Grand Test - 2 - PDF Flipbook
Grand Test - 2
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Test-02Solutions
APTITUDE
One Mark Questions:
1. Choose the word or group of words which is most opposite in
meaning to the word printed in bold.
Quiescent
a) indifferent
b) weak
c) active
d) responsive
Answer: (c)
Solution:
Meaning: Quiescent is being at rest; quiet; still; inactive or
motionless.
Synonyms: Idle, Immobile, Dormant
Antonyms: Active, Awake, Lively
2. Choose the word most nearly opposite to the given word.
Dyspeptic
a) Trusting
b) Functional
c) Euphoric
d) Talented
Answer: (c)
Solution:
Meaning: Dyspeptic is a person with poor digestion, something
that relates to poor digestion or a depressed person.
1
Example: A common dyspeptic type would be an old man
shouting "get off my lawn!" to kids.
Synonyms: Crabby, Mean, Grouchy, Irritable
Antonyms: Amiable, Good-Natured
3. Choose the word most nearly opposite to the given word.
Mendacious
a) Veracious
b) Pleasing
c) Rhythmic
d) Careful
Answer: (a)
Solution:
Meaning: A mendacious person is one who tells lies habitually
and intentionally.
Example: If you catch someone intentionally manipulating you
with a falsehood, that person is just plain mendacious.
Synonyms: Dishonest, Deceitful, Fallacious, Fraudulent
Antonyms: Frank, Honest, Truthful, Sincere
4. Choose the word which is most similar in meaning to the word
printed in bold.
Surreptitious
a) Threaten
b) Broken
c) Secret
d) Reveal
2
Answer: (c)
Solution:
Meaning: The meaning of surreptitious is something that is
done in secret or that is kept quiet.
When someone behaves in a surreptitious way, they're being
secretive.
Example: They were surreptitious about how they organized the
surprise party.
Synonyms: Clandestine, Covert, Furtive, Unauthorized, Hidden
Antonyms: Forthright, Open, Honest, Truthful
5. Father is aged three times more than his son Sunil. After 8 years,
he would be two and a half times of Sunil's age. After further 8
years, how many times would he be of Sunil's age?
a) 4 times
b) 5 times
c) 2 times
d) 3 times
Answer: (c)
Solution:
Assume that Sunil's present age = x.
Then, father's present age = 3x + x = 4x
After 8 years, father's age = 212 times of Sunil’s' age
⇒ (4x + 8) = 212(x + 8)
⇒ 4x + 8 = 52(x + 8)
⇒ 8x + 16 = 5x + 40
3
⇒ 3x = 40 – 16 = 24
⇒ x = 24/3 = 8
After further 8 years,
Sunil's age = x + 8 + 8 = 8 + 8 + 8 = 24
Father's age = 4x + 8 + 8 = 4 × 8 + 8 + 8 = 48
Father's age/Sunil's age = 48/24 = 2
4
Two Marks Questions:
6. One who does not marry, especially as a religious obligation
a) Bachelor
b) Celibate
c) Virgin
d) Recluse
Answer: (b)
7. In the question below, some part of the sentence has been left
blank. Select the best option which completes the sentence
properly, taking care of semantics, syntax and meaning.
Have you ever been to wild museum? Not sure but the name
_________.
a) reminds me of something
b) rings a bell
c) sounds known
d) sounds interesting
Answer: (b)
8. A vessel is filled with liquid, 3 parts of which are water and 5
parts syrup. How much of the mixture must be drawn off and
replaced with water so that the mixture may be half water and
half syrup?
a) 1/3
b) 1/4
c) 1/5
d) 1/7
5
Answer: (c)
Solution:
Suppose the vessel initially contains 8 liters of liquid.
Let x liters of this liquid be replaced with water.
Quantity of water in new mixture = �3 − 3 + � litres
8
Quantity of syrup in new mixture = �5 − 58 � litres
∴ �3 − 3 + � = �5 − 58 �
8
⇒ 5x + 24 = 40 – 5x
⇒ 10x = 26
⇒ x = 8
5
So, part of the mixture replaced = �58 × 18� = 1
5
9. What day of the week will 22 Apr 2222 be?
a) Monday
b) Tuesday
c) Sunday
d) Thursday
Answer: (a)
Solution:
22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-
Apr-2222)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is
a perfect multiple of 400)
6
Number of odd days in the period 2001 - 2200
= Number of odd days in 200 years
= 5 × 2 = 10 = 3
(As we can reduce perfect multiples of 7 from odd days without
affecting anything)
Number of odd days in the period 2201 - 2221
= 16 normal years + 5 leap years
= 16 × 1 + 5 × 2
= 16 + 10
= 26 = 5 odd days
Number of days from 1-Jan-2222 to 22 Apr 2222
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) =112
= 112 days = 0 odd day
Total number of odd days = (0 + 3 + 5 + 0)
= 8 = 1 odd day
1 odd day = Monday
Hence 22 Apr 2222 is Monday.
