memory based questions of gate 2020 · 2020. 4. 1. · mvi a, 21 h → 21 a h inx h → hl ... h(s)...
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Memory Based Questions of
GATE 2020Electrical Engineering
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Date of Exam : 8/2/2020Forenoon Session
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
GENERAL APTITUDE
Q.1Q.1Q.1Q.1Q.1 NPA is the asset that a customer borrows and holds it for a period of time without payingany interest. RBI has reduced the holding period for NPA thrice during the period 1993-2004. In 1993 it was four quarters (360 days) how many days is the holding period in2004?(a) 90 (b) 180(c) 45 (d) 270
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
End of Solution
Q.2Q.2Q.2Q.2Q.2 Number between 1001 to 9999 how many times 37 occurs in same sequence?(a) 279 (b) 280(c) 270 (d) 299
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
Q.3Q.3Q.3Q.3Q.3 The revenue and expenditure of four companies P, Q, R, S as shown in the figure belowcompany Q earns a profit of 10% on expenditure of 2014, the revenue of Q in 2015 is________.
P Q R S45
35
ExpenditureRevenue
Ans.Ans.Ans.Ans.Ans. ()()()()()Data insufficient
End of Solution
Q.4Q.4Q.4Q.4Q.4 Stock markets _______ at the news of the coup.(a) plugged (b) plunged(c) probed (d) poised
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.5Q.5Q.5Q.5Q.5 This book, including all its chapters, ________ interesting. The students as well asinstructor ________ in agreement with it.(a) is, was (b) is, are(c) were, was (d) are, are
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
Q.6Q.6Q.6Q.6Q.6 People were prohibited ________ there vehicles near the entrance of the main administrativebuilding.(a) to park (b) to have parked(c) from parking (d) parking
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
End of Solution
Q.7Q.7Q.7Q.7Q.7 Select the word do : undo : : trust :(a) distrust (b) entrust(c) intrust (d) untrust
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
End of Solution
Q.8Q.8Q.8Q.8Q.8 Find the ratio of AC BC
AB
+ where O is center of the circle shown below.
BA
C
O
,AB AC and BC are chords.
Ans.Ans.Ans.Ans.Ans. (1.414)(1.414)(1.414)(1.414)(1.414)
End of Solution
Q.9Q.9Q.9Q.9Q.9 Z : WV : RQP : ?(a) KIJH (b) JIHG(c) HIJK (d) KJIH
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.10Q.10Q.10Q.10Q.10 If P, Q, R, S are four individual, how many team of size exceeding one can be formedwith Q as a member?(a) 5 (b) 7(c) 8 (d) 6
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
ELECTRIC CIRCUITS
Q.11Q.11Q.11Q.11Q.11 A resistor and a capacitor are connected in series to a 10 V dc supply through a switchis closed at t = 0, and the capacitor voltage is found to cross ‘0’ voltage at t = 0.4τ(τ = time constant). The absolute value of % change required in the initial capacitorvoltage if the zero crossing has to happen at t = 0.2τ is _____.
Ans.Ans.Ans.Ans.Ans. (54.989)(54.989)(54.989)(54.989)(54.989)If initial charge polarities on the capacitor is opposite to the supply voltage then onlythe capacitor voltage crosses the zero line.Vc(t) ⇒ Final value + (Initial value – Final value) e–t/τ
0 = 10 + (–V0 – 10) e–0.4
+–
t = 0
R
C10 V10 = (V0 + 10) e–0.4
V0 = 4.918 VNow, t = 0.2τ
0 = 0.2010 ( 10)V e−′+ − −
0V ′ = 2.214
%change in voltage =4.918 2.214
100% 54.989%4.918
− × =
End of Solution
Q.12Q.12Q.12Q.12Q.12 To ensure the maximum power transfer across Vth the values of R1 and R2 will be (Diodein figure is silicon diode)
R1
Vth
+
–
R2Is
I
I < 0
(a) R1 high, R2 high (b) R1 low, R2 low(c) R1 low, R2 high (d) R1 high, R2 low
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.13Q.13Q.13Q.13Q.13 For the given circuit the value of Vth is ______.
+
–
Vth
3 Ω2 Ω
3 Ω 5 A+–4 V
Ans.Ans.Ans.Ans.Ans. (14)(14)(14)(14)(14)
+
–
Vth3 Ω2 Ω
3 Ω 5 A+–4 V
4 V
Vth
th 42
V −= 5
Vth = 14 V
End of Solution
Q.14Q.14Q.14Q.14Q.14 In the given circuit rms value of I1 is _____.
A2
A3
A1
1 10°∠
1 70°∠
Ans.Ans.Ans.Ans.Ans. (1.732)(1.732)(1.732)(1.732)(1.732)From KCL, I1 = 1∠10 + 1∠70
I1 = 1.732∠40
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
MICROPROCESSORS
Q.15Q.15Q.15Q.15Q.15 If 2001 → 98H2002 → B1H
LXI H, 2001HMVI A, 21HINX HADD MINX HMOV M, AHLTThe content of 2003H is (_____)10.
