Grand Test - 4 - PDF Flipbook
Grand Test - 4
484 Views
20 Downloads
PDF 11,492,742 Bytes
GATE
EEE
GrandTests
Test-04Solutions
APTITUDE
One Mark Questions:
1. Choose the word or group of words which is most opposite in
meaning to the word in bold.
Ecstasy
a) Hate
b) Agony
c) Languor
d) Fatigue
Answer: (b)
Solution:
Meaning: The meaning of ecstasy is a feeling of intense joy,
happiness or passion.
Example: Actors are typically in ecstasy upon winning an Oscar.
Synonyms: Bliss, Elation, Euphoria, Happiness, Joy
Antonyms: Depression, Misery, Sadness, Apathy
2. Choose the word which is most similar in meaning to the word
printed in bold.
Magniloquent
a) Amusing
b) Humorous
c) Intelligent
d) Boastful
Answer: (d)
1
Solution:
Meaning: Magniloquent is using high-flown or bombastic
language.
Synonyms: Grandiloquent, High-sounding, Lofty
Antonyms: Terse, Crisp
3. Unfortunately for the young hire, the amiable, gregarious air of
his boss during the initial interview belied a vastly more
________ style on the job, a fact which he learned to his chagrin
within the first few days of employment.
a) Draconian
b) Friendly
c) Fatuous
d) Disconcerting
Answer: (a)
4. The largest measuring cylinder that can accurately fill 3 tanks of
capacity 98, 182 and 266 litres each, is of capacity?
a) 7 lts
b) 14 lts
c) 98 lts
d) 42 lts
Answer: (b)
Solution:
To know the measuring cylinder that can fill all the given
capacities, they must be divisible by the required number.
98, 182, 266 all are divisible by 14
2
So 14 litres is the largest cylinder that can fill all the given
cylinders
(Or)
The other method is take HCF of all given capacities i.e. 98, 182
and 266.
5. In how many different ways can the letters of the word
'CORPORATION' be arranged so that the vowels always
come together?
a) 810
b) 1440
c) 2880
d) 50400
Answer: (d)
Solution:
In the word 'CORPORATION', we treat the vowels OOAIO as
one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest
are different.
Number of ways arranging these letters = 7!/2! = 2520
Now, 5 vowels in which O occurs 3 times and the rest are
different, can be arranged in 5!/3! = 20 ways
∴ required number of ways = (2520 × 20) = 50400.
3
Two Marks Questions:
6. One who believes that all things and events in life are
predetermined is a
a) Fatalist
b) Puritan
c) Egoist
d) Tyrant
Answer: (a)
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.
When it comes to success, a majority of people assume that
making it to the top ________.
a) needs hard work and maturity.
b) requires ethical compromises.
c) demands over time.
d) Requires good business knowledge.
Answer: (b)
8. 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.
Who have not seen a Bond movie and fantasied of becoming a
secret agent, going to exotic places in pursuit of dangerous
enemies assisted by beautiful damsels, wiping out the evil and
their empires?
4
a) Who have not seen Bond movie and fantasied of becoming a
secret agent, going to exotic places in pursuit of dangerous
enemies assisted by beautiful damsels, wiping out the evil
and their empires?
b) Who has not seen a Bond movie and had the fantasy of
becoming a secret agent, going to exotic places in pursuit of
dangerous enemies assisted by beautiful damsels, wiping out
the evil and their empires?
c) Who has not seen a Bond movie and fantasied becoming a
secret agent, going to exotic places in pursuit of dangerous
enemies, assisted by beautiful damsels and wiping out the
evil and their empires?
d) Who have not seen Bond movie and fantasied becoming a
secret agent, going to exotic places in pursuit of dangerous
enemies’ beautiful damsels assisting, wiping out the evil and
their empires?
Answer: (c)
9. 8 litres are drawn from a cask full of wine and is then filled with
water. This operation is performed three more times. The ratio
of the quantity of wine now left in cask to that of water is 16:
65. How much wine did the cask hold originally?
a) 18 litres
b) 24 litres
c) 32 litres
d) 42 litres
5
Answer: (b)
Solution:
Let the quantity of the wine in the cask originally be x liters.
