AC/DC
ELECTRICAL
SYSTEMS
LEARNING
ACTIVITY
PACKET
INDUCTANCE AND
CAPACITANCE
BB227-BC04UEN
LEARNING ACTIVITY PACKET 4
INDUCTANCE AND CAPACITANCE
INTRODUCTION
Up to this point, the electrical circuits that have been covered have only contained
resistive-type load characteristics. This LAP will discuss two new types of load
characteristics: inductance and capacitance. These two new characteristics play an
important role in many electrical systems that are worked with every day. Examples
include electric motors, power supplies, lighting systems, car ignitions, and flash cameras.
This LAP will also explain electromagnetism. Electromagnetism is an electrical
principle that contributes to inductance. It is the principle that allows motors to turn and
relays to operate. Therefore, it is important to first understand electromagnetism before
studying inductance.
ITEMS NEEDED
Amatrol Supplied
1
T7017 AC/DC Electrical Learning System
FIRST EDITION, LAP 4, REV. E
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 © 2013, 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
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TABLE OF CONTENTS
SEGMENT
1 ELECTROMAGNETISM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
OBJECTIVE 1 Define electromagnetism and give an application
Activity 1 Test an electromagnetic field
OBJECTIVE 2 Describe the functions of four electromagnetic devices
Activity 2 Electromagnetic device operation
SKILL 1 Connect and operate a relay in a circuit
SEGMENT
2 INDUCTANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
OBJECTIVE 3 Define inductance and give its units of measurement
OBJECTIVE 4 Describe the operation of an inductor and give its schematic symbol
OBJECTIVE 5 Describe the effect of an inductor in a DC circuit and give an application
Activity 3 Effect of inductance in a DC circuit
OBJECTIVE 6 Describe the effect of an inductor in an AC circuit and give an application
Activity 4 Effect of inductive reactance in an AC circuit
OBJECTIVE 7 State the formulas for calculating total series inductance and inductive reactance
OBJECTIVE 8 State the formulas for calculating total parallel inductance and inductive reactance
SKILL 2 Calculate the total load on an AC circuit with inductors
SEGMENT
3 CAPACITANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
OBJECTIVE 9
OBJECTIVE 10
OBJECTIVE 11
SKILL 3
SKILL 4
SEGMENT
Define capacitance and give its units of measurement
Describe the operation of a capacitor and give its schematic symbol
Describe the functions of three types of capacitors
Discharge a capacitor
Test a capacitor with a DMM
4 CHARACTERISTICS OF CAPACITANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
OBJECTIVE 12 Describe the effect of a capacitor in a DC circuit and give an application
SKILL 5 Measure the voltage across a charged capacitor
Activity 5 Effect of a capacitor in a DC circuit
OBJECTIVE 13 Describe the effect of a capacitor in an AC circuit and give an application
Activity 6 Effect of a capacitor in an AC circuit
OBJECTIVE 14 State the formulas for calculating total series capacitance and capacitive reactance
OBJECTIVE 15 State the formulas for calculating total parallel capacitance and capacitive reactance
SKILL 6 Calculate the total load on an AC circuit with capacitors
SEGMENT
5 INDUCTOR AND CAPACITOR APPLICATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
OBJECTIVE 16
OBJECTIVE 17
SKILL 7
OBJECTIVE 18
Describe the function of a fluorescent light fixture
Describe the function of an RC timing circuit in a time-delay relay
Calculate the time to charge and discharge a capacitor
Describe the function of capacitor and inductors in an electric power supply
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SEGMENT 1
ELECTROMAGNETISM
OBJECTIVE 1
DEFINE ELECTROMAGNETISM AND GIVE AN APPLICATION
When electrons flow through a wire, an invisible force is created around the
wire, as shown in figure 1. This invisible force is called a magnetic field and can
be used to do many useful tasks such as operate an electric motor. The creation of
a magnetic field by the flow of electric current is referred to as electromagnetism.
MAGNETIC
FIELD
CONDUCTOR
ELECTRON
FLOW
Figure 1. Electron Flow Through a Conductor Creates a Magnetic Field
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Winding the conductor into a coil, as shown in figure 2, causes the magnetic
field to become stronger. The strength of the field increases as the number of windings (turns) increases. For example, a coil with 5 turns has a magnetic field that is
5 times stronger than a conductor with only one turn.
S
N
MAGNETIC
FIELD
CONDUCTOR
Figure 2. Winding the conductor into a coil increases the magnetic field strength
An electromagnetic field creates a force that can be used to move metal objects.
Electromagnets are devices that use this force to produce some type of output,
usually mechanical motion. You will learn more about electromagnetic devices and
how they operate later in this segment.
Coiling a conductor also affects the current that flows through the conductor.
You will learn more about the effects of a coil on current later.
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Activity 1. Test an Electromagnetic Field
Procedure Overview
In this procedure, you will show that an electromagnetic field is generated
by a coil when current passes through it by bringing a compass near the coil.
You will also observe that by inserting an iron core into the coil, the strength
of the field is increased.

1. Remove the armature from the solenoid and connect the circuit shown in
figure 3.

2. Hold the compass 12 to 16 inches above the trainer work surface. The
compass needle will align itself with the earth’s magnetic field.

3. While looking down through the compass from above, rotate the solenoid
module until its axis is perpendicular to the compass needle as shown in
figure 3.
REMOVE
SOLENOID
ARMATURE
SOURCE SELECT
AC
DC
24V
12V
12V
W
N
S
E
EARTH'S
MAGNETIC
FIELD
SOLENOID
AXIS
Figure 3. Verifying an Electromagnetic Field with a Compass

4. Place the AC/DC selector switch in the DC position.
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
5. Turn on the power supply and slowly lower the compass toward the solenoid
until the needle rotates 90°. Then measure the distance of the compass from
the trainer surface with a rule.
Distance = ___________________________________ (Inches/Centimeters)

This should occur when the compass is approximately 2-3 inches over
the solenoid. This is the point where the compass is close enough that the
electromagnetic field from the solenoid attracts the compass needle.

6. Turn off the power supply.

7. Insert the armature into the solenoid and turn on the power supply.

We will now observe the effect the armature has on the strength of a magnetic
field. The stronger the field is, the farther away from the coil the magnetic
field lines reach.

There are three factors that determine the strength of an electromagnetic
field. Changing any one of these changes the electromagnetic field. These
factors are:
• The amount of current flowing through the conductor - As you increase
the current flow through a coil, the strength of the magnetic field increases
as well. Decreasing the current causes the field to weaken. If you stop
the flow of current, the field collapses.
• The number of turns in the coil - The more turns a coil of wire has, the
stronger the magnetic field is going to be. This happens because the
magnetic field lines, called flux lines, can interact and produce an adding
effect.
• The type of core material - Coils can be wrapped around any type of
material to form an electromagnet. If it is wrapped around a nonmagnetic material, it is called an air core magnet. If it is wrapped around
a magnetic material, usually soft iron or steel, it is called an iron core
magnet. An iron core magnet will cause a greater increase in the electromagnetic field than an air core magnet.
8. Turn on the power supply and re-position the compass 12 to 16 inches above
the solenoid. Slowly lower the compass towards the solenoid until the needle
rotates 90°. Then measure the distance of the compass from the work surface.

Distance = __________________________________ (Inches/Centimeters)



It should be farther away than in step 6 (approximately 6-7 inches) because the
armature acts as an iron core and strengthens the magnetic field. Therefore,
the magnetic flux lines will reach out farther away from the solenoid.
9. Turn off the power supply.
10. Disconnect the circuit and store all components.
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OBJECTIVE 2
DESCRIBE THE FUNCTIONS OF FOUR ELECTROMAGNETIC
DEVICES
The functions of four common electromagnetic devices are:
•Solenoid
- A solenoid is an electromagnetic device that is used most often to
produce linear motion. This motion is created by an armature (or plunger)
that is inserted inside the coil. When a magnetic field is created, the armature
moves.
•Buzzer
- A buzzer is an electromagnetic device that vibrates an armature to
produce a buzzing sound.
•Motor
- A motor is an electromagnetic device that produces rotary motion.
The motor transfers that motion through a round shaft that rotates inside the
magnetic field. Motors come in various sizes from very small to very large.
They can operate on either DC or AC electricity.
•Relay
- A relay is an electromagnetic device that is basically an electricallycontrolled switch. A relay consists of two parts: a coil and one or more sets
of switch contacts. Relays are used extensively in electrical control circuits.
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Activity 2. Electromagnetic Device Operation
Procedure Overview
In this procedure, you will connect and operate the solenoid, the buzzer,
and the fan in a circuit. This activity will help you see how electromagnetic
force can be used to perform useful functions.

1. Connect the solenoid in the circuit shown in figure 4.

Notice the two bars across the top of the solenoid schematic symbol. This
indicates an iron core. If there are no lines above the solenoid symbol, it is an
air core.

In the case of the solenoid that is included with the T7017, the two bars
indicate that the armature is inserted. If the two bars are missing, the armature
should not be inserted.
+
24V
PUSHBUTTON
SWITCH
SOLENOID
Figure 4. The Solenoid Connected in a Circuit

2. Place the AC/DC selector switch in the DC position.

3. Turn on the power supply.

4. Pull the armature out of the solenoid approximately half way.
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
5. Push and hold the pushbutton. Observe the armature.
Armature status _________________________________________ (In/Out)

You should observe that the armature is pulled inside the coil.

When current is applied to the coil, an electromagnetic field is generated, and
the armature is pulled inside the coil, as shown in figure 5. When the current
is stopped, the electromagnetic field collapses.

If the armature has a spring attached to it, the armature is retracted by the
spring because the magnetic pull is no longer present. If it does not have a
spring, it must be manually retracted.
SOLENOID DEENERGIZED
SOLENOID ENERGIZED
ARMATURE
(PLUNGER)
"C" FRAME
WIRE COIL
PLUNGER
MOVEMENT
Figure 5.
MAGNETIC
FLUX
Basic Operation of a Solenoid

6. Remove the armature.

7. Now push and hold the pushbutton.

The magnetic field is still being created. However, since the armature is not
inserted into the solenoid, the field has nothing to act on. A solenoid needs an
armature to perform a useful function.

