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Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: GATE Electrical Engineering (EE) Previous Paper 2020

Test: GATE Electrical Engineering (EE) Previous Paper 2020 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: GATE Electrical Engineering (EE) Previous Paper 2020 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: GATE Electrical Engineering (EE) Previous Paper 2020 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 2020 below.
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Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 1

Select the next element of the series: Z, WV, RQP, _______ .

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 1

Let's break down the pattern in the series: Z, WV, RQP, _______.

  • The first element is Z, which is the last letter of the alphabet.
  • The second element is WV, where both letters are moving backward from the alphabet:
    • W is the 23rd letter, and V is the 22nd letter.
  • The third element is RQP, where:
    • R is the 18th letter, Q is the 17th, and P is the 16th, all moving backward from the alphabet.

So, the pattern suggests that with each step, we are taking consecutive letters in reverse order, and the number of letters is increasing by one.

Now, following this logic, the next group should consist of four consecutive letters in reverse order from the next position after R, which would start from M (the 13th letter):

  • M (13th), L (12th), K (11th), J (10th)

Thus, the next element is MLKJ.

However, this option isn't listed, but the closest matching option based on the reverse alphabetical order is KJIH.

The correct answer is:

2. KJIH.

 

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 2

If P, Q, R, S are four individuals. How many teams of size exceeding one can be formed, with Q as a member?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 2

3C1 + 3C2 + 3C3
3 + 3 + 1 = 7
Possible combinations,
PQ, RQ, SQ, PRQ, PSQ, RSQ, PQRS

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Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 3

In four-digit integer numbers from 1001 to 9999, the digit group “37” (in the same sequence) appears ______ times.

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 3

10 × 10 + 9 × 10 + 9 × 10 = 280

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 4

Non-performing Assets (MPAs) of a bank in India is defined as an asset, which remains unpaid by a borrower for a certain period of time in terms of interest, principal, or both.
Reserve Bank of India (RBI) has changed the definition of NPA thrice during 1993-2004. in terms of the holding period of loans. The holding period was reduced by one quarter each time. In 1993, the holding period was four quarters (360 days). Based on the above paragraph, the holding period of loans in 2004 after the third revision was ______ days.

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 4

As given in question holding period was reduced by one quarter each time.
Therefore, after third division holding period remains 90 days.

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 5

Stock markets _________ at the news of the coup.

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 6

People were prohibited ________ their vehicles near the entrance of the main administrative building.

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 7

Given a semicircle with O as the centre; as shown in the figure, the ratio  is
_________ . Where  are chords.

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 7


The ratio, 

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 8

The revenue and expenditure of four different companies P, Q, R and S in 2015 are shown in the figure. If the revenue of company Q in 2015 was 20% more than that in 2014, and company Q had earned a profit of 10% on expenditure in 2014, then its expenditure (in million rupees) in 2014 was _______.

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 8

Let the revenue of company Q in 2010 = x in million rupees.
Then the revenue of company Q in 2015 = 1.2 x in million rupees.
Given in graph, 1.2x =45
x = 37.5 in million rupees
As given in question Q had earned a profit of 10%.
∵ Expenditure of company Q in 2014
= 37.5/1.1 = 34.09 ≈ 34.1

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 9

Select the word that fits the analogy:
Do : Undo : Trust : ______

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 9

Do : Undo : Trust : Distrust

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 10

This book, including all its chapters _______ interesting. The students as well as the instructor _______ in agreement about it.

*Multiple options can be correct
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 11

xR and xA are, respectively, the rms and average values of x(t) = x(t – T ), and similarly, yR and yA are, respectively, the rms and average values of y(t) = k x(t) ⋅ k, T are independent of t. Which of the following is true?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 11

Given, y(t)= Kx(t) ...(1)
then, Average of y (t)= K × Average of x(t)
⇒ YA = KXA
From equation (1),
Power of y (t)= |K|2 ⋅ power of x(t)
⇒  [∵ Power = Rms2]
⇒ YR = |K| ⋅XR
Case (I) : When K is real and positive then,
|K| = K
and
YR = KXR
Thus option (a) is satisfied.
Case (II) : When K is imaginary or complex or real and negative then,
|K| ≠ K
and YR ≠ KXR
Thus option (b) is satisfied, option (a) and (b) both satisfies the given condition.

