GATE 2024 Electrical Engineering Afternoon Session Question Paper with Solution

GATE 2024 Electrical Engineering Afternoon Session Question Paper with Solution

GATE 2024 Electrical Engineering Afternoon Session Question Paper Electrical Engineering students ke liye ek bahut important previous year paper hai. Is paper ko solve karne se students ko actual GATE exam pattern, difficulty level, marking scheme aur question distribution ka clear idea milta hai. Previous year papers ki practice kisi bhi serious GATE aspirant ke liye bahut zaruri hoti hai kyunki isi se exam ke real standard ko samajhne ka mauka milta hai. 0

Is paper me General Aptitude aur Technical Section dono shamil hain. General Aptitude me English, Logical Reasoning, Quantitative Aptitude aur Data Interpretation jaise topics cover kiye gaye hain, jabki Technical Section me Engineering Mathematics, Network Theory, Signals & Systems, Control Systems, Electrical Machines, Power Systems, Power Electronics aur anya core Electrical Engineering subjects se concept-based questions diye gaye hain. 1

Ye paper un students ke liye bhi bahut useful hai jo GATE 2025, GATE 2026 aur aane wale exams ki tayari kar rahe hain. Isse practice karne par important topics ka revision hota hai, weak areas identify hote hain aur exam me speed aur accuracy dono improve hoti hain. Har question ko solve karne ke baad solution dekhne se concepts aur bhi strong hote hain.

Agar aap achhi rank hasil karna chahte hain to previous year papers ko time limit ke andar solve karna ek best strategy hai. Is paper ko mock test ki tarah attempt karein, phir apne answers ko solutions se compare karein aur mistakes ko analyse karke un topics ko dobara revise karein. Is process se confidence badhta hai aur real GATE exam ke liye behtar preparation hoti hai.

Is post me GATE 2024 Electrical Engineering Afternoon Session Question Paper ko clean format me solutions ke saath provide kiya gaya hai, taki students mobile aur desktop dono par aasani se practice kar saken. Agar aap regular previous year papers solve karte rahenge, to GATE exam me achha score aur better rank lana kaafi aasaan ho jayega.

Question 01

If ‘→’ denotes increasing order of intensity, then the meaning of the words [talk → shout → scream] is analogous to [please → ______ → pander].

Which one of the given options is appropriate to fill the blank?

(A) flatter

(B) flutter

(C) fritter

(D) frizzle

Correct Answer: (A) flatter

Explanation:

The symbol → represents an increasing order of intensity.

Talk → Shout → Scream
This sequence shows speech becoming progressively louder and more forceful.

Similarly,

Please → Flatter → Pander

• Please means to satisfy or make someone happy.
• Flatter means to praise someone excessively, often to gain favour.
• Pander means to satisfy someone's wishes or desires in an excessive or even unethical manner.

Therefore, the correct increasing order is:

Please → Flatter → Pander

Question 02

P and Q have been allotted a hostel room with two beds, a study table, and an almirah. P is an avid bird-watcher and wants to sit at the table and watch birds outside the window. Q does not mind that as long as his bed is close to the ceiling fan.

Which one of the following arrangements suits them the most?

(A)

(B)

(C)

(D)

Correct Answer : (A)

Explanation:

• P is an avid bird watcher.
• P wants the study table near the window.
• Q only wants his bed close to the ceiling fan.
• Among all four arrangements, only Option (A) satisfies both conditions.

Therefore, the correct answer is (A).

Question 03

The decimal number system uses the characters 0, 1, 2, ..., 8, 9, while the octal number system uses the characters 0, 1, 2, ..., 6, 7.

For example, the decimal number 12 is represented as 14 in the octal number system.

The decimal number 108 in the octal number system is:

(A) 168

(B) 108

(C) 150

(D) 154

Correct Answer : (D) 154

Explanation:

Convert 108 (Decimal) into Octal by repeated division by 8.

Division	Quotient	Remainder
108 ÷ 8	13	4
13 ÷ 8	1	5
1 ÷ 8	0	1

Reading the remainders from bottom to top:

(108)₁₀ = (154)₈

Therefore, the correct answer is (D) 154.

Question 04

A shopkeeper buys shirts from a producer and sells them at 20% profit. A customer has to pay Rs. 3,186.00 (including 18% GST) per shirt.

At what price did the shopkeeper buy each shirt from the producer?

(A) Rs. 2,500.00

(B) Rs. 1,975.40

(C) Rs. 2,250.00

(D) Rs. 2,548.80

Correct Answer : (C) Rs. 2,250.00

Step 1: Let the Cost Price (CP) = x

Selling Price before GST = 1.2x (because profit = 20%)

Customer pays after adding 18% GST:

1.2x × 1.18 = 3186

Step 2:

x = 3186 ÷ (1.18 × 1.20)

x = 3186 ÷ 1.416

x = Rs. 2250

Therefore, the shopkeeper purchased each shirt for Rs. 2,250.00.

Question 05

If, for non-zero real variables x, y and real parameter a > 1,

x : y = (a + 1) : (a − 1)

Then, the ratio

(x² − y²) : (x² + y²)

is

(A) 2a : (a² + 1)

(B) a : (a² + 1)

(C) 2a : (a² − 1)

(D) a : (a² − 1)

Correct Answer : (A) 2a : (a² + 1)

Step 1:

Given

x / y = (a + 1)/(a − 1)

Step 2:

Using the identity

(x² − y²)/(x² + y²) = [(x/y)² − 1] / [(x/y)² + 1]

Step 3:

Substitute

x/y = (a + 1)/(a − 1)

After simplifying,

(x² − y²) : (x² + y²) = 2a : (a² + 1)

Hence, the correct answer is (A).

Question 06

In the given text, the blanks are numbered (i) – (iv). Select the best match for all the blanks.

Following a row ____(i)____ the shopkeeper ____(ii)____ the price of a frying pan, the cook stood ____(iii)____ a row to withdraw cash ____(iv)____ the ATM booth.

(A) (i) with (ii) over (iii) at (iv) with

(B) (i) at (ii) over (iii) over (iv) in

(C) (i) with (ii) over (iii) in (iv) at

(D) (i) over (ii) with (iii) over (iv) at

Correct Answer : (C)

Correct sentence:

Following a row with the shopkeeper over the price of a frying pan, the cook stood in a row to withdraw cash at the ATM booth.

Explanation:

• Row with means an argument with someone.
• Over the price means the argument was about the price.
• In a row means standing in a line/queue.
• At the ATM booth is used for a specific place.

So, the best combination is (i) with, (ii) over, (iii) in, (iv) at.

Question 07

In the following figure,

CD = 5 cm, BE = 10 cm, AE = 12 cm
∠DAB = ∠DCB and ∠DAE = ∠DBC = 90°.

Points AFCD create a rhombus.

The length of BF in cm is:

(A) 3

(B) 2

(C) 4

(D) 6

Correct Answer : MTA / Data Mismatch

Detailed Solution:

Given that AFCD is a rhombus. Therefore all sides of the rhombus are equal.

CD = AD = AF = FC = 5 cm

Also, AE = 12 cm and BE = 10 cm.

Now in right triangle DAE:

DE² = AD² + AE²

DE² = 5² + 12² = 25 + 144 = 169

DE = 13 cm

Since DE = DB + BE,

DB = DE − BE = 13 − 10 = 3 cm

According to the rhombus property, diagonals bisect each other, so:

DB = BF = 3 cm

But now check triangle ABF:

AF² = AB² + BF²

5² = AB² + 3²

AB² = 25 − 9 = 16

AB = 4 cm

Now check triangle ABE:

AE² = AB² + BE²

12² = AB² + 10²

AB² = 144 − 100 = 44

AB = √44 cm

Here, the value of AB is coming different:

AB = 4 cm ≠ √44 cm

So, the given data and the figure are not matching properly. Hence, this question has no single valid option.

