Gate ee previous year papers 1992 2010

273

Transcript of Gate ee previous year papers 1992 2010

Page 1: Gate ee previous year papers 1992   2010
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• Kolkata GATE FORUM 4A, ELGIN Road Next to Bhavanipur College Kolkata-700020 Tel: (033) 30947075, 30947160

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• Dhanbad GATE FORUM Flat NO: 3 B, HEM Tower, Luby Circular Road, Dhanbad – 826001 Tel: (0326) 3108848

• Durgapur GATE FORUM 2nd floor, Nachan Road Opposite Bank of India, Benachity. Durgapur - 713213

• Pune GATE FORUM 5, Kalpana Building opp. Hotel Surya Off Ghole Road Pune - 411004 Tel : (020) 25538396 / 25510078

• Guwahati GATE FORUM Maniram Dewan Road Opposite to Regalia Marriage Hall,Chandmari Guwahati 781003 Tel. 91 98640 75835

• Hubli / Dharwad

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• Mumbai

GATE FORUM 110, Shopper's Point 1st Floor , S. V. Road Opp. Andheri Railway Station. Andheri (W) Mumbai. Tel: (022) 2623 7471 / 72

• Kolkata GATE FORUM 4A, ELGIN Road Next to Bhavanipur College Kolkata-700020 Tel: (033) 30947075, 30947160

• Jaipur GATE FORUM C-16 , Greater Kailash Colony, Behind New Vidhan Sabha, Lal Kothi, Jaipur, Rajasthan. PIN : 302001,Tel:(0141) 5103580

• Trivandrum GATE FORUM TC 14/ 1679 , Behind Sanskrit College, Palayam, Trivandrum Kerala PIN : 695034. Tel : (471) 2322914,

• Nagpur GATE FORUM 3rd Floor, Samarth Chambers W.H.C Road, Opp. Chauhan Traders, Nagpur - 440 010, Maharashtra

• Dhanbad GATE FORUM Flat NO: 3 B, HEM Tower, Luby Circular Road, Dhanbad – 826001 Tel: (0326) 3108848

• Durgapur GATE FORUM 2nd floor, Nachan Road Opposite Bank of India, Benachity. Durgapur - 713213

• Pune GATE FORUM 5, Kalpana Building opp. Hotel Surya Off Ghole Road Pune - 411004 Tel : (020) 25538396 / 25510078

• Guwahati GATE FORUM Maniram Dewan Road Opposite to Regalia Marriage Hall,Chandmari Guwahati 781003 Tel. 91 98640 75835

• Hubli / Dharwad

GATE FORUMPlot no. 101, Shri Venktesh Krupa,

Shiv Basav Nagar,BELGAUM Pin 591 010

(Land mark: Naganoor swami Kalyan Mantap

• Hyderabad GATE FORUM

Suite Number 515,Model House

Hyderabad - 82Ph: +91-40-5583 3454 +91-40-5583 3242

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• Mumbai

GATE FORUM 110, Shopper's Point 1st Floor , S. V. Road Opp. Andheri Railway Station. Andheri (W) Mumbai. Tel: (022) 2623 7471 / 72

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• Guwahati GATE FORUM Maniram Dewan Road Opposite to Regalia Marriage Hall,Chandmari Guwahati 781003 Tel. 91 98640 75835

• Hubli / Dharwad

GATE FORUMPlot no. 101, Shri Venktesh Krupa,

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(Land mark: Naganoor swami Kalyan Mantap

• Hyderabad GATE FORUM

Suite Number 515,Model House

Hyderabad - 82Ph: +91-40-5583 3454 +91-40-5583 3242

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"

#4 & & & & ( ) = $

+ −

+ −

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&&, ( )

=

+& ≥ $

( ) ( )

δ− ( ) ( ) δ − − ( ) ( )

− ( ) ( )

− −

Page 159: Gate ee previous year papers 1992   2010

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) ' . && , & & )&. , ! !

,! ! ,! !⊗

,! ! ,! !⊗

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) > & & ) .&,& &

, & ) . & & ) ))' & & . ,&& ))&-, .

*>*> *>*>

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& @G , , ' &&

,

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@ ) &1 φω.

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# ω.

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Page 160: Gate ee previous year papers 1992   2010

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&& ' - ))' &

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& &&& &'2, &&

)#T & & ) &1

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# ## # 2

2 ) ,&& & &

& &&

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. & /

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# !"

$ ; & %6 & ,& ,-,,

) &1 & , & ' & ) (

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# $ , -, , & .

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Page 161: Gate ee previous year papers 1992   2010

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4 )&, &&) & &, & $

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) 1 & / & 1 &

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7 3&('# >)& %3

, )' && 7 )& & )' && ,

4 4 7

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, , ,&- # '&*$

(R74 =R =R (R77

2 3 & 1&, ' &, &L

L ' 2 & & && .

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!

= + +

= + +

' & $%& & >$%& ) .&,

, &, " >$%& ) .&

2 ' '&* & ' 2 >

1 3 N G @ & L

# & = ∠ ° $

"∠ °Ω

# #∠ − °Ω

Ω Ω

(R(R (R

=R =R

M M

H

L L

" = ∠ °Ω

Page 162: Gate ee previous year papers 1992   2010

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# !!$ # % !&'% (!!$ % "#$%& %

%

%

2 Ω# Ω

2" 3 &&' ) &, )) )' & &

)()&&, &)& (&

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. )' &, . , 2 ) (& &)) & . .)' &,)'

)

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& ) & ) . ) )& & & ((4

).8& ( ) ( ) # " = + ( ) 1

( ) # " − ( ) + ( ) 4 2 + ( ) # " +

2 3@G,, && Ω&& & S3 & - S 3 , && 1& '

& $

Ω Ω Ω Ω

22 () 22 , & & ,, &

, && & ΩΩ& ) . && & 3 4% & ) . , & & 7Ω ) & &&& , & , $

T "T

#T #T

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&& && # Ω2 Ω& ) .'23, &'& KΩ , & > ' ) & & , &&

4% 2% #% #%

M

%

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Page 163: Gate ee previous year papers 1992   2010

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24 '', &' & , & )' & & (

) . , )). & *> S *>

& ) . )' & )' &&& ) .&

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.

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.)& 1 .& ) '( & ) ,) &

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ν

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=

−ν

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& 1 .

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Page 164: Gate ee previous year papers 1992   2010

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ν

=

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Page 165: Gate ee previous year papers 1992   2010

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Page 166: Gate ee previous year papers 1992   2010

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B

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= = )& ,

2 2' &)' & ,& &&

& & ' & & & AI&&&A&D> &.& &D

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+

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+

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A ''& 1& . & &AI '

> '& & / '!D

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Page 167: Gate ee previous year papers 1992   2010

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% & $'

( % & ) $ 3 F

& #Ω ' 4 3 & ' ! ')& &., , )0

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Ω Ω

#Ω 2Ω

*,+& , * & 7T &

&.)

%

,F

Ω('

=

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=

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Page 168: Gate ee previous year papers 1992   2010

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( % & )$

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( )

−=

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Page 169: Gate ee previous year papers 1992   2010

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*,+ / & 1 &)& $

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Q. No. 1 – 20 Carry One Mark Each

1. The pressure coil of a dynamometer type wattmeter is

(A) highly inductive (B) highly resistive

(C) purely resistive (D) purely inductive

2. The measurement system shown in the figure uses three sub-systems in cascade

whose gains are specified as 1 23

1G , G and

G. The relative small errors associated

with each respective subsystem 1 2 3 1 2 3G , G and G are , andε ε ε . The error

associated with the output is:

(A) 1 23

1ε + ε +

ε (B) 1 2

3

.ε εε

(C) 1 2 3ε + ε − ε (D) 1 2 3ε + ε + ε

3. The following circuit has a source voltage Vs as shown in the graph. The current through the circuit is also shown.

The element connected between a and b could be

(A) (B)

(C) (D)

Input1G 2G

3

1

GOutput

sV+

a b

R 10k

15

10

5

0

5

10

15

−0 100 200 300 400

Vs (volts)

Time (ms)

1.5

1

0.5

0

0.5

1

1.5

−0 100 200 300 400

Current (m

A)

Time (ms)

a b

a b

a b

a b

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4. The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. The following inference can be made from this.

