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GATE-2020

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ELECTRICAL ENGINEERING

GATE - 2020ELECTRICAL ENGINEERING

Question with Detailed Solutions08/02/20

SUBJECTWISE WEIGHTAGE

S.No. Name of the Subject 1 Mark 2 Marks Total

01 Electric Circuits 2 4 6

02 EM Theory 0 8 8

03 Signals and Systems 2 4 6

04 Electrical Machines 3 8 11

05 Power Systems 3 6 9

06 Control Systems 3 10 13

07 Electrical & Electronic Measurements 1 1 3

08 Digital Electronics & Microprocessors 1 2 3

09 Analog Electronics 2 6 8

10 Power Electronics 4 4 8

11 Engineering Mathematics 4 6 10

12 Aptitude 4 4 8

13 English 1 6 7

Total Marks 100

2 Electrical Engineering

ACE Engineering Publications Hyderabad Delhi Pune Bhubaneswar Bengaluru Lucknow Chennai Vijayawada Vizag Tirupati Kolkata Ahmedabad

Section : General AptitudeQ. 01 to Q. 05 carry ONE mark:

01. In four-digit integer numbers from 1001 to 9999 the digit group “37” (in the same sequence) appears _______ times.

(a) 299 (b) 270 (c) 279 (d) 28001. Ans: (c)Sol: Four digit integer numbers from 1001 to 9999, with 37 as sequence appears in

Case-1: Case-2:

egersint

3 7

10 10 100#

. .

. . egerint

3 7

9 90

. .

10#

Case-3:

ways

3 7

9 90

. .

10#

Total = 100 + 90 + 90 = 280 ways 3737 repeated two times ∴ Total number of ways = 280 − 1 = 279

02. Given a semicircle with ‘O’ as the centre, as shown in the figure, the ratio AB

AC CB+ is _______ . Where

,AC CB and AB are chords. C

A BO

(a) 2 (b) 2 (c) 3 (d) 3

02. Ans: (a)Sol: AB = Diameter = 2r OC = radius = r = AO=OB Consider is ∆ AOC Using pythagerous theorem AC2 = r2 +r2 ⇒AC2 =2r 2

A r 0

C

rr2

AC r 2& =

Similarly ∆ COB BC r 2& =

So, ABAC CB

rr r

22 2

2=

3 GATE_2020_Questions with Solutions




03. 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 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) 45 (b) 180 (c) 135 (d) 9003. Ans: (d)Sol: Based on the given paragraph the holding period of loans in 2004 after the third revision equal to quarter days

of the year.

= 41

360# = 90 days

04. 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) 8 (c) 5 (d) 6

04. Ans: (a)Sol: Team size more than 1. Case 1: Team size = 2 As Q is fixed, only 1 member to be selected among P,R & S 3C1 = 3 ways Case 2: Team size = 3 As Q is fixed, only 2 members to be selected among P,R & S 3C2 = 3 ways Case 3: Team size = 4 As Q is fixed, only 3 members to be selected among P,R & S 3C2 = 1 ways Case 4: Team size ≥ 5 Not possible So, total possible ways = 3+3+1 =7 ways

05. This book, including all its chapters, _______ interesting. The students as well as the instructor ________ in agreement about it.

(a) is, are (b) is, was (c) are, are (d) were, was05. Ans: (a)Sol: Subject + Verb + agreement



Q. 06 to Q. 10 carry TWO marks:

06. Select the word that fits the analogy Do : Undo :: Trust : ________ . (a) Distrust (b) Untrust (c) Intruct (d) Entrust06. Ans: (a)Sol: The obvious analogical relationship is do and undo are antonyms just as trust and distrust are antonyms

07. People were prohibited _______ their vehicles near the entrance of the main administrative building (a) to have parked (b) to park (c) parking (d) from parking07. Ans: (d)Sol: Prohibited from parking means “to officially refuse allow something”

08. 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 that in 2014, and company Q had earned profit of 10% on expenditure in 2014, then its expenditure (in million rupees) in 2014 was _________

35

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

ExpenditureRevenue

30

4540

50

55

252015

5

0

10

Rev

enue

/Exp

endi

ture

(in m

illio

n ru

pees

)

Company P Company Q Company R Company S

(a) 35.1 (b) 34.1 (c) 32.7 (d) 33.7

08. Ans: (b)

Sol: Revenue of company Q in 2015 = 45 revenue of company Q in 2015 was 20% more than that in 2014.

R2015 = 120% R2014 ( R-Revenue)

R45 100120

2014=

R2014 = 37.5

Revenue of company Q in 2014 =37.5



10% profit in 2014 for company Q.

⇒ Revenue Q2014 = 110% Expenditure Q2014

.37 5 100110= Expenditure Q2014

Expenditure Q2014 .11375

34 1= =

(Note: profit & loss is calculated on Expenditure)

09. Select the next element of the series Z, WV, RQP, ______ . (a) JIHG (b) LKJJ (c) NMLK (d) KJIH

09. Ans: (d)

Sol: , ,Z WV RQP KJIH

YX UTS ONML. . .

10. Stock markets ________ at the news of the coup (a) probed (b) poised (c) plunged (d) plugged

10. Ans: (c)Sol: Plunged means plummet especially (prices, temperature, etc.) to decrease suddenly and quickly.

Section : Electrical Engineering

01. xR, xA are, respectively, the rms and average values of x(t) = x( t − T). and similarly, 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

01. Ans: (a)Sol: X(t) = X(t − T) ↓ Represents a periodic waveform with time period ‘T’. XR → RMS value XA → Average value

x(t)

T/2 T 2Tt

0

Now y(t) = Kx(t) = Kx(t −T) Ex: So, YR = K

2

y(t)

T/2 T 3T/2 2Tt

0

K

So, YA = KXA, YR = KXR



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

(a) base power of the generator (b) rated power output of the generator (c) rated voltage of the generator (d) solution of economic load dispatch 02. Ans: (d)

03. 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 mA/V, the voltage gain of the amplifier is closest to _______

(a) 1.22 (b) −2.44 (c) −1.22 (d) 2.44

03 Ans:(b)

Sol: The voltage gain of a common source amplifier is Av = −gm Rd

= .520 10 4 7 106 3# # #− −

= −2.44

04. 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 per unit current for operation of the distance relay for a solid three-phase to ground fault on the transmission line is closest to ______

(a) 6.25 (b) 5.00 (c) 3.61 (d) 1.0004. Ans: (a)Sol: Reactance of transmission line, x, = 0.2 p.u

Generator

Distance Relay

Protective zone

load

Xl

The distance relay will be protecting upto 80% length of line. The terminal voltage of generator, |Vt| = 1 p.u.

