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Sample GATE Mock Test - Electrical Engineering (EE)


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65 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Sample GATE Mock Test - Electrical Engineering (EE)

Sample GATE Mock Test - Electrical Engineering (EE) for GATE 2022 is part of GATE Electrical Engineering (EE) 2023 Mock Test Series preparation. The Sample GATE Mock Test - Electrical Engineering (EE) questions and answers have been prepared according to the GATE exam syllabus.The Sample GATE Mock Test - Electrical Engineering (EE) MCQs are made for GATE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Sample GATE Mock Test - Electrical Engineering (EE) below.
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Sample GATE Mock Test - Electrical Engineering (EE) - Question 1

Identify the correct spelling of the word.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 1

The correct spelling of the word is 'definitive' which means '(of a conclusion or agreement) done or reached decisively and with authority.' Thus option 2 is the correct answer.

Sample GATE Mock Test - Electrical Engineering (EE) - Question 2

This is the place that _______

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 2

The preposition 'about' is mandatory here thus option 1 and 4 are eliminated. Option 2 is correct as the tense present perfect continuous fits here. It conveys the meaning that the person usually talked about the place. 

Sample GATE Mock Test - Electrical Engineering (EE) - Question 3

She is brave. Her brother is more brave.

Select the most suitable sentence with respect to grammar and usage.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 3

Options 1 and 2 are incorrect as they change the meaning of what is mentioned. Option 3 is incorrect as with 'less' we use the adjective in positive form and not comparative form. The correct word here should be 'brave.' Option 4 is thus the correct answer.

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 4

When a four digit number is divided by 65, it leaves a remainder of 29. If the same number is divided by 13, the remainder would be______


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 4

Let number is N

N = 65 k +29

∴ Remainder = 3

Sample GATE Mock Test - Electrical Engineering (EE) - Question 5

Complete the following sentence.

I was ___ ___  for the bus and then I ___ sight of Craig passing by.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 5

The word 'here' and 'there' both can be used here but should be followed with 'waiting' as no other word can fit here. The word 'caught' is correct here as 'catch a sight' means 'to see something.' Option 2 thus has the correct combination of words. The word 'cot' means 'a small bed with high barred sides for a baby or very young child.'

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 6

4 – digit number greater than 5000 are randomly formed from the digits 0, 2, 3, 5 and 7. The probability of forming a number divisible by 5 when the digits are repeated is ______


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 6

For a number to be greater than 5000, d1 should be filled with either 5 or 7

∴ Total numbers formed when the digits are repeated = 2 × 5 × 5 × 5 = 250

total cases = 250 -1 = 249 ( case of 5000 is not included)

Now, For the number to be divisible by 5, unit digit d4 should be either 0 or 5.

∴ Total no. of ways = 2 × 5 × 5 × 2 = 100

favorable cases = 100 - 1=9 ( 5000 is not included))

∴Required Probability = 

Sample GATE Mock Test - Electrical Engineering (EE) - Question 7

It is theoretically possible that bacteria developed on Venus early in its history and that some were carried to Earth by a meteorite. However, strains of bacteria from different planets would probably have substantial differences in protein structure that would persist over time, and no two bacterial strains on Earth are different enough to have arisen on different planets. So, even if bacteria did arrive on Earth from Venus, they must have died out. 

The argument is most vulnerable to which of the following criticisms?

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 7

The question asks which of the statements given in the options can weaken the argument put by the author that all bacteria from Venus must have died out.

The passage states that since there is a single strain of bacteria which exists on the Earth, they all must be belonging to the Earth or let's say a single planet. But here the author does not take into consideration (as can be argued from his theory) the fact that may be all the bacteria came from Venus and there are none which originally belong to the Earth. So this criticism as mentioned in option 3 makes the argument of the author vulnerable. 

Options 1, 2 and 4 are completely irrelevant criticisms as they do not address the main argument. The argument claims that if there were Venusian bacteria on Earth, then they must have died out by now. Whether there are bacteria originally from Earth that have also disappeared from Earth is irrelevant to the question and has no effect on the given argument.

Sample GATE Mock Test - Electrical Engineering (EE) - Question 8

A man sells three articles A, B, C and gains 10% on A, 20% on B and loses 10% on C. He breaks even when combined selling prices of A and C are considered, whereas he gains 5% when combined selling prices of B and C are considered. What is his net loss or gain on the sale of all the articles?