(A) The signals are not sinusoidal

(B) The amplitudes of the signals are very close but not equal

(C) The signals are sinusoidal with their frequencies very close but not equal

(D) There is a constant but small phase difference between the signals

5. The increasing order of speed of data access for the following devices is

(i) Cache Memory

(ii) CDROM

(iii) Dynamic RAM

(iv) Processor Registers

(v) Magnetic Tape

(A) ( ) ( ) ( ) ( ) ( )v , ii , iii , iv , i (B) ( ) ( ) ( ) ( ) ( )v , ii , iii , i , iv

(C) ( ) ( ) ( ) ( ) ( )ii , i , iii , iv , v (D) ( ) ( ) ( ) ( ) ( )v , ii , i , iii , iv

6. A field excitation of 20 A in a certain alternator results in an armature current of 400A in short circuit and a terminal voltage of 2000V on open circuit. The magnitude of the internal voltage drop within the machine at a load current of 200A is

(A) 1V (B) 10V (C) 100V (D) 1000V

7. The current through the 2 kΩ resistance in the circuit shown is

(A) 0mA (B) 1mA (C) 2mA (D) 6mA

8. Out of the following plant categories

(i) Nuclear (ii) Run-of-river (iii) Pump Storage (iv) Diesel

The base load power plants are

(A) (i) and (ii) (B) (ii) and (iii) (C) (i), (ii) and (iv) (D) (i), (iii) & (iv)

9. For a fixed value of complex power flow in a transmission line having a sending end voltage V, the real power loss will be proportional to

(A) V (B) V2 (C) 1/V2 (D) 1/V

A B

6V

1kΩ 1kΩ

1kΩ1kΩ

2kΩ

C

D

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10. How many 200W/220V incandescent lamps connected in series would consume the same total power as a single 100W/220V incandescent lamp?

(A) not possible (B) 4 (C) 3 (D) 2

11. A Linear Time Invariant system with an impulse response h(t) produces output

y(t) when input x(t) is applied. When the input ( )x t − τ is applied to a system

with impulse response ( )h t − τ , the output will be

(A) ( )y t (B) ( )( )y 2 t − τ (C) ( )y t − τ (D) ( )y t 2− τ

12. The nature of feedback in the opamp circuit shown is

(A) Current - Current feedback

(B) Voltage - Voltage feedback

(C) Current - Voltage feedback

(D) Voltage - Current feedback

13. The complete set of only those Logic Gates designated as Universal Gates is

(A) NOT, OR and AND Gates (B) XNOR, NOR and NAND Gate

(C) NOR and NAND Gates (D) XOR, NOR and NAND Gates

14. The single phase, 50Hz, iron core transformer in the circuit has both the vertical arms of cross sectional area 20cm2 and both the horizontal arms of cross sectional area 10cm2. If the two windings shown were wound instead on opposite horizontal arms, the mutual inductance will

(A) double

(B) remain same

(C) be halved

(D) become one quarter

15. A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor (solid curve) and of the load (dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B?

(A) A is stable B is unstable

(B) A is unstable B is stable

(C) Both are stable

(D) Both are unstable

1kΩ

inV

2kΩ

outV

6V+

6V−

−+

sync

NN0

A

B

1.0

Torque •

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16. An SCR is considered to be a semi-controlled device because

(A) it can be turned OFF but not ON with a gate pulse

(B) it conducts only during one half-cycle of an alternating current wave

(C) it can be turned ON but not OFF with a gate pulse

(D) it can be turned ON only during one half-cycle of an alternating voltage wave

17. The polar plot of an open loop stable system is shown below. The closed loop system is

(A) Always stable

(B) Marginally stable

(C) Unstable with one pole on the RH s-plane

(D) Unstable with two poles on the RH s-plane

18. The first two rows of Routh's tabulation of a third order equation are as follows.

3

2

2 2s

4 4s. This means there are

(A) two roots at s j= ± and one root in right half s-plane

(B) two roots at s j2= ± and one root in left half s-plane

(C) two roots at s j2= ± and one root in right half s-plane

(D) two roots at s j= ± and one root in left half s-plane

19. The asymptotic approximation of the log-magnitude vs frequency plot of a system containing only real poles and zeros is shown. Its transfer function is

(A) ( )

( ) ( )10 s 5

s s 2 s 25

+

+ + (B)

( )( ) ( )2

1000 s 5

s s 2 s 25

+

+ + (C)

( )( ) ( )100 s 5

s s 2 s 25

+

+ + (D)

( )( ) ( )2

80 s 5

s s 2 s 25

+

+ +

0.1 2 5 2.5

80dB

40dB /dec−

60dB /dec−

rad /sω

Imaginary

Real

ω = ∞

0ω =

1.42−

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20. The trace and determinant of a 2×2 matrix are known to be -2 and -35 respectively. Its eigen values are

(A) -30 and -5 (B) -37 and -1 (C) -7 and 5 (D) 17.5 and -2

Q. No. 21 – 56 Carry Two Marks Each

21. The following circuit has R 10k , C 10 F= Ω = µ R. The input voltage is a sinusoid at

50Hz with an rms value of 10V. Under ideal conditions, the current is from the source is

(A) 010 mA leading by 90π

(B) 020 mA leading by 90π

(C) 010mA leading by 90

(D) 010 mA lagging by 90π

22. In the figure shown, all elements used are ideal. For time t<0, S1 remained closed and S2 open. At t=0, S1 is opened and S2 is closed. If the voltage Vc2 across the capacitor C2 at t=0 is zero, the voltage across the capacitor combination at t=0+ will be

(A) 1V (B) 2 V (C) 1.5 V (D) 3 V

23. Transformer and emitter follower can both be used for impedance matching at the output of an audio amplifier. The basic relationship between the input power Pin and output power Pout in both the cases is

(A) Pin = Pout for both transformer and emitter follower

(B) Pin > Pout for both transformer and emitter follower

(C) Pin < Pout for transformer and Pin = Pout for emitter follower

(D) Pin = Pout for transformer and Pin < Pout for emitter follower

C10 Fµ

R

10kΩ

OPAMP+

10k

R

Ω

si

Vs=10V rms, 50Hz

2S1S

3V1C 1F 2C 2F

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24. The equivalent capacitance of the input loop of the circuit shown is

(A) 2 Fµ (B) 100 Fµ (C) 200 Fµ (D) 4 Fµ

25. In an 8085 microprocessor, the contents of the Accumulator, after the following

instructions are executed will become

XRA A

MVIB F0H

SUB B

(A) 01 H (B) 0F H (C) F0 H (D) 10 H

26. For the Y-bus matrix of a 4-bus system given in per unit, the buses having shunt

elements are BUS

5 2 2.5 0

2 10 2.5 4Y j

2.5 2.5 9 4

0 4 4 8

− − = −

(A) 3 and 4 (B) 2 and 3 (C) 1 and 2 (D) 1,2 and 4

27. The unit-step response of a unity feedback system with open loop transfer function G(s) = K/ ((s + l) (s + 2)) is shown in the figure. The value of K is

(A) 0.5 (B) 2 (C) 4 (D) 6

28. The open loop transfer function of a unity feedback system is given by

( ) ( )0.1sG s e / s−= . The gain margin of this system is

(A) 11.95dB (B) 17.67dB (C) 21.33dB (D 23.9dB

1

0.75

0.5

0.25

0

0 1 2 3 4( )times s

response

input

loop

1i1kΩ

1kΩ

1kΩ

100 Fµ

100 Fµ

149i

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29. Match the items in List-I with the items in List-II and select the correct answer using the codes given below the lists.

List I List II

To Use

a. improve power factor 1. shunt reactor

b. reduce the current ripples 2. shunt capacitor

c. increase the power flow in line 3. series capacitor

d. reduce the Ferranti effect 4. series reactor

(A) a 2 b 3 c 4 d 1→ → → → (B) a 2 b 4 c 3 d 1→ → → →

(C) a 4 b 3 c 1 d 2→ → → → (D) a 4 b 1 c 3 d 2→ → → →

30. Match the items in List-I with the items in List-II and select the correct answer using the codes given below the lists.

List I List II

Type of transmission line Type of distance relay preferred

a. Short Line 1. Ohm Relay

b. Medium Line 2. Reactance Relay

c. Long Line 3. Mho Relay

(A) a 2 b 1 c 3→ → → (B) a 3 b 2 c 1→ → →

(C) a 1 b 2 c 3→ → → (D) a 1 b 3 c 2→ → →

31. Three generators are feeding a load of 100MW. The details of the generators are

Rating(MW) Efficiency (%) Regulation (p.u.)

on 100 MVA base

Generator-1 100 20 0.02

Generator-2 100 30 0.04

Generator-3 100 40 0.03

In the event of increased load power demand, which of the following will happen?

