GATE

EEE

GrandTests

Test-12Solutions

APTITUDE

One Mark Questions:

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

meaning to the word in bold.

Anachronism

a) Obsession

b) Vogue

c) Radicalism

d) Pre-emption

Answer: (b)

Solution:

Meaning: The meaning of anachronism is a person or thing that

is placed in a time where it does not fit.

Example: He avoids anachronism but does not shrink from

suggesting lessons for our own time.

Synonyms: Misplacement, prolepsis, solecism

2. Choose the word most nearly opposite to the given word.

Ensconce

a) Captive

b) Subjugate

c) Castigate

d) Expose

Answer: (d)

1

Solution:

Meaning: Ensconce is defined as to become firmly settled in or

to settle yourself or someone else in a place of comfort or safety.

Example: If you ensconce yourself somewhere, you settle in for

quite a while.

Synonyms: Hide, Situate, Stash, Conceal

Antonyms: Expose, Depart, Uncover, Disorder, Open

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

printed in bold.

Ersatz

a) Tumultuous

b) Artificial

c) Unstudied

d) Vague

Answer: (b)

Solution:

Meaning: The ersatz version of something is an inferior

substitute or imitation.

Serving as a substitute, synthetic, artificial

Example: You might speak in an ersatz French accent, but you

cannot fool the officer.

Synonyms: Counterfeit, Fake, Imitation, Phony, Sham

Antonyms: Genuine, Real, True

2

4. Even those who disagreed with Carmen’s views rarely faulted

her for expressing them, for the positions she took were as

________ as they were controversial.

a) complicated

b) political

c) subjective

d) thoughtful

Answer: (d)

5. Ajay’s age is 125% of what it was 10 years ago, but 831/3 % of

what it will be after ten years. What is his present age?

a) 45 years

b) 50 years

c) 55 years

d) 60 years

Answer: (b)

Solution:

Let the present age be x years.

Then, 125% of (x – 10) = x

And 831/3% of (x + 10) = x

Therefore, 125% of (x – 10) = 831/3% of (x + 10)

5x/12 = 250/12

x = 50 years

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.

A broad road bordered with trees

a) Avenue

b) Boudoir

c) Boulevard

d) Façade

Answer: (c)

Solution:

Boulevard

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.

Some countries are so self-respecting and even during disastrous

natural calamities they refuse any foreign aid, prefer to deal

with the crises themselves.

a) so self-respecting and even during disastrous natural

calamities they refuse any foreign aid, prefer to deal with

b) as self-respecting and even during disastrous natural

calamities will refuse any foreign aid, preferring to deal with

c) so self-respecting that even during disastrous natural

calamities they refuse any foreign aid, preferring to deal with

d) so much self-respecting that even during disastrous natural

calamities refusing any foreign aid, preferring to deal with

4

Answer: (c)

Solution:

'So-that' is the correct idiomatic usage.

8. The ratio, in which tea costing Rs. 192 per kg is to be mixed

with tea costing Rs. 150 per kg so that the mixed tea when sold

for Rs. 194.40 per kg, gives a profit of 20%.

a) 2:5

b) 3:5

c) 5:3

d) 5:2

Answer: (a)

Solution:

CP of first tea = Rs. 192 per kg.

CP of Second tea = Rs. 150 per kg.

Mixture is to be sold in Rs. 194.40 per kg, which has included

20% profit. So,

SP of Mixture = Rs. 194.40 per kg.

Let the CP of Mixture be Rs. X per kg. Therefore,

X + 20% of X = SP

6X /5 = 194.40

6X = 194.40 *5

X = Rs. 162 per kg.

Let N kg of first tea and M kg of second tea to be added.

Now, Using Allegations,

5

We get,

N/M = 12/30

N: M = 2: 5

9. A box contains 6 bottles of variety 1 drink, 3 bottles of variety 2

drink and 4 bottles of variety 3 drink. Three bottles of them are

drawn at random, what is the probability that the three are not of

a) 632/71

b) 752/833

c) 833/858

d) none of these

Answer: (c)

Solution:

Total number of drink bottles = 6 + 3 + 4 = 13.

Let S be the sample space.

Then, n(S) = number of ways of taking 3 drink bottles out of 13.

Therefore, n(S) = 13C3 = (13 × 12 × 11)/(1 × 2 × 3) = 66 × 13 =

858.

Let E be the event of taking 3 bottles of the same variety.

Then, E = event of taking (3 bottles out of 6) or (3 bottles out of

3) or (3 bottles out of 4)

6

n(E) = 6C3 + 3C3 + 4C3

= 6 × 5 × 4/1 × 2 × 3 + 1 + 4 × 3 × 2/1 × 2 × 3

= 20 + 1 + 4 = 25.