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 8

Let a, b and c be the cost prices of the three articles A, B and C.

SP = CP + Profit (or) SP = CP – Loss

⇒ SP of A = 1.1a; SP of B = 1.2b; SP of C = 0.9c

By question,

1.1a + 0.9c = a + c ⇒ 0.1a = 0.1c ⇒ a = c

1.2b + 0.9c = 1.05(b + c) ⇒ 0.15b = 0.15c ⇒ b = c = a

Gain% = {(SP – CP)/CP} × 100

⇒ Net gain on the sale of all the articles =

∴ Net gain on the sale of all the articles = 6.66%

Sample GATE Mock Test - Electrical Engineering (EE) - Question 9

Which of the following inferences can be drawn from the above graph?

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 9

Option 1 is false as graph says there is decrease in students qualifying in Physics in 2015 compared to 2014.

Option 2

Let no. of students qualifying in Biology in 2013 be 100

⇒ No. of students qualifying in Biology in 2014 = 100 – 10% of 100 = 90

⇒ No. of students qualifying in Biology in 2015 = 90 + 10% of 100 = 99

∴ The number of students qualifying in Biology in 2015 is less than that in 2013

Option 3 and option 4 are incorrect since no detail is given regarding how many students qualified the subject in 2013.

Sample GATE Mock Test - Electrical Engineering (EE) - Question 10

DRQP is a small square of side a in the corner of a big square ABCD of side A. What is the ratio of the area of the quadrilateral PBRQ to that of the square ABCD, given A/a = 3?

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 10

Area of triangle PAB = Area of triangle RCB (By symmetry)

∴ Area of Δ PAB = ½ × PA × AB

= ½ × 2A/3 × A = A2/3

Area of ΔRCB = A2/3

Now, Area of ABCD = Area of DRQP + Area of PAB + Area of RCB + Area of PBRQ

A2 = a2 + A2/3 + A2/3 + Area of PBRQ

As, A/a = 3 ⇒ a = A/3

 Area of PBRQ

⇒ Area of PBRQ = 

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 11

If the Laplace transform of y(t) is given by Y(s) = L(y(t))     then y(0) + y'(0) = _____.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 11

Y(s) = L(y(t))

Apply inverse Laplace transform,

Differentiate with respect to ‘t’.

⇒ y(0) + y'(0) = 1

Sample GATE Mock Test - Electrical Engineering (EE) - Question 12

Let f(z) = (x2 + y2) + i2xy and g(z) = 2xy + i(y2 – x2) for z = x + iy ϵ C. Then, in the complex plane C.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 12

Given that f(z) = (x2 + y2) + i 2xy

g(z) = 2xy + i (y2 – x2)

To check analyticity of a function, we need to check CR equations.

ux = vy, uy = - vx

f(z) = (x2 + y2) + i 2xy

u = x2 + y2, v = 2xy

u= 2x

uy = 2y

vx = 2y

vy = 2x

ux = vy but uy ≠ -vx

Hence, f(z) is not analytic

g(z) = 2xy + i (y2 – x2)

u = 2xy, v = y– x2

ux = 2y

uy = 2x

vx = -2x

vy = 2y

ux = vy and uy = -vx

Hence g(z) is analytic.

Sample GATE Mock Test - Electrical Engineering (EE) - Question 13

Consider the following statements P and Q:

(P) : IfM =   then M is singular.

(Q): Let S be a diagonalizable matrix. If T is a matrix such that S + 5 T = I, then T is diagonalizable.

Which of the above statements hold TRUE?

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 13

A matrix is said to be singular, if determinant of that matrix is zero.

= 1 (18 – 12) - 1 (9 – 4) + 1 (3 – 2)

= 6 – 5 + 1 = 2 ≠ 0

M is non singular

(Q) A matrix can be diagonalizable when it has distinct eigen values

S is a diagonalizable matrix. Hence, has distinct eigen values.

Let S be a 3 × 3 matrix and the eigen values of s are λ1, λ2, λ3

Given that, S + 5T = I

From the properties of Eigen values,

(a) If λ1 is an eigen value of matrix A, then -λ1 ­will be on eigen value of matrix -A.

