AC Waveforms
Electronics
Lesson Plan
Performance Objective
Upon completion of this lesson, the student will be able to explain what alternating current (AC) electricity is
and how it is generated. The student will be able to demonstrate how to perform calculations to convert one
value of AC voltage into other values by completing the AC Waveforms Exam.
Specific Objectives
 Understand how a sine wave of alternating voltage is generated.
 Explain the three ways to express the amplitude of a sinusoidal waveform and the relationship
between them.
 Define the following values for a sine wave: peak, peak-to-peak, root mean square, average, and
instantaneous.
 Calculate the RMS, average, and peak-to-peak values of a sine wave when the peak value is known.
 Calculate the instantaneous value of a sine wave.
 Convert peak, peak-to-peak, average, and RMS voltage and current values from one value to another.
 Explain the importance of the .707 constant and how it is derived.
 Define frequency and period and list the units of each.
 Calculate the period when the frequency is known and frequency when the period is known.
 Explain the sine, cosine, and tangent trigonometric functions.
 Understand the concept of phase angles.
 Calculate the value of the sine of any angle between 0and 360.
Terms
 Alternating current (AC)- a flow of electric charge that periodically changes direction.
 Direct current (DC)- the unidirectional flow of electric charge.
 Transformer- a device that uses electromagnetic induction to change one value of AC voltage into a
different value of AC voltage.
 Generator- a device that changes mechanical energy into electrical energy.
 Sine wave- a type of repeating curved wave shape created by rotational motion through a magnetic
field and described by the mathematical sine function.
 Wave- a disturbance traveling through a medium.
 Waveform- a graphic representation of a wave.
 Period (P)- the time required to complete one cycle of a waveform.
 Frequency (f)- the number of cycles of a waveform that occur in one second of time.
 Amplitude- height of a wave.
 Peak- the maximum positive or negative deviation of a waveform from its zero reference level.
 Peak-to-peak- the measurement from the highest amplitude peak to the lowest peak.
 Root mean square (RMS)- the value of AC voltage that creates the same heat as the same numerical
value of DC voltage.
Copyright © Texas Education Agency, 2014. All rights reserved.
1


Average voltage- the DC equivalent voltage over a half cycle of an AC voltage wave.
Instantaneous voltage- the voltage at a single point or instant of time.
Time
It should take approximately four, 45-minute class periods to teach the lesson and three, 45-minute class
periods to practice calculations, complete the lab and worksheets, and take the quiz.
Preparation
TEKS Correlations
This lesson, as published, correlates to the following TEKS. Any changes/alterations to the activities may result
in the elimination of any or all of the TEKS listed.
Electronics
 130.368 (c)
o (5) The student implements the concepts and skills that form the technical knowledge of
electronics using project-based assessments. The student is expected to:
(A) apply Ohm's law, Kirchoff's laws, and power laws;
(B) demonstrate an understanding of magnetism and induction as they relate to
electronic circuits; and
(C) demonstrate knowledge of the fundamentals of electronics theory.
o (6) The student applies the concepts and skills to simulated and actual work situations. The
student is expected to:
(A) measure and calculate resistance, current, voltage, and power in series, parallel, and
complex circuits; and
(B) apply electronic theory to generators, electric motors, and transformers.
o (8) The student learns the function and application of the tools, equipment, and materials used
in electronics through project-based assignments. The student is expected to:
(A) safely use tools and laboratory equipment to construct and repair circuits; and
(B) use precision measuring instruments to analyze circuits and prototypes.
Interdisciplinary Correlations
Precalculus
 111.35 (c)
o (1) The student defines functions, describes characteristics of functions, and translates among
verbal, numerical, graphical, and symbolic representations of functions, including polynomial,
rational, power (including radical), exponential, logarithmic, trigonometric, and piecewisedefined functions. The student is expected to:
(A) describe parent functions symbolically and graphically, including f(x) = xn, f(x) = 1n x,
f(x) = loga x, f(x) = 1/x, f(x) = ex, f(x) = |x|, f(x) = ax, f(x) = sin x, f(x) = arcsin x, etc.; and
Copyright © Texas Education Agency, 2014. All rights reserved.
2
(D) recognize and use connections among significant values of a function (zeros,
maximum values, minimum values, etc.), points on the graph of a function, and the
symbolic representation of a function.
o (3) The student uses functions and their properties, tools and technology, to model and solve
meaningful problems. The student is expected to:
(A) investigate properties of trigonometric and polynomial functions;
(B) use functions such as logarithmic, exponential, trigonometric, polynomial, etc. to
model real-life data; and
(E) solve problems from physical situations using trigonometry, including the use of Law
of Sines, Law of Cosines, and area formulas and incorporate radian measure where
needed.
