Official Paper

GATE EE 2023 Official Paper (Previous Year Paper)

65 questions · 180 minutes · with answers · free

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General Aptitude (10 questions)

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.

((a))
thought
((b))
is thinking
((c))
am thinking
((d))
was thinking

Show Answer

Answer: ((d))

was thinking

The correct answer is option (4)

Solution:

Rafi told Mary that he was thinking of watching a film that weekend.

Permit : _______ :: Enforce : Relax

(By word meaning)

((a))
Allow
((b))
Forbid
((c))
License
((d))
Reinforce

Show Answer

Answer: ((b))

Forbid

Concept:

Enforce is related to relax in a special manner, enforce and relax are opposite in meaning. In the same way

of relationship opposite meaning of permit will forbid.

Hence, the correct option is (2).

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?

((a))
$\frac{1}{36}$
((b))
$\frac{1}{6}$
((c))
$\frac{5}{6}$
((d))
$\frac{1}{3}$

Show Answer

Answer: ((a))

$\frac{1}{36}$

Probability of getting "1" on 1^stroll = $\frac{1}{6}$

Probability of getting "4" on 2^nd roll = $\frac{1}{6}$

Since, both events are independent

Probability of getting "1" on first roll and the "4" on the second roll = $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

Hence, the correct option is (1).

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?

((a))
A majority of tobacco users who were advised to stop consuming tobacco made an attempt to do so.
((b))
A majority of tobacco users who were advised to stop consuming tobacco did not attempt to do so.
((c))
Approximately 30% of tobacco users successfully stopped consuming tobacco.
((d))
Approximately 65% of tobacco users successfully stopped consuming tobacco.

Show Answer

Answer: ((b))

A majority of tobacco users who were advised to stop consuming tobacco did not attempt to do so.

The correct answer is 'Option 2' i.e. 'A majority of tobacco users who were advised to stop consuming tobacco did not attempt to do so.'

Concept:

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.

Option (C) cannot be inferred with certainty, as according to the given information 3 out of 10 numerical tobacco users attempted to stop using tobacco, which is not a majority.

Option (A) and option (D) cannot be inferred with certainty, as there is no information about how much tobacco users successfully stop consuming tobacco.

Option (B) can be inferred with certainty, as 3 out of 10 tobacco users attempted to stop using tobacco; which means 7 users did not attempt to do this, and 7 out of 10 is a majority.

Hence, the correct option is (2).

How many triangles are present in the given figure?

((a))
12
((b))
16
((c))
20
((d))
24

Show Answer

Answer: ((c))

No. of triangles = ΔABC, ΔACD, ΔBDC, ΔABD, ΔCDK, ΔBCK, ΔADK, ΔABK, ΔAHJ, ΔАЕН, ΔАЕЈ, ΔВЕМ, ΔBFM, ΔCFL, ΔCGL, ΔCGF, ΔDGI, ΔDHI, ΔDGH.

∴ Total no. of triangles = 20

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.

((a))
(i) and (ii)
((b))
(i) and (iii)
((c))
only (i)
((d))
only (iii)

Show Answer

Answer: ((d))

only (iii)

Concept:

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

(i) All those students who would not be eligible to vote in college elections would certainly belong to the Department of Human Science, which cannot be inferred, as it is not a necessary condition or situation according to the given information.

(ii) None of the students from departments other than human science failed to complete the registration process within the due time, which cannot be inferred, as we have no information about the departments other than human science.

(iii) All the eligible voters would certainly be students who are not from human science, which can be inferred logically with certainty, as according to the given information “it was found that surprisingly none of the students from the department of human science had completed the registration process.” So, all the eligible voters would not be from the Department of Human Science.

Hence, the correct option is (4).

Which one of the following options represents the given graph?

((a))
f(x) = x²2^-^|x|
((b))
f(x) = x 2-|x|
((c))
f(x) = |x| 2-x
((d))
f(x) = x 2-x

Show Answer

Answer: ((b))

f(x) = x 2-|x|

The given function is odd function.

Property of odd function.

F(-x) = -f(x)

Therefore, option (b) is an odd function.

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.

((a))
It is a reflection of why cancer seems so modern to most of us.
((b))
It tells us that modern cell biology uses and promotes metaphors of machinery.
((c))
Modern cell biology encourages metaphors of machinery, and cancer is often imagined as a machine.
((d))
Modern cell biology never uses figurative language, such as metaphors, to describe or explain anything.

Show Answer

Answer: ((d))

Modern cell biology never uses figurative language, such as metaphors, to describe or explain anything.

Concept:

Modern cell biology encourages us to imagine the cell as a molecular machine. Cancer is that machine unable to quench its initial command (to grow) and thus transform into an indestructible, self-propelled automaton.

From the given information we can clearly see that modern cell biology uses figurative language, such as metaphors to describe modern illnesses like cancer.

Hence, the correct option is (D).

The digit in the unit’s place of the product 3⁹⁹⁹ × 7¹⁰⁰⁰ is _______.

((a))
7
((b))
1
((c))
3
((d))
9

Show Answer

Answer: ((a))

Given: 3⁹⁹⁹ × 7¹⁰⁰⁰ = (3⁴)²⁴⁹ × 3³ × (7⁴)²⁵ = 1 × 27 × 1

Unit digit value = 7

Hence, the correct option is (A).

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 _______ cm² .