(b) If λ1 is an eigen value of matrix A, then (λ1 + 1) will be an eigen value of matrix (A + I)

(c) If λ is an eigen value of matrix A, then  will be an eigen value of matrix  where K is a scalar.

From the above properties, eigen values of T are,


As λ1, λ2, λ3 are distinct values, λ′1,λ′2,λ′3 will be distinct.

Hence, matrix T is diagonalizable

So, only Q is true.

Sample GATE Mock Test - Electrical Engineering (EE) - Question 14

Consider the differential equation

If x = 0 at t = 0 and x = 1 at t = 1, the value of x at t = 2 is 

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 14

 = 0

A.E is, (D2 – 3D + 2) = 0

⇒ (D – 1) (D – 2) = 0

⇒ D = 1, 2

x(t) = C1et + C2e2t

x = 0 at t = 0

⇒ 0 = C1 + C2 ⇒ C1 – C2

x = 1 at t = 1

⇒ 1 = C1e1 + C2e2

⇒ 1 = C1e1 + (-C1) e2

⇒ C1(e1 – e2) = 1

x(2) = C1e2 + C2e4

= e + e2

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 15

If for two vectors  and , sum   is perpendicular to the difference . The ratio of their magnitude is


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 15

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 16

A 400 V, 180° conduction mode 3 ϕ bridge inverter has star connected resistive load the RMS value of load current is 15 A. Find the value of per phase resistive load _______ Ω.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 16

R.M.S value of phase voltage

Sample GATE Mock Test - Electrical Engineering (EE) - Question 17

In the circuit I is DC current and capacitors are very large. Using small signal model which of the following is correct?

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 17

In DC equivalent circuit capacitor is open

A small signal equivalent circuit is shown in fig.

Put rd in equation (2)

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 18

how many clock pulses it will take to come back to same state If the initial state is D2D1D0 = 001


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 18

The sequence is 000 to 001 to 011 to 111 to 110 to 100 and back to 001, etc. 

Sample GATE Mock Test - Electrical Engineering (EE) - Question 19

A 3 ϕ power system has ZBUS matrix   if an impedance j0.8 is connected b/w bus-2 and ground. The modified value of Z22 is –

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 19

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 20

 

The Nyquist sampling rate for the signal   is ______ KHz


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 20

Multiplication in the domain become convolution in frequency domain therefore individual.

Frequency will add up

f1 = 400 Hz, f2 = 500 Hz

Sampling frequency fs = 2 fnet = 2 × 900 = 1.8 KHz

Sample GATE Mock Test - Electrical Engineering (EE) - Question 21

 

Three current carry conductors are shown in the figure. the value of  around the closed curve is

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 21

Ampere law: 

for isotropic medium

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 22

The open loop transfer function of unity negative feedback control system is given by​  The gain margin of the system is _______ dB


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 22

Phase crossover frequency is given as 

G.M = 20 log 164 = 44.3 db

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 23

The Lissajous figure obtained on the CRO show in the figure. Find the phase difference between the two waves applied.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 23

It can be observed from the Lissajous figure

y1 = 9 units, y2 = 10 units

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 24

An UJT shown in figure has the following parameters η = 0.67, VD = 0.7 V, IV = 3 mA, VV = 1 V, IP = 12 μA, VBB = 20 V. Find the value of VEE so as turn an UJT if RE = 1 KΩ

 

 


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 24

VEE = η VBB + VD + IP . RE

= 0.67 × 20 + 0.7 + 12 × 10-6 × 1 × 103

= 13.4 + .7 + 0.012

= 14.112

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 25

A 10 μF capacitor is initially charged with 1000 μC. At t = 0, the switch K is closed find the voltage drop across the resistor after 100 μs _______V.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 25

Here, 3 capacitor are connected in parallel.

