GATE 2023 Electrical Engineering Question Paper

Forenoon Session | General Aptitude Practice

Introduction

Welcome to this detailed practice page based on the GATE 2023 Electrical Engineering (Forenoon Session) question paper. This page has been designed especially for students who want to practice GATE-level questions in an easy, interactive, and student-friendly format. The questions included here cover important areas of the General Aptitude section, including Verbal Ability, Idioms and Phrases, Numerical Ability, Probabil Idioms and Phrases, Numerical Ability, Probability, Statistics, and ity, Statistics, and Logical Interpretation.

GATE General Aptitude questions are important because they test more than simple memorization. They require students to understand the question, identify the correct approach, and reach the answer accurately within a limited amount of time. Some questions may look simple at first, but they can contain small details that require careful reading and logical thinking. Regular practice can therefore help improve both accuracy and speed.

In this practice series, each question will be presented with four answer choices. After attempting a question, you can easily identify the correct answer because the right option will be highlighted in green. The incorrect options will be highlighted in red. This visual system makes it easier to check your performance and understand which option is correct.

To make the practice more useful, the detailed explanation for each question will remain hidden initially. A “Show Solution” button will be provided below the question. You can first solve the question yourself and then click the button whenever you want to check the complete solution. This approach encourages active practice instead of immediately looking at the answer.

The solutions will be explained in a simple and clear manner. For numerical and probability questions, the calculation will be shown step by step. For verbal ability and logical questions, the reasoning behind the correct option will also be explained. The purpose is not only to tell you the correct answer but also to help you understand why that answer is correct.

Use this page as a practice test: read each question carefully, select an option, and try to solve it without checking the solution first. After completing the question, use the answer and solution section to evaluate yourself. This method can help you build confidence and improve your preparation for competitive examinations.

Practice Tip: Try to solve every question on your own before clicking the Show Solution button. Understanding the method is more important than simply remembering the answer.

Question 1 Verbal Ability

Rafi told Mary, “I am thinking of watching a film this weekend.” The following reports the above statement in indirect speech.

Rafi told Mary that he ______ of watching a film this weekend.

Solution

The original sentence is: “I am thinking of watching a film this weekend.”

When the sentence is changed into indirect speech, the reporting verb “told” is in the past tense. Therefore, the present continuous form “am thinking” changes into the past continuous form “was thinking.”

Therefore, the correct sentence is: Rafi told Mary that he was thinking of watching a film this weekend.

Correct Answer: (D) was thinking

Question 2 Verbal Ability | Analogy

Permit : Allow :: Enforce : ______

Choose the option that has the same relationship as the given pair.

Solution

Permit means to allow something. Therefore, the relationship is based on meaning.

Similarly, Enforce means to make sure that a rule or restriction is obeyed. In this context, its opposite is Forbid, which means to prohibit or not allow something.

Therefore, the best answer is: Enforce : Forbid.

✔ Correct Answer: (B) Forbid

Question 3 Numerical Ability | Probability

Given a fair six-faced dice where the faces are labelled 1, 2, 3, 4, 5 and 6, what is the probability of getting a “1” on the first roll of dice and the “4” on the second roll?

Solution

A fair dice has 6 possible outcomes: 1, 2, 3, 4, 5 and 6.

The probability of getting 1 on the first roll is:

P(1) = 1/6

Similarly, the probability of getting 4 on the second roll is:

P(4) = 1/6

The two rolls are independent events, so we multiply their probabilities:

P(1 on first roll and 4 on second roll) = 1/6 × 1/6 = 1/36

✔ Correct Answer: (A) 1/36

Question 4 Verbal Ability | Case Studies

A recent survey shows that 65% of tobacco users were advised to stop consuming tobacco. The survey also shows that 3 out of 10 tobacco users attempted to stop using tobacco.

Based only on the interpretation of the above passage, which one of the following options can be logically inferred with certainty?

Solution

We are given that 3 out of 10 tobacco users attempted to stop using tobacco.

Therefore, the remaining users who did not attempt to stop are:

10 - 3 = 7

Thus, 7 out of 10 tobacco users did not attempt to stop using tobacco. This represents 70% of the users, which is a majority.

The information that 65% of users were advised to stop consuming tobacco does not tell us that 65% successfully stopped. Similarly, we cannot conclude that 30% successfully stopped merely because 30% attempted to stop.

Therefore, the statement that can be logically inferred with certainty is that a majority of tobacco users who were advised to stop consuming tobacco did not attempt to do so.

✔ Correct Answer: (B) A majority of tobacco users who were advised to stop consuming tobacco did not attempt to do so.

Question 5 Logical Reasoning | Numerical Computation

How many triangles are present in the given figure?

Solution

We can count the triangles systematically instead of counting them randomly.

First, consider the main rectangular/parallelogram region. The intersecting lines create 8 basic triangles.

Next, larger triangles can be formed by combining the smaller regions. These additional combinations give another 8 triangles.

Finally, there are 4 larger triangles formed by the main diagonal structure of the figure.

Total = 8 + 8 + 4

Total = 20

✔ Correct Answer: (C) 20

Question 6 Verbal Ability | Case Studies

Students of all the departments of a college who have successfully completed the registration process are eligible to vote in the upcoming college elections. However, by the time the due date of registration was over, it was found that surprisingly none of the students from the Department of Human Sciences had completed the registration process.

Based only on the information provided above, which one of the following sets of statement(s) can be logically inferred with certainty?

(i) All those students who would not be eligible to vote in college elections would certainly belong to the Department of Human Sciences.

(ii) None of the students from departments other than Human Sciences failed to complete the registration process within the due time.

(iii) All the eligible voters would certainly be students who are not from the Department of Human Sciences.

Solution

We know that students who successfully completed the registration process are eligible to vote.

The passage specifically tells us that none of the students from the Department of Human Sciences completed the registration process.

Therefore, no student from the Department of Human Sciences can be an eligible voter.

Hence, every eligible voter must be a student who is not from the Department of Human Sciences.

Therefore, statement (iii) can be logically inferred with certainty.

Statement (i) cannot be inferred because the passage does not say that every student who is not eligible to vote belongs to Human Sciences.

Statement (ii) also cannot be inferred because we are not given information that every student from other departments completed registration.

✔ Correct Answer: (D) only (iii)

Question 7 Logical Reasoning | Numerical Computation

Which one of the following options represents the given graph?

Solution

First, observe the important features of the graph.

The graph passes through the origin and is positive for x > 0, while it is negative for x < 0.

The graph also has the required symmetry for an odd function, meaning:

f(-x) = -f(x)

Consider option (B):

f(x) = x2^-|x|

Since 2^-|x| is always positive, the sign of f(x) is determined by x. Therefore, the function is negative for negative x and positive for positive x.

Also, 2^-|x| becomes very small as |x| becomes large, which matches the graph approaching the x-axis at both ends.

Therefore, option (B) matches the important features of the given graph.

✔ Correct Answer: (B) f(x) = x2^-|x|

Question 8 Verbal Ability | Case Studies

Which one of the options does NOT describe the passage below or follow from it?

We tend to think of cancer as a “modern” illness because many of the metaphors used to describe it are modern. It is associated with uncontrolled overproduction and sudden growth. Modern cell biology encourages us to imagine the cell as a kind of molecular machine.

Cancer can be viewed as a machine that cannot stop its initial growth command and consequently becomes an uncontrolled, self-propelled system.

Based on the passage, select the statement that does NOT describe or follow from it.

Solution

The passage explains that modern cell biology uses the idea of a machine as a metaphor for understanding cells and cancer.

It specifically describes the cell using a machine-like comparison. Therefore, statements about modern cell biology using metaphors of machinery are supported by the passage.

However, option (D) says that modern cell biology never uses figurative language or metaphors. This is directly opposite to the idea presented in the passage.

Therefore, option (D) is the statement that does NOT describe or follow from the passage.

✔ Correct Answer: (D) Modern cell biology never uses figurative language, such as metaphors, to describe or explain anything.

Question 9 Numerical Ability | Number System

The digit in the unit’s place of the product

3⁹⁹⁹ × 7¹⁰⁰⁰

is ______.

Solution

We only need the unit digit, so we can use the cyclic pattern of the last digits.

The unit digits of powers of 3 follow the cycle:

3, 9, 7, 1

The cycle length is 4. Since 999 ÷ 4 leaves remainder 3, the unit digit of 3⁹⁹⁹ is:

3⁹⁹⁹ → 7

Similarly, the unit digits of powers of 7 follow the cycle:

7, 9, 3, 1

Since 1000 is divisible by 4, the unit digit of 7¹⁰⁰⁰ is 1.

Therefore:

Unit digit = 7 × 1 = 7

✔ Correct Answer: (A) 7

Question 10 Logical Reasoning | Numerical Computation

A square with sides of length 6 cm is given. The boundary of the shaded region is defined by two semicircles whose diameters are the sides of the square, as shown.

The area of the shaded region is ______ cm².

Solution

The square has side length 6 cm. Therefore, each semicircle has radius:

r = 6 ÷ 2 = 3 cm

The area of each semicircle is:

½πr² = ½ × π × 3² = 9π/2

The two semicircles together have area:

9π/2 + 9π/2 = 9π

The required shaded region can be obtained by considering the portions of the two semicircles that do not overlap. Using the geometry of the two equal semicircles, the required shaded area simplifies to:

Area = 18 cm²

✔ Correct Answer: (B) 18

Question 11 Mathematics | Linear Algebra

For the vector W = [1, 2, 3]^T, determine the vector normal to the plane represented by

W^Tx = 1

Solution

The given plane is represented in the form:

W^Tx = 1

For a plane written as ax + by + cz = d, the vector [a, b, c]^T is normal to the plane.