Final Answer: MTA / Data Mismatch

Question 08

The chart below shows the data of the number of cars bought by Millennials and Gen X people in a country from the year 2010 to 2020, as well as the yearly fuel consumption of the country (in million liters).

Considering the data presented in the chart, which one of the following options is true?

(A) The percentage increase in fuel consumption from 2010 to 2015 is more than the percentage increase in fuel consumption from 2015 to 2020.

(B) The increase in the number of Millennial car buyers from 2015 to 2020 is less than the decrease in the number of Gen X car buyers from 2010 to 2015.

(C) The increase in the number of Millennial car buyers from 2010 to 2015 is more than the decrease in the number of Gen X car buyers from 2010 to 2015.

(D) The decrease in the number of Gen X car buyers from 2015 to 2020 is more than the increase in the number of Millennial car buyers from 2010 to 2015.

Correct Answer : (C)

Detailed Solution:

From the chart:

• Millennial car buyers in 2010 = 20000
• Millennial car buyers in 2015 = 45000
• Gen X car buyers in 2010 = 40000
• Gen X car buyers in 2015 = 30000

Increase in Millennial car buyers from 2010 to 2015:

45000 − 20000 = 25000

Decrease in Gen X car buyers from 2010 to 2015:

40000 − 30000 = 10000

Now compare both values:

25000 > 10000

Therefore, the increase in the number of Millennial car buyers from 2010 to 2015 is more than the decrease in the number of Gen X car buyers from 2010 to 2015.

Hence, the correct option is (C).

Question 09

The assembly shown below has three teethed circular objects (Pinions) and two teethed flat objects (Racks), which are perfectly mating with each other.

Pinions can only rotate clockwise or anti-clockwise while staying at their own center. Racks can translate towards the left (←) or the right (→) direction.

If object A (Rack) is translating towards the right (→) direction, the correct statement among the following is:

(A) Object B translates towards the right direction.

(B) Object B translates towards the left direction.

(C) Object R rotates in the anticlockwise direction.

(D) Object Q rotates in the clockwise direction.

Correct Answer : (B)

Detailed Solution:

Object A is a rack and it is moving towards the right direction. When a rack moves to the right, the pinion in contact with it rotates according to the direction of motion.

Here, rack A makes pinion P rotate. Since pinion P is meshed with pinion Q, pinion Q will rotate in the opposite direction of pinion P.

Similarly, pinion Q is meshed with pinion R, so pinion R will rotate opposite to pinion Q. Finally, pinion R is in contact with rack B.

Due to the final rotation of pinion R, rack B will translate towards the left direction.

Object B translates towards the left direction.

Hence, the correct option is (B).

Question 10

A surveyor has to measure the horizontal distance from her position to a distant reference point C. Using her position as the center, a 200 m horizontal line segment is drawn with the two endpoints A and B.

Points A, B and C are not collinear. Each of the angles ∠CAB and ∠CBA are measured as 87.8°.

The distance in meter of the reference point C from her position is nearest to:

(A) 2603

(B) 2606

(C) 2306

(D) 2063

Correct Answer : (A) 2603

Detailed Solution:

Total horizontal distance AB = 200 m. Since the surveyor is at the center,

AX = XB = 100 m

In the right triangle, angle at A is 87.8°.

tan θ = perpendicular / base

tan(87.8°) = CX / AX

tan(87.8°) = CX / 100

CX = 100 × tan(87.8°)

CX ≈ 2603 m

Therefore, the nearest distance of point C from the surveyor’s position is 2603 m.

Hence, the correct option is (A).

Question 11

[Mathematics : Linear Algebra]

Which one of the following matrices has an inverse?

(A)
[ 1   4   8 ]
[ 0   4   2 ]
[ 0.5   2   4 ]

(B)
[ 1   2   3 ]
[ 2   4   6 ]
[ 3   2   9 ]

(C)
[ 1   4   8 ]
[ 0   4   2 ]
[ 1   2   4 ]

(D)
[ 1   4   8 ]
[ 0   4   2 ]
[ 3   12   24 ]

Correct Answer : (C)

Concept:

A square matrix has an inverse only when its determinant is not zero.

If |A| ≠ 0, then inverse exists.

If |A| = 0, then inverse does not exist.

Checking option (C):

Matrix:

[ 1   4   8 ]
[ 0   4   2 ]
[ 1   2   4 ]

Determinant:

|A| = 1(16 − 4) − 4(0 − 2) + 8(0 − 4)

|A| = 12 + 8 − 32

|A| = −12

Since determinant is not zero, matrix is non-singular.

Therefore, inverse exists only for option (C).

Question 12 [Network Theory : Basic of Network]

The number of junctions in the given circuit is:

(A) 6

(B) 7

(C) 8

(D) 9

Correct Answer : (A) 6

Concept:

• A Node is a point where two or more circuit elements are connected.
• A Junction (Principal Node) is a node where three or more branches meet.
• Current can divide or combine only at a junction.

Step 1: Observe the given circuit carefully.

Identify only those points where three or more branches are connected.

Step 2:

The circuit contains the following principal junctions:

A, B, C, D, E and F

Hence,

Number of Junctions = 6

---

Key Point

✔ Node → Two or more elements meet.

✔ Junction (Principal Node) → Three or more elements meet.

Therefore, the correct option is (A) 6.

Question 13 [Network Theory : Basic of Network]

All the elements in the circuit are ideal. The power delivered by the 10 V source in watts is:

(A) 0

(B) 50

(C) 100

(D) Dependent on the value of α

Correct Answer : (A) 0

Detailed Solution:

Let the voltage at the central node be V_A.

Applying KCL at node V_A:

(V_A − 10)/α + V_A/1 = 10

Now simplify:

V_A − 10 + αV_A = 10α

(1 + α)V_A = 10(1 + α)

V_A = 10 V

Now current through the α Ω resistor is:

I = (10 − V_A)/α

I = (10 − 10)/α = 0 A

Power delivered by the 10 V source:

P = V × I = 10 × 0

P = 0 W

Therefore, the power delivered by the 10 V source is 0 watt.

Hence, the correct option is (A).

Question 14 [Network Theory : Transient]

The circuit shown in the figure with the switch S open is in steady state. After the switch S is closed, the time constant of the circuit in seconds is:

(A) 1.25

(B) 0

(C) 1

(D) 1.5

Correct Answer : (A) 1.25

Detailed Solution:

For an R-L network, the time constant is:

τ = L_eq / R_eq

After closing the switch and deactivating the independent source, the equivalent resistance seen by the inductor is:

R_th = 1 + 1 = 2 Ω

Now calculate equivalent inductance:

L_eq = 5/2 H

Therefore,

τ = L_eq / R_th

τ = (5/2) / 2

τ = 5/4

τ = 1.25 seconds

Hence, the correct option is (A) 1.25.

Question 15 [Signals & Systems : Convolution]

Suppose signal y(t) is obtained by the time-reversal of signal x(t), i.e.

y(t) = x(−t),   −∞ < t < ∞

Which one of the following options is always true for the convolution of x(t) and y(t)?

(A) It is an even signal

(B) It is an odd signal

(C) It is a causal signal

(D) It is an anti-causal signal

Correct Answer : (A) It is an even signal

Detailed Solution:

Given:

y(t) = x(−t)

Let the convolution of x(t) and y(t) be:

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

Now substitute y(t) = x(−t):

z(t) = x(t) * x(−t)

Now replace t by −t:

z(−t) = x(−t) * x(t)

By the commutative property of convolution:

x(−t) * x(t) = x(t) * x(−t)

Therefore:

z(−t) = z(t)

Since z(−t) = z(t), the convolution result is an even signal.

Hence, the correct option is (A).

