Source: https://www.scribd.com/document/315012453/EE-Objective-Paper-1
Timestamp: 2019-04-24 04:04:02+00:00

Document:
those students will be evaluated who will qualify the screening cut offs.
and EE branches are average but E&T papers are easier than last year.
take guarantee if any variation is found in actual cutoffs.
MADE EASY owes no responsibility for any kind of error due to data insufficiency/misprint/human errors etc.
2. It loses all its magnetism when there is no current flow.
3. It does not saturate easily.
field lines to be concentrated in the core material.
a current. So if there is no current then it should loose all its magnetism.
The capacitance per metre will be least in which of the above transmission lines?
presentation will be provided to score well in Paper-I.
project cost control etc. electronic materials. Information and Communication Technologies : Introduction to ICT. ECBC. elements of ethical dilemmas. orthographic projections. biodiversity. Watch Video Scroll down For Answer Key of ESE-2016 Page 2 of 3 . classification of magnetic materials. electronic mail. Watch Video 7. drawing standard. TQM : Most Importance. global warming. need of ICT in education. origin and development of ICT. ethical theories. Drawing instruments. Concept of System Software. Drawing. meaning of cloud computing. wrought iron. maintenance and services: ISO Standards.4. hexagonal closed packing. fibre optic cable etc. professional codes. crystal imperfections. photo conductors. projection of points. Application of computer. safety. thermosetting materials. rotation methods. Kaizer Tools-3m. biomagnification. GPS navigation system. project organization. Basics of Project Management: Project characteristics and types. polar molecules. IS. e-learning. Ethical dilemma. aluminum. Standards and quality practices in production. indian ethics. Watch Video 6. Basics of Energy and Environment: Renewable and non renewable energy resources. copper. Basics of Material Science and Engineering: Introduction of material science. project financing and financial appraisal. Six Sigma. virtual classroom. human values. Sampling. TPM. human progress. evidences of climate change. 7 Quality Control Tools . benefits of EIA etc. EDUSAT (Education satellite). ISHIKAWAS -7QC Tools. Watch Video 9. ethics and sustainability. network configuration of EDUSAT. classification of materials. tungsten etc. Watch Video 8. steel. cloud computing architecture. environmental laws for controlling pollution. Deming's: 14 Principles. profile planes side views. Ethics and values in engineering profession: ethics for engineers. polymers. ISO-9000 Quality Management. multimedia systems. digital libraries. history and development of internet. greenhouse gases. Quality Control. national mission on education through ICT. cubic crystal structures. semi conductor materials. Lean Manufacturing ME. thermistors. conventions on climate change. Components of ICT. methods of projection. positions of a straight line with respect to HP and VP. e-governance. ICT in networking. uses of EDUSAT. ecology. smart classes. ferrous and non ferrous metals. ISO-14000 other. trace of a line. risk analysis. construction. compounding materials. General Principles of Design. carbon credit. Watch Video 5. PDSA. miller indices. BIS Codes. environmental pollution. accidents. network topologies. Project appraisal and project cost estimations. 5S System. ceramics. Chemical bonding. crystallography. determining true length and true inclinations of a straight line. project evaluation and post project evaluation. hall effect. cast iron. plastics. projection of planes. Quality Circles. responsibilities of engineers etc. ozone depletion. Importance of Safety : Engineering Drawing. environmental degradation. Watch Video 10. insulators. projection of straight lines. climate change. wireless transmission. thermoplastic materials. special alloys steels. TQM ME. environmental ethics. titanium. importance of safety etc. acid rain. dielectrics. geometric construction and curves. fracture. risks. energy conservation. PDCA.
ESE-2016 : Electrical Engg. The Poynting vector on the surface of a long straight conductor of radius a and conductivity σ0. TV broadcast Which of the above applications is/are correct? (a) 1 only (b) 1 and 2 only (c) 2 and 3 only (d) 1. New Delhi-16 | Email : info@madeeasy. which carries current I in the z-direction. (a) 10. Radio astronomy 2. The variation of | B | with distance r from a very long straight conductor carrying a current I is correctly represented by (a) (b) |B | |B | r r (c) (d) |B | r Ans.in .madeeasy. Consider the following applications in respect of a square corner reflector: 1. |B | r (d) Corporate Office: 44-A/1. is I2 (a) (c) σ0 πb a 3 r I2 σ0 π2a (b) a 3 r (d) −I 2 2σ 0 π2 a 2 −I 2 2σ 0 π2 a 3 ar ar Ans. 2 and 3 Ans. Kalu Sarai. 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 5 Three equal point charges are located at the vertices of an equilateral triangle on the circumference of a circle of radius r. Solutions of Objective Paper-I | Set-A 9.in | Visit: www. The total electric field intensity at the centre of the circle would be (a) zero (b) q (d) (c) 12 ε 0r 2 3q 4πε 0r 2 q 3π ε 0r 2 Ans. (d) 11. Point-to-point communication 3. (d) 12.
3 and 4 Ans.madeeasy.9 × 10–6 Ω cm.ESE-2016 : Electrical Engg. 2. by passing a current greater than the critical current (IC) as per Silsbee rule. higher resistivity) than annealed copper.in . New Delhi-16 | Email : info@madeeasy.. The resistivity of hard drawn copper at 20°C is 1. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 6 13. 14. A hard drawn copper wire thus has lower conductivity (i. 2 and 3 only (d) 1. 3. 2. (a) The mechanical treatment such as cold working produces localized strain in the material which results in the increase in resistivity of material.e.e (EF −EV )/ kT n ∝ e − ΔE / kT 15. by increasing the temperature above a critical temperature (TC). ⎛ ΔE ⎞ (a) exp ⎜⎝ ⎟ kT ⎠ ⎛ 2ΔE ⎞ (b) exp ⎜⎝ ⎟ kT ⎠ ⎛ ΔE ⎞ (c) exp ⎜ − ⎝ kT ⎠⎟ ⎛ 2 ΔE ⎞ (d) exp ⎜ − ⎝ kT ⎠⎟ (c) n = Nce −(EC −EF )/ kT = Nce −(EC −EV )+ (EF −EV )/ kT = Nce − ΔE / kT .in | Visit: www. Superconductivity in a material can be destroyed by (a) increasing the temperature above a certain limit (b) applying a magnetic field above a certain limit (c) passing a current above a certain limit (d) decreasing the temperature to a point below the critical temperature Which of the above are correct? (a) 1 and 2 only (b) 2 and 3 only (c) 1. (c) Superconductivity can be destroyed– 1. Corporate Office: 44-A/1. Kalu Sarai. The number of electrons excited into the conduction band from valence band (with ΔE = forbidden energy gap and k = Boltzmann’s constant) is proportional to Ans. by applying magnetic field greater than the critical magnetic field (HC). The resistivity of annealed copper compared to hard drawn copper is (a) lesser (b) slightly larger (c) same (d) much larger Ans.
madeeasy. 18.83 mm (b) 8.in | Visit: www.3 cm (d) 0.83 mm Corporate Office: 44-A/1. nearly equal to liquid helium temperature). Kalu Sarai..ESE-2016 : Electrical Engg. The dielectric strength of rubber is 40000 V/mm at frequency of 50 Hz.083 mm Ans. Such materials (metals and alloys) are known as (a) piezoelectric materials (b) diamagnetic materials (c) superconductors (d) high-energy hard magnetic materials Ans.825 × 10–3 m 40000 = 0. (a) Dielectric strength of rubber is = 40000 V/mm Applied voltage = 33 kV Let us assume thickness is ‘t’ then 40000V 10−3 m ⇒ = t= 33 × 103V t 33 × 103 × 10−3 = 8. (c) Superconductors are the materials whose resistivity becomes very small or zero below a critical temperature. Solutions of Objective Paper-I | Set-A 16.3 mm (c) 8. 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 M H (b) 1 − M H (d) 1 − M H M H (a) We have susceptibility HM = M = μr – 1 H so relative permeability μr = 1 + M H 17. New Delhi-16 | Email : info@madeeasy.e. Page 7 The relative permeability of a medium is equal to (with M = magnetization of the medium and H = magnetic field strength) (a) 1 + (c) 1 + Ans.in . The electrical resistivity of many metals and alloys drops suddenly to zero when they are cooled to a low temperature (i.25 × 10–4 m = 0. What is the thickness of insulation required on an electrical conductor at 33 kV to sustain the breakdown? (a) 0.
However those subjects of technical syllabus which are added in ESE-2017 will be supplemented by study material.m 2nd July.500/- Rs.110016.Rank Improvement Batches India’s Best Institute for IES. These batches will be focusing on solving problems and doubt clearing sessions. 41. These batches are designed for repeater students who have already taken regular classroom coaching or prepared themselves and already attempted GATE/ESE Exams .500/- Rs. Study Material/ Books will be provided only for the technical syllabus which are newly added in ESE-2017. 22.in . it is decided that the technical subjects which are newly added in ESE 2017 syllabus over ESE 2016 syllabus will be taught from basics and comprehensively . 09958995830 www.0 CGPA Comprehensive problem solving sessions Smart techniques to solve problems Techniques to improve accuracy & speed Systematic & cyclic revision of all subjects Doubt clearing sessions Weekly class tests Interview Guidance Course Duration : Approximately 25 weeks (400 teaching hours) Syllabus Covered : Technical Syllabus of GATE-2017 & ESE-2017 Streams Timing Batch Type Date Venue nd CE. 26. 2. MADE EASY students are best bene tted by new set of questions. Looking at the importance and requirements of repeater students. GATE score card.S. study material and taxes Note: 1. 2016 Saket (Delhi) EC.m to 11:30 a. Features : Eligibility : • • • • • • • • Old students who have undergone classroom course from any centre of MADE EASY or any other Institute • Top 6000 rank in GATE Exam • Quali ed in ESE written exam • Quali ed in any PSU written exam • M.m to 5:00 p. 2016 Lado Sarai (Delhi) ME Regular th 4 July. Conventional. Tech from IIT/NIT/DTU with minimum 7. The selection of questions will be such that the Ex. MADE EASY I-card • 2 photos + ID proof Corp.500/- Rs. Post GATE batches Those students who were enrolled in long term classroom programs only Rs. ADMISSIONS OPEN Documents required : M. 31. Ph: 011-45124612. Office : 44 . Therefore if a student is weak in basic concepts & fundamentals then he/she is recommended to join regular classroom course.Tech marksheet. ME Weekend Sat & Sun : 8:00 a. G.m to 5:00 p. EE Weekend Sat & Sun : 8:00 a. New Delhi .madeeasy. General Studies and Online Test Series (OTS) .500/- Rs.500/- Fee is inclusive of classes. 24. The course fee is designed without Study Material/Books. GATE & PSUs MADE EASY oﬀers rank improvement batches for ESE 2017 & GATE 2017. Kalu Sarai. 36.m 2 July. PSUs/IES Interview call le er.A/1.500/- Rs. The content of Rank Improvement Batch is designed to give exposure for solving diﬀerent types of questions within xed time frame. 2016 8:00 a. Rank Improvement. 3. but want to give next attempt for better result. Rank Improvement Batches will be conducted at Delhi Centre only.m (Mon-Fri) Saket (Delhi) Fee Structure Ex. MADE EASY Students Batch Non-MADE EASY Students Rank Improvement Batch Rank Improvement Batch + General Studies & Engineering Aptitude Batch Those students who were enrolled in Postal Study Course.