Here,

W = [1, 2, 3]^T

Therefore, the coefficients of x, y and z are 1, 2 and 3.

Normal vector = [1, 2, 3]^T

✔ Correct Answer: (D) [1, 2, 3]^T

Question 12 Control System | Block Diagram & Signal

For the block diagram shown in the figure, the transfer function

Y(s)
R(s)

is:

Solution

Let the output of the first summing point be X(s).

At the first summing point, both inputs are positive:

X(s) = 2R(s) + Y(s)

The signal then passes through the block 1/s. Therefore, the signal entering the second summing point is:

X(s)/s

The second summing point also receives the direct input 3R(s).

Y(s) = X(s)/s + 3R(s)

Substitute X(s) = 2R(s) + Y(s):

Y(s) = [2R(s) + Y(s)]/s + 3R(s)

Multiplying both sides by s:

sY(s) = 2R(s) + Y(s) + 3sR(s)

Rearranging:

(s − 1)Y(s) = (3s + 2)R(s)

Hence,

Y(s)
R(s)  =  3s + 2
s − 1

✔ Correct Answer: (B) (3s + 2)/(s − 1)

Question 13 Control System | Nyquist Stability Criteria

In the Nyquist plot of the open-loop transfer function G(s)H(s) corresponding to the feedback loop shown in the figure,

G(s)H(s) = 3s + 5
s − 1

the infinite semicircular arc of the Nyquist contour in the s-plane is mapped into a point at:

Solution

Given open-loop transfer function:

G(s)H(s) = 3s + 5
s − 1

For the infinite semicircular arc of the Nyquist contour, the magnitude of s tends to infinity:

s → ∞

Therefore,

G(s)H(s) = 3s + 5
s − 1

Divide the numerator and denominator by s:

G(s)H(s) = 3 + 5/s
1 − 1/s

As s → ∞, 5/s → 0 and 1/s → 0.

G(s)H(s) = 3 + 0
1 − 0 = 3

✔ Correct Answer: (C) G(s)H(s) = 3

Question 14 Control System | Controller & Compensator

Consider a unity gain negative feedback system consisting of the plant G(s) and a proportional-integral controller. Let the proportional gain and integral gain be 3 and 1, respectively.

The plant is given by:

G(s) = 1
s − 1

For a unit step reference input, the final values of the controller output and the plant output, respectively, are:

Solution

The PI controller is:

G_c(s) = 3 + 1/s

Therefore,

G_c(s) = 3s + 1
s

The plant is:

G(s) = 1
s − 1

For unity negative feedback, the closed-loop transfer function is:

C(s)/R(s) = G_c(s)G(s)
1 + G_c(s)G(s)

Substituting the values:

C(s)/R(s) = 3s + 1
(s + 1)²

For a unit step input:

R(s) = 1/s

Hence,

C(s) = 3s + 1
s(s + 1)²

Using the Final Value Theorem:

c(∞) = lim_s→0 sC(s) = 1

Therefore, the plant output is:

Plant output = 1

---

Now let X(s) be the controller output. The error signal is:

E(s) = R(s) − C(s)

E(s) = s − 1
(s + 1)²

Controller output:

X(s) = 3s + 1
s E(s)

Therefore,

x(∞) = lim_s→0 sX(s) = −1

✔ Correct Answer: (D) −1, 1

यानी Controller output = −1 और Plant output = 1.

Question 15 Electrical Machine | DC Machine

The following columns present various modes of induction machine operation and the ranges of slip:

Column A Mode of operation	Column B Range of slip
(a) Running in generator mode	(p) From 0.0 to 1.0
(b) Running in motor mode	(q) From 1.0 to 2.0
(c) Plugging in motor mode	(r) From −1.0 to 0.0

The correct matching between the elements in column A with those of column B is:

Solution

The slip of an induction machine is defined as:

s = N_s − N
N_s

where N_s is synchronous speed and N is rotor speed.

1. Generator Mode

In generating operation, the rotor runs faster than synchronous speed:

N > N_s

Therefore:

−1 < s < 0

Hence, (a) → (r).

2. Motor Mode

In normal motoring operation, the rotor speed is less than synchronous speed:

0 < s < 1

Hence, (b) → (p).

3. Plugging in Motor Mode

During plugging, the direction of the rotating field is reversed while the rotor is still rotating in its original direction. Therefore, the slip lies in the range:

1 < s < 2

Hence, (c) → (q).

✔ Final Matching:

(a) → (r)
(b) → (p)
(c) → (q)

✔ Correct Answer: (A) a-r, b-p and c-q

Question 16 Electrical Machine | Single Phase Induction Motor

A 10 pole, 50 Hz, 240 V, single phase induction motor runs at 540 RPM while driving rated load. The frequency of induced rotor currents due to the backward field is:

Solution

Given:

• Number of poles, P = 10
• Supply frequency, f = 50 Hz
• Rotor speed, N_r = 540 RPM

First, calculate the synchronous speed:

N_s = 120f
P

N_s = 120 × 50
10 = 600 RPM

Forward-field slip

s_f = N_s − N_r
N_s

s_f = 600 − 540
600 = 0.1

Backward-field slip

The backward rotating field moves in the opposite direction to the rotor. Therefore, the relative speed is:

N_s + N_r

s_b = N_s + N_r
N_s

s_b = 600 + 540
600 = 1.9

Frequency of rotor current due to backward field

The rotor-current frequency due to the backward field is:

f_b = s_bf

f_b = 1.9 × 50 = 95 Hz

✔ Correct Answer: (B) 95 Hz

Question 17 Signals & Systems | Laplace Transform

A continuous time system that is initially at rest is described by the differential equation

dy(t)/dt + 3y(t) = 2x(t)

where x(t) is the input voltage and y(t) is the output voltage. The impulse response of the system is:

Solution

Given differential equation:

dy(t)/dt + 3y(t) = 2x(t)

Since the system is initially at rest, the initial condition is zero.

Taking the Laplace transform on both sides:

sY(s) + 3Y(s) = 2X(s)

Therefore:

(s + 3)Y(s) = 2X(s)

Hence, the transfer function is:

H(s) = Y(s)
X(s) = 2
s + 3

The impulse response is the inverse Laplace transform of H(s):

h(t) = L⁻¹ { 2/(s + 3) }

h(t) = 2e^−3tu(t)

✔ Correct Answer: (C) 2e^−3tu(t)

Question 18 Signals & Systems | Fourier Transform

The Fourier transform X(ω) of the signal x(t) is given by:

X(ω) = 1,   for |ω| < ω₀
0,   for |ω| > ω₀

Which one of the following statements is true?

Solution

Given:

X(ω) = 1,   for |ω| < ω₀
0,   for |ω| > ω₀

This is a rectangular spectrum. Taking the inverse Fourier transform:

x(t) = sin(ω₀t)
πt

Check x(0)

At t = 0, using the standard limit sin θ / θ → 1:

x(0) = ω₀/π

Therefore, as ω₀ increases, x(0) also increases. Hence option (B) is wrong.

Check options (C) and (D)

Put:

t = π/(2ω₀)

Then:

x(t) = sin(π/2)
π × π/(2ω₀) = 2ω₀
π²

This is not equal to either −1/π or 1/π. Therefore, (C) and (D) are also wrong.

Check option (A)

As ω₀ → ∞, the rectangular spectrum becomes wider and wider. In the limit, it approaches a constant spectrum over the entire frequency axis.

The inverse Fourier transform of a constant spectrum is an impulse (Dirac delta).

ω₀ → ∞  ⇒  x(t) → δ(t)

✔ Correct Answer: (A) x(t) tends to be an impulse as ω₀ → ∞

Question 19 Signals & Systems | Z-Transform

The Z-transform of a discrete signal x[n] is given by:

X(z) = 4z
(z − 1/5)(z − 2/3)(z − 3)

with ROC = R.

Which of the following statements is true?

Solution

Given:

X(z) = 4z
(z − 1/5)(z − 2/3)(z − 3)

The poles of X(z) are located at:

z = 1/5,   z = 2/3,   z = 3

Important Concept

The DTFT exists when the ROC of the Z-transform contains the unit circle.

The unit circle is:

|z| = 1

For the ROC to contain the unit circle, it must lie between the poles at |z| = 2/3 and |z| = 3.

2/3 < |z| < 3

Why the other options are wrong

(A) |z| > 3 does not contain the unit circle, so DTFT does not converge there.

(C) A left-sided sequence generally has an ROC inside the innermost pole. That ROC does not contain |z| = 1 in this case.

(D) A right-sided sequence has an ROC outside the outermost pole, i.e. |z| > 3, which again does not contain the unit circle.

✔ Correct Answer: (B) 2/3 < |z| < 3

Question 20 Power System | Cable and Insulator

For the three-bus power system shown in the figure, the trip signals to the circuit breakers B₁ to B₉ are provided by overcurrent relays R₁ to R₉, respectively, some of which have directional properties also.

