Revision 1
December 2014
Basic Electricity
Part 2
Instructor Guide
Reviewed by:
Cassandra Bitler
Project Manager, OGF
Approved by:
Robert Coovert
Manager, INPO Learning Development
Approved by:
Kevin Kowalik
Chairperson, Industry OGF Working Group
11/13/2014
Date
11/13/2014
Date
11/13/2014
Date
NOTE: Signature also satisfies approval of associated student guide and PowerPoint presentation.
GENERAL DISTRIBUTION
GENERAL DISTRIBUTION: Copyright © 2014 by the National Academy for Nuclear Training. Not
for sale or for commercial use. This document may be used or reproduced by Academy members
and participants. Not for public distribution, delivery to, or reproduction by any third party without
the prior agreement of the Academy. All other rights reserved.
NOTICE: This information was prepared in connection with work sponsored by the Institute of
Nuclear Power Operations (INPO). Neither INPO, INPO members, INPO participants, nor any
person acting on behalf of them (a) makes any warranty or representation, expressed or implied,
with respect to the accuracy, completeness, or usefulness of the information contained in this
document, or that the use of any information, apparatus, method, or process disclosed in this
document may not infringe on privately owned rights, or (b) assumes any liabilities with respect to
the use of, or for damages resulting from the use of any information, apparatus, method, or process
disclosed in this document.
ii
Table of Contents
INTRODUCTION....................................................................................................................... 1
TLO 1 GENERATING AC VOLTAGE ........................................................................................ 2
Overview ........................................................................................................................... 2
ELO 1.1 Simple AC Generator ......................................................................................... 2
ELO 1.2 Sine Wave Output .............................................................................................. 3
ELO 1.3 AC Generation Terminology .............................................................................. 5
ELO 1.4 Calculating RMS Voltage and Current .............................................................. 7
ELO 1.5 Phase Relationships ............................................................................................ 9
TLO 1 Summary ............................................................................................................. 11
TLO 2 INDUCTORS, CAPACITORS, AND POWER IN AC CIRCUITS .......................................... 11
Overview ......................................................................................................................... 11
ELO 2.1 Inductors ........................................................................................................... 12
ELO 2.2 Capacitors ......................................................................................................... 16
ELO 2.3 Inductive Reactance ......................................................................................... 20
ELO 2.4 Capacitive Reactance ....................................................................................... 23
ELO 2.5 Impedance ........................................................................................................ 25
ELO 2.6 Apparent, True, and Reactive Power................................................................ 27
ELO 2.7 Power Factor..................................................................................................... 30
TLO 2 Summary ............................................................................................................. 33
TLO 3 TRANSFORMERS ........................................................................................................ 34
Overview ......................................................................................................................... 34
ELO 3.1 Transformer Terminology ................................................................................ 35
ELO 3.2 Transformers Components ............................................................................... 36
ELO 3.3 Voltage, Current and Power Relationships ...................................................... 38
ELO 3.4 Transformer Applications................................................................................. 40
TLO 3 Summary ............................................................................................................. 43
TLO 4 ELECTRICAL DISTRIBUTION SYSTEMS ...................................................................... 44
Overview ......................................................................................................................... 44
ELO 4.1 Basic Electrical Distribution System ................................................................ 44
ELO 4.2 Electrical Distribution Terminology ................................................................ 47
ELO 4.3 Single-Phase Load Connections and Three-Phase Systems ............................. 50
ELO 4.4 Wye and Delta Systems .................................................................................... 52
ELO 4.5 Unbalanced Loads ............................................................................................ 56
ELO 4.6 Power Distribution Schemes ............................................................................ 57
TLO 4 Summary ............................................................................................................. 62
TLO 5 ELECTRICAL TEST EQUIPMENT ................................................................................. 62
Overview ......................................................................................................................... 62
ELO 5.1 Use of Common Test Equipment ..................................................................... 62
ELO 5.2 Operation of Common Test Equipment ........................................................... 63
TLO 5 Summary ............................................................................................................. 70
BASIC ELECTRICITY PART 2 SUMMARY ............................................................................... 70
iii
This page is intentionally blank.
iv
Basic Electricity Part 2
Revision History
Revision
Date
Version
Number
Purpose for Revision
Performed
By
11/6/2014
0
New Module
OGF Team
12/10/2014
1
Added signature of OGF
Working Group Chair
OGF Team
Duration
 11 hours
Logistics
Ensure that the presentation space is properly equipped with the following:
 Projector
 Internet access, if needed
 Whiteboard or equivalent
 Space for notes, parking lot, mockups, or materials
 Sufficient space for all students
Ensure that the following course materials are prepared and staged:
 All student materials
 Instructor materials
 Media, photos, and illustrations
 Props, lab equipment, or simulator time, as applicable
Ensure that all students have fulfilled the course prerequisites, if applicable.
Instructor preparation:
 Review the course material prior to beginning the class.
 Review the NRC exam bank and as many new exams as are available
prior to the class to ensure that you are prepared to address those items.
Ensure that all students have access to the training material for self-study
purposes.
Introduction
This module will review methods of generating, transforming, and
distributing AC power, including use of common electrical test equipment.
Rev 1
Logistics
 Use PowerPoint slides
1–3 and the instructor
guide (IG) to introduce
the Basic Electricity Part
2 module.
1
Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of 80 percent
or higher on the following Terminal Learning Objectives (TLOs):
1. Describe the theory of operation and operating characteristics of an
AC generator.
2. Describe the construction and theory of operation of Inductors and
Capacitors, their effects on AC electrical circuits, and relationship to
power factor.
3. Describe the construction, operation, and applications of transformers.
4. Describe basic industrial electrical distribution, including typical
wiring schemes used and the advantages of three-phase systems.
5. Given an electrical measuring device or piece of test equipment,
describe the use of that equipment including the electrical parameter
measured.
TLO 1 Generating AC Voltage
Overview
Duration
 5 minutes
Logistics
 Use PowerPoint slides
4–5 and the IG to
introduce TLO 1.
In this section, you will learn methods of AC voltage generation. AC
voltage generation is the core business of a power plant, and an operator
must know how generators work in order to monitor and control them.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the operation of a simple AC generator.
2. Describe the development of a sine-wave output in an AC generator.
3. Define common terms in relation to AC generation.
4. Describe the relationship between peak, average, and RMS values of
voltage in an AC power source.
5. Given a diagram of two sine waves, describe the phase relationship
between the two waves.
ELO 1.1 Simple AC Generator
Duration
 15 minutes
Logistics
 Use PowerPoint slides
6–8 and the IG to present
ELO 1.1.
Introduction
In this section, you will learn how an AC generator creates AC voltage.
Simple AC Generator
A simple AC generator consists of a conductor or loop of wire in a
magnetic field produced by an electromagnet. The two ends of the loop
connect to slip rings. The slip rings make contact with two brushes. When
the loop rotates, it cuts magnetic lines of force, first in one direction, and
then the other.
2
Rev 1
As the conductor passes through the magnetic field, the magnetic field
induces a voltage in the conductor and the slip rings transfer this voltage in
the conductor as voltage output.
Figure: Simple AC Generator
Knowledge Check
An AC generator has all of the following except:
A.
a commutator
B.
a magnetic field
C.
slip rings
D.
a conductor in relative motion with the magnetic field
ELO 1.2 Sine Wave Output
Duration
 15 minutes
Logistics
 Use PowerPoint slides
9–13 and the IG to
present ELO 1.2.
Inform
 Emphasize the sine wave
output is through the slip
rings and brushes of the
generator.
Introduction
In this section, you will learn how an AC generator develops a sine wave
output.
Developing an AC Sine Wave Voltage
When the loop is in the vertical position, at 0º, (0 degrees of rotation in the
figure below) the coils are moving parallel to the magnetic field and do not
cut magnetic lines of force. At that instant, there is no voltage induced in
the loop. As the coil rotates in a clockwise direction, each side of the coil
cuts the magnetic lines of force in opposite directions. The direction
(polarity) of the induced voltages depends on the direction of movement of
the coil.
The induced voltages are additive, making slip ring X positive (+) and slip
ring Y negative (-). The potential across resistor R causes a current to flow
from Y to X through the resistor. This current increases until it reaches a
maximum value when the coil is horizontal to the magnetic lines of force at
Rev 1
3
90º. At that instant, the horizontal coil is moving perpendicular to the
magnetic field and cutting the greatest number of magnetic lines of force.
As the coil continues to turn, the induced voltage and current decrease until
both reach zero, when the coil is again in the vertical position (180º). The
next half revolution produces an equal voltage, except with reversed
polarity (270º and 360º). The current flow through R is now from X to Y.
Figure: Developing an AC Sine Wave Voltage
The alternating reversal of polarity results in the generation of a voltage, as
shown above. As the coil rotates through 360º, voltage output in the shape
of a sine wave results.
Knowledge Check
In a simple AC generator, ______________________
causes the AC sine wave output.
4
A.
the changes in relative motion of the conductor and the
magnetic field
B.
commutation
C.
changing speed of the rotating element
D.
the pre-programmed oscillation of the field
Rev 1
ELO 1.3 AC Generation Terminology
Duration
 15 minutes
Logistics
 Use PowerPoint slides
14–17 and the IG to
present ELO 1.3.
Introduction
In this section, you will learn terminology necessary to describe AC power
and measure its effect.
Period and Frequency
When an AC generator produces a voltage, the resulting current varies in
step with the voltage. As the generator coil rotates 360°, the output voltage
goes through one complete cycle. In one complete cycle, the voltage
increases from zero to Emax in one direction, decreases to zero, increases to
Emax in the opposite direction (negative Emax), and then decreases to zero
again.
The period is the time required for the generator to complete one cycle. The
frequency (measured in hertz) is the number of cycles completed per
second.
Peak Voltage and Current
In the figure below, Emax occurs at 90°. This value is termed the peak
voltage. One way to quantify AC voltage or current is by peak value: peak
voltage (Ep) or peak current (Ip). Peak means the maximum voltage or
current appearing on an AC sine wave.
Peak to Peak Voltage and Current
Another commonly used term associated with AC is peak-to-peak value
(Ep-p or Ip-p). Peak to peak refers to the magnitude of voltage, or current
range, spanned by the sine wave.
Figure: AC Sine Wave Voltage
Rev 1
5
Effective Value of AC
A sine wave like the one shown graphically in the figure below presents the
AC generator output.
Figure: AC Voltage Sine Wave
Effective value is the value most commonly used for quantifying AC. The
effective value of AC is the amount of AC that produces the same heating
effect as an equal amount of DC. An one ampere effective value of AC will
produce the same amount of heat in a conductor, in a given time, as one
ampere of DC.
The heating effect of a given AC current is proportional to the square of the
current. It is possible to calculate the effective value of AC by squaring all
the amplitudes of the sine wave over one period, taking the average of these
values, and then taking the square root of the average. The effective value,
because it is the root of the mean (average) square of the currents, and is
termed the root-mean-square, or RMS value.
Knowledge Check
Match the terms to their appropriate definitions.
6
1. The time required for the generator
to complete one cycle
A. Period
2. The number of cycles completed
per second
B. Peak-to-peak
3. The magnitude of voltage or
current range spanned by the sine
wave.
C. Frequency
4. The root of the mean (average)
square of the currents or voltages
D. RMS
Rev 1
Knowledge Check Answer
1. A – Period
2. C – Frequency
3. B – Peak-to-peak
4. D – RMS
ELO 1.4 Calculating RMS Voltage and Current
Introduction
In this section, you will learn how to calculate RMS current and voltage.
Effective Value of AC
The effective value of current or voltage of an AC signal is equal to the root
mean square (RMS) of the signal. In an AC circuit, the current value
changes continuously over the period (or frequency). To calculate the
effective value of AC, square the average amplitudes of the sine wave over
one period, and then take the square root of the result.
The following figure illustrates the effective current applied to a sine wave,
Figure: Effective Value of AC Current
The RMS and average values are calculated in the same manner for either
curve even though they do not have the same frequency or period. The
upper curve shows a plot of the values of I over time and the effective value
of I; the lower curve shows a plot of the values of I2 over time, and the
average current.
Rev 1
7
Duration
 20 minutes
Logistics
 Use PowerPoint slides
18–20 and the IG to
present ELO 1.4.
