Grand Test - 3

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EEE

GrandTests

Test-03Solutions

APTITUDE
One Mark Questions:

1. Choose the word or group of words which is most opposite in
meaning to the word in bold.
Hirsute
a) Shaggy
b) bald
c) erudite
d) glorious
Answer: (b)
Solution:
Meaning: Hirsute is covered with or as if with hair.
Synonyms: Brushy, Fleecy
Antonyms: Bald, Furless, Smooth

2. Choose the word or group of words which is most similar in
meaning to the word printed in bold.
Abrogate
a) Usurp
b) Treasure
c) Retain
d) Abolish
Answer: (d)
Solution:
Meaning: Abrogate is defined as to end something, especially
when some formal step is needed to end it.

1

Example: Our rights cannot be denied or abrogated by the
Government.
Synonyms: Annul, Nullify, Undo, Revoke, Quash
Antonyms: Approve, Enact, Pass, Permit
3. Jean Jacques Rousseau, whose social philosophy was often
austere, was in his personal life a surprisingly _________ man:
he attended fashionable Parisian parties, wore flashy clothing,
and dated other men’s wives.
a) Flamboyant
b) Indisputable
c) Pristine
d) Astute
Answer: (a)
4. Now known as Administrative Professionals’ Day, Secretaries’
Day was created in 1952 by Harry F. Klemfuss, a public
relations professional who _______ the value and significance
of administrative assistants in order to attract more women to
the profession.
a) proscribed
b) touted
c) refuted
d) undermined
Answer: (b)

2

Solution:
The clue is that Klemfuss created Secretaries’ Day, so he must
have appreciated the value and significance of administrative
assistants. None of proscribed, refuted, undermined means
appreciated, so eliminate choices (A), (C), (D). Although touted
does not-strictly speaking-mean appreciated, a person touts only
something that is appreciated, so select choice (B).
5. Two varieties of wheat - A and B costing Rs. 9 per kg and Rs.
15 per kg were mixed in the ratio 3: 7. If 5 kg of the mixture is
sold at 25% profit, find the profit made?
a) Rs. 13.50
b) Rs. 14.50
c) Rs. 15.50
d) Rs. 16.50
e) None of these
Answer: (d)
Explanation:
Let the quantities of A and B mixed be 3x kg and 7x kg.
Cost of 3x kg of A = 9(3x) = Rs. 27x
Cost of 7x kg of B = 15(7x) = Rs. 105x
Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x
Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66
Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg
of the mixture) = 25/100 * 66 = Rs. 16.50

3

Two Marks Questions:
6. One who dabbles in fine arts for the love of it and not for
monetary gains
a) Connoisseur
b) Amateur
c) Professional
d) Dilettante
Answer: (b)
7. In the question below, complete the sentences with the
appropriate choice.
Public sector banks recorded better performance, despite
_________
a) The banking sector’s profitability being under pressure
b) The banking sector and its profitability is coming under
pressure
c) The fact that the profitability is achieved
d) It’s coming under pressure due to profitability
Answer: (a)
8. Three beakers namely A, B and C each contain 100 ml of spirit
in water solution. The ratio of spirit to water in the beakers A, B
and C is 1: 3, 1: 4 and 2: 3 respectively. 40 ml of solution is
transferred from beaker A to beaker C and then 28 ml of
solution is transferred from beaker C to beaker B. Find the final
ratio of spirit in the beakers A, B and C.

4

a) 3: 6: 8
b) 6: 15: 20
c) 15: 28: 42
d) None of these
Answer: (a)
Solution:
Initial quantity of spirit and water in the beakers

Beaker A: spirit = 25 ml and Water = 75 ml
Beaker B: spirit = 20 ml and Water = 80 ml
Beaker C: spirit = 40 ml and Water = 60 ml
After 40 ml is transferred from beaker A to beaker C, the
quantity of spirit and water in the beakers is as follows:
Beaker A: spirit = 25 – 10 = 15 ml and
Water = 75 – 30 = 45 ml
Beaker B: spirit = 20 ml and Water = 80 ml
Beaker C: spirit = 40 + 10 = 50ml and
Water = 60 + 30 = 90 ml
Now, spirit: Water in Beaker C = 5:9
After 28 ml is transferred from beaker C to beaker B:
Beaker A: spirit = 15 and Water = 40mI
Beaker B: spirit = 20 + l0 = 30ml and
Water = 80 + 18 = 98 ml
Beaker C: spirit = 50 - 10 = 40ml and
Water = 90 – 18 = 72 ml
Required Ratio = 15 : 30 : 40 = 3 : 6 : 8

