Gate Practice Test: Electrical Engineering(EE)- 14

Thus, option C is the correct answer.

=1 / 12×12 Work done by 12 men in 6 days = 1/2 Remaining work = 1−(1/2) = 1/2 After 6 men leave the work, time taken to complete the remaining work = Remaining Work/Remaining Men

∴ Time taken

= 12 × 12 / 6 × 2 =12days

So,

so the extra time taken =12 - 6 = 6 days

Thus, option A is the correct answer.

Let 'M' and 'C' be the marked price and cost price per lunch box. We can form an equation based on the given information as

i.e.

i.e. C = 8M / 10

or M:C = 10:8 or 5: 4

Let t be the time taken by the machines when they work simultaneously.

∴ 1 / t = 1 / 4 + ½

∴ 1 / t = 3 / 4

∴ t = 4 / 3

Answer C is corrected as not only will come after the sentence.

For sum = 16, we have the following cases:

(i) 6, 6, 4

(ii) 6, 4, 6

(iii) 4, 6, 6

(iv) 5, 5, 6

(v) 5, 6, 5

(vi) 6, 5, 5

Total cases of sum (16) = 6

Favourable cases for 5 appears in 3^rd throw = 2

p = 2 / 6 = 1 / 3

Let the number of pages in each volume be ′x′. The first page of the 3 volumes will be 1,1+x and 1+2x respectively. 1473 = 1 + (1 + x) + (1 + 2x) = 3 + 3xgiving

x = 1470/3 = 490

First page of volume 3 will be having pages starting from 1 + 980 = 981

R₁R₃ = R₂R₄

R₃ = R₂R₄/R₁

R₃ = 600 × 400/1000 = 240Ω

L₃ = CR2R4

L₃ = 0.5 × 10⁻⁶ × 400 × 600

L₃ = 12 × 10⁻²

L₃ = 0.122H

= 0.91I_dc

Now the rms value of supply fundamental component of input current

Harmonic factor (HF) of input current

=

=

= 0.30

⊤ ∝ la² ⇒ T = kla²

For const torque armature I must be const.

E^b = 200 − 20 × (0.6 + 0.4) = 180V

l_a = 20A for 1000rpm. E_a = kϕw

180E_a = 1500 / 1000

E_a = 120V = 200 − 20(0.4 + 0.6 + R_ext)

⇒ R_ext = 200 − 120 / 20 −1 = 3Ω

So

= 0.75 + j3.75

=

x(5) = 0.5

Load current

In A 3ϕ fully enthralled bridge inverter input rms current I or the

correct in each supply phase exist for 120∘ in every 180∘

Therefore rms value of input current

Input apparent power

= √3 × V_SI_S

= √3 × 400 × 1.15

= 796.72vA

∴ Input apparent power = Output power

796.72cosθ = 400

cos⁡θ = 0.5 lagging

z11 = v1I1∣∣t2 = 0 (Open circuit output terminal)

z11 = v1I1∣∣t2 = 0 (Open circuit output terminal)

V1 = I1R1

z11 = V1I1 = R1

z11 = v2I1∣∣I2 = 0

V2 =−αV1

V2 = −α(I1R1)

z21 = V2I1 = −αR1

z12 = v1I2∣∣I1 = 0 (Open circuit input terminal)

Since V1 = I1R1 = 0

50,Z12 = 0

z22=V2I2∣∣I1 = 0

since V1 = I1R1 = 0

Thus, α∨1 = 0

V2 = I2R2

z22 = V2I2 = R2

Alternative Method:

We have the equations for voltage V1 and V2 as

∨1 = |1R1

V2 − I2R2 + αV1=O

V2 = I2R2−αV1

From eq. (1) V2 = I2R2 − α(1R1) = (−αR1)I1 + R2I2

Comparing equation (1) and (2) to the general equation of z- parameter,

(Node equation at the top left node)

(Node equation at the top right node)

Comparing equation (1) and (2) with general equations, we get z-

parameters as

[y] =

1. A says that strategies are now available for eliminating psychiatric illnesses but it is mentioned in the very first line that the geneticists are very close but the strategy is not available till now. So, A is incorrect.