8. Release the pushbutton.

9. Turn off the power supply.

10. Disconnect the circuit.
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

11. Now connect the buzzer in the circuit shown in figure 6.
Be sure to observe the polarity of the buzzer.
+
12V
PUSHBUTTON
SWITCH
BUZZER
+
-
Figure 6. A Buzzer Connected in a Circuit

12. Turn on the power supply.

13. Press the pushbutton and listen to the buzzer.
Buzzer status _____________________________ (Sounding/Not Sounding)



The buzzer should be sounding.
14. Release the pushbutton.
Figure 7 shows the parts and operation of a buzzer.
CONTACT POINT
CONTACT STRIP
COIL
SPRING
ARMATURE
SOURCE
VOLTAGE
ES
CONTACT POINT
CONTACT STRIP
Figure 7. The Parts and Operation of a Buzzer
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
When no voltage is applied to the buzzer, the buzzer’s contact points and
contact strip make contact because of the spring. When a voltage, Es,
is supplied to the buzzer, the coil forms a magnetic field that attracts the
armature. Because of this, the contact strip is no longer making contact with
the contact points and the circuit opens. When this happens, the magnetic
field collapses and the armature returns to its original position. This starts the
process over again.

This process happens very rapidly, thus producing the buzzing sound. It will
continue until the voltage is turned off.

15. Turn off the power supply.

16. Disconnect the circuit.

17. Now connect the fan in the circuit shown in figure 8.

Be sure to observe the polarity of the fan.
+
24V
SELECTOR
SWITCH
+
M
-
Figure 8. A Fan Connected in a Circuit
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
18. Turn on the power supply.

19. Activate the selector switch. Observe the status of the fan blades.
Fan blade status ______________________________(Turning/Not Turning)

The fan blades should be turning.

A typical electric motor has at least two sets of magnetic coils. Figure 9
shows the basic components of an electric motor. The field coils remain
stationary. However, the armature coils can freely rotate within the magnetic
field created by the field coils. The motor shaft is connected to the armature.
Therefore, as the armature rotates, so does the shaft.

In a DC motor, current is supplied to both the field coils and the armature.
The opposing magnetic fields created by the field coils and armature coils is
what actually causes the rotation.
CAUTION
DC motors and AC motors are not generally interchangeable. Using the
wrong type of power source could severely damage or destroy the motor.
STATOR
(FIELD COILS)
ARMATURE
(ARMATURE COILS)
SHAFT
Figure 9. Basic Construction of a Motor

20. Turn off the selector switch.

21. Turn off the power supply.

22. Disconnect the circuit and store all components.
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SKILL 1
CONNECT AND OPERATE A RELAY IN A CIRCUIT
Procedure Overview
In this procedure, you will use a relay to control the operation of various
loads in a circuit. You will control single loads and dual loads with the relay.

1. Connect the circuit shown in figure 10.
Make sure that the output circuit is connected to the normally open and
common terminals of the relay.

SOURCE SELECT
AC
1PB
NC
DC
24V
12V
PUSH BUTTON
SWITCH
MODULE
12V
COIL
CONNECTORS
24V
C
LAMP
NO
24V
NO
LAMP
MODULE
C
NC
RELAY
MODULE
Figure 10. A Basic Relay Control Circuit
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
Figure 11 shows the basic components and operation of a relay.
FIXED
CONTACT
MOVABLE
CONTACT
(COMMON)
NORMALLY CLOSED
CONTACTS
(N.C.)
ARMATURE
NORMALLY OPEN
CONTACTS
(N.O.)
TERMINAL
FOR
COMMON
TERMINAL FOR
N.C. CONTACT
TERMINAL FOR
N.O. CONTACT
SOLENOID
COIL
SWITCH
POWER SOURCE
Figure 11. The Basic Components of a Relay

When current is not supplied to the coil, the relay is in its off or deenergized
state, as shown in figure 11. In this condition, the common terminal is
connected through the armature to the upper contact, which is the normally
closed (N.C.) contact. These contacts are said to be N.C. because they are
closed when the relay is in its de-energized or normal state.

A typical relay also has a set of normally open (N.O.) contacts. As you might
expect, this set of contacts is open while the relay is in its normal state. The
N.O. contacts are the lower contacts shown in figure 11.

When current is applied to the coil of the relay, a magnetic field is created.
This causes the armature to be pulled away from its de-energized position
and toward the coil. The movable contact on the armature then comes in
contact with the lower set of contacts (N.O.). The armature remains in this
position until the current to the coil is turned off. When the current to the coil
is removed, a spring causes the armature to return to its de-energized state.

It only takes a small amount of current to activate the coil of the relay. This
makes a relay a very important control device in many machines because you
can control high amounts of current through the contacts of a relay with a
small amount of input current to the coil. You will learn more about a relay’s
function in machine control later.

2. Perform the following substeps to operate the circuit.
A. Place the AC/DC selector switch in the DC position.
B. Turn on the power supply.
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C. Press and hold the pushbutton and observe the status of the lamp.
Lamp status _________________________________________(On/Off)
You should hear the relay click when you press the pushbutton because
the magnetic field from the coil has pulled the armature toward it. The
lamp should be on.
D. Release the pushbutton and observe the status of the lamp.
Lamp status _________________________________________(On/Off)
The lamp should now be off because the relay contacts have returned to
their normal states.
E. Press the pushbutton again and observe the motion of the relay’s armature
through the transparent housing of the relay.
Armature status __________________________ (Stays up/Pulled down)
The armature should be pulled down to make contact with the N.O.
contacts, completing the lamp circuit.
F. Release the pushbutton and observe the armature.
Armature status(Pushed up/Pulled down)
The armature should be pushed up by the spring to its normal state, making
contact with the N.C. contacts.

3. Perform the following substeps to test the other output devices in the circuit.
This will show that a relay can switch circuits with any of these output
devices.
A. Replace the lamp with a fan and repeat step 2.
B. Replace the fan with a solenoid and repeat step 2.


4. Turn off the power supply and connect the circuit shown in figure 12.
This circuit controls two different circuits. Notice that one output circuit is
connected to the normally open contacts, while the other output circuit is
connected to the normally closed contacts.
LAMP 1
NC
24V
C
NO
24V
LAMP 2
Figure 12. Relay Circuit
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A switch’s or relay’s contacts can be described by the number of poles and
throws they have. A pole refers to an independent or isolated circuit in the switch.
The circuit is isolated because current can pass through its contacts without
affecting other contacts in the switch. A throw is a specific closed contact position
to which a pole can be set (N.O. or N.C.).
Most of the switches that you have used so far have had only one set of
contacts with one pole and either one or two throws. Relays can have multiple sets
of contacts with two or more poles and throws. This allows them to control several
different circuits at one time.
Two common types of switch contacts used in relays are:
•Single-pole, double-throw (SPDT) - This type has a single pole that switches
back and forth between contacts (N.O. and N.C.), such as the one shown in
figure 10.
•Double-pole,
double-throw (DPDT) - This type has two poles that switch
back and forth between contacts (N.O. and N.C.). This type has two individual sets of N.O. contacts and two individual sets of N.C. contacts. The
schematic symbols for both types are shown in figure 13.
SINGLE-POLE
DOUBLE-THROW (SPDT)
RELAY
DOUBLE-POLE
DOUBLE-THROW (DPDT)
RELAY
NC
THROW
C
POLE
NO
NC
NC
C
COIL
CONTACTS
C
NO
NO
Figure 13. Schematic Symbols for an SPDT and DPDT Relay
NOTE
The relay that is supplied with the T7017 has double-pole, double-throw
contacts.
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
5. Perform the following substeps to operate the relay control circuit.
A. Turn on the power supply.
B. Push and hold the pushbutton and observe the status of both lamps.
Lamp 1 status ________________________________________(On/Off)
Lamp 2 status ________________________________________(On/Off)
Lamp 2 should be on because it is connected to the normally open
contacts. Lamp 1 should be off because it is connected to the normally
closed contacts, which are now open.
C. Release the pushbutton and observe the status of the lamps.
Lamp 1 status ________________________________________(On/Off)
Lamp 2 status ________________________________________(On/Off)
Lamp 2 should now be off and lamp 1 should be on.