5

9. A contract is to be completed in 56 days if 104 persons work,

each working at 8 hours a day. After 30 days, 2/5 of the work is

completed. How many additional persons should be deployed so

that the work will be completed in the scheduled time, each

person now working 9 hours a day?

a) 160

b) 150

c) 24

d) 56

Answer: (d)

Solution:

Persons worked = 104

Number of hours each person worked per day = 8

Number of days they worked = 30

Work completed = 2/5

Remaining days = 56 - 30 = 26

Remaining Work to be completed = 1 – 2/5 = 3/5

Let the total number of persons who do the remaining work = x

Number of hours each person needs to be work per day = 9

More days, less persons (indirect proportion)

More hours, less persons (indirect proportion)

More work, more persons (direct proportion)

Days 30: 26

Hours 8∶ 9 � ∷ x: 104
Work 3
5 : 2
5

6

⇒ 30 × 8 × 3 × 104 = 26 × 9 × 2 × x
5 5

⇒ x = 30×8×35×104 = 30×8×3×104 = 30×8×104
26×9×52 26×9×2 26×3×2

= 30×8×4 = 5 × 8 × 4 = 160
3×2

Number of additional persons required = 160 – 104 = 56

10. The minute hand of a clock overtakes the hour hand at

intervals of 65 minutes of the correct time. How much a day

does the clock gain or lose?

a) The clock gains 10 10/143 minutes in 24 hours.

b) The clock gains 9 10/143 minutes in 24 hours.

c) The clock gains 10 8/43 minutes in 24 hours.

d) The clock gains 10 10/43 minutes in 24 hours.

Answer: (a)

Solution:

In a correct clock, the minute hand gains 55-minute space over

the hour hand in 60 minutes.

To be together again, the minute hand must gain 60 minutes

over the hour hand 55 min are gained in 60 min.

60 min. are gained in �5605 × 60� . = 65 5 .
11

But, they are together after 65 min.

∴ Gain in 65 min. = �65 5 − 65� = 5 .
11 11

Gain in 24 hours = �151 × 646×524� = 10 10
143

∴The clock gains 10 10 minutes in 24 hours.
143

7

TECHNICAL

One Mark Questions:

11. A 5 × 7 matrix has all its entries equal to 1. Then the rank of a

matrix is

a) 7

b) 5

c) 1

d) Zero

Answer: (c)

Solution:

Let A = ⎢⎡⎢⎢1111 1 1 1 1 1 1111⎤⎥⎥⎥
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
⎣1 1 1 1 1 1 1⎦5×7

Be the given matrix

Here all rows/ all columns are same.

Applying R2 – R1, R3 – R1, R4 – R1, R5 – R1

⎡⎢⎢⎢0001 1 1 1 1 1 0001⎥⎤⎥⎥
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
⎣0 0 0 0 0 0 0⎦

∴ ρ(A) = 1 = number of non − zero rows.

8

12. If a vector R�(t) has a constant magnitude then

a) R�. dR� = 0
dt

b) R� × dR� = 0
dt

c) R�. R� = dR�
dt

d) R� × R� = dR�
dt

Answer: (a)

Solution:

R(t) is a vector with constant magnitude

⇒ �R� = R = constant

We know that R. R = R2

⇒ d �R. R� = d (R2) = 0
dt dt

⇒ dR . R + R. dR = 0
dt dt

⇒ 2R. dR = 0 ⇒ R. dR = 0
dt dt

13. The maximum value of the determinant among all 2×2 real

symmetric matrices with trace 14 is ______.

Answer: 49

Solution:

Let A = �yx y x�
14 −

Det A = x (14 - x) – y2

For maximum value of Det A, y = 0

∴A = �0x 0 x�
14 −

9

|A| = x (14 - x) = (14x - x2) = f(x) (say)
⇒ f '(x) = 14 – 2x
f '(x) = Q ⇒ X = 7
f "(x) = -2 < 0
∴At x = 7, we get f(x) as maximum and is equal to 49.
14. Consider the following circuit:

The circuit shown above is in steady state before closing the
switch What is the current i(0+) through the switch if the circuit
is closed at t = 0?
a) -4 A
b) 0 A
c) 4 A
d) 12 A
Answer: (b)
Solution:

iL(0−) = 12 = 4A
2+1

10

vc(0−) = 1 × iL(0−) = 4V
Circuit at t = 0+

Current, i1 = 4 = 4A
1

Applying KCL at node A

i(0+) + 4 − i1 = 0
⇒ i(0+) + 4 = 0

i(0+) = 0A

15. If the Z-parameters for the T-network as shown below are Z11 =

40 Ω, Z22 = 50 Ω and Z12 = Z21 = 30 Ω, then what are the values

of Z1, Z2 and Z3?

