GATE 2026 EE Official Question Paper with Answer Key & Detailed Solutions

 

GATE 2026 Electrical Engineering Question Paper with Detailed Solutions

This post provides the GATE 2026 Electrical Engineering (EE) question paper with detailed solutions in a clear and student-friendly format. The purpose of this paper analysis is to help GATE aspirants understand the type of questions asked, the difficulty level of the paper, and the correct method of solving each problem. Every question is presented with options, correct answer marking, and detailed explanation so that students can practice and revise effectively.

The GATE Electrical Engineering paper includes questions from General Aptitude, Engineering Mathematics, Electric Circuits, Signals and Systems, Electrical Machines, Power Systems, Control Systems, Power Electronics, Analog and Digital Electronics, and Measurements. This makes the paper useful for complete revision before the exam.

Why This Paper Is Important?

Previous year and memory-based GATE papers are very important for preparation because they show the actual exam pattern. By solving this paper, students can understand how concepts are applied in numerical and theoretical questions. Many questions are based on fundamentals, so a strong understanding of basic concepts is necessary. This paper also helps students identify repeated topics and important areas for revision.

In this paper, some questions are direct concept-based, while others require step-by-step calculation. For example, questions from network theory may test Norton or Thevenin equivalent concepts, while questions from machines may involve transformer vector groups, induction motor behavior, and synchronous machine operation. Power electronics questions may include diode characteristics, MOSFET switching, and converter behavior.

Question Format Used in This Post

Each question in this post is designed in an interactive format. When students click on an option, the correct answer is highlighted in green and the wrong answer is highlighted in red. If a student selects a wrong option, the correct option is also shown automatically. This makes the practice experience better and helps students learn from mistakes immediately.

A Show / Hide Solution button is also provided below every question. Students can first try the question on their own and then check the detailed solution. This method is useful because it improves problem-solving ability and builds exam confidence.

How to Use This Paper for Preparation?

First, attempt each question without looking at the solution. Note down the questions where you make mistakes. After that, read the solution carefully and understand the concept behind the answer. Do not only memorize the final answer; focus on the method. In GATE, similar concepts are often asked in different forms, so understanding the logic is more important than remembering one specific question.

For numerical questions, revise formulas and practice calculations. For theory-based questions, revise definitions, conditions, and standard results. For circuit-based and diagram-based questions, redraw the circuit or figure on paper and solve step by step. This habit will help during the actual exam.

Benefits for GATE Aspirants

• Helps in understanding the latest GATE EE exam pattern.
• Improves speed and accuracy.
• Useful for revision of important concepts.
• Provides detailed solutions for better understanding.
• Helps identify weak topics before the exam.

This GATE 2026 Electrical Engineering paper is especially useful for students preparing seriously for the upcoming GATE examination. Regular practice with such papers can improve confidence and reduce exam fear. Students should revise the solutions multiple times and make short notes of important formulas and concepts.

Final Note:
Use this paper as a complete practice set. Try to solve all questions first, then check the detailed solutions. Consistent practice, concept clarity, and proper revision are the keys to scoring well in GATE Electrical Engineering.

Question 1

'The shopkeeper sells lemons.'

In this sentence, the word 'lemons' is the ________.

A) Object

B) Subject

C) Predicate

D) Verb

Detailed Solution

The sentence is:

The shopkeeper → Subject
Sells → Verb
Lemons → Object

The word "lemons" receives the action performed by the verb "sells". Therefore, it is the object of the sentence.

Why the other options are incorrect?

• Subject → "The shopkeeper" performs the action.
• Predicate → "Sells lemons" is the predicate.
• Verb → "Sells" is the verb.

Correct Answer: A) Object

Question 2

The figure below is supposed to show three non-overlapping shapes—one oval and two triangles. Which one of the following figures P, Q, R, or S fits the missing portion indicated by '?' and completes the oval and the two triangles?

A) P

B) Q

C) R

D) S

Detailed Solution

This is a visual reasoning question.

The missing piece must simultaneously complete:

• The curved boundary of the oval.
• The upper triangle.
• The lower triangle.

Checking all four options shows that only Option Q aligns perfectly with all surrounding shapes without any overlap or mismatch.

The remaining options leave gaps or create incorrect alignments.

Correct Answer: B) Q

Question 3

At how many points will the curves y = x² and y = -x² - 2x - 1 intersect in the real (x, y) plane?

A) 0

B) 1

C) 2

D) 3

Detailed Solution

For the curves to intersect, their equations must be equal.

x² = -x² - 2x - 1

Bringing all terms to one side:

2x² + 2x + 1 = 0

Using the discriminant formula:

Δ = b² − 4ac

= 2² − 4(2)(1)

= 4 − 8 = −4

Since the discriminant is negative (Δ < 0), the quadratic equation has no real roots.

Therefore, the two curves do not intersect anywhere in the real (x, y) plane.

Correct Answer: A) 0

Question 4

'If Anish had scored hundred runs in today's match, he would have been made the captain of his team. He would have then become the youngest captain in his team's history. Unfortunately, he got out without scoring any runs. Hence, there won't be any change in the captaincy for now.'

Based on the paragraph above, which one of the following statements is true?

A) Anish made hundred runs but was denied captaincy.

B) Anish was the captain of his team before the game today.

C) The current captain is older than Anish.

D) Anish is the youngest player in his team.

Detailed Solution

Anish did not score a hundred. Therefore, he was not made captain, and the captaincy remains unchanged.

The paragraph states that if Anish had become captain, he would have been the youngest captain in the team's history. Since there is no change in captaincy, the current captain is one of the previous captains and must be older than Anish.

Why the other options are incorrect?

• A) False. Anish scored zero runs, not a hundred.
• B) False. If he had already been captain, there would be no question of making him captain.
• D) False. The paragraph only mentions that Anish would have been the youngest captain, not the youngest player.

Correct Answer: C) The current captain is older than Anish.

Question 5

Which one of the following figures P, Q, R, or S, correctly shows the 45° clockwise-rotated version of figure (I)?

A) P

B) Q

C) R

D) S

Detailed Solution

In this question, figure (I) has to be rotated by 45° clockwise.

A clockwise rotation means the upper part of the figure moves towards the right side. The shape and internal pattern should remain the same; only the orientation changes.

After rotating figure (I) by 45° clockwise, the arrangement matches with figure S.

Options P, Q, and R do not match the correct rotated position because their pattern or direction is different.

Correct Answer: D) S

Question 6

Match the words in Column I with their synonyms in Column II.

A) (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r)

B) (i)-(q), (ii)-(s), (iii)-(r), (iv)-(p)

C) (i)-(s), (ii)-(p), (iii)-(q), (iv)-(r)

D) (i)-(r), (ii)-(s), (iii)-(p), (iv)-(q)

Detailed Solution

To solve this question, match each word with its correct synonym.

Column I	Correct Synonym
Lonely	Solitary (q)
Literal	Verbatim (p)
Lousy	Terrible (s)
Lethal	Deadly (r)

Therefore, the correct matching is:

(i)-(q), (ii)-(p), (iii)-(s), (iv)-(r)

Correct Answer: A

Question 7

In the given figure, PQ is the diameter of a circle with center O. Two points R and S are chosen on the circle such that ∠ROS = 80°. When PR and QS are extended, they meet at T. The value of ∠RTS is ______.

A) 40°

B) 50°

C) 60°

D) 80°

Detailed Solution

Here, PQ is the diameter of the circle.

So, the arc PQ of the semicircle is:

Arc PQ = 180°

Also, the central angle is given:

∠ROS = 80°

Therefore, the minor arc RS is:

Arc RS = 80°

When two secants meet outside a circle, the angle formed is half the difference of the intercepted arcs.

∠RTS = 1/2 × (Arc PQ − Arc RS)

∠RTS = 1/2 × (180° − 80°)

∠RTS = 1/2 × 100° = 50°

Correct Answer: B) 50°

Question 8

Based on the relationship between each polygon and the number inside it, the value of X is ______.

A) 720

B) 596

C) 24

D) 240

Detailed Solution

The number inside each polygon follows the factorial of the number of its sides.

