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Electronics
AC Waveforms
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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.
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.
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Specific Objectives (Continued) 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 period is known. Explain the sine, cosine, and tangent
trigonometric functions. Calculate the value of the sine of any angle
between 0° and 360°.
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Introduction AC and the characteristics of a sinusoidal
waveform Time and frequency measurement of a
waveform Trigonometric functions
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What is AC?
Alternating Current (AC) is a useful form of voltage. AC is a flow of electric charge that periodically
changes direction. Direct Current (DC) is another useful form of
voltage. DC is the unidirectional flow of electric charge.
AC and DC are both useful power sources for heating, lights, and motors.
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AC vs DC
DC can power the same things AC can power. The devices may have to be constructed
differently, but they would work the same. AC cannot power some of the things that DC
can power. Example: electronic devices AC motors were not invented at the time Thomas
Edison built his first DC power station in 1882.
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Why Use AC Instead of DC?
AC is relatively easy to produce. It is created through rotational motion.
AC generation can produce large amounts of power economically.
AC voltage can be changed from one value to another relatively easily. DC voltage cannot be changed from one value to
another easily.
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Why Change Voltage?
AC voltage can be stepped up to a higher value using a transformer (up to 760 kV).
The basic concept is that “power in” equals “power out” of a transformer.
Power is voltage times current, so if voltage is made to increase, then current will decrease.
After the voltage is increased, AC is much more efficient to transmit over long distances.
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Why is AC More Efficient?
Electrical power (voltage and current) is sent over transmission wires from a generator to users.
Lower current sent through those transmission wires will produce less heat (Joule’s Law). Fewer moving electrons reduce the amount of friction. Voltage does not create heat.
Less heat means less power is lost during delivery.
We can use thinner wires for power transmission. Thin wires are more economical.
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How Do You Make AC?
Most electricity is produced by induction. In a generator, induction occurs when a
conductor moves through a magnetic field. AC is the type of electricity generated by a
conductor moving in a circle through a static magnetic field. Circular motion is easy to produce. Like a wheel going around, it is a simple and efficient
process.Copyright © Texas Education Agency, 2014. All rights reserved.
Rotary Motion
The process of using a water wheel as a mechanical power source for milling and sawing has been used for thousands of years.
A water wheel produces circular motion. This is also called a water-driven turbine.
This process was applied to creating electricity in the late 1800s.
Two more things were necessary to make AC power viable: the alternator and the transformer.
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Mechanical Power Source An electrical power plant has a capacity, but
the actual amount of power produced is a function of user demand. Higher user demand creates a larger load on the
generator. The generator then needs to draw more
mechanical power from the prime mover. If the prime mover cannot provide the additional
mechanical power, the plant will shut down.
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Hydroelectric Power The Ames Hydroelectric Generating Plant was
the world's first commercial system to produce and transmit AC electricity for industrial use.
In 1890, Westinghouse Electric supplied the station's generator and motor.
The AC was proven to be effective as it was transmitted two miles (3 km) at a loss of less than 5%. The maximum distance for DC was about a mile.
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AC Generation
In an AC generator, one rotation of the rotor shaft creates one cycle of voltage.
This voltage is not steady over the cycle, it changes and reverses polarity depending on the direction of motion of the conductor through the magnetic field. The magnetic field goes in a line from the north pole to
the south pole. The rotor that contains the conductors move in a circle
through the linear magnetic field.Copyright © Texas Education Agency, 2014. All rights reserved.
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AC Generation
Electricity is produced by induction. Induction occurs when a conductor cuts
through a magnetic field line. A conductor must move perpendicular to the
magnetic field line to cut through it. Conductor motion parallel to the magnetic
field does not cut through any magnetic field lines.
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Sine Wave Generation
During a rotation, the motion of a conductor changes from perpendicular to parallel. Perpendicular is a 90° angle. Parallel is a 0° (or 360°) angle.
Between these examples, the conductor motion is at some other angle relative to the direction of the magnetic field. Rotation goes through 0 to 360 degrees and then
repeats.Copyright © Texas Education Agency, 2014. All rights reserved.