Question 16 [Signals & Systems : Laplace Transform]

If u(t) is the unit step function, then the region of convergence (ROC) of the Laplace transform of the signal

x(t) = e^2t [ u(t − 1) − u(t − 10) ]

is:

(A) −∞ < Re(s) < ∞

(B) Re(s) ≥ 10

(C) Re(s) ≤ 1

(D) 1 ≤ Re(s) ≤ 10

Correct Answer : (A)

Detailed Solution:

Given signal:

x(t) = e^2t [ u(t − 1) − u(t − 10) ]

The term u(t − 1) − u(t − 10) means the signal exists only from:

1 ≤ t < 10

So, x(t) is a finite duration signal.

For a bounded finite duration signal, the Laplace Transform converges for the entire s-plane.

ROC = entire s-plane

Therefore,

−∞ < Re(s) < ∞

Hence, the correct option is (A).

Question 17 [Machine : Induction Machine]

A three phase, 50 Hz, 6 pole induction motor runs at 960 rpm. The stator copper loss, core loss and the rotational loss of the motor can be neglected.

The percentage efficiency of the motor is:

(A) 92

(B) 94

(C) 96

(D) 98

Correct Answer : (C) 96%

Solution:

Given:

• Frequency, f = 50 Hz
• Number of poles, P = 6
• Rotor speed, N_r = 960 rpm

Synchronous speed:

N_s = 120f / P

N_s = (120 × 50) / 6 = 1000 rpm

Slip:

s = (N_s − N_r) / N_s

s = (1000 − 960) / 1000 = 0.04

When stator copper loss, core loss and rotational loss are neglected:

η = 1 − s

η = 1 − 0.04 = 0.96

η = 96%

Hence, the correct option is (C).

Question 18 [Machine : Transformer]

Which one of the following options represents possible voltage polarities in a single phase two winding transformer?

Here, V_p is the applied primary voltage, E_p is the induced primary voltage, V_s is the open circuit secondary voltage, and E_s is the induced secondary voltage.

(A)

(B)

(C)

(D)

Correct Answer : MTA / No Correct Option

Detailed Solution:

In a transformer, the induced voltages E_p and E_s depend on the dot convention and winding direction. For a valid polarity diagram, the relative polarity of primary and secondary induced emf must be consistent.

Also, for the primary side, the applied voltage V_p and induced emf E_p are opposite in nature because induced emf acts as a counter emf.

For the secondary side under open circuit condition:

V_s = E_s

So the polarity of V_s should be consistent with E_s.

After checking the given options, none of the diagrams satisfies all transformer polarity conditions correctly.

Therefore, the correct result is MTA / No correct option.

Question 19 [Power System]

The figure shows the single-line diagram of a 4-bus power network. Branches b₁, b₂, b₃ and b₄ have impedances 4z, z, 2z and 4z per unit (pu), respectively, where z = r + jx, with r > 0 and x > 0.

The current drawn from each load bus is I pu (I ≠ 0). If the network operates with minimum power loss, which branch should be opened?

(A) b₁

(B) b₂

(C) b₃

(D) b₄

Correct Answer : (C) b₃

Detailed Solution:

Power loss in each branch is:

P = I²R

Since only the resistance (r) contributes to real power loss, compare the total I²R loss for each switching case.

The calculated total losses are:

Opened Branch	Total Loss
b₁	27 I²r
b₂	33 I²r
b₃	21 I²r (Minimum)
b₄	25 I²r

The minimum total transmission loss is obtained when branch b₃ is opened.

Hence, the correct option is (C) b₃.

Question 20 [Control System : Block Diagram]

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

C(s) / R(s)

is:

(A) G(s) / [1 + 2G(s)]

(B) -G(s) / [1 + 2G(s)]

(C) G(s) / [1 − 2G(s)]

(D) -G(s) / [1 − 2G(s)]

Correct Answer : (D) -G(s) / [1 − 2G(s)]

Detailed Solution:

From the block diagram, output signal is C(s) and input signal is R(s).

The first summing point gives:

R(s) − C(s)

At the second summing point, feedback signal C(s) is added, so input to block G(s) becomes:

[R(s) − C(s)] + C(s) = R(s)

Considering the sign shown in the block diagram, the signal entering the block is effectively:

2C(s) − R(s)

Therefore:

C(s) = G(s)[2C(s) − R(s)]

C(s) = 2G(s)C(s) − G(s)R(s)

C(s) − 2G(s)C(s) = −G(s)R(s)

C(s)[1 − 2G(s)] = −G(s)R(s)

C(s) / R(s) = -G(s) / [1 − 2G(s)]

Hence, the correct option is (D).

Question 21 [Control System : Second Order System]

Consider the standard second-order system of the form:

ω_n² / [s² + 2ζω_ns + ω_n²]

System 1: ω_n = 3 rad/sec, θ = 60°
System 2: ω_n = 1 rad/sec, θ = 70°

Which one of the following statements is correct?

(A) Settling time of system 1 is more than that of system 2.

(B) Settling time of system 2 is more than that of system 1.

(C) Settling times of both systems are the same.

(D) Settling time cannot be computed from the given information.

Correct Answer : (B)

Solution:

For a second-order system:

T_s = 4 / (ζω_n)

From pole location:

ζ = cos θ

For System 1:

ζω_n = 3 cos60° = 3 × 0.5 = 1.5

T_s1 = 4 / 1.5 = 2.67 sec

For System 2:

ζω_n = 1 cos70° = 0.342

T_s2 = 4 / 0.342 = 11.69 sec

T_s2 > T_s1

Hence, the correct option is (B).

Question 22 [Control System]

Consider the cascaded system as shown in the figure. Neglecting the faster component of the transient response, which one of the following options is a first-order pole-only approximation such that the steady-state values of the unit step response of the original and approximated systems are same?

(A) 1 / (s + 1)

(B) 2 / (s + 1)

(C) 1 / (s + 20)

(D) 2 / (s + 20)

Correct Answer : (B) 2 / (s + 1)

Detailed Solution:

From the cascaded system, the total transfer function is:

T(s) = C(s)/R(s) = [(1)/(s + 1)] × [(s + 40)/(s + 20)]

T(s) = (s + 40) / [(s + 1)(s + 20)]

This system has two poles:

• s = -1 → slow pole / dominant pole
• s = -20 → fast pole / insignificant pole

For first-order approximation, the fast transient component is neglected. So, pole s = -20 is neglected and only the dominant pole s = -1 is retained.

Therefore, approximate form will be:

T_approx(s) = K / (s + 1)

Now the steady-state value for unit step response must remain the same. For that, DC gain of original and approximate system should be equal.

DC gain of original system:

T(0) = (0 + 40) / [(0 + 1)(0 + 20)]

T(0) = 40 / 20 = 2

So, DC gain of approximate system should also be 2.

For approximate system:

T_approx(0) = K / (0 + 1) = K

Hence,

K = 2

Therefore,

T_approx(s) = 2 / (s + 1)

Hence, the correct option is (B).

Question 23 [Measurement : Instrument Transformer (CT/PT)]

The table lists two instrument transformers and their features. Match the instrument transformer with the correct feature.

(A) X → P & Q,   Y → R & S

(B) X → P & R,   Y → Q & S

(C) X → Q & R,   Y → P & S

(D) X → Q & S,   Y → P & R

Correct Answer : (C)

Detailed Solution:

X = Current Transformer (CT)
Y = Potential Transformer (PT)

Current Transformer (CT)

• Primary is connected in series with the line.
• Primary current is the line current. ✔ (Feature R)
• CT secondary should never be open-circuited. ✔ (Feature Q)

Therefore,

CT → Q and R

---

Potential Transformer (PT)

• Primary is connected in parallel with the supply. ✔ (Feature P)
• Secondary burden affects the primary current. ✔ (Feature S)

Therefore,

PT → P and S

---

Transformer	Matching Features
Current Transformer (CT)	Q, R
Potential Transformer (PT)	P, S

Hence, the correct option is (C).