Occupational Correlation (O*Net – www.onetonline.org/)
Job Title: Electricians
O*Net Number: 47-2111.00
Reported Job Titles: Chief Electrician; Control Electrician; Electrician; Industrial Electrician; Inside Wireman;
Journeyman Electrician; Journeyman Wireman; Maintenance Electrician; Mechanical Trades Specialist,
Electrician; Qualified Craft Worker, Electrician (QCW, Electrician)
Tasks
 Plan layout and installation of electrical wiring, equipment, or fixtures, based on job specifications and
local codes.
 Connect wires to circuit breakers, transformers, or other components.
 Test electrical systems or continuity of circuits in electrical wiring, equipment, or fixtures, using testing
devices, such as ohmmeters, voltmeters, or oscilloscopes, to ensure compatibility and safety of system.
 Use a variety of tools or equipment, such as power construction equipment, measuring devices, power
tools, and testing equipment, such as oscilloscopes, ammeters, or test lamps.
 Inspect electrical systems, equipment, or components to identify hazards, defects, or the need for
adjustment or repair, and to ensure compliance with codes.
 Prepare sketches or follow blueprints to determine the location of wiring or equipment and to ensure
conformance to building and safety codes.
 Diagnose malfunctioning systems, apparatus, or components, using test equipment and hand tools to
locate the cause of a breakdown and correct the problem.
 Work from ladders, scaffolds, or roofs to install, maintain, or repair electrical wiring, equipment, or
fixtures.
 Advise management on whether continued operation of equipment could be hazardous.
 Maintain current electrician's license or identification card to meet governmental regulations.
Soft Skills
 Troubleshooting
 Repairing
 Active Listening
 Critical Thinking
 Judgment and Decision making
 Installation
Copyright © Texas Education Agency, 2014. All rights reserved.
3




Active Learning
Complex Problem Solving
Equipment Maintenance
Instructing
Accommodations for Learning Differences
These lessons accommodate the needs of every learner. Modify the lessons to accommodate your students
with learning differences by referring to the files found on the Special Populations page of this website.
Preparation
 Know the proper use of a calculator
References
 Gerrish, Dugger, Roberts. (2008). Electricity & electronics. Tinley Park, Illinois: Goodheart-Willcox
Company.
 Mitchel E. Schultz. (2007). Grob’s basic electronics fundamentals of DC and AC circuits. Columbus, Ohio:
McGraw Hill.
Instructional Aids
 AC Waveforms (with Exercise Key) slide presentation and notes
 Study Guide 1
 Study Guide 1 Answer Key
 Study Guide 2
 Study Guide 2 Answer Key
 AC Waveforms Exam
 AC Waveforms Exam Answer Key
Introduction
This lesson discusses Alternating Current and various ways to describe the amplitude of a sinusoidal
waveform. The time and frequency measurement of a waveform are reviewed, and an introduction to the
trigonometric function are presented.

Show
o An electrical outlet.

Ask
o Most of us take for granted the huge infrastructure necessary to bring electricity into our homes.
Does anyone really know the characteristics of this common energy source?

Say
o Today we are going to start learning about this important energy source. Let’s get started.
Copyright © Texas Education Agency, 2014. All rights reserved.
4
Outline
OUTLINE
MI
I.
Introduction and objectives
A. Overview the lesson objectives.
B. Introduce important terms.
C. Discuss how AC is an important technology that
people tend to take for granted.
D. Discuss AC and the use of common trigonometric
functions in real world examples.
II.
What is AC?
A. Contrast/compare alternating current (AC) and
direct current (DC).
B. Explain that movement of electrons can perform
work whether the movement is always in the
same direction or back and forth.
C. Discuss motors that are used in common
household items (fans, air conditioners, washing
machines, dryers, refrigerators, and pumps).
D. Electronic devices can create AC, but electronics
does not work on AC.
E. Electronic devices typically use a power supply to
convert AC into DC for use.
III.
Why change voltage?
A. Discuss the importance of transformers in power
distribution and increased efficiency over long
distances with AC voltage.
B. A transformer can step voltage up (or down) but
power in always equals power out.
C. As voltage is stepped up, current goes down and
current is what creates heat as it travels through
wires in a power distribution system.
D. AC is the only type of voltage that works in a
transformer.
E. Until very recently there was no good method to
step up or down DC voltages.
NOTES TO TEACHER
Slides 1-4
Talk about things you
are going to be
covering in this
lesson and be careful
to not make it sound
more difficult than it
is.
Slides 5-7
Slides 8-9
In the late 1800s
there was a huge
battle between the
advocates of DC
(Thomas Edison) and
the advocates of AC
(Charles
Westinghouse).