Ans.Ans.Ans.Ans.Ans. (210)(210)(210)(210)(210)
LXI H, 2001 H → 20 01H L
MVI A, 21 H → 21AH
INX H → HL + 1 20 02H L
ADD M → [A] + data → reference of HL pair21 H + B1H = D2H → [A]
INX H → [HL] + 1 → 2002H + 1H → 2003HMOV M, A → A → [HL]
D2H → 2003H D2HLT → Stop
∴ content in the 2003 H is D2HD × 161 + 2 × 16° ⇒ 13 × 16 + 2 = 210
End of Solution
CONTROL SYSTEMS
Q.16Q.16Q.16Q.16Q.16 A differential equation with y(t) → output and x(t) → input is2
2
( )4 ( )
d y ty t
d+
x = 6x(t)
The poles of the input(a) –2j, +2j (b) +4, –4(c) –2, +2 (d) +4j, –4j
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)2
2
( )4 ( )
d y ty t
dt+ = 6x(t)
Taking Laplace transform,s2Y(s) + 4Y(s) = 6X(s)
( )( )
Y sX s
=2
6
4s +∴ Poles are at s2 + 4 = 0⇒ s = ±2j
End of Solution
Q.17Q.17Q.17Q.17Q.17 A PMDC motor is connected to 5 V at t = 0. Its speed increases from 0 to 6.32 rad/sec monotonically from t = 0 to t = 0.5s and finally settles down to 10 rad/sec. Assumenegligible armature inductance. Find the transfer function?
(a)10
1 0.5s+ (b)100.5s s+
(c)2
1 0.5s+ (d)20.5s +
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)When the armature inductance is neglected the transfer function of PMDC motorbecomes,
( )( )
m
a
sV sω
=+1
m
m
ksT
...(i)
⇒ ωm(s) =+
( )·1
ma
m
kV ssT
As input is 5 V i.e. Va(s)=5s
∴ ωm(s) = ( )+5 1
m
m
ks sT
...(ii)
As the final value of wm = 10 rad/sec
Using final value theorem,
0lim ( )ms s→
ωx
=→
=+0
5lim 10(1 )
m
m
kss sTs
⇒ km = 2
From (ii) equation, ωm(s) = = ++ +
0 110(1 ) 1m m
a as sT s sT
ωm(s) = −+
10 101
m
m
Ts sT
ωm(t) = 10 – 10–t /Tm
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
It is given that at t = 0.5 sec,
ωm = 6.32 rad/sec
∴ 6.32 = 10 – 10e–0.5/Tm
⇒ e–0.5/Tm = 0.368
−0.5
mT= ln(0.368)
−12 mT
= –0.999
Tm = =�1 1
0.51.999 2
Substituting the value of km and Tm in T.F. in equation (i)
( )( )
m
a
sV sω
=2
1 0.5s+
By option checking we can directly get the answer.
End of Solution
Q.18Q.18Q.18Q.18Q.18 For the given open loop transfer function + − +( )( )( )
Ks a s b s c
. If 1 + G(s)H(s) plane
encircles the origin once in counter clockwise direction then number of closed loop polesin right side of s-plane will be
Ans.Ans.Ans.Ans.Ans. (0)(0)(0)(0)(0)
−+N = P+ – Z+
N+ = +1 (as CCW direction)P+ = 1
∴ Z+ = P+ – N+
= 1 – 1 = 0
End of Solution
Q.19Q.19Q.19Q.19Q.19 For the given open loop transfer function with feed back gain unity,
G(s)H(s) = + +
+ + +
2
3 2
1
2 2
s ss s s K
The value of gain K. For which closed loop system is marginally stable _________ .
Ans.Ans.Ans.Ans.Ans. (8)(8)(8)(8)(8)CE is 1 + G(s)H(s) = 0
⇒ + +++ + +
2
3 2
11
2 2
s ss s s K
= 0
⇒ s3 + 3s2 + 3s + (1 + K) = 0
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
R.H. criteria
+
− +
3
2
1
1 3
3 (1 )
9 (1 ) 0
s
s K
s K
for marginal stability9 – (1 + K) = 0
⇒ K = 8
End of Solution
Q.20Q.20Q.20Q.20Q.20 A stable LTI system with single pole P, has a transfer function G(s)H(s) = +−
2 100ss P
with
DC gain of 5; the smallest possible frequency in radian/sec at only gain is(a) 11.08 (b) 70.13(c) 122.07 (d) 8.84
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)
G(s) H(s) =2 100ss p
+−
Given, K = 5
2 100
1
sS
PP
+⎛ ⎞− −⎜ ⎟⎝ ⎠
=
2
1100
1001
s
sP
P
⎛ ⎞+⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞− −⎜ ⎟⎝ ⎠
DC gain =100
5P
− =
⇒ P = –20
∴ G(s) H(s) =2 100
20ss
++
G(jω) H(jω) =2100
20 j− ω+ ω
( ) ( )G j H jω ω =2
2 2
1001
20
− ω =+ ω
⇒ ω = 8.84 rad/sec
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.21Q.21Q.21Q.21Q.21 For the given block diagram of the system
s1
1+ s21R s( ) C s( )
s20
20+
which of the following is true regarding order and stability of the system(a) 4th order and stable (b) 4th order and unstable(c) 3rd order and unstable (d) 3rd order and stable
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
1s2
1 + 1s
20 + 20s
C s( )R s( )
CE is 1 + G(s) H(s) = 0
2
1 201
( 20)( 1) ss s+ ×
++= 0
⇒ (s3 + s2) (s + 20) + 20 = 0s4 + 21s3 + 20s2 + 20 = 0s4 + 21s3 + 20s2 + 20 = 0
As coefficient of s′ is missing system is of 4th order and unstable.
