Gate Mock Test: Electrical Engineering(EE)- 15


65 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Gate Mock Test: Electrical Engineering(EE)- 15


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Attempt Gate Mock Test: Electrical Engineering(EE)- 15 | 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

The sentences given with blanks are to be filled with an appropriate word(s). Four alternatives are suggested for each question. For each question, choose the correct alternative and click the button corresponding to it.

They abandoned their comrades ______the wolves.

Solution: When something is left in between more than 2 things or persons, ‘among’ is used. There are several wolves in the sentence hence the correct answer is option D.
QUESTION: 2

Direction: Select the one which is different from the other three responses.

Solution:

49 - 33 = 16

62 - 46 = 16

83 - 67 = 16

70 - 55 = 15

All options have a difference of 16 except the D option.

Hence, option D is the answer.

QUESTION: 3

Amit rows a boat 9 kilometres in 2 hours down-stream and returns upstream in 6 hours. The speed of the boat (in kmph) is:

Solution:

Down-stream rate = 9/2 = 4.5 kmph

Upstream rate = 9/6 = 1.5 kmph

The speed of the boat = (4.5 – 1.5) kmph = 3 kmph

QUESTION: 4

Directions: In the following question, out of the four alternatives, select the word opposite in meaning to the given word.

Turgid

Solution:

Turgid = swollen and distended or congested; pompous

Bloated = inflated

Humble = plain, simple

Puffy = inflated

Tumescent = swollen

Hence, humble is the correct answer.

QUESTION: 5

Direction: Study the following information carefully and answer the given questions:

The following pie chart shows the percentage distribution of total number of Apples (Dry + Wet) sold in different days of a week:

The pie chart2 shows the percentage distribution of the total number of Apples (Wet) sold on different days of the week.

Find the ratio of Apples (Dry + Wet) sold on Tuesday to that of the Apples (Wet) sold on Thursday?

Solution:

Required ratio = 13% of 7000: 23% of 4500 = 182:207

QUESTION: 6

Directions: In each of the questions below, some statements are given followed by some conclusions. You must consider the statements to be true even if they seem to be at variance with commonly known facts. You must decide which of the following conclusions logically follows from the given statements. Give answer.

Statements:

Some jacket is shirt.

Some shirts are trouser.

No shoes are t-shirt.

All trousers are shoes.

All pants are t-shirt.

Conclusions:

I. All shirts being t-shirt is a possibility.

II. Some shoes are trouser.

III. Some jackets are t-shirt.

IV. Some shirts being t-shirt is a possibility.

Solution:

From the above diagram-

• Some shoes are trouser.

• Some shirts being t-shirt is a possibility.

Hence option B is correct.

QUESTION: 7

Direction: Which of the following is the MOST SIMILAR in meaning to the given word?

Remnant

Solution:

‘Remnant’ means ‘remainder’ which is the same as ‘residue’.

QUESTION: 8

Direction: Read the information carefully and give the answer of the following questions

Total number of children = 2000

If two-ninths of the children who play football are females, then the number of male football children is approximately what percent of the total number of children who play cricket?

Solution:

57.4% (If 2/9 th of children play football are female, then

male children = 1 - 2/9 = 7/9th of children play football;

Let’s take z% of male football children equal to cricket children, then

⇒ z% of cricket children = 7/9 th of football;

⇒ z% of (23% of 2000) = 7/9 th of (17% of 2000);

⇒ z% of 23 = 7/9 th of 17

⇒ z = 7× 17 × 100 = 57.4%;)

Short-cut : Z% of 23 = 7/9 of 17

(7x17x100)/ (23x9) =57.4%

Required percentage = (7/9)×17×100/ 23 = 57.4%

QUESTION: 9

Direction: Study the following information carefully and answer the given questions.

Seven people, P, Q, R, S, T, W and X sitting in a straight line facing north, not necessarily in the same order. R sits at one of the extreme ends of the line. T has as many people sitting on his right, as on his left. S sits third to the left of X. Q sits on the immediate left of W. Q does not sit at any of the extreme ends of the line.

If all the people are made to sit in alphabetical order from right to left, the positions of how many people will remain unchanged?

Solution:

From the given conditions, we can conclude:

After arranging in alphabetical order,

Hence, only Q′ s position will remain unchanged.

QUESTION: 10

A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in total from both as interest. The rate of interest per annum is.

