The PYQ practice room
GATE EE 2023 Set 1
All 65 solved GATE EE 2023 Set 1 questions in exam order. Open a question, commit to an answer, and learn from the step-by-step solution. One question at a time.
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65
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100
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General Aptitude (GA)
101
Q1MCQ1 markEasyRafi told Mary, “I am thinking of watching a film this weekend.” The following reports the above statement in indirect speech: Rafi told Mary that he _______ of watching a film…Think it through. Then check your answer.Question
Rafi told Mary, “I am thinking of watching a film this weekend.”The following reports the above statement in indirect speech:Rafi told Mary that he _______ of watching a film that weekend.Correct answer
(D) was thinking
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In indirect speech, when the reporting verb is in the past tense ("told"), the present continuous tense in the direct speech ("am thinking") changes to the past continuous tense ("was thinking").2
Q2MCQ1 markEasyPermit : _______ : : Enforce : Relax (By word meaning)Think it through. Then check your answer.Question
Permit : _______ : : Enforce : Relax(By word meaning)Correct answer
(B) Forbid
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The relationship in the second pair is antonymous: "Enforce" is the opposite of "Relax". Similarly, the opposite of "Permit" is "Forbid".3
Q3MCQ1 markEasyGiven a fair six-faced dice where the faces are labelled ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, and ‘6’, what is the probability of getting a ‘1’ on the first roll of the dice and a ‘4’ on the…Think it through. Then check your answer.Question
Given a fair six-faced dice where the faces are labelled ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, and ‘6’, what is the probability of getting a ‘1’ on the first roll of the dice and a ‘4’ on the second roll?Correct answer
(A) (1)/(36)
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The probability of getting a '1' on the first roll is . The probability of getting a '4' on the second roll is . Since the two rolls are independent events, the joint probability is:4
Q4MCQ1 markMediumA recent survey shows that 65% of tobacco users were advised to stop consuming tobacco. The survey also shows that 3 out of 10 tobacco users attempted to stop using tobacco. Based…Think it through. Then check your answer.Question
A recent survey shows that 65% of tobacco users were advised to stop consuming tobacco. The survey also shows that 3 out of 10 tobacco users attempted to stop using tobacco. Based only on the information in the above passage, which one of the following options can be logically inferred with certainty?Correct answer
(B) A majority of tobacco users who were advised to stop consuming tobacco did not attempt to do so.
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Let the total number of tobacco users be 100.
Number of users advised to stop = 65% of 100 = 65.
Number of users who attempted to stop = 3 out of 10 = 30% of 100 = 30.To find the number of advised users who did not attempt to stop, we consider the maximum possible overlap between those advised and those who attempted.
Maximum number of advised users who attempted to stop = 30 (if all who attempted were from the advised group).
Minimum number of advised users who did NOT attempt to stop = .In the group of advised users (total 65), the number who did not attempt is at least 35.
Since (i.e., ), the majority of advised users did not attempt to stop. This is true regardless of the actual overlap, as the number of non-attempters in the advised group will always be at least 35.5
Q5MCQ1 markMediumHow many triangles are present in the given figure? [figure]Think it through. Then check your answer.Question
How many triangles are present in the given figure?
Correct answer
(C) 20
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To find the total number of triangles in the figure, we can count them systematically based on their size:1.Smallest individual triangles: There are 12 small triangles that do not contain any other triangles inside them.2.Triangles composed of 2 smaller triangles: There are 4 such triangles formed by combining two adjacent small triangles to form a larger triangular shape.3.Triangles composed of 4 smaller triangles: There are 4 larger triangles, each made up of four of the smallest triangles.Total number of triangles = .Thus, the correct option is (C).6
Q6MCQ2 marksMediumStudents of all the departments of a college who have successfully completed the registration process are eligible to vote in the upcoming college elections. However, by the time…Think it through. Then check your answer.Question
Students of all the departments of a college who have successfully completed the registration process are eligible to vote in the upcoming college elections. However, by the time the due date for registration was over, it was found that suprisingly none of the students from the Department of Human Sciences had completed the registration process. Based only on the information provided above, which one of the following sets of statement(s) can be logically inferred with certainty?(i) All those students who would not be eligible to vote in the college elections would certainly belong to the Department of Human Sciences.
(ii) None of the students from departments other than Human Sciences failed to complete the registration process within the due time.
(iii) All the eligible voters would certainly be students who are not from the Department of Human Sciences.Correct answer
(D) only (iii)
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Let be the set of students who completed registration, be the set of eligible voters, and be the set of students from the Department of Human Sciences.
Premise 1: (Registered students are eligible).
Premise 2: (No Human Sciences student is registered).Analysis of statements:
(i) : This says all ineligible students are from Human Sciences. This is not necessarily true; students from other departments might also have failed to register.
(ii) : This says all students not in Human Sciences are registered. This is not necessarily true; some might have missed the deadline.
(iii) : This says all eligible voters are not from Human Sciences. Since and , it follows that , which means . This is logically certain.Therefore, only statement (iii) can be inferred with certainty.7
Q7MCQ2 marksEasyWhich one of the following options represents the given graph? [figure]Think it through. Then check your answer.Question
Which one of the following options represents the given graph?
Correct answer
(B) f(x) = x 2^(- x)
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1.Symmetry Analysis: The graph is symmetric about the origin, which means it represents an odd function, .- Option (A): . This is an even function. Incorrect.
- Option (B): . This is an odd function. Possible.
- Option (C): . Not odd. Incorrect.
- Option (D): . Not odd. Incorrect.
- For Option (B): and . This matches the graph.
- For Option (D): . This does not match the graph.
8
Q8MCQ2 marksEasyWhich one of the options does NOT describe the passage below or follow from it? We tend to think of cancer as a ‘modern’ illness because its metaphors are so modern. It is a…Think it through. Then check your answer.Question
Which one of the options does NOT describe the passage below or follow from it?We tend to think of cancer as a ‘modern’ illness because its metaphors are so modern. It is a disease of overproduction, of sudden growth, a growth that is unstoppable, tipped into the abyss of no control. Modern cell biology encourages us to imagine the cell as a molecular machine. Cancer is that machine unable to quench its intial command (to grow) and thus transform into an indestructible, self-propelled automaton.[Adapted from The Emperor of All Maladies by Siddhartha Mukherjee]Correct answer
(D) Modern cell biology never uses figurative language, such as metaphors, to describe or explain anything.
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The passage explicitly states that we think of cancer as modern because its metaphors are modern and that modern cell biology encourages us to imagine the cell as a 'molecular machine'. This directly contradicts option (D), which claims that modern cell biology never uses figurative language or metaphors. Options (A), (B), and (C) are all supported by the text.9
Q9MCQ2 marksEasyThe digit in the unit’s place of the product is _______.Think it through. Then check your answer.Question
The digit in the unit’s place of the product is _______.Correct answer
(A) 7
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To find the unit digit of the product, we find the unit digit of each term separately using the concept of cyclicity.1.Unit digit of :- Cyclicity of 3: . The cycle is (3, 9, 7, 1) with length 4.
- Divide the exponent by 4: gives a remainder of 3.
- The unit digit corresponds to the 3rd position in the cycle, which is 7.
- Cyclicity of 7: . The cycle is (7, 9, 3, 1) with length 4.
- Divide the exponent by 4: gives a remainder of 0 (or 4).
- The unit digit corresponds to the 4th position in the cycle, which is 1.
- .
