electrical engineering date of exam: 08/02/2020 · gate-2020 original paper – ee/1 general...

GATE-2020 ORIGINAL PAPER – EE/0

Electrical Engineering

Date of Exam: 08/02/2020

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GENERAL APTITUDE

1. Non-performing Assets (NPAs) of a bank in India is defined as an asset, which remains unpaid by

a borrower for a certain period of time in terms of interest, principal, or both. Reserve Bank of India

(RBI) has changed the definition of NPA thrice during 1993-2004, in terms of the holding period of

loans. The holding period was reduced by one quarter each time. In 1993, the holding period was

four quarters (360 days).

Based on the above paragraph, the holding period of loans in 2004 after the third revision was

______ days.

(A) 135

(B) 45

(C) 90

(D) 180

Sol. Since the holding period was reduced by one quarter every time, after the third revision, the holding

period would have reduced by 3 quarters, i.e., 270 days. Hence, the holding period in 2004 would have

been 360 – 270, i.e., 90 days. Choice (C)

2. Stock markets _____ at the news of the coup.

(A) poised

(B) plugged

(C) probed

(D) plunged

Sol. A coup could only make matters worse. Hence, ‘plunged’ or crashed is the right answer.

Choice (D)

3. In four-digit integer numbers from 1001 to 9999, the digit group “37” (in the same sequence)

appears ______ times.

(A) 299

(B) 280

(C) 270

(D) 279

Sol. We have three possible cases:

Case (1): When the first two digits are 37, the next two digits can be chosen in 10 × 10, i.e., 100 such

numbers are there.

Case (2): When the two middle digits are 37, the first digit can be chosen in 9 ways and the last digit in

10 ways. Hence, 90 such numbers are there.

Case (3): When the last two digits are 37, the first digit can be chosen in 9 ways and the second digit in

10 ways. Hence, a total of 90 numbers are there.

The number 3737 is counted twice, once in case (1) and again in case (3). Hence, a total of (100 + 90 +

90) – 1, i.e. 279 numbers are there. Choice (D)

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4. This book, including all its chapters _____ interesting. The students as well as the instructor

______ in agreement about it.

(A) is, was

(B) were, was

(C) are, are

(D) is, are

Sol. The primary subject being a book, both ‘were’ and ‘are’ are ruled out. Since the latter subject is plural

‘students’, the corresponding verb is ‘are’. Choice (D)

5. The revenue and expenditure of four different companies P, Q, R and S in 2015 are shown in the

figure. If the revenue of company Q in 2015 was 20% more than in 2014, and company Q had earned

a profit of 10% on expenditure in 2014, then its expenditure (in million rupees) in 2014 was ______.

(A) 34.1

(B) 33.7

(C) 32.7

(D) 35.1

Sol. Let the revenue and expenditure of Q in 2014 be R and E respectively.

The revenue of Q in 2015 = ₹45 million = 1.2R R = ₹37.5 million

Given, R = 1.1 E

37.5 = 1.1 E

E = 34.1 Choice (A)

6. If P, Q, R, S are four individuals, how many teams of size exceeding one can be formed, with Q as

a member?

(A) 7

(B) 6

(C) 5

(D) 8

Sol. Given, Q is included in the team. Of the remaining 3 people, at least one member should be included,

which can be done in 23 – 1, i.e., 7 ways. Choice (A)

05

10152025303540455055

Company P Company Q Company R Company S

Rev

en

ue/E

xp

en

dit

ure

(in

millio

n

rup

ees)

Revenue and Expenditure (in million rupees) of four comapnies P, Q, R and S in 2015

Revenue Expenditure

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7. People were prohibited _____ their vehicles near the entrance of the main administrative building. (A) to have parked (B) parking (C) from parking (D) to park Sol. The expression ‘prohibited from’ is correct, as per grammatical conventions. Choice (C)

8. Given a semicircle with O as the centre, as shown in the figure, the ratio 𝑨𝑪 + 𝑪𝑩

𝑨𝑩 is _____.

(A) √𝟐

(B) √𝟑 (C) 3 (D) 2

Sol. Let OA = OB = OC = r (Radius)ACB = 90°

(Angle in a semicircle)OCB = OBC = 45° = OAC

= OCAAC = BC

= √2R AC + CB

AB=

2√2R

2R

= √2 Choice (A)

9. Select the next element of the series Z, WV, RQP, _____

(A) JIHG

(B) NMLK

(C) LKJI

(D) KJIH

Sol. Between Z and WV, there is a gap of two alphabets i.e. XY; between WV and RPQ, there is a gap of

three alphabets i.e. STU; between RQP and the missing element, there should be a gap of four i.e.

LMNO. Hence the missing element should be KJIH. Hence the answer is KJIH. Choice (D)

10. Select the word that fits the analogy: Do : Undo :: Trust : ______ (A) Intrust (B) Untrust (C) Entrust (D) Distrust Sol. To ‘distrust’ is to doubt the honesty or reliability of someone; or to regard someone with suspicion. Choice (D)

A O B

C

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ELECTRICAL ENGINEERING

1. Which of the following statement is true about the two sided Laplace transform?

(A) It exists for every signal that may or may not have a Fourier transform.

(B) It has no poles for any bounded signal that is non-zero only inside a finite time interval.

(C) If a signal can be expressed as a weighted sum of shifted one sided exponentials, then its

Laplace Transform will have no poles.

(D) The number of finite poles and finite zeroes must be equal.

Sol. Let us consider finite interval signal

For finite interval signal, ROC is entire S – plane.

ROC does not include any poles.

Choice (B)

2. A common-source amplifier with a drain resistance, RD = 4.7 kΩ, is powered using a 10 V power

supply. Assuming that the transconductance, gm, is 520 µA/V, the voltage gain of the amplifier is

closest to”

(A) 2.44

(B) –2.44

(C) 1.22

(D) –1.22

Sol. RD = 4.7 K

VDD = 10 V

gm = 520 A/V

A = –gmRD

= –520 × 4.7 × 103

= –2.44 Choice (B)

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50

49.75

49.25

P

Q

R

f

0 100 110 x P(MN)

3. A single-phase inverter is fed from a 100 V dc source and is controlled using a quasi-square wave modulation scheme to produce an output waveform, v(t), as shown. The angle σ is adjusted to entirely eliminate 3rd harmonic component from the output voltage. Under this condition, for v(t), the magnitude of the 5th harmonic component as a percentage of the magnitude of the fundamental component is ______ (rounded off to 2 decimal places).

Sol. Let width of the pulse = 2d Given, output voltage is free from 3rd harmonic.

∴width of the pulse, 2d = 2

3

Magnitude of 5th harmonic component as Percentage of magnitude of fundamental component,

|𝑉05

𝑉01| =

4𝑉𝑠5

sin (5𝑑) sin(5

2)

4𝑉𝑠

sin 𝑑 sin

2

= 1

5 sin 5 ×

3

5 ×

3

= 0.2 (or) 20% Ans: 20

4. A sequence detector is designed to detect precisely 3 digital inputs, with overlapping sequences detectable. For the sequence (1,0,1) and input data (1,1,0,1,0,0,1,1,0,1,0,1,1,0), what is the output of this detector?

