Electrical Engineering (EE) : Mock Test 9 For GATE

‘complaisant’- willing to please or oblige others
‘complacent’- self-satisfied, unconcerned
Thus, the sequence in option 3 must be used to make the sentence contextually sensible.

‘vitiate’- to make faulty or defective, hurt
‘ratify’- to approve and sanction formally
The sentence implies that the ever-increasing population of the country is posing a hindrance to all the developmental programmes initiated by the government.
Hence, option 1 is the correct answer.

D = ABC + BCA + CAB
= (100 A + 10 B + C) + (100 B + 10 C + A) + (100 C + 10 A + B)
= 111 (A + B + C)
So, D is a multiple of 111.
111 is divisible by 37 but not divisible by 29. Thus, D is divisible by 37 but not divisible by 29.

In an examination 100 questions of 1 mark each are given.
After the examination, 20 questions are deleted from evaluation.
Now, total number of questions considered for evaluation = 80
Marks for each correct question = 100/80 = 1.25

Student A:
Total marks secured = 40
Number of questions answered correctly = 40/1.25 = 32
Student A had answered 4 of the deleted questions correctly
Number of questions answered correctly before the deletion of 20 questions = 32 + 4 = 36
Marks as per 100 question paper = 36

Student B:
Total marks secured = 35
Number of questions answered correctly = 35/1.25 = 28
Student A had answered 10 of the deleted questions correctly
Number of questions answered correctly before the deletion of 20 questions = 28 + 10 = 38
Marks as per 100 question paper = 38

Conclusion: B lost more than A.

The ratio of ingredients A, B and C = a : b : c
Let x be the factor.
The quantity of A in the solution X = ax
The quantity of B in the solution X = bx
The quantity of C in the solution X = cx

Solution - Y:
The ratio of ingredients A, B and C = c : b: a
Let x be the factor.
The quantity of A in the solution Y = cx
The quantity of B in the solution Y = bx
The quantity of C in the solution Y = ax

Mixture Solution:
The quantity of A in the mixture solution = (a + c)x
The quantity of B in the mixture solution = 2bx
The quantity of C in the mixture solution = (a + c)x
The ratio of ingredients A, B and C = (a + c)x : 2bx : (a + c)x = (a + c) : 2b : (a + c)
It is given that, the resultant mixture to have A, B and C in equal proportion.
⇒ 2b = a + c

Total number of people in content = total number of people in rectangle
Total number of people in rectangle = 18 + 50 + 25 = 93
Total number of people in Testbook team = 152

Hence the answer the is 61 to 75

Since m and n are both positive and their sum is equal to
Hence the value of both there fractions must be less than
Therefore n-can take values > 48 and odd
N = {49, 51, 53, 55, 57, 59}

Now we have to check which values of ‘n’ will give integral values of ‘m’.
Equation:

For m to be on integer (n - 48) should be a multiple of 12n
The above condition is satisfied for
N = 49, 51, 57
Corresponding values of m
N = 49
M = 12 × 49
N = 51

According to the second sentence of the passage, a necessary condition for being a successful professional football player is not ignoring the advice of coaches.

So all of the successful professional football players mentioned in the passage’s first sentence, including those who played the sport as youths, do not ignore the advice of coaches. That’s what correct answer states.

Rotating the smaller triangle by 180°, we get the following figure.

From symmetry we can say R is the midpoint of BC.
So P and Q are also mid points of AB and AC
From the midpoint theorem:

Since two equilateral tringles are also similar triangle we know

The region above the infiltration curve represents the excess of rainfall.
∴ Excess rainfall = Area of the hydrograph above infiltration curve.
Area of hydrograph above infiltration curve (A) is given by 


∴ Excess rainfall is 11 mm.

= A̅ B̅ C + ABC
⇒ Y = (A̅ B̅ + AB) C
⇒ Y = C(A⊙B) = C(A̅⊙B̅)

We see that both current shown in the figure enter their respective winding at dots. so, for ampere turn balance

This question is based on load sharing

By using current division rule

However, if the impedance is in PU then their PU value should be converted on a common base to ensure that the ohmic ratio remain unchanged.

From the property of unit pulse function,

F(s) = F₁(s) . F₂(s)

By using partial fractions,

By applying inverse Laplace transform,

1 – e-100t = 0.015
-e-100t = 0.985
-100t = -0.01511
t = 151.136 μsec

Calculation:
V_o = I_oR_L = 3 × 10 = 30 V
V_b = V_o (1 – D)

On time = T_ON = DT = 0.6 × 20 × 10^-6 = 12 μs 

 

When the transistor is switched on, current ramps up in the inductor and energy is stored.

