GATE  >  GATE Electrical Engineering (EE) 2023 Mock Test Series  >  Test: GATE Electrical Engineering (EE) Previous Paper 2019 Download as PDF

Test: GATE Electrical Engineering (EE) Previous Paper 2019


Test Description

65 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Test: GATE Electrical Engineering (EE) Previous Paper 2019

Test: GATE Electrical Engineering (EE) Previous Paper 2019 for GATE 2022 is part of GATE Electrical Engineering (EE) 2023 Mock Test Series preparation. The Test: GATE Electrical Engineering (EE) Previous Paper 2019 questions and answers have been prepared according to the GATE exam syllabus.The Test: GATE Electrical Engineering (EE) Previous Paper 2019 MCQs are made for GATE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: GATE Electrical Engineering (EE) Previous Paper 2019 below.
Solutions of Test: GATE Electrical Engineering (EE) Previous Paper 2019 questions in English are available as part of our GATE Electrical Engineering (EE) 2023 Mock Test Series for GATE & Test: GATE Electrical Engineering (EE) Previous Paper 2019 solutions in Hindi for GATE Electrical Engineering (EE) 2023 Mock Test Series course. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free. Attempt Test: GATE Electrical Engineering (EE) Previous Paper 2019 | 65 questions in 180 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE Electrical Engineering (EE) 2023 Mock Test Series for GATE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 1

Newspapers are a constant source of delight and recreation for me. The ________ trouble is that I read _______ many of them

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 1

Newspapers are a constant source of delight and recreation for me. The only (what bother’s) trouble is that I read too (a lot/ large) many of them.

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 2

The missing number in the given sequence 343,1331,______, 4913 is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 2

343 = 73
1331 = 113
4913 = 173
All numbers given are cube of prime numbers so 133 = 2197 satisfy the missing number.

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 3

The passengers were angry _______ the airline staff about the delay.

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 3

The passengers were angry with the airline staff about the delay.

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 4

It takes two hours for a person X to mow the lawn. Y can move the same lawn in four hours. How long (in minutes) will it take X and Y, if they work together to move the lawn?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 4

Time taken by X to now the lawn = 2 hrs.
∴ Work done by X in 
Similarly, Work done by 4 in hr = ¼
Work done by x + 4 in 1 hr = 
∴ Total time taken by X & 4 together =

= 80 Minutes

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 5

I am not sure if the bus that has been booked will be able to ____ all the students.

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 5

 I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 6

Given two sets X = {1, 2, 3} and Y = {2, 3, 4}, we construct a set Z of all possible fractions where the numerators belong to set X and the denominators belong to set Y. The product of element having minimum and maximum values in the set Z is __________.

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 6

Given that X = {1, 2, 3}
4 = {2, 3, 4}

Minimum value in 
Maximum value in  

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 7

The ratio of the number of boys and girls who participated in an examination is 4 : 3. The total percentage of candidates who passed the examination is 80 and the percentage of girls who passed is 90. The percentage of boys who passed is __________

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 7

Let number of boys participated = 4x Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates = 
Girls candidate who passed =  
Boys candidate who passed = Total passed candidate – Girls candidate who passed

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 8

An award-winning study by a group of researchers suggests that men are as prone to buying on impulse as women feel more guilty about shopping.

Which one of the following statements can be inferred from the given text?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 8

The correct statement can be concluded from Venn diagram or using the Syllogism.

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 9

How many integers are there between 100 and 1000 all of whose digits are even?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 9

For all digits of a number which lie between 100 and 1000 are even, Unit and tens digits can be filled from the set {0, 2, 4, 6, 8} But hundred’s digit does not include 0 as it will not remain a number which lie between 100 and 1000
∴ Hundreds digit set is {2, 4, 6, 8}

Total integer = 100 numbers

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 10

Consider five people – Mita, Ganga, Rekha, Lakshmi and Sana. Ganga is taller than both Rekha and Lakshmi. Lakshmi is taller than Sana. Mita is taller than Ganga.
Which of the following conclusions are true?
1. Lakshmi is taller than Rekha
2. Rekha is shorter than Mita
3. Rekha is taller than Sana
4. Sana is shorter than Ganga

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 10

Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
∴ Mita > Ganga > Rekha, Lakshmi > Sana
∴ 2 and statement 4 are correct 

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 11

The mean-square of a zero-mean random process is  where k is Boltzmann’s constant, T is the absolute temperature, and C is capacitance. The standard deviation of the random process is 