Solution:

Let the rate % = R

According to the question,

= 2200

100R + 120R = 2200

220R = 2200

R = 10%

Hence required rate % = 10%

QUESTION: 11

A 10 KVA, 400 V, 3 phase, star connected synchronous generator has an armature resistance of 0.8 Ω per phase and synchronous reactance of 1.5 Ω per phase. The voltage regulation at a power factor 0.6 lagging is _______.

Solution:

Ra = 0.80Xs = 1.5Ω

Vph = 230.94V

Eg = 255.19V

Voltage Regulation =

=

V.R. = 10.50%

QUESTION: 12

How many essential prime implicants are there in the given k-map?

Solution:

AB and are essential prime implicants because only these two have at least single 1′ ' which can have only one pairing.

QUESTION: 13

The various resistance of a wheatstone bridge are shown in figure. The battery has an internal emf of 8 V with negligible internal resistance. If the sensitivity of the galvanometer is 12 mm/µA with an internal resistance of 150 Ω.

What will be the sensitivity of the bridge in terms of deflection per unit change in resistance.

Solution:

Resistance of unknown resistor required for balance,

Therefore, the deviation in unknown resistance from the balance condition,

ΔR = 1650 - 1600

= 50Ω

So now Thevenin source emf,

ε0 = 0.039V Now the internal resistance of bridge looking into terminal C & d−

= 481.95Ω

Hence the current through the galvanometer

then deflection of galvanometer

θ = Si. Ig

= 12 × 61.7mm

θ =740.4mm

Therefore, sensitivity of bridge

QUESTION: 14

The circuit shown below has C = 20 μF and L = 5 μH. Initial voltage across the capacitor is Vs = 230 V. The load current is constant at 300 A . Let ‘A’ be the conduction time period of the auxiliary thyristor and ‘B’ is the circuit turn off time of the main thyristor. Than A + B will be _________.

Solution:

The above circuit is class 'B' commutation.

So, the conduction time of auxiliary thyristor is =

A = 31.4 \musec

The peak value of resonant current,

Voltage across the main thyristor, when it gets turned off will be

Circuit turn off time for main thyristor is

So, A + B = 31.4 + 11.62

=

QUESTION: 15

In the circuit shown in figure the switch is closed at time (t=0). The rate of change of current

across inductor will be -

Solution:

At (t = 0+) as the circuit was unenergized at (t = 0- ) ⇒

Inductor L is open circuit (O.C)

Capacitor C is short circuit (S.C)

QUESTION: 16

In the figure shown below,

Assume |Vth| = 0.3 V, the transistor is operating in

Solution:

Since the device is NMOS,

Vth = 0.3 V

VS = 0.4 V

VD = 2 V

VG = 0 V

For cutoff region, V­GS < |Vth|

Since, VGS = –0.4 V

So, the device is in the cutoff region.

QUESTION: 17

If a finite length of wire having a total length QR = 80 m, a current of 10 A is flowing in it.

A point P from the origin point or reference point is away at a distance of 4 mm. If the point P is making an angle of 75° with the point Q in clockwise direction & 60° with the point R in anticlockwise direction. What will be the magnetic field density B, at point P.

Solution: As we know that due to a finite length of wire in which I current is flowing, at a point P where the point P makes angle θ1 with point θ in clockwise direction &θ2 with R in anticlockwise direction is calculated as-

B =

where the sign of angle is tve, is being taken, a is the distance from a reference point.

S0 B at point P

B =

= 2.5 x 10-4(0.966 + 0.866)

= 0.458 x 10-3Wb/m2

B = 0.458 m-Wb/m2

*Answer can only contain numeric values
QUESTION: 18

A chopper circuit shown in the figure has an input voltage of Vs = 5 V. Duty ratio is 0.6 . The load resistance is 10 Ω. The chopper is operating at 25 kHz switching frequency. If L = 150 μH and C = 220 μF. Find the ratio of peak value of capacitor voltage to peak value of inductor current?


Solution:

The given circuit is a boost regulator.

For the boost regulator,

When the switch is ON.

Peak value of inductor current is

The ripple voltage at capacitor is

Peak voltage at capacitor,

Required ratio is = =

= 3.56

*Answer can only contain numeric values
QUESTION: 19

Consider the circuit given in the figure below, if voltage VAB and VCD in the circuit are 200 V each, then the value of the capacitor in the circuit, C is approximately ________ (in μF).