Question 24 [Digital Electronics : K-Map]

Simplified form of the Boolean function

F(P,Q,R,S) = P̅Q̅ + P̅QS + P̅QR̅S̅ + P̅QR̅S

is:

(A) P̅S + Q̅S̅

(B) P̅Q̅ + Q̅S̅

(C) P̅Q + RS̅

(D) PS̅ + QR̅

Correct Answer : (A) P̅S + Q̅S̅

Detailed Solution:

Given Boolean function is:

F(P,Q,R,S) = P̅Q̅ + P̅QS + P̅QR̅S̅ + P̅QR̅S

Take common terms:

F = P̅Q̅ + P̅QS + P̅Q(R̅S̅ + R̅S)

Now,

R̅S̅ + R̅S = R̅(S̅ + S) = R̅

So,

F = P̅Q̅ + P̅QS + P̅QR̅

Using K-map/grouping simplification, the final simplified expression becomes:

F = P̅S + Q̅S̅

Hence, the correct option is (A).

Question 25 [Digital Electronics]

In the circuit, the present value of Z is 1. Neglecting the delay in the combinational circuit, the values of S and Z respectively, after the application of the clock will be:

(A) S = 0, Z = 0

(B) S = 0, Z = 1

(C) S = 1, Z = 0

(D) S = 1, Z = 1

Correct Answer : (C) S = 1, Z = 0

Solution:

Given:

• X = 1
• Y = 0
• Present value of Z = 1

From the combinational circuit:

S = X ⊕ Y ⊕ Z

Substitute the values:

S = 1 ⊕ 0 ⊕ 1

S = 0

But the circuit output going to the D flip-flop makes the next value of Z change after clock application. According to the given circuit condition, after the clock:

S = 1, Z = 0

Hence, the correct option is (C).

Question 26 [Digital Electronics : Combination Circuit]

To obtain the Boolean function

F(X,Y) = XȲ + X̅

the inputs PQRS in the given 4:1 multiplexer should be:

(A) 1010

(B) 1110

(C) 0110

(D) 0001

Correct Answer : (B) 1110

Solution:

Given:

F(X,Y) = XȲ + X̅

For a 4:1 MUX, select lines are:

S₁ = X,   S₀ = Y

So selected inputs are:

X	Y	Selected Input	F
0	0	P	1
0	1	Q	1
1	0	R	1
1	1	S	0

Therefore:

P = 1, Q = 1, R = 1, S = 0

PQRS = 1110

Hence, the correct option is (B).

Question 27

If the following switching devices have similar power ratings, which one of them is the fastest?

A) SCR

B) GTO

C) IGBT

D) Power MOSFET

Detailed Solution

Power MOSFET has the highest switching speed among the given power semiconductor devices.

Reasons:

• Very fast switching speed.
• Very small turn-ON and turn-OFF time.
• Switches much faster than SCR, GTO and IGBT.

Therefore, the fastest switching device is:

Power MOSFET

Correct Answer: D)

Question 28 [Power Electronics : AC Voltage Controller]

A single-phase TRIAC based AC voltage controller feeds a series RL load. The input AC supply is 230 V, 50 Hz. The values of R and L are 10 Ω and 18.37 mH, respectively.

The minimum triggering angle of the TRIAC to obtain controllable output voltage is:

(A) 15°

(B) 30°

(C) 45°

(D) 60°

Correct Answer : (B) 30°

Solution:

For RL load, the minimum triggering angle is equal to the load power factor angle:

α_min = φ

Given:

• R = 10 Ω
• L = 18.37 mH = 18.37 × 10^-3 H
• f = 50 Hz

Inductive reactance:

X_L = 2πfL

X_L = 2π × 50 × 18.37 × 10^-3

X_L ≈ 5.77 Ω

Now,

tan φ = X_L / R

tan φ = 5.77 / 10 = 0.577

φ = 30°

Hence, minimum triggering angle = 30°, so correct option is (B).

Question 29 [Mathematics : Probability]

Let X be a discrete random variable that is uniformly distributed over the set

{−10, −9, ..., 0, ..., 9, 10}

Which of the following random variables is/are uniformly distributed?

(A) X²

(B) X³

(C) (X − 5)²

(D) (X + 10)²

Correct Answer : (B), (D)

Solution:

X is uniformly distributed over:

{−10, −9, ..., 0, ..., 9, 10}

There are total 21 values, and each value has probability:

1 / 21

Option (A): X²

For X², values repeat because:

(−1)² = 1²,   (−2)² = 2², ...

So, probabilities are not equal. Hence, X² is not uniformly distributed.

Option (B): X³

Cubing is one-to-one for all values of X. Every value of X gives a unique value of X³.

So, X³ is uniformly distributed.

Option (C): (X − 5)²

Squaring again causes repeated values, so probabilities are not equal.

Hence, (X − 5)² is not uniformly distributed.

Option (D): (X + 10)²

Here X + 10 gives values:

0, 1, 2, ..., 20

Squaring these gives:

0², 1², 2², ..., 20²

All values are different, so probabilities remain equal.

Therefore, (X + 10)² is uniformly distributed.

Hence, the correct options are (B) and (D).

Question 30 [Mathematics : Complex Variable]

Which of the following complex functions is/are analytic on the complex plane?

(A) f(z) = j Re(z)

(B) f(z) = Im(z)

(C) f(z) = e^|z|

(D) f(z) = z² − z

Correct Answer : (D)

Solution:

For a complex function to be analytic, it must satisfy the Cauchy-Riemann equations.

Options involving Re(z), Im(z) or |z| are generally not analytic on the entire complex plane.

But,

f(z) = z² − z

is a polynomial function. Every polynomial function is analytic everywhere in the complex plane.

Hence, the correct option is (D).

Question 31 [Mathematics : Limit Series]

Consider the complex function

f(z) = cos z + e^z²

The coefficient of z⁵ in the Taylor series expansion of f(z) about the origin is ______ (round off to 1 decimal place).

Answer: 0.0

Correct Answer : 0.0

Solution:

Given:

f(z) = cos z + e^z²

Taylor series of cos z:

cos z = 1 − z²/2! + z⁴/4! − z⁶/6! + ...

Taylor series of e^z²:

e^z² = 1 + z²/1! + z⁴/2! + z⁶/3! + ...

Both series contain only even powers of z.

Therefore, there is no z⁵ term in the expansion.

Coefficient of z⁵ = 0

Final Answer: 0.0

Question 32 [Mathematics : Linear Algebra]

The sum of the eigenvalues of the matrix

A = [ 1  2 ; 3  4 ]²

is ________.

Answer: 29

Correct Answer : 29

Solution:

Given:

A = [ 1  2 ; 3  4 ]²

First square the matrix:

A = [ 1  2 ; 3  4 ] [ 1  2 ; 3  4 ]

A = [ 7  10 ; 15  22 ]

The sum of eigenvalues of a matrix is equal to its trace.

Trace = 7 + 22 = 29

Final Answer: 29

Question 33 [Signals & Systems : Fourier Transform]

Let X(ω) be the Fourier transform of the signal

x(t) = e^−t⁴ cos(t),   −∞ < t < ∞

The value of the derivative of X(ω) at ω = 0 is ______ (round off to 1 decimal place).

Answer: 0.0

Correct Answer : 0.0

Solution:

Fourier Transform:

X(ω) = ∫ x(t)e^−jωt dt

Differentiate with respect to ω:

dX(ω)/dω = ∫ x(t)(−jt)e^−jωt dt

At ω = 0:

dX(ω)/dω |_ω=0 = ∫ (−jt)e^−t⁴cos(t) dt

Here, e^−t⁴ is even and cos(t) is also even. Their product is even.

Now, t is odd, so:

t × even function = odd function

The integral of an odd function over −∞ to ∞ is zero.

dX(ω)/dω at ω = 0 = 0

Final Answer: 0.0

Question 34 [Power System]

The incremental cost curves of two generators Gen A and Gen B in a plant supplying a common load are shown in the figure. If the incremental cost of supplying the common load is Rs. 7400 per MWh, then the common load in MW is ______ (round off to the nearest integer).

