GATE

EEE

GrandTests

Test-13Solutions

APTITUDE

One Mark Questions:

1. Choose the word or group of words which is most similar in

meaning to the word printed in bold.

Gallant

a) Quaggy

b) Snug

c) Miser

d) Brave

Answer: (d)

Solution:

Meaning: The meaning of gallant is brave, noble or polite.

Gallant is defined as a man who is polite and attentive to

women.

Example: A gallant firefighter saved the people.

Synonyms: Bold, Courageous, Fearless, Valiant

Antonyms: Coward, Fearful, Meek, Unheroic

2. Choose the word which is most similar in meaning to the word

printed in bold.

Paucity

a) Fewness

b) Pace

c) Sufficiency

d) Abundance

Answer: (b)

1

Solution:

Meaning: Paucity is defined as a small number of something.

Example: Paucity of evidence has made his first case difficult.

Synonyms: Dearth, Smallness, Deficiency, Scarceness

Antonyms: Abundance, Affluence, Plenty, Enough

3. When she first came to France from Bulgaria, she was hardly

the _________ student she later made herself out to be, since

she had access to considerable family wealth.

a) naive

b) precocious

c) impecunious

d) ambitious

Answer: (c)

4. The belief that politicians might become ________ after their

election to office led to the appointment of ethics officers at

various levels of government.

a) scrupulous

b) entrenched

c) venal

d) puzzled

Answer: (c)

Solution:

If a certain belief led to the appointment of ethics officers, that

belief must concern some ethical issue. Of the choices provided,

only “venal” fits that context. Although several of the other

2

choices are not necessarily positive characteristics, none of them

involves ethics.

5. Seats for Mathematics, Physics and Biology in a school are in

the ratio 5: 7: 8. There is a proposal to increase these seats by

40%, 50% and 75% respectively. What will be the ratio of

increased seats?

a) 2 : 3 : 4

b) 6 : 7 : 8

c) 6 : 8 : 9

d) None of these

Answer: (a)

Solution:

Originally, let the number of seats for Mathematics, Physics and

Biology be 5x, 7x and 8x respectively.

Number of increased seats are (140% of 5x), (150% of 7x) and

(175% of 8x).

3

Two Marks Questions:

6. In question given below out of four alternatives, choose the one

which can be substituted for the given word/sentence.

Commencement of adjacent words with the same letter

a) Alliteration

b) Oxymoron

c) Pun

d) Transferred epithet

Answer: (a)

Solution:

Alliteration

7. In the following question, a sentence is given which has an

underlined part. Select the best replacement for the underlined

phrase from the given options.

One of the reasons for employers giving preference to

employees from India is that they are speaking at least two

languages, one of those their mother tongue.

a) India is that they are speaking at least two languages, one of

those

b) India is that they speak at least two languages, one being

c) India is that they speak at least two languages, one including

d) India is that they are to speak at least two languages, one

being

Answer: (b)

4

Solution:

Present indefinite tense is used in denoting an ability (speaking

languages).

8. X and Y started a business with capitals Rs. 20000 and Rs.

25000. After few months Z joined them with a capital of Rs.

30000. If the share of Z in the annual profit of Rs. 50000 is Rs.

14000, then after how many months from the beginning did Z

join?

a) 7

b) 6

c) 3

d) None of these

Answer: (d)

Solution:

Investments of X, Y and Z respectively are Rs. 20000, Rs.

25000 and Rs. 30000

Let investment period of Z be x months.

Ratio of annual investments of X, Y and Z is (20000 × 12):

(25000 × 12) : (30000 × x)

= 240: 300: 30x = 8: 10: x

The share of Z in the annual profit of Rs. 50000 is Rs. 14000.

=> [x/(18 + x)] 50000 = 14000 => [x/(18 + x)] 25 = 7

=> 25x = 7x + (18 × 7) ⇒ x = 7 months.

5

Z joined the business after (12 – 7) months. i.e., 5 months.

received by A per annum for being a working partner = 1500 ×

12 = Rs.18000.

Let 'P' be the part of the remaining profit that A receives as his

share. Total income of A = (2500 + 18000 + P)

Total income of B = only his share from the remaining profit =

'P', as A and B share the remaining profit equally.

Income of A = Twice the income of B

(2500 + 18000 + P) = 2(P)

P = 20500

Total profit = 2P + 18000

= 2 × 20500 + 18000 = 59000

9. The number of ways of arranging n students in a row such that

no two boys sit together and no two girls sit together is m(m >

100). If one more student is added, then number of ways of

arranging as above increases by 200%. The value of n is:

a) 12

b) 8

c) 9

d) 10

Answer: (d)

Solution:

If n is even, then the number of boys should be equal to number

of girls, let each be a.