Then, quantity of wine left in cask after 4 operations
= � �1 − 8 �4�
∴ � �1−( 8 / )�4� = 16
81
⇒ �1 − 8 �4 = �32�4
⇒ � −8 8� = 2
3
⇒ 3 − 24 = 2
⇒ = 24
10. Two persons A and B take a field on rent. A puts on it 21
horses for 3 months and 15 cows for 2 months; B puts 15 cows
for 6 months and 40 sheep for 7 1/2 months. If one day, 3 horses
eat as much as 5 cows and 6 cows as much as 10 sheep, what
part of the rent should A pay?
a) 1/3 rd
b) 2/5 th
c) 2/3 rd
d) 1/5 th
Answer: (a)
Solution:
3h = 5c
6c = 10s
6
A = 21h*3 + 15c*2
= 63h + 30c
= 105c + 30c = 135c
B = 15c*6 + 40s*7 1/2
= 90c + 300s
= 90c + 180c = 270c
A: B = 135:270
27:52
A = 27/79 = 1/3
7
TECHNICAL
One Mark Questions: 0 � compute under
11. The matrices � − � and � 0
multiplication.
a) If a = b (or) θ = nπ, n is an integer
b) Always
c) Never
d) If acosθ ≠ bsinθ
Answer: (a)
Solution:
= � − � & = � 0 0 �
⇒ = � − � � 0 0 �
= � − �
= � 0 0 � � − �
= � − �
= ( ) = , for an integer n then AB = BA
. . commute when a = b (or) = , n is an integer.
12. What values of x, y, z satisfy the following system of linear
equations?
1 2 3 6
�1 3 4� � � = � 8 �
2 2 3 12
8
a) = 6, = 3, = 2
b) = 12, = 3, = −4
c) = 6, = 6, = −4
d) = 12, = −3, = 4
Answer: (c)
Solution:
1 2 36
[ | ] = �1 3 4� 8 �
2 2 3 12
2 → 2 − 1; 3 → 3 − 2 1
1 2 36
�0 1 1 � 2�
0 −2 −3 0
3 → 3 + 2 2
1 2 36
�0 1 1 � 2� − − − − − (1)
0 0 −1 4
( ) = 3, ( | ) = 3,
= = 3
⇒ ( ) = ( | ) =
∴ .
From equation (1), we have
+ 2 + 3 = 6,
+ = 2
− = 4
⇒ = −4, = 6 = 6
9
13. ∫− [ 6 + 7 ] is equal to
a) 2 ∫0 6
b) 2 ∫0 7
c) 2 ∫0 [ 6 + 7 ]
d) Zero
Answer: (a)
Solution:
∫− [ 6 + 7 ] = 2 ∫0 6
Sin6x is even function,
Sin7x is odd function,
∫− ( ) = �2 ∫0 ( ) ( )
0 ( )
14. If ⃗ = � + 2 � then ∮ ⃗ . � � � over the path shown in the
figure is
a) 0
b) 2
√3
10
c) 1
d) 2√3
Answer: (c)
Solution:
̅ = � + 2 �
∮ ̅. � = ∮ + 2
Using green’s Theorem,
∮ + = ∫ ∫ � − �
22
∫ 3 =1(2
= ∫ √ =3√13 − ) = ∫1√3 2 = 1
√3
15. In the given RC circuit, the current i(t) = 2cos 5000t A.
The applied voltage v (t) is
a) 28.28 cos (5000t – 450) V
b) 28.28 cos (5000 + 450) V
c) 28.28 sin (5000t – 450) V
d) 28.28 sin (5000t + 450) V
Answer: (a)
11
Solution:
=
2 5000 = 20 × 10−6 ×
⇒ = ∫ 2 5000
20×10−6
= −20 sin 5000
= = 20 5000
( ) = +
= 20 × √2 �√12 5000 − 1 5000 �
√2
= 28.28 (5000 − 450)
16. In the circuit shown in the figure below, for what value of C
will the Current-I be in phase with the sinusoidal source voltage
Vs = sin2t?
12
a) 1
4
b) 1
2
c) 1
√2
d) 1F
Answer: (a)
Solution:
= 2
⇒ ( 2) = (1+ )×� 21 �
1+ + 21
= 1+
2 +1−2
( 2) = (1+ )×{(1−2 )− 2 }
(1−2 )2+(2 )2
= (1−2 )− 2 + (1−2 )+2
(1−2 )2+(2 )2
Putting imaginary term to be zero
-2C + 1 – 2C = 0
= 1
4
17. For the below given circuit if supply frequency, ω = 2 rad/sec
and V2 = 2∠00 volts, then what is the lead angle of VL with V2?