Solution: since

Let assume angle between VAB and VBC is θ

So, taking magnitude both sides

By solving, θ = 120∘

Current through inductor hL=10A

since, we know that inductor current lags 90 by its voltage and for capacitor its current leads 90 by its voltage

C =

C =

QUESTION: 20

A uniform volume charge density ρv= 5 µC/m3 is present in the spherical shell of 0.9 < r < 1 m and ρv = 0 elsewhere. The charge present in the shell approximately is

Solution:

Q =

=

=

Q =

=

Q = 5.675µC

QUESTION: 21

A 15 MVA, 11 kV, alternator has positive, negative and zero sequence reactance are 0.3 p.u., 0.2 p.u. and 0.05 p.u. respectively. Find the value of resistance to be added in the generator neutral so that the fault current of LG fault does not exceed 1.5 times of rated line current.

Solution:

For L-G fault Current

Rn= 0.641 p.u.

R = 5.17 Ω

*Answer can only contain numeric values
QUESTION: 22

Consider a circuit shown in figure below, switch is closed at t = 0 in the circuit for V = 10 V, R = 5 Ω L = 3 H. Find the value of maximum current ‘I’ in the circuit. (in Amperes)


Solution:

Expression for current 11 for t≥0 is

Where τ=LR

Similarly, for current I2

Current 1 = 11 -I2

The value of current at this time is

I = 0.5

QUESTION: 23

A causal system has input x[n] = δ[n] + 14δ[n − 1] − 18δ[n − 2]

and output y[n] = δ[n] − 34δ[n − 1]

The impulse response of this system is

Solution:

*Answer can only contain numeric values
QUESTION: 24

Two identical unloaded generators are connected in parallel as shown below. Both the generators are having positive, negative and zero sequence impedance of j0.8 p.u., j0.2 p.u. and j0.3 p.u. respectively. If pre-fault voltage is 1 p.u. for a line to ground fault at the terminals of the generator, the fault current magnitude (in p.u.) is ___________. (Assume neutral impedance, Zn = j0.02 p.u.)


Solution: The positive sequence reactance,

Similarly, the equivalent negative sequence reactance is

For a single line to ground fault, fault current is

*Answer can only contain numeric values
QUESTION: 25

Given an assembly language program in 8085 microprocessor,

LXI B, 2384 H

LOOP: DCX B

MOV A, C

ORA B

JNZ LOO

HLT

If the system clock frequency is 2 MHz. The time required for complete execution of the program is _________ msec.


Solution:

Number of T-states required = Count x Number of T states inside loop + Number of T states outside loop Number of T states when jump is invalid.

[Here, count in decimal (2384)16=(9092)10]

=(9092 × 24 )+ 16 − 3 = 218221 T-states

Clock frequency is 2MHz

T=1 / 2 ×106 = 0.5μsec

Total execution time =218221 × 0.5 × 10 − 6 = 0.10911sec

Hence, the time required for execution of the program is 109.11 msec.

QUESTION: 26

Let x(t) be a signal that has a rational Laplace transform with exactly 2 poles located at S = –1 & S = –3. If g(t) equal to e2t x(t) & G(ω) converge. The conclusion about g(t) is

Solution:

One pole in RHS & another pole in LHS to make the system converging i.e. stable, the signal must be non-causal.

Hence Double-sided signal.

*Answer can only contain numeric values
QUESTION: 27

If the crystal frequency of 8085 microprocessor is 6MHz then total time (in millisecond) required to execute the program, before HLT instruction is _____msec.


Solution:

TTotal = ⅓ x 60 = 20ߎsec = 0.02msec

*Answer can only contain numeric values
QUESTION: 28

Consider a feedback system with characteristics equation .

The point of intersection of the asymptotes of the root locl with real axls on


Solution:

Characteristic equation is equation as: 1+G(s)H(s) = 0

On comparing this characteristics equation with the given equation

The point of intersection of the asymptotes = centroid(o)

*Answer can only contain numeric values
QUESTION: 29

In a current cumulated chopper, peak commutating current is twice that of the maximum possible load current. The source voltage is 250 V dc and main SCR take off time is 50 μs. Voltage is 250 V dc and main SCR taken off time is 50 μs. If maximum load current is 250 A, then peak capacitor voltage is………V.


Solution:

Peak Capacitor Voltage Vc = Vs + Io

Peak Current = Ip + 2Io = 2(250) = 500

= 2

= 0.5

VCP 250 + 250(0.5) = 375V

QUESTION: 30

Which of the following is incorrect:

Solution:

For inverting amplifier,

(Av)1= −RF / RL

if R1 > RF then it will be less than 1

for Non-Inverting amplifier,

(AV)NI = (1 + RF / RL)

when RF = 0, AV = 1

it can be one but cannot be less than one in any case.