The dashed line is the average of the I2 values, and the square root of that
value is the RMS, or effective value. (Square root of the mean of the
squared deviation of a waveform).
The average value is ½ Imax2.
The RMS value is
√2
𝐼
2 𝑚𝑎𝑥
then which is approximately equal to 0.707 Imax.
Calculate the effective value of voltage or current for an AC sine wave
using:
𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑣𝑎𝑙𝑢𝑒 (𝑅𝑀𝑆) = 𝑝𝑒𝑎𝑘 𝑣𝑎𝑙𝑢𝑒 × 0.707
Normal convention is that the values of AC current (I), and voltage (E) are
RMS values.
Calculating RMS Voltage and Current
Step
Action
1.
Determine the peak current or voltage.
2.
Multiply the peak current or voltage by 0.707.
3.
This is the effective (RMS) value for voltage or current.
RMS Voltage Demonstration
The peak value of voltage in an AC circuit is 200 V. What is the RMS
value of the voltage?
Solution:
E = 0.707 Emax
E = 0.707 (200 V)
E = 141.4 V
RMS Current Demonstration
The peak current in an AC circuit is 10 amps. What is the RMS value of
current?
8
Rev 1
Solution:
I = 0.707 Imax
I = 0.707 (10 amps)
I = 7.07 amps
Knowledge Check
The peak value of voltage in an AC circuit is 250 volts.
Calculate the effective voltage.
A.
120 volts
B.
125 volts
C.
177 volts
D.
353 volts
ELO 1.5 Phase Relationships
Duration
 15 minutes
Logistics
 Use PowerPoint slides
21–24 and the IG to
present ELO 1.5.
Introduction
In this section, you will learn definitions of phase relationships in AC
power.
Phase Angle Guidelines
Phase angle is the fraction of a cycle, in degrees, that has gone by since a
voltage or current has passed through a given value. The given value is
normally zero. From the figure below, take point 1 on the sine wave as the
starting point or zero phase. The phase angle at Point 2 is 30°, Point 3 is
60°, Point 4 is 90°, and so on, until Point 13 where the phase angle is 360°,
or zero once again.
Figure: AC Voltage Sine Wave
Rev 1
9
Phase Example
Phase difference is another common term for phase angle. Phase difference
describes two different voltages that have the same frequency, which pass
through zero values in the same direction, but at different times. In the
figure below, the angles along the axis indicate the phases of voltages e1
and e2. At 120°, e1 passes through the zero value, which is 60° ahead of e2
(e2 equals zero at 180°). We describe this as voltage e1 leads e2 by 60
electrical degrees, or voltage e2 lags e1 by 60 electrical degrees.
Figure: Phase Relationship
Phase difference can also compare two different currents or a current and a
voltage. If the phase difference between two currents, two voltages, or a
voltage and a current is zero degrees, they are termed to be in phase. If the
phase difference is any amount other than zero, they are out of phase.
Knowledge Check
When two AC voltages reach their peak voltage at the
same time, the voltages are said to be
__________________.
10
A.
leading
B.
lagging
C.
in phase
D.
out of phase
Rev 1
TLO 1 Summary
In this section, you learned the basic theory behind generating an AC
voltage, AC power terminology, and phase relationships. The module on
AC Motors and Generators provides more detail about AC generators.
1. A simple AC generator requires a magnetic field, a conductor, and
relative motion to generate a voltage.
2. A sine wave output results as the conductor rotates through the
magnetic field with the conductor connected through slip rings and
brushes to an output source. The output varies in proportion to the
strength of the magnetic field that the conductor passes through.
3. Common terms used in relation to AC generators are period and
frequency, peak voltage and current, effective value of AC, and peak
to peak voltage and current.
4. Measured voltage in AC circuits is usually the effective voltage
(RMS), that is equal to peak voltage times 0.707.
5. Phase angle of sine waves is the fraction of a cycle in degrees that a
voltage or current has passed through a given value of a sine wave,
usually zero. A three-phase output has each phase out of phase by
120 degrees; a later chapter will cover three-phase output in more
detail.
Duration
 15 minutes
Logistics
 Use PowerPoint slides
25–27 and the IG to
review TLO 1 material.
Use directed and nondirected questions to
students, check for
understanding of ELO
content, and review any
material where student
understanding of ELOs
is inadequate.
Figure: Three-Phase Power Output
Now that you have completed this lesson, you should be able to:
1. Describe the operation of a simple AC generator.
2. Describe the development of a sine-wave output in an AC generator.
3. Define common terms in relation to AC generation.
4. Describe the relationship between peak, average, and RMS values of
voltage in an AC power source.
5. Given a diagram of two sine waves, describe the phase relationship
between the two waves.
TLO 2 Inductors, Capacitors, and Power in AC Circuits
Overview
In this section, you will learn the effects of inductors and capacitors on AC
circuits.
Operators must understand the effects of inductors and capacitors to
monitor and anticipate the response of electrical machines and systems.
Rev 1
11
Duration
 5 minutes
Logistics
 Use PowerPoint slides
28–29 and the IG to
introduce TLO 2.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe how current flow, magnetic field, and stored energy in an
inductor relate to one another, and how an inductor opposes a change
in current flow.
2. Describe the construction of a capacitor, explain how it stores energy,
and explain how it opposes a change in voltage.
3. Describe inductive reactance (XL) and the phase relationship between
current and voltage in an inductive circuit.
4. Define capacitive reactance (XC) and the phase relationship between
current and voltage in a capacitive circuit.
5. Define impedance (Z).
6. Define apparent, true, and reactive power using a power triangle.
7. Define power factor as it relates to true power and apparent power,
and define leading and lagging power factors.
ELO 2.1 Inductors
Duration
 30 minutes
Logistics
 Use PowerPoint slides
30–38 and the IG to
present ELO 2.1.
Introduction
An inductor is a circuit element that stores electrical energy in the form of a
magnetic field. It is usually a coil of wire wrapped around a core of
permeable material. Circuits containing inductors behave differently than a
purely resistive circuit.
Counter-Electromotive Force (CEMF)
In the figure below, when the DC current is flowing through Wire A, it
generates a magnetic field around Wire A, but does not induce an
electromotive force (EMF) into Wire B because there is no relative motion
between the magnetic field and Wire B (DC circuit).
If we open the switch, the current stops flowing in Wire A, and the
magnetic field collapses. As the field collapses, it moves relative to Wire
B. The field movement induces an EMF or voltage in Wire B.
12
Rev 1
Figure: Induced EMF
This is an example of Faraday’s Law, which states that when a conductor
moves through a magnetic field, or when the magnetic field moves past a
conductor, a voltage is induced in the conductor. The EMF induced in Wire
B causes a current to flow whose magnetic field opposes the change in the
magnetic field that produced it. For this reason, an induced EMF is termed
Counter Electromotive Force or CEMF. This is an example of Lenz’s Law,
which states that the induced EMF opposes the EMF that caused it.
Inducing Electromotive Force
The three requirements for inducing an EMF are:
 A conductor
 A magnetic field
 Relative motion between the two
The faster the movement between the two, or the faster the magnetic field
collapses or expands, the greater the induced EMF. Coiling the wire in
either Circuit A or Circuit B, or both, as shown in the following figure,
increases the induction. Increasing the DC voltage will increase the current
through wire A and result in a stronger magnetic field, which will also
increase the voltage induced in wire B.
Rev 1
13
Figure: Induced EMF in Coils
Self-Induced Electromotive Force (EMF)
Self-induced EMF is another phenomenon of induction. The circuit shown
in the figure below contains a coil of wire called an inductor (L). As
current flows through the circuit, a large magnetic field sets up around the
coil. Since the current is not changing, there is no EMF produced. If we
open the switch, current flow stops and the field around the inductor
collapses. This collapsing magnetic field produces a voltage in the coil.
This is a self-induced EMF.
Figure: Self-Induced EMF
Lenz’s Law gives the polarity of self-induced EMF. The polarity is in the
direction that opposes the change in the magnetic field that induced the
EMF. The resulting current caused by the induced EMF tends to maintain
the same amount of current that existed in the circuit before opening the
switch. The inductor maintains current flow until the magnetic field has
collapsed entirely. For this reason, an inductor tends to oppose a change in
current flow.
Inductance
The induced EMF, or counter EMF (CEMF), is proportional to the time rate
of change of the current. The proportionality constant is the inductance (L).
Inductance is a measure of an inductor’s ability to induce CEMF, measured
in henries (H).
14
Rev 1
An inductor has an inductance of one henry when a one amp per second
change in current produces one volt of CEMF, as shown in the equation
below.
𝐶𝐸𝑀𝐹 = −𝐿
∆𝐼
∆𝑡
Where:
CEMF = Induced voltage (volts)
L = Inductance (henries)
∆𝐼
∆𝑡
= Time rate of change of current (amp/sec)
The minus sign shows that the CEMF is opposite in polarity to the applied
voltage.
Inductors in Series Example
To calculate the equivalence of inductors in series, simply add the
inductance values, like resistors in series. Equivalent inductance (Leq) of
multiple inductors in series shown below is:
𝐿𝑒𝑞 = 𝐿1 + 𝐿2 + . . . 𝐿𝑛
The figure below shows two inductors in series.
Figure: Inductors in Series
Inductors in Parallel Example
The figure below shows two inductors in parallel.
To calculate the equivalence of inductors in parallel, combine the values
like resistors in parallel as shown below.
1
1
1
1
= + + …
𝐿𝑒𝑞 𝐿1 𝐿2
𝐿𝑁
Rev 1
15
Figure: Inductors in Parallel
Knowledge Check
Select all of the statements about inductors that are
true.
A.
An inductor is a circuit element that will store electrical
energy in the form of a magnetic field.
B.
An inductor stores energy as a stored charge between
two plates.
C.
An inductor is usually a coil of wire wrapped around a
core of permeable material.
D.
Inductors oppose a change in voltage.
ELO 2.2 Capacitors
Duration
 30 minutes
Logistics
 Use PowerPoint slides
39–47 and the IG to
present ELO 2.2.
Introduction
In this section, you will learn how capacitors store energy and their effects
on AC circuits.
Capacitors
Capacitors are electrical devices constructed of two metal plates separated
by an insulating material, called a dielectric (shown below). The schematic
symbols shown in (b) and (c) below apply to all capacitors.
16
Rev 1
Figure: Capacitor Construction and Symbols
Charging a Capacitor
The two conductor plates of the capacitor, shown as (a) in the figure below,
are electrically neutral, because there are as many positive as negative
charges on each plate. The capacitor, therefore, has no charge.
Now, suppose we connect a battery across the plates, shown in (b) below.
When the switch is closed (c), the negative charges on Plate A are attracted
to the positive side of the battery, while the positive charges on Plate B are
attracted to the negative side of the battery. This movement of charges will
continue until the difference in charge between Plate A and Plate B is equal
to the voltage of the battery. This is now a charged capacitor. Capacitors
store energy as an electric field between the two plates.
Figure: Charging a Capacitor
Because very few of the charges can cross between the plates (the insulating
dielectric prevents this), the capacitor will remain in the charged state even
when the battery is disconnected. Because the opposite charges on the
opposing plates attract each other, they will tend to oppose any changes in
Rev 1
17
charge. In this manner, a capacitor will oppose any change in voltage felt
across it.
Discharging a Capacitor
If we place a conductor across the plates (b) below, electrons will flow
through the conductor back to Plate A, and the charges will be neutralized.
This is now a discharged capacitor.
Figure: Discharging a Capacitor
Capacitance
Capacitance is the ability to store an electrical charge. Capacitance is equal
to the amount of charge that can be stored divided by the applied voltage, as
shown in the equation below.
𝐶=
𝑄
𝑉
Where:
C = Capacitance (F)
Q = Amount of charge (Coulombs)
V = Voltage (V)
The unit of capacitance is the farad (F). A farad is the capacitance that will
store one coulomb of charge when one volt acts across the plates of the
capacitor.
The dielectric constant (K) describes the ability of the dielectric to store
electrical energy, as compared to air. Air is used as a reference and is given
a dielectric constant of 1. Therefore, the dielectric constant is
dimensionless. Some other dielectric materials are paper, Teflon, Bakelite,
mica, and ceramic.
18
Rev 1
The capacitance of a capacitor depends on three things.
