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GATE Practice Test: Electrical Engineering(EE)- 2 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - GATE Practice Test: Electrical Engineering(EE)- 2

GATE Practice Test: Electrical Engineering(EE)- 2 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The GATE Practice Test: Electrical Engineering(EE)- 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The GATE Practice Test: Electrical Engineering(EE)- 2 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Practice Test: Electrical Engineering(EE)- 2 below.
Solutions of GATE Practice Test: Electrical Engineering(EE)- 2 questions in English are available as part of our GATE Electrical Engineering (EE) Mock Test Series 2025 for Electrical Engineering (EE) & GATE Practice Test: Electrical Engineering(EE)- 2 solutions in Hindi for GATE Electrical Engineering (EE) Mock Test Series 2025 course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt GATE Practice Test: Electrical Engineering(EE)- 2 | 65 questions in 180 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study GATE Electrical Engineering (EE) Mock Test Series 2025 for Electrical Engineering (EE) Exam | Download free PDF with solutions
GATE Practice Test: Electrical Engineering(EE)- 2 - Question 1

A number of Indian goods face a _____ competition from Chinese goods in terms of prices and looks.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 1
The sentence implies that a powerful competition is faced by Indian goods from Chinese goods. ‘Fierce’ meaning ‘powerful and destructive’ is the apt fit for the blank.

Hence, option C is the correct answer.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 2

In the following question, out of the five alternatives, select the word similar in meaning to the given word.

Appease

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 2
Appease = to bring to a state of peace, quiet, ease, calm, or contentment; pacify; soothe.

Agitate = make (someone) troubled or nervous.

Pacify = to cause someone who is angry or upset to be calm and satisfied.

Interrupt = stop the continuous progress of (an activity or process).

So, option B is the correct answer.

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GATE Practice Test: Electrical Engineering(EE)- 2 - Question 3

Select the most appropriate option to substitute the underlined segment in the given sentence. If there is no need to substitute it, select No improvement.

It is not wise relying to anybody too much.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 3
‘Rely on’ is the phrase which means to be dependent as for support or maintenance. All other choices fail to provide the correct meaning to the sentence both contextually as well as grammatically.

Therefore, we can say option C is the correct answer.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 4

Select the most appropriate option to substitute the underlined segment in the given sentence. If no substitution is required, select No improvement.

His miserable condition made us wept.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 4
The English verbs let, make, have, Get, and Help are called causative verbs

because they cause something else to happen. The grammatical structure for make is given below:

MAKE + PERSON + VERB (base form).

The structure given in option A follows the above rule.

So, it is the correct answer.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 5

Select the most appropriate word to fill in the blank.

Many items made of ivory were _____ from a dealer in antiques by the custom authorities at the Delhi airport.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 5
The sentence implies that the custom authorities seized many items made of ivory.

‘Confiscate’ means ‘take or seize (someone's property) with authority’. This word fits best in the sentence.

Therefore, option A is the correct answer.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 6

Select the most appropriate option to substitute the underlined segment in the given sentence. If there is no need to substitute it, select No improvement

The man who seen the accident to occur telephoned the police.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 6
The sentence shows two actions occurring in past with a considerable gap between them. The act of the man seeing the occurrence of the accident happened first and then the man telephoned the police. In such situations, we need to keep the first occurring action in past perfect tense and second event will be kept in simple past tense.

Therefore, option B is the correct response.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 7

Select the most appropriate option to substitute the underlined segment in the given sentence. If no substitution is required, select No improvement.

If you listen to English news, it improve your English.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 7
The given sentence is an example of first conditional sentences. The first conditional is a structure we use when we want to talk about possibilities in the present or in the future. The first conditional has simple present after 'if' or ‘when’, then the simple future in the other clause.

Hence, option B is the correct answer.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 8

Select the most appropriate word to fill in the blank.

Scientists at Cambridge University are ______ how plants can give us sustainable energy.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 8

The sentence implies that the scientists at Cambridge University are carrying out the research to find how plants can give us sustainable energy.

‘Investigating’ meaning ‘carry out research or study into (a subject or problem, typically one in a scientific or academic field)’ perfectly fits in the blank.

Hence, option B is the correct answer.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 9

In the following question, out of the five alternatives, select the word similar in meaning to the given word.