(A) All the generators will share equal power

(B) Generator-3 will share more power compared to Generator-1

(C) Generator-1 will share more power compared to Generator-2

(D) Generator-2 will share more power compared to Generator-3

32. A 500MW, 21kV,, 50Hz, 3-phase, 2-pole synchronous generator having a rated p.f=0.9, has a moment of inertia of 27.5 x 103 kg-m2. The inertia constant (H) will be

(A) 2.44s (B) 2.71s (C) 4.88s (D) 5.42s

33. f(x,y) is a continuous function defined over ( )x, y 0,1 0,1∈ × . Given the two

constraints, x>y2 and y>x2, the volume under f(x,y) is

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(A) ( )2

x yy 1

y 0 x y

f x,y dxdy==

= =∫ ∫ (B) ( )

2 2

y 1 x 1

y x x y

f x,y dxdy= =

= =∫ ∫

(C) ( )y 1 x 1

y 0 x 0

f x,y dxdy= =

= =∫ ∫ (D) ( )

x yy x

y 0 x 0

f x,y dxdy==

= =∫ ∫

34. Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of at least two people in the room being born on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5?

(A) 20 (B) 7 (C) 15 (D) 16

35. A cascade of 3 Linear Time Invariant systems is causal and unstable. From this, we conclude that

(A) Each system in the cascade is individually causal and unstable

(B) At least one system is unstable and at least one system is causal

(C) At least one system is causal and all systems are unstable

(D) The majority are unstable and the majority are causal

36. The Fourier Series coefficient, of a periodic signal x(t), expressed as

( ) j2 kt / Tkk

x t a e∞ π

=−∞= ∑ are given by

2 1 0 1 2a 2 j1; a 0.5 j0.2; a j2; a 0.5 j0.2; a 2 j1;− −= − = + = = − = + and

ka 0; for k 2= > . Which of the following is true?

(A) x(t) has finite energy because only finitely many coefficients are non-zero

(B) x(t) has zero average value because it is periodic

(C) The imaginary part of x(t) is constant

(D) The real part of x(t) is even

37. The z-transform of a signal x n is given by 3 1 2 34z 3z 2 6z 2z .− −+ + − + It is

applied to a system, with a transfer function ( ) 1H z 3z 2−= − . Let the output be

y(n). Which of the following is true?

(A) y(n) is non causal with finite support

(B) y(n) is causal with infinite support

(C) y(n) = 0;|n|>3

(D) ( ) ( ) ( ) ( )j j j jz e z e z e z eRe Y z Re Y z ; Im Y z Im Y z ;θ − θ θ − θ= = = =

= − = − π ≤ θ < π

38. A cubic polynomial with real coefficients

(A) can possibly have no extrema and no zero crossings

(B) may have up to three extrema and up to 2 zero crossings

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(C) cannot have more than two extrema and more than three zero crossings

(D) will always have an equal number of extrema and zero crossings

39. Let x2 -117 = 0. The iterative steps for the solution using Newton-Raphson's method is given by

(A) k 1 kk

1 117x x

2 x+

= +

(B) k 1 kk

117x x

x+ = −

(C) kk 1 k

xx x

117+ = − (D) k 1 k kk

1 117x x x

2 x+

= − +

40. ( ) ( ) ( )2 2x yˆ ˆF x,y x xy a y xy a .= + + + It's line integral over the straight line from

(x,y)= (0,2) to (x,y) = (2,0) evaluates to

(A) -8 (B) 4 (C) 8 (D) 0

41. An ideal opamp circuit and its input waveform are shown in the figures. The output waveform of this circuit will be

(A) (B)

(C) (D)

3

2

1

0

1

2

3

1t 2t 3t4t 5t 6t

tV

V↑

6

0

3−

3t 6t

t →

V↑

6

0

3−

3t 6t

t →

V↑

6

0

3−

2t6t

t →4t

V↑

6

0

3−

2t 6tt →

4t

outV

6V

3V−

inV1kΩ

1kΩ

2kΩ

−+

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42. A 220V, 50Hz, single-phase induction motor has the following connection diagram and winding orientations shown. MM' is the axis of the main stator winding (M1M2) and AA' is that of the auxiliary winding (A1A2). Directions of the winding axes indicate direction of flux when currents in the windings are in the directions shown. Parameters of each winding are indicated. When switch S is closed, the motor

(A) rotates clockwise

(B) rotates anticlockwise

(C) does not rotate

(D) rotates momentarily and comes to a halt

43. The circuit shows an ideal diode connected to a pure inductor and is connected to a purely sinusoidal 50Hz voltage source. Under ideal conditions the current waveform through the inductor will look like

(A)

(B)

(C)

220V

50Hz

S1A 2A

a

a

r 1

L 10 / H

= Ω

= π

1M

2M

m

m

r 0.1

L 0.1 / H

= Ω

= π

A A′

M

M′

( )L 0.1/ H= π

D+ −

sv 10sin100 t

+= π

1.5

1

0.5

00 10 20 30 40 50

current

( )time ms

1.5

1

0.5

00 10 20 30 40 50

current

( )time ms

1.5

1

0.5

00 10 20 30 40 50

current

( )time ms

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(D)

44. The Current Source Inverter shown in figure, is operated by alternately turning on thyristor pairs (T1, T2) and (T3, T4). If the load is purely resistive, the theoretical maximum output frequency obtainable will be

(A) 125kHz (B) 250kHz (C) 500kHz (D) 50kHz

45. In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio of 0.8, which is much larger the commutation interval. If the maximum allowable reapplied dv/dt on TM is 50 V/µs, what should be the theoretical minimum value of C1? Assume current ripple through Lo to be negligible.

(A) 0.2 Fµ (B) 0.02 Fµ (C) 2 Fµ (D) 20 Fµ

46. Match the switch arrangements on the top row to the steady-state V-I characteristics on the lower row. The steady state operating points are shown by large black dots

1.5

1

0.5

00 10 20 30 40 50

current

( )time ms

10A

T1

D1

D4

T 4

T3

D3

D2

T2

0.1 Fµ

− +

0.1 F

− +

µ

10Ω

8Ω0C

0L

0D

1D

1C− +

1LMT

100V

+

AT

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(A) A I B II C III D IV− − − − (B) A II B IV C I D III− − − −

(C) A IV B III C I D II− − − − (D) A IV B III C II D I− − − −

47. For the circuit shown, find out the current flowing through the 2Ω resistance. Also identify the changes to be made to double the current through the 2Ω resistance

(A) ( )s5A;Put V 20V=

(B) ( )s2A; Put V 8V=

(C) ( )s5A;Put I 10A=

(D) ( )s7A; Put I 12A=

48. The figure shows a three-phase delta connected load supplied from a 400V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be

(A) 0 (B) 1600 Watt (C) 800 Watt (D) 400 Watt

49. An average-reading digital multimeter reads 10V when fed with a triangular wave, symmetric about the time-axis. For the same input an rms-reading meter will read.

(A) 20 / 3 (B) 10 / 3 (C) 20 3 (D) 10 3

+ −( )A

si

( )Isv

+ −( )B

si

( )IIsv

+ −

( )C

si

( )III

sv

+ −

( )D

si

( )IV

sv

± 2ΩSi 5A=

SV 4V=

3-Phase

Balanced

Supply

400Volts

50Hz cb

a

2Z1Z

CC

PC

( )1Z 100 j0= + Ω ( )2Z 100 j0= + Ω

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50. Figure shows the extended view of a 2 pole dc machine with 10 armature conductors. Normal brush positions are shown by A and B, placed at the interpolar axis. If the brushes are now shifted, in the direction of rotation, to A' and B' as shown, the voltage waveform A BV ′ ′ will resemble

(A)

(B)

(C)

(D)

Common Data Questions: 51 & 52

N S

A'+A

+

B'−B

1 2 3 4 5 1' 2' 3' 4' 5'

rotation at speed rad/secω

A 'B 'V

0 0.2π 0.4π 0.6π 0.8π πtω

A 'B 'V

0 0.2π 0.4π 0.6π 0.8π πtω

A 'B 'V

0 0.2π 0.4π 0.6π 0.8π πtω

A 'B 'V

0 0.2π 0.4π 0.6π 0.8π πtω

A

B

C

N S1 S2

a

b

c

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The star-delta transformer shown above is excited on the star side with a balanced, 4-wire, 3-phase, sinusoidal voltage supply of rated magnitude. The transformer is under no load condition.

51. With both S1 and S2 open, the core flux waveform will be

(A) A sinusoid at fundamental frequency (B) Flat topped with third harmonic

(C) Peaky with third harmonic (D) None of these

52. With S2 closed and S1 open, the current waveform in the delta winding will be

(A) a sinusoid at fundamental frequency (B) flat topped with third harmonic

(C) only third harmonic (D) none of these

Common Data Questions: 53 & 54

The circuit diagram shows a two winding, lossless transformer with no leakage flux, excited from a current source, i(t), whose waveform is also shown. The transformer has a magnetizing inductance of 400/πmH.