Polygon	Sides	Value
Triangle	3	3! = 6
Square	4	4! = 24
Pentagon	5	5! = 120
Hexagon	6	6! = 720

Therefore, the value inside the hexagon is:

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Correct Answer: A) 720

Question 9

Consider a linear arrangement of seven bulbs, each of which can be in the ON or OFF states. The number of bulbs which are ON at the end of Step 8 is ______.

A) 5

B) 4

C) 3

D) 0

Detailed Solution

Let ON = 1 and OFF = 0.

Initial state is:

0 0 0 1 0 0 0

Now applying the given rules step by step:

Step 1: 0 0 1 1 1 0 0

Step 2: 0 1 1 0 1 1 0

Step 3: 1 1 1 0 1 1 1

Step 4: 1 0 1 0 1 0 1

After Step 4, the pattern becomes stable, so it remains the same till Step 8:

Step 8: 1 0 1 0 1 0 1

Number of ON bulbs = 4

Correct Answer: B) 4

Question 10

P and Q are two positive integers such that

P² = Q² + 13

The product of the numbers P and Q is ______.

A) 13

B) 26

C) 39

D) 42

Detailed Solution

Given:

P² = Q² + 13

Rearranging,

P² − Q² = 13

Using the identity:

(P − Q)(P + Q) = 13

Since 13 is a prime number and P and Q are positive integers, the only possible factorization is:

P − Q = 1 and P + Q = 13

Adding the two equations:

2P = 14 ⇒ P = 7

Therefore,

Q = 6

Hence,

P × Q = 7 × 6 = 42

Correct Answer: D) 42

Question 11

Consider the infinite-length, discrete-time sequence x[n] = 0.9^|n|, where n is an integer. The region of convergence of its Z-transform X(z) is given by:

Note: z is a complex variable.

A) |z| > 0.9

B) |z| < 0.9

C) 0.9 < |z| < 1/0.9

D) { z such that |z| < 0.9 } ∪ { z such that |z| > 1/0.9 }

Detailed Solution

Given sequence:

x[n] = 0.9^|n|

This is a two-sided sequence because it exists for both positive and negative values of n.

For n ≥ 0:

x[n] = 0.9ⁿ

The ROC for the right-sided part is:

|z| > 0.9

For n < 0:

x[n] = 0.9^-n

This gives a left-sided part whose ROC is:

|z| < 1/0.9

For the complete two-sided sequence, the ROC is the common region of both:

0.9 < |z| < 1/0.9

Correct Answer: C) 0.9 < |z| < 1/0.9

Question 12

Let x_c(t) be any continuous-time periodic signal with period T. It is sampled uniformly with a sampling period T_s (where T_s ≠ T) resulting in the discrete-time sequence:

x[n] = x_c(nT_s)

Which one of the following statements is correct about x[n]?

A) x[n] will always be periodic with period T/T_s for all values of T/T_s

B) x[n] will always be periodic with period 1 for all values of T/T_s

C) x[n] will never be periodic

D) x[n] will be periodic if and only if T/T_s is a rational number

Detailed Solution

For a discrete-time sequence to be periodic, there must exist a positive integer N such that:

x[n+N] = x[n]

Since the original continuous-time signal has period T, we require:

NT_s = kT

where k is an integer.

Rearranging,

T/T_s = N/k

This is possible only when T/T_s is a rational number.

If T/T_s is irrational, no integer values of N and k satisfy the equation, so the sampled sequence is not periodic.

Therefore, x[n] is periodic if and only if T/T_s is a rational number.

Correct Answer: D

Question 13

The Laplace transform of the step response of a system is given by

Y(s) = 100 / [s(s + 100)]

The rise time is defined as the time taken for the response to go from 0.1 to 0.9 of its final value. The settling time is defined as the time taken for the response to reach 0.98 of its final value.

For this system, the rise time (T_r), settling time (T_s), and time constant (T_c), all expressed in seconds, are:

A) T_r = 0.022, T_s = 0.04, T_c = 0.01

B) T_r = 0.22, T_s = 0.404, T_c = 0.01

C) T_r = 2.2, T_s = 4.04, T_c = 1.01

D) T_r = 22, T_s = 40.4, T_c = 10.1

Detailed Solution

Given:

Y(s)=100/[s(s+100)]

This is the step response of a first-order system.

The standard form is:

G(s)=100/(s+100)

Therefore,

Time Constant:

τ = 1/100 = 0.01 s

Rise Time (10%–90%)

T_r ≈ 2.2τ

T_r = 2.2 × 0.01 = 0.022 s

Settling Time (98%)

T_s ≈ 4τ

T_s = 4 × 0.01 = 0.04 s

Hence,

• T_r = 0.022 s
• T_s = 0.04 s
• T_c = 0.01 s

Correct Answer: A

Question 14

Consider the following differential equation:

t²(d²y/dt²) + 7t(dy/dt) + 8ty = 10 sin(t)

Which one of the following options is correct?

A) It is a linear differential equation

B) It is a nonlinear differential equation

C) It is a time-invariant differential equation

D) It is a second-order partial differential equation

Detailed Solution

The given equation is:

t²(d²y/dt²) + 7t(dy/dt) + 8ty = 10 sin(t)

All derivatives of y appear only to the first power and are not multiplied together.

Also, the coefficients (t², 7t, and 8t) depend only on the independent variable t, not on y or its derivatives.

Hence, the equation satisfies the definition of a linear ordinary differential equation (ODE).

Why the other options are incorrect?

• B) Incorrect. There are no nonlinear terms such as y², (dy/dt)², or y(dy/dt).
• C) Incorrect. The coefficients depend explicitly on t, so the system is time-varying, not time-invariant.
• D) Incorrect. The equation contains derivatives with respect to only one variable (t), so it is an ODE, not a partial differential equation.

Correct Answer: A) It is a linear differential equation.

Question 15

A uniform ring of radius R carries a total charge Q. Which one of the following options correctly quantifies the magnitude of the force on a point charge of strength q kept at the center of the ring?

(ε is the permittivity of the medium)

A) Qq / (4πεR)

B) Qq / (4πεR²)

C) 0

D) (q / 4πεR) × (Q / 2πR)

Detailed Solution

A uniformly charged ring is perfectly symmetric about its center.

Each small charge element dq on the ring produces an electric field at the center.

For every charge element, there exists an opposite charge element producing an electric field of equal magnitude but opposite direction.

Hence, all electric field vectors cancel each other.

Therefore, the net electric field at the center of the ring is:

E = 0

The force on the point charge is:

F = qE = q × 0 = 0

Correct Answer: C) 0

Question 16

A positive point charge with velocity v = 5x̂ enters a region having electric field E = 4ŷ and magnetic field B = −6ẑ. Which one of the following statements is correct for the force on the charge as it enters the region?

A) The force will be along the magnetic field but perpendicular to the electric field.

B) The force will be along the electric field but perpendicular to the magnetic field.

C) The force will be perpendicular to both electric and magnetic fields.

D) The magnetic field does not exert any force on the charge.

Detailed Solution

The force on a moving charge is given by the Lorentz force equation:

F = q(E + v × B)

Given:

• v = 5x̂
• E = 4ŷ
• B = −6ẑ

Calculate the magnetic force direction:

v × B = x̂ × (−ẑ)

Using the right-hand rule:

x̂ × (−ẑ) = +ŷ

Thus, the magnetic force is also along the +ŷ direction.

The electric force is:

F_e = qE, which is also along +ŷ.

Hence, both electric and magnetic forces act in the same direction (+ŷ).

The total force is therefore:

• ✔ Along the electric field.
• ✔ Perpendicular to the magnetic field (ẑ).

Correct Answer: B) The force will be along the electric field but perpendicular to the magnetic field.

Question 17

A 15 kVA, 1100 V/220 V, single-phase two-winding transformer is configured as a 1.32 kV/1.1 kV autotransformer.

What will be the rating of the autotransformer?

A) 60 kVA

B) 75 kVA

C) 90 kVA

D) 100 kVA

Detailed Solution

Given:

• Two-winding transformer rating = 15 kVA
• Voltage rating = 1100 V / 220 V
• Autotransformer voltage = 1320 V / 1100 V

The current rating of the series winding remains the same as in the original transformer.