10. A clock is set at 5 am. If the clock loses 16 minutes in 24
hours, what will be the true time when the clock indicates 10 pm
on 4th day?
a) 9.30 pm
b) 10 pm
c) 10.30 pm
d) 11 pm
Answer: (d)
7
Solution:
Time from 5 am to 10 pm on the 4th day = 3 days 17 hours = 3
× 24 + 17 = 89 hours
Given that clock loses 16 minutes in 24 hours
⇒ > 23 hour 44 minutes of the given clock = 24 hours in a
normal clock
⇒ 23 44 hours of the given clock = 24 hours in a normal clock
60
⇒ 23 11 hours of the given clock = 24 hours in a normal clock
15
⇒ 356 hours of the given clock = 24 hours in a normal clock
15
⇒ 89 hours of the given clock = 24× 15 × 89 hours in a normal
356
clock = 24 × 15 = 6 × 15 = 90 hours
4
So the correct time is 90 hours after 5 am = 3 days 18 hours
after 5 am = 11 pm on the 4th day
8
TECHNICAL
One Mark Questions:
11. The number of linearly independent solutions of the system of
1 0 2 1
equations �1 −1 0� � 2� = 0 is equal to
2 −2 0 3
a) 1
b) 2
c) 3
d) 0
Answer: (a)
Solution:
1 0 2 1 0
�1 −1 0� � 2� = �0�
2 −2 0 3 0
2 ⟶ 2 − 1; 3 ⟶ 3 − 2 1
1 0 2 1 0
�0 −1 −2� � 2� = �0�
0 −2 −4 3 0
3 ⟶ 3 − 2 2
1 0 −2 1 0
�0 −1 −2� � 2� = �0�
0 0 0 3 0
= ( ) = 2, = = 3
∴
= − = 3 − 2
=1
9
12. The following systems of equations
x1 + x2 + x3 = 3
x1 – x3 = 0
x1 – x2 + x3 = 3 has
a) Unique solution
b) No solution
c) Infinite number of solutions
d) Only one solution
Answer: (b)
Solution:
Given AX = B
1 1 1 1 3
�1 0 1� � 2� = �0�
1 −1 1 3 1
Consider the augmented matrix [ | ]
1 1 13
[ | ] = �1 0 1 �0�
1 −1 1 1
2 ⟶ 2 − 1; 3 ⟶ 3 − 1
1 1 13
∽ �0 −1 0 �−3�
0 −2 0 −2
3 ⟶ 3 − 2 2
1 1 13
∽ �0 −1 0 �−3�
0 0 04
( ) = 2, ( | ) = 3
, ( ) ≠ 2, ( | ) ∴ Solutions does not exist
10
13. By reversing the order of integration ∫02 ∫ 2 2 ( , ) may
be represented as
a) ∫02 ∫ 2 2 ( , )
b) ∫02 ∫ √ ( , )
c) ∫04 ∫ √ / 2 ( , )
d) ∫ 2 2 ∫02 ( , )
Answer: (c)
Solution:
Given limits are
= 2 = 2 �
= 0 =2
To change the order of integration, write the variable limits for x
x = y/2 to X = �
and y = 0 to y = 4
∴ ∫04 ∫ √ / 2 ( , )
11
14. A scalar field is given by f = x2/3 + y2/3, where x and y are the
Cartesian coordinates. The derivative of 'f' along the line y = x
directed away from the origin at the point (8, 8) is
a) √2
3
b) √3
2
c) 2
√3
d) 3
√2
Answer: (a)
Solution:
= 2/3 + 2/3
A unit vector along the line y = x is � = cos + sin , where
= /4
� = cos + sin = 1 + 1 ̅
4 4 √2 √2
∇ = 2 −1 ̅ + 2 −1 ̅
3 3 3 3
At (8, 8) ∇ = 1 + 1
3 3
∴ Directional derivative = ∇ . �
= 1 + 1 = √2
3√2 3√2 3
12
15. The value of ( , l )i→m(0,0) 2− is _______.
√ −√
Answer: 0
Solution:
Along the path y = mx
( , l )i→m(0,0) 2−
√ −√
= l i →m0 2− 2
√ −√
= l i →m0 2(1− )
√ (1−√ )
3
2(1− )
= l i →m0 (1−√ ) = 0
16. The figure shows a network in which the diode is an ideal one.
the terminal v-i characteristics of the network is given by
13
Answer: (c)
Solution:
By applying KVL
The terminal v-I characteristic is v = 2i + 5
17. Consider the following circuit:
For the circuit shown above, by how much the voltage across
the inductor leads the voltage across the capacitor?
