(PAPER II) ELECTRICAL ENGINEERING 1

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1

(PAPER –II)

ELECTRICAL ENGINEERING

1. If is eigen values of A, and A is idempotent matrix, then

(A) 0

(B) 1

(C) Either 0 or 1 = =

(D) 0and 1 =

Answer: (C)

Solution: Eigen values of Idempotent matrix, i.e., ( )2A A= are 0 or 1 or both.

So, option (C) is correct.

2. The eigen values of the matrix 5 4

are1 2

(A) 5 and 2

(B) 1 and 4

(C) 1 and 6

(D) 7 and 5

Answer: (C)

Solution: Characteristic equation is 5 4

01 2

−=

−

( )( )

( )( )

2

5 2 4 0

7 6 0

1 6 0

1, 6 are the eigen values of A

− − − =

− + =

− − =

=

So, option (C) is the correct answer.

3. Using Runge’s formula of order 2, when x = 1.1 given 2dy

3x ydx

= + and y = 1.2 when x = 1, taking

h = 0.1. The value of y will be nearly

(A) 1.3

(B) 1.5

(C) 1.7



2

(D) 1.9

Answer: (C)

Solution: ( ) 2

0 0f x,y 3x y ; x 1; y 1.2; h 0.1= + = = =

Runge’s formula of order 2 is

1 0 1 2

1y y k k for i 0

2= + + =

Where ( ) ( ) ( )

( )

( ) ( ) ( )

( )

1 0 0

2 0 0 1

k hf x , y 0.1 f 1, 1.2

0.1 3 1.44 0.444

k hf x h, y k 0.1 f 1.1, 1.644

0.1 3.3 2.702736 0.6002736

= =

= + =

= + + =

= + =

1y 1.2 0.5221368 1.7 = +

Hence, option (C) is the correct answer.

4. The expression 2 x

x

2 x

Eee .

E e

(the interval of differencing being h) is

(A) x he −

(B) x he +

(C) xe

(D) x2e

Answer: (C)

Solution: ( )

( )22

x x 1 xE 1

e e E 2 E eE E

− − = = − +

( ) ( ) ( )( ) ( )( )

( )

( ) ( ) ( )

( )

( )

x x 1 x

xx h x x h 2h h

h

2x x h 2 x x

2 x x 2h x h x

x 2h h

2 xx x

2 x

E e 2e E e E f x f x h and E 1

ee 2e e e 2e 1

e

and E e e ; e E 1 e

E 2E 1 e e 2e e

e e 2e 1

Eee . e

E e

−

+ −

+

+ +

= − + = + = +

= − + = − +

= = −

− + = − +

= − +

=



3

5. The solution of differential equation ( ) ( )2 2 3 2x y 2xy dx x 3x y dy 0,− − − = is

(A) x

2log x 3log y cy− + =

(B) y

2log y 3log x cx− + =

(C) x

2log x 3log y cy+ − =

(D) y

2log y 3log x cx+ − =

Answer: (A)

Solution: Differential Equation is ( ) ( ) ( )2 2 3 2

M N

x y 2xy dx x 3x y dy 0 ... 1

− + − + =

2 2M N

x 4xy; 3x 6xyy x

= − = − +

∴ Not exact but it is homogeneous

3 2 2 3 2 2 2 2

1 1 1I.F

M.x N.y x y 2x y x y 3x y x y = = =

+ − − +

Multiplying equation (1) both sides by 2 2

1

x y, we get

( )

2 2 3 2

2 2 2 2 2 2 2 2

2

x y 2xy x 3x ydx dy 0

x y x y x y x y

1 2 x 3dx dy 0 ... 2 is exact equation

y x y y

−− + + =

− − + + =

∴ General solution is

( )yconstant

1 2 3dx dy 0

y x y

1.x 2 n x 3 ny C

y

− + =

− + =

Hence, option (A) is correct.

6. If u logxy,= where 3 3x y 3xy 1,+ + = then

duis

dx

(A) ( )( )

2

2

x x y1 log xy

y y x

+− +

+

(B) ( )( )

2

2

y x y1 log xy

x y x

+− −

+



4

(C) ( )( )

2

2

x x y1 log xy

y y x

++ −

+

(D) ( )( )

2

2

x x y1 log xy

y y x

−− +

+

Answer: (C)

Solution: u x log x x log y

u 1x. log x log y 1 log xy

x x

u x

y y

= +

= + + = +

=

Let ( ) 3 3f x,y x y 3xy 1 0= + + − = be implicit function

( )

( )( )

( )( )

2

2

2

2

2

2

f3x 3ydy x

fdx 3y 3x

y

x y

x y

du u u dy

dx x y dx

x yx1 log xy

y x y

−− + = −

+

− +=

+

= +

+= + −

+

Hence, the correct option is (C).

7. The solution of differential equation 3 3 3

x 2y

3 2 3

z z z3 4 e

x x y y

− − + =

(A) ( ) ( ) ( )x 2y

1 2 3

ez f y x f y 2x xf y 2x

27

+

= − + + + + +

(B) ( ) ( ) ( )x 2y

1 2 3


23

+

= − + + + + +

(C) ( ) ( ) ( )x 2y

1 2 3


27

−

= + + + + + +

(D) ( ) ( ) ( )x 2y

1 2 3


23

+

= − + − + + +

Answer: (A)



5

Solution: Differential Equation is ( )3 2 3 x 2yD 3D D' 4D' z e whereD and D'x y

+ − + = = =

3 2AE is m 3m 4 0

m 1 satisfies

− + =

= −

( )( )

( ) ( ) ( )

2

1 2 3

m 1 m 4m 4 0

m 1, 2, 2

CF f y x f y 2x x.f y 2x

+ − + =

= −

= − + + + +

( )

( ) ( ) ( ) ( )( ) ( )

( )

x 2y

3 2 3

x 2y

3 2 3

x 2y

1PI e

D 3D D' 4D'

1e using Rule 1 proceedure

1 3 1 2 4 2

e

27

+

+

+

=− +

=− +

=

GS is z CF PI opiton(A) = + is correct

Note: Right hand side term is not cleared in the question. So we considered x 2ye + then option (a) is

answer. If we consider x 2ye − then no option is matching.

8. If the imaginary part xv e= (x sin y + y cos y) is part of analytic function f(z) = u + iv, then f(z) is

(A) ( ) z1 z e c+ +

(B) zze c+

(C) 2zze c+

(D) ( ) z1 z e c− +

Answer: (B)

Solution: ( ) ( ) ( )

( )

( )

( )

x x x

x

vv e xsin y ycos y e x sin y ycos y e sin y

x

vand e x cos y cos y ysin y

y

u vf ' z i

x x

v vi. using Cauchy Riemann equations

y x

= + = + +

= + −

= +

= + −

1 3 0 4

0 1 4 4

1 4 4 0

−

− −

−

1−



6

Replacing x by z and y by 0, we get

( ) ( ) ( )z z

z z

f ' z e z 1 i. e 0

e .z e

= + +

= +

Integrating

( )

( ) ( )

z z

z z

z

f z e .zdz e dz c

f z e . z 1 e c

e .z c

= + +

= − + +

= +

∴ Option (B) is correct.

9. The first four terms of the Taylor series expansion of ( )( )( )

z 1f z ,

z 3 z 4

+=

− −when z = 2 is

(A) ( ) ( ) ( )2 311 27 59

z 2 z 2 z 2 ...4 8 16

− + − + − +

(B) ( ) ( ) ( )2 311 27 59

z 2 z 2 z 2 ...4 8 16

+ − − + + +

(C) ( ) ( ) ( )2 33 11 27 59

z 2 z 2 z 2 ...2 4 8 16+ − + − + − +

(D) ( ) ( ) ( )2 33 11 27 59

z 2 z 2 z 2 ...2 4 8 16− − − − + − +

Answer: (C)

Solution:

( )( )( )

( )

( )

z 2 t z t 2

z 1 A Bf z Partial fractions

z 3 z 4 z 3 z 4

A 4 and B 5

4 5 4 5f z

z 3 z 4 t 1 t 2

− = = +

+= = +

− − − −

= − =

− − = + = +

− − − −

( )

( )

( ) ( ) ( )

11

2

2

2 3

2 3

5 t4 1 t 1

2 2

5 t t4 1 t t ...... 1 ......

2 2 2

5 5 5 54 4 t 4 t 4 t .....

2 4 8 16

3 11 27 59z 2 z 2 z 2

2 4 8 16

−−

= − − −

= + + + − + + +

= − + − + − + − +

= + − + − + −

∴ Option (C) is correct.



7

10. The mean deviation about mean of a normal distribution is nearly

(A) 3

5

(B) 5

3

(C) 4

5

(D) 5

4

Answer: (C)

Solution: The mean deviation about mean μ of a normal distribution is nearly

( ) ( )4 2

or Using propertiesof Normal distribution5 5

Hence the correct option is (C).

11. Consider the following regression equations obtained from a correlation table:

y 0.516x 33.73

x 0.512y 32.52

= +

= +

The value of the correlation will be

(A) 0.514

(B) 0.586

(C) 0.616

(D) 0.684

Answer: (A)

Solution: Let the regression line of y on x is y 0.516x 33.73= +

yxb 0.516 = is regression coefficient of y on x and

regression line of x on y is x 0.512y 32.52= +

xyb 0.512 = is regression coefficient of x on y

We know that ( )

( )( )

yx xyr b .b both coefficient are positive

0.516 0.512 0.5139 0.514

=

= =

Hence the correct option is (A).



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12. If the probability of a bad reaction from a certain injection is 0.001, the chance that out of 2000

individuals, more than two will get a bad reaction will be

(A) 0.72

(B) 0.54

(C) 0.32

(D) 0.14

Answer: (C)

Solution: Let rP P= [bad reaction from certain injection] ( )0.001 very small=

( )n 2000 very large=

So, Apply Poisson distribution n.p 2= =

Let X denote number of individuals gets bad reaction ( )xe

P xx!

− =

∴ Required probability is ( ) ( )

( ) ( ) ( )0 1 2

2

2

P X 2 1 P X 2

1 P X 0 P X 1 P X 2

2 2 21 e

0! 1! 2!

51 0.32

e

−

= −

= − = − = − =

= − + +

= −


13. As per de Broglie’s relationship, the wavelength related to its mass m and velocity v is

(A) h

mv

(B) hv

m

(C) hm

v

(D) mv

h

Where: h = Planck’s constant

Answer: (A)

Solution: 2E mc h= =

from planck's theory

from Einstein equation



9

2

v

Speed of light(c) is replaced by velocity of lig

hv cmv h

h

m

ht(v)

= = =

=

Hence, the correct option is (A)

14. Which of the following statements regarding an atom are correct?

1. If two atoms with similar ionization potential form a bond, then this bond will most probably be

either covalent or metallic.

