EE Objective Paper 2 1009


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Director’s Message

UPSC has introduced the sectional cutoffs of each paper and screening cut off in

three objective papers (out of 600 marks). The conventional answer sheets of only

those students will be evaluated who will qualify the screening cut offs.

In my opinion the General Ability Paper was easier than last year but Civil

Engineering objective Paper-I and objective Paper-II both are little tougher/

lengthier. Hence the cut off may be less than last year. The objective papers of ME

and EE branches are average but E&T papers are easier than last year.

Category

Percentage

Marks

Expected Minimum Qualifying Marks in Each Paper (out of 200 Marks)OBJECTIVE

GEN

15%

30

OBC

15%

30

SC

15%

30

ST

15%

30

PH

10%

20

Category

Percentage

Marks

Expected Minimum Qualifying Marks in Each Paper (out of 200 Marks)CONVENTIONAL

GEN

15%

30

OBC

15%

30

SC

15%

30

ST

15%

30

PH

10%

20

Branch

CE

ME

EE

E&T

Expected Screening cut off out of 600 Marks (ESE 2016)

GEN

225

280

310

335

OBC

210

260

290

320

SC

160

220

260

290

ST

150

200

230

260

Note: These are expected screening cut offs for ESE 2016. MADE EASY does not

take guarantee if any variation is found in actual cutoffs.

B. Singh (Ex. IES)CMD , MADE EASY Group

MADE EASY team has tried to provide the best possible/closest answers, however if

you find any discrepancy then contest your answer at www.madeeasy.in or write your

query/doubts to MADE EASY at: [email protected]

MADE EASY owes no responsibility for any kind of error due to data insufficiency/misprint/human errors etc.


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ESE-2016 : Electrical Engg.Solutions of Objective Paper-II | Set-A

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Page3

Paper-II (Electrical Engineering)

1.1.1.1.1. Compared to the salient-pole Hydroelectric generators, the steam and the gas turbine

generators have cylindrical rotors for

(a) Better air-circulation in the machine

(b) Reducing the eddy-current losses in the rotor

(c) Accommodating larger number terms in the field winding

(d) Providing higher mechanical strength against the centrifugal stress

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)

2.2.2.2.2. Consider the following losses for short circuit test on a transformer:

1. Copper loss

2. Copper and iron losses

3. Eddy current and hysteresis losses

4. Friction and windage losses

(a) 1 only (b) 2 only

(c) 3 only (d) 2, 3 and 4

Ans.Ans.Ans.Ans.Ans. (a )(a )(a )(a )(a )

3.3.3.3.3. A 2000 V/200 V, 20 kVA, two winding single-phase transformer is reconnected as a step

up auto-transformer having 200 V/2200 V ratings. Then the power rating for the auto

transformer will be

(a) 160 kVA (b) 180 kVA

(c) 200 kVA (d) 220 kVA

Ans.Ans.Ans.Ans.Ans. (d*)(d*)(d*)(d*)(d*)

2000 V/200 V, 20 kVA, two winding T / F

Reconnected as Auto T / F → 200/2200V

200 V2000 V reconnectedas

200 V

100 V

200 V

2000 V

2200

For current distribution....

V 1 I 1 = V 2 I 2

⇒ 200 × I 1 = 110 A∴ Auto transformer rating should be

V 1

I 1

or V 2 I

2

⇒200 × 110 (or) 2200 × 10= 22000 (or) 22 kVA


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Roadmap forESE 2017 Prelims

Paper-I General Studies &Engineering Aptitude

India’s Best Institute for IES, GATE PSUs

1. Current Affairs: Current National and International issues, bilateral issues, current economic affairs,

Defence, Science and Technology, Current Government Schemes, Persons in news, Awards &

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2. Reasoning and Aptitude : Algebra and Geometry, Reasoning and Data Interpretation, Arithmetic,

coding and decoding, Venn diagram, number system, ratio & proportion, percentage, profit & loss,

simple interest & compound interest, time & work, time & distance, blood relationship, direction

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3. Engineering Mathematics : Differential equations, complex functions, calculus, linear algebra,

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Paper-I : General Studies & Engineering Aptitude

P.T.O. (Page 1 of 3)

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4. General Principles of Design, Drawing, Importance of Safety : Engineering Drawing,

Drawing instruments, drawing standard, geometric construction and curves, orthographic

projections, methods of projection, profile planes side views, projection of points, projection of

straight lines, positions of a straight line with respect to HP and VP, determining true length and

true inclinations of a straight line, rotation methods, trace of a line, projection of planes,

importance of safety etc.

5. Standards and quality practices in production, construction, maintenance and services:

ISO Standards, ISO-9000 Quality Management, ISO-14000 other, BIS Codes, ECBC, IS, TQM ME,

TPM, PDCA, PDSA, Six Sigma, 5S System, 7 Quality Control Tools , ISHIKAWAS -7QC Tools, Kaizer

Tools-3m, TQM : Most Importance, Deming's: 14 Principles, Lean Manufacturing ME, Quality

Circles, Quality Control, Sampling.

6. Basics of Energy and Environment: Renewable and non renewable energy resources, energy

conservation, ecology, biodiversity, environmental degradation, environmental pollution,

climate change, conventions on climate change, evidences of climate change, global warming,greenhouse gases, environmental laws for controlling pollution, ozone depletion, acid rain,

biomagnification, carbon credit, benefits of EIA etc.

7. Basics of Project Management: Project characteristics and types, Project appraisal and project

cost estimations, project organization, project evaluation and post project evaluation, risk

analysis, project financing and financial appraisal, project cost control etc.

8. Basics of Material Science and Engineering: Introduction of material science, classification of

materials, Chemical bonding, electronic materials, insulators, polar molecules, semi conductor

materials, photo conductors, classification of magnetic materials, ceramics, polymers, ferrous

and non ferrous metals, crystallography, cubic crystal structures, miller indices, crystal

imperfections, hexagonal closed packing, dielectrics, hall effect, thermistors, plastics,

thermoplastic materials, thermosetting materials, compounding materials, fracture, cast iron,

wrought iron, steel, special alloys steels, aluminum, copper, titanium, tungsten etc.