∴ Total capacitance = C1 + C2 + C3 = 4 + 4 + 2 = 10 μF

Now find time constant of the circuit

T = RCeq = 20 × 5 = 100 μsec

The initial voltage across the

With closing of k, the capacitor co will start discharging 

Sample GATE Mock Test - Electrical Engineering (EE) - Question 26

Match List – I with List – II and select the correct answer

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 26

Sample GATE Mock Test - Electrical Engineering (EE) - Question 27

A multimer having sensitivity of 2000 Ω/v is used for the measurement of voltage across a circuit having an output resistance 5 kΩ. The open circuit voltage of the circuit is 6 V. Find the percentage error when it is set to its 10 V scale.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 27

Input resistance of voltmeter ZL = 2000 × 10 = 20 kΩ

Output resistance of circuit Z0 = 5 kΩ

Open circuit voltage of circuit under measurement E0 = 6 V

Reading of voltmeter

∴ Percentage error in voltage reading

- 20%

Or 20% low

Sample GATE Mock Test - Electrical Engineering (EE) - Question 28

For power transformer of larger ratings, the tapping’s are located in the middle portion of the winding to

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 28

For power transformer of large rating the tapping are located in the middle portion of the winding to reduce the mechanical force affecting the winding during short-circuits.

Sample GATE Mock Test - Electrical Engineering (EE) - Question 29

The number of forward path and the number of non-toughing loop pair for the signal flow graph given in the figure below are respectively

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 29

P1 = 1, 3, 5, 8, 10

P= 12, 8, 9

P3 = 12, 8, 10

P4 = 11, 9

P5 = 11, 10

P6 = 1, 3, 5, 8, 9

Non touching loop

L1 = 2, 7

L2 = 2, (5, 6)

L3 = (3, 4), 7

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 30

 

A discrete time signal is given as.The energy of the signal is


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 30

Signal energy

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 31

The number of chords in the graph of the given circuit will be


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 31

No. of branches b = 6

nodes = 4

chords = b – (n – 1) = 3

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 32

In the conductor shown below if diameter of each conductor is 4 cm then self GMD is


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 32

Self G.M.D of bundle of 4 conductor

= 1.09 (r's3)1/4

S = Distance b/w 2 conductor

G.M.D = 1.09 (0.7788 × 2 × 10-2 × 23)1/4

G.M.D = 0.6496 m

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 33

Calculate the magnitude of line current in the circuit shown in figure.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 33

Impedance/ph = 1 + 1j + 7 + 5j = (8 + 6j)

Sample GATE Mock Test - Electrical Engineering (EE) - Question 34

An absolutely integrable signal x(t) is known to have Laplace transform with only one pole at s = 4 then x(t) is

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 34

Since the pole is at s = 4 the function is absolutely integrable thus it must contain jω axis

Hence the signal is left sided

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 35

A 3 phase full converter delivers a ripple free load current of 10 A with firing angle delay of 30° the input voltage is 3 phase, 400 V, 50 Hz. Find the R.M.S value of source current _____ in A.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 35

R.M.S value of source current

Sample GATE Mock Test - Electrical Engineering (EE) - Question 36

The meter constant of 5 A, 220 V, dc watthour meter is 3275 revolution per kwh. Calculate the speed of the disc at full load. In a test at half load the meter takes 59.5 sec to complete 30 revolutions. Calculate the error of the metre.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 36

Energy consumed in one miute

= 0.01833 kwh

∴ Revolution in one minute = E × K = 0.01833 × 3275 = 60.04 rpm

∴ Speed of disc = 60.04 RPM = 1 r.p.s

At half load 

= 2.5A,t = 59.5sec

Et = VIcosϕ×t = 2220×2.5×1×59.5 = 0.00909027kwh 

N = 30 revolutions

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 37

Circuit shown in figure. Assume L to be large enough to ensure linear growth and decay of the current through it and have continuous current find the max value of load current. (Vs = 100 V, Vo = 50 V, L = 2mH, F = 20 kHz, R = 10 Ω)


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 37

 

The function of capacitor C across-load resistance R is to make the output voltage continuous.

Ripple current 

= 5.3125 A

Sample GATE Mock Test - Electrical Engineering (EE) - Question 38

The following circuit shown in figure. the input voltage is a sinusoidal at 50 Hz with an R.M.S value of 20 V. Find the current is from the source is

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 38

= 1 + j 100 π × 20 × 103 × 20 × 10-6 = 1 + j40 π

V0 = Vs (1 + j40 π) = 20 (1 + j40 π)

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 39

Consider the system shown below'

The controllability matrix for this system is    then the sum of a, b and c is – 


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 39

= -x1 + x2

2 = -x2 + 2u

= -2x3 + 2u

 

controllability QC = [B AB A2B]

∴ A + B + C = -4 – 2 + 8 = 2

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 40

A single line diagram of 3ϕ system shown in figure. Find the short circuit current flow into 3ϕ circuit at F, is ____ KA

 


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 40

Let base MVA = 50 MVA

% reactance of alternator A at the base mVA

XA and XB is connected in parallel.