Rated current of the 220 V winding:

I = 15000 / 220 = 68.18 A

The output voltage of the autotransformer is:

V = 1100 V

Hence, the autotransformer rating is:

S = V × I

= 1100 × 68.18

≈ 75,000 VA = 75 kVA

Correct Answer: B) 75 kVA

Question 18

Consider a power system with N buses, of which P are generator buses and the remaining Q are load buses, where there is no generation.

Assume that there are no reactive power-limit violations at the generator buses.

What is the size of the Jacobian matrix in the Newton-Raphson load flow method?

A) 2N × 2N

B) (2N − 1 − P) × (2N − 1 − P)

C) (2N − Q) × (2N − Q)

D) (P + Q) × (P + Q)

Detailed Solution

In Newton-Raphson load flow, the unknown variables are:

• Voltage angles at all buses except the slack bus.
• Voltage magnitudes at load buses only.

Total buses = N

Number of generator buses = P

Number of load buses = Q

Since one bus is the slack bus:

Number of unknown voltage angles = N − 1

Voltage magnitudes are unknown only for load buses:

Number of unknown voltage magnitudes = Q

Therefore, total unknowns are:

(N − 1) + Q

Since the remaining Q buses are load buses and the other P buses are generator buses,

N = P + Q

So,

(N − 1) + Q = N − 1 + (N − P) = 2N − 1 − P

Hence, the Jacobian matrix size is:

(2N − 1 − P) × (2N − 1 − P)

Correct Answer: B

Question 19

The initial three-phase voltage phasors V_A, V_B, and V_C at a bus of a power network are as shown in Case-1. Due to a disturbance, the bus voltage phasors changed in phase by a small angle Δθ, and the magnitudes remained the same as depicted in Case-2.

Which one of the following statements is correct about the zero sequence components?

A) The zero sequence components in Case-1 and Case-2 have the same phase angle and magnitude

B) The magnitude of the zero sequence component in Case-1 is greater than that in Case-2

C) The magnitude of the zero sequence component in Case-2 is greater than that in Case-1

D) The zero sequence components in Case-1 and Case-2 have the same magnitude but different phase angles

Detailed Solution

The zero sequence component is given by:

V₀ = (V_A + V_B + V_C) / 3

In Case-2, all three phasors are shifted by the same angle Δθ, while their magnitudes remain unchanged.

So, each phasor is multiplied by the same rotation factor:

e^jΔθ

Therefore, the new zero sequence component becomes:

V_0,new = e^jΔθ V_0,old

This means:

• The magnitude remains the same.
• The phase angle changes by Δθ.

Correct Answer: D) The zero sequence components in Case-1 and Case-2 have the same magnitude but different phase angles.

Question 20

A certain application requires power at a frequency of 16.67 Hz, while the available grid frequency is 50 Hz. A 3-phase synchronous motor connected to the 50 Hz, 3-phase grid, driving a synchronous generator, is used for this application.

Which one of the following combinations is a suitable choice for the number of poles in the motor and generator, respectively?

A) 2, 6

B) 6, 2

C) 4, 6

D) 6, 4

Detailed Solution

For a synchronous machine:

N_s = 120f / P

The synchronous motor is connected to a 50 Hz supply.

If the motor has 6 poles, then:

N = 120 × 50 / 6 = 1000 rpm

The generator is driven at the same speed. If the generator has 2 poles, then output frequency is:

f = NP / 120

f = 1000 × 2 / 120 = 16.67 Hz

Hence, the suitable combination is:

Motor poles = 6, Generator poles = 2

Correct Answer: B) 6, 2

Question 21

Which one of the following options is correct regarding the typical double-squirrel-cage structure used in induction motors?

A) At starting, a larger portion of the rotor current flows in the outer cage

B) At starting, a larger portion of the rotor current flows in the inner cage

C) At rated speed, most of the rotor current flows in the outer cage

D) The purpose of the double-squirrel-cage structure is to lower the effective rotor resistance at starting

Detailed Solution

In a double-squirrel-cage induction motor, the rotor has two cages:

• Outer cage: High resistance and low leakage reactance.
• Inner cage: Low resistance and high leakage reactance.

At starting, slip is high and rotor frequency is high. Due to high leakage reactance of the inner cage, more current flows through the outer cage.

The outer cage has higher resistance, which improves starting torque.

At normal running speed, slip is low and rotor frequency is low. Then, most current flows through the inner cage because it has lower resistance.

Correct Answer: A) At starting, a larger portion of the rotor current flows in the outer cage.

Question 22

The figure shows the single-line diagram of a synchronous generator delivering P = 50 MW of power at unity power factor to an infinite bus.

I_s denotes the stator current phasor. If the field excitation is increased, which one of the following options correctly describes its effect on the stator current, power factor, and load angle of the machine?

A) Stator current increases, power factor becomes lagging, load angle remains the same

B) Stator current decreases, power factor becomes leading, load angle remains the same

C) Stator current increases, power factor becomes lagging, load angle decreases

D) Stator current increases, power factor becomes leading, load angle increases

Detailed Solution

The generator is delivering the same real power P = 50 MW to an infinite bus.

For a synchronous generator connected to an infinite bus, terminal voltage and frequency remain constant.

When field excitation is increased:

• The internal generated emf increases.
• The generator supplies more reactive power.
• The stator current increases.
• The power factor becomes lagging.

Since the real power output is maintained at the same value, the load angle is taken to remain the same for this operating condition.

Correct Answer: A) Stator current increases, power factor becomes lagging, load angle remains the same.

Question 23

A circuit with ideal elements is shown.

Which one of the following options correctly identifies all the linear elements in the circuit?

A) R only

B) R, L, and C only

C) D only

D) L, C, and D only

Detailed Solution

For ideal circuit elements:

• Resistor (R): Linear because V = IR.
• Inductor (L): Linear because V = L(di/dt).
• Capacitor (C): Linear because I = C(dV/dt).
• Diode (D): Nonlinear because its V-I characteristic is nonlinear and depends on its operating region.

Hence, the linear elements in the circuit are:

R, L and C only.

Correct Answer: B) R, L, and C only.

Question 24

For the circuit shown, which one of the following options correctly identifies the Thevenin’s equivalent parameters between nodes Y and Z?

A) V_TH = 100 V, R_TH = 10 kΩ

B) V_TH = 140 V, R_TH = 0 Ω

C) V_TH = 100 V, R_TH = 0 Ω

D) V_TH = 140 V, R_TH = 10 kΩ

Detailed Solution

Take node W as reference.

From the 100 V source:

V_X = 100 V

From the 20 V source between X and Y:

V_Y = V_X + 20 = 120 V

From the 20 V source between W and Z:

V_Z = 20 V

Therefore,

V_TH = V_Y − V_Z = 120 − 20 = 100 V

For finding R_TH, all independent voltage sources are replaced by short circuits.

After short-circuiting the voltage sources, nodes Y and Z get directly shorted through ideal voltage sources.

Hence,

R_TH = 0 Ω

Correct Answer: C) V_TH = 100 V, R_TH = 0 Ω

Question 25

Refer to the four circuits shown.

 Which one of the following options for k₁, k₂, k₃, and k₄ makes all of them realizable?

A) k₁ = 1, k₄ = −1/3, for all values of k₂ and k₃

B) k₂ = −2, k₃ = +1/3, for all values of k₁ and k₄

C) k₁ = 2, k₂ = 0.5, k₃ = −2/3, k₄ = −3

D) k₁ = 2, k₂ = −0.5, k₃ = −2/3, k₄ = +3

Detailed Solution

For ideal sources connected together, the source constraints must not contradict each other.

First circuit: Two voltage sources are connected in parallel. Hence, their voltages must be equal.

V_X = V_D

But, V_X = k₁V_D. Therefore:

k₁ = 1

Second circuit: A voltage source and a current source are in parallel. This is realizable for any value of k₂.

Third circuit: A current source and a voltage source are in parallel. This is realizable for any value of k₃.

Fourth circuit: Two current sources are connected in parallel in the same loop. For the circuit to be realizable, the net current must be zero:

I_D + I_X = 0

Given:

I_X = 3k₄I_D

So,

I_D + 3k₄I_D = 0

1 + 3k₄ = 0

k₄ = −1/3

Correct Answer: A

Question 26

A single-phase voltage source

v_s = 325 sin(2π50t) V

delivers a current

i = 12 sin(2π50t) + 9 sin(2π150t) A

The load power factor, correct up to two decimal places, is:

A) 1.00

B) 0.80

C) 0.65

D) 0.57

Detailed Solution

The voltage contains only the fundamental frequency (50 Hz).

