Chapter 1 – Introduction to Electrical Measurement
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Transcript of Chapter 1 – Introduction to Electrical Measurement
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Chapter 1 – Introduction to Electrical
Measurement
ESE 122Group 1 (a)
AdliAsif
helmi Shauqi
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INTRODUCTION TO ELECTRICAL MEASUREMENT
ESE 122
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INTRODUCTION
• Measurement is the process of obtaining the magnitude of a quantity, such as length or mass, relative to a unit of measurement, such as a meter or a kilogram.
• Measurement is basically about counting process.
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Unit, Dimensions & Standards
• International System Unit : S.I. Unit
• 4 basic unit SI second (s) : time meter (m) : length kelvin (K) : temperature kilogram (kg) : mass
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Measurement Standards
• International Standards
• Primary Standards
• Secondary Standard
• Working Standard
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International Standards
• Defined by international agreements
• Maintained at the International Bureau of Weight & Measures, Pairs
• Periodically evaluated & checked by absolute measurements in term of the fundamental units of physics.
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Primary Standards
• Maintained at institution in various countries around the world.
• Eg: SIRIM (Malaysia)SISIR (Singapore)KIRDI (Kenya)
• Use to calibrate & verify the secondary standards.
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Secondary Standard
• Used by measurement & calibration lab in the industry as basic reference standards
• Secondary standards be responsible by its own industrial lab
• Each lab sends its secondary standards to the national standards lab for calibration
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Working Standard
• Used to check & calibrate the instrument used in the lab or to make comparison measurement in industrial application
• Eg : capasitors inductors standard resistors
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Errors
• Gross Errors
• Systematic Errors
• Random Erros
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Gross Errors
• Mistakes resulting from grave lack of proper consideration, such as stupidity, confusion, carelessness, or culpable ignorance.
• To reduce it, two or more readings should be taken by different experiments
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Systematic Errors
• Instrumental
• Observational
• Environmental
• Simplification
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Random Errors
• Shows variation of results from one to another even after all systematic & gross errors have been accounted for.
• Can be determined by statistical analysis
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Error in Measurement• Absolute error e=Yn-Xn• Percent Error = I e/Yn I x 100% Accuracy = 1- I Yn-Xn/Yn I Percent Accuracy = Ax100% Precision = 1- I Xn-xn/xn I Resolution = the smallest change in a measured variable to which an instrument will respond
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• Eg : the measurement for the power across the resistor is 98W, but the expected value is 100W. Calculate the absolute error,% error, relative error, and % of accuracy
Absolute error : 100-98 = 2W% error : I 100-98/100 I x 100 = 2%Accuracy : 1- I100-98/100 I = 0.98%of accuracy : 0.98 x 100 = 98%
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• Eg : table 1 gives the set of 5 measurement. Calculate the precision of the 3rd measurement
Average value : 235/5 = 47Precision for the 3rd
reading 1- I 56-47/47 I = 0.8085
No. measurement
1 45
2 34
3 56
4 32
5 68
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Statistical Analysis of Measurement• Arithmetic mean X=(x1+x2+x3…..xn)/ n Deviation d1 = x1-X d2 = x2-X dn = xn=X Algebraic sum of the deviation dt= d1+d2+d3+…..dn Average deviation D= Id1I+Id2I+Id3I+….IdnI n Standard Deviation √( d1+d2+d3+…..dn) (reading less than 30, the denominator will n-1 minus by 1) Probable Error r = 0.6745 x standard deviation
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• Eg : the power in a electric circuit was measured by 5 different students and been recorded in the table below. Calculate the arithmetic mean, deviation, algebraic sum of the deviation, average deviation, standard deviation, probable error.
No. Measured power (W)
1 42
2 44
3 50
4 48
5 44
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• Mean = 228 5 = 45.6
Deviation Standard Deviation d1 = 42 - 45.6 = -3.6 √ 3.6²+1.6²+4.4²+2.4²+1.6² d2 = 44 - 45.6 = -1.6 4 d3 = 50 - 45.6 = 4.4 = 3.286 d4 = 48 - 45.6 = 2.4 d5 = 44 - 45.6 = -1.6 Probable error 0.6745 x 3.286 = 2.217 Sum = -3.6-1.6+4.4+2.4-1.6 = 0
o Average = 3.6+1.6+4.4+2.4+1.6 = 13.6
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Question 1
a) The terms accuracy and precision are two key parameters in measurement. Define each of them and state available standards for measurement.Solution:Accuracy: Degree of exactness of a measurement compared to the expected value.Precision: A measure of the consistency of repeatability of measurement.
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Question 1
b) Systematical error can be divided into four categories. State all of them.Solution:1) Instrumental2) Observational3) Simplification4) Environmental
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Question 1
c) The expected value of the voltage across a resistor is 90V. However, the measurement gives a value of 88V. Calculatei) absolute error
90V-88V=2Vii) percentage error
iii) relative accuracy
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Question 1
d) Time Temperature (⁰C)
1 p.m. 30.0
2 p.m. 29.5
3 p.m. 29.0
4 p.m. 28.5
5 p.m. 27.0
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Question 1
d) The temperature of measurement without air conditioning is recorded each hour from 1 pm to 5 pm and tabulated in Table Q1d. Determine:i) The average temperature within the recorded period.
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Question 1
ii) Find the average deviation and deviation for each hour.
Average Deviation
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Question 1
iii) Calculate the probable error for the data.
= 1.151
Probable error= 0.6745 x 1.151= 0.776
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Question 1
e) A system of units is required before one can make a quantitative evaluation of parameters measured. State four (4) fundamental quantities of S.I. units.1) metre2) kilogram3) second4) kelvin
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LIMITING ERRORSESE 122
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Limiting Errors
• Limiting error can be defined as the specification that any instrument can be guaranteed to be accurate within that specified limit.
• Most manufacturers of measuring instruments specify accuracy within a certain percent of a full scale reading.
• If the reading less than full scale, the limiting error increases.
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Total Measurement System Errors
• A measurement system often consists of several separate components, each of which is subject to systematic & random errors.
• Mechanism have now been presented for quantifying the errors arising from each of these sources and therefore the total error at the output of each measurement system can be calculated.
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Error in a Sum
• If
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Error in a Multiplication/Division
• For
• For
• For
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Error in a Power Factor
• For
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Question C
• Based on the circuit shown in Figure Q1c, calculate the magnitude and percentage error for total resistance and power dissipated by resistor R1. Given that R1=12kΩ R2=5kΩ ±10% R3=10KΩ ±20% and supply voltage Vs is equal to 12V ±10%.
Figure Q1c
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Solution
• R1= 12 kΩ ± 5%• R2= 5kΩ ± 10%• R3= 10 kΩ ± 20% = 9.391V
• Magnitude• R2//R3+R1• = 5k//10k+12k• = 15.3 kΩ
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Solution
• Total Error R2 + R3
= 30%
= 3.3334 + 13.333 = 16.666%
30+16.666 = 46.666%
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Solution
• Total error
= 3.922 + 10.0652 percentage power R1 error= 13.9875 % (28.9872 x 2) + 5
• voltage error = 62.9744%= 5 + 10 + 13.9872= 28.9872%
Answer:= 15.3k Ω ± 13.9875%
= ± 62.9744%
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