End of Solution
Q.22Q.22Q.22Q.22Q.22 Which of the given below signal flow is analogous to given below system?
de
a c1 1
Ans.Ans.Ans.Ans.Ans. ()()()()()
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
ELECTROMAGNETIC THEORY
Q.23Q.23Q.23Q.23Q.23 Given 2
Cˆ ˆ ˆ15 2 3mzD a Pa Pzaρ φ
⎛ ⎞= + − ⎜ ⎟⎝ ⎠
��, find electric flux crossing the cylinder ρ = 3 m;
3 varying from 0 to 5.
Ans.Ans.Ans.Ans.Ans. (180(180(180(180(180πππππ)))))
ψ/Crossing closed surface = ( )D ds D dv⋅ = ∇ ⋅∫∫ ∫∫∫�� ��� �� ��
� ...(i)
D∇ ⋅�� ��
= 1 1( ) zD D
Dz
φρ
∂∂ ∂ρ + +ρ ∂ρ ρ ∂φ ∂
= 1 1 1( 15) (2 ) ( 3 ) 15 3z Pz
∂ ∂ ∂ρ + ρ + − ρ = −ρ ∂ρ ρ ∂φ ∂ ρ
( )D dv∇ ⋅∫∫∫�� ��
=15
3 d d dz⎛ ⎞− ρ ρ ρ φ⎜ ⎟ρ⎝ ⎠∫∫∫
= 215 3d d dz d d dzρ φ − ρ ρ φ∫∫∫ ∫∫∫
=3 2 5 3 2 5
2
0 0 0 0 0 0
15 3z z
d d d d dzπ π
ρ = φ = = ρ = φ = =
ρ φ − ρ ρ φ∫ ∫ ∫ ∫ ∫ ∫
=3315(3 0) (2 ) (5) 3 ( ) (5)
3⎛ ⎞
− π − ∈π⎜ ⎟⎜ ⎟⎝ ⎠= 45(10π) – 27 (10π) = 180π = 565.2 C
End of Solution
Q.24Q.24Q.24Q.24Q.24 The vector function expressed byF = ax(5y – k1 z) + ay(3z + k2 x) + az(k3 y – 4x)
represent a conservative field, where ax, ay, az are unit vector along x, y, z directions,respectively the value of constant k1, k2 and k3 ________.(a) 4, 5, 3 (b) 8, 3, 7(c) 0, 0, 0 (d) 3, 8, 5
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
F = − + + + −i x x1 2 3ˆ ˆˆ(5 ) (3 ) ( 4 )y k z z k j k y k
is conservative field
F is irrotational, ∇ × F = 0
∂ ∂ ∂∂ ∂ ∂− + −
i
xx x1 2 3
ˆ ˆˆ
5 3 4
j k
y zy k z z k k y
= 0
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
− − − + + −i 3 1 2ˆ ˆˆ( 3) ( 4 ) ( 5)k j k k k = 0
k3 – 3 = 0 4 – k1 = 0 k2 – 5 = 0k3 = 3 k1 = 4 k2 = 5k1 = 4 k2 = 5 k3 = 3
End of Solution
Q.25Q.25Q.25Q.25Q.25 The value of mutual inductance figure shows long wire carrying current 2 A placed inaway from square coil as shown in figure. The value mutual inductance will be ____. nH.
∞
–∞
1 m
1 m
1 m
1 m2 A
Square coil( = 1)N
∞
–∞
1 m
1 m
1 m
1 m2 A
Square coil( = 1)N
Ans.Ans.Ans.Ans.Ans. (138.6)(138.6)(138.6)(138.6)(138.6)
φ ∝ Iφ = MI
B��
=µ ˆ ( due to infinite long line)2
o a BφπρI ��
Magnetic flux crossing square loop is
φ = B ds⋅∫∫�� ���
=2 1
o
1 0
µ µˆ ˆ( )2 2
o
z
da d dz a dzφ φ
ρ = =
ρ⋅ ρ =πρ π ρ∫∫ ∫ ∫
I I
φ = 2 11 0
µ(ln ) ( )
2o
zzρ = =ρπI
φ =µ
(ln2)2o
πI
m =φI
m =7
2µ (ln2) 4 10 (ln )2 2
o−π ×=
π πm = 1.386 × 10–7 Henry
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.26Q.26Q.26Q.26Q.26 In a dielectric medium εr = 2.25 and 3ˆ ˆ ˆ2 6r zE r a a ar φ= + +
�� in cylindrical, then find volume
charge density(a) 2∈o (b) 3∈o
(c) 4∈o (d) 9∈o
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)
D�
= o rE E∈ = ∈ ∈�� ��
D��
=3ˆ ˆ ˆ2.25 2 6o r zr a a ar φ
⎛ ⎞∈ + +⎜ ⎟⎝ ⎠
D��
= 6.75ˆ ˆ ˆ4.5 13.5oo r o zr a a a
r φ∈∈ + + ∈
Volume charge density
ρ v = D∇ ⋅�� ��
ρ v =1 1
( ) zr
D DrD
r r r zφ∂∂ ∂+ +
∂ ∂φ ∂
ρ v = 6.751 1( 4.5 ) (13.5 )oo or r
r r r r z∈∂ ∂ ∂⎛ ⎞∈ + + ∈⎜ ⎟∂ ∂φ ∂⎝ ⎠
= 21(4.5 ) 0 0o r
r r∂ ∈ + +∂
=1
(4.5 ) (2 ) 9o orr
∈ = ∈
End of Solution
ELECTRIC MACHINES
Q.27Q.27Q.27Q.27Q.27 In the given figure, the three windings having polarity as shown above, are connectedto 3-φ, balanced supply. The number of turns in the supply winding is 20, the voltageseen in winding X having N = 2 turns will be
N = 20
N = 20 N = 20
Van = 230 0°∠
Vbn = 230 120°∠
Vcn = 230 –120°∠
winding
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Ans.Ans.Ans.Ans.Ans. (46)(46)(46)(46)(46)
V4 = ∠ ° − ∠ ° − ∠ − °2(230 0 230 120 230 120 )
20= 46∠0° V
End of Solution
Q.28Q.28Q.28Q.28Q.28 A cylindrical rotor synchronous generator delivering constant active power at a constantterminal voltage, a current of 100 A at a 0.9 p.f. lagging. A shunt reactor is connectedso that the reactive power demand doubles then the new value of armature current is______ A.