Kalu Sarai.in | Visit: www. which is a conductor. Hence because of the thermal energy equal number electrons and holes are generated in an intrinsic semiconductor. This is due to (a) doping (b) free electrons (c) thermal energy (d) valence electrons Ans. 22. The conductivity of insulating materials (a very small value) is called as (a) residual conductivity (b) dielectric conductivity (c) ionic conductivity (d) bipolar conductivity Ans. (c) In an intrinsic semiconductor if we increase temperature then few of the electrons in valence band acquire sufficient energy so that they overcome forbidden gap and reach into the conduction band. and a corresponding hole is generated in valence band. 21. An electrically balanced atom has 30 protons in its nucleus and 2 electrons in its outermost shell. Corporate Office: 44-A/1. (a) There are 30 protons in electrically balanced atom of material. An intrinsic semiconductor has equal number of electrons and holes in it. and 2 electrons in its outermost shell).ESE-2016 : Electrical Engg. (b) When we add a metal with another metal this makes an alloy. (c) 20.in . The material made of such atom is (a) a conductor (b) an insulator (c) a semiconductor (d) a superconductor Ans. New Delhi-16 | Email : info@madeeasy. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 8 19.madeeasy. So this material is Zn (having atomic number 30. its conductivity will (a) increase (b) decrease (c) remain the same (d) increase or decrease depending on the impurity Ans. When a very small amount of higher conducting metal is added to a conductor. Conductivity of alloy is less than the conductivity of metal as alloy has less regular structure than metal. hence atomic number of material is 30.
the electron spins (like tiny magnets) become more likely to be in high energy state. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 9 23. New Delhi-16 | Email : info@madeeasy.in . In the slice processing of an integrated circuit (a) components are formed in the areas where silicon dioxide remains (b) components are formed in the areas where silicon dioxide has been removed (c) the diffusing elements diffuse through silicon dioxide (d) only on diffusion process is used Ans. ⇒ As we further heat the individual dipole spin (dipole vibration) within domain become more likely to point opposite to their neighbour which results in losing permanent magnetic behaviour. So the temperature coefficient of resistance for a doped semiconductor can be positive or negative depending upon level of doping. 2 and 3 Ans. Permanent magnet loses the magnetic (a) atomic vibration (c) realignment of dipoles Which of the above are correct? (a) 1 and 2 only (c) 1. 25. ⇒ Heating means providing extra thermal energy because of which it becomes easy for domain walls (the boundaries between regions that are lined up pointing different directions) to slide around. degenerate semiconductor) has metal like properties.ESE-2016 : Electrical Engg. Kalu Sarai. The temperature coefficient of resistance of a doped semiconductor is (a) always positive (b) always negative (c) zero (d) positive or negative depending upon the level of doping Ans. which leads to atomic vibration. But a heavily doped semiconductor (i. hence permanent magnet loses magnetic behaviour.madeeasy. hence it has positive temperature coefficient of resistance. behaviour when heated because of (b) dipole vibration (b) 1 and 3 only (d) 2 and 3 only (c) ⇒ As a permanent magnet is heated.in | Visit: www. Corporate Office: 44-A/1. (d) For a normally doped semiconductor temperature coefficient of resistance is negative. (b) In the slice processing of an integrated circuit the components are formed in selective areas known as “windows” where silicon dioxide has been etched. 24.e. That means that are less lined up so the total magnetism is reduced. Hence the domain walls will rearrange so that they reduce the large scale field energy by pointing different directions.
Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 10 26.in | Visit: www. Corporate Office: 44-A/1. then the deflection of the fluxmeter will be (a) 87. the change in flux linking with search coil is Δφ = 2 × 1 × 10–4 = 2 × 10–4 Wb Δφ = 2 × 10–4 = Gθ N 1500 × 10−6 × θ 10 θ = 1.madeeasy. 4. 3.5° (c) 65.33 × 28.7° Ans.05 × 750 × 10–6 = 1500 × 10–6 Flux linkage with search coil = 0. 27. The net magnetic moment in ferrimagnetic material is higher than that in ferromagnetic material. 180 = 76. Both ferromagnetic and ferrimagnetic materials make those domains that have favourable orientation to the applied field grow in size.5 × 200 × 10–6 = 1 × 10–4 wb As the flux is reversed. The net magnetic moment in ferromagnetic material is higher than that in ferrimagnetic material. (b) Constant of Fluxmeter G = Nc Bc Ac = 40 × 0. The magnetic field required to reduce the residual magnetization to zero is called (a) retentivity (b) coercivity (c) hysteresis (d) saturation Ans.2° π Consider the following statements: 1.in . Both ferromagnetic and ferrimagnetic materials have domain structures.05 Wb/m2 Number of turns on moving coil = 40 Area of moving coil = 750 mm2 If the flux linking 10 turns of a search coil of 200 mm2 area connected to the fluxmeter is reversed in a uniform field of 0. each domain has randomly oriented magnetic moments when no external field is applied.6° (d) 54. Kalu Sarai. 2. (b) The magnetic field required to reduce the residual magnetization (or spontaneous magnetization) to zero value is applied in reverse direction and is called coercive field.33 rad = 1.4° (b) 76. New Delhi-16 | Email : info@madeeasy.ESE-2016 : Electrical Engg. A certain fluxmeter has the following specifications: Air gap flux density = 0.5 Wb/m2.
GATE & PSUs Classroom Course is designed for comprehensive preparation of ESE.madeeasy. 2016 EC : 30th May & 9th June. 2016 MADE EASY Centres at all MADE EASY centres. ME. Classes are taken by highly experienced professors and ESE qualified toppers. Fee Structure. IN. The main feature of the course is that all the subjects are taught from basic level to advance level. 2016 CS from 29th May’16 Visit : www. visit : www.in ADMISSIONS OPEN ME from 28th May’16 CS : 30th May.Regular and Weekend Classroom Courses For ESE. High quality study material is provided during the classroom course with sufficient theory and practice test papers for objective and conventional questions alongwith regular assignments for practice. 2016 ME : 5th June. MADE EASY team has developed very effective methodology of teaching and advance techniques and shortcuts to solve objective questions in limited time. 2016 CE from 28th May’16 EE : 30th May & 5th June. GATE & PSUs. CS. EE. Course Features : • Timely coverage of technical & non-technical syllabus • Books & Reading References • Regular classroom tests followed by discussion • Doubt clearing sessions • GATE counseling session • Interview Guidance Program • All India ESE Classroom Test Series Syllabus Covered : • All Technical Subjects alongwith 10 subjects of paper-I (as per revised syllabus of ESE 2017) • Engineering Mathematics • Reasoning & Aptitude Books & Reading References : • Technical Subjects (Theory Book + Work Book) • Engineering Mathematics • Reasoning & Aptitude • Previous Years GATE Solved Papers • General English • Previous Years IES Solved Papers (Objective & Conventional) Difference between Regular and Weekend Course : In Regular Course. GATE & PSUs 2017 (On revised syllabus of ESE-2017 & GATE-2017) India’s Best Institute for IES. classes are conducted for 4 to 6 hours per day in a week for 8 to 9 months where as in Weekend Courses take 10 to 11 months for completion of syllabus as classes run nearly 8 to 9 hrs/day on every weekends and public holidays. PI New Batches Commencing at Delhi Centres Regular Batches Schedule Weekend Batches Schedule CE : 30th May & 7th June. EC.madeeasy. Streams Offered : CE.in Delhi Hyderabad Noida Jaipur Bhopal Lucknow Indore Bhubaneswar Pune Kolkata Patna 011-45124612 09958995830 040-24652324 09160002324 0120-6524612 08860378009 0141-4024612 09166811228 0755-4004612 08120035652 09919111168 08400029422 0731-4029612 07566669612 0674-6999888 09040999888 020-26058612 09168884343 033-68888880 08282888880 0612-2356615 0612-2356616 . timing & other details. There is due emphasis on solving objective and numerical questions in the class. EC from 29th May’16 EE from 29th May’16 IN : 16th June. 2016 Online Admissions Available To know more about upcoming batches.
Kalu Sarai. 29. is (a) 148 μV (c) 74 μV Ans. while the load varies between 140 Ω and 10 kΩ.madeeasy. The Hall voltage. 2 and 3 Ans.3 Ω Corporate Office: 44-A/1. VH .6 Ω (c) 36. A Zener regulator has an input voltage varying between 20 V and 30 V. (b) 111 μV (d) 37 μV (c) VH = t= I= B= RH = ? 0.1 × 10 −3 VH = 74 μ volt VH = RH ⇒ 30.1 mm 100 A 1 Wb/m2 7. The maximum resistance in series with the unregulated source and Zener diode would be (a) 3. Bz = 1 Wb/m2 and the Hall coefficient.in | Visit: www.4 × 10–11 m3/c BI 7.in .ESE-2016 : Electrical Engg. Hence statement 4 is wrong. so answer will be option (d). The desired regulated voltage is 12 V. for a thin copper plate of 0. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 11 Which of the above statements are correct? (a) 1 and 4 only (b) 1.4 × 10 −11 × 1 × 100 = 7.6 Ω (d) 93.1 mm carrying a current of 100 A with the flux density in the z-direction.4 x 10–11 m3/C.3 Ω (b) 6. New Delhi-16 | Email : info@madeeasy. RH = 7. 2 and 4 (c) 2 and 4 only (d) 1. (d) In ferrimagnetic materials dipoles are oriented in opposite direction but having different magnitude as shown below: While in ferromagnetic materials in a domain all the dipoles are oriented in same direction as shown below: So net magnetic moment in ferromagnetic materials is higher than that in ferrimagnetic materials.4 × 10 −5 V = t 0.