6. Turn off the power supply.

7. Disconnect the circuit and store all components.

You have now successfully operated a relay in a circuit.
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SEGMENT 1
SELF REVIEW
1. When electrons flow through a conductor, a(n) _________________ is
created around the conductor.
2. Winding a conductor into a(n) ____________________ will cause the
magnetic field to become stronger.
3. ______________ devices use a magnetic field to produce a desired
output.
4. A(n) _________________ is most often used to produce linear motion.
5. The buzzing sound produced by a buzzer is caused by the ___________
of the armature as the magnetic field is turned on and off rapidly.
6. A motor produces _________ motion.
7. A(n) ____________________ is basically an electrically-controlled
switch.
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SEGMENT 2
INDUCTANCE
OBJECTIVE 3
DEFINE INDUCTANCE AND GIVE ITS UNITS OF MEASUREMENT
Inductance is a characteristic of electromagnetic devices that opposes any
changes in current flow.
Inductance is measured in the unit henrys and is abbreviated with a capital h
(H). A henry is a fairly large unit, so you will usually see inductance listed in millihenrys (1/1000 Henrys). The abbreviation for millihenrys is mH.
OBJECTIVE 4
DESCRIBE THE OPERATION OF AN INDUCTOR AND GIVE
ITS SCHEMATIC SYMBOL
An inductor is created by winding an electrical wire into a coil. This means all
of the electrical devices discussed in Segment 1 are inductors. However, there is
also a specific component called an inductor, as shown in figure 14.
SCHEMATIC
SYMBOL
Figure 14. Typical Inductor and its Schematic Symbol
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The purpose of an inductor is to resist changes in current flow. Because of
this, inductors are often referred to as chokes. This feature is very useful in many
applications, as you will learn.
To understand how the inductor can be used, it is first useful to understand how
it works. The reason an inductor resists changes in current flow is that whenever
current through an inductor is changing, a voltage is induced in the inductor. The
faster the current is changing, the larger the magnitude of the induced voltage. This
induced voltage opposes the source voltage that is creating the current flow (this is
why the induced voltage is often referred to as counter emf or back emf).
It is important to note that the induced voltage is only present when the current
is changing. Each level of current flow creates an electromagnetic field around the
inductor. The strength of the field is proportional to that particular level of current.
When current is increasing, the induced voltage slows the current change and the
electromagnetic field builds. When current is decreasing, the induced voltage tries
to keep current constant by using the energy stored in the electromagnetic field
while it is collapsing.
Inductors have different effects in DC and AC circuits, as you will see in the
following objectives. Inductors are mathematically represented by L.
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OBJECTIVE 5
DESCRIBE THE EFFECT OF AN INDUCTOR IN A DC CIRCUIT
AND GIVE AN APPLICATION
Inductance affects a DC circuit only when the current is first turned on, turned
off, or when there is a change in the load resistance. These three conditions are the
only times the current is changing in a DC circuit.
As an example, when voltage is first applied to the circuit of figure 15, the
rising current level creates an opposing (induced) voltage in the inductor that
slows the increase in current. This opposing effect lasts only while the current
is rising (changing) to its steady state value. After the current has stabilized (no
longer changing), the current through the inductor is determined solely by the DC
resistance of the inductor. An electromagnetic field exists around the inductor, the
strength of which is proportional to the steady state current level.
INDUCTOR
R
12V
L
Figure 15. An Inductor in a DC Circuit
If the resistance of the circuit load in figure 15 increases for some reason, the
inductor will react. The electromagnetic field around the inductor collapses trying
to maintain the current at a constant level. After the field’s energy is gone, the
current remains at this new level until the circuit again changes.
A common application of inductance in a DC circuit is in an automobile’s ignition system. In this application, inductance is used to create a high voltage across
the spark plugs, which causes the spark to occur.
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This is accomplished by changing the resistance in the circuit suddenly to a
much higher value. To understand how this works, examine the schematic diagram
of an ignition circuit shown in figure 16.
A
R1 = 100
B
POINTS
C
SPARK
PLUG
STARTER
COIL
12V
BATTERY
L=2H
CONTACTS
R2 = 250K
Figure 16. A Car’s Ignition System
In this circuit, a switch is used to represent the “points” in the ignition system
that open and close to allow the coil to charge and then to discharge through the
spark plug contacts. R1 represents the resistance of the starter coil windings. R2
represents the resistance of gap between the contacts of the spark plug. Actually,
the resistance between the spark plug contacts is infinite since a spark plug is an
open circuit. But, we will assign to it a very high resistance of 250 K W to do the
calculations.
When the switch is in the A position, the current level is determined by the
resistance of R1 and the voltage of the battery. In this example, the current is 0.12
Amps (I = E /R1 = 12/100) using Ohm’s Law. This is the charge condition.
When the switch is changed to the B position, the resistance in the circuit is
suddenly much higher. This would normally cause the current to be reduced except
that the inductor tries to oppose this change and keep the current the same.
To do this, the inductor discharges, causing a voltage across the spark plug
contacts. The voltage caused by the inductor in figure 16, for example, will be
30,000 V. This is calculated by Ohm’s Law with the present current level, 0.12A,
and the new resistance of 250,000 ohms (E = 0.12 × 250,000 = 30,000 V).
This very high voltage causes a spark to jump the gap of the spark plug. This
spark is used to ignite the fuel/air mixture in the cylinder of the car’s engine.
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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Activity 3. Effect of Inductance in a DC Circuit
Procedure Overview
In this procedure, you will use Ohm’s law to calculate the steady state
current of an inductor in a DC circuit. You will then connect an inductor to a
DC source and verify that the steady state current in an inductor is determined
solely by its DC resistance. You will also vary the inductance of the inductor to
see what effect it has on the current.
NOTE
The change in current through the inductor happens so quickly that the
T7017 test equipment will not be able to show this concept. An oscilloscope
would be required to see this actually happening.

1. Measure the DC resistance of the solenoid with the DMM. Record this value
below.
DC resistance of solenoid = _______________________________ (Ohms)


You should find that the resistance is approximately 90 ohms.
2. Use Ohm’s Law to calculate the theoretical current through the solenoid
using a 24VDC source. Record this value below.
Theoretical current = ____________________________________ (Amps)

You should find a theoretical current of approximately 0.267 A.
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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
3. Connect the circuit shown in figure 17.
A
SOURCE SELECT
DC
+
AC
24V
-
24V
12V
12V
SOLENOID
30XR
DC
AMPS
NON
CONTACT
VOLTAGE
MIN MAX
V
200
HOLD
600 OFF 600
V
200
20
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
V
COM
10A
CAT
CAT
A
600V
300V
BATT 9V
200mA
MAX
FUSED
10A MAX
FUSED
SOLENOID
MODULE
MAX
600V
600V
Figure 17. DC Current Measurement
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
4. Perform the following substeps to operate the circuit.
A. Place the AC/DC selector switch in the DC position.
B. Set the DMM to measure DC current.
C. Turn on the T7017 power supply and observe the current reading.
Actual current = ______________________________________ (Amps)
Compare this value to the one calculated in step 2. They should be close.
Any difference may be due to a slightly lower actual 24VDC source.
In activity 1, we proved that the electromagnetic field was stronger when
the solenoid armature was in the solenoid. Since the magnetic field was
stronger, this means that we can actually vary the inductance of this solenoid depending on how much of the armature is actually inside the coil.
In the next two steps, you will vary the inductance of the solenoid to see
what effect this has on the current.

5. Turn off the power supply, grasp the armature and pull it out of the solenoid
coil.

6. Turn the power supply back on and allow the solenoid coil to slowly pull the
armature into the coil while you observe the current displayed on the DMM.
Record below whether the current changes significantly as you do this.
Current changes significantly _____________________________ (Yes/No)



You should find that the current does not change significantly. Remember that
the current in the circuit must be changing before a change in inductance will
effect the circuit. Since the source is DC and the current level has stabilized,
changing the inductance has no effect. Steady state current in a DC circuit is
solely dependent on the DC resistance of the coil.
7. Turn off the power supply.
Leave the circuit connected, it will be used in the next activity.
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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OBJECTIVE 6
DESCRIBE THE EFFECT OF AN INDUCTOR IN AN AC CIRCUIT
AND GIVE AN APPLICATION
Because the current in an AC circuit is continually changing, an inductor has
a different effect in an AC circuit than it does in a DC circuit. In an AC circuit,
the induced voltage across the inductor always resists this change. This constant
resistance is called inductive reactance. It is expressed in ohms and can be found
with the following formula:
INDUCTIVE REACTANCE FORMULA
XL = 2πfL
where
XL =
f =
π =
L =
inductive reactance (ohms)
AC frequency (hertz)
pi (3.1416)
inductance (henrys)
There are five factors that determine the amount of inductive reactance an
inductor will introduce into a circuit. Changing any of these factors will change the
inductive reactance of the inductor.
These five factors are:
•The
number of turns in the coil - The more turns the coil has, the greater
the inductive reactance.
•The
diameter of the coil - The larger the diameter, the greater the inductive
reactance.
•The length of the coil - The longer the coil, the greater the inductive reactance.
•The
type of core - An iron core produces a greater inductive reactance than
an air core.
•The
frequency of the alternating current - The higher the frequency of the
AC, the greater the inductive reactance. In this respect, inductive reactance
differs from resistance because the resistance of the resistor is the same at
any frequency.
A common application of an inductor in an AC circuit is a fluorescent light
fixture. The fluorescent light fixture uses a special inductor called a ballast. You
will learn more about this later.
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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Activity 4. Effect of Inductive Reactance in an AC Circuit
Procedure Overview
In this procedure, you will operate a solenoid coil in an AC circuit and
verify the effects of inductive reactance by measuring the current. You will then
calculate the actual inductance of the solenoid using the inductive reactance
formula you just learned. You will also observe the effects of varying the
inductance of the solenoid and its effect on current in an AC circuit.

1. Connect the circuit shown in figure 18.
A
SOURCE SELECT
AC
DC
24
VAC
24V
12V
12V
SOLENOID
30XR
AC
AMPS
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
V
COM
10A
CAT
CAT
A
600V
300V
SOLENOID
MODULE
BATT 9V
200mA
MAX
FUSED
10A MAX
FUSED
MAX
600V
600V
Figure 18. AC Current Measurement
NOTE
Verify that the armature is inserted all the way inside of the solenoid,
pointed end first.
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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

Ohm’s Law was used in the previous activity to calculate a theoretical current
of approximately 0.267Amps. If it were not for inductive reactance, the AC
circuit you just connected would give you the same result. In the next step,
you will see that the inductive reactance of the coil dramatically changes the
value of current which will flow in the circuit.
2. Perform the following substeps to operate the circuit.
A. Place the AC/DC selector switch in the AC position.
B. Set the DMM to measure AC current. You should use the mA input.
C. Turn on the power supply and observe the current reading.
Actual Current = _______________________________________ (mA)
You should read approximately 23mA of actual current. This is much less
than the theoretical value which we calculated in the previous activity.
This added resistance is the result of the induced voltage in the solenoid
opposing the source. Remember, the current in an AC circuit is constantly
changing, which means the opposing voltage (or back emf) is always
present.
In the next few steps, you will use values you measured with the formula
for inductive reactance to calculate the actual inductance of the solenoid
coil.

3. Measure the voltage across the solenoid using the analog voltmeter on the
trainer.
Voltage across the solenoid = _______________________________ (VAC)
You should measure approximately 28VAC.