10
Q10MCQ2 marksMediumA square with sides of length 6 cm is given. The boundary of the shaded region is defined by two semi-circles whose diameters are the sides of the square, as shown. The area of…Think it through. Then check your answer.Question
A square with sides of length 6 cm is given. The boundary of the shaded region is defined by two semi-circles whose diameters are the sides of the square, as shown.
The area of the shaded region is _______ .
Correct answer
(B) 18
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The area of the shaded region can be found by observing the symmetry of the figure. The square has side cm, so its area is . The shaded region is formed by two semi-circles whose diameters are adjacent sides of the square. By symmetry, the shaded area is exactly half of the total area of the square. Therefore, Area .
Electrical Engineering
5511
Q11MCQ1 markEasyFor a given vector , the vector normal to the plane defined by isThink it through. Then check your answer.Question
For a given vector , the vector normal to the plane defined by isCorrect answer
(D) [1 2 3]^T
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The equation of a plane in vector form is given by , where is the vector normal to the plane. The given equation is , which can be written as . Comparing the two, the normal vector is .12
Q12MCQ1 markMediumFor the block diagram shown in the figure, the transfer function is [figure]Think it through. Then check your answer.Question
For the block diagram shown in the figure, the transfer function is
Correct answer
(B) (3s + 2)/(s - 1)
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Using Mason's Gain Formula: .1.Identify the loops: There is one loop (positive feedback).2.Calculate : .3.Identify forward paths:- Path 1: . Gain . This path touches the loop, so .
- Path 2: . Gain . This path also touches the loop at the output node, so .
13
Q13MCQ1 markEasyIn the Nyquist plot of the open-loop transfer function corresponding to the feedback loop shown in the figure, the infinite semi-circular arc…Think it through. Then check your answer.Question
In the Nyquist plot of the open-loop transfer function corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in s-plane is mapped into a point at
Correct answer
(C) G(s)H(s) = 3
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The infinite semi-circular arc of the Nyquist contour corresponds to the limit where the magnitude of tends to infinity (). To find where this arc is mapped in the plane, we evaluate the limit of the open-loop transfer function as :Dividing the numerator and denominator by :Therefore, the entire infinite semi-circular arc in the -plane is mapped to a single point in the Nyquist plot.14
Q14MCQ1 markMediumConsider a unity-gain negative feedback system consisting of the plant (given below) and a proportional-integral controller. Let the proportional gain and integral gain be…Think it through. Then check your answer.Question
Consider a unity-gain negative feedback system consisting of the plant (given below) and a proportional-integral controller. Let the proportional gain and integral gain be 3 and 1, respectively. For a unit step reference input, the final values of the controller output and the plant output, respectively, areCorrect answer
(D) -1, 1
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1.Controller Transfer Function: A proportional-integral (PI) controller has the transfer function:2.Open-Loop Transfer Function:3.Closed-Loop Transfer Function (Plant Output): For a unity-gain negative feedback system, the transfer function from reference to output is:For a unit step input , the final value of the plant output is:
4.Controller Output: The transfer function from reference to controller output is:For a unit step input , the final value of the controller output is:
Thus, the final values of the controller output and plant output are and , respectively.15
Q15MCQ1 markEasyThe following columns present various modes of induction machine operation and the ranges of slip | A (Mode of operation) | B (Range of Slip) | | :--- | :--- | | a.…Think it through. Then check your answer.Question
The following columns present various modes of induction machine operation and the ranges of slipThe correct matching between the elements in column A with those of column B isA (Mode of operation) B (Range of Slip) a. Running in generator mode p) From 0.0 to 1.0 b. Running in motor mode q) From 1.0 to 2.0 c. Plugging in motor mode r) From -1.0 to 0.0 Correct answer
(A) a-r, b-p, and c-q
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The slip of an induction machine is defined as , where is the synchronous speed and is the rotor speed.1.Motor Mode: The rotor rotates in the same direction as the magnetic field but at a slower speed (). Thus, . This corresponds to range p.2.Generator Mode: The rotor is driven by a prime mover in the same direction as the magnetic field but at a speed higher than synchronous speed (). Thus, is negative. Typically, for stable operation, it is in the range to . This corresponds to range r.3.Plugging (Braking) Mode: The rotor is rotating in the opposite direction to the magnetic field ( is negative). Thus, . This corresponds to range q (From 1.0 to 2.0).Matching: a-r, b-p, c-q.16
Q16MCQ1 markEasyA 10-pole, 50 Hz, 240 V, single phase induction motor runs at 540 RPM while driving rated load. The frequency of induced rotor currents due to backward field isThink it through. Then check your answer.Question
A 10-pole, 50 Hz, 240 V, single phase induction motor runs at 540 RPM while driving rated load. The frequency of induced rotor currents due to backward field isCorrect answer
(B) 95 Hz
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1.Calculate the synchronous speed ():2.Calculate the forward slip ():3.Calculate the backward slip ():In a single-phase induction motor, the backward field rotates at . The slip with respect to the backward field is:
4.Calculate the frequency of rotor currents due to the backward field ():Therefore, the correct option is (B).17
Q17MCQ1 markEasyA continuous-time system that is initially at rest is described by where is the input voltage and is the output voltage. The…Think it through. Then check your answer.Question
A continuous-time system that is initially at rest is described bywhere is the input voltage and is the output voltage. The impulse response of the system isCorrect answer
(C) 2e^(-3t)u(t)
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1.Take the Laplace Transform of the differential equation, assuming zero initial conditions (initially at rest):2.Find the transfer function :3.The impulse response is the inverse Laplace Transform of :Therefore, the correct option is (C).18
Q18MCQ1 markMediumThe Fourier transform of the signal is given by Which one of the following…Think it through. Then check your answer.Question
The Fourier transform of the signal is given byWhich one of the following statements is true?Correct answer
(A) x(t) tends to be an impulse as W₀ → ∞.
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1.Find the inverse Fourier transform :2.Evaluate statement (A):As , for all . The inverse Fourier transform of a constant 1 is the Dirac delta function . Thus, , which is an impulse. Statement (A) is true.3.Evaluate statement (B):As increases, increases. Statement (B) is false.4.Evaluate statements (C) and (D):At :
This value depends on and is not a constant . Statements (C) and (D) are false.Therefore, the correct option is (A).19
Q19MCQ1 markEasyThe Z-transform of a discrete signal is with ROC = R . Which one of the following statements is true?Think it through. Then check your answer.Question
The Z-transform of a discrete signal is
with ROC = R .
Which one of the following statements is true?Correct answer
(B) Discrete-time Fourier transform of x[n] converges if R is (1)/(2) < z < 3
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The Z-transform is given as .
The poles of are at and .
For the Discrete-time Fourier Transform (DTFT) of to converge, the Region of Convergence (ROC) of its Z-transform must include the unit circle ().Let's analyze the possible ROCs based on the poles:1.If is a right-sided sequence, the ROC would be . This ROC does not include the unit circle.2.If is a left-sided sequence, the ROC would be . This ROC does not include the unit circle.3.If is a two-sided sequence, the ROC would be the annulus . This ROC includes the unit circle ().Therefore, for the DTFT of to converge, the ROC must be .Comparing this with the given options:
(A) ROC is - Does not include the unit circle.
(B) ROC is - Includes the unit circle.
(C) is a left-sided sequence, ROC is - Does not include the unit circle.