XAB = (40 × 50)/90 =22.22 

Now XAB is connected parallel with XC

∴ PU short circuit current 

= 8.4996

ISC actual = IPU × IBase

= 8.4996 × 1443.38

= 12268.15 A

= 12.268 KV

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 41

A motor running at 1500 rpm has hysteresis and eddy current loss of 180 W and 70 W respectively. If flux remain constant, the total loss will be one one-fourth at _____ rpm.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 41

Speed α F

Hysteresis loss α N

⇒ Wh = K1N

Eddy current loss α N2

⇒ We = K2N2

180 = K1 × 1500

The speed N at which total loss become 1/4

62.5 = 0.12 N + (0.03 × 10-3)N2

N ≃ 467 R.P.M

Sample GATE Mock Test - Electrical Engineering (EE) - Question 42

A 50 Hz, synchronous generator capable of supplying 400 MW of power is connected to a large power system and is delivering 80 MW when a three phase fault occurs at its terminals find the time in which the fault must be cleared if the maximum power angle is to be 85°. Assume H = 8 MJ/MVA on 100 MVA base.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 42

Pi = Pmax sin δ0

δ0 = 11.54° = 0.2 radian

and δ1 = 85° = 1.48 radian

cos δc = cos δ1 + (δ1 – δ0) sin δ0

cos δc = cos (85) + (1.48 – 0.2) sin 11.54

δc = 1.22 radian

= 36.046 ms

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 43

A 50 Hz, 3ϕ core type star-delta transformer has a line valtage ratio of 11,000/440 volts. The cross-section of the core is square with a circumscribing circle of 0.5 m diameter. If maximum flux density of 1.30 ωb/m2 then find the number of turns per phase on high voltage windings. Assume insulation occupies 8% of the total core area.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 43

We known that diameter of the circumscribing is same as diagonal of the square.

Area of square core = l × b

∴ Net cross section area = 0.125 × 0.92 = 0.115

E.M.F/Turn = 4.44 fBA = 4.44 × 50 × 1.3 × 0.115 = 33.19

Phase turn ratio =​ 

∴ Number of turn per phase on high voltage side =

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 44

After how many clock pulses counter locks up in the 1010 and 0101 states. If the initial state is Q3Q2Q1Q= 0000


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 44

Tog = toggle, NC = no change

The counter locks up in the 1010 and 0101 states, alternating between them.  

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 45

A single-phase full bridge inverter Is fed from a dc source such that fundamental. Component of output voltage is 230 V. Find the R.M.S value of diode current. For the following loads: R = 2 Ω, XL= 8 Ω, XC = 6 Ω


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 45

 

= 81.317 A

R.M.S value of diode current

= 0.1507 × √2 × 81.317 = 17.328 A

Sample GATE Mock Test - Electrical Engineering (EE) - Question 46

Two generators rated 250 MW and 400 MW are operating in parallel. The droop characteristics of the governors are 4% and 6% respectively.if a load of 450 MW be share between them. What will be the system frequency? Assume nominal system frequency is 60 Hz and no governing action.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 46

Let, Load on generator 1 = x MW

Load on generator 2 = (450 – x) MW

Reduction in frequency = Δf

Now,

From equations (i) & (ii), we get

∴ x = 217.81 MW (load on generator 1)

450 – x = 232.18 MW (load on generator 2)

and Δf = 2.09Hz

∴ System frequency = (60 – 2.09) = 57.909 Hz.

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 47

The synchronous speed of induction motor is 900 RPM. Under blocked-rotor condition the input power to the motor is 45 KW at 193.6 A. The stator resistance per phase is 0.2 Ω and the transformation ratio is a = 2. Calculate the motor starting torque (in N-m)


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 47

R2 = 0.05 Ω

Referred to the rotor resistance per phase is

R′= a2R= 0.2Ω

∴      Starting torque 

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 48

The is transistor shown below is biased for constant base current if  the value RB is ______ KΩ.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 48

VCC = ICQRC + VCEQ

If leakage current neglected

Transistor is Si device VBEQ = 0.7V

Sample GATE Mock Test - Electrical Engineering (EE) - Question 49

Find the Norton equivalent circuit for the following circuit.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 49