(A) 0,1,0,0,0,0,0,0,1,0,0,0 (B) 0,1,0,0,0,0,0,1,0,1,0,0 (C) 1,1,0,0,0,0,1,1,0,1,0,0 (D) 0,1,0,0,0,0,0,1,0,1,1,0 Sol. For sequence (1, 0, 1) the output is 1 for 3 times hence the correct option is (B). Choice (B)

5. A single 50Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to be increased by 10 MW, as a result of which, system frequency dropped to 49.75 Hz. Further increase in load in the system resulted in a frequency of 49.25 Hz. At this condition, the power in MW supplied by the generator is ______ (rounded off to 2 decimal places).

Sol.

Slope = −0.25

10=

−0.5

𝑥 −110

x = 130 MW. Ans: 130

–100V

100V

0

/2

2

5/2

7/2 4

3/2

v(t)

t 3

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6. Out of the following options, the most relevant information needed to specify the real power (P) at the PV buses in a load flow analysis is

(A) rated voltage of the generator (B) rated power output of the generator (C) solution of economic load dispatch (D) base power of the generator Sol. To specify ‘p’ at pv buses load forecasting is required, i.e, solution of economic load dispatch is most

relevant information needed. Choice (C)

7. Consider a linear time-invariant system whose input r(t) and output y(t) are related by the following differential equation:

𝒅𝟐𝒚(𝒕)

𝒅𝒕𝟐 + 4y(t) = 6r(t)

(A) +4, –4 (B) +2, –2 (C) +4j, –4j (D) +2j, –2j

Sol. d2y(t)

dt2 + 4y(t) = 6r(t)

S2y(s) + 4y(s) = 6R(s)

Y(s)

R(s) =

6

s2+4

C.E = S2 + 4 = 0 S1,2 = +j2, – j2 Choice (D)

8. A lossless transmission line with 0.2 pu reactance per phase uniformly distributed along the length

of the line, connecting a generator bus to a load bus, is protected up to 80% of its length by a distance relay placed at the generator bus. The generator terminal voltage is 1 pu. There is no generation at the load bus. The threshold pu current for the operation of the distance relay for a solid three phase-to-ground fault on the transmission line is closest to:

(A) 6.25 (B) 1.00 (C) 3.61 (D) 5.00 Sol. Z = 0.8 × 0.2 = 0.16 pu

If = 1

0.16 = 6.25 pu. Choice (A)

9. 𝒙𝑹 and 𝒙𝑨 are, respectively, the rms and average values of 𝒙(𝒕) = 𝒙(𝒕 – 𝑻), and similary, 𝒚𝑹 and 𝒚𝑨

are, respectively, the rms and average values of 𝒚(𝒕) = 𝒌𝒙(𝒕). 𝒌, 𝑻 are independent of 𝒕. Which of the following is true?

(A) 𝒚𝑨 = 𝒌𝒙𝑨 ; 𝒚𝑹 ≠ 𝒌𝒙𝑹 (B) 𝒚𝑨 ≠ 𝒌𝒙𝑨 ; 𝒚𝑹 ≠ 𝒌𝒙𝑹 (C) 𝒚𝑨 = 𝒌𝒙𝑨 ; 𝒚𝑹 = 𝒌𝒙𝑹 (D) 𝒚𝑨 ≠ 𝒌𝒙𝑨 ; 𝒚𝑹 = 𝒌𝒙𝑹

Sol. For Periodic signal x(t),

Average value , xA = 1

𝑇 ∫ 𝑥(𝑡)𝑑𝑡

𝑇

y (t) = K(t)

Average value, yA = 1

𝑇 ∫ 𝐾𝑥(𝑡)𝑑𝑡 = 𝐾𝑥𝐴

𝑇

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Rms value of x(t), xR = √1

𝑇∫ 𝑥2(𝑡)

𝑇𝑑𝑡

Rms value of y(t), yR = √1

𝑇∫ [𝐾𝑥(𝑡)]2

𝑇𝑑𝑡

yR = KxR Choice (C)

10. A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source supplies a series combination of finite resistance, R, and a very large inductance, L. The two most dominant frequency components in the source current are:

(A) 50 Hz, 0Hz (B) 150 Hz, 250 Hz (C) 50 Hz, 100 Hz (D) 50Hz, 150 Hz Sol. For highly inductive load, the source current, Is:

Fourier series representation of source current is given by:

Is ∑4𝐼0

𝑛𝑛=1,3,5….. sin (nwt)

Most dominant frequency components are fundamental and 3rd harmonic

∴ f & 3f

50Hz & 150Hz Choice (D)

11. A three-phase, 50 Hz, 4-pole induction motor runs at no-load with a slip of 1%. With full load, the slip increases to 5%. The % speed regulation of the motor (rounded off to 2 decimal places) is _____.

Sol. P = 4, f = 50Hz, NS = 120𝑓

𝑃 =

120 ×50

4

NS = 1500 r.p.m. No load Slip, So = 0.01 No load Speed, No = NS (1 – So) No = 1500 (1 – 0.01) = 1485 r.p.m. Full load slip, Sfl = 0.05 Full Load speed, Nfl = NS(1 – Sfl) Nfl = 1500 (1 – 0.05) Nfl = 1425 r.p.m.

% Speed Regulation = 𝑁𝑂 – 𝑁𝑓𝑙

𝑁𝑓𝑙 100%

= 1485 – 1425

1425 100%

= 4.21%. Ans: 4.21

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12. Thyristor T1 is trigged at an angle 𝜶 (in degree), and T2 at angle 𝟏𝟖𝟎° + 𝜶, in each cycle of the sinusoidal input voltage. Assume both thyristors to be ideal. To control the load power over the range 0 to 2 kW, the minimum range of variation in 𝜶 is:

(A) 60° to 180° (B) 60° to 120° (C) 0° to 120° (D) 0° to 60°

Sol. Given, load impedance, z = 10 –60°

= 60°

The range of firing angle () of AC Voltage regulator for which output can be controlled is:

= < 180°

∴firing angle : 60° to 180° Choice (A)

13. Which of the options is an equivalent representation of the signal flow graph shown here?

(A) (B) (C) (D)

Sol. T.F = 𝑎𝑑

1–𝑎𝑑𝑒–𝑑𝑐

From option (1)

T.F = 𝑎 (

𝑑

1–𝑐𝑑)

1–𝑎𝑒𝑑

(1–𝑐𝑑)

= 𝑎𝑑

1–𝑐𝑑–𝑎𝑒𝑑 Choice (A)

200V 50 Hz

+

–

T2

T1

10–60°

1 a

c

d

1

e 1 a(𝑑

1 –𝑐𝑑)

1

e

1 (a + c)d 1

e

1 a(d + c) 1

e

1 a(𝑐

1 – 𝑐𝑑) 1

e

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14. Consider the initial value problem below. The value of y at x = ℓn 2, (rounded off to 3 decimal

places) is ______.