9. Information and Communication Technologies : Introduction to ICT, Components of ICT,

Concept of System Software, Application of computer, origin and development of ICT, virtual

classroom, digital libraries, multimedia systems, e-learning, e-governance, network topologies,

ICT in networking, history and development of internet, electronic mail, GPS navigation system,smart classes, meaning of cloud computing, cloud computing architecture, need of ICT in

education, national mission on education through ICT, EDUSAT (Education satellite), network

configuration of EDUSAT, uses of EDUSAT, wireless transmission, fibre optic cable etc.

10. Ethics and values in engineering profession: ethics for engineers, Ethical dilemma, elements

of ethical dilemmas, indian ethics, ethics and sustainability, ethical theories, environmental

ethics, human values, safety, risks, accidents, human progress, professional codes,

responsibilities of engineers etc.

Page 2 of 3Scroll down

or Answer Key of ESE 2016

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Page5

6.6.6.6.6. Consider the following advantages of a distributed winding in a rotating machine:

1. Better utilization of core as number of evenly placed small slots are used.

2. Improved waveform as harmonic emf’s are reduced.

3. Dimensional armature reaction and efficient cooling.

Which of the above advantages are correct?(a) 1 and 2 only (b) 1 and 3 only

(c) 2 and 3 only (d) 1, 2 and 3


7.7.7.7.7 . The breadth factor for 3rd harmonic emf of a 3-phase, 4-pole, synchronous machine having

36 stator slots is

(a) 0.47 (b) 0.53

(c) 0.67 (d) 0.73

Ans.Ans.Ans.Ans.Ans. (c )(c )(c )(c )(c )Breadth factor for r th harmonic:

=

α

α

α = slot angle =

° °= = °

n = slots / pole / phase= 36 / 4 / 3 = 3

∴ Breadth factor for 3rd harmonic is

=

× ×

=× ×

8.8.8.8.8. Consider the following factors for a dc machine:

1. Interpole

2. Armature resistance

3. Reduction in field current

Which of the above factors are responsible for decrease in terminal voltage of a shunt

generator?

(a) 1 and 2 only (b) 2 and 3 only

(c) 1 and 3 only (d) 1, 2 and 3

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)


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Page6

9.9.9.9.9. A dc motor develops an electromagnetic torque of 150 N-m in a certain operation

condition. From this operating condition, a 10% reduction in field flux and 50% increase

in armature current is made. What will be new value of electromagnetic torque.

(a) 225 N-m (b) 202.5 N-m

(c) 22.5 N-m (d) 20.25 N-m


T = 150 N-m

Field reduced by 10% I a increased by 50%

∴ φ2 = 0.9 φ1and I

a 2= 1.5 I

a 1

T α φ I a

=

φ×

φ I

I

∴

= 0.9 × 1.5

⇒ T 2 = 1.35 (T 1)∴ T 2 = 1.35 × 150

= 202.5 N-m

10.10.10.10.10. A dc machine having a symmetrical closed-circuit armature winding and a sinusoidal

air gap flux density distribution, will have a sinusoidal voltage induced in the individual

coils. The resultant brush-to-brush voltage will have a waveform

(a) Sinusoidal with the negative half(b) Unidirectional and constant without

(c) Unidirectional and constant with ripples superimposed

(d) Sinusoidal positive half and zero negative half, in each cycle

Ans.Ans.Ans.Ans.Ans. (c )(c )(c )(c )(c )

11.11.11.11.11. A 3-phase, induction motor operating at a slip of 5% develop 20 kW rotor power output.

What is the corresponding rotor copper loss in this operating condition?

(a) 750 W (b) 900 W

(c) 1050 W (d) 1200 W


Slip = 0.05

Rotor power output = 20 kW

Rotor copper loss = Slip × Rotor power input

∴ Rotor power output = (1 – s ) Rotor power input


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Page7

∴ Rotor power input = rotor power input / 1 – s = 20 kW/ 1 – 0.05

= 21.0526 kW

∴ Rotor cu. loss = (0.05) (21.05) kW

≈ 1050 watts

12.12.12.12.12. What are the signs of load angle in an alternator during generator and motor operations,

respectively?

(a) Negative, negative (b) Positive, negative

(c) Negative, positive (d) Positive, positive


13.13.13.13.13. In an alternator the armature winding is kept stationary while the field winding is keptrotating for the following reasons:

1. Armature handles very large current and high voltage.

2. Armature fabrication, involving deep slots to accommodate large coils, is easy if

armature is kept stationary.

3. It is easier to cool the stator than the rotor.


(c) 2 and 3 only (d) 1, 2 and 3


14.14.14.14.14. Increasing the air gap of a squirrel-cage induction motor would result in

(a) increasing in no-load speed

(b) increase in full-load power factor

(c) increase in magnetizing current

(d) maximum available torque


15.15.15.15.15. A cumulative compound dc motor runs at 1500 rpm on full load. If its series field is

short circuited, its speed(a) becomes zero (b) remains same

(c) increases (d) decreases


\


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Batches MADE EASY offers rank improvement batches for ESE 2017 & GATE 2017. These batches are designed for repeater students who havealready taken regular classroom coaching or prepared themselves and already attempted GATE/ESE Exams , but want to give next attempt

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time frame. The selection of questions will be such that the Ex. MADE EASY students are best benetted by new set of questions.

Eligibility :Features :

Syllabus Covered : Technical Syllabus of GATE-2017 & ESE-2017 Course Duration : Approximately 25 weeks (400 teaching hours)

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16.16.16.16.16. If the capacitor of a capacitor start single-phase motor fails to open when the motor picks

up speed.

(a) The motor will stop.

(b) The auxiliary winding will be damaged.

