ACE - Cloud Object Storage · ACE Engineering Academy Hyderabad |Delhi |Bhopal |Pune |Bhubaneswar |Bengaluru |Lucknow |Patna |Chennai | Vijayawada |Visakhapatnam | Tirupati 25 k z

Hyderabad |Delhi |Bhopal |Pune |Bhubaneswar |Bengaluru |Lucknow |Patna |Chennai | Vijayawada |Visakhapatnam | TirupatiACE Engineering Academy

ACEEngineering Academy

Hyderabad | Delhi | Bhopal |Pune | Bhubaneswar | Bengaluru | Lucknow | Patna | Chennai | Vijayawada | Visakhapatnam | Tirupati

H.O: 204, II Floor, Rahman Plaza, Opp. Methodist School, Abids, Hyderabad-500001.Ph: 040-23234418, 040 -23234419 , 040-23234420, 040-24750437

EE: Electrical Engineering

General Aptitude:One Mark Questions (Q01 – 05)

01. The cost of 7 pens, 8 pencils and 3sharpeners is Rs 20. The cost of 3 pencils, 4sharpeners and 5 erasers is Rs 21. The costof 4 pens, 4 sharpeners and 6 erasers is Rs25. The cost of 1 pen, 1 pencil, 1 sharpenerand 1 eraser is ________ (Rs)

Ans: 6Sol: Let the costs of pens, pencil, eraser and

sharpener be pn, pp, e and s respectivelyGiven7pn + 8pp + 3s = 203pp + 4s + 5e = 214pn + 4s + 6e = 25Adding all three equations11pn + 11pp + 11s + 11e = 66 1pn + 1 pp + 1s + 1e = 6

02. Sentence Completion:Although some think the terms "bug" and"insect" are -------, the former term actuallyrefers to ------- group of insects.(a) parallel - an identical(b) precise - an exact(c) interchangeable - particular(d) exclusive - a separate.

Ans: (c)Sol: The word "although" indicates that the two

parts of the sentence contrast with eachother: although most people think about theterms "bug" and "insect" one way,something else is actually true about theterms. Choice (c) logically completes thesentence, indicating that while most peoplethink the terms are "interchangeable," the

term "bug" actually refers to a "particular"group of insects.

03. Sentence improvement:Underestimating its value, breakfast is ameal many people skip.(a) Underestimating its value, breakfast is a

meal many people skip(b) Breakfast is skipped by many people

because of their underestimating itsvalue

(c) Many people, underestimating the valueof breakfast, and skipping it.

(d) Many people skip breakfast because theyunderestimate its value.

Ans: (d)Sol: The problem with this sentence is that the

opening phrase "underestimating its value"modifies "breakfast," not "people." Theorder of the words in the sentence in choice(d) does not have this problem of amisplaced modifying phrase. Choice (d) alsoclarifies the causal relationship between thetwo clauses in the sentence. None of theother choices convey the informationpresented in the sentence as effectively anddirectly as choice (d).

04. Spot the error, if any:If I were her / I would accept / his offer(a) If I were her,(b) I would accept(c) his offer(d) No error

Ans: (a)

: 2 : Electrical Engg.


Sol: Rule we should use Subjective case ofpronoun after BE forms…am, is, are waswere.,, has been, have been, had been.Her is an objective case ---If I were she. is correct.

05. Kishenkant walks 10 kilometres towardsNorth. From there, he walks 6 kilometrestowards south. Then, he walks 3 kilometrestowards east. How far and in which directionis he with reference to his starting point?(a) 5 kilometres, West Direction(b) 5 kilometres, North-East Direction(c) 7 kilometres, East Direction(d) 7 kilometres, West Direction

Ans: (b)Sol: The movements of Kishenkant are as shown

in figure

A to B, B to C and C to DAC = (AB – BC) = (10 – 6) km = 4 km

Clearly, D is to the North-East of A Kishenkant’s distance from starting point A

AD = 22 CDAC

22 )3()4( 25 = 5 km

So, Kishenkant is 5 km to the North-East ofhis starting point

Two Mark Questions (Q06 – 10)

06. The infinite sum 1+7

4+

27

9+

37

16+

47

25+ - - - -

- equals

Ans: 1.8 to 2Sol: We have to find the sum of the series

1+7

4+

27

9+

37

16+

47

25+ - - - - -

Putting x =7

1 we get

1 + 22x + 32x2 + 42x3 + 52x4 + - - - - - s = 1 + 4x + 9x2 + 16x3 + 25x4

s.x = x + 4x2 + 9x3 + 16x4 + - - - - -s – sx = 1 + 3x + 5x2 + 7x3 + 9x4 + - - - - - -x(s – sx) = x + 3x2 + 5x3 + 7x4 + - - - - - -

(s – sx) –x(s – sx) = 1 + 2x + 2x2 + 2x3 + - -- - - - + to

(1 – x)2 s = 1+x1

x2

; since 1x

s =3)x1(

x1

We may use it as direct formula for solvingthis type of problem

Substituting x =7

1 we get

s = 3

7

11

7

11

=

2749

21673438

=1.81

07. If 5a2c3

z

c2b3

y

b2a3

x

and a, b

and c are in continued proportion and b, c, aare in continued proportion, then

c3

z

b2

y

a

x is _______ ( a, b and c are in

continued proportion means b2 = ac)

(a)5

155 (b) 25

(c) 46

1(d) 45

6

5

Ans: (d)Sol: Given that a, b, c are in continued proportion

b2 = ac -------- (1)Also b, c, a are in continued proportion c2 = ab ------- (2)From (1) and (2)b2c2 = a2bc a2 = bc ------- (3)Conditions (1), (2) and (3) can only besatisfied when a = b = c = k (say)

10 km

6 km 3 kmD

A

4 km

B

C

: 3 : Pre GATE Question Paper with Sol.