This battle was called
“the war of currents”
and makes for a good
research paper.
Copyright © Texas Education Agency, 2014. All rights reserved.
5
MI
OUTLINE
NOTES TO TEACHER
IV.
How do you make AC?
A. Discuss the concept of rotary motion and
generators.
B. Discuss mechanical power and electrical power
plants.
C. Electrical power generation takes work; more
power out requires more work in.
D. The work in comes from a prime mover, such as
water in a hydroelectric plant or steam pressure in
a conventional power plant.
Slides 10-12
V.
Hydroelectric power
A. Dams (creating water pressure for hydroelectric
power) were the first huge source of mechanical
energy for power plants.
B. Niagara Falls was one of the first major
hydroelectric sources.
C. Hydroelectric power creates rotary motion where
a rotor spins through a magnetic field.
D. The key concept in electrical AC generation is that
voltage is produced when a conductor cuts
through a magnetic field line.
E. The changing angle of relative motion creates the
sine wave.
F. Voltage amplitude is calculated using the sine
function.
Slides 13-17
VI.
Voltage is a function of angle
A. The graphics show the position of a conductor as
it rotates from a 90 degree angle to a 270 degree
angle.
B. Start at the 90 degree angle to show students that
maximum voltage is created when motion is
perpendicular.
C. A zero-degree angle would start on the negative Y
axis and increase going clockwise from there.
D. The animated graphics visually show how the
voltage produced is constantly changing as the
conductor rotates through the magnetic field.
Slides 18-33
Copyright © Texas Education Agency, 2014. All rights reserved.
6
MI
OUTLINE
NOTES TO TEACHER
VII.
Waveform values
A. A waveform is a plot of voltage vs time.
B. A waveform is the shape of a signal.
C. There are several ways to characterize the amount
of voltage given by a changing waveform.
D. Peak-to-peak is the value seen by an oscilloscope.
E. The peak value is the absolute value of the
maximum amplitude.
F. RMS is a heating value.
G. Average is the DC equivalent voltage.
H. Angles are given in either degrees or radians.
Slides 34-38
VIII.
Time-based waveform terms (frequency and period)
A. Frequency and period have nothing to do with
amplitude.
B. Frequency and period are time-based terms.
C. Wavelength is another term that often gets used in
the same context as frequency and period, but
wavelength is not a time-based term.
D. Wavelength is a distance that relates to the size of
the wave.
E. Inverse means one over something (there is a
button on a calculator that performs this function:
).
Slides 39-41
IX.
Waveform terminology
A. Each of these terms has a different value
associated with it.
B. The ability to convert from one value to any other
value is important to know.
C. We give the descriptions first then the formulas
needed to perform conversions.
D. There are three terms in each of the formulas, and
in most cases, the third term is a constant.
E. For instantaneous voltage, all three terms are
variables, where any two terms are necessary to
calculate the third.
F. Regular trigonometric functions take the sine of an
angle to get a value, and inverse functions give the
angle from the ratio value.
Slides 42-51
Have students
practice calculating
each of these
values, particularly
those using the
inverse function.
Copyright © Texas Education Agency, 2014. All rights reserved.
7
MI
OUTLINE
NOTES TO TEACHER
X.
Calculation examples
A. Use the formula relationship circles as a memory
aide.
B. Have students work example problems in slide
presentation.
C. Create more problems for students to practice
working.
D. Students will complete the table on the last page of
Study Guide 1.
I. Problems 1 and 2 as guided practice
II. Problems 3-5 as independent practice
XI.
Trigonometric functions
A. The motion of a conductor through the magnetic
field can be resolved into a right triangle.
B. The sine function is used because motion
perpendicular to the magnetic field is the opposite
side of the triangle.
C. Right-triangle side and angle relationships
(hypotenuse, opposite, adjacent).
D. Basic trigonometric functions (sine, cosine, and
tangent).
E. Motion parallel to the magnetic field would be the
adjacent side of the triangle.
F. Trigonometric exercise - Study Guide 2
i. Triangles 1 and 2 are teacher-guided
practice
ii. Triangles 3 and 4 are students’
independent practice
XII.
Slides 52-61
Pass out Study
Guide 1 and have
students complete
the table for guided
and independent
practice.
Slides 62-69
Upon completion of
slides, pass out
student Study Guide
2 for guided and
independent
practice.
Slides 70-76
Review answer keys
with students.
Administer the AC
Waveforms Exam
and grade with
answer key.