The probability of taking 3 bottles of the same variety =

n(E)/n(S) = 25/858.

Then, the probability of taking 3 bottles are not of the same

variety = 1 – 25/858 = 833/858.

10. A sum of Rs. 725 is lent in the beginning of a year at a certain

rate of interest. After 8 months, a sum of Rs. 362.50 more is lent

but at the rate twice the former. At the end of the year, Rs. 33.50

is earned as interest from both the loans. What was the original

rate of interest?

a) 3.46%

b) None of these

c) 4.5%

d) 5%

Answer: (a)

Solution:

Let the sum of Rs.725 is lent out at rate R% for 1 year

Then, at the end of 8 months, add additional sum of 362.50

more is lent out at rate 2R% for remaining 4 months (1/3 year)

Total Simple Interest = 33.50

⇒ 725×R×1 + 362.50×2R×13 = 33.50

100 100

⇒ 725×R×1 + 362.50×2R = 33.50

100 300

7

⇒ 725R + 725R = 33.50

100 100

⇒ 725R �1100 + 3100� = 33.50

⇒ 725R �1400� = 33.50

⇒ 725R × 4 = 10050

⇒ 725R = 2512.5

⇒ 145R = 502.5

⇒ R = 502.5 = 3.46%

145

8

TECHNICAL

One Mark Questions:

11. The solution to the differential equation f11(x) + 4f1(x) + 4f(x)

=0

a) f1(x) = e-2x

b) f1(x) = e2x, f2(x) = e-2x

c) f1(x) = e-2x, f2(x) = xe-2x

d) f1(x) = e-2x, f2(x) = e-x

Answer: (c)

Solution:

Given f11(x) + 4f1(x) + 4f(x) = 0

⇒ (D2 + 4D + 4) f(x) = 0

Consider D2 + 4D + 4 = 0

⇒ (D + 2)2 = 0

⇒ D = –2 , –2

∴ f(x) = e-2x (C1 + C2x)

and f1(x) = e-2x, f2(x) = xe-2x

12. If L{f(t)} = ss2++21, L{g(t)} = (s+s32)+(s1+2), h(t) = ∫0t f(T)g(t −

T)dT then L{h(t)} is ________.

a) s2+1

(s+3)

b) 1

(s+3)

c) s2+1 + s+2

(s+3)(s+2) s2+1

d) None

9

Answer: (b)

Solution:

h(t) = ∫0t f(T)g(t − T)dT = f(t) ∗ g(t)

L{h(t)} = L{f(t) × g(t)} = F(s) G(s) = 1

s+3

13. For the function of a complex variable w = 1nz (where w = u +

jv and z = x + jy) the u = constant lines get mapped in the z –

plane as

a) set of radial straight lines

b) set of concentric circles

c) set of confocal hyperbolas

d) set of confocal ellipses

Answer: (b)

Solution:

Given function is w = lnz where w = u + iv & z = x + iy

⇒ w = 1 log(x2 + y2) + i tan−1 �yx�

2

⇒ u = 1 log(x2 + y2) and v = tan−1 �yx�

2

But given that u = constant (c)

⇒ 1 log(x2 + y2) = c

2

⇒ log(x2 + y2) = 2c

⇒ x2 + y2 = e2c

⇒ x2 + y2 = c12 where c1 = ec

Which represents the set of concentric circles with centre as

(0, 0) and radius c1.

10

14. The Newton-Raphson method is used to find the root of the

equation x2 – 2 = 0. If the iterations are started from -1, then the

iteration will

a) converge to -1

b) converge to √2

c) converge to -√2

d) not converge

Answer: (c)

Solution:

Let f(x) = x2 – 2 and x0 = –1

xn+1 = xn − f(xn) = xn − �x2n−2� = �x2n+2�

f1(xn) 2xn 2xn

xn+1 = �x2n+2�

2xn

x1 = x02+2 = 1+2 = −3 = −1.5

2x0 −2 2

x2 = x12+2 = (−1.5)2+2 = 94+2

2x1 2(−1.5) −3

= 17/4 = −1.4166

−3

x3 = x22+2 = (−1.4166)2+2

2x2 2(−1.4166)

= 2.0067+2 = −1.4141

−2.8332

x4 = x23+2 = (−1.4141)2+2

2x3 2(−1.4141)

= 1.9996+2 = −1.4141

−2.8282

∴ The iteration will converge to −√2 = −1.4141

11

15. xl→im∞ x−Sinx = ______.

x+Cosx

Answer: 1

Solution:

xl→im∞ 1−Sixnx = 1−0 = 1

1+Coxsx 1+0

16. For the circuit shown in the figure below, the voltage across

the 1-ohm resistor is given by

a) 7 V/4

b) 5 V/4

c) 7 V/3

d) 2 V/3

Answer: (a)