New Delhi-16 | Email : info@madeeasy.628 mH (c) 0. An air-cored solenoid of 250 turns has a cross-sectional area A = 80 cm2 and length l = 100 cm. The current in a coil changes uniformly from 10 A to 1 A in half a second.ESE-2016 : Electrical Engg. The self-inductance of the coil is (a) 0. A short in any type of circuit (series. power to decrease 3.3 Ω 12/140 Imin 31. 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 12 (d) Imin = Iz min + Iz max 12 0 + Vz = 140 RL min R= 20 − 12 Vmin − Vz = = 93. (c) V= L 36 = L di dt 9 = L = 2H 1/ 2 Corporate Office: 44-A/1. (b) L= N 2 μa = 250 × 250 × 4π × 10 −7 × 80 × 10 −4 l μ = μ0μr as air μr  1 100 × 10−2 = 0.in | Visit: www.751 mH (d) 0. resistance to decrease 2. parallel or combination) causes the total circuit 1. current to increase 4. 32.425 mH (b) 0. Kalu Sarai.madeeasy.904 mH Ans.5 H (b) 1 H (c) 2 H (d) 4 H Ans. A voltmeter connected across the coil gives a reading of 36 V. Solutions of Objective Paper-I | Set-A Ans. (d) A short in any type of circuit causes the total circuit resistance to decrease and thereby the current to increase. The value of its inductance is (a) 0.in . voltage to increase Which of the above are correct? (a) 2 and 3 (b) 2 and 4 (c) 1 and 4 (d) 1 and 3 Ans.628 mH 33.
(c) V= M di 4 ⇒ 4000 = M dt 10 × 10−6 M = 10–1 ⇒ M = 0. Kalu Sarai.5 Ω (d) 10 Ω Ans. the primary current is reduced from 4 A to zero in 10 μs.1 H (d) 0.in | Visit: www. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 13 34.madeeasy.. New Delhi-16 (b) 3 Ω (d) 3 3 Ω | Email : info@madeeasy.. N resistors each of resistance R when connected in series offer an equivalent resistance of 50 Ω and when reconnected in parallel the effective resistance is 2 Ω.. A voltage of 40000 V is observed across the secondary..in .(2) Multiplying equation (1) and (2) 50 × 2n = 100 ⇒ R = 100 n R = 10 Ω R2 = 36. (d) RP RP = R n ⇒ n ⋅ R = 50 2 = R n 50 n ⇒ R = 2n RS = nR R = . The value of R is (a) 2.ESE-2016 : Electrical Engg..1 H 35.01 H Ans. For a series R-L circuit i(t) = 2 sin(ωt − 45°) If ωL = 1 Ω. the value of R is (a) 1 Ω (c) 3Ω Corporate Office: 44-A/1.5 Ω (b) 5 Ω (c) 7.(1) . In a mutually coupled circuit. The mutual inductance between the coils is (a) 100 H (b) 10 H (c) 0..
AIR Total 47 selections Stream Batch Commencement Civil / Mechanical 30th May (Morning) & 15th June (Evening) Electrical / Electronics 1st June (Morning) & 1st July (Evening) Final year or Pass out engineering students are eligible. JMI-Delhi. HBTI-Kanpur. SGSITS-Indore. NSIT-Delhi. Punjab Engg. 3. GATE & PSUs Super Talent batches are designed for students with good academic records and who have secured good ranks in GATE/ESE or other national level competitive examinations. Kalu Sarai.in Note: Super Talent batches are conducted only at Delhi Centre Note: 1. College Why most brilliant students prefer Super Talent Batches! • Highly competitive environment • Meritorious student’s group • Opportunity to solve more problems • More number of tests • In-depth coverage of the syllabus • Motivational sessions GATE-2016 : Top Rankers from Super talent batches 1 1 2 AIR AIR AIR AIR 3 • Classes by senior faculty members • Discussion & doubt clearing classes • Special attention for better performance ESE-2015 : Top Rankers from Super talent batches AIR 4 AIR 3 AIR 6 7 AIR Agam Kumar Garg Nishant Bansal Arvind Biswal Udit Agarwal Piyush Kumar Amit Kumar Mishra Anas Feroz Kirti Kaushik EC ME EE EE CE CE EE CE AIR 8 10 10 AIR 8 10 AIR AIR AIR 9 10 AIR Sumit Kumar Stuti Arya Brahmanand Rahul Jalan Aman Gupta Mangal Yadav Arun Kumar ME EE CE CE CE CE ME 9 in Top 10 58 in Top 100 6 in Top 10 ADMISSIONS OPEN Online admission facility available To enroll. tests. But due to eligibility criteria.Tech from IIT’s/NIT’s/DTU • GATE Qualified MADE EASY old students • Cleared ESE written exam • 70% marks in B. BIT-Mesra.Tech from reputed colleges (See below mentioned colleges) BITS-Pilani. BIT-Sindri. ESE/PSUs Selection Proof.110016. visit : www. Corp.Tech from private engineering colleges • 65% marks in B. Office : 44 . the composition of students in this batch is homogeneous and better than other batches. study material. (Mark sheet. 2.-Pune. Jabalpur Engg. Thapar University-Punjab. College of Engg. GATE & PSUs India’s Best Institute for IES. College of Engg. Admissions in Super Talent Batch is subjected to verification of above mentioned documents.-Roorkee.Tech • Cleared any 3 PSUs written exam • GATE rank upto 2000 • 60% marks in B. Madan Mohan Malviya-Gorakhpur. GATE Score card. 09958995830 www. Eligibility (any one of the following) • MADE EASY repeater students with 65% Marks in B. MBM-Jodhpur. teaching pedagogy is similar to other batches.A/1.madeeasy.in . Candidate should bring original documents at the time of admission. New Delhi . Ph: 011-45124612. College. Here students will get a chance to face healthy competitive environment and it is very advantageous for ambitious aspirants. Super Talent batches are a kind of regular batches in which faculty.S UPER T ALENT B ATCHES for ESE.madeeasy. 2 Photographs & ID Proof).
32 Vm π The potential difference VAB in the circuit 1A VAB + VA 1Ω 4Ω is (a) 0.32 Vm (b) 0.8 V (c) 1. Kalu Sarai. Solutions of Objective Paper-I | Set-A Ans.8 V | Email : info@madeeasy.71 Vm Ans.8 V (d) –1. A single-phase full-wave rectifier is constructed using thyristors.8 V Corporate Office: 44-A/1.54 Vm (d) 0.in . (a) The average output voltage of single phase full wave rectifier is 38.48 Vm (c) 0. V0 = 2Vm 2Vm cos ∝ = cos60° π π V0 = Vm = 0. If the peak value of π the sinusoidal input voltage is Vm and the delay angle is radian.in | Visit: www. 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 14 (a) i (t) = 2 sin(ωt − 45°) ωL = 1 Ω XL Z θ R tan θ = XL XL ⇒ R = tan θ R R= 1 ⇒ R = 1Ω tan45° 37. New Delhi-16 – 3Ω 5V 3Ω VB (b) –0.madeeasy. then the average 3 value of output voltage is (a) 0.ESE-2016 : Electrical Engg.
1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 15 (b) 1A VA 1Ω VC = 5 V 3Ω VB 5V 4Ω 3Ω V C – 0 = 5 ⇒ VC = 5 V At node A: 1+ VA − 5 VA + =0 1 4 5 VA = 16 ⇒ VA = 16 V 5 At node B: −1 + ∴ ∴ VB − 5 VB + = 0 ⇒ 2VB = 8 3 3 VB = 4 V VAB = VA – VB = 16 − 4 = −0.ESE-2016 : Electrical Engg. Two bulbs of 100 W/250 V and 150 W/250 V are connected in series across a supply of 250 V.67 Ω P= V2 250 × 250 = .67 Ω P2 150 Req = R1 + R2 = 1041.madeeasy. Solutions of Objective Paper-I | Set-A Ans. 60 W Req 1041. Kalu Sarai.8 V 5 39.66 Corporate Office: 44-A/1.in . (b) 100W/250 V 150W/250 V R1 = V2 250 × 250 ⇒ R1 = ⇒ R1 = 625 Ω P1 100 R2 = V2 250 × 250 ⇒ R2 = = 416. New Delhi-16 | Email : info@madeeasy.in | Visit: www. The power consumed by the circuit is (a) 30 W (b) 60 W (c) 100 W (d) 250 W Ans.
madeeasy. Kalu Sarai.667 ∴ Z = R – j XC or Z = R – j (X C– X L) but ⎥ X C⎥ > ⎥ X L⎥ ∴ Thevenin impedance can be realised by using either R and C or by using R.71 ∠ –15.in | Visit: www.38 – j 0. a capacitor and an inductor (b) a resistor and a capacitor (c) a resistor and an inductor (d) a capacitor and an inductor Ans.38 – j 0.in . Solutions of Objective Paper-I | Set-A 40. Thevenin’s equivalent of a circuit. operating at ω = 5 rad/s. Analog-to-digital converter with the minimum number of bits that will convert analog input signals in the range of 0-5 V to an accuracy of 10 mV is (a) 6 (b) 9 (c) 12 (d) 15 Ans. New Delhi-16 | Email : info@madeeasy.667 Ω At this frequency. Three 30 Ω resistors are connected in parallel across an ideal 40 V source. (b) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 16 Z = 2. the minimal realization of the Thevenin’s impedance will have (a) a resistor. has VOC = 3. What would be the equivalent resistance seen by the load connected across this circuit? (a) 0 Ω (b) 10 Ω (c) 20 Ω (d) 30 Ω Corporate Office: 44-A/1.9° V ZO = 2.ESE-2016 : Electrical Engg. 41. L and C but minimal realisation will be with R and C. (b) ∵ In A to D converter Accuracy = Resolution = Error ∴ VFS 2n − 1 5V n 2 −1 ≤ 10 mV ≤ 10 mV 1 ≤ 2 × 10–3 2 −1 2n – 1 ≥ 500 2n ≥ 501 n≈9 n 42.
madeeasy. IN & PI 53 Selections in Top 10 96 Selections in Top 20 368 Selections in Top 100 ME Top 20 17 Selections CE Top 100 68 Selections Top 20 16 Selections EE Top 100 65 Selections Top 20 19 Selections EC Top 100 76 Selections Top 20 10 Selections IN CS & PI Top 100 45 Selections Top 20 17 Selections Top 100 53 Selections Top 20 17 Selections Top 100 61 Selections Detailed results are available at www. EE. M. Aakash Tayal Amarjeet Kumar Tushar Sumit Kumar Suman Dutta 9in ME Top 10 A I R Nishant Bansal AIR-2 AIR-3 AIR-4 AIR-5 AIR-6 Udit Keshav Tanuj Sharma Arvind AIR-7 AIR-8 AIR-10 AIR-9 10 in EE Top 10 AIR CE Rajat Chaudhary Sudarshan Rohit Agarwal Stuti Arya AIR-3 AIR-4 AIR-6 AIR-8 AIR-8 AIR-10 AIR-10 Rahul SIngh Piyush Kumar Srivastav Roopak Jain Jatin Kumar Lakhmani Vikas Bijarniya Brahmanand Rahul Jalan 8 in Top 10 AIR EC Anupam Samantaray Arvind Biswal Kumar Chitransh AIR-2 AIR-3 AIR-5 AIR-6 AIR-10 K K Sri Nivas Jayanta Kumar Deka Amit Rawat Pillai Muthuraj Saurabh Chakraberty 6 in Top 10 AIR Agam Kumar Garg AIR-2 AIR-3 AIR-4 AIR-7 AIR-8 AIR-8 AIR-10 AIR-10 Avinash Kumar Shobhit Mishra Ali Zafar Rajesh Chaitanya Shubham Tiwari Palak Bansal Saket Saurabh 9 IN in Top 10 AIR CS AIR-2 AIR-4 AIR-6 AIR-9 AIR-10 Debangshu Chatterjee Himanshu Agarwal Jain Ujjwal Omprakash Sreyans Nahata Nilesh Agrawal 6 in Top 10 AIR PI Harshvardhan Sinha Ankita Jain AIR-4 AIR-7 AIR-8 AIR-8 Akash Ghosh Niklank Kumar Jain Shree Namah Sharma Agniwesh Pratap Maurya 5 in Top 10 AIR Gaurav Sharma 1st Ranks in ME.MADE EASY Students Top in GATE-2016 AIR-2 AIR-2 AIR-4 AIR-5 AIR-6 AIR-7 AIR-8 AIR-10 Gaurav Sharma Manpreet Singh Sayeesh T. CS. EC.in “MADE EASY is the only institute which has consistently produced toppers in ESE & GATE” .
