# Gate Mock Test: Electrical Engineering(EE)- 14

## 65 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Gate Mock Test: Electrical Engineering(EE)- 14

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Attempt Gate Mock Test: Electrical Engineering(EE)- 14 | 65 questions in 180 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE Electrical Engineering (EE) 2023 Mock Test Series for GATE Exam | Download free PDF with solutions
QUESTION: 1

### Given 4 words labelled as (P), (Q), (R) and (S). One of the pairs given below has words which have similar meaning or have opposite meaning. Identify the correct option.(P) instigate (Q) enquire(R) construe (S) interpret

Solution: Construe and interpret have identical meaning (a word or action).

Thus, option C is the correct answer.

QUESTION: 2

### A job can be completed by 12 men in 12 days. How many extra days will be needed to complete the remaining job, if 6 men leave after working for 6 days?

Solution: work done by one man in 1 day

=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

QUESTION: 3

### Which of the following phrases should be placed in the blank spaces so as to make a grammatically correct and meaningful sentence.Even though the institute had lot of space.______

Solution: Answer b would be a grammatically correct sentence.
QUESTION: 4

In the question, there are 4 statements followed by 4 conclusions numbered as (a), (b), (c) and (d), assume the given statements to be true even if they are at variance with commonly known facts, identify the conclusion which following from the given 3 statements

S1: Some guitars are posters.

S2: All posters are doors.

S3: Some doors are tablets.

S4: All tablets are books

Conclusions:

C1: Some doors are guitars.

C2: Some books are posters.

C3: Some tablets are guitars.

Solution: S1 is ‘particular affirmative’ type of statement whereas S2 is a universal affirmative. S1 & S2 combined lead to ‘particular affirmative’ i.e. some guitars are doors ‘and’ some doors are guitars’ i.e. C3 is not possible. C2 is also NOT possible since no conclusion is possible regarding books and posters; hence only C1 following leading to option (a).
QUESTION: 5

Choose one word out of the given option to replace the phrase ‘person who insists on adherence to formal rules of the literal meaning’.

Solution: Pedant is a person who strictly goes by the rule of the book of the literal meaning of the written text.

Thus, option A is the correct answer.

QUESTION: 6

A shopkeeper gives a discount on the marked price based on the quantity bought by a customer. Raja bought 10 pieces of a lunch box and was given a 10% discount by the shopkeeper. When Radhika bought 20 pieces of the same lunch box, she was given a 15% discount by the shopkeeper. In both the cases, the net amount of profit for the shopkeeper was identical. Which of the following is the ratio of marked price to the cost price for the lunch box?

Solution:

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

QUESTION: 7

It would take one machine 4 hours to complete a production order and another machine 2 hours to complete the same order. If both machines work simultaneously at their respective constant rates, the time taken to complete the same order is __________ hours.

Solution:

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

QUESTION: 8

Which of the options given below should replace the part of the sentence written in BOLD so as to get a meaningful and grammatically correct sentence?

We had not only helped them with money but also with new machinery and raw material.

Solution:

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

QUESTION: 9

A fair die is thrown three times and the sum of three numbers is found to be 16. Find the probability that 5 appears on the third throw.

Solution:

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 3rd throw = 2

p = 2 / 6 = 1 / 3

QUESTION: 10

A work of classics is split up in 3 volumes-each volume having equal number of pages. It is also known that the page numbers are running across the 3 volumes. If the sum of the first page number of the 3 volumes is 1473, the number printed on the first page of volume 3 is ________.

Solution:

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

QUESTION: 11

Find the value of the resistance and the inductance of the branch CD if the balance is obtained under these conditions. if the arms of an ac. maxwell bridge are given as: AB is a non-inductive resistance of 1000 Ω in parallel with a capacitor of capacitance 0.5 μF, BC is a non inductive resistance of 600Ω and CD is an inductive impedance which is unknown and DA is a non-inductive resistance of 400 Ω?

Solution: The balanced condition is

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

QUESTION: 12

A single phase semi converter is operated from 120V, 50Hz ac supply. The load current with an average value IDC is continuous and ripple free firing angle α=π/6. Determine the harmonic factor of input current?