For a 3-φ to ground fault at the end of protective zone, the fault current will be, .I xV0 8f

t=,

. .0 8 0 21#

= = 6.25 p.u So, the relay will operate for a threshold current of 6.25 A



05. Consider the initial value problem below. The value of y at x = ln(2) is _______ (rounded off to 3 decimal places)

dx

dyx y2 , y(0) = 1

05. Ans: 0.886

Sol: dx

dyx y2

dx

dyy x2

.I F e e.dx x1= =#

ye xe dx2x x` = # = 2ex(x−1) + C y(0) = 1 ⇒ 1 = − 2 + C ∴ C = 3 ∴ yex = 2ex (x −1) +3 at x = ln2 → 2y = 4 (ln2 −1) +3 = 4 (0.693 −1) + 3 ∴ y = 0.886

06. A single phase, 4 kVA, 200V/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 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)

06. Ans: 162.5 WSol: At 200 V, 50 Hz coreloss, Wi1 = 450 W At 160 V, 40 Hz coreloss, Wi1 = 320 W What is Iron loss at 100V, 25 Hz

Verification of fV ratio

50

20040160=

fV = constant, Bm = constant

Then Wi = Af+Bf2

450 = A (50)+B(50)2....... (1) 320 = A (40)+B(40)2....... (2) By solving (1) & (2) A = 4 & B = 0.1 Wi at 00 V, 25 Hz =Af+Bf2

= 4 (25)+0.1(25)2 = 162.5 W



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

c

11

e

a d

11

e

a(d+c)(a) 11

e

(a)

− cd1ca

11

e

(c)

− cd1da

(a+c)d 11

e

(d)

07. Ans: (c)Sol: Apply the Mason’s gain formula to the

given SFG T.F = ade dcad

1 − − .........(1)

apply the Mason’s gain formula to the option (c) T.F =

cdade

acdd

ade dcad

11

11

.........(2)

(1) = (2)

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

equation ( )

( ) ( )dt

d y ty t r t4 6

2

2

, the poles of this systems are at

(a) + 4j, −4j (b) + 2, −2 (c) +4, −4 (d) +2j, −2j08. Ans: (d)Sol: Applying the laplace transform s2Y(s) + 4Y(s) = 6R(s)

.( )

( )T F

R s

Y s

s 462

s2 + 4 = 0 ∴poles are s = ± j2

09. 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, 150 Hz (b) 50 Hz, 100 Hz (c) 50 Hz, 0 Hz (d) 150 Hz, 250 Hz



09. Ans: (a) Sol: 1-φ full wave diode bridge rectifier is fed from 230 V, 50 Hz sinusoidal source having supplied with large value of

inductance, then shape of supply current would be square wave. For the square wave, if fourier analysis is carried out, it will wave

sini nI

n t4

,s

n

0

1 3

3

=

& 0/ The frequency components, they are present in wave form are 1,3,5,7,9, ......etc. and having frequencies of 50 Hz,

150 Hz, 250 Hz, 350 Hz.... etc.10. The Thevenin equivalent voltage VTh, in V ( rounded off to 2 decimal places) of the network shown below is

_______ .

5 A4 V Vth3 Ω

2 Ω 3 Ω +

−

+−

10. Ans: 14 Sol: Thevenin;s voltage (VTh) VTh

5 A4 V Vth3 Ω

2 Ω 3 Ω

+−

( )V

2

45

Th −− = 0 ⇒ VTh = 14 volts

11. 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) ( )sin sinm n d1

00

# (b) ( ) ( )sin sinp q d2

10

/

/

2

2

#

(c) sin sinLim p q d2

10

"3

# (d) ( ) ( )sin cosp q d2

10

#

11. Ans: (d)

Sol: ( ) ( )sin cosp q d2

1

# , the integrand is an odd function its value is zero.



12. A 3-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 motor (rounded off to 2 decimal places) is ________ .

12. Ans: 4.21Sol: Poles P = 4, frequency (f) = 50 Hz

synchronous speed Ns = Pf120 = 1500 rpm

No load slip = 1% No load speed ( )N N S1rn s n , , = 1500 (1 − 0.01) = 1485 rpm full load slip ( Sf, ) = 5% full load speed (Nrfl) = ( )N S1s f− , = 1500 (1 −0.05) = 1425 rpm

Speed Regulation = NN N

100rf

rn rf #−,

, ,

= 4.21%

13. Consider a negative unity feedback system with forward path transfer function G(s) = ( ) ( ) ( )s a s b s c

K

,

where K, a, b, c are positive real numbers. For a Nyquist plot enclosing the entire imaginary axis and right half

of s-plane in clockwise direction, the Nyquist plot of [1 + G(s)], encircles the origin of [1 + G(s)]-plane once

in clockwise direction and never passes through this origin for a certain value of ‘K’. The number of poles of

( )

( )

G s

G s

1 + lying in the open right half of the s-plane is _______ .

13. Ans: 2 Sol: P = 1 N = −1 N = P − Z Z = P − N = 1 − (−1) Z = 2

GHG

1 + has two poles in the right half of s-plane

14. 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 statements is true?