𝒅𝒚

𝒅𝒙 = 2x – y , y(0) = 1

Sol. 𝑑𝑦

𝑑𝑥 = 2𝑥 – y

sy(s) – y(0) = 2

𝑠2 – y(s)

y(s) [s + 1] = 2

𝑠2 + y(0)

y(s) = 1

𝑠+1 [

2

𝑠2 + 𝑦(0)]

y(s) = 2

𝑠2(𝑠+1) +

𝑦(0)

𝑠+1

y(s) = 2

𝑠2(𝑠+1) +

1

𝑠+1

y(𝑥) = L–1 [2

𝑠2 +–2

𝑠+

2

𝑠+1] + L–1 [

1

𝑠+1]

y(𝑥) = 2𝑥 – 2u(𝑥) + 2𝑒–𝑥 + 𝑒–𝑥

y(𝑥)|x=ln2 = 2ln2 – 2u(ln2) + 3e– ln2

= 0.8863. Ans: 0.88000 to 0.88630

15. The Thevenin equivalent voltage, VTH, in V (rounded off to 2 decimal places) of the network shown

below, is ______. Sol. By KVL, +Vth – 10 – 4 = 0

Vth = 14V Ans: 14

4V 3

2 3

5A VTH

+

–

– +

4V

+

2 5A

5A VTH

+

–

– +

–

4V

+ – 10V

4V

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16. Consider a signal 𝒙[𝒏] = (𝟏

𝟐)

𝒏 𝟏[𝒏], where 𝟏[𝒏] = 𝒊𝒇 𝒏 < 𝟎, 𝒂𝒏𝒅 𝟏[𝒏] = 𝟏 𝒊𝒇 𝒏 ≥ 𝟎. The z-

transform of 𝒙[𝒏 – 𝒌], 𝒌 > 𝟎 𝒊𝒔 𝒛−𝒌

𝟏−𝟏

𝟐 𝒛−𝟏

with region of convergence being

(A)|𝒛| < 𝟐

(B)|𝒛| > 𝟐

(C) |𝒛| > 𝟏

𝟐

(D) |𝒛| < 𝟏

𝟐

Sol. Given, x[n] = (1

2)

𝑛𝑢 [𝑛]

x[n] causal signal

x[n – k] causal signal (∵K > 0)

z x[n –k] = 𝑧–𝐾

1 – 1

2 𝑧 –1

Region of coverage; |𝑧| > 1

2

Choice(C)

17. A three-phase cylindrical rotor synchronous generator has a synchronous reactance XS and a

negligible armature resistance. The magnitude of per phase terminal voltage is VA and the

magnitude of per phase induced emf is EA. Considering the following two statements, P and Q.

P: For any three-phase balanced leading load connected across the terminals of this synchronous

generator, VA is always more than EA

Q: For any three-phase balanced lagging load connected across the terminals of this synchronous

generator, VA is always less than EA

(A) P is false and Q is true.

(B) P is true and Q is false.

(C) P is false and Q is false.

(D) P is true and Q is true.

Sol. Given, Per phase induced e.m.f = EA

Per phase terminal voltage = VA

% Voltage Regulation = 𝐸𝐴 – 𝑉𝐴

𝑉𝐴 100%

For lagging P.F loads, always voltage regulation is(+)ve i.e., EA > VA

For lagging P.F loads, voltage regulation may be (+)ve, Zero (or) (–) ve i.e.,

EA = VA V. R = 0

EA > VA V. R is(+)ve

EA < VA V. R is (–)ve

P is false, Q is true Choice (A)

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VFD forward voltage drop

VD Voltage across diode

18. A double pulse measurement for an inductively loaded circuit controlled by IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode. D, represented approximately as a piecewise linear plot of current vs time at diode turn-off Lpar is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1, 2, 3 or 4) at which the IGBT experiences the highest current stress is ______.

Sol. Reverse recover characteristics of diode are as follows:

Till point 3, there is no appreciable change in voltage across diode.

Due to stray inductance, reverse transient voltage appears across diode at point 3

∴The load will carry maximum current. (That current is through switch)

Ans: 3

1

Diode current

Vsource –

D

Lpar Rload

IGBT

Lload

4 2

3

+

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19. The cross-section of a metal-oxide-semiconductor structure is shown schematically. Starting from an uncharged condition, a bias of +3 V is applied to the gate contact with respect to the body contact. The charge inside the silicon dioxide layer is then measured to be +Q. The total charge contained with in the dashed box shown, upon application of bias, expressed as a multiple of Q (absolute value in Coulombs, rounded off to the nearest integer) is ______.

Sol. It is a MoS capacitor where charge stored in the oxide region (or) at the interface oxide region only

Charge stored in the dotted region is zero. Ans: 0

20. Currents through ammeters A2 and A3 in the figure are 1 10° and 1 70°, respectively. The

reading of the ammeter A1 (rounded off to 3 decimal places) is _____A.

Sol.

𝐼1 = 𝐼2 + 𝐼3

Given, 𝐼2 = 1 10°

𝐼3 = 1 70°

𝐼1 = 1 10° + 170°

= √12 + 12 + 2 × 1 × 1cos (60)

|𝐼1| = 1.732 Ans: 1.732

21. Consider a negative unity feedback system with forward path transfer function G(s) = 𝑲

(𝒔+𝒂)(𝒔 –𝒃)(𝒔+𝒄),

where K, a, b, c are positive real numbers. For a Nyquist path enclosing the entire imaginary axis and right half of the s-plane in the clockwise direction, the Nyquist plot of (1 + G(s)), encircles the origin of (1 + G(s))-plane once in the clockwise direction and never passes through this origin for

a certain value of K. Then, the number of poles of 𝑮(𝒔)

𝟏+𝑮(𝒔) lying in the open right half of the s-plane

is ______.

Sol. N = P – Z

–1 = 1– Z

Z = 2 Ans: 2

GATE

Silicon Dioxide

Si

BODY

DASHED BOX

I1

I2

I3 A3

A1

A2

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Im

Re O

C

2

22. The value of the following complex integral, with C representing the unit circle centered at origin in the counter clockwise sense, is:

∫𝒛𝟐+𝟏

𝒛𝟐−𝟐𝒛

𝒄 dz

(A) –𝛑I (B) 8𝛑i (C) 𝛑I (D) –8𝛑i

Sol. Let I = ∫Z2+1

Z2 –2Z dz

c

where C is the unit circle, |Z| = 1 Z = 0,2 are the singularities of the integrand, of which only Z = 0 lies inside C

By Cauchy's integral formula, we have

∫Z2+1

Z2 –2Z

cdz = ∫ [

(Z2+1

𝑍 – 2)

Z]

cdz

= 2πi (Z2+1

Z –2) at Z = 0

= – πi Choice (A)

23. A single-phase, 4 kVA, 200 V/100 V, 50 Hz transformer with laminated CRGO steel core has rated

no-load loss of 450 W. When the high-voltage winding is excited with 160 V, 40 Hz sinusoidal ac supply, the no-load losses are found to be 320 W. When the high-voltage winding of the same transformer is supplied from a 100 V, 25 Hz sinusoidal ac source, the no-load losses will be _____ W (rounded off to 2 decimal places).

