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GATE Mock Test: Electrical Engineering (EE)- 1


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65 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | GATE Mock Test: Electrical Engineering (EE)- 1

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

Choose the most approximate word from the options given below to complete the following sentence.

If I had known that you were coming, I _______ you at the airport

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

The sentence talks about a past incident and a condition has been mentioned. Thus option 2 fits here correctly. 'Would have' is the correct tense to be used here. Option 1 is illogical. 'Would' must be used here and not 'will' as it is in the past tense. Past perfect tense is incorrect here.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 2

Choose the most approximate word from the options given below to complete the following sentence.

I believe in the _____ of positive thinking thus always recommend these books to my clients.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 2

The word that fits here is a noun thus options 3 and 4 are eliminated. The form should be singular as for abstract nouns we do not use the plural form. Option 1 is thus the correct answer.

The word powerful is an adjective and 'empowering' is a verb.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 3

Consider a circle of radius r. Fit the largest possible square inside it and the largest possible circle inside the square. What is the radius of the innermost circle?

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 3

The radius of outer circle = r

∴ Diagonal of square = 2r

Side of square (a) = 

Now, the radius of the inner circle = 

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 4

A bird files along the three sides of a field in the shape of an equilateral triangle at speeds of 3, 6, 8 km/hr respectively. The average speed of the bird is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 4

Average speed = Total distance/Total time

Total distance = a + a + a = 3a

Total time = 

 

Average speed  = 

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 5

Choose the most appropriate words from the options given below to fill in the blanks.

She was ______ to travel abroad and _____ in the field of commerce as per her wishes.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 5

The sentence mentions a person doing something as per her wishes thus the first word must be a positive reaction. 'Restive' which means 'restless' and 'jinxed' which means 'cursed' are incorrect here. Between options 1 and 3,the word 'elated' which means 'too happy' and the word 'venture' which means 'undertake a risky or daring journey or course of action' fit here correctly. The word 'cease' means 'stop' and does not convey a proper meaning. Thus option 3 is the correct answer.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 6

In a 1600 m race around a circular track of length 400 m, the faster runner and the slowest runner meet at the end of the sixth minute, for the first time after the start of the race. All the runners maintain uniform speed throughout the race. If the faster runner runs at twice the speed of the slowest runner. Find the time taken by the faster runner to finish the race.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 6

As, the faster runner is twice as fast as the slowest runner, the faster runner would have complete two rounds by the time the slowest runner completes one round.

Their first meeting takes place after the fastest runner takes 6 min to complete two rounds and the slowest runner completes one round.

Time taken to complete one round by fastest runner = 6/2 = 3 minutes

Rounds needed to complete the race = 1600/400 = 4

Time taken to complete the race = 4 × 3 = 12 min

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 7

Select the pair that best expresses a relationship similar to that expressed in the pair:

Horse: Foal

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 7

The young one of a horse is called a foal. Similarly, the young one of a goose is known as gosling. Thus, option 2 contains the pair that best expresses a relationship similar to that expressed in the given pair. In all the other pairs, the first word expresses the male form of the latter.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 8

Read the following passage and find out the inference stated through passage.

Juvenile delinquency is also termed as Teenage Crime. Basically, juvenile delinquency refers to the crimes committed by minors. These crimes are committed by teenagers without any prior knowledge of how it affects the society. These kind of crimes are committed when children do not know much about outside world.

Which of the following Inferences is correct with respect to above passage? 

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 8

Since teenagers here have tender age and commit crimes unknown to its consequences, clearly they should not be treated as any criminal. They need to be taught so that they can make difference between right and wrong and do not commit these crimes in future. Hence inference in option 1 is correct.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 9

If n and y are positive integers and 450 y = n³, which of the following must be an integer? 

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 9

450 y = n³ implies that 450 y is the cube of an integer. 

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor.

8 is the cube of an integer because 8 = 2³ = 2*2*2.

Thus, when we prime-factorize 450 y, we need to get at least 3 of every prime factor. 

Here's the prime factorization of 450 y:

450 y = 2 * 3² * 5² * y 

Since 450 provides only one 2, two 3's, and two 5's, and we need at least 3 of every prime factor, the missing prime factors must be provided by y. Thus, y must provide at at least two more 2's, one more 3, and one more 5. 