a) 450
b) 900
c) 1350
d) 1800
Answer: (c)
Solution:
Let current through inductor, ∠
14
∴ ∠ . = … … … . ( )
1
+
and = 2∠00 … … … . ( )
1
from equation (i) and (ii),
∠ . = 2∠
1+
⇒ − tan−1( ) = 0
⇒ = tan−1(2 × 0.5 × 1) = 450
As we know, in inductor current lags the voltage by 900 so
inductor voltage phase angle = 900 + 450 = 1350 i.e., inductor
voltage leads the capacitor voltage by 1350.
18. Match List-I (Metal-semiconductor band diagram under
equilibrium) with List-II (Type of contact) and select the
correct answer using the code given below the lists:
List-I List-II
15
Codes:
ABC D
a) 1 4 2 3
b) 1 4 3 2
c) 4 1 3 2
d) 4 1 2 3
Answer: (a)
Solution:
In case of rectifier gap between Ec to Ev will be large in case of
ohmic contact gap between Ec to Ev will be small and in n-type
case Ef will be close to Ec and in p-type EF will be close to Ev.
19. In Boolean Algebra
If = ( + )� + �, then
a) = +
b) = +
c) = +
d) = +
Answer: (c)
16
Solution:
= ( + )( ̅ + )
= ̅ + + ̅ +
= + ̅ +
= ̅ +
20. What is the simplified form of the Boolean expression =
( + )� + �� + �?
a)
b)
c)
d)
Answer: (c)
Solution:
= ( + )( + � )( ̅ + )
= ( + � + + � )( ̅ + )
= ( + � + + � )( ̅ + � )
(∵ = � = 0)
= ( + � + )( ̅ + )
= (1 + � + )( ̅ + )
= (1 + 1)( ̅ + ) (∵ 1 + 1 = 1)
= ( ̅ + ) = � +
17
= 0 + (∵ � = 0)
=
21. Which one of the following pairs is NOT correctly matched?
(input x(t) and output y(t)).
a) Unstable system: ( ) − 0.1 ( ) = ( )
b) Nonlinear system: ( ) + 2 2 ( ) = ( )
c) Non-causal system y(t) = x (t + 2)
d) Non-dynamic System, y(t) = 3x2(t)
Answer: (b)
Solution:
(i) For the system to be stable, the roots of the characteristic
equation should lie in the LHS of s-plane.
( ) − 0.1 ( ) = ( )
⇒ ( ) = 1
( ) −0.1
⇒ s = 0.1 which is in RHS of s-plane, hence the system is
unstable.
(ii) The system is linear if the response to ax1(t) + bx2(t) is ay1(t)
+ by2(t).
( ) + 2 2 ( ) = ( )
The condition of linearity is satisfied here, hence the system is
linear.
(iii) A system is casual if the output at any time depends only on
values of the input at the present time and in the past.
18
(iv) A system is said to be memoryless or static if its output for
each value of the independent variable at a given time is
dependent only on the input at the same time.
22. Match List-I (Characteristic of f(t)) with List-II (Functions)
and select the correct answer using the codes given below the
lists:
List-I
A. f(t) (1 – u(t)) = 0
B. f(t) + k ( )⁄ = 0; k is a positive constant
C. f(t) + k 2 (2 ) = 0; k is a positive constant
D. f(t) (g(t) – g (0) = 0; for any arbitrary g(t)
List-II
1. Decaying exponential
2. Growing exponential
3. Impulse
4. Causal
5. Sinusoid
Codes:
A BCD
a) 4 1 5 3
b) 1 4 5 3
c) 4 2 5 1
d) 2 5 4 1
Answer: (a)
19
Solution:
i) ( ) + ( ) = 0
Taking Laplace transform, (1 + Ks) F(s) = 0
Characteristic equation is 1 + Ks = 0
Complementary function C.F = e-t/K
which is decaying exponential
ii) ( ) + 2 ( ) = 0
2
Its Laplace transform is (1 + Ks2) F(s) = 0
Characteristic equation is 1 + Ks2 = 0
Roots are s = ±j√1
C.F = cos 1 + sin 1
√ √
which is a sinusoid function
iii) ( ){1 − ( )} = 0
∵ ( ) = 1 > 0
0 < 0
So for t < 0, f(t) = 0
Therefore, f(t) is a causal function
23. Match List-I (Block Diagram) with List-II (Transformed
Block Diagram) and select the correct answer:
List-I List-II
20
Codes:
AB C D
a) 3 4 2 1
b) 4 3 1 2
c) 3 4 1 2
d) 4 3 2 1
Answer: (c)
24. Consider the following diagram:
For the multiple gear system shown above, which one of the
following gives the equivalent inert a referred to shaft 1?