2. The given data says that the geneticists are working on the genetic roots of psychiatric illnesses such as depression and schizophrenia, which implies that these two diseases have genetic basis. Thus, B is the correct option.

3. C says that all human diseases can be traced back to genes and how they are expressed, but the data given in the question talks about psychiatric illnesses such as depression and schizophrenia only. So, C is also incorrect.

4. D says that in future, genetics will be the only relevant field for identifying psychiatric illnesses, which cannot be inferred from the given data. So, D is also incorrect.

So, B is the correct option

Considering input loop, since, I_G = 0A and I_S = I_D

V_GS = −I_DR_S

⇒

Since,

Substituting values,

⇒ 2.5mA =

⇒ 2.5mA=IDSS(2564)

⇒ I_DSS = 64/25 × 2.5mA

⇒ I_DSS = 6.4mA

the meter uses half wave rectifier and input is 10 V r.m.s

E_av = 1/2 (Eav over a cycle of input)

Now Ep = √2 Erms = 14.14V

E_av = 0.6 E_p = 8.99 ≈ 9V

Therefore, Eav (output) = 1/29 = 4.5V

E_dc = 0.45Erms

R_s = E_dc/l_dc − R_m = 0.45E_rms/ldc − R_m = 0.45×10/1×10⁻³ − 200

R_s = 4.3KΩ

f(x) = x³ − 3x² + 2x

f′(x) = 3x² − 6x + 2

f′(c) = 3c² − 6c + 2

by mean value through, we have

∴ 3c² − 6c + 2 = 3/4

12c² − 24c + 5=0

c = 6 ± √21/6

C = 6 − √21/6 = 0.24↔(0, 1/2)

Stages

It is a gray counter.

Initial value = 2

Final value = 6/8 = ¾

Given data

Overall efficiency of the power station is n overall =η_thermal ×n_electrical

= 0.4 × 0.7 = 0.28

Units generated /hr = (70×10³)×1

= 70kwh

Heat produced / hr

H= electrical output in heat units /n overall

= 70 × 10³ × 860/0.28 = 215 × 10⁶Kcal(1kwh = 860Kcal)

Coal consumption 1hr=H/ calorific value

= 215 × 10⁶/5000

= 43 × 10³kg

=

Putting

= -0.94

Multiplication in the time domain results in convolution in the frequency domain.

The range of convolution in frequency domain is [−1500Hz, 1500Hz]

So, the maximum frequency present in z(t) is 1500Hz Nyquist rate is 3000Hz or 3 kHz.

kW = kVA cosϕ

K_va ∝ 1/cos⁡ϕ

If the power factor increases the kVA demand will decrease.

=

=

let G(s) = 2/(s + c)

⇒ g(t) = L⁻¹{G(s)} = 2e^−ct

f(t) = L⁻¹{G(s) e^−bs}

=2e^{− k(t − b)}u(t − b)

For a discrete time signal with input as x[n] has impulse response h[n] =kδ[n], the output y[n] of the system is given as:

y[n] = x[n] × h[n]

y[n] = x[n] ×kδ[n]

y[n] = k(x × δ)[n]

y[n] = kx[n]

Taking Fourier transform on both sides

Y(ω) = X(ω) x H(ω)

Therefore H(ω) = Y(ω)/X(ω)

This is known as the transfer function of the system.

h[n] = F^-1[H(ω)].

Here, we have to use Poisson's Equation,

, because

20 × 3 × 2x + 10 × 4 × 3y² = −ρ/ε

At point (2,0)f

ρ = −240ε⁰

N = = 1500rpm

I = 9.5 x 10³ kg-m²

K.E. = = 117.2MJ

The loop equations are:

V₁ = 3I₁ + 5(I₁ + I₂) = 8I₁ + 5I₂

V₂ = 4I₂ + 5(I₁ + I₂) = 5I₁ + 9I₂

When I₂ = 0,

Z₁₁ = V₁I₁ = 8

Z₂₁ = V₂I₁ = 5 When I₁ = 0,

Z₁₂ = V₁I₂=5

Z₂₂ = V₂/I₂ = 9

Since Z11 ≠ Z₂₂, the network is not symmetrical. But, Z₁₂ = Z₂₁, so the network is reciprocal.

Thus, the fundamental frequency of the output voltage is 6ω.