2. When atoms with different ionization potentials form a bond, the bond will be mainly ionic.

3. If the atoms or molecule already has its outer shells completely full, then the bonding between the

atoms or molecules will be a secondary bond when it solidifies.

(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (D)

Solution: If two atoms with similar ionisation, then covalent bonds are most likely to occur, since both atoms

have the same affinity for electrons and after full filling the octet configuration and become stable

for different ionisation, Ionic bonds are most likely to occur.

Since, molecules already have its outer shells completely full, hence there won’t be any

modifications in electronic structure. Hence, Bonding between molecules will be secondary bond.

15. A barium titanate crystal is inserted in a parallel plate condenser of area 10mm × 10 mm. The plates

having a separation of 2mm, give a capacitance of 910 F.− If the value of 12 1

0 8.854 10 Fm ,− − = the

relative dielectric constant of the crystal will be nearly

(A) 2640

(B) 2450

(C) 2260

(D) 2080

Answer: (C)



10

Solution: ( )o r

12 69 r

3

r

AC Basic concept

d

8.854 10 100 1010 F

2 10

2260

− −−

−

=

=

Hence, option (C) is correct.

16. A transformer core is wound with a coil carrying an alternating current at a frequency of 50 Hz. The

hysteresis loop has an area of 60,000 units when the axes are drawn in units of 4 210 Wbm− − and 2 110 Am .−

If the magnetization is uniform throughout the core volume of 0.01 m3, the power loss due to hysteresis

will be

(A) 300W

(B) 350W

(C) 400W

(D) 450W

Answer: (A)

Solution: Given: f 50Hz, A 60,000 units= =

4 2 2 3

h

B 10 wb m , H 10 A m, V 0.01m

P ?

−= = =

=

Area of hysteresis loop 4 260,000 10 10−=

Energy lost in per unit volume 4 2 3 3Area Volume 60,000 10 10 1m 600Jm−= = =

Power loss due to hysteresis, ( )h

Energy lost in per unit Volume of core 600 0.01P

time 1/ f

= =

h

600 0.1P 300W

1

50

= =

17. When ferromagnetic or ferrimagnetic materials are magnetized, the direction of magnetization in any

domain will be rotated from its preferential direction. This will show an anisotropic behaviour. On

removal of the magnetizing force, the total magnetization will in general have a non-zero value. This

behavior is due to

(A) Crystal anisotropic

(B) Stress anisotropic



11

(C) Shape anisotropics

(D) Crystal, stress and shape anisotropics

Answer: (D)

Solution: The ferromagnetic or ferrimagnetic materials shows anisotropic behaviour due to crystal, stress and

shape of anisotropics.

18. The paramagnetic susceptibility varies inversely with the absolute temperature for ordinary fields and

temperature. It is given by the relation

C

T =

The relation is known as

(A) Phenomenon of magnetostriction

(B) Curie law of paramagnetism

(C) Hall Effect

(D) Diamagnetism

Answer: (B)

Solution: Magnetic susceptibility of paramagnetic material depends on temperature, as

m

C

T =

19. If the interaction between the atomic permanent dipole moments is zero or negligible and the individual

dipole moments are oriented at random, the material will be a

(A) Ferromagentic material

(B) Ferrimagnetic material

(C) Paramagnetic material

(D) Antiferromagnetic material

Answer: (C)

Solution: Magnetic material Diploe arrangement

Paramagnetic →


Randomly



12

20. The magnetic moments of diamagnetic materials are mainly due to

(A) Electron spin angular momentum

(B) Nuclear spin angular momentum

(C) Orbital angular momentum of the electrons

(D) Centrifugal angular momentum

Answer: (C)

Solution: The magnetic moments of diamagnetic materials are mainly due to orbital angular momentum of

the electrons and due to this diamagnetic material repel magnetic lines.

21. The inductance of an air-cored coil is proportional

1. The square of the number of turns

2. The diameter of the coil

3. A form factor, F, dependent on the ratio of coil radius to coil length plus winding depth.

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

Answer: (B)

Solution: Since, 2

2

2 2

N AL where A d

N dL

= → =

Hence, the inductance is proportional to the square of diameter of coil.

Hence, statement 2 is wrong and the correct option is (B).

22. Light is capable of transferring electrons to the free-state inside a material thus increasing the electrical

conductivity of the materials. When the energy imparted to the electrons is quite large, the latter may be

emitted from the materials into the surrounding medium. This phenomenon is known as

(A) Photoemissive effect

(B) Photovoltaic effect

(C) Photoconductive effect

(D) Photo absorptive effect



13

Answer: (A)

Solution: Photo emissive effect: Electrons or charge carrier absorb light from light energy. Electron or

charge carriers get excited will be ejected out of the material.

Hence, the correct option is (A).

23. Which of the following statements is/are correct

1. Conductor contains a large number of electrons in the conduction band at room temperature. No

energy gaps exist and the valence and conduction bands overlap.

2. A semiconductor is a material in which the energy gap is so large that practically no electron can be

given enough energy to jump this gap.

3. An insulator is a solid with an energy gap small enough for electrons to cross rather easily from the

valence band to the conduction band.

(A) 1 only

(B) 2 only

(C) 3 only

(D) 1, 2 and 3

Answer: (A)

Solution:

Hence the correct answer is option (A).

24. Which of the following statements regarding superconducting materials are correct, when a large number

of metals become superconducting below a temperature?

1. The resistivity of the superconductor is zero.

2. The magnetic flux density B vanishes

3. Ferromagnetic and Antiferromagnetic metals are good examples of superconducting materials.

(A) 1, 2 and 3

(B) 1 and 3 only

Semi conductor

CB

VB

( )1 2eV gap−

Insulator

( )huge gap

CB

VB

Conductor

CB

VB

Overlapping



14

(C) 1 and 2 only

(D) 2 and 3 only

Answer: (C)

Solution: A super conductor characteristics are:

Hence statement 3 is wrong and the correct option is (C).

25. A voltage source-series resistance combinations is equivalent to a current source-parallel resistance

combination if and only if their

1. Respective open-circuit voltages are equal

2. Respective short-circuit currents are equal

3. Resistance remains same in both cases


(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (D)

Solution: Based on given information we can say, we are supposed to find a condition under what a practical

voltage source is equivalent to practical current source.

In circuit 1 In circuit 2

O.C voltage is sV O.C voltage is s sI R

S.C current is s sV R S.C current is sI

0 =

Perfect diamagnetic

C C CB , J & I are limits

+−

sR

circuit 1

sV LV

LI

+

−

sR

circuit 2

sI LV

+

−LI



15

We can say both circuits are equivalent if there O.C voltage, S.C current and resistance same, more over

from the load terminal we know the Thevenin model and Norton model are equivalent, this implies O.C

voltage , short circuit current and resistance should be same.

Hence the 3 given statements are true.

Hence the correct answer is option (D).

26. For a network graph having its fundamental loop matrix fB and its sub-matrices tB and B corresponding

to twigs links, which of the following statements are correct?

1. B is always an identity matrix.

2. tB is an identity matrix

3. fB has a rank of ( )b n 1 ,− − where b is the number of branches and n is the number of nodes of the

graph.

(A) 1 and 2 only

(B) 2 and 3 only

(C) 1 and 3 only

(D) 1, 2 and 3

Answer: (C)

Solution: It is given that a network graph having fundamental loop matrix fB and submatrices tB and B

f tB I B=

So tB is not an identity matrix.

B :link sub matrix is always identity matrix (by definition it is true).

fB has rank b-(n-1) (by definition it is true).

Hence statements 1,3 are true & 2 is false

Hence the correct answer is option (C).

27. The resistance R of a conductor is

(A) EA

J

(B) EJ

A

(C) E

JA



16

(D) JA

E

Where:

E = Electric field intensity

A = Cross-sectional area

J = Current density

=Length of conductor.

Answer: (C)

Solution: We know that from Ohm’s law

( )

( )

J E

E 1J where resistivity

E... 1

J

Resistance RA

E E

JA J

=

= =

=

=

= =


28. Which of the following statements are correct for an ideal constant voltage source?

1. Its output voltage remains absolutely constant whatever the change in load current.

2. It possesses zero internal resistance so that internal voltage drop in the source is zero

3. Output voltage provided by the source would remain constant irrespective of the amount of current

drawn from it.

4. Output voltage provided by the source varies with the amount of current drawn from it

(A) 1, 2 and 4 only

(B) 1, 3 and 4 only

(C) 2, 3 and 4 only

(D) 1, 2 and 3 only

Answer: (D)

Solution: The given statements for an ideal voltage source are

1. Its output voltage is constant irrespective of load current (True statement by definition)

2. It possesses zero internal resistance so that internal voltage drop in source is zero (True

statement by definition)



17

3. Output voltage provided by same would remain constant irrespective of current drawn (True

statement by definition).

4. output voltage provided by source varies with amount of current drawn from it.

(False statement as it contradicts statement 3).

So, statements 1,2,3 are true & 4th is false


29. Which of the following statements are correct?

1. A lowpass filter passes low frequencies and stops high frequencies.

2. A highpass filter passes high frequencies and rejects low frequencies.

3. A bandpass filter passes frequencies within a frequency band and attenuates frequencies outside the

band.

4. A bandstop filter passes frequencies within the band and blocks/attenuates frequencies outside a

frequency band.

(A) 1, 2 and 4 only

(B) 1, 3 and 4 only

(C) 2, 3 and 4 only

(D) 1, 2 and 3 only

Answer: (D)

Solution: The given statements are

1. a low pass filter passes low frequencies and stops high frequencies (True statement, by definition

of LPF)

2. a high pass filter passes high frequencies and stops low frequencies (True statement by definition

of HPF)

3. a band pass filter passes frequencies within a frequency band and attenuates frequencies outside

the band (True statement, by definition of band pass filter).

4. A band stop filter passes frequencies within the band and blocks frequencies outside a frequency

band (false statement). The true statement for Band stop filter is it blocks frequencies within the

band and passes frequencies outside a frequency band.

So statements 1,2,3 are true and statement 4 is false




18

30. A point charge of 910− C is placed at a point A in the free space. The potential difference between the two

points 20cm and 10cm away from the charge at A will be

(A) 40V

(B) 45V

(C) 50V

(D) 55V

Answer: (B)

Solution: Given, Charge9Q 10 C−=

( )

BC B C

1 2 1 2

9 9

V V V

Q Q Q 1 1

4 r 4 r 4 r r

100 10010 9 10

10 20

9 10 5 9 5 45 volts

−

= −

= − = −

= −

= − = =

Hence the correct option is (B)

31. According to Gauss’s theorem, the surface integral of the normal component of electric flux density D

over a closed surface, containing free charge is

(A) Q

(B) 0

Q

(C) 0Q

(D) 2

0

Q

Answer: (A)

Solution: From Gauss’s law

D.dA Q=


Q B C

A

1r 10cm=

2r 20cm=



19

32. A unit magnetic pole may be defined as that pole which when placed in vacuum at distance of one metre

from a similar and equal pole repels it with a force of

(A) 1

Newtons4

(B) 0 Newtons4

(C) 0

Newtons4

(D) 0

1Newtons

4

Answer: (B)

Solution: In practice magnetic monopoles doesn’t exist. But if we assume that if we have two magnetic

monopoles of magnetic charges 1 2m mq and q separated by a distance “r”. Then the magnetic force

exerted by magnetic monopole 1 on magnetic monopole 2 is

1 2m m0

2

q qF

4 r

=

If we consider 1 2m mq , q as unit magnetic monopoles separated by unit distance, then we get

( )oF N4

=

Hence the correct option is (B).

