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Gate Practice Test: Electrical Engineering(EE)- 15 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - Gate Practice Test: Electrical Engineering(EE)- 15

Gate Practice Test: Electrical Engineering(EE)- 15 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Gate Practice Test: Electrical Engineering(EE)- 15 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Gate Practice Test: Electrical Engineering(EE)- 15 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)- 15 below.
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Gate Practice Test: Electrical Engineering(EE)- 15 - Question 1

The sentences given with blanks are to be filled with an appropriate word(s). Four alternatives are suggested for each question. For each question, choose the correct alternative and click the button corresponding to it.

They abandoned their comrades ______the wolves.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 1
When something is left in between more than 2 things or persons, ‘among’ is used. There are several wolves in the sentence hence the correct answer is option D.
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 2

Direction: Select the one which is different from the other three responses.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 2

49 - 33 = 16

62 - 46 = 16

83 - 67 = 16

70 - 55 = 15

All options have a difference of 16 except the D option.

Hence, option D is the answer.

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Gate Practice Test: Electrical Engineering(EE)- 15 - Question 3

Amit rows a boat 9 kilometres in 2 hours down-stream and returns upstream in 6 hours. The speed of the boat (in kmph) is:

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 3

Down-stream rate = 9/2 = 4.5 kmph

Upstream rate = 9/6 = 1.5 kmph

The speed of the boat = (4.5 – 1.5) kmph = 3 kmph

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 4

Directions: In the following question, out of the four alternatives, select the word opposite in meaning to the given word.

Turgid

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 4

Turgid = swollen and distended or congested; pompous

Bloated = inflated

Humble = plain, simple

Puffy = inflated

Tumescent = swollen

Hence, humble is the correct answer.

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 5

Direction: Study the following information carefully and answer the given questions:

The following pie chart shows the percentage distribution of total number of Apples (Dry + Wet) sold in different days of a week:

The pie chart2 shows the percentage distribution of the total number of Apples (Wet) sold on different days of the week.

Find the ratio of Apples (Dry + Wet) sold on Tuesday to that of the Apples (Wet) sold on Thursday?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 5

Required ratio = 13% of 7000: 23% of 4500 = 182:207

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 6

Directions: In each of the questions below, some statements are given followed by some conclusions. You must consider the statements to be true even if they seem to be at variance with commonly known facts. You must decide which of the following conclusions logically follows from the given statements. Give answer.

Statements:

Some jacket is shirt.

Some shirts are trouser.

No shoes are t-shirt.

All trousers are shoes.

All pants are t-shirt.

Conclusions:

I. All shirts being t-shirt is a possibility.

II. Some shoes are trouser.

III. Some jackets are t-shirt.

IV. Some shirts being t-shirt is a possibility.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 6

From the above diagram-

• Some shoes are trouser.

• Some shirts being t-shirt is a possibility.

Hence option B is correct.

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 7

Direction: Which of the following is the MOST SIMILAR in meaning to the given word?

Remnant

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 7

‘Remnant’ means ‘remainder’ which is the same as ‘residue’.

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 8

Direction: Read the information carefully and give the answer of the following questions

Total number of children = 2000

If two-ninths of the children who play football are females, then the number of male football children is approximately what percent of the total number of children who play cricket?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 8

57.4% (If 2/9 th of children play football are female, then

male children = 1 - 2/9 = 7/9th of children play football;

Let’s take z% of male football children equal to cricket children, then

⇒ z% of cricket children = 7/9 th of football;

⇒ z% of (23% of 2000) = 7/9 th of (17% of 2000);

⇒ z% of 23 = 7/9 th of 17

⇒ z = 7× 17 × 100 = 57.4%;)

Short-cut : Z% of 23 = 7/9 of 17

(7x17x100)/ (23x9) =57.4%

Required percentage = (7/9)×17×100/ 23 = 57.4%

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 9

Direction: Study the following information carefully and answer the given questions.

Seven people, P, Q, R, S, T, W and X sitting in a straight line facing north, not necessarily in the same order. R sits at one of the extreme ends of the line. T has as many people sitting on his right, as on his left. S sits third to the left of X. Q sits on the immediate left of W. Q does not sit at any of the extreme ends of the line.

If all the people are made to sit in alphabetical order from right to left, the positions of how many people will remain unchanged?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 9

From the given conditions, we can conclude:

After arranging in alphabetical order,

Hence, only Q′ s position will remain unchanged.

