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


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30 Questions MCQ Test - Test: GATE Electrical Engineering (EE) Previous Paper 2019

Test: GATE Electrical Engineering (EE) Previous Paper 2019 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: GATE Electrical Engineering (EE) Previous Paper 2019 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: GATE Electrical Engineering (EE) Previous Paper 2019 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: GATE Electrical Engineering (EE) Previous Paper 2019 below.
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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.

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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 

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