a) 1 + 2 � 12�2 + 3 � 21 43�2
b) 1 + 2 � 12�2 + 3 � 21 43�2
c) 1 + 2 � 21�2 + 3 � 13 24�2
d) 1 + 2 � 12�2 + 3 � 31 24�2
21
Answer: (a)
25. A d.c shunt motor is excited from an alternating power
frequency voltage source. Its brush-axis is rotated by an angle
from the geometrical neutral axis. The torque developed will be
proportional to which one of the following?
a) sin α
b) cos α
c) tan α
d) cos 2α
Answer: (b)
Solution:
When the brush currents are alternating the effect of shifting the
brushes is to shift the space phase of the magnetic field
produced by armature currents by a like amount with respect to
the stator field.
26. A 50 kW d.c shunt motor is loaded to draw rated armature
current at any given speed. When driven (i) at half the rated
speed by armature voltage control and (ii) at 1.5 times the rated
speed by field control, what are the approximate output powers
delivered by the motor?
a) 25 kW in (i) and 75 kW in (ii)
b) 25 kW in (i) and 50 kW in (ii)
c) 50 kW in (i) and 75 kW in (ii)
d) 50 kW in (i) and 50 kW in (ii)
Answer: (b)
22
Solution:
Armature voltage control has constant torque mode so
Power = torque × speed
⇒ P ∝ speed
⇒ P (at half the rated speed) = 50 = 25
2
Field control has constant power mode.
So, P (at 1.5 times the rated speed) = 50 kW
27. Neglecting all losses, how is the developed torque (T) of a dc
separately excited motor, operating under constant terminal
voltage, related to its output power (P)?
a) T ∝√P
b) T ∝ P
c) T2 ∝ P3
d) T is independent of P
Answer: (b)
Solution:
=
Output power ‘P’ = Vt ×
Vt = constant and separately excited motor so, ϕ is also constant.
Hence, P ∝
and ∝
So, ∝
23
28. Consider the following statements:
In the figure below, the curves pertain to the dc motor.
1. Speed Vs armature-current characteristic of a dc shunt motor.
2. Torque Vs armature-current characteristic of dc shunt motor.
3. Speed Vs armature current characteristic of dc series motor.
4. Torque Vs armature-current characteristic of dc series motor.
What is the correct sequence of characteristics?
a) 4132
b) 2134
c) 4312
d) 2314
Answer: (c)
Solution:
For constant voltage supply Eb drops slightly
For shunt motor,
∝
∝ ∝ ∴
which is almost constant.
24
For series motor,
∝ ∝ 2
∝ ∝ ∝ 1
29. For some given transmission line the expression for voltage
regulation is given by | |−| | × 100%. Hence,
| |
a) this must be a 'short' line
b) this may either be a 'medium line' or a 'short line
c) this expression is true for any line
d) this may either be a 'medium line' or a ‘long line’
Answer: (a)
Solution:
In case of a short line, when load is removed, the voltage at the
receiving end is equal to the voltage at the sending end. At full
load � , � = | | and at no load, � , � = | |.
30. An equipment has an impedance 0.9 p.u. to a base of 20 MVA,
33 kV. To the base of 50 MVA, 11 kV the p.u. impedance will
be ______
Answer: 20.25
Solution:
. = 0.9 .
( ) = 20 ;
( ) = 33
( ) = 50 ;
( ) = 11
25
( ) = .( ) × ( ) × ( )2( )
( ) ( )2( )
= 0.9 × 50 × (33)2
20 (11)2
( ) = 20.25
31. When cathode of a thyristor is made more positive than its
anode
a) all the junctions are reverse biased
b) outer junctions are reverse biased and central one is forward
biased
c) outer junctions are forward biased and central one is reverse
biased
d) all the junctions are forward biased
Answer: (b)
Solution:
When cathode is made more positive with respect to anode,
thyristor operates in reverse blocking mode, and the device
behaves as it two diodes are connected in series with reverse
voltage applied across them.
32. In single pulse modulation of PWM inverters, the pulse width
is 1200. For an input voltage of 220 V dc, what is the rms value
at the fundamental component of the output voltage is ______V.