((a))
6π
((b))
18
((c))
20
((d))
9π

Show Answer

Answer: ((b))

Area of region (A + B + C + D) = 2[36 - (π.3²)] = 72 - 18π

Area of region (I + II + III + IV) = 36 - [72 - 18π] = 18π - 36

Area of region "I" = (18π - 36)/4 =

Area of shared region = Area of = (A + B + I)

= 9π - (9π - 18) = 18

Hence, the correct option is (B).

Electrical Engineering (55 questions)

For a given vector w = [1 2 3]^T , the vector normal to the plane defined by w^Tx = 1 is

((a))
[−2 −2 2]^T
((b))
[3 0 −1]^T
((c))
[3 2 1]^T
((d))
[1 2 3]^T

Show Answer

Answer: ((d))

[1 2 3]^T

Given: W = [1, 2, 3]

$W_x^T=1$

$\left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right]\left[\begin{array}{l} x \ y \ z \end{array}\right]=1$

ϕ : x + 2y + 3z = 1

Normal vector or gradient of ϕ = î + 2ĵ + 3k̂

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

Hence, the correct option is (D).

For the block diagram shown in the figure, the transfer function $\rm \frac{Y(s)}{R(s)}$ is

((a))
$\rm \frac{2s+3}{s+1}$
((b))
$\rm \frac{3s+2}{s-1}$
((c))
$\rm \frac{s+1}{3s+2}$
((d))
$\rm \frac{3s+2}{s+1}$

Show Answer

Answer: ((b))

$\rm \frac{3s+2}{s-1}$

Signal flow graph:

Forward paths,

P₁ = 3, Δ₁ = 1

P2 = $\frac{2}{S}$ , Δ₂ = 1

Loops : L₁ = $\frac{1}{S}$

Using Masson's graph formula,

$\frac{\mathrm{Y}(\mathrm{s})}{\mathrm{R}(\mathrm{s})}=\frac{\mathrm{P}_1 \Delta_1+\mathrm{P}_2 \Delta_2}{1-\mathrm{L}_1}$

$=\frac{3+\frac{2}{\mathrm{~S}}}{1-\frac{1}{\mathrm{~S}}}$

$=\frac{3 \mathrm{~S}+2}{\mathrm{~S}-1}$

In the Nyquist plot of the open-loop transfer function

$\rm G(s)H(s)=\frac{3s+5}{s-1}$

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

((a))
𝐺(𝑠)𝐻(𝑠) = ∞
((b))
𝐺(𝑠)𝐻(𝑠) = 0
((c))
𝐺(𝑠)𝐻(𝑠) = 3
((d))
𝐺(𝑠)𝐻(𝑠) = −5

Show Answer

Answer: ((c))

𝐺(𝑠)𝐻(𝑠) = 3

Nyquist Contour:

Given:

$\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) =\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1}$

Put s = Re^jθ

$\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) =\operatorname{Lim}_{\mathrm{R} \rightarrow \infty} \frac{3 \operatorname{Re}^{j \theta}+5}{\operatorname{Re}^{j \theta}-1}$

G(s)H(s) = 3

Consider a unity-gain negative feedback system consisting of the plant G(s) (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, are

$\rm G(s)=\frac{1}{s-1}$

((a))
∞, ∞
((b))
1, 0
((c))
1, -1
((d))
-1, 1

Show Answer

Answer: ((d))

-1, 1

Given system can be drawn as shown below,

Here,

X(s) → Controller output

C(s) → Plant output

$C(s)=R(s) \frac{\left(3+\frac{1}{s}\right)\left(\frac{1}{s-1}\right)}{1+\left(3+\frac{1}{s}\right) \frac{1}{s-1}}$ = $R(s) \frac{(3 s+1)}{s(s-1)+3 s+1}$ = $\frac{(3 s+1)}{s(s-1)+3 s+1} \times \frac{1}{s}$

∴ c(∞) = $\lim _{s \rightarrow 0} s C(s)=1$

$X(s)=[R(s)-C(s)]\left(3+\frac{1}{s}\right)$

$X(s)=[R(s)-C(s)]\left(\frac{3 s+1}{s}\right)$

$\rm x(\infty) =\lim s_{s \rightarrow 0}X(s)=\lim _{s \rightarrow 0}R(s)-C(s)$

$x(\infty) =\lim _{s \rightarrow 0}\left(\frac{1}{s}-\frac{1}{s} \frac{(3 s+1)}{\left(s^2+2 s+1\right)}\right)(3 s+1)$

$=\lim _{s \rightarrow 0} \frac{1}{s}\left[\frac{s^2+2 s+1-3 s-1}{s^2+2 s+1}\right](3 s+1)$

$=\lim _{s \rightarrow 0} \frac{1}{s}\left[\frac{s^2-s}{s^2+2 s+1}\right](3 s+1)$ = $\lim _{s \rightarrow 0} \frac{(s-1)(3 s+1)}{s^2+2 s+1}=-1$

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

A 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

<br>

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

((a))
a-r, b-p, and c-q
((b))
a-r, b-q, and c-p
((c))
a-p, b-r, and c-q
((d))
a-q, b-p, and c-r

Show Answer

Answer: ((a))

a-r, b-p, and c-q

The torque slip characteristics are shown in the figure below.

