Gate ee 2008 with solutions

GATE EE2008

Brought to you by: Nodia and Company Visit us at: www.nodia.co.in

PUBLISHING FOR GATE

Q.1 - Q.20 carry one mark each.

MCQ 1.1 The number of chords in the graph of the given circuit will be

(A) 3 (B) 4

(C) 5 (D) 6

SOL 1.1 No. of chords is given as l 1b n= − + b " no. of branches n " no. of nodes l " no. of chords

6b = , n 4= l 6 4 1= − + 3=Hence (A) is correct option.

MCQ 1.2 The Thevenin’s equivalent of a circuit operation at 5ω = rads/s, has 3.71 15.9V Voc += − % and 2.38 0.667Z j0 Ω= − . At this frequency, the minimal

realization of the Thevenin’s impedance will have a(A) resistor and a capacitor and an inductor

(B) resistor and a capacitor

(C) resistor and an inductor

(D) capacitor and an inductor

SOL 1.2 Hence (A) is correct option.Impedance Zo 2.38 0.667j Ω= −

Page 2 GATE EE 2008 www.gatehelp.com


PUBLISHING FOR GATE

Constant term in impedance indicates that there is a resistance in the circuit.Assume that only a resistance and capacitor are in the circuit, phase difference in thevenin voltage is given as θ ( )tan CR1 ω=− − (Due to capacitor)

Zo R Cj

ω= −

So, C1

ω 0.667=

and R 2.38 Ω=

θ ..tan 0 667

1 2 381 #=− −b l

74.34 15.9c c=− =−[

given Voc 3.71 15.9c+= −So, there is an inductor also connected in the circuit

MCQ 1.3 A signal ( )sine tt ωα- is the input to a real Linear Time Invariant system. Given K and φ are constants, the output of the system will be of the form ( )sinKe vtt φ+β- where(A) β need not be equal to α but v equal to ω

(B) v need not be equal to ω but β equal to α

(C) β equal to α and v equal to ω

(D) β need not be equal to α and v need not be equal to ω

SOL 1.3 Hence (D) is correct option.Let ( ) ( )x t X sL

( ) ( )y t Y sL

( ) ( )h t H sL

So output of the system is given as ( )Y s ( ) ( )X s H s=Now for input ( )x t τ− ( )e X s (shifting property)sL τ-

( )h t τ− ( )e H ssL τ−

So now output is ' ( )Y s ( ) ( )e X s e H ss s$= τ τ- -

' ( )Y s ( ) ( )e X s H ss2= τ-

' ( )Y s ( )e Y ss2= τ-

Or ' ( )y t ( 2 )y t τ= −

MCQ 1.4 X is a uniformly distributed random variable that takes values between 0 and 1. The value of { }E X3 will be(A) 0 (B) 1/8

(C) 1/4 (D) 1/2



PUBLISHING FOR GATE

SOL 1.4 X is uniformly distributed between 0 and 1So probability density function

( )f XX 1, 0 1

0,

x

otherwise

< <= )

So,

{ }E X3 ( )X f X dxX3

0

1= #

( )X dx13

0

1= #

X4

4

0

1

= : D

41=

Hence (C) is correct option

MCQ 1.5 The characteristic equation of a (3 3# ) matrix P is defined as ( )a I P 2 1 03 2λ λ λ λ λ= − = + + + =If I denotes identity matrix, then the inverse of matrix P will be(A) ( )P P I22 + + (B) ( )P P I2 + +

(C) ( )P P I2− + + (D) ( )P P I22− + +

SOL 1.5 According to CAYLEY-HAMILTON Theorem every non-singular square matrix satisfies its own characteristic equation.Characteristic equation ( )a I Pλ λ= − 2 1 03 2λ λ λ= + + + =Matrix P satisfies above equation P P P I23 2+ + + 0= I ( )P P P23 2=− + +

Multiply both sides by P 1−

P 1− ( 2 )P P I2=− + +Hence (D) is correct option.