Ans.Ans.Ans.Ans.Ans. (125.29)(125.29)(125.29)(125.29)(125.29)At Pconstant, Ia1 cos φ1 = Ia2 cos φ2
cos φ1 = 0.9
tan φ1 = 0.484 = QP
⇒2QP
= 0.9686 = tan φ2
cos φ2 = 0.7182∴ 100 × 0.9 = Ia2 × 0.7182⇒ Ia2 = 125.29 A
End of Solution
Q.29Q.29Q.29Q.29Q.29 A cylindrical rotor generator having internal emf 1 + j 0.7 V and terminal voltages(1 + j 00 V). The synchronous reactance is 0.7 pu whereas subtransient reactance is0.2 pu for 3-φ bolted short circuit at generator the value of subtransient generated internalemf is ______.
Ans.Ans.Ans.Ans.Ans. (1.019)(1.019)(1.019)(1.019)(1.019)
Prefault current, I0 =− + −= =1 0.7 1
10.7
f t
d
E V jjX j
Subtransient induced emf,
′′fE = + ′′ I0 0dV jX = 1 + j0.2 × 1 = 1 + j0.2
End of Solution
Q.30Q.30Q.30Q.30Q.30 4 kVA, 200/100 V, 50 Hz single phase transformer has no load core loss of 450 W. Ifhigh voltage side is energized by 160 V, 40 Hz, the core loss will be 320 W. Find thecore loss if the high voltage side is energized by 100 V, 25 Hz.
Ans.Ans.Ans.Ans.Ans. (162.5)(162.5)(162.5)(162.5)(162.5)200 V, 50 Hz, Pc = 450 Watt160 V, 40 Hz, Pc = 320 Watt100 V, 25 Hz, Pc = ? Watt
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Vf
= const. = 20050
= 16040
= 10025
So, Pc = Af + Bf2
450 = A × (50) + B × (50)2 ...(i)320 = A × (40) + B × (40)2 ...(ii)
From (i) and (ii),
45050
= A + B(50) ...(iii)
32040
= A + B(40) ...(iv)
(iii) – (iv),(9 – 8) = B(10)
B =1
10
and A = − ×19 50
10A = 4
Now at 100 V, 25 Hz,
Pc = × + × 214 25 (25)
10= 100 + 62.5 = 162.5 Watt
End of Solution
Q.31Q.31Q.31Q.31Q.31 A 3-φ, 50 Hz, 4 pole induction motor runs at no load with a slip of 1%, at full load themotor runs at a slip of 5%. The percentage speed regulation of the motor is _____.
Ans.Ans.Ans.Ans.Ans. (4.02)(4.02)(4.02)(4.02)(4.02)4 pole, 50 Hz I.M has no load slip 1%4 pole, 50 Hz I.M has full load slip 5%
NS = 1500 rpmN0 = Ns(1 – s) = 1500(1 – 0.01) = 1485N = Ns(1 – s) = 1500(1 – 0.01) = 1425
Speed regulation is
%S.R. =−
×0 100N N
N =
− × =1485 1425100 4.02%
1425
End of Solution
Q.32Q.32Q.32Q.32Q.32 A sequence detector is designed to detect precisely 3 digital inputs, with overlappingsequence detectable. For the sequence (1, 0, 1) and input data (1,1,0,1,0,0,1,1,0,1,0,1,1,0)the output is
Ans.Ans.Ans.Ans.Ans. ()()()()()
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.33Q.33Q.33Q.33Q.33 A synchronous generator has lossless reactance Xs. Then VA is terminal voltage and Ef
is internal emf voltage.P :P :P :P :P : If power factor is leading then always VA > Ef
Q :Q :Q :Q :Q : If power factor is lagging then always VA < Ef
Which of the statement is true?(a) P is false, Q is true (b) P is false, Q is false(c) P is true, Q is true (d) P is true, Q is false
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
End of Solution
Q.34Q.34Q.34Q.34Q.34 A 250 DC shunt motor having armature resistance of 0.2 Ω and field resistance of 100 Ω.It draws a no load current of 5 A at 1200 rpm. The brush drop is 1 V per brush at alloperating conditions. If motor draws 50 A at fulled load and flux per pole is decreasedby 5% because of armature reaction. The speed of the motor at full load is ____ rpm.(a) 900 (b) 1000(c) 1200 (d) 1220
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)No load current 5 A
B.R.D = 1 V per brushloaded, IL = 50 A
Rsh = 100 Ω
Ish = =2502.5 A
100Ia0 = 2.5 AIaL = 47.5 AV = Eb + IaRa + B.R.D
0.2 Ω 250 V
Eb no load = V – Ia0Ra – B.R.D= 250 – 2.5(0.2) – 1 × 2= 247.5 Volts
Eb load = 250 – 47.5(0.2) – 1 × 2= 238.5 volts
2
1
NN =
φ×
φ2 1
1 2
b
b
EE
2
1200N
=φ
×φ
1
1
238.5247.5 0.95
N2 = 1217.22 rpm
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.35Q.35Q.35Q.35Q.35 For cylindrical rotor synchronous generators which of the following statements is/arecorrect.1. VT always greater then Eg at loading load.2. VT always less than Eg at lagging load.(a) only 1 (b) only 2(c) both 1 and 2 (d) None
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Only 2 option correct.