⇒ n = 2a

6

Then the number of arrangements = 2 × a! × a!

If one more students is added, then number of arrangements,

= a! × (a + 1)!

But this is 200% more than the earlier

⇒ 3 × (2 × a! × a!) = a! × (a + 1)!

⇒ a + 1 = 6 and a = 5

⇒ n = 10

But if n is odd, then number of arrangements, = a!(a + 1)!

Where, n = 2a + 1

When one student is included, number of arrangements,

= 2(a + 1)!(a + 1)!

⇒ By the given condition, 2(a + 1) = 3, which is not possible.

10. Two parallel chords on the same side of the center of a circle

are 5 cm apart. If the chords are 20 and 28 cm long, what is the

radius of the circle?

a) 14.69 cm

b) 15.69 cm

c) 18.65 cm

d) 16.42 cm

Answer: (b)

Solution:

Draw the two chords as shown in the figure. Let O be the center

of the circle. Draw OC perpendicular to both chords. That

divides the two chords in half.

7

So CD = 10 and AB = 14. Draw radii OA and OD, both equal to

radius r.

We are given that BC = 5, the distance between the two chords.

Let OB = x.

We use the Pythagorean Theorem on right triangle ABO

AO² = AB² + OB²

r² = 14² + x²

We use the Pythagorean Theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB + BC = x + 5, so

r² = 10² + (x + 5)²

So we have a system of two equations:

r² = 14² + x²

r² = 10² + (x + 5)²

Since both left sides equal r², set the right sides equal to each

other.

14² + x² = 10² + (x + 5)²

196 + x² = 100 + x² + 10x + 25

196 = 125 + 10x

71 = 10x ⇒ 7.1 = x

r² = 14² + x²

r² = 196 + (7.1)²

r² = 196 + 50.41 = 246.41

r = √246.41

r = 15.69745202 cm

8

TECHNICAL

One Mark Questions:

11. Biotransformation of an organic compound having

concentration (x) can be modeled using an ordinary differential

equation dx + kx2 = 0, where k is the reaction rate constant. If x

dt

= a at t = 0 then solution of the equation is

a) x = ae–kt

b) 1 = 1 + kt

x a

c) x = a(1 – e–kt)

d) x = a + kt

Answer: (b)

Solution:

Given dx = −kx2…….. (1)

dt

and x = a at t = 0 ……... (2)

⇒ ∫ dx = −k ∫ dt − c

x2

⇒ − 1 = −kt − c

x

⇒ 1 = kt + c

x

By (2), (3) becomes, 1 = 0 + c

a

∴ The solution of (1) is 1 = kt + 1

x a

9

12. Using Cauchy's integral theorem, the value of the integral

(integration being taken in contour clockwise direction)

∫c z3−6 dz is where C is |z| = 1

3z−i

a) 2π − 4πi

81

b) π − 6πi

8

c) 4π − 6πi

81

d) 1

Answer: (a)

Solution:

I = ∫C z3−6 dz = 1 ∫|z|=1 z3−6 dz

3z−i 3 z−i/3

The integrand is not analytic at z = i/3 which lies insides C.

By Cauchy integral formula = 1 2πi ��3i �3 − 6� = 2π − 4πi

3 81

13. A particle, starting from origin at t = 0 s, is traveling along x -

axis with velocity v = π cos �π2 t�m/s. At t = 3 s, the difference

2

between the distance covered by the particle and the magnitude

of displacement from the origin is _______.

Answer: 2

Solution:

Displacement x = ∫ vdt

= sin �π2t� + C

At t = 0 ⇒ x = 0

∴C=0

10

⇒ x = sin �π2t�

After t = 3s ⇒ x = −1

i.e., the particle is moved '1' unit left of the origin.

But x = sin �π2t� is oscillating

After t = 3s, the total distance moved by the particle is '3' units

Since At t = 0 ⇒ x = 0

At t = 1 ⇒ x = 1

At t = 2 ⇒ x = 0

At t = 3 ⇒ x = −1

The required difference = 3 − |−1|

=3–1=2

14. A parallel circuit consists of two branches. One branch has RL

and L connected in series and the other branch has RC and C

connected in series.

Consider the following statements:

1. The two branch currents will be in quadrature if RLRC = L/C.

2. The impedance of the whole circuit is independent of

frequency, if RL = RC and ω = 1/√LC.