MCQ 1.6 If the rank of a ( )5 6# matrix Q is 4, then which one of the following statement is correct ?(A) Q will have four linearly independent rows and four linearly independent

columns

(B) Q will have four linearly independent rows and five linearly independent columns

(C) QQT will be invertible

(D) Q QT will be invertible

SOL 1.6 Rank of a matrix is no. of linearly independent rows and columns of the matrix.Here Rank ( )Q 4ρ =



PUBLISHING FOR GATE

So Q will have 4 linearly independent rows and flour independent columns.Hence (A) is correct option

MCQ 1.7 A function ( )y t satisfies the following differential equation :

( )

( )dt

dy ty t+ ( )tδ=

where ( )tδ is the delta function. Assuming zero initial condition, and denoting the unit step function by ( ), ( )u t y t can be of the form(A) et (B) e t-

(C) ( )e u tt (D) ( )e u tt-

SOL 1.7 Given differential equation for the function

( )

( )dtdy t

y t+ ( )tδ=

Taking Laplace on both the sides we have, ( ) ( )sY s Y s+ 1= ( 1) ( )s Y s+ 1=

( )Y s s 11= +

Taking inverse Laplace of ( )Y s ( )y t ( )e u tt= − , t 0>Hence (D) is correct option.

MCQ 1.8 The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure.

(I)

(II)

If such a diode is used in clipper circuit of figure given above, the output voltage V0 of the circuit will be



PUBLISHING FOR GATE

SOL 1.8 Assume the diode is in reverse bias so equivalent circuit is

Output voltage V0 10sin t10 1010 #ω=

+ 5 sin tω=

Due to resistor divider, voltage across diode 0V <D (always). So it in reverse bias for given input.Output, V0 5 sin tω=Hence (A) is correct option.

MCQ 1.9 Two 8-bit ADCs, one of single slope integrating type and other of successive approximate type, take TA and TB times to convert 5 V analog input signal to equivalent digital output. If the input analog signal is reduced to 2.5 V, the approximate time taken by the two ADCs will respectively, be(A) ,T TA B (B) /2,T TA B

(C) , /T T 2A B (D) /2, /2T TA B

SOL 1.9 Conversion time does not depend on input voltage so it remains same for both type of ADCs.Hence (A) is correct option



PUBLISHING FOR GATE

MCQ 1.10 An input device is interfaced with Intel 8085A microprocessor as memory mapped I/O. The address of the device is 2500H. In order to input data from the device to accumulator, the sequence of instructions will be(A) LXI H, 2500H (B) LXI H, 2500H

MOV A, M MOV M, A

(C) LHLD 2500H (D) LHLD 2500H

MOV A, M MOV M, A

SOL 1.10 Hence ( ) is Correct Option

MCQ 1.11 Distributed winding and short chording employed in AC machines will result in(A) increase in emf and reduction in harmonics

(B) reduction in emf and increase in harmonics

(C) increase in both emf and harmonics

(D) reduction in both emf and harmonics

SOL 1.11 Distributed winding and short chording employed in AC machine will result in reduction of emf and harmonics.Hence (D) is correct option.

MCQ 1.12 Three single-phase transformer are connected to form a 3-phase transformer bank. The transformers are connected in the following manner :

The transformer connecting will be represented by(A) Y d0 (B) Y d1

(C) Y d6 (D) Y d11

SOL 1.12 Transformer connection will be represented by Y d1.Hence (B) is correct option.

MCQ 1.13 In a stepper motor, the detent torque means(A) minimum of the static torque with the phase winding excited

(B) maximum of the static torque with the phase winding excited

(C) minimum of the static torque with the phase winding unexcited

(D) maximum of the static torque with the phase winding unexcited

SOL 1.13 Detent torque/Restraining toque:The residual magnetism in the permanent magnetic material produced.



PUBLISHING FOR GATE

The detent torque is defined as the maximum load torque that can be applied to the shaft of an unexcited motor without causing continuous rotation. In case the motor is unexcited.Hence (D) is correct option.