(c) The capacitor will be damaged.(d) The winding will be damaged.


17.17.17.17.17. For a 3-phase induction motor, what fraction/multiple of supply voltage is required for

a direct on line starting method such that starting current is limited to 5 times the full

load current and motor develops 1.5 times full load torque at starting time?

(a) 1.632 (b) 1.226

(c) 0.816 (d) 0.456


Let ‘ x’ be the fraction of supply votlage

∴ T st α ( xv )2 and T f α v

2

T st → starting torque

T f → full load torque

=

x

⇒ x =

=

∴ x = 1.22

18.18.18.18.18. What is the material of slip rings in an induction machine?

(a) Carbon (b) Nickel

(c) Phosphor bronze (d) Manganese


19.19.19.19.19. The stator loss of a 3-phase induction motor is 2 kW. If the motor is running with a slip

of 4% and power input of 90 kW, then what is the rotor mechanical power developed?

(a) 84.48 kW (b) 86.35 kW

(c) 89.72 kW (d) 90.52 kW


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Page9


Stator loss = 2 kWSlip = 0.04

Power input – stator loss = Stator ouput = Rotor input∴ Rotor input = 90 kW – 2 kW = 88 kW

Rotor power output = (1 – s ) Rotor power input∴ Rotor power input = (1 – 0.04) (88)

= 84.48 kW

20.20.20.20.20. In a single-phase capacitor start induction motor, the direction of rotation

(a) can be changed by reversing the main winding terminals.

(b) cannnot be changed.

(c) is dependent on the size of the capacitor.

(d) can be changed only in large capacitor motors.


21.21.21.21.21. Air pollution due to smoke around a thermal power station can be reduced by installing

(a) Induced draft fan (b) Super heater

(c) Economizer (d) Electrostatic precipator


22.22.22.22.22. The load curve is useful in deciding.

1. The operating schedule of generating units.

2. The total installed capacity.

Which of the above statements is/are correct?


(c) Both 1 and 2 (d) Neither 1 nor 2


23.23.23.23.23. The maximum demand on a steam power station is 480 MW. If the annual load factor

is 40%, then the total energy generated annually is

(a) 19819.2 × 105 kWh (b) 18819.2 × 105 kWh

(c) 17819.2 × 105 kWh (d) 16819.2 × 105 kWh


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Page10


Load factor =

Average power = 0.4 × 480 = 192 mW

= 192×103 kWTotal energy generated in one year

= 192 × 103 × 365 × 24 kW hr

= 16819.2 × 106 kWh

1.1.1.1.1.

24.24.24.24.24. To equalize the sending and receiving end voltages, impedance is connected at the

receiving and of a transmission line having the following ABCD parameters:

A = D = 0.9∠0° B = 200∠90° ΩThe impedance so connected would be

(a) 1000∠0° Ω (b) 1000∠90° Ω(c) 2000∠90° Ω (d) 2000∠0° Ω


z

T L⋅

Equivalent T-parameter =

Aeq

=

β+ =

∠∠ +

for V S = V R

Aeq

=

∠= ∠ +

⇒ z = 2000∠90°

25.25.25.25.25. The maximum efficiency in the transmission of bulk ac power will be achieved when the

power factor of the load is

(a) slightly less than unity lagging

(b) slightly less than unity leading

(c) unity

(d) considerably less than unity


For max. Power Transfer Condition load should be of leading nature.

1.1.1.1.1.


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1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

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1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

Page11

26.26.26.26.26. A speed of a dc motor is

(a) directly proportional to back emf and inversely proportional to flux.

(b) inversely proportional to back emf and directly proportional to flux.

(c) directly proportional to back emf as well as to flux.

(d) inversely proportional to back emf as well as to flux.


α φ

27.27.27.27.27. When the sending end voltage and current are numerically equal to the receiving end

voltage and current respectively, then the line is called

(a) a tuned line (b) a transposed line

(c) a long line (d) a short line


V S = V R , S S = S R For a tuned line.

1.1.1.1.1.

28.28.28.28.28. If V m is the peak value of an applied voltage in a half wave rectifier with a large capacitor

across the load, then the peak inverse voltage will be

(a) 0.5 V m (b) V m (c) 1.5 V

m (d) 2.0 V

m

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d) A

V M

V S +

–

K

LO AD

+

V M

–

The steady state voltage across load V M [due to large capacitor]

∴ PIV = 2 V M

29.29.29.29.29. A 100 MVA generator operates load of 50 Hz frequency. The load is suddenly reduced

to 50 MW. The steam valve begins to close only after 0.4 and if the value of the inertia

constant H is 5s, then the frequency at 0.4s is nearly

(a) 38 Hz (b) 44 Hz

(c) 51 Hz (d) 62 Hz


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Page12


f n =

+ ∆

=

× + × ×

= 51 Hz

1.1.1.1.1.

30.30.30.30.30. A 25 MVA, 33 kV transformer has a pu impedance of 0.9. The pu impedance at a new

base 50 MVA at 11 kV would be

(a) 10.4 (b) 12.2

(c) 14.4 (d) 16.2


Z pun

=

× ×

=

× × = 16.2 pu

31.31.31.31.31. Symmetrical components are used in power system for the analysis of

(a) balanced 3-phase fault

(b) unbalanced 3-phase fault

(c) normal power system under steady conditions

(d) stability of system under disturbance


1.1.1.1.1.

32.32.32.32.32. For V-curves for a synchronous motor the graph is drawn between

(a) terminal voltage and load factor

(b) power factor and field current

(c) field current and armature current

(d) armature current and power factor


33.33.33.33.33. Critical clearing angle is related to(a) stability study of power system

(b) power flow study of power system

(c) regulation of transmission line

(d) power factor improvement of the system


1.1.1.1.1.