33. An analogous of magnetic circuit ‘permeability’ in electrical circuit is

(A) Reluctivity

(B) Conductance

(C) Conductivity

(D) Resistivity

Answer: (C)

Solution: Magnetic permeability is analogous to conductivity.


34. The magnetizing force at the centre of a circular coil varies

1. Directly as the number of its turns



20

2. Directly as the current

3. Directly as its radius

4. Inversely as its radius


(A) 1, 2 and 3 only

(B) 1 and 4 only

(C) 1, 2 and 4 only

(D) 2 and 3 only

Answer: (C)

Solution:

From Ampere’s law,

0H d NI

NIF H

2 r

F H N

I

1

r

=

=


35. An uncharged capacitor of 0.01 F is charged first by a current of 2 mA of 30s and then by a current of 4

mA for 30s. The final voltage in it will be

(A) 12V

(B) 18V

(C) 24V

(D) 30V

Answer: (B)

Solution: Based on given information we can say C 0.01= Farad, cV (0 ) 0V− = ,

We need to obtain final voltage cV .

Magnetizing force H

r I



21

( ) ( )

t

c c

0

3 3

1V i dt

C

130 2 10 30 4 10

0.01

6 12 18

− −

=

= +

= + =

Hence the correct answer is (B).

36. A capacitor of 10 pF is connected to a voltage source of 100V. If the distance between the capacitor plates

is reduced to 50%, while it remains connected to the 100V supply, the value of potential gradient in the

second case will be

(A) Half of earlier value

(B) Same as earlier value

(C) Twice of earlier value

(D) One fourth of earlier value

Answer: (C)

Solution: Given

Capacitance C=10pF

Voltage source V=100V

Capacitor is connected to the 100V supply implies that the voltage across the capacitor is fixed.

We know that potential across the capacitor is given by fixV Ed=

1

V is fixed implies Ed

As distance between the plates reduces to half, E-field (or) potential gradient becomes double.

Hence the correct option is (C)


1. Accuracy is the closeness with which an instrument approaches the true value of the quantity being

measured.

2. Precision is a measure of the reproducibility of the measurement.

3. Precision of an instrument can be improved upon by calibration.

4. Accuracy may be specified in terms of limits of errors.

(A) 1, 2 and 3 only

(B) 1, 3 and 4 only

2 mA

ci

4mA

0 30 60 ( )t sec



22

(C) 1, 2 and 4 only

(D) 2, 3 and 4 only s

Answer: (C)

Solution: i. Accuracy is the closeness with which an instrument reading approaches the true value of the

quantity being measured. Thus, accuracy of a measurement means confirmity to the truth. The

accuracy may be specified in terms of inaccuracy or limits of errors. Hence statements (i) & (iv)

are correct

ii. Precision is a reproducibility of the measurements that is given a fixed value of quantity,

precision is a measure of degree of agreement within a group of measurements.

Hence statement (ii) is correct

iii. The process of calibration improves accuracy but not the precision. Hence statement (iii) is not

correct.

38. An electrodynamometer instrument can be used as

1. Wattmeter and VAR meter.

2. Power factor meter and frequency meter.

3. Transfer instrument

(A) 1 and 3 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (D)

Solution: ● Electrodynamometer type of instruments are used as wattmeter’s, varmeters and with some

modification as power factor meters and frequency meters.

● A transfer instrument is one that may be calibrated with a dc source and then used without

modification to measure AC. Electro dynamometer type instruments are transfer type which

have same accuracy for both DC and AC.

39. The moving iron instruments when measuring voltages or currents.

(A) Indicating the same value of the measuring for both ascending and descending values of current

(B) Indicate higher values of the measurement for ascending values of current

(C) Indicate higher values of the measurement for descending values of current

(D) Indicate lower values of the measurement for both ascending values of current



23

Answer: (C)

Solution: Hysteresis error in moving iron instruments:

This error occurs as the value of flux density is different for the same current for ascending &

descending values.

The values of flux density is higher for descending values of current, therefore, the instrument tends

to indicate higher values of measurement for descending values of current (and voltage) than for

ascending values

40. True RMS-reading voltmeter

1. Measures the RMS value of voltage accurately.

2. Eliminates the error due to waveform.

3. Uses the thermocouple for heating.


(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (D)

Solution: 1. The voltage measured by the DC voltmeter is approximately equal to the rms value of input

signal.

Hence it measures RMS value of voltage accurately

2. The true rms value is measured independently of the waveform of AC signal. Hence eliminates

the error due to waveform.

3. To measure the rms value of an arbitrary waveform, we may feed an input signal to a heating

element in close proximity to a thermocouple as shown in figure.

41. Instrument transformers are

(A) Used to extend the range of the AC measuring instruments only

(B) Used to isolate the measuring instruments from the high voltage only

rmsV Heater Thermocouple

+

−

P



24

(C) Used to extend the range and isolate the measuring instruments

(D) Not used at generating stations and transformer stations

Answer: (C)

Solution: Transformers used in conjunction with measuring instruments for measurements purposes are called

“Instrument transformers”

Uses:

1. The range of instrument can be extended, so that current, voltage, power and energy can be

measured with instruments of moderate size.

2. The measuring circuit is isolated from the power circuit.Hence option (C) is correct.

3. In power systems, very large currents and voltages are handled by stepping down them with the

help of instrument transformers.

42. The power in a 3-phase circuit is measured with the help of 2-wattmeters; the readings of one of the

wattmeter is positive and that of the other is negative. The magnitude of readings is different. It can be

concluded that the power factor of the circuit will be

(A) Unity

(B) Zero

(C) 0.5

(D) Less than 0.5

Answer: (D)

Solution: The readings of two wattmeter’s

( )

( )1 L L

2 L L

W V I cos 30

W V I cos 30

= −

= +

Given that, one wattmeter reading is positive & other is negative & magnitude of readings of both

meters are different.

Let say 1W is positive & 2W is negative.

( )

( )

( )

2 2

L 1

So, W 0 W negative

V I cos 30 0

cos 30 0

30 90

60

cos 0.5

=

+

+

+

Power factor should be less than 0.5 in order to satisfy the given condition.



25

43. In a Q-meter, distributed capacitance of a coil is measured by changing the capacitance of the tuning

capacitor. The values of tuning capacitor are 1 2C and C for resonant frequencies 1 1f and 2f respectively.

The value of distributed capacitance will be

(A) 1 2C C

2

−

(B) 1 2C 2C

3

−

(C) 1 2C 4C

3

−

(D) 1 2C 3C

2

−

Answer: (C)

Solution: Given: Tuning capacitance is 1C at resonant frequency 1f & tuning capacitance is 2C at resonant

frequency 12f

2 1

2

1

f 2f

fn 2

f

=

= =

Distributed capacitance, 2

1 2d 2

C n CC

n 1

−=

−

( )2

1 2

2

1 2 1 2d

C 2 C

2 1

C 4C C 4CC

4 1 3

−

−

− −= =

−

44. In a digital voltmeter, during start of conversion, zero indication is displayed and is called auto zeroing.

This is achieved by

(A) Using a positive reference voltage

(B) Using a negative reference voltage

(C) Properly charging the differentiator circuit capacitance to ground

(D) Properly discharging the integrator circuit capacitance to ground

Answer: (D)

Solution: During start of conversion, the zero indication is found on the display. It is called Auto zeroing.

This is achieved by properly discharging the integrator circuit capacitance to ground.



26

During Auto-zeroing, S switch from inV to ground and capacitor C of dual slope converter discharges to

the ground.

45. A CRT has an anode voltage of 2000V and parallel deflecting plates 2 cm long and 5 mm apart. The screen

is 30 cm from the centre of the plates. If the input voltage is applied to the deflecting plates through

amplifiers having an overall gain of 100, the input voltage required to deflect the beam through 3cm will be

(A) 1V

(B) 3V

(C) 5V

(D) 7V

Answer: (A)

Solution: Given: Anode voltage, aE 2000V;= length of deflecting plates, 2

d 2cm 2 10 m−= = distance

between plates, d 35mm 5 10 m−= =

Screen-distance, L=30 cm 230 10 m−=

2

Gain 100

D 3cm 3 10 m.−

=

= =

Voltage applied to deflecting plates,

ad

d

2d.E .DE

L.=

Substituting all values, we get

3 2

d 2 2

2 5 10 2000 3 10E 100

30 10 2 10

− −

− −

= =

Input voltage for deflection, di

EV

Gain=

i

100V 1V

100= =

−

+

C

SinV

R



27

46. An aquadag is used in a CRO to collect

(A) Primary electrons only

(B) Secondary emission electrons

(C) Both primary electrons and secondary emission electrons

(D) Heat emission electrons

Answer: (B)

Solution: The electrons when bombards & strikes the screen release secondary emission electrons. These

secondary electrons are collected by an aqueous solution of graphite is called Aquadag & connected

to the second anode.

47. A resistance wire strain gauge with a gauge factor of 2 is bonded to a steel structural member subjected to

a stress of 100 MN/m2. The modulus of elasticity of steel is 200 GN/m2. The percentage change in the

value of the gauge resistance due to the applied stress will be

(A) 0.1%

(B) 0.3%

(C) 0.5%

(D) 0.7%

Answer: (A)

Solution: Gives: Gauge factor; fG 2=

2 6 2

2 9 2

Stress 100MN m 100 10 N m

r 200 GN m 200 10 N m

R?

R

= =

= =

=

We have Guage factor,

( )

( )

f

R R R RG ... 1

strain

stressmodulus of elasticity, ... 2

strain

= =

=

From (1) & (2),

Strain, stress

=



28

( )

63

9

f

3 3

100 100.5 10

200 10

RNow, G

R

2 0.5 10 10

−

− −

= =

=

= =

Percentage change in the value of guage resistance due to applied stress will be

3R

100 10 100 0.1%R

− = =

48. Capacitive transducers can be used for the measurement of liquid level. The principle of operation used in

this case is the change of capacitance with change of

(A) Distance between plates

(B) Area of plates

(C) Dielectric

(D) Resonance

Answer: (C)

Solution: 1. The principle of operation of capacitive transducers is based on equation

o r. AAC

d d

= =

The capacitive transducer works on the principle of change of capacitance which may be caused by

(i) change in overlapping area A,

(ii) change in distance between the plates, d and

(iii) change in dielectric constant

2. The capacitive transducers used for measurement of level of non-conducting liquid uses the

principle of change of capacitance with change of dielectric.

