ee objective paper 1


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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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Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 | Email : [email protected] | Visit: www.madeeasy.in

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

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Paper-I (Electrical Engineering)

1.1.1.1.1. Permeance is inversely related to

(a) resistance (b) conductance

(c) reluctance (d) capacitance

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

2.2.2.2.2. Consider the following statements regarding an ideal core material:

1. It has very high permeability

2. It loses all its magnetism when there is no current flow.

3. It does not saturate easily.

Which of the above statements are correct?

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

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

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

⇒ An ideal core material should have high permeability which causes the magneticfield lines to be concentrated in the core material.

⇒ The magnetic field is often created by a coil of wire around the core that carriesa current. So if there is no current then it should loose all its magnetism.

⇒ Core saturation means the core flux stops increasing with increase in magnetizingcurrent, which is undersirable. Hence for proper operation the core should not

saturate easily.

3.3.3.3.3. The capacitance of a conducting sphere of radius r with a total charge of q uniformly

distributed on its surface is

(a) proportional to qr (b) independent of r

(c) proportional to

(d) independent of q


4.4.4.4.4. The characteristic impedance of a transmission line depends upon

(a) shape of the conductor (b) surface treatment of the conductor

(c) conductivity of the material (d) geometric configuration of the conductor



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Page4

5.5.5.5.5. In a series R-L-C circuit supplied by a source of 125 V at a resonant frequency of 220 Hz,

the magnitudes of the voltages across the capacitor and the inductor are found to be

4150 V. If the resistance of the circuit is 1Ω, then the selectivity of the circuit is

(a) 33.20 (b) 3.32

(c) 0.0301 (d) 0.301

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

+

125 V

f 0

= 220 Hz

V C

= V L = 4150 volts

R = 1 Ω

Q =

= =

Selectivity = Q = 33.2

6.6.6.6.6. The value of characteristic impedance in free space is equal to

(a)

µ

ε(b)

µ ε

(c)

µ ε(d)

ε

µ


7.7.7.7.7. The magnitude of magnetic field strength H is independent of

(a) current only (b) distance only

(c) permeability of the medium only (d) both current and distance


8.8.8.8.8. Consider the following types of transmission lines:

1. Open-wire line

2. Twin-lead wire

3. Coaxial cable

The capacitance per metre will be least in which of the above transmission lines?

(a) 1 only (b) 2 only

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



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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

sense test, permutation & combinations etc.

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,

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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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9.9.9.9.9. Three equal point charges are located at the vertices of an equilateral triangle on the

circumference of a circle of radius r . The total electric field intensity at the centre of the

circle would be

(a) zero (b)

πε

(c)

ε(d)

π ε


10.10.10.10.10. The Poynting vector on the surface of a long straight conductor of radius a and

conductivity σ0, which carries current I in the z -direction, is

(a)

σ π

I

(b)

−σ π

I

(c)

σ π

I (d)

−σ π

I


11.11.11.11.11. Consider the following applications in respect of a square corner reflector:

1. Radio astronomy

2. Point-to-point communication

3. 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, 2 and 3


12.12.12.12.12 . The variation of |B | with distance r from a very long straight conductor carrying a current

I is correctly represented by

(a) | |B

r

(b) | |B

r

(c) | |B

r

(d) | |B

r



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13.13.13.13.13. The resistivity of hard drawn copper at 20°C is 1.9 × 10–6 Ω cm. The resistivity of annealedcopper compared to hard drawn copper is

(a) lesser (b) slightly larger

(c) same (d) much larger


The mechanical treatment such as cold working produces localized strain in the material

which results in the increase in resistivity of material. A hard drawn copper wire thus

has lower conductivity (i.e., higher resistivity) than annealed copper.

14.14.14.14.14. The number of electrons excited into the conduction band from valence band (with

∆E = forbidden energy gap and k = Boltzmann’s constant) is proportional to

(a)

∆ (b)

∆

(c)

∆ − (d)

∆ −


n =

− −

=

− − + −

= −−∆

n ∝ −∆

15.15.15.15.15. 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?


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


Superconductivity can be destroyed–

1. by increasing the temperature above a critical temperature (T C ).

2. by applying magnetic field greater than the critical magnetic field (H C ).

3. by passing a current greater than the critical current ( I C ) as per Silsbee rule.