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Voltage Amplitude
The amount of voltage produced at any point is proportional to the sine of the angle of motion relative to the direction of the magnetic field lines.
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Voltage is a Function of Angle
There is a magnet in a generator. Field lines go from north to south.
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Voltage is a Function of Angle
A conductor is placed in the magnetic field.
This is actually a single conductor that loops back through the magnetic field.
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Voltage is a Function of Angle
Arrows show the direction of motion for the conductor in this position.
This motion is perpendicular to the direction of the field lines (at 90°).
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Voltage is a Function of Angle
The sine of a 90° angle equals 1.
This represents the maximum or peak voltage out of the generator.
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Voltage is a Function of Angle
We are here on the sine wave.
This represents the maximum or peak voltage out of the generator
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Voltage is a Function of Angle
Here is the position of a conductor after a rotation of 45° (actual angle = 135°).
The motion is at an angle of 45° to the direction of the magnetic field lines.
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Voltage is a Function of Angle
The sine of 45° (or 135°) is 0.707. We are here on the sine wave.
The amount of voltage produced at this angle is 0.707 of the peak voltage.
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Voltage is a Function of Angle
This shows a conductor after another rotation of 45° from the previous example.
This motion is parallel to the magnetic field lines and represents an angle of 180°.
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Voltage is a Function of Angle
The sine of 180° (or 0°) is zero. We are here on the sine wave.
At this instant the voltage produced by the generator is zero volts.
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Voltage is a Function of Angle
The conductor rotates another 45°. Polarity starts to reverse.
This motion is now down through the magnetic field lines (the opposite direction).
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Voltage is a Function of Angle
The Sine of 225° is 0.707. We are here on the sine wave.
Polarity is opposite because the direction of motion is going the opposite way.
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Voltage is a Function of Angle
Here is the position of the conductor after another 45° rotation.
This more clearly shows that the direction of motion (red circle) is down.
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Voltage is a Function of Angle
The sine of 270° is negative one. We are here on the sine wave.
This is the negative peak with equal but opposite amplitude of the positive peak.
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AC Generation Example
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To see animation, copy the link into your computer browser. http://commons.wikimedia.org/wiki/File:Dynamo.wechsel.wiki.v.1.00.gif
Waveform Plot At each point in the rotating cycle, the amount of voltage
produced is equal to the sine of the angle created by the direction of motion of the conductor relative to the direction of the magnetic field.
This produces the changing voltage used for electricity known as the sine wave.
As the angle between the conductor and the magnetic field changes, the voltage changes.
Peak
RMS
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To see animation, copy link into your computer browser.https://phet.colorado.edu/en/simulation/generator
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Common Terms A plot of the voltage vs time produces a wave
form in the shape of a sine wave. This wave form has many terms we need to learn. There are terms for the amplitude or size of the
wave: Peak, peak to peak, RMS, instantaneous, and average
voltage There are terms for how fast the wave changes:
Frequency, period
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Waveform Values Graphically
The angle represents time. An angle is used because a time measurement would change with a change in frequency.
0.637a
VAVG
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Angular Measurement
We use an angle to measure instantaneous voltage values at an instant of time.
At a given angle, the relative amplitude at that point is the same for any sine wave regardless of the frequency.
A sine wave is created due to a conductor moving in a circle through a magnetic field. A circle always has 360 degrees. A sine wave always has 360 degrees for one cycle.
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Radians
There is another way to measure the angle of a circle or sine wave.
It is called the radian measurement because of the relationship between the radius and the circumference of a circle.
There are 2π radians in 360 degrees.C = 2πr
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Degrees to Radians
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Time-based Waveform Terms Wave- a disturbance traveling through a
medium. For AC electricity, the movement of the electrons
back and forth in the wire Waveform- a graphic representation of a
wave. Waveform depends on both movement and time. Example: ripple on the surface of a pond A change in the vertical dimension of a signal is a
result of a change in the amount of voltage.