Sol. V1 = 200V, f1 = 50HZ; Wi1 = 450W V2 = 160V, f2 = 40HZ; Wi2 = 320W V3 = 100V, f3 = 25HZ; Wi3 = ?

𝑉1

𝑓1 =

𝑉2

𝑓2 =

𝑉3

𝑓3

200

50 =

160

40 =

100

25

𝑉

𝑓 = Constant

Wi = Af + Bf2 450 = A(50) + B(50)2

320 = A(40) + B(40)2

A = 4, B = 1

10 = 0.1

Wi3 = Af3 + Bf32

= (4 25) + (0.1 252) Wi3 = 162.5 W Ans: 162.5

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24. Which of the following is true for all possible non-zero choices of integers m, n ; m ≠ n, or all possible non-zero choices of real numbers p, q ; p ≠ q, as applicable?

(A) 𝟏

𝝅∫ 𝒔𝒊𝒏 𝒎𝜽 𝐬𝐢𝐧 𝒏𝜽 𝒅𝜽

𝝅

𝟎 = 0

(B) 𝟏

𝟐𝝅∫ 𝒔𝒊𝒏 𝒑𝜽 𝐜𝐨𝐬 𝒒𝜽 𝒅𝜽

𝝅

−𝝅 = 0

(C) 𝟏

𝟐𝝅∫ 𝒔𝒊𝒏 𝒑𝜽 𝐬𝐢𝐧 𝒒𝜽 𝒅𝜽

𝝅/𝟐

−𝝅/𝟐 = 0

(D) 𝐥𝐢𝐦𝜶 → ∞

𝟏

𝟐𝜶∫ 𝒔𝒊𝒏 𝒑𝜽 𝐬𝐢𝐧 𝒒𝜽 𝒅𝜽

𝜶

−𝜶 = 0

Sol. We know that ∫ f(x)dxa

–a =

2 ∫ f(x)dx; if (x) is even; a

0

0; if f(x)is odd

Consider option B

sinpθ.cosqθ is an odd function when p ≠ q

1

2π ∫ sinpθ. cosqθ dθ

π

–π = 0 Choice (B)

25. ax3 + bx2 + cx + d is a polynomial on real x over real coefficients a, b, c, d where in a ≠ 0. Which of the following statement is true?

(A) d can be chosen to ensure that x = 0 is a root for any given set a, b, c. (B) No choice of coefficients can make all roots identical. (C) a, b, c, d can be chosen to ensure that all roots are complex. (D) c alone cannot ensure that all roots are real.

Sol. Let f(x) = ax3 + bx2 + cx + d; a≠ 0

When d = 0, f(x) becomes ax3 + bx2 + cx

x = 0 becomes a root for any values of a, b, and c Choice (A)

26. A 250 V dc shunt motor has an armature resistance of 0.2 Ω and a field resistance of 100 Ω. When the motor is operated on no-load at rated voltage, it draws an armature current of 5 A and runs at 1200 rpm. When a load is coupled to the motor, it draws total line current of 50 A at rated voltage, with a 5% reduction in the air-gap flux due to armature reaction. Voltage drop across the brushes can be taken as 1 V per brush under all operating conditions. The speed of the motor, in rpm, under this loaded condition, is closest to:

(A) 1000 (B) 1200 (C) 1220 (D) 900

Sol. Given;

Vt = 250V, Ra = 0.2, Rsh = 100 No load: 𝐼𝑎𝑜 = 5A, No = 1200 r.p.m

Ish = 𝑉𝑡

𝑅𝑠ℎ =

250

100 = 2.5A

𝐸𝑏𝑜 = Vt – 𝐼𝑎𝑜 Ra – Total Brush drop

= 250 – (5 0.2) – 2 = 247V

Full load: IL = 50A, Ia(F.L) = 50 – 2.5 = 47.5A

Ebfl = 250 – (47.5 0.2) – 2 = 238.5V

Eb = Kaω 𝐸𝑏𝑓ℓ

𝐸𝑏𝑜 =

∅𝐹𝐿

∅𝑜 𝑁𝐹𝐿

𝑁𝑜

Given: fl = 0.95o

238.5

247 = 0.95

𝑁𝑓𝑙

1200

Nfl = 1219.68 r.p.m

1220 r.p.m. Choice (C)

`


27. Let ax and ay be unit vectors along x and y directions, respectively. A vector function is given by F = ax y – ay x

The line integral of the above function ∫ 𝑭. 𝒅𝒍

𝒄

Along the curve C, which follows the parabola y = x2 as shown below is ______ (rounded off to 2 decimal places).

Sol.

Given F = ax y –ay x

We have to evaluate

∫ F. dl

c

where C is y = x2 as shown in the figure

y = x2 dy = 2xdx

∫ 𝐅. 𝐝𝐥

c = ∫ (y dx – x dy)

c

= ∫ [x2dx – x(2x dx)]2

x= –1

= ∫ [x2 – 2x2 ]2

–1dx

=∫ (– x2)2

–1dx

=–𝐱𝟑

𝟑]

−𝟏

𝟐

= −8

3 –

1

3 = – 3 Ans: –3.05 to –2.95

28. The number of purely real elements in a lower triangular representation of the given 3 × 3 matrix, obtained through the given decomposition is _____.

[𝟐 𝟑 𝟑𝟑 𝟐 𝟏𝟑 𝟏 𝟕

] = [𝒂𝟏𝟏 𝟎 𝟎𝒂𝟏𝟐 𝒂𝟐𝟐 𝟎𝒂𝟏𝟑 𝒂𝟐𝟑 𝒂𝟑𝟑

] [𝒂𝟏𝟏 𝟎 𝟎𝒂𝟏𝟐 𝒂𝟐𝟐 𝟎𝒂𝟏𝟑 𝒂𝟐𝟑 𝒂𝟑𝟑

]

𝑻

(A) 6 (B) 9 (C) 5 (D) 8

C

2 x

y

1

–1

4

C

2 x

y

1

–1

4

`


Sol. Given [2 3 33 2 13 1 7

] = [a11 0 0a12 a22 0a13 a23 a33

] [a11 0 0a12 a22 0a13 a23 a33

]

𝑇

= [a11 0 0a12 a22 0a13 a23 a33

] [

a11 a12 a13

0 a22 a23

0 0 a33

]

[2 3 33 2 13 1 7

] = [

a112 a11a12 a11a13

a12a11 a122 + a22

2 a12a13 + a22a23

a13a11 a13a12 + a23a22 a132 + a23

2 + a332

]

Comparing the corresponding elements on both sides,

a112 = 2 a11 = ± √2 a11 is real

a11a12 = 3 a12 = 3

a11 = ±

3

√2 a12 is real

a11a13 = 3 a13 = 3

a11 = a13= ±

3

√2 a13 is real

a122 + a22

2 = 2 9

4 + a22

2 = 2 a222 = –

1

4 a22 = ± i

1

2 a22 is imaginary

a12a13 + a22a23 = 1 a23 = 1 – a12a13

a22 = ± (2 ±

9

2) i a23 is imaginary

a132 + a23

2 + a332 = 7

9

4 – (2 ±

9

2)

2 + a33

2 = 7 a332 = 7 –

9

4 + (2 ±

9

2)

2 > 0 a33 is real.