Smallest possible case: 

y = 2² * 3 * 5

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 10

The graph shows cumulative frequency % of research scholars and the number of papers published by them. Which of the following statements is true?

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 10

In this problem, cumulative frequency is given, so converting into frequency

From the table, it is clear that option b, i.e.

60% of the scholars published at least 2 papers is correct.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 11

For which value of x, the following matrix A is a singular matrix.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 11

A matrix is singular if and only if its determinant is zero.

 

x(3+1)−2(x2+1) = 0−2x2+4x−2 = 0

x2−2x+1 = 0

(x−1)2 = 0

x = 1

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 12

Find the laplace transform of t2sin (2t).

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 12

According to formula, L(tf(t)) =​

Here, f(t) = sin (2t) ⇒ F(s) =

∴ L(t2sin (2t)) =  = 

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 13

Consider the following function:

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 13

ii) (4 - x)x=2 = 2

Hence at x = 2, f(x) is continuous

∴ f(x) is not differentiable.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 14

 value of b = ?

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 14

The given limit is in 0/0 form, So

Applying L hospitals rule

⇒ log b - 1 = -1 - log b

⇒ 2 log b = 0

⇒ b = 1

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 15

The integrating factor of equation y log y dx + (x – log y) dy = 0 is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 15

y log y dx + (x – log y) dy = 0

It is a Leibnitz’s equation in x.

Integrating factor =

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 16

Relationship between input x(t) and output y(t) of a system is given as

The transfer function of this system is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 16

By applying Laplace transform,

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 17

For the two square inputs in figure, the PMMC meter will read, maximum when the phase difference between them is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 17

PMMC meter reads average value if the output of the XOR gate.

The output of XOR gate will be high when both the inputs are different.

For a phase difference of T/2 both X and Y will be different as shown below.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 18
  1. For the circuit shown the hybrid parameter matrix [h] is

 

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 18

Equations of hybrid parameter are,

When, V2 = 0

V1 = -2I2

By applying KCL at 0,

⇒ h21 = -2

V1 = -2I2 = -2 (-2I1) = 4I1

⇒ h11 = 4

When, I1 = 0

⇒ h22 = 0.5

By applying KVL,

⇒ -V1 + 2V2 – I2 = 0

⇒ -V1 + 2V2 – 0.5 V2 = 0

⇒ h12 = 1.5

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 19

H(z) is discrete rational transfer function. To ensure that both H(z) and its inverse are stable its

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 19

For H(z) to be stable the poles of H(z) must be inside the unit circle.

For the inverse of H(z) to be stable the poles inverse of H(z) must be inside the unit circle.

The poles of inverse of H(z) are the zeros of H(z)

Hence, both poles and zeros of H(z) must be inside the unit circle.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 20

In the dc circuit shown in the below figure, the node voltage V2 at steady state is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 20

By voltage division

By voltage division,

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 21

A DC-DC converter is used to feed the resistive load and its output is always inverted. An alternating square wave voltage is getting produced across in inductor having rms value of 200 V and frequency of 50 Hz. the duty cycle of the switch is______


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 21

Given circuit is Buck Boost converter

Voltage across inductor would be, V= 200 V

Vo = 200 V

⇒ δ = 1 – δ

⇒ δ = 0.5

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 22

If the multiplexer is controlled such that the channels one sequenced every 5 μs as 1, 2, 1, 3, 1, 4, 1, 2, 1, 3, 1, 4, 1, … , the input connected to channel 1 will be sampled at the rate of

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 22

Sampling period, Ts = 2 × 5 = 10 μs

Sampling rate, 100k samples/sec

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 23

What percentage of current IDSS is the drain current for a JFET, if the gate to source voltage is 65% of the pinch-off voltage.


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 23

VGS = 0.65 Vp

The drain current of JFET is given as

 

 (1−0.65)2 = 0.1225 = 12.25% 

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 24

A uniform line charge of infinite length with PL = 15 nC/m lies along the z-axis.