AC Waveforms Exam
Multiple Intelligences Guide
Existentialist
Interpersonal
Intrapersonal
Kinesthetic/
Bodily
Logical/
Mathematical
Musical/Rhythmic
Naturalist
Verbal/Linguistic
Visual/Spatial
Copyright © Texas Education Agency, 2014. All rights reserved.
8
Application
Guided Practice
 Study Guide 1 - problems 1 and 2
 Study Guide 2 - triangles 1 and 2
Independent Practice
 Study Guide 1 - problems 3 and 4
 Study Guide 2 - triangles 3 and 4
Summary
Review
Have students restate lesson objectives.
Evaluation
Informal Assessment
The teacher monitors students during guided and independent practice activities and may assign a grade for
the independent activities.
Formal Assessment
Students will complete the AC Waveforms Exam.
Enrichment
Extension
To add an interdisciplinary component, students will write a short research paper on “the war of the currents.”
Students will create their own problems for independent practice. See lesson outline (page 5/Notes to Teacher).
Copyright © Texas Education Agency, 2014. All rights reserved.
9
Student Study Guide 1
AC Waveforms
Waveform
A wave is a disturbance traveling through a medium. A waveform is a graphic representation of a wave. Like a
wave, a waveform depends on movement and on time. The ripple on the surface of a pond is a movement of
water in time. Wave shapes tell you a great deal about the signal. Any time you see a change in the vertical
dimension of a signal, you know that this amplitude change represents a change in voltage. Wave shapes
alone are not the whole story. To completely describe a waveform, you will need to find its particular
parameters. Depending on the signal, these parameters might be frequency, period, amplitude, width, rise
time, or phase.
Frequency
The frequency of a waveform is the number of cycles of the waveform that occur in one second of time. The
common unit of measurement is hertz (Hz).
Period
The period of a waveform, which sometimes is called its time, is the time required to complete one cycle of a
waveform. It is measured in units of seconds, such as seconds, tenths of seconds, milliseconds, or
microseconds.
Figure 2.1 Sample of Waveform
If a waveform is to be properly described in terms of its period or frequency, it must be a repetitious
waveform. A repetitious waveform is one in which each following cycle is identical to the previous cycle.
Waveform Amplitude Specifications
In addition to frequency and period values, a third major specification of a waveform is the amplitude or
height of the wave. There are three possible ways to express the amplitude of a sinusoidal waveform: peak,
peak-to-peak, and root-mean-square (RMS).
Copyright © Texas Education Agency, 2014. All rights reserved.
10
1. Peak
The peak amplitude of a sinusoidal waveform is the maximum positive or negative deviation of a waveform
from its zero reference level. The sinusoidal waveform is a symmetrical waveform, so the positive peak value
is the same as the negative peak value as shown in Figure 2.2. If the positive peak has a value of 10 volts, then
the negative peak will also have a value of 10 volts. When measuring the peak value of a waveform, either
positive or negative peaks can be used.
Figure 2.2 Positive Peak and Negative Peak Value
2. Peak–to-Peak
The peak-to-peak amplitude is simply a measurement of the amplitude of a waveform taken from its positive
peak to its negative peak as shown in Figure 2.3.
Figure 2.3 Peak-to-Peak Amplitude
Figure 2.4 Peak-to-Peak Amplitude
For sinusoidal waveform, if the positive peak value is 10 volts in magnitude, then the negative peak value of
the same waveform is also 10 volts. Measuring from peak-to-peak, there is a total of 20 volts. Therefore, the
value of the sinusoidal waveform in Figure 2.2 can be specified as either 10 volts peak or 20 volts peak-topeak.
Copyright © Texas Education Agency, 2014. All rights reserved.
11
For the non-sinusoidal waveform shown in Figure 2.4, the peak-to-peak value of the voltage can be
determined by adding the magnitude of the positive and the negative peak. In this example, the peak-to-peak
amplitude is 18 volts plus 2 volts for a total of 20 volts, peak-to-peak.
3. Root-Mean-Square
The third specification for AC waveform is called root-mean-square, abbreviated RMS. This term allows the
comparison of AC and DC circuit values. Root-mean-square values are the most common methods of
specifying sinusoidal waveforms. In fact, almost all AC voltmeter and ammeters are calibrated so that they
measure AC values in terms of RMS amplitude.
RMS Relations to DC Heating Effect
The RMS value is also known as the effective value and is defined in terms of the equivalent heating effect of
direct current. The RMS value of a sinusoidal voltage is equivalent to the value of a DC voltage, which causes
an equal amount of heat due to the circuit current flowing through a resistance. The RMS value of a sinusoidal
voltage or current waveform is 70.7 percent or 0.707 of its peak amplitude value.