Acrimonious

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 9

Acrimonious (adj.): (typically of speech or discussion) angry and bitter.

Tawdry (adj.): showy but cheap and of poor quality.

Gaudy (adj.): extravagantly bright or showy, typically so as to be tasteless.

Vitriolic (adj.): violent hate and anger expressed through severe criticism.

Ameliorate (v.):

make (something bad or unsatisfactory) better.

Courteous (adj.): polite, respectful, or considerate in manner.

So, the correct answer is option B.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 10

Direction: In the following question, a part of the sentence is bold. Below the sentence alternatives to the bold part are given at (A), (B) and (C) which may improve the sentence. Choose the correct alternative. In case ‘No improvement’ is needed, your answer is (D).

The professor has agreed to take remediable classes for the weaker students.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 10

'Remedial' should be used to make the sentence grammatically correct. It means giving or intended as a remedy or cure while remediable means capable of being cured; treatable. The classes don't need to be cured.

Thus, option B is the correct answer.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 11

The open loop transfer function of a unity negative feedback system is given by Find the value of 'K' such that the phase margin is 60

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 11

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 12

The starting current of a star-connected, 3-phase induction motor at rated voltage is 6 times the full load current and full load slip of 5%. Auto-transformer is used to limit the starting current from mains to 4 times the full load current. Determine the ratio of starting torque to the full load torque.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 12
In case of auto transformer,

As we known

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 13

A DC voltage of 50 V is applied to a coil having R = 10 Ω, L = 10 H. The time taken by the current to reach 75% of its final value is –

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 13
In a coil having R and L.

The charging current will be

i = 1 (1 - e - t/T)

i = 5 [ l - e-7]

Time required to reach 75% of its final value,

0.75 x 5 = 5 [1 - e-]

e-t = 0.25

t = 1.386 sec

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 14

For the operational amplifier circuit shown in the figure below, what is the maximum possible value of R1, if the voltage gain required is between – 10 and – 25? (The upper limit on RF is 1 MΩ)

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 14

So lesser the gain higher will be R1

So fo r AV = - 10

→ R1 will be maximum

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 15

In the circuit of figure, Vs = 2 cos3t volts the VL(t) will be-

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 15
Current across capacitor

Where,

V0 = [2 cos 3f] volts

Voltage across inductor

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 16

The power absorbed by 0.5 Ω resistance in the given circuit is:

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 16
The circuit can be redrawn as

Since, bridge is in the balanced condition.

So, power absorbed by 0.5Ω resistance is

P = 12R = ((3.43)2 × 0.5) = 5.88W

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 17

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 GATE Practice Test: Electrical Engineering(EE)- 2 - Question 17
Given data L = 50 mH, C = 0.05 μF

Critical resistance to have transient free oscillations is

R = 500 Ω

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 18

A signal x(t) is denoted by a function which is periodic and continuous in its time domain representation. Then, which of the following would be true for the fourier transform of the signal

(i) Discrete and Periodic

(ii) Discrete and Aperiodic

(iii) Real{x(t)} transforms in to Real{X(jω)}

(iv) Real{x(t)} transforms in to Imag{X(jω)}

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 18
It is a well-established thumb rule that is tabulated below

Also, Even part of a signal gets transformed into real part of Fourier representation and Odd part of signal gets transformed into Imaginary part of Fourier representation.

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 19

Find the differential equation of the system described by the transfer function given as:

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 19
We have,

⇒ Y (s) [2s2 + 5s] = X(s) • (s + 3)

⇒ 2s2 • y(s) + 5 • y(s) = s • X(s) + 3. X(s)

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 20

Which of the following is correct for 1110100 ÷ 1010?

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 20
By simple divide method,

1010) 1110100(1011

(1) 1) 1) 1) 1) 1) 1010

10000

1010

110

Therefore, Quotient= 1011 and remainder= 110

*Answer can only contain numeric values
GATE Practice Test: Electrical Engineering(EE)- 2 - Question 21

Fault current of a line to ground fault and LLL fault for an unloaded generator is same. If X1 =0.3 pu, X2 = 0.2 pu and Xo = 0.04 pu. Then calculate value of inductive reactance (Xn) required for natural grounding (in pu)


Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 21

for LG fault:-

IfLG = 3 ER1 / (X1 eq + X 2eq + X oeq)

For LLL fault:-

ILL = E R1 / X 1eq

ILL = 1/0.3

Both faults currents are same

IFLG = LFLL

0 .54 + 3 X n = 0 .9

3 Xn = 0.9 - 0.54

Grounding reactance (Xn) = 0.12 pu

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 22

Determine the value of V and I in the given network.