Solution: Supply rms current

= 0.91Idc

Now the rms value of supply fundamental component of input current

Harmonic factor (HF) of input current

=

=

= 0.30

QUESTION: 13

A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added is

Solution:

⊤ ∝ la2 ⇒ T = kla2

For const torque armature I must be const.

Eb = 200 − 20 × (0.6 + 0.4) = 180V

la = 20A for 1000rpm. Ea = kϕw

180Ea = 1500 / 1000

Ea = 120V = 200 − 20(0.4 + 0.6 + Rext)

⇒ Rext = 200 − 120 / 20 −1 = 3Ω

QUESTION: 14

The value of impedance of short transmission line having supply of 6.5 kV with the rating of 10 kVA is (1+j5) p.u. What is the impedance at the new base of 13 kV and 30kVA?

Solution: As we know that by the formula that

So

= 0.75 + j3.75

QUESTION: 15

A Signal x(t) has Fourier transform x(ω) phase and magnitude of x(ω) are shown below

Determine x(t) at t = 5sec

Solution:

=

x(5) = 0.5

QUESTION: 16

A 3-ɸ bridge converter is provided with line-line voltage of 400 V. A load which is resistive in nature of value 100 ohm takes 400 W of power from the converter, the input power factor will be

Solution:

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 × VSIS

= √3 × 400 × 1.15

= 796.72vA

∴ Input apparent power = Output power

796.72cosθ = 400

cos⁡θ = 0.5 lagging

QUESTION: 17

The z-parameters of the network shown in figure are given by

Solution:

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,

QUESTION: 18

The admittance parameter matrix [y] is

Solution:

(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] =

QUESTION: 19

Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses such as depression and schizophrenia, and consequently, that doctors will be able to eradicate these diseases through early identification and gene therapy. On which of the following assumptions does the statement above rely?

Solution:
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

QUESTION: 20

Determine the maximum drain current in mA for the JFET in the given network, if VGS = −3V and Vp = −8V

Solution:

Considering input loop, since, IG = 0A and IS = ID

VGS = −IDRS

Since,

Substituting values,

⇒ 2.5mA =

⇒ 2.5mA=IDSS(2564)

⇒ IDSS = 64/25 × 2.5mA

⇒ IDSS = 6.4mA

QUESTION: 21

An ac voltmeter uses half wave rectifier and the basic meter with full scale deflection current of 1 mA and the meter resistance of 200 Ω. Calculate the multiplier resistance for a 10 V r.m.s range on the voltmeter.

Solution:

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

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

Now Ep = √2 Erms = 14.14V

Eav = 0.6 Ep = 8.99 ≈ 9V

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

Edc = 0.45Erms

Rs = Edc/ldc − Rm = 0.45Erms/ldc − Rm = 0.45×10/1×10−3 − 200

Rs = 4.3KΩ

QUESTION: 22

Let f(x) = x(x - 1)(x - 2) be defined in [0,0.5]. Then the value of c of the mean value theorem is

Solution:

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)

QUESTION: 23

For the flip flop configuration shown below, find the operation carried out by this configuration.

Solution:

Stages

It is a gray counter.

QUESTION: 24

If Laplace transform of a function is given by

Find the value of f(0+) and f(∞) respectively.

Solution:

Initial value = 2

Final value = 6/8 = ¾

QUESTION: 25

A 70 MW steam station uses coal of calorific value 5000 Kcal/kg. Thermal efficiency of the station is 40% and electrical efficiency 70 % . calculate the coal consumption per hour when the station is delivering its fuel rated output is ___x 103 kg?

Solution:

Given data

Overall efficiency of the power station is n overall =ηthermal ×nelectrical

= 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

QUESTION: 26

The value of is ____

Solution:

=

Putting

= -0.94

QUESTION: 27

Consider two real valued signals, x(t) band-limited to [−500 Hz, 500Hz] and y(t) bandlimited to [−1kHz, 1kHz]. For z (t) = x(t). y(t), the Nyquist sampling frequency (in kHz) is __________

Solution: x(t ) is band limited to [−500Hz, 500Hz] y(t )is band limited to [−1000Hz, 1000Hz] z(t ) = x (t ).y(t )

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.

QUESTION: 28

Keeping the kW demand constant if the load power factor increases the kVA demand –

Solution:

kW = kVA cosϕ

Kva ∝ 1/cos⁡ϕ

If the power factor increases the kVA demand will decrease.