MCQ 1.14 A two machine power system is shown below. The Transmission line XY has positive sequence impedance of Z1 Ω and zero sequence impedance of Z0 Ω

An ‘a’ phase to ground fault with zero fault impedance occurs at the centre of the transmission line. Bus voltage at X and line current from X to F for the phase ‘a’, are given by Va Volts and Ia amperes, respectively. Then, the impedance measured by the ground distance relay located at the terminal X of line XY will be given by(A) /2Z1 Ω^ h (B) /2Z0 Ω^ h

(C) ( )/2Z Z0 1 Ω+ (D) /V Ia a Ω^ h

SOL 1.14 Given for X to F section of phase ‘a’Va -Phase voltage and Ia -phase current.Impedance measured by ground distance,

Relay at X Current from phase 'a'Bus voltage=

IV

a

a Ω=

Hence (D) is correct option.

MCQ 1.15 An extra high voltage transmission line of length 300 km can be approximate by a lossless line having propagation constant .0 00127β = radians per km. Then the percentage ratio of line length to wavelength will be given by(A) 24.24 % (B) 12.12 %

(C) 19.05 % (D) 6.06 %

SOL 1.15 For EHV line given data isLength of line 300= km and .0 00127β = S rad/km

wavelength λ .2 .2

0 00127 4947 39βπ π= = = km

So %lλ . .4947 39

300 100 0 06063 100# #= =

%lλ 6.063=


MCQ 1.16 A-3-phase transmission line is shown in figure :



PUBLISHING FOR GATE

Voltage drop across the transmission line is given by the following equation :

VVV

ZZZ

ZZZ

ZZZ

III

a

b

c

s

m

m

m

s

m

m

m

s

a

b

c

3

3

3

=

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

Shunt capacitance of the line can be neglect. If the has positive sequence impedance of 15 Ω and zero sequence impedance of 48 Ω, then the values of Zs and Zm will be(A) 31.5 ; 16.5Z Zs mΩ Ω= =

(B) 26 ; 11Z Zs mΩ Ω= =

(C) 16.5 ; 31.5Z Zs mΩ Ω= =

(D) 11 ; 26Z Zs mΩ Ω= =

SOL 1.16 For three phase transmission line by solving the given equation

We get, VVV

a

b

c

ΔΔΔ

R

T

SSSS

V

X

WWWW

( )

( )( )

X XX X

X X

III

00

0

0

002

s m

s m

s m

a

b

c

=−

−+

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Zero sequence Impedance 2 48X Xs m= + = ...(1) and Positive Sequence Impedance Negative Sequence Impedance= ( )X Xs m= − 15= ...(2)By solving equation (1) and (2) Z Xors s 26= and 11Z Xorm m =Hence (B) is correct option.

MCQ 1.17 In the single phase voltage controller circuit shown in the figure, for what range of triggering angle ( )α , the input voltage ( )V0 is not controllable ?

(A) 0 45< <c cα (B) 45 135< <c cα

(C) 90 180< <c cα (D) 135 180< <c cα



PUBLISHING FOR GATE

SOL 1.17 Hence (A) is correct option.

R jXL+ j50 50= +

tanφ RL

5050 1ω= = =

φ 45c=so, firing angle ‘α ’ must be higher the 45c, Thus for 0 45< < cα , V0 is uncontrollable.

MCQ 1.18 A 3-phase voltage source inverter is operated in 180c conduction mode. Which one of the following statements is true ?(A) Both pole-voltage and line-voltage will have 3rd harmonic components

(B) Pole-voltage will have 3rd harmonic component but line-voltage will be free from 3rd harmonic

(C) Line-voltage will have 3rd harmonic component but pole-voltage will be free from 3rd harmonic

(D) Both pole-voltage and line-voltage will be free from 3rd harmonic components

SOL 1.18 A 3-φ voltage source inverter is operated in 180c mode in that case third harmonics are absent in pole voltage and line voltage due to the factor ( / )cos n 6π . so both are free from 3rd harmonic components.Hence (D) is correct option.

MCQ 1.19 The impulse response of a causal linear time-invariant system is given as ( )h t . Now consider the following two statements :Statement (I): Principle of superposition holdsStatement (II): ( )h t 0= for t 0<Which one of the following statements is correct ?(A) Statements (I) is correct and statement (II) is wrong

(B) Statements (II) is correct and statement (I) is wrong

(C) Both Statement (I) and Statement (II) are wrong

(D) Both Statement (I) and Statement (II) are correct

SOL 1.19 Since the given system is LTI, So principal of Superposition holds due to linearity.For causal system ( ) 0h t = , 0t <Both statement are correct.Hence (D) is correct option.