VRMS = 0.707 Vpeak
IRMS = 0.707 Ipeak
A sinusoidal voltage with peak amplitude of 1 volt has the same effect as a DC voltage of 0.707 volts as far as
its ability to reproduce the same amount of heat in a resistance. Because the AC voltage of 1 volt peak or
0.707 volts RMS is as effective as a DC voltage of 0.707 volts, the RMS value of voltage is also referred to as
the effective value.
Determining the 0.707 Constant
How is the 70.7 percent of peak-value constant derived? Essentially, the words root-mean-square define the
mathematical procedure used to determine the constant.
Copyright © Texas Education Agency, 2014. All rights reserved.
12
Study Guide 1 (Continued)
P-P
2
RMS
PK
0.707
INST.
Sine 
PK
AVG.
PK
0.637
PK
EXAMPLES
120 VAC = 170 Vpk
Formula: PK = RMS  0.707
120  0.707 = 169.7
(round off to 170 Vpk)
18 V @ 72  = 19 Vpk
Formula: PK = Instantaneous  Sine
18  72 Sine = 18.9
(round off to 19)
30 Vpk = 21.2 VAC
Formula: RMS = 0.707 X PK
0.707 X 30 = 21.2
350 V @ 23.5 = 30 Vpk
Formula: PK = Instantaneous  Sine
350  23.5 Sine = 877.7
(round off to 878)
50 Vpp = 17.7 Vrms
Step 1: Need to find PK
Formula: PK = P-P  2
50  2 = 25m
Step 2: Find RMS
Formula: RMS = 0.707 X PK
0.707 X 25 = 17.675
(round off to 17.7 Vrms)
Find the angle with 454 V instantaneous and a
PK of 908 V
Formula: Sine (θ) = Instantaneous  PK
454  908 = 0.5
nd
2 Sine (.5) = 30
20 V Average = 22.2 Vrms
= 31.4 Vpk
= 62.8 Vp-p
Step 1: Find PK, Formula: PK = Average  0.637
= 20 0.637 = 31.39 (round off to 31.4)
Step 2: Find RMS, Formula: RMS = 0.707 X PK
= 0.707 X 31.4 = 22.19 (round off to 22.2)
Step 3: Find P-P, Formula: P-P = 2 X PK
= 2 X 31.4 = 62.8
Copyright © Texas Education Agency, 2014. All rights reserved.
13
Study Guide 1 (Continued)
#
rms
peak
pk-to-pk
average
200mV
instantaneous
________ V @ 72º
113V
________ V @ 90º
96.4
________ V @ 235º
1.5V @ 122º
689V
________ V @ 35º
Copyright © Texas Education Agency, 2014. All rights reserved.
14
Student Study Guide 2
The Sine Wave and Sine Trigonometric Function
The term sinusoidal has been used to describe a waveform produced by an AC generator. The term sinusoidal
comes from a trigonometric function called the sine function.
Right-Triangle Side and Angle Relationships
Trigonometry is the study of triangles and their relationships involving lengths and angles of triangles. The
basic triangle studied in trigonometry is a right triangle, which is a triangle that has a 90◦ angle as one of its
three angles. A 90◦ triangle has a unique set of relationships from which the rules for trigonometry are
derived.
To help distinguish the sides of a right triangle from one another, a name is given to each side. The sides of the
triangle are named with respect to the angle theta. The side of the triangle across from or opposite to the
angle theta is called the opposite side.
The longest side of a right triangle is called the hypotenuse. The remaining side is called the adjacent side
because it lies beside or adjacent to the angle. These three names are commonly abbreviated to their first
initials, O, H, and A.
Basic Trigonometric Functions
In trigonometry, these ratios have specific names. The three most commonly-used ratios in the study of right
triangles are called sine, cosine, and tangent. The sine of the angle theta is equal to the ratio formed by the
length of the opposite side divided by the length of the hypotenuse:

Sine ө = opposite
hypotenuse
The cosine of the angle theta is equal to the ratio formed by length of the adjacent side divided by the length
of the hypotenuse:

Cosine ө = adjacent
hypotenuse
The tangent of the angle theta is equal to the ratio formed by length of the opposite side divided by the length
of the adjacent side:

Tangent ө = opposite
adjacent
Copyright © Texas Education Agency, 2014. All rights reserved.
15
Student Study Guide 2 (Continued)
Right Triangle
Hypotenuse
Opposite
Adjacent
Formulas
Opposite
Sine
Adjacent
Hypotenuse
Cosine
Hypotenuse
Opposite
Tangent
Adjacent
Copyright © Texas Education Agency, 2014. All rights reserved.
16
Name________________________________________Date_______________________Class______________
Student Study Guide 2 (Continued)
Calculate the value of the missing side.
