Chapter 1 – Introduction to Electrical

Measurement

ESE 122Group 1 (a)

AdliAsif

helmi Shauqi

INTRODUCTION TO ELECTRICAL MEASUREMENT

ESE 122

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.

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

Measurement Standards

• International Standards

• Primary Standards

• Secondary Standard

• Working Standard

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.

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.

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

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

Errors

• Gross Errors

• Systematic Errors

• Random Erros

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

Systematic Errors

• Instrumental

• Observational

• Environmental

• Simplification

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

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

• 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%

• 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

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

• 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

• 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

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.

Question 1

b) Systematical error can be divided into four categories. State all of them.Solution:1) Instrumental2) Observational3) Simplification4) Environmental

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

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

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.

Question 1

ii) Find the average deviation and deviation for each hour.

Average Deviation

Question 1

iii) Calculate the probable error for the data.

= 1.151

Probable error= 0.6745 x 1.151= 0.776

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

LIMITING ERRORSESE 122

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.

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.

Error in a Sum

• If

Error in a Multiplication/Division

• For

• For

• For

Error in a Power Factor

• For

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

Solution

• R1= 12 kΩ ± 5%• R2= 5kΩ ± 10%• R3= 10 kΩ ± 20% = 9.391V

• Magnitude• R2//R3+R1• = 5k//10k+12k• = 15.3 kΩ

Solution

• Total Error R2 + R3

= 30%

= 3.3334 + 13.333 = 16.666%

30+16.666 = 46.666%

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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PercyDustfinger

Karl Komodo