13
a) 150
b) 450
c) 900
d) 1350
Answer: (d)
Solution:
= 2 = 2∠00 = 2∠00
1
= 2 = 2∠00
� 1 �
= 2∠00 = 2∠900
� 2×10.5�
= +
= 2∠00 + 2∠900 = 2√2∠450
= ×
= 2√2∠450 × ( )
= 4√2∠1350
18. A specimen of intrinsic germanium with the density of charge
carriers of 2.5×1013/cm3, is doped with donor impurity atoms
such that there is one donor impurity atom for every 106
germanium atoms. The density of germanium atoms is
4.4×1022/cm3. The hole density would be
a) 4.4 × 1016/cm3
b) 1.4 × 1010/cm3
c) 4.4 × 1010/cm3
d) 1.4 × 1016/cm3
14
Answer: (b)
Solution:
In an n-type material ≃ ,
= = 2.5 × 1013/ 3
= 4.4×1022 / 3
106
. = 2
= �2.5×1013�2
4.4×1016
= 1.4 × 1010/ 3
19. Match List-I (Type of Device) with List-II (Characteristics/
Application) and select the correct answer using the codes
given below the lists:
List-I List-II
A. Zener diode 1. Display panel
B. Tunnel diode 2. Voltage reference
C. Schottky diode 3. Light detection
D. Photo diode 4. Negative resistance
5. High frequency switching
Codes:
AB C D
a) 3 4 5 2
b) 2 5 1 3
c) 3 5 1 2
d) 2 4 5 3
Answer: (d)
15
Solution:
Zener diode is used as voltage reference in voltage regular
circuit.
Tunnel diodes are -ve resistance devices and used as high speed
switching in microwave oscillators.
Photo diodes respond to light falls on it as photo current changes
due to change in intensity of light.
Schottky diode is used for high speed switching.
20. The Boolean expression = + + + + + +
+ + + ABC reduces to
a) A
b) B
c) C
d) A + B + C
Answer: (b)
Solution:
= + + + + + + + +
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ +
= � + � ⋅ + � + �
= ⋅ + ⋅
= � + � =
16
21. Consider the following systems:
1. Y[K] = x[k] + a1x [k – 1] – b1y [k – 1] – b2y [k – 2]
2. Y[K] = x[k] + a1x [k – 1] – a2x [k – 2]
3. Y[K] = x[k+1] + a1x [k] + a2x [k – 1]
4. Y[K] = a1x [k] + a2x [k +1] – b1y [k – 2]
Which of the systems given above represent recursive discrete
system?
a) 1 and 4
b) 1 and 2
c) 1, 2 and 3
d) 2, 3 and 4
Answer: (a)
Solution:
System 2 & 3 are not recursive because it is not in the form y(n
– k) = Σx(n – k)
22. Match List-I (Equation Connecting input x(n) and Output
y(n)) with List-II (System Category) and select the correct
answer using the codes given below the lists:
List-I
A. Y(n + 2) + y(n +1) + y(n) = 2x(n + 1) + x(n)
B. N2y2(n) + y(n) = x2(n)
C. Y(n +1) + ny (n) = 4nx(n)
D. Y(n +1) y(n) = 4x(n)
17
List-II
1. Linear, time-variable, dynamic
2. Linear, time-invariant, dynamic
3. Non-linear, time-variable, dynamic
4. Non-linear, time invariant, dynamic
5. Non-linear, time-variable, memory less
Codes:
ABC D
a) 3 5 2 1
b) 3 2 5 4
a) 2 3 5 1
b) 2 5 1 4
Answer: (d)
23. x[n] is defined as x[n] =�10, for n ≤ −2 or n > 4 determine
otherwise
the value of n for which x [– n – 2] is guaranteed to be zero.
a) N < 1 and n > 7
b) N < – 4 and n > 2
c) N < – 6 and n > 0
d) N < – 2 and n > 4
Answer: (c)
18
Solution:
24. Consider the following three cases of block diagram algebra A
B and C:
Which of the above relations are correct?
a) A and B
b) B and C
c) A and C
d) A, B and C
Answer: (b)
19
25. Which one of the following block diagrams is equivalent to the
below shown block diagram?
Answer: (b)
Solution:
E = RG – C
Which is satisfied by (b) option.