3. The circuit is in resonance for all the frequencies if RL= RC.

4. The two branch currents will be in phase at ω = 1/√LC.

Which of the above statements are correct?

a) 1 and 2

b) 2 and 3

c) 1 and 3

11

d) 3 and 4

Answer: (a)

Solution:

Calculate Yin(jω). The impedance of the whole circuit is

independent of frequency when imaginary Yin(jω) = 0 which

gives RL = RC and ω = 1/√LC. In other words, circuit is in

resonance for RL and RC and ω = 1/√LC. i.e., not for all the

frequencies.

Further, two branch currents will be in quadrature when

tan−1 �ωC1RC� + tan−1 �ωRLL� = 900 which gives RLRC = L

C

15. The circuit given below is constituted by an iron cored coil and

a capacitor. At resonance, the circuit behaves like

a) an open circuit

b) a short circuit

c) a pure resistor of value R

d) a pure resistor of value much higher than R

Answer: (d)

12

Solution:

Resonant frequency, ω0 = 1 �1 − R2C

√LC L

R < �CL

Input impedance, Zin = R2+ω02L2

R

= R + 1 �1 − RL2C� L2

RLC

= R + 1 [L − R2C]

RC

= R + L − R

RC

Zin = L

RC

as R2 < L

C

⇒ R < L

RC

⇒ Circuit behaves like a pure resistor of value much higher than

R.

13

16. The power dissipated in the controlled source of the network

shown below is _______ W.

Answer: 14

Solution:

Let current through the circuit = I

So that, VA = 7I

∴ Using KVL,

36 – 15I – 2VA = 0

36 – 15I – 3 × (7I) = 0 ⇒ I = 1 A

and VA = 7V

∴ Power dissipated in the controlled source = 2VAI = 14 W

17. The power consumed by a coil is 300 W when connected to a

30 V dc source and 108 W when connected to a 30 V ac source.

The reactance of the coil is _______ Ω.

Answer: 4

Solution:

14

I2R = 300

⇒ V2 = 300 ⇒ 302 = 300

R R

⇒ 900 = 300 ⇒ R = 3Ω

R

I = V = 30

√R2+X2 √R2+X2

�√323+0X2�2 × 3 = 108

5 = √32 + X2

X=4Ω

18. The cut-in voltage of Zener diode Dz and diode D shown in the

figure below is 0.7 V. Breakdown voltage of Dz is 3.3 V and

reverse breakdown voltage of D is 50 V. The other parameters

can be assumed to be the same as those of an ideal diode. Then

what are the values of the peak output voltage V0?

15

Positive Half Negative Half

cycle in V cycle in V

a) 3.3 1.4

b) 4 5

c) 3.3 3.3

d) 4 4

Answer: (b)

Solution:

In positive half cycle when Vi > 4 volt

Diode DZ will be break down and diode D will be ON

So, output voltage = VZ + Vcut in of D

= 3.3 + 0.7 = 4V

And for Vi < 4V

DZ will be open circuited so output

V0 = Vi sin ωt × 1 = Vi sin ωt

1+1

So maximum output in positive half cycle = 4 V in negative half

cycle diode D will be open circuited as it is in reverse biased

condition and reverse voltage is < 50 V.

So, output = 1 × Vi sin ωt

1+1

= Vi sin ωt

2

So maximum (magnitude) in negative half = 10 = 5V

2

As, Vi = 10V (since maximum of Vi sin ωt = 10)

16

19. The modified work function of an n-channel MOSFET is -0.85

V. If the interface charge is 3 × 10-4 C/m2 and the oxide

capacitance is 300 μF/m2, the flat band voltage is ________V.

Answer: 1.85 V

Solution:

Given that, WC = −0.85

Interface charge q = 3 × 10-4 c/m2

Oxide capacitance,

C = 300 × μF/m2

= 300 × 10-6 F/m2

We know that flat band voltage

V = q (1 − WC)

C

V = 3×10−4 [1 − (−0.85)]

300×10−6

= 1.85V

20. The JFET in the below, circuit has a loss = 10 mA, VP = 5 V.

The value of the resistance RS for a drain current of IDS = 6.4

mA is _____ kΩ.