25k

z

k

y

k

x5

k5

z

k5

y

k5

x

k

z

3

1

k

y

2

1

k

x

c3

z

b2

y

a

x

=3

25

2

2525

=6

545

6

275

6

1125

08. Rasputin was born in 3233 B.C. The year ofbirth of Nicholas when successively dividedby 25, 21 and 23 leaves remainder of 2, 3and 6 respectively. If the ages of Nicholas,Vladimir and Rasputin are in arithmeticprogression, when was Vladimir born?(a) 3227 B.C (b) 3229 B.C(c) 3230 B.C (d) 3231 B.C

Ans: (c)Solution: The year of birth of Nicholas

25 21 23

2 3 6 3227The ages of Nicholas, Vladimir and Rasputinare in A.PThe ages of Nicholas Vladimir Rasputin

3227 ? 3233

Vladimir age =2

RasputinNicholas

=2

32333227= 3230 B.C

09. Recent studies have highlighted the harmfuleffects of additives in food (colors,preservatives, flavor enhancers etc.). Thereare no synthetic substances in the foods weproduce at Munchon Foods - we use onlynatural ingredients. Hence you can be sureyou are safeguarding your family’s healthwhen you buy our products, says MunchonFoods. Which of the following, if true,would most weaken the contention ofMunchon Foods?(a) Some synthetic substances are not

harmful

(b) Some natural substances found in foodscan be harmful

(c) Food without additives is unlikely totaste good

(d) Munchon Foods produces only breakfastcereals

Ans: (b)Sol: Munchon’s contention is that buying their

products safeguards health. To weaken thatargument we can show that, for some reason,their foods might not be healthy.So think about an alternative cause

10. To open a lock, a key is taken out of acollection of n keys at random. If the lock isnot opened with this key, it is put back intothe collection and another key is tried. Theprocess is repeated again and again. It isgiven that with only one key in thecollection, the lock can be opened. Theprobability that the lock will open in ‘nth’trail is _____

(a)n

n

1

(b)

n

n

1n

(c) 1 –n

n

1n

(d) 1–

n

n

1

Ans: (c)Sol: Probability that the lock is opened in a trail

isn

1 (since there is exactly one key, which

opens the lock) The chance that the lock is not opened in

a particular trail = 1 –n

1

P(lock is opened in nth trial) = 1– P(lock isnot opened in n trials)

= 1 –nn

n

1n1

n

11



– 10 V

10 V

3k

3V

Vi

2k

–4V

+–

2k

5V

3k

V0

3k

+

–

Vi

3k 2k3k

3V 5V

Vx

3k

I1

I2I3

I42k

I5

– 4V

V0+

–

One Mark Questions (Q11 – 35)

11. For the op-amp circuit shown, the op-amp isideal. The breakdown voltage of zenerdiodes is 7 V and its V = 0.7 V whenforward biased. Then the value of Hysteresisvoltage is ______ (in volts).

Ans: 2.57 range(2.45 to 2.65)

Sol: The saturated voltages of the op-amp areV0 = (Vz + V)

= 7.7 VWhen V0 = + 7.7 V, the voltage at noninverting terminal(Vx) is Vx = VUTP

When V0 = – 7.7 V, Vx = VLTP

Apply KCL at non inverting terminalI1 = I2 + I3 + I4 + I5

k3

4V

k2

V

k3

3V

k2

5V

k3

VV xxxxx0

0 3 15 2 6 3 2 8

3 6x x x x xV V V V V V

2V0 – 2Vx = 10 Vx – 1312 Vx = 2 V0 + 13

02 13

12x

VV

When V0 = 7.7 V, Vx = VUTP

V366.2

12

137.72VUTP

When V0 = – 7.7 V, Vx = VLTP

2( 7.7) 13

12LTPV

= – 0.2 V

Hysteresis Voltage = VUTP – VLTP

= 2.566 V 2.57 V

12. Characteristic equation of a feedbackcontrol system is s3 + ks2 + 8s + 8 = 0, toobtain fixed amplitude oscillations, the valueof ‘k’ is _____.

Ans: k = 1 (Range = 1)Sol:

s3 1 8s2 k 8s1 8k 8

k

0

s0 8 0

8k – 8 = 0 k = 1

13. A 400 kV, 3-, 200 km length losslesstransmission line operated at 50 Hz. If surgeimpedance of transmission line is 400 thenline sending end current at no load is _____kA.

Ans: 0.123 (Range from 0.10 to 0.13)Sol: At no load

sphs

VI C.

A



D QX

QClock

D Q

Q

D Q

Q

Z

1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 11 2 3 4

s sphc

1 1| I | sin . .V

Z cos

50

103 8

sphc

1Tan .V

Z

2

= 31 180 400tan 1.047 10 200 kA

400 3

= 0.1227 kA.

14. A volt meter placed across 3 resistor,shown in above figure reads 45 volts ,thenthe magnitude of current reading in ammeteris ________ (Amperes)

Ans: 19.34 (Range 19 to 20)Sol:

V = I Z = 15 (3 + j3) = o45245 current through resitive branch

= o45568.5

7/80

45245

Total current through Ammeter= 15 + 5.56845

= 19.3411.74Magnitude of current = 19.34

15. If the serial data applied at input x in thefollowing circuit serially from left to right,then the number of times the 1 occurs at z is______.

Data: 10 0 1 1 1 0 0 1 1 1 1 0 1 1 1

Ans: 4 (Range: 4)Sol: If the input x is 1 in the consecutive clock

cycles then z becomes 1, means finding thebit pattern ‘111’ and it is findingoverlapping pattern.

