Rony Parvej s EEE. Lecture 5 & 6: MIST Special. Update: 30 April, fecabook.com/ronyiut

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1 Rony Parvej s EEE Lecture 5 & 6: MIST Special Update: 30 April, 2015 fecabook.com/ronyiut

2 M.Sc. in EECE Admission Test Question of Military Institute of Science and Technology (MIST) Compiled by: Rony Parvej (IUT, EEE 07) Time: 90 minutes Full Marks: 100 Exam Date: Venue: MIST 1. MCQ (i) A-t/m is the unit of (a) Flux density (b) Magnetic field intensity (c) Permeability (d) Flux (ii) What is the unit of permeability? (a) A-t (b) A-t/m (c) H/m (d) Wb-t (iii) In series RLC circuit when the frequency of the circuit is less than resonance frequency? (a) X L >X C (b) X L < X C (c) X L =X C (d) None (iv) What happens when power factor is increased? (a) Reactive power is increased (b) Active power is increased (c) Apparent power is increased (d) None (v) What are linear circuit elements? (a) the circuit elements whose parameters are not changed with change of voltage & current. (b) (c) (d) (vi) Voltage Regulation depends on (a) Load current (b) Power pactor (c) both (d) none. (vii) - (x). মন ই 2. Elaborate following words: PSM, IGCTs, DCPSK, CSDN, GTO, DGPS, CDMA, IDMT, SCADA 3. What is ideal transformer? 4. Write down the characteristics of an ideal Op-Amp. 5. What are the losses in an compound dc generator? 6. What are linear circuit element & bilateral linear circuit element? 7. What do you mean by coupling co-efficient between two coils is 0.6? 8. A series RLC circuit has R=1000 Ω, L=100mH & C=10 µµf. Find: (a) Resonance frequency (b) Q-factor (c) 3dB point 9. A 3-phase delta connected load has Z=20 60 and line voltage is 400 V. Find: (a) Total power (b) Power consumed in each phase. 10. Write down the conditions of parallel operation of two alternators. 11. Write down four main differences between CT and PT. 12. Find the output of a full wave rectifier if the input is 13. Draw the following signals: (b) x(t) = 2 u(t) + δ (t-1)

3 14. Find Vo 15. Find I E, I B and I C. ( β was not given ) 16. Find I 1, I 2 and V A power station has following loading: Lighting load: 150 Kw for 7.00pm to 10.00pm Residential load: 50KW for 7.00pm to 12.00pm Pumping Load: 75KW for 3.00pm to 7.00pm Find diversity factor. 18. Take 150 MVA, 15KV as base and draw the reactance diagram:

4 1. (i) A-t/m is the unit of (a) Flux density (b) Magnetic field intensity (c) Permeability (d) Flux (ii) What is the unit of permeability? (a) A-t (b) A-t/m (c) H/m (d) Wb-t Reference: Course material of University of Minnesota

5 (iii) In series RLC circuit when the frequency of the circuit is less than resonance frequency? (a) X L >X C (b) X L < X C (c) X L =X C (d) None Before resonance frequency, f r capacitive reactance dominates and after resonance, inductive reactance dominates and at resonance the circuit acts purely as resistive circuit causing a large amount of current to circulate through the circuit. Reference: & So, series resonant circuit is inductive if it operates at a frequency higher than resonant frequency. iv What happens when power factor is increased? (a) Reactive power is increased (b) Active power is increased (c) Apparent power is increased (d) None Ans: Active power is increased v What are linear circuit elements? (a) the circuit elements whose parameters are not changed with change of voltage & current. (b) (c) (d) Ans: the circuit elements whose parameters are not changed with change of voltage & current vi Voltage Regulation depends on Load current (b) Power pactor (c) both (d) none. voltage regulation depends not only on load current, but also on its power factor. Reference: jnrengineer.co.za/doc/electricity%20basics.pdf