4. Use the current value obtained in step 2 and the voltage value you just
measured in step 3 with Ohm’s Law to calculate the actual inductive reactance
of the solenoid.
XL =
V
I
Inductive Reactance, XL = ________________________________ (Ohms)

You should obtain a value of approximately 1217 ohms.
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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

5. Use the inductive reactance value you just calculated in step 4 to determine
the actual inductance of the solenoid coil. Record your answer below. Use the
appropriate frequency (50 Hz or 60 Hz) for your location.
The inductive reactance formula can be re-arranged as shown to solve for
inductance.
L=
where
XL
f
π
L
=
=
=
=
XL
2π f
inductive reactance (ohms)
AC frequency (hertz)
pi (3.1416)
inductance (henrys)
Actual inductance of solenoid coil = _______________________ (henrys)



You should obtain an answer of approximately 3.8 henrys for 50 Hz or 3.2
henrys for 60 Hz.
6. You will now observe the effects of varying the inductance of the solenoid in
an AC circuit.
Grasp the solenoid armature and slowly pull it out of the coil while you
observe the current displayed by the DMM. Does it change?
Current Change _______________________________________ (Yes/No)



The current should vary with the position of the armature. In the previous
activity, you saw that varying the inductance in a DC circuit had no effect
because the current was not changing. Since you are now using AC (the
current is continuously changing), changing the inductance affects the circuit.
7. Turn off the power supply.
In the next activity, you will be analyzing series and parallel combinations
of inductors. Complete the following steps to obtain the actual inductance of
one of the primary coils of the transformer module. You will need this value
to perform the next activity.
NOTE
The transformer will be covered in a later LAP. You will only be using it
as an inductor in this LAP.
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
8. Replace the solenoid in your circuit with one of the primary coils of the
transformer module as shown in figure 19.
SOURCE SELECT
AC
DC
24V
12V
12V
STEP DOWN
TRANSFORMER
MODULE
30XR
AC
AMPS
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
V
COM
10A
CAT
CAT
A
600V
300V
BATT 9V
SIDE LABELED
"PRIMARY"
200mA
MAX
FUSED
10A MAX
FUSED
MAX
600V
600V
Figure 19. Circuit for Step 8

9. Repeat steps 2 through 5 with the transformer module, record your data
below.
Current through transformer primary = ________________________ (mA)
Voltage across transformer primary = _________________________ (VAC)
Inductive reactance, XL = _________________________________ (Ohms)
Actual inductance of transformer coil = _____________________ (henrys)
NOTE
The answer you obtain for an actual inductance will vary slightly from
approximately 2.3-2.5 henrys @ 60 Hz or 2.8-3.0 henrys @ 50 Hz. The
transformer is manufactured with the turns ratio as the primary concern. Actual
inductance values will vary.

10. Turn off the power.

11. Disconnect and store all components.
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OBJECTIVE 7
STATE THE FORMULAS FOR CALCULATING TOTAL SERIES
INDUCTANCE AND INDUCTIVE REACTANCE
Like resistors, inductors are usually manufactured in certain standard values.
Sometimes an application may require an inductance value that is not a standard
value. When this occurs, it may be necessary to combine inductors in series or
parallel to achieve the desired value.
Since inductors are similar to resistors, in that they tend to oppose or resist
current, connecting inductors in series or parallel will have similar results to
connecting resistors in series. The inductive reactance that the inductors introduce
in the circuit adds to the total load of the circuit in the same manner as adding more
resistors.
When inductors are connected in series, the total inductance equals the sum of
the individual inductances.
TOTAL SERIES INDUCTANCE FORMULA
LT = L1 + L2 + L3 …
where
LT = total inductance in Henrys
L1, L2, L3 = individual inductances in Henrys
The same is also true for the total inductive reactance.
TOTAL SERIES INDUCTIVE REACTANCE FORMULA
XLT = XL1 + XL2 + XL3 …
where
XLT = total inductive reactance in Ohms
XL1, XL2, XL3 = individual inductive reactances in Ohms
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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OBJECTIVE 8
STATE THE FORMULAS FOR CALCULATING TOTAL PARALLEL
INDUCTANCE AND INDUCTIVE REACTANCE
When inductors are connected in parallel, the total inductance and inductive
reactance of the circuit are reduced according to the following formulas.
TOTAL PARALLEL INDUCTANCE FORMULA
1
1
1
+
+
L1 L2 L3
where
LT = total inductance in Henrys
L1, L2, L3 = individual inductances in Henrys
TOTAL PARALLEL INDUCTIVE REACTANCE FORMULA
X LT =
1
1
1
1
+
+
X L1 X L 2 X L3
where
XLT = total inductive reactance in Ohms
XL1, XL2, XL3 = individual inductive reactances in Ohms
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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SKILL 2
CALCULATE THE TOTAL LOAD ON AN AC CIRCUIT
WITH INDUCTORS
Procedure Overview
In this procedure, you will calculate the total inductance and inductive
reactance in both series and parallel circuits given the individual inductance
values.
You will also use the inductance values you obtained in the previous
activity for the solenoid and transformer to calculate the theoretical current
that will flow when they are connected in series and parallel. You will then
measure actual current to verify your calculations.

1. Calculate the inductive reactance for the circuit shown in figure 20.
XL = __________________________________________________ (Ohms)

The solution is as follows:

Step 1 - Calculate the total inductance as follows:
LT = L1 + L2 + L3
LT = 6H + 4H + 5H
LT = 15H

The total inductance is 15 henrys.

Step 2 - Using the total inductance found in step 1, calculate the total inductive
reactance as follows:
XL = 2πfL
XL = 2 × 3.1416 × 60 × 15
XL = 5655 Ohms
L1 =6H
24 VAC
60Hz
L2 =4H
L3 =5H
Figure 20. Series Inductors
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

The total inductive reactance is 5,655 ohms.
2. Calculate the total inductance and total inductive reactance of the circuit in
figure 20 if the inductor values are 50H, 10H and 25H.
The total inductance= ____________________________________ (henrys)
The total inductive reactance = ____________________________ (Ohms)


Total inductance is 85 henrys and reactance is 32,044 ohms.
3. Calculate the total inductance and total inductive reactance of the circuit in
figure 20 if the frequency is 50 Hertz.
The total inductance = ___________________________________ (henrys)
The total inductive reactance = _____________________________ (Ohms)


The total inductance is still 15 henrys but the reactance is now 4,712 ohms.
4. Calculate the total inductance and total inductive reactance of the circuit
in figure 21. Use the actual inductance values you obtained in the previous
activity for the solenoid and one primary winding of the transformer. Use the
appropriate frequency for your location.
The total inductance @ 50 Hz = ____________________________ (henrys)
The total inductance @ 60 Hz = ___________________________ (henrys)
The total inductive reactance @ 50 Hz = _____________________ (Ohms)
The total inductive reactance @ 60 Hz = _____________________ (Ohms)
A
24 VAC
60Hz
L1
SOLENOID
L2
TRANSFORMER
PRIMARY
Figure 21. Two Inductors in Series


In the next few steps, you will verify the calculations you made in step 4 by
connecting the two inductors in series and measuring the resulting current.
5. Connect the circuit shown in figure 21.
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
6. Use Ohm’s law and the total inductive reactance you calculated in step 4 to
now calculate the theoretical current that should be displayed by the ammeter
when you energize the circuit.
NOTE
The T7017 supply is rated at 24 VAC when it is under full load. Since this
circuit will not fully load the supply, use 28 VAC as a source voltage for this
calculation.
I=
V
XL
Theoretical current @ 50 Hz = _______________________________ (mA)
Theoretical current @ 60 Hz = ______________________________ (mA)

7. Perform the following substeps to operate the circuit.
A. Turn on the DMM and set it to measure AC amps. Use the mA input.
B. Place the AC/DC selector switch in the AC position.
C. Turn on the power supply and observe the displayed current value on the
DMM.
Current = _____________________________________________ (mA)


8. Compare the theoretical value calculated in step 6 with the value now
displayed on the DMM.
Are the two values close ?
_____________________________________________________ (Yes/No)



The two values should be close.
9. Turn off the power supply.
10. Calculate the total inductive reactance of the parallel inductance circuit
shown in figure 22, using the appropriate frequency for your location.
XL = __________________________________________________ (Ohms)
24 VAC
60 Hz
L1 =2.5H
L2 =1.8H
L3 =1.2H
Figure 22. Inductors in Parallel
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
The solution is as follows:

Step 1 - Calculate the total inductance as follows:
LT =
1
1
L1
LT =
+
+
L2
1
L3
1
1
2.5
LT =
1
+
1
1.8
+
1
1.2
1
0.4 + 0.55 + 0.83
LT = 0.56 henry

The total inductance is 0.56 henry.

Step 2 - Using the total inductance found in step 1, calculate the total inductive
reactance as follows:
For 50 Hz
XL = 2πfL
XL = 2 × 3.1416 × 50 × 0.56
XL = 176 ohms
For 60 Hz
XL = 2πfL
XL = 2 × 3.1416 × 60 × 0.56
XL = 211 ohms


The total inductive reactance is 176 ohms for 50 Hz and 211 ohms for 60 Hz.
11. Calculate the total inductance and total inductive reactance of the circuit in
figure 22 if the inductor values are 5H, 25H and 7H.
The total inductance = ___________________________________ (henrys)
The total inductive reactance @ 50 Hz = _____________________ (Ohms)
The total inductive reactance @ 60 Hz = _____________________ (Ohms)

The total inductance is 2.63 henrys and the inductive reactance is 826.24
ohms at 50 Hz and 991.49 at 60 Hz.
NOTE
A good application of this calculation is to determine the power load on
a control circuit that has many solenoids controlling fluid power values or
electric motor starters. These solenoids are always connected in parallel. Most
manufacturers, however, make this easy for you by specifying the current draw
for the solenoid at a certain voltage level.
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
12. Calculate the total inductance and total inductive reactance of the circuit
in figure 23. Use the actual inductor values you obtained in the previous
activity for the solenoid and one primary winding of the transformer. Use the
appropriate frequency for your location.
The total inductance @ 50 Hz = ____________________________ (henrys)
The total inductance @ 60 Hz = ___________________________ (henrys)
The total inductive reactance @ 50 Hz = _____________________ (Ohms)
The total inductive reactance @ 60 Hz = _____________________ (Ohms)
A
24 VAC
60 Hz
L1
SOLENOID
L2
TRANSFORMER
PRIMARY
Figure 23. Two Inductors in Parallel

In the next few steps, you will verify the calculations you made in step 12 by
connecting the two inductors in parallel and measuring the resulting current.