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 12

A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode. D, represented approximately as a piecewise linear plot of current vs time at diode turn-off.
Lpar is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1. 2, 3 or 4) at which the IGBT experiences the highest current stress is ______.


Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 13

The value of the following complex integral, with C representing the unit circle centered at origin in the counterclockwise sense, is:

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 13





By integral formula,
2πi f (0) where, 

*Multiple options can be correct
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 14

Which of the following is true for all possible non-zero choices of integers m, n; m ≠ n. or all possible non-zero choices of real numbers p, q ; p ≠ q, as applicable?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 14



(A): 
Put l = π in rule 4

Given that, m ≠ n

(B): 
Put l = π in rule 5

Given that, m ≠ n

(C) 
When, α → ∞,

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 15

A common-source amplifier with a drain resistance, RD = 4.7 kΩ, is powered using a 10 V power supply. Assuming that the transconductance, gm, is 520 µA/V, the voltage gain of the amplifier is closest to:

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 15

Given data: RD = 4.7 kΩ, gm = 520 μA/V
Voltage gain of CS amplifier
=–gmRD
= –520 μAV × 4.7 kΩ = –2.44

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 16

The Thevenin equivalent voltage, VTH, in V (rounded off to 2 decimal places) of the network shown below, is _______ .


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 16



Vth = 14 V

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 17

Consider a signal x[n] = (1/2)n 1[n], where 1[n] = 0 if n < 0, and 1[n] = 1 if n ≥ 0. The z-transform of x[n – k], k > 0 is  with region of convergence being

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 17



For x(n – k) ROC will be |z| > 1/2.

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 18

A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source supplies a series combination of finite resistance. R, and a very large inductance, L, The two most dominant frequency components in the source current are:

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 19

Consider the initial value problem below. The value of y at x = ln 2. (rounded off to 3 decimal places) is ________ .
dy/dx = 2x - y, y(0) = 1


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 19

dy/dx = 2x - y, y(0) = 1, y at x = ln 2
dy/dx + y = 2x
P =1, Q =2x

Solution, y(I.F)= ∫Q (I.F .)dx
yex = ∫ 2x ⋅exdx = 2(xex −ex)+ C
y =2x – 2 + ce–x
y(0) = 1
1 = 0 – 2 + C
C = 3
∴ y =2x – 2 + 3e–x
At x = ln 2
y = 2(ln 2) – 2 + 3e–ln2​
1.386 - 2 + 3/2 = 0.886

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 20

A single-phase, 4 kVA, 200 V/100 V, 50 Hz transformer with laminated CRGO steel core has rated no-load loss of 450 W. When the high-voltage winding is excited with 160 V, 40 Hz sinusoidal ac supply, the no-load losses are found to be 320 W. When the high-voltage winding of the same transformer is supplied from a 100 V, 25 Hz sinusoidal ac source, the no-load losses will be _________W (rounded off to 2 decimal places).


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 20

200 V, 50 Hz, Pc = 450 Watt
160 V, 40 Hz, Pc = 320 Watt
100 V, 25 Hz, Pc = ? Watt

So, Pc = Af + Bf2
450 = A × (50) + B × (50)2 ...(i)
320 = A × (40) + B × (40)2 ...(ii)
From (i) and (ii),  ...(iii)
320/40 = A + B(40) ...(iv)
Equation (iii) – (iv),
(9 – 8) = B(10)
B = 1/10
and A = 9 - (1/10) x 50 = 4
Now at 100 V, 25 Hz, Pc = 4 x 25 + (1/10) x (25)2
= 100 + 62.5 = 162.50 Watt

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 21

Currents through ammeters A 2 and A 3 in the figure are 1∠10° and 1∠70° respectively. The reading of the ammeter A1 (rounded off to 3 decimal places) is ________ A.