So, option B is an incorrect statement and correct Answer.

QUESTION: 31

A single-phase transformer, with secondary voltage of 250 V, 50 Hz delivers power to R load of 15 Ω through a half wave-controlled rectifier circuit. For firing angle of 45°, the form factor is____

Solution:

Average value of output voltage

RMS value of output voltage,

Vorms =168.55V

Form factor, FF = = 1.75

QUESTION: 32

Time period of the function X(n) is ejm2 will be


Solution:

x(n) = ejm2 = cosπn2 + jsinπn2

x(n) = (-1)r2 + 0

x(n) = (-1)n2

x(o0 = (-1)02 = 1

x(1) = (-1)2 = -1

x(2) = (-1)22 = 1

x(3) = (-1)32 = -1

Hence time period will be 2

*Answer can only contain numeric values
QUESTION: 33

A 200 V, unsaturated shunt motor has an armature resistance (including brushes and interpolar winding) of 0.05 Ω and field resistance of 200 Ω.

The value of resistance in Ω to be added in the field circuits increases the speed from 1500 RPM to 1800 RPM. When the supply current is 300 A.


Solution:

As, armature current is nearly equal to supply current. The value of back emf will not change much

As we known,

is nearly constant

Renva = 240 - 200 = 40Ω

QUESTION: 34

Two resistors R1 and R2 are connected in series. The voltage across the series combination is V. A voltmeter V1 is used to measure the voltage across the resistor R1 and a voltmeter V2 is used to measure the voltage across the resistor R2.

The voltmeter V1 is ±X% and the voltmeter V2 is +y% accurate. The voltmeter V1 reads 10 V and V2 reads 5 V. The absolute error of two voltmeters are the same and absolute error in total voltage V is 0.2 V.

The values of X and Y (in %) respectively are:

Solution:

The voltmeter V1 is ±x% and voltmeter V2 is +y% accurate

V1 = 10V, V2 = 5

Absolute error,

∨1 = x100 × 10 = 0.1x

V2 = y100 × 5 = 0.05y

Absolute error of two voltmeter are same

0.1x = 0.05y

y = 2x −(i)

Total voltage, V = V1 + V2

Absolute error in total voltage,

V = 0.1 x + 0.05y

0.2 = 0.1 x + 0.05y

Put equation (i) in above expression

0.2 = 0.1 x + (0.05)2x

⇒x=1%, y=2%

*Answer can only contain numeric values
QUESTION: 35

Independent trials consisting of a rolling of a fair die are performed, the probability that 2 appears before 3 or 5 is __________ (Up to three decimal)


Solution:

(0.2, 0.4)

Probability =

=

=

= 3 / 8

= 0.375

*Answer can only contain numeric values
QUESTION: 36

In the circuit shown all the values are up on 100MV, 400KV base (L-L) If a (solid 3 phase vault) occurs at the cantro of the transmission line at an instant to = 9.735msec. What is the magnitude of initial DC offset current? (in kA) (Nominal system frequency f = 50Hz). The voltage in phase a is V = Vm sin ωt.


Solution:

Magnitude of DC offset current =

At switching instant Ioc =

= 23.57 pu

IDC= 3.402KA

QUESTION: 37

In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)

Solution:

The peak value of phase voltage

= Vmph = 132 √3 × 2 = 107.77KV

Active Recovery voltage

= K1K2Vmphsin⁡ϕ

K1 = 0.95

K2 = 1 (grounded fault)

cos⁡ϕ = 0.4 ⇒ ϕ = cos−1⁡(0.4)=66.42

Sin⁡ϕ = 0.916

Active recovery voltage

= 0.95 × 1 × 107.77 × 0.916

= 93.85KV

Restriking Voltage

= 93.85k(1 − cos⁡ωt)

RRRV = d / dt [93.85 × 103(1 − cos⁡ωt)]

RRRV =93.85 ×103 × ω. sinwt (RRRV) average = 2π(93.85 × 103 × ω)

=2/π × 93.85 × 103 × 2πf

= 4f x 93.85 x 103

= 4 x 16 x 93.85 x 106

= 4f x 93.85 x 10–3 KV/μsec

Average RRRV = 6 KV/μsec

*Answer can only contain numeric values
QUESTION: 38

The transformation ratio of an auto transformer for a 30 kW, 400, 3-ϕ induction motor is______. If the starting torque is to be 75% of full load torque. Assume the full load slip to the 5% and the short circuit current to be 6 times the full load current.