13. Connect the circuit shown in figure 23.

14. Use Ohm’s law and the appropriate total inductive reactance you calculated
in step 12 to now calculate the theoretical current that should be displayed by
the ammeter when you energize the circuit.
NOTE
The T7017 supply is rated at 24 VAC when it is under full load. Since this
circuit will not fully load the supply, use 26.5 VAC as a source voltage for this
calculation.
I=
V
XL
Theoretical current @ 50 Hz = _______________________________ (mA)
Theoretical current @ 60 Hz = ______________________________ (mA)
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
15. Perform the following substeps to operate the circuit. Use the mA input.
A. Turn on the DMM and set it to measure AC amps.
B. Place the AC/DC selector switch in the AC position.
C. Turn on the power supply and observe the displayed current value on the
DMM.
Current = _____________________________________________ (mA)


16. Compare the theoretical value calculated in step 12 with the value now
displayed on the DMM.
Are the two values close?
_____________________________________________________ (Yes/No)

The two values should be close.

17. Turn off the power supply.

18. Disconnect and store all components.
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SEGMENT 2
SELF REVIEW
1. __________ is the characteristic of an electromagnetic device that
opposes any change in current flow.
2. Inductance is measured in units called _______________.
3. There are __________ factors that determine the amount of inductive
reactance an inductor will introduce into a circuit.
4. When inductors are connected in ______________, the total inductance
equals the sum of the individual inductances.
5. A(n) ______________ in a fluorescent light fixture is a special type of
inductor.
6. The resistance of a car’s spark plug is _____________ since a spark
plug is an open circuit.
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40
SEGMENT 3
CAPACITANCE
OBJECTIVE 9
DEFINE CAPACITANCE AND GIVE ITS UNITS OF
MEASUREMENT
Capacitance is the measure of a device’s ability to store electrical energy. The
basic unit of capacitance is the farad, which is usually abbreviated with a capital
f (F).
Since a farad is such a large unit, you will see most values listed in microfarads, which means millionths of a farad (0.000001 F).
OBJECTIVE 10
DESCRIBE THE OPERATION OF A CAPACITOR AND
GIVE ITS SCHEMATIC SYMBOL
A capacitor is a device that stores electrical energy. When a voltage source is
applied across a capacitor, the capacitor begins storing energy. As it does this, the
voltage across its leads increases until it reaches the same voltage as the source
voltage. This stored energy is called an electrostatic charge.
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Figure 24 shows a typical capacitor and its schematic symbol. Capacitors are
mathematically represented by C.
Figure 24. A Typical Capacitor and its Schematic Symbol
As shown in figure 25, a capacitor is constructed by separating two metal
plates with an insulating material. This insulating material is called the dielectric.
When voltage is applied to the capacitor, the electrostatic charge is stored in the
dielectric.
WIRE
METAL
PLATES
DIELECTRIC
INSULATING
MATERIAL
WIRE
Figure 25. Construction of a Capacitor
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OBJECTIVE 11
DESCRIBE THE FUNCTIONS OF THREE TYPES OF
CAPACITORS
Capacitors are available in three basic styles:
• Nonpolarized (used in both AC and DC circuits)
• Polarized or electrolytic (polarity must be observed, used mainly in DC
circuits)
• Variable (can be adjusted)
Nonpolarized
Nonpolarized capacitors are not polarity sensitive. It does not matter which
way they are connected in a circuit. Nonpolarized capacitors have no polarity
markings and usually have one lead on each end. Figure 26 shows some typical
nonpolarized capacitors. The schematic symbol in figure 24 is the symbol for a
nonpolarized capacitor.
Figure 26. Nonpolarized Capacitors
There are several types of nonpolarized capacitors including paper, ceramic,
polystyrene and mica. Most of these are less than 1mF and are used in filtering and
coupling applications. They are available in a wide range of tolerances depending
on the application. Timing applications, which require a very accurate value of
capacitance use special close- tolerance versions of 1% or less.
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Electrolytic
Polarized or electrolytic capacitors usually have markings to let you know the
polarity of their leads. The negative lead is usually marked with a minus (-). The
positive lead will always be the longer of the two leads.
Care must be used when installing these capacitors. If installed backwards, the
capacitor will become very hot and may even explode! Most electrolytic capacitors
have both leads on the same end. Figure 27 shows some typical electrolytic capacitors and the schematic symbol for an electrolytic capacitor.
SCHEMATIC
SYMBOL
+
Figure 27. Electrolytic Capacitors and their Schematic Symbol
Electrolytic capacitors provide the most capacitance in the smallest space and
at the least cost. They are available with capacitance values ranging from 10μF to
thousands of m farads. Electrolytics are used in most DC power supplies to provide
a smooth DC output level. Since they are most often used in filtering applications,
their tolerances fluctuate widely. Larger electrolytic capacitors have typical tolerances of -10 to +50%, meaning that a 10,000μF capacitor has an actual value that
can range from 9,000 to 15,000μF.
Special non-polarized versions of these capacitors are available which allow
you to use them in AC applications. The motor start capacitor is the most common
of these types and is the type supplied with the T7017 trainer. Motor start capacitors are manufactured with a value range rather than a tolerance. The capacitors
supplied with the T7017 have a range of 88 to 106μF.
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Variable
Variable capacitors are much like variable resistors in that their value can be
adjusted by turning a knob. Small variable capacitors, called trimmer capacitors,
are adjusted by using a screwdriver to turn the adjustment screw. Figure 28 shows
some variable capacitors along with the variable capacitor’s schematic symbol.
SCHEMATIC
SYMBOL
+
Figure 28. Variable Capacitors and Their Schematic Symbol
Regardless of the type of capacitor, the higher the capacitance, the larger the
amount of charge the capacitor can store. The three factors that determine the
capacitance value of a capacitor are:
•The
area of the plates - The larger the surface area is, the larger the capacitance is.
•The
distance between the plates - The farther apart the plates are, the less
the capacitance will be.
•The
type of dielectric used - Some materials have the ability to create a
stronger electrical field than others, which causes higher capacitance.
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SKILL 3
DISCHARGE A CAPACITOR
Procedure Overview
Capacitors have the ability to deliver an extremely high level of current
when charged. If you touch both leads at the same time, you could receive
a serious electrical shock. Therefore, before connecting or disconnecting a
capacitor, it is a good safety practice to discharge the capacitor. Large value
capacitors should be discharged through a resistor to prevent damaging the
dielectric or welding of the discharging device. Most large capacitors are
usually installed with a bleeder resistor connected across their terminals for
this reason. The capacitors you will be using are relatively small and can be
discharged by shorting the terminals with an insulated test lead.
In this procedure, you will use one of the lead wires supplied with the
T7017 to discharge a capacitor. Since the capacitors are mounted on module
blocks, you can discharge the capacitor by connecting the wire across the
connection posts on the module.

1. Discharge the capacitor by touching the connection posts with the lead wire,
as shown in figure 29. If there is any charge, you should notice a spark when
the wire touches the posts.
CAUTION
Be sure that you are holding the wire by the insulation when you discharge
the capacitor to avoid getting shocked.
Spark _______________________________________________ (Yes/No)

There may or may not be a spark depending on whether or not the previous
user left a charge on the capacitor.
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CAPACITOR
MODULE
HOLD THE WIRE
BY THE INSULATION
Figure 29. Discharging a Capacitor with a Test Lead Wire

2. Connect the capacitor to the power supply through a knife switch, as shown
in figure 30.
CAPACITOR
MODULE
24V
12V
12V
KNIFE
SWITCH
MODULE
Figure 30. Connecting a Capacitor

3. Place the AC/DC selector switch in the DC position.

4. Close the switch.

5. Turn on the power supply and leave it on for a few seconds to allow the
capacitor to charge.

6. Open the switch.

7. Turn off the power supply.

8. Repeat step 1 and observe if there is a spark.
Spark ________________________________________________ (Yes/No)


There should be a spark this time since we know the capacitor was charged.
9. Disconnect the capacitor and store it.
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SKILL 4
TEST A CAPACITOR WITH A DMM
Procedure Overview
A capacitor can be tested by measuring its resistance with an ohmmeter.
This allows you to determine if the dielectric is shorted. If the capacitor is
good, the display will drop to a fairly low resistance reading and then start to
climb rapidly until it reaches overload. If the resistance value is close to zero,
and does not change, the dielectric is shorted and the capacitor is bad.
In this procedure, you will test a capacitor to determine if it is good using
a DMM.