The necessary condition for the system to be protected for short circuit fault at any part of the system between bus 1 and the R-L loads with isolation of minimum portion of the network using minimum number of directional relays is:

Solution

In this three-bus power system, the main objective of the protection scheme is to isolate the faulty portion of the network while keeping the healthy part in service.

The directional relays are required where fault current can flow in more than one possible direction. They prevent the relay from unnecessarily tripping for faults occurring in the adjacent section.

For the given arrangement, the required minimum directional relays are R₃ and R₄.

These relays are made directional so that they block faults towards Bus 2. This provides proper discrimination between faults on the connected lines and faults associated with Bus 2.

Required directional relays:

R₃ → Directional overcurrent relay
R₄ → Directional overcurrent relay

Both block faults towards Bus 2.

Why the other options are incorrect

(B) introduces an additional directional relay R₇. Since the question asks for the minimum number of directional relays, this is unnecessary.

(C) uses several directional relays and therefore does not satisfy the minimum-relay requirement.

(D) specifies R₃ and R₄ as blocking faults towards Line 1 and Line 2, respectively. The required blocking direction is towards Bus 2.

✔ Correct Answer: (A)

R₃ and R₄ are directional overcurrent relays blocking faults towards Bus 2.

Question 21 Power System | Per Unit System

The expressions of fuel cost of two thermal generating units as a function of the respective power generation P_G1 and P_G2 are given as:

F₁(P_G1) = 0.1P_G1² + 40P_G1 + 120 Rs/hr
0 MW ≤ P_G1 ≤ 350 MW

F₂(P_G2) = 0.2P_G2² + 30P_G2 + 100 Rs/hr
0 MW ≤ P_G2 ≤ 300 MW

Where a is a constant. For a given value of a, optimal dispatch requires the total load of 290 MW to be shared as P_G1 = 175 MW and P_G2 = 115 MW.

With the load remaining unchanged, the value of a is increased by 10% and optimal dispatch is carried out. The changes in P_G1 and the total cost of generation, F = (F₁ + F₂) in Rs/hr, will be as follows:

Solution

For economic load dispatch, the optimal operating condition is obtained by making the incremental costs of the two generating units equal:

IC₁ = IC₂

For Unit 1:

IC₁ = dF₁/dP_G1 = 2aP_G1 + 40

For Unit 2:

IC₂ = dF₂/dP_G2 = 0.4P_G2 + 30

Effect of increasing a by 10%

When a increases, the incremental cost of Unit 1 increases for the same value of P_G1.

Therefore, to restore the condition IC₁ = IC₂, the generation from Unit 1 must be reduced.

P_G1 ↓

Since the total load is fixed at 290 MW:

P_G1 + P_G2 = 290 MW

If P_G1 decreases, P_G2 must increase to maintain the same total load.

Effect on total generation cost

Increasing a directly increases the quadratic fuel-cost term of Unit 1:

aP_G1²

Thus, after the new optimal dispatch is obtained, the total fuel cost of generation increases.

✔ Correct Answer: (A)

P_G1 will decrease and the total generation cost F will increase.

Question 22 Electrical Machine | DC Machine

The four stator conductors (A, A′, B and B′) of a rotating machine are carrying DC currents of the same value, the directions of which are shown in the figure. The rotor coils a-a′ and b-b′ are formed by connecting the back ends of conductors a′ and a, and b′ and b, respectively, as shown in the figure.

The emf induced in coil a-a′ and coil b-b′ are denoted by E_a-a′ and E_b-b′, respectively. If the rotor is rotated at uniform angular speed ω rad/s in the clockwise direction, which of the following correctly describes E_a-a′ and E_b-b′?

Solution

The stator conductors carry DC currents and therefore produce a stationary magnetic field in space.

From the given arrangement, the rotor coils a-a′ and b-b′ are positioned along the q-axis at the instant shown.

The emf induced in a rotating coil depends on the rate of change of magnetic flux linkage with respect to the rotor position.

At the instant shown, both rotor coils are aligned along the q-axis. Hence, the flux linkage does not change with respect to the rotor position at this instant.

E_a-a′ = 0     E_b-b′ = 0

Therefore, both induced emfs are zero at the instant shown.

✔ Correct Answer: (D)

E_a-a′ = E_b-b′ = 0.

Question 23 Power Electronics | Chopper & Commutation

The chopper circuit shown in the figure feeds power to a 5 A DC constant current source. The switching frequency of the chopper is 100 kHz. All the components can be assumed to be ideal.

The gate signals of switches S₁ and S₂ are shown in the given waveform. The average voltage across the 5 A current source is:

Question 24

Mathematics (Linear Algebra)

In the figure, the vectors u and v are related as Au = v by a transformation matrix A. The correct choice of A is:

✓ Correct Answer: (A)

Solution

Given:

u = (4, 3)
v = (5, 0)

We need a transformation matrix A such that:

Au = v

Consider option (A):

A = [ 4/5    3/5
-3/5    4/5 ]

Now multiply A with u:

Au = [ 4/5    3/5
-3/5    4/5 ] [ 4
3 ]

First component:
(4/5 × 4) + (3/5 × 3) = 16/5 + 9/5 = 25/5 = 5

Second component:
(-3/5 × 4) + (4/5 × 3) = -12/5 + 12/5 = 0

Therefore,

Au = (5, 0) = v

✓ Hence, the correct option is (A).

Question 25

Mathematics (Numerical Methods)

Suppose 10,000 observations are independently obtained from a statistically stationary Gaussian process having mean zero and standard deviation σ₀.

Two estimates of the variance are defined as:

σ̂₁² = (1/10000) Σ_n=1¹⁰⁰⁰⁰ x_n²

σ̂₂² = (1/9999) Σ_n=1¹⁰⁰⁰⁰ x_n²

Which of the following statements is true?

Detailed Solution

Given that the random samples have mean = 0 and standard deviation σ₀.

Therefore, the variance is:

E(x_n²) = σ₀²

Now consider the first estimator:

σ̂₁² = (1/10000) Σ_n=1¹⁰⁰⁰⁰ x_n²

Taking expectation on both sides:

E(σ̂₁²) = (1/10000) Σ E(x_n²)

Since every sample has the same variance, E(x_n²) = σ₀².

E(σ̂₁²) = (1/10000)(10000σ₀²) = σ₀²

Hence the stated relation for the first estimator is satisfied.

✅ Correct Answer: (C)

Concept: For zero-mean samples, the expected value of x² is equal to the variance. Therefore, E(x²) = σ₀².

Question 26

Power Electronics (Power Semiconductor Device)

A semiconductor switch needs to block voltage V of only one polarity (V > 0) during OFF state as shown in Fig. (i) and carry current in both directions during ON state as shown in Fig. (ii).

Which of the following switch combination(s) will realize the same?

Detailed Solution

The required semiconductor switch must satisfy two conditions:

1. OFF state:
It must block voltage of only one polarity (V > 0).

2. ON state:
It must allow current to flow in both directions.

Therefore, the required switch configuration should provide bidirectional current conduction while maintaining the required voltage-blocking property in the OFF state.

Among the given combinations, the configurations in (A) and (D) satisfy these requirements.

✅ Correct Answer: (A) and (D)

Concept: A switch that conducts current in both directions needs a suitable bidirectional current path. The selected semiconductor arrangement must also retain the required one-polarity voltage blocking capability when OFF.

Question 27

Signals & Systems (Classification of System)

Which of the following statement(s) is/are True?

Question 28

Power System (Per Unit System)

The bus admittance matrix (Y_bus) of a 3-bus power system is given below.

Y_bus = ⎡ −j15    −j10    j5 ⎤
⎢ j10    −j13.5    j4 ⎥
⎣ j5      j4     −j8 ⎦

Considering that there is no shunt inductor connected to any of the buses, which of the following cannot be true?

Solution

The given bus admittance matrix is

Y_bus = ⎡ −j15   −j10   j5 ⎤
⎢ j10   −j13.5   j4 ⎥
⎣ j5     j4    −j8 ⎦

For a power system bus admittance matrix, the diagonal element of each bus represents the total admittance connected to that bus, including the contributions from the transmission lines and any shunt elements.

The off-diagonal elements represent the mutual admittances between the corresponding buses.

From the given matrix, the relevant elements are:

• Y₁₂ = −j10
• Y₁₃ = j5
• Y₂₃ = j4

Using the diagonal-element condition of the bus admittance matrix, the shunt elements required at the buses can be identified by comparing each diagonal element with the sum of the corresponding line admittances.

For the given matrix, the resulting configuration shows that a shunt element is required at buses 2 and 3, while the stated condition does not permit the configurations represented by options (A) and (C).

✅ Correct Options: (A) and (C)

Therefore, the configurations described in (A) and (C) cannot be true.

Question 29 – Network Theory

Topic: Steady State Analysis

The values of parameters of the circuit shown in the figure are:

For time t < 0, the circuit is at steady state with switch K in closed condition. If the switch is opened at t = 0, the value of the voltage across the inductor V_L at t = 0⁺, in volts, is ________.

Round off your answer to 1 decimal place.

Solution

At t < 0, the switch is closed and the circuit is in DC steady state.

At steady state:

• Inductor behaves as a short circuit.
• Capacitor behaves as an open circuit.

Therefore, the current through the inductor branch is obtained using current division between R₁ and R₃.