 Area of conductor plates
 Separation between the plates
 Dielectric constant of insulation material
The equation below illustrates the formula to find the capacitance of a
capacitor with two parallel plates.
𝐶=𝐾
𝐴
(8.85 × 10−12 )
𝑑
Where:
C = Capacitance
K = Dielectric constant
A = Area
d = Distance between the plates
8.85 x 10-12 = Constant of proportionality
Types of Capacitors
Dielectric material serves to classify all commercial capacitors. The most
common dielectrics are air, mica, paper, and ceramic capacitors, plus the
electrolytic type.
Capacitors in Series Example
Capacitors in series are combined like resistors in parallel. The figure
below shows the total capacitance, CT, of capacitors connected in series and
equation follows.
Figure: Capacitors Connected in Series
1
1
1
1
1
= + + + …
𝐶𝑇 𝐶1 𝐶2 𝐶3
𝐶𝑁
When only two capacitors are in series, the equation simplifies as shown
below.
𝐶𝑇 =
𝐶1 𝐶2
𝐶1 + 𝐶2
When all the capacitors in series are the same value, compute the total
capacitance by dividing the capacitor’s value by the number of capacitors in
series as shown below.
𝐶𝑇 =
Rev 1
𝐶
𝑁
19
Where:
C = Value of any capacitor in series
N = The number of capacitors in series with the same value.
Capacitors in Parallel Example
Capacitors in parallel are combined like resistors in series. When connected
in parallel, the total capacitance, CT, is the sum of the individual
capacitances as given below.
𝐶𝑇 = 𝐶1 + 𝐶2 + 𝐶3 + … + 𝐶𝑁
Figure: Capacitors Connected in Parallel
Knowledge Check
A capacitor is ________________________________.
A.
two metal plates separated by an insulating material
that opposes a change in current flow
B.
two metal plates separated by an insulating material
that opposes a change in voltage
C.
a coil of wire around a magnetic core that opposes a
change in voltage
D.
a coil of wire around a magnetic core that opposes a
change in current flow
ELO 2.3 Inductive Reactance
Duration
 20 minutes
Logistics
 Use PowerPoint slides
48–55 and the IG to
present ELO 2.3.
Introduction
In AC circuits, inductors present a resistance to current flow that is termed
inductive reactance. In order to understand the behavior of AC circuits, it is
necessary to understand inductive reactance and its effects on voltage and
current.
Inductive Reactance
Any device relying on magnetism or magnetic fields to operate is a form of
inductor. Motors, generators, transformers, and coils are all inductors.
20
Rev 1
In an inductive AC circuit, the current is continually changing and is
continuously inducing an EMF. Because this EMF opposes the continuous
change in the flowing current, we measure its effect in ohms. This
opposition of the inductance to the flow of an alternating current is
inductive reactance (XL). The equation below is the mathematical
representation of the current flowing in a circuit that contains only inductive
reactance.
𝐼=
𝐸
𝑋𝐿
Where:
I = Effective current (A)
XL = Inductive reactance (Ω)
E = Effective voltage across the reactance (V)
The value of XL in any circuit is dependent on the inductance of the circuit
and on the rate at which the current is changing through the circuit. This
rate of change depends on the frequency of the applied voltage. The
equation below is the mathematical representation for XL.
𝑋𝐿 = 2𝜋𝑓𝐿
Where:
π ≈ 3.14
f = Frequency (Hertz)
L = Inductance (Henries)
The magnitude of an induced EMF in a circuit depends on how fast the flux
that links the circuit is changing. In the case of self-induced EMF (such as
in a coil), a counter EMF is induced in the coil due to a change in current
and flux in the coil. This CEMF opposes any change in current, and its
value at any time will depend on the rate at which the current and flux are
changing at that time. Inductors in AC circuits expand and collapse their
magnetic fields in an attempt to keep current in the circuit constant.
In a purely inductive circuit, the resistance is negligible in comparison to
the inductive reactance.
Voltage and Current Relationship in an Inductive Circuit
As previously stated, any change in current in a coil (either a rise or a fall)
causes a corresponding change of the magnetic flux around the coil.
Because the current changes at its maximum rate when it is going through
its zero value at 90° (point b I n the figure below) and 270° (point d), the
flux change is also the greatest at those times. Consequently, the selfinduced EMF in the coil is at its maximum value at these points. Because
the current is not changing at the point when it is going through its peak
value at 0° (point a), 180° (point c), and 360° (point e), the flux change is
Rev 1
21
zero at those times. Therefore, the self-induced EMF in the coil is at its
zero value at these points.
Figure: Current, Self-Induced EMF, and Voltage in an Inductive Circuit
According to Lenz’s Law, the induced voltage always opposes the change
in current flow. Referring to the figure above, with the current at its
maximum negative value (point a), the induced EMF is at a zero value and
falling. Thus, when the current rises in a positive direction (point a to point
c), the induced EMF is of opposite polarity to the applied voltage and
opposes the rise in current.
Notice that as the current passes through its zero value (point b) the induced
voltage reaches its maximum negative value.
When the current is at its maximum positive value (point c), the induced
EMF is at a zero value and rising. As the current is falling toward its zero
value at 180° (point c to point d), the induced EMF is of the same polarity
as the current and tends to keep the current from falling. When the current
reaches a zero value, the induced EMF is at its maximum positive value.
Later, when the current is increasing from zero to its maximum negative
value at 360° (point d to point e), the induced voltage is of the opposite
polarity as the current and tends to keep the current from increasing in the
negative direction. Thus, the induced EMF lags the current by 90°.
The value of the self-induced EMF varies as a sine wave and lags the
current by 90°, as shown in the figure above. The applied voltage must be
equal and opposite to the self-induced EMF at all times.
Therefore, the current lags the applied voltage by 90° in a purely inductive
AC circuit.
22
Rev 1
Knowledge Check
Inductive reactance is caused
by__________________________.
A.
the induced EMF in inductors
B.
stored electrical charge in circuit components
C.
hysteresis losses
D.
resistance in the conductors
ELO 2.4 Capacitive Reactance
Duration
 20 minutes
Logistics
 Use PowerPoint slides
56–61 and the IG to
present ELO 2.4.
Introduction
In AC circuits, capacitors present a resistance to current flow known as
capacitive reactance. In order to understand the behavior of AC circuits,
you must understand capacitive reactance and its effects on voltage and
current.
Capacitive Reactance
There are many natural forms of capacitance in AC power circuits, such as
transmission lines, fluorescent lighting, and computer monitors. Normally,
the inductors counteract the effects of capacitance in an electrical
distribution system. However, where capacitors outnumber inductive
devices, capacitive reactance will affect the amount of current flowing in an
AC electrical circuit.
Capacitors in an AC circuit charge and discharge in an attempt to keep
voltage constant. Capacitive reactance is the opposition by a capacitor (or a
capacitive circuit) to the flow of AC current. The capacitance of the circuit
and the rate at which the applied voltage is changing affect the current
flowing in a capacitive circuit. The frequency of the voltage supply
determines the rate at which the applied voltage is changing. If the supply
voltage frequency or the capacitance of a given circuit is increased, the
current flow will increase. If the frequency or capacitance is increased, the
opposition to current flow decreases; therefore, capacitive reactance, which
is the opposition to current flow, is inversely proportional to frequency and
capacitance.
The units of capacitive reactance (XC) are ohms, just like inductive
reactance. The equation below is a mathematical representation for
capacitive reactance.
𝑋𝐶 =
Rev 1
1
2𝜋𝑓𝐶
23
Where:
f = Frequency (Hz)
π ≈ 3.14
C = Capacitance (farads)
The equation below is the mathematical representation for the current that
flows in a circuit with only capacitive reactance.
𝐼=
𝐸
𝑋𝐶
Where:
I = Effective current (A)
E = Effective voltage across the capacitive reactance (V)
XC = Capacitive reactance (Ω)
Voltage and Current Relationships in a Capacitive Circuit
The figure below shows a plot of the variation of an alternating voltage
applied to a capacitor, the charge on the capacitor, and the current flowing
through the capacitor. The current flow in an AC circuit containing
capacitance depends on the rate at which the voltage changes.
The current flow in the figure below is greatest at points a, c, and e. At
these points, the voltage is changing at its maximum rate (i.e., passing
through zero). Between point a and point b, the voltage and charge are
increasing, and the current flow is into the capacitor, but decreasing in
value. At point b, the capacitor is fully charged, and the current is zero.
From point b to point c, the voltage and charge are decreasing as the
capacitor discharges, and its current flows in a direction opposite to the
voltage. From point c to point d, the capacitor begins to charge in the
opposite direction, and the voltage and current are again in the same
direction.
At point d, the capacitor is fully charged, and the current flow is again zero.
From point d to point e, the capacitor discharges, and the flow of current is
opposite to the voltage. The figure shows the current leading the applied
voltage by 90°.
In any purely capacitive AC circuit, current leads applied voltage by 90°.
24
Rev 1
Figure: Voltage, Charge, and Current in a Capacitive Circuit
Knowledge Check
Capacitive reactance is dependent on all of the
following except ____________________.
A.
applied voltage
B.
frequency
C.
area of the conducting plates
D.
dielectric constant
ELO 2.5 Impedance
Introduction
No circuit is without some resistance, whether desired or not. Both resistive
and reactive components in an AC circuit oppose current flow. The total
opposition to current flow in an AC circuit depends on its resistance, its
reactance, and the phase relationships between them.
Rev 1
25
Duration
 20 minutes
Logistics
 Use PowerPoint slides
62–64 and the IG to
present ELO 2.5.
Impedance
Impedance is the total opposition to current flow in an AC circuit. The
equation below is the mathematical representation for the magnitude of
impedance in an AC circuit.
𝑍 = √𝑅 2 + 𝑋 2
Where:
Z = Impedance (Ω)
R = Resistance (Ω)
X = Net reactance (Ω)
Resistance, Reactance and Impedance
The figure below shows the relationship between resistance, reactance, and
impedance in an AC circuit.
Figure: Relationship between Resistance, Reactance,
and Impedance
The current through a certain resistance is always in phase with the applied
voltage. Resistance (R) plots on the zero axis.
The current through an inductor lags applied voltage by 90°; inductive
reactance (XL) plots along the 90° axis.
Current through a capacitor leads applied voltage by 90°; capacitive
reactance (XC) plots along the -90° axis.
26
Rev 1
Knowledge Check
Adding a capacitor to an inductive circuit will
_______________________.
A.
reduce the impedance, because the capacitive reactance
counteracts some of the inductive reactance
B.
increase the impedance, because all reactance adds to
the impedance
C.
reduce the resistance of the circuit
D.
cause no change to the circuit at all, since it is
primarily inductive
ELO 2.6 Apparent, True, and Reactive Power
Introduction
Whereas a DC electrical circuit has one form of power, power in an AC
electrical circuit is comprised of three separate components, each uniquely
related to the other. This chapter will discuss true power, apparent power,
reactive power, and their behavior in AC electrical circuits.
The Power Triangle
In AC circuits, current and voltage are normally out of phase due to the
effects of inductive and capacitive reactance. As a result, not all the power
produced by a generator in an AC application can accomplish work.
Power calculations differ in AC circuits from DC circuits. The power
triangle, shown below, equates AC power to DC power by showing the
relationship between generator output (Apparent Power - S) in volt-amperes
(VA), usable power (True Power - P) in watts, and wasted or stored power
(Reactive Power - Q) in volt-amperes-reactive (VAR). The phase angle (θ)
represents the inefficiency of the AC circuit and corresponds to the total
reactive impedance (Z) to current flow in the circuit.
Rev 1
27
Duration
 30 minutes
Logistics
 Use PowerPoint slides
65–70 and the IG to
present ELO 2.6.
Figure: Power Triangle
The power triangle represents comparable values that can be used directly
to find the efficiency level of generated power to usable power, which is
expressed as the power factor (discussed later). You can calculate Apparent
Power, Reactive Power, and True Power by using the DC equivalent (RMS
value) of the AC voltage and current components, along with the power
factor.
Apparent Power
Apparent Power (S) is the power delivered to an electrical circuit. The
equation below is a mathematical representation of Apparent Power. Voltamperes (VA) are the units for Apparent Power.