26. Match List-I (Types of electrical loads) with List-II
(Torque-speed characteristics) and select the correct answer:
List-I
A. Hoist
B. Fans
C. Machine Tools (Lathe, Milling machine etc.)
D. Loads with fluid friction
20
List-II
1. Torque ∝ (speed)2
2. Torque ∝ (speed)
3. Constant Torque
4. Torque ∝ 1/(speed)
Codes:
AB C D
a) 1 3 2 4
b) 1 3 4 2
c) 4 1 3 2
d) 3 1 2 4
Answer: (c)
27. Assertion (A): A d.c motor draws high current at the time of
starting.
Reason (R): While starting a d.c motor, it takes some time to
develop a non-zero value of back e.m.f.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is NOT the correct explanation
of A.
c) A is true but R is false.
d) A is false but R is true.
Answer: (a)
21
Solution:
Back EMF depend on speed (Eb = k.ϕ.ω)
Hence reason as well as assertion is correct and R is correct
explanation of A.
28. The maximum efficiency occurs in a separately excited d.c
generator when the terminal voltage is 220 V and the induced
emf is 240 V, the stray losses, if the armature resistance is 0.2Ω,
will be _______ W.
Answer: 2000
Solution:
= − = 240−220 = 100
0.2
For maximum efficiency,
Copper loss = Ia2Ra
Steady losses = 1002 × 0.2 = 2000 W
Hence, option (b) is correct.
29. A dc series motor is running at rated speed and rated voltage,
feeding a constant power load. If the speed has to be reduced to
0.25 p.u., the supply voltage should be reduced to _____ p.u.
Answer: 0.5
22
Solution:
Power =
=
∝
= 2
∴ 1 = 2 2
2 2 1
⇒ 1 = 2 2
0.25 2 1
⇒ 2 2 = 4 2 1
⇒ 2 = 2 1
⇒ 2 = 2 1
As =
1 = 1 1 = 1 = 1
2 2 2 2×0.25 0.5
⇒ 2 = 0.5 1
30. Consider the following statements:
Surge impedance loading of a transmission line can be increased
by
1. increasing its voltage level
2. addition of lumped inductance in parallel
3. addition of lumped capacitance in series
4. reducing the length of the line
Which of these statements are correct?
a) 1 and 3
b) 1 and 4
23
c) 2 and 4
d) 3 and 4
Answer: (a)
Solution:
Surge impedance loading = ( )2
0
SIL can be increased by increasing its voltage level.
Z0 = Surge impedance can be decreased by addition of lumped
capacitance is series which reduces the effective reactance of the
line resulting into higher SIL.
31. A 100 km long transmission line is loaded at 110 kV. If the
loss of line is 15 MW and the load is 150 MVA, the resistance
of the line is
a) 8.06 ohms per phase
b) 0.806 ohm per phase
c) 0.0806 ohm per phase
d) 80.6 ohms per phase
Answer: (a)
Solution:
100 km long line is a short transmission line so, Z = R + jX; and
assuming single phase line.
= 2 = 15 × 106
= �111500××110063�2 × ⇒ = 8.06 / ℎ
24
32. Consider the following statement
Assertion (A): The critical rate of change of forward-voltage is
the value of dv/dt, at which the device just goes into conduction
without a gate pulse
Reason (R): Thyristors go to a state of conduction with the
application of sharp rate of change of forward-voltage in the
absence of gate pulse, even before the break forward voltage
limit is reached
Of these statements
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of
A.
c) A is true but R is false.
d) A is false but R is true.
Answer: (a)
Solution:
With forward voltage across the anode and cathode of a
thyristor, the two outer junctions J1, J3 are forward biased, but
inner junction J2 is reverse biased. This reverse biased junction
behaves like a capacitance. If forward voltage is suddenly
applied, a charging current through junction capacitance may
turn on the SCR.
25
33. The latching current in the below circuit is 4 mA. The
minimum width of the gate pulse required to turn on the
thyristor is ______ μs.