16. For the circuit shown in figure the diode hasV = 0.7 V, Rf = 20 . The approximatevalue of current through the diode is

(a) 25 mA (b) 21 mA(c) 18 mA (d) None

Ans: (c)Sol: The equivalent circuit of the diode is

V

A

10 5

2

j3 3

10 10/7

3 j3

45 –+

15 A

V+ –

A

+ 100 100 10 V

150

100 100 10 V

20

150 0.7 V

Vx

0.7 V

20



0Ω0.33 k

6.2 V0.1 kΩ

B

E

C

‘Si’

10 V

q

CR

surface ‘S’ of thespherical conductor

spherical conducting volume

Apply KCL at Vx

120

7.0V

100

V

150

V10 xxx

120

7V10V12

15

V10 xxx

80 – 8Vx = 22Vx – 730Vx = 87 Vx= 2.9VThen current through diode is

ID = mA33.18120

7.0Vx

17. The zener diode in the figure shown istemperature compensated and value of Sitransistor is very high. If the currentthrough 0.1 k resistor is 55 mA at 25o C.What is the approximate current through 0.1k at 65 C?

(a) 100 mA (b) 56 mA(c) 55 mA (d) 54 mA

Ans: (b)Sol: VB = 6.2 V

VE = VB 0.7 = 5.5 V

IE =5.5

0.1k = 55 mA at 25o C

For 1o C rise in temperature the value of VBE

by 2.5 mVT = 65 25 = 40o C

BEV = 0.7 40 2.5 103 = 0.6 V

B EV V 0.6

EV 6.2 0.6 5.6 V

EI =

k1.0

6.5 = 56 mA

18. In a standard 2nd order system steady stateerror to a unit ramp input can be reduced by(a) Increase both n and (b) Increase n and decrease (c) Decrease n and increase (d) Decrease both n and

Ans: (b)

Sol:2n

n

G(s)s (s 2 )

nv

s 0k t G(s)

2

ssv n

1 2e

k

ess if n

19. If the numbers A = 1 1 1 0 1 1 0 0 and B = 11 1 1 0 1 1 1 are two signed numbers in 2‘scomplement representation, then A – B in2’s complement representation is ________.(a) 1 1 1 0 0 0 1 1 (b) 1 1 1 1 0 1 0 1(c) 0 0 0 1 1 1 0 1 (d) 1 1 1 1 0 1 0 0

Ans: (b)Sol: A + 2’s of B

1 1 1 0 1 1 0 00 0 0 0 1 0 0 11 1 1 1 0 1 0 1

20. ‘C’ is a cavity in the spherical conductingvolume of radius R m. There is a charge ‘q’Coulomb in the cavity as shown in thefigure.



The charge density on the surface ‘S’ is(a) 0(b) Data insufficient

(c) 22

qC / m

4 R

(d) 22

qC / m

4 R

Ans: (c)Sol: Under static conditions, the electric field at

any point in the conductor body must bezero.But q in cavity, by itself, will certainlyproduce electric field at all points in theconducting body (and outside).

Hence a charge –q coulomb, drawn from theconductor body, appears on the cavitysurface C. (This is called induced charge).This leaves charge +q in the conductorbody, which appears on surface S.

q in cavity and –q on C together producezero field at all points outside C and hencethe conductor body.

q on S must independently produce zerofield at all points in the conductor body. Forthis reason, q must distribute uniformly onsurface S. Hence charge density on S (q/4R2) c/m2.

21. A long solenoid has 1000 turns/m and aradius greater than (80/ ) cm. A shortsolenoid of 2000 turns and radius (80/ ) cmis located along the axis of the longsolenoid. For a current of 2 A in the shortsolenoid, the total flux linkages of the longsolenoid is.(a) (b) data insufficient(c) 0 (d) 1.024 Wb-turns

Ans: (d)Sol: Assume a current I through the long

solenoid, and let it have N1 turns/m. Then B

at any point inside the long solenoid is 0 N1

I Wb/m2, along its axis.Since the short solenoid is along the axis ofthe long one, if it has a radius R m, total fluxlinkages of the short solenoid = 0 N1 I R2

N2 Wb-turns (where N2 is the number ofturns of the short solenoid). Hence mutualinductance between the two solenoids = 0 N1 N2 R

2 H.Now, if a current ‘I’ flows through the shortsolenoid, flux linkages of the long solenoidwill be 0 N1 N2 R

2 Wb-turns.Substituting given numerical values, fluxlinkages of the long solenoid = 1.024 Wb-turns.

22. When primary and secondary windings arenot placed symmetrically with respect to thetransformer core, the mechanical forcesdeveloped are(a) External radial forces(b) Internal axial forces(c) Internal radial forces(d) External axial forces

Ans: (d)

23. For which value of the following systemof equations is inconsistent?

3x + 2y + z = 10 2x + 3y + 2z = 10 x + 2 y + z = 10

(a)5

7(b)

5

7

(c)7

5(d)

7

5

Ans: (a)

Sol: 0

21

232

123

3(3 – 4) – 2(2–2) + (4–3) = 0 5 – 7 = 0

=5

7



24. If r r rX cos isin

3 3

for r = 1, 2, 3,

.... Then X1. X2. X3 .... (upto infinity) equalto

(Where, i = 1 )(a)1 + i (b) 1

(c) i (d)1 3

i2 2

Ans: (c)

Sol:r

i3

r r rX cos sin e ,

3 3

r = 1, 2, 3, ...