The number of purely real elements = 4 + 3 = 7 Ans: 7

29. A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and

subtransient reactance of 0.2 pu. It is operating at (1 + j0) pu terminal voltage with an internal emf

of (1 + j0.7) pu. Following a three-phase solid short circuit fault at the terminal of the generator,

the magnitude of the subtransient internal emf (rounded off to 2 decimal places) is ______ pu.

Sol.

𝐼 = 1 + 𝑗 0.7 1

𝐽 0.7

𝐼 = 10°

𝑔11 = (j 0.2) 𝐼 + 1 0°

= (j 0.2)(10°) + 1 0°

= 1 + j 0.2

|𝐸𝑔11| = 1.0198 pu. Ans: 1.01

`


30. Which of the following options is correct for the system shown below?

(A) 4th order and stable

(B) 3rd order and stable

(C) 3rd order and unstable

(D) 4th order and unstable

Sol. T.F =

1

𝑠2(𝑠+1)

1+ 20

𝑠2(𝑠+1)(𝑠+20)

T.F = 𝑠+20

𝑠2(𝑠+1)(𝑠+20)+20

= 𝑠+20

(𝑠3+ 𝑠2) (𝑠+20)+20

T.F = 𝑠+20

𝑠4 + 20𝑠3 + 𝑠3 + 20𝑠2+20

C.E. = s4 + 21s3 + 20s2 + 20 = 0

4th order and unstable

∵ s1 coefficient is missing Choice (D)

31. A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω

resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz

sinusoidal source through cables which have negligible internal resistance and

a total inductance of 2.28 mH. If the thyristors are triggered at an angle

𝜶 = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______.

Sol. Given, Vs = 230 V

f = 50 Hz

R= 10

Source inductance, Ls = 2.228 mH

45°

Overlap angle, = ?

R(s) +

–

1

𝑠 + 1

1

𝑠2

20

𝑠 + 20

`


The average output voltage, VOA = 2 𝑉𝑚

cos –

2𝐼𝑜𝑤𝐿𝑠

VOA = IO R

Io R = 2 𝑉𝑚

cos –

2𝐼𝑜𝑤𝐿𝑠

Io = (

2 𝑣𝑚

cos)

𝑅+ 2 𝑤𝐿𝑠

Io =

2 ×230 × √2

cos 45°

10 + (2 ×100 ×2.28 ×10–3

)

Io = 14A

To find overlap angle ‘’

Cos – Cos ( + ) = 2𝐼𝑜𝑤𝐿𝑠

𝑉𝑚

Cos 45 – Cos (45° + ) = 2 ×14 ×100 2.28 × 10–3

230 × √2

Cos (45° + ) = 0.6454

45° + = 49.80

= 4.80° Ans: 4.80

`


`


32. Windings ‘A’, ‘B’ and ‘C’ have 20 turns each and are wound on the same iron core as shown, along

with winding ‘X’ which has 2 turns. The figure shows the sense (clockwise/anti-clockwise) of each

of the windings only and does not reflect the exact number of turns. If windings ‘A’, ‘B’ and ‘C’ are

supplied with balanced 3-phase voltages at 50 Hz and there is no core saturation, the no-load RMS

voltage (in V, rounded off to 2 decimal places) across winding ‘X’ is ______.

Sol. 𝑉𝑥

𝑉𝐴 =

2

20 Vx =

2

20 VA

𝑉𝑥

𝑉𝐵 =

2

20 Vx =

2

20 VB, Vx =

2

20 VC

Vx = 2

20[2300° – 230 – 120° – 230120°]

Vx = 46 V Ans:46

33. A conducting square loop of side length 1 m is placed at a distance of 1 m from a long straight

wire carrying a current I = 2 A as shown below. The mutual inductance, in nH (rounded off to 2

decimal places), between the conducting loop and the long wire is _____.

Sol. M =

I =

N∅

I

N = 1, I = 2A

M = ∅

I =

∅

2 =

1

2 ∬ B d S

The Flux density at a radial distance ‘r’ from the wire and between the wire and loop,

B = μ0I

2πr (Assume the infinitely long wire)

M = 1

2 ∫

μ0I

2πr

𝑟=2

𝑟=1 1dr

= μ0I

4π∫

dr

r

2

𝑟=1 =

4π × 10−7 ×2

4π ln2

M = 138. 63 nH. Ans: 138.63

a = 1m

a = 1m d = 1m

I = 2A

z

`


34. The figure below shows the per-phase Open Circuit Characteristics (measured in V) and Short

Circuit Characteristics (measured in A) of a 14 KVA, 400 V, 50 Hz, 4-pole, 3-phase, delta connected

alternator, driven at 1500 rpm. The field current, If is measured in A. Readings taken are marked

as respective (x, y) coordinates in the figure. Ratio of the unsaturated and saturated synchronous

impedances (Zs(unsat)/Zs(sat)) of the alternator is closest to:

(A) 1.000 (B) 2.000 (C) 2.025 (D) 2.100

Sol. ZS(unsat) = 𝑉𝑜𝑐

𝐼𝑆𝐶| 𝐼𝑓= 𝑅𝑎𝑡𝑒𝑑

Equation of airgap line,

VOC = (210−110

2 – 1) If + 10

VOC = 100 If + 10

At If = 8A Voc = (100 8) + 10 = 810

∴ ZS(unsat) =

810

40 = 20.25

At If = 8A ISC = 40 A

ZS(sat) = 400

40 = 10

∴ZS(unsat)

ZS(sat) =

20.25

10 = 2.025 Choice (C)

35. Bus 1 with voltage magnitude V1 = 1.1 pu is sending reactive power Q12 towards bus 2 with voltage

magnitude V2 = 1 pu through a lossless transmission line of reactance X. Keeping the voltage at

bus 2 fixed at 1 pu, magnitude of voltage at bus 1 is changed, so that the reactive power Q12 sent

from bus 1 is increased by 20%. Real power flow through the line under both the condition is zero.

The new value of the voltage magnitude. V1, in pu (rounded off to 2 decimal places), at bus 1 is

––––––––.

VOC

OCC (8,400

)

(4, 20)

SCC

(0,0)

(0,10)

(1,110)

(2,210)

ISC

If

V1 V2

Bus2 Bus1 Q12

`


Sol.