Answer: 171.5
Solution:
0 = ∑ ∞ =1,3,5 4 sin sin sin
2
26
( 01) = 4 sin sin
2
and = 120 = 600
2
( 01) = 1 �4 sin 600� = 171.5
√2
33. The value of a quantity and its uncertainty are given as 26455
± 3754 without rounding off. Only two significant digits are
relevant for error. Value of error rounded off to two significant
figures is
a) 26500 ± 3800
b) 26400 ± 3800
c) 26460 ± 3750
d) 26400 ± 3700
Answer: (a)
Solution:
Significant figures convey actual information regarding the
magnitude and the measurement precision of a quantity.
Hence, option (a) is correct.
34. What is the value of the integral ∮ � � � ⃗ along the curve c (c is
the curve ABCD in the direction of the arrow)?
27
a) 2 � � + � �/√2
b) −2 � � + � �/√2
c) 2R �
d) −2R �
Answer: (d)
Solution:
∫ � � � ⃗ = ∫0 /2 . � = �
2
∫ � � � ⃗ = ∫+− (− � ) = −2 �
∫ � � � ⃗ = ∫−− / 2 . � = − �
2
∴ ∫ � � � ⃗ = ∫ � � � ⃗ + ∫ � � � ⃗ + ∫ � � � ⃗ = −2 �
35. A plane slab of dielectric having dielectric constant 5, placed
normal to a uniform field with a flux density of 2 C/m2, is
uniformly polarized. The polarization of the slab is ______C/m2.
Answer: 1.6
Solution:
Polarization, �⃗ = � ⃗ �1 − 1 � = 2 �1 − 51� = 1.6 / 2
28
Two Marks Questions:
36. The value of the integral of the function g(x, y) = 4x3 + 10y4
along the straight line segment from the point (0, 0) to the
point(1, 2) in the xy-plane is
a) 33
b) 35
c) 40
d) 56
Answer: (a)
Solution:
Equation of straight line segment from (0, 0) to (1, 2) is
y = 2(x - 1)
∫ ( , ) = ∫01(4 3 + 10 4)
= ∫01[4 3 + 10 × 16( − 1)4]
= 33
37. The complex number z = x + jy which satisfy the equation │z
+1│= 1 lie on
a) a circle with (1, 0) as the centre and radius 1
b) a circle with (-1, 0) as the centre and radius 1
c) y – axis
d) x – axis
Answer: (b)
Solution:
The general equation of circle is given by │z + z0│= r
Where z0 is centre and radius is r
29
Now the equation of the given circle is │z + 1│= 1 (or) │z – (-
1) │= 1
∴ centre = (-1, 0) and radius = 1
38. The lowest Eigen value of the 2 × 2 matrix �41 23� is ____
Answer: 2
Solution:
| − | = 0
⇒ (4 − )(3 − ) − 2 = 0
⇒ ( 2 − 7 + 10 = 0)
∴ = 2, 5
39. The value of l i →m0 �2 − + � is _______.
Answer: -0.333
Solution:
l i →m0 �2 − + � �00�
Applying L’ Hospital rule
l i →m0 �2 − + � = l i →m0 �2 +− − �
= l i →m0 �3 − − �
= − cos(0)
3 0−0
= − 1
3
= −0.333
30
40. The equivalent inductance seen at terminals A-B in Fig. is
_____ H.
Answer: 8H
Solution:
For the given coupled circuit shown in Fig
L1 = L2 = L3 = 4H
The coupling between,
L1 and L2 is opposing (M = 2H)
L2 and L3 is aiding (M = 1H)
L3 and L1 is opposing (M = 1H)
LAB = Leq
= L1 + L2 + L3 – (2×2) + (2×1) – (2×1)
= 12 – 4 + 2 – 2 = 8H
31
41. The voltage and current waveforms for an element are
shown in Figure. The circuit element is inductor and its
value is _____ H.
Answer: 2H
Solution:
For the given i(t) and v(t),
( ) = ( ) , ℎ = 2
Compare with ( ) =
Therefore, the circuit element is an inductor with L = 2H
(Or)
From the given i(t) ~ t and v(t) ~ t waveforms
( ) = �0 0 ≤ ≤ 2
2 ≤
∴ = �10 0 ≤ ≤ 2
2 ≤
And, 2 = �02 0 ≤ ≤ 2
2 ≤
32
But this is the given V (t)
∴ ( ) = 2 , = 2
Hence the element must be an inductance of 2H.