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Page13

34.34.34.34.34. A 2-pole, 50 Hz, 11 kV, 100 MW alternator has a moment of inertia of 10,000 kg.m2.

The value of inertia constant, H is

(a) 3.9s (b) 4.3s

(c) 4.6s (d) 4.9s


M =

− ω ⋅ × = π

− × π × × × =

×π ×

H = 4.93

35.35.35.35.35. Stability of a power system can be improved by:

1. Using series compensations2. Using parallel transmission lines

3. Reducing voltage of transmission

Which of the above statements are correct?


(c) 2 and 3 (d) 1 and 2


1.1.1.1.1.

36.36.36.36.36. Equal-area criterion is employed to determine:

(a) The steady-state stability(b) The transient stability

(c) The reactive power limit

(d) The rating of a circuit breaker


37.37.37.37.37. Consider the following advantages with respect to HVDC transmission:

1. Long distance transmission

2. Low cost of transmission

3. Higher efficiencyWhich of the above advantages are correct?


(c) 2 and 3 only (d) 1, 2 and 3


If the line length is more than 600 km HVDC is more economical than HVAC. So, HVDC

is economical to transmit Bulkpower for long distance.

1.1.1.1.1.


19/48


SUPER T ALENT B ATCHESSUPER T ALENT B ATCHESfor ESE, GATE & PSUs

Super Talent batches are designed for students with good academic records and who have secured good ranks in GATE/ESE or other nationallevel competitive examinations. Super Talent batches are a kind of regular batches in which faculty, study material, tests, teaching pedagogy is

similar to other batches. But due to eligibility criteria, the composition of students in this batch is homogeneous and better than other batches.

Here students will get a chance to face healthy competitive environment and it is very advantageous for ambitious aspirants.

Eligibility (any one of the following)

BITS-Pilani, BIT-Sindri, HBTI-Kanpur, JMI-Delhi, NSIT-Delhi, MBM-Jodhpur, Madan Mohan Malviya-Gorakhpur, College of Engg.-Roorkee,

BIT-Mesra, College of Engg.-Pune, SGSITS-Indore, Jabalpur Engg. College, Thapar University-Punjab, Punjab Engg. College

• 65% marks in B.Tech from reputedcolleges (See below mentioned colleges)

• 70% marks in B.Tech fromprivate engineering colleges

• 60% marks in B.Tech from IIT’s/NIT’s/DTU

• Cleared any 3 PSUs written exam

• Cleared ESE written exam• GATE Qualified MADE EASY old students

• MADE EASY repeater students with 65% Marks in B.Tech

• GATE rank upto 2000

Why most brilliant students prefer Super Talent Batches!

• Highly competitive environment • Meritorious student’s group • Classes by senior faculty members

• Opportunity to solve more problems • In-depth coverage of the syllabus • Discussion & doubt clearing classes

• More number of tests • Motivational sessions • Special attention for better performance

Note: 1. Final year or Pass out engineering students are eligible.

Candidate should bring original documents at the time of admission. (Mark sheet, GATE Score card, ESE/PSUs Selection Proof, 2 Photographs & ID Proof). 2.

Admissions in Super Talent Batch is subjected to verification of above mentioned documents. 3.

GATE-2016 : Top Rankersfrom Super talent batches

Agam Kumar Garg

EC

1AIR

Nishant Bansal

ME

1AIR

Arvind Biswal

EE

2AIR

Udit Agarwal

EE

3AIR

Piyush Kumar

CE

4AIR

Sumit Kumar

ME

8AIR

Stuti Arya

EE

10AIR

Brahmanand

CE

10AIR

Rahul Jalan

CE

10AIR

9 in Top 10 58 in Top 100

ESE-2015 : Top Rankersfrom Super talent batches

Amit Kumar Mishra

CE

3AIR

Anas Feroz

EE

6AIR

Kirti Kaushik

CE

7AIR

Aman Gupta

CE

8AIR

Mangal Yadav

CE

9AIR

Arun Kumar

ME

10AIR

6 in Top 10 Total 47 selections

ADMISSIONS OPENOnline admission facility available

To enroll, visit : www.madeeasy.inNote: Super Talent batches are conducted only at Delhi Centre

Civil /Mechanical

Electrical /Electronics

Stream

30th May (Morning) &

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www.madeeasy.inCorp. Office : 44 - A/1, Kalu Sarai, New Delhi - 110016; Ph: 011-45124612, 09958995830


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Page14

38.38.38.38.38. The three sequence voltages at the point of fault in a power system are found to be

equal. The nature of the fault is

(a) L – G (b) L – L – L

(c) L – L (d) L – L – G


39.39.39.39.39. A distance relay with inherent directional property is known as

(a) Buchholtz relay

(b) Admittance relay

(c) Directional over current relay

(d) Directional switched relay


40.40.40.40.40. Consider the following circuit breakers for 220 kV substations:

1. Air

2. SF6

3. Vaccum

Which of the above circuit breakers can be used in an indoor substation?

(a) 1, 2 and 3 (b) 1 only

(c) 2 only (d) 3 only


41.41.41.41.41. A semiconductor differs from a conductor in that it has

(a) only one path for the free electrons in the valence band.

(b) only one path for holes in the conductance band.

(c) two paths followed by free electrons and holes, one an ordinary path in the conduction

band and the other one an extraordinary path in the valence band, respectively.

(d) Two paths followed by free electrons and holes, one an extraordinary path in the

conduction band and the other one an ordinary path in valence band, respectively.


42.42.42.42.42. Which of the following circuits is used for converting a sine wave into a square wave?

(a) Monostable multivibrator (b) Bistable multivibrator

(c) Schmitt trigger circuit (d) Darlington complementary pair



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1 2 3 4 5 6 7 8 9 0

1 2 3 4 5 6 7 8 9 0

Page15

43.43.43.43.43. What is the type of breakdown that occurs in a Zener diode having breakdown voltage

(6 V)?

