Hyderabad Delhi Bhopal Pune Bhubaneswar Bengaluru Lucknow Patna Chennai Vijayawada Visakhapatnam Tirupati Kukatpally Kolkata ESE - 19

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1 Website : www. aceengineeringpublications.com ACE Engineering Publications (A Sister Concern of ACE Engineering Academy, Hyderabad) Hyderabad Delhi Bhopal Pune Bhubaneswar Bengaluru Lucknow Patna Chennai Vijayawada Visakhapatnam Tirupati Kukatpally Kolkata ESE - 19 (P ) Electrical Engineering (Volume - II) (Analog and Digital Electronics, Systems and Signal Processing, Control Sytems, Electrical Machines, Power Systems and Power Electronics and Drives ) Previous years Objective Questions with Solutions, Subjectwise & Chapterwise ( ) ACE is the leading institute for coaching in ESE, GATE & PSUs H O: Sree Sindhi Guru Sangat Sabha Association, # /1/A, King Koti, Abids, Hyderabad Ph: / 19 / 20 / 21, All India 1 st Ranks in ESE 43 All India 1 st Ranks in GATE

2 Copyright 2018 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, digital, recording or otherwise, without the prior permission of the publishers. Published at : Sree Sindhi Guru Sangat Sabha Association, # /1/A, King Koti, Abids, Hyderabad , Telangana, India. Phones : / 20 / info@aceenggpublications.com hyderabad@aceenggacademy.com Authors : Subject experts of ACE Engineering Academy, Hyderabad While every effort has been made to avoid any mistake or omission, the publishers do not owe any responsibility for any damage or loss to any person on account of error or omission in this publication. The publishers will be obliged if mistakes are brought to their notice through , for correction in further editions. info@aceenggpublications.com First Edition : 2011 Revised Edition : 2018 Printed at : Rowshni Graphics, Hyderabad. Price :. 950/- ISBN :

3 Foreword UPSC Engineering Services in Electrical Engineering Volume - II Objective Questions: From The Stage-I (Prelims) of ESE 2018 consists of two objective papers. Paper-I is of General Studies & Engineering Aptitude. Paper-II is of Electrical Engineering of 300 Marks and 3 hours duration. In stage-ii (Mains), the technical syllabus is divided into two papers. The subjects included in this volume are: 1. Analog and Digital Electronics 2. Systems and Signal Processing 3. Control Systems 4. Electrical Machines 5. Power Systems 6. Power Electronics and Drives Keeping in view of the above topics, the present Volume-II for prelims of new pattern is redesigned using the previous questions from 1992 onwards. The style, quality and content of the Solutions for previous ESE Questions of Electrical Engineering, will encourage the reader, especially the student whether above average, average or below average to learn the concept and answer the question in the subject without any tension. However it is the reader who should confirm this and any comments and suggestions would be pleasantly received by the Academy. It is observed that majority of ESE objective Questions are being asked as it is in many PSUs, state service commission, state electricity boards and even in GATE exam. Hence we strongly recommend all students who are competing for various competitive exams to use this book according to the syllabus of the exam concerned. This book can also be used by fresh Teachers in engineering colleges to improve their Concepts. We proudly say that questions and solutions right from 1992 onwards are given in this book. The questions which appeared early 90 s are most conceptual oriented and these are being repeated in the recent exams in a different way. Hence we advise the students to practice these questions compulsorily. The student is also advised to analyze why only a particular option is correct and why not others. Evaluate yourself, in which case, these other options are correct. With this approach you yourself can develop four questions out of one question. The student is advised to solve the problems without referring to the solutions. The student has to analyze the given question carefully, identify the concept on which the question is framed, recall the relevant equations, find out the desired answer, verify the answer with the final key such as (a), (b), (c), (d), then go through the hints to clarify his answer. The student is advised to have a standard text book ready for reference to strengthen the related concepts, if necessary. The student is advised not to write the solution steps in the space around the question. By doing so, he loses an opportunity of effective revision. With best wishes to all those who wish to go through the following pages. Y.V. Gopala Krishna Murthy, M Tech. MIE, Chairman & Managing Director, ACE Engineering Academy,.