6 2 Elaborate following words: PSM, IGCTs, DCPSK, CSDN, GTO, DGPS, CDMA, IDMT, SCADA PSM : Phase Shift Modulation IGCTs : Integrated Gate-Commutated Thyristors DCPSK: Differentially Coherent Phase Shift Keying CSDN : Circuit Switched Data Network GTO : Gate Turn Off (thyristor) DGPS : Differential Global Positioning System CDMA : Code Division Multiple Access IDMT : Inverse Definite Minimum Time SCADA: Supervisory Control and Data Acquisition 2. What is ideal transformer? An ideal transformer is an imaginary transformer which does not have any loss in it, means no core losses, copper losses and any other losses in transformer. Efficiency of this transformer is considered as 100%. Write down the characteristics of an ideal Op-Amp. 1. Infinite Open-Loop Gain 2. Infinite Input Impedance 3. Infinite Bandwidth 4. Zero Output Impedance 5. Zero Noise Contribution 6. Zero output Offset 7. Differential Inputs Stick Together Explanation:

7 What are the losses in an compound dc generator? 1. Copper losses Armature Cu loss Shunt Cu loss Series Cu loss 2. Iron losses Hysteresis loss Eddy current loss 3. Mechanical losses Friction loss Windage lo

8 What are linear circuit element & bilateral linear circuit element? Linear bilateral circuit element: A linear circuit element is an electrical element with a linear relationship between current and voltage. Resistors are the most common example of a linear element; other examples include capacitors, inductors, and transformers. Linear bilateral circuit element: It is a circuit element that behaves the same way if it is connected in the opposite direction. For example, a resistor behaves the same way no matter if it's connected left-to-right or right-to-left. In contrast, a diode is not a bilateral component, because it conducts current in one direction, and does not conduct in the other. What do you mean by coupling coefficient between two coils is 0.6? Coupling coefficient is a measure of the strength of interaction between two coils. coupling coefficient between two coils is 0.6 means 60% of the flux set-up in one coil links the other coil. (Reference: Dictionary + Objective Electrical Technology By Mehta) A series RLC circuit has R=1000 Ω, L=100mH & C=10 µµf. Find: (a) Resonance frequency (b) Q-factor (c) 3dB point Bandwidth, The upper and lower -3dB frequency points, ƒ H and ƒ L Ans.

9 Similar: Bandwidth of a Series Resonance Circuit Series Resonance Example No1 A series resonance network consisting of a resistor of 30Ω, a capacitor of 2uF and an inductor of 20mH is connected across a sinusoidal supply voltage which has a constant output of 9 volts at all frequencies. Calculate, the resonant frequency, the current at resonance, the voltage across the inductor and capacitor at resonance, the quality factor and the bandwidth of the circuit. Also sketch the corresponding current waveform for all frequencies. Resonant Frequency, ƒ r Circuit Current at Resonance, I m

10 Inductive Reactance at Resonance, X L Voltages across the inductor and the capacitor, V L, V C ( Note: the supply voltage is only 9 volts, but at resonance the reactive voltages are 30 volts peak! ) Quality factor, Q Bandwidth, BW The upper and lower -3dB frequency points, ƒ H and ƒ L Current Waveform

11 Series Resonance Example No2 A series circuit consists of a resistance of 4Ω, an inductance of 500mH and a variable capacitance connected across a 100V, 50Hz supply. Calculate the capacitance require to give series resonance and the voltages generated across both the inductor and the capacitor. Resonant Frequency, ƒ r Voltages across the inductor and the capacitor, V L, V C Reference: 9 A 3-phase delta connected load has Z=20 60 and line voltage is 400 V. Find: (a) Total power (b) Power consumed in each phase. (a) P total = 3 V ϕ I ϕ cos θ =3*400*20* cos 60 = W=12KW Ans. (b) P ϕ = P total / 3 = 12/3 KW =4 KW Ans. Z ϕ = V ϕ = V L = 400 V. I ϕ = V ϕ / Z ϕ = 400 / = Write down the conditions of parallel operation of two alternators. There are five conditions that must be met before the synchronization process takes place. The alternator must have * Equal line voltage, * Frequency, *Phase sequence, *Phase angle and * Waveform to that of the system to which it is being synchronized.

