GATE EE 2023 Official Paper (Previous Year Paper)

The correct answer is option (4)

Solution:

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

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).

Probability of getting "1" on 1^st roll = \(\frac{1}{6}\)

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).

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).

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

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).

The given function is odd function.

Property of odd function.

F(-x) = -f(x)

Therefore, option (b) is an odd function.

Concept:

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 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).

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

Unit digit value = 7

Hence, the correct option is (A).

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).

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).

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,

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

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

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

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

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 torque slip characteristics are shown in the figure below.

S > 1 ⇒ Plugging mode

0 < S < 1 ⇒ Mtoring mode

S < 0 ⇒ Generating mode

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).

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).

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).

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).

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).

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

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

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).

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).

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).

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).

\(\sigma_0^2\) 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), (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).

1. No information about amplitude of h(n) is given. So, option (C) is wrong.
2. h(n) can have any amplitude less than infinity for the LTI system to be stable. So, option (D) is wrong.
3. A causal LTI system can also be unstable. So, option (A) is wrong.
4. 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).

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).

At t = 0-, switch is closed and circuit is in steady state condition (inductor acts as short circuit and capacitor acts as open circuit).

By CDR, I_L(0^-) = \(10 × \frac{3}{3+2}=6 \mathrm{~A}\)

V_C(0^-) = 6 × 2 = 12 V

For t = 0⁺, switch is opened

Apply KVL,

6 × 2 + V_L - 12 - 4 × 2 = 0

V_L = 20 - 12

V_L = 8V

Hence, the correct answer is 8V.

Given torque is constant, T = KϕI_a = Constant

Field is also unchanged, ϕ = Constant

So, I_a = Constant

The back emf with rated voltage

E_b1 = V_s - I_aR_a = 400 - 1.2 × 15

Eb1 = 382 V

and back emf with 10% drop in supply voltage

E_b2 = V_s - I_a2R_a = 400 × 0.90 - 1.2 × 15

Eb2 = 342 V

We know E_b ∝ N (for separately excited motor)

So, 

\( \frac{E_{b 2}}{E_{b 1}} =\frac{N_2}{N_1} \)

N₂ = \(\frac{342}{382} \times 1500=1342.93\)

N₂ = 1343 rpm

Given signal y(t) and x(t) is shown in figure below,

Given: z(t) = x(t)*y(t) = \(\int_{-\infty}^{\infty} y(τ) x(t-τ) d τ\)

x(t - τ) can be plotted as shown below,

Maximum value of convolution will be obtained when positive portion of y(τ) is multiplied with x(t - τ)

∴ 1 + t = 5

t = 4 sec

At t = 4 sec, maximum value of convolution will be obtained.

 

Circuit can be redrawn

By applying voltage division rule, we can get

\(V_{a b}=8 \times \frac{1.5}{1.5+0.5}=6 \mathrm{~V}\)

For nominal π model:

B = Z = [R + j(2π + L)]l

= [0.035 + j(2π × 50 × 1 × 10^-3)] × 400

= 126.44 ∠83.642° = |B|∠β

\(P_R=\left|\frac{V_S V_R}{B}\right| \cos (β-δ)-\left|\frac{A V_R^2}{B}\right| \cos (β-\alpha)\)

\(\frac{2 P_R}{2 δ}=0\) ⇒ δ = β

At δ = β ⇒ \(\left(P_R\right)_{\max }=\left|\frac{V_S V_R}{B}\right|-\left|\frac{A V_R^2}{B}\right| \cos (\beta-\alpha)\)

For maximum PR occur at

δ = β = 83.64° = θ

Given: \(\frac{d y}{d t}=\frac{e^{-α t}}{2+α t}\)    ...(i)

α = 0.01

y(0) = 0 (initial)

\(f(t, y)=\frac{d y}{d t}=\frac{e^{-0.01 t}}{2+0.01 t}\)

Euler's method: Step size h = t (given)

y₁ = y₀ + hf(t, y) (at t₀ = 0, y₀ = 0)

\(=0+1\left[\frac{e^{-0.01 \times 0}}{2+0.01 \times 0}\right] \)

\(=0+1 \times \frac{1}{2}=\frac{1}{2}=0.5\)

So, at t = 1, y₁ = 0.5

Given that:

 

Best fit of straight line (y = ax + b) by least square.