QUESTION: 29

A delayed unit step function as

Its Laplace Transform is

Solution:

=

=

QUESTION: 30

Find the inverse Laplace transform of the function F(s) =

Solution:

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)

QUESTION: 31

A discrete time signal with input as x[n] has impulse response h[n]=kδ[n] where k is constant. The output y[n] of the system is given as:

Solution:

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)].

QUESTION: 32

The Potential (scalar) distribution in free space is given as V=10y4+20x3. if ε0 is permittivity of free space then the charge density ρ at the point (2,0) is

Solution:

Here, we have to use Poisson's Equation,

, because

20 × 3 × 2x + 10 × 4 × 3y2 = −ρ/ε

At point (2,0)f

ρ = −240ε0

QUESTION: 33

A 4-pole, 50Hz, 100MVA turbo generator has a moment of inertia 9.5 x 103 kg-m2. What is the kinetic energy stored in the machine?

Solution:

N = = 1500rpm

I = 9.5 x 103 kg-m2

K.E. = = 117.2MJ

QUESTION: 34

Determine the reciprocity and symmetry of the given two-port network.

Solution:

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.

QUESTION: 35

If ω is the frequency of the 3-phase source supplying to a 3-phase full-wave bridge rectifier, what is the fundamental frequency of the output voltage?

Solution: In a 3-phase full-wave bridge rectifier, there are six combinations of line-to-line voltages, taking 2 phases at a time. If the period of the source is 360 °, then the transition will take place every 360/6 = 60 °. Hence, there will be six transitions that will occur for each period of the source voltage.

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

QUESTION: 36

Find the complete solution of the state equation given below

Given the initial condition as

Solution:

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

We know that,

Finding

=

=

State transition matrix

=

Now,

And

=

=

=

Thus,

=

QUESTION: 37

In a 3-phase line operating at 50 Hz, the conductors, each of 0.96 cm diameter, are arranged as given below.

Determine the inductance of the line.

Solution:

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

QUESTION: 38

The divide by N counter is shown below. If initially Q0 = 0, Q1 = 1, Q2 = 0 the value of N is _______.

Solution:

State is repeating after 5 clock pulses. N = 5

*Answer can only contain numeric values
QUESTION: 39

Find the difference between rank of A and columns of null space of A?

A =

Solution:

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

QUESTION: 40

If an electric field is given as:

And,

Determine the work (in nJ) that is to be done in moving a 3μC charge along this path if the path is located at P(-0.3, -4, 0.6).

Solution:

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

QUESTION: 41

DC shunt motor is coupled to the identical DC shunt generator. The field of the generator is also connected to the same supply source as the motor. The armature resistance is 0.02 pu and the mechanical losses are 0.05 pu. Armature reactions can be neglected. The armature of the generator is connected to a load resistance RL. With rated voltage across the motor, the load resistance across the generator is adjusted to obtain rated armature current in both motor and generator. The pu value of RL is

Solution:

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

Eb = V − ImRa = 1 − 1 × 0.02 = 0.98pu

Mechanical output power of motor

= EbIm -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

EgIg = 0.93 − 0.05 = 0.88pu ⇒ Eg = 0.88puTerminal voltage of generator

= Eg − lgRa = 0.88 − 0.02 = 0.86pu

RL= V9/I9 = 0.86/1 = 0.86pu

QUESTION: 42

Consider a 3-bit number A and 2 bit number B are given to a multiplier. The output of the multiplier is realized using AND gate and one bit full adders. If minimum number of AND gates required are X and one bit full adders required are Y, then X + Y =

Solution:

Number of AND gates required X = 6

Number of one bit full adders required Y = 3

X + Y = 6 + 3 = 9

X1, = X2

QUESTION: 43

If z = sinh⁡ ucosv + icoshusinv then for what values of z, the function w = f(z) = u + iv is not analytic for

Solution:

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

QUESTION: 44

In the below parallel adder, A4 A3 A2 A1 is a BCD number and B4 B3 B2 B1 = 0011, then the circuit acts as

Solution:

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.

QUESTION: 45

Let = αβγ = 1,α,β,γ ∈ R and X = Then MX=0 has infinitely many solutions if trace (M) is ……………………..