X1. X2. X3 ......2 3

1 1 1 ii .....3 3 3 2e e = i

25. Inverse Z–Transform of

)2Z(3Z

Z2

2

(a) (3n+1 + 2n+1) u(n)(b) (3n + 2n) u(n)(c) ((n–2)3n + 2n+1) u(n)(d) None of the above

Ans: (c)

Sol: Z–1 1n2

1n2

Z)2Z()3Z(

)Z(Z

Res/Z=3 +Res/Z=2

= ((n–2)3n + 2n+1) u(n)

26. The best resolution of three and half digitdigital multimeter is______. The operatingVoltage Ranges are: 0-2V,0-20V,0-200V.(a) 2mV (b) 1mV(c) 3 mV (d) 0.5mV

Ans: (b)

Sol: For 31

2

R =N 3

1 11mV

10 10

N = No. Of full digits

27. A 3, 10 kVA load has a p.f of 0.342leading. The power is measured by twowattmeter method. The reading of eachwattmeter will be(a) – 1.732 kW & 7.655 kW(b)– 1kW & 4.42 kW(c) – 0.577 kW & 2.55 kW(d) 1.973 kW & 1.973 kW

Ans: (b)Sol: Given power is Apparent power

L LS 3 V I3

L L10 10 3 V I 3

L L

10 10V I

3

= 5773.5 VAcos = 0.342 = 70

W1 = VL IL cos (30 + ) = – 1 kWW2 = VL IL cos (30 – ) = 4.42 kW

28.

In above circuit, the values of Z22 & Z21

respectively are

(a)2R

1R,

2R

1

(b)2R

1R,

2R

1

(c)

2R

1,

2R

1R

(d)2R

1,

2R

1R

Ans: (d)

+

+

1

1 R

I1

I2I1

V1 V2

V2

+

–

+

–



LTI System(a)

System(b)

(t) outputx(t)

Input

y(t)

h(t)x(t) y(t)

Sol:

By applying KVL to mesh ①–V1–V2 + R(I1 – I3) = 0 ........(a)

By applying KVL to mesh②I3 + (I3 + I2) – I1 + R(I3 – I1) = 0 .........(b)

By applying KVL to mesh③–V2+(I2 + I3) – I1 = 0V2 = I2 – I1 + I3 .........(c)From equation (b)

2R

II1RI 21

3

Substitute value of I3 in equation (c)

Then, V2 = I2 – I1 +

2R

II1R 21

V2 = 21 I2R

1RI

2R

1

Z21 =2R

1RZ,

2R

122

29. The short circuit faults are commonly takingplace in power system. Which of thefollowing statements are correct.(i) Severe over voltages are taking place an

unbalanced fault with ground(ii) In case of unsymmetrical faults, the

subtransient currents are same aspositive sequence subtransient current

(iii)The pickup value of the relay is selectedbased on single line to ground fault

(iv)Positive sequence network terminalvoltage is zero in case of symmetricalfaults.

(a) (i), (iii) and (iv) (b) (iii) and (iv)(c) (ii), (iii) and (iv) (d) (i) and (iii)

Ans: (b)

30. Consider the cascade of two systems shownin the figure below system ‘B’ is the inverseof system ‘A’

Suppose the input is (t). What is the output(t)?

(a) (t) = u(t); where u(t) is unit stepfunction

(b) (t) = s(t); where s(t); where s(t) is stepresponse.

(c) (t) = (t)(d) (t) = 2u(t) – (t)

Ans: (c)Sol: Because the two systmes are inverse, the

2nd system should produce the output whichis same as the initial input.

31. Consider the following LTI systems

Let x(t) and h(t) be the odd signals. Theoutput y(t) is defined as y(t) = x(t) * h(t).Then the output y(t) is.(a) Even signal(b) Odd signal(c) Can not be determined(d) None of the above

Ans: (a)Sol: y(t) = x(t) * h(t) and y(t) = x(–t) * h(–t)

Because x(t) & h(t) are odd functionsx(t) = – x(–t) & h(t) = – h(– t).Substituting eqation (2)y(–t) = [– x(t)] * [– h(t)] = x(t) * h(t) = y(t) y(t) is even signal.

+

+

1

1

R

I2I1

V1 V2

V2

+

–

+

–

I1 I3 I2

I1

③②①



32. A practical diode bridge rectifier with afilter capacitor is shown below.

The above circuit can be described by thefollowing equations in every half cycle ofline frequency.

d

dS S

dd

di 1idt LA Vvdv

0dt

Then the state matrix A is given by

(a)

load

d

d

SS

S

R

C

C

1L

1

R

L

(b)

loaddd

SS

S

RC

1

C

1L

1

L

R

(c) d load d

S

S S

1 1

C R C

R1

L L

(d)

loaddS

dS

S

RC

1

L

1C

1

L

R

Ans: (b)Sol: From circuit,

sddsd

s VviRdt

diL ………(1)

Cd dLoad

dd iR

v

dt

dv ………..(2)

ssd

d

Loaddd

ss

s

d

d

v0L

1

v

i

Rc

1

c

1L

1

L

R

dt

dv

dt

di

33. A 3-phase squirrel-cage induction motor canbe connected to have either 4 or 8 poles. Afan operated by this motor takes 15 kW atlow speed. At higher operating speed, thefan would take _________ kW.(a) 120 kW (b) 60 kW(c) 30 kW (d) None

Ans: (a)Sol: For fan load

T 2rN ; P 3

rN

Nr sok

1N

P

3

ok

1P

P

15 (8)3 = P2 (4)3

P2 = 120 kW

34. The slip test results of a 3-phase, starconnected synchronous machine are givenbelowVmax (rms, line-to-line) = 200 VVmin (rms, line-to-line) = 180 VIa max (rms, phase value) = 20 AIa min = (rms, phase value) = 15 AThe direct axis reactance and Quadratureaxis reactance per phase respectively(a) 13.33 , 9 (b) 5.77 , 6.92 (c) 10 , 12 (d) 7.7 , 5.19

Ans: (d)

Sol: Xd = max

min

V / Ph 200 / 37.7

I / Ph 15

Xq = min

a max

V / Ph 180 / 35.19

I / Ph 20

+

_

Rload

id

~ Cd Vd

+

s

Ls Rs



+

–

+–

20 Ω

Vi 100 Ω

20 Ω

20 Ω

V0

+

–

+–

20 Ω

Vi 100 Ω

20 Ω

20 Ω

V0

I4

I1

I3

I2

0 V

Vx

35.