The current contains:

• Fundamental component = 12 sin(2π50t)
• Third harmonic component = 9 sin(2π150t)

The harmonic current does not contribute to real power, because the source voltage has no third harmonic component.

Real Power:

P = V_rms × I_1,rms

Total RMS Current:

I = √(I₁² + I₃²)

Using peak values:

I₁ = 12/√2 A, I₃ = 9/√2 A

I = √[(12² + 9²)/2] = √112.5 = 10.61 A

Power Factor:

PF = I₁ / I = (12/√2) / 10.61 ≈ 0.80

Correct Answer: B) 0.80

Question 27

The figure shows a straight-line approximation for the forward characteristics of a power diode. A continuous on-state current of 15 A is flowing through the diode.

What is the power loss in the diode?

A) 32.8 W

B) 21.2 W

C) 18.6 W

D) 23.1 W

Detailed Solution

From the graph:

• Threshold voltage, V₀ = 1.0 V
• At I = 50 A, V = 1.8 V

The diode's dynamic resistance is

r_d = (1.8 − 1.0) / 50 = 0.8 / 50 = 0.016 Ω

For I = 15 A, the forward voltage is

V_D = V₀ + I r_d

= 1 + (15 × 0.016)

= 1.24 V

The power dissipated is

P = V_D × I

= 1.24 × 15

P = 18.6 W

Correct Answer: C) 18.6 W

Question 28

Consider the circuit shown in Figure (a). A gate pulse V_g is applied between time instants t₀ and t₁. After t₁, during the MOSFET turn OFF process, it experiences a voltage overshoot.

Based on the V_ds waveforms shown in Figure (b), which one of the following options is correct?

A) R₁ > R₂ > R₃

B) R₁ > R₃ > R₂

C) R₃ > R₂ > R₁

D) R₂ > R₃ > R₁

Detailed Solution

During MOSFET turn OFF, the rate of fall of current depends on the gate resistance R_g.

A larger gate resistance causes slower turn OFF, which reduces di/dt and hence reduces the voltage overshoot.

A smaller gate resistance causes faster turn OFF, larger di/dt, and therefore larger voltage overshoot.

From the waveform:

• With R_f = R₁, voltage overshoot is the highest.
• With R_f = R₃, voltage overshoot is intermediate.
• With R_f = R₂, voltage overshoot is the lowest.

Therefore, resistance is lowest for highest overshoot and highest for lowest overshoot:

R₁ > R₃ > R₂

Correct Answer: B) R₁ > R₃ > R₂

Question 29

The asymptotic Bode magnitude plot of a system is shown.

Which one of the following options best represents the transfer function of the system?

A) G(s)= (1+s/ω₀)/(s/ω₀)

B) G(s)= (s/ω₀)/(1+s/ω₀)

C) G(s)= (1+s/ω₀)/(1−s/ω₀)

D) G(s)= (1−s/ω₀)/(1+s/ω₀)

Detailed Solution

From the Bode magnitude plot:

• At low frequencies, the slope is −20 dB/decade.
• At ω = ω₀, the slope changes from −20 dB/decade to 0 dB/decade.
• At high frequencies, the magnitude becomes constant at 0 dB.

A slope of −20 dB/decade at low frequency indicates a zero at the origin in the denominator (or equivalently, a differentiator in the numerator).

The change of slope by +20 dB/decade at ω₀ indicates the presence of a pole at ω₀.

Thus the transfer function is

G(s)= (s/ω₀)/(1+s/ω₀)

This gives:

• Low frequency: |G(jω)| ∝ ω → −20 dB/decade.
• High frequency: |G(jω)| → 1 (0 dB).

Correct Answer: B

Question 30

Two n × n matrices A and B have a common eigenvalue 2, and the same corresponding nonzero eigenvector.

Which of the following options is/are correct?

Note: I is the n × n identity matrix.

A) Determinant (A − 2I) = 0

B) Determinant (B − 2I) = 0

C) Determinant (A + B − 2I) = 0

D) Determinant (A + B − 4I) = 0

Detailed Solution

Let the common nonzero eigenvector be x.

Since 2 is an eigenvalue of A:

Ax = 2x

So,

(A − 2I)x = 0

Since x ≠ 0, the matrix A − 2I is singular.

det(A − 2I) = 0

Similarly, since 2 is also an eigenvalue of B:

det(B − 2I) = 0

Now,

(A + B)x = Ax + Bx = 2x + 2x = 4x

So, 4 is an eigenvalue of A + B.

Therefore,

det(A + B − 4I) = 0

But det(A + B − 2I) = 0 is not necessarily true.

Correct Answers: A, B and D

Question 31

A 220V/12V single-phase transformer is designed for use in India and rated 100 VA at 50 Hz. Later, this unit is shipped to the USA where it is used as a 110V/6V transformer at 60 Hz.

Which of the following statements is/are correct?

A) No-load current drawn would be smaller for operation in the USA compared to that in India

B) For the same load current, the losses would be higher in the USA compared to that in India

C) The peak magnetic flux density in the core would be higher while operating in the USA compared to that in India

D) The eddy current losses in the core would be approximately 44% higher while operating in the USA compared to that in India

Detailed Solution

For a transformer:

B_max ∝ V / f

In India:

V/f = 220/50 = 4.4

In USA:

V/f = 110/60 = 1.83

Since V/f is much smaller in the USA, the flux density is smaller. Hence, the magnetizing current or no-load current is also smaller.

So, statement A is correct.

Statement B is incorrect because for the same load current, copper loss remains nearly the same, and core losses do not become higher.

Statement C is incorrect because peak flux density is lower in the USA.

Statement D is incorrect because eddy current loss is not 44% higher; with lower voltage, it actually reduces.

Correct Answer: A only

Question 32

Given that F(x,y,z) = sin(y)x̂ + cos(x)ŷ + 5ẑ, the integral ∯_S F(x,y,z) · dS over the unit sphere S centered at the origin evaluates to ______.

Round off to one decimal place.

Detailed Solution

Using the divergence theorem:

∯ F · dS = ∭ (∇ · F) dV

Given:

F = sin(y)x̂ + cos(x)ŷ + 5ẑ

Now calculate the divergence:

∇ · F = ∂/∂x[sin(y)] + ∂/∂y[cos(x)] + ∂/∂z[5]

= 0 + 0 + 0 = 0

Hence,

∯ F · dS = ∭ 0 dV = 0

Final Answer: 0.0

Question 33

In the linear regulator circuit shown, the base to emitter voltage V_BE of the BJT is 0.6 V. The Zener diode clamps the base voltage to 5.4 V. Ignore the biasing current of the Zener diode and BJT.

The maximum possible efficiency of the regulator circuit is ______ %.

Round off to one decimal place.

Detailed Solution

The Zener diode clamps the transistor base voltage at:

V_B = 5.4 V

Given:

V_BE = 0.6 V

Therefore, the regulated output voltage is:

V_out = V_B − V_BE

V_out = 5.4 − 0.6 = 4.8 V

Ignoring the bias currents, the input current is approximately equal to the load current.

Hence, the maximum efficiency is:

η = (P_out/P_in) × 100

= (V_out/V_in) × 100

= (4.8/10) × 100 = 48.0%

Final Answer: 48.0%

Question 34

Consider the circuit shown. Assume that the diode D is ideal. The supply voltage is v_s = 325 sin(2π50t) V, L = 500 μH, and R = 10 Ω.

The peak diode current is ______ A.

Round off to one decimal place.

Detailed Solution

Given:

• Peak voltage, V_m = 325 V
• Resistance, R = 10 Ω
• Inductance, L = 500 μH = 0.0005 H
• Frequency, f = 50 Hz

Angular frequency:

ω = 2πf = 2π × 50 = 314.16 rad/s

Inductive reactance:

X_L = ωL = 314.16 × 0.0005 = 0.157 Ω

Total impedance:

Z = √(R² + X_L²)

Z = √(10² + 0.157²) ≈ 10.001 Ω ≈ 10 Ω

Peak diode current:

I_peak = V_m / Z

I_peak = 325 / 10 = 32.5 A

Final Answer: 32.5 A

Question 35

A is an m × m skew-symmetric matrix with real-valued entries, and x is an m-dimensional column vector with real-valued entries such that x^Tx = 1.

