Gate Mock Test: Electrical Engineering(EE)- 12


65 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Gate Mock Test: Electrical Engineering(EE)- 12


Description
Attempt Gate Mock Test: Electrical Engineering(EE)- 12 | 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

Select the most appropriate synonym of the given word.

Solicit

Solution: The meaning of 'solicit' is to ask for or try to obtain something from someone. The word “sanction,” which means official permission or approval for action is similar to this.

Sanction means official permission or approval for action.

Scum means a layer of dirt or froth on the surface of a liquid.

Disguise means to give someone or oneself a different appearance in order to conceal one's identity.

Thus, option A is the correct answer.

QUESTION: 2

Select the most appropriate option to fill in the blank.

Although his brother is blind, he is very fast _________ calculations.

Solution: Let us understand the meaning of the given words :-

"In" = used for expressing the situation of something that is or appears to be enclosed or surrounded by something else.

"With" = accompanied by another person or thing.

"About" = concerning.

"At" = used to show the activity in which someone's ability is being judged.

Since doing fast calculations is an ability, so the preposition "at" is used here.

Hence, option C is the correct answer.

QUESTION: 3

Two trains one 160 m and the other 140 m long are running in opposite directions on parallel rails, the first at 77 km an hour and the other at 67 km an hour. How long will they take to cross each other?

Solution: If two trains be moving in opposite directions at rates wand vkm/h respectively, then their relative speed = (u+ v) km/h

Further, if their lengths are x and y km.

Then, time is taken to cross each other =

Here, total length = 160 + 140 = 300 m

Relative speed = (77 + 67) km/h = 144 km/h

QUESTION: 4

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

Conjecture

Solution: Conjecture (noun) means a guess or idea which is not based on definite knowledge. So, option A is the correct answer.
QUESTION: 5

In the following question, a sentence is given with blanks to be filled in with an appropriate word(s). Select the correct alternative out of the four and indicate it by selecting the appropriate option.

He is _____ rich, yet he _____ about high taxes.

Solution: The use of "yet" indicates that the sentence talks about two contradictory things. Now, let's put the words given in options in the blank fields and see if they make a proper sentence

Option A: Obscenely means morally wrong, often describing something that is wrong because it is too large. It fits in the first blank as the sentence means to say that a person is very much rich. The verb "whine" means to make a sound similar to such a cry. It also fits in the second blank and gives a meaning to the sentence. The complete sentence means to say that he is very rich but still he does not want to pay taxes and cry about it.

Option B: The use of adjective "very" seems correct in the first blank field. However, the second word "careless" makes the option an incorrect choice. It is because the sentence requires a verb while "careless" is an adjective.

Option C: This option is also incorrect as both the words does not give any meaning to the sentence and are grammatically incorrect too.

Option D: Although, both the words fits in the blanks grammatically, they fail to convey the correct meaning in the sentence.

Thus, option A is the correct answer.

QUESTION: 6

If a landlord sell a portion of land for Rs. 40000, one-fourth of the outlay was gained. What should be the selling price in order to have loss of 25%?

Solution: Given that, if a landlord sell a portion of land for Rs. 40000, one-fourth of the outlay was gained. We know that, CP =

Now, SP in order to have loss of 25% =

= Rs. 24000

QUESTION: 7

In the following question, a sentence is given with a blank to be filled in with an appropriate word. Select the correct alternative out of the four and indicate it by selecting the appropriate option.

The little boy ran ____ fast that he was ____ for breath.

Solution: Option B has the correct fillers. Running too fast is less likely to result in a fight for breath but rather a gasp for breath.

Hence, ‘so, gasping’ is the most suitable response.

QUESTION: 8

The product of two numbers is 2028 and their H.C.F is 13. The numbers of such pair is:

Solution:

Let the numbers be 13a and 13b.

Then, 13a × 13b = 2028

ab = 2028/169 = 12

Now, the co-primes with product 12 are (1, 12) and (3, 4).