Z = 0 + j1 VA = 12030VVB = 1200VIn the figure shown above, which of thefollowing statement is correct(a) Power delivered by network A is –

7200.4 W(b) Power delivered by network B is 7200.4

W(c) Power delivered by network A is 7200.4

W(d) Power absorbed by network A is 6800.4

WAns: (c)

Sol: I =Z

VV BA

=1j

012030120 oo = 62.1215 A

The power delivered by network ‘A’ isPA = IVAA cos|I||V| = (120) (62.12) cos(30 – 15)

= 7200.4 W Power absorbed by network ‘B’ isPB = |VB||I|cos(VB – I) = (120) (62.12) cos(0 – 15) = 7200.4 W Option ‘C’ is correct

Two Mark Questions (Q36 – 65)

36. For the op-amp circuit shown, the op ampassumed to be ideal. The value of voltage

gain 0

i

V

V is _________.

Ans: 11 range(10.9 to 11.1)Sol: The given circuit is

Apply KCL at inverting terminal

i xV 0 V

20 100

Vx = 5Vi

Apply KCL at Vx

20

V

20

VV

100

V xx0x

x 0 x xV 5V 5V V

100 20

5V0 6Vx = 5Vx

5V0 = 11Vx

5V0 = 11(5Vi)

0

i

V11

V

The magnitude of voltage gain 0

i

V11

V

37. Consider the system shown in figure below

The value of ‘k1’ to obtain the steady stateerror of – 0.25 to a unit step disturbance

input1

D(s)s

is _______.

Ans: 6 (Range: 6)

NetworkA

NetworkB

I

VA VB

+ +

– –

Z

+–1k (s 2)

s 3

+

2k

s s 4

D(s)C(s)R(s)

+



8q

N8q

C

Spherical electron cloud ofradius R m and total charge – 8q C.

N: Nucleus with charge 8q C.at center of electron cloud.

C: Center of the spherical electron cloud,which shifts (with no distortion in shape)to the left by 1 nano – meter while the nucleusshifts to the right by the same amount.

10 ay N/C

q = 1.6 10–19 Coul

(b)

(a)

N

Py = 2m

0

z

dz

I = 1A

az

z 5 m2

z 5 m2

2 2y z

Sol:

2

1 2

kE(s) s(s 4)

k k (s 2)D(s) 1s(s 3)(s 4)

2

1 2

k (s 3)E(s) D(s)

s(s 3)(s 4) k k (s 2)

Error 2

s 01 2

( k )(3)Lt sE(s)

2k k

= – 0.25

k1 = 6

38. Open loop transfer function of a unity

feedback control system is4

k

s 16, the

value of ‘k’ at the break in /away point is__________.

Ans: 16 (Range : 16)Sol:

S = 0 is a break point4

s 0k (2) k = 16.

39.

Fig. (a) shows a dielectric atom when thereis no external electric field. Fig. (b) showsthe same atom when there is a uniformexternal electric field of 10 ay V/m. Thedipole moment of the atom in fig (b) is_________ 1028 C-m.

Ans: 25.6 (range 25 to 26)Sol: In fig. (a), there is a charge 8q at point N.

The electron cloud appears, for pointsoutside the sphere, as a charge ‘–8q’ at N.Thus, net charge at N (for points outside theatom) is zero. The atom is electricallyneutral. It has no dipole moment.In fig. (b) again for points outside the atom,there is a charge ‘– 8q’ at C and 8q at N. Ifthe distance between C and N is y, thedipole moment is 8qy. Here y = 2 10–9 mand 8q = 12.810–19 C. Dipole momentnumerically is 25.6 10 – 28 C-m.

40. In figure, the vector magnetic potential at Pis _________ 10 –10 Wb/m.

Ans: 1924 (Range: 1923 to 1925)Sol:

j2

–j2

2–2

s = 0

j

z 5 m

y = 2 m P0

1 A

z

z 5 m



700W1300 VAR

800W600 VAR

I

100VAR

80W 120W1200 VAR

+

–

(Inductive) (Capacitive)

120 V(Capacitive)

(Capacitive)

z /20 z

2 2

z2

Idz aA(P)

4 y z

z /20 z

2 2

z2

I a dz.

4 y z

From figure, z = y tan , dz = y sec2 d2 2 2 2y z y (1 tan ) = y sec .

z2

0 z

z2

IaA(P) sec d

4

= z

0 z 2e

z2

Ialog sec tan

4

2 2y ( / 4)z / 2; sec

y

, tan

2y

2 2y ( / 4)z ; sec

2 y

,

tan2y

22

0 ze 2

2

yI a 4 2A(P) log4

y4 2

Substituting given numerical values,7

z e

4 10 9 5A(P) a log

4 9 5

= 1924 10 – 10 Wb/m.

41. A 240 V, 50 A, 800 rpm dc shunt motor hasarmature circuit resistanc of 0.2 . If loadtorque is reduced to 60% of its full loadvalue and a resistance of 2 is inserted inseries with armature circuit, the motor speedwill be _________ rpm. Armature reactionweakens the field flux by 4% at full load andby 2% at 60% of full load.