42. An incandescent lamp is marked 40 W, 240 V. If resistance
at room temperature (260C) is 120 Ω, and temperature
coefficient of resistance is 4.5 × 10-3/0c, then its 'ON' state
filament temperature in 0C is approximately ____
Answer: 2470-2471
Solution:
ON state,
= [ ]2 = (240)2 = 1440Ω
40
= 0[1 + ]
1440 = 120[1 + 4.5 × 10−3( − 26)]
1 + 4.5 × 10−3( − 26) = 12
4.5 × 10−3( − 26) = 11
⇒ ( − 26) = 11/4.5 × 10−3
⇒ = 2470. 40
43. The octal equivalent of the HEX number AB.CD is
a) 253.314
b) 253.632
c) 526.314
d) 526.632
Answer: (b)
33
Solution:
44. A, B, C and D are input bits, and Y is the output bit in the
XOR gate circuit of the figure below. Which of the following
statements about the sum S of A, B, C, D and Y is correct?
a) S is always either zero or odd
b) S is always either zero or even
c) S = 1 only if the sum of A, B, C and D is even
d) S = 1 only if the sum of A, B, C and D is odd
Answer: (c)
Solution:
For the given figure in the question;
S=A+B+C+D+Y
If A = B = C = D = 0 ⇒ Y = 0 ⇒ S = 0
If A, B, C, D inputs has added No. of one’s,
Y = 1 ⇒ S = even
34
45. If u(t), r(t) denote the unit step and unit ramp functions
respectively and u(t) * r(t) their convolution, then the function
u(t + 1) * r(t – 2) is given by
a) (1/2) (t – 1) (t – 2)
b) (1/2) (t – 1) (t – 2)
c) (1/2)(t – 1)2 u (t – 1)
d) none of the above
Answer: (c)
Solution:
( ) ↔ 1
( + 1) ↔
( ) ↔ 1
2
( − 2) ↔ −2
2
( + 1) ∗ ( − 2)
↔ . −2
2
↔ −
3
2 ( ) ↔ 1
2 3
( −1)2 ( − 1) ↔ −
2 3
35
46. x [n] = 0; n < -1, n > 0, x[-1] = -1, x[0]
= 2is the input and
y [n] = 0; n < -1, n > 2, y [-1] = -1
= y [1], y [0] = 3, y [2] = -2 is the output of a discrete-time
LTI system. The system impulse response h[n] will be
a) h [n] = 0; n < 0, n > 2,
h [0] = 1, h [1] = h [2] = -1
b) h [n] = 0; n < -1, n > 1,
h [-1] = 1, h [0] = h[1] = 2
c) h [n] = 0; n < 0, n ≥ 3, h [0] = -1
h [1] = 2, h [2] = 1
d) h [n] = 0; n < -2, n > 1,
h [-2] = h[1] = -2, h [-1] = -h [0] = 3
Answer: (a)
Solution:
( ) ≠ 0, from n = -1 to n = 0
( ) ≠ 0, from n = -1 to n = 2
As y(n) = x(n) * h(n),
h(n) ≠ 0, from n = 0 to n = 2 from width property of
convolution
This is satisfied only for option in (a).
Detailed analysis follows:
From the given data
x(n) = -1 δ(n + 1) + 2 δ(n)
and y(n) = -δ (n + 1) + 3δ(n) – 1 δ(n – 1) - 2 δ(n – 2)
36
X(z) = -z-1 + 2
Y(z) = -z-1 + 3 – z-1 – 2 z-2
( ) = ( ) = − −1+3− −1−2 −2
( ) 1− −1− −2
∴ ℎ( ) = 1 ( ) − 1 ( − 1) − 1 ( − 2)
ℎ(0) = 1, ℎ(1) = ℎ(2) = −1
ℎ( ) = 0, < 0
ℎ( ) = 0, > 2
47. The signal flow graph of the system is shown in the given
figure. The transfer function C(s)/D(s) of the system is
a) 1( ) 2( )
1+ 1( ) ( )
b) 1( ) 2( )
1− 1( ) 2( ) ( )
c) − 2( )
1+ 1( ) 2( ) ( )
d) − 2( )
1− 1( ) 2( ) ( )
Answer: (c)
Solution:
( ) =?
( )
Put, R(s) = 0, graph reduces to
37
Mason gain formula,
= 1 ∑ ∆
∆
There is only one forward path,
1 = (−1)� 2( )�(1) = − 2( )
One loop → − 1( ) ( ) 2( )
∆ = 1 − �− 1( )� ( ) 2( )
= 1 + 1( ) 2( ) ( )
∆1= 1 ( − ℎ )
( ) = − 2( )
1+ 1( ) 2( ) ( )
= ( )
( )
48. Consider a system having forward path and feedback path
transfer functions 16 and (1 + 2s) respectively. Which is the
s(s+0.8)
characteristic polynomial of the system?
a) s2 + 0.8s + 16
b) s2 + 2.8s + 16
c) s2 + 32.8s + 16
d) s2 + 32s + 16
Answer: (c)
38
Solution:
( ) = 16
s(s+0.8)
( ) = (1 + 2 )
( ) 16
1+ ( ) ( )
. = = s(s+0.8)
1+(1s+(s2+ )0(.81)6)
. = 16
2+32.8 +16
Characteristic polynomial = s2 + 32.8s + 16
49. A unity feedback control system has forward path transfer
function G(s) given by
( ) = 10(1+ )
2( +12)( +5)
The steady-state error due to unit parabolic input r (t) = t2⁄2 u
(t) is ______.