a) 10 Ω, 20 Ω and 30 Ω
b) 20 Ω, 30 Ω and 20 Ω
c) 30 Ω, 40 Ω and 10 Ω
d) 40 Ω, 50 Ω and 10 Ω
Answer: (a)

11

Solution:

V1 = Z11I1 + Z12I2
V2 = Z21I1 + Z22I2
From the given network

V1 = Z1I1 + Z3(I1 + I2)
V1 = (Z1 + Z3)I1 + Z3I2 … (i)
and V2 = Z2I2 + Z3(I1 + I2)
V2 = Z3I1 + (Z3 + Z2)I2 … ( )
Comparing equation (i) and (ii) with standard equations

Z11 = Z1 + Z3
40 = Z1 + Z3 and Z12 = Z3

∴ Z3 = 30 Ω
∴ Z1 = 10Ω and Z3 + Z2 = Z22

Z3 + Z2 = 50
∴ Z2 = 20 Ω
16. For the circuit as shown below, if E = E1 and I is removed, then

V = 5 volts. If E = 0 and I = 1 A, then V = 5 volts. For E = E1

and I replaced by a resistor of 5 Ω, what is the value of V in

volts is ______.

Answer: 2.5

12

Solution:

When I is removed I2 = 0
∴ V = E = E1 = 5V
∴ E1 = 5V
5 = 1(R2 + 1)
∴ R2 = 4Ω

When I is replaced by 5Ω

E1 = (R2 + 1 + 5)I2

5 = (4 + 1 + 5)I2

∴ I2 = 1 A and V = 1 × 5
2 2

V = 2.5 V

17. In the circuit shown below, the maximum power absorbed by

the load resistance RL is ______ W.

Answer: 2.25

13

Solution:

Resistance seen across the terminal ab is

= 12×6 = 4
12+6

For maximum power transfer,

= 4

Now, current passing through RL,

I = 18 × 6
(6‖4+12) 6+4

= 18 × 6 = 0.75
2.4+12 10

Now maximum power absorbed by RL is

= I2RL = (0.75)2 × 4 = 2.25 Ω

18. Find the break region (voltage range) over which the dynamic

resistance of a diode is multiplied by a factor of 1000. Let this

region be contained between v1 and v2, then is │v1 – v2│given

by

a) logC (1000VT)

b) 1000 VT
c) (loge103)VT

d) The value cannot be computed with the given data

Answer: (c)

14

Solution:

ηVT rd,2 ev1.η/VT
I0ev.η/VT rd,1 ev2.η/VT
r = ⇒ =d

∴ 1000.rd,1 = e|v1−v2|.VηT
rd,1

∴ loge103 = |v1 − v2|. η
VT

⇒ |v1 − v2| = VT(loge103) for η = 1

19. An op-amp has a differential gain of 103 and a CMRR of 100.

The output voltage of the op-amp with inputs of 120μV and 80

μV will be _______ mV.

Answer: 41

20. The 6V Zener diode shown in figure, has zero Zener resistance

and a knee current of 5 mA. The minimum value of R, so that

the voltage across it does not fall below 6 V is _____ kΩ.

Answer: 80
Solution:

= 10−6 = 80
50

15

IL = Is − 5 = 80 − 5 = 75 mA

∴ R = 6 = 80Ω
75×10−3

21. The simplified form of a logic function = � � ∙ � � is

a) A + B
b) AB
c) +
d) +
Answer: (d)
Solution:

= � � ∙ � �

Using Demorgan’s theorem

= � + � ∙ � + �
= �� +�� + �� +��

= +

22. The state models

X(k + 1) = �−0β −1α� x(k) + �01�u(k)

Y(k) = [0 1] �xx12((kk))� is represented in the difference equation
as

a) C (k + 2) + αc (k + 1) + βc (k) = u(k)

b) C (k + 1) + αc (k) + βc (k – 1) = u (k – 1)

c) C (k – 2) + αc (k – 1) + βc (k) = u(k)

16

d) C (k – 1) + αc (k) + βc (k + 1) = u(k + 1)

Answer: (a)

Solution:

x1(k + 1) = x2(k) … (i)
x2(k + 1) = −βx1(k) − αx2(k) + u(k) … (ii)
y(k) = x2(k) … (iii)
From equation (i) and (ii)

x1(k + 2) = −βx1(k) − αx1(k + 1) + u(k)
⇒ c(k + 2) = −βc(k) − αc(k + 1) + u(k)

⇒ c(k + 2) + αc(k + 1) + βc(k) = u(k)

23. The signal x(t) = A cos (ωt + ϕ) is

a) an energy signals

b) a power signals

c) an energy as well as a power signal

d) neither an energy nor a power signal

Answer: (b)

Solution:

Energy signal: A signal is said to be an energy signal if the

energy of the signal is finite (i.e., E < ∞) and power is zero.