Approximation given by normal equations

∑y_i = a ∑x_i + bn   ...(i)

∑x_iy_i = a ∑\(x_i^2\) + b ∑x_i   ...(ii)

 

Substituting values in eqn. (1) and (2)

5.8 = a(0) + 3b   ...(iii)

2 = a(2) + 0(b)    ...(iv)

Solving we get, a = 1, \(b = \frac{5.8}{3}\) = 1.93

 ∴ Slope of best fit y = ax + b is 'a' = 1

The transfer function, TF = Ke^-sT_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

 

 

 

 

 

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).

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

\( \mathrm{TF}=\frac{s+α}{(s+α β)} \)

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 \(\alpha = \frac{4}{\sqrt 4} = 2\)

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

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

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.

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)

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".

By Geometrical distribution,

\(E(x) =\frac{1}{P} \)

\(V(x) =\frac{q}{P^2} \)

\(E(x) =4=\frac{1}{P} \)

\(q =1-\frac{1}{4}=\frac{3}{4} \)

\(P(H) =\frac{1}{4}, \)

\(P(T) =\frac{3}{4}\)

Required probability, P(E) = P(T H)

= \(\frac{3}{4} \times \frac{1}{4}=\frac{3}{16}=0.1875\)

Given:

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

Transfer function is given by,

T.F. = C[sI - A]-1B + D

\( {[s I-A]=\left[\begin{array}{cc} s & -1 \ 1 & s+2 \end{array}\right]} \)

\([s I-A]^{-1}=\left[\begin{array}{cc} s & -1 \ 1 & s+2 \end{array}\right]^{-1} \)

\( {[s I-A]^{-1}=\frac{1}{s(s+2)+1}\left[\begin{array}{cc} s+2 & 1 \ -1 & s \end{array}\right]}\)

\(T . F .=\left[\begin{array}{ll} 3 & -2 \end{array}\right] \frac{1}{s(s+2)+1}\left[\begin{array}{cc} s+2 & 1 \ -1 & s \end{array}\right]\left[\begin{array}{l} 0 \ 1 \end{array}\right]+1 \)

\(T . F .=\frac{1}{s^2+2 s+1}\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{ll} s+2 & 1 \ -1 & s \end{array}\right]\left[\begin{array}{l} 0 \ 1 \end{array}\right]+1 \)

\( T . F .=\frac{1}{s^2+2 s+1}\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{l} 1 \ s \end{array}\right]+1 \)

\(T . F .=\frac{3-2 s}{s^2+2 s+1}+1 \)

\(T . F=\frac{s^2+4}{s^2+2 s+1} \)

\(H(s)=\frac{s^2+4}{s^2+2 s+1}\)

s = jω

\(H(\mathrm{j} ω)=\frac{4-ω^2}{1+2 j ω-ω^2}\)

Steady state output of system is zero

4 - ω² = 0

4 = ω²

ω = 2 rad/sec

Hence, the correct answer is 2.

Given:

Z̅_s = 0.1 + j0.3pu = 0.316∠71.56

P_em = 2.5 pu

E = ?

Let V = 1 pu

Power output of synchronous motor

\( P_{0=}=\frac{E V}{Z_s}-\frac{E^2}{Z_s^2} R_G \)

\(2.5=\frac{E \times 1}{0.316}-\frac{E^2}{0.316^2} \times 0.1 \), E = 1.58 pu

\(N_s=120 × \frac{50}{6}=1000 \mathrm{rpm}\)

⇒ Slip speed = 40 rpm

Here, VIf = Constant. So, slip speed remains constant.