1. Locate both of the capacitor modules in the T7017 and discharge them both.

2. Set the DMM to measure resistance and test each capacitor as shown in figure
31.
Capacitor 1 _________________________________________ (Good/Bad)
Capacitor 2 _________________________________________ (Good/Bad)

If the resistance goes to a low value and then climbs to overload, the capacitor
is good. If the resistance goes to a value near zero and stays, the dielectric is
shorted and the capacitor is bad.
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CAPACITOR
MODULE
30XR
NON
CONTACT
VOLTAGE
MIN MAX
V
200
HOLD
600 OFF 600
V
200
20
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
CAT
CAT
V
COM
A
600V
300V
BATT 9V
200mA
MAX
FUSED
10A MAX
FUSED
MAX
600V
600V
Figure 31. Testing a Capacitor with a DMM


3. Repeat step 2 again to confirm your results.
If you find a bad capacitor, inform your instructor immediately.

4. Discharge the capacitors.

5. Store the capacitors.
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SEGMENT 3
SELF REVIEW
1. A(n) _________ is a device that can store electrical energy.
2. A(n) _____________ capacitor can be used in AC or DC circuits.
3. ____________ capacitors provide the most capacitance in the smallest
space and at the least cost.
4. The measure of a capacitor’s ability to store energy is called
____________.
5. Capacitance is measured in units called ________.
6. The insulating material between the plates in a capacitor is called a(n)
___________.
7. It is good practice to ___________ a capacitor to avoid a possible
severe shock.
8. A capacitor can be tested by measuring its resistance with an
ohmmeter, which allows the user to determine if the dielectric is
____________________.
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SEGMENT 4
CHARACTERISTICS OF CAPACITANCE
OBJECTIVE 12
DESCRIBE THE EFFECT OF A CAPACITOR IN A DC CIRCUIT
AND GIVE AN APPLICATION
There are two effects a capacitor has in a DC circuit. First, a capacitor charges
to a voltage that is equal to the source voltage.
Second, since the dielectric is an insulator and the voltage potential across the
capacitor is equal to the source, a charged capacitor acts like an open and will not
allow current to flow in the circuit.
Capacitor Voltage Effect
If a DC voltage is applied to a capacitor, as shown in figure 32, current is
not able to cross the dielectric. A surplus of electrons then builds up on the plate
connected to the negative terminal of the voltage source. On the plate connected to
the positive terminal, there will be a shortage of electrons.
SHORTAGE OF
ELECTRONS
+
12V
-
+
-
CAPACITOR
SURPLUS OF
ELECTRONS
Figure 32. A Capacitor in a DC Circuit
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These electrons continue to build up on the negative side of the capacitor until
the voltage across the plates of the capacitor is equal to the source voltage. This
means the capacitor has stored a charge.
Because of this fact, a charged capacitor can be used as a voltage source for
other circuits. The only problem is that it usually can supply only a very short
burst of voltage. Once the capacitor is fully discharged, it has no more voltage.
One common application of a capacitor in a DC circuit is in the electronic flash
of a camera. The capacitor stores a charge and then discharges when the button is
pressed.
Capacitor Current Effect
If you construct a DC circuit with a capacitor and lamp, such as the one in
figure 33, current flows to the lamp for only a split second and then stops. The
current flows only long enough to charge the capacitor. When the capacitor is fully
charged, the current stops because the charge of the capacitor is equal to the source
voltage. Therefore, the charged capacitor acts like an open circuit in a DC circuit.
CAPACITOR WILL NOT
LET CURRENT FLOW
+
LAMP WILL NOT
LIGHT
Figure 33. The Effect of a Capacitor in a DC Circuit
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SKILL 5
MEASURE THE VOLTAGE ACROSS A CHARGED CAPACITOR
Procedure Overview
In this procedure, you will connect a capacitor to the power supply of the
T7017 and allow it to charge. You will then disconnect the capacitor from the
source and measure the voltage across the capacitor plates with the DMM.

1. Locate one of the capacitor modules and discharge it with a test lead wire.

2. Connect the capacitor to the T7017 power supply, as shown in figure 34.
SOURCE SELECT
AC
DC
24V
12V
12V
CAPACITOR
MODULE
Figure 34. The Effect of a Capacitor in a DC Circuit

3. Select DC with the AC/DC selector switch and turn on the power supply.

4. Wait a few seconds for the capacitor to charge and then turn off the power
supply.

5. Carefully disconnect the capacitor from the power supply.

6. Set the DMM to measure DC volts and then measure the voltage across the
capacitor.
Capacitor full charge voltage = ______________________________ (Volts)


The voltage should initially be very close to the source voltage and then start
to drop off as the capacitor discharges through the DMM.
7. Discharge the capacitor with a lead wire.
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Activity 5. Effect of a Capacitor in a DC Circuit
Procedure Overview
In this procedure, you will connect a capacitor in various DC circuits.
In the first circuit, the capacitor will be connected to serve as a momentary
voltage source. In the second circuit, the capacitor will be connected in series
with a lamp to show how a charged capacitor will block current.

1. Connect the circuit shown in figure 35.
SOURCE SELECT
AC
DC
24V
12V
12V
SELECTOR
SWITCH
PUSHBUTTON
SWITCH
+
12V
SELECTOR
SWITCH
PUSH BUTTON
SWITCH
MODULE
C
-
+
BUZZER
+
CAPACITOR
MODULE
-
BUZZER
MODULE
Figure 35. Pictorial and Schematic of Experiment Setup

2. Place the AC/DC selector switch in the DC position and turn on the power
supply.

3. Rotate the selector switch to the right (ON) to allow the charge circuit to
charge the capacitor.

4. Rotate the selector switch back to the OFF position and then press and hold
the pushbutton. Note the status of the buzzer.
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Buzzer status _____________________________ (Sounds/Does not sound)

It should sound for a short time as the capacitor discharges through it. It should
start out fairly loud and then decrease in volume as the voltage decreases.

5. Release the pushbutton.

6. Repeat steps 3-5 to become more familiar with the operation of this circuit.
NOTE
The DMM shown in figure 36 is a generic representation of a DC ammeter.
A
SOURCE SELECT
+
AC
12V
DC
C
24V
12V
12V
30XR
DC
AMPS
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
V
COM
10A
CAT
CAT
A
600V
300V
BATT 9V
200mA
MAX
FUSED
10A MAX
FUSED
MAX
600V
600V
CAPACITOR
MODULE
LAMP
MODULE
Figure 36. Pictorial and Schematic of Setup

7. Turn off the power supply.

8. Discharge the capacitor.

9. Set the DMM to measure DC current.

10. Connect the circuit shown in figure 36.

11. Turn on the power supply and observe the lamp.
Lamp status ____________________________________________(On/Off)

The lamp should not light. There is a small amount of current flow until the
capacitor is charged, but it is not enough to cause the lamp to light.
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
12. Observe the current value displayed on the DMM display.
Current reading _________________________________________ (Amps)

OBJECTIVE 13
It should be zero because the capacitor is charged and has blocked the current
flow.

13. Turn off the power supply.

14. Discharge the capacitor.

15. Disconnect the circuit.

16. Store all components.
DESCRIBE THE EFFECT OF A CAPACITOR IN AN AC CIRCUIT
AND GIVE AN APPLICATION
A capacitor has a different effect in an AC circuit than it does in a DC circuit.
The current in a DC circuit is blocked because the capacitor charges to the level of
the source voltage and current cannot flow through the dielectric.
In an AC circuit, the voltage and current are continually changing (alternating),
so the capacitor is constantly charging and discharging. The load is then driven by
the charge and discharge current from the capacitor because current cannot flow
through the dielectric. The amount of charge and discharge current depends on
how much the capacitor can store (its capacitance) and therefore appears as a resistance to the AC source. This apparent resistance to AC current is called capacitive
reactance which is expressed in ohms and can be found with the following formula:
CAPACITIVE REACTANCE FORMULA
Xc =
where
Xc =
f =
π =
C =
1
2π fC
capacitive reactance (ohms)
AC frequency (hertz)
pi (3.1416)
capacitance (farads)
There are two factors that affect the amount of capacitive reactance a capacitor
will display in a circuit. These are:
•The
capacitance value - The higher the capacitance (more charge is stored
and released), the lower the capacitive reactance. The lower the capacitance
(less charge is stored and released), the higher the capacitive reactance.
•The
frequency of the current-The higher the frequency (the number of
times per second that the charge is stored and released), the lower the capacitive reactance. The lower the frequency, the higher the capacitive reactance.
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RF (radio frequency) and AF (audio frequency) filters are common applications of capacitors in AC circuits. These filters take advantage of the ability of
the capacitor to block unwanted DC signals while allowing the AC signal to pass
through. The reactance of the capacitor at different frequencies is also used to
separate unwanted AC signals from those of the correct frequency.
Activity 6. Effect of a Capacitor in an AC Circuit
Procedure Overview
In this procedure, you will connect and operate an AC circuit containing a
capacitor. You will then measure the current and voltage in the circuit and use
the capacitive reactance formula to calculate the actual capacitance value of
the capacitors used in your T7017.

1. Discharge the capacitors.

2. Connect the circuit shown in figure 37.
CAUTION
Make sure you have the test leads for the DMM connected to the correct
input terminals to measure AC current. Use the 10A input.
A
SOURCE SELECT
AC
DC
C
24 VAC
24V
12V
12V
30XR
AC
AMPS
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
V
COM
10A
CAT
CAT
A
600V
300V
BATT 9V
200mA
MAX
FUSED
10A MAX
FUSED
MAX
600V
600V
CAPACITOR
MODULE
LAMP
MODULE
Figure 37. A Capacitor in an AC Circuit
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

This is the same circuit you used in the previous activity with DC voltage.
When operated on DC, the current was blocked and the lamp would not light.
You will now observe the operation of a capacitor in an AC circuit.
3. Perform the following substeps to operate the circuit.
A. Turn on the DMM and set it to measure AC amps.
B. Place the AC/DC selector switch in the AC position.
C. Turn on the T7017 power supply.

4. Observe the lamp. Does it light?
Lamp On _____________________________________________ (Yes/No)


The lamp should be on.
5. Observe the ammeter and record its reading below.
Actual current = ________________________________________ (Amps)

The ammeter should indicate approximately 0.15 Amps. This will vary
depending on the actual capacitance of your capacitor. The lamp is actually
being illuminated by the charge and discharge current from the capacitor, the
capacitor blocks DC but will pass AC.

In the next few steps, you will take actual readings and use the capacitive
reactance formula to calculate the actual capacitance value of the capacitor.

6. Use the T7017 voltmeter to measure the voltage across the capacitor and
record below.
Actual voltage = _________________________________________ (VAC)


You should read approximately 5.5 VAC.
7. Use Ohm’s law and the actual voltage and current values obtained in steps
5 and 6 to calculate the actual capacitive reactance of the capacitor in the
circuit. Record your answer below.
Xc =
V
I
Actual capacitive reactance = ______________________________ (Ohms)

Your answer should be approximately 36.7 ohms.
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
8. The capacitive reactance formula can be re-arranged as shown to solve for
capacitance.
C=
where

Xc =
f =
π =
C =
1
2π fX c
capacitive reactance (ohms)
AC frequency (hertz)
pi (3.1416)
capacitance (farads)
Use the actual capacitive reactance value you calculated in step 7 to now
calculate this capacitor’s actual capacitance. Record your answer below. Use
the appropriate frequency for your location.
NOTE
The answer you obtain will be in farads, multiply your answer by 1 million
to convert to micro-farads.
Actual capacitance = ________________________________________ (μF)


You should obtain an answer of approximately 87μF for 50 Hz and 72μF for
60 Hz.
9. Turn off the power supply.