Power signal: A signal is said to be a power signal if its power

contained is finite. So, energy of power signal is infinite.

x(t) = A cos(ωt + ∅)
Energy of x(t) is E = ∫−∞∞|x(t)|2dt

E = ∫−∞∞ A2dt = ∞

17

Power of x(t) is P = T lim 1 ∫−TT//22|x(t)|2dt
→∞ T

= T lim 1 ∫−TT//22 A2dt
→∞ T

= T lim 1 . A2T = A2
→∞ T

Since the power of the signal is finite, it is a power signal

Note: Signal having infinity duration with their value tending to
a non-zero constant as t → ∞ are in general a power signal.

24.Consider the following block diagrams:

Which of these block diagrams can be reduced to transfer

function C(s) = 1−GG11G2?
R(s)

a) 1 and 3

b) 2 and 4

c) 1 and 4

d) 2 and 3

18

Answer: (b)
25. Select the correct transfer function v0(s)/v1(s) from the

following for the given network.

a) 2
s(s+1)

b) s
(s+2)

c) s
(2s+1)

d) 2s
(s+1)

Answer: (c)

Solution:

Applying nodal analysis,

V1 − V0 = V0 + V0
s

V1 = 2V0 + V0
s

V1 = V0 �2ss+1�

∴ V0 = s
V1 2s+1

19

26. Match List-I (dc motors) with List-II (Characteristics
labeled 1, 2, 3 and 4) and select the correct answer:
List-I
A. Shunt motor
B. Series motor
C. Cumulative compound motor
D. Differential compound motor
List- II

Codes: D
ABC 4
3
a) 3 2 1 4
b) 4 1 2 3
c) 3 1 2
d) 4 2 1
Answer: (c)

20

27.What is the increase in the torque expressed as percentage of

initial torque, if the current drawn by a d.c series motor is

increased from 10 A to 12 A (Neglect saturation)?

a) 21 %

b) 25 %

c) 41 %

d) 44 %

Answer: (d)

Solution:

Assuming for the motor to be a dc series motor as saturation has

to be neglected.

T ∝ Ia2

∴ T2 = 1.44T1

∴ % increase in the torque = 0.44T1 × 100 = 44%
T1

28.The capacitance of an overhead transmission line increases with

1. increase in mutual geometrical mean distance

2. increase in height of conductors above ground

Select the correct answer from the following:

a) Both 1 and 2 are true

b) Both 1 and 2 are false

c) Only 1 is true

d) Only 2 is true

Answer: (b)

21

Solution:

Capacitance of a transmission line = πϵ0 as mutual geometric
ln�hr�

mean distance increases, capacitance decreases.

29. The charging reactance of 50 km. length of the line is 1500 Ω.

What is the charging reactance for 100 km length of the line?

a) 1500 Ω

b) 3000 Ω

c) 750 Ω

d) 600 Ω

Answer: (c)

Solution:

Xc = 1 ∝ 1 as C ∝ length
ωC length

Xc = 50
1500 100

⇒ Xc = 750 Ω

30.Match List-I (Thyristors) with List-II (Symbols) and select

the correct answer:

List-I

A. Silicon-controlled rectifier (SCR)

B. Silicon-controlled switch (SCS)

C. Silicon-unilateral switch (SUS)

D. Light-activated SCR (LASCR)

22

List-II

Codes:
ABC D

a) 3 4 1 2
b) 4 3 1 2
c) 3 4 2 1
d) 4 3 2 1
Answer: (d)
31. A 0 to 300 V voltmeter has an error of ± 2% of fsd. What is the
range of readings if true voltage is 30 V?
a) 24 V – 36 V
b) 20 V – 40 V
c) 29.4 V – 30.6 V

23

d) 20 V – 30 V

Answer: (a)

Solution:

Error = 2 ×
100

= 2 × 300
100

= 6

∴ Range for 30 V = 30 ± 6 = 24 V – 36 V

32.Match List-I (Error parameters) with List-II (Values) and

select the correct answer; (σ is the standard deviation of

Gaussian error)

List-I List-II

A. Precision index 1. 0.67σ

B. Probable error 2. 3σ

C. Error limit 3. 0.39/σ

D. Peak probability density of error 4. 0.71/σ

Codes:

ABC D

a) 4 2 1 3

b) 4 1 2 3

c) 3 1 2 4

d) 3 2 1 4

Answer: (b)

24

33. Which of the following equations is correct?
a) � × � = | |2
b) � � × � � + � � × � � = 0
c) � × � � × � � = � × � � × � �
d) � . � + � . � = 0
Answer: (b)
Solution:
( � × � ) = 0
Since cross product with same vector is zero because θ = 0
so sin θ = 0
� × � = 0
� × � = − �
� � × � � + � � × � � = � + (− � ) = 0

34. Consider the following statements relating to Laplace's
equation:
1. Solution of Laplace's equation with two different approved
methods lead to different answers.
2. Every physical problem satisfying Laplace's equation must
contain at least two conducting boundaries.
3. Every field (if ρv = 0) satisfies Laplace's equation.
4. Every conceivable configuration of electrodes or conductors
produces a field for which Δ2V = 0.

25

Which of these statements are correct?
a) 1, 3 and 4
b) 3 and 4
c) 1 and 2
d) 2, 3 and 4
Answer: (b)
Solution:
• Every physical problem must contain atleast one conducting

boundary and usually contains two or more.
• Solution of Laplace’s equation with two different approved

methods lead to same answer.
• Laplace’s equation is all-embracing for applying as it does

wherever volume charge density ( ) is zero, it states that
every conceivable configuration of electrodes or conductors
produces a field for which Δ2V = 0.
35. A flat slab of dielectric, εr = 5 is placed normal to a uniform
field with a flux density D = 1 C/m2. The slab is uniformly
polarized. What is the polarization P of the slab in ______
Coulomb/m2?
Answer: 0.8
Solution:

P�⃗ = �ϵrϵ−r 1� �D�⃗ = �5−51� . 1 = 0.8C/m2

26

Two Marks Questions:

36. The number of terms in the expansion of general determinant

of order of order n is

a) n2

b) n!

c) N

d) (n +1)2

Answer: (b)

Solution:

If = [ 11]1×1 then | | = | 11| = 11

If A = � 1211 12 � then | | = 11 22 − 12 21
22

11 12 13
If A = � 21 22 23� then

31 32 33

| | = 11 22 33 − 11 23 32 + 12 31 23 − 12 21 33

+ 13 22 32 − 13 31 22

∴ The number of terms in expansion of 1st order

determinant is 1!

The number of terms in expansion of 2nd order determinant is 2!

The number of terms in expansion of 3rd order determinant is 3!

And so on the number of terms in expansion of nth order

determinant is n!.

27

37. For the equation ̈( ) + 3 ̇( ) + 2 ( ) = 5, the solution x(t)
approaches the following values as → ∞

a) 0

b) 5/2

c) 5

d) 10

Answer: (b)

Solution:
Given ̈( ) + 3 ̇( ) + 2 ( ) = 5

⇒ ( 2 + 3 + 2) = 5 0

The Auxiliary equation is
D2 + 3D + 2 = 0

⇒ (D + 1)(D + 2) = 0

⇒ D = −1, −2

∴ xc = C1e−t + C2e−2t

P. I = xp = 1 (5e0t )
D2+3D+2

= 5 (∵ f(0) ≠ 0)
2

Solution is x = C1e−t + C2e−2t + 5
2

But, As t → ∞, x → 5
2

38. Consider a die with the property that the probability of a face

with 'n' dots showing up is proportional to 'n'. The probability of

the face with three dots showing up is_______.

Answer: 0.14

28

Solution:

Let X be the random variable, which denotes the number of dots

on a face of the die. Then probability distribution table is shown

below.

X 123 4 56

P(X) K 2K 3K 4K 5K 6K
Σ ( ) = 1 (where ‘K’ is the proportionality constant)

i. e. , 21K = 1

⇒ = 1
21

∴ The required probability = 3K = 3 = 1 = 0.14
21 7

39. In Fig., the value of R is ______ Ω.

Answer: 12
Solution:

29

The relevant circuit is shown in Fig. Write loop equation:
�10 − 6 0 � 14 − �5 + 6 0 � 1 = 60
Solving, R = 12 Ω

40. Two ac sources feed a common variable resistive load as
shown in Fig. Under the maximum power transfer condition, the
power absorbed by the load resistance RL is ____ W.

Answer: 625 W
Solution:
Vth across the terminals of RL is found from the circuit shown in
Fig. (1).