Solution:

1. By O.C. the 1A current source.

12

V1′Ω = 4×1 = 1V

4

2. By S.C. the voltage source.

V1′′Ω = 3×1 = 3 V

4 4

By superposition theorem

V1tΩotal = V1′Ω + V1′′Ω = 1 + 3 = 7 V

4 4

17. The currents Ix and Vx in the below circuit are respectively.

a) 5A; 10V

b) 10 A; 20 V

c) 6 A; 12 V

d) 4 A; 8 V

Answer: (a)

Solution:

13

By applying source transformation

35 − 20 = 18 1 − 3 2

15 = 18 1 − 3 2 ……… (i)

20 = 5 2 − 3 1 ……… (ii)

By equation (i) and (ii),

i1 = 1.66A

i2 = 5A

So, ix = 5A

Vx = 10V

18. In the circuit given below, D1 and D2 are ideal.

Which one of the following represents the transfer

characteristics of the circuit?

14

Answer: (b)

Solution:

Now get VD1 and VD2 to know about forward or reverse biasing

condition of diodes.

V �16 + 14� = 10

6

⇒ V = 4 volt

VD1 = Vi − 4 (for Vi < 4 V it is off initially)

VD2 = 1volt, hence

VD2 = forward biased so, V0 = 5V

for Vi > 5V, V0 = Vi (∵ D1 gets on)

15

19. Match List-I with List-II and select the correct answer using

the codes given below the lists:

List-I List-II

Codes: D

ABC 2

3

a) 3 1 4 2

b) 2 1 4 2

c) 3 4 1

d) 2 4 1

Answer: (d)

Solution:

16

20. In a ripple counter, the state whose output has a frequency

equal to 1/8th that of the clock signal applied to the first stage,

also has an output periodicity equal to 1/8th that of the output

signal obtained from the last stage. The counter is

a) Modulo-8

b) Modulo-6

c) Modulo-64

d) Modulo-16

Answer: (c)

Solution:

In a ripple counter at output of each flip-flop frequency becomes

half of that of its previous flip flops.

So if input frequency of clock is fc, then at output of first flip

flop becomes fc/2 then at output of second flip flop becomes

(fc/2)/2 = fc/4 then at output of third flip-flop becomes (fc/4)/2 =

fc/8.

So at the output of third flip flop frequency becomes 1/8 of input

frequency.

So period at output of third flip-flop

T = 1 = 8

fc/8 fc

at the output of each flip flop time period becomes double to

that of its previous one.

∵ Given that periodicity T at third stage is 1/8 of that of last

state.

So, last stage period,

17

T′ = 8T

T′ = 64

fc

So, frequency at output of last stage

1 = 1 = fc

T′ (64/fc) 64

∵ we know that at last stage of ripple counter frequency

f = Input frequency

Modulo number

fc = fc

64 Modulo number

Modulo number = 64

21. A system is characterized by the input-output relation y(t) =

x(2t) + x(3t) for all f, where y(t) is the output and x(t) is the

input. It is

a) linear and causal

b) linear and non-causal

c) non-linear and causal

d) non-linear and non-causal

Answer: (b)

Solution:

Given y(t) = x(2t) + x(3t)

Scaling operation is always a non-causal operation.

Now check for linearity:

y1(t) = x1(2t) + x2(3t)

also, y2(t) = x2(2t) + x2(3t)

y3(t) = x3(2t) + x3(3t) ……... (i)

where x3(3t) = x1(t) + x2(t) ……... (ii)

18

from (i) and (ii)

y3(t) = x1(2t) + x2(2t) + x1(3t) + x2(3t)

y3(t) = y1(t) + y2(t)

Thus, the given system is a linear and non-causal system.

22. Statement (I): The total energy of an energy signal falls

between the limits 0 and ∞.

Statement (II): The average power of an energy signal is zero.

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: (b)

Solution:

Energy of any signal is given by

E = ∫−∞∞|x2(t)|dt

and power of a signal is given by

P = Tli→m∞ ∫−TT//22 1 |x2(t)|dt

T

For energy signal, Energy is finite

∴ P = Tli→m∞ E

T

P = E = 0

∞

19

→ All the finite duration and bounded signals are energy

signals.

Hence statements (I) and (II) are correct but statement (II) is not

correct explanation of statement (I).

23. The below shown feedback control system has to be reduced to

equivalent unity feedback system. Which one of the following is

equivalent?

