electrical measurement & interface

8/10/2019 ELECTRICAL MEASUREMENT & Interface

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Pharos University

EE212_ELECTRICAL MEASUREMENT & Interface

Prepared By:

Dr. Sahar Abd El Moneim Moussa

Dr. Sahar Abd El Moneim Moussa 1


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CH1_ ERRORS IN MEASUREMENTS


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INTRODUCTION

Measurement involves using an instrument as a physicalmeans of determining a quantity or variable.

A measuring device: a device used for obtaining a valuewhich nearly represents an unknown variable or quantity.

The advancement of science and technology leads to the

progress in the measuring techniques.



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Classification of Measuring Instruments:

1- Mechanical Instruments:

These are the first type of measuring instruments.

Since they have moving parts then they are useful for stableoperation but due the inertia of the moving elements are

slow then they are not suitable for measuring dynamic

variables and transient operations.



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Classification of Measuring Instruments:( cont.)

2- Electrical Instruments:

Since 1801. By discovering electricity, the electrical

measuring instruments appear.

Electrical instruments have in general electric circuit

consisting of some resistors and a magnetic circuit consisting

of either a permanent or a coil with an iron core.

It depends upon the interaction between the electric circuit

and the magnetic circuit.



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Classification of Measuring Instruments:( cont.)

3- Electronic Instruments:

Then the electronic measuring instruments appear which

depends upon electronic components such as diodes and

transistors.

They are fast and can be used in transient measurements.



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Methods of Measurements:

1- Direct Method:

The unknown quantity is directly compared against a

standard. Which are common for measurements of physical

quantities like length, mass and time.

Disadvantages are:

i- limited accuracy due to human factors.

Ii- Less accuracy



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Methods of Measurements: (cont.)

1- Indirect Method: The unknown quantity under measurement is determined

via the use of measurement systems as follows:

Advantages are:

i- High accuracy & sensitivity with lower cost

ii- Measurement of non-physical quantities



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Basic definitions:

1- True Value: AtIt is the actual value of a variable to be measured

2- Measured Value: Am

It is the value measured by a specified instrument

3- Static (Absolute) Error:(

A)

It is the difference between measured value (Am ) and true

value ( At ). It is the deviation of the measured value from the true value.


A= Am-At


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Basic definitions: (cont.)

4- Static Correction:(

C)

It is the correction to be added to the measured

value to give the true value


C =-

At= At-Am


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5- Relative Error:(

r)

The Static Error is expressed as a fraction of the full scale

deflection or the true value


r (as a fraction of true vale) =( Am - At ) / At

r (as a fraction of true vale) =( Am - At ) / f.s.d

r %(as a fraction of true vale) = (A / At) x 100%

r %(as a fraction of true vale) = (A / f.s.d) x 100%


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Example 1:

A meter reads 127.5 V and the true value of the voltage is

127.43 V, determine the:

i- Static error

ii- Static correction for this instrument.

Solution:

The Static Error: A = AmAt = 127.5127.43 = + 0.07 V

Static correction: C =A =0.07 V



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Example 2:

A voltmeter has a true value of 1.5V. An analog indicating

instrument with a scale 0-2.5 shows a voltage of 1.46V. Whatare the values of absolute error and correction. Express the

error as a fraction of the true value and f.s.d.

Solution:


It will Solved in the lecture


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6- Limiting Error:

It is the specification given to an instrument which

guarantees that a full scale deflection will be within a certain

percentage of a full-scale reading. However, the readings thatare less than full scale has a larger limiting error.

Therefore it is the limit of the deviation from the specified

value specified by manufacturer



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Example 3:

A 300 V voltmeter is specified to be accurate within 2% at

full scale. Calculate the limiting error when the instrument isused to measure a 120 V source and 60 V. Comment.

Solution:

The limiting error at full scale is 2%

The magnitude of the limiting error at full scale is

= 300 x ( 2/100 ) = 6 V

The limiting error at 120 V = 6/120 x 100 = 5%

The limiting error at 60 V = 6/60 x 100 = 10%

The results of the previous example give us an important

conclusion, which is:


Measurements of low values has higher limiting errors, so readings

should be made near full scale to decrease error.