Answer: 156 kΩ

17

Solution:

iDSS = 10mA

VP = 5V

iDS = 6.4 mA

iDS = iDSS �1 − VVGPS�2

6.4 = 10 �1 − VGS �2

VP

VGS = 1V

By applying KVL,

VGS + RSiDS = 0

1 − 6.4 RS = 0

RS = 156.25 Ω

21. Match List-I (Circuit Symbols) with List-II (Nomenclature)

and select the correct answer using the codes given below:

List-I List-II

18

Codes:

ABC D

a) 4 3 1 2

b) 3 4 2 1

c) 4 3 2 1

d) 3 4 1 2

Answer: (c)

22. Which one of the following is the correct statement?

The system characterized by the equation y(t) = ax(t) + b is

a) linear for any value of b

b) linear if b > 0

c) linear if b < 0

d) non-linear

Answer: (d)

Solution:

y(t) = ax(t) + b

Let x(t) = x1(t) + x2(t)

Then y(t) = ax1(t) + ax2(t) + b

y(t) ≠ {ax1(t) + b} + {ax2(t) + b}

y(t) ≠ y1(t) + y2(t)

Hence the system is non-linear

19

1; n = 1

23. A signal V(n) is defined by V[n] = � −1; n = −1

0; n = 0 and |n| > 1

which is the value of composite signal defined as V[n] +V [– n]?

a) 0 for all integer values of n

b) 2 for all integer values of n

c) 1 for all integer values of n

d) –1 for all integer values of n

Answer: (a)

Solution:

v[n] + v[-n]

for all integer values of n

20

24. A mechanical system is as shown in the figure below. The

system is set into motion by applying a unit impulse force δ(t).

Assuming that the system is initially at rest and ignoring friction

what is the displacement x(t) of mass?

a) 1 exp(−m ∙ t)

√k

b) 1 sin(t)

√mk

c) 1 sin ��mk ∙ t�

√mk

d) 1 ��mk . t�

√mk

Answer: (c)

Solution:

δ(t) = md2x(t) + k x(t)

dt2

Taking laplace transform

1 = ms2X(s) + k[X(s)]

∴ X(s) = 1

ms2+k

X(s) = 1

m�s2+mk �

X(t) = 1 sin ��mk . t�

√mK

21

25. Consider the following statements with regards to signal flow

graph:

1. The number of loops are 3.

2. The number of loops are 2.

3. The number of forward paths are 3.

4. C/R ratio is 40/81

5. C/R ratio is 28/81

Which of these statements are correct?

a) 1, 3, 4 and 5

b) 1, 3 and 4

c) 2, 3 and 4

d) 3, 4 and 5

Answer: (b)

Solution:

C = P1∆1+P2∆2+P3∆3

R ∆

= 5×4+2×4+3×4

1−{−5×4×2−2×4×2−3×4×2}

⇒ C = 40

R 81

Hence, option (b) is correct

22

26. When is the mechanical power developed by a d.c motor

maximum?

a) Back e.m.f is equal to applied voltage

b) Back e.m.f is equal to zero

c) Back e.m.f is equal to half the applied voltage

d) None of the above

Answer: (c)

Solution:

P = Eb × Ia; Ia = Vt−Eb

Ra

dP = (Vt−2Eb) = 0 for max

dEb Ra

Vt = 2Eb or Eb = Vt

2

27. A separately excited dc generator is feeding a dc shunt motor.

If the load torque on the motor is reduced to half then

a) the armature current of both motor and generator are reduced

to half.

b) the armature current of motor is halved and that of generator

remains unchanged.

c) the armature current of generator is halved and that of motor

remains unchanged.

d) the armature current of both machines remains unchanged.

Answer: (a)

Solution:

For separately excited dc generator

23

Load torque, T = KϕIam

When load torque is reduced to half,

T = KϕIa′ m ⇒ Ia′ m = Iam

2 2

28. For an extra-high voltage overhead transmission line, four

conductors are used per phase (in a bundle) at the corners of a

square of side‘s’ meter. The GMR (Geometric Mean Radius) of

each conductor is rm’ meter.

a) �rm′ × s2 × √2s�1/4

b) (rm′ × s3)1/4

c) (rm′ × 3s3)1/4

d) �rm′ × �√2s�3 1/4

�

Answer: (a)

Solution:

GMR of bundle conductor

24

= 4×4��rm′ × s × s × s√2�4

= 4×4�rm′ × s2 × s√2

29. Statement (I): The expression for the value of inductance L

per conductor of an unsymmetrically spaced 3-phase overhead

transmission line contains an imaginary term.

Statement (II): The presence of the imaginary term is due to

the mutual inductance between the phase conductors and can be

eliminated by symmetrically transposing the three-line

conductors along the length of the line

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.