2 Ω (d) 6 V and 5. (ii) (b) 6 V and 1. (i) are 10 V and 2 Ω.in | Visit: www.2 Ω | Email : info@madeeasy..ESE-2016 : Electrical Engg. what are Thevenin's equivalents? A A 3Ω B Fig.... 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 17 (a) 40 V +  30 Ω 30 Ω 30 Ω Req Replacing voltage source by a short circuit Req = 0 Ω 43.2 Ω Corporate Office: 44-A/1. (ii).madeeasy. Thevenin’s equivalents of the network in Fig.. Solutions of Objective Paper-I | Set-A Ans. 2 ⎛ 4 ⎞ ⎛ 4 ⎞ ⎟ ⎜ ⎟ 2⎠ ⎝ 2⎠ (3)2 + ⎜⎝ 2 = 25 = 5A 2 P = Irms ⋅ R = (5)2 ⋅ (10) = 250 W 44.2 Ω (c) 10 V and 5. New Delhi-16 B Fig. The current i(t) through a 10 Ω resistor in series with an inductance is given by i(t) = 3 + 4 sin (100t + 45°) + 4 sin (300t + 60°) A The RMS value of the current and the power dissipated in the circuit are respectively (a) 5 A and 150 W (b) 11 A and 250 W (c) 5 A and 250 W (d) 11 A and 150 W Ans..in . If a resistance of 3 Ω is connected across terminals AB as shown in Fig. Kalu Sarai. (i) (a) 10 V and 1.. (c) Irms = = 2 2 Irms 1 + Irms 2 + .
01 W Ans.in | Visit: www. ZL = Z∗TH ZL = 500 – j100 Irms = Vrms z Z = ZL + ZTH ⇒ Z = 500 + j100 + 500 – j100 = 1000 ∴ Irms = Vrms 10 1 ⇒ Irms = = A Z 1000 100 2 Pmax = Irms ⋅ RL 2 ⎛ 1 ⎞ × 500 = 0. connected to a load.2 W (b) 0.in .ESE-2016 : Electrical Engg. has an e.1 W (c) 0. Solutions of Objective Paper-I | Set-A Ans.madeeasy.m.05 W (d) 0.f. A voltage source. 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 18 (b) 2Ω 2Ω A 10 V 10 V + +  3 Ω VTh – B 10 × 3 = 6V VTH = 3+2 RTH: Deactivate independent source 2Ω 3Ω Rth RTH = 3×2 6 = = 1. New Delhi-16 | Email : info@madeeasy.2 Ω 3+2 5 45. The maximum power that can be transferred to the load is (a) 0.05 W = ⎜ ⎝ 100 ⎟⎠ Corporate Office: 44-A/1. of 10 V and an impedance of (500 + j 100)Ω. Kalu Sarai. (c) 500 + j100 + ZL 10 V I For maximum power transfer.
madeeasy. L = 0. I = 2A 3.5 Ω Ans. A source of 10 V and internal impedance of 2 Ω is connected to the primary. 4 ZL = 2 ⇒ ZL = 1 = 0. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 19 46. VR = 100 2 V 2. An ideal transformer is rated 220/110 V. The power transferred to a load ZL connected across the secondary would be a maximum.in . Zi = ZL (2)2 ⇒ Zi = 4ZL And for maximum power transfer.. Kalu Sarai. Zi = 2 Ω ∴ 47. when | ZL | is (a) 4 Ω (b) 2 Ω (c) 1 Ω (d) 0.(1) V1 N1 N 220 N1 = ⇒ 1 =2 = ⇒ V2 N2 110 N2 N2 As From equation (1). 2 and 3 Corporate Office: 44-A/1..ESE-2016 : Electrical Engg.25 H Which of the above values are correct? (a) 2 and 3 only (b) 1 and 2 only (c) 1 and 3 only (d) 1.in | Visit: www. (d) In an ideal transformer N1 ⎛N ⎞ Zi = Z L ⎜ 1 ⎟ ⎝ N2 ⎠ N2 2 .5 Ω 2 Consider the following values for the circuit shown below : 100 Ω VR u(t) = 250 2 sin 600t L 150 V 1. New Delhi-16 | Email : info@madeeasy.
Siddhikh Hussain AIR 2 AIR 3 AIR 4 AIR 5 AIR 7 AIR 8 AIR 9 AIR 10 Saurabh Pratap Siddharth S.madeeasy. Nikki Bansal Nagendra Tiwari Anas Feroz Amal Sebastian Dharmini Sachin Sudhakar Kumar Vishal Rathi 9 EE Selections in Top 10 AIR EE E&T S.in “MADE EASY is the only institute which has consistently produced toppers in ESE & GATE” .MADE EASY Students TOP in ESE-2015 AIR 2 AIR 3 AIR 4 AIR 5 AIR 6 AIR 7 AIR 8 Amit Sharma Dhiraj Agarwal Pawan Jeph Kirti Kaushik Aman Gupta AIR 8 AIR 9 AIR 10 10 CE Selections in AIR Top 10 CE Palash Pagaria Piyush Pathak Amit Kumar Mishra AIR 2 AIR 3 AIR 4 AIR 5 AIR 6 AIR 7 Ashok Bansal Aarish Bansal Vikalp Yadav Naveen Kumar Raju R N Sudhir Jain Mangal Yadav Aishwarya Alok AIR 9 AIR 10 10 ME Selections in AIR Top 10 ME Pratap Bandi Sreenihar Kotnana Krishna Arun Kr. Maurya AIR 2 AIR 3 AIR 5 AIR 6 AIR 7 AIR 8 AIR 9 AIR 10 Partha Sarathi T. Piyush Vijay Manas Kumar Panda Kumbhar Piyush Nidhi Shruti Kushwaha Anurag Rawat 9 Selections in Top 10 AIR E&T Ijaz M Yousuf 4 Streams 4 Toppers All 4 MADE EASY Students 38 in Top 10 352 73 in Selections out of total 434 Top 20 MADE EASY selections in ESE-2015 : 82% of Total Vacancies CE Selections in Top 10 10 Selections in Top 20 20 MADE EASY Selections 120 Out of 151 Vacancies ME Selections in Top 10 10 Selections in Top 20 18 MADE EASY Selections 83 Out of 99 Vacancies MADE EASY Percentage 83% EE Selections in Top 10 9 Selections in Top 20 16 MADE EASY Selections 67 Out of 86 Vacancies MADE EASY Percentage 78% E &T Selections in Top 10 9 Selections in Top 20 19 MADE EASY Selections 82 Out of 98 Vacancies MADE EASY Percentage MADE EASY Percentage 79% 84% Detailed results are available at www.
Solutions of Objective Paper-I | Set-A Ans.madeeasy.in . 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 20 (*) 100 Ω + VR – + 150 V – v(t) V 2 = VR2 + VL2 Vrms = 250 2 2 = 250 (250)2 = VR2 + (150)2 ⇒ VR = 200V I= VR 200 = ⇒ I = 2A 100 100 VL = IXL ⇒ XL = ωL = 75 ⇒ L = 150 = 75 2 75 = 0. Kalu Sarai.in | Visit: www.125 H 600 L = 0. The response of a series R-C circuit is given by 2V q0 − π C I(s) = 1 ⎞ ⎛ R ⎜s + ⎟ ⎝ RC ⎠ where q0 is the initial charge on the capacitor.ESE-2016 : Electrical Engg. New Delhi-16 | Email : info@madeeasy.125 H 48. et / RC ⎛ 2V q0 ⎞ − ⎟ ⎜ R ⎝ π C⎠ (d) Zero (d) lim i(t ) = lim s I (s) t →∞ s →0 ⎛ 2V q0 ⎞ ⎜⎝ π − C ⎟⎠ =0 = lim s ⋅ s →0 1 ⎞ ⎛ R ⎜s + ⎟ ⎝ RC ⎠ Corporate Office: 44-A/1. What is the final value of the current? (a) 1 ⎛ 2V q0 ⎞ − ⎟ ⎜ R⎝ π C⎠ (b) (c) Infinity Ans.
(d) V +  + V1 – Req = R 3 P= + V2 – + V3 – R V2 V2 3V 2 = = Req R /3 R V1 = V2 = V3 = V V +  + V1 – + V2 – R Req = R 2 V2 2V 2 = R /2 R V1 = V2 = V As we can see from above circuits. • Total voltage and branch voltage remain unaffected We have taken equal resistances even unequal values give the same analysis.in . Kalu Sarai. What should be done to find the initial values of the circuit variables in a first-order R-C circuit excited by only initial conditions? (a) To replace the capacitor by a short circuit (b) To replace the capacitor by an open circuit (c) To replace the capacitor by a voltage source (d) To replace the capacitor by a current source Ans. no change in total voltage and branch voltage Which of the above is/are correct? (a) 1 only (b) 2 only (c) 3 only (d) 1. (c) As capacitor does not allow the sudden change in voltage ∴ for initial values we replace capacitor by a voltage source. 50. • Total power decreases.madeeasy. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 21 49. • Total resistance increases.in | Visit: www. opening a branch results in 1. 2 and 3 Ans. increase in total resistance 2. In a parallel resistive circuit.ESE-2016 : Electrical Engg. P= Corporate Office: 44-A/1. decrease in total power 3. New Delhi-16 | Email : info@madeeasy.