Triangle #1
Triangle #2
Hypotenuse?
Hypotenuse?
8'
Opposite
10' Adjacent
5.3 Rods
Opposite
6.8 Rods Adjacent
Hypotenuse:
Hypotenuse:
Triangle #3
Triangle #4
125 miles
Hypotenuse
56'
Hypotenuse
Opposite?
23.2'
Opposite
Adjacent?
85 miles Adjacent
Opposite:
Adjacent:
Copyright © Texas Education Agency, 2014. All rights reserved.
17
Study Guide 1 Answer Key
AC Waveforms
Waveform
A wave is a disturbance traveling through a medium. A waveform is a graphic representation of a wave. Like a
wave, a waveform depends on movement and on time. The ripple on the surface of a pond is a movement of
water in time. Wave shapes tell you a great deal about the signal. Any time you see a change in the vertical
dimension of a signal, you know that this amplitude change represents a change in voltage. Wave shapes
alone are not the whole story. To completely describe a waveform, you will need to find its particular
parameters. Depending on the signal, these parameters might be frequency, period, amplitude, width, rise
time, or phase.
Frequency
The frequency of a waveform is the number of cycles of the waveform that occur in one second of time. The
common unit of measurement is hertz (Hz).
Period
The period of a waveform, which sometimes is called its time, is the time required to complete one cycle of a
waveform. It is measured in units of seconds, such as seconds, tenths of seconds, milliseconds, or
microseconds.
Figure 2.1 Sample of Waveform
If a waveform is to be properly described in terms of its period or frequency, it must be a repetitious
waveform. A repetitious waveform is one in which each following cycle is identical to the previous cycle.
Waveform Amplitude Specifications
In addition to frequency and period values, a third major specification of a waveform is the amplitude or
height of the wave. There are three possible ways to express the amplitude of a sinusoidal waveform: peak,
peak-to-peak, and root-mean-square (RMS).
Copyright © Texas Education Agency, 2014. All rights reserved.
18
1. Peak
The peak amplitude of a sinusoidal waveform is the maximum positive or negative deviation of a waveform
from its zero reference level. The sinusoidal waveform is a symmetrical waveform, so the positive peak value
is the same as the negative peak value as shown in figure 2.2. If the positive peak has a value of 10 volts, then
the negative peak will also have a value of 10 volts. When measuring the peak value of a waveform, either
positive or negative peaks can be used.
Figure 2.2 Positive Peak and Negative Peak Value
2. Peak–to-Peak
The peak-to-peak amplitude is simply a measurement of the amplitude of a waveform taken from its positive
peak to its negative peak as shown in figure 2.3.
Figure 2.3 Peak-to-Peak Amplitude
Figure 2.4 Peak-to-Peak Amplitude
For sinusoidal waveform, if the positive peak value is 10 volts in magnitude, then the negative peak value of
the same waveform is also 10 volts. Measuring from peak-to-peak, there is a total of 20 volts. Therefore, the
value of the sinusoidal waveform in Figure 2.2 can be specified as either 10 volts peak or 20 volts peak-topeak.
Copyright © Texas Education Agency, 2014. All rights reserved.
19
For the non-sinusoidal waveform shown in Figure 2.4, the peak-to-peak value of the voltage can be
determined by adding the magnitude of the positive and the negative peak. In this example, the peak-to-peak
amplitude is 18 volts plus 2 volts for a total of 20 volts, peak-to-peak.
3. Root-Mean-Square
The third specification for AC waveform is called root-mean-square, abbreviated RMS. This term allows the
comparison of AC and DC circuit values. Root-mean-square values are the most common methods of
specifying sinusoidal waveforms. In fact, almost all AC voltmeter and ammeters are calibrated so that they
measure AC values in terms of RMS amplitude.
RMS Relations to DC Heating Effect
The RMS value is also known as the effective value and is defined in terms of the equivalent heating effect of
direct current. The RMS value of a sinusoidal voltage is equivalent to the value of a DC voltage, which causes
an equal amount of heat due to the circuit current flowing through a resistance. The RMS value of a sinusoidal
voltage or current waveform is 70.7 percent or 0.707 of its peak amplitude value.
VRMS = 0.707 Vpeak
IRMS = 0.707 Ipeak
A sinusoidal voltage with peak amplitude of 1 volt has the same effect as a DC voltage of 0.707 volts as far as
its ability to reproduce the same amount of heat in a resistance. Because the AC voltage of 1 volt peak or
0.707 volts RMS is as effective as a DC voltage of 0.707 volts, the RMS value of voltage is also referred to as
the effective value.