Ans: 597.44 (Range from 597 to 598)

Sol: At rated load, Ea1 = V1 Ia1ra

= 240 50 0.2 = 230 VAlso Ea1 N11

Rated torque TL1 = Ka1Ia1

New torque TL2 = 0.6TL1= Ka2Ia2

Given 1 = 0.96 and 2 = 0.98. [when ismainfield flux at no load]

2aa

a

1L

1L

2L

1L

I98.0K5096.0K

6.01

T6.0T

TT

Ia2 = 29.39 ANew counter Emf = Ea2 = 240 – 29.32 (2 +0.2) = 175.34 V

a1 1 1

a 2 2 2 2

E N 230 800 0.96

E N 175.342 N 0.98

Motor speed, N2 = 597.44 rpm.

42. A single channel Analog Cathode RayOscilloscope is operated in X-Y DisplayMode.An ellipse with major axis in 2nd and4th Quadrant and rotating in anti-colckwisedirection displayed on the screen.The Y-intercept is at 2 div and Ymax is at 4 div. Thephase difference between Vertical andHorizontal input signals is_______ (degrees)

Ans: 210 (Range: 210)Sol:

= 180 + 1 in

max

Ysin

Y

= 180+30 = 210.

43.

The magnitude of ‘I’ in the above figure is________ Ampere

Yin=2 Ymax=4



00 01 11 10

00 1 1

01 1 1 1 1

11 1 1 1

10 1 1

CDAB

Ans: 15.025 (Range 14.9 to 15.1)Sol: PT = 700 + 800 + 80 + 120 = 1700 W

QT = 1300 – 600 – 100 – 1200 = – 600 VARST = PT + jQT =1700–j600=1803–19.4VA

I = A025.15120

1803

V

ST

44.

In the RC circuit shown above, the swich isclosed on position 1 at t = 0 and after 1 timeconstant is moved to position 2.Then the value of the current just afterswitch is moved to position 2 is _______Ampere

Ans: – 0.1053 (Range – 0.1050 to – 0.1060)Sol: When switch is at position 1:

i = C1e–t/RC = C1e

–4000t

RC = 500 0.5 10–6 = 250 10–6

At t = 0, i0 = A04.0500

20

R

V

i = 0.04 e–4000t

One time constant = 1RC = 500 0.5 10–6

= 250 micro secondsAt this point i = 0.04e–1 = 0.0147 AmpWhen switch is moved to position 2:Voltage across capacitor c = 20(1– e-1) =12.65 V 0V)0(V cc

This voltage and the 40 Volt source bothdrive current in the opposite direction fromthe current caused by 20 volt source.The current equation for the second transientis i = C2 e–4000(t – t)

At t = t, i =

500

65.1240

= –0.1053 Amp

i = – 0.1053 e–4000(t –t)

45. The number of minimal expressions of thefunction f (A, B, C, D) = m(0, 2, 4, 5, 6, 7,8, 10, 12, 13, 15) in sum of products form is_______.

Ans: 4 (Range: 4)Sol:

BACBBDDB (or)

BD BD BC A D (Or)

BD BD CD A B (Or)

BD BD CD A D

500

0.5 F40 V

–

+

i2

+

–

1

20 V

i

– 0.1053– 0.12

– 0.10

– 0.08

– 0.06

– 0.04

– 0.02

0

0.02

0.04

tt

i



46. A 3-phase WRIM develops rated torque at aslip of 0.05 with slip ring short circuited. Ifthe rotor circuit resistance is increased tofour times by inserting external resistance inthe rotor circuit, then at rated load torque,the slip would be ______.

Ans: 0.2 (Range: 0.2)Sol: For fan load

T 2

2

SV

R;

2

2

R

KSVT

For the same torque,R

S= Constant

1 2

21 22

S S

R R

2S0.05

R 4R

S2 = 0.05 4 = 0.2

47. The value of 2

2

4 dx|x1| ________

Ans: 12 (Range: 12)

Sol: 2

2

4 dx|x1| 2

0

4 dx|x1|2 ( |x1| 4 is

even function)

= dxx1dxx121

0

2

1

44

= 12

48. Let f (x,y) = k xy – x3y – xy3 for (x, y) R2,where ‘k’ is a real constant. The directionalderivative of f(x,y) at the point (1,2) int the

direction of unit vectori j

2 2

is15

2.

Then the value of k is ________________.

Ans: 4 (Range: 4)

Sol: (grad f).15

a2

[ky – 3x2y – y3] i + [kx – x3 – 3xy2] j .

2

15

2

j

2

i

i j2k 6 8 i k 1 12 j .

2 2

=15

2 k = 4

49. An inverter employs output voltage matchesto reduce it’s lower order harmonics. Theoutput wave form is as shown in thefollowing figure. The rms value of outputvoltage V0 at fundamental frequency is 70 Vfor 1 = 30. Then 2 ________ degrees.

Ans: 41 (Range: 39 to 43)Sol: The fourier analysis of the wave form is to

be carried out to find out fundamentalcomponent. The wave form is havingquarter wave symmetry.Hence An = 0

Bn =

tdtnsin4

2

1

1 2/

2/0

tdtnsin100tdtnsin100tdtnsin1004

n

ncos2ncos211004 21

B1 = 21 cos2cos21400

21 cos230cos21400

B270

2 = 41

2/2 t

1

2

V0

100 V

–2

–1



SD1 = 1 + j1 SD2 = 1 + j1

Bus - 1 Bus - 2Z = j 0.2 p.u

Bus-2Bus-1Z = j0.2

I2 I2V1

SD1 = 1 + j1 SD2 = 1 + j1

Bus-2Bus-1Z = j0.2I2I2V1SD1 = 1 + j1SD2 = 1 + j1I0

I1

I2

I3

4 1mux

y f (A, B, C, D)

S1 S0

C D

50. Consider the following two bus powersystem network with complex powerdemands at buses 1 and 2 as SD1= 1 + j1p.u; and SD2 = 1 + j1 p.u respectively. Bus-2voltage is given as 1 p.u in magnitude.