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16.16.16.16.16. The relative permeability of a medium is equal to (with M = magnetization of the medium

and H = magnetic field strength)

(a)

+ (b)

−

(c)

+ (d)

−


We have susceptibility

H M

=

= µ

r – 1

so relative permeability

µr =

+

17.17.17.17.17. The electrical resistivity of many metals and alloys drops suddenly to zero when they

are cooled to a low temperature (i.e., nearly equal to liquid helium temperature). Such

materials (metals and alloys) are known as

(a) piezoelectric materials (b) diamagnetic materials

(c) superconductors (d) high-energy hard magnetic materials


Superconductors are the materials whose resistivity becomes very small or zero below

a critical temperature.

18.18.18.18.18. The dielectric strength of rubber is 40000 V/mm at frequency of 50 Hz. What is the

thickness of insulation required on an electrical conductor at 33 kV to sustain the

breakdown?

(a) 0.83 mm (b) 8.3 mm

(c) 8.3 cm (d) 0.083 mm


Dielectric strength of rubber is = 40000 V/mm

Applied voltage = 33 kV

Let us assume thickness is ‘t ’ then

− =

×

⇒ t =

−× × = 8.25 × 10–4 m = 0.825 × 10–3 m

= 0.83 mm


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RankImprovement

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

for better result. The content of Rank Improvement Batch is designed to give exposure for solving different types of questions within xed

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)

• Old students who have undergone classroom course from anycentre of MADE EASY or any other Institute

• Top 6000 rank in GATE Exam

Qualied in ESE written exam•

Qualied in any PSU written exam•

M. Tech from IIT/NIT/DTU with minimum 7.0 CGPA•

• Comprehensive problem solving sessions

Smart techniques to solve problems•

Techniques to improve accuracy & speed•

Systematic & cyclic revision of all subjects•

Doubt clearing sessions•

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Ex. MADE EASY Students

Those students who were enrolled inPostal Study Course, Rank Improvement,

Conventional, G.S, Post GATE batches

Non-MADE EASY

Students

Fee is inclusive of classes, study material and taxes

Fee Structure

Those students whowere enrolled in long termclassroom programs only

Batch

Rank Improvement Batch

Rank Improvement Batch

+ General Studies &

Engineering Aptitude Batch

Rs. 26,500/- Rs. 24,500/- Rs. 22,500/-

Rs. 41,500/- Rs. 36,500/- Rs. 31,500/-

Streams Batch Type Timing Date Venue

CE, ME Weekend Sat & Sun : 8:00 a.m to 5:00 p.mnd

2 July, 2016 Saket (Delhi)

EC, EE Weekend Sat & Sun : 8:00 a.m to 5:00 p.mnd

2 July, 2016 Lado Sarai (Delhi)

ME Regular 8:00 a.m to 11:30 a.m (Mon-Fri)th

4 July, 2016 Saket (Delhi)

Documents required : M.Tech marksheet, PSUs/IES Interview call leer, GATE score card, MADE EASY I-card 2 photos + ID proof •

Rank Improvement Batches will be conducted at Delhi Centre only.

Note: 1. These batches will be focusing on solving problems and doubt clearing sessions. Therefore if a student is weak in basic concepts& fundamentals then he/she is recommended to join regular classroom course.

2. Looking at the importance and requirements of repeater students, it is decided that the technical subjects which are newly

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3. The course fee is designed without Study Material/Books, General Studies and Online Test Series (OTS) . However those subjects

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19.19.19.19.19. The conductivity of insulating materials (a very small value) is called as

(a) residual conductivity (b) dielectric conductivity

(c) ionic conductivity (d) bipolar conductivity


20.20.20.20.20. An intrinsic semiconductor has equal number of electrons and holes in it. This is due

to

(a) doping (b) free electrons

(c) thermal energy (d) valence electrons


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. Hence

because of the thermal energy equal number electrons and holes are generated in an

intrinsic semiconductor.

21.21.21.21.21. When a very small amount of higher conducting metal is added to a conductor, its

conductivity will

(a) increase

(b) decrease

(c) remain the same

(d) increase or decrease depending on the impurity

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

When we add a metal with another metal this makes an alloy. Conductivity of alloy is

less than the conductivity of metal as alloy has less regular structure than metal.