I_L(0⁻) = 10 × 3 / (2 + 3) = 6 A

Since inductor current cannot change instantaneously:

I_L(0⁺) = 6 A

Also, the capacitor voltage cannot change instantaneously. Since the capacitor was initially open-circuited in DC steady state, its initial voltage is obtained from the circuit.

After the switch is opened, applying KCL to the parallel network gives the required instantaneous voltage across the inductor.

V_L(0⁺) = 8.0 V

Hence, the correct option is (B).

Question 30

Electrical Machine (DC Machine)

A separately excited DC motor rated 400 V, 15 A, 1500 rpm drives a constant torque load at rated speed operating from 400 V DC supply drawing rated current. The armature resistance is 1.2 Ω. If the supply voltage drops by 10% with field circuit unaltered, the resultant armature speed of the motor in RPM is ______. (Round off to the nearest integer.)

Correct Answer: 1343 RPM

Solution

Given:

Supply voltage, V₁ = 400 V
Armature current, Iₐ = 15 A
Armature resistance, Rₐ = 1.2 Ω
Initial speed, N₁ = 1500 rpm

Since the motor is separately excited and the field circuit is unaltered:

Φ = constant

The supply voltage drops by 10%.

V₂ = 0.9 × 400
V₂ = 360 V

For a DC motor:

Eₐ = V - IₐRₐ

Also, for constant flux:

Eₐ ∝ N

Therefore,

N₁ / N₂ = (V₁ - IₐRₐ) / (V₂ - IₐRₐ)

1500 / N₂ = (400 - 15 × 1.2) / (360 - 15 × 1.2)

1500 / N₂ = 382 / 342

N₂ = 1500 × 342 / 382

N₂ ≈ 1343 rpm

Final Answer = 1343 RPM

Question 31

Signals & Systems (Basic of Signal)

For the signals x(t) and y(t) shown in the figure, z(t) = x(t) * y(t) is maximum at t = T₁. Then T₁ in seconds is ______. (Round off to the nearest integer.)

Correct Answer: 4

Solution

Given:

z(t) = x(t) * y(t)

Here, * denotes convolution.

From the graph, x(t) is a rectangular pulse:

x(t) = 1,   -1 ≤ t ≤ 1

The signal y(t) starts from y(1) = -2 and reaches y(5) = 2.

Therefore its equation in the interval 1 ≤ t ≤ 5 is:

y(t) = t - 3

Convolution with the rectangular pulse means that z(t) is obtained by taking the area of a moving 2-second portion of y(t).

For 2 ≤ t ≤ 4, the complete interval [t-1, t+1] lies inside the ramp.

z(t) = ∫_t-1^t+1 (τ - 3)dτ

z(t) = 2(t - 3)

Hence, in this interval, z(t) continuously increases.

For 4 ≤ t ≤ 6, the upper limit reaches t = 5, so:

z(t) = ∫_t-1⁵ (τ - 3)dτ

z(t) = 2 - (t - 4)²/2

This expression decreases after t = 4. Therefore, the maximum occurs exactly at:

T₁ = 4 seconds

Hence, the correct answer is 4.

Question 32

Network Theory (Network Theorem)

For the circuit shown in the figure, V₁ = 8 V, DC and I₁ = 8 A, DC. The voltage V_ab in Volts is ______. (Round off to 1 decimal place.)

 

Solution

Given:

V₁ = 8 V
I₁ = 8 A

From the given circuit, the two 3 Ω resistors are connected in parallel. Therefore, their equivalent resistance is:

R = 3 ∥ 3

R = (3 × 3) / (3 + 3)

R = 1.5 Ω

Now, using the Voltage Divider Rule (VDR):

V_ab = (8 × 1.5) / (1.5 + 0.5)

V_ab = 12 / 2
V_ab = 6 V

✅ Correct Answer: (B) 6 V

Question 33

Power System (Performance of Transmission Line)

A 50 Hz, 275 kV line of length 400 km has the following parameters:

Resistance: R = 0.035 Ω/km
Inductance: L = 1 mH/km
Capacitance: C = 0.001 μF/km

The line is represented by the nominal-π model. The magnitudes of sending-end and receiving-end voltages (denoted by V_S and V_R, respectively) are maintained at 275 kV.

The phase angle difference θ between V_S and V_R required for maximum possible active power to be delivered to the receiving end is ______ degrees.
(Round off to 2 decimal places.)

Solution

Given:

f = 50 Hz
Length = 400 km
R = 0.035 Ω/km
L = 1 mH/km
C = 0.001 μF/km

Total series resistance of the line:

R = 0.035 × 400

R = 14 Ω

Total inductance:

L = 1 × 10⁻³ × 400

L = 0.4 H

Therefore, the series impedance is:

Z = R + jωL

Z = 14 + j(2π × 50 × 0.4)

Z ≈ 14 + j125.66 Ω

For the nominal-π transmission-line model, the ABCD parameters are used. The receiving-end power can be written in terms of the phase difference θ as:

P_R = (|V_S||V_R| / |B|) cos(θ − β)

− (|A||V_R|² / |B|) cos(β − α)

For maximum possible active power delivered to the receiving end, the first cosine term must be maximum.

For maximum power:

cos(θ − β) = 1

Therefore,
θ = β

On calculating the angle of the transmission-line parameter B for the given nominal-π model:

θ ≈ 83.64°

✅ Correct Answer: (C) 83.64°

Question 34

Mathematics (Differential Equations)

In the following differential equation, the numerically obtained value of y(t), at t = 1, is ______.
(Round off to 2 decimal places.)

dy/dt = e^−αt / (2 + αt)

α = 0.01      y(0) = 0

Solution

Given differential equation:

dy/dt = e^−αt / (2 + αt)

α = 0.01
y(0) = 0

Using Euler's Method with step size:

h = 1

Euler's method is:

y₁ = y₀ + h f(t₀, y₀)

At t₀ = 0 and y₀ = 0:

y₁ = 0 + 1 × [ e^−0.01×0 / (2 + 0.01×0) ]

= 1 × (1 / 2)

y₁ = 0.5

Therefore, at t = 1:

✅ Correct Answer: (B) 0.50

Question 35

Mathematics (Numerical Methods)

Three points in the x-y plane are (−1, 0.8), (0, 2.2) and (1, 2.8). The value of the slope of the best fit straight line in the least square sense is ________.

(Round off to two decimal places.)

Enter your answer:

Solution

Given three points:

(−1, 0.8),   (0, 2.2),   (1, 2.8)

Let the best-fit straight line be:

y = ax + b

For the least-square approximation, the normal equations are:

Σy = aΣx + nb

Σxy = aΣx² + bΣx

x	y	x²	xy
−1	0.8	1	−0.8
0	2.2	0	0
1	2.8	1	2.8

Σx = 0
Σy = 5.8
Σx² = 2
Σxy = 2

Substituting these values in the normal equations:

5.8 = a(0) + 3b

2 = a(2) + b(0)

From the second equation:

2 = 2a

a = 1

Since a is the slope of the best-fit straight line, therefore:

✅ Correct Answer: 1.00

Question 36

Control System (Bode Plot)

Magnitude and phase plots of an LTI system are shown in the figure. The transfer function of the system is:

 

Solution

From the given Bode plot, let the transfer function be:

T(s) = Ke^−Ts

Here, T represents the transportation lag.

From the magnitude plot:

20 log K = 8 dB

K = 10^8/20

K ≈ 2.51

From the phase plot, at:

ω = 1 rad/sec

φ = −60°

For a pure transportation lag:

φ = −ωT × 180°/π

Substituting the values:

−60° = −1 × T × 180°/π

T = 60π/180

T ≈ 1.047 sec

Therefore:

T(s) = Ke^−Ts

T(s) = 2.51e^−1.047s

✅ Correct Answer: (D) 2.51e^−1.047s

Question 37

Consider the Op-Amp based circuit shown in the figure. Ignore the conduction drops of diodes D₁ and D₂. All the components are ideal and the breakdown voltage of the Zener is 5 V. Which of the following statements is true?

Solution

The input voltage is:

V_in = 10 sin(1000t) V

The input resistor and feedback resistor are both 1 kΩ. Therefore, when the feedback path is active, the magnitude of the closed-loop gain is:

Gain = − R_f / R_in

= −1 kΩ / 1 kΩ

= −1

Hence, without limiting, the output would be:

V_o = −10 sin(1000t) V

However, the Zener diode network limits the output voltage. The Zener breakdown voltage is 5 V.

During the positive half-cycle, the output is limited to:

V_o,max = +5 V

During the negative half-cycle, the corresponding limiting level is:

V_o,min = −10 V

Therefore, the maximum and minimum output voltages are:

✅ Correct Answer: (D)

Maximum value = +5 V
Minimum value = −10 V

Question 38

Consider a lead compensator of the form

K(s) = 1 + s/α 1 + s/(αβ)   , β > 1, α > 0

The frequency at which this compensator produces maximum phase lead is 4 rad/sec. At this frequency, the gain amplification provided by the controller, assuming asymptotic Bode magnitude plot of K(s), is 6 dB.

The values of α, β respectively are:

Solution

Given:

K(s) = 1 + s/α 1 + s/(αβ)

Rewriting the compensator:

K(s) = s + α s + αβ × αβ/α

K(s) = β(s + α)/(s + αβ)

For a lead compensator, the frequency at which maximum phase lead occurs is:

ω_m = √(α × αβ)

= α√β

Given that the maximum phase lead occurs at ω_m = 4 rad/sec:

α√β = 4

Therefore:

α√β = 4    ...(1)

The gain amplification at this frequency is given as 6 dB.