They have high power rating and low resistance value. 54. which is initially not a relaxed system 3. 2 and 3 Ans. ground 2. the preferred reference node is a node that is connected to 1. In nodal analysis. The precision resistors are (a) carbon composition resistors (b) wire-wound resistors (c) resistors with a negative temperature coefficient (d) resistors with a positive temperature coefficient Ans. 52. (a) Reciprocity theorem is applicable in case of linear and bilateral networks containig only one independent source.in | Visit: www. Two networks are said to be dual when (a) their node equations are the same (b) the loop equations of one network are analogous to the node equations of the other (c) their loop equations are the same (d) the voltage sources of one networks are the current sources of the other Ans. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 22 51. containing R.in . (a) We always prefer to take that node as the reference node which is at ground potential. Corporate Office: 44-A/1.madeeasy. (b) Wire wound resistors are used as precision resistors. (b) Duality means mathematical representation of both the networks should be identical (kVL and kCL) ∴ Loop equations of one network are analogous to the node euqations of the other. New Delhi-16 | Email : info@madeeasy. having both dependent and independent sources Which of the above is/are correct? (a) 1 only (b) 1 and 2 only (c) 2 and 3 only (d) 1. many parts of the network 3. Kalu Sarai. 53. L and C elements 2. Reciprocity theorem is applicable to a network 1. the highest voltage source .ESE-2016 : Electrical Engg. Which of the above is/are correct? (a) 1 only (b) 2 only (c) 3 only (d) 1. 2 and 3 Ans.
solved examples.madeeasy. • Many students are selected every year in ESE. considering the syllabus and standards of the competitive examinations. • Postal Study course is also available at all centres of MADE EASY.in MADE EASY Centres Delhi Hyderabad Noida Jaipur Bhopal Lucknow Indore Bhubaneswar Pune Kolkata Patna 011-45124612 09958995830 040-24652324 09160002324 0120-6524612 08860378009 0141-4024612 09166811228 0755-4004612 08120035652 09919111168 08400029422 0731-4029612 07566669612 0674-6999888 09040999888 020-26058612 09168884343 033-68888880 08282888880 0612-2356615 0612-2356616 .madeeasy. • Conventional practice booklets (for ESE). The content is self sufficient and there will be no need to refer several text books for any subject. effective and easy to understand. Note : • Postal Study Course for only General Studies and Engg. objective & conventional practice questions. The study material is authored by experienced faculties supported by research and development wing of MADE EASY. GATE & PSUs by reading Postal Study Course only. • Previous exam solved papers (15 to 25 years). GATE & PSUs On the revised pattern and syllabus of ESE-2017 and GATE-2017 Postal Study Course is the distance learning program designed to meet the needs of the college going students and working professionals who are unable to join our classroom courses. Aptitude paper (Paper-I of ESE-2017) is also available. The study material is compact. GATE and PSUs Streams Offered : CE • ME/PI • EE • EC • CS • IN www. • General studies and Engineering Aptitude booklets (As per ESE-2017 syllabus). MADE EASY has made all efforts to provide an error free material introducing smart and shortcut techniques for better understanding. • Objective practice booklets.in Courses Offered : • GATE • GATE + PSU • ESE Buy Now • ESE.Postal Study Course For ESE-2017 & GATE-2017 India’s Best Institute for IES. • Online purchase facility available at : www. Content includes : • Theory books with solved examples. • The content is very much student friendly expressed in lucid language such that an average student can also understand the concepts easily. Features of Postal Study Course : • The content of new MADE EASY Postal Course 2017 covers all the basic fundamentals.
80 mJ (d) 2. Kalu Sarai.19 mJ (c) 4.25 (b) 2.madeeasy.5) × 1 Pmax = 6.in | Visit: www.38 mJ (b) 22. How much electrical energy must be applied to produce output energy of 7.06 x 10–3 J? (a) 25. then the voltage across the current generator and P are (a) 5 V and 6. delivers maximum power P in watts to its load of RL Ω.5 × 1 = 2.32.in . A piezoelectric crystal has a coupling coefficient K of 0. (c) Total Response = Forced Response + Natural Response i(t) = i(∞) + [i(0+) – i(∞)] e–t /τ v (t) = v (∞) + [v (0+) – v (∞) e–t /τ So it can be either in the form or K + K1 e–at K1 ⋅ e −at ↓ if final valueis zero 56.5 (d) 2.25 W Corporate Office: 44-A/1. (d) + 5A 1Ω VL – RL = 1 Ω For maximum power transfer RL = 1 Ω IL 5 × 1 = 2.5 (c) 5 V and 12.5 V IL = ∴ 2 Pmax = IL2 ⋅ R = (2.26 mJ Ans. shunted by its own resistance of 1 Ω.ESE-2016 : Electrical Engg. Which of the following is true for the complete response of any network voltage or current variables for a step excitation to a first-order circuit? (a) It has the form k1e–at (b) It has the form k (c) It may have either the form (a) or the form of (a) plus (b) (d) It has the form e+at Ans.5 A 1+ 1 VL = 2.25 Ans. (b) 57. New Delhi-16 | Email : info@madeeasy. If a constant current generator of 5 A.5 V and 12. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 23 55.5 V and 6.
Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 24 58. when its current coil is connected in R phase and the potential coil is connected across R and neutral of a balanced 400 V (RYB sequence) supply. New Delhi-16 | Email : info@madeeasy. The line current is 54 A. Three star-connected loads of 3∠60° Ω each and three delta-connected loads of 9∠60° Ω each are connected in parallel and fed from a three-phase balanced source having lineto-neutral voltage of 120 V. (c) R 3∠60° 9∠60° Y B Converting Δ-Load into star R 80 A 3∠60° 120 V 120 V 3∠60° Y B 120 = 40A 3 = 40 + 40 = 80 A I= Itotal 59.ESE-2016 : Electrical Engg. If the potential coil reconnected across B-Y phases with the current coil in R phase. the new reading of the wattmeter will be nearly (a) 10 kW (b) 13 kW (c) 16 kW (d) 19 kW Ans. A wattmeter reads 10 kW. Kalu Sarai. The line currents drawn from the supply will be (a) 10 A each (b) 20 A each (c) 80 A each (d) 160 A each Ans. (b) R W N B Y Corporate Office: 44-A/1.in .in | Visit: www.madeeasy.
the connection of R phase got reversed. (a) 61. It increases the rise time. Consider the following statements regarding the effect of adding a pole in the openloop transfer function on the closed-loop step response: 1.in | Visit: www. 2.as well as delta-connected loads Ans. The new line voltages will have a relationship VYB (a) VRY = VBR = VBR (b) VRY = VYB = VRY (c) VYB = VBR = (d) VRY = VYB = VBR 3 3 3 Ans. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 25 P = Vph Iph cos φ 10 × 103 = R 400 × 54 × cos φ 3 W B Y cos φ = 0. Which of the above statements are correct? (a) 1.6 = 12. It increases the maximum overshoot.8 P = VLIL sin φ = 400 × 54 × 0. star-connected alternator is V. By mistake. New Delhi-16 | Email : info@madeeasy.madeeasy. Kalu Sarai.46 kW  13 kW 60. The phase voltage of a three-phase. 3.ESE-2016 : Electrical Engg.in . It reduces the bandwidth. (d) 62. Two-wattmeter method of power measurement in three-phase system is valid for (a) balanced star-connected load only (b) unbalanced star-connected load only (c) balanced delta-connected load only (d) balanced or unbalanced star. 2 and 3 (b) 1 and 2 only (c) 2 and 3 only (d) 1 and 3 only Corporate Office: 44-A/1.
A series R-L-C circuit is connected to a 25 V source of variable frequency.5 cycles (c) 5 cycles (d) 10 cycles Ans. the number of signals displayed on the screen will be (a) 1.25 cycles (b) 2.madeeasy. The circuit current is found to be a maximum of 0. (b) ∵ Time period of I/P signal 5 sin (314 t + 45°) 2π = 314 T T = 20 m sec ∵ CRO horizontal scale has 10 divisions with base setting 5 ms/div ∴ Total time period of horizontal scale = 10 × 5 = 50 m sec No. A CRO screen has 10 divisions on the horizontal scale.ESE-2016 : Electrical Engg.5 A . fr = 400 Hz. ∴ ⇒ VL = VC and VR = VS VR = 25 V 25 ⇒ R = 50 Ω 0.in | Visit: www. (a) We know that adding a pole i.5 Ω and 0. the values of R and L are respectively (a) 50 Ω and 300 mH (b) 12. Assuming ideal components. VC = 150 V As the circuit is in resonance.e.5 64. of cycle = 50 msec 20 msec = 2.5 VR = IR ⇒ R = XL = ωL ⇒ L = Corporate Office: 44-A/1. the order of denominator increase then (i) maximum overshoot increase (ii) rise time increases (iii) unstability increases (iv) reduces band width so answer is (a) 63. New Delhi-16 300 ⇒ L = 0.119 H (c) 50 Ω and 0.119 H 2π × 400 | Email : info@madeeasy. (c) Imax = 0. If a voltage signal 5 sin (314t + 45°) is examined with a line base setting of 5 ms/div.5 150 = 300 VL = I XL ⇒ XL = 0.5 A at a frequency of 400 Hz and the voltage across C is 150 V.in .5 Ω and 300 mH Ans.119 H (d) 12.. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 26 Ans. Kalu Sarai.
Aptitude (Paper-I) 17 Part Syllabus Tests : Engineering Discipline (Paper-II) 4 Full Syllabus Tests : GS & Engg. 2016 For online admission & other details.India’s Best Institute for IES.madeeasy. • Subjectwise analysis to know your weak areas.2017 Preliminary Examination (Paper-I & Paper-II) 15 Part Syllabus Tests : GS & Engg. GATE & PSUs All India ONLINE For TEST SERIES ESE . • Compete with thousands of ESE aspirants. • Test papers are designed as per ESE 2017 standard and pattern. • Series of part syllabus and full syllabus tests for better practicing. Aptitude (Paper-I) 4 Full Syllabus Tests : Engineering Discipline (Paper-II) Total 40 Tests Features of ESE Online Test Series : QUALITY test papers COMPREHENSIVE VIDEO SOLUTIONS with explanations COMPETITION WITH thousands of aspirants 40 TOTAL 40 test papers android & IOS AVAILABLE ON FAST TECHNICAL • Questions are newly developed by R & D wing of MADE EASY. • High level of accuracy to make question papers error free. visit : www. Test series is likely to start in October.in . • Fully explained & well illustrated video solutions for all questions. • Opportunity to win rewards/prizes for securing position in ESE 2017. • Comprehensive and detailed analysis report of test performance.