12 [Waveform and phase sequence are fixed by the construction of the generator and its connections to the system, but voltage, frequency and phase angle must be controlled each time a generator is to be connected to a grid.] 11 Write down four main differences between CT and PT. Current Transformer A current transformer (CT) is a series connected measurement device designed to provide a current in its secondary coil proportional to the current flowing in its primary. The secondary of a C.T can not be open circuited on any circumstance when it is under service. A CT may be considered as a series transformer. The primary current in a C.T is independent of the secondary circuit conditions (burden). The primary winding of the CT is connected in series with the line carrying the current to be measured. Hence it carries of the full line current. Potential transformer Potential transformers (PT) are a parallel connected type of instrument transformer, used for metering and protection in high-voltage circuits or phasor phase shift isolation. The secondary of a P.T can be open circuited without any damage being caused either to the operator or the transformer. P.T may be considered as a parallel transformer. The primary current of a P.T depends upon the secondary circuit conditions (burden). The primary winding P.T is connected across the line of voltage to be measured. Hence the full line voltage is impressed across its terminal. Find the output of a full wave rectifier if the input is Draw the following signals: (b) x(t) = 2 u(t) + δ (t-1)

13 Find Vo Find I E, I B and I C. ( β was not given ) Solution: β was not given so, let β = 90. Taking loop, I E I B - 2.5= I E I B = (β+1)I B I B = 6.8 [ য ন ত I E = (β+1)ib ] 6800*91 I B I B = I B I B = I B = 6.8 I B = 6.8/ = µa Ans. I C = β I B = 90*9.460 = µa Ans. I E = I C + I B = = µa Ans.

14 Find I 1, I 2 and V Considering the short circuits, the circuit can be redrawn as So, I 1 =20/4K = 5mA Ans. V 2 is the voltage across open circuit So, V 2 =20 V Ans. And I 2 =0A Ans. Take 150 MVA, 15KV as base and draw the reactance diagram: 18

15 Similar: From Stevenson Junior s Book Answer:

16

17 17 A power station has following loading: Lighting load: 150 KW for 7.00pm to 10.00pm Residential load: 50KW for 7.00pm to 12.00pm Pumping Load: 75KW for 3.00pm to 7.00pm Find diversity factor. Type 3.00pm to 7.00pm 7.00pm to 10.00pm 10.00pm to 12.00pm Lighting load 150 KW Residential load 50KW 50KW Pumping Load 75KW Total 200KW Sum of Individual Maximum Demand Diversity factor = Maximum demand of the circuit at the time of system peak 150 KW + 50 KW + 75 KW = = Ans. 200KW

18 Rony Parvej s EEE Job Preparation Recruitment Test Question of Rural Power Company Limited (RPCL) Compiled by: Rony Parvej (IUT, EEE 07) Post: Assistant Engineer (Electrical) Time: 90 minutes Full Marks: 100 Exam Date: Venue: MIST 1. Calculate the values of i 2, i 3, V 2, V 3, V L, V C of the network shown in figure at the following times: (a)at time, t = 0+ immediately after the switch Sis closed. 10 (b) At time, t i.e. the steady state. 2. In a series circuit containing pure resistance and a pure inductance, the current and the voltage are expressed as i = 5 Sin (314t + 2π / 3 ) and v = 15 Sin ( 314t + 5π / 6 ). Find: 10 (a) What is the impedance of the circuit? (b) What is the value of the resistance? (c) What is the average power drawn by the circuit. (d) What is the power factor? 3. Draw the circuit diagram of a push-pull amplifier. Explain why it is called so? In the following circuit, I 1 = 2 A, I 2 = - 3 A, I 3 = 1.5 A, I 5 = 5 A. Find the value of I 0 and I In the following circuit find: 10 (a) Voltage across R L (b) Current in the secondary coil (c) Will the power remain same both in primary and secondary?