𝑆 = 𝐼 2 𝑍 = 𝐼𝐸
Where:
S = Apparent Power (VA)
I = RMS current (A)
E = RMS voltage (V)
Z = Impedance (Ω)
True Power
True Power (P) is the power consumed by the resistive loads in an electrical
circuit. The equation below is a mathematical representation of true power.
Watts are the units for true power.
𝑃 = 𝐼 2 𝑅 = 𝐸𝐼 cos 𝜃
28
Rev 1
Where:
P = True Power (watts)
I = RMS current (A)
E = RMS voltage (V)
R = Resistance (Ω)
θ = Angle between E and I sine waves
Reactive Power
Reactive Power (Q) is the power component in an AC circuit necessary for
the expansion and collapse of magnetic (inductive) and electrostatic
(capacitive) fields. Volt-amperes-reactive (VAR) are the units for Reactive
Power.
The equation below is a mathematical representation for reactive power.
𝑄 = 𝐼 2 𝑋 = 𝐸𝐼 sin 𝜃
Where:
Q = Reactive power (VAR)
I = RMS current (A)
X = Net reactance (Ω)
E = RMS voltage (V)
θ = Angle between the E and I sine waves
Unlike True Power, Reactive Power is unusable power because it is stored
in the circuit itself. This power is stored by inductors, because they expand
and collapse their magnetic fields in an attempt to keep current constant,
and by capacitors, because they charge and discharge in an attempt to keep
voltage constant. In AC electrical circuits, inductance and capacitance
alternately exchange reactive power (consume and give back) with the
circuit’s AC source.
Reactive Power is a function of a system’s amperage. The power delivered
to the inductance is stored in the magnetic field during field expansion, and
returned to the source when the field collapses. The power delivered to the
capacitance is stored in the electrostatic field when the capacitor is
charging, and returned to the source when the capacitor discharges. The
circuit conserves reactive power, since none of the reactive power delivered
to the circuit by the source is consumed, but it is all returned to the source.
Additionally, these reactive loads consume no True Power in order to
maintain their magnetic and electrostatic fields. Alternating current
constantly changes; thus, the cycle of expansion and collapse of the
magnetic and electrostatic fields constantly occurs.
Circulating current is the term for this current that is constantly flowing
between the source and the inductive and capacitive loads in an AC circuit
in order to maintain magnetic fields. Circulating currents account for no
real work in the circuit.
Rev 1
29
Total Power
The Total Power delivered by the source is the same as Apparent Power.
Part of this Apparent Power, called True Power, dissipates by the circuit
resistance in the form of heat. The rest of the Apparent Power returns to the
source by the circuit inductance and capacitance (Reactive Power).
Knowledge Check
Match the power terms with their appropriate location
on the power triangle.
A.
Apparent power
B.
Reactive power
C.
True power
Knowledge Check Answer
A – Apparent power
C – True power
B – Reactive power
ELO 2.7 Power Factor
Duration
 20 minutes
Logistics
 Use PowerPoint slides
71–75 and the IG to
present ELO 2.7.
Introduction
In this section, you will learn the definition of power factor, and understand
how it affects AC power systems.
Power Factor
Power factor (pf) is the ratio between True Power and Apparent Power.
True Power is the power consumed by an AC circuit, whereas Apparent
Power is a representation of the total power delivered to an AC circuit.
Reactive Power accounts for a portion of the Apparent Power, which is
30
Rev 1
power that is stored in an AC circuit and accomplishes no real work in the
circuit.
Power factor is represented by cos θ in an AC circuit. It is the ratio of True
Power to Apparent Power, where θ is the phase angle between the applied
voltage and current sine waves and is the angle between P and S on a power
triangle.
Figure: Power Triangle
The equation below is a mathematical representation of power factor.
cos 𝜃 =
𝑃
𝑆
Where:
cos θ = Power Factor (pf)
P = True Power (watts)
S = Apparent Power (VA)
Lagging Power Factor
Power factor also determines what part of the Apparent Power is True
Power. It can vary from 1, when the phase angle is 0°, to 0, when the phase
angle is 90°. In an inductive circuit, the current lags the voltage. This type
of circuit has a lagging power factor, as shown in the figure below.
Rev 1
31
Figure: Lagging Power Factor
Leading Power Factor
In a capacitive circuit, the current leads the voltage. This type of circuit has
a leading power factor, as shown in the figure below.
Figure: Leading Power Factor
An electrical circuit that supplies power to loads such as motors, will
exhibit a lagging power factor. An electrical circuit that supplies power to
loads such as fluorescent lighting will exhibit a leading power factor. Most
industrial electrical distribution systems exhibit a lagging power factor
because inductive loads normally account for a larger percentage of the
reactance seen in these types of circuits.
ELI the ICE Man
You may want to use a mnemonic memory device, "ELI the ICE man," to
remember the voltage/current relationship in AC circuits. ELI refers to an
inductive circuit (L) where current (I) lags voltage (E). ICE refers to a
capacitive circuit (C) where current (I) leads voltage (E).
32
Rev 1
Knowledge Check
Select all of the statements about power factor (pf) that
are true.
A.
Power factor is the ratio of true power and reactive
power.
B.
Power factor cannot be greater than one.
C.
Power factor is the sine of the power triangle.
D.
Power factor is the ratio between true power and
apparent power.
TLO 2 Summary
In this section, you learned about inductors, capacitors, and about how they
affect AC power system operation. We also reviewed inductive and
capacitive reactance, and their phase relationship to current and voltage.
We reviewed the concept of total circuit resistance, termed impedance, and
the relationship between true, apparent and reactive power.
1. An inductor stores electrical energy in the form of a magnetic field.
The three requirements for inducing an EFM are: a conductor, a
magnetic field, and relative motion between them.
2. Capacitors are electrical devices constructed of two metal plates
separated by an insulating material, called a dielectric. Capacitors
store energy as an electric field between the two plates, as they
charge.
3. Inductors present a resistance to current flow that is termed inductive
reactance. Because this EMF opposes the continuous change in the
flowing current, we measure its effect in ohms. Any device relying
on magnetism or magnetic fields to operate is a form of inductor;
motors, generators, transformers, and coils are all inductors.
XL = 2πf
There are many natural forms of capacitance in AC power circuits,
such as transmission lines, fluorescent lighting, and computer
monitors. When capacitors outnumber inductive devices, capacitive
reactance will affect the amount of current flowing in an AC electrical
circuit. The units of capacitive reactance (XC) are ohms, just like
inductive reactance (XL).
1
𝑋𝐶 =
2𝜋𝑓𝐶
4. Impedance is the total opposition to current flow in an AC circuit.
Both resistive and reactive components in an AC circuit oppose
current flow. The total opposition to current flow in an AC circuit
depends on its resistance, its reactance, and the phase relationships
between them.
Rev 1
33
Duration
 15 minutes
Logistics
 Use PowerPoint slides
76–79 and the IG to
review TLO 2 material.
Use directed and nondirected questions to
students, check for
understanding of ELO
content, and review any
material where student
understanding of ELOs
is inadequate.
5. In AC circuits, current and voltage are normally out of phase due to
the effects of inductive and capacitive reactance. The power triangle
equates AC power to DC power by showing the relationship between
generator output (Apparent Power - S) in volt-amperes (VA), usable
power (True Power - P) in watts, and wasted or stored power
(Reactive Power - Q) in volt-amperes-reactive (VAR). The phase
angle (θ) represents the inefficiency of the AC circuit and corresponds
to the total reactive impedance (Z) to current flow in the circuit.
6. Power factor (pf) is the ratio between True Power and Apparent
Power. True Power is the power consumed by an AC circuit, whereas
Apparent Power is a representation of the total power delivered to an
AC circuit.
Now that you have completed this lesson, you should be able to:
1. Describe how current flow, magnetic field, and stored energy in an
inductor relate to one another, and how an inductor opposes a change
in current flow.
2. Describe the construction of a capacitor, explain how it stores energy,
and explain how it opposes a change in voltage.
3. Describe inductive reactance (XL) and the phase relationship between
current and voltage in an inductive circuit.
4. Define capacitive reactance (XC) and the phase relationship between
current and voltage in a capacitive circuit.
5. Define impedance (Z).
6. Define apparent, true, and reactive power using a power triangle.
7. Define power factor as it relates to true power and apparent power,
and define leading and lagging power factors.
TLO 3 Transformers
Overview
Duration
 5 minutes
Logistics
 Use PowerPoint slides
80–81 and the IG to
introduce TLO 3.
In this section, you will learn how transformers operate.
Transformers are necessary for power plant operation. Operators must
understand how they work in order to monitor and control them.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Define the common terms as they pertain to transformers: mutual
induction, turns ratio, impedance ratio, and efficiency.
2. Describe the construction of the following components of a
transformer: primary coil, secondary coil, and iron core.
3. Describe the voltage, current and power relationships between the
primary and secondary windings of transformers.
4. State the applications of each of the types of transformers: distribution
transformers, power transformers, control transformers, auto
transformers, isolation transformers, instrument potential
transformers, and instrument current transformers.
34
Rev 1
ELO 3.1 Transformer Terminology
Introduction
A transformer is a device that transfers electrical energy from one circuit to
another by electromagnetic induction. This energy always transfers without
a change in frequency, but usually with changes in current and voltage. In
this section, you will learn terminology frequently used in discussion of
transformer operations.
Transformer Terminology
Here are important terms for discussing transformer operation.
 Mutual Induction
If flux lines from the expanding and contracting magnetic field of one
coil cut the windings of another nearby coil, a voltage will be induced
in that coil. Inducing an EMF in a coil by magnetic lines of flux
generated in another coil is mutual induction. The amount of
electromotive force (EMF) induced by this method depends on the
relative positions of the two coils.

Turns Ratio
Each winding of a transformer (primary and secondary) contains a
certain number of turns of wire. The turns ratio is the ratio of the
number of turns of wire in the primary winding to the number of turns
of wire in the secondary winding, as shown in the equation below.
𝑇𝑢𝑟𝑛𝑠 𝑅𝑎𝑡𝑖𝑜 =
𝑁𝑃
𝑁𝑆
Where:
NP = Number of turns on the primary coil
NS = Number of turns on the secondary coil
 Impedance Ratio
Maximum power transfers from one circuit to another through a
transformer when the impedances are equal, or matched. A
transformer winding constructed with a definite turns ratio can
perform an impedance matching function. The turns ratio will
establish the proper relationship between the primary and secondary
winding impedances. The ratio between the two impedances is the
impedance ratio, as shown in the equation below.
𝑁𝑃 2 𝑍𝑝
( ) =
𝑁𝑆
𝑍𝑆

Efficiency
Efficiency of a transformer is the ratio of the power output to the
power input, as illustrated by the equation below.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑃𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑆
=
× 100
𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡
𝑃𝑃
Where:
Rev 1
35
Duration
 20 minutes
Logistics
 Use PowerPoint slides
82–85 and the IG to
present ELO 3.1.
PS = Power of secondary
PP = Power of primary
Knowledge Check
Match the terms with their appropriate definitions.
1. Inducing an EMF in a coil by magnetic
lines of flux generated in another coil
A. Efficiency
2. The ratio of the number of turns of wire in
the primary winding to the number of turns
of wire in the secondary winding
B. Mutual
induction
3. The ratio between the two impedances
C. Impedance
ratio
4. The ratio of the power output to the power
input
D. Turns
ratio
Knowledge Check Answer
1. B – Mutual induction
2. D – Turns ratio
3. C – Impedance ratio
4. A – Efficiency
ELO 3.2 Transformers Components
Duration
 20 minutes
Logistics
 Use PowerPoint slides
86–90 and the IG to
present ELO 3.2.
Introduction
In this section, you will learn the components of a transformer and how
each function.
Transformer Components
Every transformer has a primary winding and one or more secondary
windings. The primary winding receives electrical power from an AC
source and induces electrical energy into the secondary winding(s). The
energy appears as an electromotive force (EMF) across the secondary
winding, and if a load connects to the secondary, energy in the form of
current transfers to the load.
Transformers provide a means of transferring electrical energy from one
circuit to another with no direct electrical connections between the circuits.
They are used extensively for AC power transmission, various control, and
indication functions and for isolating electrical circuits from one another.
The most important application of a transformer is for raising (stepping-up)
or lowering (stepping down) the source of voltage to a desired level.
Transformers have the ability to convert power at a given current and
voltage to an equivalent power at a different current and voltage.