When the transistor is switched off, the inductor voltage reverses and acts together with the battery voltage to forward bias the diode, transferring energy to the capacitor.

When the transistor is switched on again, the load current is maintained by the capacitor, energy is stored in the inductor and the cycle can start again.

The value of the load voltage is increased by increasing the duty cycle or the on-time of the transistor.

During the transistor on-period, assuming an ideal inductor and transistor
V_b = V_L = Ldi/dt
V_b/s = (sL) i(s)
i(s) = V_b/s²L
i(t) = (V_b/L)t = kt

Over an on-time of T_ON, the change of battery current will be
Δi = k T_ON

During the off-period
V_o = V_b + V_L = V_b + L(di/dt) = V_b + L(Δi/Δt)
= V_b + L(V_b/L)T_ON /T_off = V_b(1 + T_ON /T_off)

Let the duty cycle
D = T_ON T, and T_off = T – T_ON = T(1 - D). Now
V_o = V_b (1 – DT/T(1 – D))

which simplifies to
V_o = V_b/(1 - D)

In an ideal circuit, V_bI_b = V₀I₀. Therefore, from eq.
I_b = (V_o/V_b)I_b = I_o/(1 – D)
I_b = I_b + Δi/2
I_bo = I_b – Δi/2

When the steady variation of battery current has been reached, the variation will be between l_o and I₁, as shown in.

By applying voltage division in the input loop,

i₀ will be zero, as there is no closed path.
By applying KCL in the right-side circuit,

As 10 Ω and 5 Ω are connected in parallel, voltage across them are equal.
⇒ 10 i₁ = 5 i₂ ⇒ i₂ = 2i₁ ----(2)
From equation (1) and (2)
i₁ + 2i₁ = -3 ⇒ i₁ = -1 A

The s-domain circuit for the given circuit is

When V₂ = 0

By applying KVL in the second loop
sI₂(s) + 4 (I₁(s) + I₂(s)) = 0
⇒ 4 I₁ (s) + (s + 4) I₂(s) = 0

Type of the system indicates the number of poles at origin.
Order of the system indicates the total number of poles.

Characteristic equation,
1 + G(s)H(s) = 0

⇒ s(s + 1)(s + 6) = 0
⇒ s³ + 7s² + 6s + K = 0
Routh array:

For the system to be stable, there should be no sign change in the first column of Routh array.

Given that, current through meter (I_m) = 60 μA
Total required current (I) = 100 mA
Meter resistance (R_m) = 100 Ω

Current flows though shunt resistance
I_sh = I - I_m
= 100 × 10^‑3 – 60 × 10^-6
= (100 – 0.06) × 10^-3
= 99.94 mA

Power consumed in the load,
PL = V₃ I_L cos θ
V₁ = I_LR

Given:
V_GS = -6V

V_DS = V_DD – I_DR_D = 40 – 3.2 × 3 = 30.4 V

Voltage at A point V_A = 4 × 4 = 16 V
By using virtual short circuit condition V_A = V_B = 16 V
Voltage droop across feedback resistance V_4kΩ = 4 × 4 = 16 V
Output voltage V_o = 16 + 16 = 32 V

Given that, Z₁ = 0.2 pu
Z₂ = 0.15 pu
Z₀ = 0.05 pu
Z_n = 0.05 pu
For a G fault, fault current is given by

Directional relays are not required at supply side and at radial feeders as the power flow unidirectional but not bidirectional.
so, the directional over-current relays will be required at 2 and 3 only.

y_22(old) = -j 14 p.u
Z₂₃ = j 0.2 p.u
y’₂₃ = -j5 p.u

y₂₂ = -j 14 – j5 + j 0.25
= -j 18.75 p.u.

Let X is a random variable represents an event of rolling a die and P(X = x_i) represents the probability of getting sum as x_i.

While rolling a die, the probability of getting different numbers as follows.

Greatest rate of increase of function f is directional derivative at that point.
∇f = (2xyz + 3y²) i + (x²yz + 6xy) j + x²y k
At the given point (2, 1, -1), ∇f = -i + 8 j + 4 k
Greatest rate of increase 

Given that, Eigen values of a matrix A = 3, 4 and 5
From the properties of Eigen values, determinant of a matrix is equal to the product of their Eigen values.
⇒ |A| = 3 × 4 × 5 = 60
|Adj A| = |A|^n-1 = |A|² = 60² = 3600

Probability that solenoid is defective 
The probability that solenoid is not defective = 1 – 0.05 = 0.95
Since 24 samples are there, the probability of zero defectives will be