(b) L C R Zeq ∴ Zeq = jωL + Z 1 R 1 − j ωRC j ωc = × Z= 1 1 + j ωRC 1 − j ωRC R+ j ωc R× Z= R − j ωR 2C 1 + ω2R 2C 2 Zeq = j ωL + = R − j ω R 2C 1 + ω 2R 2C 2 ( ) j ω L 1 + ω 2 R 2C 2 + R − j ω R 2C 2 1 + ω R 2C 2 ⎡ ωL + ω 3R 2C 2L − ωR 2C ⎤ ⎦ = +j⎣ 1 + ω 2 R 2C 2 1 + ω 2 R 2C 2 At resonance impedance is purely resistive in nature.21 1 ω2 = 5 – 1 ⇒ ω2 = ω = 2 rad/sec ω2 = Corporate Office: 44-A/1. is (a) 1 rad/s (b) 2 rad/s (c) 3 rad/s (d) 4 rad/s Ans.madeeasy. ∴ Equating Imaginary part to zero. Solutions of Objective Paper-I | Set-A 65. Kalu Sarai. ωL + ω3R2C2L – ωR2C = 0 ω3R2C2L = ω[R2C – L] R ω2 = R 2C 2 2 R C L − 1 2 R C 2L 1 1 1 1 − 2 2 ⇒ ω2 = − LC R C 10. R = 1 Ω and C = 1 F. The resonant frequency for the circuit L R C for L = 0.in | Visit: www. New Delhi-16 | Email : info@madeeasy.2 H.ESE-2016 : Electrical Engg.in 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 27 .
(1) . Which one of the following conditions will be correct. V1 = Z11 I1 + Z12 I2 V2 = Z21I1 + Z22 I2 Z11 = 60 Z12 = 20 Z21 = 20 Z22 = 40 • For symmetry Z11 = Z22 but as in the given case 60 ≠ 40 ∴ Network is not symmetrical • For reciprocity Z12 = Z21 and 20 = 20 ∴ Network is reciprocal. From equation (2) 20I1 = V2 – 40I2 ⇒ I1 = 1 V2 − 2 I2 20 Corporate Office: 44-A/1.(2) .madeeasy..(3) | Email : info@madeeasy. when three identical bulbs forming a star are connected to a three-phase balanced supply? (a) The bulb in R phase will be the brightest (b) The bulb in Y phase will be the brightest (c) The bulb in B phase will be the brightest (d) All the bulbs will be equally bright Ans. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 28 66. (d) 67..e. A = D 1 4. The network is reciprocal.. New Delhi-16 . y11 = 50 Which of the above is/are correct? (a) 2 only (b) 2 and 4 (c) 1 only (d) 1 and 3 Ans. (b) V1 = 60I1 + 20 I2 V2 = 20I1 + 40I2 Comparing (1) and (2) with standard Z-parameters equations i.in ...ESE-2016 : Electrical Engg. 3..in | Visit: www. Kalu Sarai. The network is both symmetrical and reciprocal. For the two-port network shown in the figure + I1 I2 V1 + V2 – – V1 = 60I1 + 20I2 and V2 = 20I1 + 40I2 Consider the following for the above network : 1. 2.
the required shunt resistance is (a) 10 Ω (b) 15 Ω (c) 20 Ω (c) 25 Ω Ans. D=2 Z 22 Z 22 = | Z | Z 11Z 22 − Z 12 Z 21 40 40 1 = = 60 × 40 − 20 × 20 2000 50 1 50 If the total powers consumed by three identical phase loads connected in delta and star configurations are W1 and W2 respectively. V1 = AV2 – BI2 I1 = CV2 – DI2 A=3 B = 100 1 20 A≠ D C= ∴ Y11 = = Y11 = 68. Kalu Sarai.madeeasy.. (d) Rsh = Rm ⎛ I ⎞ ⎜⎝ I − 1⎟⎠ m Corporate Office: 44-A/1.in | Visit: www. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 29 Substituting (3) in (1): ⎡ 1 ⎤ V1 = 60 ⎢ V2 − 2 I 2 ⎥ + 20 I 2 ⎣ 20 ⎦ V1 = 3V2 – 120 I2 + 20I2 V1 = 3V2 – 100I2 Comparing (3) and (4) with standard ABCD parameters equation i. For extending its range to measure 500 μA.e. .ESE-2016 : Electrical Engg. New Delhi-16 = 100 = 25 Ω 500 ⎛ ⎞ − 1⎟ ⎜⎝ 100 ⎠ | Email : info@madeeasy..in . A 100 μAammeter has an internal resistance of 100 Ω.(4) 3W2 (b) W2 3 (d) W2 3 (a) PΔ = 3PY W1 = 3 W2 69. then W1 is (a) 3W2 (c) Ans.
Test series is likely to start in August. • Post GATE exam counseling & support for M. • Opportunity to win rewards and prizes for GATE-2017 top rankers. 2016 For online admission & other details. GATE & PSUs now available at all platforms 12 + Basic Level 4 + Basic Level 12 + Advanced Level Total 4 + 4 + 30 Advanced Level GATE Mock Tests 66 Tests Features of GATE Online Test Series : • Video solutions by senior faculty. • Comprehensive and detailed analysis report of test performance. • High level of accuracy to make question papers error free.madeeasy. visit : www. • Feel of GATE exam i.All India ONLINE For TEST SERIES GATE . • Fully explained and well illustrated solutions to all the questions. Tech & PSUs. • Questions are newly designed by senior faculty and not taken from previous competitive exams.in .e it is going to be a kind of mini GATE exam. • Opportunity to compare your performance among thousands of students including MADE EASY quality students. • Quality questions exactly similar to GATE standard and patten. • Basic level and Advanced level questions are designed having subjectwise and full syllabus tests.2017 India’s Best Institute for IES.
A 200 V PMMC voltmeter is specified to be accurate within ±2% of full scale.in . (b) % Error at any reading 71. 72.in | Visit: www.ESE-2016 : Electrical Engg. New Delhi-16 | Email : info@madeeasy. (c) Corporate Office: 44-A/1. The degree to which an instrument indicates the changes in measured variable without dynamic error is (a) repeatability (b) hysteresis (c) precision (d) fidelity Ans. (d) 73. is (a) ± 8% (b) ± 4% (c) ± 2% (d) ± 1% Ans. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 30 70. s 3 + 2s + 1 s 2 + 3s + 2 (b) 1 (d) 3 (c) ⎛ s 3 + 2s + 1⎞ Since function is f(s) = ⎜ ⎝ s 2 + 3s + 2 ⎟⎠ Nothing is common between numerator and denominator so number of pole is 2. = Full scale error × Full scale value Full scale reading value = 2 × 200 = 4% 100 How many poles does the following function have? F(s) = (a) 0 (c) 2 Ans. Kalu Sarai. Which of the following devices gives the most accurate result? (a) PMMC (b) Hot-wire (c) CRO (d) Electrodynamic Ans. Loading by the measuring instruments introduces an error in the measured parameter. when the instrument is used to measure a voltage of 100 V. The limiting error.madeeasy.
Air-friction damping 2. New Delhi-16 | Email : info@madeeasy. while digital data acquisition system is used when narrow frequency width is to be monitored (c) when quantity to be monitored varies slowly.in | Visit: www. while digital data acquisition system is preferred when quantity is time-invariant Ans. 2 and 3 Ans.f.m. no error is introduced due to 1. analog data acquisition system is used (a) for narrow frequency width.in . (b) 77. During the measurement of resistance by Carey Foster bridge. In data acquisition system. while its counterpart is preferred if the quantity to be monitored varies very fast (d) when quantity to be monitored is time-variant. while digital data acquisition system is used when wide frequency width is to be monitored (b) for wide frequency width.madeeasy. Eddy-current damping PMMC type instruments use which of the above? (a) 1 only (b) 2 only (c) 3 only (d) 1. contact resistance 2. Kalu Sarai. thermoelectric e. connecting leads 3. (c) 76. (d) Corporate Office: 44-A/1. Which of the above are correct? (a) 1 and 2 only (b) 1 and 3 only (c) 2 and 3 only (d) 1. Fluid-friction damping 3. (c) 75. 2 and 3 Ans. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 31 74.ESE-2016 : Electrical Engg. Consider the following types of damping : 1. A moving-coil galvanometer can be used as a DC ammeter by connecting (a) a high resistance in series with the meter (b) a high resistance across the meter (c) a low resistance across the meter (d) a low resistance in series with the meter Ans.
Kalu Sarai. The error in the meter reading is (a) 4% (b) 8% (c) 12% (d) 16% Ans.in | Visit: www. (d) % Er = Pm − PT × 100 = tan φ × tan β × 100 PT ⎛ 2πfLP ⎞ = tan φ ⋅ ⎜⎝ R ⎟⎠ × 100 P ⎛ 2π × 50 × 0.ESE-2016 : Electrical Engg. MI instrument 2.7. loss angle measurement 3. Electrodynamometer instrument Which of the above instruments is/are free from hysteresis and eddy-current losses? (a) 1 only (b) 2 only (c) 3 only (d) 1. 3 and 4 Ans.madeeasy. The frequency of the supply is 50 c/s.5 H. and inductance 0. A wattmeter is measuring the power supplied to a circuit whose power factor is 0.in . Consider the following instruments : 1. Electrostatic instrument 3. (b) 80.5 ⎞ = tan(cos −1 0. 2 and 4 only (d) 1. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 32 78. New Delhi-16 | Email : info@madeeasy. Dummy strain gauges are used for (a) compensation of temperature ' changes (b) increasing the sensitivity of bridge (c) compensating for different expansions (d) calibration of strain gauge Ans. 2. simple balance detector like PMMC instrument 4. (c) 79. 2 and 3 Ans. capacitance value (magnitude) 2. providing safety to operators by incorporating Wagner earthing device Which of the above are correct? (a) 1 and 3 only (b) 3 and 4 only (c) 1. (a) 81.7) ⋅ ⎜ ⎟⎠ × 100 = 16% ⎝ 1000 Corporate Office: 44-A/1. Schering bridge is a very versatile AC bridge and is used for capacitor testing in terms of 1. The wattmeter has a potential coil circuit of resistance 1000 Ω.