Since:

20 log₁₀|K(jω)| = 6 dB

|K(jω)| = 10^6/20

|K(jω)| ≈ 2

At ω = 4 rad/sec:

|K(jω)| = β √(ω² + α²) / √(ω² + (αβ)²)

= β√(16 + α²) / √(16 + (αβ)²)

= 2

Squaring both sides:

β²(16 + α²) = 4[16 + (αβ)²]

On simplifying:

16β² = 64 + 3(αβ)²

From equation (1):

α√β = 4

α²β = 16

The values satisfying the given conditions are:

✅ Correct Answer: (B)

α = 2
β = 4

Question 39

A 3-phase, star connected, balanced load is supplied from a 3-phase, 400 V (rms), balanced voltage source with phase sequence R-Y-B, as shown in the figure. If the wattmeter reading is −400 W and the line current is I_R = 2 A (rms), then the power factor of the load per phase is

The power factor is:

Solution

Given:

V_L = 400 V
I_R = 2 A
Wattmeter reading = −400 W

For the given wattmeter connection:

W = V_pcI_cc cos φ_pc-cc

Since the phase sequence is R-Y-B:

V_pc = V_YB
I_cc = I_R = 2 A

For a balanced lagging load:

W = V_YBI_R cos(90° − θ)

For a balanced leading load:

W = V_YBI_R cos(90° + θ)

The wattmeter reading is negative:

W = −400 W

Therefore, the load power factor is leading.

For the leading case:

−400 = 400 × 2 × cos(90° + θ)

cos(90° + θ) = −0.5

θ = 30°

Hence, the power factor is:

Power Factor = cos 30° = 0.866 leading

✅ Correct Answer: (C)

0.866 leading

Question 40

An 8-bit ADC converts analog voltage in the range of 0 to +5 V to the corresponding digital code as per the conversion characteristics shown in the figure. For V_in = 1.9922 V, which of the following digital output, given in hex, is true?

The digital output in hexadecimal is:

Solution

Given:

n = 8 bits
V_max = 5 V
V_in = 1.9922 V

For an 8-bit ADC, the number of quantization levels is:

2⁸ = 256

Therefore, the step size is:

Step size = V_max 2ⁿ − 1 = 5 255 V

Hence, the digital code corresponding to the input voltage is obtained from:

Digital value = V_in Step size

= 1.9922 5/255

Therefore:

Digital value ≈ 101.6

The corresponding ADC code is approximately 102 in decimal.

Convert decimal 102 into hexadecimal:

102 ÷ 16 = 6 remainder 6

Therefore,
102 = 66 H

✅ Correct Answer: (C)

66 H

Question 41

The three-bus power system shown in the figure has one alternator connected to bus 2 which supplies 200 MW and 40 MVAr power. Bus 3 is an infinite bus having a voltage of |V₃| = 1.0 p.u. and an angle of −15°. A variable current source |I₁|∠φ is connected at bus 1 and controlled such that the magnitude of the bus 1 voltage is maintained at 1.05 p.u. and the phase angle of the source current is φ = θ₁ + π/2 where θ₁ is the phase angle of the bus 1 voltage. The three buses can be categorized for load flow analysis as:

Solution

For load flow analysis, the type of a bus is determined from the quantities specified at that bus.

Bus 1

At bus 1, the magnitude of voltage is maintained at |V₁| = 1.05 p.u.

The current source angle is given by:

φ = θ₁ + π/2

Therefore, the current is in quadrature with the bus voltage. Hence the real power supplied by the current source is zero:

P₁ = 0

Thus, P and |V| are specified at bus 1. Therefore, bus 1 is a:

P − |V| bus

Bus 2

The alternator connected to bus 2 supplies:

P₂ = 200 MW
Q₂ = 40 MVAr

Both active power P and reactive power Q are specified.

Therefore, Bus 2 = P − Q bus

Bus 3

Bus 3 is an infinite bus.

Its voltage magnitude and phase angle are specified:

|V₃| = 1.0 p.u.
θ₃ = −15°

Hence, bus 3 is the reference/slack bus.

Therefore, Bus 3 = Slack bus

✅ Correct Answer: (D)

Bus 1 → P − |V| bus
Bus 2 → P − Q bus
Bus 3 → Slack bus

Question 42

Consider the following equation in a 2-D real-space:

|x₁|^p + |x₂|^p = 1,    p > 0

Which of the following statement(s) is/are true?

Solution

Given:

|x₁|^p + |x₂|^p = 1

Case 1: p = 1

The equation becomes:

|x₁| + |x₂| = 1

This represents a diamond-shaped square with vertices:

(1,0), (0,1), (−1,0), (0,−1)

Its diagonals are both equal to 2.

Area = ½ × 2 × 2 = 2

Therefore, (D) is true.

Case 2: p = 2

The equation becomes:

x₁² + x₂² = 1

This is a circle of radius 1.

Area = πr² = π

Therefore, (A) is true.

Case 3: p → ∞

As p becomes very large, the equation |x₁|^p + |x₂|^p = 1 approaches:

max(|x₁|, |x₂|) = 1

Hence the curve approaches a square whose sides extend from −1 to +1 along both axes.

Area = 2 × 2 = 4

Therefore, (B) is true.

Case 4: p → 0

For non-zero x, as p → 0:

|x|^p → 1

Hence the limiting equation does not give a normal enclosed region with area 1. Therefore statement (C) is false.

✅ Correct Options: (A), (B) and (D)

(A) Area = π when p = 2
(B) Area → 4 when p → ∞
(D) Area = 2 when p = 1

Question 43

In the figure, the electric field E and magnetic field B point in the z and x directions, respectively, and have constant magnitude. A positive charge q is released from rest at the origin. Which of the following statement(s) is/are true?

Solution

The force acting on a moving charge in electric and magnetic fields is given by the Lorentz force:

F = qE + q(v × B)

Initially the charge is released from rest, so v = 0. Therefore, initially the magnetic force is zero and only the electric force acts on the charge.

Motion of the charge

Since the electric field is in the z-direction, the charge initially accelerates in the z-direction.

As the charge gains velocity in the z-direction, the magnetic force starts acting because:

F_m = q(v × B)

The magnetic field is along the x-direction. Hence, for velocity along z:

z × x = y

Therefore, the magnetic force produces motion in the y-direction.

Option (A)

The charge does not move with constant velocity in the z-direction because the electric field continuously produces acceleration.

Therefore, (A) is false.

Option (B)

There is no force component that produces motion in the x-direction.

Therefore, the x-coordinate remains zero and the charge always remains in the y-z plane.

Hence, (B) is true.

Option (C)

The electric field continuously accelerates the charge in the z-direction, while the magnetic force causes the motion to curve in the y-direction.

The resulting trajectory is not a circle. It is a cycloidal-type trajectory.

Therefore, (C) is false.

Option (D)

The magnetic force causes the charge to acquire motion in the y-direction.

Hence, the charge progresses in the positive y-direction.

Therefore, (D) is true.

✅ Correct Options: (B) and (D)

(B) The charge remains in the y-z plane.
(D) The charge progresses in the y-direction.

Question 44

All the elements in the circuit shown in the figure are ideal. Which of the following statements is/are true?

(All diodes and circuit elements are ideal.)

Solution

Since all the circuit elements are ideal, each diode can be treated as either a short circuit when forward biased or an open circuit when reverse biased.

Case 1: Switch S is ON

When S is ON, the corresponding node is connected to the reference potential. Considering the ideal diode conditions and the directions of the current sources, D₁ and D₂ are forward biased, while D₃ is reverse biased.

Therefore, statement (A) is TRUE.

Case 2: Switch S is OFF

When S is OFF, the circuit conditions change. Under this condition, D₁ is reverse biased, whereas D₂ and D₃ are forward biased.

Therefore, statement (C) is TRUE.

Hence, the correct statements are:

✅ Correct Options: (A) and (C)

Thus, (A) and (C) are the correct statements.

Question 45

The expected number of trials for first occurrence of a “head” in a biased coin is known to be 4. The probability of first occurrence of a “head” in the second trial is __________.
(Round off to 3 decimal places)

Solution

Let the probability of getting a Head be p.

Therefore, the probability of getting a Tail is q = 1 − p.

The number of trials required for the first occurrence of a Head follows a Geometric Distribution.

E(X) = 1 / p

Given that the expected number of trials is 4:

4 = 1 / p
p = 1/4

Hence,

q = 1 − p
q = 1 − 1/4
q = 3/4

For the first occurrence of Head in the second trial, the first trial must be Tail and the second trial must be Head.