Solution:

As we know,

x2 = 0.41667

x = 0.645

*Answer can only contain numeric values
QUESTION: 39

In resonant pulse commutation, C = 30 μF and L = 5 μH. Initial voltage across capacitor = VS = VC = 240 V. If the constant load current of 240 A, then circuit then of time for main thyristor is______μsec.


Solution:

Circuit turn off time,

Where, Vab = Vscos⁡ωt

Vab is voltage across the main thyristor when it gets turned off.

Vab = 240cos24.09o

Vab = 219.09V

QUESTION: 40

Consider the circuit shown below:

The value of capacitance C required to make circuit power factor unity is:

Solution:

The equivalent impedance seeing from source side is

Yeq = (0.035 - j0.0136) + jwC

For unit power factor, the imaginary part should b zero

j(wC - 0.0136) = 0

QUESTION: 41

Let x (t) = rect (t−1/2) (where rect (x)=1 for −1/2 ≤ x ≤ 1/2 and zero otherwise). Then if

sinc⁡(x) = sin⁡(πx) / πx, the Fourier Transform of x(t) + x(−t) will be given by

Solution:

x (t) = rect (t-1/2)

y (t) = x (t) + x (-t) = rect (t-1/2) + rect (-t-1/2) = rect(t/2)

Fourier Transform of y (t) is

y(ω) = ∫−11I⋅e−jωdt

y(ω) = [e−jω/−jω]−11

y(ω) = e−e−jω/jω

y(ω) = 2/ω[e−e−jω/2j

y(ω) = 2/ωsin⁡ω

y(w) = −2sin⁡(n−ω/n)/(π−ω/n)

y(ω) = 2sin⁡(ω/n)

But none of the four options match this Answer The result can be modified as follows.

y(ω) = 2/ωsin⁡ω

y(ω) = 2/ω⋅2sin⁡(ω/2)⋅cos⁡(ω/2)

y(ω) = 2sin⁡(n−ω/2π)/(Π−ω/2π)⋅cos⁡(ω/2)

y(ω) = 2sinc⁡(ω/2π)⋅cos⁡(ω/2)

which matches option C.

QUESTION: 42

A 100 MVA 50 Hz two alternator operates at no load at 3000 rpm. A Load of 27 mw is suddenly applied to the machine and the steam values to the turbine commence to open after 0.6 sec due to time – log in the governor system. Calculate the frequency to which the generator voltage drops before the steam flow to increase to meet the steam flow to increase to meet the new load. Inertia constant H = 4.5 KW – sec / KVA.

Solution:

We know the swing equation

= sync honours speed

Integrate on both side

Wr = rotor speed

So ds/dt = -9.42t

Now at t = 0.6

Wr - ws= -5.652

Wr = ws - 5.652

= 308.348 rad/sec

f’ = 49.1 Hz

QUESTION: 43

In the given circuit, the silicon transistor has β = 75 and a collector voltage VC=9V.

Then the ratio of RB and RC is

Solution:

Correct answer is 105.1

Given β = 75

VC = 9 V

We redraw the given circuit as

From the circuit, we have

15 − RC(IC + IB) = VC

15−VC = RC(IC + IB)

RC(IC + IB) = 6V……….(1)

Applying Kirchoff's law across B-E Junction,

VC − RBIB − VBC = 0

or 9 − 0.7 = RBIB

or RBIB = 8.3......(2)

Also,IC = βIB = 75lB…….(3)

Solving equations (1),(2), and (3), we get

RC(75 + 1)IB = 6……….(4)

RB⋅IB = 8.3……….(5)

Hence, dividing equation (5) by (4), we obtain the desired ratio as

RB/RC × (1/76)= 8.3/6

RB/RC = {76 x 8.3} / 6 = 105.13

QUESTION: 44

The output characteristics for a BJT are given below.

Determine the hybrid parameter hfe, if VCEQ = 6V and ICEQ = 3.5mA

Solution:

Firstly, the given Q-point is plotted on the transfer characteristics.

Keeping VCE Constant, the change in ib is chosen along the straight line at VCEQ = 6V

Next, the corresponding change in ic by drawing horizontal lines intersecting at IB to the vertical axis.