3. Change in dielectric constant.

49. The hexadecimal of the binary number (11010011)2 is

(A) 16D3

(B) 16D4

(C) 16C3

(D) 16C4



29

Answer: (A)

Solution: Given binary Number (1101 0011)2

We know that, 1 Hexa decimal digit = 4 binary bits

So

0 to 9 as Binary

1010 A

1011 B

1100 C

1101 D

→

→

→

→

50. Which of one of the following relations from the Boolean algebra pertaining to ‘AND’ operation cannot

be verified when A and B can take only the value 0 or 1?

(A) AB = BA

(B) AA = A

(C) A1 = 1

(D) A0 = 0

Answer: (C)

Solution: We need to obtain which of the following can’t be verified

AB=BA (True, commutative law)

A. A=A (True, by property)

A. 1 =1 (False, A .1=A)

A. 0=0 (True, by properly)

So, A.1=1 cannot be verified

Hence the correct answer is (C).

51. Which of the following design levels of a computer are widely used in computer design?

1. Gate level

2. Processor level

3. Register level

4. User level

( ) ( )2 16

1101 0011 D 3=



30

(A) 1 and 3 only

(B) 2 and 4 only

(C) 3 and 4 only

(D) 1, 2 and 3 only

Answer: (D)

Solution: A computer can be designed with respect to gate level, register level as well as processor level but

not with respect to user level.

52. Which one of the following is a powerful web platform for web applications and web services, built-in

virtualization technologies, variety of new security tools, enhancements and streamlined configuration and

management tools?

(A) Internet Explorer

(B) Internet Information Services

(C) Web Matrix

(D) Visual Web Developer

Answer: (C)

Solution: Web matrix is an integrated environment offered by many famous web hosting providers, which is

used for developing ASP. NET applications. It helps us by providing us with a very easy access to

the websites which are windows azure

53. Which one of the following is the correct sequence of steps for executing an instruction during CPU's

processing ?

(A) Fetch instruction, Read data, Decode instruction, Store data and Execute instruction

(B) Decode instruction, Read data, Execute instruction, Fetch Next instruction and Store data

(C) Decode instruction, Decode next operands, Fetch Next instruction and Store data

(D) Fetch instruction, Decode instruction, Read operands, Execute instruction and Store data

Answer: (D)

Solution:

Fetch Next

instructions

Execute Next

instructionsStart Halt

Fetch cycle Execute cycle



31

In a basic computer, each instruction cycle consists of the following phases:

1. Fetch instruction from memory

2. Decode instruction from memory

3. Read the effective address from memory

4. Execute the instruction

54. Which one of the following is the correct combination of registers in DMA controller?

(A) Data register, Stack pointer and Data counter

(B) Data register, Address register and Data counter

(C) Data register, Stack pointer and Address register

(D) Data register, Program counter and Data counter

Answer: (B)

Solution: DMA controller contains the following

Registers. They are

i. Data counter

ii. Data register

iii. Address register

55. A multiprocessing technology which enables software to treat a single processor as two processors to

utilize the processing power in the chip that would otherwise go unused and lets the chip operate more

efficiently resulting in faster processing is called

(A) Systematic multiprocessing

(B) Massively parallel processing

(C) Co–processing

(D) Hyper threading

Data counter

Data register

Address register

Control logic

Data lines

Address lines

Read

write Interrupt

DMA ACK

DMA REQ



32

Answer: (D)

Solution: Hyper threading is a technology used by same intel microprocessors that allows a single

microprocessor to act like two separate processors to the OS and the app programs that use it.

It is a feature of intel’s 1A-32 processor architecture. With hyper threading a microprocessor’s

“core” processor can execute two (rather than one) concurrent streams of instructions sent by the

OS. Having two streams of execution units to work on allows more work to be done by the

processor during each clock cycle. To the OS, the hyper threading microprocessor appears to be two

separate processors.

56. A physical implementation of the type declaration in high-level programming languages where major

information types should be assigned formats for identification is called

(A) Storage order

(B) Tag

(C) Error correction

(D) Error detection

Answer: (A)

57. Which of the following factors are to be considered while selecting number representations to be used in a

computer?

1. Number types to be represented

2. Range of values to be encountered

3. Cost of the hardware to store and process the numbers

4. Positional notation with fixed weight

(A) 1,2 and 4 only

(B) 1,3 and 4 only

(C) 2, 3 and 4 only

(D) 1,2 and 3 only

Answer: (A)

Solution: While selecting under representations to be used in computer, the following factors to be

considered, they are

i. number types to be represented

ii. range of values to be encountered

iii. positional notation with fined weight



33

58. If a negative binary number is to be represented by n–bits, then the standard format will be

(A) Sign bit '0' on left and magnitude right

(B) Sign bit ‘1’ on left and magnitude on right

(C) Sign bit ‘0’ on right and magnitude on left

(D) Sign bit '1' on right and magnitude on left

Answer: (B)

Solution: The standard pattern for a negative binary number is to be represented by n-bits, is

Sign bit ‘1’ on left and magnitude on right.

59. The physical address translation in virtual memory address with Memory Management Unit (MMU) is

done by which one of the following mechanisms?

(A) Multiply virtual, address by some constant

(B) Translation lookaside buffer (TLB)

(C) Encryption key

(D) Using general purpose register in CPU

Answer: (B)

Solution: While converting the PA to VA with mmv is done by TLB.

It reduces EMAT.

60. Which one of the following satellite systems is most often used for Global Positioning System (GPS) ?

(A) Geosynchronous

(B) Geostationary

(C) Low Earth Orbit

(D) Medium Earth Orbit

Answer: (D)

Solution: In GPS system, medium earth orbit satellite system is used.

61. In a tunnel diode, the width of the junction barrier is

(A) Directly proportional as the square root of impurity concentration

(B) Inversely proportional as the square root of impurity concentration

(C) Directly proportional as square of impurity concentration



34

(D) Inversely proportional as square of impurity concentration

Answer: (B)

Solution: Width of the junction = dd

1 2w W

q NN

∴ W is inversely proportional to the square root of impurity concentration.


62. In a grounded-emitter transistor, when emitter current becomes zero in cut-off region the emitter potential

is called

(A) Floating Emitter Potential

(B) Breaking Emitter Potential

(C) Cascading Emitter Potential

(D) Cut-off Emitter Potential

Answer: (A)

63. When maximum reverse-biasing voltage is applied between the collector and base terminals of the

transistor and emitter is open circuited, breakdown occurs due to

(A) Avalanche breakdown

(B) Avalanche multiplication

(C) Punch–through

(D) Reach–through

Answer: (B)

Solution: Junction between emitter and base region is prone to Zener break down and junction between base

and collector is prone to Avalanche break down.

64. In a Field Effect Transistor (FET) the maximum voltage that can be applied between any two terminals is

given by

(A) Low DSV causing avalanche breakdown

(B) Low GSV causing avalanche breakdown

(C) DSV = 0 when gate is reverse–biased



35

(D) GSV = 0 when gate is reverse–biased

Answer: (A)

Solution: In FET, the two terminals voltage VDS is causing avalanche break down.

65. A depletion–type MOSFET can be operated in an enhancement mode where negative charges are induced

into n–type channel by applying

(A) Positive Gate Voltage

(B) Negative Gate Voltage

(C) Positive Drain Voltage

(D) Negative Drain Voltage

Answer: (A)

Solution: Depletion type MOSFET are operating in enhancement mode to induce negative charge, we need to

apply positive Gate voltage.

66. The double base diode which is operated with the emitter forward biased and a smaller emitter junction is

called

(A) Field Effect Transistor (FET)

(B) Uni-Junction Transistor (UJT)

(C) Bipolar Junction Transistor (BJT)

(D) Metal Oxide Semiconductor Field Effect Transistor (MOSFET)

Answer: (B)

Solution:


P N

1B

2BUJT



36

67. Which one of the following is not a distortion type that exists either separately or simultaneously in

amplifiers?

(A) Linear distortion

(B) Non-linear distortion

(C) Frequency distortion

(D) Delay distortion

Answer: (A)

Solution: When the transistor is operating in the linear region of the transfer characteristics, then only the

output will be exact replica of the input so, we don’t have any distortion when it operates in the

linear region of the transfer characteristic.

Hence there will be no distortion like linear distortion

Hence the correct option is (A)

68. Transistor noise caused by the recombination and generation of carriers on the surface of the crystal is

called

(A) Thermal noise

(B) Excess noise

(C) White noise

(D) Shot noise

Answer: (*)

Solution: Flicker noise (or) 1

fnoise is exhibited by all normal amplifying devices and transistors are no

exception.

Flicker noise is known to arise from the generation (or) recombination of carriers on the surface.

69. During a low frequency response of an amplifier which is invariably of RC–coupled type, there is a range

of frequency characteristics over which the amplification is constant and delay is also constant, called

(A) Low band frequency

(B) Mid band frequency

(C) High band frequency

(D) Hyper band frequency



37

Answer: (B)

Solution: RC coupled amplifier frequency response is as shown in figure (1) & figure (2)

In the mid band frequency ( Vs frequency)

d

Time delay T constant.d

− = =


70. In a crystal oscillator, especially when piezoelectric crystal like quartz is applied, then the inductor L,

capacitor C and resistor R are the analogs of the mechanical system as

( )phase

( )fig : 2

frequency→

T = −

0A

Lf Hf

mid band frequency

High frequency

frequency

low frequency

( )fig : 1

Gain



38

(A) Mass, compliance, viscous-damping factor

(B) Mass, spring constant, viscous-damping

(C) Mass, momentum, viscous-damping factor

(D) Mass, displacement, viscous-damping factor

Answer: (A)

Solution: L Mass

C Compliance

R Viscous damping

=

=

=

71. In a phase shift oscillator using an FET, at a certain frequency if the phase shift introduced by the RC

network is 180°, then the total phase shift from gate around the circuit and back to the gate will be

(A) 0°

(B) 90°

(C) 180°

(D) 270°

Answer: (A)

Solution: In a phase shift oscillator using FET, FET amplifier provides a phase shift of 180° and it is given

that RC network provides another 180° phase shift.

Overall phase shift is 360° (or) 0°


72. In a feedback amplifier, which configuration increases bandwidth, decreases non-linear distortion and

improves transconductance with negative feedback?

(A) Voltage-series

(B) Current-series

(C) Voltage-shunt

(D) Current-shunt

Answer: (B)

Solution: If we have negative feedback then in general it increases. Band width and decreases non-linear

distortion. But to increase transconductance Current series is good.




39

73. In order to balance the offset voltage of an operational amplifier, a small DC voltage is applied to input

terminals where the connection is

(A) Series with both inverting as well as non–inverting input

(B) Series with non–inverting input

(C) Shunt with inverting input

(D) Shunt with non–inverting input

Answer: (B)

Solution:

osV is connected in series with non-inverting i/p.