S > 1 ⇒ Plugging mode

0 < S < 1 ⇒ Mtoring mode

S < 0 ⇒ Generating mode

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 is

((a))
100 Hz
((b))
95 Hz
((c))
10 Hz
((d))
5 Hz

Show Answer

Answer: ((b))

95 Hz

Given : P = 10, f = 50 Hz, V_f = 240 V, N_r = 540 pm

The frequency of induced rotor current due to backward field is given by,

$N_s=\frac{120 f}{P}=\frac{120 \times 50}{10}=600 \mathrm{rpm}$

$S_f=\frac{N_s-N_r}{N_s}$ = $\frac{600-540}{600}=\frac{60}{600}=0.1$

As we know, backward field will rotate opposite to forward field so it will try to make the rotor in opposite direction so slip will be

$S_b=\frac{N_s+N_r}{N_s}$ $=\frac{600+540}{600}=\frac{1140}{600}=1.9$

As we know, frequency in the rotor is slip frequency.

So, frequency due to backward slip will be S_bf_s = 1.9 × 50 = 95 Hz

Hence, the correct option is (B).

A continuous-time system that is initially at rest is described by

$\rm \frac{dy(t)}{dt}+3y(t)=2x(t)$ ,

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

((a))
3e^-2t
((b))
$\rm \frac{1}{3}e^{-2t}u(t)$
((c))
2 e^-3t u(t)
((d))
2e^-3t

Show Answer

Answer: ((c))

2 e^-3t u(t)

Given: $\frac{d y(t)}{d t}+3 y(t)=2 x(t)$

Taking Laplace transform on both sides, we get

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

⇒ (s + 3)Y(s) = 2X(s)

⇒ $\frac{Y(s)}{X(s)}=\frac{2}{s+3}$

⇒ $H(s)=\frac{2}{s+3}$

∴ Impulse response will be, h(t) = L^-1(H(s)) = 2e^-3tu(t)

Hence, the correct option is (C).

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

𝑋(𝜔) = 1, for |𝜔| < 𝑊₀

= 0, for |𝜔| > 𝑊₀

((a))
𝑥(𝑡) tends to be an impulse as 𝑊₀ → ∞
((b))
𝑥(0) decreases as 𝑊₀ increases.
((c))
At $\rm t=\frac{\pi}{2W_0}, x(t)=-\frac{1}{\pi}$
((d))
At $\rm t=\frac{\pi}{2W_0}, x(t)=\frac{1}{\pi}$

Show Answer

Answer: ((a))

𝑥(𝑡) tends to be an impulse as 𝑊₀ → ∞

Given:

$X(ω)= \begin{cases}1, & \text { for }|ω|<ω_0 \ 0, & \text { for }|ω|>ω_0\end{cases}$

By taking inverse Fourier transform,

$x(t)=\frac{\sin ω_0 t}{\pi t}$

$x\left(\frac{\pi}{2 ω_0}\right)=\frac{2 ω_0}{\pi \times \pi} \sin ω_0 \times \frac{\pi}{2 ω_0}$ $=\frac{2 ω_0}{\pi^2} \sin \frac{\pi}{2}=\frac{2 ω_0}{\pi^2}$

So, option (C) and (D) are wrong.

$x(0)=\underset{t \rightarrow 0}{L t} \frac{\sin ω_0 t}{\pi t}=$ $\underset{t \rightarrow 0}{L t} \frac{ω_0 \cos ω_0 t}{\pi}=\frac{ω_0}{\pi}$

So, x(0) ∝ ω₀ Option (B) is wrong.

When ω0 → ∞, X(ω) will be a D.C signal and inverse Fourier transform of a D.C signal will be impulse signal.

So, option (A) is correct.

Hence, the correct option is (A).

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

$\rm X(z)=\frac{4z}{\left(z-\frac{1}{5}\right)\left(z-\frac{2}{3}\right)(z-3)}$ with ROC = R.

Which one of the following statements is true?

((a))
Discrete-time Fourier transform of x[n] converges if R is |𝑧| > 3
((b))
Discrete-time Fourier transform of x[n] converges if R is $\rm \frac{2}{3}<|z|<3$
((c))
Discrete-time Fourier transform of x[n] converges if R is such that x[n] is a left-sided sequence
((d))
Discrete-time Fourier transform of x[n] converges if R is such that x[n] is a right-sided sequence

Show Answer

Answer: ((b))

Discrete-time Fourier transform of x[n] converges if R is $\rm \frac{2}{3}<|z|<3$

Given:

$X(z)=\frac{4 z}{\left(z-\frac{1}{5}\right)\left(z-\frac{2}{3}\right)(z-3)}$

Poles of X(z) are located at z = $\frac{1}{5}$ , z = $\frac{2}{3}$ and z = 3.

For DTFT to converge, the ROC of Z-transform of x() should contain unit circle.

If x(n) is a right sided sequence then the ROC is |z|>3 which does not include unit circle. So, option (D) and (A) are wrong.

If R.O.C. is $\frac{2}{3}$ < |z| < 3, the R.O.C. includes unit circle. So, option (B) is correct.

If x(n) is a left sided then R.O.C will be |z| < $\frac{1}{5}$ which does not include unit circle. So, option (C) is wrong.

Hence, the correct option is (B).