10. Remove the capacitor from the circuit and replace it with the other capacitor
provided with the T7017.

11. Turn on the power supply and repeat steps 5 thru 8 to determine the actual
capacitance of the 2nd capacitor. Record your data below.
Actual current = ________________________________________ (Amps)
Actual voltage = _________________________________________ (VAC)
Actual capacitive reactance = _____________________________ (Ohms)
Actual capacitance = ________________________________________ (μF)

The values obtained should be very close to those obtained with the first
capacitor. Any difference will be from a slightly different capacitance value
for the second capacitor.

12. Turn off the power supply.

13. Disconnect and store all components.
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OBJECTIVE 14
STATE THE FORMULAS FOR CALCULATING TOTAL SERIES
CAPACITANCE AND CAPACITIVE REACTANCE
The formulas for calculating series and parallel capacitance and capacitive
reactance are different than for inductance and inductive reactance.
Total series capacitance can be calculated in the same manner as parallel
inductance. This is shown in the following formula:
TOTAL SERIES CAPACITANCE FORMULA
CT =
where
CT =
C1 =
C2 =
C3 =
1
1
1
1
+
+
C1 C2 C3
total capacitance (Farads or Micro farads)
capacitance of capacitor 1 (Farads or Micro farads)
capacitance of capacitor 2 (Farads or Micro farads)
capacitance of capacitor 3 (Farads or Micro farads)
The total capacitive reactance of series capacitors is the sum of the individual
reactances.
TOTAL SERIES CAPACITIVE REACTANCE FORMULA
XCT = XC1 + XC2 = XC3 …
where
XCT
XC1
XC2
XC3
=
=
=
=
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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total capacitive reactance (Ohms)
capacitive reactance of C1 (Ohms)
capacitive reactance of C2 (Ohms)
capacitive reactance of C3 (Ohms)
60
OBJECTIVE 15
STATE THE FORMULAS FOR CALCULATING TOTAL PARALLEL
CAPACITANCE AND CAPACITIVE REACTANCE
The total capacitance of parallel capacitors is calculated as follows:
TOTAL PARALLEL CAPACITANCE FORMULA
CT = C1 + C2 + C3 …
where
CT = total capacitance (Farads or Micro farads)
C1, C2, C3 = individual capacitances (Farads or Micro farads)
The total capacitive reactance of parallel capacitors is calculated as follows:
TOTAL PARALLEL CAPACITIVE REACTANCE FORMULA
X CT =
1
1
1
1
+
+
X C1 X C 2 X C 3
where
XCT
XC1
XC2
XC3
=
=
=
=
BB227-BC04UEN INDUCTANCE AND CAPACITANCE
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total capacitance reactance (Ohms)
capacitive reactance of C1 (Ohms)
capacitive reactance of C2 (Ohms)
capacitive reactance of C3 (Ohms)
61
SKILL 6
CALCULATE THE TOTAL LOAD ON AN AC CIRCUIT
WITH CAPACITORS
Procedure Overview
In this procedure, you will calculate the total load on an AC circuit by
first calculating the total capacitive reactance. You will then use that value to
calculate the total resistance of the circuit. You will also make measurements
to verify your calculations.

1. Calculate the total capacitive reactance of the circuit in figure 38.
XC = __________________________________________________ (Ohms)

The solution is as follows:

Step 1 - Calculate the total capacitance as follows:
CT =
1
1
C1
CT =
+
1
C2
CT =
1
C3
1
1
10 μ F
CT =
+
+
1
5μF
+
1
8μ F
1
0.1 + 0.2 + 0.125
1
0.425
CT = 2.4 μ F

The total capacitance is 2.4μF.
C1=10 F
24 VAC
60 Hz
C2=5 F
C3=8 F
Figure 38. Capacitors in Series
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
Step 2 - Using the total capacitance found in step 1, calculate the total
capacitive reactance at 60 Hz as follows:
Xc 
Xc 
1
2 fC
1
23.1416600.0000024
X c  1105 Ohms


The total capacitive reactance at 60 Hz is 1105 ohms.
2. Calculate the total capacitance and total capacitive reactance of the circuit in
figure 38 if the capacitor values are 100μF, 50μF and 40μF.
The total capacitance = _____________________________________ (μF)
The total capacitive reactance = ____________________________ (Ohms)


Total capacitance is 18.2mF and reactance is 146 ohms.
3. Calculate total capacitance and total capacitive reactance of the circuit in
figure 38 if used with 50 Hertz.
The total capacitance = ______________________________________ (μF)
The total capacitive reactance = ____________________________ (Ohms)

The total capacitance is still 2.4mF but the reactance is now 1326 ohms.
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
4. Calculate the total capacitance and total capacitive reactance of the circuit
in figure 39. Use the actual capacitance values you obtained in the previous
activity for the two capacitors supplied with your T7017. Use the appropriate
frequency for your location.
The total capacitance= ______________________________________ (μF)
The total capacitive reactance @ 50 Hz = ____________________ (Ohms)
The total capacitive reactance @ 60 Hz = ____________________ (Ohms)
A
24 VAC
60 Hz
C1
C2
Figure 39. Two Capacitors in Series

In the next few steps, you will verify the calculations you made in step 4 by
connecting the two capacitors in series and measuring the resulting current.

5. Connect the circuit shown in figure 39.

6. Use Ohm’s law and the total capacitive reactance you calculated in step 4
to calculate the theoretical current that should be displayed by the ammeter
when you energize the circuit.
NOTE
The T7017 supply is rated at 24 VAC when it is under full load. Since this
circuit will not fully load the supply, use 26.5 VAC as a source voltage for this
calculation.
Theoretical current = _____________________________________ (Amps)
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
7. Perform the following substeps to operate the circuit.
A. Turn on the DMM and set it to measure AC amps. Use the 10A input.
B. Place the AC/DC selector switch in the AC position.
C. Turn on the power supply and observe the displayed current value on the
DMM.
Current = ___________________________________________ (Amps)

8. Compare the theoretical value calculated in step 6 with the value now
displayed on the DMM. Are the two values close ?
_____________________________________________________ (Yes/No)



The two values should be close.
9. Turn off the power supply.
10. Calculate the total capacitive reactance of the circuit shown in figure 40.
XCT = _________________________________________________ (Ohms)

The solution is as follows:

Step 1 - Calculate the total capacitance as follows:
CT = CC1 + CC2 + CC3
CT = 25μF + 5μF + 10μF
CT = 40μF

The total capacitance is 40μF.
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
Step 2 - Using the total capacitance found in step 1, calculate the total
capacitive reactance as follows:
1
X 
c 2 fC
1
X 
c 2  3.1416  60  0.00004
X  66.3 Ohms
c

The total capacitive reactance is 79.58 ohms at 50 Hz and 66.3 ohms at 60
Hz.
24 VAC
60 Hz
C =25 F
C =5 F
C =10 F
Figure 40. Capacitors in Parallel
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
11. Calculate the total capacitance and total capacitive reactance of the circuit
in figure 40 if the capacitor values are 100μF, 50μF and 40μF. Use the
appropriate frequency for your location.
The total capacitance= ______________________________________ (μF)
The total capacitive reactance @ 50 Hz= _____________________ (Ohms)
The total capacitive reactance @ 60 Hz= _____________________ (Ohms)


Total capacitance is 190μF and the total capacitive reactance is 16.75 ohms
at 50Hz or 13.96 ohms at 60Hz.
12. Calculate the total capacitance and total capacitive reactance of the circuit
in figure 41. Use the actual capacitance values you obtained in the previous
activity for the two capacitors supplied with your T7017. Use the appropriate
frequency for your location.
The total capacitance = _____________________________________ (μF)
The total capacitive reactance @ 50 Hz = ____________________ (Ohms)
The total capacitive reactance @ 60 Hz = ____________________ (Ohms)
A
24 VAC
60 Hz
C1
C2
Figure 41. Two Capacitors in Parallel


In the next few steps, you will verify the calculations you made in step 12 by
connecting the two capacitors in parallel and measuring the resulting current.
13. Connect the circuit shown in figure 41.
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
14. Use Ohm’s law and the total capacitive reactance you calculated in step 4
to calculate the theoretical current that should be displayed by the ammeter
when you energize the circuit.
NOTE
The T7017 supply is rated at 24 VAC when it is under full load. Since this
circuit will not fully load the supply, use 26.5 VAC as a source voltage for this
calculation.
I=
V
Xc
Theoretical current, I = ___________________________________ (Amps)

15. Perform the following substeps to operate the circuit.
A. Turn on the DMM and set it to measure AC amps. Use the 10A input.
B. Place the AC/DC selector switch in the AC position.
C. Turn on the power supply and observe the displayed current value on the
DMM.
Actual Current = _____________________________________ (Amps)

16. Compare the theoretical value calculated in step 14 with the value now
displayed on the DMM. Are the two values close ?
_____________________________________________________ (Yes/No)

The two values should be close.