(a) a, b, c, d can be chosen to ensure that all roots are complex (b) c alone can’t ensure that all roots are real (c) d can be chosen to ensure that x = 0 is a root for any given set a, b, c (d) no choice of coefficients can make all roots identical



14 Ans: (c)Sol: Given that ax3 + bx2 + cx + d = 0 Suppose d = 0 and atleast one of a,b,c ≠ 0 then x = 0 is a root . ‘C’ is true.

15. Consider a signal x(n) = ( )n2

11n

d n , where 1(n) = 0 if n < 0 and 1(n) = 1 if n ≥ 0.

The Z-Transform of x(n − k ), k > 0 is z

z

12

1

k

1− −

−

with region of convergence being

(a) |z| < 2 (b) |z| > 2

1 (c) |z| < 2

1 (d) |z| > 215. Ans: (b)

Sol: ( ) ( )x n u n21 n

= b l

( ) ; | |x zz

z

zz

21

1 211

21

1&

-

( ) . ( )x n k z x zk)− -

zz z

z

1 211

1 21

kk

1 1)

-

- -

-

|z| > 21

Time shifting property will not effect ROC

16. The value of the following complex integral, with ‘c’ representing the unit circle centered at origin in the counter

clockwise sense, is: Z Z

ZdZ

2

1

c

2

2

#

(a) −8pi (b) pi (c) 8pi (d) −pi

16. Ans: (d)

Sol: z zz

dz21

C

2

2

] g# ( C: |z| = 1)

z zz

dz21

C

2

] g#

z = 0 is inside It is rotating in anti-clockwise direction

i Lt zz zz

221

z 0

2

"d ] g n

= i221 b l

= −pi



17. A single 50 Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to increase 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)

17. Ans: 130Sol: A generator supplying 100 MW power to a system at 50 Hz frequency. The demand was increased by 10 MW and the frequency was reduced to 49.75 Hz. Change in demand, ∆P = 10 MW Change in frequency, ∆f = 0.25 Hz

Speed regulation, R = .

pf

10

0 25

OO = Hz/MW

= 0.025 Hz/MW The power supplied by the generator at 49.75 Hz is 100 + 10 = 110 MW

10MW 20MW100MW Power

49.75 Hz50 Hz

Frequency

49.25 Hz

At this present operation again load was increases by a value PnewO for which the new frequency was 49.25 Hz. that in . . .f Hz49 75 49 25 0 5newO

As pf

Rnew

new

OO =

..

p 0 0250 5

newO = = 20 MW

The power supplied by the generator at 49.25 Hz is 110 + 20 = 130 MW

18. A 3- phase cylindrical rotor synchronous generator has a synchronous reactance XS and a negligible armature resistance. The magnitude of per phase terminal voltage VA and magnitude of per phase induced emf is EA.

Consider the following statements, P and Q P: For any 3-phase balanced leading load connected across the terminal of this synchronous generator, VA is

always more than EA

Q: For any 3-phase balanced lagging load connected across the terminals of this synchronous generator VA is always less than EA

Which of the following options are correct. (a) P is false Q is true (b) P is true Q is false



(c) P is false Q is false (d) P is true Q is true18. Ans: (a)Sol: P: For any 3-phase balanced leading load, VA is always more than EA. For leading loads VA is depends on the magnitude of load and load power factor for a given excitation. The regu-

lation may be positive, zero or negative for leading PF loads.

Regulation = | |

| | | |

V

E V−

If EA > VA ⇒ regulation in positive EA = VA ⇒ regulation zero EA < VA ⇒ regulation negative This can be observed from the phasor diagram

EAEA

Ia

EA

jIaXs

VA VA

VA < EAlead pf

VA = EAlead pf

VA > EAlead pf

VA

Ia

jIaXsIa

Therefore statement ‘P’ is false. Q: For any 3-phase balanced lagging load VA is always is less than EA

The voltage regulation is always positive for lag PF loads ∴VA is always less than EA. This can be observed from the phasor diagram also.

E

V

jIa ∝s

Ia

∴Statement Q is true. ∴P is false, Q is correct.

19. A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode, D, represented approximately as a piecewise linear plot of current versus time at diode turn-off. Lpar is the parasitic inductance due to 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 IGBT experiences the highest current stress is _______ .

time

1

Diodecurrent

2

3

4+− Rload

Lload

Lpar

D

Vsource

IGBT

19. Ans: 3Sol: At point 3 current through switch subject abrupt change in

dt

di. Hence at that point IGBT subject to the hight

current stress.



20. Current through ammeters A2 and A3 in the figure are 1∠10° and 1∠70° respectively. The reading of ammeter A1 (rounded off to 3 decimal places) is _______ A.

A2

A3

A1

I1

I2

I3

20. 1.732Sol: I I I1 2 3 r r r

= [1∠10°] + [1∠70°] = [0.985 + j0.173] + [0.342 + j0.939] = [1.327 + j1.113] I1r = 1.732∠40° A

21. The cross section of a metal-oxide semiconductor structure is shown schematically. Staring from an uncharged condition a bias of +3 V is applied to the gate contact with respect to body contact. The charge inside the silicon dioxide layer is the 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 _____

Silicon Dioxide

Si

DASHED BOX

GATE

BODY

21. Ans: 0

Sol: The given figure is MOSFET, it behaves like a capacitor. The total charge inside the dotted line is zero.

22. Which of the following statements is true about the two-sided Laplace transform?(a) It has no poles for any bound signal that is non-zero only inside a finite time interval.(b) It exists for every signal that may or may not have a Fourier Transform(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 zeros must be equal.