Using superposition theorem,

Vth = 110∠00 + 90∠00 = 100∠00
2 2

Rth is found by replacing voltage sources by short-circuit

Zth = (6 + j8)‖(6 + j8) = (3 + j4)Ω

Thevenin's equivalent about A, B is shown in Fig. (2)

30

For maximum power transfer,

RL = √32 + 42 = 5Ω

⃗I = 100 , I = 100 , I2 = 104
8+j4 √80 80

Power absorbed by load,

(RL) = I2RL = 104 × 5 = 625W
80

41. An 11 V pulse of 10 µs duration is applied to the circuit shown

in Fig. Assuming that the capacitor is completely discharged

prior to applying the pulse, the peak value of the capacitor

voltage is ______V.

Answer: 6.32
Solution:

31

The relevant circuit is shown in Fig.

The Time constant T fall all responses in the circuit

= 10 × 103 × 11 × 10−9 = 10
11

If the input is 11 u (t),

VC(t) = F. V + (F. V − I. V) �1 − e−Ʈt �

Where, F. V = 11 × 10 = 10V and I. V = 0
11

(F.V and I.V are the final and initial values of VC (t))

When the pulse of 10 µsis applied, VC (t) as given in (i) is true

for 0 < t < 10 µs
∴ Peak value of VC (t) occurs at t = 10 µ sec
Peak value = 10 (1 − −1) = 6.32

42. A voltage signals 10 sin ωt is applied to the circuit with ideal

diodes as shown in figure. The max and minimum values of the

output waveform of the circuit are respectively.

a) +10 V and -10 V
b) +4 V and -4 V
c) +7 V and -4 V
d) +4 V and -7 V

32

Answer: (d)
Solution:
For positive half cycle

i) (when Vi > 4V) D2 ON, D1 = off then
ii) (when Vi < 4V) D1 and D2 = off

⇒ 0 =
iii) For negative half cycle,

D1 – ON & D2 = OFF
Vi – 10I + 4 – 10 I = 0

= � 2 +04�

When Vi = -10 V (MaxValue)

= −6
20

0 + 4 − 10 = 0

0 = 10 − 4 = 10 �−206� − 4

= −3 − 4 ⇒ 0 = −7

33

43. For the circuit shown in Fig. the Boolean expression for the
output Y in terms of inputs P, Q, R and S is

a) + + +
b) + + +
c) ( + )( + )
d) ( + )( + )
Answer: (b)
Solution:

In Two level logic ‘NAND – NAND’
Logic = ‘AND – OR’ Logic

44. Which of the following circuits is a realization of the above
function F?

34

Answer: (d)
Solution:
The above SOP expression, which can be implemented using
AND – OR logic obtaining � and � using NAND gates which
acts as inverters.
45. The figure shows a digital circuit constructed using negative
edge triggered J - K flip flops. Assume a starting state of Q2 Q1
Q0 = 000. This state Q2 Q1 Q0 = 000 will repeat after ____
number of cycles of the clock CLK.

Answer: 6
Solution:
The flip-flops 1 and 2 synchronously clocked by Q0 which is
CLK/2

35

After 3 clock cycles of CLK/2 Q2Q1 is becoming 00 again

So after 6th clock cycle of CLK, we see Q2Q1Q0 = 000 again
46. A continuous-time system is described by y(t) = e-│x(t)│, where

y (t) is the output and x (t) is the input, y (t) is bounded
a) only when x (t) is bounded
b) only when x (t) is non-negative
c) only for t ≥ 0 if x(t) is bounded for t ≥ 0
d) even when x (t) is not bounded
Answer: (d)
Solution:
y(t) = e−|x(t)|, |x(t)| is always positive for positive as well as
negative values of x(t).
∴ When x(t) is bounded, y(t) is bounded.
Even if x(t) is not bounded i.e., x(t) = ±∞ for any t
y(t) = e-∞ = 0
∴ y(t) is bounded even when x(t) is not bounded.
47. G (z) = αz-1 + βz-3 is a low-pass digital filter with a phase
characteristic same as that of the above equation if
a) α = β
b) α = - β
c) α = β(1/3)
d) α = β-(1/3)

36

Answer: (a)
Solution:
For G(z) = αz-1 + βz-3
G(ejω) = αe-jω + βe-3jω
The frequency response is given by
G(ejω) = e-j2ω (αejω + βe-jω)

= 2α cos (ω) e-j2ω, for α = β
48. The signal flow graph for a system is given below. The transfer

function Y(s)/U(s) for this system is

a) +1
5 2+6 +2

b) +1
2+6 +2

c) +1
2+4 +2

d) 1
5 2+6 +2

Answer: (a)