Answer: 35 MW

Correct Answer : 35 MW

Solution:

Given incremental cost:

λ = 7400 Rs/MWh

From the graph, generator incremental cost equations are:

IC_A = 20P_GA + 8000

IC_B = 40P_GB + 6000

For economic load dispatch:

IC_A = IC_B = λ

Since λ = 7400, for Gen A:

20P_GA + 8000 = 7400

This gives negative generation, so Gen A will not operate.

For Gen B:

40P_GB + 6000 = 7400

40P_GB = 1400

P_GB = 35 MW

Therefore, common load:

P_L = 35 MW

Final Answer: 35

Question 35 [Power Electronics : Chopper]

A forced commutated thyristorized step-down chopper is shown in the figure. Neglect the ON-state drop across the power devices.

The capacitor is initially charged to 50 V with the polarity shown in the figure. The load current I_L is constant at 10 A.

Initially, Th_M is ON and Th_A is OFF. The turn-off time available to Th_M in microseconds, when Th_A is triggered, is ______.

Answer: 50 μs

Correct Answer : 50 μs

Solution:

Given:

• Capacitance, C = 10 μF
• Initial capacitor voltage, V = 50 V
• Load current, I₀ = 10 A

The turn-off time available is:

t_off = CV / I₀

Substitute values:

t_off = (10 × 10^-6 × 50) / 10

t_off = 50 × 10^-6 seconds

t_off = 50 μs

Final Answer: 50

Question 36 [Mathematics : Vector Calculus]

Consider a vector u = 2x̂ + ŷ + 2ẑ. The directional derivative of the function

f(x,y,z) = 2ln(xy) + ln(yz) + 3ln(xz)

at point (1,1,1) in the direction of vector u is:

(A) 0

(B) 7 / 5√2

(C) 7

(D) 21

Correct Answer : (C) 7

Solution:

Given:

f(x,y,z) = 2ln(xy) + ln(yz) + 3ln(xz)

Gradient of function:

∇f = x̂ ∂f/∂x + ŷ ∂f/∂y + ẑ ∂f/∂z

After differentiation:

∇f = x̂(5/x) + ŷ(3/y) + ẑ(4/z)

At point (1,1,1):

∇f = 5x̂ + 3ŷ + 4ẑ

Given direction vector:

u = 2x̂ + ŷ + 2ẑ

Magnitude of u:

|u| = √(2² + 1² + 2²) = 3

Directional derivative:

D_uf = ∇f · (u/|u|)

D_uf = [(5×2) + (3×1) + (4×2)] / 3

D_uf = (10 + 3 + 8) / 3 = 21/3

D_uf = 7

Hence, the correct option is (C).

Question 37 [Signals & Systems]

The input x(t) and output y(t) of a system are related as:

y(t) = e^−t ∫_−∞^t e^τx(τ)dτ

The system is:

(A) Nonlinear

(B) Linear and time-invariant

(C) Linear but not time-invariant

(D) Noncausal

Correct Answer : (B)

Solution:

Given system:

y(t) = e^−t ∫_−∞^t e^τx(τ)dτ

Rewrite it as:

y(t) = ∫_−∞^t e^−(t−τ)x(τ)dτ

This is convolution form:

y(t) = x(t) * h(t)

where:

h(t) = e^−tu(t)

Since the output is convolution of input with impulse response, the system is linear and time-invariant.

Hence, the correct option is (B).

Question 38 [Signals & Systems]

Consider the discrete-time systems T₁ and T₂ defined as follows:

{T₁x}[n] = x[0] + x[1] + ... + x[n]
{T₂x}[n] = x[0] + (1/2)x[1] + ... + (1/2)ⁿx[n]

Which of the following statements is true?

(A) T₁ and T₂ are BIBO stable

(B) T₁ and T₂ are not BIBO stable

(C) T₁ is BIBO stable but T₂ is not BIBO stable

(D) T₁ is not BIBO stable but T₂ is BIBO stable

Correct Answer : (D)

Solution:

For BIBO stability, a bounded input must produce a bounded output.

For T₁:

{T₁x}[n] = x[0] + x[1] + ... + x[n]

Take bounded input x[k] = 1.

{T₁x}[n] = 1 + 1 + ... + 1 = n + 1

As n increases, output becomes unbounded. So, T₁ is not BIBO stable.

---

For T₂:

{T₂x}[n] = x[0] + (1/2)x[1] + ... + (1/2)ⁿx[n]

Again take bounded input x[k] = 1.

{T₂x}[n] = 1 + 1/2 + 1/4 + ... + (1/2)ⁿ

This is a geometric series whose sum is finite:

Sum < 2

So, T₂ is BIBO stable.

Hence, the correct option is (D).

Question 39 [Signals & Systems : Z-Transform]

If the Z-transform of a finite-duration discrete-time signal x[n] is X(z), then the Z-transform of the signal

y[n] = x[2n]

is:

(A) Y(z) = X(z²)

(B) Y(z) = 1/2 [X(z^−1/2) + X(−z^−1/2)]

(C) Y(z) = 1/2 [X(z^1/2) + X(−z^1/2)]

(D) Y(z) = 1/2 [X(z²) + X(−z²)]

Correct Answer : (C)

Solution:

For the signal:

y[n] = x[2n]

This means only the even-indexed samples of x[n] are selected.

For extracting even samples, the Z-transform relation is:

Y(z) = 1/2 [ X(z^1/2) + X(−z^1/2) ]

This formula cancels the odd-indexed terms and keeps only the even-indexed terms.

Hence, the correct option is (C).

Question 40 [Electrical Machine : Synchronous Machine]

A 3-phase, 11 kV, 10 MVA synchronous generator is connected to an inductive load of power factor √3/2 via a lossless line with per-phase inductive reactance of 5 Ω. The per-phase synchronous reactance of the generator is 30 Ω with negligible armature resistance.

If the generator is producing the rated current at the rated voltage, then the power factor at the terminal of the generator is:

(A) 0.63 lagging

(B) 0.87 lagging

(C) 0.63 leading

(D) 0.87 leading

Correct Answer : (A) 0.63 lagging

Solution:

Given load power factor:

cos φ = √3 / 2

So,

φ = 30° lagging

The load is inductive, so current lags the load voltage by 30°.

Because the line has inductive reactance, there will be an additional voltage drop of jIX in the line. This changes the terminal voltage angle of the generator.

On solving the phasor relation using rated current and rated voltage condition, the terminal power factor of the generator comes out to be:

Power Factor = 0.63 lagging

Hence, the correct option is (A).

Question 41

For the three-bus lossless power network shown in the figure, the voltage magnitudes at all the buses are equal to 1 per unit (pu), and the differences of the voltage phase angles are very small. The line reactances are represented in the figure, where α, β and γ are strictly positive. The bus injections P₁ and P₂ are in pu. If P₁ = mP₂, where m > 0, and the real power flow from bus 1 to bus 2 is 0 pu, then which one of the following options is correct?

A) γ = mβ

B) β = mγ

C) α = mγ

D) α = mβ

Detailed Solution

Given:

• All bus voltage magnitudes are 1 pu.
• Voltage angle differences are very small.
• P₁ = mP₂, where m > 0.
• Real power flow from bus 1 to bus 2 is zero.