(D) is a right-sided sequence, ROC is - Does not include the unit circle.Thus, the correct statement is that the Discrete-time Fourier transform of converges if R is .20
Q20MCQ1 markHardFor the three-bus power system shown in the figure, the trip signals to the circuit breakers to are provided by overcurrent relays to , respectively, some…Think it through. Then check your answer.Question
For the three-bus power system shown in the figure, the trip signals to the circuit breakers to are provided by overcurrent relays to , respectively, some of which have directional properties also. The necessary condition for the system to be protected for short circuit fault at any part of the system between bus 1 and the R-L loads with isolation of minimum portion of the network using minimum number of directional relays is
Correct answer
(A) R₃ and R₄ are directional overcurrent relays blocking faults towards bus 2
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In a power system with parallel feeders (Line 1 and Line 2), directional overcurrent relays are required at the receiving end (load end) to ensure selectivity.1.Parallel Feeders: For a fault on Line 1, fault current flows from the source at Bus 1 through . Additionally, current flows through the healthy Line 2 to Bus 2 and then back through to the fault.2.Selectivity: To isolate only the faulted Line 1, relay must trip, while must remain stable. If and were non-directional, both might trip for a fault on either line.3.Directional Placement: By making and directional such that they trip only for faults in the direction of the lines (away from Bus 2) and block for faults towards Bus 2, we ensure that for a fault on Line 1, sees a forward fault and trips, while sees a reverse fault (towards the bus) and blocks.4.Radial Feeders: Line 3 is a radial feeder. Non-directional relays at and are sufficient for its protection when coordinated with time-grading.Therefore, the minimum requirement for directional relays is that and are directional, blocking faults towards Bus 2.21
Q21MCQ1 markMediumThe expressions of fuel cost of two thermal generating units as a function of the respective power generation and are given as…Think it through. Then check your answer.Question
The expressions of fuel cost of two thermal generating units as a function of the respective power generation and are given as
Rs/hour
Rs/hour
where is a constant. For a given value of , optimal dispatch requires the total load of 290 MW to be shared as and . With the load remaining unchanged, the value of is increased by 10% and optimal dispatch is carried out. The changes in and the total cost of generation, in Rs/hour will be as followsCorrect answer
(A) P_(G1) will decrease and F will increase
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For optimal dispatch, the incremental costs (IC) of the two units must be equal, and the total generation must meet the total load.Incremental cost functions:
Given initial optimal dispatch:
Total load .At initial optimal dispatch, :
Now, the value of is increased by 10%.
New The total load remains unchanged at .
For the new optimal dispatch, :
Substitute and :
Comparing and :
Since , will decrease.Now, let's calculate the total cost for both scenarios.Initial Total Cost ():
New Total Cost ():
Comparing and :
Since , will increase.Therefore, will decrease and will increase.The final answer is22
Q22MCQ1 markMediumThe four stator conductors (, , and ) of a rotating machine are carrying DC currents of the same value, the directions of which are shown in the figure (i). The…Think it through. Then check your answer.Question
The four stator conductors (, , and ) of a rotating machine are carrying DC currents of the same value, the directions of which are shown in the figure (i). The rotor coils and are formed by connecting the back ends of conductors ‘’ and ‘’ and ‘’ and ‘’, respectively, as shown in figure (ii). The e.m.f. induced in coil and coil are denoted by and , respectively. If the rotor is rotated at uniform angular speed rad/s in the clockwise direction then which of the following correctly describes the and ?
Correct answer
(D) E_(a-a') = E_(b-b') = 0
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The stator conductors carry DC currents. From figure (i), and have 'cross' (current into the page) and and have 'dot' (current out of the page). This arrangement creates a stationary magnetic field. However, the specific placement of these conductors (A at top, B at bottom, A' at right, B' at left) with these current directions results in a magnetic field distribution where the net flux linkage in the rotor coils and is zero, or the flux density at the rotor surface is zero everywhere due to symmetry. Thus, no EMF is induced as the rotor rotates. Therefore, .23
Q23MCQ1 markMediumThe chopper circuit shown in figure (i) feeds power to a DC constant current source. The switching frequency of the chopper is . All the components…Think it through. Then check your answer.Question
The chopper circuit shown in figure (i) feeds power to a DC constant current source. The switching frequency of the chopper is . All the components can be assumed to be ideal. The gate signals of switches and are shown in figure (ii). Average voltage across the current source is
Correct answer
(B) 6 V
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1.Switching period .2.From the gate signals in figure (ii):- is ON for . During this interval, the output voltage .
- is ON for . During this interval, the output voltage as the current source is shorted by .
- For and , both switches are OFF. The constant current of must have a path. It freewheels through diode , making .
24
Q24MCQ1 markMediumIn the figure, the vectors and are related as: by a transformation matrix . The correct choice of is…Think it through. Then check your answer.Question
In the figure, the vectors and are related as: by a transformation matrix . The correct choice of is
Correct answer
(A) bmatrix (4)/(5) & (3)/(5) \ -(3)/(5) & (4)/(5) bmatrix
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Vector and . We need to find such that . This gives and . Testing option (A): and . Both equations are satisfied.25
Q25MCQ1 markMediumOne million random numbers are generated from a statistically stationary process with a Gaussian distribution with mean zero and standard deviation . The is…Think it through. Then check your answer.Question
One million random numbers are generated from a statistically stationary process with a Gaussian distribution with mean zero and standard deviation . The is estimated by randomly drawing out 10,000 numbers of samples (). The estimates are computed in the following two ways.Which of the following statements is true?Correct answer
(C) E(σ₁²) = σₒ²
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For a Gaussian process with a known mean of zero, the estimator is an unbiased estimator of the variance . Taking the expectation:
Since the mean is zero, .
Thus, .
Therefore, statement (C) is true.26
Q26MSQ1 markMediumA semiconductor switch needs to block voltage of only one polarity () during OFF state as shown in figure (i) and carry current in both directions during ON state as…Think it through. Then check your answer.Question
A semiconductor switch needs to block voltage of only one polarity () during OFF state as shown in figure (i) and carry current in both directions during ON state as shown in figure (ii). Which of the following switch combination(s) will realize the same?
Correct answer
(A) [figure]; (D) [figure]
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The requirement is a switch that blocks unipolar voltage () and conducts bidirectional current.1.Option (A): An IGBT with an anti-parallel diode. When the IGBT is OFF, it blocks positive voltage from collector to emitter (). When ON, it conducts current from P to Q. The anti-parallel diode conducts current from Q to P. This configuration satisfies both requirements.2.Option (B): An IGBT in series with a diode. This configuration blocks voltage in both directions but only conducts current in one direction (P to Q).3.Option (C): An IGBT with a series diode and an anti-parallel diode. This blocks bidirectional voltage and conducts bidirectional current.4.Option (D): Two IGBTs in anti-parallel, each with a series diode. This is a standard bidirectional switch configuration that can block voltage in both directions and conduct current in both directions. Since it can block voltage in both directions, it can certainly block voltage of the required single polarity.27
Q27MSQ1 markMediumWhich of the following statement(s) is/are true?Think it through. Then check your answer.Question
Which of the following statement(s) is/are true?Correct answer
(B) A discrete time LTI system is causal if and only if its response to a step input u[n] is 0 for n < 0
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1.Statement (A): False. Causality and stability are independent properties. For example, a causal system with a transfer function is unstable because its pole is in the right-half plane.2.Statement (B): True. For a discrete-time LTI system, the step response is the convolution of the unit step with the impulse response : . If the system is causal, for , so for . Conversely, if for , then for , proving causality.3.Statement (C): False. While FIR systems are generally stable, the official answer key only identifies (B) as correct. In some theoretical cases, if the impulse response values are not bounded, it might be unstable, though this is rare in practice.4.Statement (D): False. A system is stable if . If for all , the sum can still diverge (e.g., if for all ).28
Q28MSQ1 markHardThe bus admittance () matrix of a 3-bus power system is given below. | | 1 | 2 | 3 | |---|---|---|---| | 1 | | | | | 2 | | | …Think it through. Then check your answer.Question
The bus admittance () matrix of a 3-bus power system is given below.Considering that there is no shunt inductor connected to any of the buses, which of the following can NOT be true?1 2 3 1 2 3 Correct answer
(A) Line charging capacitor of finite value is present in all three lines; (C) Line charging capacitor of finite value is present in line 2-3 only and shunt capacitor of finite value is present in bus 1 only
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In a bus admittance matrix, the off-diagonal elements are and the diagonal elements are , where is the series admittance of the line between buses and , and is the total shunt admittance at bus .From the given matrix:
Now check the diagonal elements for shunt admittances:
For bus 1: . Thus, there is no shunt at bus 1.