The probability that at least one is defective = 1 - the probability of zero defectives
= 1 – 0.292 = 0.708
The answer is in between 0.65 and 0.75

Angle of departure is given by,
θ_D = 180° + ∠G(s)H(s) (or) 180° - ϕ
ϕ = (∠ poles - ∠ zeros) of G(s)H(s)

θ_D = 180° - 90 – (π – tan^-1 (2))
= -90 + tan^-1(2) = -26.56

We need to find e_ss due to D(s).
So, R(s) = 0, the reduced block diagram will be

Transfer function for a given state space representation is, C[sI - A]^-1 B + D


Impulse response = L^-1[T/F] = 
At t = 1sec, impulse response = 1

V_avg = V_b T_on f = 100 × 6 × 10^-3 × 130 = 78 V
E = V_a – I_avg × R = 78 – (100 × 0.1) = 68 V
E = k_V I_avg ω

For quasi – square wave
2d = 0.5 × 180° = 90°
d = 45°
RMS value of fundamental voltage

Load impedance at third harmonic frequency is

RMS value of third harmonic

J₀ = K₀ = 1
J₁ = K₁ = Q₀
J₂ = K₂ = Q₀Q₁
Initial value of Q₀Q₁Q₂ = 010

The state just before the third clock pulse is the state at the end of second clock pulse.
At the end of second clock pulse
= Q₀Q₁Q₂ = 001

Pole = 6, conductor/slot = 8, flux = 0.05
Winding factor K_W = 0.96, speed = 1000 RPM

Number of conductor z = slot × conductor/slot = 90 × 8 = 720

E_ph = 4.44 K_c K_d f N = 4.44 × 0.96 × 0.05 × 50 × 120 = 1278.72 V

Back emf E_b = V – I_aR_a = 480 – 110 × 0.2 = 458 V
where Z = 860
For lap winding A = P = 6
Flux ϕ  = 0.05

Let the signal represents ramp function
r(t) = t u(t)
signal is sampled at t = T
s(n) = r(nT)
= nT u(nT) = nT u(n)
By applying z-transform

By applying Delta to star conversion as shown below:


Now the given circuit becomes,



Power supplied by 40 V source = 40 × 0.5 = 20 W

Maximum average power delivered to the load, when 

R = 40 Ω
X_L = jωL = j (5 × 10³) (8 × 10^-3) = 40 j Ω

Now the Thevenin circuit will be

|I_rms| = 1 A
Average power delivered = (1)² (20) = 20 W

Circuit under no-load condition

By voltage division

⇒ 5R₂ = 40 + R₂ ⇒ R₂ = 10 Ω
The circuit with load R_L,

Let the parallel combination of R₂ and R_L is R_eq.

By voltage division,

The Thevenin equivalent circuit for DC model is

Since B is very large, therefore I_B is small and can be ignored.

Force due to weight = Wg
Force due to electromagnet = 2F_e

Magnetic field in air gap,
B_g = μ₀H_g
Ampere turn, NI = H_g. x

Given that, μ_r = 60
N₁ = 1200
N₂ = 1000
d₁ = 5 cm ⇒ r₁ = 2.5 cm
l = 100 cm

Let x% be the percentage of winding which remains unprotected
The voltage of unprotected winding 

Reactive power absorbed by load
Q = P tan ϕ
= 1200 × tan (cos^-1 (0.6))
= 1600 kVAR
At unity power factor, reactive power requirement = 0
Rating of synchronous condenser = 1600 kVAR
Rating of alternator = 1200 kVA

 [Sum of residues at poles]
Poles: z = 0, 3
Both the poles are inside the |z| = 4
Residue at z = 0,

The given integral is in the form of M dx + N dy
By using Green’s theorem,

Given that, y
y(1) = 2 ⇒ y₀ = 2, x₀ = 1
h = 0.1
According to Euler’s method, y

First iteration:
y₀ = 2, x₀ = 1, h = 0.1
y
y

Second iteration:
y₁ = 2.22, x₁ = x₀ + h = 1 + 0.1 = 1.1, h = 0.1
y
y

⇒ (2 – λ) (4 – λ) – 3x = 0
⇒ λ² – 6λ +8 – 3x = 0

Roots of the above equation are Eigen values and let the Eigen values be λ₁ and λ₂
From the properties of Eigen values,

For the Eigen values to be positive, λ₁λ₂ > 0
⇒ 8 – 3x > 0
⇒ x < 8/3
For roots to be real, 
⇒ 36 – 32 + 12x > 0
⇒ x > -1/3
The range of for which Eigen values of matrix are real and positive is: 
Maximum value of x = 8/3 = 2.67

Particular integral is,