Solution:

37

Forward paths:

P1 = U − x1 − x2 − x3 − Y = S−2
P2 = U − x1 − x2 − x3 − Y = S−1

Loops:

L1 = x1 − x2 − x3 − x1 = −2S−2
L2 = x1 − x2 − x3 − x1 = −2S−1

L3 = x2 − x3 − x2 = −4S−1

L4 = x2 − x3 − x2 = −4

Δ1 = 1 Δ2 = 1

T. F = S−2+S−1
1−(−2s−2−2s−1−4s−1−4)

= S−2+S−1
s+2s−2+6s−1

= s+1
5s2+6s+2

49. In the system whose signal flow graph is shown in the figure.

U1(s) and U2(s) are inputs. The transfer function Y(s) is
U1(s)

a) k1
JLs2+JRs+k1k2

b) k1
JLs2−JRs−k1k2

c) k1−U2(R+sL)
JLs2+(JR−U2L)s+k1k2−U2R

d) k1−U2(sL−R)
JLs2−(JR+U2L)s+k1k2−U2R

38

Answer: (a)

Solution:

Y(s) = L1.1S.K1.1J .S1
U1(s) 1−��L1��1S�(K1)�1J ��1S�(−K2)�−�−LSR�

= K1
JLS2+JRS+K1K2

50. The roots of the closed loop characteristic equation of the

system shown in fig is

a) -1 and -15

b) 6 and 10

c) -4 and -15

d) -6 and -10

Answer: (d)

Solution:

( ) = (3)(15) & ( ) = 1
( +15)( +1)

Characteristic equation = 1 + G(s)H(s)

= 1 + 3(15) = 0
( +15)( +1)

2 + 16 + 15 + 45 = 0

2 + 16 + 60 = 0

( + 6)( + 10) = 0

Roots are -6 & -10

39

51. The loop gain GH of a closed loop system is given by the
following expression ( +2 ) ( +4). The value of K for which the

system just becomes unstable is

a) K = 6

b) K = 8

c) K = 48

d) K = 96

Answer: (c)

Solution:

Characteristic equation = 1 + G(s)H(s) = 0

1 + = 0
( +2)( +4)

( 2 + 2 )( + 4) + = 0

3 + 2 2 + 4 2 + 8 + = 0

3 + 2 2 + 8 + = 0

52. The figure as shown below represents the sinusoidal flux-
density distribution in the air-gap of a dc machine
If l = the axial length of the armature
r = the radius of the armature and
P = the number of poles in the machine
Then, what is the total flux per pole Q, in terms of the peak flux-
density Bpeak?

40

a) 2 BPeak .

b) 4 BPeak .

c) 2 BPeak .

d) 4 BPeak .

Answer: (b)

Solution:

Total flux per pole = 4 � �

53. A permanent magnet dc commutator motor has a no load speed

of 6000 rpm when connected to an 120V dc supply. The

armature resistance is 2.5 ohms and other losses may be

neglected. The speed of the motor with supply voltage of 60V

developing a torque 0.5 Nm, is _______ rpm.

Answer: 2673

Solution:

= ×
60

Under no load condition =

120 = × 6000
60

= 1.2 … … . (1)

41

Power developed (P) = 2 =
60

2 ×0.5 = × ×
60 60

(from equation 1)

2 × 0.5 = 1.2

= 2 ×0.5 = 2.62
1.2

2 = 60 − 2.62 × 2.5 = 53.45

2 = 2
1 1

2 = 53.45 × 6000 = 2672.75
120

≅ 2673

Data for Question: 54

A separately excited DC motor runs at 1500 rpm under no-load

with 200 V applied to the armature. The field voltage is

maintained at its rated value. The speed of the motor, when it

delivers a torque of 5 Nm, is 1400 rpm as shown in the figure.

The rotational losses and armature reaction are neglected.

42

54. For the motor to deliver a torque of 2.5 Nm at 1400 rpm, the

armature voltage to be applied is ______ V.

Answer: 193.3

Solution:

T = 2.5 N-m, N = 1400 rpm, V =?

2 = 2
1 1

2.5 = 2
5 3.92

2 = 1.96

∝

As speed is same Eb also same

= 186.67

= −

= +

= 186.67 + 1.96 × 3.4

= 193.3

55. A 220 V, 10 kW, 900 rpm separately excited DC motor has an

armature resistance Ra = 0.02 Ω. When the motor operates at

rated speed and with rated terminal voltage, the electromagnetic

torque developed by the motor is 70 Nm. Neglecting the

rotational losses of the machine, the current drawn by the motor

from the 220 V supply is _______ A.