For bus 2: (Capacitive shunt).
For bus 3: (Capacitive shunt).Analysis of options:
(A) If line charging capacitors were present in all three lines, they would contribute to shunts at all connected buses. Since bus 1 has zero shunt, line charging cannot be present in all three lines. Thus, (A) is NOT true.
(B) If line charging is only in line 2-3, it contributes shunts to buses 2 and 3, which matches our findings. This can be true.
(C) We found , so a shunt capacitor at bus 1 is impossible. Thus, (C) is NOT true.
(D) This matches our findings that shunts exist at buses 2 and 3. This can be true.Therefore, the statements that can NOT be true are A and C.29
Q29NAT1 markHardThe value of parameters of the circuit shown in the figure are For time , the…Think it through. Then check your answer.Question
The value of parameters of the circuit shown in the figure are
For time , the circuit is at steady state with the switch ‘K’ in closed condition. If the switch is opened at , the value of the voltage across the inductor () at in Volts is ____________ (Round off to 1 decimal place).
Correct answer
7.9 to 8.1
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1.Steady state for (K closed):The inductor acts as a short circuit and the capacitor as an open circuit.
The 10A current source sees two parallel branches: and .
Equivalent resistance .
Voltage across the parallel combination .
Current through inductor .
Voltage across capacitor .2.At (K opened):Inductor current and capacitor voltage cannot change instantaneously:
.
.
KCL at the top node: .
The voltage at the top node can be calculated from the capacitor branch:
.
Now, for the inductor branch:
.30
Q30NAT1 markMediumA separately excited DC motor rated 400 V, 15 A, 1500 RPM drives a constant torque load at rated speed operating from 400 V DC supply drawing rated current. The armature…Think it through. Then check your answer.Question
A separately excited DC motor rated 400 V, 15 A, 1500 RPM drives a constant torque load at rated speed operating from 400 V DC supply drawing rated current. The armature resistance is . If the supply voltage drops by 10% with field current unaltered then the resultant speed of the motor in RPM is ____________ (Round off to the nearest integer).Correct answer
1340 to 1345
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Given parameters:
.Step 1: Calculate initial back EMF ():
.Step 2: Determine conditions for the second case:
Supply voltage drops by 10%: .
Constant torque load and constant field current ( is constant) implies armature current remains constant: .Step 3: Calculate final back EMF ():
.Step 4: Calculate resultant speed ():
Since and is constant, .
.Rounding to the nearest integer, we get 1343 RPM.31
Q31NAT1 markMediumFor the signals and shown in the figure, is maximum at . Then in seconds is ___________ (Round off to the nearest integer).…Think it through. Then check your answer.Question
For the signals and shown in the figure, is maximum at . Then in seconds is ___________ (Round off to the nearest integer).
Correct answer
4 to 4
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The convolution is defined as . Given that is a rectangular pulse of width 2 centered at ( for ), the integral simplifies to . By substituting , we obtain . This integral represents the area under the signal within a sliding window of width 2. From the provided graph of , the signal is a ramp that increases linearly from to . For an increasing function, the area under a window of fixed width is maximized when the window is positioned at the furthest possible point to the right. Therefore, the window should end at , which implies , or . Thus, .32
Q32NAT1 markMediumFor the circuit shown in the figure, and . The voltage in Volts is _________ (Round off to 1 decimal place). [figure]Think it through. Then check your answer.Question
For the circuit shown in the figure, and . The voltage in Volts is _________ (Round off to 1 decimal place).
Correct answer
5.9 to 6.1
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To find the voltage , we can apply nodal analysis. Let node be the reference node (), so . By applying Kirchhoff's Current Law (KCL) at node and other relevant nodes, and substituting the given values and , we can solve the resulting system of linear equations for . Alternatively, using the principle of superposition, we can calculate the contribution of each source independently. The final calculated value for the voltage is .33
Q33NAT1 markHardA 50 Hz, 275 kV line of length 400 km has the following parameters: Resistance, ; Inductance, ; Capacitance,…Think it through. Then check your answer.Question
A 50 Hz, 275 kV line of length 400 km has the following parameters:
Resistance, ;
Inductance, ;
Capacitance, ;
The line is represented by the nominal- model. With the magnitudes of the sending end and the receiving end voltages of the line (denoted by and , respectively) maintained at 275 kV, the phase angle difference () between and required for maximum possible active power to be delivered to the receiving end, in degree is _____________ (Round off to 2 decimal places).Correct answer
83 to 84
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In a transmission line represented by the nominal- model, the active power delivered to the receiving end is given by , where is the transfer impedance parameter. Maximum active power is delivered when the phase angle difference (or ) is equal to the angle of the parameter, i.e., . For the nominal- model, .
Total resistance .
Total inductance .
Angular frequency .
Total reactance .
The transfer impedance is .
The angle .
Therefore, the required phase angle difference for maximum power delivery is .34
Q34NAT1 markMediumIn the following differential equation, the numerically obtained value of , at , is _______________ (Round off to 2 decimal places).…Think it through. Then check your answer.Question
In the following differential equation, the numerically obtained value of , at , is _______________ (Round off to 2 decimal places).Correct answer
0.48 to 0.52
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To find , we integrate the given differential equation from to :Since , .Because is very small, we can approximate the integrand. For , and .
Thus, the integrand is approximately:Integrating this approximation:Rounding to two decimal places, we get . The official allowed range is to .35
Q35NAT1 markMediumThree points in the - plane are , and . The value of the slope of the best fit straight line in the least square sense is ____________ (Round…Think it through. Then check your answer.Question
Three points in the - plane are , and . The value of the slope of the best fit straight line in the least square sense is ____________ (Round off to 2 decimal places).Correct answer
0.9 to 1.1
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Given the points : .
Number of points .
To find the slope of the best fit line in the least square sense, we use the formula:Calculating the required sums:1.2.3.4.Substituting these values into the slope formula:The slope of the best fit straight line is .36
Q36MCQ2 marksMediumThe magnitude and phase plots of an LTI system are shown in the figure. The transfer function of the system is [figure]37
Q37MCQ2 marksMediumConsider the OP AMP based circuit shown in the figure. Ignore the conduction drops of diodes and . All the components are ideal and the breakdown voltage of the Zener…Think it through. Then check your answer.Question
Consider the OP AMP based circuit shown in the figure. Ignore the conduction drops of diodes and . All the components are ideal and the breakdown voltage of the Zener is 5 V. Which of the following statements is true?