Answer: (a)

Solution:

The expression for the value of inductance L, when the

conductors of a 3-ϕ transmission line are unsymmetrically

spaced consists imaginary terms because the flux linkages and

inductance of various phase are different which causes unequal

voltage drops in three-phases and unequal transfer of power

between phases due to mutual inductances even if the currents in

25

the conductors are balanced. This is avoided by transposition of

conductors.

30. Match List-I (Limiting factor) with List-II (Safe Operating

Area Portion) and select the correct answer using the code

given below the Lists:

List-I

A. The peak voltage limit

B. Secondary breakdown limit

C. Power dissipation limit

D. Peak current limit

List-II

Codes: D

ABC 3

1

a) 2 1 4 1

b) 4 3 2 3

c) 2 3 4

d) 4 1 2

Answer: (b)

26

Solution:

Safe operating area of a power transistor specifies the safe

operating limits of collector current IC versus collector emitter

voltage VCE, for reliable operation of the transistor, the IC and

VCE must always lie within this area.

31. Statement (I): For random error with normal distribution,

probable error = ±0.6745 , where is the standard deviation.

Statement (II): Probable error ξρ is that error value where there

is a 50% chance that any observation has a random error no

greater than ± ξρ.

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.

Answer: (a)

27

32. Three d.c voltmeters are connected in series across a 120 V d.c

supply. The voltmeters are specified as follows:

Voltmeter A: 100 V, 5 mA

Voltmeter B: 100 V, 250 ohms/V

Voltmeter C: 10 mA, 15,000 ohms

The voltages read by the meters A, B and C are respectively

a) 40, 50 and 30 V

b) 40, 40 and 40 V

c) 60, 30 and 30 V

d) 30, 60 and 30 V

Answer: (a)

Solution:

R1 = 100 = 20000Ω

5×10−3

R2 = 250 × 100 = 25000Ω; R3 = 15000Ω

∴ I = 120 = 2mA

R1+R2+R3

V1 = 2 × 10−3 × 20000 = 40V

V2 = 2 × 10−3 × 25000 = 50V

V3 = 2 × 10−3 × 15000 = 30V

33. A point charge +Q is brought near a corner of two right angle

conducting planes which are at zero potential as shown in the

given Figure. Which one of the following configurations

describes the total effect of the charges for calculating the actual

field in the first quadrant?

28

Answer: (d)

Solution:

We know that on the surface of conductor tangential component

of electric field is zero only normal component exist in order to

find this we use the concept of method of images i.e. by

replacing the surface by placing a charge of a magnitude and

29

polarity so that the electric field component existing remains

normal to surface and for this we replace surfaces by placing

equal and opposite charges at a distance same as existing

charge's distance from surface on the other side of surface so

from the options, option D is correct.

34. A potential field is given by V = 3x2y – yz. Which of the

following is NOT true?

a) At the point (1, 0, –1), V and the electric field vanish.

b) x2y = 1 is an equipotential plane in the xy-plane.

c) The equipotential surface V = – 8 passes through the point P

(2, – 1, 4).

d) A unit normal to the equipotential surface

Answer: (a)

Solution:

V = 3x2y − yz

E� = −∇V = − �∂∂Vx a� x + ∂V a� y + ∂V a� z �

∂y ∂z

E� = −�6xy a�x + (3x2 − z)a�y − ya�z�

So, E� at (+1, 0, −1) = −�0 + 3a�y − 0�

= −3a�y ≠ 0

35. If the magnetic flux through each turn of the coil consisting of

200 turns is (t2 – 3t) milli-Webers, where t is in seconds, then

the induced emf in the coil at t = 4 sec is _______V.

Answer: –1 V

30

Solution:

e = −N d∅

dt

= −200 × �ddt (t2 − 3t)� t=4 sec

at

= −200 × [2t − 3]at t=4 sec

= −1000mV

= − 1V

31

Two Marks Questions:

36. A sphere of unit radius is centered at the origin. The unit

normal at a point (x, y, z) on the surface of the sphere is the

vector.

a) (x, y, z)

b) �√13 , 1 , √13�

√3

c) �√ 3 , , √ 3 �

√3

d) �√ 2 , , √ 2 �

√2

Answer: (a)

Solution:

Equation of sphere is x2 + y2 + z2 = 1

Litϕ(x, y, z) = x2 + y2 + z2

Normal to the surface is

∇ϕ = 2xı̅ + 2yȷ̅ + 2zk�

Unit normal is ∇ϕ = 2�xı+̅ yȷ+̅ zk��

|∇ϕ| �4x2+4y2+4z2

= xı̅ + yȷ̅ + zk� [Hence x2 + y2 + z2 = 1]

37. A fair coin is tossed 10 times. What is the probability that only

the first two tosses will yield heads?

a) �12�2

b) 10c2 �21�2

c) �12�10

32

d) 10c2 �12�10

Answer: (c)

Solution:

Probability for first two tosses to yield heads is (1/2)2, so

remaining tosses must be tails.