53. The peak voltage across A and B, with S open is

(A) 400/πV (B) 800V (C) 4000/πV (D) 800/πV

54. If the waveform of i(t) is changed to ( ) ( )i t 10sin 100 t A,= π the peak voltage

across A and B with S closed is

(A) 400V (B) 240V (C) 320V (D) 160V

Common Data Questions: 55 & 56

A system is described by the following state and output equations

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )( ) ( )

1

1 2

2

2

1

dx t3x t x t 2u t

dt

dx t2x t u t

dt

y t x t

where u t is the input and y t is the output

= − + +

= − +

=

( )i t

•A

•B

S

30Ω

1:1

5ms 10ms 15ms 20ms 25ms 30ms t

10A

0

( )i t

10A

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55. The system transfer function is

(A) 2

s 2

s 5s 6

++ −

(B) 2

s 3

s 5s 6

++ +

(C) 2

s 5

s 5s 6

++ +

(D) 2

2s 5

s 5s 6

−+ −

56. The state transition matrix of the above system is

(A) 3t

2t 3t 2t

e 0

e e e

− − −

+ (B)

3t 2t 3t

2t

e e e

0 e

− − −

(C) 3t 2t 3t

2t

e e e

0 e

− − −

+

(D) 3t 2t 3t

2t

e e e

0 e

− −

Linked Answer Questions: Q.57 to Q.60 Carry Two Marks Each

Statement for Linked Answer Questions: 57 & 58

The figure above shows coils 1 and 2, with dot markings as shown, having 4000 and 6000 turns respectively. Both coils have a rated current of 25A. Coil 1 is excited with single phase, 400V, 50Hz supply

57. The coils are to be connected to obtain a single phase, 400/1000V, auto transformer to drive a load of 10kVA. Which of the options given should be exercised to realize the required auto transformer?

(A) Connect A and D; Common B (B) Connect B and D; Common C

(C) Connect A and C; Common B (D) Connect A and C; Common D

58. In the autotransformer obtained in Question 57, the current in each coil is

(A) Coil-1 is 25 A and Coil-2 is 10 A (B) Coil-1 is 10 A and Coil-2 is 25 A

(C) Coil-1 is 10 A and Coil-2 is 15 A (D) Coil-1 is 15 A and Coil-2 is 10 A

Statement for Linked Answer Questions: 59 & 60

Coil 2

C

D

••Coil 1

A

B

+ −

+−5V

2kΩ 1kΩ

2kΩ3VAB

B

A

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59. For the circuit given above, the Thevenin’s resistance across the terminals A and B is

(A) 0.5kΩ (B) 0.2kΩ (C) 1kΩ (D) 0.11kΩ

60. For the circuit given above, the Thevenin’s voltage across the terminals A and B is

(A) 1.25V (B) 0.25V (C) 1V (D) 0.5V

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Q. No. 1 – 25 Carry One Mark Each

1. The value of the quantity P, where 1

x

0

P xe dx,is equal to= ∫

(A) 0 (B) 1 (C) e (D) 1/e

2. Divergence of the three-dimensional radial vector field r is

(A) 3 (B) 1/r (C) ˆˆ ˆi j k+ + (D) ( )ˆˆ ˆ3 i j k+ +

3. The period of the signal ( )x t 8sin 0.8 t4π⎛ ⎞= π +⎜ ⎟

⎝ ⎠ is

(A) 0.4 sπ (B) 0.8 sπ (C) 1.25s (D) 2.5s

4. The system represented by the input-output relationship ( ) ( )5t

y t x d , t 0−∞

= τ τ >∫ is

(A) Linear and causal (B) Linear but not causal

(C) Causal but not linear (D) Neither linear nor causal

5. The switch in the circuit has been closed for a long time. It is opened at t = 0. At t = 0+, the current through the 1 Fμ capacitor is

(A) 0A (B) 1A (C) 1.25A (D) 5A

6. The second harmonic component of the periodic waveform given in the figure has an amplitude of

(A) 0

(B) 1

(C) 2 / π

(D) 5

7. As shown in the figure, a 1Ω resistance is connected across a source that has a load line v + i = 100. The current through the resistance is

(A) 25A (B) 50A

(C) 100A (D) 200A

5V

t 0=

1 Fμ 4Ω

1+

0

1−

T /2 T t

Source v+

i

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8. A wattmeter is connected as shown in the figure. The wattmeter reads

(A) Zero always

(B) Total power consumed by Z1 and Z2

(C) Power consumed by Z1

(D) Power consumed by Z2

9. An ammeter has a current range of 0 - 5 A, and its internal resistance is 0.2Ω . In order to change the range to 0 - 25 A, we need to add a resistance of

(A)0.8Ω in series with the meter (B) 1.0Ω in series with the meter

(C)0.04Ω in parallel with the meter (D) 0.05Ω in parallel with the meter

10. As shown in the figure, a negative feedback system has an amplifier of gain 100 with ±10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately:

(A) 10±1%

(B) 10 ±2%

(C) 10 ±5%

(D) 10 ±10%

11. For the system ( )2 ,

s 1+ the approximate time taken for a step response to reach 98%

of its final value is

(A) 1s (B) 2s (C) 4s (D) 8s

12. If the electrical circuit of figure (b) is an equivalent of the coupled tank system of figure (a), then

(A) A, B are resistances and C, D capacitances

(B) A, C are resistances and B, D capacitances

(C) A, B are capacitances and C, D resistances

(D) A, C are capacitances and B, D resistances

Current coil

Potential coil

Wattmeter

1Z

2Z

100 10%±

9100

+

1h2h A C

B D

( )a Coupled tank ( )b Electrical equivalent

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13. A single-phase transformer has a turns ratio of 1:2, and is connected to a purely resistive load as shown in the figure. The magnetizing current drawn is 1A, and the secondary current is 1A. If core losses and leakage reactance’s are neglected, the primary current is

(A) 1.41A (B) 2A (C) 2.24 A (D) 3 A

14. Power is transferred from system A to system B by an HVDC link as shown in the figure. If the voltages VAB and VCD are as indicated in the figure, and I > 0, then

(A) VAB<0, VCD<0, VAB>VCD (B) VAB>0, VCD>0, VAB>VCD

(C) VAB>0, VCD>0, VAB<VCD (D) VAB>0, VCD<0

15 A balanced three-phase voltage is applied to a star-connected induction motor, the phase to neutral voltage being V. The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance referred to the stator, and the magnetizing reactance are denoted by s r s r mr ,r ,x ,x and X , respectively. The magnitude of the starting current of the motor is given by

(A) ( ) ( )2 2

s r s r

V

r r x x+ + + (B)

( )22s s m

V

r x X+ +

(C) ( ) ( )2 2

s r m r

V

r r X x+ + + (D)

( )22s m r

V

r X x+ +

16. Consider a step voltage wave of magnitude 1pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant the travelling wave reaches the reactor is

(A) –1pu (B) 1pu (C) 2pu (D) 3pu

1.21A

ACSystem A

Rectifier

A

B

ABVl

CDV

Inverter

ACSystem B

C

D

Power Flow⎯⎯⎯⎯⎯⎯→

A

Reactor

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17. Consider two buses connected by an impedance of (0+j5)Ω. The bus 1 voltage is o100 30 V∠ , and bus 2 voltage is o100 0 V.∠ The real and reactive power supplied by

bus 1, respectively, are

(A) 1000W, 268VAr (B) –1000W, –134Var

(C) 276.9W, –56.7Var (D) –276.9W, 56.7Var

18. A three-phase, 33kV oil circuit breaker is rated 1200A, 2000MVA, 3s. The symmetrical breaking current is

(A) 1200 A (B) 3600 A (C) 35 kA (D) 104.8 kA

19. Consider a stator winding of an alternator with an internal high-resistance ground fault. The currents under the fault condition are as shown in the figure. The winding is protected using a differential current scheme with current transformers of ratio 400/5 A as shown. The current through the operating coil is

(A) 0.17875 A (B) 0.2A (C) 0.375A (D) 60 kA

20. The zero-sequence circuit of the three phase transformer shown in the figure is

(A) (B)

(C) (D)

CT ratio 400 /5 CT ratio 400 /5

( )220 j0 A+ ( )250 j0 A+

Operating coil

R

Y

B

r

b

y

R r

G

R r

G

R r

G

R r

G

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21. Given that the op-amp is ideal, the output voltage V0 is

(A) 4V

(B) 6V

(C) 7.5V

(D) 12.12V

22. Assuming that the diodes in the given circuit are ideal, the voltage V0 is

(A) 4V

(B) 5V

(C) 7.5V

(D) 12.12V

23. The power electronic converter shown in the figure has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a

(A) step-down chopper (buck converter)

(B) half-wave rectifier

(C) step-up chopper (boost converter)

(D) full-wave rectifier

24. Figure shows a composite switch consisting of a power transistor (BJT) in series with a diode. Assuming that the transistor switch and the diode are ideal, the I-V characteristic of the composite switch is

(A) (B)

(C) (D)

0V10V 15V10kΩ

10kΩ

10kΩ

INV

AP

B

L

+

OUTV

+ V

I

V

I

V

I

V

I

V

I

R

2V+

2R

10V+

10V−0V

−+

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25. The fully controlled thyristor converter in the figure is fed from a single-phase source. When the firing angle is 0°, the dc output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60°, assuming continuous conduction?