P(TH) = P(T) × P(H)

= (3/4) × (1/4)

= 3/16

= 0.1875

Rounding off to 3 decimal places:

✅ Correct Answer: 0.187

Question 46

Consider the state-space description of an LTI system with matrices

A = ⎡ 0   1 ⎤
⎣ −1  −2 ⎦      B = ⎡ 0 ⎤
⎣ 1 ⎦      C = [ 3  −2 ]      D = 1

For the input sin(ωt), ω > 0, the value of ω for which the steady-state output of the system will be zero is ________.
(Round off to the nearest integer)

Solution

For a state-space system, the transfer function is

G(s) = C(sI − A)⁻¹B + D

Given:

sI − A = ⎡ s  −1 ⎤
⎣ 1  s+2 ⎦

(sI − A)⁻¹ = 1/(s²+2s+1) ⎡ s+2  1 ⎤
⎣ −1  s ⎦

Therefore,

C(sI − A)⁻¹B + D

= [3  −2] · 1/(s+1)² ⎡1⎤
          ⎣s⎦ + 1

= (3 − 2s)/(s+1)² + 1

G(s) = (s² + 4)/(s+1)²

For sinusoidal input sin(ωt), the steady-state output becomes zero when the magnitude of the frequency response is zero. Hence,

G(jω) = ((jω)² + 4)/(jω + 1)²

= (4 − ω²)/(1 + jω)²

For zero output:
4 − ω² = 0

ω² = 4
ω = 2 rad/sec

✅ Correct Answer: (B) 2

Hence, the required value of ω is 2 rad/sec.

Question 47

A 3-phase synchronous motor with synchronous impedance of 0.1 + j0.3 p.u. per phase has a static stability limit of 2.5 p.u. The corresponding excitation voltage in per unit is ________.
(Round off to 2 decimal places)

Solution

Given synchronous impedance per phase:

Z_s = 0.1 + j0.3 p.u.

Therefore, the magnitude of synchronous impedance is:

|Z_s| = √(0.1² + 0.3²) ≈ 0.316 p.u.

At the static stability limit, the maximum power output is:

P_max = EV / X_s − E²R_s / X_s²

Here, P_max = 2.5 p.u., V = 1 p.u., R_s = 0.1 and |Z_s| ≈ 0.316.

Hence,

2.5 = E/0.316 − (E² × 0.1)/(0.316²)

Solving the above equation gives:

E ≈ 1.58 p.u.

✅ Correct Answer: (B) 1.58 p.u.

Question 48

A 3-phase, 415 V, 50 Hz, 6 pole, 960 RPM squirrel cage induction motor drives a constant torque load at rated speed operating from rated supply and delivering rated output. If the supply voltage and frequency are reduced by 20%, the resultant speed of the motor in RPM (neglecting the stator leakage impedance and rotational losses) is ________.
(Round off to the nearest integer)

Solution

Given:

• Supply voltage, V₁ = 415 V
• Frequency, f₁ = 50 Hz
• Number of poles, P = 6
• Rated speed, N_r1 = 960 RPM

Synchronous speed is

N_s1 = 120f₁/P = (120 × 50)/6 = 1000 RPM

Therefore, the initial slip is

s₁ = (N_s1 − N_r1)/N_s1 = (1000 − 960)/1000 = 0.04

Both voltage and frequency are reduced by 20%. Hence,

V₂ = 0.8V₁
f₂ = 0.8f₁ = 40 Hz

Since V/f remains constant and the motor operates in the low-slip region, for constant torque:

s₂f₂ = s₁f₁

Therefore,

s₂ = (0.04 × 50)/40 = 0.05

New synchronous speed:

N_s2 = (120 × 40)/6 = 800 RPM

Hence, the new motor speed is

N_r2 = N_s2(1 − s₂)
= 800 × 0.95
= 760 RPM

✅ Correct Answer: (C) 760 RPM

Question 49

The period of the discrete-time signal x[n] described by the equation below is N = ______.
x[n] = 1 + 3sin (15πn/8 + 3π/4) − 5sin (πn/3 − π/4)
Round off to the nearest integer.

Solution

Given,

x[n] = 1 + 3sin(15πn/8 + 3π/4) − 5sin(πn/3 − π/4)

For a discrete-time sinusoidal signal sin(ωn + φ), the signal is periodic if ω/(2π) is rational.

For the first sinusoidal component:

ω₁ = 15π/8

N₁ = 2π / gcd(15π/8, 2π)

N₁ = 16

For the second sinusoidal component:

ω₂ = π/3

N₂ = 6

The overall period of the signal is the LCM of the individual periods:

N = LCM(N₁, N₂)

N = LCM(16, 6)

N = 48

✅ Correct Answer: (C) 48

Therefore, the fundamental period of the given discrete-time signal is 48.

Question 50

The discrete-time Fourier transform of a signal x[n] is

X(Ω) = (1 + cosΩ)e^−jΩ

Consider that x_p[n] is a periodic signal of period N = 5 such that

x_p[n] = { x[n],   for n = 0, 1, 2
       0,   for n = 3, 4 }

If x_p[n] = Σ a_ke^j2πkn/N , then the magnitude of the Fourier series coefficient a₃ is ______.

Round off to 3 decimal places.

Solution

Given,

X(Ω) = (1 + cosΩ)e^−jΩ

Using cosΩ = (e^jΩ + e^−jΩ)/2 , we get

X(Ω) = [1 + (e^jΩ + e^−jΩ)/2] e^−jΩ
= e^−jΩ + 1/2 + 1/2 e^−j2Ω

Comparing with the DTFT relation, the corresponding sequence is

x[n] = { 1/2, 1, 1/2 }
for n = 0, 1, 2

Since the periodic sequence has N = 5,

x_p[n] = { 1/2, 1, 1/2, 0, 0 }

The Fourier series coefficient is

a_k = (1/N) Σ^N−1_n=0 x_p[n] e^−j2πkn/N

For k = 3 and N = 5,

a₃ = 1/5 [ (1/2) + e^−j6π/5 + (1/2)e^−j12π/5 ]

|a₃| ≈ 0.038

✅ Correct Answer: (C) 0.038

Therefore, the magnitude of the Fourier series coefficient a₃ is 0.038.

Question 51

For the circuit shown, if i = sin 1000t, the instantaneous value of the Thevenin's equivalent voltage (in Volts) across the terminal a-b at time t = 5 ms is ______.

Round off to 2 decimal places.

✏️ Enter your answer

V

Solution

Step 1: Source Transformation

Given current source:

i = sin 1000t

By source transformation, the current source with impedance (10 + j10) Ω is converted into an equivalent voltage source.

V = (10 + j10) sin 1000t

Step 2: Apply KVL

Applying KVL in the loop:

(10 + j10)sin 1000t + (10 + j10)i_x − 4i_x + (10 − j10)i_x = 0

(10 + j10)sin 1000t − 16i_x = 0

Step 3: Find i_x

i_x = (10 + j10)sin 1000t / 16

Step 4: Find V_th

V_th = i_x(10 − j10)

V_th = [(10 − j10)(10 + j10)] sin 1000t / 16

(10 − j10)(10 + j10) = 100 + 100

V_th = 200 sin 1000t / 16

V_th = 12.5 sin 1000t

Step 5: Put t = 5 ms

t = 5 ms = 0.005 s

V_th = 12.5 sin(1000 × 0.005)

V_th = 12.5 sin 5

sin 5 ≈ −0.958924

V_th = 12.5 × (−0.958924)

V_th ≈ −11.9865 V

✅ Correct Answer: −11.98 V

Therefore, the instantaneous value of the Thevenin's equivalent voltage across terminal a-b is −11.98 V.

Question 52

The admittance parameters of the passive resistive two-port network shown in the figure are

The power delivered to the load resistor R_L in Watt is ______.

Round off to 2 decimal places.

✏️ Enter your answer

W

Solution

Step 1: Given Y-parameters

The Y-parameter matrix of the given two-port network is

[Y₂] = [ 5    −2.5
−2.5    1 ] S

Step 2: Y-parameters of 3 Ω resistor

The 3 Ω resistor connected between port 1 and port 2 has the Y-parameter matrix

[Y₁] = [ 1/3    −1/3
−1/3    1/3 ] S

Step 3: Overall Y-parameter matrix

Since the 3 Ω resistor is connected in parallel with the two-port network, their Y-parameter matrices are added.

[Y] = [Y₁] + [Y₂]

= [ 1/3    −1/3
−1/3    1/3 ] + [ 5    −2.5
−2.5    1 ]

[Y] = [ 16/3    −8.5/3
−8.5/3    4/3 ] S

Step 4: Find Thevenin Voltage

The Y-parameter equations are

I₁ = (16/3)V₁ − (8.5/3)V₂

I₂ = −(8.5/3)V₁ + (4/3)V₂

For finding the Thevenin voltage, the output is open-circuited. Therefore, I₂ = 0.

0 = −(8.5/3)V₁ + (4/3)V₂

4V₂ = 8.5V₁

V₁ = 20 V

V₂ = (8.5/4)(20)

V_th = 42.5 V

Step 5: Find Thevenin Resistance

From the Y-parameter equation,

I₂ = −(8.5/3)V₁ + (4/3)V₂

R_th = 3/4 Ω

R_th = 0.75 Ω

Step 6: Load Current

Given:

R_L = 6 Ω

Therefore,

I_L = V_th / (R_th + R_L)

I_L = 42.5 / (0.75 + 6)

I_L ≈ 6.29 A

Step 7: Power Delivered to Load

P_L = I_L²R_L

P_L = (6.29)² × 6

P_L ≈ 237.38 W

✅ Correct Answer: 237.38 W

Therefore, the power delivered to the load resistor R_L = 6 Ω is 237.38 W.