Thus, hfe is computed as,

⇒ hfe =

⇒ hfe = = 200

QUESTION: 45

Find the resistance and inductance of the coil if the arms of an ac Maxwell Bridge are given as: AB and BC are non-reactive resistors of value 100 Ω each. Let DA is a standard variable reactor and value of L1 of resistance 32.7Ω and CD comprises a standard variable resistor R connected in series with a coil of unknown impedance. Balance was obtained with L1=47.8mH and R=1.36Ω?

Solution:

Figure shows Ac Bridge; in the figure we observe the products of

the resistances of opposite arms are equal

32.7 × 100 = (1.36 + R4)100

32.7 = 1.36 + R4

R4 = 32.7 − 1.36 = 31.34Ω

L1 × 100 = L4 × 100

L4 = L1 = 47.8 mH

QUESTION: 46

A 3 phase full converter charges a battery from a three phase supply of 230V, 50Hz. The battery emf is 200V and its internal resistance is 0.5Ω. On account of inductance connected with the battery charging current is constant at 20A. Compute the supply power factor.

_______ lag.

Solution:

We know that

√3vssscos⁡ϕ = volo…(i)

Battery terminal voltage Vo is

V0 = E + IOr

= 200 + 20 × 0.5

= 210V

IO = 2OA (given)

We know that supply current is question square wave of amplitude

Io = 20A and for 120 over every half cycle

Using (i)

√3 × 230 × 16.33cos⁡ϕ = 210 × 20

cos⁡ϕ = 0.646lag

QUESTION: 47

A non-salient pole, 3-phase, synchronous machine with negligible resistance is on an infinite bus. Its synchronizing power coefficient will be maximum when δ = _______ degrees.

Solution:

Synchronizing power coefficient at any,

This will be maximum when δ=0.

QUESTION: 48

The current coil in a wattmeter is connected in series with an ammeter & an inductive load. The ammeter reading is 5.5 A. A voltmeter & the voltage coil are connected across 200 Hz supply. The reading of voltmeter & wattmeter is 220V & 25 watt respectively. The inductance of the voltage circuit is 12 mH & its resistance R = 2400 Ω. If the voltage drop across the ammeter & the current coil are negligible, then what will be the approximate percentage error in the wattmeter reading.

Solution:

Inductive reactance of pressure coil =2π × 200 × 12 × 10−13 = 15Ω

Resistance of pressure coil =2400Ω

Phase angle of pressure coil

Since,

tan⁡β = 152 / 400 = 6.25 × 10−3

So, β = tan−1⁡(6.25 × 10−3)

=0.36 or β = 0.36

We know that for inductive load,

Reading of the wattmeter αcos⁡β.cos⁡(ϕ−β)

True power αcos⁡ϕ

Where cos ϕ load p.f.

Therefore,

True power

= cos⁡ϕ / cos⁡β.cos⁡(ϕ − β) × Re ading of wattmeter

Here true power = 12R

= (5.5)2R

Which is equal to = 12Zcos⁡ϕ

Impedance

Z = 220 / 5.5 = 40Ω

So, true power = (5.5)2 × 40cos⁡ϕ

Reading of wattmeter is given as = 25 watt

So,

(5.5)2 × 40 cos⁡ϕ = cos⁡ϕ / cos⁡β⋅cos⁡(ϕ − β) × 5

Or,cos⁡β⋅cos⁡(ϕ − β) = 25(5.5)2 × 40 = 0.0206

since β = 0.36

So, cos⁡(0.36)⋅cos⁡(ϕ − 0.36) = 0.0206

or cos⁡(ϕ − β) ≃ 0.0206

(ϕ−β) ≃ cos−1⁡(0.0206)

ϕ−β ≃ 88.81

so, ϕ ≃ 88.81 + 0.36 ≃ 89.17

therefore, percentage error

=tan⁡ϕ⋅tan⁡β × 100

=tan⁡(89.17)⋅tan⁡(0.36) × 100

=43.37%

QUESTION: 49

A synchronous motor is drawing 40 A from 440 V, 3-ϕ supply at unity p.f. with a field current of 0.9 A. The motor has synchronous reactance of 1.4 Ω. If mechanical load on motor remains the same. Find the % change in the field current which would result in 0.85 p.f. leading is _________? (Assuming linear magnetization)

Solution:

Since it is given that mechanical load is constant i.e. P= constant

V/cos⁡ϕ = constant

I Cos ϕ = constant

I1Cos⁡ϕ1 = I2Cos⁡ϕ2

40 × 1 = 12×0.85

I2 = 47.05A

When power factor is 0.85 leading motor takes the current 1 = 47.05

∠31.78∘

In that condition the generated emf is

Ef = 254.03∠0− (47.05 < 31.78)(1.4="" />< />)

Ef = 254.03 − 65.87 < />

Ef = 294.09∨

Since given that linear magnetization

Ef ∝ If

Ef2 / Ef1 = If2 / If1

294.09 / 260.12 = If2 / 0.9

|lf2| = 1.017A

% change in the field current

= 1.017 − 0.9 / 0.9 × 100

= 13% increases

QUESTION: 50

Two identical 1 − ϕ transformer of 15kVA, gave the following results, when tested by back to back method.

W1 = 1000 watt (Primary)

W2 = 1500watt( Secondary )

Calculate the percentage efficiency of the transformer at 1/4th load 0.8 P.F. lag

Solution:

In the back to back test the primary wattmeter gives the iron loss.

Each transformer has iron loss

=W1 / 2 = 1000 / 2 = 500W and secondary wattmeter reading gives the full load copper loss.

Each transformer has full load copper loss

=W2 / 2 = 1500 / 2 = 750W

Full load loss at 1/4th of load is

=

Efficiency η=

η=0.8458×100%

η=84.58%

QUESTION: 51

A long transmission line has R = 6 Ω, G = 240 mS L = 2 H and C = 80 mF is operating at 50 Hz, 400 kV. If this line is terminated with 5 Ω, the voltage across the load as ______ kV.

Solution:

Zc =

=

= 5Ω

Zc = ZL

Hence, VL = VS = 400 kV

QUESTION: 52

The value of is Where [x] is the greatest integer function.

Solution:

[x] = 0; 0 ≤ x ≤ 1

[x] = 1; 1 ≤ x ≤ 2

[x] = 2; 2 ≤ x ≤ 3

So.

= 3 /2 + 5

= 6.5

QUESTION: 53

The Cauchy’s mean value of the functions ex & e−x between (0, 1) is

Solution:

f(x) = ex, g(x) = e−x

a = 0, b = 1

Cauchy's mean value a < c="" />< b="" />

f′(x) = ex & g(x) = −e−x

⇒ −e2c =

⇒ −e2c = −e

⇒ 2c = 1

⇒ c = 0.5

QUESTION: 54

The daily variation of load on a 100 KVA transformer is as follows:

8:00 AM to 1 PM – 65 kW, 45 KVAR

1 PM to 6 PM – 80 kW, 50 KVAR

6 PM to 1 AM – 30 kW, 30 KVAR

1 AM to 8 AM – No load

This transformer has no load area loss of 270 W and a full load ohmic loss of 1200 W. The all-day efficiency of the transformer is

Solution:

No load core losses are constant throughout 24 hrs. core losses = 270 × 24 = 6.48 kW

8 AM to 1 PM

kW output = 65 kW, kWh output = 325 kWh

Cu loss is proportional to S2

Culoss = 749.86 x 5 = 3.75kWh

1PM to 6PM

KWh output = x 5 = 400kWh

Culoss = 1068 x 5 = 5.34kWh

Culoss = 216 x 7 = 1.512kWh

= 97.96%

QUESTION: 55

The output of a continuous-time, linear time-invariant system is represented by T{x(t)} where x(t) is the input signal. A signal z(t) is called eigen-signal of the system T, when T{z(t)} = Y z(t), where Y is a complex number, in general, and is called an eigen-value of T. Assume the impulse response of the system T is real and even. Then which of the following statements is TRUE?

Solution:

Given impulse response is real and even Thus H(jω0) will also be real and even Therefore H(jω0) = H(−jω0)

If cos⁡(t) is input then

Output will be =

If sin⁡(t) is input then

Output will be

So sin⁡(t) and cos⁡(t) are Eigen signal with same Eigen values.

*Answer can only contain numeric values
QUESTION: 56

The open-circuit and short-circuit characteristic of a single-phase synchronous generator are shown on the same graph below.

A field current of 60 A produces the rated voltage. If the effective armature resistance of the generator is 0.3 Ω, determine the synchronous reactance of the machine [in Ohm].