Since power flow between bus 1 and bus 2 is zero,

δ₁ − δ₂ = 0

Therefore,

δ₁ − δ₃ = δ₂ − δ₃

Using the DC power flow approximation,

P₁ = 1/(βx) sin(δ₁−δ₃)

P₂ = 1/(γx) sin(δ₂−δ₃)

Since the angle differences are equal,

P₁/P₂ = γ/β

But,

P₁ = mP₂

Hence,

γ = mβ

Correct Answer: A)

Question 42

A BJT biasing circuit is shown in the figure, where V_BE = 0.7 V and β = 100. The Quiescent Point values of V_CE and I_C are respectively:

A) 4.6 V and 2.46 mA

B) 3.5 V and 2.46 mA

C) 2.61 V and 3.13 mA

D) 4.6 V and 3.13 mA

Detailed Solution

The voltage divider gives the Thevenin equivalent values:

V_TH = 12 × 50/(100+50) = 4 V

R_TH = 100kΩ || 50kΩ = 33.33kΩ

Applying KVL in the base-emitter loop,

I_B = (V_TH−V_BE) / (R_TH+(β+1)R_E)

I_B = 3.3 / (33.33k+101k) = 24.5 μA

Therefore,

I_C = βI_B = 100×24.5μA = 2.46 mA

Emitter current,

I_E ≈ 2.48 mA

Emitter voltage,

V_E = I_ER_E = 2.48 V

Collector voltage,

V_C = 12−(2.46mA×2kΩ) = 7.08 V

Hence,

V_CE = V_C−V_E = 7.08−2.48 = 4.6 V

Correct Answer: A)

Question 43 [MSQ]

Let f(t) be a real-valued function whose second derivative is positive for −∞ < t < ∞. Which of the following is/are always true?

A) f(t) has at least one local minimum

B) f(t) cannot have two distinct local minima

C) f(t) has at least one local maximum

D) The minimum value of f(t) cannot be negative

Detailed Solution

Given that,

f''(t) > 0,   −∞ < t < ∞

This means the function is strictly convex.

For a strictly convex function, there cannot be two distinct local minima. If a local minimum exists, it will be unique.

Checking option A:

f(t) has at least one local minimum is not always true. For example, f(t)=e^t has f''(t)>0 but no local minimum.

Checking option B:

f(t) cannot have two distinct local minima. This is always true for a function with positive second derivative.

Checking option C:

f(t) has at least one local maximum is not true. Since f''(t)>0, the function is convex, so local maximum is not possible.

Checking option D:

The minimum value of f(t) cannot be negative is not true. A function can have negative minimum value.

Correct Answer: B)

Question 44 [MSQ]

Consider the function f(t) = [max(0,t)]² for −∞ < t < ∞, where max(a,b) denotes the maximum of a and b. Which of the following statements is/are true?

A) f(t) is not differentiable

B) f(t) is differentiable and its derivative is continuous

C) f(t) is differentiable but its derivative is not continuous

D) f(t) and its derivative are differentiable

Detailed Solution

Given:

f(t) = [max(0,t)]²

So,

f(t) = { t²,   t ≥ 0
0,   t < 0 }

Now derivative is,

f'(t) = { 2t,   t > 0
0,   t < 0 }

At t = 0,

LHD = 0 and RHD = 0

Therefore, f(t) is differentiable at t = 0.

Also,

f'(0) = 0

The derivative function is continuous at t = 0 because both side limits of f'(t) are equal to 0.

Hence, f(t) is differentiable and its derivative is continuous.

Correct Answer: B)

Question 45 [MSQ]

Which of the following differential equations is/are non-linear?

A) t x(t) + t(dx(t)/dt) = t²eᵗ,   x(0)=0

B) ½eᵗ + x(t) dx(t)/dt = 0,   x(0)=0

C) x(t)cos t − (dx(t)/dt)sin t = 1,   x(0)=0

D) x(t) + e^(dx(t)/dt) = 1,   x(0)=0

Detailed Solution

A differential equation is linear if:

• The dependent variable and its derivatives appear only to the first power.
• No product of the dependent variable and its derivatives is present.
• No transcendental function of the dependent variable or its derivatives is present.

Option (A):

tx(t)+t(dx/dt)=t²eᵗ

This is linear because both x(t) and dx/dt appear only in first degree.

Option (B):

½eᵗ+x(t)(dx/dt)=0

This is non-linear because it contains the product x(t)(dx/dt).

Option (C):

x(t)cos t−(dx/dt)sin t=1

This is linear since the coefficients depend only on t.

Option (D):

x(t)+e^(dx/dt)=1

This is non-linear because a transcendental function of the derivative is present.

Correct Answers: B and D

Question 46 [MSQ]

For a two-phase network, the phase voltages V_p and V_q are to be expressed in terms of sequence voltages V_α and V_β as

[ V_p
V_q ] = S [ V_α
V_β ]

The possible option(s) for matrix S is/are:

A) [ 1  1
1  −1 ]

B) [ 1  1
1  1 ]

C) [ 1  1
1  0 ]

D) [ −1  1
1  1 ]

Detailed Solution

For a valid sequence transformation matrix, the phase quantities must be expressible as independent combinations of the sequence components.

Option (A): The matrix is non-singular and satisfies the required transformation.

Option (B): Both rows are identical, making the determinant zero. Hence it is singular and not valid.

Option (C): This matrix does not satisfy the required sequence transformation.

Option (D): The determinant is non-zero and it also provides a valid transformation.

Therefore, the valid matrices are:

A and D

Correct Answers: A and D

Question 47 [MSQ]

Which of the following options is/are correct for the Automatic Generation Control (AGC) and Automatic Voltage Regulator (AVR) installed with synchronous generators?

A) AGC response has a local effect on frequency while AVR response has a global effect on voltage.

B) AGC response has a global effect on frequency while AVR response has a local effect on voltage.

C) AGC regulates the field current of the synchronous generator while AVR regulates the generator’s mechanical power input.

D) AGC regulates the generator’s mechanical power input while AVR regulates the field current of the synchronous generator.

Detailed Solution

AGC stands for Automatic Generation Control. It mainly controls the real power generation and hence controls the system frequency.

Frequency is a system-wide quantity, so AGC has a global effect on frequency.

AVR stands for Automatic Voltage Regulator. It controls the field current of the synchronous generator and hence regulates the terminal voltage.

Voltage control is more local in nature, so AVR has a local effect on voltage.

Checking option B:

AGC response has a global effect on frequency while AVR response has a local effect on voltage. This is correct.

Checking option D:

AGC regulates the generator’s mechanical power input while AVR regulates the field current of the synchronous generator. This is correct.

Correct Answers: B and D

Question 48

Two passive two-port networks P and Q are connected as shown in the figure. The impedance matrix of network P is

Z_P = [ 40Ω  60Ω
80Ω  100Ω ]

The admittance matrix of network Q is

Y_Q = [ 5S  −2.5S
−2.5S  1S ]

Let the ABCD matrix of the two-port network R in the figure be

[ α  β
γ  δ ]

The value of β in Ω is ________. Rounded off to 2 decimal places.

Detailed Solution

For network P, using Z-parameters:

A_P = Z₁₁/Z₂₁ = 40/80 = 0.5

B_P = (Z₁₁Z₂₂ − Z₁₂Z₂₁)/Z₂₁

B_P = (40×100 − 60×80)/80 = −10 Ω

For network Q, using Y-parameters:

A_Q = −Y₂₂/Y₂₁ = −1/(−2.5)=0.4

B_Q = −1/Y₂₁ = −1/(−2.5)=0.4 Ω

D_Q = −Y₁₁/Y₂₁ = −5/(−2.5)=2

For cascade connection:

β = A_PB_Q + B_PD_Q

β = (0.5)(0.4) + (−10)(2)

β = −19.8 Ω

Correct Answer: −19.80 Ω

Question 49

For the circuit shown in the figure, the source frequency is 5000 rad/sec. The mutual inductance between the magnetically coupled inductors is 5 mH with their self inductances being 125 mH and 1 mH. The Thevenin’s impedance, Z_th, between the terminals P and Q in Ω is ________. Rounded off to 2 decimal places.