(a) Avalanche breakdown only

(b) Zener breakdown only

(c) Avalanche breakdown where breakdown voltage is below 6 V and Zener breakdownotherwise

(d) Zener breakdown where breakdown voltage is below 6 V and Avalanche breakdown

otherwise


44.44.44.44.44. Consider the following statements:

A power supply uses bridge rectifier with a capacitor input filter. If one of the diodes

is defective, then

1. The dc load voltage will be lower than its expected value.2. Ripple frequency will be lower than its expected value.

3. The surge current will increase considerably.



(c) 2 and 3 only (d) 1, 2 and 3


45.45.45.45.45. The lowest frequency of ac components in the outputs of half-wave and full-wave rectifiers

are, respectively, (where ω is the input frequency)(a) 0.5 ω and ω (b) ω and 2 ω

(c) 2 ω and ω (d) ω and 3 ω


1-1-1-1-1-φφφφφ HWR:HWR:HWR:HWR:HWR: V 0↑

π

2π

Fourier Series

V 0(t ) =

+ ⋅ ω − ω − ω

π π π....

∴ Lowest frequency of ac componetns is ‘ω ’


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Page16

1-1-1-1-1-φφφφφ FWR:FWR:FWR:FWR:FWR: V 0↑

π 2π

ω t

Fourier Series:

V 0(ω t ) =

− ω − ω

π π π......

∴ Lowest frequency of ac component is ‘2ω ’

46 .46.46.46.46. A half-wave rectifier circuit using ideal diode has an input voltage of 20 sinω t volt. Thenaverage and rms value of output voltage respectively are

(a)

π (b)

π

(c)

π

(d)

π


1-1-1-1-1-φφφφφ HWR:HWR:HWR:HWR:HWR: V 0↑

π2π

ω t

V 0

=

+ ∝π

α = 0, V 0 = =

π π

V or

=

= =

47.47.47.47.47. For a BJT; I C = 5 mA, I B = 50 µA and I CBO = 0.5 µA, then the value of β is

(a) 99 (b) 91(c) 79 (d) 61


I C

= β I B + β I CBO

β =

−

−×

= =+ ×

I

I I


23/48

MADE EASY Students Top in GATE-2016

METop 10

9in

A I R Nishant Bansal Gaurav Sharma

AIR-2

Manpreet Singh

AIR-2

Sayeesh T. M.

AIR-4

Aakash Tayal

AIR-5

Tushar

AIR-7

Sumit Kumar

AIR-8

Amarjeet Kumar

AIR-6

Suman Dutta

AIR-10

EETop 10

10in

Arvind Biswal

AIR-2

Udit

AIR-3

Keshav

AIR-4

Tanuj Sharma

AIR-5

Arvind

AIR-6

Rajat Chaudhary

AIR-7

Sudarshan

AIR-8

Rohit Agarwal

AIR-9

Stuti Arya

AIR-10

A I R Anupam Samantaray

CETop 10

8in

Piyush Kumar Srivastav

AIR-4

Roopak Jain

AIR-6

Jatin Kumar Lakhmani

AIR-8

Vikas Bijarniya

AIR-8

Brahmanand

AIR-10

Rahul Jalan

AIR-10

Rahul SIngh

AIR-3

Kumar Chitransh A I R

ECTop 10

6in

K K Sri Nivas

AIR-2

Jayanta Kumar Deka

AIR-3

Amit Rawat

AIR-5

Pillai Muthuraj

AIR-6

Saurabh Chakraberty

AIR-10

A I R Agam Kumar Garg

INTop 10

9in

A I R Harshvardhan Sinha Avinash Kumar

AIR-2

Shobhit Mishra

AIR-3

Ali Zafar

AIR-4

Rajesh

AIR-7

Chaitanya

AIR-8

Palak Bansal

AIR-10

Saket Saurabh

AIR-10

Shubham Tiwari

AIR-8

53Selections in Top 101 Ranks inst

ME, EE, EC, CS, IN & PI 368 Selections in Top 10096 Selections in Top 20

METop 20 Top 100

17Selections Selections

68

CETop 20 Top 100


65

ECTop 20 Top 100


45

EETop 20 Top 100


76

CS & PITop 20 Top 100


53

INTop 20 Top 100


61

CSTop 10

6in

Himanshu Agarwal

AIR-4

Jain Ujjwal Omprakash

AIR-6

Nilesh Agrawal

AIR-10

Sreyans Nahata

AIR-9

Debangshu Chatterjee

AIR-2

A I R Ankita Jain

Akash Ghosh

AIR-4

Niklank Kumar Jain

AIR-7

Shree Namah Sharma

AIR-8

Agniwesh Pratap Maurya

AIR-8

PITop 10

5

A I R Gaurav Sharma

in

etailed results are available at

www m dee sy in

“MADE EASY is the only institute which has consistently produced toppers in ESE & GATE”


24/48


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Page18

52.52.52.52.52 . A clamper circuit

1. Adds or subtracts a dc voltage to or from a waveform.

2. Does not change the shape of the waveform.


(a) 1 only (b) 2 only(c) Both 1 and 2 (d) Neither 1 nor 2

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)

53.53.53.53.53. The operational amplifier circuit shown in figure having a voltage gain of unity has

(a) High input impedance and high output impedance

(b) High input impedance and low output impedance

(c) Low input impedance and low output impedance

+

–v

iv 0

(d) Low input impedance and high output impedance



1. Race-around condition occurs in a JK flip-flop when the inputs are 1, 1.

2. A flip-flop is used to store one bit of information.

3. A transparent latch consists of D-type flip-flops.

4. Master-slave configuration is used in a flip-flop to store 2-bits of information.


(a) 1, 2 and 3 only (b) 1, 2 and 4 only

(c) 3 and 4 only (d) 1, 2, 3 and 4

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)

A Master-slave configuration stores only 1 bit of information.

55.55.55.55.55. An operational amplifier has a slew rate of 2 V/ µ sec. If the peak output is 12 V, what

will be the power bandwidth?

