately 3.6V.
NOTE
Actual values will vary depending on the exact values of the resistors and
the source voltage.
E. Turn off the power supply.
In the next two steps you will measure the unloaded voltage produced by
the divider network.

4. Disconnect RL from the circuit and turn on the power supply.

5. Now measure the voltage across R2 (the series combination of the two 25
ohm resistors) to determine if unloaded voltage is close to the 3.6 V value
produced when the load was attached?
Unloaded voltage close to loaded voltage? __________________ (Yes/No)

You should find that the unloaded voltage is much greater than 3.6 V. This
is because the loaded divider network requires that the load be constant for
proper operation.

6. Turn off the power supply.

7. Disconnect the circuit and store all components.
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SEGMENT 3
SELF REVIEW
1. A(n) _____________________ network creates a voltage that is lower
than the source voltage.
2. A loaded voltage divider network looks exactly like a firm or stiff
network, except the bleeder resistor is _______ times the resistance of
the load resistor.
3. To deal with the effects of changes in load, firm and stiff voltage
dividers are designed so that the ____________ resistor is much less
than the load resistance.
4. The main application of voltage divider networks is with electronic
circuits that need ________ voltage levels to operate.
5. The firm and stiff voltage dividers are commonly used to supply voltage
to devices that have relatively high ______________.
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SEGMENT 4
TROUBLESHOOTING
OBJECTIVE 8
EXPLAIN THE EFFECT OF A SHORT CIRCUIT
A short circuit across any branch of a parallel circuit causes the entire circuit
to be short circuited, as shown in figure 34. All current flows through the short
circuited branch and bypasses the other branches. This causes an excessive amount
of current to flow since there is practically no resistance.
IT
+
R1
20V
R2
R3
IT
SHORT
Figure 34. A Short in a Parallel Circuit
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If a short circuit occurs in a series circuit, excessive current flow may or may
not occur. For example, if a single component is shorted, as shown in figure 35, the
current continues to flow through the rest of the circuit. Only the shorted component is bypassed. This means the circuit flow will be higher, but it may not be
excessive.
SERIES CIRCUIT
IT
+
SHORT
Figure 35. A Shorted Componet in Series Circuit
When excessive current occurs, something in the circuit is usually damaged. It
might damage the power supply or it may cause a wire to burn up. In most cases,
a short eventually creates an open, such as a wire burning up. The open can be in
the main line, in the branch containing the short, or anywhere along the path to the
short.
Adding a fuse or circuit breaker to the main line of a parallel circuit helps to
protect the components should a short develop in any branch. If this happens, the
fuse or circuit breaker opens, as shown in figure 36.
IT
+
FUSE
BLOWS
SHORT
Figure 36. A Fuse in a Parallel Circuit
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OBJECTIVE 9
DESCRIBE THE FOUR STEPS FOR TROUBLESHOOTING A SHORT
CIRCUIT
The four steps for troubleshooting a circuit to find a short are:

Step 1. Look for visible signs of damage - There may be visible signs, such
as smoke or soot that will indicate where the short occurred. If there are no
visible signs, you will have to continue on to the next step.

Step 2. Disconnect the power supply - Do not replace or reset any circuit
protection devices at this time. Wait until the problem has been resolved.

Step 3. Connect an ohmmeter across the main line - This is shown in
figure 37.
OHMMETER ACROSS
MAIN LINE
RT=?
R1
R2
R3
DISCONNECTED
POWER SUPPLY
DISCONNECT
BRANCHES
ONE AT A TIME
Figure 37. Connecting an Ohmeter to Locate a Short

Step 4. Disconnect each branch one at a time - If the resistance stays
close to zero when the branch is disconnected, the short is in another branch.
Reconnect that branch and move on to the next one.

If the resistance suddenly rises when a branch is disconnected, that is the
branch where the short occurred. If disconnecting all branches still results in
a zero ohm reading, the short has developed directly across the main line and
not in any branch.
Once you have located the short, you can correct the problem and then return
the circuit to operation. You should make sure that all the branches are reconnected, the circuit protection device is replaced or reset, and the power supply is
reconnected.
This 4-step procedure is one of the most common techniques used by electricians and electronic technicians to troubleshoot for shorts.
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SKILL 8
LOCATE A SHORT CIRCUIT
Procedure Overview
In this procedure, you will troubleshoot a circuit to locate a short circuit
using the four-step process. You will then repair the short and return the circuit
to operation.
NOTE
Consult your instructor as to whether you should continue with this
procedure or not. You will need an extra 3-amp fuse.

1. Connect the circuit shown in figure 38 and turn on the AC power supply.
Observe the fuse.
Fuse status __________________________________ (Blown/Not blown)

The fuse should blow immediately indicating a short.
3A
FUSE
12V
SHORTING
LEAD
R1=
10
R2=
25
R3=
25
Figure 38. Parallel Circuit with a Short
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
2. Perform the following substeps to troubleshoot the circuit.
A. Look at the circuit to see if there are any visible signs of a short (smoke or
soot).
B. Turn off and disconnect the power supply from the circuit.
C. Set the DMM to measure resistance and connect it across the circuit, as
shown in figure 39.
12V
DMM
R1=10
R2=25
R3=25
DISCONNECT HERE
ONE BRANCH AT A TIME
Figure 39. Troubleshooting with a DMM
D. Disconnect Branch 1 (R1) of the circuit and observe the resistance reading.
Resistance - Branch 1 disconnected = ____________________ (Ohms)
E. Reconnect branch 1.
F. Repeat substeps D and E for the other two branches.
Resistance - Branch 2 (R2) disconnected = _________________ (Ohms)
Resistance - Branch 3 (R3) disconnected = _________________ (Ohms)
The resistance readings when branches 1 and 2 are disconnected should
be zero. When branch 3 is disconnected the resistance reading should rise.
That means this is the branch where the short occurred.