Answer: 30

43

Solution:

= 220
= 900

= 70 − = = 70 −
∴Developed power = =
(70) �2 6×0900� =
⇒ ( − ) = 6597.3
[220 − (0.02)] = 6597.3
220 − 0.02 2 = 6597.3
0.02 2 − 220 + 6597.3 = 0

Solving above equation

= 30.06
56. A lightning stroke discharges impulse current of 10 kA (peak)

on a 400 kV transmission line having surge impedance of 250

Ω. The magnitude of transient overvoltage traveling waves in

either direction assuming equal distribution from the point of

lightning strike will be

a) 1250 kV

b) 1650 kV

c) 2500 kV

d) 2900 kV

Answer: (a)

Solution:

As the current is distributed equally current on each side =

5000A

44

Surge impedance of transmission line = 250 Ω
.'. The magnitude of transient over voltage = 250 × 5000

= 1250 kV
57. A lossless transmission line having Surge Impedance Loading

(SIL) of 2280 MW. A Series capacitive compensation of 30% is
emplaced. Then SIL of the compensated transmission line will
be
a) 1835 MW
b) 2280 MW
c) 2725 MW
d) 3257 MW
Answer: (c)
Solution:
Let characteristic impedance

( ) = � = �11..00 = 1 .

= � // = �

Given that for a given line 30% series capacitive compensation
is provided. Hence the series impedance of line is 0.7 or (70%)
of original value.

∴ = �10..07 = 0.836 .

Surge impedance loading (SIL) = 2

⇒ ∝ 1

45

( )2 = 1
( )1 2

( )2 = 1.0 × 2280 × 106
0.836

= 2725 × 106

= 2725

58. A 110 kV, single core coaxial, XLPE insulated power cable

delivering power at 50 Hz, has a capacitance of 125 nF/km. If

the dielectric loss tangent of XLPE is 2 ×10–4, then dielectric

power loss in this cable in W/km is ______.

Answer: 31.7

Solution:

Dielectric power loss = V2 ωC Tanδ
= (110 × 103)2 × 2 × 50 × 125 × 10−9 × 2 × 10−4

= 95 /

As we are using a single core cable

Dielectric loss in each core = 95 = 31.7 /
3

59. The A, B, C, D constant of a 220 kV line are: A = D =

0.94∠10, B = 130∠730, C = 0.001∠900. If the sending end

voltage of the line for a given load delivered at nominal voltage

is 240 kV, the % voltage regulation of the line is ______ %.

Answer: 16

Solution:

Voltage regulation = ( . )− ( . ) × 100
( . )

= +

46

Under no load condition IR = 0

∴ ( . ) = 255.3

∴ % = 255.3−220 × 100 = 16%
220

60. Thyristor T in the figure below is initially off and is triggered

with a single pulse of width 10 μs. It is given that L = �1 0 0� H
and L = �1 0 0� F. Assuming latching and holding currents of

the thyristor are both zero and the initial charge on C is zero, T

conducts for

a) 10 μs
b) 50 μs
c) 100 μs
d) 200 μs

Answer: (c)
Solution:
DC supply given to LC circuit with thyristor in the circuit, the
nature of current is sinusoidal. Thyristor does not allow current
in opposite direction.
Conduction time f circuit is one half cycle.

= √ = �1 0 0 × 100 ⇒ 100

47

61. In the single phase diode bridge rectifier shown in fig, the load
resistor is R = 50 Ω. The source voltage is V = 200 sin ωt,
Where ω = 2 × 50 radians per second. The power dissipated in
the load resistor R is _____W.

Answer: 400

Solution:

Given R = 50 Ω, V = 200 sin ωt

The power dissipated in load is 2 2

The output waveform is shown in figure below

In a full wave rectifier with R load

=
√2

= 2 = � /√2�2

= �2√020�2 × 1
50

= 400

48

Data for Question: 62
The input voltage given to a converter is

= 100√2 sin(100 )
The current drawn by the converter is

= 10√2 sin �100 − 3 � + 5√2
sin �300 + 4 � + 2√2 sin �500 − 6 �
62. The active power drawn by the converter is ____ W.
Answer: 500
Solution:
Active power will be drawn by converter only due to
fundamental component.
Therefore, Active power = Vs Is1 cosϕ1

= 100 × 10 cos 600
Active power = 500 watts
63. A chopper is employed to charge a battery as shown in figure.
The charging current is 5A. The duty ratio is 0.2. The chopper
output voltage is also shown in figure. The peak to peak ripple
current in the charging current is _____ A.

Answer: 0.48

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