End of Solution
Q.36Q.36Q.36Q.36Q.36 Find (unsaturated)
(Saturated)
ZZ = ?
(0, 10)
(8,400)
(2,210)
(1,110)(4,20)
(8,40)
If = 4If
Ans.Ans.Ans.Ans.Ans. (2.05)(2.05)(2.05)(2.05)(2.05)
Xs unsat =41020
= 20.5 V
Xsat =40040
= 10 Ω
(0, 10)
(8,400)
(2,210)
(1,110)(4,20)
(8,40)
If = 4If⇒ un sat
sat
s
s
XX
= 2.05
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
ANALOG & DIGITAL ELECTRONICS
Q.37Q.37Q.37Q.37Q.37 DC bias voltage of 13 V is applied between gate and body time. The charge measuredin the silicon dioxide layer is +Q,
SiO2 Oxide
Si
GATE
Body
3 V
Box
The total charge in the box region is ________ multiple of +Q. (Give answer to nearestinteger)
Ans.Ans.Ans.Ans.Ans. (#)(#)(#)(#)(#)A
End of Solution
Q.38Q.38Q.38Q.38Q.38 A diode is biased of –0.03 V having an ideality factor of 1513
; and VT = 26 mV; if the
current has to be rased to 1.5 times of the than required voltage will be(a) –4.5 V (b) –0.09 V(c) –0.02 V (d) None of above
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Given data: VD1 = 0.03 V = –30 mV
η =1513
; VT = 26 mV
ID2 = 1.5 ID1
ID2 = 2 /D TV Vs e ηI ...(i)
ID1 = 1 /D TV Vs e ηI ...(ii)
Equation (i) and (ii),
2
1
D
D
II
= 2 1 /D D TV V Ve − η
VD2 – VD1 = 2
1
ln DT
DV
⎛ ⎞η ⎜ ⎟
⎝ ⎠
II
= 1
1
1.51526 mV ln
13D
D
⎛ ⎞× ⎜ ⎟
⎝ ⎠
II
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
= 15 × 2 mV ln (1.5)= 30 mV × 0.40 = 12 mV
VD2 – (–30 mV) = 12 mVVD2 = 12 mV – 30 mV
= –18 mV = –0.018 V
End of Solution
Q.39Q.39Q.39Q.39Q.39 For a common source amplifier gm = 520 μA/V2 and rd = 4.7 kΩ calculate gain of amplifier.(a) –2.447 (b)(c) (d)
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)gm = 520 μA/V2
rd = 4.7 kΩAv = –gm rd
= –520 μA/V2 × 4.7 kΩ= –2.447
End of Solution
SIGNALS & SYSTEMS
Q.40Q.40Q.40Q.40Q.40 A causal control system having poles at (–2, 1), (2, –1) and zeros at (2, 1) and (–2, –1).Identify the nature of transfer function.(a) Unstable, complex, all pass (b) Stable, real, all pass(c) Unstable, complex, HP (d) Stable, real, HPF
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)All pass filter because poles and zeros are mirror-image of eachother (i..e for each polethere is a mirror-image zero).Unstable because one pole in the RHS of s-plane.
⇒ H(s) = [ ( 2 ) [ (2 )](2 )] [ ( 2 )]
s j s js j s j
− − + − −− + − − − σ
–1
1
–2 2
j
=[ (2 ) [ (2 )][ (2 )] [ (2 )]s j s js j s j
+ − − −− + + +
=2 2 2
2 2 2(2 ) [4 1 4 ](2 ) [4 1 4 ]
s j s js j s j
− − − −=− + − − +
= 2
2(3 4 )(3 4 )
s js j
− −− +
Transfer function is complex.h(t) or impulse response will be also complex because poles are not occuring inconjugate pair as well as zeros are also not occuring in conjugate pair.
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.41Q.41Q.41Q.41Q.41 x(t) ∗ h(t) = y(t), where h(t) is impulse response.