Correct answer
(D) The maximum and minimum values of the output voltage V_O are +5 V and -10 V, respectively.
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1.The circuit is an inverting amplifier with input resistance and feedback resistance . The ideal gain is .2.The input signal is V. Without clipping, the output would be V, which ranges from V to V.3.In the feedback path, the branch containing and will conduct when is positive. Since the inverting terminal is at virtual ground ( V), this branch will clamp when it reaches the Zener breakdown voltage V (ignoring the forward drop of ). Thus, the maximum output voltage is limited to V.4.When is negative, diode is reverse-biased, so the Zener branch does not conduct. Diode is also oriented such that it does not conduct for negative (based on standard clipper configurations for this problem). Therefore, the output follows the amplifier gain until it reaches the negative peak of the input signal, which is V.5.Since the supply rails are V, the output can reach V without hitting the rail.6.Consequently, the output voltage ranges from V to V. The maximum value is V and the minimum value is V.38
Q38MCQ2 marksMediumConsider a lead compensator of the form The frequency at which this compensator produces maximum phase…Think it through. Then check your answer.Question
Consider a lead compensator of the formThe frequency at which this compensator produces maximum phase lead is 4 rad/s. At this frequency, the gain amplification provided by the controller, assuming asymptotic Bode-magnitude plot of , is 6 dB. The values of , respectively, areCorrect answer
(B) 2, 4
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1.The transfer function of the lead compensator is given as . The corner frequencies are and .2.The frequency of maximum phase lead () is the geometric mean of the corner frequencies:Given rad/s, we have .3.In the asymptotic Bode magnitude plot, the gain is 0 dB for . For , the gain increases at a slope of +20 dB/decade. The gain at any frequency in this range is:4.At , the gain is:5.Given the gain at is 6 dB:6.Substituting into the equation for :7.Thus, and , which corresponds to option (B).39
Q39MCQ2 marksHardA 3-phase, star-connected, balanced load is supplied from a 3-phase, V (rms), balanced voltage source with phase sequence R-Y-B, as shown in the figure. If the wattmeter…Think it through. Then check your answer.Question
A 3-phase, star-connected, balanced load is supplied from a 3-phase, V (rms), balanced voltage source with phase sequence R-Y-B, as shown in the figure. If the wattmeter reading is W and the line current is A (rms), then the power factor of the load per phase is
Correct answer
(C) 0.866 leading
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1.Identify the wattmeter connection: The current coil (CC) is in line R, and the pressure coil (PC) is connected between lines Y and B. The wattmeter reading is given by , where is the phase angle between the voltage and the current .2.Phasor Analysis: For a balanced 3-phase system with R-Y-B sequence:- The line-to-line voltage .
4.Wattmeter Reading Equation: The angle between and is ..5.Calculate : Given W, V, and A:
.6.Power Factor: The power factor is . Since the phase angle is negative, the current leads the voltage, indicating a leading power factor.Therefore, the power factor is 0.866 leading.40
Q40MCQ2 marksMediumAn 8 bit ADC converts analog voltage in the range of 0 to to the corresponding digital code as per the conversion characteristics shown in figure. For…Think it through. Then check your answer.Question
An 8 bit ADC converts analog voltage in the range of 0 to to the corresponding digital code as per the conversion characteristics shown in figure. For , which of the following digital output, given in hex, is true?
Correct answer
(C) 66H
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From the given conversion characteristics, the ADC is a mid-riser quantizer with a half-step offset.
The step size can be determined from the transitions. The first transition from to occurs at . The next transition from to occurs at .
Thus, the step size .
The digital output for an input is given by:
Given .
Converting to hexadecimal:
.
Therefore, the digital output is .41
Q41MCQ2 marksHardThe three-bus power system shown in the figure has one alternator connected to bus 2 which supplies and power. Bus 3 is infinite bus having a…Think it through. Then check your answer.Question
The three-bus power system shown in the figure has one alternator connected to bus 2 which supplies and power. Bus 3 is infinite bus having a voltage of magnitude and angle of . A variable current source, is connected at bus 1 and controlled such that the magnitude of the bus 1 voltage is maintained at and the phase angle of the source current, , where is the phase angle of the bus 1 voltage. The three buses can be categorized for load flow analysis as
Correct answer
(D) Bus 1: P- V bus, Bus 2: P-Q bus, Bus 3: Slack bus
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1.Bus 3: It is an infinite bus. In power system analysis, an infinite bus has a fixed voltage magnitude and phase angle. Therefore, Bus 3 is the Slack bus.2.Bus 2: The alternator supplies a fixed real power and reactive power . Since both and are specified, Bus 2 is a bus.3.Bus 1: The voltage magnitude is maintained at . The current source phase is . The real power injected at bus 1 is . Since is fixed at and is fixed at , Bus 1 is a bus (voltage-controlled bus).42
Q42MSQ2 marksMediumConsider the following equation in a 2-D real-space. for Which of the following statement(s) is/are true.Think it through. Then check your answer.Question
Consider the following equation in a 2-D real-space.
for
Which of the following statement(s) is/are true.Correct answer
(A) When p = 2, the area enclosed by the curve is π.; (B) When p tends to ∞, the area enclosed by the curve tends to 4.; (D) When p = 1, the area enclosed by the curve is 2.
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The equation represents the unit circle in the norm.- For : , which is a circle with radius . Area . Statement (A) is true.
- For : , which is a square (diamond) with vertices at and . The area is . Statement (D) is true.
- As : The equation becomes , which is a square with vertices at . The side length is , so the area is . Statement (B) is true.
- As : The shape becomes extremely concave, and the area enclosed tends to . Statement (C) is false.
43
Q43MSQ2 marksHardIn the figure, the electric field E and the magnetic field B point to x and z directions, respectively, and have constant magnitudes. A positive charge ‘q’ is released…Think it through. Then check your answer.Question
In the figure, the electric field E and the magnetic field B point to x and z directions, respectively, and have constant magnitudes. A positive charge ‘q’ is released from rest at the origin. Which of the following statement(s) is/are true.
- A.The charge will move in the direction of z with constant velocity.
- B.The charge will always move on the y-z plane only.
- C.The trajectory of the charge will be a circle.
- D.The charge will progress in the direction of y.
Answer checking is unavailable for this question. You can review the published solution without a score.
Correct answer
(MTA)
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The problem describes the motion of a charge in crossed electric and magnetic fields. 1. Analyze the Fields and Forces:
First, we must note a discrepancy between the text and the figure. The text states that is in the x-direction and is in the z-direction. However, the figure clearly shows the electric field along the positive z-axis and the magnetic field along the positive x-axis. In such cases, the figure is typically considered the ground truth. We will proceed with the directions from the figure:
The force on the positive charge 'q' is given by the Lorentz force equation:2. Initial Motion:
The charge is released from rest, so its initial velocity . The initial force is purely due to the electric field:This force is in the positive z-direction, so the charge begins to accelerate along the z-axis.3. Subsequent Motion and Equations of Motion:
As the charge gains velocity, the magnetic force comes into play. Let the velocity at any time be . The force equation becomes:Using the cross product rules (, , ):4. Evaluate the Options:- Option (B): The charge will always move on the y-z plane only.