Q12 new = 1.2 Q12 old

Q12 = |𝑉1|

𝑋 [|𝑉1| |𝑉2| 𝐶𝑜𝑠 𝛿]

Cos 𝛿 = 1 ∵ P = 0

|𝑉1 𝑛𝑒𝑤|

𝑋 [|𝑉1 𝑛𝑒𝑤| |𝑉2|] = 1.2 ×

1.1

𝑋 [1.1 1]

|𝑉1 𝑛𝑒𝑤|2 = |𝑉1 𝑛𝑒𝑤| = 1.2 × 1.1 × 0.1 |𝑉1 𝑛𝑒𝑤| = 1.11806 0.11806 |𝑉1 𝑛𝑒𝑤| = 1.11806 pu Ans: 1.11

36. The temperature of the coolant oil bath for a transformer is monitored using the circuit shown. It

contains a thermistor with a temperature-dependent resistance, Rthermistor = 2 (1 + T) k, where T

is the temperature in °C. The temperature coefficient, , is – (4 0.25) %/°C. Circuit paramerers: R1

= 1 k, R2 = 1.3 k, R3 = 2.6 k. The error in the output signal (in V, rounded off to 2 decimal places) at 150°C is ________.

Sol.

= (4 ± 0.25)% /°C

Vout

Rthermistor

3V R1 1 K

R3 = 2.6 K

R2 = 1.3 K

+

+

–

0.1 V

–

+ –

Vout

Rthermistor

3V R1

R3

R2

+

+

–

0.1 V

–

+ –

`


Case (i)

∝1= −4.25

100/°C

RTM = 2[1 + ∝1 T] K

= 2[1 −4.25

100 × 150] K

= –10.75K

VR1 = Vx = 3×1K

(–10.75+1)K = –0.3676V

𝑉𝑜𝑢𝑡1−(−0.3076)

2.6𝐾 =

−0.3076−0.1

1.3K

Vout1 = –0.8152 -0.3076 = –1.1228

Case (ii) ∝1 = – (4 – .25)% / °C

= −3.75

100/°C

RTM = 2[1 + ∝2 T] K

= 2[1 −3.75

100× 150] K

= –9.25 K

VR1 = Vy = 3×1K

(−9.25+1)K= – 0.3636V

VOut2−(−0.3636)

2.6K=

−0.3636−0.1

1.3K

Vout2 = –0.9292–0.3636 = –1.2928Volt

V0 = Vout1 –Vout2 = –1.1228 + 1.2928 = 0.17 volt Ans: 0.17

37. The causal realization of a system transfer function H (s) having poles at (2, –1), (–2, 1) and zeroes

at (2, 1), (–2, 1) will be

(A) unstable, complex. allpass

(B) stable, real, allpass

(C) stable, complex, lowpass

(D) unstable, real, highpass

Sol. T.F =k(s–2–j1)(𝑠+2+j1)

(s–2+j1)(𝑠+2−j1)

|T.F| = K (for = s = j)

The system is unstable, complex, all pass Choice (A)

`


38. A benchtop dc power supply acts as an ideal 4 A current source as long as its terminal voltage is

below 10 V. Beyond this point, it begins to behave as an ideal 10 V voltage source for all load

currents going down to 0 A. When connected to an ideal rheostat, find the load resistance value

at which maximum power is transferred, and the corresponding load voltage and current.

(A) Short, A, 10V

(B) 2.5 , 4 A, 10 V

(C) 2.5 , 4 A, 5 V

(D) Open, 4 A, 0 V

Sol. As per the given data:

If V 10V It acts as 4A ideas current source.

If I 4A It acts as 10V idea voltage source.

In order to transfer maximum power,

I should be maximum i.e., RL should be minimum. But to maintain 4A of current, 10V across RL, RL should

be 2.5.

RL = 2.5; I = 4A; V = 10V. Choice (B)

39. Consider a permanent magnet dc (PMDC) motor which is initially at rest. At t = 0, a dc voltage of 5 V is applied to the motor. Its speed monotonically increases form 0 rad/s to 6.32 rad/s in 0.5 s and finally settles to 10 rad/s. Assuming that the armature inductance of the motor is negligible, the transfer function for the motor is

(A) 𝟐

𝒔 + 𝟎.𝟓

(B) 𝟏𝟎

𝒔 + 𝟎.𝟓

(C) 𝟏𝟎

𝟎.𝟓𝒔 + 𝟏

(D) 𝟐

𝟎.𝟓𝒔 + 𝟏

Sol. 0 rad/sec to 6.32 rad/sec means 63.2 percent of its final value (final value is 10 rad/sec), in 0.5 seconds

means time constant () = 0.5 sec

From the given options, option (D)

i.e., 2

0.5𝑠+1 satisfies

= 0.5, final value

= Lt𝑠0

𝑆.5

𝑠

2

0.5𝑠+1 = 10

Choice (D)

RL

4A

= 2.5 4A

+

–

10V

4A

RL I = 10

RL 10V ±

`


40. Which of the following options is true for a linear time-invariant discrete time system that obeys

the difference equations: y[n] – ay[n – 1] = b0x[n] – b1 x[n – 1] (A) The system is necessarily causal. (B) y[n] is unaffected by the values of x[n – k]; k > 2. (C) The system impulse response is non-zero at infinitely many instants. (D) When x[n] = 0, n < 0, the function y[n]; n > 0 is solely determined by the function x[n].

Sol. Given y[n] = a y[n – 1] + b0 x[n] – b1 x[n – 1] The above system represents a recursive system , i.e., output depends on previous output, present and

past inputs. Such system will have infinite impulse response. Option C is correct. Choice (C)

41. A non-ideal Si-based pn junction diode is tested by sweeping the bias applied across its terminals

form –5 V to +5 V. The effective thermal voltage, VT, for the diode is measured to be (29 2) Mv. The resolution of the voltage source in the measurement range is 1 Mv. The percentage uncertainty (rounded off to 2 decimal places) in the measured current at a bias voltage of 0.02 V is ________.

Sol. VT = (29 2) mv = 27 mv to 31 mv

wKTI = I0 exp (𝑉

𝑉𝑇) = I0 exp (

𝑉

𝑉𝑇) (n = 1)

at VT = 27 mv, I1 = I0 exp (0.02

27 × 10–3) = 2.09 I0

at VT = 31 mv, I2 = I0 exp (0.02

31 × 10–3) = 1.906 I0

at mean value of VT = 27 + 31

2 = 29 mv, I3 = I0 exp (

0.02

29 × 10–3)

= 1.998 I0

% Uncertainty = 𝑟𝑎𝑛𝑔𝑒 / 2

𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 × 100 %

=

(2.097 – 1.906)

2

1.998 × 100% = 4.77% Ans: 4.77

42. A non-ideal diode is based with a voltage of –0.03 V, and a diode current of I1 is measured. The thermal voltage is 26 mV and the ideality factor for the diode is 15/13. The voltage, in V, at which the measured current increases to 1.5I1 is closest to: (A) –4.50 (B) –1.50 (C) –0.02 (D) –0.09

Sol. = 15

13 = 1.154

VT = 26 mV V1 = –0.03 V at I = I1 V2 = ? at I2 = 1.5 I1

We know that I = I0 [𝑒𝑥𝑝 (𝑉

𝑉𝑇) – 1]

𝐼2

𝐼1 =

[exp (V2VT

) – 1]

[exp (V1VT

) – 1]

1.5 = 𝑒𝑥𝑝 (

𝑉21.154 × 10–3 × 26

) – 1

𝑒𝑥𝑝 (–0.03

1.154 × 26 × 10–3)

`


exp [𝑉2

1.154 × 26 × 10–3) – 1 = 1.5 × (–0.632)

= –0.948

𝑒𝑥𝑝 (𝑉2

1.154 × 26 × 10–3) = 1 – 0.948 = 0.052

V2 = 1.154 × 26 × 10–3 /n (0.052) = –88.7 mV = –0.088 Volt

– 0.09 Volt Choice (D)

43. In the dc-dc converter circuit shown, switch Q is switched at a frequency of 10 kHz with a duty ratio of 0.6. All components of the circuit are ideal, and the initial current in the inductor is zero. Energy stored in the inductor in mJ (rounded off to 2 decimal places) at the end of 10 complete switching cycles is ________.