So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4).

Clearly, there are 2 such pairs.

Hence, option B is correct.

QUESTION: 9

Numbers of boys and girls are 'x' and 'y' respectively; ages of one girl and one boy are 'a' years and 'b' years respectively. The average age (in years) of all the boys and girls collectively is-

Solution:

According to the question: Numbers of boys and girls are 'x' and 'y' respectively; ages of one boy and one girl are 'b' years and 'a' years respectively.

Average = sum of observations/no of observations

QUESTION: 10

Directions: Read the following information carefully and answer the questions that follow:

The bar graph given below shows the percentage distribution of the total expenditures of a company under various expense heads during 2013.

Percentage Distribution of Total Expenditure of a Company

The total amount of expenditure of the company is how many times of the expenditure on Advertisements?

Solution:

Let the total Expenditure be Rs. X Then, the expenditure on Advertisements = Rs. (15% of x)

Ratio of the total exp excitiore to the expenditure on Advertisements

Hence the total expenditure is 20/3 times the expenditure on Advertisement.

QUESTION: 11

Given the feedback as unity and open loop transfer function as

For k=1, the gain margin is 18db. When the gain margin is 10dB,k is

Solution: Open laap transfer function of given unity feedback system is G ( s ) =

For k = 1

Gain margin = 18 ab

For k =?

Gain margin = 10ab

We knaw Gm = 20 log 10

QUESTION: 12

A 50 Hz, 8 pole generator with H = 6 MJ/MVA delivers 1 p.u. to an infinite bus has and supply 45% of peak power capacity as shown. A 3 – Φ fault occur at the point F. find the critical clearing time (in seconds)?

Solution: Given that Pm sin O0=0.45Pm=1

0.45Pm = 1

Applying equal area criteria

Integrating on both sides

Again integrating,

QUESTION: 13

Determine the input and output impedances of the given network.

Solution: Neglecting ro , the ac equivalent circuit for the network is drawn as,

zb=βre+(β+1)RE

⇒Zb=150×3.8+151×1×103=151.57kΩ

The input impedance of the network is given as,

Similarly, the output impedance of the network is given as, Zo = Rc = 3.5 Ω

QUESTION: 14

Find the Laplacian of the scalar field

V = 10r sin2 θ cosø

Solution:

QUESTION: 15

lf

And Laplace transform of f(t) = F (s) = s~3/2

Then Laplace transform of g(t) if g(t) =

Solution:

QUESTION: 16

A 50 Hz, 4 pole Turbogenerator rated at 110 MVA, 112 KV has an inertia consultant of 10 MJ/MVA. Mechanical input is suddenly rise to 100 MW from 70 MW then the rotor acceleration is

Solution: Since we know that swing equation

M = 10M/MM/AW5 = 2лf = 2 x lEo x 50 elect degree /sec

Rating of = S = 110MVAPmech = 100mw

Machine P = 70mw

QUESTION: 17

A 11 kV, 3-ϕ core 1 km long underground cable has the capacitance measured between any two cores with the third core connected to the sheath is 0.45 μF/km. The reactive delivered by the 3-ϕ cable for 1 km length is __________ kVAr/km.

Solution: C/ph = 2 × 0.45 μF/km

= 0.9 μF/km

Qc (3-ϕ) = 3Vph2 2πf C/ph

QUESTION: 18

A-3-ϕ bridge inverter delivers power to a resistive load from a 450 V dc source. For a star connected load of 15 Ω per phase, the rms value of thyristor current is__________(Assume 180° CM)

Solution: In 180° CM, waveform of thyristor current is

QUESTION: 19

A 100kVA, 415V, Y-connected synchronous machine generates rated open-circuit voltage of 415V at a field current of 15A. The short circuit armature current at a field current of 10A is equal to its rated armature current. The saturated synchronous impedance in p.u., if armature resistance is 0.1pu is

Solution:

*Answer can only contain numeric values
QUESTION: 20

In the oscillator circuit shown in figure, Find the value of frequency of oscillation (in MHz).