Answer: 4
Solution:
=
∫0 . = ∫0
⇒ . =
⇒ =
∴ = 4×10−3×0.1 = 4
100
34. An analog voltage signal whose highest significant frequency
is 1 kHz is to be coded with a resolution of 0.01 percent for a
voltage range of 0 – 10 V. The minimum sampling frequency
and the minimum number of bits should respectively be
a) 1 kHz and 12
b) 1 kHz and 14
c) 2 kHz and 12
d) 2 kHz and 14
Answer: (d)
Solution:
26
Minimum sampling frequency = Nyquist sampling rate = 2 kHz
and resolution = 0.01 = 1 × 100 ⇒ ≈ 14
2 −1
35. Match List-I (Term) with List-II (Type) and select the correct
answer:
List-I List-II
A. Curl � � = 0 1. Laplace Transform
B. Div � � = 0 2. Irrotational
C. Div Grad (∅) = 0 3. Solenoidal
D. Div Div (∅) = 0 4. Not defined
Codes:
A B CD
a) 2 3 1 4
b) 4 1 3 2
c) 2 1 3 4
d) 4 3 1 2
Answer: (a)
Solution:
⃗ = 0 that means vector ⃗ is irrotational
⃗ = 0 that means vector ⃗ is solenoidal
Since flux coming out from a solenoid is zero.
( ) = ∇. (∇ ) = 0 →
( ) = not defined because
→ scalar quantity and div of a scalar quantity is not defined.
27
Two Marks Questions:
36. For a scalar function f(x, y, z) = x2 + 3y2 + 2z2, the directional
derivative at the point P(1, 2, -1) in the direction of a vector ̅ −
̅ + 2 � is
a) −18
b) −3√6
c) 3√6
d) 18
Answer: (b)
Solution:
f(x, y, z) = x2+ 3y2 + 2z2, P(1, 2, -1)
� = ̅ − ̅ + 2 �
∇ = 2 ̅ + 6 ̅ + 4 �
∇ ] = 2 ̅ + 12 ̅ − 4 �
Directional derivative = ∇ . � = 2−12−8 = − 18 = −3√6
| � | √1+1+4 √6
37. The real root of the equation xex = 2 is evaluated using
Newton-Raphson's method. If the first approximation of the
value of x is 0.8679, the 2nd approximation of the value of x
correct to three decimal places is
a) 0.865
b) 0.853
c) 0.849
d) 0.838
Answer: (b)
28
Solution:
Given f(x) = xex – 2 = 0 and x1 = 0.8679
(∵ 1( ) = + )
Newton-Raphson’s method is given by 2 = 1 − ( 1)
1( 1)
= 0.8679 − 0.8679 0.8679−2 = 0.853
(0.8679 0.8679+ 0.8679)
38. The product of the non-zero eigenvalues of the matrix
⎢⎢⎢⎡0001 0 0 0 0001⎥⎥⎥⎤ is ________.
1 1 1
1 1 1
1 1 1
⎣1 0 0 0 1⎦
Answer: 6
Solution:
Let the given matrix be ‘A’
∴ | − | = 0
1 − 0 0 0 1
1 −
�� 0 1 1 0 �� = 0
0 1 1 − 1 0
1 −
0 1 1 0 0
1 −
1 0 0
Expanding by 1st row
1 − 1 1 0
1 −
(1 − ) � 1 1 0 �
1 1 1 − 0
1 −
0 0 0
0 1 − 1 1
+1 �00 1 1 −
1 1 � = 0
1 1 1 −
0 0 0
29
1 − 1 1
= (1 − )2 � 1 1 − 1 �
1 1 1 −
1 − 1 1
−1 � 1 1 − 1 � = 0
1 1 1 −
1 − 1 1
� 1 1 − 1 � [(1 − )2 − 1] = 0
1 1 1 −
3( − 2)( − 3) = 0
= 0,0,0,2,3
∴ = 2 × 3 = 6
39. For the network given in figure below, the Thevenin's voltage
Vab is _____ V.
Answer: −1.5
Solution:
+ ( −16) + ( +30) = 0
10 10 15
3 +3 −48+2 +60 = 0
6
30
8 = −12
= −3 = −1.5
2
40. Consider the circuit shown in Fig., If the frequency of the
source is 50 Hz, then a value of t0 which results in a transient
free response is _______ ms.
Answer: 1.78
Solution:
The complete response i(t) in the given RL circuit shown in Fig.
is given by
( ) = + = − ( − 0 ) + 1 sin( + )
Where, = √ 2 + 2 2
And = − tan−1 � �
For ( 0) = 0, = − 1 sin( 0 + )
∴ = 0, if K = 0, or 0 = −
0 = tan−1 � �
31
= 2× ×50×0.01 = 0.2
5
0 = 0.561, 0 = 0.561
2 ×50
= 561 = 1.78
2 ×50
Statement for Linked Answer Q. 41:
A coil of inductance 10 H and resistance 40 Ω is connected as shown
in the Fig. After the switch S has been in contact with point 1 for a
very long time, it is moved to point 2 at t = 0.