For f = 50 × 0.80= 40 Hz, the synchronous speed

\(N_{S 2}=120 \times \frac{40}{6}\)

N_s2 = 800 rpm

So, motor speed N₂ = N_s2 - slip speed

= 800 - 40 = 760 rpm

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

Frequency components of x(n) are: \(f_1=\frac{15 \pi}{16 \pi}=\frac{15}{16} \) and \( f_2=\frac{\pi}{6 \pi}=\frac{1}{6}\)

∴ Time period, N₁ = 16 and N₂ = 6

Time period of x(n) will be,

N = L.C.M [N₁, N₂] = L.C.M[16, 6] = 48

Hence, the correct answer is 48

We know that, \(a_k=\frac{1}{N} \sum_{n=0}^{N-1} x(n) e^{\frac{-j 2 \pi}{N} k n}\)

\( a_k=\left.\frac{1}{N} X(\Omega)\right|_{\Omega=\frac{2 \pi k}{N}} \)

\(a_k=\left.\frac{1}{N}(1+\cos \Omega) e^{-j \Omega}\right|_{\Omega=\frac{2 \pi k}{N}}\) = \(\frac{1}{N}\left(1+\cos \frac{2 \pi k}{N}\right) e^{\frac{-j 2 \pi k}{N}} \)

\( a_3=\frac{1}{5}\left(1+\cos \frac{2 \pi \times 3}{5}\right) e^{\frac{-j 2 \pi \times 3}{5}} \)

\( a_3=\frac{1}{5}\left(1+\cos \frac{6 \pi}{5}\right) e^{\frac{-j 6 \pi}{5}} \)

\( \left|a_3\right|=\frac{1}{5}\left(1+\cos \frac{6 \pi}{5}\right)=0.038\)

Hence, the correct answer is 0.038

Applying source transformation

V_th = i_x(40 - j10)

Applying KVL

10 + j10 = (10 + j10)i_x - 4i_x + (10 - j10)i_x

\(i_x =\frac{10+j 10}{16}\)

V_th = i_x(10 - j10)

\(=\frac{100+100}{16}=12.5 \text { Volts }\)

V_th = 12.5 sin 1000t

= 12.5 sin 1000 × 5 × 10^-3

= -11.98 V

Y = [Y]A + [Y]B

\( Y_A =\left[\begin{array}{cc} \frac{1}{3} & -\frac{1}{3} \ -\frac{1}{3} & \frac{1}{3} \end{array}\right] \)

\(Y_B =\left[\begin{array}{cc} 5 & -2.5 \ -2.5 & 1 \end{array}\right] S \)

\(Y =\left[\begin{array}{cc} \frac{1}{3} & -\frac{1}{3} \ -\frac{1}{3} & \frac{1}{3} \end{array}\right]+\left[\begin{array}{cc} 5 & -2.5 \ -2.5 & 1 \end{array}\right] \)

\(=\left[\begin{array}{cc} \frac{16}{3} & -\frac{8.5}{3} \ -\frac{8.5}{3} & \frac{4}{3} \end{array}\right] \)

\(I_1 =\frac{16}{3} V_1-\frac{8.5}{3} V_2 \)

\(I_2 =-\frac{8.5}{3} V_1+\frac{4}{3} V_2\)

Put I2 = 0 and V1 = 20 V

\(0=-\frac{8.5}{3} × 20+\frac{4}{3} V_2\)

\( V_{\text {th }} =V_2=\frac{8.5 × 20}{4}=42.5 \mathrm{~V} \)

\(R_{\text {th }} =\frac{V_2}{I_2} \)

\(I_2 =\frac{4}{3} V_2 \)

\(\frac{V_2}{I_2} =\frac{3}{4} \Omega\)

Equivalent circuit

\(I=\frac{42.5}{\frac{3}{4}+6}=6.296 \mathrm{~A}\)

P = I²R_L = (6.296)² × 6

= 238 W

l = 500 mm = 0.5 m

A = 5000 mm² = 5 × 10^-3 m²

μ_r = 5000

E = 200 V

So, \(R=\frac{l}{\mu_o \mu_R A}\) = \(\frac{0.5}{4 \pi \times 10^{-7} \times 5000 \times 5 \times 10^{-3}}\) =15915.49

and E = 4.44 fNB_mA

\(N=\frac{E}{4.44 f N B_m A}\) = \(\frac{200}{4.44 \times 50 \times 1 \times 5 \times 10^{-3}}\)

N ≅ 181

\(L=\frac{N^R}{R}=\frac{181}{15915.49}=2058.43 \mathrm{mH}\)

The self-inductance of the winding a - b = L + 0.16 H = (2058.43 + 160) mH

= 2218.43

Given circuit is as shown below, 

 