Answer: (d)

Solution:

C = R �1+GGH�

Which is satisfied by (d) option

C = R × �1+GGHH� = R �1+GGH�

H

20

24. Consider the following relations with regard to the below

shown gear trains:

1. θ1 = N2

θ2 N1

2. T2 = J2 d2θ2 + B2 dθ2

dt2 dt

3. T1 = J2 �NN21�2 d2θ1 + B2 �NN21�2 dθ1

dt2 dt

Which of these relations are correct?

a) 1, 2 and 3

b) 1 and 2 only

c) 2 and 3 only

d) 1 and 3 only

Answer: (a)

Solution:

Number of teeth is proportional to the radius r1 = NN21.

r2

Distance travelled on the surface of the gear is the same for both

r1θ1 = r2θ2 ⇒ r1 = θθ21.

r2

Work done by one gear is equal to the other T1θ1 = T2θ2

⇒ T1 = θ2

T2 θ1

21

Combining,

T1 = θ2 = N1 = r1 = ω2

T2 θ1 N2 r2 ω1

Torque on gear can be transferred to other gear similar to

transformers transferred impedance with ratio N1/N2.

Hence, option (a) is correct.

25. Statement (I): In large dc motors which drive reversing rolls

of a steel mill and in the generators which supply them, large

fluctuations of currents because higher voltages being induced

in the coils located between adjacent commutator segments.

This may result in the so called 'flash-over' between adjacent

brushes, causing a heavy short-circuit.

Statement (II): To counteract the possibilities of a 'flash-over',

a 'compensating winding', which is embedded in the slots on the

pole-faces, is connected in series with the armature winding and

which supplies an mmf that magnetizes in the opposite direction

to that of the armature mmf.

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)

22

26. A two-winding transformer is used as an auto- transformer.

The kVA rating of the autotransformer compared to the two-

winding transformer will be

a) 3 times

b) 2 times

c) 1.5 times

d) Same

Answer: (b)

Solution:

Answer would depend upon the voltage ratio a = V1 and type of

V2

auto-transformer i.e., step-up or step-down.

(KVA)Auto = �1−1a� (KVA)two−wdg

Where a < 1

For step-down transformer

(KVA)Auto = �a−a1� (KVA)two−wdg

Where a > 1

Let, N1 = N2 (Ideal two winding transformer for step-up

transformer

a = 1 then (KVA)Auto = 2 × (KVA)two−wdg

2

For step-down transformer

a = 2 then (KVA)Auto = 2 × (KVA)two−wdg

23

27. The exact equivalent circuit of a two-winding transformer is

given in the figure given below. For affecting simplification, the

parallel magnetizing branch, consisting of RC and XQ is shifted

to the left of the primary leakage impedance (r1 + jx1). This

simplification introduces the inaccuracy, in the neglect of:

a) Voltage-drop in the primary impedance due to the secondary

current.

b) Voltage-drop in the primary impedance due to the exciting

current.

c) Voltage-drop in the secondary impedance due to the exciting

current.

d) Reduction in values of RC and XQ of the exciting circuit.

Answer: (b)

Solution:

In approximate equivalent circuit, the parallel magnetizing

branch is shifted to the left of the primary leakage impedance

with the assumption that voltage drop in the primary impedance

due to the exciting current can be neglected.

24

28. Percentage resistance and percentage reactance of a

transformer are 1% and 4%, respectively. What is voltage

regulation at power factor 0.8 lagging and 0.8 leading?

a) 2.4% and -0.8%, respectively

b) 3.2% and -1.6% respectively

c) 3.2% and -3.2%, respectively

d) 4.8% and 1.6%, respectively

Answer: (b)

Solution:

Percentage voltage regulation for lagging p.f

= (εr cos θ + εx sin θ) × 100

as εr = 0.01 and εx = 0.04

∴ % VR = (0.01 × 0.8 + 0.04 × 0.6) × 100 = 3.2%

∴ VR for leading p.f = (0.01 ×0.8 – 0.04 × 0.6) × 100

= –1.6%

29. Equivalent π model is quite suitable for analyzing the

performance of trans-mission line of

a) 50 kms length

b) 150 kms length

c) 250 kms length

d) All of the above

Answer: (c)

Solution:

Lines over 250 km are considered to be long lines. For such

long lines, equivalent π model is used.

25

30. The earthing electrodes should be placed within what distance

in meter from the building whose installation system is being

earthed?

Answer: 1.5

Solution:

The earthing electrodes should be placed within 1.5 m from the

building whose installation system is being earthed.

31. Which one of the following is the main advantage of SMPS

over linear power supply?

a) No transformer is required

b) Only one stage of conversion

c) No filter is required

d) Low power-dissipation

Answer: (d)

Solution:

In SMPS filter required is easy to design there are multiple stage

in SMPS transformer is also used in SMPS small physical size

and less weight is main reason to wide spread use of SMPS's.

32. The transformer utilization factor of a single phase full-wave

rectifier is _______.