Solution: Since it contains two capacitor and one inductor, the given oscillator circuit is Colpitts oscillator.

Frequency of oscillation of this oscillator is given by

QUESTION: 21

For a random variable x(-∞

Solution:

P(x≥60)=α⇒P(x≤40)=αP(40≤x≤60)=1-P(x≤40)-P(x≥60)P(40≤x≤60)=1-α-αP(40≤x≤60)=1-2α

QUESTION: 22

The state variable of a system is given by the following equation

If the given system is said to be unobservable, the value of a should be


Solution:

The given system is said to be unobservable

*Answer can only contain numeric values
QUESTION: 23

Determine the value of the difference between backward and forward slip in case of a 230V, 50Hz, 4 pole single phase induction machine if synchronous speed = 2372rpm, rotor speed = 1969 rpm [Write the answer upto two decimal point]


Solution: The slip of the rotor w.r.t. the forward rotating field,

sf = (ns-n)/ns ……………..(1)

Since the backward rotating flux rotates opposite to that of the stator, the sign of n must be changed.

The slip of the rotor w.r.t. the backward rotating field,

sb = [ns-(-n)]/ns = (ns+n)/ns …… (2)

Subtracting 1 from 2, we get,

sb-sf = [(ns+n)/ns]- [(ns-n)/ns] = 2n/ns = 2x1969/2372 = 1.67

QUESTION: 24

Consider the program given below :

1. MOV A, C

2. STC

3. CMC

4. DCR C

Let, the content of register C is 00 H initially and the above instructions are executed, then content of register C and flag register is respectively

Solution:

1. MOV A, C

2. STC

3. CMC

4. DCR C

QUESTION: 25

A 3-∅ SRIM with star connected rotor is having standstill impedance of (0.4 + j 4)Ω and 400V is the peak induced emf across the slip rings. At max torque, what is the rotor current per phase in Amperes? [write the answer upto two decimal points]


Solution: R2 =0.4X2 =4

Peak induced line voltage in the rotor = 400V

So, rms line voltage = (400/√2) V = 282.84 V

rms Phase voltage = (282.84)/√3 V = 163.3 V

Rotor current per phase =

QUESTION: 26

For the circuit shown in figure, which of the following represents i(t) for (t > 0).

Solution: At t < 0,="" the="" l="" acts="" as="" short="" />

⇒ i(0) = 3 A

At t > 0, the circuit becomes source free as

i(0−) = i(0+) = 3A

Time constant =

Time constant =

The current equation;

QUESTION: 27

A Stable non-minimum phase transfer function containing one zero & one pole, if the transfer function is pole dominating. The type of compensator, the given transfer function is:

Solution: It is mention, Non-minimum phase that means either zero is right Side or pole is right side. Since transfer function is stable. Hence, zero is Right side.

For compensator: System must stable also System should be minimum phase type. That’s means poles & zero must lie on the left side of the plane.

QUESTION: 28

An alternator supplying 500 kW at 0.8 pf lagging has its pf raised to unity by means of over-excited synchronous motor. At a constant armature current the reactive power supplied by motor and the power factor of operation are

Solution: Load =500kW at 0.8 pf lagging

Load

Load

When a synchronous motor is connected to improve the pf operating in over-excited mode. It will supply 375 KVAR and operate at zero power factor leading.

QUESTION: 29

Two power plants interconnected by a tie-line as shown in the below figure have loss formula coefficient B11 = 10–3 MW–1. Power is being dispatched economically with plant ‘1’ as 100 MW and plant ‘2’ as 125 MW. The penality factor for plants 1 and 2 are respectively.

Solution: since the load is at the bus of plant -2, therefore B12=B21=0 and B22=0 for a 2 -plant system

Penality factor of Plant-1

Similarly,

Penality factor of Plant-2 L2

QUESTION: 30

What will be the minimum value that can be measured for 8-bit output if the input is 5 V.