MCQ 1.20 It is desired to measure parameters of 230 V/115 V, 2 kVA,



PUBLISHING FOR GATE

single-phase transformer. The following wattmeters are available in a laboratory:W1 : 250 V, 10 A, Low Power FactorW2 : 250 V, 5 A, Low Power FactorW3 : 150 V, 10 A, High Power FactorW4 : 150 V, 5 A, High Power FactorThe Wattmeters used in open circuit test and short circuit test of the transformer will respectively be(A) W1 and W2 (B) W2 and W4

(C) W1 and W4 (D) W2 and W3

SOL 1.20 Given: 1-φ transformer, 230 V/115 V, 2 kVA W1 : 250 V, 10 A, Low Power Factor W2 : 250 V, 5 A, Low Power Factor W3 : 150 V, 10 A, High Power Factor W4 : 150 V, 5 A, High Power FactorIn one circuit test the wattmeter W2 is used and in short circuit test of transformer W3 is used.Hence (D) is correct option.

Q.21 to Q.75 carry two marks each.

MCQ 1.21 The time constant for the given circuit will be

(A) 1/9 s (B) 1/4 s

(C) 4 s (D) 9 s

SOL 1.21 Time constant of the circuit can be calculated by simplifying the circuit as follows



PUBLISHING FOR GATE

Ceq 32= F

Equivalent Resistance

Req 3 3 6 Ω= + =Time constant τ R Ceq eq=

6 32

#= 4= sec


MCQ 1.22 The resonant frequency for the given circuit will be

(A) 1 rad/s (B) 2 rad/s

(C) 3 rad/s (D) 4 rad/s

SOL 1.22 Impedance of the circuit is

Z j LR

R

j C

j C1

1

ω= ++ω

ω

j L j CRR

j CRj CR

1 11

#ω ω ωω= + + −

−

( )

j LC R

R j CR1

12 2 2ω

ωω= +

+−

( )

C Rj L C R R j CR

11

2 2 2

2 2 2 2

ωω ω ω=

++ + −

[ ( ) ]

C RR

C Rj L C R CR

1 11

2 2 2 2 2 2

2 2 2 2

ω ωω ω ω=

++

++ −

For resonance ( ) 0Im Z =So, (1 )L C R2 2 2ω ω+ CR2ω=



PUBLISHING FOR GATE

0.1L = H, 1C = F, 1R Ω= So, 0.1[1 (1)(1)]2

#ω ω+ (1)(1)2ω= 1 2ω+ 10=& ω 39= = rad/secHence (C) is correct option.

MCQ 1.23 Assuming ideal elements in the circuit shown below, the voltage Vab will be

(A) 3 V− (B) 0 V

(C) 3 V (D) 5 V

SOL 1.23 By applying KVL in the circuit 2 5V iab − + 0=

1i = A, Vab 2 1 5#= − 3=− VoltHence (A) is correct option

MCQ 1.24 A capacitor consists of two metal plates each 500 500# mm2 and spaced 6 mm apart. The space between the metal plates is filled with a glass plate of 4 mm thickness and a layer of paper of 2 mm thickness. The relative primitivities of the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the capacitance will be (Given that .8 85 100

12#ε = - F/m )(A) 983.3 pF (B) 1475 pF

(C) 637.7 pF (D) 9956.25 pF

SOL 1.24 Here two capacitance C1 and C2 are connected in series, so equivalent capacitance is

Ceq C CC C1 2

1 2= +

C1 dAr

1

0 1ε ε=

.4 10

8 85 10 8 500 500 103

12 6

#

# # # # #= −

− −

.442 5 10 11#= − F



PUBLISHING FOR GATE

C2 dAr

2

0 2ε ε=

.2 10

8 85 10 2 500 500 103

12 6

#

# # # # #= −

− −

.221 25 10 11#= − F

Ceq . .. .

442 5 10 221 25 10442 5 10 221 25 10

11 11

11 11

# #

# # #=+− −

− −

.147 6 10 11#= −

1476- pFHence (B) is correct option.