The quantity x^TAx evaluates to ______.

Answer in integer.

Detailed Solution

A skew-symmetric matrix satisfies:

A^T = −A

Let

k = x^TAx

Since k is a scalar, its transpose is equal to itself:

k = k^T

Taking transpose:

(x^TAx)^T = x^TA^Tx

Substituting A^T = −A:

k = -x^TAx = -k

Hence,

2k = 0

k = 0

Therefore, for every real skew-symmetric matrix A and every real vector x,

x^TAx = 0

Final Answer: 0

Question 36

A time-limited waveform g(x) is specified as:

g(x) = -k, for -π < x ≤ 0
g(x) = +k, for 0 < x ≤ π
g(x) = 0, otherwise

A new waveform f(x) is constructed from g(x) as a periodic function with period 2π. The sum of the coefficients of the third harmonics of the sine and cosine terms in the trigonometric Fourier series expansion of f(x) is 2/(3π). What is the value of k?

A) 1

B) 1/2

C) 1/3

D) 1/4

Detailed Solution

The periodic function is odd, so cosine coefficients are zero.

For sine coefficient:

bₙ = (2k/π) ∫₀ᴨ sin(nx) dx

bₙ = (2k/π) × [1 − (-1)ⁿ]/n

For third harmonic, n = 3:

b₃ = 4k/(3π)

Given sum of third harmonic sine and cosine coefficients:

4k/(3π) = 2/(3π)

k = 1/2

Correct Answer: B) 1/2

Question 37

In the circuit shown, the open-loop gain of the operational amplifier is A₀ = 105.

Find the voltage gain of the circuit.

Round off to two decimal places.

A) −16.67

B) −20.00

C) −21.00

D) −12.67

Detailed Solution

This is an inverting amplifier with:

• Input resistor, R_in = 5 kΩ
• Feedback resistor, R_f = 100 kΩ
• Open-loop gain, A = 10⁵

Ideal inverting gain is:

−R_f/R_in = −100/5 = −20

For finite open-loop gain:

A_v = −20 / (1 + 21/A)

Substituting A = 10⁵:

A_v = −20 / (1 + 21/100000)

A_v ≈ −19.996 ≈ −20.00

Correct Answer: B) −20.00

Question 38

Three single-phase 11 kV / 3.3 kV transformers are connected to form a three-phase transformer bank as shown.

Considering ABC phase sequence, determine the vector group of the transformer.

A) Dd0

B) Dd4

C) Dd6

D) Dd10

Detailed Solution

From the given winding connections:

• HV side is connected in Delta (D).
• LV side is also connected in Delta (d).
• The winding polarities are corresponding, producing zero phase displacement between primary and secondary line voltages.

Therefore, the transformer belongs to vector group:

Dd0

Correct Answer: A) Dd0

Question 39

In the circuit shown, the phase currents are 

Ia = 572.812 + j50.115A

Ib = -254.525 - j459.175A

Ic = -207.803 + j444.091A

Given that the CTs are ideal (300:5) and the auxiliary transformer (YNd) ratio is 2:1, determine the value of I_AR (rounded to three decimal places).

A) 0 A

B) 0.653 ∠17.556° A

C) 537.240 ∠4.105° A

D) 8.954 ∠4.105° A

Detailed Solution

Sum of the primary phase currents:

I_A+I_B+I_C =(572.812+j50.115)+(-254.525-j459.175)+(-207.083+j444.091)

=111.204+j35.031 =116.59∠17.556° A

Main CT ratio = 300:5 = 60:1

I_CT=116.59/60=1.943∠17.556° A

Considering the 2:1 YNd auxiliary transformer connection, the relay current becomes:

I_AR=1.943/3=0.653∠17.556° A

Correct Answer: B) 0.653 ∠17.556° A

Question 40

The operating characteristic of a reactance relay is given by X ≤ 1 Ω, where X is the reactance calculated by the relay. Its operating characteristic in the admittance plane (G-B plane), where G and B denote conductance and susceptance respectively, is:

A) G² + (B + 0.5)² ≥ 1/4

B) B ≥ 1

C) (G − 1)² + B² ≤ 1/2

D) G² + (B − 1)² ≥ 1/2

Detailed Solution

Admittance is

Y = G + jB

Therefore, the impedance is

Z = 1/Y = (G − jB)/(G² + B²)

The reactance is the imaginary part of Z:

X = −B/(G² + B²)

Applying the relay operating condition:

−B/(G² + B²) ≤ 1

Rearranging,

G² + B² + B ≥ 0

Completing the square,

G² + (B + 0.5)² ≥ 1/4

Correct Answer: A) G² + (B + 0.5)² ≥ 1/4

Question 41

A three-phase two-winding transformer has a voltage transformation ratio Vp/Vs = 0.866 + j0.5. The admittance corresponding to the leakage impedance is y_t p.u. (Magnetizing branch neglected). Which of the following is the correct Y-bus representation?

A) [ y_t/(0.866+j0.5)    -y_t/(0.866+j0.5) ;
    -y_t/(0.866+j0.5)    y_t/(0.866+j0.5) ]

B) [ y_t -y_t ;
    -y_t y_t ]

C) [ y_t -y_t/(0.866+j0.5) ;
    -y_t/(0.866+j0.5) y_t ]

D) [ y_t -y_t/(0.866−j0.5) ;
    -y_t/(0.866+j0.5) y_t ]

Detailed Solution

Voltage transformation ratio is

t = V_p/V_s = 0.866 + j0.5

Its complex conjugate is

t* = 0.866 − j0.5

For an off-nominal transformer, the Y-bus matrix is

[ y_t/|t|²    −y_t/t*
−y_t/t    y_t ]

Since

|t|² = (0.866)² + (0.5)² = 1

Therefore,

[ y_t    −y_t/(0.866−j0.5)
−y_t/(0.866+j0.5)    y_t ]

Correct Answer: D

Question 42

An electrical component has voltage drop v = V_m sin(ωt), when the current through it is i = I_m sin(ωt − θ). What is the average power dissipated over a half cycle corresponding to ω?

A) 0

B) V_mI_m cosθ

C) (V_mI_m/2) cosθ

D) (V_mI_m/4) cosθ

Detailed Solution

The instantaneous power is:

p(t)=v(t)i(t) =V_msin(ωt) × I_msin(ωt−θ)

Using the trigonometric identity:

sinA·sinB = ½[cos(A−B) − cos(A+B)]

Therefore,

p(t)= (V_mI_m/2) [cosθ − cos(2ωt−θ)]

Over one half cycle, the average value of cos(2ωt−θ) is zero.

Hence, the average power over a half cycle is:

P_avg = (V_mI_m/2) cosθ

Correct Answer: C) (V_mI_m/2) cosθ

Question 43

The electrical network shown has an independent voltage source of 10 V and a current source of 1u(t) mA. 

The voltage across the capacitor at time instants t = 0⁺, t = 0.50, and t = ∞, respectively, is:

A) 8.00 V, 28.00 V, 26.36 V

B) 8.00 V, 26.36 V, 28.00 V

C) 10.00 V, 26.36 V, 28.00 V

D) 10.00 V, 28.00 V, 26.36 V

Detailed Solution

Before switching, the current source is zero. So the capacitor voltage is obtained using voltage division:

V_C(0^-) = 10 × 100k/(25k + 100k) = 8 V

Capacitor voltage cannot change instantly, therefore:

V_C(0⁺) = 8 V

At steady state after switching, the capacitor acts as open circuit. Applying KCL:

(V − 10)/25k + V/100k = 1 mA

V = 28 V

Equivalent resistance seen by capacitor:

R_eq = 25k || 100k = 20kΩ

Time constant:

τ = R_eqC = 20k × 10μF = 0.2 s

Now,

V_C(t) = V∞ + [V(0⁺) − V∞]e^−t/τ

At t = 0.50 s:

V_C(0.5) = 28 + (8 − 28)e^−0.5/0.2

V_C(0.5) = 26.36 V

Correct Answer: B) 8.00 V, 26.36 V, 28.00 V

Question 44

The terminal voltage and current of a linear electrical network are given in the table. The correct choice for the Norton equivalent parameters (I_N, R_N) is:

A) I_N = 3.0 A, R_N = 24.0 Ω

B) I_N = 12.0 A, R_N = 2.0 Ω

C) I_N = 2.0 A, R_N = 12.0 Ω

D) I_N = 2.0 A, R_N = 24.0 Ω

Detailed Solution

From the given table:

(i_t, v_t) = (-0.5, 18), (0.5, 30), (1.0, 36)

Since the network is linear, the relation between voltage and current is a straight line.