36
Rev 1
A transformer works on the principle that varying magnetic flux transfers
energy by magnetic induction from one set of coils to another. An AC
source produces this magnetic flux.
The coil of a transformer that is energized from an AC source is called the
primary winding (coil), and the coil that delivers the induced AC to the load
is called the secondary winding (coil).
In the figure below, the primary and secondary coils are on separate legs of
the magnetic circuit for ease of understanding. In actual construction, half
of the primary and secondary coils wind on each of the two legs, with
sufficient insulation between the two coils and the core to insulate the
windings from one another and the core.
A transformer wound such as the one below will operate at a greatly
reduced efficiency due to magnetic leakage. Magnetic leakage is the part of
the magnetic flux that passes through either one of the coils, but not through
both. As the distance between the primary and secondary windings
increases, the magnetic circuit lengthens, and the magnetic leakage
increases.
Figure: Basic Core Type Transformer
When alternating (AC) voltage is applied to the primary winding, an
alternating current will flow through the primary winding that will
magnetize the iron core, first in one direction and then in the other direction.
This alternating flux flowing around the entire length of the magnetic circuit
induces a voltage in both the primary and secondary windings. The induced
voltage will be at the same frequency as that of the AC source.
Since the same flux links both windings, the voltage induced per turn of the
primary and secondary windings must be the same value and same
direction. In the primary winding, this voltage opposes the voltage applied
to the primary winding and is counter-electromotive force (CEMF).
Rev 1
37
Knowledge Check
Match the terms to the appropriate description.
1. The coil that is energized by the AC
source
A. Leakage
2. The coil that is connected to the load
B. Primary
3. Magnetic material that the coils are
wound around
C. Secondary
Magnetic flux that passes through only
4. one of the coils
D. Core
Knowledge Check Answer
1. B - Primary
2. C - Secondary
3. D - Core
4. A - Leakage
ELO 3.3 Voltage, Current and Power Relationships
Duration
 20 minutes
Logistics
 Use PowerPoint slides
91–95 and the IG to
present ELO 3.3.
Introduction
One of the most important functions associated with transformers is their
ability to step-up or step-down voltage. The voltage induced in the
secondary windings of a transformer is dependent on the ratio of turns of
the primary winding to turns of the secondary winding.
Primary to Secondary Voltage Relationship
If the primary winding has more turns than the secondary winding, the
output voltage from the secondary winding will be lower than the voltage
applied to the primary winding (step-down transformer). Conversely, if the
secondary winding has a greater number of turns than the primary winding,
the output voltage from the secondary will be greater than the voltage
applied to the primary winding (step-up transformer).
The voltages induced in the windings of a transformer are directly
proportional to the number of turns of the coils in the transformer. The
equation below expresses this relationship.
𝑉𝑃 𝑁𝑃
=
𝑉𝑆 𝑁𝑆
Where:
VP = Voltage on primary coil
VS = Voltage on secondary coil
38
Rev 1
NP = Number of turns on the primary coil
NS = Number of turns on the secondary coil
The ratio of primary voltage to secondary voltage is the voltage ratio (VR).
The ratio of primary turns of wire to secondary turns of wire is the turns
ratio (TR). By substituting into the equation above, we find that the voltage
ratio is equal to the turns ratio.
𝑉𝑅 = 𝑇𝑅
A voltage ratio of 1:5 (read 1 to 5) means that for each volt on the primary,
there will be 5 volts on the secondary. If the secondary voltage of a
transformer is greater than the primary voltage, the transformer is a step-up
transformer. A ratio of 5:1 means that for every 5 volts on the primary,
there will only be 1 volt on the secondary. When secondary voltage is less
than primary voltage, the transformer is a step-down transformer.
The current ratio is inversely proportional to the voltage ratio in a
1
1
transformer, therefore, the CR =
= .
𝑉𝑅
𝑇𝑅
Primary to Secondary Current Relationship
The current in the windings of a transformer is inversely proportional to the
voltage in the windings. A transformer that steps-up voltage, will stepdown current, and vice-versa. The equation below expresses this
relationship.
𝑉𝑃 𝐼𝑆
=
𝑉𝑆 𝐼𝑃
Where:
IP = Primary coil current
IS = Secondary coil current
Since the voltage ratio is equal to the turns ratio, we can express the current
ratio in terms of the turns ratio as well, as in the equation below.
𝑁𝑃 𝐼𝑆
=
𝑁𝑆 𝐼𝑃
Primary to Secondary Power Relationship
It is important to remember that regardless of whether a transformer is
acting to step-up or step-down voltage, ideally, the input power of the
transformer remains equal to the output power of the transformer (minus
any internal losses associated with the transformer). The equation below
demonstrates this relationship.
Rev 1
39
𝑃𝑃𝑟𝑖𝑚𝑎𝑟𝑦 = 𝑃𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦
𝑉𝑃 𝐼𝑃 = 𝑉𝑆 𝐼𝑆
Where:
PPrimary = Input power
PSecondary = Output power
VP = Primary voltage
VS = Secondary voltage
IP = Primary current
IS = Secondary current
Therefore, if voltage increases by a factor of 5 (VR = 5), the current is
reduced by a factor of 5 (CR = 1/5).
Knowledge Check
Match the terms with their appropriate definitions.
1. Voltage on the primary side divided by
voltage on the secondary side
A. Step-up
transformer
2. Ratio of primary turns to secondary
turns
B. Step-down
transformer
3. A transformer with higher secondary
voltage than primary voltage
C. Voltage
ratio
4. A transformer with lower secondary
voltage than primary voltage
D. Turns ratio
Knowledge Check Answer
1. C – Voltage ratio
2. D – Turns ratio
3. A – Step-up transformer
4. B – Step-down transformer
ELO 3.4 Transformer Applications
Duration
 20 minutes
Logistics
 Use PowerPoint slides
96–103 and the IG to
present ELO 3.4.
Introduction
Transformers construction varies so that a transformer’s characteristics
match its intended application. Differences in construction may involve the
size of the windings or the relationship between the primary and secondary
windings. The function the transformer serves in a circuit, i.e. an isolation
transformer, is another transformer designation.
40
Rev 1
Distribution Transformer
Electrical power distribution and transmission systems make extensive use
of distribution transformers. This class of transformer has the highest
power, or volt-ampere ratings, and the highest continuous voltage rating.
The cooling method associated with the transformer normally determines
the power rating for a distribution transformer. Some transformers use oil
or some other heat-conducting material to remove heat. Others use forced
air cooling (fans).
In a distribution transformer, increasing the size of the primary and
secondary windings increases the ampere rating. Increasing the voltage
rating of the insulation used in manufacturing the transformer increases the
voltage rating.
Power Transformers
Electronic circuits use power transformers; these come in many different
types and applications. Electronics or power transformers generally have a
rating of 300 volt-amperes and below. These transformers normally
provide power to the power supply circuit of an electronic device. The
power amplifier used in an audio receiver is an example of this type of
transformer.
Control Transformers
Electronic circuits that require constant voltage or constant current with a
low power or volt-amp rating generally use control transformers. Various
filtering devices, such as capacitors, minimize variations in the output of
these types of transformers, resulting in a more constant voltage or current.
Auto Transformers
Low power applications where a variable voltage is required generally use
an auto transformer. The auto transformer is a special type of power
transformer, because it consists of only one winding. Tapping or
connecting at differing points along the winding will yield different
voltages. The figure below shows a schematic of an auto transformer.
Figure: Auto Transformer Schematic
Rev 1
41
Isolation Transformers
Isolation transformers are normally low power transformers used to isolate
noise from or to ground electronic circuits. Since a transformer cannot pass
DC voltage from primary to secondary, any DC voltage present in the
circuit (such as noise) will not pass through; therefore, the transformer acts
to isolate this noise.
Isolation transformers have application in electrical circuits to ensure that a
fault developed in one portion of the circuit will not affect another portion
of the circuit. This is possible because there is no direct electrical
connection between the primary and secondary windings in a transformer.
The primary and secondary winding connect by magnetic flux only.
Instrument Potential Transformers
The instrument potential transformer (PT) steps down the voltage of an
electrical circuit to a low value that can be effectively and safely used for
the operation of instruments such as ammeters, voltmeters, watt meters and
relays used for various protective purposes.
Instrument Current Transformers
The instrument current transformer (CT) steps down the current of a circuit
to a lower value. The same types of equipment as the potential transformer
described above use this type of transformer. In an instrument current
transformer, the secondary winding is a coil consisting of many turns of
wire, wound around the primary coil, which contains only a few turns of
wire. This allows measurements of high values of current.
Because of the design of this type of transformer, it is necessary to follow a
special procedure when not being operated under load. A current
transformer should always be short-circuited when not connected to an
external load. The magnetic circuit design of a current transformer is for
low magnetizing current when under load; because of this, a large increase
in magnetizing current will result in the build up a large flux in the
magnetic circuit. This will cause the transformer to act as a step-up
transformer, inducing an excessively high voltage in the secondary when
under no load.
42
Rev 1
Knowledge Check
Match the terms with the appropriate description.
1. Used in electrical power distribution
and transmission systems
A. Isolation
transformer
2. Sometimes used in electrical circuits
to ensure that a fault developed in
one portion of the circuit will not
affect another portion of the circuit
B. Auto
transformer
3. Consists of only one winding
C. Distribution
transformer
4. Steps down the voltage of an
electrical circuit to a low value that
can be effectively and safely used for
the operation of instruments
D. Instrument
potential
transformer
Knowledge Check Answer
1. C – Distribution transformer
2. A – Isolation transformer
3. B – Auto transformer
4. D – Instrument potential transformer
TLO 3 Summary
In this section, you learned the different types of transformers and their
applications.
Now that you have completed this lesson, you should be able to:
1. Define the common terms as they pertain to transformers: mutual
induction, turns ratio, impedance ratio, and efficiency.
2. Describe the construction of the following components of a
transformer: primary coil, secondary coil, and iron core.
3. Describe the voltage, current and power relationships between the
primary and secondary windings of transformers.
4. State the applications of each of the types of transformers:
Distribution Transformer, Power Transformers, Control
Transformers, Auto Transformers, Isolation Transformers, Instrument
Potential Transformers, and Instrument Current Transformers.
Rev 1
43
Duration
 15 minutes
Logistics
 Use PowerPoint slides
104–105 and the IG to
review TLO 3 material.
Use directed and nondirected questions to
students, check for
understanding of ELO
content, and review any
material where student
understanding of ELOs
is inadequate.
TLO 4 Electrical Distribution Systems
Overview
Duration
 5 minutes
Logistics
 Use PowerPoint slides
106–107 and the IG to
introduce TLO 4.
In this section, you will learn the principles and operation of electrical
distribution systems.
Operators must understand how electrical distribution systems work in
order to monitor and control plant electrical equipment.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the design of a basic industrial electrical distribution system.
2. Define common terms associated with electrical distribution systems
and wiring schemes used in these systems.
3. Describe the two methods of connecting single-phase loads to a threephase power source and the advantages of three-phase systems.
4. Given a diagram of a wye or delta-connected three-phase system,
describe the voltage/current relationships of the circuit.
5. State the indications of an unbalanced load in a three-phase power
system.
6. Describe the purpose of common power distribution schemes.
ELO 4.1 Basic Electrical Distribution System
Duration
 20 minutes
Logistics
 Use PowerPoint slides
108–116 and the IG to
present ELO 4.1.
Introduction
An industrial facility’s electrical distribution system comprises all of the
electrical hardware that is located between the site-generated electrical
power or commercially purchased electrical power, and facility loads.
Electrical Distribution System
The electrical distribution system provides the power for process loads such
as pumps, fans, compressors, lighting, instrumentation, heating ventilation
and air conditioning, and operating and control circuits. It also provides the
means to connect and disconnect electrical power to these loads via cables,
wires, and various types of protective devices.
Because the electrical system interfaces with almost all of the equipment in
an industrial facility, it is vital that at least part of the electrical distribution
system remain powered at all times. Sources of backup power such as
emergency diesel generators, batteries, and uninterruptible power supplies
(UPS) help ensure constant power to vital components.
A basic electrical distribution system consists of three parts:
1. A generating system
2. A transmission system
3. A distribution system
44
Rev 1
The figure below shows key components of a typical industrial electrical
distribution system.