4 Syllabus for Electrical Engineering (EE) 1. Analog and Digital Electronics: Operational ampli iers characteristics and applications, combinational and sequential logic circuits, multiplexers, multi-vibrators, sample and hold circuits, A/D and D/A converters, basics of ilter circuits and applications, simple active ilters; Microprocessor basics- interfaces and applications, basics of linear integrated circuits; Analog communication basics, Modulation and demodulation, noise and bandwidth, transmitters and receivers, signal to noise ratio, digital communication basics, sampling, quantizing, coding, frequency and time domain multiplexing, power line carrier communication systems. 2. Systems and Signal Processing: Representation of continuous and discrete-time signals, shifting and scaling operations, linear, time-invariant and causal systems, Fourier series representation of continuous periodic signals, sampling theorem, Fourier and Laplace transforms, Z transforms, Discrete Fourier transform, FFT, linear convolution, discrete cosine transform, FIR ilter, IIR ilter, bilinear transformation. 3. Control Systems: Principles of feedback, transfer function, block diagrams and signal low graphs, steady-state errors, transforms and their applications; Routh-hurwitz criterion, Nyquist techniques, Bode plots, root loci, lag, lead and lead-lag compensation, stability analysis, transient and frequency response analysis, state space model, state transition matrix, controllability and observability, linear state variable feedback, PID and industrial controllers. 4. Electrical Machines: Single phase transformers, three phase transformers - connections, parallel operation, auto-transformer, energy conversion principles, DC machines - types, windings, generator characteristics, armature reaction and commutation, starting and speed control of motors, Induction motors - principles, types, performance characteristics, starting and speed control, Synchronous machines - performance, regulation, parallel operation of generators, motor starting, characteristics and applications, servo and stepper motors. 5. Power Systems: Basic power generation concepts, steam, gas and water turbines, transmission line models and performance, cable performance, insulation, corona and radio interference, power factor correction, symmetrical components, fault analysis, principles of protection systems, basics of solid state relays and digital protection; Circuit breakers, Radial and ringmain distribution systems, Matrix representation of power systems, load low analysis, voltage control and economic operation, System stability concepts, Swing curves and equal area criterion. HVDC transmission and FACTS concepts, Concepts of power system dynamics, distributed generation, solar and wind power, smart grid concepts, environmental implications, fundamentals of power economics. 6. Power Electronics and Drives: Semiconductor power diodes, transistors, thyristors, triacs, GTOs, MOSFETs and IGBTs - static characteristics and principles of operation, triggering circuits, phase control recti iers, bridge converters - fully controlled and half controlled, principles of choppers and inverters, basis concepts of adjustable speed dc and ac drives, DC-DC switched mode converters, DC-AC switched mode converters, resonant converters, high frequency inductors and transformers, power supplies.

5 CONTENTS I. Analog and Digital Electronics 01.Operational Amplifiers Number system and Coding Conversion Boolean Algebra Logic Gates Combinational Circuits Sequential Circuits Logic Gate Families AD and DA Conversion Semiconductor Memories Architecture of Pin Details and Interrupt of Interfacing with Instruction set of Programing with Microprocessor Amplitude Modulation Angle Modulation Digital Communication Channel Capacity and Information Theory Receivers Noise

6 22. Radar Communication Sattellite and Television Communication Miscellaneous II. Systems and Signal Processing 01. Classification & Modeling Fourier Series & Application Fourier Transforms & Application Laplace Transforms & Application Z-Transforms & Application III Control Systems 01. Basics of Control Systems Block Diagram and Signal Flow Graph Time Response Analysis Stability Root Locus Diagram Frequency Response Analysis Controllers and Compensators State Space Analysis Mathematical Modeling of Physical Systems Descrete Data Control Systems Non-linear Control Systems

7 IV Electrical Machines 01. Transformers DC Machines Synchronous Machines Induction Machines V Power Systems 01. Generating Stations Performance of Transmission Lines and Cables Insulators Load Frequency Control Fundamentals of Power Economics Load Flow Studies Faults Stability Circuit Breakers Protection HVDC Transmission VI Power Electronics and Drives 01. Basics and Power Semiconductor Devices AC-DC Converters DC-DC Converters DC-AC Converters AC-AC Converters Fundamentals of Drives