Determining the 0.707 Constant
How is the 70.7 percent of peak-value constant derived? Essentially, the words root-mean-square define the
mathematical procedure used to determine the constant.
Copyright © Texas Education Agency, 2014. All rights reserved.
20
Study Guide 1 Answer Key (Continued)
P-P
2
PK
RMS
0.707
INST.
Sine 
PK
PK
AVG
0.637
PK
EXAMPLES
120 VAC = 170 Vpk
Formula: PK = RMS  0.707
120  0.707 = 169.7
(round off to 170 Vpk)
18 V @ 72  = 19 Vpk
Formula: PK = Instantaneous  Sine
18  72 Sine = 18.9
(round off to 19)
30 Vpk = 21.2 VAC
Formula: RMS = 0.707 X PK
0.707 X 30 = 21.2
350 V @ 23.5 = 30 Vpk
Formula: PK = Instantaneous  Sine
350  23.5 Sine = 877.7
(round off to 878)
50 Vpp = 17.7 Vrms
Step 1: Need to find PK
Formula: PK = P-P  2
50  2 = 25m
Step 2: Find RMS
Formula: RMS = 0.707 X PK
0.707 X 25 = 17.675
(round off to 17.7 Vrms)
Find the angle with 454 V instantaneous and a
PK of 908 V
Formula: Sine (θ) = Instantaneous  PK
454  908 = 0.5
nd
2 Sine (.5) = 30
20 V Average = 22.2 Vrms
= 31.4 Vpk
= 62.8 Vp-p
Step 1: Find PK, Formula: PK = Average  0.637
= 20 0.637 = 31.39 (round off to 31.4)
Step 2: Find RMS, Formula: RMS = 0.707 X PK
= 0.707 X 31.4 = 22.19 (round off to 22.2)
Step 3: Find P-P, Formula: P-P = 2 X PK
= 2 X 31.4 = 62.8
Copyright © Texas Education Agency, 2014. All rights reserved.
21
Study Guide 1 Answer Key (Continued)
AC Waveforms
#
RMS
peak
pk-to-pk
average
instantaneous
200mV
283mV
566mV
180mV
269mV @ 72º
80v
113V
226V
72V
113V @ 90º
96.4
136V
272V
87V
111V @ 235º
1.25V
1.77V
3.54V
1.13V
1.5V @ 122º
764µV
1.08mV
2.16mV
689V
619µV @ 35º
Copyright © Texas Education Agency, 2014. All rights reserved.
22
Study Guide 2 Answer Key
The Sine Wave and Sine Trigonometric Function
The term sinusoidal has been used to describe a waveform produced by an AC generator. The term sinusoidal
comes from a trigonometric function called the sine function.
Right-Triangle: Side and Angle Relationships
Trigonometry is the study of triangles and their relationship involving lengths and angles of triangles. The basic
triangle studied in trigonometry is a right triangle, which is a triangle that has a 90◦ angle as one of its three
angles. A 90◦ triangle has a unique set of relationships from which the rules for trigonometry are derived.
To help distinguish the sides of a right triangle from one another, a name is given to each side. The sides of the
triangle are named with respect to the angle theta. The side of the triangle across from or opposite to the
angle theta is called the opposite side.
The longest side of a right triangle is called the hypotenuse. The remaining side is called the adjacent side
because it lies beside or adjacent to the angle. These three names are commonly abbreviated to their first
initials, O, H, and A.
Basic Trigonometric Functions
In trigonometry, these ratios have specific names. The three most commonly-used ratios in the study of right
triangles are called sine, cosine, and tangent.
1. The sine of the angle theta is equal to the ratio formed by the length of the opposite side divided by
the length of the hypotenuse.

Sine ө = opposite
hypotenuse
2. The cosine of the angle theta is equal to the ratio formed by length of the adjacent side divided by the
length of the hypotenuse.

Cosine ө = adjacent
hypotenuse
3. The tangent of the angle theta is equal to the ratio formed by length of the opposite side divided by
the length of the adjacent side.

Tangent ө = opposite
adjacent
Copyright © Texas Education Agency, 2014. All rights reserved.
23
Study Guide 2 Answer Key (Continued)
Right Triangle
Hypotenuse
Opposite
Adjacent
Formulas
Opposite
Sine
Adjacent
Hypotenuse
Cosine
Hypotenuse
Opposite
Tangent
Adjacent
Copyright © Texas Education Agency, 2014. All rights reserved.
24
Study Guide 2 Answer Key (Continued)
Calculate the value of the missing side.
Triangle #1
Triangle #2
Hypotenuse?
Hypotenuse?