22.22.22.22.22. An electrically balanced atom has 30 protons in its nucleus and 2 electrons in its

outermost shell. The material made of such atom is

(a) a conductor (b) an insulator

(c) a semiconductor (d) a superconductor


There are 30 protons in electrically balanced atom of material, hence atomic number

of material is 30. So this material is Z n (having atomic number 30, and 2 electrons in

its outermost shell), which is a conductor.


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Page9

23.23.23.23.23. 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


For a normally doped semiconductor temperature coefficient of resistance is negative.

But a heavily doped semiconductor (i.e. degenerate semiconductor) has metal like

properties, hence it has positive temperature coefficient of resistance.

So the temperature coefficient of resistance for a doped semiconductor can be positive

or negative depending upon level of doping.

24.24.24.24.24. 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


In the slice processing of an integrated circuit the components are formed in selective

areas known as “windows” where silicon dioxide has been etched.

25.25.25.25.25. Permanent magnet loses the magnetic behaviour when heated because of

(a) atomic vibration (b) dipole vibration(c) realignment of dipoles



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


⇒ As a permanent magnet is heated, the electron spins (like tiny magnets) becomemore likely to be in high energy state, which leads to atomic vibration. That means

that are less lined up so the total magnetism is reduced.

⇒ Heating means providing extra thermal energy because of which it becomes easy

for domain walls (the boundaries between regions that are lined up pointing differentdirections) to slide around. Hence the domain walls will rearrange so that they reduce

the large scale field energy by pointing different directions, hence permanent magnet

loses magnetic behaviour.

⇒ As we further heat the individual dipole spin (dipole vibration) within domain becomemore likely to point opposite to their neighbour which results in losing permanent

magnetic behaviour.


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Page10

26.26.26.26.26. The magnetic field required to reduce the residual magnetization to zero is called

(a) retentivity (b) coercivity

(c) hysteresis (d) saturation


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.

27.27.27.27.27. A certain fluxmeter has the following specifications:

Air gap flux density = 0.05 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 mm 2 area connected to the fluxmeter

is reversed in a uniform field of 0.5 Wb/m2

, then the deflection of the fluxmeter will be(a) 87.4° (b) 76.5°

(c) 65.6° (d) 54.7°


Constant of Fluxmeter

G = N c B c Ac = 40 × 0.05 × 750 × 10–6 = 1500 × 10–6

Flux linkage with search coil = 0.5 × 200 × 10–6

= 1 × 10–4 wb

As the flux is reversed, the change in flux linking with search coil is

∆φ = 2 × 1 × 10–4 = 2 × 10–4 Wb

∆φ =

θ

2 × 10–4 =

−× × θ

θ = 1.33 rad =

× = °π

28.28.28.28.28. Consider the following statements:

1. Both ferromagnetic and ferrimagnetic materials have domain structures; each domain

has randomly oriented magnetic moments when no external field is applied.

2. Both ferromagnetic and ferrimagnetic materials make those domains that have

favourable orientation to the applied field grow in size.

3. The net magnetic moment in ferromagnetic material is higher than that in ferrimagnetic

material.

4. The net magnetic moment in ferrimagnetic material is higher than that in ferromagnetic

material.


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Page11


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

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


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. Hence statement 4 is wrong, so answer will be option (d).

29.29.29.29.29 . The Hall voltage, V H , for a thin copper plate of 0.1 mm carrying a current of 100 A with

the flux density in the z -direction, B z = 1 Wb/m2 and the Hall coefficient,

R H = 7.4 x 10–11 m3 /C, is

(a) 148 µV (b) 111 µV(c) 74 µV (d) 37 µV


V H = ?

t = 0.1 mm

I = 100 A

B = 1 Wb/m2

R H = 7.4 × 10–11 m3 /c

V H =

I =

−−

−× × ×

= ××

⇒ V H = 74 µ volt

30.30.30.30.30. A Zener regulator has an input voltage varying between 20 V and 30 V. The desired

regulated voltage is 12 V, while the load varies between 140 Ω and 10 kΩ. The maximumresistance in series with the unregulated source and Zener diode would be

(a) 3.3 Ω (b) 6.6 Ω(c) 36.6 Ω (d) 93.3 Ω


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Page12


I min = I z min + I z max

+=

R =

−

I =

− = 93.3 Ω

31.31.31.31.31. A short in any type of circuit (series, parallel or combination) causes the total circuit

1. resistance to decrease

2. power to decrease

3. current to increase

4. voltage to increase


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

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


A short in any type of circuit causes the total circuit resistance to decrease and thereby

the current to increase.