8 Electrical Engineering Previous years ESE Objective questions (Prelims) Weightage (2014 to 2018) S.No. Name of the Subject Engineering Mathematics Electrical Materials Electric Circuits & Fields Electrical & Electronic Measurements Computer Fundamentals Basic Electronics Engineering Analog & Digital Electronics Systems & Signal Processing Control Systems Electrical Machines Power Systems Power Electronics & Drives Total No.of Questions:

9 Chapter 1 Transformers 01. The desirable properties of transformer core material are (IES-92) (a) low permeability and low hysteresis loss (b) high permeability and high hysteresis loss (c) high permeability and low hysteresis loss (d) low permeability and high hysteresis loss 02. The efficiency of two identical transformers under load conditions can be determined by (IES-92) (a) back to back test (b) open circuit test (c) short circuit test (d) any of the above 03. For an ideal transformer the windings should have (IES-92) (a) maximum resistance on primary side and least resistance on secondary side (b) least resistance on primary side and maximum resistance on secondary side. (c) equal resistance on primary and secondary side. (d) no ohmic resistance on either side 04. Two single-phase 100 kva transformers, each having different leakage impedances are connected in parallel. When a load of 150 kva at 0.8 power factor lagging is applied (IES-92/13) (a) both transformers will operate at power factor more than 0.8 lagging (b) both transformers will operate at power factor less than 0.8 lagging (c) one of the transformers will operate at power factor more than 0.8 lagging and other will operate at power factor less than 0.8 lagging (d) both transformers will operate at identical power factors 05. Scott connections are used for (IES-92) (a) single phase to three phase transformation (b) three phase to single phase transformation (c) three phase to two phase transformation (d) any of the above. 06. Two single phase transformers with equal turns have impedance of (0.5 + j 0.3) ohms and (0.6 + j 1) ohms with respect to the secondary, If they operate in parallel, how will they share a load of 100 kw at 0.8 p.f. lagging? (IES-92) (a) 50 kw, 50 kw (b) 62 kw, 38 kw (c) 78.2 kw, 21.8 kw (d) 85.5 kw, 14.5 kw

10 ACE Engineering Publications Transformers The losses on a transformer are I. Copper losses II. Eddy current losses III Hysteresis losses The constant power loss of a transformer loss is given by (IES-92) (a) I only (b) I and II only (c) II and III only (d) I, II and III. 08. Which of the following will improve the mutual coupling between primary and secondary circuits? (IES-92) (a) Transformer oil of high break down voltage. (b) High reluctance magnetic core (c) Winding material of high resistivity (d) Low reluctance magnetic core 09. The secondary of a current transformer under operating conditions is short-circuited to avoid (IES-92) (a) break in primary winding (b) insulation break-down (c) core saturation and high voltage induction (d) high voltage surge. 10. The inductive reactance of a transformer depends on (IES-92) (a) electromotive force (b) magneto motive force (c) magnetic flux (d) leakage flux 11. Which of the following connection of transformer will give the highest secondary voltage? (IES-92) (a) delta primary, delta secondary (b) delta primary, star secondary (c) star primary, star secondary (d) star primary, delta secondary 12. In a transformer, if the iron losses and copper losses are 40.5 kw and 50 kw respectively, then at what fraction of load will the efficiency be maximum? (IES-93/14) (a) 0.8 (b) 0.57 (c) 0.70 (d) Can a 50 Hz transformer be used as 25 Hz, if the input voltage is maintained constant at the rated value corresponding to 50 Hz? (IES-93) (a) Yes, since the voltage is constant,current levels will not change (b) No, flux will be doubled which will drive the core to excessive saturation (c) No, owing to decreased reactance of transformer, input current will be doubled at load (d) Yes, at constant voltage, insulation will not be overstressed. 14. Short-circuit test is performed on a transformer with a certain impressed voltage at rated frequency. If the short-circuited test is now performed with the same magnitude impressed voltage, but at a frequency higher than the rated frequency, then (IES-93)