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Frequency (f)- the number of cycles of the waveform that occur in one second of time. Measured in hertz (Hz)
Period (P)- the time required to complete one cycle of a waveform. Measured in seconds, tenths
of seconds, milliseconds, or microseconds
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Time-based Waveform Terms
Frequency and Period
There is an inverse relationship between frequency and period.
Example: A frequency of 100 Hz gives a period of 0.01 sec (10 ms)
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f=1P
P=1fand
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Waveform Terminology Amplitude- height of a wave
Expressed in one of the following methods Peak (PK, pk, or Pk) Peak-to-peak (P-P, PP, or pp) Root-mean-square (RMS) Average (AVG)
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Waveform Amplitude Specification Peak- the maximum positive or negative
deviation of a waveform from its zero reference level. Sinusoidal waveforms
are symmetrical. The positive peak value
of sinusoidal will be equal to the value of the negative peak.
Measured at an instant of time
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Waveform Amplitude Specification (continued Peak-to-peak is the measurement from the
highest amplitude peak to the lowest peak. Sinusoidal waveform
If the positive peak value is 10 volts in magnitude, then the negative peak is also 10; therefore, peak-to- peak is 20 volts.
Non-sinusoidal waveform It is determined by adding
magnitude of positive and negative peaks.
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Waveform Amplitude Specification (Cont’d.) Root-mean-square (RMS)
Measured over one (or more) cycles of the wave. Allows the comparison of AC and DC circuit values. RMS values of AC create the same heat as that
same numerical voltage value of DC.
RMS is most common method of specifying the value of sinusoidal waveforms.
Almost all voltmeter and ammeters are calibrated so that they measure AC values in terms of RMS amplitude.
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VRMS = VDC = heat
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RMS Heating Effect VRMS = 0.707 Vpeak
IRMS = 0.707 Ipeak
A sinusoidal voltage with peak amplitude of 1 volt has the same heating effect as a DC voltage of 0.707 volts. AC creates slightly more heat than DC. Comparing RMS value to average voltage.
Due to this, the RMS value of voltage is also referred to as the effective value.
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Determining the 0.707 Constant
To determine the 0.707 constant, you must use the mathematical procedure suggested by the name, root-mean-square.
This has nothing to do with the sine function even though this is the same value as sin 45°.
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VRMS =
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Average Voltage
Average voltage is the DC equivalent voltage. The average voltage of AC over a full sine wave
equals zero. The positive half cycle is equal to the negative half
cycle. Because of this we only look at half the wave
(the positive half cycle). This means we are looking at an angle of π radians.
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Average Voltage
Average voltage over a half cycle:
Or, VAVG = .637 VPK VAVG is produced by rectifying then filtering the
AC in a voltage regulator. This is the process used in a DC power supply.
VAVG = =
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Instantaneous Voltage
This is an important value when sampling a wave at several points. Example: analog to digital conversion
Different types of waves might have the same peak or average voltage but different instantaneous voltages Examples: square wave, triangle wave
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Instantaneous Voltage
Instantaneous voltage is the voltage at a single point or instant of time.
To find the angle, take the inverse sine. Usually a 2nd function on a calculator
= sin θ
( )
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Relationship Circles
P – P RMS
2 PK 0.707 PK
INST. AVG.
Sine° PK 0.637 PK
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Relationship CalculationsEXAMPLES120 VAC = 170 Vpk
Formula: PK = RMS 0.707120 0.707 = 169.7 (round off to 170 Vpk)
18 V @ 72 = 19 VpkFormula: PK = Instantaneous Sine18 Sine(72°) = 18.9 (round off to 19)
30 Vpk = 21.2 VACFormula: RMS = 0.707 X PK0.707 X 30 = 21.2 350 V @ 23.5 = 30 Vpk
Formula PK = Instantaneous Sine350 Sine(23.5°) = 877.7(round off to 878)
50 Vpp = 17.7 VrmsStep 1: Need to find PKFormula: PK = P-P 250 2 = 25mStep 2: Find RMSFormula: RMS = 0.707 X PK0.707 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 PK454 908 = 0.5 2nd Sine (.5) = 30
20 V Average = 22.2 Vrms= 31.4 Vpk = 62.8 Vp-pStep 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: PK = 2 X PK= 2 X 31.4 = 62.8
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Calculate peak voltage given RMS voltage
VRMS = 120 V Formula: PK = RMS 0.707 120 0.707 = 169.7
(round up to 170 Vpk) 120 VRMS = 170 Vpk
P – P RMS
2 PK 0.707 PK
INST. AVG.