(A) 150V

(B) 210V

(C) 300V

(D) 100 Vπ

Q. No. 26 – 51 Carry Two Marks Each

26. At t = 0, the function ( ) sin tf t has

t=

(A) a minimum (B) a discontinuity

(C) a point of inflection (D) a maximum

27. A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is

(A) 1/3 (B) 3/7 (C) 1/2 (D) 4/7

28. An eigenvector of 1 1 0

P 0 2 2 is0 0 3

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

(A) T

1 1 1−⎡ ⎤⎣ ⎦ (B) T

1 2 1⎡ ⎤⎣ ⎦ (C) T

1 1 2−⎡ ⎤⎣ ⎦ (D) T

2 1 1−⎡ ⎤⎣ ⎦

29. For the differential equation 2

2

d x dx6 8x 0

dtdt+ + = with initial conditions

( )t 0

dxx 0 1 and 0

dt =

= = , the solution is

(A) ( ) 6t 2tx t 2e e− −= − (B) ( ) 2t 4tx t 2e e− −= −

(C) ( ) 6t 4tx t e 2e− −= − + (D) ( ) 2t 4tx t e 2e− −= +

30. For the set of equations, 1 2 3 4x 2x x 4x 2+ + + = and 1 2 2 43x 6x 3x 12x 6+ + + = .

The following statement is true (A) Only the trivial solution 1 2 3 4x x x x 0 exists= = = =

(B) There are no solutions

(C) A unique non-trivial solution exists (D) Multiple non-trivial solutions exist

+

dcV

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31 x(t) is a positive rectangular pulse from t = -1 to t = +1 with unit height as shown

in the figure. The value of ( ) ( )2| X | d where X∞

−∞

ω ω ω∫ is the Fourier transform of

x(t) is

(A) 2

(B) 2π

(C) 4

(D) 4π

32. Given the finite length input x[n] and the corresponding finite length output y[n] of an LTI system as shown below, the impulse response h[n] of the system is

(A) h n 1,0,0,1=⎡ ⎤⎣ ⎦↑

(B) h n 1,0,1=⎡ ⎤⎣ ⎦↑

(C) h n 1,1,1,1=⎡ ⎤⎣ ⎦↑

(D) h n 1,1,1=⎡ ⎤⎣ ⎦↑

33. If the 12Ω resistor draws a current of 1A as shown in the figure, the value of resistance R is

(A) 4Ω (B) 6Ω (C) 8Ω (D) 18Ω

34. The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a, b) and (c, d), respectively. It has an impedance matrix Z with parameters denoted by ijz .A 1Ω resistor is connected in series with the network at

port 1 as shown in the figure. The impedance matrix of the modified two-port network (shown as a dashed box) is

(A) 11 12

21 22

z 1 z 1

z z 1

+ +⎛ ⎞⎜ ⎟

+⎝ ⎠ (B) 11 12

21 22

z 1 z

z z 1

+⎛ ⎞⎜ ⎟

+⎝ ⎠ (C) 11 12

21 22

z 1 z

z z

+⎛ ⎞⎜ ⎟⎝ ⎠

(D) 11 12

21 22

z 1 z

z 1 z

+⎛ ⎞⎜ ⎟

+⎝ ⎠

( )x t

1

1− 0 1 t

[ ]h n[ ] x n 1, 1= −

[ ] x n 1,0,0,0, 1= −

2A

R

6V12Ω1A

1Ω a

b

c

d

e

P

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35. The Maxwell's bridge shown in the figure is at balance. The parameters of the inductive coil are

(A) 2 3 4 4 2 3R R R /R , L C R R= = (B) 2 3 4 4 2 3L R R /R , R C R R= =

(C) ( )4 2 3 4 2 3R R /R R , L 1 C R R= = (D) ( )4 2 3 4 2 3L R /R R ,R 1 / C R R= =

36. The frequency response of ( ) ( ) ( )G s 1 / s s 1 s 2⎡ ⎤= + +⎣ ⎦ plotted in the complex

( )G jω plane (for 0 ) is< ω < ∞

(A) (B)

(C) (D)

37. The system 1 2 0

x Ax Bu with A , B is0 2 1−⎡ ⎤ ⎡ ⎤

= + = =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

(A) stable and controllable (B) stable but uncontrollable

(C) unstable but controllable (D) unstable and uncontrollable

38. The characteristic equation of a closed-loop system is

s(s+1)(s+3)+k(s+2)=0, k>0 . Which of the following statements is true?

(A) Its roots are always real

(B) It cannot have a breakaway point in the range 1 Re s 0− < <⎡ ⎤⎣ ⎦

(C) Two of its roots tend to infinity along the asymptotes Re[s] = -1

(D) It may have complex roots in the right half plane

R j L+ ω 3R

4R

2R( )4j / C− ω

3 / 4−Im

Re

0ω =3 / 4−

Im

Re

0ω =

1/6−

Im

Re

0ω =

1/6−

Im

Re

0ω =

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39. A 50 Hz synchronous generator is i n i t i a l l y connected to a long lossless transmission l i n e which is open circuited at the receiving end. With the f i e l d voltage held constant, the generator is disconnected from the transmission l ine. Which of the following may be sa id about the steady state terminal voltage and field current of the generator?

(A) The magnitude of terminal voltage decreases, and the field current does not change

(B) The magnitude of terminal voltage increases, and the field current does not change

(C) The magnitude of terminal voltage increases, and the field current increases

(D) The magnitude of terminal voltage does not change, and the field current decreases

40. A separately excited dc machine is coupled to a 50Hz, three-phase, 4-pole induction machine as shown in the figure. The dc machine is energized first and the machines rotate at 1600 rpm. Subsequently the induction machine is also connected to a 50Hz, three-phase source, the phase sequence being consistent with the direction of rotation. In steady state,

(A) Both machines act as generators

(B) The dc machine acts as a generator, and the induction machine acts as a motor

(C) The dc machine acts as a motor, and the induction machine acts as a generator

(D) Both machines act as motors

41. A balanced star-connected and purely resistive load is connected at the secondary of a star-delta transformer as shown in the figure. The line-to-line voltage rating of the transformer is 110V/220V. Neglecting the non-idealities of the transformer, the impedance 'Z' of the equivalent star-connected load, referred to the primary side of the transformer, is

Long Transmission Line

receiving end

DC machine Induction machine4 pole, 50Hz

50Hz, balancedthree phase supply−

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(A) ( )3 j0+ Ω (B) ( )0.866 j0.5− Ω (C) ( )0.866 j0.5+ Ω (D) ( )1 j0+ Ω

42. Consider a three-phase, 50Hz, 11kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figure. The voltage across the two insulators is

(A) e1 3.74kV, e2 2.61kV= =

(B) e1 3.46kV, e2 2.89kV= =

(C) e1 6.0kV, e2 4.23kV= =

(D) e1 5.5kV, e2 5.5kV= =

43. Consider a three-core, three-phase, 50Hz, 11kV cable whose conductors are denoted as R, Y and B in the figure. The inter-phase capacitance (C1) between each pair of conductors is 0.2 Fμ and the capacitance between each l ine conductor and the sheath is0.4 Fμ . The per-phase charging current is

(A) 2.0A (B) 2.4A (C) 2.7A (D) 3.5A

R

Y

B

R

Y

B

Z

Z Z

r

b

y

110 /220V

C

5C

5C

e2

e1

Conductor

R

B Y

C1 C1

C2

C2C1

C2

Outer Sheath

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44. For the power system shown in the figure below, the specifications of the components are the following:

G1: 25 kV, 100 MVA, X=9%

G2: 25'kV, 100MVA, X=9%

T1: 25 kV/220 kV, 90 MVA, X=12%

T2: 220kV/ 25 kV, 90 MVA, X=12%

Line1: 220 kV, X= 150 ohms

Choose 25 kV as the base voltage at the generator G1, and 200 MVA as the MVA base. The impedance diagram is

(A)

(B)

(C)

(D)

G1

T1

Line 1

T2

G2

Bus 1 Bus 2

j0.27

j0.18

G1

j0.42 j0.27

j0.18

G2

j0.27

j0.18

G1

j0.62 j0.27

j0.18

G2

j0.27

j0.21

G1

j0.42 j0.27

j0.21

G2

j0.3

j0.21

G1

j0.42 j0.3

j0.21

G2

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45. The transistor circuit shown uses a s i l i c o n transistor with BE C EV 0.7 V, I I= ≈ and a dc current gain of 100. The value of V0 is