The electric field at (5, 12, 20) will be – v/m


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 24

Electric field at a point P(x, y, z) is given as

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 25

A three phase, 50 Hz transmission line of length 120 km has a capacitance of  It is represented as a π – model. The shunt admittance at each end of the transmission line will be

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 25

In π model, the shunt admittance at each end of the line is 

= j300×10−6  

= 300×10−6 ∠90∘ mho

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 26

The fundamental period of the sequence x[n] = 3 sin (1.3 πn + 0.5 π) + 5 sin (1.2 πn) is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 26

x[n] = 3 sin (1.3 πn + 0.5 π) + 5 sin (1.2 π n)

Time period = LCM (20, 5) = 20

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 27

What should be the input voltage to turn off the SCR in the circuit shown in figure below when holding current of the SCR is 4 mA and load resistance is 4 kΩ.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 27

To turn off the SCR anode current should go less than holding current

⇒ V = 4 × 10-3 × 4 × 103

⇒ V = 16 V

Form the options, V = 15 V is correct. for the remaining options, holding current is less than anode current.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 28

 

The inverse Fourier transform of a function given 

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 28

Fourier transform of

⇒ g(t) = δ(t) + 2.e-3|t|

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 29

Figure shows the polar plot of a system. The transfer function of the system is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 29

From the given polar plot,

From the options, 5(1 + 0.1s) satisfies this condition.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 30

In the circuit of figure A, B, C are the inputs and P, Q are the two outputs. The circuit is a

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 30

From the given circuit diagram,

P = A⊕B⊕C

Q = AB + BC + CA

The above two expressions presents sum (P) of a full adder and carry (Q) of a full adder

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 31

A 6 pole slip ring induction motor operating on 50 Hz supply is driven by a variable speed prime mover as a frequency changer. If it is operated at 1500 rpm is opposite direction then the supply frequency generated is – (in Hz)


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 31

General equation for the frequency of an induction frequency changer is

+ve for opposite direction of rotation

-ve for same direction of rotation

1000 rpm

 = 125 Hz

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 32

The two-wattmeter method is sued to measure power in 3-phase, 3-wire balanced inductive circuit. The line voltage and line current are 400 V and 10 A respectively. If the load power factor is 0.866 lagging, then readings of the two wattmeter’s are

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 32

Given that, line voltage (VL) = 400 V

Line current (IL) = 10 A

Cos ϕ = 0.866

⇒ ϕ = cos-1(0.866) = 30°

W1 = VLIL cos (30 – ϕ)

= 400 × 10 × cos (30 – 10) = 4000 W

W2 = VLIL cos (30 + ϕ)

= 400 × 10 × cos (30 + 30) = 2000 W

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 33

A three phase squirrel cage induction motor has maximum torque equal to twice the full load torque. The ratio of motor starting torque to its full load torque, if it is started by start-delta starter given slip at maximum torque is 0.2 is –


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 33

When using, Direct on line starter,

Starting torque of star delta starter is one third of that obtained by DOL starter.

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 34

The rated making current of a circuit breaker is given as 90 kA (peak). The braking capacity of circuit breaker if it is rated at 22 kV


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 34

Rated symmetrical braking current = 

Braking capacity =  ×22×35.29 = 1344.88MVA

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 35

Consider the schmit trigger circuit shown below, A triangular wave which goes from -15 V to 15 V is applied to the inverting input of op-amp. Assume that the output of the op-amp swing from +20V to -20V the voltage at the non-inverting input switch between.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 35

Apply KCL at the non-inverting input end

V – 20 + V + 20 + V – V0 = 0

3V = V0

Since V0 swings from -20 to +20

∴ V switch between 

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 36

In the circuit shown below the average power consumed by the 1 Ω resistor is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 36

By applying superposition theorem,

= 10 sin (1000 t – 45°)

I = I1 + I2

Power dissipated 

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 37

An unknown voltage source Eu (with negligible internal resistance) is connected to a potentiometer circuit as shown in the following figure. If the galvanometer current is 10 μA with the direction as indicated, then value of Eu is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 37

By applying KVL,

-1.6 + 100 Iw + 500 Iw + 1000 (Iw + Ig) = 0

⇒ 1600 Iw + 1000 Ig = 1.6

⇒ 1600 Iw + 1000 (10 × 10-6) = 1.6

⇒ 1600 Iw + 0.01 = 1.6

By applying KVL,

-Eu + Ig(100) + (Iw + Ig) (1000) = 0

⇒ Eu = 1000 Iw + 1100 Ig

= 0.99375 + 0.011 = 1.00475 V

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 38

A 100 MVA synchronous generator operates on full load at a frequency of 50 Hz. Inertia constant is 8 MJ/MVA. The load is suddenly reduced 100 MW. Due to time lag in governor system, the steam valve begins to close after 0.4 seconds. The change in frequency that occurs in this time is_____


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 38

The rating of the machine (S) = 100 MVA

Inertia constant = H = 8 MJ/MVA

Kinetic energy stored in the rotating parts of the generator = SH = 800 MJ

Energy transferred in 0.4 sec = 100 × 0.4 = 40 MJ

 

⇒ f2 = 51.23Hz

Change in frequency = 1.23 Hz

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 39

In the circuit shown below the steady-state in reached with the switch K open. Subsequently the switch is closed at time t = 0.