For the three-bus power system shown in the figure, the trip signals to the circuit breakers B₁ to B₉ are provided by overcurrent relays R₁ to R₉, respectively, some of which have directional properties also. The necessary condition for the system to be protected for short circuit fault at any part of the system between bus 1 and the R-L loads with isolation of minimum portion of the network using minimum number of directional relays is

((a))
R₃ and R₄ are directional overcurrent relays blocking faults towards bus 2
((b))
R₃ and R₄ are directional overcurrent relays blocking faults towards bus 2 and R₇ is directional overcurrent relay blocking faults towards bus 3
((c))
R₃ and R₄ are directional overcurrent relays blocking faults towards Line 1 and Line 2, respectively, R₇ is directional overcurrent relay blocking faults towards Line 3 and R₅ is directional overcurrent relay blocking faults towards bus 2
((d))
R₃ and R₄ are directional overcurrent relays blocking faults towards Line 1 and Line 2, respectively.

Show Answer

Answer: ((a))

R₃ and R₄ are directional overcurrent relays blocking faults towards bus 2

Hence, R₃ and R₄ are directional over current relays, which only operates when fault occurs in line-1 and line-2 respectively. But blocking faults towards bus-2.

Hence, the correct option is (A).

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

𝐹₁ (𝑃_𝐺1 ) = 0.1 $\rm aP_{G1}^2$ + 40 𝑃_𝐺1 + 120 𝑅𝑠/ℎ𝑜𝑢𝑟 0 𝑀𝑊 ≤ 𝑃_𝐺1 ≤ 350 MW

𝐹₂ (𝑃_𝐺2) = 0.2 $\rm P_{G2}^2$ + 30 𝑃_𝐺2 + 100 𝑅𝑠/ℎ𝑜𝑢𝑟 0 𝑀𝑊 ≤ 𝑃_𝐺2 ≤ 300 MW

where a is a constant. For a given value of a, optimal dispatch requires the total load of 290 MW to be shared as 𝑃_𝐺1 = 175 𝑀𝑊 and 𝑃_𝐺2 = 115 𝑀𝑊. With the load remaining unchanged, the value of a is increased by 10% and optimal dispatch is carried out. The changes in 𝑃_𝐺1 and the total cost of generation, F (= F₁ + F₂) in Rs/hour will be as follows

((a))
𝑃_𝐺1 will decrease and F will increase
((b))
Both 𝑃_𝐺1 and F will increase
((c))
𝑃_𝐺1 will increase and F will decrease
((d))
Both 𝑃_𝐺1 and F will decrease

Show Answer

Answer: ((a))

𝑃_𝐺1 will decrease and F will increase

$F_1\left(P_{g_1}\right)=0.1 a P_{G_1}^2+40 P_{G_1}+120 \mathrm{Rs} / \mathrm{hr}$

$F_2\left(P_{g_2}\right)=0.2 P_{G_2}^2+30 P_{G_2}+100 \mathrm{Rs} / \mathrm{hr}$

Incremental cost, $I_{C_1}=\frac{d F_1}{d P_{G_1}}=0.2 a P_{G_1}+40$

$I_{C_2}=\frac{d F_2}{d P_{G_2}}=0.4 P_{G_2}+30$

0.2a × 175 + 40 = 0.4 × 115 + 30

a = 1.028

Now, $P_1^{\prime}+P_2^{\prime}$ = 175 + 115 = 290 mW

a' = 1.1 × 1.028 = 1.1308

F1 = 0.1 × 1.1308 $P_1^{2{\prime}}$ + 40 $p_1^{\prime}$ + 120

F2 = 0.2 $P_1^{2{\prime}}$ + 30P2 + 100

$I_{C_{\mathrm{1}}}$ = 0.226 $p_1^{\prime}$ + 40

$I_{C_{\mathrm{2}}}$ = 0.4 $P_2^{{\prime}}$ + 30

For optimal load, $I_{C_{\mathrm{1}}}$ = $I_{C_{\mathrm{2}}}$

0.226 $p_1^{\prime}$ + 40 = 0.4 $p_2^{\prime}$ + 30

0.226 $p_1^{\prime}$ - 0.4 $p_2^{\prime}$ = -10 ...(i)

$P_1^{\prime}+P_2^{\prime}$ = 290

$p_1^{\prime}$ = 169 MW

$p_2^{\prime}$ = 121 MW

So, P1 decreases

F when P₁ = 175 MW and P₂ = 115 MW

F₁ = 0.1 × 1.028 × 175² + 40 × 175 + 120 = 10268.2 Rs/hr

F₂ = 0.2 × 115² + 30 × 115 + 100 = 6195 Rs/hr

F = F₁ + F₂ = 16463 Rs/hr

$F^{\prime}_1$ = 0.1 × 1.1308 × 169² + 40 × 169 + 120 = 10109.6

$F^{\prime}_2$ = 0.2 × 121² + 30 × 121 + 100 = 6658.2

F' = $F^{\prime}_1$ + $F^{\prime}_2$ = 16767.8 Rs/hr

So, F increase

Hence, the correct option is (A).