32.32.32.32.32. An air-cored solenoid of 250 turns has a cross-sectional area A = 80 cm2 and length

l = 100 cm. The value of its inductance is

(a) 0.425 mH (b) 0.628 mH

(c) 0.751 mH (d) 0.904 mH


L =

µ

l =

− −

−

× × π × × ×=

×

µ = µ0

µr as air µ

r 1

33.33.33.33.33. The current in a coil changes uniformly from 10 A to 1 A in half a second. A voltmeter

connected across the coil gives a reading of 36 V. The self-inductance of the coil is

(a) 0.5 H (b) 1 H

(c) 2 H (d) 4 H


V =

i

36 =

= =


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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


i (t ) = ω − °

ω L = 1 Ω

θ

Z

R

X L

tan θ =

⇒ R =

θ

R =

⇒ = Ω°

37.37.37.37.37. A single-phase full-wave rectifier is constructed using thyristors. If the peak value of

the sinusoidal input voltage is V m and the delay angle is

π radian, then the average

value of output voltage is

(a) 0.32 V m (b) 0.48 V m (c) 0.54 V

m (d) 0.71 V

m


The average output voltage of single phase full wave rectifier is

V 0

=

∝π

=

°π

V 0 = =π

38.38.38.38.38. The potential difference V AB in the circuit

V AB

1 A

1 Ω 3 Ω5 V

4 Ω 3 Ω

V A

+ –

V B

is

(a) 0.8 V (b) –0.8 V

(c) 1.8 V (d) –1.8 V


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Page15


5 V

1 Ω 3 Ω

V BV A

V C

= 5 V

1 A

4 Ω 3 Ω

V C – 0 = 5 ⇒ V C = 5 V

At node A:

−+ + = 0

5 V A

= 16 ⇒

=

At node B :

−− + + = 0 ⇒ 2V B = 8

∴ V B = 4 V

∴ V AB

= V A – V

B =

− = −

39.39.39.39.39. Two bulbs of 100 W/250 V and 150 W/250 V are connected in series across a supply

of 250 V. The power consumed by the circuit is

(a) 30 W (b) 60 W(c) 100 W (d) 250 W


100W/250 V 150W/250 V

R 1 =

×⇒ = ⇒ = Ω

R 2 =

×

⇒ = = ΩR

eq = R 1 + R 2= 1041.67 Ω

P =

×=


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Page16

40.40.40.40.40. Thevenin’s equivalent of a circuit, operating at ω = 5 rad/s, hasV OC = 3.71 ∠ –15.9° VZ O

= 2.38 – j 0.667 ΩAt this frequency, the minimal realization of the Thevenin’s impedance will have

(a) a resistor, a capacitor and an inductor(b) a resistor and a capacitor

(c) a resistor and an inductor

(d) a capacitor and an inductor


Z = 2.38 – j 0.667

∴ Z = R – jX C 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 , L andC but minimal realisation will be with R and C .

41.41.41.41.41. 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


∵ In A to D converter

Accuracy = Resolution = Error

∴

− ≤ 10 mV

− ≤ 10 mV

− ≤ 2 × 10–3

2n – 1 ≥ 5002n ≥ 501n ≈ 9

42.42.42.42.42. Three 30 Ω resistors are connected in parallel across an ideal 40 V source. What wouldbe the equivalent resistance seen by the load connected across this circuit?

(a) 0 Ω (b) 10 Ω(c) 20 Ω (d) 30 Ω


23/57

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”


24/57



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Page17


40 V 30 Ω 30 Ω 30 Ω

R eq

Replacing voltage source by a short circuit

R eq= 0 Ω

43.43.43.43.43. The current i(t ) through a 10 Ω resistor in series with an inductance is given byi(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


I rms =

+ + I I

= ( )

+ = =

P = ⋅ = ⋅ = I

44.44.44.44.44. Thevenin’s equivalents of the network in Fig. (i) are 10 V and 2 Ω. If a resistance of3 Ω is connected across terminals AB as shown in Fig. (ii), what are Thevenin'sequivalents?