MCQ 1.25 A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross-sectional area of 300 mm2. The inductance of the coil corresponding to a magnetizing current of 3 A will be(Given that 4 100

7#μ π= - H/m)(A) 37.68 Hμ (B) 113.04 Hμ

(C) 3.768 Hμ (D) 1.1304 Hμ

SOL 1.25 Hence (B) is correct option.Circumference l 300= mm no. of turns n 300= Cross sectional area A 300= mm2

Inductance of coil L ln A0

2μ=

( )

4 10 (300) 300 10300 10 3

7 2 6

#

# # # #π= −

− −

.113 04= Hμ

MCQ 1.26 In the circuit shown in the figure, the value of the current i will be given by

(A) 0.31 A (B) 1.25 A

(C) 1.75 A (D) 2.5 A

SOL 1.26 By writing node equations at node A and B

V V1

51

0a a− + − 0=

2 5Va − 0=



PUBLISHING FOR GATE

& Va 2.5= VSimilarly

V V V34

10b ab b− ++ − 0=

( )V V V

V34b a b

b− − + 0=

4(2.5 ) 3V V Vb b b− − + 0= 8 10Vb − 0=& Vb 1.25= V

Current i 1.25V1b= = A

Hence (B) is correct option.

MCQ 1.27 Two point charges 10Q C1 μ= and 20Q2 = mC are placed at coordinates (1,1,0) and ( , , )1 1 0− − respectively. The total electric flux passing through a plane 0z 2= will be(A) 7.5 Cμ (B) 13.5 Cμ

(C) 15.0 Cμ (D) 22.5 Cμ


MCQ 1.28 Given a sequence [ ]x n , to generate the sequence [ ] [ ]y n x n3 4= − , which one of the following procedures would be correct ?(A) First delay ( )x n by 3 samples to generate [ ]z n1 , then pick every 4th sample of

[ ]z n1 to generate [ ]z n2 , and than finally time reverse [ ]z n2 to obtain [ ]y n .

(B) First advance [ ]x n by 3 samples to generate [ ]z n1 , then pick every 4th sample of [ ]z n1 to generate [ ]z n2 , and then finally time reverse [ ]z n2 to obtain [ ]y n

(C) First pick every fourth sample of [ ]x n to generate [ ]v n1 , time-reverse [ ]v n1 to obtain [ ]v n2 , and finally advance [ ]v n2 by 3 samples to obtain [ ]y n

(D) First pick every fourth sample of [ ]x n to generate [ ]v n1 , time-reverse [ ]v n1 to obtain [ ]v n2 , and finally delay [ ]v n2 by 3 samples to obtain [ ]y n

SOL 1.28 In option (A) [ ]z n1 [ ]x n 3= − [ ]z n2 [4 ] [4 3]z n x n1= = − [ ]y n [ ] [ 4 3] [3 4 ]z n x n x n2= − = − − = −Y

In option (B) [ ]z n1 [ ]x n 3= + [ ]z n2 [4 ] [4 3]z n x n1= = + [ ]y n [ ] [ 4 3]z n x n2= − = − +

In option (C) [ ]v n1 [ ]x n4=



PUBLISHING FOR GATE

[ ]v n2 [ ] [ 4 ]v n x n1= − = − [ ]y n [ 3] [ 4( 3)] [3 4 ]v n x n x n2= + = − + = −Y

In option (D) [ ]v n1 [ ]x n4= [ ]v n2 [ ] [ 4 ]v n x n1= − = − [ ]y n [ 3] [ 4( 3)] [3 4 ]v n x n x n2= − = − − = −Y


MCQ 1.29 A system with ( )x t and output ( )y t is defined by the input-output relation :

( ) ( )y t x t dt2

τ=3-

-#

The system will be(A) Casual, time-invariant and unstable

(B) Casual, time-invariant and stable

(C) non-casual, time-invariant and unstable

(D) non-casual, time-variant and unstable

SOL 1.29 Input-output relation

( )y t ( )x dt2τ τ=

3-

-#

Causality :Since ( )y t depends on ( )x t2− , So it is non-causal.Time-variance :