Therefore the probability for remaining tosses to be tails is

(1/2)8

Hence required probability = (1/2)2. (1/2)8

= (1/2)10

38. The surface integral ∬s 1 (9xi − 3yj). nds over the sphere

π

given by x2 + y2 + z2 = 9 is _______.

Answer: 216

Solution:

Using Gauss-divergence theorem

1 ∬s(9xi − 3yj) nds = 1 ∭v div[9xi − 3yj]dv

π π

= 1 ∭v(9 − 3)dxdydz

π

= 6 × volume of the sphere

π

= 6 × 4 × π(3)3 = 216

π 3

39. A 400V, 50 Hz, three phase balanced source supplies power to

a star connected load whose rating is 12√3 kVA, 0.8pf (lag).

The rating (in kVAR) of the delta connected (capacitive)

reactive power bank necessary to bring the pf to unity is

________ KVAR.

33

Answer: 12.47

Solution:

The rating of star connected load is given as 12√3 kVA, 0.8 p.f

(lag)

Active power consumed by the load,

P = 12√3 × 0.8×103

= 16.627 kW

Reactive power consumed by the load

= 12√3 × sin (cos-1 0.8) × 103

Q1 = 12.47 kVAR

Reactive power consumed by the load at unity power factor is

Q2 = × sin (cos-1 1) = 0

(1)

∴ kVAR to be supplied by the delta connected capacitor bank

= Q1 – Q2

Qc = 12.47 kVAR

40. For the below circuit, if the current I = 3A and 1.5 A for RL = 0

and 2Ω respectively, then what is the value of I for RL = 1 Ω?

Answer: 2.0 A

34

Solution:

V = I(R + RL)

V = 3(R + 0)

V = 1.5(R + 2)

3R = 1.5 R + 3

1.5 R = 3

⇒ R = 2Ω ⇒ V = 3 R = 6 V

I = V = 6 = 2A

R+RL 2+1

41. In the circuit given below when R is infinite V = 4V and when

R = 0, the current through R is 4 A. lf R = 3Ω what is the

current through it?

Answer: 1 A

Solution:

R is infinite, implies open circuited.

So, 'Vth' is Thevenin’s equivalent voltage a cross resistor. Let

Rth Thevenin’s equivalent resistance of network.

For R = ∞, Network

35

R = 0, I = 4 A implies Norton current of network.

Norton equivalent,

From above networks,

Rth = V = 4 = 1Ω

I 4

R = 3 Ω,

I = 4 = 4 = 1A

Rth+R 1+3

42. For the n-channel enhancement MOSFET shown in figure, the

threshold voltage Vth = 2V. The drain current ID of the MOSFET

is 4 mA when the drain resistance is 1 kΩ. If the value of RD is

increased to 4 kΩ, the drain current ID will become

36

a) 2.8 mA

b) 2.0 mA

c) 1.4 mA

d) 1.0 mA

Answer: (c)

Solution:

Given that Vth = 2V,

VD = VG, VS = 0

⇒ VDS = VGS

We know ID = 4 mA, RD = 1 kΩ

10 – IDRD – VD = 0

VD = 10 – 4 = 6V

VD = 6V = VGS

then ID = k�VGS − VGS(th)�2

4mA = k × [6 − 2]2 ⇒ k = 1 mA/v2

4

But when RD changed to 4 kΩ then

VGS = VDS = 10 − 4ID (ID in mA)

Then ID = k�VGS − VGS(th)�2

ID = 1 [10 − 4ID − 2]2 = 1 [8 − 4ID]2

4 4

37

ID = 4[2 − ID]2

ID = 4(ID2 + 4 − 4ID)

⇒ ID2 − 4.25ID + 4 = 0

ID = 4.25±�(4.25)2−16 = 4.25±1.436

2 2

= 2.843 mA (or) 1.407 mA

⇒ ID = 2.843 mA (or) 1.407 mA

For N-channel E-MOSFET VGS > Vth

VGS = 10 − 4(2.843) VGS = 10 − 4(1.407)

= −1.372 V = 4.372 V

∴ ID = 1.407 mA

43. The decimal number 4097 is represented in four forms as

shown below. Match List-I (Type of Representation) with

List-II (Number) and select the correct answer:

List I List II

A. Binary 1. 0000 0000 0000 1001

B. BCD 2. 0000 0000 0001 0001

C. Octal 3. 0001 0000 0000 0001

D. Hexadecimal 4. 0100 0000 1001 0111

Codes:

ABC D

a) 3 1 2 4

b) 2 4 3 1

c) 3 4 2 1

d) 2 1 3 4

38

Answer: (c)

Solution:

Binary: 0001 0000 0000 0001

212 + 20 = 4097

BCD: 0100 0000 1001 0111

4 0 97

Octal: 0000 0000 0001 0001

→ 1 × 84 + 1 × 80

→ 4097

Hexa decimal: 0000 0000 0000 1001

= 1 × 163 + 1 × 160

= 4097

44. Which one of the following is correct?

A binary number with n digits all which are unity has the value

a) 2n

b) 2n-1

c) 2n – 1

d) 2(n-1) – 1

Answer: (c)

Solution:

Highest value of n-bit number → 2n – 1

e.g., n = 3

23 – 1 = 7 → (111)2

39

45. Given: �(224)r = (13)r.

What is the value of the radix r is _______.

Answer: 5

Solution:

Converting (224)r into decimal equivalent

(224)r = 2 × r2 + 2 × r1 + 4 × r0

(13)r = 1 × r1 + 3 × r0 = r + 3

As per question; √2r2 + 2r + 4 = (r + 3)

Squaring on both sides,

2r2 + 2r + 4 = r2 + 6r + 9

⇒ r2 − 4r − 5 = 0

(r – 5)(r + 1) = 0

⇒ r = 5, -1

r cannot take negative values.

Hence, r = 5

46. x(t) is a positive rectangular pulse from t = –1 to t = +1 with

unit height as shown in the figure. The value of ∫−∞∞|X(ω)|2 dω

{where X (ω) is the Fourier transform of x (t)} is

a) 2

b) 2π

c) 4

d) 4π

40

Answer: (d)

Solution:

Recall the Parseval’s theorem for F.T pair:

If x(t) → X(ω), X(f)

Energy of the signal, x(t) is

E = ∫−∞∞|x(t)|2 dt = 1 ∫=∞−∞|X(ω)|2 dω

2π

= ∫f∞=−∞|X(f)|2 df

= Area under x2(t) = 2 for the given signal, x(t)

shown in fig.

∴ ∫−∞∞|X(ω)|2 dω = 2π ∫−∞∞|x(t)|2 dt

= 2π × 2 = 4π

47. A sinusoid x(t) of unknown frequency is sampled by an

impulse train of period 20 ms. The resulting sample train is next

applied to an ideal low pass filter with a cutoff at 25 Hz. The

filter output is seen to be a sinusoid of frequency 20 Hz. This

means that x(t) has a frequency of

a) 10 Hz

b) 60 Hz

c) 30 Hz

d) 90 Hz

41

Answer: (c)

Solution:

Ts = 20 msec

fs = 1 = 1

Ts 20×10−3

= 1000 = 50 Hz

20

Assume the frequency of the input signal is ‘fm’ then

n = 0 fm1 − fm

n = 1 fs − fm, fs + fm

n = 2 2fs − fm1, 2fs + fm

Consider ‘a’ option: fm = 10 then output frequencies are = 10, -

10, 40, 60, 90, 110… Output is 10 Hz.

Consider ‘b’ option: fm = 60 Hz then output frequencies are =

60, -60, -10, 110, 40, 160… No output.

Consider ‘c’ option: fm = 30 Hz then output frequencies are =

30, -30, 20, 80, 70, 130… The output of low pass filter = 20 Hz.

42

48. The feedback system shown below oscillates at 2 rad/s when

a) K = 2 and a = 0.75

b) K = 3 and a = 0.75

c) K = 4 and a = 0.5

d) K = 2 and a = 0.5

Answer: (a)

Solution:

Characteristic equation,

1 + k(s+1) = 0

s3+as2+2s+1

s3 + as2 + 2s + 1 + ks + k = 0

s3 + as2 + s(k + 2) + k + 1 = 0

RH tabulation

For oscillations a(k+2)−(k+1) = 0

a

a(k + 2) − (k + 1) = 0

a = k+1 ……… (i)

k+2

43

AE as2 + k + 1 = 0 ……… (ii)

Using eq(1) in (2)

s2 + k + 2 = 0

s = ±√k + 2 = ±j2 (2rad/sec given)

Gives k = 2, Using eq(1), a = 0.75

49. A unity feedback system has an open loop transfer function,

(s) = sK2. The root locus plot is

Answer: (b)