A

B

A

B

3 Ω

Fig. (i) Fig. (ii)

(a) 10 V and 1.2 Ω (b) 6 V and 1.2 Ω(c) 10 V and 5.2 Ω (d) 6 V and 5.2 Ω


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Page18


A

B

10 V

2 Ω

10 V

2 Ω

3 Ω V Th

+

–

V TH =

× =

+R

TH: Deactivate independent source

2 Ω

R th

3 Ω

R TH =

×=

+ = 1.2 Ω

45.45.45.45.45. A voltage source, connected to a load, has an e.m.f. of 10 V and an impedance of

(500 + j 100)Ω. The maximum power that can be transferred to the load is(a) 0.2 W (b) 0.1 W

(c) 0.05 W (d) 0.01 W


+

10 V

500 + 100 j

Z L

I

For maximum power transfer;

Z L

= Z ∗TH

Z L

= 500 – j 100

I rms =

Z = Z L + Z TH ⇒ Z = 500 + j 100 + 500 – j 100 = 1000∴ I

rms=

⇒ = = I

P max

= ⋅ I

=

× =


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Page19

46.46.46.46.46. An ideal transformer is rated 220/110 V. A source of 10 V and internal impedance of

2 Ω is connected to the primary. The power transferred to a load Z L connected across

the secondary would be a maximum, when |Z L| is

(a) 4 Ω (b) 2 Ω

(c) 1 Ω (d) 0.5 ΩAns.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)

In an ideal transformer

N 1

N 2

Z i=

...(1)

As

=

⇒

= ⇒ =

From equation (1), Z i= Z

L (2)2 ⇒ Z

i = 4Z

L

And for maximum power transfer,

Z i= 2 Ω

∴ 4 Z L

= 2 ⇒ Z L =

= Ω

47.47.47.47.47. Consider the following values for the circuit shown below :

V R

100 Ω

150 VLu t t ( ) = 250 2 sin 600

1. V R =

2. I = 2A

3. L = 0.25 HWhich of the above values are correct?


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


27/57

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

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

+ V R

–

150 V

+

–

v t ( )

100 Ω

V 2 = +

V rms

=

=

(250)2 = + ⇒ =

I =

= ⇒ = I

V L = I X L ⇒ X L =

=

ω L

= 75 ⇒ L =

=

L = 0.125 H

48.48.48.48.48. The response of a series R-C circuit is given by

I (s ) =

−π +

where q 0 is the initial charge on the capacitor. What is the final value of the current?

(a)

− π

(b)

− π

(c) Infinity (d) Zero


→∞

i =

→

I

=

→

− π⋅ = +


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1 2 3 4 5 6 7 8 9 0

Page21

49.49.49.49.49. 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


As capacitor does not allow the sudden change in voltage

∴ for initial values we replace capacitor by a voltage source.

50.50.50.50.50. In a parallel resistive circuit, opening a branch results in

1. increase in total resistance

2. decrease in total power

3. no change in total voltage and branch voltageWhich of the above is/are correct?


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


V V 1 V 2 V 3

+

–

+

–

+

–

R

R eq =

P =

= =

V 1 = V 2 = V 3 = V

V V 1 V 2

+

–

+

–

R R eq =

P =

=

V 1

= V 2 = V

As we can see from above circuits, Total resistance increases.

• Total power decreases.

• Total voltage and branch voltage remain unaffected

We have taken equal resistances even unequal values give the same analysis.


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Page22

51.51.51.51.51. 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


Wire wound resistors are used as precision resistors. They have high power rating and

low resistance value.

52.52.52.52.52. In nodal analysis, the preferred reference node is a node that is connected to

1. ground

2. many parts of the network

3. the highest voltage source .

Which of the above is/are correct?


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


We always prefer to take that node as the reference node which is at ground potential.

53.53.53.53.53. 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


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.

54.54.54.54.54. Reciprocity theorem is applicable to a network

1. containing R , L and C elements

2. which is initially not a relaxed system

3. having both dependent and independent sourcesWhich of the above is/are correct?

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

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


Reciprocity theorem is applicable in case of linear and bilateral networks containig only

one independent source.