Slope gives Norton resistance:

R_N = Δv/Δi = (30 − 18)/(0.5 − (-0.5))

R_N = 12/1 = 12 Ω

Now using:

v_t = R_N(I_N + i_t)

Using v_t = 30 V and i_t = 0.5 A:

30 = 12(I_N + 0.5)

30 = 12I_N + 6

I_N = 2 A

Correct Answer: C) I_N = 2.0 A, R_N = 12.0 Ω

Question 45

The digital circuit shown has 3 inputs (x, y, and z). 

The simplified logical expression for the output (OUT) is:

A) x̄yz̄

B) 0

C) x̄(y + z)

D) 1

Detailed Solution

The top gate is a NOR gate. Its output becomes 1 only when all inputs are 0.

NOR output = (x + y + z)' = x̄ȳz̄

The lower logic network output is combined with the NOR output using the final AND gate.

But whenever the NOR output is 1, we have:

x = 0, y = 0, z = 0

For these input values, the lower logic network output becomes 0.

Therefore, final output is:

OUT = 0

Correct Answer: B) 0

Question 46

The MOSFET switches shown in the circuit are ideal. 

Which of the following is the correct Boolean logical expression of the output (OUT), and the maximum possible power (P) consumed by the circuit?

A) OUT = AB + C̄, P = 5 mW

B) OUT = (Ā + B̄)C̄, P = 5 mW

C) OUT = ABC̄ + C̄, P = 7.5 mW

D) OUT = ABC, P = 7.5 mW

Detailed Solution

The left pull-down path conducts when both switches controlled by A and B are ON.

So, one condition for OUT to become LOW is:

AB = 1

The right 10 kΩ branch controls another MOSFET. When C = 0, the node is pulled HIGH through 10 kΩ, so this MOSFET turns ON and pulls OUT LOW.

Hence, OUT becomes LOW when:

AB + C̄ = 1

Therefore, the output logic is:

OUT = AB + C̄

Using De Morgan’s theorem:

OUT = ABC

Now, for maximum power consumption:

When C = 1, current flows through the 10 kΩ resistor:

P₁ = V²/R = 5²/10000 = 2.5 mW

When A = 1 and B = 1, current also flows through the 5 kΩ resistor:

P₂ = 5²/5000 = 5 mW

Maximum power:

P = 5 mW + 2.5 mW = 7.5 mW

Correct Answer: D) OUT = ABC, P = 7.5 mW

Question 47

Consider the single-phase voltage source inverter circuit feeding an inductive load (L). Assume that the power MOSFET switches are ideal. S1 and S2 are switched on during the first 10 μs, and S3 and S4 are switched on during the next 10 μs in a switching cycle. The switches in the same leg are switched in a complementary fashion. Neglect the dead time. The waveform of the inductor current iL in steady state is triangular with a peak value of 5 A. 

The rms value of the current through the switch S1 is:

A) 2.88 A

B) 2.04 A

C) 3.54 A

D) 2.50 A

Detailed Solution

In one switching cycle, S1 and S2 conduct for 10 μs, and S3 and S4 conduct for the next 10 μs.

Therefore, the switching period is:

T = 20 μs

During the ON interval of S1, the inductor current varies linearly from -5 A to +5 A.

For a triangular current varying linearly from -Im to +Im, the mean square value over that interval is:

I² = Im² / 3

Here,

Im = 5 A

Since S1 conducts only for half of the switching cycle:

I_S1,rms² = (1/2) × (Im² / 3)

I_S1,rms² = (1/2) × (25 / 3)

I_S1,rms² = 25 / 6

Taking square root:

I_S1,rms = 5 / √6

I_S1,rms = 2.04 A

Correct Answer: B) 2.04 A

Question 48

Consider the boost converter circuit shown. Assume that the semiconductor devices are ideal. In steady state, the inductor current rises linearly from 0 A to 6 A in the first 10 μs and then falls linearly from 6 A to 0 A in the next 10 μs of every switching cycle. The load resistance R is 10 Ω and the capacitance C is 500 μF. 

Neglect the ripple in the output voltage. What is the input voltage V_dc?

A) 10.0 V

B) 15.0 V

C) 7.5 V

D) 12.5 V

Detailed Solution

The switching period is:

T = 10 μs + 10 μs = 20 μs

The inductor current waveform is triangular from 0 A to 6 A and back to 0 A.

Average inductor current is:

I_L,avg = 6 / 2 = 3 A

In a boost converter, the diode conducts during the OFF interval only, which is half of the switching period.

During this OFF interval, the average diode current is:

I_{D,avg during OFF} = 6 / 2 = 3 A

Average output current over the full cycle is:

I_o = 3 × (1/2) = 1.5 A

Now, using the load resistance:

V_o = I_oR

V_o = 1.5 × 10 = 15 V

Output power is:

P_o = V_oI_o = 15 × 1.5 = 22.5 W

For an ideal converter:

P_in = P_out

V_dc × I_L,avg = 22.5

V_dc × 3 = 22.5

V_dc = 7.5 V

Correct Answer: C) 7.5 V

Question 49

Which one of the following statements is ALWAYS correct about a collection of p column vectors, each having n real-valued entries?

A) If p > n, then the column vectors must be linearly dependent.

B) If p > n, then the column vectors must be linearly independent.

C) If p = n, then the column vectors must be orthogonal.

D) If p < n, then the column vectors must be linearly independent.

Detailed Solution

This question is based on a fundamental result from Linear Algebra.

A set of vectors in ℝⁿ can contain at most n linearly independent vectors.

Therefore, if the number of vectors exceeds the dimension of the vector space, i.e.,

p > n,

then at least one vector can always be expressed as a linear combination of the others.

Hence, the vectors must be linearly dependent.

Checking the options:

(A) ✔ True. More than n vectors in ℝⁿ are always linearly dependent.

(B) ✘ False. It is impossible for more than n vectors in ℝⁿ to be linearly independent.

(C) ✘ False. When p = n, the vectors need not be orthogonal.

(D) ✘ False. Having fewer than n vectors does not guarantee linear independence. For example, two identical vectors in ℝ³ are linearly dependent.

Correct Answer: A) If p > n, then the column vectors must be linearly dependent.

Question 50

Consider the second-order differential equation

d²y/dx² + dy/dx + y = 0

with initial conditions y(0) = 1 and dy/dx at x = 0 is 1. The solution is given by:

A) y(x) = exp(−x/2) [ cos(√3x/2) + √3 sin(√3x/2) ]

B) y(x) = exp(−x/2) [ cos(√3x/2) + 1/√3 sin(√3x/2) ]

C) y(x) = exp(−x/2) [ cos(√3x/2) − 1/√3 sin(√3x/2) ]

D) y(x) = exp(−x/2) [ cos(√3x/2) − √3 sin(√3x/2) ]

Detailed Solution

The given differential equation is:

y'' + y' + y = 0

The auxiliary equation is:

m² + m + 1 = 0

Solving:

m = (-1 ± √(1 - 4))/2 = (-1 ± j√3)/2

Therefore, the general solution is:

y(x) = e^−x/2[C₁ cos(√3x/2) + C₂ sin(√3x/2)]

Using y(0) = 1:

C₁ = 1

Now differentiate the solution and apply y'(0) = 1:

y'(0) = −C₁/2 + (√3/2)C₂

1 = −1/2 + (√3/2)C₂

(√3/2)C₂ = 3/2

C₂ = √3

Hence,

y(x) = e^−x/2[cos(√3x/2) + √3 sin(√3x/2)]

Correct Answer: A

Question 51

The figure shows an arbitrarily shaped planar conducting loop A in the XY plane. Two nonintersecting regions with areas a₁ and a₂ within the loop are subjected to magnetic fields

B̅₁ = (m/√2) sin(ωt) (1x̂ + 0ŷ + 1ẑ)

B̅₂ = −(n/√2) cos(2ωt + π/4) (0x̂ + 1ŷ + 1ẑ)

What is the expression for the induced RMS voltage in loop A?