Answer: 0.693

33. The value of the multiplier resistance for a dc voltmeter,

having 50 V range with 5 kΩ/V sensitivity, employing a 200 μA

meter movement and having internal resistance of 100Ω, is

given by _______ kΩ.

26

Answer: 249.9

Solution:

V = 50 V, SV = 5 kΩ/V

IFS = 200 μA, Rm = 100 Ω

Vm = IFS × Rm = 20 mV

m = V = 2500

Vm

Rs = Rm(m − 1) = 249.9kΩ

34. Consider the arrangement of two equal and opposite charges of

magnitude q separated by an infinitesimal distance l as shown in

the figure given below. If ar is the unit vector in the direction r

and a∅ is the unit vector in the direction θ, the electric field at

the point P is

a) −2q cos θ ar − q sin θ aθ

4πεr3 4πεr3

b) −2q cos θ ar + q sin θ aθ

4πεr3 4πεr3

c) 2q cos θ ar + q sin θ aθ

4πεr3 4πεr3

d) 2q cos θ ar − q sin θ aθ

4πεr3 4πεr3

Answer: (c)

27

Solution:

Electric field � ⃗ at a distance r from centre of a dipole is given by

E�⃗ = p [2 cos θ . a� r + sin θ . a� θ ]

4πεr3

Where, p = q × l = ql

35. Point charges of –10 nC and 10 nC are located in free space at

(–1, 0, 0) m and (1, 0, 0) m respectively. What is the energy

stored in the field is _________ nJ.

Answer: –450

Solution:

Energy of the system = q1q2

4πϵ0r

= −10×10−9×10×10−9

4×π×8.85×10−12×2

= −449.59nJ ≃ −450nJ

28

Two Marks Questions:

36. Considered the following system of equations in three real

variables x1, x2 and x3:

2x1 – x2 + 3x3 = 1

3x1 – 2x2 + 5x3 = 2

– x1 + 4x2 + x3 = 3

This system of equations has

a) No solution

b) A unique solution

c) More than one but a finite number of solutions.

d) An infinite number of solutions.

Answer: (b)

Solution:

1 −4 −1 −3

(A|B) = �2 −1 3 � 1 �

3 2 52

R2 → R2 − 2R1; R3 → R3 − 3R1

1 −4 −1 −3

�0 7 5 � 7 �

0 14 8 11

R3 → R3 − 2R2

1 −4 −1 −3

�0 7 5 � 7 �

0 0 −2 −3

ρ(A) = 3 = ρ(A|B) = 3 = number of unknowns

∴ Unique solution exists.

29

37. The vector field F = xı̅ − yȷ̅ (where i and j are unit vectors) is

a) divergence free, but not irrotational

b) irrotational, but not divergence free

c) divergence free and irrotational

d) neither divergence free nor irrotational

Answer: (c)

Solution:

F� = xi − yj

div F� = 1 − 1 = 0 ⇒ F� is solenoidal (divergence free)

i j K�

curl F� = �∂∂x ∂ ∂∂z� = 0� ⇒ F� is irrotational

∂y

x −y 0

38. A simple random sample of 100 observations was taken form a

large population. The sample mean and the standard deviation

were determined to be 80 and 12, respectively. The standard

error of mean is ________.

Answer: 1.2

Solution:

Standard error of mean = σ

√n

(σ = S.D of sample, n = size of the sample) = 12 = 1.2

√100

39. In Newton-Raphson iterative method, the initial guess value

(Xini) is considered as zero while finding the roots of the

equation: f(x) = –2 + 6x – 4x2 + 0.5x3. The correction, Δx, to be

added to Xini in the first iteration is ________.

30

Answer: 0.33

Solution:

f(x) = (–2 + 6x – 4x2 + 0.5x3)

f ′(x) = (6 − 8x − 1.5x2)

x0 = 0

⇒ x1 = x0 − f(x0) = 0 − (−2) = 1

f′(x0) 6 3

∴ ∆x = (x1 − x0) = �31 − 0� = 0.33

40. In the given circuit, the parameter k is positive, and the

power dissipated in the 2Ω resistor is 12.5 W. The value of

k is ______.

Answer: 0.5

Solution:

Given that

Voltage across the 2Ω resistor = V0

Power dissipated in 2Ω resistor = 12.5W

V02 = 12.5W

R

V02 = 12.5 × 2 = 25

V0 = 5V

Apply KCL at the junction of 1Ω, RΩ and 10Ω in the circuit,

31

V0 + KV0 = 5

2

5 + 5K = 5

2

5K = 5 − 2.5 = 2.5

K = 2.5 = 0.5

5

41. In the circuit shown below, the node voltage VA is ________V.

Answer: 11.42

Solution:

VA − 5 + (VA+10I1) + (VA−10) = 0

5 5 10

2VA − 50 + 2VA + 20I1 + VA − 10 = 0

5VA + 20I1 = 60 ………… (1)