3. Correct the short circuit by removing the wire that is shorting the resistor.
NOTE
This wire is used to simulate what happens when a component, such as a
resistor, actually shorts out. In an actual situation, you would need to replace
the shorted component. In this case, you will simply remove the wire.
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
4. Make sure all the branches are reconnected and observe the resistance
reading.
Resistance reading = ____________________________________ (Ohms)

The reading should be 5.56 ohms, which is the effective resistance of the
circuit. Therefore, the short circuit has been eliminated. If the reading is still
zero, another short is present and must be found.

5. Replace the blown fuse with a 3-amp fuse.

6. Reconnect the power supply to the circuit.

7. Operate the circuit and observe the fuse.
Fuse status __________________________________ (Blown/Not blown)

OBJECTIVE 10
The fuse should not blow since the short has been repaired.

8. Turn off the power supply.

9. Disconnect the circuit and store all components.
DESCRIBE THE THREE BASIC STEPS FOR TROUBLESHOOTING AN
OPEN CIRCUIT
In many cases, an open is caused by a short. This happens when a short
continues for a while, causing a wire or a component to burn and open. It also
causes an open in a fuse or circuit breaker.
If one branch of a parallel circuit has opened, the rest of the circuit will still
operate but with a reduced total current. This occurs because the effective resistance of the circuit has been increased.
If an open occurs in a series circuit, current will not flow.
The method for troubleshooting a circuit to locate an open is much the same as
for locating a short. The three basic steps are:
Step 1. Disconnect the power supply from the circuit
Step 2. Connect the ohmmeter across the main line - This allows you
measure the total resistance of the circuit.
Step 3. Disconnect each branch, one at a time - The total resistance should
increase as each branch is disconnected. If disconnecting a branch has no effect
on the total resistance, the branch you are testing has an open somewhere.
More than one branch may have an open. Therefore, you should continue to
check the rest of the branches to be sure that another branch doesnt have an
open.
NOTE
If the open occurs in the main line, the circuit will not operate at all.
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Once a branch with an open has been found, it must be determined if a short
caused the open. If this is the case, you will have to repair or replace the component
to correct the short and open. You should then check the whole circuit for any other
shorts.
SKILL 9
LOCATE AN OPEN CIRCUIT
Procedure Overview
In this procedure, you will troubleshoot a parallel circuit to determine
where an open has occurred. You will then repair the open and return the
circuit to operation.

1. Connect the circuit shown in figure 40 and then turn on the power supply.
Note the status of the lamp in branch 2.
Lamp status __________________________________________ (On/Off)

The lamp should be off because there is no path for current to flow in branch
2.
DISCONNECTED
WIRE
+
12V
R1=10
R2=3.1
R3=25
Figure 40. A Parallel Circuit with an Open

2. Perform the following three substeps to troubleshoot the circuit for an open.
A. Turn off and disconnect the power supply from the circuit.
B. Set the DMM to measure resistance, connect it across the main line of the
circuit and observe the reading.
Total resistance = ____________________________________ (Ohms)
The total resistance should be approximately 7.2 ohms.
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C. Disconnect branch 1 and record the resistance reading.
Resistance - Branch 1 disconnected = ____________________ (Ohms)
D. Reconnect branch 1.
E. Repeat substeps C and D for the other two branches.
Resistance - Branch 2 disconnected = ____________________ (Ohms)
Resistance - Branch 3 disconnected = ____________________ (Ohms)
When branches 1 and 3 are disconnected, the total resistance reading
should increase. However, when branch 2 is disconnected, the total resistance reading does not change. This means branch 2 contains the open.
Also, since branch 2 contains a lamp, we could have determined that
branch had an open since the lamp did not come on when the circuit was
operated.

3. Repair the open by connecting the loose wire to the component in branch 2,
as shown in figure 41.
RECONNECT
WIRE
+
12V
R1=10
R2=3.1
R3=25
Figure 41. Connecting a Loose Wire

4. Perform the four steps to troubleshoot the circuit to determine if there is a
short present in the circuit.
Short ________________________________________________ (Yes/No)


You should not find a short present.
5. Measure the total resistance of the circuit.
Total resistance = _______________________________________ (Ohms)

The total resistance should now be lower than it was in step 2B of Skill 7.
This is because the resistance of the lamp is now affecting the circuit.
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
6. Reconnect the power supply to the circuit and turn on the circuit. Observe the
lamp status.
Lamp status __________________________________________ (On/Off)

The lamp should now be on since the open has been repaired.

7. Turn off the power supply.

8. Disconnect the circuit and store all components.
NOTE
The same results you saw in this skill would have occurred if the lamp had
been burned out, which happens frequently with lamps. When a lamp burns
out, the filaments inside the lamp open.
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SEGMENT 4
SELF REVIEW
1. A short circuit across any branch of a(n) ____________ circuit causes
the entire circuit to be short circuited.
2. Adding a fuse or circuit breaker to the _________ of a parallel circuit
will help protect the components from damage.
3. Eventually, a short in a parallel circuit will cause a(n) ________________.
4. There are ____ basic steps in troubleshooting a parallel circuit for
opens.
5. An open in a branch of a parallel circuit will cause the effective
resistance to _________.
6. In a series circuit, if a single component is ____________, the current
continues to flow through the rest of the circuit.
7. In most cases, repairing an open in a circuit requires _________ a
component that has opened.
8. If an open occurs in the main line, the ___________ will not operate at
all.
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