At time t = 0+, current I, is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 39

As the steady state is reached with the switch open, at t= 0- the capacitor will be open and inductor will be short.

At = t = 0-

I(0−1) = 0A

V= (0−1) = 10V

 

At t = 0+

At t = 0+

By applying KVL

-5 + I + 2I + 10 = 0

 

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 40

Two single phase transformers rated 1000 kVA and 500 kVA have per unit leakage impedance of (0.01 + j0.04) pu and (0.02 + j0.06) pu respectively. The largest kVA load that can be delivered by the parallel combination of these two transformers without loading any transformer is – (in kVA)


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 40

Z1 = (0.01 + j0.04) pu, kvA1 = 1000 kVA

Z2 = (0.02 + j0.06) pu, kvA2 = 500 kVA

Load shared by transformer 2,

Largest kVA load on both transformers,

Largest kVA = 1000 + 326.2 = 1326.2 kVA

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 41

A three phase full converter bridge is connected to supply voltage of 240 V per phase and a frequency of 50 Hz. The source inductance is 5 mH and load current an dc side is constant at 18 A at 30° firing angle. The percentage voltage regulation due to source inductance is _____


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 41

Voltage at no load = 

Voltage regulation due to source inductance

= 5.55 %

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 42

The signal x(t) and h(t) shown in the figures are convolved to yield y(t)

 

Which one of the following figures represent the output y(t)?

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 42

x(t) = δ(t + 1) – δ(t – 1)

x(t) × h(t) = y(t)

= h(t + 1) – h(t – 1)

 

 

y(t) = h(t + 1) – h(t – 1)

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 43

For the bridge circuit in figure, the voltage are:

V2 = 2 cos (1000 t + 45°) and

Vd = 0

The value of R = 100 Ω. Then Zx will be

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 43

By applying KVL,

-V1 + IR - Vd  = 0

⇒ V1  = IR

By applying KVL,

-V2 + Vd + I Zx  = 0

⇒ I Zx  = V2

= 2cos(1000t+45 )

⇒ Zxcos(1000t) = (1000t+45 )

 

⇒Zx =

= 100 + 100 j

So, Zx is a 100 Ω resistance in series with 100 Ω inductive reactance

ωL = 100

So, Zx is a 100 Ω resistance in series with 100 mH inductor

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 44

A coil having an inductance (L) of 10 mH and resistance R is connected in series with an ideal 100 μF capacitor (C). When excited by a voltage source of value of (1000t) V, the series RLC circuit draws 20 W of power. The value of the coil resistance R is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 44

Z = R + j10 – j10 = R

Power drawn (P) = 20

⇒ R = 5Ω

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 45

Consider the given circuit and the cut in voltage of each diode is 0.6 V. (take a value of all resistance = 20K)

What is the value of ID, ID2, ID3 respectively?

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 45

Assume that three diodes are OFF

As seen by biasing of diode, for each diode the p terminal of diode is greater than n terminal so, our assumption is incorrect.

Step-2

Assume that 3 diode are ON. So, equivalent circuit is

VC = 0 – 0.7 = -0.7 V

VB = -0.7 + 0.7 = 0 V

VA = VB – 0.7 = -0.7 V

I1 = ID1 + ID2

1 mA = 1.97 + ID2

ID2 = -0.97 mA < 0

I2 = ID2 + ID3

0.97 = - 0.97 + ID3

ID3 = 1.94 mA > 0

From the result ID3 and ID1 are greater then zero and ID2 which is less then zero. So, D3 will be aff.

Step-3

Apply KVL in loop 1

20 – 20 KI1 – 0.7 – 20 ID1 – (-4) = 0

I1 = ID1

ID2 = 0

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 46

The state diagram corresponding to the following circuit is

 

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 46

Hence, the below state diagram represents the given circuit.