22. Ans: (a)



Sol: It has no poles for any bounded signal i.e., non-zero only inside a finite time interval.23. Thyristor T1 is triggered at angle a (in degree) and T2 at angle 180° + a in each cycle of the sinusoidal input

voltage. Assume both thyristor to be ideal. To control the load power over the range 0 to 2 kW. The minimum range of variation in a is

T1

T2200 V50 Hz

+10∠−60° Ω

−

(a) 0° to 120° (b) 60° to 120° (c) 60° to 180° (d) 0° to 60°23. Ans: (a)Sol:

ωt

ωt

ι

V

120°

A 1-φ full wave AC voltage controller feeding capacitive load 10∠−60° means current loads voltage by 60°. It is possible to have control for a range of 0°−120°. In the other range thyristor gets commutated. Hence range of firing angle is 0°−120°

24. 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, 1, 0, 1, 1, 0 (b) 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0 (c) 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0 (d) 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 024. Ans: (b)Sol: The given input data 1 1 0 1 0 0 1 1 0 1 0 1 1 0 The sequence detector is for 101 with overlapping sequence. Input : 1 1 0 1 0 0 1 1 0 1 0 1 1 0 Output : 0 0 0 1 0 0 0 0 0 1 0 1 0 0 No correct option. But 101 sequence is there for 3 times in the given data. So the nearest option is (b)



25. 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 wave form n(t) as shown. The angle ‘s’ is adjusted to entirely eliminate 3rd harmonic component from the output voltage. Under this condition, for n(t), the magnitude of 5th harmonic component as a percentage of the magnitude of the fundamental component is ______ . (rounded off to two decimal places)

σ

σ

σ

σ σπ/2

3π/2

π 2π 3π

σ

σ σ

7π/2

4π5π/2ωt

σ

ν(t)

0

−100 V

25. −20Sol: The wave for for m is produced in 1-φ inverter using single pulse modulation the value of s is so selected that it

is free from 3rd harmonic. It is only for a pulse width of 120° i.e s = 30° Fourier analysis expression of output voltage wave form is

( ) sin sin sinV t nV n

nd n t4

2,

s

n 1 3

3

=

& 0/

. ( )

. ( )

sin sin

sin sin

VV

V

V

42 60

2

1

54

25

5 602

1

i

s

s

s #

b

b l

l

//

%sinsin

sinsin

51

25 2

60300

100# # #

100V

2dπ

ωt120°

σ=30° σ=30°

2d=120°V(t)d=60°

%V

V20

1

5

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

Q12Bus-1 Bus-2

V1 V2

26. Ans: 1.12Sol: Real power flow,

| | | |. sinP X

V V1 2

As P = 0, sinδ = 0, δ = 0°

Reactive power, | |

| | | |Q XV

V V121

1 2 7 A

X

Q12

|V2|∠0°|V1|∠0°



|V2| is fixed at 1 p.u.

| |

| |Q XV

V 1121

1 7 A

With |V1|old = 1.1 p.u, assume that Q12 old = Q Q12 was increased by 20%, that is Q12new = 1.2 Q with |V2|new

| |

| |

| || |

QQ

XV

V

XV

V

1

1

old

new

oldold

newnew

12

12

11

11

77 A

A

.. ( . )

| | | |V V1 2

1 1 1 1 1

1new new1 1 7 A

|V1|new [|V1|new−1] = 0.132 |V1|

2new |V1|new− 0.132 = 0

| |

.V

2

1 1 4 0 132new1

2! #

= .

2

1 1 236!

|V1|new = 1.118 , −0.118 Suitable value of |V1|new = 1.12

27. 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 statement is true?

(a) for all t∈ (−∞, ∞) z(t) ≥ y(t) (b) for all t∈(−∞, ∞) z(t) ≤ y(t) (c) for some but not all t∈(−∞, ∞) z(t) ≥ y(t) (d) for some but not all t∈(−∞, ∞) z(t) ≤ y(t)

27. Ans: (a)Sol: x(t) ∗ h(t) = y(t) |x(t)| ∗ |h(t)| = z(t) For any two signals amplitudes A.B ≤ |A|. |B|

28. 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 and K3 are given by (a) K1 = 3, K2 = 3, K3 = 7 (b) K1 = 0, K2 = 0, K3 = 0 (c) K1 = 4, K2 = 5, K3 = 3 (d) K1 = 3, K2 = 8, K3 = 5



28. Ans: (c)Sol: Given F a y k z a z k x a k y x5 3 4x y z1 2 3 v t t t_ ^ _i h i As Fv is conservative (or) irrotational and hence F 0#d =v (or) A null vector. . . .

a

x

y k z

a

y

z k x

a

z

k y x

a a a

5 3 4

0 0 0

x y z

x y z

1 2 3

22

22

22

t t t

t t t

_ ^ _i h i k3 − 3 = 0 ⇒ k3 = 3 − 4 + k1 = 0 ⇒ k1 = 4 k2 − 5 = 0 ⇒ k2 = 5 ∴ k1 = 4, k2 = 5 & k3 = 3

29. A cylindrical rotor synchronous generator with constant real power output and constant terminal voltage is supplying 100 A current to a 0.9 lagging pf 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)

29. Ans: 125.29Sol: Cylindrical rotor synchronous generator with constant real power output and constant terminal voltage i.,e P = constant; V = constant Supplying 100 A to 0.9 lag PF load i.,e φ = 25.84°

I=100∠−25.84°

I = 100 cos25.84 − j100 sin25.84 = 90 − j43.58 An ideal reactor is connected in parallel with the load as the result, total Q is doubled, keeping ‘P’ constant. ∴Active component of current is same; where as reactive component is doubled

I =90−j43.58−j43.58

I = 90 − j43.58 − j43.58 = 90 − j87.16 = .90 87 162 2+ = 125.29 A



30. For real number x and y with y = 3x2 + 3x + 1, the maximum and minimum value of y for x ∈ [−2, 0] are respectively

(a) 1 and 4

1 (b) −2 and 2

1− (c) 7 and 4

1 (d) 7 and 1

30. Ans: (c)Sol: Let f(x) = 3x2+3x+1, x ∈(2, 0) f ‘ (x) = 0 ⇒(6x+3) = 0

( , )x 21

2 0!

f ‘’(x) = 6 > 0 ∴f (−2) = 7 f(0) =1

f 21

43

23

1 41 b l

∴ maximum value = 7 & minimum value = 1/4

31. Consider the diode circuit shown below. The diode, D, obeys current voltage characteristic , ,I I e where V1 1 0D S V

VTT

D h7 A . 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, the value of R1 and R2 would you choose?

