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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


End of Solution

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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


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


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



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


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


End of Solution



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



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.


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



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



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


−+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



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


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



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


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



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


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



− − − + + −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


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



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


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




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 ______.


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



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


End of Solution



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


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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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


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



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



= 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



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



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 <


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



⇒ 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.


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



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


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



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°#


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



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


End of Solution



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



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


End of Solution



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


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



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


+−∫

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



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



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



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 ∫


End of Solution

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