The four stator conductors (A, A', B and B') of a rotating machine are carrying DC currents of the same value, the directions of which are shown in the figure (i). The rotor coils a-a' and b-b' are formed by connecting the back ends of conductors ‘a’ and ‘a'’ and ‘b’ and ‘b'’, respectively, as shown in figure (ii). The e.m.f. induced in coil a-a' and coil b-b' are denoted by E_a-a' and E_b-b', respectively. If the rotor is rotated at uniform angular speed ω rad/s in the clockwise direction then which of the following correctly describes the E_a-a' and E_b-b' ?

figure (i): cross-sectional view figure (ii): rotor winding connection diagram

((a))
Ea-a' and Eb-b' have finite magnitudes and are in the same phase
((b))
Ea-a' and Eb-b' have finite magnitudes with Eb-b' leading Ea-a'
((c))
Ea-a' and Eb-b' have finite magnitudes with Ea-a' leading Eb-b'
((d))
Ea-a' = Eb-b' = 0

Show Answer

Answer: ((d))

Ea-a' = Eb-b' = 0

At this instant, coil aa' and bb' are along q-axis, so there is no induced emf in both coils

E_aa' = E_bb' = 0.

Hence, the correct option is (D).

The chopper circuit shown in figure (i) feeds power to a 5 A DC constant current source. The switching frequency of the chopper is 100 kHz. All the components can be assumed to be ideal. The gate signals of switches S₁ and S₂ are shown in figure (ii). Average voltage across the 5 A current source is

((a))
10 V
((b))
6 V
((c))
12 V
((d))
20 V

Show Answer

Answer: ((b))

6 V

when switch S₁ ON → ν₀ = ν_s = 20V

D₂ ON → ν₀ = 0 volt

S₂ ON → ν₀ = 0 volt (no energy stored)

$V_{0(\text { ang })}=\frac{20 \times 3}{10}=6 \mathrm{Volt}$

Hence, the correct option is (B).

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

((a))
$\begin{bmatrix}\frac{4}{5}&\frac{3}{5}\\ -\frac{3}{5}&\frac{4}{5}\end{bmatrix}$
((b))
$\begin{bmatrix}\frac{4}{5}&-\frac{3}{5}\\ \frac{3}{5}&\frac{4}{5}\end{bmatrix}$
((c))
$\begin{bmatrix}\frac{4}{5}&\frac{3}{5}\\ \frac{3}{5}&\frac{4}{5}\end{bmatrix}$
((d))
$\begin{bmatrix}\frac{4}{5}&-\frac{3}{5}\\ \frac{3}{5}&-\frac{4}{5}\end{bmatrix}$

Show Answer

Answer: ((a))

$\begin{bmatrix}\frac{4}{5}&\frac{3}{5}\\ -\frac{3}{5}&\frac{4}{5}\end{bmatrix}$

Given: Au = v

Considering option (A),

$\left[\begin{array}{cc} \frac{4}{5} & \frac{3}{5} \ \frac{-3}{5} & \frac{4}{5} \end{array}\right]\left[\begin{array}{l} 4 \ 3 \end{array}\right]=\left[\begin{array}{l} 5 \ 0 \end{array}\right]$

Hence, option (A) satisfies the relation Au = v.

Considering option (B),

$\left[\begin{array}{cc} \frac{4}{5} & -\frac{3}{5} \ \frac{3}{5} & \frac{4}{3} \end{array}\right]\left[\begin{array}{l} 4 \ 3 \end{array}\right]=\left[\begin{array}{c} \frac{7}{5} \ \frac{24}{5} \end{array}\right]$

Hence, option (B) does not satisfy the relation Au = v.

Considering option (C).

One million random numbers are generated from a statistically stationary process with a Gaussian distribution with mean zero and standard deviation 𝜎₀.

The 𝜎₀ is estimated by randomly drawing out 10,000 numbers of samples (𝑥_𝑛). The estimates 𝜎̂₁, 𝜎̂₂ are computed in the following two ways.

$\rm \hat {\sigma}1^2=\frac{1}{10000}\Sigma{n=1}^{10000}x_n^2$

$\rm \hat {\sigma}2^2=\frac{1}{9999}\Sigma{n=1}^{10000}x_n^2$

((a))
$E(\hat {\sigma}_2^2)=\sigma_0^2$
((b))
$E(\hat {\sigma}_2)=\sigma_0$
((c))
$E(\hat {\sigma}_1^2)=\sigma_0^2$
((d))
$E(\hat {\sigma}_1)=\hat \sigma_2$

Show Answer

Answer: ((c))

$E(\hat {\sigma}_1^2)=\sigma_0^2$

$\sigma_0^2$ is the variance of population of 1000 samples given by

$\sigma_o^2=\frac{1}{10000} \sum_{n=1}^{10000}(X-{\bar X})^2$

Given X̅ = 0

$\sigma_o^2=\frac{1}{10000} \sum X_n^2$

We know that $E\left[\sigma_1^2\right]=\sigma_1^2=\sigma_0^2$

Hence, option (c) is the correct answer.

A semiconductor switch needs to block voltage V of only one polarity (V > 0) 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?

((a))
((b))
((c))
((d))

Show Answer

Answer: ((a))

(A), (D)

Concept:

I-V characteristic of given switch :

From the given configuration, the current flows in both directions (Bidirectional). The switch also has on drop voltage. The switch configuration in options (A) and (D) can provide bidirectional current. Hence, the correct options are (A) and (D).

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

((a))
If an LTI system is causal, it is stable
((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
((c))
If a discrete time LTI system has an impulse response ℎ[𝑛] of finite duration the system is stable
((d))
If the impulse response 0 < |ℎ[𝑛]| < 1 for all 𝑛, then the LTI system is stable.