I1

I

I

V

+

−

D

R1

R2

(a) Small R1 and large R2 (b) Large R1 and small R2

(c) Small R1 and small R2 (d) Large R1 and large R2

31. Ans: (a)Sol: Step (1): h > 1 ⇒ Let diode is silicon diode with VD = 0.7 V ∵ I < 0 (given), the diode is ON (∵ The current through R1 is to be reverse).

I1

I< 0

V>0

+

−

VD=0.7V

IDR1

R2 R

Step (2): For I to be less than zero, V to be more than zero (OR) for example say V ≈ VD , R2 should be very high and R1 to should be very low.



32. A non-ideal diode is biased with a voltage of −0.03 V, and a diode current of I1 is measured. The thermal voltageis 26 mV and the ideality factor for the diode is 15/13. The voltage, in V, at which the measured current increasesto 1.5I1 is closest to(a) − 4.50 (b) −1.50 (c) − 0.09 (d) − 0.02

32. Ans: (c)

Sol: Diode current ID = I e V

V

0 1T

D

8 B

I

I

e

e

1

1/

D

D

V

V V

/D V

D T

1

2

T1

2

Given . , ,V V mV0 03 1315

26D T1 ,

, .I I I I1 5D D1 11 2= =

V13

1526 10T

3# # #

= 0.03

.

I

I

e

e1 5

1

1/ .V

1

10 03

. / .

D

0 03 0 03

2#

⇒ −0.9481 = e 1/ .V 0 03D2 −

By applying logarithm and simplify we will getVD2 = −0.09

33. Let ar , aφ and az be unit vectors along r, φ and z directions respectively in the cylindrical coordinate system. Forthe electric flux density given by D = ( a a r a rz15 2 3r z zt t t ) Coulomb/m2, the total electric flux, in coulomb,emanating from the volume enclosed by a solid cylinder of radius 3m and height 5m oriented along the z-axiswith its base at the origin is _________(a) 108 p (b) 90 p (c) 180 p (d) 54 p

33. Ans: (c)Sol: Given electric flux density, /D a ra rz a c m15 2 3r z

2 zv t t t

top

bottom

sidez = 5

z = 0



From Gauss’s law

net . .D ds D d[ ]net electric flux

Method Methodols 1 2

d - -

o

v v v##

Method 1: . . . .D ds D ds D ds D ds

( ) ( ) ( )net

side bottomz

topzssss 3 0 5

t= = =

v v v v v v v v####

where ρ (or) r = 3 is the radians of cylinder

. ( ) . ( )

( ) . ( )

a rd dz a r z a rdrd a

rz a rdr d a

15 3

3

( ) ( )

( )

r rr

z zzrz

z z

zr

3 00

2

0

3

0

5

0

2

0

2

50

3

z

r

z

r

z

r

= =====

==

=

t t t t

t t

####

##

( ) ( )z r15 3 0 3

350

205

3

0

3

02 r r

= 450p − 270p C180net

34. A benchtop dc power supply acts as an ideal 4A 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) Open, 4 A, 10 V (b) 2.5 Ω, 4 A, 10 V (c) 2.5 Ω, 4 A, 5 V (d) Short, ∞ A, 10 V34. Ans: (b)Sol: +

V4V0 < V < 10V

RL

−

+I

V0 < I < 4 A

RL10V

−

Characteristics

So, 42(RḺ) = 40 W (OR) R

W10

40L

2

⇒RL = 2.5 Ω

Pmax = 40 W4A

0 10VV

i



35. In the dc-dc converter shown in figure, 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 _______ .

+−

Q

50 V

50 V

10 mHD

35. Ans: 5Sol: The given DC-DC converter is match with buck boost converter.

+−

Q

50 V

50 V

10 mHD

In DC-DC buck boost converter, for continuous conduction.

..

VDD

V V1 1 0 6

0 650 75s0 #

But given otuput voltage is 50V, so it is not continuous conduction mode. At the same time, it is not discontinuous conduction mode. (DC.m) because, in D.C m output voltage will be more than 50V.

so in every cycle inductor getting emerges in 0.6 T and deliver during 0.4 T duration. The slope of current during ON time (TON) and OFF time (TOFF) are as follows.

...........LV

t TS

ON#

........LV

t TOFF0 #

L = 10 mH T s

10 101

1003#

D = 0.6

200 mA 300 mA

100 mA tTTON 2T

3T

iL

L t50 # L t50 #−

During 1st cycle, at the end of TON,

.I mA10 1050

100 10 0 6 300L1 36

## # #= =-

-



at the end of T; IL2 = L T30050

OFF#−

.I 300 1010 1050

0 4 100 10L23

36#

## # # - -

-

= 100 mA At the end of 1st cycle it gets a current of 100 mA. It continuous so on at the end of 10th cycle, current through

inducted is 1000 mA. i.e., 1A ∴ Stored emerges in induced = LI2

1 2 = ( )21

10 10 13 2# # #- = 5 mJ

36. Consider a negative unity feedback system with the forward path transfer function s s s Ks s2 2

13 2

2

+ + ++ + , where K is

a positive real number. The value of K for which the system will have some of its poles in the imaginary axis is

_______ . (a) 7 (b) 9 (c) 8 (d) 636. Ans: (c)Sol: Characteristic equation

s s s ks s

12 2

103 2

2

s3 + 3s2 + 3s + k + 1 = 0

k 1

k 1

3

( )

s

s

s

s

k

1

3

39 1

3

2

1

0

9 −(k + 1) = 0 k = 8 3s2 + 9 = 0 s = j 3! poles are on the imaginary axis

37. A stable real linear time-invariant system with single pole at P, has a transfer function ( )H ss ps 1002

with a dc

gain of 5. The smallest positive frequency, in rad/s, at unity gain is closed to (a) 8.84 (b) 122.87 (c) 11.08 (d) 78.1337. Ans: (a)