Given that K is opened and K̅ is closed initially. So given circuit becomes,

Figure (A) can be drawn in phase form as shown below,

∴ \( I_C=\frac{1 \angle 0^{\circ} \times 10}{10-j 10}=\frac{1}{\sqrt{2}} \angle 45^{\circ} \ \)

\(V_C=-j 10 \times I_C=-j 10 \times \frac{1}{\sqrt{2}} \angle 45^{\circ}=\frac{10}{\sqrt{2}} \angle-45^{\circ} \)

∴ \(V_C(t)=\frac{10}{\sqrt{2}} \sin \left(1000 t-45^{\circ}\right) \mathrm{V}\)

Now, at t = t1, K is closed and K̅ is opened. So, the circuit becomes,

∴ Steady state voltage across capacitor will be, VC(∞) = 5

Now, for transient free condition at t = t1.

VC(t1) = VC(∞)

\( \frac{10}{\sqrt{2}} \sin \left(1000 t_1-45^{\circ}\right)=5 \)

\(\sin \left(1000 t_1-45^{\circ}\right)=\frac{5 \sqrt{2}}{10}=\frac{1}{\sqrt{2}} \)

\(1000 t_1-45^{\circ}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)

1000t1 - 45° = 45°

1000t1 - 90° = \(\frac{\pi}{2}\)

t1 = \(\frac{\pi}{2}\) ms = 1.57 ms

Hence, the correct answer is 1.57 ms.

At steady state,

V_C(0) = V_S = 100 V

i_L(0) = 0A

For t > 0,

\( i_C=V_S \sqrt{\frac{C}{L}} \sin ω_0 t=100 \sqrt{\frac{1}{4}} \sin ω_0 t \)

i_C = 50 sin ω₀t

\( I_0=\frac{100}{4}=25 \mathrm{~A}\)

Now, i_T = i_C + I₀

When i_C = -i₀

i_T = 0 A

 

\(t_c^*=\frac{\pi}{\omega_0}+\theta \)

\(\theta=\frac{1}{\omega_0} \sin ^{-1}\left[\frac{I_0}{I_{c p}}\right] \)

\( \theta=\frac{1}{\omega_0} \times \sin ^{-1}\left[\frac{25}{50}\right]\)

\( \theta=\frac{\pi}{6 \omega_0} \)

\( t_C^*=\frac{\pi}{\omega_0}+\frac{\pi}{6 \omega_0} \) = \(=\frac{1}{\omega_0}\left[\frac{7 \pi}{6}\right]=\frac{7 \pi}{6} \times \sqrt{L C}\)

\(t_c^*= \) 7.33 μ sec

From circuit,

D₁ = B ⊕ C = Q₂ ⊕ Q₁

D₂ = Q₁

Let, Q₂Q₁ =00

Sequence:

ABCD

Q̅₂Q₂Q₁(Q₁ ⊕ Q₂)

1000

The decimal equivalent of 1000 is (8)₁₀

R0 ← 60H

R1 ← 46H

ADD R0, R1 ⇒ \(\begin{array}{r} 0110 \quad 0000 \ +0100 \quad 0110 \ \hline 1010 \quad 0110 \ \hline \end{array}\)

∴ Carry flag = 0

Over flow flag = 1

Auxiliary carry flag = 0

Parity flag = 0

CAOP = (0010)₂ = (2)₁₀

The average output voltage,

\( V_0 =\frac{1}{2 \pi}\left[\int_\alpha^\pi V_m \sin \omega t d(\omega t) \right.\) + \(\left. \int_{\pi+\alpha}^{2 \pi+\alpha}-V_m \sin \omega t d(\omega t)\right] \)

\(V_0 =\frac{V_m}{2 \pi}[1+3 \cos \alpha] \)

\(\frac{100}{2 \pi}\left[1+3 \cos 60^{\circ}\right]=39.78 \mathrm{~V}\)

Assume that Zenor diode is OFF.

Applying KVL in loop shown,

25 - I_B × 7 kΩ - I_E × (10Ω + 20 Ω) = 0

25 - IB × 7 kΩ - (β + 1)I_B × 30Ω = 0

25 - IB × 7 kΩ - 100I_B × 30Ω = 0

\(I_B=\frac{25}{10000}=2.5 \mathrm{~mA}\)

∴ I_E = 100 × 2.5 = 250 mA

V = 250 mA × 10 = 2.5 V < 5 V

∴ Zener diode is not in breakdown and assumption is correct.