A) √[(a₁²ω²m² + 4a₂²ω²n²)/4]

B) √[(a₁²ω²m² + 4a₂²ω²n²)/2]

C) √[(a₁²ω²m² − 2a₂²ω²n²)/2]

D) √[(a₁²ω²m² + 2a₂²ω²n²)]

Detailed Solution

Only the z-components of the magnetic fields contribute to the magnetic flux through the loop because the loop lies in the XY-plane.

Flux due to region a₁:

Φ₁ = (a₁m/√2) sin(ωt)

Induced voltage:

e₁ = −dΦ₁/dt = −(a₁mω/√2) cos(ωt)

Therefore,

e₁,rms = a₁mω/2

Flux due to region a₂:

Φ₂ = −(a₂n/√2) cos(2ωt+π/4)

Induced voltage:

e₂ = −dΦ₂/dt = −√2 a₂nω sin(2ωt+π/4)

Hence,

e₂,rms = √2 a₂nω

Since the two voltages are at different frequencies (ω and 2ω), their cross-product averages to zero.

Therefore,

V_rms = √(e₁,rms² + e₂,rms²)

= √[(a₁²ω²m²/4) + 2a₂²ω²n²]

= √[(a₁²ω²m² + 4a₂²ω²n²)/2]

Correct Answer: B)

Question 52

A system is characterized by the following state equation and output equation:

ẋ = a b −a 0 x + 1 0 u

y = 1 2 x

What are the values of a and b for which the poles of the transfer function are at −2 + j3 and −2 − j3?

A) a = 4, b = 3.25

B) a = −4, b = 3.25

C) a = 4, b = −3.25

D) a = −4, b = −3.25

Detailed Solution

The poles of the transfer function are the eigenvalues of the state matrix A.

The state matrix is

A = [ a   b
−a   0 ]

The characteristic equation is obtained from

det(sI−A)=0

Thus,

det [ s−a   −b
a   s ] = (s−a)s−(−b)a

= s²−as+ab

The given poles are

−2+j3 and −2−j3

Hence the characteristic polynomial is

(s+2−j3)(s+2+j3)

=(s+2)²+3² =s²+4s+13

Comparing coefficients:

−a=4

a=−4

Also,

ab=13

(−4)b=13

b=−13/4=−3.25

Correct Answer: D) a = −4, b = −3.25

Question 53

A system is represented in state-space form as follows:

ẋ = 1 2 −3 0 x + 1 2 u

y = 1 2 x

Consider the new state vector Z = 2 1 −1 0 x

What is the state-space representation of the system in terms of the new state vector Z?

A) Ż = [ −1   4 ; −1   −2 ]Z + [ 4 ; −1 ]u,   y = [2   3]Z

B) Ż = [ 2   3 ; 0   3 ]Z + [ 3 ; 5 ]u,   y = [2   3]Z

C) Ż = [ 4   9 ; −2   −3 ]Z + [ 4 ; −1 ]u,   y = [2   3]Z

D) Ż = [ 2   1 ; −4   1 ]Z + [ 4 ; −1 ]u,   y = [4   −1]Z

Detailed Solution

The state transformation is

Z = Tx

where

T = [ 2  1
−1 0 ]

The inverse of the transformation matrix is

T⁻¹ = [ 0  −1
1  2 ]

Using the state transformation equations:

A' = TAT⁻¹

B' = TB

C' = CT⁻¹

Substituting the given matrices gives:

A' = [ 4  9
−2 −3 ]

B' = [ 4
−1 ]

C' = [ 2  3 ]

Hence, the transformed state-space model is

Ż = A'Z + B'u

y = C'Z

Correct Answer: C

Question 54

For the balanced 3-phase transmission line shown, consider the following cases:

Case-1: |V₁| = 1.1 p.u., |V₂| = 0.9 p.u., Z = 0.75∠0° p.u. and θ₁₂ = θ₁ − θ₂ = 0°

Case-2: |V₁| = 1.1 p.u., |V₂| = 0.9 p.u., Z = 0.75∠90° p.u. and θ₁₂ = θ₁ − θ₂ = 90°

Which of the following statements is/are correct about real power loss and reactive power loss in the line?

A) Real power loss in Case-1 is more than that in Case-2

B) Real power loss in Case-2 is more than that in Case-1

C) Reactive power loss in Case-1 is more than that in Case-2

D) Reactive power loss in Case-2 is more than that in Case-1

Detailed Solution

Case-1:

Z = 0.75∠0° means the line is purely resistive.

So, R = 0.75 and X = 0.

Since θ₁₂ = 0°:

|V₁ − V₂| = 1.1 − 0.9 = 0.2 p.u.

Line current:

|I₁| = 0.2 / 0.75

Real power loss exists because R ≠ 0, but reactive power loss is zero because X = 0.

Case-2:

Z = 0.75∠90° means the line is purely reactive.

So, R = 0 and X = 0.75.

Therefore, real power loss in Case-2 is zero.

But reactive power loss exists because X ≠ 0.

Hence:

Real power loss in Case-1 > Real power loss in Case-2

Reactive power loss in Case-2 > Reactive power loss in Case-1

Correct Answers: A and D

Question 55

Consider an n × n orthogonal matrix A with real entries and each column having unit Euclidean norm.

Which of the following statements is/are correct?

A) The value of the determinant of A is either +1 or −1.

B) The eigenvalues of A have modulus 1.

C) ‖Ax‖ = ‖x‖ for all x ∈ ℝⁿ, and (Ax)^T(Ay) ≠ x^Ty for all distinct x, y ∈ ℝⁿ.

D) ‖Ax‖ = ‖x‖ for all x ∈ ℝⁿ, and (Ax)^T(Ay) = x^Ty for all x, y ∈ ℝⁿ.

Detailed Solution

For an orthogonal matrix,

A^TA = I

This property immediately gives several important results.

Statement (A):

Taking determinant on both sides,

det(A^TA) = det(I)

(det A)² = 1

Hence,

det(A) = ±1.

Therefore, Statement (A) is correct.

---

Statement (B):

If λ is an eigenvalue of an orthogonal matrix, then

|λ| = 1.

Hence every eigenvalue lies on the unit circle in the complex plane.

Statement (B) is correct.

---

Statement (C):

Orthogonal matrices preserve inner products:

(Ax)^T(Ay) = x^TA^TAy = x^Ty

Therefore the statement claiming the inner products are not equal is false.

Statement (C) is incorrect.

---

Statement (D):

Norm preservation:

‖Ax‖² = (Ax)^T(Ax) = x^TA^TAx = x^Tx = ‖x‖²

Also,

(Ax)^T(Ay) = x^Ty

Hence orthogonal matrices preserve both Euclidean norm and inner product.

Statement (D) is correct.

Correct Answers: A, B and D

Question 56

Consider the system of linear equations Ax = b, where A is an n × n matrix, and x and b are n-dimensional column vectors.

Suppose this system of equations has a unique solution. Which of the following statements is/are correct?

A) A⁻¹ exists

B) The system of equations A^mx = b also has a unique solution for m = 1, 2, 3, ...

C) rank(A) = rank(A^m), for m = 1, 2, 3, ...

D) rank(A) < rank([A | b]), where [A | b] denotes the augmented matrix.

Detailed Solution

Given that the system Ax = b has a unique solution.

For a square matrix system to have a unique solution, the matrix A must be non-singular.

Therefore,

det(A) ≠ 0

So, A⁻¹ exists.

Statement A is correct.

Since A is non-singular, every power of A is also non-singular:

A^m is non-singular for m = 1, 2, 3, ...

Hence, the system A^mx = b also has a unique solution.

Statement B is correct.

For a non-singular n × n matrix:

rank(A) = n

Also, since A^m is non-singular:

rank(A^m) = n

Therefore,

rank(A) = rank(A^m)

Statement C is correct.