Sol: j p

j 1005

2

^ h

At ω = 0, p = −20 TF =

ss

201002

++

j

j

20

1002

^ h =

20

10012 2

2

∴ω = 8.84 rad/sec



38. Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends for which there are ideal transformers with turns ratios as shown bus admittance matrix for the system is

Vi

Bus-i Bus-j tj : 1

Y

1 : ti

Vj

(a) t t Yt Y

t Y

t t Y

i j

i

j

i j2

2

−−= G (b) t t Y

t Y

t Y

t t Y

i j

i

j

i j2

2−−

= G

(c) ( )

( )t t Y

t t Y

t t Y

t t Y

i j

i j

i j

i j2

2

− −− −= G (d) t Y

t t Y

t t Y

t Y

i

i j

i j

j

2

2−−= G

38. Ans: (d)Sol: The figure given in the question might be as follows

Bus-i Bus-jline

y

1 : ti

Vi Vj

tj : 1

The per phase electrical equivalent circuit will be

Vi

YIiIj

Vi.ti

1 : ti tj : 1

Ideal transformerIdeal transformer Line

Ii/tiIj/tj

Vj

+ + + +

−−−−

Vj.tj

i j

The relation between Bus-i and Bus-j voltages and currents, I

I

Y

Y

Y

Y

V

V

i

j

ii

ji

ij

jj

i

j== = =G G G

From the equivalent circuit shown above,

Equation (1) tI

V t V t Yi

ii i j j ^ h

. .I t Y V t t Y Vi i i i j j2 ^ ^h h ...... (1)

Equation (2) tI

V t V t Yj

jj j i i ^ h

. .I t t Y V t Y Vj i j i j j2 ^ ^h h ......... (2)

From equation (1) & (2),

.

.

,

.Y

t Y

t t Y

t t Y

t YBus

i

i j

i j

j

2

2

6 =@ G



39. The causal realization of a system transfer function H(s) having poles at (2, −1) and (−2, 1) and zeros at (2, 1) and (−2, −1) will be

(a) Unstable, complex, all pass (b) Stable, complex, low pass (c) Stable, real, all pass (d) Unstable, real, high pass39. Ans: (a)Sol: Poles − 2 + j 2 − j Zeros 2 + j − 2 − j .

( ) ( )

( ) ( )T F

s j s j

s j s j

2 2

2 2

|TF| = 1 ∴ All pass system Complex system Pole is in the right half of s-plane ∴Unstable system.

40. The figure below shows as per phase open circuit characteristics (measured in volt) and short circuit characteristics (measured in A) of a 14 kVA, 400 V, 50 Hz, 4-pole, 3-ph, delta connected alternator driven at 1500 rpm. The field current If is measured in A. Readings are marked as respective (x, y) coordinates in the figure. Ratio of the unsaturated and saturated synchronous impedances (ZS unsaturated/ZS saturated) of the alternator is closest to:

VOCISC

If

(0, 10)

(0, 0)SCC

(4, 20)

(2, 21

0)

(1, 110)

(8, 400)OCC

(a) 2.025 (b) 1.00 (c) 2.100 (d) 2.0040. Ans: (c)Sol: A 14 kVA, 400 V, 50 Hz, 4-pole, 3-φ, ∆−connected alternator From SCC ⇒ If = 4A, ISC = 20 A Then If = 2A ⇒ ISC = 10A If = 8A ⇒ ISC = 40 A ∵ SCC linear Unsaturated synchronous impedance is belongs to linear portion of OCC i.e., If = 2A

Zs(unsat) = IVsc

oc If belongs to unsaturated portion

i.e., If = 2A

= 10210 = 21 Ω

Saturated synchronous impedance is belongs to non-linear portion of OCC i.e., If = 8A



Zs(sat) = IVsc

oc If belongs to unsaturated portion = 40

400 = 10 Ω

i.e., If = 8A ∴ The ratio of unsaturated synchronous impedance to saturated synchronous impedance is

( )

.Z satZ unsat

1021

2 1s

s = =] g

41. A 250 V DC shunt motor has an armature resistance of 0.2 Ω and 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 a line current of 50A 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 1V per brush drop. The speed of the motor in rpm under loaded condition is closest to:

(a) 1200 (b) 1220 (c) 1000 (d) 90041. Ans: (b)Sol: No-load Condition: Ia1 = 5A, N1 = 1200 rpm Brush drop = 1V/brush Total brush drop = 2V Eb1 = V − Ia1Ra − B.D = 250 − 5 × 0.1 − 2

V=250 VRa=0.1Ω

5A2.5A

rf =100Ω

= 247.5 V Load condition: Ia2 = 47.5 A φ2 = 0.95 φ1

N2 = ____ Eb2 = V − Ia2Ra − B.D = 250 − 47.5 × 0.1 − 2 = 238.5 V

NN

EEb

b

1

2

2

1

1

2 #

V=250 VRa=0.1Ω

47.5 A

50 A

2.5A

..

.N1200 247 5

238 50 95

2

1

1#

= 1.0164

⇒ N2 = 1220 rpm

42. Consider a permanent magnet dc (PMDC) motor which is initially at rest. At t = 0, a dc voltage of 5V is applied to the motor. Its speed monotonically increases from 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 of the motor is

(a) . s0 5 110

+ (b) . s0 5 12

+ (c) .s 0 5

2+ (d)

.s 0 510+



42. Ans: (b)Sol: Given that the final value to a step of 5V is 10 and T = 0.5s

Let T.F = . s1 0 52

+ Final value ( )

..Lts TF s ss s

51 0 52 5

10s 0

"

T.F =

. s1 0 52

+

43. 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 has negligible internal resistance and total inductance of 2.28 mH. If the thyristors are triggered at a = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______.