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55.55.55.55.55. 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 k 1e –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


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 + K 1 e –at

56.56.56.56.56. A piezoelectric crystal has a coupling coefficient K of 0.32. How much electrical energy

must be applied to produce output energy of 7.06 x 10–3 J?

(a) 25.38 mJ (b) 22.19 mJ

(c) 4.80 mJ (d) 2.26 mJ


57.57.57.57.57. If a constant current generator of 5 A, shunted by its own resistance of 1 Ω, deliversmaximum power P in watts to its load of R L Ω, then the voltage across the currentgenerator and P are

(a) 5 V and 6.25 (b) 2.5 V and 12.5

(c) 5 V and 12.5 (d) 2.5 V and 6.25


5 A V L

+

–

RL

= 1 Ω1 Ω

I L

For maximum power transfer R L = 1 Ω

I L =

× =

+∴ V L = 2.5 × 1 = 2.5 V

P max = ( )

⋅ = × I

P max

= 6.25 W


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58.58.58.58.58. Three star-connected loads of 3∠60° Ω each and three delta-connected loads of 9∠60° Ωeach are connected in parallel and fed from a three-phase balanced source having line-

to-neutral voltage of 120 V. 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


Y

9 60°∠

3 60°∠

B

R

Converting ∆-Load into star

Y

3 60°∠ 3 60°∠

B

R

120 V120 V80 A

I =

=

I total = 40 + 40 = 80 A

59.59.59.59.59. A wattmeter reads 10 kW, when its current coil is connected inR phase and the potential

coil is connected across R and neutral of a balanced 400 V (RYB sequence) supply.

The line current is 54 A. 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


R

B

Y

W

N


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P = V ph I ph cos φ

10 × 103 =

× × φ

R

B

Y

W

cos φ = 0.8P = V L I L sin φ

= 400 × 54 × 0.6 = 12.46 kW 13 kW

60.60.60.60.60. The phase voltage of a three-phase, star-connected alternator is V . By mistake, the

connection of R phase got reversed. The new line voltages will have a relationship

(a)

= = (b)

= =

(c)

= = (d) V

RY = V

YB = V

BR


61.61.61.61.61. 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- as well as delta-connected loads


62.62.62.62.62. Consider the following statements regarding the effect of adding a pole in the open-

loop transfer function on the closed-loop step response:

1. It increases the maximum overshoot.

2. It increases the rise time.

3. It reduces the bandwidth.


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

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


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Page26


We know that adding a pole i.e., 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)

63.63.63.63.63. A CRO screen has 10 divisions on the horizontal scale. If a voltage signal 5 sin (314t + 45°)

is examined with a line base setting of 5 ms/div, the number of signals displayed on

the screen will be

(a) 1.25 cycles (b) 2.5 cycles

(c) 5 cycles (d) 10 cycles


∵ Time period of I / P signal 5 sin (314 t + 45°)

π = 314

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. of cycle =

= 2.5

64.64.64.64.64. A series R-L-C circuit is connected to a 25 V source of variable frequency. The circuit

current is found to be a maximum of 0.5 A at a frequency of 400 Hz and the voltageacross C is 150 V. Assuming ideal components, the values of R and L are respectively

(a) 50 Ω and 300 mH (b) 12.5 Ω and 0.119 H(c) 50 Ω and 0.119 H (d) 12.5 Ω and 300 mH


I max = 0.5 A ; f r = 400 Hz, V C = 150 V

As the circuit is in resonance,

∴ = and V R = V S ⇒ V R = 25 V

V R = I R ⇒ R =

⇒ = Ω

V L = I X L ⇒ X L =

=

X L = ω L ⇒ L =

⇒ =

π ×


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65.65.65.65.65. The resonant frequency for the circuit

L

R C

for L = 0.2 H, R = 1 Ω and C = 1 F, is(a) 1 rad/s (b) 2 rad/s

(c) 3 rad/s (d) 4 rad/s


C R

L

Z eq

∴ Z eq = j ω L + Z

Z =

×− ω ω

= ×+ ω − ω +

ω

Z =

− ω

+ ω

Z eq =

− ω ω +

+ ω

=( )

ω + ω + − ω

+ ω

=

ω + ω − ω ++ ω + ω

At resonance impedance is purely resistive in nature.