Also, I1 = (VA1−010)………… (2)

So, 5VA + 20 �VA1−010� = 60

7VA = 80 → VA = 80

7

⇒ VA = 11.42 Volts

32

42. In the given circuit, the voltage V��L� has a phase angle of

_________ deg with respect to V��S�.

Answer: 60

Solution:

For the given circuit shown in Fig

V����L⃗ = V�����S⃗j10 = 10�V��S�∠900

17.32+j10 z∠ tan−1�171.032�

�V���L⃗ , leads �V���S⃗ by θ = 900 − tan−1 �171.032�

= (900 − 300) = 600

43. Two monoshot Multivibrator, one positive edge triggered (M1)

and another negative edge triggered (M2) are connected as

shown in figure.

33

The monoshots M1 and M2 when triggered produce pulses of

width T1 and T2 respectively, where T1 > T2. The steady state

output voltage V0 of the circuit is

Answer: (c)

Solution:

Given that M1 is positive edge triggered monoshot

Multivibrator.

• The one input for AND gate always ‘1’ because capacitor is

charged to ‘+5V’ i.e., logic ‘1’.

34

• Let initially V0 = 0, then Q�2 = 1, which is fed to another

input of AND gate, when both the inputs of AND gate are

‘1’, M1 gives a pulse for a time ‘T1’.

• M2 gets triggered, when T1 is falling from ‘1’ to ‘0’ (∴ M2 is

negative edge triggered).

• Then M2 gives a pulse for a time ‘T2’

• After T2 V0 = 0, Q�2 = 1,

This process repeats as explained above, which gives output

V0 as

44. The shift register shown in fig. is initially loaded with the bit

pattern 1010 Sequential acts. Subsequently the shift register is

clocked, and with each clock pulse the pattern gets shifted by

one-bit position to the right. With each shift, the bit at the serial

input is pushed to the left most position (MSB). After how may

clock pulses will the content of the shift register become 1010

again?

35

a) 3

b) 7

c) 11

d) 15

Answer: (b)

Solution:

1, −T/4 < t ≤ 3T/4

45. A signal x(t) is given by x(t) = �−1,3T/4 < t ≤ 7T/4, which

−x(t + T)

among the following gives the fundamental Fourier term of x(t)?

a) 4 cos �πTt − π4�

π

b) π cos �2πTt + π4�

4

c) 4 sin �πTt − π4�

π

d) π sin �2πTt + π4�

4

Answer: (a)

36

Solution:

The given signal x(t) is a periodic waveform with period

T0 = 2T and satisfies half-wave symmetry:

x(t) = −x �t ± 22T�

= −x(t ± T) as shown in Fig.1

Fundamental frequency = ω0

= 2π = π/T

2T

x1(t) = x �t + T4� shown in Fig.2

Possesses even symmetry and also half wave symmetry.

∴ Fundamental Fourier term of

x1(t) = 2×M cos �πT t�,

π

Where, M = peak to peak value of x1(t) = 2

As x(t) = x1�t − (T/4)�

37

Fundamental component of

x(t) = 4 cos �πT �t − T4�� = 4 cos �πT t − π4�

π π

46. Let the single x(t) = ∑+k=∞−∞(−1)k δ �t − 20k00�. Be passed

through an LTI system with frequency response H(ω), as given

in the figure below

The Fourier series representation of the output is given as

a) 4000 + 4000cos (2000πt) + 4000cos (4000πt)

b) 2000 + 2000cos (2000πt) + 2000cos (4000πt)

c) 4000cos (2000πt)

d) 2000cos (2000πt)

Answer: (c)

Solution:

Cn = 1 ∫0T0 x(t) e−jnω0tdt

T0

Cn = 1 �∫01 x(t) e−jnω0tdt�

1×10−3

Cn = 103�1 − e−jnω0�0.5×10−3��

38

Cn = 103 �1 − e−jn�1×21π0−3��0.5×10−3��

Cn = 103�1 − e−jnπ� = 1000⌊1 − (−1)n⌋

X(ω) = 2π ∑n∞=−∞ Cnδ(ω − nωs)

X(ω) = 2000π ∑∞n=−∞(1 − (−1)n)δ(ω − 2000nπ)

X(ω) = 2000π[… 2δ(ω + 2000π) + 2δ(ω − 2000π) + ⋯ ]

X(ω) = 4000[δ(ω + 2000π) + δ(ω − 2000π) + ⋯ ]

X(ω) = 4000[cos(2000πt) + cos(6000πt) + ⋯ ]

The output of low part filter is y(t) = 4000 cos(2000πt)

47. Match List-I (Characteristic Equation) with List-II (Nature

of Unit Step Response) and select the correct answer using the

code given below:

List – I

A. s2 + 8s + 16

B. s2 + 24s + 225

C. s2 + 20.25

D. s2 + 20s + 100

List – II

1. Undamped

2. Underdamped

3. Critical damped

4. Overdamped

39

Codes:

ABC D

a) 1 3 2 4

b) 4 3 1 2

c) 1 4 3 2

d) 4 2 1 3

Answer: (d)

Solution:

Comparing with 2 + 2 + 2

• s2 + 8s + 15, ω2n = 15, 2ξωn = 8

⇒ ξ > 1 → over damped

• s2 + 24s + 225, ωn = 15, 2ξωn = 24

⇒ ξ < 1 → under damped

• s2 + 20.25 = 0, ξ = 0 under damped

• s2 + 20s + 100 = 0, ωn = 10, 2ξωn = 20

⇒ ξ = 1 → critically damped

• s2 + 20.25, 2ξωn = 0

⇒ ξ = 1 → un damped

48. The unit impulse response of a second-order under damped

system starting from rest is given by:

C(t) = 12.5e-6t sin 8t, t ≥ 0

What is the natural frequency in rad/sec and the damping factor?

a) 10 and 0.6

b) 10 and 0.8

c) 8 and 0.6

40

d) 8 and 0.8

Answer: (a)

Solution:

G(t) = 12.5e-6t sin 8t

T. F = (12.5)(8) = 100

s2+12s+100 s2+12s+100

Characteristic equation = s2 + 12s + 100

Comparing characteristic equation with

s2 + 2ξωn + ω2n = 0

ω2n = 100 ⇒ ωn = 10

2ξωn = 12 ⇒ ωn = 0.6

49. Consider the equation: 2s4 +s3 + 3s2 + 5s + 10 = 0.

The number of roots this equation has in the right half of s-plane

is ________.

Answer: 2

Solution:

Routh array

There are two sign changes, so two pole in RHS

41

50. Consider the system described by the following state space

equations

�xx12̇̇ � = �−01 −11� �xx12� + �01� u;

y = [1 0] �xx12�

If u is unit step input, then the steady state error of the system is

______.

Answer: 0

Solution:

TF = C[SI − A]−1B

SI − A = �1S S−+11�

[1 0]�S−+11 1S��10�

S(S+1)+1

TF =

= 1

S2+S+1

Find value of the output = lsi→m0 STF = 1

Input = Unit step = 1

Error = tl→im∞[Input − Output] = 1 − 1 = 0

∴ Error = 0

51. A 220 V, 15 kW, 1000 rpm shunt motor with armature

resistance of 0.25Ω, has a rated line current of 68A and a rated

field current of 2.2 A. The change in field flux required to obtain

a speed of 1600 rpm while drawing a line current of 52.8 A and

a field current of 1.8 A is

42

a) 18.18% increase

b) 18.18% decrease

c) 36.36% increase

d) 36.36% decrease

Answer: (d)

Solution:

N1 = 1000 rpm

Eb1 = 220 − 65.8 × 0.25 = 203.55V

N2 = 1600 rpm

Eb2 = 220 − 51 × 0.25

= 207.25V

2 = Eb2 × ∅1

1 Eb1 ∅2

1600 = 207.25 × ∅1

1000 203.55 ∅2

⇒ ∅1 = 1600×203.55 = 1.571

∅2 1000×207.25

∅1−∅2 = 1 − ∅2 = 1 − 1 = 0.3636

∅1 ∅1 1.571

% Change in flux = 36.36% decrease

43

52. A 8 pole dc generator has a simplex wave wound armature

containing 32 coils of 6 turns each. Its flux per pole is 0.06 Wb.

The machine is running at 250 rpm. The induced armature

voltage is ______ V.

Answer: 384

Solution:

Induced emf in a DC generator,

E = ∅ZN × P

60 A

= 0.06×(32×6×2)×250 × 8

60 2

E = 384V

53. A 3-phase delta/star transformer is supplied 6000 V on delta

connected side. The terminal voltage on the secondary side

when supplying full load at 0.8 lagging power factor is 415 V,

the equivalent resistance and reactance drops for the transformer

are 1% & 5% respectively. The turns ratio of the transformer is

_______.

Answer: 24

Solution:

Turns ratio = primary induced voltage

secondary induc ed voltage

Secondary phase voltage = terminal phase voltage

(1−%Reg)

% Reg = % R cos∅ + % Xsin∅ [∵ Lagging Load]

= 1 × 0.8 + 5 × 0.6 = 3.8%

44

E2 = V2 ph = 415 = 249.06

1−0.038 √3×0.962

∴ Turns ratio = V1ph = 6000 = 24

V2ph 249.06

54. A 25 kVA, 400 V, Δ-connected, 3-phase, cylindrical rotor

synchronous generator requires a field current of 5 A to

maintain the rated armature current under short-circuit

condition. For the same field current, the open-circuit voltage is

360 V. Neglecting the armature resistance and magnetic

saturation, its voltage regulation (in % with respect to terminal

voltage), when the generator delivers the rated load at 0.8 pf

leading at rated terminal voltage is _________%.