Sol.

f = 10 kHz

D = 0.6

Q is ON: D OFF

(0 to 0.6 T)

VL = + 50 V

Inductor starts charging:

IL: 0 Imax

Q is OFF: D ON

(0.6T – T)

Inductor starts charging:

IL : Imax Imin & VL = – 50 V

50 V

10 mH

50 V

D

Q

`


∫ 𝑑𝐼𝐿𝐼𝑚𝑎𝑥

𝑜 = ∫

𝑉𝐿

𝐿

0.6 𝑇

𝑜 𝑑𝑡

Imax = 50

10 × 10–3 0.6 T = 30

10 ×10–3 10 ×10–3 (∵ T =

1

𝑓)

= 0.3

The time available for discharging is 0.4 T& Voltage across inductor is – 50 V

∴∫ 𝑑𝐼𝐿𝐼𝑚𝑖𝑛

𝐼𝑚𝑎𝑥 ∫

𝑉𝐿

𝐿

0.4 𝑇 dt

Imin1 – Imax = –50

10 ×10–3 × 0.4 T = –20

10 ×10–310 ×10–3 (∵ T = 1

𝑓)

Imin1 – 0.3 = – 0.2

I min1 =0.1 A

Next cycle, Inductor current will be:

Imax2 = 0.4 A (ON - Period)

Imin2 = 0.2 A (OFF - Period)

End of 10th cycle, Imin10 = 10 × 0.1

= 1A

Energy stored in inductor, E = 1

2 LI2

= 1

2 × 10 × 10–3 × 1 = 5mJ Ans: 5

44. An 8085 microprocessor accesses two memory locations (2001H) and (2002H), that contain 8-bit numbers 98H and B1H, respectively. The following program is executed:

LXI H, 2001H MVI A, 21H INX H ADD M INX H MOV M, A HLT At the end of this program, the memory location 2003H contains the number in decimal (base 10) form ________.

`


Sol. LXI H, 2001 H

MVI A, 21 H

INX H HL + 1

ADD M [A] + data @ reference of HL pair

21H + B1H = D2H [A]

INX H [HL] + 1 2002H + 1H 2003H

Mov M, A [A] to memory

2003H D2

D 161 + 2 160 = (210)10 Ans: 210

45. A cylindrical rotor synchronous generator with constant real power output and constant terminal voltage is supplying 100 A current to a 0.9 lagging power factor load. An ideal reactor is now connected in parallel with the load, as a result of which the total lagging reactive power requirement of the load is twice the previous value while the real power remains unchanged. The armature current is now ________ A (rounded off to 2 decimal places).

Sol. (P, Vt) Constant, Cos = 0.9

P = Vt Ia Cos

Vt Ia1 Cos1 = Vt Ia2 Cos2

Ia1 Cos1 = Ia2 Cos2 ––––– (1)

tan = 𝑄

𝑃

tan1 = 𝑄

𝑃; tan2 =

2𝑄

𝑃;

tan2

tan1

= 2

tan2 = 2 tan1= 2 tan (cos –1(0.9))

tan2 = 0.9686

cos2 = 0.7183 From equation (1)

100 0.9 = Ia2 0.7183

Ia2 = 125.2958

125.3A Ans: 125.3

46. A resistor and a capacitor are connected in series to a 10 V dc supply through a switch. The switch

is closed at t = 0, and the capacitor voltage is found to cross 0 V at t = . 𝟒 𝝉 , where 𝝉 is the circuit

time constant. The absolute value of percentage change required in the initial capacitor voltage if

the zero crossing has to happen at t = 𝟎. 𝟐 𝝉 is __________ (rounded off to 2 decimal places).

Sol.

20 H L

01

21H A

20 H L

02

D2 A

10 V

t = 0 R

C

`


Initial voltage across capacitor is negative.

Vc (t) = 0 at t = 0.4

if zero crossing required at t = 0.2 then % change in initial voltage across capacitor is ______.

Vc(t) = 10 + (Vi – 10) e–t/ = 0

t = 0.4

10 + 0.67 Vi – 10 × 0.67 = 0

Vi = –4.9253 –––– (1)

10 + (Vi – 10)e–0.2 = 0

10 + Vi – 8.1873 = 0

Vi = –2.214 –––– (2)

% change = −4.9253+2.214

−4.9253 × 100 = 55% Ans: 55

47. For real numbers, x and y, with y = 3x2 + 3x + 1, the maximum and minimum value of y for x [–2, 0] are respectively, ________. (A) –2 and –1/2. (B) 1 and 1/4. (C) 7 and 1. (D) 7 and 1/4.

Sol. Given y = 3x2 + 3x + 1

yI = 6x + 3

yI = 0 6x + 3 = 0 x = –1

2

The maximum value of y in [–2, 0] = Maxy (–1

2) , y(– 2), y(0)

= Max 1

4, 7,1

= 7

The minimum value of y in [–2, 0] = Miny (–1

2) , y(– 2), y(0)

= Min 1

4, 7,1

= 1

4 Choice (D)

48. Consider a negative unity feedback system with the forward path transfer function 𝒔𝟐 + 𝒔 + 𝟏

𝒔𝟑 + 𝟐𝒔𝟐 + 𝟐𝒔 + 𝒌,

where K is a positive real number. The value of K for which the system will have some of its poles on the imaginary axis is ________. (A) 9 (B) 7 (C) 8 (D) 6

Sol. G(s) = 𝑠2+𝑠+1

𝑠3+2𝑠2+2𝑠 +𝑘

C.E = s3 + 2s2 + 2s + k + s2 + s + 1= 0

C.E = s3 + 3s2 + 3s + k + 1 = 0

9 = k +1

K = 8 Choice (C)

`


49. Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of

which there are ideal transformers with turns ratios as shown. Bus admittance matrix for the system is:

(A) [𝒕𝒊𝒕𝒋𝒀 −(𝒕𝒊 − 𝒕𝒋)

𝟐𝒀

−(𝒕𝒊 − 𝒕𝒋)𝟐

𝒀 𝒕𝒊𝒕𝒋𝒀]

(B) [𝒕𝒊𝒕𝒋𝒀 −𝒕𝒋

𝟐𝒀

−𝒕𝒋𝟐𝒀 𝒕𝒊𝒕𝒋𝒀

]

(C) [−𝒕𝒊𝒕𝒋𝒀 𝒕𝒋

𝟐𝒀

𝒕𝒋𝟐𝒀 −𝒕𝒊𝒕𝒋𝒀

]

(D) [𝒕𝒋

𝟐𝒀 −𝒕𝒊𝒕𝒋𝒀

−𝒕𝒊𝒕𝒋𝒀 𝒕𝒋𝟐𝒀

]

Sol.