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 10

A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in total from both as interest. The rate of interest per annum is.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 10

Let the rate % = R

According to the question,

= 2200

100R + 120R = 2200

220R = 2200

R = 10%

Hence required rate % = 10%

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 11

A 10 KVA, 400 V, 3 phase, star connected synchronous generator has an armature resistance of 0.8 Ω per phase and synchronous reactance of 1.5 Ω per phase. The voltage regulation at a power factor 0.6 lagging is _______.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 11

Ra = 0.80Xs = 1.5Ω

Vph = 230.94V

Eg = 255.19V

Voltage Regulation =

=

V.R. = 10.50%

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 12

How many essential prime implicants are there in the given k-map?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 12

AB and are essential prime implicants because only these two have at least single 1′ ' which can have only one pairing.

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 13

The various resistance of a wheatstone bridge are shown in figure. The battery has an internal emf of 8 V with negligible internal resistance. If the sensitivity of the galvanometer is 12 mm/µA with an internal resistance of 150 Ω.

What will be the sensitivity of the bridge in terms of deflection per unit change in resistance.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 13

Resistance of unknown resistor required for balance,

Therefore, the deviation in unknown resistance from the balance condition,

ΔR = 1650 - 1600

= 50Ω

So now Thevenin source emf,

ε0 = 0.039V Now the internal resistance of bridge looking into terminal C & d−

= 481.95Ω

Hence the current through the galvanometer

then deflection of galvanometer

θ = Si. Ig

= 12 × 61.7mm

θ =740.4mm

Therefore, sensitivity of bridge

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 14

The circuit shown below has C = 20 μF and L = 5 μH. Initial voltage across the capacitor is Vs = 230 V. The load current is constant at 300 A . Let ‘A’ be the conduction time period of the auxiliary thyristor and ‘B’ is the circuit turn off time of the main thyristor. Than A + B will be _________.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 14

The above circuit is class 'B' commutation.

So, the conduction time of auxiliary thyristor is =

A = 31.4 \musec

The peak value of resonant current,

Voltage across the main thyristor, when it gets turned off will be

Circuit turn off time for main thyristor is

So, A + B = 31.4 + 11.62

=

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 15

In the circuit shown in figure the switch is closed at time (t=0). The rate of change of current

across inductor will be -

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 15

At (t = 0+) as the circuit was unenergized at (t = 0- ) ⇒

Inductor L is open circuit (O.C)

Capacitor C is short circuit (S.C)

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 16

In the figure shown below,

Assume |Vth| = 0.3 V, the transistor is operating in

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 16

Since the device is NMOS,

Vth = 0.3 V

VS = 0.4 V

VD = 2 V

VG = 0 V

For cutoff region, V­GS < |Vth|

Since, VGS = –0.4 V

So, the device is in the cutoff region.

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 17

If a finite length of wire having a total length QR = 80 m, a current of 10 A is flowing in it.

A point P from the origin point or reference point is away at a distance of 4 mm. If the point P is making an angle of 75° with the point Q in clockwise direction & 60° with the point R in anticlockwise direction. What will be the magnetic field density B, at point P.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 17
As we know that due to a finite length of wire in which I current is flowing, at a point P where the point P makes angle θ1 with point θ in clockwise direction &θ2 with R in anticlockwise direction is calculated as-

B =

where the sign of angle is tve, is being taken, a is the distance from a reference point.

S0 B at point P

B =

= 2.5 x 10-4(0.966 + 0.866)

= 0.458 x 10-3Wb/m2

B = 0.458 m-Wb/m2

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 18

A chopper circuit shown in the figure has an input voltage of Vs = 5 V. Duty ratio is 0.6 . The load resistance is 10 Ω. The chopper is operating at 25 kHz switching frequency. If L = 150 μH and C = 220 μF. Find the ratio of peak value of capacitor voltage to peak value of inductor current?


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 18

The given circuit is a boost regulator.

For the boost regulator,

When the switch is ON.

Peak value of inductor current is

The ripple voltage at capacitor is

Peak voltage at capacitor,

Required ratio is = =

= 3.56

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 19

Consider the circuit given in the figure below, if voltage VAB and VCD in the circuit are 200 V each, then the value of the capacitor in the circuit, C is approximately ________ (in μF).