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 11

Given that
Mean square of random process 
Mean is given zero ⇒ E (x) = 0
We know that E(x2) – [E(x)]2 = variance

Standard deviation = 

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 12

The characteristic equation of a linear timeinvariant (LTI) system is given by 
∆(s) = s4 + 3s3 + 3s2 + s + k = 0.
The system is BIBO stable if

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 12

Applying R.H criteria for stability
Δ(S) = S4 + 3S3 + 3S2 + S + K = 0

For stability, first column should be greater than zero

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 13

The inverse Laplace transform of   for ≥ 0 is 

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 13


H (t) = L–1 [H(S)]

H (t) = e–t+2te–t

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 14

A 5 kVA, 50 V/100 V, single-phase transformer has a secondary terminal voltage of 95 V when loaded. The regulation of the transformer is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 14

We know that
Voltage Regulation = 
Given that VFL = 95V
VNL = 100 V

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 15

A three-phase synchronous motor draws 200 A from the line at unity power factor at rated load. Considering the same line voltage and load, the line current at power factor of 0.5 leading is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 15

We know that P = VI cos φ, as load and voltage are same
∴ I cos φ = constant
l1 cos φ1 = I2 cos φ2
I1 = 200A
Cos φ1 = 1
Cos φ2 = 0.5
 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 16

A cv-axial cylindrical capacitor shown in Figure (i) has dielectric with relative permittivity εr1 = 2. When one-fourth portion of the dielectric is replaced with another dielectric of relative permittivity εr2, as shown to Figure (ii), the capacitance is doubled. The value of εr2 is _________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 16

We know that




Total portion cover 2π

Both are connected in parallel

c2= cr1+ cr2


Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 17

The parameter of an equivalent circuit of a three-phase induction motor affected by reducing the rms value of the supply voltage at the rated frequency is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 17



By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which changes the magnetizing reactance

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 18

The output response of a system is denoted as y(t), and its Laplace transform is given by

The steady state value of y(t) is 

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 18


For finding steady state value, we will apply final value theorem

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 19

The open loop transfer function of a unity feedback system is given by 
In G(s) plane, the Nyquist plot of G(s) passes through the negative real axis at the point 

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 19


Nyquist plot cut the negative real
Axis at w = phase cross over frequency



Magnitude at cutting point

Then, the co-ordinates becomes (-0.5, j0).

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 20

A current controlled current source (CCCS) has an input impedance of 10 Ω and output impedance of 100 kΩ. When this CCCS is used in a negative feedback closed loop with a loop gain of 9, the closed loop output impedance is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 20

CCCS is current-shunt-negative feedback amplifier.
The output impedance Rof=R0(1+βA),Given Aβ = 9
= 100 × 103 (1 + 9)
= 1000 kΩ

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 21

A six-pulse thyristor bridge rectifier is connected to a balanced three-phase, 50 Hz AC source. Assuming that the DC output current of the rectifier is constant, the lowest harmonic component in the AC input current is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 21

We know that,
For 6-pulse converter harmonic present in AC current are 6K ± 1
General expression NK ± 1  [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 × 50 = 250 Hz

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 22

The current I flowing in the circuit shown below in amperes (round off to one decimal place) is ___________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 22


Applying nodal analysis at point 1 whose voltage is assumed as V1.

Solving (1) and (2)

8I = V1 – 6 
8I = 20 – 2I – 6
10I = 14
I = 1.4 A

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 23

The partial differential equation

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 23

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 24

Which one of the following function is analytic in the region |z| ≤ 1?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 24

 the singularity z = –2 lies outside the |Z| < | 
∴ By Cauchy’s integral theorem

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 25

If f = 2x3 + 3y2 + 4z, the value of line integral  C grade f . dr evaluated over contour C formed by the segments (-3, -3, 2) → (2, -3, 2) → (2, 6, 2) → (1, 6, -1) is


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 25

Given that
y = 2x3 + 3y2 + 4z

Applying the limits 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 26

Five alternators each rated 5 MVA, 13.2 kV with 25% of reactance on its own base are connected in parallel to a busbar. The short-circuit level in MVA at the busbar is ___________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 26


Net reactance of generator

Short Circuit MVA = ISC ×Base MVA = 20 × 5 = 100 MVA

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 27

Given Vgs is the gate-source voltage, Vds is the drain voltage, and Vth is threshold voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in saturation are

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 27

For NMOS transistor to be in saturation the condition will be
VGS > Vth And VDS ≥ VGS – VTh