Detailed Solution

Given:

ω = 5000 rad/sec

L₁ = 125 mH,   L₂ = 1 mH,   M = 5 mH

Using the coupled inductor equivalent circuit,

L₁ − M = 125mH − 5mH = 120mH

L₂ − M = 1mH − 5mH = −4mH

Converting inductors into reactance:

X_L1 = jω(120×10⁻³) = j600 Ω

X_M = jω(5×10⁻³) = j25 Ω

X_L2 = jω(−4×10⁻³) = −j20 Ω

Capacitive reactance:

X_C = 1/(jωC) = 1/(j×5000×50×10⁻⁶) = −j4 Ω

After simplifying the equivalent network and finding impedance seen from terminals P and Q:

Z_th ≈ 5.33 Ω

Correct Answer: 5.33 Ω

Question 50

In the circuit shown, Z₁ = 50∠−90° Ω and Z₂ = 200∠−30° Ω. It is supplied by a three phase 400 V source with the phase sequence being R-Y-B. Assume the watt meters W₁ and W₂ to be ideal. The magnitude of the difference between the readings of W₁ and W₂ in watts is ________. Rounded off to 2 decimal places.

Detailed Solution

Given:

Z₁ = 50∠−90° Ω

Z₂ = 200∠−30° Ω

Line voltage = 400 V

For R-Y-B phase sequence, take:

V_RY = 400∠0°

V_YB = 400∠−120°

V_BR = 400∠−240°

Using the given circuit and ideal wattmeter connections, the branch currents are calculated from the corresponding phase voltages and impedances.

The wattmeter readings are obtained using:

W = Re[V I*]

After phasor calculation,

|W₁ − W₂| ≈ 692.82 W

Correct Answer Range: 692 to 693 W

Question 51

In the (x, y, z) coordinate system, three point-charges Q, Q, and αQ are located in free space at (−1,0,0), (1,0,0), and (0,−1,0), respectively. The value of α for the electric field to be zero at (0,0.5,0) is ________. Rounded off to 1 decimal place.

Detailed Solution

At point (0,0.5,0), the x-components of electric fields due to the two charges Q cancel each other.

Only y-components will remain.

E_Q,y = 0.357 kQ

Since there are two equal charges Q,

E_{total from Q} = 2 × 0.357 kQ = 0.715 kQ

Electric field due to αQ is:

E_αQ = 0.444 k(αQ)

For net electric field to be zero:

0.715kQ + 0.444k(αQ) = 0

α = −0.715 / 0.444 = −1.61

Correct Answer: −1.6

Question 52

The given equation represents a magnetic field strength H̅(r,θ,ϕ) in the spherical coordinate system, in free space. Here, r̂ and θ̂ represent the unit vectors along r and θ, respectively. The value of P in the equation should be ________. Rounded off to the nearest integer.

H̅(r,θ,ϕ) = 1/r³ ( r̂ P cosθ + θ̂ sinθ )

Detailed Solution

Given magnetic field:

H̅ = 1/r³ ( r̂ P cosθ + θ̂ sinθ )

In free space, magnetic flux density must satisfy:

∇ · B̅ = 0

Since B̅ = μ₀H̅, therefore:

∇ · H̅ = 0

For spherical coordinates:

∇ · H̅ = 1/r² ∂(r²H_r)/∂r + 1/(r sinθ) ∂(H_θ sinθ)/∂θ

Here,

H_r = P cosθ/r³

H_θ = sinθ/r³

Now,

1/r² ∂(r²H_r)/∂r = 1/r² ∂(P cosθ/r)/∂r = −P cosθ/r⁴

And,

1/(r sinθ) ∂(H_θsinθ)/∂θ = 1/(r sinθ) ∂(sin²θ/r³)/∂θ = 2cosθ/r⁴

So,

∇ · H̅ = (−P cosθ + 2cosθ)/r⁴

For ∇ · H̅ = 0:

−P + 2 = 0

P = 2

Correct Answer: 2

Question 53

If the energy of a continuous-time signal x(t) is E and the energy of the signal 2x(2t−1) is cE, then c is ________. Rounded off to 1 decimal place.

Detailed Solution

Energy of signal x(t) is:

E = ∫ |x(t)|² dt

For time scaling:

x(at) → Energy = E/|a|

For amplitude scaling:

kx(t) → Energy = k²E

Given signal:

2x(2t−1)

Here amplitude factor is 2 and time scaling factor is 2.

Energy = 2² × E/2 = 2E

Therefore,

cE = 2E

c = 2.0

Correct Answer: 2.0

Question 54

A 3-phase star connected slip ring induction motor has the following parameters referred to the stator:

R_s = 3Ω,   X_s = 2Ω,   X′_r = 2Ω,   R′_r = 2.5Ω

The per phase stator to rotor effective turns ratio is 3 : 1. The rotor winding is also star connected. The magnetizing reactance and core loss of the motor can be neglected. To have maximum torque at starting, the value of the extra resistance in ohms, effective to the rotor side, to be connected in series with each phase of the rotor winding is ________. Rounded off to 2 decimal places.

Detailed Solution

For maximum torque at starting,

R′_2,total = √[R_s² + (X_s + X′_r)²]

Substituting the values:

R′_2,total = √[3² + (2+2)²]

R′_2,total = √(9+16) = 5Ω

Already rotor resistance referred to stator side is:

R′_r = 2.5Ω

Extra resistance referred to stator side:

R′_ext = 5 − 2.5 = 2.5Ω

Turns ratio:

N_s : N_r = 3 : 1

So, rotor side extra resistance:

R_ext = R′_ext / 3²

R_ext = 2.5 / 9 = 0.277Ω

Correct Answer: 0.28 Ω

Question 55

A 5 kW, 220 V DC shunt motor has 0.5Ω armature resistance including brushes. The motor draws a no-load current of 3 A. The field current is constant at 1 A. Assuming that the core and rotational losses are constant and independent of the load, the current drawn by the motor while delivering the rated load, for the best possible efficiency, is ________. Rounded off to 2 decimal places.

Detailed Solution

Given:

V = 220 V,   R_a = 0.5Ω,   I_sh = 1 A

P_out = 5 kW,   I_L,no-load = 3 A

At no-load, armature current is:

I_a0 = I_L − I_sh = 3 − 1 = 2 A

Back emf at no-load:

E_b0 = V − I_a0R_a = 220 − (2)(0.5) = 219 V

Constant rotational and core loss:

P_loss = E_b0I_a0 = 219 × 2 = 438 W

At rated load:

Mechanical power developed = P_out + P_loss

E_bI_a = 5000 + 438 = 5438 W

Also,

E_b = V − I_aR_a

Therefore,

(220 − 0.5I_a)I_a = 5438

0.5I_a² − 220I_a + 5438 = 0

Solving,

I_a = 26.28 A

Total current drawn by motor:

I_L = I_a + I_sh = 26.28 + 1 = 27.28 A

Correct Answer: 27.28 A

Question 56

The signal line diagram of a lossless system is shown in the figure. The system is operating in steady-state at a stable equilibrium point with the power output of the generator being P_max sinδ, where δ is the load angle and the mechanical power input is 0.5P_max. A fault occurs on line 2 such that the power output of the generator is less than 0.5P_max during the fault. After the fault is cleared by opening line 2, the power output of the generator is (P_max/√2) sinδ. If the critical fault clearing angle is π/2 radians, the accelerating area over the power angle curve is ________ P_max. Rounded off to 2 decimal places.

Detailed Solution

Initial operating condition:

P_m = P_max sinδ₀

0.5P_max = P_max sinδ₀

δ₀ = π/6

After fault clearing:

P_e = P_max/√2 sinδ

Maximum rotor angle:

0.5P_max = P_max/√2 sinδ_m

sinδ_m = 1/√2

δ_m = 3π/4

Critical clearing angle:

δ_c = π/2

By equal area criterion:

A_acc = A_dec

A_dec = ∫_π/2^3π/4 [(P_max/√2)sinδ − 0.5P_max] dδ

A_dec = 0.5P_max − (π/8)P_max

A_dec = 0.107P_max

Correct Answer: 0.11 P_max

Question 57

Consider the closed-loop system shown in the figure with

G(s)= K(s²−2s+2)/(s²+2s+5)

The root locus for the closed-loop system is to be drawn for 0 ≤ K ≤ ∞. The angle of departure from the pole (−1+j2), in degrees, is ________. Rounded off to the nearest integer.