Question 53

When the winding c-d of the single-phase, 50 Hz, two-winding transformer is supplied from an AC current source of frequency 50 Hz, the rated voltage of winding c-d is 200 V (rms), 50 Hz is obtained at the open-circuited terminals a-b.

The cross sectional area of the core is 5000 mm² and the average core length traversed by the mutual flux is 500 mm. The maximum allowable flux density in the core is B_max = 1 Wb/m² and the relative permeability of the core material is μ_r = 5000.

The leakage impedance of the winding a-b and winding c-d at 50 Hz are

Considering the magnetizing characteristics to be linear and neglecting core loss, the self-inductance of the winding a-b in millihenry is ______.

Round off to 1 decimal place.

✏️ Enter Your Answer

Enter your answer in millihenry (mH):

Solution

Given:

A = 5000 mm² = 5000 × 10⁻⁶ m²
l = 500 mm = 0.5 m
B_max = 1 Wb/m²
μ_r = 5000
f = 50 Hz
V = 200 V

For the transformer winding, the induced RMS voltage is

V = 4.44 f Φ_mN₁

The maximum flux is

Φ_m = B_maxA
= 1 × (5000 × 10⁻⁶)
= 0.005 Wb

Therefore,

200 = 4.44 × 50 × 0.005 × N₁
N₁ ≈ 180.16 turns

The magnetizing inductance is

L_m = N₁² / Reluctance

L_m = N₁² μ₀ μ_r A / l

L_m = (180.16)² × 4π × 10⁻⁷ × 5000 × 5000 × 10⁻⁶ / 0.5

L_m ≈ 2.039 H
= 2039 mH

The leakage inductance of winding a-b is obtained from

Z_leakage = jωL_l
j100πL_l = j100π × 0.16
L_l = 0.16 H
= 160 mH

Hence, the total self-inductance of winding a-b is

L₁ = L_m + L_l
= 2.039 + 0.16
= 2.199 H
= 2199 mH ≈ 2200 mH

✅ Correct Answer: 2200 mH

Therefore, the self-inductance of winding a-b is 2200 mH.

Question 54

The circuit shown in the figure is initially in the steady state with the switch K in open condition and K̄ in closed condition. The switch K is closed and K̄ is opened simultaneously at the instant t = t₁.

If the current source is i(t) = sin(1000t), the minimum value of t₁, in milliseconds, such that there is no transient in the voltage across the 100 μF capacitor is ______.

Round off to 2 decimal places.

✏️ Enter Your Answer

Enter your answer in milliseconds (ms):

Solution

Given:

i(t) = sin(1000t)
C = 100 μF
R = 10 Ω
V = 5 V

Step 1: Initial capacitor voltage

Before switching, K is open and K̄ is closed. In the steady state, the capacitor behaves as an open circuit. Therefore, there is no current through the 10 Ω resistor in the lower branch.

V_C(t₁⁻) = 5 V

Step 2: Condition for no transient

For the capacitor voltage to have no transient component, its initial voltage must be equal to the forced steady-state voltage immediately after switching.

V_C(t₁⁻) = V_C(t₁⁺)
= V_forced(t₁)

Step 3: Find the forced capacitor voltage

After switching, the current source is connected to the 10 Ω resistor and capacitor. For the steady-state voltage across the capacitor:

V_C(t) = 10 × i(t)
= 10 sin(1000t)

Step 4: Apply the no-transient condition

5 = 10 sin(1000t₁)
sin(1000t₁) = 1/2

The minimum positive value occurs for

1000t₁ = π/6
t₁ = π/(6000) second

Converting into milliseconds:

t₁ = π/(6000) × 1000
t₁ ≈ 0.5236 ms
≈ 0.52 ms

✅ Correct Answer: 0.52 ms

Therefore, the minimum value of t₁ is 0.52 ms.

Question 55

The circuit shown in the figure has reached steady state with thyristor T in OFF condition. Assume that the latching and holding currents of the thyristor are zero.

The thyristor is turned ON at t = 0 sec. The duration in μsec for which the thyristor would conduct, before it turns OFF, is ______

Round off to 2 decimal places.

✏️ Enter your answer:

Solution

Given:

V_s = 100 V
R = 4 Ω
L = 4 μH
C = 1 μF

Before the thyristor is turned ON, the circuit is in steady state. Therefore, the capacitor is charged to the source voltage and the inductor current is zero.

V_C(0) = V_s = 100 V
i_L(0) = 0 A

After the thyristor is turned ON, the LC branch forms a source-free oscillatory circuit.

The natural angular frequency is

ω₀ = 1/√(LC)

= 1/√[(4 × 10⁻⁶)(1 × 10⁻⁶)]

ω₀ = 5 × 10⁵ rad/s

The current through the LC branch is

i_C = V_s√(C/L) sin(ω₀t)

= 100√(1/4) sin(ω₀t)

i_C = 50 sin(ω₀t) A

The current through the resistor is

I₀ = V_s/R = 100/4

I₀ = 25 A

The thyristor current is the sum of the two currents:

i_T = i_C + I₀

The thyristor turns OFF when its current becomes zero. Therefore,

i_T = 0
i_C + I₀ = 0
50 sin(ω₀t) + 25 = 0
sin(ω₀t) = −1/2

The first positive value of ω₀t satisfying this condition is

ω₀t = 7π/6

Hence,

t = (7π/6)/(5 × 10⁵)

t = 7.33 × 10⁻⁶ s

t = 7.33 μs

✅ Correct Answer: 7.33 μs

Therefore, the thyristor conducts for 7.33 μs before turning OFF.

Question 56

Neglecting the delays due to the logic gates in the circuit shown in the figure, the decimal equivalent of the binary sequence [ABCD] of initial logic states, which will not change with clock, is ______.

Enter the decimal equivalent.

✏️ Enter your answer:

Solution

From the given logic circuit, the input signals can be written in terms of the flip-flop outputs.

A = Q̅₁
B = A̅ = Q₁
C = Q₂
D = B ⊕ C

For the initial state Q₁Q₂ = 00, we get:

Q₁ = 0,   Q₂ = 0
A = Q̅₁ = 1
B = A̅ = 0
C = Q₂ = 0
D = B ⊕ C = 0 ⊕ 0 = 0

Therefore, the binary sequence is

[ABCD] = [1000]

The next-state inputs of the D flip-flops are

D₁ = Q₂ = 0
D₂ = D = 0

Hence, the next state is again Q₁Q₂ = 00. Therefore, the initial state remains unchanged with every clock.

Now convert 1000 from binary to decimal:

(1000)₂ = 1×2³ + 0×2² + 0×2¹ + 0×2⁰
= 8

✅ Correct Answer: 8

Therefore, the decimal equivalent of the binary sequence [ABCD] = 1000 is 8.

Question 57

In a given 8-bit general purpose micro-controller there are the following flags:

C - Carry, A - Auxiliary Carry, O - Overflow flag, P - Parity
P = 0 for even parity and P = 1 for odd parity.

R0 and R1 are the two general purpose registers of the micro-controller.

After execution of the following instructions, the decimal equivalent of the binary sequence of the flag pattern [CAOP] will be ______.

MOV R0, +0x60
MOV R1, +0x46
ADD R0, R1

Solution

Given instructions are:

MOV R0, +0x60
MOV R1, +0x46
ADD R0, R1

Therefore,

R0 = 60H
R1 = 46H

Convert the hexadecimal values into binary:

60H = 0110 0000
46H = 0100 0110

0110 0000
+ 0100 0110
─────────
1010 0110

Hence the result is:

Result = A6H

Carry flag (C) = 0
Overflow flag (O) = 1
Auxiliary Carry flag (A) = 0
Parity flag (P) = 0

Therefore, the flag pattern [CAOP] is:

[CAOP] = [0010]

Converting the binary number 0010 into decimal:

(0010)₂ = 2₁₀

✅ Correct Answer: (C) 2

Therefore, the decimal equivalent of the flag pattern [CAOP] is 2.

Question 58

The single-phase rectifier consisting of three thyristors T₁, T₂, T₃ and a diode D₁ feeds power to a 10 A constant current load.

T₁ and T₃ are fired at angle α = 60° and T₂ is fired at α = 240°.

The reference for α is the positive zero crossing of V_in.

The average voltage V_o across the load in volts is ________.

Round off to 2 decimal places.

Solution

Given input voltage is:

V_in = V_m sin(ωt)

V_m = 100 V

The firing angle is:

α = 60°

For the given rectifier and constant current load, the output voltage waveform contains the following two conducting intervals:

Positive output: V_o = V_m sin(ωt)
for α ≤ ωt ≤ π

Negative input portion: V_o = −V_m sin(ωt)
for π + α ≤ ωt ≤ 2π

Therefore, the average output voltage is

V_o(avg) = 1/(2π) [ ∫_α^π V_m sin(θ)dθ + ∫_π+α^2π −V_m sin(θ)dθ ]

On evaluating the integrals:

V_o(avg) = V_m/(2π) [1 + 3 cos α]

Substituting V_m = 100 V and α = 60°:

V_o(avg) = 100/(2π) [1 + 3 cos 60°]

= 100/(2π) [1 + 3(0.5)]

= 100/(2π) × 2.5

= 39.79 V

✅ Correct Answer: (C) 39.79 V

Therefore, the average voltage across the load is 39.79 V.

Question 59

The Zener diode in circuit has a breakdown voltage of 5 V. The current gain β of the transistor in the active region is 99. Ignore base-emitter voltage drop V_BE.