74. Multivibrator circuit that remains in stable state until a triggering signal causes transition to quasistable

state and returns to stable state after certain time is called

(A) Astable multivibrator

(B) Monostable multivibrator

(D) Bistable multivibrator

(D) Unstable multivibrator

Answer: (B)

Solution: The monostable multivibrator has one stable state in which it can remain indefinitely.

It also has a quasi-stable state to which it can be triggered and in which it stays for a predetermined

interval equal to the desired width of the output pulse. When this interval expires, the monostable

multivibrator returns to its stable state and remains there awaiting another triggering signal.


−

+

1R

−+

2R

osV

0V

i p offset voltage.



40

75. In a paraphase amplifier, where two amplifiers are connected in cascade, the output from second stage

(A) Equals signal input without change of sign

(B) Equals signal input with change of sign

(C) Does not equal to signal input and has no sign change

(D) Does not equal to signal input but has sign change

Answer: (A)

Solution:

Paraphase amplifiers are also called as phase splitters. They produce from a single input wave

form, two output wave forms that have exactly opposite instantaneous polarities.

If these two wave forms were produced as the result of a single sine-wave input they might be

considered 180° out of phase.

One wave form appearing to have been displaced 180° along the time axis.

76. Which one of the following statements is not correct for an active filter used in the field of

communications and signal processing?

(A) It is more economical

(B) It does not cause loading of the source or load.

(C) It is easier to tune or adjust

(D) It exhibits insertion loss

Answer: (D)

Solution: Active filter doesn’t exhibit insertion loss.

Hence the correct option is (D).

77. A two–step procedure in a typical diffusion apparatus to obtain the complementary–error–function

Gaussian distribution involves the first step and second step respectively as

(A) Predeposition and Drive–in

(B) Predeposition and Drive–out

(C) Drive–in and Postdeposition

(D) Drive–out and Postdeposition

+

−

InputParaphase

Amplifier

1

+−

2

+−



41

Answer: (A)

Solution: Predeposition also known as infinite source diffusion is done before drive in diffusion.

78. In AM modulation, the equation of the modulating signal is given by f(t) = Am cos ωm t. If the amplitude

of the carrier wave is A and there is no over–modulation, the modulation efficiency will be

(A) 33.3%

(B) 38.6%

(C) 43.3%

(D) 48.6%

Answer: (A)

Solution: For the Modulation ( ) of AM signal

( )

2

2

1100 100 33.3%

2 2 1

1 No over modulation

= = =

+ +

79. For a binary phase–shift keying modulator with a carrier frequency of 70 MHz and input bit rate of 10

Mbps, the maximum Upper Side Frequency (USF) and minimum Lower Side Frequency (LSF) are

respectively

(A) 85 MHz and 65 MHz

(B) 75 MHz and 65 MHz

(C) 55 MHz and 45 MHz

(D) 55 MHz and 45 MHz

Answer: (B)

Solution: In BPSK

c

b

f 70mHz

R 10mbps

=

=

Maximum fundamental frequency of binary input (fm) bR 105mbps

2 2= = =

Lower side band frequency, LSF 70 5 65MHz= − =

Upper side band frequency, USF 70 5 75MHz= + =

So option (B) is correct.



42

80. In modulation system, the energy per bit-to-noise power density ratio b 0E N is

(A) bfC

N B

(B) b

N B

C f

(C) b

C B

N f

(D) bfN

C B

Where, N = Noise power of thermal (W)

B = Bandwidth (Hz)

C = Carrier power (W)

fb = Bit rate (bps)

Answer: (C)

Solution: Signal to Noise (SNR)

b b

o

b

0 b

E fC

N N B

E C B

N N f

=

=

Hence option (C) is the correct option.

81. Which one of the following is not a transmission parameter of a private line data circuit that utilizes public

telephone network?

(A) Geographical parameter

(B) Bandwidth parameter

(C) Interface parameter

(D) Facility parameter

Answer: (A)

Solution: Geographical parameter is not a transmission parameter of a private line data circuit that utilizes

public telephone network.



43

82. The Shannon limit for information capacity I is

(A) 2

SBlog 1

N

−

(B) 2

SBlog 1

N

+

(C) 10

SBlog 1

N

−

(D) 10

SBlog 1

N

+

Where N = Noise power (W)

B = Bandwidth (Hz)

S = Signal power (W)

Answer: (B)

Solution: Shannon capacity formula (or) information capacity of Shannon limit

2

SI B log 1

N

= +

Where N=Noise

B=Bandwidths

S=Signal

83. In a time division multiplexing, there are 8000 samples for a digital signal-0 channel that uses 8 kHz

sample rate and 8 bit PCM code. The line speed will be

(A) 56 kbps

(B) 64 kbps

(C) 76 kbps

(D) 84 kbps

Answer: (B)

Solution: sLine speed nf

8 8k 64k

=

= =

84. For an 8–PSK system, operating with an information bit rate of 24 kbps, the baud rate will be

(A) 16,000

(B) 12,000



44

(C) 8,000

(D) 6,000

Answer: (C)

Solution: Baud rate for M-Ary PSK b

2

R

Log M=

For 8-PSK, Baud rate 3

2

24 108000

1og 8

= =


85. During transformation of independent variable, if two signals identical in shape are displaced relative to

each other, then the difference in propagation time from point of origin of transmitted signal results in

(A) Time shift

(B) Time reversal

(C) Time scaling

(D) Time reduction

Answer: (A)

Solution: Difference between propagation time is Time shift

86. Which of the following statements is/are correct?

1. A continuous-time system is a system in which, continuous-time input signals are applied; resulting

in continuous-time output signals.

2. A system is said to be linear if it follows the superposition theorem.

3. A system is said to be non-linear if it follows the superposition theorem.

(A) 1 only

(B) 1 and 2 only

(C) 2 only

(D) 1 and 3 only

Answer: (B)


1. A continuous-time system is a system in which, continuous-time input signals are applied;

resulting in continuous-time output signals (True statements by definition of continuous time

system).



45

2. A system is said to be linear if it follows the superposition theorem (True statement by definition

of linear system)

3. A system is said to be non-linear if it follows the superposition theorem (False statement),the

true statement for non linear system is “it does not follow super position theorem.

So, statement 1,2 are correct & statement 3 is false

Hence the correct answer is option (B).

87. A discrete time signal is said to be unit sample sequence if

(A) ( )n 1for n 0

0 for n 0

= =

=

(B) ( )n 2 for n 0

0 for n 0

= =

=

(C) ( )n 1for n 0

0 for n 0

= − =

=

(C) ( )n 2 for n 0

0 for n 0

= − =

=

Answer: (A)

Solution: In this case we need to find expression of ( )n .

( )

By defination

n 1for n 0

0 for n 0

= =

=


88. A signal is said to be

1. Deterministic if there is no uncertainty over the signal at any instant of time.

2. Deterministic if it is expressible through a mathematical equation.

3. Random or non–deterministic if there is uncertainty over the signal at the instant of time.

4. Random or non–deterministic if it is not expressible through a mathematical equation.


(A) 1,2 and 3 only

(B) 1,2 and 4 only

(C) 3 and 4 only

(D) 1, 2, 3 and 4



46

Answer: (D)


1. A signal is said to be deterministic if there is no uncertainty over the signal at any instant of time

(True statement by definition)

2. A signal is said to be deterministic if it is expressible through a mathematical equation (True

statement by definition).

3. A signal is random or non–deterministic if there is uncertainty over the signal at the instant of

time. (True statement by definition).

4. A signal is random or non–deterministic if it is not expressible through a mathematical equation.

(True statement by definition).

So all the given 4 statement are true

Hence the correct answer is option D

89. Linear time-invariant systems that, are designed to pass some frequencies essentially undistorted and

significantly attenuate or eliminate others are

(A) Frequency-shaping filters

(B) Frequency-selective filters

(C) Time-shaping filters

(D) Time-selective filters

Answer: (B)

Solution: LTI systems that are designed to pass same frequencies essentially undistorted and significantly

alternate other are frequency selective filters (by definition)


90. If the input signal x(t) and impulse response h(t) of a continuous-time system are described as

( ) ( ) ( ) ( )3tx t e u t and h t u t 1 ,−= = − the output y(t) will be

(A) ( )3 t 11

1 e3

− − −

(B) 3t1

1 e3

− −

(C) ( )3 t 11

1 e3

− + +



47

(D) 3t1

1 e3

− +

Answer: (A)

Solution: For a continuous time system the input is ( ) ( )3tx t e u t−=

Impulse response, ( ) ( )h t u t 1= −

We need to obtain y(t)

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )0

at at

a t tat

0 0

y t x t *h t1

e u t * u t 1 e u t Standard resulta

1e u t *u t t 1 e u t t

a

− −

− −−

=

= −

− = − −

If we take a = 3, t0 = 1 then

( ) ( ) ( )

( ) ( )

( ) ( )

( )( )

3t

3 t 1

3 t 1

y t x t *h t

e u t * u t 1

11 e u t 1

3

1or 1 e for t 1

3

−

− −

− −

=

= −

= − −

= −


91. Consider the LTI system whose response to the input ( ) ( )t 3tx t e e u t− − = + is ( ) ( )t 4ty t 2e 2e u t .− − = −

The system is impulse response will be

(A) ( )2t 4t3e e u t

2

− − +

(B) ( )2t 4t3e e u t

2

− − −

(C) ( )2t 4t1e e u t

2

− − +

(D) ( )2t 4t1e e u t

2

− − −

Answer: (A)

Solution: It is given that for a LTI system

( ) ( )

( ) ( )

t 3t

t 4t

input x t e e u t

output y t 2e 2e u t

− −

− −

= +

= −

We need to obtain impulse response



48

( ) ( ) ( )

( )( )

( )

( )( )

y t x t *h t

2 2 s 11Y s s 1 s 4 s 4H s 2

1 1 s 1X s1

s 1 s 3 s 3

s 4 s 1 3

s 4 s 42 22s 4s 3 s 1

s 3s 3

3 s 3 s 32 3

s 4 2s 4 s 4 s 2

1/ 2 1/ 2 33

s 4 s 2 2

=

+ − − + + + = = = +

+ ++ + +

+ − − + + = = ++ + + + +

+ + = = + + + +

= + = + +

( ) ( )2t 4t

1 1

s 4 s 2

3h t e e u t

2

− −

+ + +

= +


92. Consider an LTI system with a system function

( )1

1H z

11 z

4

−

=

−

If difference equation will be

(A) ( ) ( ) ( )1

y n y n 1 x n2

− − =

(B) ( ) ( ) ( )1

y n y n 1 x n4

− − =

(C) ( ) ( ) ( )1

y n y n 1 x n2

+ − =

(D) ( ) ( ) ( )1

y n y n 1 x n4

− + =

Answer: (B)

Solution: The transfer function of a LT I system is

( )1

1H z

11 z

4

−

=

−

we need to obtain its time domain equation



49

( )( )

( )

( ) ( )

( ) ( ) ( )

1

1

Y z 1H z

1X z1 z

4

1Y z z Y(z) X z

4

1y n y n 1 x n

4

−

−

= =

−

− =

− − =

Hence the correct answer option (B).