Figure: Typical Industrial Electrical Distribution System
Generating System
The generating system for a typical industrial facility’s electrical system
consists of sources of power, together with associated step-up and stepdown transformers. The local commercial (purchased) power system or onsite electrical generation are the sources of generating facilities.
Power to industrial facilities normally falls into two general categories;
normal power and standby power. Normal power is the power normally
supplied to the facility’s electrical equipment. It generally comes from a
normal distribution system. Standby power supplies power to electrical
loads at the facility when normal power is lost. Standby power sources are
generally from an on-site diesel generator or from backup batteries.
Transmission System
The transmission system includes the high-voltage cables, power poles, and
switching stations that connect the generating system to the distribution
system. The transmission system forms a grid that extends across the
industrial facility.
Transmission system voltage can be very high (115 kVAC is common).
Because of power (I2R) losses over long runs of electrical cable, electrical
transmission systems transform voltage to a high value in order to lower the
current flowing through transmission lines. High voltage/low current
allows for transmission of electricity over great distances with minimal
power loss due to the resistance of the electrical cables.
Rev 1
45
Distribution System
The distribution system for a typical industrial facility usually consists of
one or more electrical substations, where the incoming high voltage from
the transmission system goes through transformers down to a lower voltage
(13.8 kVAC or 480 VAC) for use by the facility’s electrical loads.
Substations
Electrical substations transform incoming high voltage down to a value
usable by the facility and distribute it to electrical equipment throughout the
facility. This transformation often takes two steps. First, substation
distribution transformers transform voltage from 115 kVAC to an
intermediate value of voltage such as 13.8 kVAC. This intermediate
voltage may power some large facility loads directly.
In order to obtain the lower voltage (480 VAC) required by most facility
electrical loads, a second substation transforms the 13.8 kVAC again, to
480 VAC. From the 480 VAC substation, a switchgear lineup distributes
electrical power to various facility load centers, motor control centers, and
individual electrical loads.
Load Centers
Switchgear is a term used to describe a group of circuit breakers tied to a
common electrical bus. An electrical bus is a copper bar (single-phase) or
set of copper bars (three-phase) used to connect the switchgear to the
distribution transformer and to its associated breakers.
This type of switchgear is a load center. A load center is typically equipped
with a main supply breaker to receive power from a distribution
transformer, and several feeder breakers to supply power to various motor
control centers in the facility and larger electrical loads.
Electrical switchgear is often provided with meters and indicating devices
which can be used by facility operators to monitor distribution system
performance.
Motor Control Centers
Motor control centers (MCCs) act as centralized distribution and control
points for various 480 VAC components and loads. MCCs receive power
from a load center and distribute this power to individual electrical loads via
breakers, fuses, motor controllers, etc. MCCs may also feed lighting and
instrumentation loads via transformers (480 VAC to 208/120 VAC).
Like load centers, MCCs are equipped with monitoring and instrumentation
devices such as voltmeters, ammeters, etc.
46
Rev 1
Knowledge Check
Match the sub-system of the electrical distribution system with the
appropriate definition.
1. One or more electrical substations,
where the incoming high voltage is
transformed down to a lower
voltage
A. Substations
2. Includes the high-voltage cables,
power poles, and switching stations
forming a “grid” which extends
across the industrial facility.
B. Load centers
3. Transform incoming high voltage
down to a value usable by the
facility and distribute it to electrical
equipment throughout the facility.
C. Distribution
system
4. Equipped with a main supply
breaker to receive power and
several feeder breakers to supply
power to various motor control
centers in the facility and larger
electrical loads
D. Transmission
system
Knowledge Check Answer
1. C – Distribution system
2. D – Transmission system
3. A – Substations
4. B – Load centers
ELO 4.2 Electrical Distribution Terminology
Introduction
In this section, you will learn the meaning of terms that are used frequently
in describing electrical systems.
Electrical Distribution Terminology
Electrical distribution systems include numerous types of devices. Each of
these devices performs a specific function within the distribution system.
The following is a list of common devices found in electrical distribution
systems and their definitions.
 Area Substation – receives power from the site distribution system for
use in a particular facility. A substation typically contains a switch, a
transformer, and a circuit breaker.
Rev 1
47
Duration
 20 minutes
Logistics
 Use PowerPoint slides
117–122 and the IG to
present ELO 4.2.












Distribution Substation – a substation that receives high voltage
transforms it down to a lower voltage and distributes it to electrical loads
via circuit breakers.
Load Center – receives power from a substation and supplies power to
facility motor control centers and larger electrical loads.
Motor Control Center – receives 480 VAC power and distributes it to
individual process loads and other electrical panels via circuit breakers,
controllers, fuses, etc.
Contactor – an electro-mechanical device that controls power to a piece
of equipment, normally associated with motor controllers.
Panel boards – small distribution panels that contain numerous moldedcase circuit breakers, usually provides power to 208 VAC or 120 VAC
loads.
Tie-Breaker – circuit breaker used to tie two electrical busses together.
Main Breaker – circuit breaker used to connect bus bars of switchgear
assemblies to the output of transformers.
Feeder Breaker – circuit breaker used to receive power from a
switchgear bus bar, and direct the power to downstream electrical loads.
Switchgear – an assembly of circuit breakers electrically connected to a
system of electrical busses (solid copper conductors).
Normal Power – power received from the normal supply source at an
industrial facility; normal power may be generated on-site or purchased
from a commercial utility.
Standby Power – power source that comes on line upon a loss of normal
power; a diesel generator or by battery backup power via a UPS may
supply standby power.
Vital/Essential Loads – loads requiring constant power to ensure that a
facility can operate safely.
Wiring Scheme Terminology
In this section, you will learn terms commonly used to describe wiring
schemes.
To understand wiring schemes used in power distribution systems,
familiarity with the following terms is required.
 Ampacity – the maximum sustained current (in amperes) that a
conductor can carry while remaining within its temperature rating.
 Bond – the permanent joining of metallic parts or circuits assuring
electrical continuity, and safe current conductance for any expected
current.
 Conductor – any wire, cable, or substance that is capable of carrying an
electrical current.
 Ground – a conducting connection, whether intentional or accidental,
between a circuit or piece of equipment and the earth, or a body serving
as earth, which has zero electrical potential.
 Leg – a current-carrying conductor intended to deliver power to or from
a load normally at an electrical potential other than ground.
 Neutral – a current-carrying conductor normally tied to ground so that
the electrical potential is zero.
48
Rev 1

Phase voltage – the greatest root mean square (effective) difference of
potential between any two legs of the circuit.
Knowledge Check
Match the terms with their appropriate description.
1. An electro-mechanical device that
controls power to a piece of
equipment
A. Contactor
2. Receives power from a substation
and supplies power to facility
motor control centers and larger
electrical loads
B. Feeder
breaker
3. Small distribution panels, which
contain numerous molded-case
circuit breakers. Usually provide
power to 208 VAC or 120 VAC
loads.
C. Load center
4. Circuit breaker, which receives
power from a switchgear buss bar
and directs the power to
downstream electrical loads
D. Panel boards
Knowledge Check Answer
1. A – Contactor
2. C – Load center
3. D – Panel boards
4. B – Feeder breaker
Rev 1
49
Knowledge Check
Match the terms with the appropriate definitions.
1. The current in amperes that a
conductor can carry continuously
under the conditions of use without
exceeding its temperature rating
A. Ampacity
2. A conducting connection, whether
intentional or accidental, between a
circuit or piece of equipment and the
earth, or some body serving as earth;
a place of zero electrical potential
B. Bond
3. A current-carrying conductor
normally tied to ground so that the
electrical potential is zero
C. Ground
4. Permanent joining of metallic parts
or circuits assuring electrical
continuity and capacity to safely
conduct any current
D. Neutral
Knowledge Check Answer
1. A – Ampacity
2. C – Ground
3. D – Neutral
4. B – Bond
ELO 4.3 Single-Phase Load Connections and Three-Phase
Systems
Duration
 30 minutes
Logistics
 Use PowerPoint slides
123–127 and the IG to
present ELO 4.3.
Introduction
This section explains how single-phase power connects to the power
system.
Single-Phase Load Connections
The source of single-phase power in all facilities is by generation from a
single-phase generator or by utilization of one phase of a three-phase power
source. Each phase of the three-phase distribution system is a single-phase
generator electrically spaced 120 degrees from the other two; therefore, a
three-phase power source is convenient and practical to use as a source of
single-phase power.
Single-phase loads can connect to three-phase systems utilizing two
methods. The diagram shown in the figure below illustrates these
connections.
50
Rev 1
Figure: Three-Phase to Single-Phase Connection
The first scheme (A) in the figure above provides connection of the load
from a phase leg to a ground point, referred to as a phase-to-ground scheme.
The second scheme (B) in the figure above connects the single-phase load
between any two legs of the three-phase source, referred to as a phase-tophase connection.
The choice of schemes, phase-to-phase or phase-to-ground, allows several
voltage options depending on whether the source system is a three-phase
delta or wye configuration. The three-phase segment of this chapter will
discuss this topic.
Advantages of Three-Phase Systems
The design of three-phase AC circuits lends itself to a more efficient
method of producing and utilizing an AC voltage.
A three-phase electrical system is a combination of three single-phase
electrical systems. In a three-phase balanced system, power comes from a
three-phase AC generator that produces three separate and equal voltages,
each of which is 120° out of phase with the other two voltages.
Figure: Three-Phase AC
Three-phase equipment (motors, transformers, etc.) weighs less and is more
efficient than single-phase equipment of the same power rating. ThreeRev 1
51
phase generation has a wide range of voltages, and can power single-phase
loads.
Knowledge Check
Which of the following are common three-phase to
single-phase connection methods? (select all that are
correct)
A.
Phase to line
B.
Phase to ground
C.
Phase to phase
D.
Neutral to ground
Knowledge Check
Which of the following is not an advantage of threephase systems?
A.
Three-phase equipment weighs less than single-phase
equipment with the same rating.
B.
Three-phase equipment is smaller than single-phase
equipment of the same rating.
C.
Three-phase equipment is more efficient than singlephase equipment.
D.
Three-phase equipment is DC, so there is no need for
transformers.
ELO 4.4 Wye and Delta Systems
Duration
 30 minutes
Logistics
 Use PowerPoint slides
128–136 and the IG to
present ELO 4.4.
Introduction
Three-phase systems can connect in two different ways. These connections
are Wye (Y) and Delta (Δ) connections.
Wye-Connected
If the three common ends of each phase connect at a common point
(neutral) and the other three ends connect to a three-phase line, it is a wye,
or Y, connection.
52
Rev 1
Figure: Wye-Connected
Delta-Connected
If the three phases connect in series to form a closed loop, it is a delta, or Δ,
connection.
Figure: Delta-Connected
Balanced Three-Phase System
A three-phase system, that has identical impedance in each secondary
winding, has balanced loads. See the figure below for balanced ∆ and
balanced Y systems. The impedance of each winding in a delta load is
shown as Z∆ (a), and the impedance in a wye load is shown as ZY (b). For
either the delta or the wye connection, the lines A, B, and C supply a threephase system of voltages.
Rev 1
53
Figure: Three-Phase Balanced Loads
Voltage and Current in Delta-Connected Systems
The table below contains the formulas for calculating line and phase voltage
and current for delta-connected systems.
Delta-Connected Systems
Formula
Line Current
𝐼𝐿 = √3𝐼𝑝ℎ𝑎𝑠𝑒
Line Voltage
𝑉𝐿 = 𝑉𝑝ℎ𝑎𝑠𝑒
Voltage and Current in a Wye-Connected System
The table below contains the formulas for calculating line and phase voltage
and current in a Wye-connected system.
Wye-Connected Systems
Formula
Line Current
𝐼𝐿 = 𝐼𝑝ℎ𝑎𝑠𝑒
Line Voltage
𝑉𝐿 = √3𝑉𝑝ℎ𝑎𝑠𝑒
Phase Power
Because the impedance of each phase of a balanced delta or wye system has
equal current, phase power is one third of the Total Power. The equation
below is the mathematical representation for phase power (Pphase) in a
balanced delta or wye system.
𝑃𝛷 = 𝑉𝛷 𝐼𝛷 cos 𝜃
54
Rev 1
Total Power
Total Power (PT) is equal to three times the single-phase power. The
equation below is the mathematical representation for Total Power in a
balanced delta or wye system.