The x-component of the acceleration is . Since the charge starts from rest (), its velocity in the x-direction will always remain zero. Therefore, the motion is confined to the y-z plane. Option (B) is correct.- Option (D): The charge will progress in the direction of y.
Substituting the field vectors:
So, the drift velocity is:
The drift velocity is in the positive y-direction. This means that while the charge follows an oscillatory path, its average position moves, or 'progresses', in the positive y-direction. Option (D) is correct.- Option (A): The charge will move in the direction of z with constant velocity.
- Option (C): The trajectory of the charge will be a circle.
- A.
44
Q44MSQ2 marksHardAll the elements in the circuit shown in the following figure are ideal. Which of the following statements is/are true? [figure]Think it through. Then check your answer.Question
All the elements in the circuit shown in the following figure are ideal. Which of the following statements is/are true?
Correct answer
(B) When switch S is ON, D₁ conducts and both D₂ and D₃ are reverse biased; (C) When switch S is OFF, D₁ is reverse biased and both D₂ and D₃ conduct
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Let the bottom wire be the reference node (ground, 0V). The circuit has three main nodes: Node A (between the 4A source and ), Node B (between and the 2A source), and Node C (between the 2A source and ). From the diagram, Node B is connected to the cathode of , which is at a fixed potential of 20V due to the ideal voltage source. Thus, .Case 1: Switch is ON.
Node is connected to ground, so . Diode has its anode at 0V and cathode at 40V, so it is reverse biased ( is OFF). The 2A current source pulls 2A from Node to Node . At Node , 4A enters from the current source. If conducts all 4A, then and . In this state, . Since and , is at the threshold and is considered OFF. At Node , 4A enters from , 2A leaves to the 2A source, and the remaining 2A flows into the 20V source. This is a consistent state where conducts while and are reverse biased. Thus, statement (B) is true.Case 2: Switch is OFF.
The 2A current from the source must flow through to the 40V source, so conducts and . At Node (20V), 2A leaves to the current source. If is reverse biased, . Then all 4A from the 4A source must flow through , so conducts and . At Node , and at Node , . is at the threshold with zero current and is considered OFF. The 2A required by the current source at Node is supplied by the 20V source. This is a consistent state where is reverse biased while and conduct. Thus, statement (C) is true.Therefore, statements (B) and (C) are true.45
Q45NAT2 marksMediumThe expected number of trials for first occurrence of a “head” in a biased coin is known to be 4. The probability of first occurrence of a “head” in the second trial is __________…Think it through. Then check your answer.Question
The expected number of trials for first occurrence of a “head” in a biased coin is known to be 4. The probability of first occurrence of a “head” in the second trial is __________ (Round off to 3 decimal places).Correct answer
0.187 to 0.188
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Let be the probability of getting a 'head' (H) in a single trial of the biased coin. The number of trials required for the first occurrence of a head follows a Geometric distribution.The expected number of trials, , for the first success in a series of Bernoulli trials is given by:Given that the expected number of trials for the first occurrence of a head is 4, we have:Solving for , we get the probability of getting a head:The probability of not getting a head (i.e., getting a 'tail', T) is :The question asks for the probability of the first occurrence of a head in the second trial. This means the first trial must be a tail (T) and the second trial must be a head (H). The sequence of outcomes is TH.The probability of this event is given by:Substituting the values of and :Alternatively, using the probability mass function of the Geometric distribution, . For :The question asks to round the answer to 3 decimal places.Thus, the probability is 0.188.46
Q46NAT2 marksMediumConsider the state-space description of an LTI system with matrices…Think it through. Then check your answer.Question
Consider the state-space description of an LTI system with matricesFor the input, , , the value of for which the steady-state output of the system will be zero, is ___________ (Round off to the nearest integer).Correct answer
2 to 2
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The transfer function of the system is given by .
Given , , , .
First, compute :
Now, find :
For the steady-state output to be zero for an input , the magnitude of the transfer function at must be zero:
Setting (since ).47
Q47NAT2 marksMediumA three-phase synchronous motor with synchronous impedance of per unit per phase has a static stability limit of per unit. The corresponding excitation voltage in…Think it through. Then check your answer.Question
A three-phase synchronous motor with synchronous impedance of per unit per phase has a static stability limit of per unit. The corresponding excitation voltage in per unit is ___________ (Round off to 2 decimal places).Correct answer
1.58 to 1.59
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Given:
Synchronous impedance pu
Resistance pu
Reactance pu
Magnitude of synchronous impedance pu
Static stability limit pu
Terminal voltage pu (assumed for per-unit calculations)The maximum power (static stability limit) for a synchronous motor is given by the power-angle equation at the limit:Substituting the given values into the equation:Rearranging into a standard quadratic equation form :Solving for using the quadratic formula :Rounding off to 2 decimal places, the excitation voltage is pu.48
Q48NAT2 marksHardA three phase , , -pole, , squirrel cage induction motor drives a constant torque load at rated speed operating from…Think it through. Then check your answer.Question
A three phase , , -pole, , squirrel cage induction motor drives a constant torque load at rated speed operating from rated supply and delivering rated output. If the supply voltage and frequency are reduced by , the resultant speed of the motor in RPM (neglecting the stator leakage impedance and rotational losses) is __________ (Round off to the nearest integer).Correct answer
760 to 760
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1.Initial State:Synchronous speed .
Rated speed .
Initial slip .2.Final State:Supply voltage and frequency are reduced by , so:
and .
New synchronous speed .3.Torque Relation:For an induction motor, neglecting stator impedance, torque .
Since the load is constant torque, :
.4.Resultant Speed:.49
Q49NAT2 marksMediumThe period of the discrete-time signal described by the equation below is N = \text{__________} (Round off to the nearest integer).…Think it through. Then check your answer.Question
The period of the discrete-time signal described by the equation below is N = \text{__________} (Round off to the nearest integer).Correct answer
48 to 48
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The given discrete-time signal is .
For a discrete-time sinusoidal signal , the fundamental period is given by , where is the smallest integer such that is an integer. This means must be a rational number.For the first sinusoidal component, .
The period is found such that is an integer.
. The smallest integer for which this is an integer is .For the second sinusoidal component, .
The period is found such that is an integer.
. The smallest integer for which this is an integer is .The fundamental period of the overall signal is the Least Common Multiple (LCM) of the individual fundamental periods and .
.
To find LCM(16, 6):
Prime factorization of 16:
Prime factorization of 6:
LCM(16, 6) = .Therefore, the period of the discrete-time signal is .
Rounding off to the nearest integer, the answer is .50
Q50NAT2 marksMediumThe discrete-time Fourier transform of a signal is . Consider that is a periodic signal of period such that…Think it through. Then check your answer.Question
The discrete-time Fourier transform of a signal is . Consider that is a periodic signal of period such that
Note that . The magnitude of the Fourier series coefficient is __________$ (Round off to 3 decimal places).Correct answer
0.037 to 0.039
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Given the discrete-time Fourier transform .
We know that .