Show Answer

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

No information about amplitude of h(n) is given. So, option (C) is wrong.
h(n) can have any amplitude less than infinity for the LTI system to be stable. So, option (D) is wrong.
A causal LTI system can also be unstable. So, option (A) is wrong.
If the response to a step input u[n] is 0 for n < 0, then the discrete time LTI system will be causal.

So, option (B) is true.

Hence, the correct option is (B).

The bus admittance (Y_bus) 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?

((a))
Line charging capacitor of finite value is present in all three lines
((b))
Line charging capacitor of finite value is present in line 2-3 only
((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
((d))
Line charging capacitor of finite value is present in line 2-3 only and shunt capacitor of finite value is present in bus 3 only

Show Answer

Answer: ((a))

Line charging capacitor of finite value is present in all three lines

Given:

∑R₁ = 0, so no shunt element at bus-1 is present.

Where as ∑R₂ = j0.5 and ∑R₃ = j1

so, shunt element are present at bus-2 and 3.

Hence, the correct option are (A) & (C).

The value of parameters of the circuit shown in the figure are

R1 = 2Ω, R₂ = 2Ω, R₃ = 3Ω, L = 10 mH, C = 100 μF

For time t < 0, the circuit is at steady state with the switch ‘K’ in closed condition. If the switch is opened at t = 0, the value of the voltage across the inductor (V_L) at t = 0⁺ in Volts is ____________ (Round off to 1 decimal place).

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

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

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

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

Resistance, R = 0.035 Ω/km;

Inductance, L = 1 mH /km;

Capacitance, C = 0.01 μF/km;

The line is represented by the nominal-π model. With the magnitudes of the sending end and the receiving end voltages of the line (denoted by V_S and V_R, respectively) maintained at 275 kV, the phase angle difference (θ) between V_S and V_R required for maximum possible active power to be delivered to the receiving end, in degree is _____________ (Round off to 2 decimal places).

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

$\rm \frac{dy}{dt}=\frac{e^{-at}}{2+at}$ , 𝛼 = 0.01 and 𝑦(0) = 0

Three points in the x-y plane are (-1, 0.8), (0, 2.2) and (1, 2.8). The value of the slope of the best fit straight line in the least square sense is ____________ (Round off to 2 decimal places).

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

((a))
2.51 e^-0.032s
((b))
$\rm \frac{e^{-2.514s}}{s+1}$
((c))
1.04 e^-2.514s
((d))
2.51 e^-1.047s

Show Answer

Answer: ((d))

2.51 e^-1.047s

The transfer function, TF = Ke^-s^T_d (tranportation lag)

Given magnitude, M = 8 dB = 20 log [K]

K = 2.511

and angle at ω = 1 rad/sec = -60°

Angle, $\phi=-\omega T_\alpha \times \frac{180^{\circ}}{\pi}$

$60^{\circ}=-1 \times T_d \times \frac{180^{\circ}}{\pi}$

T_d = 1.047

So, required transfer function

TF = 2.511e^-1.047s

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

((a))
The maximum and minimum values of the output voltage V_O are +15 V and -10 V, respectively.
((b))
The maximum and minimum values of the output voltage V_O are +5 V and -15 V, respectively.
((c))
The maximum and minimum values of the output voltage V_O are +10 V and -5 V, respectively
((d))
The maximum and minimum values of the output voltage V_O are +5 V and -10 V, respectively.

Show Answer

Answer: ((d))

The maximum and minimum values of the output voltage V_O are +5 V and -10 V, respectively.

<br>

For positive half cycle the diodes D1 and Dz are forward bias and D2 is reverse biased and the circuit is shown below,

When V_in = 10V,

$V_0=\frac{-R}{R} \times 10=-10 \mathrm{~V}$

∴ V₀ will be -10 V.

For negative half cycle, diode D₂ is forward bias and D₁, D_z is reverse bias. Zenor diode is in breakdown region and the circuit is shown below,

∴ V₀ = 5 V

∴ $V_{0_{\max }}$ = 5 V and $V_{0_{\max }}$ = -10 V

Hence, the correct option is (D).

Consider a lead compensator of the form

$\rm K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{β a}}$ , β > 1, α > 0

The 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 𝐾(𝑠), is 6 dB. The values of 𝑎, 𝛽, respectively, are

((a))
1, 16
((b))
2, 4
((c))
3, 5
((d))
2.66, 2.25

Show Answer

Answer: ((b))

2, 4

$\mathrm{TF}=\frac{1+\frac{s}{α}}{1+\frac{s}{α β}}, α>0, β>1$

$\mathrm{TF}=\frac{s+α}{(s+α β)}$

Given: $ω_m=√{α(α β)}=4$

α√β = 4 ...(1)

Given at ω = ω_m

$x =\frac{1}{β}$

Amplification, $M =10 \log _{10} \frac{1}{x}=6$

10 log₁₀(β) = 6

β = 4

and $\alpha = \frac{4}{\sqrt 4} = 2$

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

((a))
Unity
((b))
0.5 leading
((c))
0.866 leading
((d))
0.707 lagging

Show Answer

Answer: ((c))

0.866 leading

Given: V_line = 400 Volt ⇒ Y - connected

V_phase = $\frac{V_{\text {ine }}}{√{3}}=\frac{400}{√{3}}$

I_line = I_phase = 2 Amp

Wattmeter reading = -400 Watt

This Wattmeter connection is related to reactive power measurement by single wattmeter method.