The angle made by Bus-1 voltage withrespect to Bus-2 voltage will be ______degrees.

Ans: 9.46 (Range: 9 to 10)Sol:

Let V2 = 1.00 p.uas S2 = V2. . 2I

2

1 j1I

1 0

I2 = 1 – j1 punow Bus-1 voltage, V1 = V2 + I2 Z

V1 = 10 + (1 – j1) (j0.2) = 1.2169.46

51. Unit step response of a system is 2(1 – e–4t)u(t) then the final value of the output of thesystem to a unit impulse is _____.(a) 0 (b) 2(c) 4 (d) None

Ans: (a)

Sol: 4tdIR 2(1 e )u(t)

dt = 8 e–4(t) u(t)

tt IR 0

52. The system described by the differential

equation2

2

d y dy du7 12y 5 17u

dt dt dt . If

the state space representation is in diagonalcanonical form, given byx(t) Ax(t) Bu(t) and Y(t) = C X(t). If

matrix C [0.5 0.5] . Then matrix ‘B’ is

(a)2

6

(b)3

6

(c)4

6

(d)6

3

Ans: (c)

Sol: T.F =Y(s)

U(s) =

2

5s 17

s 7s 12

=5s 17

(s 3)(s 4)

1 2b bT.F

s 3 s 4

b1 = 2 , b2 = 3

4

B6

53. The minimum number of two input NORgates required to implement functionf (A, B, C, D) = m (1, 3, 4, 7, 8, 9, 11, 12,14, 15) additionally with the following 4 1mux are _______ (assuming onlyuncomplemented variables are available).

(a) 6 (b) 5(c) 3 (d) None



I0

I1

I2

I3

4 1mux

y f (A, B, C, D)

S1 S0

C D

AB

A1

B

Ib

R

R

R

V = ?

Ia

Ans: (b)Sol:

54. A 3- , 100 kVA, 3000 V star connectedalternator has an effective armatureresistance is 0.2 . A field current of 40 Aproduces a short circuit current of 200 A andopen circuit emf of 1040 V (line to line).The full load regulation for 0.8 leadingpower factor is(a) 2.2% (b) 1.8%(c) 4.833% (d) 81.66%

Ans: (b)

Sol: VL = 1040 V Vph =1040

3 = 600 V

Ifull =3100 10

19.2A3 3000

;

Vph =3000

1730 V3

Zs = OC

sc

E 6003

I 200 , Ra = 0.2

Xs =2 2s aZ R = 2 23 0.2 = 2.99

E = 2 2a a a sV cos I R (Vsin I X )

=2

2

(1732 0.8 19.2 0.2)

(1732 0.6 19.2 2.99)

= 1701.3 VRegulation =E V 1699 1730

100 100 1.8%V 1730

55. A balanced three – phase AC voltage sourceis connected to a balanced star connectedload through a star-delta transformer asshown in the figure. If the magnetizingcurrent is neglected and Ia = 25 A and Ib =500. The line-to-line voltage is 20060on the star side, then what is the value of ‘V’in figure (in volts).

(a) 10060

(b) o603

100

(c) o303

100

(d) 10030

C D I0 I1 I2 I3

S1 S0 0 0 0 1 1 0 1 1A B(00) 0 2

A B(01) 5 6

AB(10) 10

AB (1 1) 13

A+B B AB 1



20060

2530

o303

200 R

R

R

50 o30R3

50R0

50R–120

500

50–120

o303

50

+

+

–

–

Ans: (d)Sol:

253

50

I

I

V

V

1ph

ph

2ph

1ph 2

Vph2 = 10030

56. A 3-phase load of 1500 kW, 0.8 powerfactor lagging is supplied from a tap-changertransformer, 11/33 kV, delta on LV and staron HV side. The leakage impedance of LVwinding is (0.5 + j6.5) per phase and thatof H.V. windings is (1.5 + j24) . If the lowvoltage side is supplied at 11 kV, find thesetting of tap-changer to give 33 kV at fullload.(a) 5.05% (b) 6.05%(c) 4.05% (d) 3.05%

Ans: (a)Sol:

o

3

3

86.3680.328.010333

101500I

86.3680.323

k33x

3

k33

×(1.5+j24+3(0.5+j6.5)

100.346.20016x3

k33

x = 1.0505

%tap = %05.51001

1x

57. If the probability of hitting a target is5

1 and

if 10 shots are fired, what is the conditionalprobability that the target being hit atleasttwice assuming that atleast one hit is alreadyscored?(a) 0.6999 (b) 0.624(c) 0.892 (d) 0.268

Ans: (a)

Sol: P(x 2| x 1) = 1xP

2xP

=n

1nn

q1

npqq1

=

10

910

5

41

5

4

5

110

5

41

= 0.6999

58. Let w (y1, y2) be the wronskian of twolinearly independent solutions y1 and y2 ofthe equation, y p(x) y Q(x) y 0 . Theproduct of w(y1, y2).p(x) =(a) 2 1 1 2y y y y (b) 1 1 2 2y y y y (c) 1 2 2 1y y y y (d) 1 2 1 2y y y y

Ans: (a)Sol: 1 1 1y p(x)y Q(x)y 0.......(1)

2 2 2y p(x)y Q(x)y 0.......(2) (1) y2 – (2) y1 p(x)

1 2 1 2 2 1 1 2y y y y y y y y

w (y1, y2) p(x) = 2 1 1 2y y y y

1.5 j24 (0.5+j6.5)k2

3

kv33x

3

k33+ +

––



59. Given the differential equation y1 = x – ywith initial condition y(0) = 0. The value ofy(0.1) calculated numerically upto the thirdplace of decimal by the 2nd order Runge-Kutta method with step size h = 0.1 is_____________

Ans: 0.005 (Range: no need)Sol: Given y1 = x – y

Also given y(0) = 0 and h = 0.1y(0.1) = ?