∴ Equating Imaginary part to zero.ω L + ω 3R 2C2L – ω R 2C = 0

ω 3R 2C 2L = ω [R 2C – L]

ω 2 =

−

ω 2 =

− ⇒ ω = −

ω 2 = 5 – 1 ⇒ ω 2 = ω = 2 rad/sec


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66.66.66.66.66. Which one of the following conditions will be correct, 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


67.67.67.67.67. For the two-port network shown in the figure

I 1

I 2 ++

V 2

V 1

– –

V 1 = 60 I 1 + 20 I 2 and V 2 = 20 I 1 + 40 I 2

Consider the following for the above network :

1. The network is both symmetrical and reciprocal.

2. The network is reciprocal.

3. A = D

4. y 11

=

Which of the above is/are correct?

(a) 2 only (b) 2 and 4

(c) 1 only (d) 1 and 3


V 1 = 60 I 1 + 20 I 2 ...(1)

V 2 = 20 I 1 + 40 I 2 ...(2)

Comparing (1) and (2) with standard Z -parameters equations i.e.

V 1

= Z 11

I 1 + Z

12 I

2

V 2

= Z 21

I 1 + Z

22 I

2

Z 11 = 60 Z 12 = 20

Z 21 = 20 Z 22 = 40

• For symmetry

Z 11 = Z 22 but as in the given case60 ≠ 40 ∴ Network is not symmetrical

• For reciprocity

Z 12

= Z 21

and 20 = 20 ∴ Network is reciprocal.From equation (2)

20 I 1 = V

2 – 40 I

2 ⇒

= − I I ...(3)


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Page29

Substituting (3) in (1):

V 1 =

− + I I

V 1 = 3V 2 – 120 I 2 + 20 I 2

V 1 = 3V 2 – 100 I 2 ...(4)

Comparing (3) and (4) with standard ABCD parameters equation i.e.

V 1 = AV 2 – B I 2

I 1 = CV 2 – D I 2A = 3 B = 100

C =

D = 2

∴ A ≠ D

Y 11 =

=

−

=

= =

× − ×

Y 11 =

68.68.68.68.68. If the total powers consumed by three identical phase loads connected in delta and star

configurations are W 1 and W 2 respectively, then W 1 is

(a) 3W 2

(b)

(c) (d)


P ∆ = 3P YYYYYW 1 = 3W 2

69.69.69.69.69. A 100 µAammeter has an internal resistance of 100 Ω. For extending its range to measure500 µA, the required shunt resistance is

(a) 10 Ω (b) 15 Ω(c) 20 Ω (c) 25 Ω


R sh

=

= = Ω

− − I

I


40/57

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70.70.70.70.70. A 200 V PMMC voltmeter is specified to be accurate within ±2% of full scale. The limitingerror, when the instrument is used to measure a voltage of 100 V, is

(a) ±8% (b) ±4%(c) ±2% (d) ±1%


% Error at any reading

=

=

×=

71.71.71.71.71. How many poles does the following function have?

F (s ) =

+ ++ +(a) 0 (b) 1

(c) 2 (d) 3


Since function is f (s ) =

+ + + +

Nothing is common between numerator and denominator so number of pole is 2.

72.72.72.72.72. The degree to which an instrument indicates the changes in measured variable without

dynamic error is

(a) repeatability (b) hysteresis

(c) precision (d) fidelity


73.73.73.73.73. Loading by the measuring instruments introduces an error in the measured parameter.

Which of the following devices gives the most accurate result?

(a) PMMC (b) Hot-wire

(c) CRO (d) Electrodynamic



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74.74.74.74.74. 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


75.75.75.75.75. Consider the following types of damping :

1. Air-friction damping

2. Fluid-friction damping

3. Eddy-current damping

PMMC type instruments use which of the above?


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


76.76.76.76.76. In data acquisition system, analog data acquisition system is used

(a) for narrow frequency width, while digital data acquisition system is used when wide

frequency width is to be monitored

(b) for wide frequency width, while digital data acquisition system is used when narrow

frequency width is to be monitored

(c) when quantity to be monitored varies slowly, 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 systemis preferred when quantity is time-invariant


77.77.77.77.77. During the measurement of resistance by Carey Foster bridge, no error is introduced

due to

1. contact resistance

2. connecting leads

3. thermoelectric e.m.f.

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

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



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78.78.78.78.78. Scheri