41. If, at t = 0+, the voltage across the coil is 120 V, the value of
resistance R is _______ Ω.
Answer: 40
Solution:
Coil R & L in series
32
(0− ) = 120 = 2
60
Circuit at t =0+
KVL [Polarity of the voltage across the coil is not specified in
the problem to set a positive value of R, it is selected as shown
in the figure]
2R + 20(2) – 120 = 0
2R = 80
R = 40Ω
33
42. In the circuit shown below, the knee current of the ideal Zener
diode is l0 mA. To maintain 5V across RL, the minimum value
of RL in 1Ω and the minimum power rating of the Zener diode
in mW respectively are
a) 125 and 125
b) 125 and 250
c) 250 and 125
d) 250 and 250
Answer: (b)
Solution:
= 10−5 = 50
100
= 10
= −
34
= 50 − 10 − 40
= = 5 = 125
40×10−3
Power rating of the zener is =
= 50
∴ = 5 × 50 × 10−3 = 250
43. The op amp shown in the figure has a finite gain A = 1000 and
an infinite input resistance. A step voltage Vi = 1 mV is applied
at the input at time t = 0 as shown. Assuming that the
operational amplifier is not saturated, the time constant (in
millisecond) of the output voltage V0 is _____ msec.
Answer: 1001
Solution:
= ℎ
To find ℎ across C, then the equivalent circuit is given by
− = … . (1)
35
+ 1 + 1000 = 0
1001 = −1
= − 1 … . (2)
1001
From (1) and (2)
∴ = − = 1
1001
ℎ = 1 = 1001
= ℎ = 1001
R = 1KΩ, C = 10−6
= 1001 × 1000 × 10−6 = 1001
(OR)
To find Ceq equivalent circuit diagram is shown below
= [1 − (− )][ ]
= [1001]
= [ = 1 , = 1 , = 1 ]
= [1001] = 1001
36
44. For a flip-flop formed from two NAND gates as shown in
figure. The unusable state corresponds to
a) X = 0, Y = 0
b) X = 0, Y = 1
c) X = 1, Y = 0
d) X = 1, Y = 1
Answer: (a)
Solution:
When X = 0, Y = 0 both NAND gate outputs are 1
45. A Boolean function f (A, B, C, D) = Π (1, 5, 12, 15) is to be
implemented using an 8 × 1 multiplexer (A is MSB). The inputs
ABC are connected to the select inputs S2 S1 S0 of the
multiplexer respectively.
Which one of the following options gives the correct inputs to
pins 0, 1, 2, 3, 4, 5, 6, 7 in order?
a) , 0, , 0, 0, 0, 0, ,
b) , 1, , 1, 1, 1, ,
37
c) , 1, , 1, 1, 1, ,
d) , 0, , 0, 0, 0, ,
Answer: (b)
Solution:
The Boolean function ( , , , ) = Π(1,5,12,15)
It is similar to = Σ(0,2,3,4,6,7,8,9,10,11,13,14)
46. Given the relationship between the input u (t) and the output y
(t) to be ( ) = ∫0 (2 + − ) −3( − ) ( ) the transfer function
Y(s)/U(s) is
a) 2 −2
+3
b) +2
( +3)2
c) 2 +5
+3
d) 2 +7
( +3)2
Answer: (d)
Solution:
Relation between input, u(t) and output, y(t):
( ) = ∫0 (2 + − ) −3( − ) ( )
38
Observe that the above integral is a convolution operation
between the input u(t) and the impulse response of the system.