Solution:

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

QUESTION: 46

Compute the voltage gain for the following circuit with input frequency 2.5 KHz. [Mention the nearest Integer value in dB]

Solution:

= 3.9 ≈ 4db

QUESTION: 47

In the half bridge rectifier as shown below, the peak current( in A) passing through the diode is ________ (If the supply input is given as Vs = 200Sin(100πt))

Solution:

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

So, maximum value of diode current is

(tD)max = 2.09A

QUESTION: 48

For a given connected network and for the fixed tree, fundamental loop matrix is given by

The cut set matrix corresponding to the same tree is

Solution:

Relationship between tie-set matrix and cut-set matrix

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

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

Where [Ql ] = - [BT ]T

Relationship between tie-set matrix and cut-set

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

[Q]= tie-set matrix =[Q1:I]

Where [Ql ] = - [BT ]T

=

[Q] =

QUESTION: 49

The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current IR.

Solution:

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

we have

VRY = 400∠0⁰V

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

QUESTION: 50

Given . Find the value of ∮ on the closed path shown below.

Solution:

From Stokes theorem

=

=

=

=

=

QUESTION: 51

The bus admittance matrix for a power system network is given by,

A new transmission line which is represented below is connected between bus 1 and 3.

The modified bus admittance matrix is

Solution:

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

Similarly,

QUESTION: 52

A six-pole, 3−ϕ,50 Hz induction motor has a rotor resistance and rotor reactance of 0.02Ω and 0.13Ω respectively at standstill. The external resistance to be added in the rotor circuit to get 75% of maximum torque at starting is

Solution:

sm =

sm =

sm = 0.453

sm = 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

Rext = 0.058 x 0.02 = 0.038 Ω

QUESTION: 53

Consider a discrete signal,

The value of will be

Solution:

Z-transform of signal will be

Putting z = 1

Z transform of signal x(n) will be:

Putting z = 1

= 2.75

QUESTION: 54

The integral , where ′D′ denotes the disc x2 + y2 ≤ 4, evaluate to

Solution:

Let x = rcos⁡θ & y = rsin⁡θ

dxdy = rdrdθ

=

=

=

= 20

QUESTION: 55

Consider the discrete time signals h[n] and its shifted form h[n−k].

The value of &

Solution:

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)

So,A = n − 9 and B = n + 3

S0,A − B= (n − 9) − (n + 3)

= −12

QUESTION: 56

A unity negative feedback system is described by the following state model

The steady-state error of the system due to step input of strength ' 12′ is

Solution:

The transfer function is given by =

e88 = input - output

e39 = 12 - 12 = 0

QUESTION: 57

Consider the circuit shown in figure below:

Find the phase angle of current ‘I’ with respect to voltage V2.

Solution:

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.

QUESTION: 58

Find Vx for which maximum power is transferred to load.

Solution:

Apply Thevenin’s theorem in load side

To supply maximum power,

VAB = 10/2 = 5V

By KCL at node A

Vx = 9.5 Volts

*Answer can only contain numeric values
QUESTION: 59

There are 3 fair coins and 1 false coin with tails on both sides. A coin is chosen at random and tossed 4 times. If ‘tails’ occurs all 4 times, then the probability that the false coin has been chosen for tossing is ___.

Solution:

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

QUESTION: 60

Consider an asymptotic Bode plot of a minimum phase linear system as shown in the figure below. The transfer function for the above system will be

Solution:

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

QUESTION: 61

The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is _______

Solution:

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

Vrms = Vm/(√2)

Vrms = 100/(1.414)

= 70.72v

QUESTION: 62

Find the value of Ix in the given circuit.

Solution:

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

Ix = 310

Apply KVL in the loop

= 2.5√2∠105°A

QUESTION: 63

Consider the signal . The time-period of the signal will be:

Solution:

= 12

= 18

Condition for periodic function:

N1/N2 = 12/18 is rational number

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

N = 36

QUESTION: 64

If the maximum phase provided by the compensator is 30° and this is achieved at √6 rad/sec. The transfer function of the compensator is __________.

Solution:

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

QUESTION: 65

Find the value of A and B for signal, g(t) = Ay(Bt), such that y(t) = x(t) × h(t) and g(t) = x(3t) × h(3t)

Solution:

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

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