(A) 4.65A

(B) 5V

(C) 6.3V

(D) 7.23V

46. The TTL circuit shown in the figure is fed with the waveform X (also shown). All gates have equal propagation delay of 10 ns. The output Y of the circuit is

(A) (B)

(C) (D)

47. When a "CALL Addr" instruction is executed, the CPU carries out the following

sequential operations internally: Note: (R) means content of register R ((R)) means content of memory location pointed to by R PC means Program Counter SP means Stack Pointer (A) (SP) incremented (B) ( )PC Addr←

( )( )( ) ( )PC Addr

SP PC

( )( ) ( )( )

SP PC

SP incremented

(C) ( )PC Addr← (D) ( )( ) ( )SP PC←

( )( )( ) ( )SP incremented

SP PC←

( )( )SP incremented

PC Addr←

10V+

10kΩ 50kΩ

0V

100Ω

100 nsX

1

0 t

X Y

Y

1

0 t

Y

1

0 t

Y

1

0 t

Y

1

0 t

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Common Data Questions: 48 & 49

A separately excited DC motor runs at 1500 rpm under no-load with 200 V applied to the armature. The field voltage is maintained at its rated value. The speed of the motor, when it delivers a torque of 5 Nm, is 1400 rpm as shown in the figure. The rotational losses and armature reaction are neglected.

48. The armature resistance of the motor is,

(A)2Ω (B) 3.4Ω (C) 4.4Ω (D) 7.7Ω

49. For the motor to deliver a torque of 2.5 Nm at 1400 rpm the armature voltage to be applied is

(A) 125.5V (B) 193.3V (C) 200V (D) 241.7V

Common Data Questions: 50 & 51

Given f(t) and g(t)as shown below:

50. g(t)can be expressed as

(A) ( ) ( )g t f 2t 3= − (B) ( ) tg t f 3

2⎛ ⎞= −⎜ ⎟⎝ ⎠

(C) ( ) 3g t f 2t

2⎛ ⎞= −⎜ ⎟⎝ ⎠

(D) ( ) t 3g t f

2 2⎛ ⎞= −⎜ ⎟⎝ ⎠

51. The Laplace transform of g(t) is

(A) ( )3s 5s1e e

s− (B) ( )5s 3s1

e es

− −−

(C) ( )3s

2se1 e

s

−−− (D) ( )5s 3s1

e es

( )speed rpm

15001400

0 5 ( )torque Nm

( )f t

1

0 1 t

( )g t

1

0t3 5

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Linked Answer Questions: Q.52 to Q.55 Carry Two Marks Each

Statement for Linked Answer Questions: 52 & 53

The following Karnaugh map represents a function F.

52. A minimized form of the function F is

(A) F XY YZ= + (B) F XY YZ= + (C) F XY YZ= + (D) F XY YZ= +

53. Which of the following circuits is a realization of the above function F?

(A)

(B)

(C)

(D)

Statement for Linked Answer Questions: 54 & 55

The L-C circuit shown in the figure has an inductance L=1mH and a capacitanceC 10 F= μ .

X

Y

Z

F

X

Y

Z

F

X

Y

Z

F

X

YZ

F

L

t 0=

Ci

100V−

+

1 1 1 0

01001

0

00 01 11 10F

X

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54. The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t = 0. The current i through the circuit is:

(A) ( )35cos 5 10 t A× (B) ( )45sin 10 t A

(C) ( )310cos 5 10 t A× (D) ( )410sin 10 t A

55. The L-C circuit of Q54 is used to commutate a thyristor, which is initially carrying a current of 5A as shown in the figure below. The values and initial conditions of L and C are the same as in Q54. The switch is closed at t = 0. If the forward drop is negligible, the time taken for the device to turn off is

(A) 52 sμ (B) 156 sμ (C) 312 sμ (D) 26 sμ

General Aptitude (GA) Questions

Q.No. 56-60 Carry One Mark Each

56. 25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is

(A) 2 (B) 17 (C)13 (D) 3

57. The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair.

Unemployed: Worker

(A) fallow: land (B) unaware: sleeper (C) wit: jester (D) renovated: house

58. Choose the most appropriate word from the options given below to complete the following sentence

If we manage to ____________ our natural resources, we would leave a better planet for our children.

(A) uphold (B) restrain (C) cherish (D) conserve

59. Which of the following options is closest in meaning to the word: Circuitous?

(A) cyclic (B) indirect (C) confusing (D) crooked

L

t 0=

Ci

100V−

+

5A100V 20Ω

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60. Choose the most appropriate word from the options given below to the complete the following sentence:

His rather casual remarks on politics ___________ his lack of seriousness about the subject.

(A) masked (B) belied (C) betrayed (D)suppressed Q.No. 61 - 65 Carry Two Marks Each 61. Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and sisters).

All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts:

i. Hari’s age + Gita’s age > Irfan’s age + Saira’s age ii. The age difference between Gita and Saira is 1 year. However Gita is not the

oldest and Saira is not the youngest. iii. There are no twins. In what order were they born (oldest first)? (A) HSIG (B) SGHI (C) IGSH (D) IHSG 62. 5 skilled workers can build a wall in 20days; 8 semi-skilled workers can build a wall

in 25 days; 10 unskilled workers can build a wall in 30days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall?

(A) 20 (B) 18 (C) 16 (D) 15 63. Modern warfare has changed from large scale clashes of armies to suppression of

civilian populations. Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause.

Which of the following statements best sums up the meaning of the above passage: (A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in warfare would be undesirable (D) People in military establishments like to use chemical agents in war. 64. Given digits 2,2,3,3,4,4,4,4 how many distinct 4 digit numbers greater than 3000

can be formed? (A) 50 (B) 51 (C) 52 (D) 54

65. If 137+276=435 how much is 731+672? (A) 534 (B) 1403 (C) 1623 (D)1513

Page 250: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 1 of 14

= =+

!

" #$ $

+ = + =

% !

$

" & ' ()

∴*!*

( )

= +

= + + = + +

= +

#

Page 251: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 2 of 14

( ) ( ) ( )

"

"

ω

ω

+= = = ± = ±+ +

=

' + ,-+,' + !

ε.. ∞/ '. #$

=

/ +

0' !12 1ω

0' !1ω! 0' ,+ +3+ 4 ,)!5

ωω

= = ∞

.6+.!∞7

"$

" " "

"

∞ ∞−− += = = =

=

( ) ( ) ( )

( )( ) $

$ $

$

= − = −

=+

=

$

"

#

5 "5

" "

2 = = = =

7

0 6 ++ ()

∴*!*

Page 252: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 3 of 14

$ '+

&!6

"

8+98:*;!** !*

*;<* 98: +1

*-!*

#

7

$

$

$

( ) ( )

( )

#

#

## ##

$7 $ 7

$

= ∠= ∠ −= ∠

− ∠= = = ∠ −

= = ∠ −

+ + =−= − + = ∠ − + ∠ −

−= − − − −

− = ∠ − = ∠

=

$$

=

=

&

6

>

>

>

*

-

&

&

&?

?

Page 253: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 4 of 14

*+6!

$ #

+6!

#

× ×=

$" & ' () (, +(+/.ε, '+

@' !2(, +, ,+,

A!

$

$#

*0B!*

*:'!*

8+:'

:'!Ω

$

'+, + , 4 2;C!5

( )

( )

−+ =

+

>D>22!5 D>2>2!

8+ - ) 6 D>< >2<

<>5>E2 ><1

<

$

F!D)

F!$GD)HD!D$)

)!GD$)H

+

0

Ω

B

Ω

Ω

0

Ω

B

Ω

Ω "*

F

$ ?

+

122

1 F

Page 254: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 5 of 14

( )( )

"

=

+

( )

= .

( ) ( )"

=

+

() +++! ( )

"

→ →=

+

() +++!

$7

0' 3;CD71 ( )

π

7 7"

π π − − = − −

/' '+.' . + ' !1

7

"

ωπ

ω π

− − = −

=

. '

"

π πω

ππ

=

= =

,+D + , D2I

"

( ) ( ) ( )$##

+=+

' 6+.0! $## 7

ωω− − + − − −

0!#ω+(J , - >!$##

"

"

"$

""

Page 255: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 6 of 14

( ) ( ) ( )

( )

− − −

= = + − =

− =

=

"

.+( !

∂ ∂+∂

!

$ +

K$ 2I

6. (

"#

( )

( ) ( )

( )

( ) ( )

( )

$

$

$

$

→∞ → →

→∞

+ + =

+ + × =

=+ +

→ ∞ →

= = = + +

=

" ∞

( ) ( )( )

$ #$

$

+ + = =

+ +

=

" ( )

5

!