Solution: Resolution =

= 20mV

QUESTION: 31

Area enclosed between the curves y2 = 4x and x2 = 4y is

Solution:

Curve y2=4x

Curve x2=4y

Intection point of both curve

y24x=4√4y=8√y

y4−64y=0

y=0,y=4

then x = 0, x = 4

Required area = dxdy

QUESTION: 32

A 5 hp, 4-pole, 50 Hz, 3-ϕ induction motor is operating at 1440 rpm on full load. The stator copper loss 100 W and rotational loss is 250 W. Find the efficiency of motor? (Take 1 hp = 746 W)

Solution:

In induction motor given output power = 5hp = 5 x 746 = 3730W

The mechanical power developed

Pmech = Pout + rotational losses

Pmech = 3730 + 250 = 3980W

Nr = 1440rpm

Motor input power = Pmech + rotor Cu loss + Stator Cu loss

= 3980 + 165.83 + 100

= 4245.83W

Efficiency

QUESTION: 33

In the circuit given below if V = A - Bl, then what are the value of A and B respectively?

Solution:

We obtained Thevenin equivalent across the load terminal.

Thevenin Voltage: (Open Circuit Voltage)

Thevenin Impedance:

Now, the circuit becomes as

QUESTION: 34

Calculate the time of operation of an IDMT type over-current relay is used to protect a feeder through 400/1 ACT, relay has plug setting of 125٪ and TMS =0.6 and fault current is 4000 A . Use the following characteristic.

Solution:

Given, plug setting =125%

TMS=O.G

Fault current, If=4000A

Pick up current, I PK= plug setting × C.T. secondary current=1.25×1=1.25

Plug Setting multiplier,

Now, time of operation of any relay,

T=TMS x (Time calculated from IDMT table corresponding)

T = 0.6 x 3.3 = 1.98 sec

QUESTION: 35

Two alternator G1 and G2 rated as 300MW and 600MW respectively are operating in parallel. The droop characteristics of their governors are 3 and 4, respectively from no Ioad to full load. At no load the alternators are operating at a system frequency of 50Hz. Assume free governor operation, if total load of 800MW is being shared by the generator. What will be the system frequency at this load?

Solution:

Machine, G1:

Machine, G2:

By solving (i) and (ii),

We get. P1 = 480 MW, f = 48.4Hz

QUESTION: 36

What are the initial and final values of impulse response h(t) of an LTI system which is described by the differential equation 3y "(t) + 5y'(t) + 2y(t) = 4x' (t) — x(t)

Solution: Electrical Engineering

Apply Laplace transform

Initial value h(0) =

Final value

QUESTION: 37

An armature of DC machine carrying 1024 lap connected conductors where each conductor is carrying 10 A rotates at 300 rpm. If the machine has 6 poles with a flux/pole of 0.025 Wb, the power output of the armature will be

Solution:

Given data

Z = 1024, la = 10A, N = 300 rpm, P = 6

ø = 0.025wb

Ea = 128 V

Total armature current = 10 x 6 = 60 A

[As A = 6 , total current is Ia x A ]

The power output

Pa = Eala= 60 x 128 = 7.680w

QUESTION: 38

A toy company produces toys with an average of 120 toys per week. Assume production of toys as a random variable, having standard deviation = 20. What will be the maximum hope that this week production will be at least 140 toys?

Solution:

Here, μ=120 hours, σ=20 hours

For x=140,

z=(x-μ)/σ = (140-120)/20=1

Then, P(x>120)=P(z>1)=0.3413 ≈ 34%

QUESTION: 39

The rotor poles of a 6-pole synchronous generator shift by 20 mechanical degrees from no load to full load. The torque angle between extinction voltage and the terminal voltage ‘E’ at full load is _______ degrees.