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 28

The total impedance of the secondary winding, leads, and burden of a 5 A CT is 0.01 Ω. If the fault current is 20 times the rated primary current of the CT, the VA output of the CT is ___________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 28

Isec = 5 × 20 = 100 A
V = Isec R = 100 × 0.01 = 1V
VA output of CT = VIsec = 100 × 1 100 VA

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 29

The Ybus matrix of a two-bus power system having two identical parallel lines connected between them in pu is given as

The magnitude of the series reactance of each line in pu (round off up to one decimal place) is ___________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 29

Y12= – (y12) = – j20
Series admittance of each line = 
Series reactance of each line = 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 30

The rank of the matrix, is__________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 30


Determinant of M = |M|
|M| = 0 [0 – 1] – 1 [0 – 1] + 1 [1 – 0]
|M| = 2
|M| ≠ 0
∴ Rank of M = Number of Non zero rows in Echelon form
P (M) = 3 

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 31

The symbols, a and T, represent positive quantities, and u(t) is the unit step function. Which one of the following impulse responses is NOT the output of a causal linear time-invariant system?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 31

H (t) = 1 + e–at u (t)

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 32

A system transfer function is  If a1 = b1 = 0, and all other coefficients are positive, the transfer function represents a

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 32


At s = 0
H (0) = constant
At s = ∞

∴ It is a low par filter 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 33

The output voltage of a single-phase full bridge voltage source inverter is controlled by unipolar PWM with one pulse per half cycle. For the fundamental rms component of output voltage to be 75% of DC voltage, the required pulse width in degrees (round off up to one decimal place) is ________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 33

Waveform for output voltage of single phase full bridge PWM inverter


Vo1rms = fundamental rms output voltage

Given, Vo1 = 0.754 Vdc 


Pulse width = 2d = 112.88

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 34

In the circuit shown below, the switch is closed at t = 0. The value of θ in degrees which will give the maximum value of DC offset of the current at the time of switching is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 34



For Maximum value of DC offset A
θ – φ = – 90

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 35

M is a 2 × 2 matrix with eigenvalues 4 and 9. The eigenvalues of M2 are

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 35

M is a 2 × 2 Matrix with Eigen value 4 and 9 If has λ1, λ2 _ _ _ _ _ λn Eigen values
Mn → λ1n, λ2n _ _ _  λnn Eigen values
M2 → 42, 92
∴ M2 has Eigen values as 16 and 81

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 36

A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 μF/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end and energized with 400 kV at the sending end, the receiving end line voltage end line voltage in kV (round off to two decimal places) will be _____________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 36

VS = 400 KV
l = 300 km
L1 = 1 mH / km / phase
C1 = 0.01 μF / km / phase

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 37

The current I flowing in the circuit shown below in amperes is ______________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 37

According to Mill man’s Theorem, the equivalent circuit of the given circuit is



Eeq = 0V
So, the current I flowing is 0 A 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 38

A 220 V (line), three-phase, Y-connected, synchronous motor has a synchronous impedance of (0.25 + j2.5) Ω/phase. The motor draws the rated current of 10 A at 0.8 pf leading. The rms value of line-to-line internal voltage in volts (round off to two decimal places) is ____________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 38

For synchronous motor
Eg = V1 – IZ

Z = (0.25 + j 2.5)Ω 
I = 10 ∠ –36.86 A

Eg = 141.658 ∠–8.7 V (phase)
Eg = 245.36 V (line)

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 39

The closed loop line integral 

evaluated counter-clockwise, is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 39

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 40

The voltage across and the current through a load are expressed as follows

The average power in watts (round off to one decimal place) consumed by the load is _____________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 40




P = 588.89 watts 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 41

A delta-connected, 3.7 kW, 400 V(line), three-phase, 4-pole, 50-Hz squirrel-cage induction motor has the following equivalent circuit parameters per phase referred to the stator: R1 = 5.39 Ω, R2 = 5.72 Ω, X1 = X2 = 8.22 Ω. Neglect shunt branch in the equivalent circuit. The starting line current in amperes (round off to two decimal places) when it is connected to a 100 V (line), 10 Hz, three-phase AC source is __________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 41