−= = −+ +

+ L1 +3+6 =D1>5=≥"7

α (5 5

" "

= =

0/ +α

&3 (, + (- + ( (-)+ (,.3 (3

∴ ( )5#

#$ $ #

#

φ α φ

φ

=

↓ ↑

Page 256: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 7 of 14

,

0!>@

:+4 !>φ& !>φω6 0!&!>φω6&

$

"

#

( )( )

( )( )

# "

""

%

% %

%

=

= ×

=

" J

J

# J

$ J

&

&

% &

'&

'&

'&

'&

= Ω= Ω

= += Ω= − = Ω

7 0/ +3, +7.( ,, + (

∴B!0 φ!"× G,17H B!*

0!"@

&3,, +. 3 +,

B!0 φ

"

( φ

φ+= → = =

,φ!. B!2!$

#

?!IDI2I!1I7

?!1I7

#

5

"

5

"

) * ) *

=

+ = += + == =

Page 257: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 8 of 14

# &!

*

= =

M!" "∠ −

#$

6

#

δ ×

=

=

×= +

=

&

&

M*

2

1

∠$

&

M*

2

1

N

$F

F* !

&3 -

Page 258: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 9 of 14

N

$

&3 -

( )6

×= =+

#"

%3 + , 3 ' '+L / ',+

%3 + , 3 ' + , / ',+

= =

( ) ( )

)

" " $

5 "

π

ω ππ

ω

= = +

= = =

=

) !B 2

# )./ +6!

&6

= =

6!5

'

&&

= + = '+ !

"

= Ω

##

# #"Ω

+++!7 #"

"O#"

− =

#

#7 @!

@!1

@!@C@5@!=@

3)

3

Ω

Page 259: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 10 of 14

( )

$ $

"$

, $$"

φ

φφ

− −= = + −

==

777

7777

77

$

* $

*

&+

&+ +

+ +

βαβ

β

α µ

β µ µ

= = =+

=

= = =

= = ≈ =

$

C + 16 +

5(

(

= +

=

" 6-.

2

1*

A

* !

F

F

B

Page 260: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 11 of 14

5( ( = + =

#

!!5B/) ...

B -

'.' . ,+ ' ( , ,++ 3/.

( )

( , +

+.) !

+

7 *!* 3!>CL

A! !

A!6C :!6 ,

" ,

&

= =

,++ ' (,++

!

( )

" "

"

αα

− −

− × −

− − − =

− − =

,++ !"7 +≈

: !:2 ( ), % ,

ω = +

!2PP!"%6

( ) ( ) ( )6 , ,

$ $

7

&

α απ

α

= + = +

= ==

0

FF

B

(

C

*"Ω

Ω

,

Page 261: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 12 of 14

&,!

"

+=+

&(, +, '+ ,+,

0

*!2

!"Ω

: + , !"2"2!Ω!

:6 , !

,

= =

∴ ! 1

, +.) 3 !D7!O!

, ' + ( +.)++ ' 4+ 3 ' ,++ ' ,++ '

,++

,

= ×

=

= =

( )

( ) ( )

( )

( ) ( )

$

$

$

$

$

$

$

$

$

+

+

φ

φ

−−− −

−− −

= = −

− =

= − = − −

− + = = +

+

− =

C

Ω

*

Page 262: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 13 of 14

( ) ( ) ( )

( )

( )

( )

$

$ $

$

$$

$

$

$

$

+ *

−−

2−

= − +

−+ = + +

+ − + = +

− + + =

+

( )$

$$

−=

$ !>*##>* λ $1' F!Ω 3!

( )$$ , $ ##

"

$# "7"

"777

#

-. -. -.

-.

!

$

$

$

φ= × = × × ×

= −

= + × = +

=

= −

>@!>*,φ

0!>@

!

$ $ ##

"

δ

δ

× × ×

=

" @' $1φ3 ,,+ ' 3 ,++ 6 (-) ' 4 , + , , )

∴"×#!×$×&

$

×=

F !Ω

&6!

&

∂ =∂

!1.6+(

Page 263: Gate ee previous year papers 1992   2010

"! #$##&% '$ !% !'

Page 14 of 14

6. 4 ,+,

N

;

*

+(

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written permission. Discuss this questions paper at www.gateforum.com. 1

Answer keys

1 A 2 B 3 C 4 C 5 D 6 A 7 D

8 C 9 A 10 A 11 D 12 B 13 D 14 C

15 D 16 B 17 A 18 B 19 D 20 D 21 C

22 C 23 A 24 B 25 B 26 B 27 28 B

29 C 30 B 31 D 32 A 33 D 34 D 35 A

36 D 37 C 38 39 B 40 B 41 C 42 A

43 44 B 45 B 46 B 47 B 48 C 49 B

50 C 51 C 52 C 53 54 C 55 B 56 B

57 B 58 59 60 C 61 C 62 B 63 C

64 A 65 A 66 C 67 C 68 C 69 B 70 A

71 72 73 74 75 76 C 77

78 B 79 C 80 C 81 B 82 D 83 C 84 D

85 A

Explanation:-

1. ( )b n 1− +

b No of branches

n No of nodes

==

2. Vth

Ithzth

=

3.71 15.9

2.4 8.0

1.54 7.8

−=

−= −

I lead V so it is a combination of Resistance and capacitance

3. When we apply ( )dt dte sin t to LTI system O /P is of form Ke sin t ,− −ω ω + φ

so, compare this equation with ( )tKe sin t so ,−β υ + φ β = α

υ = ω

4. 4

3

0

x 1x dx 1

4 4= =∫ ∫

5. It should satisfy the characteristic equation 3 2 2 l 0λ + λ + λ + =

( )

( )3 2 2 1

2 1 1 2

P P 2P 1 0 P P P 2I P 0

P 0 and P P 2I P 0 P P P 2I

− −

⇒ + + + = ⇒ + + + =

≠ + + + = ⇒ = − + +

15.4−

7.8−ref

I

V

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6. Rank is 4, so Q will have four linearly independent columns and four linearly

independent rows.

7. ( ) ( )SY s y s 1+ =

( )

( )t

1Y s

s 1

e u t−

=+

8. Up to 5.7V, the diode is in reverse biased so, V0=Vi and diode is forward biased

after Vi > 5.7 and reaches peak value and comes to again 5.7V, in this output voltage is 5.7V, after that again diode is in reverse bias so V0=Vi

11. Theory bit

12.

13. It is a predefined statement

15. 200km=l

0.00127radias per km

2 24947.39km

0.00127

3006.06%

4947.39

β =π π

λ = = =β

= =λl

16. PZ Zs Zm 48 Zs zm= − = −

oZ Zs 2Zm 15 Zs 2zm

Zs 26

Zm 11

= + = +

= Ω= Ω

17. 12X 50Tan Tan 45º

R 50

− − φ = = =

0Below 45º of firing angle the out put voltage V is not controlable∴

18. Line voltage is free from triplex harmonics

A

BC

30º

YDI

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19. Linear system holds superposition theorem

Causal system means h(t) = 0 for t < 0

20. In O.C. test pf is low, so we use low pf meters and In S.C. test pf is high, so we

use high pf meters.

S.C.Testtest

2kvA8.69A. so weuse 10Ameter

230v

=

21. 2 1 2 6 2

R 6,C , RC 42 1 3 3

× ×= = = τ = = =

+

22. ( ) ( )1Z 0.1s 1 /s

= +

equating imaginary pout to zero

3rad / secω =

23. i 1A=

2

ab 2

ab

v 1 2 2v

v v 5 0

v 2.5 3v

= × =

− + − =

= = −

24. 12 6

0 0

3 31 2

1 2

A A 8.85 10 500 500 10C ,C 1475PF

d dd 4 10 2 10

8 2

− −

− −

∈ ∈ × × × ×= = = =

× ×+ +∈ ∈

25. ( ) ( )22 32

0 r0 r

3

300 300 10N AL 0.3m

300 10

× µ × µ × ×µ µ= ⇒ =

×Q l

l

26. 15 2i 0− + =

1

ab

ab 1

ab

ab

i 2.5

v 2.5volt

2.5 v i 0

v i 2.5

4v 3i i 0

i v

i 1 2.5

2.5i 1.25

2

=

=− + + =

+ =

− + + =

=

+ =

= =

0.15

11s

z ⇒

+

abv1A

2v+

5Vmiv

+−

1i

1

5v 1 v

+

+abv −

i1Ω

+− abv∆

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28. Hint: (i) Apply shifting, (ii) scaling and (iii) Time reversal

29. Non-causal because the signal is defined for t<0 also time invariant since y(n-k)

= [x(n-K)] where ( ) ( )x n y n→

30. sinc t convolution in time dmin is ( ) in frequencyα

( )sinc Bt so it will be max ,Bα

32. ( ) ( ) ( ) ( )n1 j y z x z H z+ =

( ) ( )