Solution:

Torque angle δ=Pα/2

Given P=6,α=20

δ = 6 x 20/2 = 600

QUESTION: 40

In the figure, the value of resistor R is (25 + I/2) ohms, where I is the current in amperes. The current I is _____.


Solution: We redraw the given circuit as

Applying KVL in the circuit, we have

300=1R

300=1(25+1/2)

300=251+12/2

12+501−600=0

∣(1+60)−10(1+60)=0

(1−10)(1+60)=0

SO,I=10A,−60A

Now, for the two obtained values of current, we get

Since the Resistance cannot be negative.

Thus, the current through the circuit is 1 = 10A

QUESTION: 41

For the given signal, x(t)=3cos⁡80πt determine the sampling frequency(in Hz) at which aliasing will take place.


Solution:

According to the sampling theorem, the sampling frequency should be atleast equal to twice the highest frequency component of the signal, below which overlapping, or aliasing tales place.

Highest frequency component,

Hence, minimum sampling frequency to avoid aliasing, fs=2×40=80Hz

Hence, aliasing will occur if the sampling frequency is less than 80

Hz, i.e., at 40 Hz

QUESTION: 42

A generator rated 100 k V A 13.8 kV and having reactance x1 = x2 = 0.1 p.u. and x0 = 0.05 p.u. its neutral being grounded through a current reactor of 0.08 pu. The generator is connected to a connected step up transformer with 0.06 p.u. reactance its neutral being solidly grounded. A double line to ground fault occurs at star side of transformer. Calculate fault current in p.u.

Solution:

+ve sequence n/w is

-ve sequence n/w is

Zero sequence n/w

Fault current

|If|=3|Ifo|

= 3 x 3.57

= 10.71 p.u.

QUESTION: 43

A barium titanate piezo-electric crystal when subjected to a pressure of 2 MN/m2 developed a voltage of 36 V. If the charge sensitivity of barium titanate is 150 pC/N and its permittivity is 12.5 × 10–9 F/m, find the thickness of the crystal.

Solution:

Given r Pressure applied, P=2MN/m2=2×106N/m2Voltage developed, E0=36V Charge density, d=150pC/N=150×10−12C/N

Permittivity, ε0εr=12.5×10−9F/m

Now, voltage sensitivity of the crystal is given as

For a piezo'electric crystal of thickness 't', the voltage developed due to a pressure 'P' is given as, E0 = gtP

QUESTION: 44

The logic function implemented by the circuit given below is

Solution:

*Answer can only contain numeric values
QUESTION: 45

The voltage applied to the primary winding of an unloaded single-phase transformer is given by

v=400cos⁡αt+100cos⁡3αt

The primary has 500 turns and frequency of the fundamental component of the applied voltage is 50Hz. What is the maximum value of flux in mWb is ________


Solution:

QUESTION: 46

A single phase full converter supplies power to RLE load. The source voltage is 230V, 50Hz and load is R = 2ohms, L= 15mH, E=100V for a firing angle delay of 300. Find the average output current in case current extinguishes at 2000

Solution:

*Answer can only contain numeric values
QUESTION: 47

A sheet of Bakelite 4.5 mm thick is tested at 50Hz between electrode 0.12 m in diameter. The Schering bridge employs a standard air capacitor C2 of 106pF capacitance a non reactive resistance R4 of 1000πΩ in parallel with a variable capacitor C4 and a non-reactive variable resistance R3 balance is obtained with C4 = 0.5μF and R2 = 260Ω.

Then the relative permittivity of sheet is _______ if permittivity of free space is = 8.84 × 10-12 F/m.


Solution: The Schering Bridge is shown And r1=

= 1.23 x 106Ω

= 130pF

Relative permittivity

ɛr = 5.9

QUESTION: 48

Consider the given figure and calculate the feedback factor of the wave generator if R1=5kΩ, R2=20kΩ, R=500Ω and C=10 µF. Also, calculate the one cycle period?