( ) = [(2 + ) −3 ( )] ∗ ( )
∴ ℎ( ) = (2 + ) −3 ( )
Use the LT pairs:
−3 ( ) → 1
+3
−3 ( ) → 1
( +3)2
( ) = 2 + 1 = (2 +7)
+3 ( +3)2 ( +3)2
= ( ) = ( ) = 2 +7
( ) ( +3)2
47. y[n] denotes the output and x[n] denotes the input of a
discrete-time system given by the difference equation y[n]-
0.8y[n – 1] = x[n] + 1.25x[n + 1]. Its right-sided impulse
response is
a) causal
b) unbounded
c) periodic
d) non-negative
Answer: (d)
Solution:
From the given difference equation
( ) − 0.8 ( − 1) = ( ) + 1.25 ( + 1)
( ) = ( ) = 1+1.25 1
( ) 1−0.8 −1
= 1 + 1.25
1−0.8 −1 1−0.8 −1
39
For right-sided impulse response, h(n) the ROC is | | > 0.8
∴ ℎ( ) = (0.8) ( ) + 1.25(0.8) +1 ( + 1)
ℎ( ) ≠ 0, < 0
So, it can be verified that h(n) is non casual, bounded and not
periodic.
ℎ( ) ≤ , < ∞; h(n) is bounded.
h(n) is not periodic as it is a monotonically decaying sequence.
→ h(n) is non negative
48. A system with input x (t) and output y (t) is defined by the
input - output relation: ( ) = ∫−−∞2 ( ) . The system will be
a) causal, time - invariant and unstable
b) casual, time - invariant and stable
c) non-causal, time invariant and unstable
d) non - causal, time -variant and unstable
Answer: (d)
Solution:
Given ( ) = ∫−−∞2 ( )
y at time t is depending on values of x(t) in the range of t = -∞ to
(-2t
i.e., y for negative values of time is depending on x at positive
values of time.
Hence system is non-casual.
For a bounded input like a step function the output is not
bounded, as the input is integrated.
40
Hence system is unstable
For input, x(t) output ( ) = ∫−−∞2 ( )
For ( − 0), = ∫ − =2− ∞ ( − 0)
= ∫ − =2− ∞− 0 ( 1) 1 … … (1)
When 1 = − 0
But = ( − 0) = ∫−−∞2( − 0) ( ) … … . . (2)
As equations (i) and (ii) are not equal, system is time-variant
∴ System is non-casual, time-variant, and unstable.
It may be noted that the system is Linear.
49. The Laplace transform of a function f(t) is ( ) = 5 2+23 +6 .
( 2+2 +2)
As → ∞, f(t) approaches is ______.
Answer: 3
Solution:
Given ( ) = 5 2+23 +6
( 2+2 +2)
l →im∞ ( ) = l i→m0 ( ) = 6 = 3
2
50. A system with transfer function ( ) = � 2+9�( +2) is
( +1)( +3)( +4)
excited by sin (ωt). The steady-state output of the system is zero
at ω = _____ rad/s.
Answer: 3
Solution:
Find the value of “ω” where | ( )| = 0
| ( )| = �( � ( + 1 ))(2 + 9 +�(3 ) ( + 2 +)4)� = 0
41
− 2 + 9 = 0
= 3 /
51. The armature resistance of permanent magnet dc motor is 0.8
ohm. At no load, the motor draws 1.5 A from a supply voltage
of 25 V and runs at 1500 rpm. The efficiency of the motor while
it is operating on load at 1500 rpm drawing a current of 3.5 A
from the same source will be
a) 48.0%
b) 57.1%
c) 59.2%
d) 88.8%
Answer: (a)
Solution:
At no load developed power (P) = EbIa
Eb = 25 – 1.5 × 0.8 = 23.8 V
P = 23.8 × 1.5 = 35.7 W
Under no load condition the developed power is useful to
overcome friction & windage losses.
∴ friction & windage losses = 35.7W
Under loaded condition armature Cu loss = (3.5)2 × 0.8 = 9.8W
∴ Total losses = 35.7 + 9.8 = 45.5 W
Input to motor = × = 25 × 3.5
= 87.5
= − × 100
42
= 87.5−45.5 × 100 = 48%
87.5
Data for Question 52:
A 240V dc shunt motor draws 15A while supplying the rated
load at a speed of 80 rad/s. The armature resistance is 0.5 ohm
and the field winding resistance is 80 ohm.
52. The net voltage across the armature resistance at the time of
plugging will be
a) 6 V
b) 234 V
c) 240 V
d) 474 V
Answer: (d)
Solution:
Motor operation before plugging:
Circuit immediately after plugging (reversing supply to the
armature, while shunt field current direction is unchanged):
43
Since field flux direction is unchanged, and speed also is
unchanged (speed cannot change instantaneously due to inertia)
the induced emf does not change in magnitude or polarity and is
as shown in fig.2. The voltage drop across the armature
resistance (by KVL) is then (240 + 234) = 474 V.