( )

( ) ( ) ( ) ( )( )

z 11 j

n

1

1

x z y n

z

z 1 j

zO is zero for x n 1 j and the z. Transform of x nP z 1 j

1The ROC of 1 z H z

2

implies it has 1 pole and 1 zero

> +

=

− +

= + =− +

Q

33. ( ) ( ) ( )( )nnz aP P

n 1z a

1 dRe sides of z . z z z where

n 1 ! dz

−−

=× = ×

( ) ( )

( )( ) ( ) ( )( )( )

( )( )

( )

n

n 1 n 1Re sides

2n 1

2

n n 1 n 1z a

z a

z a at z a n order of pole

given problem1 d

n 2 z z z z / z a2 1 ! dz

1 d zz 2 a

1! dz z a

dz nz na

dz

− −

− −= =

− = =

= × = × =−

= −−

= = =

34. ( ) ( )22f x x 4= −

( ) ( )( ) ( )( )

1 2

1 2

11 2

11

11

11 111

111

f x 2 x 4 .2x

f x 0 4x x 4 0, x 0,2, 2

f x 12x 16

x 0 f 16 max

x z f 64 max

x z f 0 f 24x

x z f 24 2 48 max

= −

= ⇒ − = ⇒ = −

= −

= = − →

= − = − →

= = =

= = × = →

( )H z

1

/ 2α/ 2−α

B / 2B / 2−

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35. ( )( )

( ) nn x

n 1 n n1n

f xx x f x e 1

f x+ = − = −

( )( )

( ) n

10 x1

1 0 n1 10

f x ex x f x e 1 0.71828

f x e

−= − = = − − =

36. ( ) ( ) ( )1 1T T 1 T T 1A A A A A A A A

− −+ − −= = =

1

A by checking options

AA A AA A A AA A A+ − + += = ∴ ≠

37. 0 2

hx y 2y

2= +

39. ( ) ( ) istF s f t e dtα

−α

= ∫

( )1

ist

1

1 e dt

2 sincs

2 sinc2

=

=

ω = π

40. The both transfer for a cont mirror

0 1 ref

1

S I I

0 0.7 0.7 5I

1

3.6mA

=

− − +=

=

41. ( ) ( )1 1v 0 v 0t 0v− = =

1c c

at the half cycle after t 0

at v half cycle v 5v

=− =

( )x t 1

0 1 t

1− 1 t

( )f t

( ) ( ) ( )f t x t x t= + −

1kΩ

1Iref

5

I

+

5v−

5vK ~kvL

+ −c1v

5v

+

+ 2 2c cv v 10v=5−

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42. Combine two waveforms

44. Here monoshot o/p’s Q1 and Q2 has pulse widths of 1 2ON 0NT and T .

So, when triggered according to truth table given Q1 responds to 20NT and Q2 for

1ONT and waveform will be

2

1 2 1 2

0N

0N 0N 0N 0N

T1f ,Duty cycle

T T T T

= = + +

45. DAD SP

PCHL

After first instruction HL 2700H

PCHL HL contes exchanged with PC

so, PC 2700 Sp 2700

= =

46. The first op-amp is a square wave generator and now after adjusting voltage at +ve terminal to 2.5v the second op-amp an follows

( )0i

i o

Now

d 2.5 vv 2.5e

R at

1v 2.5 2.5 V

RC

so triangle wave shifts upward

−−=

− = −∫

47. 1500 1425 7.8 7.8

S 0.05, effective rotor resis tance1500 2 0.05 2 S

− = = = − − Q

48. Hint: core losses 1200w are also considered

invQ

R

R

−+ 2Ov ( )... 2

( )... 1

inv

−+ 1Ov

( ) ( )1 2 So A is Answer+

1ONT

1ONT

iv

R

C

2.5v

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49. Hint: rs 2

dE N ;

dt

θ= −

t 0 to 1 the o /p voltage is ve and constant value

t 1 to 2 the o /p voltage is zero

t 2 to 2.5 the o /p voltage is ve and constant

= −== +

50. Direct

51.

( )

2

1 2

1 21 122

2 1

12

2

v

L L 1 0.3 L

C CSIL SILC0.7, SIL

LSIL v L 0.7CCL

C

= = = = =

52.

1 2

1 2

G G

dC dC

dP dP=

( )( )

( ) ( )

1 2

1 2

G G

G G

1 0.11 P 30.06 P ...... 1

P P 250 ...... 2

solve 1 & 2

+ =

+ =

54. ( ) ( ) ( )2 2 2

4Remove e then loss 5 z 3 z 2 z 25 9 4 36z= + + = + + =

( ) ( )2 223e then loss z 111 z 7 z 2 z 1 49 4 54z+ + = + + =

So, if we remove C3 then system will operate with minimum loss.

55. 2 2

Hint : Pf = ωεαπ

57. ( )1bE 220 2.5 20 170= − × =

1

2

2

b

b

b

E 1000

E 600

600E 170 102

1000

V 102 50 152

152duty cycle 0.608

250

=

= × =

= + =

= =

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60. st

2max

T 2a

T 1 a=

+

2

2

2

2 a1 a 1

1 a

R 10.666

x 1.5

V 0.66 400 266.6V

frequency sf 0.666 50 33.3Hz

×= ⇔ =

+

= =

= × == = × =

61. m2Vcos 130 129ºα = − ⇔ α =

π

62. ( )2

WD

2 10 3T 9 0.31 31% ,Rms value 7.8

9 2

π × ×= − = = = =

× π

63. ( )0 s d d

1 20V V 40V, since poweris constant 20 4 40 I , I 2A

1 0.5 0.5= × = = × = × =

64. ( )( ) ( ) ( )C S

G S R S 1R S

= =

( ) ( )x s 0Lt C t Lt SC t→∞ →

=

65. ( )

( ) ( )1 2

10 S 1 S 10C C

S 10 10 S 1

+ += =

+ +

66. At starting the stop is -40dB/decode ∴ At starting we have 2 poles and in next

portion the stop is changes from -40dB/decode to -20dB/decode (i.e. we have

one zero and next portion the stop is changes from (-20dB/decode to 0 dB/decode)(i.e. one zero)

∴ Total we have 2 poles and 2 zeros.

10− 1−10− 1−

pole zero

lead compensator lag compensator

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67. ( )( )

( ) ( )

( ) ( )

( ) ( )2

3 2

KK

C S S S 3 S 10 S 3S S 10 K

KR S K1S S 3 S 10

S 13S 30S K

=+ + + + +

=+ =+ + + + +

3 2C.E S 13S 30S K 0

1 K 30 13 390

0 K 390

+ + + =× = × =< <

68. ( )( )

2n

2 2n n

C S

R S S 2 s

ω=

+ ξω + ω

n n10 2 20

1 so critical damped

ω = ξω =

ξ =

69. fy No. of horizontol tangencies

, fx , fyfx No. of vertical tangencies 2

ω= = ω =

70. ( ) BC ADz 500 Z Z= = ×

76. Area upto 5sec i.e. ( )1 2 3

1 1 dqA A A 2 4 3 2 3 2 13GC, i q idt

2 2 dt

+ + = × × + × × + × = = =

∫Q

78. ( ) 1T.F C SI A B D

−= − +

79. ss

1e

K=

80. 0

i

A

V SCHint : put SC J

1VSC

R

= = ω+

( )

0

2i

2 22

AA

0 0

i i

V C 1

V 11 1

C CRR

V V0 0, 1 so high pass filter.

V V

ω= =

++ ω ω

ω → = ω → ∞ =

81. From the given circuit

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82. sh aI I I= +

a

a

b

b

24015 I

80

I 12A

At plugging the net voltage across armature resistance

is =V+E

E 2400 12 0.5 234 240 234 474v

= +

=

= − × = = + =

83. 2 a

234Ia 1.25 I 1.25 12 15A,R 0.5 15.1

15= × = × = = − = Ω

84. a sV E I x= +

a

s

f

v 10

I 0.6 0

x 190

10 E 0.6 0190

E 1 j0.6 1.17 30.96

1.17, 30.96 lag

==

=

= +

= − = −

=

85. a 1I 1.2Ia 1.2 0.6 0.72= = × =

( )( )

2 2 22 s f t f t

2 2 2

f tt 2

s

f

f

Ia x E U 2E V cos

0.72 1 1.17 1 2 1.17 1 cos

cos 0.79 37.73

sin 0.6120

E vsin v Ia cos cos 0.994

x

E cos 1.17x0.79 0.924

E cos V,hence PF is 0.994lagging

= + − δ

× = + − × × × δ

δ = ⇒ δ =δ =

δ = θ ⇒ θ =

δ = =

δ <