Solution: The given circuit is that af a square wave generator. Given

R1 = 5 kΩ , R2 = 20k0, R = 5000 and C = 10μF

To calculate feedback factor:

The periad T is given by,

Therefore

QUESTION: 49

For a closed-loop system whose block diagram is shown in fig. below if the system has the maximum overshoot to unit step input is 30% and time to peak is 2 sec, then find the value of K × T _______


Solution:

Transfer function =

C .E . = s2 + KTs + K = 0

Standard characteristic equation =

It is given % Mp = 30%

QUESTION: 50

In a single machine infinite bus system, generator has inertia constant H = 5 MW sec/MVA and frequency is 50Hz. Initial mechanical input is 0.75 pu and it is operating at an angle δ = 30°

If suddenly the mechanical input is raised by 33.33% then what will be the speed of rotor during first Swing, at the instant when mechanical input equals electrical output. (in rad/sec)

Solution: Using Swing equation

QUESTION: 51

A load coil is calibrated in an environment at a temperature of 25°C has the following deflection load characteristic.

When used in an environment at 40°C, its characteristic changes to following:

Calculate sensitivity drift at 40°C and zero drift coefficient in unit μm/°C?

Solution: Sensitivity at 25°C =

30/um / kg

Sensitivity at 40°C =

Sensitivity drift at 40°C = 32 = 30 = 2μm/kg

Zero drift at 40°C = 0.3 - 0 = 0.3mm = 300mm

So, zero drift coefficient

Hence, the correct option is (C).

QUESTION: 52

A parallel plate capacitor of 50 pF having an air dielectric is charged to 15 kV. It is then electrically isolated. The plates are pulled away from each other until the distance is 5 times more than before. The energy needed to pull the plates is_______ mJ.

Solution:

As capacitance is inversely proportional to distance between plates.

QUESTION: 53

The input voltage to a converter and current drawn by the converter are expressed as

The input power factor of the converter is

Solution:

QUESTION: 54

The Y-parameter presentation of the circuit shown in the figure:

Find the value of = A + C + D

Solution: The network consists of two 2 -port networks connected in series.

The Z-parameter matrix of the upper network:

The Z-parameter matrix of the lower network:

The Z-parameter matrix of overall network is

QUESTION: 55

For the given ROC in z – plane which one of the following statement is correct?

Solution: In the given figure

which means it is right sided \& causal system & also f in it e ; system is also stable

For the ROC |z| < 2="" it="" is="" left="" sided="" signal="" &="" non="" -="" />

→ From the z - plane it is also observe that the ROC contains the unit circle so we can say the system is stable but non - causal For non-causal system the ROC is inside the outermost pole.

QUESTION: 56

The average power absorbed by the network shown in the figure (a) is _____ (in W).

The output of the current source is shown in the Figure (b).

Solution: In figure (a), the circuit represents a balance bridge that’s why 18 Ω resistor will not have current across it.

As, 18 × 3 = 9 × 6

So, I = 0 (Balance bridge)

The equivalent resistor across current source is-

Req = (9 + 3)||(6 + 18)

Req = 12||24 = 8 Ω

Average power absorbed by the network:

Pavg = (Irms)2 Req … (i)

From figure (b),

From equation (i)

QUESTION: 57

A buck boost converter is supplying an average current of 2A to a load from an input supply of 25 V. What is the peak current in ampere passing through the inductor if duty cycle is 0.4, assume filter inductance is 0.2 mH and switching frequency 50 kHz? (Answer upto two decimal places).