93. It is assumed that quantization error, e(n) is a sequence of random variables where

1. The statistics do not change with time.

2. It is a sequence of uncorrelated random variables.

3. It is uncorrelated with the quantizer input x(n).

4. The probability density function is uniformly distributed over the range of values of quantization

error.


(A) 1, 2 and 3 only

(B) 1, 2 and 4 only

(C) 3 and 4 only

(D) 1, 2, 3 and 4

Answer: (B)

94. For a given differential equation,

( ) ( )

( ) ( )2

2

d y t dy t4 5y t 5x t

dt dt+ + =

With ( )( )dy t

y 0 1and 2dt −

−

= =

And input x(t) = u(t)

The output y(t) will be

(A) 2u(t) – 2e–2t sin t

(B) u(t) + 2e–2t sin t

(C) u(t) – e–t sin t

(D) 2u(t) + e–t sin t



50

Answer: (B)

Solution: It is given that

( ) ( )( ) ( )

( )

( )

( ) ( )

2

2

0

d y t dy t4 5y t 5x t

dt dt

y 0 1

dy t2

dt

x t u t

−

−

+ + =

=

=

=

We need to obtain y(t)

( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

( )( )

( )

( )( )

( )

( )

( )

( ) ( ) ( )

2 1 1

2 1

2 2

2 2

2

2

22

2 2

2t

5s Y s sy 0 y 0 4sY s 4y 0 5Y s

s

5Y s s 4s 5 s 4 y 0 y 0

s

s 4 1 2 5Y s

s 4s 5 s s 4s 5

s 6 s

s 4s 5 s s 4s 5

s s 6 5

s s 4s 5

s 6s 5 1 2

s s 4s 5s s 4s 5

1 12

s s 2 1

y t u t 2e sin t u t

− − −

− −

−

− − + − + =

+ + = + + +

+ + = +

+ + + +

++

+ + + +

+ +=

+ +

+ += = +

+ ++ +

= +

+ +

= + Hence the correct answer is option (B).

95. The Nyquist rate for the signal ( ) ( ) ( )1

x t cos 4000 t cos 100 t2

=

will be

(A) 5 kHz

(B) 10 kHz

(C) 15 kHz

(D) 20 kHz

Answer: (A)

Solution: The given signal is

( ) ( ) ( ) 1

x t cos 4000 t cos 1000 t2

=



51

We need to obtain nyquist rate of x(t)

If g(t) having maximum frequency 1f

h(t) having maximum frequency 2f then [k g(t) h(t)] will have maximum frequency 1 2f f .+

cos 4000 t have maximum frequency: 2kHz

cos1000 t have maximum frequency: 0.5 kHz

( ) ( ) ( )

m 1 2

1x t cos 4000 t cos100c t have maximum frequency

2

f f f 2 0.5 2.5kHz

=

= + = + =

So, Nyquist rate of x(t) is 2fm 2 2.5 5kHz.= =

Hence the correct answer is option is (A).

96. Which of the following statements is/are correct?

1. A system is said to be Finite Impulse Response (FIR), if the output samples of the system depend

only on the present input and a finite number of past or previous input samples.

2. If the output of a system y(n) depends only on the present input and past inputs, but not on past

outputs, then it is called a non-recursive system.

3. If the output of a system y(n) depends only on the present input and past inputs, but not on past

outputs, then it is called a recursive system.

(A) 1 only

(B) 1 and 2 only

(C) 1 and 3 only

(D) 3 only

Answer: (B)


1. In a FIR system output depends only on present input and a finite number of past or previous

input samples (True by definition of FIR filter)

( ) ( ) ( ) ( )

( ) 1 1 1

1 1 1

Ex : y n a x n b x n 1 c x n 2

h n a , b , c

= + − + −

=

2. If the output of a system y(n) depends only on present input and past input but not on past output

the it is called non recursive system. (True by definition)

( ) ( ) ( ) ( )( )Ex : y n ax n bx n 1 c x n 2 .= + − + −



52

3. If the output of a system y(n) depends only on present input and past input but not on past output

it is called recursive system (false in definition)

[Ex: ( ) ( ) ( )y n ay n 1 bx n= − + this is example of recursive system,

where present output y(n) depends on present input x(n) and past output]

So, statement 1,2 are correct & 3 is false.


97. The value of the steady state error for first order system, 1

sT 1+with Unit Ramp Function will be

(A) 1

T

(B) T

(C) t

TT 1 e−

−

(D)

t

T1

eT

−

Answer: (B)

Solution: A first order system having transfer function ( )1

G s ,sT 1

=+

we need to obtain steady state error

due to unit Ramp input.

Assuming G(s) is transfer function of 1st order unity feedback closed loop system; its corresponding

open loop transfer function is ( )1

1G s , then

sT=

( )ss

1s 0

x 0

1 1 1e T

1 1lims G slims

sT T→

→

= = = =

[Note: if we take G(s) as open loop transfer function sse = ]

Hence the correct answer is option is (B).

98. If the number of zeros are less than the number of poles, i.e. Z < P, then the value of the transfer function

becomes zero for s → ∞. Hence we say that there are zeros at infinity and the order of such zeros is

(A) P + Z

(B) P – Z

(C) Z – P

(D) Z



53

Answer: (B)

Solution: If number of zero are less than number of poles, then the value of transfer function becomes zeros

for s→ . Hence, we say that there are zeros at infinity and the order of such zeros is

P-Z.

In root locus diagram when we have Z<P, then P-Z number of zeros exist as s .→


99. The method for determination of the stability of the feedback system as a function of an adjustable gain

parameter which does not provide detailed information concerning location of closed-loop poles as a

function of gain K is called

(A) Root locus method

(B) Nyquist criterion method

(C) Bode plot method

(D) Routh-Hurwitz criterion method

Answer: (D)

Solution: In R-H criterion only stability information can be accessed however location of poles can’t be

commented.


100. Consider the sinusoidal transfer function in time-constant form

( )( )

2

2 1 jG j

j1

10

+ =

+

Asymptotic log magnitude characteristic of factor (1 + jω) is straight line of

1. 0 dB for ω ≤ 1.

2. +20 dB/decade for ω ≥ 1.

3. –20 dB/decade for ω ≥ 1.

Which of the above relations is/are correct?

(A) 1 only

(B) 1 and 2 only

(C) 1 and 3 only

(D) 2 only



54

Answer: (B)

Solution: The transfer function given is ( )( )

2

2 1 sG s

s1

10

+= +

we need to comment about (1+s) in its asymptotic

bode plot.

If we have the term ( )s1 sT+ , the corner frequency is s

1

Tso when

s

1,

T the term contribute 0db.

When s

1,

T the term contributes 20db/dec

Here sT 1, So 1scope is 0db

1 scope is 20db / dec

=

Hence statement 1, 2 are correct & 3 is false.


101. A graphical technique for plotting the closed-loop poles of a rational system functions as a function of the

value of gain for both continuous-time and a discrete-time system is

(A) Root locus method

(B) Nyquist criterion method

(C) Bode plot method

(D) Routh-Hurwitz criterion method

Answer: (A)

Solution: The graphical techniques for plotting closed loop poles of a rational system function as a function of

the value of gain for both continuous and discrete time system is Root locus method.


102. Which one of the following statements regarding ‘Root locus’ is not correct?

(A) By addition of poles to left half, the root locus shifts towards right hand side and stability of system

decreases.

(B) By addition of zero towards left, the root locus shifts towards left half, since root locus shifts toward

left half, the relative stability of control system increases.

(C) By addition of zero towards left side, the root locus shifts towards left half, the relative stability

remains same.

(D) By addition of poles to the left half, the system stability decreases, while by addition of zeros

towards left half, the stability of system increases.



55

Answer: (C)

Solution: When poles are added left half of s plane

i. Root locus shift towards right

ii. Relative stability decreases

when zeroes added in Right half of s plane

i. Root locus shift towards left.

ii. Relative stability increases

In option C it is given that when zeros added on LHS relative stability remain same, so this

statement is false


103. In time domain, the relative stability is measured by maximum overshoot and damping ratio. In frequency

domain the relative stability is measured by

(A) Steady state error

(B) Damping ratio

(C) Resonant peak

(D) Bandwidth

Answer: (C)

Solution: In time domain relative stability is measured by maximum overshoot, damping ratio similarly in

frequency domain its by resonant peak hence the correct answer is option C.

104. Consider a feedback system with the characteristic equation

( )( )

11 K 0

s s 1 s 2+ =

+ + for root locus.

The angles of asymptotes A and the centroid of the asymptotes A− are respectively

(A) 60°, 120°, 180° and –1

(B) 45°, 90°, 300° and 0

(C) 60°, 180°, 300° and –1

(D) 45°, 90°, 180° and 0

Answer: (C)



56

Solution: ( ) ( )

( )

( )

A

open loop poles open loop zeros

p z

0 1 2 01

3 0

− =

−

− − −= = −

−

( )

( )( )

A

2q 1 180, q 0,1, 2. .....

p z

2q 1 1802q 1 60

3 0

60, 180, 300 for q 0, 1, 2

+ = =

−

+= = +

−

= =


105. Which one of the following statements regarding an effect of phase lead network is not correct?

(A) The velocity constant is usually increased.

(B) The slope of the magnitude curve is reduced at the gain crossover frequency, as a result relative

stability improves.

(C) Phase margin increased.

(D) The bandwidth decreased

Answer: (D)

Solution: A phase lead network adds a zero to an existing system which increases the slope and hence

increases the frequency/bandwidth as evident from bode plot.

Hence, the correct option is (D).

106. A lead compensator

1. Speeds up the transient response.

2. Increases the margin of stability of system.

3. Helps to increases the system error constant though to a limited extent.


(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (A)

Solution: A lead compensator adds a zero to an existing system and has the following effect.



57

● Improves/speeds up to transient response

● Increases the marginal stability

● Decreases the type of the system, increases steady state error and decreases error coefficient as

Type ( ) ( )ss

ss p v a

1 1and e

e k or k or k



1. The pair (AB) is controllable implies that the pair (ATBT) is observable.

2. The pair (AB) is controllable implies that the pair (ATBT) is unobservable.

3. The pair (AC) is observable implies that the pair (ATCT) is controllable.

4. The pair (AC) is observable implies that the pair (ATCT) is uncontrollable.

(A) 1 and 3 only

(B) 1 and 4 only

(C) 2 and 3 only

(D) 2 and 4 only

where: A, B and C are having their standard meanings.

Answer: (A)

Solution: From duality properties of controllability and observability, we now

● If pair (A B) is controllable, then pair (ATBT) is observable.