So, Wattmeter reading = $=\sqrt{3} \cdot V_{\mathrm{ph}} \cdot I_{\mathrm{ph}} \sin (ϕ)$

$-400 \text { Watt } =\sqrt{3} \times \frac{400}{\sqrt{3}} \times 2 \times \sin (ϕ)$

$\sin (ϕ) =-\frac{1}{2}$

⇒ $ϕ =\sin ^{-1}\left(-\frac{1}{2}\right)=-30^{\circ}$ ⇒ Leading

P.f. of load = cos(ϕ) = cos (-30°)

P.f. = 0.866 leading

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

((a))
64H
((b))
65H
((c))
66H
((d))
67H

Show Answer

Answer: ((c))

66H

Vin = 1.992 V,

n = 8

Vfs = 5 V

We know, from graph

Step size = $\frac{V_{f s}}{2^n-1} \$

Step size = $\frac{5}{2^8-1}=\frac{5}{255}$

Analog input = Step size × (decimal equivalent of binary code)

$1.992 =\frac{5}{255} \times D$

$D =\frac{1.992 \times 255}{5}=(101.592)_{10}$

For decimal equivalent of 101,

Analog input = $\frac{5}{255} \times 101=1.980$

Here 1.98 is less than 1.992 so, we have to take decimal equivalent as 102

For (102)₁₀ Analog input = $\frac{5}{255} \times 102=2$

2 is near about 1.992 so, decimal equivalent will 102 in hexadecimal, (102)₁₀ = 66 H

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

((a))
𝐵𝑢𝑠 1 𝑆𝑙𝑎𝑐𝑘 𝑏𝑢𝑠

𝐵𝑢𝑠 2 𝑃 − |𝑉| 𝑏𝑢𝑠

𝐵𝑢𝑠 3 𝑃 − 𝑄 𝑏𝑢𝑠
((b))
𝐵𝑢𝑠 1 𝑃 − |𝑉| 𝑏𝑢𝑠

𝐵𝑢𝑠 2 𝑃 − |𝑉| 𝑏𝑢𝑠

𝐵𝑢𝑠 3 𝑆𝑙𝑎𝑐𝑘 𝑏𝑢𝑠
((c))
𝐵𝑢𝑠 1 𝑃 − 𝑄 𝑏𝑢𝑠

𝐵𝑢𝑠 2 𝑃 − 𝑄 𝑏𝑢𝑠

𝐵𝑢𝑠 3 𝑆𝑙𝑎𝑐𝑘 𝑏𝑢𝑠
((d))
𝐵𝑢𝑠 1 𝑃 − |𝑉| 𝑏𝑢𝑠

𝐵𝑢𝑠 2 𝑃 − 𝑄 𝑏𝑢𝑠

𝐵𝑢𝑠 3 𝑆𝑙𝑎𝑐𝑘 𝑏𝑢𝑠

Show Answer

Answer: ((d))

𝐵𝑢𝑠 1 𝑃 − |𝑉| 𝑏𝑢𝑠

𝐵𝑢𝑠 2 𝑃 − 𝑄 𝑏𝑢𝑠

𝐵𝑢𝑠 3 𝑆𝑙𝑎𝑐𝑘 𝑏𝑢𝑠

At Bus (1), voltage magnitude is maintained also specified active power is zero. So, it is a PV Bus.

At Bus (2), P and Q are specified and |𝑉|, δ are unknown. Hence, it is PQ Bus.

At Bus (3), |𝑉| and δ are specified. So it is slack bus.

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

|𝑥₁|^p + |𝑥₂|^𝑝 = 1 for 𝑝 > 0

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

((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.
((c))
When p tends to 0, the area enclosed by the curve is 1.
((d))
When p = 1, the area enclosed by the curve is 2.

Show Answer

Answer: ((a))

When p = 2, the area enclosed by the curve is π.

Given equation in a 2-D real space is,

|x1|p + |x2|p = 1, p > 0

Let p = 1,

|x1| + |x2| = 1 is a square

∴ Area = 4 × $\frac{1}{2}$ × 1 × 1 = 2

Let p = 2

|x1|² + |x2|² = 1

$x_1^2 + x_2^2 = 1$ which is a circle of radius 1.

∴ Area = π × 1² = π

For p = 0, the curve does not exists.

Let p = ∞

|x1|p + |x2|p = 1

If x₁ < 1, then x₂ = 1 or -1

It will form square of side 2.

Area = (Side)² = 2² = 4

Hence, the correct options are (A), (B) & (D)

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

Show Answer

Answer: ((b))

The charge will always move on the y-z plane only

Net force applied on charge, $\vec F_{Total} = \vec F_e + \vec F_m = q. \vec E + q(\vec V \times \vec B)$

Initially charge is at rest. No magnetic force is experienced, due to electric field charge moves in the z direction with increasing velocity. Now because of increasing v $\vec F_m$ increases in perpendicular of velocity vector and creates a cycloid trajectory.

"There is miss match between statement and figure. As per the figure only (B) and (D) are the correct answer. Must go for Marks To All".

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

((a))
When switch S is ON, both D₁ and D₂ conducts and D₃ is reverse biased
((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
((d))
When switch S is OFF, D₁ conducts, D₂ is reverse biased and D₃ conducts

Show Answer

Answer: ((a))

When switch S is ON, both D₁ and D₂ conducts and D₃ is reverse biased

<br>

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).