Consider the state equation be, X˙= AX + BU Where,

We know that,

Finding

=

=

State transition matrix

=

Now,

And

=

=

=

Thus,

=

Radius of the conductor = 0.96 / 2 x 100 = 0.0048m

Mutual GMD (geometric mean distance) of the conductor = = 2.296m

⇒

= 1.234 x 10^-6H/m

= 1.234 mH/km

State is repeating after 5 clock pulses. N = 5

We need to find the REF form of the matrix to find null space of the matrix

REF =

Clearly it has 3 pivot columns

Thus we have

Rank = 3

Column in null space = 2 = free variable in matrix.

Difference = 3 - 2 = 1

Here, length and work is differential in nature, and hence, no integration is required. (Differential because charge is moved through ΔL ).

Therefore,

=

Now it is given that Q = 3 × 10⁻⁶

So, we get:

Now, we have point P = (−0.3, −4,0.6)

x = −0.3, y = −4 and z = 0.6

Therefore,

= 4.89 x 10^-10J

= 0.489 nJ

Motor is connected across rated voltage V = 1 pu

Rated armature current flaws in bath the motor and generator Im=Ig=1pu

Back emf in motor

E_b = V − I_mR_a = 1 − 1 × 0.02 = 0.98pu

Mechanical output power of motor

= E_bI_m -mechanicallasses

= 0.98 × 1 − 0.05 = 0.93 pu

This power is given to the generator.

Output power of generator = Output power of motor - mechanical Iasses

E_gI_g = 0.93 − 0.05 = 0.88pu ⇒ E_g = 0.88puTerminal voltage of generator

= E_g − l_gR_a = 0.88 − 0.02 = 0.86pu

Laad resistance

R_L= V₉/I₉ = 0.86/1 = 0.86pu

Number of AND gates required X = 6

Number of one bit full adders required Y = 3

X + Y = 6 + 3 = 9

X1, = X2

z=sinh⁡ ucosv+icoshusinv

By using hyperbolic properties cosh ⁡iv = cosv; sinh ⁡iv = isinv

z = sinh ⁡u cosh ⁡iv+cosh ⁡u sinh ⁡iv Now using sinh⁡(u+iv)=sinh⁡ u cosh ⁡iv+cosh ⁡u sinh ⁡iv we get

z = sinh⁡(u + iv)u + iv=sinh_-1 zw = sinh^-1 z

∴ z is not analytic for z² = -1 ⇒ z = ±i

This binary adder simply adds 0011 i.e., 3 (in decimal) to a BCD number.

When 3 is added to the BCD then it becomes excess-3 code.

So, it is a BCD to excess-3 Code converter.

MX = 0 has infinitely many solutions if |M| = 0 i.e., det(A) = 0

α(βγ - 1) - 1(γ - 1) + (1 - β) = 0

αβγ - α - γ + 1 + 1 - β = 0

αβγ - (α + β + γ) + 2 = 0

1 - (α + β + γ) + 2 = 0

(α + β + γ)=3

tr(M) = α + β + γ = 3

= 3.9 ≈ 4db

We know that maximum value of A sincut + Bcos⁡ωt = √A²+B²

So, maximum value of diode current is

(t_D)_max = 2.09A

Relationship between tie-set matrix and cut-set matrix

[B] = tie-set matrix = [I:B_T ]

[Q] = tie-set matrix = [Q_l:I]

Where [Q_l ] = - [B_T ]^T

Relationship between tie-set matrix and cut-set

matrix [B] = tie-set matrix =[I:B_T]

[Q]= tie-set matrix =[Q₁:I]

Where [Q_l ] = - [B_T ]^T

=

[Q] =

Taking V_RY = V∠0⁰ as a reference phasor, and assuming RYB phase sequence,

we have

V_RY = 400∠0⁰V

Z₁ = 20∠30⁰Ω = (17.32+j10)Ω I_R = (400∠0o)/(20∠30o)=(17.32-j10) A.