The current through the 20 Ω resistance in milliamperes is ________.

Round off to 2 decimal places.

Solution

Given:

V_S = 25 V
β = 99
R_B = 7 kΩ
R_E = 10 Ω
R_L = 20 Ω
V_Z = 5 V

First assume that the Zener diode is OFF.

Applying KVL around the loop:

25 − I_B(7 kΩ) − I_E(10 Ω + 20 Ω) = 0

Since,

I_E = (β + 1)I_B
I_E = 100I_B

Therefore,

25 − 7kI_B − 100I_B(30) = 0

25 − 10000I_B = 0

I_B = 25/10000

I_B = 2.5 mA

Hence emitter current is:

I_E = 100 × 2.5

I_E = 250 mA

Now calculate the voltage across the 10 Ω resistance:

V = I_E × 10
V = 250 mA × 10 Ω
V = 2.5 V

Since,

2.5 V < 5 V

The voltage is less than the Zener breakdown voltage. Therefore, the Zener diode remains OFF.

✅ Correct Answer: (C) 250 mA

Therefore, the current through the 20 Ω resistance is 250.00 mA.

Question 60

The two-bus power system has one alternator supplying a synchronous motor load through a Y-Δ transformer. The positive, negative and zero-sequence diagrams of the system are given in the question.

For a bolted line-to-line fault between phases b and c at bus 1, neglecting all pre-fault currents, the magnitude of the fault current (from phase b to phase c) in p.u. is ______.

Round off to 2 decimal places.

Enter Your Answer

Solution

For a bolted line-to-line fault between phases b and c, there is no zero-sequence current.

I₀ = 0

For a line-to-line fault, the positive- and negative-sequence currents are related as

I₂ = −I₁

Positive-sequence Thevenin impedance

Looking into bus 1, the generator branch has j0.1 + j0.1 = j0.2, while the synchronous motor branch has j0.3.

Z₁ = (j0.2 || j0.3)
= j (0.2 × 0.3)/(0.2 + 0.3)
Z₁ = j0.12 p.u.

Negative-sequence impedance

Z₂ = j0.12 p.u.

Positive-sequence current

Since the pre-fault current is neglected, the internal voltage is taken as E = 1 p.u.

I₁ = E/(Z₁ + Z₂)
= 1/(j0.12 + j0.12)
= 1/j0.24
|I₁| = 4.1667 p.u.

Fault current in phase b

For a b-c line-to-line fault,

I_b = (a² − a)I₁
|a² − a| = √3
|I_b| = √3 × 4.1667
|I_b| = 7.2169 p.u.

✅ Correct Answer: 7.22 p.u.

Therefore, the magnitude of the fault current is 7.22 p.u.

Question 61

An infinite surface of linear current density K = 5a_x A/m exists on the x-y plane.

The magnitude of the magnetic field H at a point (1, 1, 1) due to the surface current in A/m is ______.

Round off to 2 decimal places.

Enter Your Answer

Solution

Given surface current density:

K = 5a_x A/m

The infinite current sheet lies in the x-y plane. Therefore, the unit normal to the surface is a_z.

Magnetic field due to an infinite current sheet

H = 1/2 (K × a_n)

Here, K = 5a_x and a_n = a_z.

H = 1/2 (5a_x × a_z)
a_x × a_z = −a_y
H = −2.5a_y A/m

Magnitude of magnetic field

|H| = 2.5 A/m

The point (1,1,1) is above the x-y plane, so the magnitude remains the same.

✅ Correct Answer: 2.50 A/m

Therefore, the magnitude of the magnetic field is 2.50 A/m.

Question 62

The closed curve shown in the figure is described by

r = 1 + cosθ
where r = √(x² + y²)

The curve is represented parametrically by

x = r cosθ
y = r sinθ

The magnitude of the line integral of the vector field F = −y i + x j around the closed curve is ______.

Round off to 2 decimal places.

Enter Your Answer

Solution

Given vector field:

F = −y i + x j

The required line integral is

∮ F · dl = ∮ (−y dx + x dy)

Using Green's theorem

By Green's theorem,

∮ (P dx + Q dy) = ∬ (∂Q/∂x − ∂P/∂y) dA
P = −y,   Q = x
∂Q/∂x = 1
∂P/∂y = −1
∴ ∂Q/∂x − ∂P/∂y = 2

Therefore,

∮ F · dl = 2 × Area

Area of the cardioid

The closed curve is

r = 1 + cosθ

Area enclosed by a polar curve is

Area = 1/2 ∫₀^2π r² dθ

Area = 1/2 ∫₀^2π (1 + cosθ)² dθ
= 1/2 ∫₀^2π (1 + 2cosθ + cos²θ) dθ
= 3π/2

Required line integral

∮ F · dl = 2 × 3π/2
= 3π
= 9.4248...
|∮ F · dl| = 9.42

✅ Correct Answer: 9.42

Therefore, the magnitude of the line integral is 9.42.

Question 63

The signal x(t) = 2cos(180πt)cos(60πt) is sampled at 200 Hz and then passed through an ideal low pass filter having cut off frequency of 100 Hz. The maximum frequency present in the filtered signal in Hz is ______.

Round off to the nearest integer.

Solution

Given signal:

x(t) = 2cos(180πt)cos(60πt)

Using the trigonometric identity

2cos(A)cos(B) = cos(A+B) + cos(A−B)

Therefore,

x(t) = cos(240πt) + cos(120πt)

f₁ = 120 Hz
f₂ = 60 Hz

The sampling frequency is f_s = 200 Hz.

Due to sampling, spectral replicas occur at integer multiples of the sampling frequency.

For 120 Hz component:
|120 − 200| = 80 Hz

For 60 Hz component:
|60 − 200| = 140 Hz

The ideal low-pass filter has cut-off frequency 100 Hz. Hence, only frequency components below 100 Hz will pass.

60 Hz → passes
80 Hz → passes
120 Hz → rejected
140 Hz → rejected

Therefore, the maximum frequency present at the output of the low-pass filter is:

f_max = 80 Hz

✅ Correct Answer: (B) 80 Hz

Hence, the maximum frequency present in the filtered signal is 80 Hz.

Question 64

A balanced delta connected load consisting of the series connection of one resistor R = 15 Ω and a capacitor C = 212.21 μF in each phase is connected to a 3-phase, 50 Hz, 415 V supply terminals through a line having an inductance of L = 31.83 mH per phase.

Considering the change in the supply terminal voltage with loading to be negligible, the magnitude of the voltage across the terminals V_AB in volts is ______.

Round off to the nearest integer.

Solution

Given:

R = 15 Ω
C = 212.21 μF
L = 31.83 mH
f = 50 Hz
V_LL = 415 V

First, calculate the inductive reactance of the line.

X_L = 2πfL

= 2π × 50 × 31.83 × 10⁻³

X_L = 10 Ω

Now calculate the capacitive reactance of the capacitor.

X_C = 1/(2πfC)

= 1/(2π × 50 × 212.21 × 10⁻⁶)

X_C ≈ 15 Ω

Therefore, the impedance of each delta-connected load phase is:

Z_Δ = R − jX_C

= 15 − j15 Ω

Its magnitude and angle are:

|Z_Δ| = √(15² + 15²)

= 15√2 Ω

∠Z_Δ = −45°

Since the load is delta connected, the phase voltage is equal to the line voltage:

V_AB = V_phase

Let the load terminal voltage be V_AB = V∠0°. The phase current is therefore:

I_AB = V/(15 − j15)

= V/(15√2) ∠45°

For a balanced delta load, the line-current relation gives the corresponding line-current magnitude as:

|I_A| = √3 |I_AB|

= √3 V/(15√2)

The line reactance is j10 Ω. For the balanced system, the line voltage drop between phases A and B is:

V_s,AB = V_AB + j10(I_A − I_B)

On substituting the balanced phase relationships, the magnitude relation becomes:

V_s,AB² = V² + 2V² − 2V²

V_s,AB = V

The supply terminal voltage is given as:

V_s,AB = 415 V

Therefore,
V_AB = 415 V

✅ Correct Answer: (C) 415 V

Hence, the magnitude of the voltage across the terminals V_AB is 415 V.

Question 65

A quadratic function of two variables is given by

f(x₁,x₂) = x₁² + 2x₂² + 3x₁ + 3x₂ + x₁x₂ + 1

The magnitude of maximum rate of change of the function at the point (1,1) is ______.

Round off to the nearest integer.

✏️ Enter Your Answer

Solution

Given,

f(x₁,x₂) = x₁² + 2x₂² + 3x₁ + 3x₂ + x₁x₂ + 1

The maximum rate of change of a scalar function is equal to the magnitude of its gradient.

∇f = î ∂f/∂x₁ + ĵ ∂f/∂x₂

∂f/∂x₁ = 2x₁ + x₂ + 3
∂f/∂x₂ = 4x₂ + x₁ + 3

At the point (1,1),

∇f_(1,1) = î(2×1 + 1 + 3) + ĵ(4×1 + 1 + 3)
= 6î + 8ĵ

Therefore, the magnitude of the gradient is

|∇f_(1,1)| = √(6² + 8²)
= √(36 + 64)
= √100
= 10

✅ Correct Answer: 10

Therefore, the magnitude of maximum rate of change at the point (1,1) is 10.