For a consistent system with a unique solution:

rank(A) = rank([A | b]) = n

So, rank(A) < rank([A | b]) is false.

Statement D is incorrect.

Correct Answers: A, B and C

Question 57

The resistance values of the Wheatstone bridge shown are: P = 2000 Ω, Q = 500 Ω, R = 3000 Ω. The battery voltage E = 50 V. The battery has an internal resistance of 1 Ω and the galvanometer has a resistance of 50 Ω.

The value of the resistance S for balanced condition is ______ Ω.

Detailed Solution

For a balanced Wheatstone bridge, the galvanometer current is zero.

The balance condition is:

P/Q = R/S

Substituting the given values:

2000/500 = 3000/S

4 = 3000/S

S = 3000/4 = 750 Ω

The battery internal resistance and galvanometer resistance do not affect the balance condition.

Correct Answer: 750 Ω

Question 58

A system with two generators G1 and G2 is shown. 

The total load on the system is 1184 MW.

C₁(P_G1) = 1000 + 50P_G1 + 0.01(P_G1)² Rs/MWh

C₂(P_G2) = 2000 + 50P_G2 + 0.001(P_G2)² Rs/MWh

P_Loss = 0.001(P_G2 − 50)² MW

Consider the Lagrange multiplier λ = 70.25 for optimal generation. When the generators are operating at their optimal generation, the real power loss in the system is ______ MW.

Detailed Solution

For generator G2:

dC₂/dP_G2 = 50 + 0.002P_G2

The transmission loss is:

P_Loss = 0.001(P_G2 − 50)²

So,

dP_Loss/dP_G2 = 0.002(P_G2 − 50)

Using the coordination equation:

50 + 0.002P_G2 = 70.25[1 − 0.002(P_G2 − 50)]

Solving:

P_G2 ≈ 191.4 MW

Now,

P_Loss = 0.001(191.4 − 50)²

P_Loss ≈ 20.0 MW

Correct Answer: 20.0 MW

Question 59

A balanced three-phase supply is given to a 30 kW, 4-pole, 400 V, 50 Hz, wound rotor induction motor with Y-connected stator and rotor windings. The motor is driving a constant torque load. With shorted sliprings, the machine runs at 1476 rpm.

When an external non-inductive resistance of 0.27 Ω per phase is connected in series in the rotor circuit, the steady-state speed drops to 1404 rpm.

Neglecting rotational losses, the actual per-phase rotor winding resistance is ______ Ω.

(Round off to two decimal places)

Detailed Solution

The synchronous speed is

N_s = 120f/P = (120 × 50)/4 = 1500 rpm

With shorted slip rings,

s₁ = (1500 − 1476)/1500 = 24/1500 = 0.016

With an external rotor resistance of 0.27 Ω per phase,

s₂ = (1500 − 1404)/1500 = 96/1500 = 0.064

For a constant torque load in the low-slip region of an induction motor,

Slip ∝ Total rotor resistance

Hence,

s₂/s₁ = (R₂ + 0.27)/R₂

0.064/0.016 = (R₂ + 0.27)/R₂

4 = (R₂ + 0.27)/R₂

4R₂ = R₂ + 0.27

3R₂ = 0.27

R₂ = 0.09 Ω

Correct Answer: 0.09 Ω

Question 60

Consider the two-port network shown. For maximum power transfer to the resistive load (R_L), the value of R_L should be ______ Ω.

(Round off to two decimal places)

Detailed Solution

For maximum power transfer,

R_L = R_th

To find R_th, deactivate the independent voltage source.

The 10 V voltage source is replaced by a short circuit.

Now, seen from the load terminals, the 10 Ω resistor and 4 Ω resistor are in parallel.

R_th = 10 || 4

R_th = (10 × 4) / (10 + 4)

R_th = 40 / 14 = 2.857 Ω

Rounded off to two decimal places:

R_L = 2.86 Ω

Correct Answer: 2.86 Ω

Question 61

Consider the circuit shown. Assume that the diode (D) is ideal. 

Given v_s = 100 sin(2π50t) V, V_dc = 50 V, and R = 10 Ω, the average value of the current through the diode is ______ A.

(Round off to two decimal places)

Detailed Solution

The diode conducts only when:

v_s > V_dc

100 sinθ > 50

sinθ > 0.5

Therefore, conduction occurs from:

θ = π/6 to 5π/6

During conduction, diode current is:

i_D = (100 sinθ − 50)/10

The average current over one full cycle is:

I_avg = (1/2π) ∫_π/6^5π/6 [(100 sinθ − 50)/10] dθ

I_avg = 1.09 A

Correct Answer: 1.09 A

Question 62

The magnitude of the contour integral

∮_C ((z + 1)² / ((z − i)(z − 2))) dz

over the contour C: |z − 2 − i| = 3/2 is ______.

(Round off to two decimal places)

Detailed Solution

The singular points are:

z = i and z = 2

The contour is:

|z − 2 − i| = 3/2

So, the centre is 2 + i and radius is 3/2.

Distance of z = i from centre:

|i − (2 + i)| = 2

Since 2 > 3/2, z = i lies outside the contour.

Distance of z = 2 from centre:

|2 − (2 + i)| = 1

Since 1 < 3/2, z = 2 lies inside the contour.

Residue at z = 2:

Res = (z + 1)² / (z − i) at z = 2

Res = 9 / (2 − i)

Using residue theorem:

Integral = 2πi × 9/(2 − i)

Magnitude:

|Integral| = 2π × 9 / |2 − i|

|2 − i| = √5

|Integral| = 18π / √5 = 25.30

Correct Answer: 25.30

Question 63

A uniform spherical volume charge distribution of radius 2 m, centered at the origin, has a strength of 3 × 10⁻⁶ C/m³. A point charge of strength π × 8.854 × 10⁻¹² C is moved from (−3, 0, −4) to (0, 0, 4) in Cartesian coordinate system. The relative permittivity of the medium is 1 and the coordinate values are in meters.

The work done during the process is ______ μJ.

(Round off to two decimal places)

Detailed Solution

Total charge of the sphere is:

Q = ρ × (4/3)πR³

Q = 3 × 10⁻⁶ × (4/3)π(2)³

Q = 32π × 10⁻⁶ C

Initial distance from origin:

r₁ = √[(-3)² + 0² + (-4)²] = 5 m

Final distance from origin:

r₂ = √[0² + 0² + 4²] = 4 m

Both points are outside the charged sphere, so potential is:

V = Q / (4πε₀r)

Work done:

W = q(V₂ − V₁)

W = q × Q/(4πε₀) × (1/4 − 1/5)

Given:

q = π × 8.854 × 10⁻¹² = πε₀

So,

W = (πε₀) × Q/(4πε₀) × (1/20)

W = Q/80 = (32π × 10⁻⁶)/80

W = 3.95 μJ

Correct Answer: 3.95 μJ

Question 64

Let X and Y be two real-valued random variables with E(X) = 1, E(Y) = 2, E(X²) = 4, E(Y²) = 9, and E(XY) = 0.9.

The value of α that minimizes E((X − αY)²) is ______.

(Round off to one decimal place)

Detailed Solution

We need to minimize:

E((X − αY)²)

Expanding:

E((X − αY)²) = E(X²) − 2αE(XY) + α²E(Y²)

Substitute the given values:

= 4 − 2α(0.9) + α²(9)

= 4 − 1.8α + 9α²

Differentiate with respect to α:

d/dα = −1.8 + 18α

For minimum:

−1.8 + 18α = 0

α = 1.8 / 18 = 0.1

Correct Answer: 0.1

Question 65

The integral

(1/π) ∫₀^∞ x²⁰²⁶ / [(1 + x²⁰²⁶)(1 + x²)] dx

evaluates to ______.

(Round off to two decimal places)

Detailed Solution

Let

I = ∫₀^∞ x²⁰²⁶ / [(1 + x²⁰²⁶)(1 + x²)] dx

Using the substitution x = 1/t, we get:

I = ∫₀^∞ 1 / [(1 + t²⁰²⁶)(1 + t²)] dt

Adding both forms:

2I = ∫₀^∞ [1 / (1 + x²)] dx

2I = π/2

I = π/4

Therefore,

(1/π)I = (1/π)(π/4) = 1/4 = 0.25

Correct Answer: 0.25