(a) 36.5 kHz (b) 26.5 kHz

(c) 22.5 kHz (d) 12.5 kHz


f m

=

×= =

π × × π ×


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Page19

56.56.56.56.56. A voltage follower is used as

1. An isolation amplifier

2. A buffer amplifier


(a) 1 only (b) 2 only(c) Both 1 and 2 (d) Neither 1 nor 2


57.57.57.57.57. If a , b , c are 3 input variables, then Boolean function y = ab + bc + ca represents

1. A 3 input majority gate

2. A 3 input minority gate

3. Carry output of a full adder

4. Product circuit for a , b and c

Which of the above statements are correct?(a) 2 and 3 (b) 2 and 4

(c) 1 and 3 (d) 1 and 4


The following truth table represents a 3-input majority gate as well as the carry output

of a full adder.

a b c y

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

0

0

1

0

1

1

1

Simplification using K-map,

bc

bc bc bc bc a

a

a 1 1 1

1

y ab bc ca= + +

58.58.58.58.58. In a 2 input CMOS logic gate, one input is left floating i.e. connected neither to ground

nor to a signal. What will be the state of that input?

(a) 1 (b) 0

(c) same as that of the other input (d) indeterminate (neither 1 nor 0)



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MADE EASY Students TOP in ESE-2015

MADE EASY selections in ESE-2015 : 82% of Total Vacancies

ME Selections in Top 10 10 18 83 99 83%Selections in Top 20 MADE EASY Selections Out of Vacancies MADE EASY Percentage

CE 10 20 120 151 79%Selections in Top 10 Selections in Top 20 MADE EASY Selections Out of Vacancies MADE EASY Percentage

EE 9 16 67 86 78%Selections in Top 10 Selections in Top 20 MADE EASY Selections Out of Vacancies MADE EASY Percentage

E T 9 19 82 98 84%& Selections in Top 10 Selections in Top 20 MADE EASY Selections Out of Vacancies MADE EASY Percentage

352Selections

out of total

434

73in

Top 20

38in

Top 10

4 Streams Toppers4

All 4 MADE EASY

Students

Ashok Bansal

AIR 2

Aarish Bansal

AIR 3

Vikalp Yadav

AIR 4

Naveen Kumar

AIR 5

Raju R N

AIR 6

Sudhir Jain

AIR 7

Bandi Sreenihar

AIR 8

Kotnana Krishna

AIR 9

Arun Kr. Maurya

AIR 10

AIR

PratapME

Partha Sarathi T.

AIR 2

Anas Feroz

AIR 6

Amal Sebastian

AIR 7

Vishal Rathi

AIR 10

Sudhakar Kumar

AIR 9 AIR 8

Dharmini SachinNagendra Tiwari

AIR 5

Nikki Bansal

AIR 3

S. Siddhikh Hussain

AIR

EE

Siddharth S.

AIR 3

Piyush Vijay

AIR 4

Kumbhar Piyush

AIR 7

Shruti Kushwaha

AIR 9

Anurag Rawat

AIR 10

Nidhi

AIR 8

Saurabh Pratap

AIR 2

Manas Kumar Panda

AIR 5

Ijaz M Yousuf

AIR

E&T

Piyush Pathak

AIR 2

Amit Kumar Mishra

AIR 3

Amit Sharma

AIR 4

Dhiraj Agarwal

AIR 5

Pawan Jeph

AIR 6

Kirti Kaushik

AIR 7

Aman Gupta

AIR 8

Mangal Yadav

AIR 9

Aishwarya Alok

AIR 10

Palash Pagaria

AIR

CE

ME Top 10Selections in

10

EETop 10

Selections in

9

E&TTop 10

Selections in

9

CETop 10

Selections in

10

etailed results are available at

www m dee sy in

“MADE EASY is the only institute which has consistently produced toppers in ESE & GATE”


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Page20

59.59.59.59.59. The expression for MOD number for a ripple counter with N flip-flop is

(a) N (b) 2 N

(c) 2N – 1 (d) 2N – 1


MOD number of a ripple counter represents the total number of states followed by it

with N flip-flops, the total possible number of states that could be followed is 2 N.

60.60.60.60.60. Why a ROM does not have data inputs?

(a) It does not have a WRITE operation.

(b) Data inputs are integrated with data outputs

(c) Data inputs are integrated with address inputs.

(d) ROM is sequentially accessed.


ROM ⇒ Read only memory, so no write operation.


1. RAM is a non-volatile memory.

2. RAM is a volatile memory whereas ROM is a non-volatile memory.

3. Both RAM and ROM are volatile memories but in ROM data is nor when power is

switched off.



(c) 3 only (d) none of the above


RAM ⇒ Volatile, ROM ⇒ Non-volatileVolatile ⇒ Data only with power supplyNon-volatile ⇒ Data is present permanently.

62.62.62.62.62. Consider the following instructions:

1. LOCK 2. STD

3. HLT 4. CLI

Which of the above are machine control instructions?(a) 1 and 4 (b) 1 and 3

(c) 2 and 3 (d) 2 and 4


HLT & LOCK are for internal machine control operations of NP.

HLT ⇒ To stop NP execution of programLOCK ⇒ Related to 8086, instruction with lock prefix is executed first.


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Page21

63.63.63.63.63. What is the assembler directive statement used to reserve an array of 100 words in

memory and initialize all 100 words with 0000 and give it a name STORAGE?

(a) STORAGE DW 100 (b) STORAGE DW 100 DUP (0)

(c) STORAGE DW 100 DUP (?) (d) STORAGE DB 100


Words so DW ⇒ Define words i.e., 2 bytes of memory locations are reserved.But for array STORAGE DW 100 DUP (0)

DUP (0)⇒Means 100 words space is reserved with data in location initialized as 0000H.