Singh (Ex.in . GATE. experienced and trained management staff • MADE EASY is the only institute which has consistently produced Toppers in ESE. IES) CMD.Crack in1 Attempt st ESE. Office: 44-A/1. GATE & PSUs Mr.Tech admissions • Full Time Interview support for ESE & PSUs Regular updation on Vacancies/Notifications Timely completion of syllabus • Display on notice board and announcement in classrooms for vacancies notified by government departments • 4-6 hrs classes per day • Well designed course curriculum • Syllabus completion much before the examination date • Notification of ESE. Kalu Sarai (Near Hauz Khas Metro Station) New Delhi-110016 Ph: 011-45124612 www. Focused and Well planned course curriculum • Thoroughly revised and updated • Focused and relevant to exam • Comprehensive so that. there is no need of any other text book • Designed by experienced & qualified R&D team of MADE EASY • Course planning and design directly under our CMD • GATE & ESE both syllabus thoroughly covered • Course coordination and execution directly monitored by our CMD Dedication and Commitment Best Infrastructure & Support • Professionally managed • Pre-planned class schedule • Subjects completed in continuity • Well equipped audio-visual classrooms • Clean and inspiring environment • In campus facility of photocopy. PSUs and state services exams Professionally Managed & Structured Organization Maximum Selections with Top Rankers • MADE EASY has pool of well qualified. Secured and Hygienic Campus Environment Corp.madeeasy. MADE EASY Group Why most of the students prefer ! Comprehensive Coverage Best Pool of Faculty • More than 1000 teaching hours • Freshers can easily understand • Emphasis on fundamental concepts • Basic level to advanced level • Coverage of whole syllabus (Technical and Non technical) • India's best brain pool • Full time and permanent • Regular brain storming sessions and training • Combination of senior professors and young energetic top rankers of ESE & GATE Focused and Comprehensive Study B ooks Consistent. B. bookshop and canteen • Best quality teaching tools • No cancellation of classes • Starting and completion of classes on time • Co-operation and discipline Complete guidance for written and personality test Regular Assessment of Performance MADE EASY has a dedicated team which provides round the year support for • Self assessment tests (SAT) • ESE all India Classroom Test Series • GATE Online Test Series • Subject-wise classroom tests with discussion • Examination environment exactly similar to GATE & UPSC exams • Interpersonal Skills • Communication Skills • GD and Psychometric Skills • Mock Interviews Motivation & Inspiration • Motivational Sessions by experts • Interaction with ESE & GATE toppers Counseling Seminars and Guidance • Expert Guidance Support • Career counseling • Techniques for efficient learning • Post GATE counseling for M. GATE & PSUs • Highest Selections in GATE • Highest Selections in ESE Audio Visual Teaching | Hostel Support | Safe.
value = –4 A MI reads RMS value = I 02 + 1 2 I1 = 2 Corporate Office: 44-A/1. rated frequency balanced source is supplying power to a balanced three-phase load carrying a line current of 5 A at an angle of 30° lagging. The two meters will read respectively (a) –4 A and –5 A (b) 4 A and –5 A (c) –4 A and 5 A (d) 4 A and 5 A Ans. (a) W1 = = W2 = = 84. New Delhi-16 (−4)2 + 1 (3 2 )2 = 5A 2 | Email : info@madeeasy.01 Ω (c) with shunt resistance of 0. when a potential difference 10 mV is applied across its terminals.01 Ω (d) with series resistance of 0. used for measuring the power drawn by the circuit. A moving-coil instrument gives full-scale deflection of 10 mA.0001 Ω Ans. VL I L 400 VL I L 400 cos (30 – φ) × 50 × cos (30 – 30) = 2000 W cos (30 + φ) × 50 × cos(30 + 30) = 1000 W A current of −4 + 3 2 sin(ωt + 30°) A is passed through a centre zero PMMC meter and a moving-iron meter.in . The readings of the two wattmeters W1 and W2.madeeasy. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 33 82. three-phase.0001 Ω 83. the same instrument can be used (a) with shunt resistance of 0.ESE-2016 : Electrical Engg. (a) Rsh = Rm V /I = m m m −1 ⎛ I ⎞ − 1⎟ ⎜ ⎝ Im = 10 10 100 ⎠ ⎛ ⎞ − 1⎟ ⎜⎝ −3 ⎠ 10 × 10 = 0. A 400 V. Kalu Sarai.0001 Ω (b) with series resistance of 0. To measure currents up to 100 A.in | Visit: www. are respectively (a) 2000 W and 1000 W (b) 1500 W and 1500 W (c) 2000 W and 1500 W (d) 1500 W and 1000 W Ans. (c) PMMC reads Avg.
The percentage error in the calculated value of R =100 Ω (voltmeter reading 200 V/ammeter reading 2 A) is nearly A R RA = 0. (b) Rm 1000 m = R + 1 = 200 + 1 = 6 sh Corporate Office: 44-A/1. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 34 85.6 mΩ (d) decrease the resistance by 49. Kalu Sarai.ESE-2016 : Electrical Engg.1 × 120 × 5 × 10–6 = –7. A structural member is compressed to produce a strain of 5 μm/m. New Delhi-16 | Email : info@madeeasy.1 Ω A RV = 2000 Ω (a) –2% (c) 2% Ans. The nickel wire strain gauge has a gauge factor of –12.6 mΩ Ans. (b) Gauge factor = Gf = ΔR /R (Δl /l) ⎛ Δl ⎞ ΔR = Gf ⋅ R ⋅ ⎜ ⎟ ⎝ l ⎠ = –12.in | Visit: www.1.26 mΩ (c) increase the resistance by 49. The change in resistance due to compressive strain will (a) increase the resistance by 7.in .26 mΩ (b) decrease the resistance by 7.madeeasy. The values of ammeter and voltmeter resistances are 0.1 Ω and 2000 Ω respectively as shown in the figure below. The pre-stress resistance of the gauge is 120 Ω.26 mΩ –ve indicates decrease of resistance. What is the multiplying power of a shunt of 200 Ω resistance when used with a galvanometer of 1000 Ω resistance? (a) 4 (b) 6 (c) 12 (d) 20 Ans. 86. (b) –5% (d) 5% (b) % Er = − V 200 × 100 = − × 100 = –5% I Rv 2 × 2000 87.
Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 35 88.in | Visit: www. An inverse z-transform x(kT) of X(z) = 1 − e −aT (z − 1)(z − e −aT ) is (a) 1 – e–akT (c) 1 – eakT Ans. works with both planar and non-planar circuits 2.ESE-2016 : Electrical Engg. (b) 1 + e–akT (d) 1 + eakT (*) Corporate Office: 44-A/1. uses Kirchhoffs voltage law Which of the above is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Ans. Its conversion time is (a) 18 μs (b) 16 μs (c) 8 μs (d) 4. Kalu Sarai.5 μs Ans. An 8-bit successive approximation A-to-D converter is driven by a 2 MHz clock. The mesh-current method 1. 89. of bits being converted) = 1 2 × 106 × 8 = 4 ms 90.in . (b) Mesh analysis is valid only for planar networks and for its application we apply kVL. (d) Convertion time = Clock period × (No. care should be taken not to open circuit the (a) primary of a voltage transformer when the secondary is connected to the rated load (b) secondary of a voltage transformer when the primary is energized with the rated voltage (c) primary of a current transformer when the secondary is connected to the rated load (d) secondary of a current transformer when the primary is carrying the rated current Ans. In using instrument transformers.madeeasy. (d) 91. New Delhi-16 | Email : info@madeeasy.
1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 36 A system has a transfer function C (s) 4 = 2 R (s) s + 1. A large bandwidth corresponds to slow response. Kalu Sarai. 2. K . For a damping s(s + 4) (d) Open loop transfer function is k s(s + 4) since the β = 1 and damping factor = 0. Which of the above statements are correct? (a) 1 and 2 (b) 2 and 3 (c) 1 and 4 (d) 3 and 4 Ans. New Delhi-16 | Email : info@madeeasy.5 k ⎛ ⎞ so closed loop function is ⎜ 2 ⎝ s + 4s + k ⎟⎠ so and so ω0 = k 2 × 0. A large resonant peak corresponds to a small overshoot in transient response. The cut-off rate indicates the ability of the system to distinguish the signal from noise.in | Visit: www. 3. 4. (a) ⎛ −4 2 21 ⎞ 4 ⎛ ⎞ ± i⎟ The transfer function of system is ⎜⎝ 2 ⎟⎠ poles are at ⎜⎝ s + 1.6s + 4 5 5 ⎠ The 2% tolerance ban setting time is 4τ ⎛ 4 ⎞ 4 =5 so 4τ ⇒ ⎜ ⇒ 4 ⎝ ξωn ⎟⎠ 5 93. (d) 94. Resonant frequency is indicative of the speed of transient response.ESE-2016 : Electrical Engg.6s + 4 For a unit-step response and 2% tolerance band.5 × k = 4 k = 16 Corporate Office: 44-A/1.5. Solutions of Objective Paper-I | Set-A 92.madeeasy. the value of the gain K must be set to (a) 1 (b) 2 (c) 4 (d) 16 Ans. the settling time will be (a) 5 seconds (b) 4 seconds (c) 3 seconds (d) 2 seconds Ans.in . The open-loop transfer function of a unity feedback system is factor of 0. Consider the following statements with reference to the response of a control system: 1.
India’s Best Institute for IES.57 31 Saurabh Singh Lodhi 140 665 23 Ishan Shrivastava 140 709.67 56 Harmandeep Singh 148 640 3 Amit Kumar Mishra 150 766.45 39 Anubhaw Mishra 142 657 59 Sandeep Singh Olla 144 678. Personality Test Total Marks Rank Personality Test Total Marks 13 Neetesh Agrawal 150 708 9 Shruti Kushwaha 144 754.21 31 Akhil Pratap Singh 134 673 49 Aman Chawla 136 709.23 7 Sudhir Jain 140 708 11 Raman Kunwar 142 732.88 12 Pankaj Fauzdar 149 712 1 Ijaz M Yousuf 142 801.36 1 Shaik Siddhikh Hussain 135 772 14 Shyam Sundar Sharma 136 745. GATE & PSUs MADE EASY will conduct INTERVIEW GUIDANCE PROGRAM FOR ESE-2016 Soon after the announcement of written results Interview is the most crucial stage which decides the selection or rejection of the candidate.57 Dhanesh Goel 140 747.46 29 Anuj Kumar Mishra 146 675 21 Nishant Kumar 144 712.24 41 Praseed Sahu 140 653 Vedant Darbari 140 642 Vinay Kumar 140 598 24 Abhishek Verma 140 705. and over the years we have noticed that only few candidates managed to score above 120 marks.57 3 Nikki Bansal 134 761 43 Anshul Agarwal 136 713.77 Delhi Hyderabad Noida Jaipur Bhopal Lucknow Indore Bhubaneswar Pune Kolkata Patna 011-45124612 09958995830 040-24652324 09160002324 0120-6524612 08860378009 0141-4024612 09166811228 0755-4004612 08120035652 09919111168 08400029422 0731-4029612 07566669612 0674-6999888 09040999888 020-26058612 09168884343 033-68888880 08282888880 0612-2356615 0612-2356616 www. the ratio of finally selected candidates to written qualified candidates is 1:2.22 11 Ankita Gupta 146 714 18 Hitesh 142 743.44 74 Electrical Engineering Electronics & Telecommunication Engg. In previous engineering services examinations.67 36 Rohit Singh 148 659 2 Piyush Pathak 150 783.5 Obtaining 120 marks in engineering services interview is considered as impressive score.22 Rank MADE EASY Centres Name Name Name 2 Partha Sarathi Tripathy 141 772 13 20 Apurva Srivastava 140 692 60 Harshit Mittal 140 705.98 9 Sudhakar Kumar 132 718 8 Nidhi 132 754. As per the analysis. numerous candidates from MADE EASY secured more than 140 marks which is an extraordinary achievement of qualitative training and sincere efforts of the aspirants.22 22 Umesh Prasad Gupta 146 687 2 Saurabh Pratap Singh 140 791. ESE-2015 MADE EASY’s Top 10 Performers of Personality Test in all 4 Streams Civil Engineering Rank Name Mechanical Engineering Personality Test Total Marks Rank Personality Test Total Marks 1 Palash Pagaria 150 783.12 54 65 Yogendra Singh 140 676.madeeasy.88 13 Kumar Sourav 140 699 6 Pawan Jeph 140 745.in .
