Module 2: AC Measurements - Rahma NajjarMeasurements and instrumentation Module 2: AC Measurements....
Transcript of Module 2: AC Measurements - Rahma NajjarMeasurements and instrumentation Module 2: AC Measurements....
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Measurements and instrumentation
Module 2: AC Measurements
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Watch the following video
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Module objectivesUpon successful completion of this module, students should be able to: • Familiarise with the definition of alternating current (AC) • Identify appliances/devices that operate on AC. • Familiarise with the laboratory instruments /equipments used to
generate and measure AC signal. • Use a function generator to generate an AC signal. • Use an Oscilloscope/multimeter to measure AC signal parameters. • Differentiate between rms value and peak value.
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Alternating current (AC) is the most common form of electrical power delivered to homes and businesses through electrical outlets and is used every day in all aspects of life.!
This is due to the fact that AC power is easy to transmit and convert.
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The AC voltage alternates between positive And negative values. By tracing the variation over time, we can plot it as a "waveform". The AC voltage alternates (varies) in polarity and AC current alternates in direction.
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The usual waveform of an AC waveform is a sine wave.
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The usual waveform of an AC waveform is a sine wave. However, in certain applications, different waveforms are used, such as triangular, square, or sawtooth waves.
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Triangular waveform
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Square waveform
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Sawtooth wave
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You can find
different shapes of waveforms
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AC signal parameters
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At the following picture and answer the question.
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What do you think the difference between the direct current and the
alternating current is? .
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AC Signal parameters
Since the DC voltage and current quantities are generally stable, it is easy to express how much voltage or current is present in any part of a circuit. While AC is a varying signal, it changes from positive to negative.
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After knowing the difference between AC and DC signals, try to tell whether the following signals are of AC of
DC type?
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Knowing that AC is a varying signal, how can we
express how large or small an AC quantity is? How can a signal
measurements value be assigned to something that is constantly
changing?
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AC Signal parameters
To measure an AC signal , it is important to know the following AC signal parameters:
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AC Signal parametersPeak value
(Vp)Peak to peak value (Vp-p)
Root mean square value (Vrms)
Cycle Period Frequency
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Peak Value (Vp)Peak value is the maximum value reached by an AC wave.
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Peak to peak Value (Vp-p)Peak to peak value is a measure of the total height between opposite peaks. It is twice the peak value.
Vp-p = 2 x Vp
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Root mean square value(Vrms)
The RMS value is the effective value of an AC signal. A measure of the peak value does not give the actual AC value. The true value can be calculated using the formula: !
(They are only true for some waves because the constant or the multiplier is not 0.707 for other shapes)
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CycleOne complete waveform is called a cycle.
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Period
The period is the time taken to complete one cycle. It is measures in second (s)
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PeriodFor example, The period for the waveform shown in the following figure is 20 ms
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Frequency
Number of cycles per second is called frequency.
F = 1/T or T = 1/F where T is the period. Frequency is measured in hertz (Hz)
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Summary
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Skill number 1 page 6:Calculate the period, frequency, and the RMS value of the sine wave:
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Skill number 1 solution:A. What is the period of the waveform shown? Period = 20 ms B. Calculate the frequency of the signal. Frequency = 1/T = 1/20ms. = 0.05 KHz C. Find the RMS amplitude of the signal? Vrms = 0.707 x Vpeak , where Vpeak = 6 volt Vrms = 0.707 x 6 = 4.242 volt.
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Try to answer the following questions.
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Question 1Study the following waveform, and calculate the following: 1. Vp 2. Vp-p 3. Vrms 4. Period 5. Frequency
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Question 1 answerStudy the following waveform, and calculate the following: 1. Vp = 2.4V 2. Vp-p= 2 xVp = 2x 2.4 = 4.8 V 3. Vrms = 0.707 xVp = 0.707 x 2.4 = 1.7 V 4. Period = 4 ms 5. Frequency = 1/T = 1/(4ms) = 0.25 KHz
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Question 2Study the following waveform, and calculate the following: 1. Vp 2. Vp-p 3. Vrms 4. Period 5. Frequency
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Question 2 answer:Study the following waveform, and calculate the following: 1. Vp = 40 V 2. Vp-p= 2 xVp = 2x 40 = 80 V 3. Vrms = 0.707 xVp = 0.707 x 40 = 28.28 V 4. Period = 0.016 s 5. Frequency = 1/T = 1/(0.016) = 62.6 Hz
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Laboratory Equipment
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Watch the following video
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Laboratory Equipment
In the laboratory, AC signals are produced using the Function Generator. !
And displayed and measured using the Oscilloscope.
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Function generatorThe function generator is a device that generates voltage signals. Specifically, a sine, a triangle, or a square wave may be created. Also, the signals parameters may be specified.
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Function generatorThe basic form of a sine waveform as a function of time (t), is as follows: V(t) = A sin(2.π.F.t) Where, A, is the amplitude. F, is the frequency.
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For example,
The function : V(t) = 3 sin(2.π.(100).t) is created by adjusting the function generator to output a sine wave with a frequency of 100 Hz, and peak value of 3 volts.
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Oscilloscopes are used to display and measure signal parameters.
Oscilloscope
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Specifically, the oscilloscope displays a signal with an axis for voltage, and an axis for time. The scales of the two axes are given in Volts/Division and Second / Division respectively.
Oscilloscope
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Skill Number (2) page 8:To calculate the peak voltage and peak-to-peak voltage from the oscilloscope reading if the Volt/div = 2 V and Time/div = 500 ms.
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Skill Number (2) solution:A. What is the peak voltage (Vp)? Peak voltage (Vp) = 7 divisions x 2V / Division = 14 volt B. What is the peak-to-peak voltage?. Peak-to-peak voltage = 2 x Vpeak = 2x 14 = 28 volts. Or, Vpp = 14 div x 2V/div. = 28 volts.
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Skill Number (3) page 9:A. To calculate the peak voltage with the given rms voltage.
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Skill Number (3) solution:A. What is the rms voltage? RMS voltage = 3.44 volts. !
B. What is the peak voltage? Vrms = 0.707 x Vp Vp = Vrms / 0.707 = 4.86 V
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Skill Number (3):B. Calculate the rms voltage with the given peak voltage. Volt/div = 5V.
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Skill Number (3) solution:A. What is the peak voltage? Vp = 1division x 5 volts /division Vp = 5 volts. B. What is the RMS voltage? Vrms = 0.707 xVp Vrms = 0.707 x 5 = 3.53 Volts
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Lab activity1
Objective: to generate AC voltage using a function generator, and to measure the voltage using the oscilloscope.
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Procedure: voltage generation and measurement: Connect the function generator to the oscilloscope as shown in the figure below:
Lab activity1
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Procedure: voltage generation and measurement: Set the function generator to output a sine wave with a frequency of 1 KHz, and peak value of 5 V.
Lab activity1
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Procedure: voltage generation and measurement: Follow the steps for measuring period, frequency, peak and rms voltages, and sketching the waveform.
A. Period and frequency
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A. Period and frequency
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B. peak voltage
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C. RMS voltage
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D. Waveform
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Don't forget to solve the homework posted on PLATO.
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