Sine° PK 0.637 PK
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Calculate RMS voltage given peak voltage
Vpk = 30 V Formula: RMS = 0.707 X PK 0.707 X 30 = 21.2 V 30 Vpk = 21.2 VRMS
P – P RMS
2 PK 0.707 PK
INST. AVG.
Sine° PK 0.637 PK
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Calculate RMS given p-p
Vpp = 50 V Step 1: Find Vpk
Formula: PK = p-p 2 50 2 = 25 V
Step 2: Find RMS Formula: RMS = 0.707 X PK 0.707 X 25 = 17.675
(round off to 17.7 Vrms) 50 Vpp = 17.7 Vrms
P – P RMS
2 PK 0.707 PK
INST. AVG.
Sine° PK 0.637 PK
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Calculate VRMS, Vpk, and Vpp given VAVG
VAVG = 20 V Step 1: Find Vpk,
Formula: PK = Average 0.637 = 20 0.637 = 31.39
(round up to 31.4) Step 2: Find RMS,
Formula: RMS = 0.707 X PK = 0.707 X 31.4 = 22.19 (round up to 22.2) Step 3: Find P-P,
Formula: P-P = 2 X PK = 2 X 31.4 = 62.8Copyright © Texas Education Agency, 2014. All rights reserved.
P – P RMS
2 PK 0.707 PK
INST. AVG.
Sine° PK 0.637 PK
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Calculate peak voltage from instantaneous voltage
VINST = 18 V @ 72° Formula: PK = Instantaneous Sine
of the angle = 18 Sine(72°) = 18 .951 = 18.9 V
(round up to 19) 18 V @ 72 = 19 Vpk
P – P RMS
2 PK 0.707 PK
INST. AVG.
Sine° PK 0.637 PK
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Calculate peak voltage from instantaneous voltage
VINST = 350 V @ 23.5° Formula: PK = Instantaneous
Sine(angle) = 350 Sine(23.5°) = 350 0.4 = 877.7
(round up to 878) 350 V @ 23.5 = 878 Vpk
P – P RMS
2 PK 0.707 PK
INST. AVG.
Sine° PK 0.637 PK
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Calculate phase angle given instantaneous voltage and peak voltage
454 VINST at unknown angle with a Vpk of 908 V Formula: Sine θ =
Instantaneous PK 454 908 = 0.5 2nd Sine (.5) = 30
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P – P RMS
2 PK 0.707 PK
INST. AVG.
Sine° PK 0.637 PK
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Relationship Exercise
# rms peak pk-to-pk average instantaneous
200mV ________ ____ V @ 72º113V ________ V @ 90º
96.4 ________ V @ 235º1.5V @ 122º
689V ________ V @ 35º Go to Answers
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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. Sines, cosines, and tangents are numbers equal
to the ratio of the lengths of the sides. The sine function is used in AC because the
opposite side is the direction of motion of the conductor through a magnetic field relative to the angle.
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Right-Triangle: Side and Angle Relationships A right triangle has a 90° angle. Each side is named with respect to the angle
you are using (called the angle theta, or θ). The side of the triangle across from 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.
Each of these sides are commonly abbreviated to their initials, O, H, and A.
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Right Triangle
Hypotenuse
Opposite
Adjacent
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θ
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Basic Trigonometric Functions
In trigonometry, there are three common ratios used to study right triangles.
1. SineA. 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.
B. Sine θ =
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Basic Trigonometric Functions(Continued)
2. CosineA. The cosine of the angle theta is equal to the ration formed by
length of the adjacent side divided by the length of the hypotenuse.