Apply KVL at node Va

13 Vb = 3 Va + 10

Now find ISC = IN

 

Apply KVL at Node Va

13 V = 3 Va + 10

Now find ISC = IN

 

V0 = 40 i2, i3 = 0.5 V0 = 20 i2

5 + 10 (i1 – 20 i2) = 0

-5 + 40 i2 + 20 (i2 – 20 i2) = 0

Sample GATE Mock Test - Electrical Engineering (EE) - Question 50

Find the Fourier series of rectangular pulse which is represented by for one time period

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 50

The waveform is given by

 

Again,   

[it may be observed here that awill be zero for even values of n].

∴ The Fourier series is

Sample GATE Mock Test - Electrical Engineering (EE) - Question 51

The admittance locus of the circuit shown in figure is

 

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 51

0 < R1 < ∞

y = y1 at R1 = ∞

Sample GATE Mock Test - Electrical Engineering (EE) - Question 52

A bridge shown in figure is at balance. The parameter r1 and c1 are

 

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 52

At balance

Z1Z4 = Z2Z3

Equating real and imaginary terms

Sample GATE Mock Test - Electrical Engineering (EE) - Question 53

A synchronous motor has 1000 KW, 3ϕ, Y connected 3.3 KV, 28 poles, 50 Hz. synchronous reactance of 4.20 Ω/ph. Its field excitation is adjusted to result in unity P.F operation at rated load. Compute the maximum power that the motor can deliver with its excitation remaining constant at this value.

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 53

Excitation emf, E = 1905∠0° - j175 ∠0 × 4.20

= 1905 – j735 = 2041.88 V

= 2778.41 KW

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 54

The power input in a three phase three wire delta connected balanced load is measured by the two wattmeter method. The reading of wattmeter A is 5000 W and wattmeter B is –1000 W (with reversal of connection) if the voltage of the circuit is 440 V, 50 Hz, what is the value of capacitance (in μF) connected in delta at the source to cause the whole of the power measured by the wattmeter A only.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 54

= 0.3592 lag

Now capacitance is connected at the source which does not affect the active power consumption

W = 5000 – 1000 = 4000 W

Before connecting capacitor

For delta Vph = VL = 440 V and IL = √3 Iph

i.e. Iph = 8.4362 A

∴ Rph = Zph cos ϕ = 18.735 Ω

Xph = Zph sin ϕ = 48.67 Ω (inductive)

After connecting capacitor

WA = W = 4000 W, WB = 0

WB = VLIL cos (30 + ϕ) = 0

30 + ϕ = 90°

ϕ = 60°

cos ϕ = cos 60° = 0.5 lagging

Due to pure capacitor, Rph remain same

Rph = 18.735 Ω

Xcph = Xph – X'ph = 48.67 – 32.45

= 16.2246 Ω

= 196.1896 μF

Sample GATE Mock Test - Electrical Engineering (EE) - Question 55

The system characteristics equation given below represents

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 55

Using Laplace transform

Where a > 1

Here Zero is dominant over pole

∴ It is lead network and act as High pass filter (or network)

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 56

A single phase full converter bridge is connected to RLE load the source voltage is 230 V, 50 Hz. The average load current of 8 A is constant over the working range for R = 0.4, L = 2 mH and E = -120 V. Find the input P.F


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 56

For E = -120 V, the full converter is operating as a line commutated inverter therefore power from dc source to AC source

Vr Ior cos ϕ = EI0 – I2or R

Ior = I0 = 8 A

= 0.508 Lag

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 57

The maximum efficiency of 550 KVA, 3300/430 V, 50 Hz 1 ϕ transformer is 98% and occurs at ¾ full load, unity power factor. If the impedance is 10%, calculate the regulation (in %) at full load at 0.8 P.F lagging.