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

$A =\rm \begin{bmatrix}0&1\\ -1&-2\end{bmatrix}, B=\rm \begin{bmatrix}0\\ 1\end{bmatrix}, C=\rm \begin{bmatrix}3&-2\end{bmatrix}, D = 1$

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

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

A three phase 415 V, 50 Hz, 6-pole, 960 RPM, 4 HP squirrel cage induction motor drives a constant torque load at rated speed operating from rated supply and delivering rated output. If the supply voltage and frequency are reduced by 20%, the resultant speed of the motor in RPM (neglecting the stator leakage impedance and rotational losses) is __________ (Round off to the nearest integer).

The period of the discrete-time signal 𝑥[𝑛] described by the equation below is 𝑁 = ___________________ (Round off to the nearest integer).

$\rm x[n]=1+3\sin\left(\frac{15\pi}{8}n+\frac{3\pi}{4}\right)-5\sin\left(\frac{\pi}{3}n-\frac{\pi}{4}\right)$

The discrete-time Fourier transform of a signal 𝑥[𝑛] is 𝑋(Ω) = (1 + 𝑐𝑜𝑠Ω)𝑒^−𝑗Ω. Consider that 𝑥_𝑝[𝑛] is a periodic signal of period N = 5 such that

𝑥_𝑝 [𝑛] = 𝑥[𝑛], for 𝑛 = 0, 1 ,2

= 0, for 𝑛 = 3, 4

Note that $\rm x_p[n]=\Sigma_{k=0}^{N-1}a_ke^{j\frac{2\pi}{N}kn}$ . The magnitude of the Fourier series coefficient 𝑎₃ is _______________ (Round off to 3 decimal places).

For the circuit shown, if 𝑖 = 𝑠𝑖𝑛 1000𝑡, the instantaneous value of the Thevenin’s equivalent voltage (in Volts) across the terminals a - b at time t = 5 ms is __________ (Round off to 2 decimal places).

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

𝑦₁₁ = 5 𝑆, 𝑦₂₂ = 1 𝑆, 𝑦₁₂ = 𝑦₂₁ = −2.5 𝑆

The power delivered to the load resistor RL in Watt is __________ (Round off to 2 decimal places).

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 mm² and the average core length traversed by the mutual flux is 500 mm. The maximum allowable flux density in the core is B_max = 1 Wb/m² and the relative permeability of the core material is 5000. The leakage impedance of the winding a-b and winding c-d at 50 Hz are (5 + j100π × 0.16) Ω and (11.25 + j100π × 0.36) Ω, 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).

The circuit shown in the figure is initially in the steady state with the switch K in open condition and K̅ in closed condition. The switch K is closed and K̅ is opened simultaneously at the instant t = t₁, where t₁ > 0. The minimum value of t₁ in milliseconds, such that there is no transient in the voltage across the 100 μF capacitor, is ____________ (Round off to 2 decimal places).

The circuit shown in the figure has reached steady state with thyristor ‘T’ in OFF condition. Assume that the latching and holding currents of the thyristor are zero. The thyristor is turned ON at t = 0 sec. The duration in microseconds for which the thyristor would conduct, before it turns off, is _____ (Round off to 2 decimal places).

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 ____________.

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, R1

The single phase rectifier consisting of three thyristors T₁, T₂, T₃ and a diode D₁ feed power to a 10 A constant current load. T₁ and T₃ are fired at α = 60° and T₂ is fired at α = 240°. The reference for α is the positive zero crossing of V_in. The average voltage V_O across the load in volts is _____ (Round off to 2 decimal places).

The Zener diode in circuit has a breakdown voltage of 5 V. The current gain β of the transistor in the active region in 99. Ignore base-emitter voltage drop V_BE. The current through the 20 Ω resistance in milliamperes is ________(Round off to 2 decimal places).

The two-bus power system shown in figure (i) has one alternator supplying a synchronous motor load through a Y-Δ 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).

An infinite surface of linear current density 𝐊 = 5𝐚̂_𝐱 A/m exists on the x-y plane, as shown in the figure. The magnitude of the magnetic field intensity (H) at a point (1,1,1) due to the surface current in Ampere/meter is _______ (Round off to 2 decimal places).

The closed curve shown in the figure is described by

r = 1 + cos θ, where r = $\sqrt{x^2+y^2}$ ; x = r cosθ, y = r sinθ

The magnitude of the line integral of the vector field $F = - y\hat{i} + x\hat{j}$ around the closed curve is ____________ (Round off to 2 decimal places).

A signal 𝑥(𝑡) = 2𝑐𝑜𝑠(180𝜋𝑡)𝑐𝑜𝑠(60𝜋𝑡) is sampled at 200 Hz and then passed through an ideal low pass filter having cut-off frequency of 100 Hz.

The maximum frequency present in the filtered signal in Hz is _____________ (Round off to the nearest integer).

A balanced delta connected load consisting of the series connection of one resistor (R = 15 Ω) and a capacitor (C = 212.21 μF) in each phase is connected to threephase, 50 Hz, 415 V supply terminals through a line having an inductance of L = 31.83 mH 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 V_AB in Volts is _____________ (Round off to the nearest integer).

A quadratic function of two variables is given as

$\rm f(x_1, x_2)=x_1^2+2x_2^2+3x_1+3x_2+x_1x_2+1$

The magnitude of the maximum rate of change of the function at the point (1,1) is _________ (Round off to the nearest integer).

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