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 19
since

Let assume angle between VAB and VBC is θ

So, taking magnitude both sides

By solving, θ = 120∘

Current through inductor hL=10A

since, we know that inductor current lags 90 by its voltage and for capacitor its current leads 90 by its voltage

C =

C =

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 20

A uniform volume charge density ρv= 5 µC/m3 is present in the spherical shell of 0.9 < r < 1 m and ρv = 0 elsewhere. The charge present in the shell approximately is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 20

Q =

=

=

Q =

=

Q = 5.675µC

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 21

A 15 MVA, 11 kV, alternator has positive, negative and zero sequence reactance are 0.3 p.u., 0.2 p.u. and 0.05 p.u. respectively. Find the value of resistance to be added in the generator neutral so that the fault current of LG fault does not exceed 1.5 times of rated line current.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 21

For L-G fault Current

Rn= 0.641 p.u.

R = 5.17 Ω

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 22

Consider a circuit shown in figure below, switch is closed at t = 0 in the circuit for V = 10 V, R = 5 Ω L = 3 H. Find the value of maximum current ‘I’ in the circuit. (in Amperes)


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 22

Expression for current 11 for t≥0 is

Where τ=LR

Similarly, for current I2

Current 1 = 11 -I2

The value of current at this time is

I = 0.5

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 23

A causal system has input x[n] = δ[n] + 14δ[n − 1] − 18δ[n − 2]

and output y[n] = δ[n] − 34δ[n − 1]

The impulse response of this system is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 23

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 24

Two identical unloaded generators are connected in parallel as shown below. Both the generators are having positive, negative and zero sequence impedance of j0.8 p.u., j0.2 p.u. and j0.3 p.u. respectively. If pre-fault voltage is 1 p.u. for a line to ground fault at the terminals of the generator, the fault current magnitude (in p.u.) is ___________. (Assume neutral impedance, Zn = j0.02 p.u.)


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 24
The positive sequence reactance,

Similarly, the equivalent negative sequence reactance is

For a single line to ground fault, fault current is

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 25

Given an assembly language program in 8085 microprocessor,

LXI B, 2384 H

LOOP: DCX B

MOV A, C

ORA B

JNZ LOO

HLT

If the system clock frequency is 2 MHz. The time required for complete execution of the program is _________ msec.


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 25

Number of T-states required = Count x Number of T states inside loop + Number of T states outside loop Number of T states when jump is invalid.

[Here, count in decimal (2384)16=(9092)10]

=(9092 × 24 )+ 16 − 3 = 218221 T-states

Clock frequency is 2MHz

T=1 / 2 ×106 = 0.5μsec

Total execution time =218221 × 0.5 × 10 − 6 = 0.10911sec

Hence, the time required for execution of the program is 109.11 msec.

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 26

Let x(t) be a signal that has a rational Laplace transform with exactly 2 poles located at S = –1 & S = –3. If g(t) equal to e2t x(t) & G(ω) converge. The conclusion about g(t) is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 26

One pole in RHS & another pole in LHS to make the system converging i.e. stable, the signal must be non-causal.

Hence Double-sided signal.

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 27

If the crystal frequency of 8085 microprocessor is 6MHz then total time (in millisecond) required to execute the program, before HLT instruction is _____msec.


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 27

TTotal = ⅓ x 60 = 20ߎsec = 0.02msec

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 28

Consider a feedback system with characteristics equation .

The point of intersection of the asymptotes of the root locl with real axls on


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 28

Characteristic equation is equation as: 1+G(s)H(s) = 0

On comparing this characteristics equation with the given equation

The point of intersection of the asymptotes = centroid(o)

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 15 - Question 29

In a current cumulated chopper, peak commutating current is twice that of the maximum possible load current. The source voltage is 250 V dc and main SCR take off time is 50 μs. Voltage is 250 V dc and main SCR taken off time is 50 μs. If maximum load current is 250 A, then peak capacitor voltage is………V.


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 29

Peak Capacitor Voltage Vc = Vs + Io

Peak Current = Ip + 2Io = 2(250) = 500

= 2

= 0.5

VCP 250 + 250(0.5) = 375V

Gate Practice Test: Electrical Engineering(EE)- 15 - Question 30

Which of the following is incorrect:

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 15 - Question 30

For inverting amplifier,

(Av)1= −RF / RL

if R1 > RF then it will be less than 1

for Non-Inverting amplifier,

(AV)NI = (1 + RF / RL)

when RF = 0, AV = 1

it can be one but cannot be less than one in any case.

So, option B is an incorrect statement and correct Answer.

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