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 57

Output at ¾ full load and unit power factor =

Total loss = 420.92 – 412.5 = 8.42 KW

Loss is equally divided b/w iron and copper loss

Copper loss = 7.48299

Percentage resistance = 

Percentage impedance = 10%

Percentage reactance = 

∴ Percentage regulation = 

= 1.74 × 0.8 + 9.84 × 0.6

= 1.392 + 5.91 = 7.30%

Sample GATE Mock Test - Electrical Engineering (EE) - Question 58

The full range of audibility in audio frequency oscillator is 

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 58

The full range of audibility in audio frequency oscillator is a. 0 to 20 Hz b. 20 Hz to 2 kHz

Sample GATE Mock Test - Electrical Engineering (EE) - Question 59

The matrix representation of the relevant two port parameters that describe the circuit

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 59

Writing KVL for input loop

V1 = 5I1 + 3I2

Writing KVL for outer loop

I2(1 + 2j – 2j) + 3I2

V2 = 4I2

∴ h parameter = ​

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 60

A shunt motor 500 V, draw 40 A while supplying rated load at speed of 120 rad/sec. The armature resistance is 2 Ω and field winding resistance 250 Ω, external resistance is inserted in series with the armature circuit so that armature current does not exceed 160% of its rated value when the motor is plugged. The breaking torque at the instant of plugging is _____ N-m


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 60

Field current 

 

I = 40 – 2 = 38 A

Ea = 500 – 38 × 2 = 424 V

At the instant of plugging

Plugging torque =

= 214.83 N-m

Load torque​ 

∴ Braking torque = 214.83 + 134.27

= 349.1 N-m

Sample GATE Mock Test - Electrical Engineering (EE) - Question 61

Consider the matrix equation

The condition for existence of a non-trivial solution, and the corresponding normalised solution (up to a sign) is

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 61

For non-trivial solution, the rank of the matrix should be less than the number of variables. i.e. r < n.

For this, |A| = 0

⇒ (4c – 3b) – (2c – 6) + (b – 4) = 0

⇒ 4c – 3b – 2c + 6 + b – 4 = 0

⇒ 2c – 2b + 2 = 0

⇒ b = c + 1

The vectors x1, x2 ….. xare said to be linearly dependent, if there exist numbers λ1, λ2 ……. λn, not all zero such that

λ1x1 + λ2x2 + ….. + λnxn = 0

Here,

λ1 + λ2 + λ= 0      ----(1)

λ1 + 2λ2 + 3λ= 0      ----(2)

1 + bλ2 + 2cλ= 0

1 + (c + 1) λ2 + 2cλ= 0      ----(3)

From (1) & (3):

λ2 = -2λ3

λ1 = λ3 = λ

λ2 = -2λ

so corresponding normalised solution:

*Answer can only contain numeric values
Sample GATE Mock Test - Electrical Engineering (EE) - Question 62

Two points are chosen randomly on a line 9 cm long. Determine the probability that the distance between them is less than 3 cm


Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 62

x: The distance of first point from the start of the line

y: distance of second point from the start of the line segment

x,y ϵ[0,9]

So sample space is Area of region bounded by

 x ≥ 0, y ≥ 0, x ≤ 9, y ≤ 9

This is square of side 9

Area = 81 cm2

The region of our interest is

|x-y| < 3

0 ≤ x ≤ 9

0 ≤ y ≤ 9

Area of shaded region = 2 (area of triangle) + area of rectangle

 = 45

Probability = 45/81 =0.55

Sample GATE Mock Test - Electrical Engineering (EE) - Question 63

The area bounded by the curve y =x (3 – x)2, the x-axis and the ordinates of the maximum and minimum points of the curve is

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 63

y = x(3 – x)2

−x.2(3−x) + (3−x)= 3(x2−4x+3) = 0

(x - 3) (x – 1) = 0 ⇒ x = 3, x = 1

Sample GATE Mock Test - Electrical Engineering (EE) - Question 64

Given N > 0, the iterative equation for finding using Newton-Raphson method is:

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 64

Sample GATE Mock Test - Electrical Engineering (EE) - Question 65

If y = 3e2x + e-2x - αx is the solution of the initial value problem

 Where, α, β ϵ R, then

Detailed Solution for Sample GATE Mock Test - Electrical Engineering (EE) - Question 65

y = 3e2x + e-2x – αx

Given that, 

⇒ 1 = 6 – 2 - α

⇒ α = 3

Complementary solution

(D2 + β) = 0      ----(1)

The given is, y = 3e2x + e-2x – αx

It indicates, 2 and -2 are the roots of auxiliary equations.

⇒ (D + 2) (D – 2) = 0

⇒ D2 – 4 = 0

By comparing this equation with equation (1)

β = -4

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