● If pair (A C) is observable, then pair (ATCT) is controllable .


108. Bounded-input bounded-output stability implies asymptotic stability for

1. Completely controllable system

2. Completely observable system

3. Uncontrollable system

4. Unobservable system


(A) 1 and 4 only

(B) 1 and 2 only

(C) 2 and 3 only

(D) 3 and 4 only



58

Answer: (B)

Solution: If a linear system is BIBO system and the state space representation is minimal i.e., both

controllable and observable, then the system is asymptotically stable.

Hence, the correct option is (B).

109. The degree of humming level of the noise caused in the transformers may be reduced by

(A) Magnetostriction

(B) High flux density in core

(C) Tightening of core by clamps

(D) Quality of transformer oil

Answer: (C)

Solution: Various ways to reduce humming sound in transformer due to magnetostriction are

● using cushion padding and oil barrier

● using innovation design like stiffeners while constructing the walls of the transformer bank

● proper fixing and clamping of the lamination frames


110. A transformer with a 10 : 1 ratio and rated at 50 kVA, 2400/240 V, 50 Hz is used to step down the voltage

of distribution system. The low tension voltage is to be kept constant at 240 V. If the transformer is fully

loaded at 0.8 power factor (lag), the load impedance connected to low-tension side will be nearly,

(A) 3.15 Ω

(B) 2.60 Ω

(C) 1.15 Ω

(D) 0.60 Ω

Answer: (C)

Solution: ( )

22

3

240VZ 1.15

VA 50 10= = =


111. A DC shunt generator supplies a load of 7.5 kW at 200 V. If the armature resistance is 0.6 Ω and field

resistance is 80 Ω, the induced emf will be

(A) 224 V

(B) 218 V

(C) 212 V



59

(D) 204 V

Answer: (A)

Given, a fV 200V, r 0.6, r 80= = =

3

f

A a f

a a

7.5 10I 37.5A

200

200I 2.5A

80

I I I 40

E V I r 200 40 0.6 200 24 224V

= =

= =

= + =

= + = + = + =


112. A 4-pole DC motor is lap-wound with 400 conductors. The pole shoe is 20 cm long and the average flux

density over one-pole-pitch is 0.4 T, the armature diameter is 30 cm. When the motor is drawing 25 A

and running at 1500 rpm, the torque developed will be nearly

(A) 30 Nm

(B) 40 Nm

(C) 50 Nm

(D) 60 Nm

Answer: (A)

Solution: av

av av

P 4, Z 400, 0.2m, B 0.4T, D 0.3m

Area DB B

Pole P

0.4 0.3 0.20.006 wb

4

= = = = =

= =

= =

We have a

ZPK constant

2 A= =

But given lap winding, so P = A

a

a

a a a

400 200K

2

For I 25A,

Torque, T K . .I

2000.006 25 30N m

= =

=

=

= = −

113. In an unloaded shunt generator, when switch is closed, a small field current is produced which leads to

generation of still larger voltages due to addition of



60

(A) Armature voltage

(B) Residual flux voltage

(C) Generated voltage

(D) Voltage drop

Answer: (B)

Solution: Voltage build up in dc shunt generator takes place when field mmf aids the residual flux.


114. The generator efficiency of a shunt generator will be maximum when its variable loss is equal to

(A) Constant loss

(B) Stray loss

(C) Iron loss

(D) Friction and windage loss

Answer: (A)

Solution: The condition of maximum efficiency in a dc machine is

“Variable loss = constant loss”


115. Induction motor can be regarded as a generalized transformer due to certain similarities except rated

(A) Frequency

(B) Flux

(C) Speed

(D) Induced emf

Answer: (C)

Solution: Induction motor is a rotating electrical device whereas transformer is static, hence speed cannot be

similar between them.


116. A 3-phase, 400/200 V, Y-Y connected wound-rotor induction motor has 0.06 Ω rotor resistance and 0.3 Ω

standstill reactance per phase. To make the starting torque equal to the maximum torque, the additional

resistance required in the rotor circuit will be



61

(A) 0.24 Ω/phase

(B) 0.34 Ω/phase

(C) 0.42 Ω/phase

(D) 0.52 Ω/phase

Answer: (A)

Solution: Given 20 20r 0.06 , x 0.3= =

For maximum starting torque

2020

r 'x and s 1

5= = where 20r ' is new rotor resistance.

Then, 20 20r ' x 0.3= = or

20, extra 20r 0.3 r

0.3 0.06 0.24 / phase

= −

= − =

117. Potier triangle method is helpful in obtaining the voltage regulation of synchronous machines by

determining the armature

(A) Leakage reactance and its reaction mmf

(B) Leakage reactance and air-gap flux

(C) Resistance and its reaction mmf

(D) Resistance and air-gap flux

Answer: (A)

Solution: Potier triangle method of determining voltage regulation of synchronous machine is based on

determination of armature leakage reactance and its reaction mmf.


118. In a synchronous motor, the magnitude of stator back emf, Eb depends on

(A) Speed of the motor

(B) Load on the motor

(C) Both the speed and rotor flux

(D) Rotor excitation only

Answer: (C)

Solution: Stator back emf, b w phE 4.44 k fN=



62

( )b

b

E rotor flux and

E f speed

So, Eb depends on both the speed and rotor flux.


119. A stepper motor has a step angle of 2.5°. If the shaft is to make 25 revolutions, the number of steps

required will be

(A) 1800

(B) 2200

(C) 2800

(D) 3600

Answer: (D)

Solution: Given, 2.5 , = Number of revolutions = 25 = n

Number of steps = n.360 2.5 360

36002.5

= = =

Hence, the correct option is (D)

120. In which of the following respects, the servomotors differ in application capabilities from large industrial

motors?

1. They produce high torque at all speeds including zero speed.

2. They are capable of holding a static position.

3. They do not overheat at standstill or lower speeds.

4. Due to low-inertia they are not able to reverse direction quickly.

(A) 1, 2, 3 and 4

(B) 1, 2 and 3 only

(C) 1, 2 and 4 only

(D) 3 and 4 only

Answer: (B)

Solution: Statement 4 is incorrect as due to low inertia reversal of direction is quick.

Hence, by option elimination, the correct option is (B).



63

121. Which of the following statements regarding steam boilers are correct?

1. The boiler must be capable of quick starting and loading.

2. The boiler should have no joints exposed to flames.

3. The boiler must be capable of burning low ash content coal efficiently.

(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (D)

Solution: Characteristics of a good steam Boiler are

● It should have maximum steam generation with minimum fuel consumption.

● It should be economical to install

● It should be capable of quick starting and quick shut down

● It should be light in weight

● It should meet rapid fluctuation of loads


122. Which of the following are the main parts of a power system?

1. Generating stations

2. Transmission systems

3. Distribution network

(A) 1 and 2 only

(B) 1 and 3 only

(C) 2 and 3 only

(D) 1, 2 and 3

Answer: (D)

Solution: A power system mainly consists of

● Generating stations

● Transmission system

● Distribution system

Hence, the correct option is (D)



64

123. Which of the following factors affect Corona?

1. Atmospheric conditions, temperature, humidity, moisture, ice and fog

2. Current of conductor

3. Waveform

4. Condition of surface of conductors, smoothness and dust

(A) 1, 2 and 3 only

(B) 1, 3 and 4 only

(C) 1, 2 and 4 only

(D) 2, 3 and 4 only

Answer: (B)

Solution: ● Power loss in Corona is given by

( )25

L ph c

f 25 rP 242 10 V V kW/km/ph

d

− + = −

● Critical disrupting voltage is given by

( )C 0

dV m g r n kV/ph rms

r

=

where,

0m = irregularity or stranding factor

g = 212kV/cm (rms)

s = air density factor 3.92b

;T 273

=+

b = barometric pressure in cm

From above expression we may conclude that corona is independent of conductor’s current.


124. Bundled conductors that are used to increase line voltage in EHV lines for raising critical corona voltage

depend on

(A) Number of conductors in the group

(B) Voltage gradient

(C) Optimum spacing

(D) Communication interference

Answer: (C)



65

125. In a 275 kV transmission line with line constants A 0.85 5 and B 200 75 ,= = if the voltage profile at

each end is to be maintained at 275 kV, the power at Unity Power Factor (UPF) will be nearly

(A) 98 MW

(B) 118 MW

(C) 144 MW

(D) 184 MW

Answer: (B)

Solution: Given, S RV V 275kV, A 0.85 5 , B 200 75 , pf 1= = = = =

We know

( ) ( )

( ) ( )

2

S R R

R

2

S R R

R

V V A VP cos cos

B B

V V A VQ sin sin

B B

= − − −

= − − −

Since pf =1 RQ 0 =

i.e., ( ) ( )

( ) ( )

2

S R RV V A Vsin sin 0

B B

sin 75 0.85 sin 75 5 0

22

− − − =

− − − =

=

Then, ( ) ( )

( )

2

S R R

R

2

V V A VP cos cos

B B

275cos53 0.85 cos70 118

200

= − − −

= −


126. When the sinusoidal steady state current is called the symmetrical short circuit current, then the

unidirectional transient component is called

(A) AC short circuit current

(B) DC short circuit current

(C) AC offset current

(D) DC offset current

Answer: (D)

Solution: Current during transients on a transmission line is given by



66

( ) ( ) ( )bRVm m

DC offset current AC steady state fault current

V Vi t e sin sin t

z z

−= − + + −

1 Ltan

R

switching angle

− =

=


127. Consider the following balanced line-to-neutral voltages with abc sequence:

an

p bn

cn

V 277 0

V V 277 120 volts

V 277 120

= = − +

The values of V0, V1 and V2 are respectively

(A) 0, 177 0 and 177 0

(B) 277 0 , 0 and 0

(C) 0, 277 0 and 0

(D) 277 0 , 0 and 177 0

Answer: (C)

Solution: Given,

an

p bn

cn

V 277 0

V V 277 120

V 277 120

= = − +

We have to find symmetrical components

AC symmetrical currentDC offset current

t

( )i t



67

a0 an

2

012 a1 bn

2

a2 cn

a0

2

a1

2

a2

a0 a1 a2

V 1 1 1 V1

V V 1 V3

V 1 V

V 1 1 1 277 01

V 1 277 1203

V 1 277 120

V 0, V 277 0, V 0

= =

= − +

= = =

NOTE: Since pV is balanced, so only positive sequence component will be non zero.


128. In an HVDC transmission, the DC output voltage can be controlled to get inverter operation when the

firing angle α is

(A) α = 0°

(B) 0° < α < 90°

(C) 90° < α < 180°

(D) 0° < α < 180°

Answer: (C)

Solution: The dc output voltage of a full wave converter proportional to cos

i.e, oV cos

where, for 0 90; cos 0 normal mode and for 90 180; cos 0 Inverter mode

Hence the correct option is (C)

129. In an HVDC transmission mode, the link which has two circuits that are almost independent of each other

is called

(A) Monopolar link

(B) Bipolar link

(C) Homopolar link

(D) Dualpolar link

Answer: (B)

Solution: In bipolar HVDC link, each direct current DC cable is connected to an independent converter as

shown below.