11 586 Electrical Machines ACE Engineering Publications 141.a 142.b 143.b 144.c 145. a 146.b 147.c 148.d 149.a 150.a 151.d 152.b 153.c 154.b 155. a 156.c 157. b 158.d 159.c 160.c 161.a 162.c 163.d 164.c 165.a 166.a 167. a a 170. b 171. a 172. d 173. c 174. a 175. c 176. b 177. b 178. a 179. b 180. c 181. a 03. Ans: (d) Sol: In ideal transformer, resistance of windings and magnetic leakage flux are zero. 04. Ans: (c) Sol: The X R ratios of transformers to be connected in parallel should be equal to avoid operation of transformers at different power factors. Solutions 01. Ans: (c) Sol: For a transformer, permeability ( ) of Transformer core should be high i.e if permeability increases, then magnetizing component of current required to create rated flux in the core decreases. Hysteresis loss in the core should be low to get more efficiency. If, X A X B, R A R B A > B cos A < cos B A B L I B I A I L V 2 I A I B 02. Ans: (a) Sol: During open circuit and short circuit tests all the losses do not occur simultaneously, therefore the exact temperature rise can not be determined. In order to get exact temperature rise Sumpners test/ back to back test should be conducted. Iron loss and copper loss are obtained from wattmeter readings, with these results we can find efficiency of transformers. Where, I A = current shared by transformer-a I B = current shared by transformer-b Therefore, load current I I I 05. Ans: (c) Sol: Scott connection is one of the special connection of the Transformer, which is used to convert 3-phase to 2-phase and vice versa. L A B

12 ACE Engineering Publications Transformers Ans: (c) Sol: Z 1 = (0.5 + j0.3) = Z 2 = (0.6+j1.0) = Given sharing of 100 kw load at 0.8 power factor lagging 100 i.e. 125 kva at = 36.6 O 0.8 For Transformer-I Sharing = S L Z2 Z Z = = kva at O lag i.e kw at lag For Transformer-II Sharing = S L 1 2 Z2 Z Z = = kva at 55.7 o lag = kw at lag 07. Ans: (c) Sol: Copper loss I 2 i.e depends on load current called variable losses. Iron loss (W h + W e ) V 2 (applied voltage), called constant losses. 08. Ans: (d) Sol: Permeability 1 reluc tan ce If the permeability of the core is increases, the mutual coupling between primary and secondary will be improved. 09. Ans: (c) Sol: A current transformer is connected in a circuit as shown. I CT Primary of the CT + ac supply A Load Secondary of the CT Primary of the CT has few turns. It can simply consist of a metal bar of negligible resistance. Secondary of the CT is virtually short circuited by a low resistance ammeter coil. Unlike in conventional transformers, the primary current is not decided by the secondary current. It is decided by the load current. If the secondary is open, there is no balancing mmf, the core flux will be very high and a high emf is induced in the secondary. This will cause severe insulation stress in the CT and could be a risk to operating persons. 10. Ans: (d) Sol: As leakage flux is more, coefficient of coupling of transformer will decrease and also the inductive reactance drop will be increased.

13 588 Electrical Machines ACE Engineering Publications 11. Ans: (b) Sol: Connections Line Voltages v If 1 is 100% x Voltage on secondary side - V 1 : v 1 100% x 14. Ans: (d) Sol: S.C test is conducted at reduced voltage (V sc ) and at rated frequency. If V sc constant and f (f > frated) is increased, the consequences are I R 01 SC X 01 V SC 1 V2 0 Y Y V 1 : v 1 100% x - Y V 1 : 3 x v % Y - V 1 : V % 3x For same rating and applied voltage, /Y connection gives 73.3% of more voltage compared to / (or) Y/Y connection. 12. Ans: (d) Sol: Efficiency is maximum at any x of full W i 40.5 load = = 0.9 Wcu 50 Maximum Efficiency occurs at 0.9 of Full Load. 13. Ans: (b) V Sol: B max f Here V constant, f decreased to half B max increased to double, which will drive the core in to deep saturation and also I is very high to create double the rated flux. (i) R 01 = constant (ii) X 01 f (iii) Equivalent circuit under S.C test z x R z Vsc cons tan t (iv) I SC = z R 01 cons tan t (v) cos sc = z Ans: (b) Sol: Open circuit test is convenient to conduct on LV side by opening H.V winding due to the following reasons: 1. If the test is conducted on LV side, LV source sufficient to conduct the test to maintain rated flux. 2. If the test is conducted on LV side, low range meters are sufficient to conduct the test. 3. As magnitude of no-load current is more on LV side, this high no-load current can be accurately measured on LV side when compared to HV side. 01