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 47

A single phase full bridge inverter controls the power in a resistance load. The nominal value of input dc voltage is Vdc = 230 V and a uniform pulse width modulation with five pulses per half cycle is used for the required control the width of each pulse is 30°, if the dc supply increased by 20%, then the pulse width (in degrees) to maintain the same load power is _____


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 47

Given that, input dc voltage Vdc = 230 V 

After increase in the dc supply

Vdc = 1.2 × 230 = 276 V

to maintain the same load power.

⇒ d = 20.83°

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 48

Given   then the total current crossing the plate z = 0.2 in the az direction, for ρ varying as 0 < ρ < 16 is ____ in (A)


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 48

t = ρ2+4  

dt = 2ρ dρ

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 49

A three phase Y-Δ transformer is rated 400 MVA, 220 kV/22 kV. The short circuit impedance of the transformer measured on low voltage side is 0.121 Ω and because of the low value it is considered equal to the leakage reactance. The per unit value of reactance to represent this transformer in a system whose base on the high voltage side of the transformer is 100 MVA, 230 kV is ___ (pu)

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 49

On its own base the pu reactance of the transformer is,

On the chosen base the reactance becomes,

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 50

The polar form of the given signal x(t) is  ​  The value of 2πk2 for the given signal is


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 50

From the given signal,

Time period (T0) = 2π

 

 

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 51

A 230V, 10 kW dc shunt generator has 1000 turns on each pole. At rated speed, a shunt field current of a 2 A produces no load voltage of 230 V, but at rated load a voltage of 230 V is produced by a field current of 3 A. The number of series field turns per pole required for long shunt connection, if it is made possible to maintain the load voltage constant without changing field current is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 51

Total mmf required at rated load

= 3 × 1000 = 3000 AT

Total mmf required at no load

= 2 × 1000 = 2000 AT

The mmf to be supplied by series field winding

= 3000 – 2000 = 1000 AT

The line current at full load,

Armature current = 43.48 + 3 = 46.48 A

Required series filed turns per pole

​turns per pole.

≈ 22 turns per pole.

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 52

A plant with a transfer function  is controlled by a PI controller with KP = 1 and Ki ≥ 0 in a unity feedback configuration. The lowest value of Ki that ensures zero steady state error for a step change in the reference input is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 52

Given transfer function = 

Kp = 1

Characteristic equation is,

1 + G(s) H(s) = 0

For system to be stable kI > 0

And  

Potion error constant 

= ∞

Steady state error is independent of kI. Lowest value of kI is 1/3

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 53

For the circuit in the figure the thyristor voltage after SCR gets self-commutated is _____ (in V)

 

Assume initial voltage across the capacitor is zero.


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 53

Vc(t) = Vs (1 – cos ωot )

the thyristor is self-commutated at ωot = π 

Vc (t) = 2Vs

⇒ -150 + V+ VC = 0

⇒ V= 150 - VC

= 150 – 2Vs = 150 – 300 = -150 V

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 54

 

Which of the following circuits implements the Boolean function F(A,B,C)=∑(1,2,4,6)?

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 54

Given that, F(A, B, C) = ∑ (1, 2, 4, 6)

From the options, selection liens are A, B

Hence, the following circuits implements the given Boolean function

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 55

A 50 kW. 240 shunt motor takes 15A when running light at 1400 rpm. The resistance of the armature and field are 0.25 Ω and 100 Ω respectively. If mechanical losses to be negligible, the efficiency (in%) of the motor when taking 150 A is


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 55

Shunt resistance = 100 Ω

Shunt Current

Armature current (Ia) = I- Ish­

= 15 – 2.4 = 12.6 A

At no load, Eb = 240 – (12.6 × 0.25) = 236.86 V

No load losses = (236.85) (12.6) + (12.6)2 (0.25) + (240) (2.4) = 3600 W

At load,

Ia = 150 – 2.4 = 147.6 A

P losses = (147.6)2 (0.25) + 3600

= 9046.44 W

Input power = 240 × 150 = 36 kW

Efficiency​ 

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 56

Consider two single input single output systems with state and output equations as given below:

System 1:

System 2:

The two systems are connected in cascade by applying the output of system 1 to system 2, that is by making u2(t) = y1(t). Determine the transfer function of the combined system.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 56

Transfer function is given by,

T|F = C(SI – A)-1 B

For the system 1:

For the system 2:

Both systems are connected in cascade,

Overall transfer function =

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 57

A three phase transmission line having conductors each of resistance 1.0 Ω/km is supplying a load of 15 MW at 0.8 power factor lagging. The receiving end voltage is 120 kV. The losses in the transmission line is found to be 15% of load. The distance over which the load supplied is ______ (in km)


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 57

P = 15 MW

Power factor = 0.8 lagging

R = 1 Ω/km

VR = 120 kV

Line losses = 

Line current 

 

Line losses = 3I2R = 2.25×10 6

Length of line = 92.16 km

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 58

For the op-amp circuit shown below, V0 is equal to _____(take β = 100)


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 58

Due to symmetry

V01 = Av(V2−V1)

V01 = −20(5.5−5) = −20×0.5 =−10V 

Due to the virtual ground concept

VA = 1V

Now, apply KVL to the base-emitter circuit of the transistor.

-10 + 20 – 33 × 103 × Ib – 0.7 – 1 = 0

= 25.25 mA

V0 = 1 – 25.25 × 10-3 × 10 = 0.74

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 59

A feedback control system with high gain k, is shown in the figure below:

 

Then the closed loop transfer function is

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 59

Transfer function 

Sensitivity of transfer function for the variation in G(s) is

Given that, k is very high

Hence  will be zero at higher values of k.

Sensitivity of the transfer function for the variation in H(s) is

For higher values of K,  will be -1.

Hence, transfer function is sensitive to perturbations in H(s) but not to perturbations in G(s).

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 60

A three phase 100 kVA, 3000 V star connected alternator has an effective armature resistance is 0.2 Ω. A field current of 40 A produces a short circuit current of 200 A and open circuit emf of 1040 V line to line. The full load regulation of 0.8 leading power factor is – (in %)


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 60

EOC = 1040 V

E = 1701.23 V

% Regulation = 

 =−1.776%

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 61


The column vector (aba) is a simultaneous Eigen vector of  and​

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 61

Given that, X =  is an Eigen vector of A.

⇒ AX = λ1X

 

For all values of a and b,  is an Eigen vector of A.

 

Given that,   is an Eigen vector of B

⇒ BX = λ2X

being an eigen vector of matrix B

a + b = λ2 a      ----(1)

2a = λ2 b      ----(2)

From (1) and (2)

⇒ ab + b2 = 2a2

⇒ 2a2 - ab - b2 = 0

⇒ 2a2 - 2ab + ab - b2 = 0

⇒ 2a (a - b) + b (a - b) = 0

⇒ b = a, b = -2a

For b = a, (or) b = -2a, X is an eigen vector of matrix B.

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 62

 

The value of         is equal to ______.


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 62

Poles are, z = 0, ± π/2, ± 3π/2, …..

The poles inside |z| = 4 are,

Residue of f(z) at 

It is in 0/0 form

From L-hospital’s rule

Residue of f(z) at 

It is in 0/0 form

From L-hospital’s rule

From Cauchy’s residue theorem,

∫f(z)dz = 2πi  (sum of all residues)

*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 63


Let y(x) be the solution of the initial value problem 2xy=xandy(0)=0. Find the value of ________. (Correct up to 2 decimal places).


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 63

Integrating on both sides:

Now given y (0) = 0

Thus,

 C = 0

Now,

ln (1 – 2y) = -x2

1−2y = e−x2​

GATE Mock Test: Electrical Engineering (EE)- 1 - Question 64

2000 cashew nuts are mixed thoroughly in flour. The entire mixture is divided into 1000 equal parts. And each part is used to make 1 biscuit. Assume that no cashews are broken in the process. A biscuit is picked at random. The probability it contents no cashew is _____.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 64

Assuming poisson distribution

Alternate:

  • Probability a particular cashew is in particular biscuit is 
  • Probability a particular cashew is not in a particular biscuit =
  • Probability that no cashew in =
*Answer can only contain numeric values
GATE Mock Test: Electrical Engineering (EE)- 1 - Question 65

Using Newton – Raphson method, find  after first iteration, assuming 2 as the initial approximation. Correct upto 2 decimal places.


Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 1 - Question 65

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