Answer: 6
Solution:
( ) = 10(1+ )
2( +12)( +5)
= l i→m0 ( )
1+ ( )
( ) = � 2 2 ( )� = 1
3
= lim ( )× 1 3
→0 1+ 2( 1 +01(12+)( ) +5)
= l i→m0 2 ( +12)( +5)
( +12)( +5)+10(1+ )
= (12)(5) = 6
10
39
50. Given a unity feedback system with G(s) = K the value of
s(s+4)
K for damping ratio of 0.5 is _________
Answer: 16
Solution:
Characteristic equation: 2 + 4 +
On comparing with 2 + 2 + 2
. = ( ) =
1+ ( ) 2+4 +
2 =
2 = 4
= 0.5
(2)(0.5)�√ � = 4
⟹ = 16
51. A DC shunt motor is running at 1200 rpm, when exited with
220 V DC. Neglecting losses and saturation, the speed of the
motor when connected to a 175 V DC supply is
a) 750 rpm
b) 900 rpm
c) 1050 rpm
d) 1200 rpm
Answer: (d)
Solution:
At V = 220 V, N = 1200 rpm
Neglecting losses & Saturation
40
∝ ∝ − ⟹ ∝ ∝
∅ ∅ ∅
∵
∴ Losses are neglected
2 1
= = × = × = 1 2 2 2 1 2
1 1 1 2 1 2
1
∴ 2 = 1 = 1200
52. A 200V, 2000 rpm, 10A, separately excited d.c motor has an
armature resistance of 2A. Rated d.c voltage is applied to both
the armature and field winding of the motor. If the armature
drawn 5A from the source, the torque developed by the motor is
_____ N-m.
Answer: 4.30 N-m
Solution:
V = 200 V, N = 2000 rpm = (100 π/s) π/s
Ia = 10 A; Ra = 2Ω
Eb = V – Ia Ra = 200 – 10 × 2 = 180 V
If now the armature draws a current of 5 A,
torque developed = 540×5 = 4.30 −
200
53. A three-phase, three winding Δ/Δ/Y (1.1 kV/6.6 kV/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.
41
Answer: 625
Solution:
Given transformer is ∆/∆/
Wdg (1) / wdg (2) / wdg (3)
Δ /Δ / Y
1.1 kV / 6.6 kV / 400 V
By applying superposition theorem
a) Load on 6.6 kV (Δ) side is 900 kVA, 0.8 pf lag
⇒ 900 × 103 = √3 × 6.6 × 103 ×
= 78.72
ℎ( ) = 78.72 = 45.46 / ℎ
√3
= 45.46∠ − 36.86 / ℎ
⇒ The current on 1.1 kV side will be
Iph(source) = 46.46 × �16..16�
= 272.76∠ − 36.86 / ℎ
b) Load on 400 V (Y) is 300 kVA, 0.6 pf lag
⇒ 300 × 103 = √3 × 400 ×
= 300×103 = 433.01
√3×400
⇒ ℎ = 433.01∠ − 53.13 / ℎ
∴ The current of source side (1.1 kV) will be
Iph(source) = 400/√3 × 433.01
[1100]
= 90.90∠ − 53.13 / ℎ
∴ Total source phase current will be
42
= 272.76∠ − 36.86 + 90.90∠ − 53.13
= 360.88 / ℎ;
pf = 0.65
∴ The line current will be = 625 A
54. A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating
of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine
at synchronous speed is 100 MJ. The machine is running
steadily at synchronous speed and delivering 60 MW power at a
power angle of 10 electrical degrees. If the load is suddenly
removed, assuming the acceleration is constant for 10 cycles,
the value of the power angle after 5 cycles is ____ electrical
degrees.