8'
Opposite
10' Adjacent
5.3 Rods
Opposite
6.8 Rods Adjacent
Hypotenuse: _____12.8’__________
Hypotenuse: _____8.62 Rods______
Triangle #3
Triangle #4
125 miles
Hypotenuse
56'
Hypotenuse
Opposite?
85 miles Adjacent
Opposite: _____91.7 miles________
23.2'
Opposite
Adjacent?
Adjacent: _____51’_____________
See the next page for detailed answers.
Copyright © Texas Education Agency, 2014. All rights reserved.
25
Triangle #1
Given the adjacent side = 10’ and the opposite side = 8’, what is the length of the
hypotenuse side?
Step 1 - Find the degree angle
Tangent = Opposite
Adjacent
= 8_ = 0.8
10
nd
= .8 enter 2 tan on your calculator
= 38.65980825 round up to 38.66
= 38.66°
Step 2 - Change the degree angle to cosine
Hypotenuse = Adjacent
Cosine 38.66
= 10’ _______ (take 38.66° enter cosine on your calculator your answer is
.780866719
0.780866719)
= 12.8’
Triangle #2
Given the adjacent side = 6.8 rods and the opposite side = 5.3 rods, what is the length of
the hypotenuse side?
Step 1 - Find the degree angle
Tangent = Opposite
Adjacent
= 5.3 rods = 0.779411765
6.8 rods
= enter 2nd tangent 0.779411765on your calculator
= 37.93°
Step 2 - Change the degree angle to sine
Hypotenuse = Opposite
Sine
= 5.3 rods
Sine 37.93 (enter 37.93, enter sine on the calculator)
= 5.3 rods
0.6146982793
= 8.62 rods
Copyright © Texas Education Agency, 2014. All rights reserved.
26
Triangle #3
Given the hypotenuse side = 125 miles and the adjacent side = 85 miles, what is the
length of the opposite side?
Step 1 - Find the angle in degrees
Cosine = Adjacent
Hypotenuse
= 85 miles_ = 0.68
125 miles
= enter 2nd function button on calculator, then enter cosine 0.68
= 47.15635696 (round off to 47.16 degrees)
Step 2 - Change the degree angle to sine
Opposite = Sine x Hypotenuse
enter sine 47.16 on calculator
= 0.7332553462
= 0.7332553462 x 125 miles
= 91.65691828 (round off to 91.7)
= 91.7 miles
Triangle #4
Given the hypotenuse = 56’ and the opposite side = 23.2’, what is the length of the
adjacent side?
Step 1 - Find the angle in degrees
Sine = Opposite__
Hypotenuse
= 23.2’ = 0.4142857143
56’
= enter 2nd sine 0.414285 (or inv sine)
= 24.47 degrees
Step 2 - Change the degree angle to cosine
Adjacent = Cosine x Hypotenuse
= enter cosine 24.47 on the calculator
= 0.910178279 x 56’
= 50.96998362 (round off to 51’)
= 51’
Copyright © Texas Education Agency, 2014. All rights reserved.
27
Name________________________________________Date_______________________Class______________
AC Waveforms Exam
Complete the table by calculating the missing or incomplete values for the RMS, peak amplitudes, the peak-topeak values, the average values and/or the instantaneous values.
#
rms
1
2
peak
pk-to-pk
average
2.63uV
1.55uV @ 36º
45V
3
________ V @ 165º
3.54pV
________ ____ @ 70º
4
5
3.75V @ 70.7º
10.0kV
2.43kV @ 190º
6
7
8
9
instantaneous
82.1mV
72.3uV
40mV @ 18º
54.1uV @ 212º
905V
7.91V @ 179º
29.6mV
________ V @ 247º
Copyright © Texas Education Agency, 2014. All rights reserved.
28
AC Waveforms Exam Key
Complete the table by calculating the missing or incomplete values for the RMS, peak amplitudes, the peak-topeak values, the average values and the instantaneous values.
#
RMS
peak
pk-to-pk
average
instantaneous
1
1.86uV
2.63uV
5.26uV
1.68uV
1.55uV @ 36º
2
45V
63V
126V
40V
16 V @ 165º
3
1.25pV
1.77pV
3.54pV
1.13pV
-1.77pV @ 70º
4
2.81V
3.97V
7.94V
2.53V
3.75V @ 70.7º
5
10.0kV
14kV
28kV
8.92kV
2.43kV @ 190º
6
91.2mV
129mV
258mV
82.1mV
40mV @ 18º
7
72.3uV
102uV
204uV
65uV
54.1uV @ 212º
8
320V
453V
905V
289V
7.91V @ 179º
9
32.9mV
46.5mV
93mV
29.6mV
-42.8mV @ 247º
Copyright © Texas Education Agency, 2014. All rights reserved.
29