𝑃𝑇 = 3𝑉𝛷 𝐼𝛷 cos 𝜃
Delta-Connected System
In a delta-connected system,
𝑉𝐿 = 𝑉𝑝ℎ𝑎𝑠𝑒 and 𝐼𝑝ℎ𝑎𝑠𝑒 =
√3𝐼𝐿
so: 𝑃𝑇 = √3𝑉𝐿 𝐿𝐿 cos 𝜃
3
Wye-Connected System
In a wye-connected load,
𝐼𝐿 = 𝐼𝑝ℎ𝑎𝑠𝑒 and 𝑉𝑝ℎ𝑎𝑠𝑒 =
√3𝑉𝐿
so: 𝑃𝑇 = √3𝑉𝐿 𝐿𝐿 cos 𝜃
3
The above equations demonstrate that Total Power formulas for delta- and
wye-connected systems are identical.
Apparent Power and Reactive Power
Total Apparent Power (ST) in volt-amperes and total Reactive Power (QT)
in volt-amperes-reactive are related to total True Power (PT) in watts as
shown in the figure below.
Figure: Power Triangle
A balanced three-phase system has True, Apparent, and Reactive powers
given by the following equations.𝑃𝑇 = √3𝑉𝑇 𝐼𝐿 cos 𝜃
Rev 1
55
𝑆𝑇 = √3𝑉𝑇 𝐼𝐿
𝑄𝑇 = √3𝑉𝑇 𝐼𝐿 sin 𝜃
Calculating Voltage and Current Demonstration
In a Wye-connected power system, phase voltage is 150 volts, and phase
current is 5 amps. Calculate line voltage and line current.
Solution:
𝐿𝑖𝑛𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = 𝑃ℎ𝑎𝑠𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 5 𝑎𝑚𝑝𝑠
𝐿𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = √3
𝑃ℎ𝑎𝑠𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 260 𝑣𝑜𝑙𝑡𝑠.
Knowledge Check
A three-phase power system has the following
parameters. Line current is 12 amps. Phase current is
12 amps. This system is a _____________________.
A.
DC system
B.
Delta-connected system
C.
Wye-connected system
D.
X-connected system
ELO 4.5 Unbalanced Loads
Duration
 20 minutes
Logistics
 Use PowerPoint slides
137–139 and the IG to
present ELO 4.5.
Introduction
In this section, you will learn the indications of an unbalanced load.
Unbalanced Loads
An important property of a three-phase balanced system is that the phasor
sum of the three line or phase voltages is zero, and the phasor sum of the
three line or phase currents is zero. When the three load impedances are not
equal to one another, the phasor sums and the neutral current (In) are not
zero, and the load is, therefore, unbalanced. An imbalance occurs when an
open or short circuit appears at the load. The figure below shows balanced
and unbalanced three-phase systems.
56
Rev 1
Figure: Three-Phase Systems
In a fault condition, the neutral connection in a wye-connected load will
carry more current than the phase under a balanced load. Abnormally high
currents in one or more of the phases indicate unbalanced three-phase
circuits. If allowed to continue, this may cause damage to equipment.
Knowledge Check
A three-phase Wye-connected system with an
unbalanced load will have a neutral current of zero.
A.
False
B.
True
ELO 4.6 Power Distribution Schemes
Introduction
In this section, you will learn about different distribution system layouts.
Three-Wire, Single-Phase Edison System
The only approved method of wiring single-phase power from a three-phase
system is the scheme commonly referred to as the 3-wire, single-phase
Edison system. The illustration below depicts the use of a center-tapped
transformer, with the center tap grounded, providing half voltage (120 V)
connections on either side or full voltage (240 V) across both sides.
Figure: Three-Wire Edison Scheme
Rev 1
57
Duration
 30 minutes
Logistics
 Use PowerPoint slides
140–149 and the IG to
present ELO 4.6.
The physical connections to the transformer secondary involve two
insulated conductors and one bare conductor. Conductors for currentcarrying legs or neutral legs require insulation. The remaining un-insulated
conductor serves as a safety ground, bonded to the ground point of the
system.
In all cases, three wires will connect to the load terminals. The safety
ground will connect to each junction box, or device, in the distribution
system. In the case of half voltage (120 V) use, the intended path of the
current is from the supply leg through the load and back to the source via
the neutral leg. The ground carries no current unless a fault occurred in the
system, in which case the current would flow safely to ground.
In the full voltage system (240 V), the insulated conductors are connected
across the full winding of the transformer, and the un-insulated conductor is
again bonded to the grounded center tap. In a balanced system, all currents
will flow on the insulated conductors, and the grounded neutral will carry
no current, acting only in a ground capacity. In the case of either an
unbalanced load or a fault in the system, the bare conductor will carry
current, but the potential will remain at zero volts because it connects to the
ground point. As in the case of the half voltage system, the un-insulated
conductor will connect to each device in the system for safety.
Three-Phase Wiring Schemes
Unlike the single-phase wiring scheme that must make a provision for a
neutral leg and separate ground, the three-phase system needs neither a
separate neutral nor a ground to operate safely. However, to prevent an
unsafe condition, all 3 and 4-wire, three-phase systems can include an
effective ground path. As with the previous single-phase discussion, only
the secondary side of the transformer and its connected load need study.
Three-Wire, Three-Phase Delta System
The simplest three-phase system is the 3-wire Delta configuration, normally
used for transmission of power in the intermediate voltage class from
approximately 15,000 volts to 600 volts. The figure below shows the two
methods of connecting the Delta secondary.
58
Rev 1
Figure: Three-Wire, Three-Phase Delta Scheme
The upper diagram depicts the ungrounded Delta, normally confined to
protected environments such as fully enclosed ducts or overhead
transmission lines where access by personnel requires extraordinary means.
Each conductor’s ground voltage is equal to the full phase voltage of the
system.
The lower diagram shows a ground point affixed to one corner of the Delta,
which effectively lowers one phase’s voltage reference to ground to zero,
but retains a phase-to-phase voltage potential. The corner-grounded phase
acts in much the same way as the grounded neutral of the single-phase
Edison system, carrying current and maintaining ground potential.
The corner-grounded Delta system has an obvious economy in wiring costs,
and the grounded phase can physically protect the other two phases from
accidental grounding or lightning strikes in outdoor settings. This system is
rarely used for low voltage (under 600 V), however, because of the absence
of a safety ground required by many facilities for circuits involving
potential worker contact.
Four-Wire, Three-Phase Delta System
The 4-wire, three-phase Delta system combines the ungrounded Delta
discussed above for three-phase loads with the convenience of the Edison
system for single-phase loads. As depicted below, one side of the Delta has
Rev 1
59
a grounded-neutral conductor connected to a center tap winding on one
phase.
Figure: Four-Wire Delta System
The single-phase voltage on each side of the half-tap is one-half the voltage
available in the normal phase-to-phase relationship. This provides the same
half-or full-voltage arrangement seen in the normal Edison scheme with a
grounded neutral.
Notice also that the legs coming from the corners of the Delta would have a
normal ungrounded appearance if it were not for the center tap of one phase.
Thus, at any given location in the system, either three-phase power at full
voltage or single-phase power with half or full voltage is equally possible.
However, several strict procedures are required in the operation of this
system.
 Carefully balance all loads on both the single-phase and three-phase legs.
 Because the voltage between one leg and the grounded neutral is
considerably higher than the rest of the single-phase system, you must
take a measurement between the neutral and the phase to identify the
high leg, or abnormal voltage.
 Never use the high leg as a single-phase source because no ground or
grounded neutral exists for this circuit.
4-Wire, Three-Phase Wye System
Until now, the voltage, the phase voltage, and the ground voltage of the
three-phase systems have been equal, with the one exception of one phase
of the corner-grounded Delta. The Wye system has completely different
voltage characteristics from the Delta system. In the Wye system, the
ground voltage or voltage available from phase to ground is the phase
voltage divided by 1.73.
In the figure below, an example of the Wye system, (or center-grounded
Wye), extends three current-carrying insulated conductors and an insulated
grounded neutral to the loads.
Depending on the selection of conductors, one of the following is available:
60
Rev 1



A reduced-voltage single-phase between a phase leg and the neutral leg
A full-voltage single-phase circuit between any two phase legs
A full-voltage three-phase power
Some precautions are necessary when balancing the single-phase loads in
the system. Size the full load ampacity of the neutral to 1.73 times the
highest phase ampacity. This is to avoid either an over-current condition if
a fault is present or the operation of single-phase loads at reduced voltage if
an accidental interruption causes severely unbalanced loads.
Figure: Four-Wire, Three-Phase Wye System
As with all other grounded systems, establish bonds between the grounded
neutral and all components of the system. This system is the safest
possible, multi-purpose distribution system for low voltage and is common
in the 208/120-volt range in many facilities.
Knowledge Check
The _________________________ is normally
confined to protected environments such as fully
enclosed ducts or overhead transmission lines that
cannot be reached by personnel without extraordinary
means.
Rev 1
A.
4-wire, three-phase delta system
B.
Ungrounded delta system
C.
Edison system
D.
4-wire, three-phase Wye system
61
TLO 4 Summary
Duration
 15 minutes
Logistics
 Use PowerPoint slides
150–154 and the IG to
review TLO 4 material.
Use directed and nondirected questions to
students, check for
understanding of ELO
content, and review any
material where student
understanding of ELOs
is inadequate.
In this section, you learned the different types of distribution systems and
the advantages of each type. You also learned the means of connecting
single-phase loads to three-phase power supplies.
Now that you have completed this lesson, you should be able to:
1. Describe the design of a basic industrial electrical distribution system.
2. Define common terms associated with electrical distribution systems
and wiring schemes used in these systems.
3. Describe the two methods of connecting single-phase loads to a threephase power source and the advantages of three-phase systems.
4. Given a diagram of a wye or delta-connected three-phase system,
describe the voltage/current relationships of the circuit.
5. State the indications of an unbalanced load in a three-phase power
system.
6. Describe the purpose of common power distribution schemes.
TLO 5 Electrical Test Equipment
Overview
Duration
 5 minutes
Logistics
 Use PowerPoint slides
155–156 and the IG to
introduce TLO 5.
 Bring sample meters, test
equipment from
Electrical lab to
demonstrate, if available.
Ensure that any required
PPE is used.
In this section, you will learn the uses of various types of electrical test
equipment, and how they must connect to the circuits they measure.
Basic knowledge of electrical test equipment allows an operator to avoid
equipment damage and personnel injury during use; it also helps ensure that
operators collect correct/accurate data.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the use of the following electrical test meters: ground
detector, multimeter, megger, and synchroscope.
2. Describe the operation and electrical parameter measured of the
following electrical test equipment: ground detector, multimeter,
megger, and synchroscope.
ELO 5.1 Use of Common Test Equipment
Duration
 10 minutes
Logistics
 Use PowerPoint slides
157–159 and the IG to
present ELO 5.1.
Introduction
In this section, you will learn about common types of electrical test meters
and equipment.
Ground Detectors
The ground detector is an instrument that detects conductor insulation
resistance to ground. An ohmmeter, or a series of lights, can detect the
insulation strength of an ungrounded distribution system.
62
Rev 1
Most power distribution systems in use today are of the grounded variety;
however, some ungrounded systems still exist.
Multimeter
The multimeter is a portable, single instrument capable of measuring
various electrical values including voltage, resistance, and current.
Meg-Ohm Meter (Megger)
The megger is a portable instrument used to measure insulation resistance.
The megger consists of a hand-driven DC generator and a direct reading
ohmmeter.
Synchroscope
A synchroscope indicates when two AC generators are in the correct phase
relation for connecting in parallel and shows whether the incoming
generator is running faster or slower than the on-line generator.
Knowledge Check
Parameters normally measured by installed meters
include _________. (select all that are true)
A.
hysteresis
B.
power (watts)
C.
voltage
D.
current (amps)
ELO 5.2 Operation of Common Test Equipment
Introduction
In this section, you will learn about the operation of the test equipment and
parameters measured.
Ground Detectors
Ground detectors measure insulation resistance to ground in ohms.
Ohmmeter Method of Ground Detection
In the ohmmeter method (shown in the figure below), a DC voltage is
applied to the conductor. If a leakage path exists between the conductor
insulator and ground, a current will flow through the ground to the
ohmmeter proportional to the insulation resistance of the conductor.