43. Ans: 4.8Sol: A single phase full bridge fully controlled rectifier having large value of inductance. The source is having finite

value of inductance of 2.28 mH. The question is given with 1-φ full wave rectifier with the effect of source inductance

R = 10 Ω Vs = 230V Ls = 2.28 mH a = 45° cos cos V

LsI2

m0 _ i

But in the given data I0 is not given

cos cosVLsI2

m0

_ i

cos cosI Lsv2m

0 _ i7 A

I RV

00` =

sinV V t d t1

m0 a n

r a

+

+

] g#

cosV

tm

a nr a++6 @

cos cosVm _ ^i h7 A

cos cosV

Vm0 _ ^i h7 A

( )cos cosI RV

RVm

00

7 A

cos cos cos cosLsV

RV

2m m` _ _i i7 7A A



.cos cos

cos cos

2 100 2 28 101

45 45

101

45 45

3&# # #

#

-

_

_

i

i

7

7

A

A

( )

( . ) ( ( )

cos cos

cos cos

45 45

102 28 10 200 45 453

&

# #

-

7 A

. .cos cos45 1 0 0456 45 1 0 0456& _ i6 6@ @

⇒ .. cos

cos1 04560 9544 45

45# _ i

( ) .cos 45 0 6454&

m = 4.8

44. A non-ideal Si-based PN junction diode is tested by sweeping the bias applied across its terminals from −5V to + 5V. 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 % uncertainty (rounded off to 2 decimal places) in measured current at a bias voltage of 0.02 V is _________.

44. Ans: 4.77Sol: Diode current expression I = I e V

V

0 T

d

(h = 1) Vd = diode bias voltage and VT = Thermal voltage Given: Vd = 0.02 V, VT = (29 ± 2) mV Diode current (I) = I e /V V

0D T

Apply logarithm on both sides ln I = ln lnI e /V V

0d T+ ^ h

ln I = ln IV

V

T

d

0 +

Partial differentiation wrt ‘VT’ only

⇒ .I V

I

V

V10

T T

d

22

2 ⇒ V

IIV

V

T T

d

22

2

Uncertainty uI = V

Iu

TV

2

T2!

2

2d n = IV

Vu

T

d

V2

2

T2! −e o = I

V

Vu

T

d

V2 T!

Percentage uncertainty = I

u100

I#

V

Vu 100

T

d

V2 T! #=

= .

29 10

0 022 10 100

3

3!#

# # #−

−

^ h = ± 4.76%



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

LXI H, 2001 H 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 ___ .45. Ans: 210Sol: LXI H, 2001 H ⇒ HL : 2001 MVI A, 21H ⇒ A : 21 INX H ⇒ HL : 2002 ADD M ⇒ A : D2 INX H ⇒ HL : 2003 MOV M, A ⇒ A : D2 HLT 21H + B1H = D2H--------------- Memory data 2001H 98H 2002H B1H 2003H D2H After execution of this program the accumulator contents is D2H. The decimal equivalent of D2H is 210.

46. A resistor and capacitor are connected in series to a 10 V dc supply through a switch. The switch closed at t = 0, and the capacitor voltage is found to cross 0 V at t = 0.4t, where t is circuit time constant. The absolute value of % change required in the initial capacitor voltage if the zero crossing has to happen at t = 0.2t is _______ . (rounded off to 2 decimal places)

46. Ans: 55Sol: To cross zero voltage let us consider ‘−ve’ initial voltage across capacitor

v(t) = v(∞) + [V(0) − V(∞)] et

( )

( ) ( ) [ ]

V V

V V

T RC

V t V e

0

10 10 10 RC

t0

03

_

`

a

bbbbbbbb

So, voltage across capacitor reach 0 volts at t = 0.4 t 0 = 10 − [V0 + 10]e−0.4

Rt = 0

10 V CV0

+

−



V0 = 4.925 V0 = 4.925 V Now voltage across capacitor has to reach ‘0’ volts at t = 0.2t only

0 = V e10 10 .01 0 2 6 @

V01 = 2.1 volts

= |

.

| . . |%

V

V V

4 925

4 925 2 21100

0

0 01

#

% change in voltage = 55% −V0

−10V

0V t

v(t)

0.2τ

0.4τ

−V′0

47. Which of the following options is true for a linear time-invariant discrete time system that obeys the difference equation y[n] − ay[n − 1] = b0x[n] − b1x[n − 1]

(a) The system is necessarily causal (b) The system impulse response is non-zero at infinitely many instants (c) y(n) is unaffected by the values of x[n − k]; k > 2 (d) When x[n] = 0, n < 0, the function y[n]; n > 0 is solely determined by the function x[n]47. Ans: (d)

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

(a) 4th order and stable (b) 3rd order and unstable (c) 3rd order and stable (d) 4th order and unstable

s 11+s 11+ s

12

s 2020+

R(s)+

−

48. Ans: (d)Sol: CE = ( ) ( )G s H s1 0

( ) ( )s s s1

11 1

2020

02

( )s s s 20 20 03 2 ^ h s s s21 20 20 04 3 2

number of roots of CE = 4 + s4 1 20 20+ s3 21 0 0+ s2 20 20 0− s1 − 21+ s0 20

Number of sign changes in the first column = 2 ∴ two roots are present in the right half of s-plane, hence system is unstable.



49. 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 _______ .

Z I=2A

d=1ma =1m

a =1m

49. Ans: 138.62Sol: The mutual inductance between two conductors is given by

M N I12

M I1212 (a N = 1, for single square turn)

where 12 is the magnetic flux on conductor 2 due to long straight wire (conductor 1).

.B dss

12 v v# .Ia d dz a2

z

0

z z

t

t t## I

d dz21

z

0

1

2

0

1

t= =

# #

( )lnI

21

0 2

1] g

( )lnM I 2 212

12 0

ln24 10

27#

- ] g

M12 = 138.629 nH

z =1m

ρ =2mρ=1m

d=1mI = 2A

z

a=1m

conductor 2

conductor 1

50. 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+aT) kΩ, where T is temperature in °C. The temperature coefficient, a is −(4 ± 0.25)% /°C circuit parameters R1 = 1kΩ , R2 = 1.3 kΩ , R3 = 2.6 kΩ . The error in the output signal (in V, rounded off to 2 decimal places) at 150°C is ________.