19 Rony Parvej s EEE Job Preparation 6. (a) How alternators are classified? 3x5=15 (b) Describe advantages and disadvantages of 3 lamp synchronization method of Alternator. (c) What is the function of protective relays? 7. (a) Which device is used to convert direct current to alternating current? 5x1=5 (b) A multimeter is connected across a dc device. It is showing 0V. What can be the best explanation behind this? (c) What is the effect of increasing frequency on Transformer rating? (d) Why Alternator stator is laminated? (e) Two resistances of different values are connected in parallel. Will the current and voltage across them be same? 8. Correct the following sentences according to instructions: 8x1=8 (a) Light alternates darkness. (Preposition) (b) He went a visitor college. (Preposition) (c) I am much happy to meet you. (Correction) (d) He ate very much and become ill. (Correction) (e) Everybody hates a liar. (Negative) (f) He is the best boy in the class. (Negative) (g) We must yield to our fate. (Negative) (h) 9. Translate into English: 7 যৎক র ২৬ অগস ট, ১৯১০ মভসডড সনয় য স কসডয় ন ভক এক ম ট ট ডযয এক ধন কয থসরক ফযফ য় ফ ডয অরডফন য় সযফ ডয এক ম ট সশ জন মগ রন কডযন ত য ন ভ একডন মগ ন ক স মফ জ সক স সউ সনডক র ও মফ জ সক সউ দম পসতয সতন ন ত ডনয ভডধয সতসন স ডরন ফ য ম ট মক জ নডত এআ ফ সরক সটআ বসফষ যডত অতত ভ নফত য মফ য় অত মসনডয় গ কযডফন, ভ নফত য মফ য় ফযয় কযডফন সনডজয য ট জ ফন এফ ভ নফত য জনন ডয় উঠডফন সতসন অয মকউ নন অভ ডদয ফ য স য় ভ দ য মতডয ১০ ধ যণ জ ঞ ন ১৫x১=১৫ ১ ফ র ডদ-ব যত ভ ন ত চ স ডত মক মক স ব ক ষয কডযন? ২ ডফত ডভ ট ম ত ফ প ত ভ স ডম দ ধ য য কত? ৩ ভসর ফ য সতডকয য কত? ত ডদয ন ভ সক? ৪ ভ স ম ডদ ধয ভয় মক ন মজর কডফ থভ ত র ভ য়? ৫ ফ র ডদডয থতননসতক ভ দ র ভ কত? ৬ ফ র ডদড মভ ফ আর ফয সক কডফ চ র য়? ৭ শ য ভডদ মক ন মদডক ফর য়? ৮ ফ র ডদডয জ ত য় ত ক ক য ক য ফযফ য কযডত ডযন? ৯ এসয় ক প টফর কয় ফ য য য ন স ত য়? ১০ জ সত ঘ কডফ ফ র ডক অন তজত সতক ভ ত ব ল সদফ সডডফ মঘ লন কডয? ১১ A Long Walk to Freedom ফআসট ক য মর? ১২ ভ স ম ডদ ধ ফ র ডদডয একভ ত র সনয়সভত মক টয মন -মক টয কয় নম বয মক টডযয ধ ডন স র? ১৩ মক ন ভ ঘর ম র ট সজন দ য ন ডভ সযসচত স ডরন? ১৪ সফড য ফডচডয় দ র ততভ ণ মক নসট? ১৫

20 Rony Parvej s EEE Job Preparation 1. Calculate the values of i 2, i 3, V 2, V 3, V L, V C of the network shown in figure at the following times: (a)at time, t = 0+ immediately after the switch Sis closed. 10 (b) At time, t i.e. the steady state. 2. In a series circuit containing pure resistance and a pure inductance, the current and the voltage are expressed as i = 5 Sin (314t + 2π / 3 ) and v = 15 Sin ( 314t + 5π / 6 ). Find: 10 (a) What is the impedance of the circuit? (b) What is the value of the resistance? (c) What is the average power drawn by the circuit. (d) What is the power factor? 3. Draw the circuit diagram of a push-pull amplifier. Explain why it is called so? 10 A Push-Pull Amplifier combines two signals to form a third signal. The amplifiers operate in phase opposition, with one transistor amplifying the positive cycle while the other amplifies the negative transition. The circuit is called a Push-Pull because one transistor pushes in one direction while the other pulls in another direction.