14 Chapter 2 Performance of Transmission Line R = Z Lines & Cables 01. As the operating voltage and consequently the electric stress on the dielectric of solid type cable is increased from a low value, the dielectric power factor cos remains almost unchanged up to a certain value of the stress beyond which cos increases very rapidly. This is due to increase in the (IES-93) (a) resistivity of the dielectric material (b) ionization in the voids present in the dielectric (c) core-to-core capacitance of the cable. (d) core-to-earth capacitance of the cable 02. The L/C ratio for 132 kv and 400 kv lines are typically and respectively. The natural 3-phase loading for the two lines will be respectively (IES-93) (a) MW and 2560 MW (b) 44 MW and 2560 MW (c) 44 MW and 640 MW (d) 640 MW and 44 MW 04. Different terminal conditions at the receiving end of a loss less overhead transmission line having surge impedance Z are sketched in List-I. The incident and the reflected current waves are sketched in List-II. Match List-I with List-II and select the correct answer using the codes given below the Lists: (IES-93) List I (Line with different terminal conditions) Line A. Short Circuit B. C. List - II (Incident and reflected current waves) I I Line R = Z No reflection I Line Open Circuit 03. A quarter wave line will behave as (a) a rectifier (b) an amplifier (c) a transformer (d) a variable capacitance (IES-93) 3. I I Codes: A B C A B C (a) (b) (c) (d) 1 3 2

15 ACE Engineering Publications Performance of Transmission Lines & Cables Assertion (A): The electric field intensity E is proportional to X, the distance from the axis of the conductor of radius, r, for x r. Reason (R): Bundled conductors increase the line capacitance from that of a single conductor of the same cross-sectional area. (IES-93) (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT a correct explanation of A (c) A is true but R is false (d) A is false but R is true 06. Assertion (A): In a high voltage cable, concentric conducting surfaces are provided in an insulating medium of same permittivity throughout the cross-section of the cable to reduce the ratio of maximum to minimum electric stress. Reason (R): The potential differences between the conducting surfaces redistribute the charge distribution (IES-93) 07. The inductance of single-phase two-wire power transmission line per kilometer gets doubled when the (IES-93) (a) distance between the wires is doubled (b) distance between the wires is increased four fold. (c) distance between the wires is increased as square of original distance (d) radius of the wire is doubled 08. Corona loss can be reduced by the use of hollow conductors, because (IES-94) (a) the current density is reduced (b) the eddy current in the conductor is eliminated (c) for a given cross section, the radius of the conductor is increased (d) of better ventilation in the conductor. 09. The insulation of modern EHV lines is designed based on (IES-94) (a) the lightning (b) corona (c) radio interference (d) switching voltage 10. A 3-phase 50 Hz transmission line has the following constants (line to neutral) Resistance = 10, Inductive reactance = 20, Capacitive susceptance = The inductance of the arc suppressor coil to be used in the system is (IES-94) (a) H (b) H (c) H (d) H 11. Consider the following statements: 1. The insulation resistance of cable will increase if the length of the cable is increased 2. For the same overall diameter of cable, the grading of cable will increase the safe working voltage.