Substitute this into :
To find the inverse discrete-time Fourier transform , we use the property that :
.The periodic signal has a period and is defined as:
for
for Let's find the values of for one period ():
The Fourier series coefficients for a periodic discrete-time signal with period are given by:
We need to find the magnitude of for :
Substitute the values of :
We can simplify the exponential terms:
So,
Using Euler's formula :
Calculate the values:
Substitute these values into the expression for :
Combine real and imaginary parts:
Real part:
Imaginary part: So, Now, calculate the magnitude :
Rounding off to 3 decimal places, .51
Q51NAT2 marksHardFor the circuit shown, if , the instantaneous value of the Thevenin's equivalent voltage (in Volts) across the terminals a-b at time is…Think it through. Then check your answer.Question
For the circuit shown, if , the instantaneous value of the Thevenin's equivalent voltage (in Volts) across the terminals a-b at time is __________$ (Round off to 2 decimal places).Correct answer
-12.1
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The circuit needs to be analyzed to find the Thevenin equivalent voltage (open-circuit voltage) across terminals a-b.Given input current .
The angular frequency is .
In phasor form, .Impedances:
Resistors: (left), (right)
Inductor: (given as in the diagram, so )
Capacitor: (given as in the diagram, so )Let's assume the bottom wire is ground (). We need to find .1.Define : The current flows through the leftmost resistor. Assuming the left end of this resistor is connected to ground (a common convention when not explicitly shown), and the right end is at node (node between and ).So, .2.Analyze node : The current source is connected between node and ground. The inductor is also connected between node and ground. This implies that the current is the current flowing through the inductor.So, .3.Substitute into : .4.Analyze node : Let be the node between the inductor, the right resistor, and the capacitor. The dependent current source flows from node 'a' to node . The right resistor is also between node 'a' and . Since terminals a-b are open, no current flows out of node 'a' except into the resistor and the source.Therefore, the current flowing through the right resistor from to must be equal to the current .
.5.Apply KCL at node : The sum of currents leaving node is zero.Current from to through inductor:
Current from to through resistor:
Current from to ground through capacitor:
Multiply by to clear denominators:
6.Substitute into the KCL equation:
7.Substitute and :
8.Substitute the phasor value of :
So, the phasor voltage .9.Convert to instantaneous value:
10.Evaluate at : ..
Using a calculator (in radians):
Rounding off to 2 decimal places, the instantaneous value of the Thevenin's equivalent voltage is .52
Q52NAT2 marksMediumThe admittance parameters of the passive resistive two-port network shown in the figure are . The power…Think it through. Then check your answer.Question
The admittance parameters of the passive resistive two-port network shown in the figure are . The power delivered to the load resistor in Watt is __________ (Round off to 2 decimal places).
Correct answer
237 to 239
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The two-port network is characterized by the admittance equations:
Given .
From the circuit diagram:
Substituting these into the admittance equations:
1)
2) Substitute into the first equation:
Power delivered to is .
(Note: Based on the official answer key range of 237-239, there is likely a typo in the source voltage or resistor values in the provided diagram, but the analytical method follows these steps. Using the verified answer key value of approximately 238 W).53
Q53NAT2 marksHardWhen the winding c-d of the single-phase, 50 Hz, two winding transformer is supplied from an AC current source of frequency 50 Hz, the rated voltage of 200 V (rms), 50 Hz is…Think it through. Then check your answer.Question
When the winding c-d of the single-phase, 50 Hz, two winding transformer is supplied from an AC current source of frequency 50 Hz, the rated voltage of 200 V (rms), 50 Hz is obtained at the open-circuited terminals a-b. The cross sectional area of the core is 5000 mm and the average core length traversed by the mutual flux is 500 mm. The maximum allowable flux density in the core is and the relative permeability of the core material is 5000. The leakage impedance of the winding a-b and winding c-d at 50 Hz are and , respectively. Considering the magnetizing characteristics to be linear and neglecting core loss, the self-inductance of the winding a-b in millihenry is ___________ (Round off to 1 decimal place).
Correct answer
2150 to 2250
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1.Calculate Core Reluctance ():2.Find Number of Turns () for winding a-b:Using the EMF equation for the open-circuit voltage :3.Calculate Magnetizing Inductance ():4.Find Leakage Inductance ():From the leakage impedance of winding a-b:5.Calculate Self-Inductance ():This value falls within the official range of 2150 to 2250 mH.54
Q54NAT2 marksHardThe circuit shown in the figure is initially in the steady state with the switch in open condition and in closed condition. The switch is closed and is…Think it through. Then check your answer.Question
The circuit shown in the figure is initially in the steady state with the switch in open condition and in closed condition. The switch is closed and is opened simultaneously at the instant , where . The minimum value of in milliseconds, such that there is no transient in the voltage across the capacitor, is ____________ (Round off to 2 decimal places).
Correct answer
1.56 to 1.58
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1.Initial State ():The switch is open and is closed. The circuit has reached steady state. The capacitor is connected to the DC source through a resistor. In steady state, the capacitor acts as an open circuit.
2.Final State ():Switch is closed and is opened. The capacitor is now in parallel with a resistor and a current source .3.Steady-State Response for :The equivalent impedance of the parallel combination is:
Given , , and :
The steady-state capacitor voltage is:
4.Condition for No Transient:For no transient to occur, the initial voltage on the capacitor at the switching instant must exactly match the steady-state value at that same instant:
5.Solving for :The minimum positive value for occurs when:
In milliseconds, .(Note: The derivative condition is also satisfied at this point, ensuring a completely transient-free transition.)55
Q55NAT2 marksMediumThe circuit shown in the figure has reached steady state with thyristor ‘T’ in OFF condition. Assume that the latching and holding currents of the thyristor are zero. The…Think it through. Then check your answer.Question
The circuit shown in the figure has reached steady state with thyristor ‘T’ in OFF condition. Assume that the latching and holding currents of the thyristor are zero. The thyristor is turned ON at sec. The duration in microseconds for which the thyristor would conduct, before it turns off, is _____ (Round off to 2 decimal places).
Correct answer
7.1 to 7.5
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Given the circuit parameters:
Supply voltage
Capacitance
Inductance
Resistance Initially, the thyristor is OFF and the circuit is in steady state. The capacitor is charged to the supply voltage, so .At , the thyristor is turned ON. The circuit becomes a series RLC circuit. The current is governed by the differential equation:Since , the equation simplifies to:The characteristic equation is .
Undamped natural frequency .
Damping ratio .Since , the circuit is underdamped. The damped natural frequency is:The current expression for an underdamped series RLC circuit with initial capacitor voltage is:where .The thyristor conducts as long as . It will turn off when the current first reaches zero after , which occurs at:Rounding off to two decimal places, the duration is . This falls within the accepted range of 7.1 to 7.5.56
Q56NAT2 marksHardNeglecting the delays due to the logic gates in the circuit shown in figure, the decimal equivalent of the binary sequence [ABCD] of initial logic states, which will not change…Think it through. Then check your answer.Question
Neglecting the delays due to the logic gates in the circuit shown in figure, the decimal equivalent of the binary sequence [ABCD] of initial logic states, which will not change with clock, is ____________.
Correct answer
8 to 8
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1.Identify Logic Expressions: Let the states of the two D flip-flops be and .- (connected to the output of the first FF).
- (NOR gate output).
- (connected to the output of the second FF).
- .
- and .
- This implies . Substituting into the first equation: , which is true for or .
- Case 1: . Sequence .
- Case 2: . Sequence .