in | Visit: www. the closed-loop system (a) may become unstable (b) stability may improve (c) stability may not be affected (d) will become highly stable Ans. The frequency of sustained oscillation for marginal stability. If G (s)H (s) ⇒ s(s + 1)(s + 5) + 2k = 0 ⇒ s 3 + 6 s 2 + 5s + 2k = 0 Corporate Office: 44-A/1. the forward path transfer function is given by G(s) = 40 s(s + 2)(s2 + 2s + 30) The steady-state error of the system for the input (b) ∞ (d) 30t 2 (a) 0 (c) 20t 2 Ans. because the poles may go to the right half of s plane. Kalu Sarai. (a) When gain k of the system is varied from 0 to ∞ then the closed loop system may became unstable.madeeasy. 97.ESE-2016 : Electrical Engg. 5t 2 is 2 (b) G(s) = 40 s(s + 2)(s 2 + 2s + 30) since type of system is 1 so steady state error for 5t 2 will be ∞ 2 96. For a unity feedback control system. for a control system G(s)H(s) = 2K s(s + 1)(s + 5) and operating with negative feedback. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 37 95. When gain K of the open-loop transfer function of order greater than unity is varied from zero to infinity. New Delhi-16 | Email : info@madeeasy.in . (a) G(s)H(s) = 2k s(s + 1)(s + 5) for marginal stability we need to find frequency of sustained oscillation. is (b) 6 r/s (d) 6 r/s (a) 5 r/s (c) 5 r/s Ans.
lag-lead compensator Which of the above is/are. Consider the following statements : 1. 2 and 3 only (b) 3 and 4 only (c) 1. 2. lag compensator 3. 2 and 3 | Email : info@madeeasy. 2. correct? (a) 1 only (c) 3 only Corporate Office: 44-A/1. 3. New Delhi-16 (b) 2 only (d) 1. Kalu Sarai.in | Visit: www. lead compensator 2.madeeasy. Which of the above statements are correct? (a) 1. (c) Complementary root locus refer to root loci with negative k. 2 and 4 only (d) 1.in . Complementary root locus (CRL) refers to root loci with positive K. (c) (a) adding a zero lead to decrease in the angle of asymptote so push root locus to left (b) adding a pole lead to increase in the angle of asymptote so push root locus to right. (d) adding of pole in forward path transfer function increase maximum overshoot and adding a zero reduces maximum overshoot. Adding a zero to the G (s)H (s) tends to push root locus to the left.ESE-2016 : Electrical Engg. Adding a pole to the G (s)H (s) tends to push root locus to the right. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 38 Now from Rough Huswitz criteria 3 s 2 s 1 s 0 s 1 5 6 2k 30 – 2k 6 2k so k = 15 Now we get that k = 15 So 6s2 + 30 = 0 ωoscillation = 5 rad/sec 98. 3 and 4 Ans. 4. An R-C network has the transfer functions Gc(s) = s 2 + 10s + 24 s 2 + 10s + 16 The network could be used as 1. Adding a zero to the forward path transfer function reduces the maximum overshoot of the system. 99.
Kalu Sarai. (a) meters constant k= 5 No.in | Visit: www.ESE-2016 : Electrical Engg. If an energy meter makes 5 revolutions in 100 seconds. 100. –8 and zero are –4.5 1. The partial fraction expansion of the function F(z) = 4z 2 − 2 z z 3 − 5z 2 + 8z − 4 is 2 12 (a) z − 1 + (z − 2)2 (c) 2 2 12 (b) z − 1 + z − 2 + (z − 2)2 1. the meter constant is (a) 800 rev/kWh (b) 222 rev/kWh (c) 147 rev/kWh (d) 13 rev/kWh Ans.5 1 (d) z − 1 + z − 2 + (z − 2)2 Ans. of revolutions = = 800 Rev/kWhr 225 100 kW hr × 1000 3600 102. (b) 101. Corporate Office: 44-A/1. (d) Since closed loop system is having a feedback so the control action depends on output.madeeasy.5 12 + z − 1 (z − 1)(z − 2) 1. New Delhi-16 | Email : info@madeeasy. Solutions of Objective Paper-I | Set-A Ans. In a closed-loop control system (a) control action is independent of output (b) output is independent of input (c) there is no feedback (d) control action is dependent on output Ans. –6 Gc(s) = so (s + 4)(s + 6) (s + 4) (s + 6) = ⋅ (s + 2)(s + 8) (s + 8) (E55 s + 2) E55F F lead lag so Gc(s) will work as lead lag or lag lead compensation.in . when a load of 225 W is connected. 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 39 (c) Gc(s) = s 2 + 10s + 24 s 2 + 10s + 16 so poles are –2.
p.in | Visit: www. (b) For critically damped system the system should have poles which are purely real. 105. equal and negative.47 (c) 5.47 57. unequal and negative G(s) = 57.? (a) 21.3K s(s + 10) What value of K will result in a steady-state error of 1°.3)k = 1° s →0 1+ s(s + 10) k= 10 × 60 = 10. For (a) (b) (c) (d) Ans. A second-order position control system has an open-loop transfer function a critically damped system.74 (b) 10. when the input shaft rotates at 10 r. 104. The characteristic polynomial of a system can be defined as (a) denominator polynomial of given transfer function (b) numerator polynomial of given transfer function (c) numerator polynomial of a closed-loop transfer function (d) denominator polynomial of a closed-loop transfer function Ans. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 40 103.m. equal and negative complex conjugate with negative real part real.523 Ans. New Delhi-16 | Email : info@madeeasy.3 Corporate Office: 44-A/1. the closed-loop poles are purely imaginary real.madeeasy. Kalu Sarai. (b) G(s) = 57.in .3k s(s + 10) Input is 10 rpm and steady state error is 1° steady state error is given by slim →0 so so X (s) 1 + G(s )H (s) 10 × 60 s lim (57. (d) Since poles are most important to determine properties of a system so denominator of closed loop system is called characteristic polynomial of a system.23 (d) 0.ESE-2016 : Electrical Engg.
New Delhi-16 0 dB line (ω) | Email : info@madeeasy. 107. For a type-I system. (b) Gain margin is the factor by which the gain of system should be increased to drive it to marginally stable condition on drive it to oscillations. Gain margin is the factor by which the system gain can be increased to drive it to (a) stability (b) oscillation (c) the verge of instability (d) critically damped state Ans. the intersection of initial slope of bode plot with 0 dB axis give error constant for example k s(s + p) 20 log (k) slope = 20 dB/decade k 109. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 41 106. Kalu Sarai.madeeasy. Nichols' chart is used to determine (a) transient response (c) open-loop frequency response (b) closed-loop frequency response (d) settling time due to step input Ans. The desirable features of a servomotor are (a) low rotor inertia and low bearing friction (b) high rotor inertia and high bearing friction (c) low rotor inertia and high bearing friction (d) high rotor inertia and low bearing friction Ans. (b) For type-1 system.in | Visit: www. the intersection of the initial slope of the Bode plot with 0 dB axis gives (a) steady-state error (b) error constant (c) phase margin (d) cross-over frequency Ans.in . (b) 108.ESE-2016 : Electrical Engg. (a) Corporate Office: 44-A/1.
(d) ΔR / R ⎞ ⎛ ⎜⎝ Q Gf = Δ ⎟⎠ l /l Corporate Office: 44-A/1. (d) Self loop can exist in signal flow graph also. Statement (II): Lead compensator increases the margin of stability. the order of a system increases and higher the order of the system. (a) 112. Kalu Sarai. Statement (I): For type-II or higher systems. Ans. 113. Statement (I): Self-loops can exist in block diagram but not in signal flow graph.in | Visit: www. Statement (I): The gauge factor of a strain gauge is the ratio of strain to per unit change in resistance. Statement (II) : Both block diagrams and signal flow graphs are applicable to linear time-invariant systems.ESE-2016 : Electrical Engg.in . Statement (II): Poisson's effect is defined as producing less strain with opposite sign on the plane perpendicular to the applied load. one labelled as 'Statement (I) and the other as ‘Statement (II)’. Ans. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 42 Directions: Each of the following eleven (11) items consists of two statements. Examine these two statements carefully and select the answers to these items using the code given below : Code : (a) Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true 110. Ans. Statement (II): With integral control action. Statement (I): Stability of a system deteriorates when integral control is incorporated into it. more the system tends to become unstable. lead compensator may be used.madeeasy. 111. (a) Since lead compensation increases the margin of stability so we use higher order lead compensation. New Delhi-16 | Email : info@madeeasy. Ans.
(c) Closed loop system has feedback to account environment changes and became stable. New Delhi-16 | Email : info@madeeasy. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 43 114. stable operation is possible if gain is suitably reduced.ESE-2016 : Electrical Engg.in . i.. Ans. Ans. Statement (I): In type-0 and type-l systems. 116. (b) As we know that voltage is the work done per unit charge dW dQ and power is the energy per unit of time i. 115. (a) Alloy has less regular structure than a metal because of which conductivity of alloy decreases with increase in alloy content and resistivity of alloy increases in comparison to metal. Statement (II): A solid solution has a less regular structure than a pure metal. Statement (II): Any one of the compensators lag.in | Visit: www. (a) 117. Ans. lead.e. Statement (II): Closed-loop system is inaccurate as it cannot account environmental or parametric changes and may become unstable. dW dt ∴ Both are correct.e. Statement (I): The electrical conductivity of a solid solution alloy drops off rapidly with increased alloy content. Kalu Sarai. Statement (II): Power is energy per unit of time. lag-lead may be used to improve the performance. Corporate Office: 44-A/1. Ans. Statement (I): Open-loop system is inaccurate and unreliable due to internal disturbances and lack of adequate calibration.madeeasy. Statement (I): Voltage is the energy per unit charge created by charge separation.
Statement (I): Optical pyrometers are used as transducers for the measurement of flame temperature in a boiler. (b)  Corporate Office: 44-A/1.in | Visit: www. Ans. New Delhi-16 | Email : info@madeeasy. (a) 119.in . (a) 120. Statement (I): A constant temperature type hot-wire anemometer is suitable for turbulent flow measurements. Statement (I): The null voltage of an LVDT cannot be reduced to an insignificant value. Solutions of Objective Paper-I | Set-A 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 44 118. Ans. Ans. Kalu Sarai. the bandwidth is increased. Statement (II): Hall effect transducers are primarily used to measure flux density.ESE-2016 : Electrical Engg. Statement (II): When the resistance of the hot wire is kept constant by incorporating current feedback. Statement (II): Non-invasive methods are suitable for flame temperature measurement in a boiler.madeeasy.

References: V. 
 V. 
 V. 
 V. 
 V. 
 V. 
 V. 
 V.