Solution:

QUESTION: 58

Given a signal If the z-transform of signal Cosωon.u[n] is X[z] then which of the following diagram can represent pole location of X[z] on the z-plane

Solution:

Given the signal

z-transform of

so the poles of z -transform of are given by

poles of z-transform of will be given by

for the given signal,

∴ poles will be given by

poles will lie at

QUESTION: 59

A 3-∅, half wave controlled rectifier is supplying an R-L load, where L⟶∞. A freewheeling diode is also connected across the load. The converter is operating with firing angle of 60°, then the RMS value of fundamental source current is

Solution:

Since a>30°, so there will be freewheeling action. So, the source current waveform will be as shown below

We have to find IS1 To find IS1 we have to find the fourier series of

QUESTION: 60

Find the solution of the Linear differential equation

Solution:

Given ( D2 - 4D + 3) y = cos x

Solution is y = yc + yp

y c : Consider (D2 - 4 D + 3)y = 0(D - 1)(D - 3) = 0

So, the solution is y = yc + yp

QUESTION: 61

Initially the switch is at position 1 & capacitor reaches to its steady state condition. Then at t=0 switch is moved from position 1 to 2 at t=0. Find the current flowing i(t) in 10 ohm resistance at t = 0+?

Solution:

When the switch 5 was at 1, i.e. at the steady state, capacitar being fully charged, io = i.

So, i = 12/(50 + 10)A = 0.2A

Voltage drop across 50 ohm resistor = (50 x 0.2 ) V = 1 0 V

Voltage drop across capacitor, VC = (12 - 10 ) V = 2 V

Vclt= o = 2 V

Following switching of S from 1 ta 2 , the voltage source is removed and the capacitor will start discharging through 10 ohm in parallel to 50 ahm.

i(t) (att ~ 0+) = 2 V / 10ahm = 0 .2 A

However,

QUESTION: 62

Determine the transfer function for the given asymptotic bode plot.

Solution:

G > 0, B < 0="" so="" the="" lacus="" will="" be="" in="" />th quadrant.

The initial slope of the plat is -20 db/decade. This indicates a term ( s ) in the denominator.

At ω = lape changes to -40 db / decade.

This indicates a term in the denominator.

At ω = lape changes to -8 0 d b / decade.

This indicates a term in the denominator, as the net slape change is -40

db/decade. Naw, computing the gain K,

we have, 201og10K = 40 ⇒ log10 K = 2

⇒ K = 102 = 100

Hence, the required transfer function is:

QUESTION: 63

Given the sequence current

Ia1= -j1.653 pu

Ia2= j0.5 pu

Ia0= j1.153 pu

Find which type of electrical system it is if the pre fault voltage is 1.2pu and the post fault voltage is 0.625pu?

Solution:

For this first we have to find the sequence impedances

As all sequence current components are not same so it must be a LLG fault.

For LLG fault

Va1=Va2=Va0= Va/3= 0.625/3=0.2083pu

Va1= Ea- Ia1Z1

Z1= (Ea-Va1)/Ia1

Z1=(1.2-0.2083)/-j1.653= j0.59pu

Z2= -Va2/Ia2= -0.2083/j0.5= j0.4166pu

Z0= -Va0/Ia0= -0.2083/j1.153= j 0.18pu

So it must be a synchronous machine because impedances are in a relation of

Z1 > Z2 > Z0.

*Answer can only contain numeric values
QUESTION: 64

A 11/6.6/0.5kV, star/star/delta, three winding transformer has a magnetizing current of 3A. At rated voltages, the secondary supplies a balanced load of 300 kVA at 0.8 pf lag and tertiary supplies a balanced load of 150 kW at 0.9 pf lead. The magnitude of primary phase current (in Amp) will be


Solution:

Magnetising current Iμ=-j3.0

Secondary phase current at 0.8pf lag Secondary phase current referred to primary

I2=15.746∠−36.87∘

Tertiary line current at 0.9 leading

Tertiary phase current at 0.9 leading

Tertiary phase current referred to primary a t 0-9 lead I3 = 8.75∠25.84°

Primary current =Iμ+I2+I3=−j3+15.746∠−36.87∘+8.75∠25.84∘

=22.21∠−22.86∘

So, magnitude of primary phase current= 22.21 A

QUESTION: 65

In the circuit shown below the value of I is

Solution:

From the figure

V1 = 20 V and V3 = 2 4 VKCL

At supernode

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code