17. Turn off the power supply.

18. Disconnect and store all components.
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SEGMENT 4
SELF REVIEW
1. A capacitor in a DC circuit will charge to a voltage equal to the
________ voltage.
2. A charged capacitor acts like an ______ in a DC circuit.
3. Current cannot cross the ___________ if a DC voltage is applied to a
capacitor.
4. The apparent resistance of a capacitor in an AC circuit is called
capacitive __________.
5. When the AC frequency is increased, the capacitive reactance of a
capacitor is ____________.
6. When capacitors are connected in series in an AC circuit, the overall
reactance of the circuit is __________.
7. Connecting capacitors in parallel in an AC circuit tends to increase the
__________ of the capacitors.
8. When capacitors are connected in parallel, the current is ___________.
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SEGMENT 5
INDUCTOR AND CAPACITOR APPLICATIONS
OBJECTIVE 16
DESCRIBE THE FUNCTION OF A FLUORESCENT
LIGHT FIXTURE
One of the components in a fluorescent light fixture is a type of inductor called
a ballast, as shown in figure 42.
Figure 42. A Typical Ballast
The ballast serves two purposes. First, it delivers a momentary voltage charge
that is higher than the source voltage to the lamp, much like the coil in a car’s
ignition system. Second, it then limits the current flowing through the lamp to a
predetermined value. To understand how it does this, figures 43 and 44 show its
operation in two stages.
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Figure 43 shows the state of the light fixture immediately after supplying power
to it by turning on the switch. The automatic switch, called a starter, is normally
closed until the light fixture is turned on. Because the starter has a low resistance, a
fairly high level of current begin to flow through it. The purpose of this is to establish a high level of current for the inductor to try to maintain.
STARTER CLOSED
CATHODE
CATHODE
LAMP, NOT LIGHTED
BALLAST
SWITCH
TURNED ON
LINE VOLTAGE
Figure 43. The Condition of a Fluorescent Light at the Instant It is Turned On
About a second after the light is turned on, the starter opens and stays open
until the light is turned off. When this happens, the only path for the current to flow
is through the gas in the fluorescent lamp. However, the resistance of the gas is
very high because it is cold and does not conduct well.
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Because the ballast is trying to maintain the current flow, it develops a very
high instantaneous voltage charge across the high resistance filaments (cathodes)
of the fluorescent lamp.
This instantaneous voltage is much higher than the source voltage and is
needed to get the current to flow through the cold gas. Once the gas is heated by
the high voltage, the current starts flowing through the gas causing it to glow, as
shown in figure 44.
STARTER OPEN
LAMP, LIGHTED
BALLAST
SWITCH
ON
LINE VOLTAGE
Figure 44. The Operation of the Ballast
The resistance is greatly reduced when the gas is heated. The gas effectively
becomes a low-resistance conductor. At this point, the ballast performs another
function.
Since the current applied is AC, the ballast introduces inductive reactance that
resists the change in current flow, thus limiting the current to a predetermined value
to keep the tube from burning out. Without this, the fluorescent lamp would not last
very long because the resistance of the gas in the tube is so low once it gets hot.
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OBJECTIVE 17
DESCRIBE THE FUNCTION OF AN RC TIMING CIRCUIT
IN A TIME-DELAY RELAY
An RC timing circuit uses a resistor and a capacitor connected in series in a DC
circuit to create an electronic timer, as shown in figure 45. This is possible because
the capacitor will charge at a specific rate. This rate depends on the amount of
capacitance and resistance and can be found by calculating the RC time constant.
However, it will take five time constants to fully charge or discharge the RC circuit.
R
+
C
Figure 45. An RC Circuit
The RC time constant can be calculated as follows:
RC TIME CONSTANT FORMULA
T=R×C
where
T = Time Constant in Seconds
R = Resistance in Ohms
C = Capacitance in Farads
Once you have determined the time constant, you can use that information
to calculate the charge and discharge time for the RC circuit. The charge time is
calculate as follows:
RC CIRCUIT CHARGE TIME FORMULA
Charge Time = T × 5
where
T = Time Constant in Seconds
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The discharge time is calculated as follows:
RC CIRCUIT DISCHARGE TIME FORMULA
Discharge Time = T × 5
where
T = Time Constant in Second
Time-Delay Relays
Time-delay relays, like the one shown in figure 46, often use these resistorcapacitor (RC) timing circuits to obtain a desired delay before they activate the
devices that are connected to it. Time-delay relays that use this type of timing
device are called solid-state, time-delay relays.
The time delay in this type of relay can be changed by adjusting the capacitance or resistance of the circuit. Usually, it is more economical to use a potentiometer to adjust the resistance than it is to use a variable capacitor to adjust
the capacitance. The adjustment is accomplished by turning a timer knob to the
desired time setting.
TIME DELAY
ADJUSTMENT KNOB
TIMER
MODULE
SUBBASE
Figure 46. A Solid-State, Time-Delay Relay
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SKILL 7
CALCULATE THE TIME TO CHARGE AND DISCHARGE
A CAPACITOR
Procedure Overview
In this procedure, you will calculate the time required to charge a capacitor.
You will then calculate the time required to discharge a capacitor through the
analog voltmeter of the T7017. Finally, you will operate the circuit and verify
your calculations.

1. Calculate the RC time constant for the circuit in figure 47.
R=100
+
12V
C=.001F
Figure 47. Calculating RC Time Constant

The resistance of the resistor is 100 ohms and the capacitance of the capacitor
is .001F.
T = R×C
T = 100 × .001
T = __________________________________________________(Seconds)


The time constant is 0.1 seconds.
2. Calculate the discharge time for the circuit.
Discharge Time = 0.1 × 5

The solution is as follows:
Discharge Time = ______________________________________(Seconds)

Since it takes 5 time constants to discharge, we need to multiply 0.1 times 5.

The discharge time is therefore 0.5 seconds.
NOTE
These time calculations are for DC circuits only.
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
3. Examine the circuit in figure 48.
MODEL 7017
SOURCE SELECT
5
10
15
20
25
0
30
A-C VOLTS
AC
DC
24V
12V
12V
VOLTMETER
-
10 RESISTOR
OHM MODULE
SELECTOR
SWITCH
+
PUSH BUTTON
SWITCH
MODULE
CAPACITOR
MODULE
R1=10
+
V
12V
CHARGE PATH
DISCHARGE
PATH
R=30k
7017
VOLTMETER
Figure 48. Circuit Setup

4. Calculate the time it takes for a full charge of the capacitor. Use the resistance
value and capacitance value shown in figure 48.
Full charge time = ______________________________________(Seconds)


It should be approximately 0.0045 seconds (4.5 milliseconds).
5. Calculate the time it takes for full discharge of the capacitor. Use the resistance
value shown in figure 48 for the analog voltmeter.
Full discharge time = ___________________________________(Seconds)

It should be approximately 13.5 seconds.

6. Connect the circuit shown in figure 48.

7. Place the AC/DC switch in the DC position.

8. Turn on the power supply.
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
9. Rotate the selector switch to the right to allow the capacitor a few seconds to
charge.

10. Rotate the selector switch back to the left and press the pushbutton switch to
discharge the capacitor.

Observe the voltmeter display and record the time it takes for the display to
fall back to zero.
Time = ______________________________________________(Seconds)

OBJECTIVE 18
It should be approximately the same as the time calculated in step 5, 13.5
seconds.

11. Repeat steps 9 and 10 to make sure of your time.

12. Turn off the power supply.

13. Discharge the capacitor.

14. Disconnect the circuit and store all components.
DESCRIBE THE FUNCTION OF CAPACITOR AND INDUCTORS IN
AN ELECTRIC POWER SUPPLY
Capacitors and inductors are often used in DC electric power supplies to help
create a constant DC output voltage. To understand how this is accomplished, let’s
first examine figure 49, which shows the first step of converting from AC to DC.
In this first step, the power supply converts the AC voltage from the wall outlet
into a pulsating DC voltage. The electronic device that accomplishes this conversion is called a rectifier.
POSITIVE
PEAK
+V
POSITIVE
PEAK
+V
0V
NO NEGATIVE CYCLE
-V
-V
NEGATIVE
PEAK
BEFORE
RECTIFIER
(AC VOLTAGE)
AFTER
RECTIFIER
(PULSATING DC)
Figure 49. Converting AC Voltage to Pulsating DC Voltage
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Basically, what the rectifier does is eliminate the negative cycle of the AC
input. This means that the current now only flows in one direction (DC). However,
the voltage is pulsating between 0 volts and the positive peak voltage. This is
where the capacitor comes in. The purpose of the capacitor is to perform step 2 of
the process by creating a voltage between the pulses.
Figure 50 shows how a capacitor would be connected in a power supply. In this
circuit, the capacitor charges as the voltage rises. When the voltage starts to fall,
the capacitor begins to discharge and opposes the fall.
RECTIFIER
120VAC
CHARGE
DISCHARGE
+
C
LOAD
Figure 50. Using a Capacitor as a Filter in a Power Supply
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The value of the capacitor is usually fairly large so that the discharge time
is long enough to keep the DC voltage close to its peak while the AC voltage
is decreasing. By the time the AC voltage nears its next peak, the capacitor has
discharged only a small amount and requires only a small amount of recharging
to get back to the peak voltage, as shown in figure 51. This results in an almost
constant DC voltage, although a small fluctuation called ripple will be present.
This technique of using a capacitor to smooth out the changes in voltage is called
filtering.
CHARGE
DISCHARGE
CHARGE
DISCHARGE
DC VOLTAGE WITH RIPPLE
Figure 51. DC Output Voltage Created
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An inductor can be added to the circuit to further reduce the ripple, as shown
in figure 52. The ripple can be almost eliminated because the inductor resists any
change in the current, even a small amount.
INDUCTOR
RECTIFIER
+
LOAD
WITH INDUCTOR
ADDED
0 VAC
WITH CAPACITOR
ONLY
Figure 52. Adding an Inductor to the Capacitive Filter Circuit
NOTE
The power supply of your T7017, shown in figure 53, uses the type of
circuit shown in figure 50. It has a rectifier and uses a capacitor as a filter. The
T7017 power supply uses a voltage regulating device to eliminate the ripple.
Figure 53. Components of the T7017 Power Supply
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SEGMENT 5
SELF REVIEW
1. The inductor used in a fluorescent light fixture is called a(n)
___________.
2. An RC timing circuit uses a resistor and a(n) ________________
connected in series.
3. __________________ relays often use RC circuits to obtain a desired
time delay.
4. Capacitors and inductors are often used in electric DC power supplies
to create a(n) __________ DC output voltage.
5. Using a capacitor to smooth out changes in voltage is called
__________.
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