Rev 1
63
Duration
 30 minutes
Logistics
 Use PowerPoint slides
160–173 and the IG to
present ELO 5.2.
Figure: Simple Ohmmeter Ground Detector
Lamp-Type Ground Detector
In the ground detector lamp method (shown below), a set of three lamps
connect through transformers to the system. To check for grounds, the
switch is closed and the brilliance of the lamps observed. If the lamps are
equally bright, no ground exists and all the lamps receive the same voltage.
If any one of the three lamps is dark, and the other two lamps are brighter,
the phase with the darkened lamp has a short to ground. In this case, the
primary winding of the transformer has a short to ground and receives no
voltage.
Figure: Lamp-Type Ground Detector Circuit
Multimeter
The volt-ohm-milliammeter (VOM) is the most commonly used multimeter.
The typical VOM has a meter movement with a full-scale current of 50 µA,
or a sensitivity of 20 KΩ/V, when used as a DC voltmeter. A single meter
movement indicates current, AC and DC voltage, and resistance. These
instruments can be analog or digital devices. Range switches may be
provided for scale selection (e.g., 0-1 V, 0-10 V, etc.) on analog-type
64
Rev 1
multimeters, however, many digital multimeters are auto-ranging and will
automatically respond to indicate the measured parameter on the correct
scale.
Voltmeters
You can construct a simple DC voltmeter by placing a resistor (RS), called a
multiplier, in series with the ammeter movement, and marking the meter
face to read voltage as shown below. Voltmeters are connected in parallel
with the load (RL) being measured.
Figure: Simple DC Voltmeter
When a circuit includes a voltmeter, the voltmeter will draw current from
that circuit. This current causes a voltage drop across the resistance of the
meter; the meter subtracts this voltage drop from the voltage measured by
the meter. This reduction in voltage is the loading effect. It can have a
serious effect on voltage measurement accuracy, especially for low current
circuits.
Normally, manufacturers counter this effect by constructing the voltmeter’s
movement of an extremely high resistance to limit the current flow through
the voltmeter.
The accuracy of a voltmeter is the ratio of measured voltage when the meter
is in the circuit to the voltage measured with the meter out of the circuit.
Ammeter Operation
The ammeter measures electric current. It may read in units of amperes,
milliamperes, or microamperes. The ammeter must be in series with the
circuit to be tested, in order to measure current, as shown below.
Rev 1
65
Figure: Ammeter
When an ammeter is in series with a circuit, it will increase the resistance of
that circuit by an amount equal to the internal resistance of the meter Rm.
The equation below is the mathematical representation of the current
without the meter installed.
𝐼𝑜 =
𝑉
𝑅𝑜
The next equation is the mathematical representation of the current with the
meter installed in the circuit.
𝐼𝑤 =
𝑉
𝑅𝑜 + 𝑅𝑚
The accuracy of the ammeter KA is the ratio of the current when the meter is
in the circuit, Iw, to the current with the meter out of the circuit, Io.
Ammeter Shunts
An ammeter with a full scale current deflection (Im) can be shunted with a
resistor (RSH) in order to measure currents in excess of full scale deflection
current (Im) as shown in the figure below. The reason for shunting an
ammeter is to extend the range of the ammeter and, thereby, measure
currents higher than the original full scale value.
Figure: Ammeter with Shunt Installed
66
Rev 1
Ohmmeters
An ohmmeter measures the resistance of a wire or a circuit. When used as a
test device, an ohmmeter aids the troubleshooter in determining if a ground
or a short exists in a circuit.
The ohmmeter is an instrument used to determine resistance. A simple
ohmmeter (shown below) consists of a battery, a meter movement
calibrated in ohms, and a variable resistor.
In order to obtain an accurate measurement of a component’s resistance,
connect the ohmmeter to a component removed from the circuit as
illustrated in the figure below. The reason for removing the component is
that measurement of current through the component determines the
resistance. If the component remains in the circuit, and a parallel path
exists in the circuit, the current will flow in the path of least resistance and
give an erroneous reading.
Figure: Simple Ohmmeter Circuit
Ro, in the figure above, is an adjustable resistor whose purpose is to zero the
ohmmeter and correct for battery aging. It also helps to limit current along
with the meter resistance Rm. Zeroing the ohmmeter is accomplished by
shorting the ohmmeter terminals a and b and adjusting Ro to give full-scale
deflection.
When the unknown resistance Rx connects across the ohmmeter terminals,
calculate the current by computing the total series resistance and applying
the equation below (Ohm’s Law).
𝐼=
𝑉
𝑅𝑜 + 𝑅𝑥
Meg-Ohm Meter (Megger)
The figure below shows a simplified circuit diagram of the instrument.
Rev 1
67
The moving element of the ohmmeter consists of two coils, A and B rigidly
mounted to a pivoted central shaft and are free to rotate over a C-shaped
core (C on figure below). These coils connect by means of flexible leads.
The moving element may point in any meter position when the generator is
not in operation.
As current provided by the hand-driven generator flows through coil B, the
coil will tend to set itself at right angles to the field of the permanent
magnet. With the test terminals open, giving an infinite resistance, no
current flows in coil A. Thereby, coil B will govern the motion of the
rotating element, causing it to move to the extreme counter-clockwise
position, which is marked as infinite resistance.
Figure: Simple Megger Circuit
Coil A has windings that produce a clockwise torque on the moving
element. With the terminals marked line and earth shorted, giving a zero
resistance, the current flow through the coil A is sufficient to produce
enough torque to overcome the torque of coil B. The pointer then moves to
the extreme clockwise position, which is marked as zero resistance.
Resistance (R2) will protect coil A from excessive current flow in this
condition.
When the test terminals, line, and earth connect across an unknown
resistance, the opposing torques of coils A and B balance each other so that
the instrument pointer comes to rest at some point on the scale. The scale is
calibrated such that the point r directly indicates the value of resistance
being measured, usually in MΩ.
Synchroscope
The synchroscope consists of a two-phase stator. The two stator windings
are at right angles to one another, and by means of a phase-splitting
network, the current in one phase leads the current of the other phase by
90°, thereby generating a rotating magnetic field. The stator windings
connect to the incoming generator, and a polarizing coil connects to the
running generator.
The rotating element is unrestrained and is free to rotate through 360°. It
consists of two iron vanes mounted in opposite directions on a shaft, one at
the top, and one at the bottom, magnetized by the polarizing coil.
68
Rev 1
If the frequencies of the incoming and running generators are different, the
synchroscope will rotate at a speed corresponding to the difference. If
incoming frequency is higher than running frequency, it will rotate in the
clockwise direction; if incoming frequency is less than running frequency, it
will rotate in the counterclockwise direction. When the synchroscope
indicates 0° phase difference, the pointer is at the 12 o’clock position and
the two AC generators are in phase.
Knowledge Check
A ground detector measures
__________________________.
A.
insulation resistance to ground
B.
difference in phase currents
C.
difference in phase voltage
D.
current on the neutral connection
Knowledge Check
Synchroscopes are used to parallel DC generators.
A.
True
B.
False
Knowledge Check
A multimeter can be used to measure all of the
following except_____________________.
Rev 1
A.
AC voltage
B.
DC voltage
C.
hysteresis
D.
resistance
69
TLO 5 Summary
Duration
 10 minutes
Logistics
 Use PowerPoint slide
174–175 and the IG to
review TLO 5 material.
Use directed and nondirected questions to
students, check for
understanding of ELO
content, and review any
material where student
understanding of ELOs
is inadequate.
In this section, you learned how common installed and portable meters
work, and what they measure.
Now that you have completed this lesson, you should be able to:
1. Describe the use of the following electrical test meters: ground
detector, multimeter, megger, and synchroscope.
2. Describe the operation and electrical parameter measured of the
following electrical test equipment: ground detector, multimeter,
megger, and synchroscope.
Basic Electricity Part 2 Summary
Duration
 30 minutes
Logistics
 Review PowerPoint
slides 176–181.
In this module, the following key concepts and definitions were covered:
1. AC Generators
a. To generate a voltage in an AC machine, a magnetic field, a
conductor and relative motion between them required.
b. Major components in a AC generator are the rotor or armature,
stator, slip rings and brushes.
c. The output of an AC generator is a sine wave generated by the
induced voltage in the coil as is rotates through the field.
Figure: Developing an AC Sine Wave Voltage
d. One rotation of the coil through the field is a period or cycle, the
number of cycles completed per second is the frequency.
70
Rev 1
e. Voltage and current output values vary as the coil cuts through the
magnetic field, with the peak values generated at the maximum
flux area, voltage at this point is termed peak voltage, and the
effective voltage (RMS) is equal to 0.707 times the peak value.
f. Phase angle describes the relationship between two or more signal,
in a three-phase output the signals are out of phase by 120 degrees.
2. Inductors, Capacitors and Power relationships
a. An inductor is a circuit element that will store electrical energy in
the form of a magnetic field.
b. An inductor tends to oppose a change in current flow, the faster
the relative motion between the field and coil, the greater the value
of induced voltage.
c. Inductance is a measure of an inductor’s ability to induce CEMF,
measured in henries (H).
d. An inductor has an inductance of one henry when a one amp per
second change in current produces one volt of CEMF, as shown
below.
e. Capacitors store energy as an electric field between two plates,
and is equal to the amount of charge that can be stored divided by
the applied voltage.
f. The unit of capacitance is the farad (F). A farad is the capacitance
that will store one coulomb of charge when one volt acts across
the plates of the capacitor.
g. Inductive reactance and capacitive reactance are present in AC
circuits due to the continuously changing output signal, both affect
the current flow in the circuit.
h. 𝐼𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 = 𝑋𝐿 = 2𝜋𝑓𝐿, and
1
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 = 𝑋𝐶 =
2𝜋𝑓𝐶
i. Impedance is the total opposition to current flow in an AC circuit.
j. Power in AC circuits is comprised of Apparent, true and reactive
power, not all power produced in the circuit can provide work due
to capacitive and inductive reactance, and current/voltage being
out of phase.
k. Power triangle represents comparable values that can be used to
find efficiency.
l. Power factor is the ratio between True Power and Apparent Power
and represented by cos θ in an AC circuit.
3. Transformers
𝑁𝑃
a. Works on Mutual induction, Turns Ratio =
, matched
𝑁𝑆
impedance is important for maximum transfer of power, efficiency
of the ration of power output to power input in a transformer.
b. Primary coil is connected to the AC supply; secondary coil is
connected to the load.
c. The voltages induced in the windings of a transformer are directly
proportional to the number of turns of the coils in the transformer.
d. Transformers are constructed to match their intended application.
Rev 1
71
4. Electrical Distribution
a. Consists of three parts (generation, transmission and distribution).
b. Substations transform from hog voltage to usable voltage e.g.: 480
V, load center is “switchgear” with main breaker and feeder
breakers, MCCs provide centralized distribution.
c. Terminology – ampacity is maximum current a conductor can
carry with overheating.
d. Single phase connections connected to 3 phase by phase to ground
or phase to phase connection
e. Three-phase advantage over single-phase: equipment weighs less
and is more efficient, provides a wide range of voltages and can
provide singe phase power.
f. Abnormally high currents in one or more phases is indicative of an
unbalanced load.
5. Electrical test equipment
a. Ground detectors measure resistance in ohms to ground of
insulation resistance.
b. Multimeters used for measurement of voltage, current and
resistance in one meter, can be analog or digital readout with
manual or automatic scaling.
c. Meggers used to check insulation resistance in ohms, through use
of hand driven DC generator and coils.
d. Synchroscope is used to monitor phase difference between two
generators to ensure they are correctly phased to allow paralleling
with excessive currents.
72
Rev 1
Summary
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a grade of
80 percent or higher on the following TLOs:
1. Describe the theory of operation and operating characteristics of an
AC generator.
2. Describe the construction and theory of operation of Inductors and
Capacitors, their effects on AC electrical circuits, and relationship to
power factor.
3. Describe the construction, operation, and applications of transformers.
4. Describe basic industrial electrical distribution, including typical
wiring schemes used and the advantages of three-phase systems.
5. Given an electrical measuring device or piece of test equipment,
describe the use of that equipment including the electrical parameter
measured.
Rev 1
73