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Example 4:

The inductance of an inductor is specified by a manufacturer as

20H5%. Determine the limits of the inductance betweenwhich it is guaranteed

Solution:




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Characteristics of the Instruments:

1- Accuracy:

Accuracy is defined as the closeness with which an instrumentreading approaches to the true value of the quantity measured.

2- Resolution:

It is the smallest value that can be detected by a certaininstrument



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Characteristics of the Instruments: (cont.)

3- Sensitivity:

It is the ratio between the output magnitude to the input

magnitude

Deflection factor = 1/Sensitivity

4- Precision:

It is how the instrument will insist on its reading for a

certain quantity for a number of readings



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Example 5:

If an Ohmmeter requires a change of 8 to produce a change

in the deflection of 2 mm. Find its sensitivity.

Solution:

sensitivity is given by:

S = 2 / 8 = 0.25 mm /

Example (6):

If an ammeter has an input of 10 A which produces 5 mm. Find

the sensitivity.

Solution:




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Example 7:

A moving coil voltmeter has a uniform scale with 100 divisions,

the full-scale reading is 200 V and 1/10 of a scale division canbe estimated with a fair degree of certainty. Determine the

resolution of the instrument.

Solution:

1 scale division = 200/100 = 2 V

Resolution = 1/10 of the scale division = (1/10) x 2 = 0.2 V



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Example (8):

A digital voltmeter has a read-out range from 0 to 9999

counts. Determine the resolution of the instrument in volts

when the full scale reading is 9.99V.

Solution:




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Errors in Measurements:

Errors may come from different sources and are usually

classified under the following headings:

1- Gross Errors:

These type of errors are mainly due to human mistakes such

as misreading of instruments, incorrect adjustment and

improper application of instruments.

Gross errors can be reduced by reading and recordingmeasurements with great care and by taking more than one

measurement to be sure that the required reading is

accurate as possible.



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Errors in Measurements: (cont.)

2- Systematic Errors:

These type of errors are either due to shortcoming of the

instrument or environmental errors due to external

conditions affecting the measurement such as temperature,humidity, pressure, loading effects, insertion effects and

magnetic or electrostatic fields.



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3- Random Errors:

These errors are due to unknown reasons.

The effect of random errors is minimized by measuring thegiven quantity many times under the same conditions and

calculating the arithmetic mean of the obtained values

Arithmetic mean=


=


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Range of possible Error: Is the largest deviation from the

arithmetic mean.

Example (9):

A set of independent current measurements were recorded as

10.03, 10.10, 10.11,and 10.08A. Calculate the range of possibleerrors.

Solution:



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Loading Effect Error:

Introducing a measuring device to a system will cause signal

distortion.

This distortion may take the form of reduction in themeasured quantity magnitude, waveform distortion, phase

shift and may be all these undesirable features put together.



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Loading Effect Error: (cont.)

1- Loading effect of a shunt connected device:

On connecting a shunt connected instrument whose

impedance is ZL. The actual voltage E0 decreases to EL.

Therefore the %loading error in measurement of E0 equals:


%

100%


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Loading Effect Error: (cont.)

2- Loading effect of a series connected device:

On connecting a shunt connected instrument whose

impedance is ZL. The actual voltage I0 decreases to IL.

Therefore the %loading error in measurement of I0 equals:


%

100%


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Example (10):

A 50 voltage range voltmeter is connected across the

terminals A and B of the circuit shown in the figure. Find the

reading of the voltmeter under open circuit and loadedconditions. Find the accuracy and the loading error. The

voltmeter has a resistance of 1000 k.

Solution:




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Propagation of Errors:

1- Sum OR Difference of two quantities:

If W is the sum or difference of two quantities:

+ :

Therefore, the relative error wwill be:


[

+

]


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Example (11):

Three resistors having the following ratings: R1=2005%,R2=10010%. Find the magnitude of the resultant

resistance and limiting errors in percentage and ohms, if theabove resistances are connected in series .

Solution:




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Propagation of Errors:(cont.)

2- Product OR Division of two quantities:

If W is the Product or division of two quantities:

: /

Therefore, the relative error wwill be:


[ + ]


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Example (12):

The measurements of potential difference and current for

certain element are V=202%, I=35%. Calculate the

percentage error in the calculated power.

Solution:


electrical measurement & interface

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