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 21

I =1∠10° + 1∠70°
I = 1.732∠40°
The ready of ammeter is 1.732 A.

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 22

A single-phase inverter is fed from a 100 V dc source and is controlled using a quasi square wave modulation scheme to produce an output waveform, v(t). as shown. The angle σ is adjusted to entirely eliminate the 3rd harmonic component from the output voltage. Under this condition, for v(t), the magnitude of the 5th harmonic component as a percentage of the magnitude of the fundamental component is _______(rounded off to 2 decimal places).


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 22

Using result,
For V3 = 0
cos3σ = 0
or 3σ = π/2
σ = π/6
Now, 

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 23

Thyristor T1 is triggered at an angle α (in degree), and T2 at angle 180° + α, in each cycle of the sinusoidal input voltage. Assume both thyristors to be ideal. To control the load power over the range 0 to 2 kW, the minimum range of variation in α is:

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 24

A sequence detector is designed to detect precisely 3 digital inputs, with overlapping sequences detectable. For the sequence (1, 0, 1) and input data (1,1, 0,1,0,0,1,1,0,1,0,1,1,0): what is the output of this detector?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 24

Sequence detector problem:
• If consider the case of non-overlapping sequence detector, then the pattern 101 is appearing 2 times in the given bit sequence.
• If we consider the case of overlapping sequence detector, then the pattern 101 is appearing 3 times in the given bit sequence.
The question says that the detector is overlapping, hence answer is (a).

*Answer can only contain numeric values
Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 25

A three-phase. 50 Hz. 4-pole induction motor runs at no-load with a slip of 1%. With full load, the slip increases to 5 %. The % speed regulation of the motor (rounded off to 2 decimal places) is _________ .


Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 25

4 pole, 50 Hz I.M has no load slip 1%
4 pole, 50 Hz I.M has full load slip 5%
NS = 1500 rpm
N0 = Ns(1 – s) = 1500(1 – 0.01) = 1485
N = Ns(1 – s ) = 1500(1 – 0.01) = 1425
Speed regulation is

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 26

A lossless transmission line with 0.2 pu reactance per phase uniformly distributed along the length of the line, connecting a generator bus to a load bus, is protected up to 80% of its length by a distance relay placed at the generator bus. The generator terminal voltage is 1 pu. There is no generation at the load bus. The threshold pu current for operation of the distance relay for a solid three phase-to-ground fault on the transmission line is closest to:

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 26


= 5 pu for 100% of line

Relay is operated for 80%
Zf =0.8 Zl ⇒ 0.8 × 0.2 = 0.16 p.u.
For 80% of line, 

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 27

ax3 + bx2 + cx + d is a polynomial on real x over real coefficients a, b, c, d wherein a ≠ 0. Which of the following statements is true?

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 27

ax3 + bx2 + c x + d =0 ; a ≠ 0
x = 0 is the root for any values of a, b, c only when d = 0.

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 28

Consider a linear time-invariant system whose input r (t ) and output y (t) are related by the following differential equation.

The poles of this system are at

Detailed Solution for Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 28


[s2 + 4] Y (s )= 6 R (s)

Poles: s2 + 4 = 0
s =±j2

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 29

A three-phase cylindrical rotor synchronous generator has a synchronous reactance Xs and a negligible armature resistance. The magnitude of per phase terminal voltage is VA and the magnitude of per phase induced emf is EA. Considering the following two statements, P and Q.
P : For any three-phase balanced  leading load connected  across the terminals of this synchronous generator, VA is always more than EA.
Q : For any  three-phase  balanced  lagging load connected across the terminals of this synchronous generator, VA is always less than EA.
Which of the following options is correct?

Test: GATE Electrical Engineering (EE) Previous Paper 2020 - Question 30

Which of the following statements is true about the two sided Laplace transform?

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