⏐x(t)⏐ ∗ ⏐h(t)⏐ = z(t ) then which of the following is correct.(a) For every –∞ < t < ∞, z(t) ≤ y(t) (b) For every –∞ < t ≤ ∞, z(t) ≥ y(t)(c) For some –∞ < t < ∞, z(t) ≤ y(t) (d) For some –∞ < t ≤ ∞, z(t) ≥ y(t)
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Since, y(t) = x(t) + h(t)
and z(t) = ( ) ( )t h t×x
Case-1:Case-1:Case-1:Case-1:Case-1:
1
–1
0
x( )t
t
1
10
h( )t
t
then, y(t) and z(t)
0
y( )t
t1 2
–1 0
z( )t
t1 2
–1
Case-2:Case-2:Case-2:Case-2:Case-2:
1
10
x( )t
t
1
10
h( )t
t
then, y(t) = z(t)
0 t1 2
1
Thus, z(t) ≥ y(t), for all ‘t’
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.42Q.42Q.42Q.42Q.42 A signal x(n) is given by 1 1( )2
n
n⎛ ⎞⎜ ⎟⎝ ⎠
where,
1 01( )
0 0
nn
n≥⎧
= ⎨ <⎩
Z-transform x(n – K) is 1,
11
2
KZ
Z
−
−− then what will be ROC of x(n – K)
(a) 12
Z > (b) Z < 2
(c) Z > 2 (d) 12
Z <
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
x(n) = 1 ( )2
n
u n⎛ ⎞⎜ ⎟⎝ ⎠
, ROC of x(n) : 12
z >
( ) ( )n K X z−x ���⇀↽��� =11
12
KZ
Z
−
−−, ROC of x(n – K) :
12
Z >
For x(n – K) ROC will be 12
Z > .
End of Solution
Q.43Q.43Q.43Q.43Q.43 If xA and xR are average and rms values of signal x(t) = x(t – T) respectively.yA and yR are average and rms value of signal y(t) = Kx(t) respectively.K and T are independent of t.(a) yA = KxA, yR = KxR (b) yA ≠ KxA, yR = KxR
(c) yA = KxA, yR ≠ KxR (d) yA ≠ KxA, yR ≠ KxR
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)Given that,⇒ Average x(t) = Xa , Rms x(t) = XR
⇒ Average y(t) = Ya , Rms y(t) = YR
⇒ x(t) = x(t – T)⇒ y(t) = Kx(t) ...(i)⇒ Ya = KXa
Since, Power y(t) = 2K power x(t)
⇒ Rms y2(t) = 2 2Rms ( )K tx
⇒ Rms y(t) = 2 Rms ( )K tx
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
⇒ YR = RK X
Assuming K as real and positive,YR = KXR
End of Solution
Q.44Q.44Q.44Q.44Q.44 Which of the following statement is true about the two sided Laplace transform?(a) It exists for a signal that may or may not have Fourier transform.(b) It has no poles for any bounded signal that is non-zero only inside a finite time interval.(c) If a signal can be expressed as a weighted sum of shifted one sided exponentials,
then its Laplace transform will have no poles.(d) The number of finite poles finite zeros must be equal.
Ans.Ans.Ans.Ans.Ans. ()()()()()
End of Solution
POWER SYSTEMS
Q.45Q.45Q.45Q.45Q.45 A 50 Hz, power system network is operated under load of 100 MW. When the load isincreases, the power input by the synchronous generator is increases by 10 MW andfrequency of the rpm fall to 49.75 Hz. What will be the load an power system for thefrequency falls to 49.25 Hz?
Ans.Ans.Ans.Ans.Ans. (130)(130)(130)(130)(130)Assumed full load frequency is 50 Hz
tan θ =−
−50 49.75110 100
= −
−x49.75 49.25
( (110))
0.2510
=−x0.5
( 150)
x – 110 =×0.5 10
0.25x = 110 + 20 = 130 MW
Increase in load = 130 – 110 = 20 MW
5049.7549.25
f(Hz)
MWx100 110
θθ
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.46Q.46Q.46Q.46Q.46 Find admittance matrix 2 bus is connected by 2 transformers having ratios 1 : Ti andTj : 1 respectively and line having Y admittance.
1 : Ti Tj : IY
(a)
2
2
j
j j
T TT
TT T
⎡ ⎤−⎢ ⎥⎢ ⎥−⎣ ⎦
i i
i(b)
2
2
j
j j
T TT
TT T
⎡ ⎤−⎢ ⎥⎢ ⎥−⎣ ⎦
i i
i
(c)2
2
j j
j
TT T
T TT
⎡ ⎤−⎢ ⎥⎢ ⎥−⎣ ⎦
i
i i(d)
2
2
j j
j
TT T
T TT
⎡ ⎤−⎢ ⎥⎢ ⎥−⎣ ⎦
i
i i
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)I = Y(TiVi – VjTj)Ii = Ti I
= Ti2YVi – TiTjYVj
Ij = –Tj I Vi
Ii
Vj
I Ij
Y1 : Tx Tj : 1
= –IiTjYVi + Tj2YVj
⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
iII j
=⎡ ⎤− ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎣ ⎦⎣ ⎦
i i i
i
2
2
j
jj j
T Y TT Y VVTT Y T Y
End of Solution
Q.47Q.47Q.47Q.47Q.47 A transmission line is being protected by distance protection. 80% of the line is beingprotected. The transfer reactance is 0.2 p.u. A solid 3-φ, fault occurred at the end ofthe transmission line. The minimum level of fault current to activate the relay(a) 6.25 (b) 5(c) 1.25 (d) 3
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
If = =1 10.2ThZ
= 5 pu for 100% of line
Zl = 0.2 p.u.