+−

+−3V

0.1V

RThermistor

R1

R2

R3

V0

+−



50. Ans: −0.09Sol: Given R1 = 1 k, R2 = 1.3 k, R3 =2.6 k, a = −(4 ± 0.25)% per °C, RT = 2(1 + aT) kΩ Note: Initially, consider a as constant.

i.e., / C100

4c = −0.04 ............... (1)

Then RT = 2(1 − 0.04 × 150) = 10 kΩ ............. (2)

.V

R

R

R R

R V

R

R V1

3 0 1

T0

2

3

1

1

2

3# #

d n= =G G .............. (3)

= .

.

.

. .

k

k

k k

k V

k

k V11 3

2 6

1 10

1 3

1 3

2 6 0 1# #

d dn n ........... (4)

V0 = −1.2 V ................. (5) (original output voltage)

+−

+−3V

V1

V2

0.1V

RThermistor

R1

R2

R3

V0

+−

Case (i) Consider a = −(4 + 0.25)% per °C = −0.0425/°C ............. (6) Then RT = 2(1 − 0.0425 × 150) = −10.75 kΩ ................... (7)

⇒V01 = .

.k k

k VV3

1 10 75

1 30 2

#

−−< F = −1.123 V ................ (8)

Case (ii): Consider a = −(4 − 0.25)% per °C = −0.0375/°C ............. (6) Then RT = 2(1 − 0.375 × 150) = −9.25 kΩ ................... (6)

⇒V02 = .

.k k

k VV3

1 9 25

1 30 2

#

−−< F = −1.29 V

Error on the output = −0.077 V to −0.09 V ............. (12) i.e., minimum error = −0.077 V Maximum error = −0.09 V.

51. Let ax and ay be unit vectors along x and y directions, respectively. A vector function is given by F = axy − ayx.

The line integral of the above function .F dc

,# along the curve C, which follows the parabola y = x2 as shown

below is _____ (rounded off to 2 decimal places).

−1

1

2x

y

c

4



51. Ans: − 3 Sol: Given F ya xax y t t

given curve is y= x2

. .Now F dr y dx xdy,

,

C 1 1

2 4

-^

^

h

h

##

.x dx x x dx2x

2

1

2

=-

] g#

.x dx2

1

2

& −-

#

4

−2−1

(−1,1) (2,4)

y = x2

x3 3

18 1

3

1

2

& &

-

^ h#

⇒ − 3 52. The number of purely real elements in a lower triangular representation of the given 3 × 3 matrix, obtained

through the given decomposition is ________ .

a

a

a

a

a a

a

a

a

a

a a

2

3

3

3

2

1

3

1

7

0 0

0

0 0

0

T11

12

13

22

23 33

11

12

13

22

23 33

=

R

T

SSSSSSSS

R

T

SSSSSSSS

R

T

SSSSSSSS

V

X

WWWWWWWW

V

X

WWWWWWWW

V

X

WWWWWWWW

(a) 8 (b) 6 (c) 9 (d) 552. Ans: (a)

Sol: a

b

d

c

e f

a

b

d

c

e f

2

3

3

3

2

1

3

1

7

0 0

0

0 0

0

T

=

J

L

KKKKKKK

J

L

KKKKKKK

J

L

KKKKKKK

N

P

OOOOOOO

N

P

OOOOOOO

N

P

OOOOOOO

a

b

d

c

e f

a b

c

d

e

f

0 0

0 0

0 0

=

J

L

KKKKKKK

J

L

KKKKKKK

N

P

OOOOOOO

N

P

OOOOOOO

a

ab

ad

ab

b c

bd ce

ad

bd ce

d c f

2

2 2

2 2 2

J

L

KKKKKKK

N

P

OOOOOOO

Comparing both sides, a2 = 2, ab = 3, ad =3, b2 + c2 = 2, bd + ce = 1 and d e f 72 2 2

By solving these, we will get a = ,2! is purely real

b = 2

3! , is also real

d =

2

3! , is also real

c = i

2

5! , is imaginary



e = i10

7! , is imaginary

f =

5

37! , is real

If we consider ± signs then the number of possible real elements can be 8 ∴ Option (a) is correct.

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

E rr

a a a23

6r z b l where ar, aφ, az ae unit vectors along r, φ, 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, e0, in SI units is given by:

(a) 4e0 (b) 3e0 (c) 5e0 (d) 9e0

53. Ans: (d)Sol: Given: Relative permittivity 4

9r

E r r r z23

6 v t v t

From Gauss’s law, the volume charge density is given by

. .D Er0d d/ ov v

Assume the medium is homogeneous

.Er0 d o v ( )r r rE

EzrEz

1r r0 22

22

22

z ] g< F . .

r rr r

r zr

12

36r0 2

222

22 ] b ]g l g; E

r r49 1

4 0 00# # 6 @ 9 0` o



54. A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and sub-transient 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 sub-transient internal emf (rounded of to 2 decimal places) is _____ pu.

54. Ans: 1.0198Sol: A cylindrical rotor generator has Xd = 0.7 p.u X’’

d = 0.2 p.u

jXdE V

Terminal voltage of generator, V j1 0

Internal emf under steady state, . .E j p u1 0 7

Prefault current in the generator, Ijx

E Vg

d

.

.I

j

j

0 7

1 0 7 1g

= 1 p.u

Sub-transient emf calculation: ( )E V I jX

''g d m

= 1+ 1(j0.2)

jX″d

Ig

E″ V

= 1+ j0.2 | | . . .E p u1 0 2 1 0198'' 2 2 ] g

55. 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 doesn’t reflect the exact number of turns. If windings A, B and C are supplied balanced 3-phase voltages at 50 Hz and there is no saturation the no-load rms voltage (in V, rounded off to 2 decimal places) across winding ‘x’ is _____

230∠120oV

230∠0oV

230∠−120oV

C

A+

+

+

XB

55. Ans: 46

Sol: Sense of B and C coils is same. Whereas sense of A is opposite

The voltage across coil X

= V20

2230 0 230 120 230 120X c c c+ + + ^ h

= 46 V

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