𝑰𝒊

𝒕𝒊 = (vi ti vj tj) Y

Ii = vi ti2 Y vj tj Y ti → (1)

𝑰𝒋

𝒕𝒋 = (vj tj vi ti)Y

Ij = Yvj tj2 vi ti tj Y→ (2)

[𝐼𝑖𝐼𝐽

] = [𝑡𝑖

2 𝑌 𝑡𝑗 𝑡𝑖𝑌

ti tj Y 𝑡𝑗2 𝑌

] [𝑉𝑖𝑉𝐽

] Choice (D)

50. A stable real linear time-invariant system with single pole at p, has a transfer function

H (s) = 𝒔𝟐 + 𝟏𝟎𝟎

𝒔 − 𝒑 with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest

to:

(A) 78.13

(B) 8.84

(C) 11.08

(D) 122.87

V1 V2

Bus2 Bus1 Y

1 : ti tj : 1

Vi

`


Sol. H(s) = 𝑠2+100

𝑠 –𝑝

H(s) = 𝑠2 + 100

𝑠 –20

|H(j)| = 1 = |–2+100|

√2+400

(–2 + 100)2 = 2 + 400

x2 + 104 – 200x = x + 400

x2 – 201x + 9600 = 0

x = 122.86, 78.133

= 11.08, 8.83

The smallest possible is 8.83 Choice (B)

51. The vector function expressed by F = ax (5y – k1z) + ay (3z + k2x) + az (k3y – 4x) represents a conservative field, where ax, ay, az are unit vectors along x, y and z directions, respectively. The values of constants k1, k2, k3 are given by: (A) k1 = 0, k2 = 0, k3 = 0 (B) k1 = 3, k2 = 8, k3 = 5 (C) k1 = 4, k2 = 5, k3 = 3 (D) k1 = 3, k2 = 3, k3 = 7

Sol. Given that F = ax(5y – k1z) + ay(3z + k2x) + az(k3y – 4x) is a conservative field.

Curl F = 0

|

𝐚𝐱 𝐚𝐲 𝐚𝐳

∂

∂x

∂

∂y

∂

∂z

5y – k1z 3z – k2x k3y − 4x

| = 0

(k3 – 3)ax – (–4 + k1)ay + (k2 – 5)az = 0

k3 – 3 = 0, – 4 + k1 = 0, k2 – 5 = 0

k3 = 3, k1 = 4, k2 = 5 Choice (C)

52. The static electric field inside a dielectric medium with relative permittivity, r = 2.25, expressed in cylindrical coordinate system is given by the following expression

E = ar 2r + a (𝟑

𝒓) + az 6 where ar, a, az are unit vectors along r, and z directions, respectively. If

the above expression represents a valid electrostatic field inside the medium, then the volume

charge density associated with this field in terms of free space permittivity, 0, in SI units is given by:

(A) 50

(B) 40 (C) 90 (D) 30

Sol. D = v, the volume charge density

(εo εv E) = v

2.25 εo [1

r

𝜕

𝜕r(𝑟 𝐸𝑟) +

1

r

𝜕𝐸∅

𝜕∅+

𝜕𝐸𝑧

𝜕𝑧] = v

v = 2.25 εo [1

r

𝜕

𝜕r(2𝑟2) +

1

r

𝜕

𝜕∅ (

3

r) +

𝜕

𝜕𝑧 (6)]

= 2.25 εo[4 + 0 + 0] =9.00 εo Choice (C)

`


53. Suppose for input x(t) a linear time-invariant system with impulse response h(t) produces output

y(t), so that x(t) * h(t) = y(t). Further, if |x(t)| * |h(t)| = z(t), which of the following statements is true?

(A) For all t (–, ), z(t) y(t)

(B) For some but not all t (–, ), z(t) y(t)

(C) For all t (–, ), z(t) y(t)

(D) For some but not all t (–, ), z(t) y(t)

Sol. Given x(t) * h(t) = y(t)

|𝑥(𝑡)| * |ℎ(𝑡)| = |𝑧(𝑡)|

x(t) (or) h(t) can be negative (i.e < 0)

y(t) can be negative

|𝑥(𝑡)| & |ℎ(𝑡)| always positive

∴z(t) always positive

∴z(t) ≥ y(t) for all ‘t’ Choice (A)

54. Let ar, a and az be unit vectors r, and z directions, respectively in the cylindrical coordinate

system. For the electric flux density given by D = (ar 15 + a 2r – az 3rz) Coulomb/m2, the total

electric flux, in Coulomb, emanating from the volume enclosed by a solid cylinder of radius 3 m

and height 5 m oriented along the z-axis with its base at the origin is:

(A) 180

(B) 90

(C) 108

(D) 54

Sol. Total flux emanating from the cylinder = ∯ D 𝑑S

∯ D 𝑑S = ∬ D 𝑑S

𝑡𝑜𝑝 ∬ D 𝑑S

𝑏𝑜𝑡𝑡𝑜𝑚+ ∬ D 𝑑S

𝑠𝑖𝑑𝑒𝑠

= ∫ ∫ −3𝑟2𝑧3

𝑟=0

2𝜋

∅=0 dr d∅ + ∫ ∫ 3𝑟2𝑧

3

𝑟=0

2𝜋

∅=0 dn d∅ + ∫ ∫ 15r d∅ dz

5

𝑧=0

2𝜋

∅=0

z = 8 z = 0 r = 3

= –270 + 0 + 450

= 180C Choice (A)

`


55. Consider the diode circuit shown below. The diode, D, obeys the current-voltage characteristic ID

= IS (𝐞𝐱𝐩 (𝒗𝒑

𝒏𝑽𝑻) – 𝟏) , where n > 1, VT > 0, VD is the voltage across the diode and ID is the current

through it. The circuit is biased so that voltage, V > 0 and current, I < 0. If you had to design this

circuit to transfer maximum power from the current source (I1) to a resistive load (not shown) at

the output, what values of R1 and R2 would you choose?

(A) Small R1 and small R2.

(B) Large R1 and small R2.

(C) Small R1 and large R2.

(D) Large R1 and large R2.

Sol. As > 1, diode is Si and Vbi = 0.7V for Si

As I < 0, I = –ve flows from cathode to anode of diode and it is ON

For I < O, and V > O i.e: V = Vbi, R2 must be very high and R1 to be very low Choice (C)

+

–

I1

D

R2

R1

I

I

V

Vbi = 0.7V I1

–

↑

R1

R2 RL

V

+ (V > 0volt)

I

`


electrical engineering date of exam: 08/02/2020 · gate-2020 original paper – ee/1 general...

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