21 Rony Parvej s EEE Job Preparation 4. In the following circuit, I 1 = 2 A, I 2 = - 3 A, I 3 = 1.5 A, I 5 = 5 A. Find the value of I 0 and I At node A, I o + I 1 + I 2 = 0 or, I o = 0 so, I o = 1 A (Ans.) At node B, x= I o + I 5 Or, x = 1+5 = 6 A At node C, Or, So, x + I 1 = I 3 + y = y y = 6.5 A At node D, I 4 = y + I 2 Or, I 4 = = 3.5 A (Ans.) 5. In the following circuit find: 10 (a) Voltage across R L (b) Current in the secondary coil (c) Will the power remain same both in primary and secondary? (a) V S = V P * N S /N P = 120*1/10= 12 V (Ans.) (b) I S = I P * N P /N S = 1*10/1 = 10 A (Ans.) (c) Primary Power = V P * I P = 120*1=120 VA Secondary Power = V S * I S = 12*10 =120VA So, the power will remain same both in primary and secondary. 6. (a) How alternators are classified? 3x5=15 Alternators can be classified by method of excitation, number of phases, the type of rotation,and their application. According to rotating part a. Revolving armature type b. Revolving field type According to its rotor construction - a. Smooth cylindrical type b. Salient pole alternator According to number of phases - a. Single-phase b. Polyphase By excitation a. Direct connected DC generator b. Transformation and rectification c. Brushless alternators According to application these machines are classified as 1. Automotive type - used in modern automobile. 2. Diesel electric locomotive type - used in diesel electric multiple unit. 3. Marine type - used in marine. 4. Brush less type - used in electrical power generation plant as main source of power. 5. Radio alternators - used for low brand radio frequency transmission.

22 Rony Parvej s EEE Job Preparation (b) Describe advantages and disadvantages of 3 dark lamp synchronization method of Alternator. Advantages: The synchronous switch using lamps is inexpensive Checking for correctness of the phase sequence can be obtained in a simple manner which is essential especially when the Alternator is connected for the first time or for fresh operation after disconnection Disadvantages: The rate of flickering of the lamps only indicates the frequency difference between the bus-bar and the incoming Alternator. The frequency of the incoming Alternator in relation to the bus-bar frequency is not available. (c) What is the function of protective relays? The function of protective relays is to trip a circuit breaker when a fault is detected. 7. (a) Which device is used to convert direct current to alternating current? Answer:Inverter (b) A multimeter is connected across a dc device. It is showing 0V. What can be the best explanation behind this? Answer: The desired mode or range is not selected appropriately. (c) What is the effect of increasing frequency on Transformer rating? Answer: KVA rating will increase. Explaination: The KVA rating of a transofrmer is limited by 1) the maximum voltage you can apply and 2) the maximum current you can allow through the windings at the end of the day: how hot can we let it get? The peak magnetic flux in the core is proportional to the volts per turn and inversely to the frequency. If we increase the frequency the peak flux in the core falls, so the magnetising current decreases, so we have less loss due to heating of the windings i.e. the copper losses. But also the iron losses increase (eddy currents and hysteresis loss). If we decrease the frequency, the peak flux increases, and the magnetising current increases. We have less losses from iron loss, but we increase copper losses. There is also a risk that decreasing the frequency too much could lead to saturation, the magnetising inductance falls very rapidly and the mag. current increases very rapidly (this can also happen if you apply too high a voltage) This can lead to very high losses and is usually the limiting factor: there is a trade off between the amount of copper (number of turns) and the maximum copper loss you want to allow. So if we decrease the frequency, but dont want the copper losses to increase, we would need to reduce the voltage. Like wise if we increase the frequency, we can afford to increase the voltage with out exceeding our maximum allowable magnetising current (remember the current in the windings will be the sum of the magnetising current and the load current). But if we increase the frequency too much, our iron losses will go up, so we cant afford as much in the way of copper loss without too much heating, so we would then need to reduce the load current (effectively the KVA rating) A small decrease in frequency from nominal will probably reduce the KVA rating more than a small increase in frequency, especially since the applied voltage would probably be fixed.