53. A voltage V = 400 sin314.16t is applied to a 1-phase
transformer on no-load. If the no load current of the transformer
is 2sin (314.16t – 850), then magnetization branch impedance
will be approximately
a) 141∠90
b) 200∠-85
c) 200∠85
d) 282∠-80
Answer: (c)
Solution:
44
Magnetization branch impedance
0 = ℎ = �4√020�∠00 = 200∠850
ℎ �√22�∠−850
54. The hysteresis loop of a magnetic material has an area of 5cm2
with the scales given as 1cm = 2AT and 1cm = 50mWb. At
50Hz the total hysteresis loss is ______ W.
Answer: 25
Solution:
Area enclosed by the hysteresis loop = energy loss due to
hysteresis per cycle.
50 Hz = 50 cycles/sec
The hysteresis loop is given to have an area of 5 cm2.
An area of 5 cm2 corresponds to 5 × 2× 50 (At) (m Wb) = 0.5
Joules.
�i d∅ : Watts; id∅: Watt − sec or Joules�
dt
With 50 cycles per sec, the hysteresis loss equals 50×0.5 = 25W
55. Consider the model shown in figure of a transmission line with
a series capacitor at its mid-point. The maximum voltage on the
line is at the location
45
a) P1
b) P2
c) P3
d) P4
Answer: (c)
Solution:
The voltage profile for the given line is shown in fig.
56. At an industrial sub-station with a 4 MW load, a capacitor of 2
MVAR is installed to maintain the load factor at 0.97 lagging. If
the capacitor goes out of service, the load power factor becomes
a) 0.85 lag
b) 1.00
c) 0.80 lag
d) 0.90 lag
Answer: (c)
Solution:
Let the initial power factor angle = ∅1
After connecting a capacitor, the power factor angle = ∅2
Given ∅2 = cos−1 0.97
= 14.070
46
(tan ∅1 − tan ∅2) = supplied by capacitor
4 × 106(tan ∅1 − tan 14.07) = 2 × 106
cos ∅1 = 0.8
Hence if the capacitor goes out of service the load power factor
becomes 0.8 lag
57. The horizontally placed conductors of a single phase line
operating at 50 Hz are having outside diameter of 1.6 cm, and
the spacing between centers of the conductors is 6 m. The
permittivity of free space is 8.854×10-12 F/m. The capacitance to
ground per kilometer of each line is
a) 4.2×10-9 F
b) 8.4×10-9 F
c) 4.2×10-12 F
d) 8.4×10-12 F
Answer: (b)
Solution:
= 0.8 , = 6 , 0 = 8.854 × 10−12 /
Capacitance to ground = 2 0 /
( / )
= 2 0 /
�0.8×610−2�
= 8.404 × 10−12 /
= 8.404 × 10−9 /
47
58. A load is supplied by a 230 V, 50 Hz source. The active power
P and the reactive power Q consumed by the load are such that 1
kW ≤ P ≤ 2 kW and 1 kVAR ≤ 2 kVAR. A capacitor connected
across the load for power factor correction generates 1 kVAR
reactive power. The worst case power factor after power factor
correction is
a) 0.447 lag
b) 0.707 lag
c) 0.894 lag
d) 1
Answer: (b)
Solution:
Worst power factor corresponding to Pmin and Qmax
Pmin = 1 kW
Qmax = 2 KVAR
Qc = 1 KVAR
P = 1 kW, Q = 1 kVAR
= �tan−1 � �� = 0.707
59. A single phase AC distributor supplies two single phase loads
as shown in figure. The voltage drop from A to C is ____ V.
Answer: 31.5
48
Solution:
∴ Total voltage drop from A to C is,
= 66.5∠ − 15.7 × 0.25∠900 + 40∠00 × 0.35∠900
= 30.34∠81.470
60. The given figure shows a step – down chopper switched at 1
kHz with a duty ratio D = 0.5. The peak-peak ripple in the load
current is close to _______ A.
Answer: 0.125
Solution:
Given α = 0.5 f = 1 kHz
L = 200 mH R = 5 Ω
Vs = 100 V
Peak to peak ripple in inductor current
∆ = (1 − )
= 100 × 0.5 × 0.5 × 10−3
0.2
= 0.125
49
Data Loading...