Detailed Solution

Given:

G(s)= K(s²−2s+2)/(s²+2s+5)

Zeros are:

s = 1 ± j

Poles are:

s = −1 ± j2

Angle of departure from pole (−1+j2) is:

θ_d = 180° + Σ angle from other poles − Σ angle from zeros

Angle from other pole:

90°

Angles from zeros:

153.43° and 123.69°

Therefore,

θ_d = 180° + 90° − (153.43° + 123.69°)

θ_d = −7.12°

Magnitude of angle of departure:

≈ 7°

Correct Answer: 7°

Question 58

Consider the stable closed-loop system shown in the figure. The asymptotic Bode magnitude plot of G(s) has a constant slope of −20 dB/decade at least till 100 rad/sec, with the gain crossover frequency 10 rad/sec. The asymptotic Bode phase plot remains constant at −90° at least till ω = 10 rad/sec. The steady-state error of the closed-loop system for a unit ramp input is ________. Rounded off to 2 decimal places.

Detailed Solution

Given that the initial slope of Bode magnitude plot is −20 dB/decade and phase is −90°.

So the open-loop transfer function behaves like:

G(s) = K/s

Gain crossover frequency is:

ω_gc = 10 rad/sec

At gain crossover frequency:

|G(jω)| = 1

K/ω = 1

K/10 = 1

K = 10

For unity feedback system, steady-state error for unit ramp input is:

e_ss = 1/K_v

Velocity error constant:

K_v = lim_s→0 sG(s)

K_v = lim_s→0 s(K/s) = K = 10

Hence,

e_ss = 1/10 = 0.10

Correct Answer: 0.10

Question 59

Consider the stable closed-loop system shown in the figure. The magnitude and phase values of the frequency response of G(s) are given in the table. The value of gain K₁(>0) for a 50° phase margin is ________. Rounded off to 2 decimal places.

ω in rad/sec	Magnitude in dB	Phase in degrees
0.5	−7	−40
1.0	−10	−80
2.0	−18	−130
10.0	−40	−200

 

Question 60

In the given circuit, the diodes are ideal. The current I through the diode D₁ in milliamperes is ________. Rounded off to 2 decimal places.

Detailed Solution

Assume both ideal diodes D₁ and D₂ are ON.

Applying nodal analysis at left node x:

(V_x−10)/1k + V_x/1k + (V_x−3)/1k = 0

3V_x − 13 = 0

V_x = 13/3 = 4.33 V

Current through horizontal 1kΩ resistor:

I_D2 = (4.33−3)/1k = 1.33 mA

Current through lower 1kΩ resistor:

I₃ = 3/1k = 3 mA

By KCL at right node:

I + I_D2 = I₃

I = 3 − 1.33 = 1.67 mA

Correct Answer: 1.67 mA

Question 61

A difference amplifier is shown in the figure. Assume the op-amp to be ideal. The CMRR in dB of the difference amplifier is ________. Rounded off to 2 decimal places.

Detailed Solution

Given resistor values:

R₁ = 10.5kΩ,   R_f = 101kΩ

R₂ = 9.5kΩ,   R_g = 101.5kΩ

Voltage at non-inverting terminal:

V₊ = V_in2 × 101.5/(101.5+9.5)

V₊ = 0.9144V_in2

Output voltage:

V_o = (1+101/10.5)V₊ − (101/10.5)V_in1

V_o = 9.711V_in2 − 9.619V_in1

Differential gain:

A_d = (9.711+9.619)/2 = 9.665

Common mode gain:

A_cm = 9.711 − 9.619 = 0.092

Therefore,

CMRR = 20log(A_d/A_cm)

CMRR ≈ 40.40 dB

Correct Answer: 40.40 dB

Question 62

A single-phase half-controlled bridge converter supplies an inductive load with ripple free load current. The triggering angle of the converter is 60°. The ratio of the rms value of the fundamental component of the input current to the rms value of total input current of the bridge is ________. Rounded off to 3 decimal places.

Detailed Solution

For a single-phase half-controlled bridge converter with continuous load current:

i_s = I_d   from   α to π

i_s = -I_d   from   π+α to 2π

Total rms input current is:

I_{rms} = I_d√[(π−α)/π]

For α = 60° = π/3:

I_{rms} = I_d√(2/3)

The rms value of fundamental component is:

I_{1,rms} = 0.7797 I_d

Required ratio:

I_{1,rms}/I_{rms} = 0.7797/0.8165 = 0.955

Correct Answer: 0.955

Question 63

A single-phase bridge voltage source inverter (VSI) feeds a purely inductive load. The inverter output voltage is a square wave in 180° conduction mode. The fundamental frequency of the output voltage is 50 Hz. If the DC input voltage of the inverter is 100 V and the value of the load inductance is 20 mH, the peak-to-peak load current in amperes is ________. Rounded off to the nearest integer.

Detailed Solution

For a purely inductive load,

V = L(di/dt)

Given:

V_s = 100 V,   L = 20 mH = 0.02 H

f = 50 Hz

Time period:

T = 1/f = 1/50 = 0.02 sec

In one half cycle, current changes from negative peak to positive peak.

ΔI = (V/L)(T/2)

ΔI = (100/0.02)(0.02/2)

ΔI = 5000 × 0.01 = 50 A

Therefore, peak-to-peak load current is:

I_pp = 50 A

Correct Answer: 50 A

Question 64

In the DC-DC converter shown in the figure, the current through the inductor is continuous. The switching frequency is 500 Hz. The voltage (V_o) across the load is assumed to be constant and ripple free. The peak inductor current in amperes is ________. Rounded off to the nearest integer.

Detailed Solution

Given:

V_s = 20 V, L = 2 mH, V_o = 40 V, f = 500 Hz, R = 10 Ω

For a boost converter,

V_o = V_s /(1-D)

40 = 20 /(1-D)

D = 0.5

Average output current:

I_o = V_o/R = 40/10 = 4 A

Average inductor current:

I_L(avg) = I_o /(1-D) = 4/0.5 = 8 A

Inductor current ripple:

ΔI_L = (V_sD)/(Lf)

ΔI_L = (20×0.5)/(0.002×500) = 10 A

Peak inductor current:

I_L(max) = I_L(avg) + ΔI_L/2

I_L(max) = 8 + 10/2 = 13 A

Correct Answer: 13 A

Question 65

A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications:

Rated armature voltage	210 A
Rated armature current	10 A
Rated speed	1200 rpm
Armature resistance	1Ω
Input to converter bridge	240 V at 50 Hz

The armature of the DC motor is fed from the full-controlled bridge and the field current is kept constant. Assume that the motor is running at 600 rpm and the armature terminals of the motor are suitably reversed for regenerative braking. If the armature current of the motor is to be maintained at the rated value, the triggering angle of the converter bridge in degrees should be ________. Rounded off to 2 decimal places.

Detailed Solution

At rated condition:

E_b = V - I_aR_a

E_b = 210 - (10×1) = 200 V

Since field current is constant, back emf is proportional to speed.

E_{b,600} = 200 × 600/1200 = 100 V

During regenerative braking, required converter output voltage is:

V_0 = -(E_b - I_aR_a)

V_0 = -(100 - 10×1) = -90 V

For single-phase full-controlled bridge:

V_0 = (2V_m/π)cosα

V_m = √2 × 240 = 339.41 V

-90 = (2×339.41/π)cosα

cosα = -0.4165

α = 114.61°

Correct Answer: 114.61°