From Stokes theorem

=

=

=

=

=

The new transmission line is connected between bus ( 1 ) and (3), so element has to be changed, are Y₁₁,Y₁₃, Y₃₁, Y₃₃

Similarly,

Modified bus admittance matrix.

s_m =

s_m =

s_m = 0.453

s_m = R/X = 0.453

R = 0.453 x 0.13

R = 0.058Ω

So, the external resistance to be connected to the rotor circuit would be

R_ext = 0.058 x 0.02 = 0.038 Ω

Z-transform of signal will be

Putting z = 1

Z transform of signal x(n) will be:

Putting z = 1

= 2.75

Let x = rcos⁡θ & y = rsin⁡θ

dxdy = rdrdθ

=

=

=

= 20

h[k] = (1/2)^k−1(u[k+3]−u[k−10])

So, h[k] is non zero only in range − 3 ≤ k ≤ 9

So, h[−k] is non zero only in range − 9 ≤ k ≤ 3

Shifting h[-k] by n, so

h[n − k] is non zero only in range, (n − 9) ≤ k ≤ (n + 3)

S_o,A = n − 9 and B = n + 3

S₀,A − B= (n − 9) − (n + 3)

= −12

The transfer function is given by =

Steady-state output is

e₈₈ = input - output

e₃₉ = 12 - 12 = 0

Equivalent voltage across terminal A-B is:

V = V₁ + V₂

V = 150(0.5 + j0.866) + 180(1 − j)

V = 255 − j50 = 259.85 < />

Impedance across the terminal A-B is

Current

Angle of voltage source V₂ = 180√2 ∠ − 45^∘

So, the angle current ' " and voltage V₂ is =11.09^∘.

Apply Thevenin’s theorem in load side

To supply maximum power,

V_AB = 10/2 = 5V

By KCL at node A

V_x = 9.5 Volts

Required Probability = Favorable Outcomes / Total possible outcomes

Favourable outcomes = A false coin is chosen and flipped every time Probability of selecting a false coin =1/4 Probability of getting a tail on every flip of false coin =1. Favourable outcome

= 1/4 × 1 = 1/4

Total possible outcomes = Favourable outcomes + Unfavorable outcomes Unfavorable outcomes = A fair coin is chosen and flipped every time to get tail Probability of selecting a fair coin = 3/4 Probability of flipping a fair coin 4 times and getting tails every time

= (1/2)⁴ = 116 ∴ Unfavourable outcomes

= 3/4 × 1/16 = 3/64

Total possible outcomes

= 1/4 + 3/64 = 19/64

∴ Required probability

=

= 16/19

= 0.84

The initial line has slope of −40dB/decadel i.e., the transfer function of initial line is K/s²

At ω = ω1, the slope has been changed. From ω = ω₁ to ω = 25, slope is −20dB/ decade

So,

−20 =

log⁡25 − log⁡ω₁=0.8

log⁡ω₁ = 0.597

ω₁ = 3.95

According to the initial line.

M(dB) = 20log⁡K − 40log⁡ω

At ω = ω₁M(dB) = 22

22 = 20log⁡K = 40×0.597

log⁡K = 2.294

K = 196.78

So, the transfer function is

It is a Wheatstone bridge. For the bridge to be balanced, the required condition is

Z₁Z₄ = Z₂Z₃

From the given circuit, we have z₁ = z₄ = j1 − Ω

z₂ = z₃ = (1/j) − Ω

Now substituting these value for bridge balance equation

j×j = (1/j) × (1/j)

j² = 1/j²

− 1 = −1

Hence, it is a balanced, Wheatstone bridge, and reading of voltmeter (rms) is

V_rms = Vm/(√2)

V_rms = 100/(1.414)

= 70.72v

Let the source current is I, (current direction is outward from the source)

I_x = 31₀

Apply KVL in the loop

= 2.5√2∠105°A

⇒ = 12

⇒ = 18

Condition for periodic function:

N₁/N₂ = 12/18 is rational number

Overall time-period, N = LCM of (12,18)

N = 36

Given, ϕm = 30^∘

1/2 = 1−α/1+α

α = 13

Let assume the compensators transfer function is

Than maximum phase is provided at

So, the required compensators transfer function

We know that,

So,

=

Given form is,

=

So, we can write it as,

Comparing both equations, we get,

Taking Inverse Fourier Transform.

Comparing with given signal, g(t) = Ay(Bt)

A = 1/3 and B = 3