Hence, the correct answer is 250.

\(\left(Z_{\text {leq }}\right)_{\text {bus } 1}=j 0.2 | j 0.3\) = \(=\frac{j 0.2 \times 0.3}{0.5}=j 0.12 \mathrm{pu}\)

\(\left(Z_{\text {2eq }}\right)_{\text {bus } 1}=j 0.2 | j 0.3=j0.12 \mathrm{pu}\)

The per phase equivalent circuit for LLG fault.

\(I_{a f}^{(1)} =\frac{1 \angle 0^{\circ}}{j 0.12+j 0.12}=4.16 \mathrm{pu} \)

So, fault current, \(I_F =\sqrt{3} \times I_{o f}^{(1)}=7.21 \mathrm{pu} \)

\(\overrightarrow{H_p}=\frac{1}{2}\left(\vec{K} \times \hat{a}_n\right)\)

\(\overrightarrow{H_p}=\frac{1}{2}\left(5 \hat{a}_x \times \hat{a}_z\right)=-2.5 \hat{a}_y \)

\( \left|\overrightarrow{H_p}\right|=2.5 \mathrm{~A} / \mathrm{m}\)

C : r = (1 + cos θ)

\(\oint_C \bar{F} \cdot \overline{d r}=\oint_C-y d x+x d y\)

By Green theorem

\(=2 \iint_R d y d x \)

= 2[Area of region bounded by curve 'C']

\(=2 × \frac{1}{2} \int_{\theta=0}^{2 π} r^2 d \theta \)

\(=\int_0^{2 π}(1+\cos \theta)^2 d \theta \)

\(=\int_0^{2 π}\left(1+\cos ^2 \theta+2 \cos \theta\right) d \theta \)

\( =(\theta)_0^{2 π}+4 \int_0^{π / 2} \cos ^2 \theta d \theta+4 \int_0^π \cos \theta d \theta \)

\(=2 π+4 × \frac{1}{2} × \frac{π}{2}+0\)

= 2π + π

= 3π = 3 × 3.14

= 9.42

Given: 

x(t) = 2cos(180πt) cos(60πt)

= 

F₁ = 120 Hz and F₂ = 60 Hz

Now, present frequency component,

= nf_s ± f₁

= f₁, f_s ±f₁, 2f_s ±f₁

= 120 Hz, 200 ± 120, ....

= 120 Hz, 80 Hz, 320 Hz, ...

and other frequency component.

= nf_s ± f₂

= f₂, f_s ±f₂, 2f_s ±f₂, ...

= 60, 200 ± 60, ....

= 120 Hz, 140 Hz, 260 Hz, ...

Given cut-off frequency of LPF = 100 Hz

∴ It is pas only 60 Hz and 80 Hz frequency component.

∴ Max. frequency = 80 Hz.

X_L = 2πfL = 2π × 50 × 31.83 × 10^-3 = 10 Ω

\(X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \pi \times 50 \times 212.21 \times 10^{-6}}\) = 15 Ω

Star to delta ⇒ \(Z_Y=\frac{Z_{\Delta}}{3}=\frac{15-j 15}{3}\) = (5 - j5) Ω

\( V_{A N} =\frac{(5-j 5)}{(5-j 5+j 10)} \times \frac{415}{\sqrt{3}}\) = \(\frac{(5-j 5)}{(5+j 5)} \times \frac{415}{\sqrt{3}} \)

\(I_{A N} =\frac{415}{\sqrt{3}} V\)

So, V_AB = 415 V

Let, \( \phi =x_1^2+2 x_2^2+3 x_1+3 x_2+x_1 x_2+1 \)

\(\vec{\nabla} \phi =i \phi_{x 1}+j \phi_{x 2} \)

= i(2x₁ + 3 + x₂) + j(4x₂ + 3 + x₁)

\( \left.\vec{\nabla} \phi\right|_{(1,1)} =6 \hat{i}+8 \hat{j} \)

\(|\vec{\nabla} \phi| =\sqrt{6^2+8^2}=10\)