16 808 Power Systems ACE Engineering Publications 3. The normal operating temperature of PVC cable is 70 o C 4. The thermal resistance of coil increases as the moisture content of coil increases. (IES-95) Of these statements (a) 1 and 2 are correct (b) 2 and 3 are correct (c) 3 and 4 are correct (d) 1 and 4 are correct 12. Compared with a solid conductor of the same radius, corona appears on a stranded conductor at a lower voltage, because stranding (IES-95) (a) assists ionization (b) makes the current flow spirally about the axis of the conductor (c) produces oblique sections to a plane perpendicular to the axis of the conductor (d) produces surfaces of smaller radius 13. Transmission lines are transposed to (IES-96) (a) reduce copper loss (b) reduce skin effect (c) prevent interference with neighboring telephone line (d) prevent short- circuit between any two lines 14. A single-phase transmission line of impedance j0.8 ohm supplies a resistive load of 500A at 300V. The sending-end power factor is (IES-96) (a) unity (b) 0.8 lagging (c) 0.8 leading (d) 0.6 lagging 15. On an infinitely long transmission line, the attenuation co-efficient is expressed in terms of the ratio K of the two voltages one kilometer apart on the line, in bels/km. The value of this co-efficient is given by (a) log e K (b) (c) 2 K (IES-96) 2log10 K (d) None of the above 16. For a transmission line with resistance R, reactance X, and negligible capacitance, the generation constant A is (IES-96) (a) 0 (b) 1 (c) R + jx (d) R + X 17. A surge voltage of 1000 kv is applied to an overhead line with its receiving end open. If the surge impedance of the line is 500, then the total surge power in the line is (a) 2000 MW (c) 2 MW (IES-96) (b) 500 MW (d) 0.5 MW

17 ACE Engineering Publications Performance of Transmission Lines & Cables 825 Solutions 01. Ans: (b) Sol: As the operating voltage increases, the stress will increase so that the di-electric strength at insulation will fail due to the presence of clouds in the di-electric medium. 02. Ans: (c) Sol: Surge Impedance or natural loading V Z 2 C, where Z For 132 kv line, C L C SIL = MW For 400 kv line, SIL = 640 MW Ans: (c) Sol: Quarter wave line will be have as a transformer for impedance matching. It can transform low impedance into high impedance and vice versa. 04. Ans: (a) Sol: For short circuit, reflection current is positive. For a open circuit, reflection current is negative. For a flat line, there is no reflection of current and voltage. 06. Ans: (a) Sol: The given method of grading is known as inters heath grading. In this method, instead of using various dielectrics and having composite dielectric to distribute the stress properly extra metallic sheaths between the conductor and the main sheath. This method improves voltage distribution in the dielectric of the cable and consequently uniform potential gradient is obtained. 07. Ans: (c) Sol: Inductance of a transmission line D H L 1 K ln r m Where, D = distance between the conductors. r = radius of the conductor. K = If the distance between the conductors is squared, Inductance will be doubled 2 D D L 2 K ln 2 K ln = 2 L 1 r r 08. Ans: (c) Sol: By using hollow conductors the diameter of the conductor is increased without materially adding to its weight, when the diameter is increased corona loss is reduced.

18 826 Power Systems ACE Engineering Publications 09. Ans: (d) Sol: Switching surge (OR) voltage is the base for the design of insulation of modern EHV lines 10. Ans: (d) Sol: The Inductive reactance of the arc suppressor coil is given as 20 X L = 20 2 C = C = 1.27 F L C = H. 11. Ans: (b) Sol: The insulation resistance of cable will decrease of the length of cable is increased and vice versa (i.e., 1 R ) in By grading of a cable is meant the distribution of dielectric material such that the difference between the maximum gradient and minimum is reduced, there by a cable of the same size could be operated at higher voltages or For same operating voltage, size of the cable could be smaller. The normal operating temperature of PVC cable is 70 0 C The thermal resistance of coil decreases as the moisture content of coil increases and vice versa. 12. Ans: (d) Sol: For stranded conductor the cross-section of conductor is a series of arcs of circles each of much smaller diameter than the conductor as a whole. The potential gradient for such a conductor will be greater than for the equivalent smooth conductor, so that the breakdown voltage for a stranded conductor will be some what less for a smooth conductor. 13. Ans: (c) Sol: Transmission Lines are transposed only to prevent interference with neighboring telephone lines 14. Ans: (d) Sol: V s = V r + Z I r = j = j400 = I s = I r = cos s = cos (53.13) = 0.6lag 15. Ans: (c) Sol: V x = V 0 e x V V x 0 e x For x = 1 km V 0 k V x e = k 1 k e