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 22
Applying nodal analysis,

⇒ 2V + 4(V - 6) + V = 0

⇒ 2V + 4V - 24 + V = 0

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 23

The instantaneous polarity of primary winding of an ideal transformer is shown in figure below. The direction of flux and polarity of instantaneous voltage induced in secondary is

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 23
For the given direction of current and polarity of induced voltage in primary, according to principle of Right Hand Thumb Rule the direction of flux will be clockwise and as the instantaneous direction of flux is clockwise, according to Lenz's law as effect opposes cause, the polarity of instantaneous induced in secondary is 'd' negative and 'c positive'.
GATE Practice Test: Electrical Engineering(EE)- 2 - Question 24

Given below is a 3-phase full-bridge rectifier. When the diode D3 is conducting, which of the following diodes cannot conduct at the same time?

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 24
KVL around any path shows that only one diode in the top half of the bridge will conduct at the same time. The diode that is conducting has its anode connected to the highest phase voltage at that moment. Hence, when

D3 is conducting, D1 and D5 cannot conduct. Similarly, only one diode can conduct in the bottom half at the same time. The diode that is conducting has its cathode connected to the lowest phase voltage at that moment. Hence, when D3 is conducting, D6 cannot conduct.

*Answer can only contain numeric values
GATE Practice Test: Electrical Engineering(EE)- 2 - Question 25

A simple slide wire potentiometer is used for measurement of current in a circuit. The voltage drop across a standard resistor of 0.5 Ω is balanced at 95 cm. find the magnitude of the current (in Ampere) if the standard cell emf of 1.95 volts is balanced at 60 cm.


Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 25
The answer is in between 6.5 and 7

For the same working current, if 60 cm corresponds to 1.95 volt.

Then 95cm of the slide wire corresponds to =1.95 / 60x10-2 x 95x10-2 = 3.421 volt

So, cross the resistance 0.5 Ω the voltage drop is 3.421 volt.

Then the value of the current is (I) = 3.421 / 0.5 = 6.84 A

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 26

Determine V0 ( in Volts) in the given network.


Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 26
Applying current-divider,

Hence,

V0 = 4 × I2 = 4 × 6.25× = 25V

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 27

In the figure shown below, ‘N’ is two port network. If I = 2 A, then the values of V1 and V2 will be

Assume the Y - parameter matrix for ' N′ as

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 27
Y- parameter of Network 'N'

I1 = 2V1 + V2........(i)

I2 = 2V1 + 2V2 ........ (ii)

Applying KCL at input node,

Applying KCL at output node,

Putting the value of I1 and l2 in equation (iii) and (iv),

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 28

Find the value of y{z), n > 0 if y(n) + (n - 1 ) - (n ~ 2) = 0 and y(-1) = y { - 2) = 1?

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 28
Given y{n) + y{n - 1) - y(n - 2) = 0

Y (-1)= y (-2 ) = 1

Apply Z-transform to the above equation, we get

GATE Practice Test: Electrical Engineering(EE)- 2 - Question 29

Let the state transition matrix of a system be given as

Then which of the following is equal to

Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 29
We have,

Hence, replace ′t′ in state transition matrix by '-t'.

*Answer can only contain numeric values
GATE Practice Test: Electrical Engineering(EE)- 2 - Question 30

The full-load torque angle of a synchronous motor at rated voltage and frequency is 30° elect. The stator resistance is negligible. ______ degree will be the torque angle if the load torque and terminal voltage remaining constant, the excitation and frequency are raised by 10%


Detailed Solution for GATE Practice Test: Electrical Engineering(EE)- 2 - Question 30
The answer is in between 33 and 34

Assuming Ra to be negligible,

sin δ = 1.1 sin 30 = 0.55

8 = 33.36

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