23 Rony Parvej s EEE Job Preparation (d) Why Alternator stator is laminated? Answer: To reduce eddy current loss. Explaination: In electrical engineering, lamination is a construction technique used to reduce unwanted heating effects due to eddy currents in components. Critical to achieving high efficiency is selecting a core material that has low core loss and high permeability. The term "core loss" relates to the total energy lost through the generation of heat. Heat, a form of energy loss, is produced by eddy currents in the core material and by a behavior called magnetic hysteresis. Eddy currents are small stray electrical currents that are generated within the core material by the magnetic field. These can be minimized, but not avoided in total. Since there is a current flowing in the core material (steel), heat is generated. This source of electrical energy loss is called "eddy current loss". The resistance of the core is increased in two ways: 1. constructing the core from a number of thin sections (light-gauge sheets) or laminations, and 2. alloying the steel with elements such as manganese, silicon, and aluminum; elements that increase the electrical resistance of steel. (e) Two resistances of different values are connected in parallel. Will the current and voltage across them be same? Answer: Voltage will remain same but the current will be different. ১০ ধ যণ জ ঞ ন ১৫x১=১৫ ১ ফ র ডদ-ব যত ভ ন ত চ স ডত মক মক স ব ক ষয কডযন? ( শ খ ম জজব - ইজ র গ ন ধ ) ২ ডফত ডভ ট ম ত ফ প ত ভ স ডম দ ধ য য কত? (৬৭৬ জন) ৩ ভসর ফ য সতডকয য কত? ত ডদয ন ভ সক? (২ জন কয ড ন ড. মত য মফগভ ও ত য ভন সফসফ) ৪ ভ স ম ডদ ধয ভয় মক ন মজর কডফ থভ ত র ভ য়? (মড য, ৬ সডডম বয, ১৯৭১) ৫ ফ র ডদডয থতননসতক ভ দ র ভ কত? (২০০ নসটকয র ভ আর) ৬ ফ র ডদড মভ ফ আর ফয সক কডফ চ র য়? (- ড চ ব ল ব ক জল. ৩১ ম চচ ২০১১ স লল) ৭ শ য ভডদ মক ন মদডক ফর য়? (থ আরয ন ড) ৮ ফ র ডদডয জ ত য় ত ক ক য ক য ফযফ য কযডত ডযন? (র ষ ট রজ ও প রধ নমন ত র ) ৯ এসয় ক প টফর কয় ফ য য য ন স ত য়? (দ ই বছর র র) ১০ জ সত ঘ কডফ ফ র ডক অন তজত সতক ভ ত ব ল সদফ সডডফ মঘ লন কডয? (১৯৯৯ স ললর ১৭ নল ম বর অন জ ইউলনলক র জরস অজধলব লন) ১১ A Long Walk to Freedom ফআসট ক য মর? (মনরন ভয ডন ডর ) ১২ ভ স ম ডদ ধ ফ র ডদডয একভ ত র সনয়সভত মক টয মন -মক টয কয় নম বয মক টডযয ধ ডন স র? (১০ ন শসক টর) ১৩ মক ন ভ ঘর ম র ট সজন দ য ন ডভ সযসচত স ডরন? (সম র ট আওরঙ গলজব) ১৪ সফড য ফডচডয় দ র ততভ ণ মক নসট? (জচ ব ঘ) ১৫

24 Rony Parvej s EEE Job Preparation Lecture-6: mist special Recruitment Test Question of Dhaka Electric Supply Company Limited (DESCO) Collected from: Ridwanur Rahman Ayon (MIST 07) Compiled by: Rony Parvej (IUT, EEE 07) Post: Assistant Engineer (Electrical) Time: unknown Full Marks: unknown Exam Date: 2013 (probable) Venue: MIST 1. Answer following questions: ১ স তজন ব রশ র শ র র ন ম ল খ ন ২ ২ জন মল ব রপ রত শ র র ন ম ল খ ন ৩ সশস ত র ব ল ন লদবস শ রব প ল ত য়? ৪ ফর শ রয়জ আশ রদ ন র ননত শ র চশ র? এর ন ম ফর শ রয়জ আশ রদ ন ন ন? ৫ লতত ম শ ররর প র ত ন ম ল? ৬ প রথম এভ শ ররস ট লবজয় ন র ন? ৭ নমগ লসল ল? ৮ লতশ র ত তম সম ভব র ন খ? ৯ ম ন র অপল ম এর উদ ভ ব ন? ১০ SPARSO ল? এর জ ল? 2. Elaborate following words: MNLP, ETP, WFP, WTO, WFO, MDG, FAO, ITU, EVM, IFAD, UNV, PABX, SSL, SOMS, OIC, FRCS, CIDA. 3. Two batteries of different emfs are connected in series with each other and with external load resistance. The current is 3A. When the polarity of one of the batteries is reversed the current is 1A. Find the ratio of emf s of the two batteries. 4. Why is power system never operated under maximum power conditions? Answer: Under the conditions of maximum power transfer, the efficiency is low (50%) and there is greater voltage drop in the lines. In a power system, the goal is higher efficiency rather than maximum power. For these reasons, maximum power transfer is not desired in a power system. 5. What are the advantages of Thevenin s Theorem? 6. Which one will produce more heat? (Answer: Square)