Answer: 12.70
Solution:
= 1000 = 4
250
δ = 100
Ps = Pe = 60 MW
δ = δ + Δδ
∆ = (∆ )2
2
43
= −
= (∆ )2
2
= 60−0 × (0.1)2
2
5
180
∆ = 60×186×50 × (0.1)2
250×4 2
6 × 180 × 5 × (0.1)2 = 2.70
2
= 10 + 2.7 = 12.70
55. A cable has the following characteristics. L = 0.201 μH/m and
C = 196.2 pF/m. The velocity of wave propagation through the
cable is
a) 32 m/s
b) 159.24 μ/ms
c) 0.0312 m/s
d) 159.24 m/(μ-s)
Answer: (d)
Solution:
Velocity of wave in a cable = 1
√
Given L = 0.201 μH/m
C = 196.2 pF/m
∴ = 1
√0.201×10−6×196.2×10−12
= 159240110 /
= 159.24 / −
44
56. A generator delivers power of 1.0 p.u. to an infinite bus
through a purely reactive network. The maximum power that
could be delivered by the generator is 2.0 p.u. A three-phase
fault occurs at the terminals of the generator which reduces the
generator output to zero. The fault is cleared after tc seconds.
The original network is then restored. The maximum swing of
the rotor angle is found to be δmax = 110 electrical degrees. Then
the rotor angle in electrical degrees at t = tc is ______ degrees.
Answer: 69.14
Solution:
= 1 = 1.0 .
Before fault = 1 = 2.0 .
0 = sin−1 � 1� = 300
= 0.523
During fault 2 = 0 ⇒ 2 = 0
= 1100 = 110×
180
= 1.919
3 = 1 � ℎ 1 = �
45
Critical clearing angle, =
cos−1 � ( − 0)+ 3 cos − 2 cos 0�
3 − 2
= cos−1 �1.0(1.919−0.5223−)0+2 cos 1100−0�
= 69.140
57. Consider a synchronous generator connected to an infinite bus
by two identical parallel transmission lines. The transient
reactance x' of the generator 0.1 pu. Due to some previous
disturbance, the rotor angle (d) is undergoing an undamped
oscillation, with the maximum value of δ(t) equal to 1300. One
of the parallel lines trip due to relay mal-operation at an instant
when δ(t) = 1300 as shown in the figure. The maximum value of
the per unit line reactance x, such that the system does not lose
synchronism subsequent to this tripping is _______ p.u.
Answer: 0.67
46
Solution:
As the alternator is already under going undamped oscillations
with δmax = 1300, at this instant is any fault occurs even then also
to maintain stability ‘δ’ value must not go beyond 1300, for a
given mechanical input. sin =
∴ 2 = ,
Where Xeq = Equivalent reactance when online is removed due
to mal operation of relay.
Xeq = 0.1 + X
∴ 1.0×1.0 sin 130 = 1.0
(0.1+ )
= 0.67 . [∵ = 1.0 .
47
58. A cylindrical rotor generator delivers 0.5 pu power in the
steady-state to an infinite bus through a transmission line of
reactance 0.5 pu. The generator no- load voltage is 1.5 pu and
the infinite bus voltage is 1 pu. The inertia constant of the
generator is 5 MW-s/ MVA and the generator reactance is 1 pu.
The critical clearing angle, in degrees, for a three-phase dead
short circuit fault at the generator terminal is ______ deg.
Answer: 79.6
Solution:
= 1 = 0.5
Before fault 1 = ⇒ 1.5×1.0 = 1.0
1.5
During fault 2 = 0,
After the fault 3 = 1.0
= sin−1 � 1� ⇒ sin−1 �10..05� = 300
0( ) = 30× ⇒ 0.52
180
= 180 − sin−1 � 3�
⇒ 180 − sin−1 �01.5� = 1500
( ) = 150× ⇒ 2.618
180
Critical clearing angle, = cos−1 � � ( − 0)+ 3 cos
3
= cos−1 �0.5(2.618−0.512.0)+1.0 cos 1500�
= 79.450
48
59. A MOSFET rated for 15 A, carries a periodic current as shown
in figure. The ON state resistance of the MOSFET is 0.15 Ω.
The average ON state loss in the MOSFET is
a) 33.8 W
b) 15.0 W
c) 7.5 W
d) 3.8 W
Answer: (c)
Solution:
Given ‘ON’ state Resistance = 0.15 Ω
From fig. MOSFET carries a current 10A
From figure, Average power loss,
= 1 ∫0 2
2
= 100 ∫0 (0.15)
= 50(0.15) ( )
= 7.5
Data for Question 60:
A 1:1 Pulse transformer (PT) is used to trigger the SCR in the
adjacent figure. The SCR is rated at 1.5 kV, 250 A with IL = 250
mA, IH = 150 mA, and IGmax = 150 mA with IL = 250 mA, IGmin
= 100 mA. The SCR is connected to an inductive load, where L
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