80%
Relay is operated for 80%Zf = 0.8 Zl ⇒ 0.8 × 0.2 = 0.16 p.u.
For 80% of line, If =1
0.16 = 6.25 p.u.
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.48Q.48Q.48Q.48Q.48 Voltage at bus 1 = 1.1 p.u.Voltage at bus 2 = 1 p.u.
Q12
1 2jX
Voltage at bus 2 is kept constant, Q12 is the sending reactive power from 1 to 2.On changing the bus 1 voltage, Q12 increases by 20%.Active power is zero in both the condition find the new value of bus 1 voltage.
Ans.Ans.Ans.Ans.Ans. (#)(#)(#)(#)(#)A
End of Solution
POWER ELECTRONICS
Q.49Q.49Q.49Q.49Q.49 In AC voltage controller shown below. Thyristor T1 is fired at α and T2 is fired at 180° + α.To control the output power over range 0 to 2 kW. The minimum range of variation inα is
10 –60° ∠ Ω
T1
T2200 V,50 Hz
(a) 0° to 60° (b) 0° to 120°(c) 60° to 120° (d) 60° to 180°#
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)
End of Solution
Q.50Q.50Q.50Q.50Q.50 A 1-φ, fully controlled rectifier is connected to highly inductive RL load with R = 10 Ωat 230 V, 50 Hz. The source inductance is 2.28 mH. If the firing angle α = 45°, thenthe overlapping angle will be
Ans.Ans.Ans.Ans.Ans. (4.25)(4.25)(4.25)(4.25)(4.25)1-φ, SCR bridge rectifier
α = 45°, R = 10 Ωsupply 230 V, 50 Hz
Ls = 2.28 mHµ = ?
ΔVd = α − α + μπ
[cos cos( )]mV = 4f LsI0
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
V0 = α −π
I02
cos 4ms
Vf L
I0R = α −π
I02
cos 4ms
Vf L
Find I0
I0 10 = −⋅ − × × ×π
I30
2.230 2 cos45 4 50 2.28 10
I0(10 + 0.456) = 146.64
I0 =146.4910.456
= 14.01 A
ΔVd0 = − +π
230 2 [cos45 cos(45 µ)] = 4 × 50 × 2.28 × 10–3 × 14
= 6324cos45° – cos(45° + µ) = 0.061
45 + µ = 49.75∴ µ = 4.25°
End of Solution
Q.51Q.51Q.51Q.51Q.51 Value of σ adjusted so that 3rd harmonic is completely eliminated. Find the percentagemagnitude of 5th harmonic w.r.t. fundamental component at this condition.
σ
σ σ
2πt
π/2σ0
V0
3 /2π
Ans.Ans.Ans.Ans.Ans. (20)(20)(20)(20)(20)
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.52Q.52Q.52Q.52Q.52 A single-phase, 50 Hz full bridge rectifier with highly inductive RL load. The two mostdominant frequency components will be(a) 50 Hz, 150 Hz (b) 50 Hz, 100 Hz(c) 150 Hz, 250 Hz (d) 50 Hz, 0 Hz
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)1-φ full bridge rectifier,
is
n = 1, 3, 5, ...f = 50, 150, 250 dominant
End of Solution
Q.53Q.53Q.53Q.53Q.53 A DC-DC converter shown below having switching frequency of 10 kHz with duty ratio0.6. All the components are ideal and initial inductor current is zero. Energy stored inthe inductor (in mJ) at the end of 10 switching cycles is
10 mH 50 V
50 V
Ans.Ans.Ans.Ans.Ans. (5)(5)(5)(5)(5)Buck boost converter
0.6T T
50L
–50L
i
t0.4T
D = 0.6 → stores energy
D = = 0.6ONTT
TON = 0.6T → store energyTOFF = 0.4 T → releasing energy
For one cycle: Rise in current for 0.2T
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
For 10 cycles: Find rise in current (0.2T) × 10 = 2T
i =50
tL
i = −×= = =
⋅ × ⋅3 3
50 50 2 100(2 ) 1A
10 10 10 10T
L LP
∴ Energy stored = −= × ⋅ ⋅i2 3 21 1(10 10 ) (1)
2 2L = 5 mJ
End of Solution
ELECTRICAL & ELECTRONICS MEASUREMENTS
Q.54Q.54Q.54Q.54Q.54 RT → ThermistorRT = 2(1 + αT)
Temperature rise = 150%Find errors in the output voltage?
V0
R3R2
0.1 V
R1
+–
3 V
RT
R1 = 1 kHz, R2 = 1.3 kΩ, R3 = 2.6 kΩ
Ans.Ans.Ans.Ans.Ans. ()()()()()
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
ENGINEERING MATHEMATICS
Q.55Q.55Q.55Q.55Q.55 A vector function is given by ˆ ˆyF y a a= −x x�
. The line integral of the above function
c
F d⋅∫ l� �
along the curve C given by y = x2 as shown below ________.