B. Cosine θ = 3. Tangent
A. 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.
B. Tangent θ =
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Simple Memory Aid
Remember the acronym SOH-CAH-TOA Pronounced “sock ah toa”
Sine equals opposite over hypotenuseCosine equals adjacent over hypotenuseTangent equals opposite over adjacent
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Basic Trigonometric Functions (Continued)
Opposite
Sine Hypotenuse
Adjacent
Cosine Hypotenuse
Opposite
Tangent Adjacent
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Trigonometric ExerciseTriangle #1
Hypotenuse? 8’ Opposite
10’ AdjacentWhat is the Hypotenuse? ___________
Triangle #2
Hypotenuse? 5.3 Rods Opposite
6.8 Rods AdjacentWhat is the Hypotenuse? ___________
Triangle #3
125 Miles Hypotenuse Opposite?
85 miles AdjacentWhat is the Opposite? ____________
Triangle #4
56’ Hypotenuse 23.2’ Opposite
Adjacent?What is the Adjacent? _____________
Go to Answers
End of
Presentation
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Answer Keys
Follow This Slide
End of
PresentationGo to Answers
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Relationship Key
# rms peak pk-to-pk average instantaneous
1. 200mV 283mV 566mV 180mV 269mV @ 72º2. 80v 113V 226V 72V 113V @ 90º3. 96.4 136V 272V 87V 111V @ 235º4. 1.25V 1.77V 3.54V 1.13V 1.5V @ 122º5. 764µV 1.08mV 2.16mV 689V 619µV @ 35º
Go to the Trigonometric
Functions
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Trigonometric Exercise KeyTriangle #1
Hypotenuse? 8’ Opposite
10’ AdjacentHypotenuse = 12.8’
Triangle #2
Hypotenuse? 5.3 Rods Opposite
6.8 Rods AdjacentHypotenuse = 8.62 Rods
Triangle #3
125 Miles Hypotenuse Opposite?
85 miles AdjacentOpposite = 91.7 miles
Triangle #4
56’ Hypotenuse 23.2’ Opposite
Adjacent?Adjacent = 51’
End of
Presentation
Go to the Calculations
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Trigonometric Exercise Calculations
Triangle #1 Given Adjacent side = 10’ and Opposite side = 8’. What is the Hypotenuse?
Step 1 - Find the degree angleTangent = Opposite Adjacent
= 8_ = 0.8 10 = enter 2nd tangent (0.8) on your calculator = 38.65980825 round up to 38.66º = 38.66°
Step 2 - Change the degree angle to cosineHypotenuse = Adjacent (take 38.66°, enter cosine on your calculator, your answer is 0.780866719)
Cosine = 10’____________ .780866719 = 12.8’
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Trigonometric Exercise Calculations (Continued)
Triangle #2Given Adjacent side = 6.8 rods and Opposite side = 5.3 rods. What is the Hypotenuse?
Step 1 - Find the degree angleTangent = Opposite
Adjacent = 5.3 rods
6.8 rod enter 2nd tangent 0.779411765 on your calculator
= 37.93°Step 2 - Change the degree angle to sineHypotenuse = Opposite Sine
= 5.3 rods sine 37.93 (enter 37.93, enter sine on the calculator ) = 5.3 rods
0.6146982793 = 8.63
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= 0.779411765
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Trigonometric Exercise Calculations (continued)
Triangle #3Given Hypotenuse side = 125 miles and Adjacent side = 8.5 miles What is the Opposite side?
Step 1 - Find the degree angleCosine = = 0.68
enter 2nd function button on calculator enter cosine 0.68 = 47.15635696 (round off to 47.16º)
Step 2 - Change the degree angle to sineOpposite = 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
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=
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Trigonometric Exercise Calculations (continued)
Triangle #4Given Hypotenuse side = 56’ and Opposite side = 23.2’ What is Adjacent side?
Step 1 - Find the degree angleSine = = 0.4142857143 = enter 2nd sin 0.4142857143 = 24.47ºStep 2 - Change the degree angle to cosineAdjacent = Cosine x Hypotenuse
enter cosine 24.47 on the calculator = 0.910178279 x 56’ = 50.96998362 (round off to 51’) = 51’
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=