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The earth power is set at a single point in each station hence it exhibit the independent control for the

positive and negative poles.

130. In a photovoltaic system, there is a thermally generated small reverse saturation current which flows even

in the absence of light, called

(A) Photon current

(B) Diode current

(C) Leakage current

(D) Dark current

Answer: (D)

Solution: Dark current:

It is the relatively small electric current that flows through photosensitive devices given, when no

photons are entering the device, it consists of the charges generated in the detector when no outside

radiation is entering the detector.


131. In a power system, due to interconnection or grid formation and transmission line redundancy, the ability

to serve all power demands without failure over long periods of time, is due to

(A) Power system quality

(B) Power system reliability

VSC1+

VSC2+

VSC1

−

VSC2

−

T1+

T1−

HVDC cable −

T2+

T2−AC grid1 AC grid 2

HVDC cable +



69

(C) Computers and microprocessors

(D) Reserve generating capacity

Answer: (B)

Solution: Power system reliability:

It refers to the ability of power system to provide adequate and stable power demands without

failure over longer periods of time.


132. In wind power, the speed which is considered as the single most important parameter is

(A) Wind speed

(B) Peripheral speed

(C) Tip speed

(D) Blade speed

Answer: (A)

Solution: Power in the wind = 31

AV2

Where, A = swept area of blades = ( )2 2r m

V = wind speed (m/s)

ρ = density of air (Kg/m3)


133. Communication circuitry is an extra circuit used to turn off

(A) Line-commutated thyristors

(B) Phase-commutated thyristors

(C) Force-commutated thyristors

(D) Reverse-commutated thyristors

Answer: (C)

Solution: Forced commutation: -

(i) Forward current of the thyristor is forced to be zero by an external circuit known as commutation

circuit

(ii) It includes class B, C and D commutation

(iii) It is usually employed in dc choppers and inverter. Hence, the correct option is (C)



70

134. TRIAC as a bidirectional triode thyristor is used to control the output voltage by varying conduction time

or firing delay angle in

(A) AC-DC converters (Controlled rectifiers)

(B) AC-AC converters (AC voltage controllers)

(C) DC-DC converters (DC choppers)

(D) DC-AC converters (Inverters)

Answer: (B)

Solution: The given options are

Option A: AC-DC converters

(i) They convert constant ac voltage to variable dc output voltage

(ii) They use thyristors

Option B: AC-AC converters

(i) They converter fixed ac input voltage into variable ac output voltage

(ii) They employ two thyristors in antiparallel i.e. triac

Option C: DC-DC converters

(i) They convert fixed dc input voltage to a controllable dc output voltage.

(ii) For higher power circuits, thyristors are used

(iii) For lower power circuits, transistors are used

Option D: DC-AC converters

(i) They convert fixed dc voltage to a variable ac voltage

(ii) For low power applications, power transistors are used

(iii) For high power applications thyristors and GTO’s are used

Hence, the correct option is (B)

135. For large power output, multiphase rectifiers are used along with filters to reduce level of harmonics by

increasing the fundamental frequency in

(A) Diode rectifier

(B) Bridge rectifier

(C) Star rectifier

(D) Delta rectifier

Answer: (C)

Solution: Multiphase star rectifier

(i) Used for larger power output



71

(ii) Because of higher frequency ripples, filtering process simplifies i.e size of filter reduces due

to higher ripple frequency

Hence, the correct option is (C)

136. In a Bipolar Junction Transistor (BJT) due to current flow to small portion of the base, hot spots are

produced causing localized excessive heating and damaging the transistor.

This switching limit is called

(A) Forward-Biased Safe Operating Area (FBSOA)

(B) Reverse-Biased Safe Operating Area (RBSOA)

(C) Power Derating

(D) Second Breakdown (SB)

Answer: (D)

Solution: The option given are

Option A: Forward biased safe operating area (FBSOA)

It pertains to the transistor operation when base emitter junction is forward to turn on the transistor

Option (B): Reverse biased safe operating area (RBSOA)

Safe operating area for transistor during turnoff is known as RBSOA.

Option C: Power Derating

It is a factor of a transistor by which the power dissipation rating of a transistor falls when the

transistor junction temperature increases

Option D: Secondary Break down

It is a failure mode in a power transistor with large junction area where under certain condition of

current and voltage, the current concentrates in small spots of the base emitter junction causing

local heating and finally resulting in a short between collector and emitter.


137. In a three-phase inverter with 180° conduction, there are six modes of operation in a cycle where duration

of each mode is

(A) 90°

(B) 75°

(C) 60°

(D) 45°



72

Answer: (C)

Solution: In 3 phase inverter with 180°, conduction mode, we have 6 modes for 1 cycle (i.e. 360°)

Therefore for 1 mode, 60° angle will be available hence the correct option is (C).

138. In a closed-loop control of squirrel cage induction motor, the field-oriented control strategy implemented is

(A) Scalar control

(B) Vector control

(C) Adaptive control

(D) Frequency control

Answer: (B)

Solution:

Vector Control: Here, the stator current of a 3− AC electric motor is identified as two orthogonal

components one is magnetic flux and the other is torque.

Instantaneous torque control is possible

Scalar control: It lacks the feature of instantaneous torque control hence vector control is preferred

as such places


139. In a DC motor drive, if the armature current is revised by keeping field current positive producing a

braking torque, then the drive is said to be operating is

(A) Motoring mode

(B) Regenerative braking mode

(C) Dynamic braking mode

(D) Plugging mode

Answer: (D)

Solution: As per the definition of plugging in motors, the correct option is (D)

Vector control Scalar control

control methodVFD



73

140. If the induction motor drive is capable of bidirectional power flow where limited range of speed control is

required for large power applications, then this arrangement is called

(A) Static conductance drive

(B) Static Scherbius drive

(C) Static compressive drive

(D) Static reluctance drive

Answer: (B)

Solution: To convert slip power to AC line and return it back to the line, two types of approach are used

1. Static Kramer Drive: only allows operation at sub-synchronous speed i.e unidirectional power

flow

2. Static Scherbius Drive: operates at both below and above synchronous speed i.e.

bidirectional power flow


141. In a DC-DC switched-mode converter if the output voltage polarity is opposite to input voltage, then this

inverting regulator is called

(A) Buck regulator

(B) Boost regulator

(C) Buck-Boost regulator

(D) Cuk regulator

Answer: (C)

Solution: Circuit of Buck boost regular is

(polarity assumed same as input)

sV

sw

+

−

0VR



74

Since the polarity developed due to flow of LI thorough load R is opposite to

0V polarity (same as input

voltage) and so ( )o s

DV V

1 D= −

−

Hence, the correct option is (C)

142. In a zero current switching resonant converter, the switching loss and noise are increased due to presence

of capacitive coupling-called

(A) Miller capacitor

(B) Series resonant capacitor

(C) Parallel resonant capacitor

(D) Switch capacitor

Answer: (A)

Solution: In zero current switching resonant converter, during turn ON, considerable rate of change of voltage

is coupled to the gate driven circuit through a capacitor called Millen capacitor thus increasing loss

and noise. It is one limitation. Another is that switches are under high current stress resulting in

higher conduction loss.

143. Which one of the following devices is not a switched-mode DC power supply?

(A) Fly back forward converter

(B) Full bridge converter

(C) Push-pull converter

(D) Resonant converter

Answer: (D)

Solution: The various configurations of SMPS converters are

● Flyback converter

● Push pull converter

LI

sw ON sw OFF

+

−

LIR



75

● Full bridge converter

● Half bridge converter

So, resonant converter is not included in it


144. The ideal core should exhibit very high permeability in case of transformers and inductor core due to

magnetic saturation caused by DC imbalance condition that can be minimized by

(A) Low permeability core only

(B) High permeability core only

(C) Low and high permeability combination core

(D) No permeability core

Answer: (C)

Directions: Each of the next four (4) items consists of two statements, one labelled as ‘Statement (I)’ and the

other as ‘Statement (II)’. You are to 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

145. Statement (I): In a substitutional semiconductor, atom is replaced by an occasional foreign atom. The

imperfections may be deliberately controlled or created in transistor material.

Statement (II): The lattice vacancies created when certain atoms in a semiconductor are missing are

known as Schottky defects.

Answer: (B)

Solution: In substitutional semiconductor, parent atom is replaced by foreign atom. It may be deliberately

controlled or altered so as to improve the semiconductor usability. Schottky defects are lattice

vacancies created when a pair of opposite charge ions are missing. It maintains the neutrality of the

atom but decreases the density.



76

Hence, both the statements are individually correct, but statement II is not the correct explanation of

Statement-I.


146. Statement (I): A cache is a memory unit placed between the CPU and main memory M and is used to

store instructions, data or both.

Statement (II): The cache's effect is to increase the average time required to access an instruction or

data word, typically to just a single-clock cycle.

Answer: (C)

Solution: A cache is a memory unit which is located between main memory and CPU. It works based on

“locality of reference” principle. It contains the most referenced instructions & data which results in

reducing the average access time of memory.

147. Statement (I): A buffer is not an area in RAM or on the hard drive designated to hold input and

output on their way in or out of the system.

Statement (II): The process of placing items in a buffer so they can be retrieved by the appropriate

device when needed is called spooling.

Answer: (D)

Solution: Buffer means that, it is temporary storage. It is part of every memory unit [main (or) secondary]

148. Statement (I): The power diodes are three-layer devices.

Statement (II): The impurity concentrations of power diodes vary layer to layer.

Answer: (B)

Solution: The structure of power diode is as shown below

n substrate+

n−

p+

Anode+

Cathode−



77

From above figure, we can see it is a three-layered device.

Also n+substrate is heavily doped and n- is lightly doped so impurity concentration of power diodes vary

from layer to layer.


149. Statement (I): Registers are used for storage of small data in the microprocessor.

Statement (II): All registers are accessible to the user through instructions.

Answer: (C)


1. Registers are used for small data storage in microprocessor (True, mostly they are

of 8 bit/16bit)

2. All register are accessible to user through instructions (false, in 8085 microprocessor w, z can’t

be accessed).

So, statement 1 is true and statement 2 is false


150. Statement (I): In a three-phase induction motor, the maximum torque is directly proportional to

standstill reactance.

Statement (II): In a three-phase induction motor, the speed or the slip at which maximum torque

occurs is determined by the rotor resistance.

Answer: (D)

Solution: In a 3− inductor motor, max

2

1T

x

i.e., maximum torque inversely proportional to standard reactance.

Hence, Statement-I is false.

2m

2

rS

x= i.e, slip at which maximum torque occurs depends upon rotor resistance.


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(PAPER II) ELECTRICAL ENGINEERING 1