–1 2
14
Ans.Ans.Ans.Ans.Ans. (–2.33)(–2.33)(–2.33)(–2.33)(–2.33)
F = −x xˆ ˆyya a
F = −i xy j
⋅∫C
F d r = +∫ x1 2C
F d F dy
y = x2
dy = 2x dx
= − ⋅∫ x x x x x2 2d d
= −∫ x x x2 2( 2 )d = − −
⎛ ⎞− = −⎜ ⎟⎝ ⎠∫
xx x22 3
2
1 13
d
=− −+ =8 1 73 3 3
= –2.33
End of Solution
Q.56Q.56Q.56Q.56Q.56dyd x
= 2x – y, y(0) = 1. Find y at x = ln2
Ans.Ans.Ans.Ans.Ans. (0.886)(0.886)(0.886)(0.886)(0.886)
xdyd
= 2x – y; y(0) = 1, y at x = ln 2
+x
dyy
d= 2x
P = 1, Q = 2x
I.F. = ∫ xPde = ∫ =x x1de e
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Solution, y(I.F) = ∫ I x( . .)Q F d
yex = ⋅∫ xx x2 e d = − +x xx2( )e e C
y = 2x – 2 + ce–x
y(0) = 11 = 0 – 2 + CC = 3
∴ y = 2x – 2 + 3e–x
At x = ln 2y = 2(ln 2) – 2 + 3e–ln2
= − + 31.386 2
2 = 0.886
End of Solution
Q.57Q.57Q.57Q.57Q.57 Value of integral 2
21
2z
z z+
−∫ along the contour ⏐z⏐ = 1 is
(a) πi (b) –π i(c) 8πi (d) –8π i
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
+−∫
2
2
1
2
zdz
z z where C is ⎪z⎪ = 1
–1 10=
+−∫
2 1( 2)z
dzz z
=
+−∫
2 12
zz dz
zBy integral formula,
2πi f(0) where, f(z) =+
−
2 12
zz
= ⎛ ⎞π ⎜ ⎟⎝ ⎠−i 12
2 = –πi
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.58Q.58Q.58Q.58Q.58 The number of purely real elements in lower triangular representation of given 3 × 3 matrixobtained through given decomposition is
11 11
21 22 21 22
31 32 33 31 32 33
2 3 3 0 0 0 03 1 2 0 03 6 5
a aa a a aa a a a a a
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Ans.Ans.Ans.Ans.Ans. (#)(#)(#)(#)(#)
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
2 3 33 1 23 6 5
=
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
ll ll l l
11 11 12 13
21 22 22 23
31 32 33 33
0 00 0
0 0
u u uu u
u
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
2 3 33 1 23 6 5
=
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
ll ll l l
11 12 13
21 22 23
31 32 33
0 0 10 0 1
0 0 1
u uu
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
2 3 33 1 23 6 5
=
⎡ ⎤⎢ ⎥+ +⎢ ⎥⎢ ⎥+ + +⎣ ⎦
l l ll l l l ll l l l l l
11 11 12 11 13
21 21 12 22 21 13 22 23
31 31 12 32 31 13 32 23 33
u uu u uu u u
l11 = 2 l11u12 = 3 l11u13 = 32u12 = 3 2u13 = 3
l21 = 3 u12 =32
u13 =32
l31 = 3
l21u12 + l22 = 1
⎛ ⎞ +⎜ ⎟⎝ ⎠l22
332 = 1
l21l13 + l22u23 = 2
u23 =−
=−
92 52
7 72
u23 =57
l22 =−− =9 7
12 2
l22 =−72
l31u12 + l32 = 6
⎛ ⎞ +⎜ ⎟⎝ ⎠l32
3(3)2
= 6
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
l32 = − =9 36
2 2
l32 =32
l31u13 + l32u23 + l33 = 5
l33 = − −9 155
2 14
l33 =− −70 63 15
14 =
− −=8 414 7
l33 =−47
End of Solution
Q.59Q.59Q.59Q.59Q.59 If y = 3x2 + 3x + 1, for x∈[–2, 0], find maximum and minimum value in the given range.
(a) 4 and 1 (b) 7 and 14
(c)14
and 1 (d) –2 and 12−
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)y = 3x2 + 3x + 1 in [–2, 0]
∂∂xy
= 6x + 3,∂∂x
2
2
y = 6
xdyd
= 0
6x + 3 = 0
x =−12
x
2
2
d yd
= 6 > 0 minimum
Maximum value of y in [–2, 0] is maximum {f(–2), f(0)}max{7, 1} = 7
Minimum value of y in [–2, 0]
⎧ ⎫⎛ ⎞−⎜ ⎟⎪ ⎪⎝ ⎠−⎪ ⎪⎪ ⎪↓ ↓ ↓⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
1( 2) (0) 2
min , ,
7 1 14
ff f
=14
Maximum value 7, minimum value 14
.
End of Solution
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Memory Based Questions of GATE 2020Electrical Engg. | Forenoon Session
Q.60Q.60Q.60Q.60Q.60 Which of the following is true?
(a)0
1 sin sin 0m t n t m nπ
ω ω = ≠π ∫ (b)
1 sin sin 0p t q tπ
−π
ω ω =π ∫
(c)0
1 cos cos 02
p t q tπ
ω ω =π ∫ (d)
1 sin cos 02
p t q tπ
−π
ω ω =π ∫
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)
End of Solution
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