Let x0 = 0, x1 = x0 + 1.h = 0 + 0.1 = 0.1The 2nd order Runge-Kutta method is givenby

y1 = y(x1) = y0 +2

1 (k1 + k2)

where k1 = h f(x0, y0) andk2 = h f(x0 + h, y0 + k1)k1 = (0.1) [x0 y0] = (0.1) (0 – 0) = 0k2 = (0.1) [(x0 + h) – (y0 + k1)] = (0.1) [0 + 0.1 – (0 + 0)] = 0.01

y1 = y(0.1) = 0 +2

1(0 + 0.01) =

2

01.0

= 0.005

60.

For the circuit shown above, which of thefollowing statement is true(a) The resonance is possible when RL = 4

only(b) The resonance is possible when RL = 10 only

(c) The resonance is possible for all positivevalues of RL

(d) The resonance is not possible with anyvalue of RL

Ans: (d)

Sol: The total admittance is

YT =5j4

1

10jR

1

L

=

100R

10

41

5j

41

4

100R

R2L

2L

L

For resonance imaginary part must bezero

i.e, 18R,100R

10

41

5 2L2

L

Thus there is no value of RL which resultsin resonacne.

61. Synchronous generator is connected in aninfinite bus through a transformer anddouble circuit line. The infinite bus voltageis 1.0 pu and the internal voltage ofgenerator is also 1.0 pu. The tansferreactance before fault is 0.5 pu. A 3-phasefault is taking place near to the bus bar onone of the double circuit. The fault iscleared by circuit breaker after few cycles.The peak power after fault cleared is 75% ofpeak power before fault. The criticalclearing angle of the rotor at the faultcleared by CB is when the generator isdelivery 1.0 pu power before fault to infinitebus.(a) 68.5o (b) 74.8o

(c) 59.2o (d) 52.6o

Ans: (c)

Sol: Pm1 = 0.25.0

0.10.1

X

EV

eq

Pm2 = 0Pm3 = 0.75 Pm1 = 1.5

0 =1

m1

PSsin

P

= 30, 0(rad) = 0.523

m = 1

m3

PS180 sin

P

= 1 1.0180 sin

1.5

= 138.2o

m(rad) = 2.41

j10

RL4

– j5



j 0.5

j 0.4-j10

j0.5 j0.8

j0.25j0.1j0.2

~

① ② ③~

c =1 m m3 m

m3

PS( ) P coscos

P

= 1 1.0 2.41 0.523) 1.5cos138.2

cos1.5

= 1 1.0 2.41 0.523 1.118

cos1.5

= 1cos 0.512 = 59.2o

62. A power system network is shown belowalong with per unit reactances. Busadmittance matrix is

(a)

7 5 10

j 5 17 10

10 10 14

(b)

14100

10175

057

j

(c)

9.45 5 0

j 5 16.95 10

0 10 15.25

(d)

9.5 5 15

j 5 17 10

15 10 15.25

Ans: (c)Sol: z10 = j0.5, y10 = j2

y11 = y10 + y10(2) + y12 +112y

2 = j9.45

z10(2) = j0.4, y10(2) = j2.5z12 = j0.2, y12 = j5

y22 = y20 +y21 +y23 +112y

2 = j16.95

112z = j10, 1

12y = j0.1

z20 = j0.5, y20 = j2y33 = y30 + y30(2) + y31 = j15.25z23 = j0.1, y23 = j10y12 = y12 = (j5) = j5z30 = j0.25, y30 = j4z30(2) = j0.8, y30(2) = j1.25y12 = y21

y13 = 0 = y31

y23 = j10 = y32

63. Find the energy of the signalx(t) = 8 sinc (4t) cos 2t(a) 12 J (b) 8 J

(c)6

J

(d) None of the above

Ans: (a)Sol: 8 sinC 4(t) 2 rect (f/4)

Using frequency shifting propertyf 1 f 1

x(f ) rect rect4 4

Energy

2|tx|E

=3

2

3

x (t)

1 1 32

3 1 1

1 4 1

= 2 + 8 + 2

= 12 J

64. A AC voltage controller fed from 230 V, 50Hz AC supply and supplying power to aninductive load of R = 100 , L = 0.318 H.The output across the load is alternating andswitching devices are given with firing angleof 30 in every half cycle. The rms, averagevoltages across load are(a) 230 V, 0 V(b) 226.65 V, 179.3 V(c) 226.65 V, 0 V(d) Data insufficient



Ans: (a)Sol: Given circuit is AC voltage controller whose

outout is AC.R = 100 XL = 2fL = 2×0.318 ×50 = 100

= 1 LXtan

R

= 1 100tan

100

= 45

Since < 45; then rms output voltage issupply voltage.Vor = 230 V ; V0 = 0

65. A DC-DC converter is used to feed theresistive load and its output is alwaysinverted. An alternating square wavevoltage is getting produced across inductorhaving rms value of 100 V and frequency of50 Hz. The duty cycle of the switch is_________

(a) 0.5(b) 0(c) Alternating voltage can not be produced

in DC- DC converter(d) Can not be determined

Ans: (a)Sol:

Given circuit is Buck Boost convertervoltage across inductor would beVS = 100 VV0 = 100 V

V0 = S

DV

1 D

100 = 100×D

1 D

D = 1 D = D =1

2 = 0.5

Vs = 100 V

V0 = 100 VTON

Tt

VL

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