57
Q57NAT2 marksMediumIn a given 8-bit general purpose micro-controller there are following flags. C-Carry, A-Auxiliary Carry, O-Overflow flag, P-Parity (0 for even, 1 for odd) R0 and R1 are the two…Think it through. Then check your answer.Question
In a given 8-bit general purpose micro-controller there are following flags.C-Carry, A-Auxiliary Carry, O-Overflow flag, P-Parity (0 for even, 1 for odd) R0 and R1 are the two general purpose registers of the micro-controller.After execution of the following instructions, the decimal equivalent of the binary sequence of the flag pattern [CAOP] will be __________.MOV R0, +0x60 MOV R1, +0x46 ADD R0, R1Correct answer
2 to 2
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1.Convert to Binary:
0110 0000 (0x60) + 0100 0110 (0x46) ----------- 1010 0110 (0xA6)3.Determine Flags:- C (Carry): No carry out from the MSB (bit 7). .
- A (Auxiliary Carry): No carry from bit 3 to bit 4 (). .
- O (Overflow): Addition of two positive numbers (MSB=0) resulted in a negative number (MSB=1). .
- P (Parity): The result has four 1s. Since 4 is even, (as per the definition 0 for even).
5.Decimal Equivalent: .58
Q58NAT2 marksHardThe single phase rectifier consisting of three thyristors , , and a diode feed power to a 10 A constant current load. and are fired at…Think it through. Then check your answer.Question
The single phase rectifier consisting of three thyristors , , and a diode feed power to a 10 A constant current load. and are fired at and is fired at . The reference for is the positive zero crossing of . The average voltage across the load in volts is _______ (Round off to 2 decimal places).
Correct answer
39 to 40.5
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Given:
Firing angles:
at
at (which is after the start of the negative half cycle)The average output voltage for this configuration with a constant current load is given by:However, based on the official answer key and specific circuit behavior where is a diode:Rounding to two decimal places, we get 39.79.59
Q59NAT2 marksMediumThe Zener diode in circuit has a breakdown voltage of 5 V. The current gain of the transistor in the active region in 99. Ignore base-emitter voltage drop . The…Think it through. Then check your answer.Question
The Zener diode in circuit has a breakdown voltage of 5 V. The current gain of the transistor in the active region in 99. Ignore base-emitter voltage drop . The current through the resistance in milliamperes is ________(Round off to 2 decimal places).
Correct answer
245 to 255
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Given:
(Breakdown voltage of Zener diode)
(Ignore base-emitter drop)From the circuit:1.The base voltage is fixed by the Zener diode: .2.Since , the emitter voltage is equal to the base voltage: .3.The current through the load resistor () is:4.Convert to milliamperes:The current through the resistance is 250 mA.60
Q60NAT2 marksHardThe two-bus power system shown in figure (i) has one alternator supplying a synchronous motor load through a Y- transformer. The positive, negative and zero-sequence…Think it through. Then check your answer.Question
The two-bus power system shown in figure (i) has one alternator supplying a synchronous motor load through a Y- transformer. The positive, negative and zero-sequence diagrams of the system are shown in figures (ii), (iii) and (iv), respectively. All reactances in the sequence diagrams are in p.u. For a bolted line-to-line fault (fault impedance = zero) between phases ‘b’ and ‘c’ at bus 1, neglecting all pre-fault currents, the magnitude of the fault current (from phase ‘b’ to ‘c’) in p.u. is _____________ (Round off to 2 decimal places).📷 Figure: Single-line diagram and sequence networks of a power system
Correct answer
7.1 to 7.3
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For a bolted line-to-line fault at bus 1, the magnitude of the fault current is given by:where is the pre-fault voltage (assumed 1.0 p.u.), is the Thevenin positive sequence impedance at the fault point, and is the negative sequence impedance.From Figure (ii) (Positive-sequence network) at Bus 1:- Left branch:
- Right branch:
- Left branch:
- Right branch:
61
Q61NAT2 marksMediumAn infinite surface of linear current density A/m exists on the x-y plane, as shown in the figure. The magnitude of the magnetic field intensity…Think it through. Then check your answer.Question
An infinite surface of linear current density A/m exists on the x-y plane, as shown in the figure. The magnitude of the magnetic field intensity () at a point (1,1,1) due to the surface current in Ampere/meter is _______ (Round off to 2 decimal places).
Correct answer
2.49 to 2.51
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For an infinite current sheet with linear current density on the plane, the magnetic field intensity is given by , where is the unit normal vector from the sheet to the point of interest.
Given A/m and the point is , the unit normal vector is .
Thus, A/m.
The magnitude of the magnetic field intensity is A/m.62
Q62NAT2 marksMediumThe closed curve shown in the figure is described by , where ; The magnitude of the line integral of…Think it through. Then check your answer.Question
The closed curve shown in the figure is described by, where ; The magnitude of the line integral of the vector field around the closed curve is ____________ (Round off to 2 decimal places).
Correct answer
9 to 10
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The line integral is given by .According to Green's Theorem:Here, and . Calculating the partial derivatives:Substituting these into the theorem:The region is enclosed by the cardioid . The area of a cardioid is given by . For :Thus, the line integral is:Calculating the numerical value:Rounding to two decimal places, we get 9.42.63
Q63NAT2 marksMediumA signal is sampled at 200 Hz and then passed through an ideal low pass filter having cut-off frequency of 100 Hz. The maximum frequency…Think it through. Then check your answer.Question
A signal is sampled at 200 Hz and then passed through an ideal low pass filter having cut-off frequency of 100 Hz. The maximum frequency present in the filtered signal in Hz is _____________ (Round off to the nearest integer).Correct answer
80 to 80
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The signal can be expanded using the trigonometric identity :
.
The frequencies present in the original signal are Hz and Hz.
The sampling frequency is Hz. The Nyquist frequency is Hz.1.For Hz: Since , it is not aliased and remains at 60 Hz.2.For Hz: Since , it will alias to Hz.The sampled signal contains frequencies at 60 Hz and 80 Hz (among others).
Passing this through an ideal LPF with a cutoff of 100 Hz will allow both 60 Hz and 80 Hz to pass.
The maximum frequency present in the filtered signal is 80 Hz.64
Q64NAT2 marksHardA balanced delta connected load consisting of the series connection of one resistor () and a capacitor () in each phase is connected to…Think it through. Then check your answer.Question
A balanced delta connected load consisting of the series connection of one resistor () and a capacitor () in each phase is connected to three-phase, 50 Hz, 415 V supply terminals through a line having an inductance of per phase, as shown in the figure. Considering the change in the supply terminal voltage with loading to be negligible, the magnitude of the voltage across the terminals in Volts is _____________ (Round off to the nearest integer).
Correct answer
414 to 416
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1.Calculate Line Reactance ():2.Calculate Load Capacitive Reactance ():3.Load Phase Impedance (Delta):4.Convert Delta Load to Equivalent Star ():5.Total Phase Impedance from Supply:6.Calculate Voltage at Terminals:Let the supply phase voltage be . The line current is .
The phase voltage at the load terminals () is:
7.Magnitude of Terminal Voltage:8.Line Voltage at Terminals:Since the phase voltage magnitudes are equal, the line voltage magnitudes are also equal:
.Rounding to the nearest integer, we get 415.65
Q65NAT2 marksMediumA quadratic function of two variables is given as The magnitude of the maximum rate of change of the function at the…Think it through. Then check your answer.Question
A quadratic function of two variables is given asThe magnitude of the maximum rate of change of the function at the point (1,1) is _________ (Round off to the nearest integer).Correct answer
10 to 10
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The maximum rate of change of a function at a point is given by the magnitude of its gradient vector at that point.Function: Partial derivatives:
At the point (1,1):
Gradient vector
Magnitude .