Answer: –14.56%

Solution:

25 kVA, 400V, ∆ − connected

∴ IL = 25×1000 = 36.08A

√3×400

⇒ Iph = 36.08 = 20.83A

√3

Isc = 20.83A when If = 5A

Voc(line) = 360V when If = 5A

Xs = VIsocc�If=given

= 360(phase voltage) = 17.28Ω

20.83(phase current)

For a given leading pf load [cos∅ = 0.8 lead]

⇒ E0 = �(Vcos∅ + IaRa)2 + (Vsin∅ + IaXs)2

= �[400 × 0.8]2 + [400 × 0.6 − 20.83 × 17.28]2

45

= 341 Volts/ph

⇒ Voltage regulation = |E|−|V| × 100

|V|

= 341−400 × 100

400

= −14.56%

55. Consider a three-phase, 50 Hz, 11 kV distribution systems.

Each of the conductors in suspended by an insulator string

having two identical porcelain insulators. The self -capacitance

of the insulator is 5 times the shunt capacitance between the link

and the ground, as shown in the figure. The voltages across the

two insulators are

a) e1 = 3.74 kV, e2 = 2.61 kV

b) e1 = 3.46 kV, e2 = 2.89 kV

c) e1 = 6.0 kV, e2 = 4.23 kV

d) e1 = 5.5 kV, e2 = 5.5 kV

Answer: (b)

46

Solution:

At “A” point I1 = I2 + I3

e1(5Cω) = e2(ωC) + e2(5Cω)

5e1 = 6e2 …….. (i)

e1 + e2 = 11k

√3

e1 + e2 = 6.35kV ……… (ii)

From equations (i) & (ii) we can get,

e2 = 2.886 kV

e1 = 3.46 kV

56. The parameters of the transposed overhead transmission line

are given as: self-reactance Xs = 0.4 Ω/km and mutual reactance

Xm = 0.1 Ω/km. The positive sequence reactance X1 = 0.3 Ω/km

and zero sequence reactance X0 in Ω/km are _______.

Answer: 0.6

Solution:

Zero sequence reactance X0 = Xs + 2Xm

= 0.4 + 2(0.1)

= 0.6 Ω/km

47

57. In an unbalanced three phase system phase current Ia = 1∠(–

900)pu, negative sequence current Ib2 = 4∠(–1500)pu, zero

sequence current Ic0 = 3∠(900)pu. The magnitude of phase

current Ib in pu is ______ pu.

Answer: 11.53

Solution:

Ia = 1∠–900; Ib2 = 4∠–1500

Ic0 = 3∠900

Ib = Ib0 + Ib1 + Ib2 as Ia = Ia0 + Ia1 + Ia2

Ia = Ic0 + Ib1 �α12� + Ib2(1/α)

Ib1 �α12� = Ia − Ic0 − Ib2(1/α)

= (1∠ − 900) − (3∠900) − (4∠ − 1500) × 1

1∠1200

Ib1 = 8∠150 pu

Ib = Ib0 + Ib1 + Ib2

= 3∠900 + 8∠1500 + 4∠ − 1500

= 11.53∠154.43 pu

58. A 50 Hz, 4-pole, 500 MVA, 22 kV turbo generator is

delivering rated mega volt-amperes at 0.8 power factor.

Suddenly fault occurs reducing its electric power output by

40%. Neglect losses and assume constant power input to the

shaft. The accelerating torque in the generator in MNm at the

time of fault will be ________.

Answer: 1.018

48

Solution:

Before fault

Mechanical input (Ps1) = electrical output

�Pe1� = 500 × 0.8 × 106

= 400 MW

During fault electrical output,

�Pe2� = 0.6 × 400 × 106

= 240 MW

During fault mechanical output,

�Ps2� = 400MW

Accelerating power,

Pa = Ps2 − Pe2 = 400 − 240 = 160MW

Ns = 120×50 = 1500 rpm

4

ωs = 2πNs = 2π×1500 = 157 rad/sec

60 60

Accelerating torque,

Ta = Pa = 160×106 = 1.018MNm

ωs 157

59. A single-phase, 230 V, 50 Hz ac mains fed step down

transformer (4:1) is supplying power to a half - wave

uncontrolled ac – dc converter used for charging a battery (12 V

dc) with the series current limiting resistor being 19.04 Ω. The

charging current is

a) 3.43 A

b) 1.65 A

49