1. Auxiliary carry flag is used only by the DAA and DAS instruction.

2. Zero flag is set to 1 if the two operands compared are equal.

3. All conditional jumps are long type jumps.

Which of the above statements are correct?(a) 1, 2 and 3 (b) 1 and 2 only

(c) 1 and 3 only (d) 2 and 3 only


AC ⇒ Auxiliary carry flags staus is used only in DAA { DASDecimal adjust after additions {subtraction}

All conditional jumps are short jumps z = 1 if data is same when compared.

65.65.65.65.65. If a 3-phase slip ring induction motor is fed from the rotor side with stator winding short

circuited, then frequency of currents flowing in the short circuited stator is(a) slip × frequency

(b) supply frequency

(c) frequency corresponding to rotor speed

(d) zero


66.66.66.66.66. The reversing of a 3-phase induction motor is achieved by

(a) Y-∆ starter

(b) DOL starter(c) Auto transformer

(d) Interchanging any two of the supply line



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67.67.67.67.67. Consider the following interrupts for 8085 microprocessor:

1. INTR 2. RST 5.5

3. RST 6.5 4. RST 7.5

5. TRAP

If the interrupt is to be vectored to any memory location then which of the above interruptis/are correct?

(a) 1 and 2 only (b) 1, 2, 3 and 4

(c) 5 only (d) 1 only


INTR is non-vectored so no spacific location.

68.68.68.68.68. The instruction JNC 16-bit refers to jump to 16-bit address if

(a) Sign flag is set (b) CY flag is reset

(c) Zero flag is set (d) Parity flag is reset


JNC 16 bit address

Jump if no carry/ if CY = 0 i.e., reset.

69.69.69.69.69. Consider the symbol shown below.

What function does the above symbol represent in a program flow chart?

(a) A process (b) Decision making

(c) A subroutine (d) Continuation


⇒ Symbol for subroutine.

70.70.70.70.70. Which one of the following statements is correct regarding the instruction CMP A?

(a) Compare accumulator with register A

(b) Compare accumulator with memory

(c) Compare accumulator with register H

(d) This instruction does not exist


CMPA ⇒ Compare accumulator with itself i.e., register ‘A’.


31/48


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71.71.71.71.71. The instruction RET executes with the following series of machine cycle

(a) Fetch, read, write (b) Fetch, write, write

(c) Fetch, read, read (d) Fetch, read


RET ⇒ Unconditional returns1 B instruction, 3 machine cycles (Fetch, Read, Read), 10 T-states.

Read operations are from state memory.

72.72.72.72.72. Consider the following circuits:

1. Full adder 2. Half adder

3. JK flip-flop 4. Counter

Which of the above circuits are classified as sequential logic circuits?

(a) 1 and 2 (b) 3 and 4

(c) 2 and 3 (d) 1 and 4


Full adder — Combinational

Half adder — combinational

JKFF — Sequential

Counter — sequential

73.73.73.73.73. When a peripheral is connected to the microprocessor in input/output mode, the data

transfer takes place between(a) any register and I/O device (b) memory and I/O device

(c) accumulator and I/O device (d) HL register and I/O device


In I/O mapped I/O data transfer takes place only between accumulator and I/O device.

74.74.74.74.74. While execution of I/O instruction takes place, the 8-bit address of the port is placed

on

(a) lower address bus

(b) higher address bus(c) data bus

(d) lower as well as higher order address bus


In IN and OUT instructions, as I/O has 8 bit address in last machine cycle address

is placed on A15 – A8 (AD 7 – AD 0).


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75.75.75.75.75. The port C of 8255 can be configured to work in

(a) mode 0, mode 1, mode 2 and BSR

(b) mode 0, mode 1 and mode 2

(c) mode 2 and BSR

(d) BSR mode only


Only port C can be used in BSR mode i.e., bit set reset in mode 0 as only port i/p

or o/p mode 1 and 2 for control signals.


1. Semiconductor memories are organized as linear array of memory locations.

2. To address a memory location out of N memory locations, at least log N bits of

address are required.

3. 8086 can address 1048576 addresses.4. Memory for an 8086 is set up as two banks to make it possible to read or write

a word will one machine cycle.



(c) 3 and 4 only (d) 1, 2, 3 and 4


All are true.

In 8086 two banks are used as data bus is 16 bits wide i.e., each bank contains

one byte.

77.77.77.77.77. The sticker over the EPROM window protects the chip from

(a) infrared light from sunlight

(b) UV light from fluorescent lights and sunlight

(c) magnetic field

(d) electrostatic field


EPROM ⇒ Erasable Programmable Read Only Memory.A crystal window is present on the chip so that data can be erased by exposing chip

to UV light after removing chip out of board.

78.78.78.78.78. A 2400/240 V, 200 kVA, single-phase transformer has a core loss of 1.8 kW at rated

voltage. Its equivalent resistance is 1.1%. Then the transfer efficiency at 0.9 power factor

and on full load is

(a) 95.60% (b) 96.71%

(c) 97.82% (d) 98.93%


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Equivalent resistance = 1.1%

% copper loss = % resistance drop

∴ copper loss = 1.1% of output

= 0.011 (200) = 2.2

η =

=

×× + +

= 97.82%

79.79.79.79.79. The 8259 A programmable interrupt controller in cascade mode can handle interrupt of

(a) 8 priority levels (b) 16 priority levels

(c) 32 priority levels (d) 64 priority levels


In 8259A i.e., pic a maximum of 64 I/O devices can be connected in interrupt mode.

By using nine 8259, 64 I/O’s are possible in cascaded mode.

80.80.80.80.80. 8259 A programmable interrupt controller uses the following initialization commands:

1. ICW1

2. ICW2

3. ICW3 4. ICW4If 8259 A is to be used in cascaded and fully nested mode, the ICW1 bits D

0 and D

1are

(a) 0 and 0 (b) 1 and 0

(c) 0 and 1 (d) 1 and 1


In ICW, bits , D0

D1

0 ICW not needed→ 4

1 ICW needed

0 Cascade mode

1 Single mode

→

→

→

4

So, D 0

and D 1

should be 1 and 0.