Solution:

From the given figure, at the field current, If = 60A

Open-circuit voltage, VOC = 500V

Short-circuit armature, ISC = 275A

Hence, the synchronous impedance,

Synchronous reactance,

QUESTION: 57

Find the inverse Z- transform of X(z)

Solution:

QUESTION: 58

A three phase to ground fault occurs at " ' in the network show in figure. Generator G1 : E1 = 1 pu. ,x1 = x2 = 0.8 pu. x0 = 0.5 pu.

Generator G2 : E2 = 1pu,x1 = x2 = 0.6pu

X0 = 0.3pu

Transformer T1 : x1 = x2 = X0 = 0.22pu

T2 : x1 = x2 = x0 = 0.15pu

Each line:

x1 = x2 = x0 = 0.67pu, x2 = 0.02pu

The current flowing through ground is :

Solution:

Three phase to ground fault occurs so we require only positive sequence network

Fault current

QUESTION: 59

For the given system, determine the overall response Y(s) of the system corresponding to both inputs.

Solution:

When multiple inputs are present in a linear system, output is computed by treating each input independently. The overall response of the system is obtained by adding the outputs corresponding to each input.

Assuming x2 (s)= o

The block diagram can be reduced to,

Reducing the inner feedback loop,

The block diagram is reduced to,

Thus,

= output due to X1(s) acting alone

Assuming X1(s)=0

The block diagram is reduced to,

thus. Yx2(s)= output due to X2(s) acting alone

Hence, the response of the system,

QUESTION: 60

Find the inverse fourier transform of X (t) =

Solution:

Given

X(w) =

We know that, X(w) =

=

=

QUESTION: 61

In sinusoidal pulse width modulation, the lowest-order harmonic in the modulated signal is the 15th harmonic. Determine the number of pulses per half cycle of the output signal.


Solution:

In sinusoidal PWM, all the harmonics less than or equal to (2p – 1) are eliminated, where p is the number of pulses per half cycle. Since the lowest-order harmonic is 15th, the harmonics till 13th are eliminated.

Thus,

2p − 1 = 13

⇒ 2p = 13 + 1 = 14

⇒ p = 14/2 = 7

Hence, there are 7 pulses per half cycle in the output signal.

QUESTION: 62

A 300V DC shunt motor is drawing a full load current of 45A. The armature resistance and shunt field resistance are 0.6Ω and 160Ω respectively. If the torque remains same throughout, by how much should the flux be reduced to increase the speed by 40%?

Solution:

T ∝ ∅Ia

since torque is constant and flux and speed are changing. SOt

1Ia1 =∅ 2Ia2

Ia2 = ∅1 / ∅2Ia1

Let

1 / ∅2 = x, then

Ia2 = 45x as Ia1 = 45A

Eb1 = 300 − (0.6 × 45) = 273V

Eb2 = 300 − (0.6 × 45x) = (300 − 27x)V

It is given that N2 = 1.4N1 since speed is increased by 40% Now, we know Eb ∝ N∅

⇒ Eb1 / Eb2 = N1∅1 / N22

382.2 = 300x − 27x2=0

27x2 − 300x + 382.2=0

x = 9.64,1.4685

For x = 9.64, the increase in speed criteria is not satisfied. Percentage change in flux

= ∅1−∅2 / ∅1

= 1 − ∅2 / ∅1 = 1 − 11.4685 = 31.9%

QUESTION: 63

A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.

Solution:

GH = 30 × 10 = 300MJ

Pm = 26800hp = 26800 × 0.746 = 20,000KW

Pe = 16000KW

Pa = Pm − Pe = 20,000 − 16,000 = 4000KW

Pa = Md2δ/dt2

M = GH/180f = 300/(180 × 50) = 1/30

4 = 1/30(d2δ/d1)

d2δ/dt2 = 120%sec2 = 120 × π/180 = 2.09rad/sec2

Multiply by 2(dδ/dt). dt on both sides

Integrating the above equation

Further integration

constant for 10 cycle which means 10 cycles = 0.2 sec δ = 1.045(0.2)2 = 0.0418rad = 1.8

QUESTION: 64

Find the value of average output voltage if a 230V, 50Hz, single-pulse SCR is feeding a RL load with α = 40 ° and β = 210 °?

Solution:

Vo = Vm/2π x (cosα-cosβ)

Where Vm = √2Vs

Vm = 84 V

QUESTION: 65

Consider a circuit as shown in fig below

Equivalent resistance across the terminals A & B is_____Ω.

Solution:

⇒ Req = 1 + ¾

⇒ 7/4 Ω

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