25 Rony Parvej s EEE Job Preparation RMS Value Lecture-6: mist special Sinusoidal Voltage: Square Wave Voltage: Triangular Wave Voltage: 7. What is an ideal transformer? 8. What do you mean by coupling co-efficient between two coils is 0.6? 9. Why do we use a multiplier with a voltmeter? 10. What are the factors on which the economic choice of transmission voltage depends on? 11. How is the efficiency of an alternator effected by load power factor? Answer: The efficiency of an alternator depends not only on kva output but also on p.f. of the load. For a given load, efficiency is maximum at unity p.f. and decreases as the p.f. falls. 12. Why higher distribution factor is needed in power plant? 13. If Q 1 Q 2 > 0, What can we say about the nature of force? Answer: If Q 1 Q 2 > 0, it means the product of the magnitude of two charges is positive (because Q 1 Q 2 is greater than zero). This implies that charges are similar. i.e., either both positive or both negative. Hence, repulsion will be the result. (Reference: ch-5, Objective electrical Technology V.K. Mehta) 14. High diversity factor should be the desirable feature of a power system. Why? Answer: Greater diversity factor means lesser maximum demand. This in turn means that lesser plant capacity is required. Thus, the capital investment on the plant is reduced.

26 Rony Parvej s EEE Job Preparation Lecture-6: mist special Recruitment Test Question of Dhaka Electric Supply Company Limited (DESCO) Compiled by: Rony Parvej (IUT, EEE 07) Post: Assistant Engineer (Electrical) Time: unknown Full Marks: unknown Exam Date: 2013 (probable) Venue: MIST [ম প রশ ন ই ররজ র ছ একজন র ক ষ ছ র এর র য ট ক মরন ছ র ছনরজর ভ আম রক জ ছনর র সভ র ই প রশ নগ র এখ রন র ছ ম ] 1. Write a short note on Padma Bridge Write a short note on IMO Write a short note on Global warming Two cells with same e.m.f. E and different internal resistances r 1 and r 2 are connected in series to an external resistance R as shown in figure. Find the value of R such that the potential difference across the terminals of first cell is zero What is unsymmetrical fault? Why it occurs? Which type of unsymmetrical fault is most severe? What is the importance of time constant of a circuit? 5 7. Write a short note on PCM What is shear stress? 3 9. What is the value of a in the following triangle? 10. Elaborate: (i) ANSI, (ii) JFET, (iii) CMOS, (iv) MOSFET, (v) Translation MCQ (5 questions) 5 x 2 = 10 (i) What will happen if heat is applied again in a turbine (FF Quota) ২০১৩ র শ ধ ম ত র ম ছ রয দ ধ সক ট DESCO এর ছরক র টরমন ট এক স ম র ছ র অ লগ রণক র একজরনর র কর ক রন ও ছক য র ছ ছছ ম কক ছ ছ ছ উল ন ম ভ নন-ছ টকরমন ট র ৩০ এ ছ টকরমন ট র ৭০ এরকম ছ নন-ছ টকরমন ট র Abbreviation (e.g. VOIP, IMF etc) ও অন য ন য ধ রণ জ ঞ রনর প রশ ন ছ ছ টকরমন ট রর প রশ ন স ল জ ছ ছক ছ লটক ক র শ চ ন আর সয প রশ ন মরন আর সগ র র 1. why maximum power theorem can not be used in transmission line? 2. Induction motor-a slip theke 1ta problem. 3. very simple 1ta ckt problem. 4. high voltage transmission er jonno required condition in system.