Given R1 = 5.39Ω, R2 = 5.72Ω, X1 = X2 = 8.22Ω  for frequency → 10 Hz

Starting phase current at 10 Hz

IPn = 8.63A
Starting line current =  

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 42

In a 132 kV system, the series inductance up to the point of circuit breaker location is 50 mH. The shunt capacitance at the circuit breaker terminal is 0.05 μF. The critical value of resistance in ohms required to be connected across the circuit breaker contacts which will give no transient oscillation is _____________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 42

Given data L = 50mH, C = 0.05 μF
Critical resistance to avoid current shopping will be given as

R = 500Ω  

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 43

In the single machine infinite bus system shown below, the generator is delivering the real power of 0.8 pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees (round off to one decimal place) is ____________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 43


Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 44

In the circuit below, the operational amplifier is ideal. If V1 = 10 mV and V2 = 50 mV, the output voltage (Vout) is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 44




 

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 45

In the circuit shown below, X and Y are digital inputs, and Z is a digital output. The equivalent circuit is a

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 45



The above expression is for XOR gate 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 46

A 0.1 μF capacitor charged to 100 V is discharged through a 1 kΩ resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1 V is _________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 46

Discharging of capacitor equation 
VC (t) = Voe–t/τ
Where τ = RC = (103) (10–7) = 10–4 sec
Vo = 100V
Vc(t) = 100 e–104t
Vc(t) = 1V
1 = 100 e–104t
T = 0.46 msec

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 47

A periodic function f(t), with a period of 2π, is represented as its Fourier series, 

the Fourier series coefficients a1 and b1 of f(t) are

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 47



*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 48

If A = 2xi + 3yj + 4zk and u = x2 + y2 + z2, then div (uA) at (1, 1, 1) is ________.
 


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 48


div (UA) = (6x2 + 2y2 + 2z2) + (3x2 + 9y2 + 3z2) + (4x2 + 4y2 + 12z2)
at (1, 1, 1)  ⇒ x = 1, y = 1, z = 1
div (UA) = 45

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 49

A moving coil instrument having a resistance of 10 Ω, gives a full-scale deflection when the current is 10 m

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 49

PMMC Instrument
Ifs = 10 mA
Rm = 10Ω

100 = Ifs (Rm + Rse)
100 = 10 × 10–3 (10 + Rsc)
Rse = 10000 – 10 = 9990Ω  

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 50

Consider a state-variable model of a system 

     
where y is the output, and r is the input. The damping ratio ξ and the undamped natural frequency ωn (rad/sec) of the system are given by

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 50


Comparing the above equation with the given problem 

C = (1          0)
Characteristic equation is 
|SI – A| = 0

s2 + 2Sβ + α = 0.........(1)
s2 + 2ξωns + ωn2 = 0.............(2)
Comparing (1) and (2)
ωn2 = α

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 51

The enhancement type MOSFET in the circuit below operates according to the square law. μnCox = 100 μA/V2, the threshold voltage (VT) is 500 mV. Ignore channel length modulation. The output voltage Vout is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 51


Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 52

The asymptotic Bode magnitude plot of a minimum phase transfer function G(s) is shown below.

Consider the following two statements.
Statement I : Transfer function G(s) has three poles and one zero.
Statement II : At very high frequency (ω → ∞), the phase angle (jω)

Q. Which one of the following options is correct?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 52


From the given Bode plot,
T(S) = Transfer function =

It has three poles and no zero
So, statement 1 is false
∠T(s) = – 90 – tan–1 w – tan–1 
∠T(jw) |w → ∞ = – 270o
So, statement 2 is true

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 53

A single-phase transformer of rating 25 kVA, supplies a 12 kW load at power factor of 0.6 lagging. The additional load at unity power factor in kW (round off to two decimal places) that may be added before this transformer exceeds its rated kVA is _____________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 53

Load supplied previously before adding extra load 12 KW at pf of 0.6
SLoad = 12 + j16
Now, Let P be extra load added   (Qextra = as unity p.f)
SLoad = 12 + P + j16

Rated KVA |Srated| = 25

So, 7.20 KW is extra load which is added

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 54

Consider a 2 × 2 matrix M = [v1 v2], where. v1 and v2 are the column vectors. suppose  where u1T and u2T are the row vectors. Consider the following statements :
Statement 1 :
Statement 2 :

Q. Which of the following options is correct?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 54

M–1 M = I

Statement 1 and 2 are both correct 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 55

A single-phase fully-controlled thyristor converter is used to obtain an average voltage of 180 V with 10 A constant current to feed a DC load. It is fed from single-phase AC supply of 230 V, 50 Hz. Neglect the source impedance. The power factor (round off to two decimal places) of AC mains is ___________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 55

