POWER II

PAST QUESTION AND ANSWER ON ELECTRICAL AND ELECTRONIC INSTRUMENTATION 1 COURSE CODE: EEC 126. CLASS:ND1


(1) The measuring system of an indicating instrument is subjected to deflecting torque, controlling torque…………………….torque (1 mark)
Answer; the indicating instrument contains three types are (1)deflecting torque (2) controlling torque (3) damping torque
(2) Dynanometer wattmeter instrument is most commonly used ……………..
Answer; dynanometre are commonly used in AC and DC
(3) Moving iron instrument are either of attraction type or …………….
Answer; It is either of repulsion type or attraction type
(4) A ……………. Resistor extends the range of an ammeter
Answer; NOTE: THE RANGE OF PMMC AMMETER CAN BE EXTENDED BY CONNECTING A LOW RESISTANCE SHUNT IN PARALLEL, WHILE THE RANGE OF A VOLTMETER CAN BE EXTENDED BY CONNECTING A HIGH RESISTANCE CALLED MULTIPLIER IN SERIES.
(5) The digital instrument count the cycles produced by the unknown frequency during a precisely controlled…………….
Answer; period of time
(6) Define the term measuring instruments
ANSWER; MEASURING INSRUMENT ARE INSTRUMENT THAT ARE MAJORLY ELECTROMECHANICAL OR ELECTRONIC DEVICE, THAT IS USED TO MEASURE ELECTRICAL OR NON ELECTRICAL DEVICE IN ORDER TO DISPLAY THEIR QUANTITY EITHER IN A DICRETE FORM(USING DIGITAL INSTRUMENTS) OR VARYING FORM BY THE USE OF A POINTER (USING ANALOG INSTRUMENT)
c) List 5 advantages of electronic instrument over electromechanical instrument.
ANSWER
I) It is used to monitor nearby signal
ii) it’s not as enormous or big as compared to electromechanical instrument
iii) it has high flexibility
iv) it also have high sensitivity
v) low cost
vi) faster response when in use i. e during usage
d) Explain with the aid of an appropriate diagram and mathematical analysis that the deflecting torque of a permanent magnetic moving coil instrument (p.m. m. c) is proportional to the current flowing through it.

ANSWER
NOTE: Before showing the mathematical analysis you are expected to list all the parameters used in the p. m. m. C and their name. Such as;
B=breadth of the, coil in Meter
N=number of turns in the coil in turns
B=The flux density of the coil, in weber per meter
L=length of the depth of the coil in meter
I=current flowing through the coil in ampere

Followed by the solving… Such as;

The force magnitudeexperienced in the coil
F=BIL (BREADTH × CURRENT × LENGTH)
Number of turns experience at each side of the coil
F NBIL
Deflecting torque( Td) =F × distance at a right angle I.e BREADTH
Td= F×B I.e NBIL×B
A=L×B. so therefore,
Td=NBIA….. (1)
But we are asked to prove that deflecting torque (Td) is at right angle to the current flowing through. Then, we say
Constant (k) =NBA
Substitute NBA as K into equation
Td=KI
Recall, that anytime you have a quantity equal to constant (k) , with a value or parameter. It can be rewritten as the quantity DIRECTLY PROPORTIONAL TO THAT VALUE
So therefore,
Td is directly proportional to the current flowing through the coil
2) a Explain the construction and operation of a permanent magnetic moving coil instrument
ANSWER
This is based on the principle of operation that When an instrument is connected a circuit ;so as to measure voltage and current. A current pass through the coil due to the current being measured, which is then placed in the magnetic field of the permanent magnet which therefore allow a mechanical force to be exerted upon it. This allow the pointer on the moving system to be deflected in a clockwise way in the graduated scale, so as to indicate increase in current and voltage.
The deflecting torque on a permanent magnetic moving coil in a moving coil instrument, is as a result of the interaction between the field set up by a permanent magnet( this is reason it is called permanent magnetic moving coil).

ii) SYSTEMATIC ERROR :This error can be caused due to two factors
(a) Faulty instrument : this deals with the short coming of instrument
(b) Environmental factors: some instrument are affected by environmental condition which can lead to error.
iii) RANDOM ERROR : This are errors that can not be directly established. This means that the caused of the error is unknown because of random variation in the parameter or system of measurementmeasurement


2b) A moving coil instrument with a resistance of 10 ohms give a full scale deflection with a potential differene of 45mV. The coil has 100 turns, an effective depth of 3cm and a width of 2.5cm. The controlling torIque exerted by the spring is 49×10^-9NM. Calculate the density in the air gap
Solution
R=10ohms
TC=49×10^-9NM
P. d=45×10^-3 V
N=100turns
A=2.5×3×10^-4M=7.5×10^-4
B=?
I=V/R100
=45 ×10^-3/10
=4.5×10^-3A
At equilibrium Tc=Td
Tc=NBIA
49×10^-9=100×B×4.5×10^-3×7.5×10^-4
B=49×10^-9/3375×10^-7
B=49×10^-2/3375
=0.145×10^-3weber/M^2
2c) Highlights two merit and two demerit of pmmc
Two merit
I) They are accurate and reliable
ii) They have uniformly scale i.e evenly divided scale
Two demerit
I) it is limited to D. C only
ii) it is quite expensive compare to moving iron instrument because of their accurate design
3a) describe three types of error
I)
GROSS ERROR :This deals majorly with human beings little mistake which can be caused by;
(a) IMPROPER ADJUSTMENT OF INSTRUMENT :Mostly when a laboratory personnel is trying to adjust the instrument to suit the measurements due to human errors some little mistake could take place which can later be ignored
(b) COMPUTATION ERROR:The operator may make mistake when putting the values in a written form
(c) CALCULATION ERROR:The operator might make mistake while calculating

3b) A voltmeter having a sensitivity of 100 ohm /V reads 100volts on it 150V scale when connected across an unknown resistor in series with milliameter when the milliameter reads 5mA, calculate
I) apparent resistance of the unknown resistor
ii) actual resistance of the unknown resistor
iii) Error due to loading effect of the voltmeter
SOLUTION
I) finding total resistance Rt=Vt/It
100/5×10^-3=20000ohms=20kohm
So therefore apparent resistance of the unknown resistor is 20Kohms


ii) voltage resistance =100×150=15000ohms(why is because we are given 100ohms per volts and 150v is given to represent one ohm)
Actual resistance =Rv×Rt/Rv-Rt
=150×20/150-20
=3000/130
=23.08Kohms


iii) % error =Actual -Apparent /Actual ×100
=23.08-20.0/23.0×100
=3.08/23.08×100
=13.34%


3c) Draw the major block diagram of a Cathode ray tube and explain each unit/stage

Cathode ray tube: it display the quantities being measured
Vertical amplifier :it amplify the signal waveform to be viewed
Trigger circuit : it produces trigger pulses to start horizontal sweep
Sweep generator :it produces saw tooth voltage waveform used for horizontal deflection to the electron beam
Horizontal amplifier: it is fed with a saw tooth voltage which is then applied to the x-plates
High and low voltage supply: it supply voltage power

4a) Highlights 4 advantages of digital frequency
i) The digital output signal can be coded as decimal BCD or binary codes
ii) digital frequency meter measures Unknown frequency by counting the number of cycles the frequency produce in a precisely contolled period of tim
iii) in a Schmitt trigger the signal is converted into square wave with a very fast rise and fall time than differentiated and chip.

Oluwamuyide Peter

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