81.81.81.81.81. The induced emf in the armature conductor of a D.C. machine is

(a) Sinusoidal (b) Trapezoidal

(c) Rectangular unidirectional (d) Triangular



35/48

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82.82.82.82.82. If a carrier of 100% modulated AM is suppressed before transmission, the power saving

is nearly

(a) 50% (b) 67%

(c) 100% (d) 125%


For AM; with µ = 1, 66.6% of AM power distributed to carrier.

83.83.83.83.83. An FM signal is represented by v = 12 sin (6 × 108t + 5 sin 1250t ). The carrier frequency

f c and frequency deviation δ, respectively, are

(a) 191 MHz and 665 Hz (b) 95.5 MHz and 995 Hz

(c) 191 MHz and 995 Hz (d) 95.5 MHz and 665 Hz


V = 12 sin (6 × 108t + 5 sin 1250t )

Standard FM expression is

s (t ) = Ac

cos (2πf c t + β sin2πf

m t )

By comparison ⇒ 2πf c

= 6 × 108

f c

= 95.5 MHz

β = 5; 2πf m

= 1250

∆f = βf m

= 995 Hz

84.84.84.84.84. When the modulation frequency is doubled the modulation index is halved and the

modulating index is halved and the modulation voltage remains constant. This happens

when the modulating system is(a) AM (b) PM

(c) FM (d) Delta Modulation


For FM; β =

For FM; if modulating frequency is doubled then modulation index is halfed.

85.85.85.85.85. v = A sin (ω c t + m sin ω m t ) is the expression for(a) Amplitude modulated signal (b) Frequency modulated signal

(c) Phase modulated signal (d) Carrier signal used for modulation


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Ans.Ans.Ans.Ans.Ans. (b or c)(b or c)(b or c)(b or c)(b or c)

Given V = A sin (ω c t + m sinω m t )Standard FM expression ⇒ s (t ) = Ac sin[ω c t + 2πk f ∫ m (t )dt ]Standard PM expression ⇒ s (t ) = Ac sin[ω c t + kpm (t )]

If m (t ) = Am cosω m t then given expression is FM.If m (t ) = Am sinω m t , then given expression is PM.

86.86.86.86.86. The four basic elements in a PLL are loop filter, loop amplifier, VCO and

(a) Up converter (b) Down converter

(c) Phase detector (d) Frequency multiplier


87.87.87.87.87. In a frequency modulated (FM) system, when the audio frequency is 500 Hz and audiofrequency voltage is 2.4 V, the frequency deviation δ is 4.8 kHz. If the audio frequencyvoltage is now increased to 7.2 V then what is the new value of deviation?

(a) 0.6 kHz (b) 3.6 kHz

(c) 12.4 kHz (d) 14.4 kHz


Given: ∆f = 4.8 kHz = k fA

m; A

m = 2.4 V

If message amplitude is 7.2 Volts

i.e., A′ m = 3Am Now ∆f ′ = 3 × 4.8 = 14.4 kHz

88.88.88.88.88. Modulation is used to

1. Separate different transmission

2. Reduce the bandwidth requirement

3. Allow the use of practicable antennas

4. Ensure that intelligence may be transmitted over long distance



(c) 2 and 4 only (d) 1, 2, 3 and 4



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Page28

89.89.89.89.89. Carson’s rule is (with symbols having their standard meaning)

(a) B = 2 DW (b) B = 2 (D + 1) W

(c) = + (d) =

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Bandwidth of FM is (β + 1) 2f m

90.90.90.90.90. Consider the following features of FM vis-a-vis AM:

1. Better noise immunity is provided

2. Lower bandwidth is required

3. The transmitted power is better utilized

4. Less modulating power is required

Which of the above are advantages of FM over AM?


(c) 2 and 4 only (d) 1, 2, 3 and 4


Bandwidth requirement of FM higher than AM.

91.91.91.91.91. The total characteristic of a stabilizer is

(a) constant output voltage with low internal resistance.

(b) constant output current with low internal resistance.

(c) constant output voltage with high internal resistance.

(d) constant internal resistance with variable output voltage.


92.92.92.92.92. For a d.c. shunt generator to self excite, the conditions to be satisfied are that there

must be some residual magnetism in the field magnet, it must be in the proper direction

and the shunt field resistance must be

(a) Above the critical field resistance (b) Equal to the critical field resistance

(c) Less than the armature resistance (d) Less than the critical field resistance


93.93.93.93.93. In an IGBT cell the collector and emitter are respectively

(a) n and p (b) n + and p +

(c) p and n (d) p + and n +


39/48

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Page30

96.96.96.96.96. NAND and NOR gates are called ‘Universal’ gates primary because

(a) they are available everywhere.

(b) they are widely used in I .C . packages.

(c) they can be combined to produce AND, OR and NOR gate.

(d) they can be manufactured easily.


A logic gate is known as universal logic gate if all other logic gates can be obtained/

implemented from it.

97.97.97.97.97. If a medium transmission line is represented by nominal T , the value of B of ABCD

constant is

(a) Z (b)

+

(c)

+ (d)

+


98.98.98.98.98. To turn-off a GTO what is required at the gate?

(a) A high amplitude (but low energy) negative current

(b) A low amplitude negative voltage

(c) A high amplitude negative voltage(d) A low amplitude negative voltage

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)

A negative gate current of 20 to 25% of Anode current is required to turn off GTO.

99.99.99.99.99. A chopper circuit is operating on TRC control mode at a frequency of 2 kHz of a 230 V

dc supply. For output voltage of 170 V, the conduction and blocking periods of a thyristor

in each cycle are respectively

(a) 0.386 ms and 0.114 ms (b) 0.369 ms and 0.131 ms

(c) 0.390 ms and 0.110 ms (d) 0.131 ms and 0.369 ms


f = 2 kHz

V S

= 230V

V 0

= 170 V

V 0

= αV S

α =

= =


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