Vsr Isr cos φ = VoIo
For single phase fully – controlled converter
Io = Isr = 10A

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 56

A DC-DC buck converter operates in continuous conduction mode. It has 48 V input voltage, and it feeds a resistive load of 24 Ω. The switching frequency of the converter is 250 Hz. If switch-on duration is 1 ms, the load power is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 56

Given that
Switch frequency, fs = 250Hz
Load resistance RL = 24Ω
Supply voltage Vs = 48V
TON = 1 msec

P = 6 watts 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 57

In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in μH) is ________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 57

Po = 120w, Vs = 24V, Vo = 48V

α = 0.5 [Duty cycle]
Po = VoIo = 120
 
VSIS = VoIo

At boundary of continuous & discontinuous

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 58

A 220 V DC shunt motor takes 3 A at noload. It draws 25 A when running at fullload at 1500 rpm. The armature and shunt resistances are 0.5 Ω and 220 Ω, respectively. The no-load speed in rpm (round off to two decimal places) is ________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 58


No load
INL = 3A 

Ia = IL – If = 2A 
Back cmf = EbN = V – IaRa
= 220 – 2 × 0.5 = 219V
Full load
IFL = 25A  Nf = 1500 rpm
If = 1A
Ia = IFL – If = 24A
EbF = V – IaRa = 220 – 24 × 0.5 = 208 V
We know E α speed (N)

(NN = speed at no load)

NN = 1579.33 rpm
 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 59

A fully-controlled three-phase bridge converter is working from a 415 V, 50 Hz AC supply. It is supplying constant current of 100 A at 400 V to a DC load. Assume large inductive smoothing and neglect overlap. The rms value of the AC line current in amperes (round off to two decimal places) is ________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 59

Ac line current rms = (Is)rms =

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 60

The output expression for the Karnuaugh map shown below is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 60



F (P, Q, R, S) = 

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 61

The probability of a resistor being defective is 0.02. There are 50 such resistors in a circuit. The probability of two or more defective resistors in the circuit (round off to two decimal places) is __________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 61

P = 0.02
n = 50
λ = np = 50 (0.02) = 1
P (x ≥ 2) = 1 – P (x < 2)
= 1 – [P(x =0) + P (x = 1)]
P (x ≥ 2) = 1 – e–1 (1 + 1) = 0.26

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 62

The magnetic circuit shown below has uniform cross-sectional area and air gap of 0.2 cm. The mean path length of the core is 40 cm. Assume that leakage and fringing fluxes are negligible. When the core relative permeability is assumed to be infinite, the magnetic flux density computed in the air gap is 1 tesla. With same Ampere-turns, if the core relative permeability is assumed to be 1000 (linear), the flux density in tesla (round off to three decimal places) calculated in the air gap is _________________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 62


Lair = 0.2 cm
Lm = 40 cm
Given Bo = 1 Tesla at µr  
Lcore = 40 – 0.2 = 39.8cm
Let a = uniform cross – sectional area
We know that

ST = Sairgap + Score

Case 1: when μr

Case 2: 
Μr = 1000
MMF = Same

Put NI1 from (1) 

BL = 0.834 Tesla

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 63

A 30 kV, 50 Hz, 50 MVA generator has the positive, negative, and zero sequence reactances of 0.25 pu, 0.15 pu, and 0.05 pu, respectively. The neutral of the generator is grounded with a reactance so that the fault current for a bolted LG fault and that of a bolted three-phase fault at the generator terminal are equal. The value of grounding reactance in ohms (round off to one decimal place) is ___________.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 63

Fault current for SLG fault

Fault current for 3φ fault

Xn = 0.1 Pu
Xn (inΩ) =
Xn (inΩ) = 1.8Ω  

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 64

The line currents of a three-phase four wire system are square waves with amplitude of 100 A. These three currents are phase shifted by 120o with respect to each other. The rms value of neutral current is

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 64

Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 65

The transfer function of a phase lead compensator is given by 

The frequency (in rad/sec), at which ∠D(jω) is maximum, is 

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2019 - Question 65



Frequency at which ∠T (jw) is maximum.......(i)

 

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about Test: GATE Electrical Engineering (EE) Previous Paper 2019 Page
In this test you can find the Exam questions for Test: GATE Electrical Engineering (EE) Previous Paper 2019 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: GATE Electrical Engineering (EE) Previous Paper 2019, EduRev gives you an ample number of Online tests for practice