Solution:

(s)H(s) = k

s2

Centroid = 0

44

Angle of asymptotes ±900

50. The open loop transfer function G(s) of a unity feedback

control system is given as, (s) = k�s+32� . From the root locus,

s2(s+2)

it can be inferred that when k tends to positive infinity

a) three roots with nearly equal real parts exist on the left half of

the s-plane

b) one real root is found on the right half of the s-plane

c) the root loci cross the jω axis for a finite value of k; k ≠ 0

d) three real roots are found on the right half of the s-plane

Answer: (a)

Solution:

Number of asymptotes = |p − z|

= |3 − 1| = 2

Centroid, σ = 0−2−�23�

3−1

σ = −2 = 0.67

3

Angle of asymptotes + 900, -900

Therefore the RLD is given below

45

k → ∞ roots are −2 , −2 ± j∞

3 3

It is clear from the RLD when k → ∞ all the three roots are left

side of s-plane and have equal real parts.

51. In the GH(s) plane, the Nyquist plot of the loop transfer

function (s)H(s) = πe−0.25s passes through the negative real

s

axis at the point

a) (–0.25, j0)

b) (–0.5, j0)

c) (–1, j0)

d) (–2, j0)

Answer: (b)

Solution:

Point of interaction with negative real axis = � �jωpc�H�jωpc��

∠πe−0.25jωpc = −π

jωpc

− �0.25ωpc + π2� = −π

0.25ωpc = π ⇒ ωpc = 2π

2

�πe−0.25jωpc � = π = π = 1

ωpc 2π 2

jωpc

∴ Nyquist plot passes through (-0.5, j0)

46

52. A 3-phase induction motor develops a torque 'T' when driven

from a balanced 3-phase supply. The terminal voltage and

frequency are halved so that the air-gap flux remains the same

while the load torque is kept constant. Then the slip

a) Is reduced to half

b) Remains same as the previous value

c) Speed is reduced to half

d) Speed remains the same

Answer: (d)

Solution:

In stable region,

Torque, T = sV2

ωsr′2

Torque ∝ V2.s (ωs = 2πf)

ωs

T = kV2s

f

T1 = kV21s1

f1

and T2 = k(V1/2)2 . s2

(f1/2)

⇒ s2 = 2s1

Slip speed = Ns − Nr

Slip, s = Ns−Nr

Nr

Ns2 = 120f2

p

= 120×f1/2 = Ns1

p 2

47

(slip speed)1 = s1Ns1

(slip speed)2 = s2Ns2

= (2s1) �N2s1� = s1Ns1

So, slip speed remain same.

53. A three-phase, three winding Δ/Δ/Y (1.1kV/6.6kV/400 V)

transformer is energized from AC mains at the 1.1 kV side. It

Supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV

winding and 300 kVA load at 0.6 power factor lag from the 400

V winding. The RMS line current in ampere drawn by the 1.1

kV winding from the mains is ________ A.

Answer: 625

Solution:

Given transformer is ∆/∆/Y

Wdg (1) / wdg (2) / wdg (3)

∆ /∆ / Y

1.1 kV / 6.6 kV / 400V

By applying superposition theorem

a) Load on 6.6kV (∆) side is 900 kVA, 0.8 pf lag

⇒ 900 × 103 = √3 × 6.6 × 103 × IL

IL = 78.72 A

Iph(load) = 78.72 = 45.46A/ph

√3

= 45.46∠ − 36.86 A/ph

⇒ The current on 1.1 kV side will be

Iph(source) = 46.46 × �16..16�

48

= 272.76∠ − 36.86 A/ph

b) Load on 400 V (Y) is 300 kVA, 0.6 pf lag

⇒ 300 × 103 = √3 × 400 × IL

IL = 300×103 = 433.01 A

√3×400

Iph = 433.01∠ − 53.13 A/ph

∴ The current of source side (1.1 kV) will be

Iph(source) = 400/√3 × 433.01

[1100]

= 90.90∠ − 53.13 A/ph

∴ Total source phase current will be

= 272.76∠ − 36.86 + 90.90∠ − 53.13

= 360.88 A/ph; Pf = 0.65 lag

∴ The line current will be = 625 A

54. The power input to a 415V, 50Hz, 6 pole, 3- phase induction

motor running at 975 rpm is 40 kW. The stator losses are 1 kW

and friction and windage losses total 2 kW. The efficiency of the

motor is ______ %.

Answer: 90%

Solution:

Power input = 40 kW,

Slip (s) = Ns−N = 1000−975 = 0.025

Ns 1000

49