( )y t ( ) ( )x d y tt

02

0τ τ τ τ= − = −3-

-Y#

So this is time-variant.Stability :Output ( )y t is unbounded for an bounded input.For exampleLet ( )x τ ( )boundede= τ-

( )y t Unboundede d e1

2 2t t$τ= =

−3 3

ττ

-

-

- -

-

-

8 B#


MCQ 1.30 A signal ( ) ( )x t tsinc α= where α is a real constant ( )xsinc ( )sinx

x= ππ

^ h is the input to a Linear Time Invariant system whose impulse response ( ) ( )h t tsinc β= , where β is a real constant. If min ( , )α β denotes the minimum of α and β and similarly, max ( , )α β denotes the maximum of α and β, and K is a constant, which one of the following statements is true about the output of the system ?(A) It will be of the form ( )tsincK γ where ( , )minγ α β=

(B) It will be of the form ( )tsincK γ where ( , )maxγ α β=

(C) It will be of the form ( )tsincK α

(D) It can not be a sinc type of signal



PUBLISHING FOR GATE

SOL 1.30 Output ( )y t of the given system is ( )y t ( ) ( )x t h t)=Or ( )Y jω ( ) ( )X j H jω ω=given that ( )x t ( )tsinc α= ( )h t ( )tsinc β=Fourier transform of ( )x t and ( )h t are

( )X jω [ ( )] ,x t rect2

< <Fαπ

αω α ω α= = −` j

( )H jω [ ( )] ,h t rect2

< <Fβπ

βω β ω β= = −` j

( )Y jω rect rect2 2

2

αβπ

αω

βω= ` `j j

So, ( )Y jω rect2

Kγ

ω= ` j

Where γ ( , )min α β=And ( )y t ( )tK sinc γ=Hence (A) is correct option.

MCQ 1.31 Let ( )x t be a periodic signal with time period T , Let ( ) ( ) ( )y t x t t x t t0 0= − + + for some t0. The Fourier Series coefficients of ( )y t are denoted by bk . If b 0k = for all odd k , then t0 can be equal to(A) /T 8 (B) /T 4

(C) /T 2 (D) T2

SOL 1.31 Let ak is the Fourier series coefficient of signal ( )x tGiven ( )y t ( ) ( )x t t x t t0 0= − + +Fourier series coefficient of ( )y t bk e a e ajk t

kjk t

k0 0= +ω ω-

bk 2 cosa k tk 0ω= bk 0= (for all odd k )

k t0ω 2π= , k odd"

kT

t20

π 2π=



PUBLISHING FOR GATE

For k 1= , t0 T4

=


MCQ 1.32 ( )H z is a transfer function of a real system. When a signal [ ] (1 )x n j n= + is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of z1 2

1 1− -^ h H(z) is the entire Z-plane (except z 0= ). It can then be

inferred that ( )H z can have a minimum of(A) one pole and one zero (B) one pole and two zeros

(C) two poles and one zero (D) two poles and two zeros

SOL 1.32 Hence ( ) is correct option.

MCQ 1.33 Given ( )( )

X zz a

z2=

− with z a> , the residue of ( )X z zn 1- at z a=

for n 0$ will be(A) an 1- (B) an

(C) nan (D) nan 1-

SOL 1.33 Hence (D) is correct option.

Given that ( )X z ( )z a

z2=

−, z a>

Residue of ( )X z zn 1- at z a= is

( ) ( )dzd z a X z zn

z a2 1= − -

=

( )( )dz

d z az a

z zn

z a

22

1= −−

-

=

dzd zn

z a=

=

nznz a

1= -= nan 1= -

MCQ 1.34 Consider function ( ) ( 4)f x x2 2= − where x is a real number. Then the function has(A) only one minimum (B) only tow minima

(C) three minima (D) three maxima

SOL 1.34 Given function ( )f x ( )x 42 2= − ' ( )f x ( )x x2 4 22= −To obtain minima and maxima '( )f x 0= ( )x x4 42 − 0= ,x x0 4 02= − = x 2& !=So, x 0, 2, 2= + − '' ( )f x ( ) ( )x x x4 2 4 42= + − x12 162= −



PUBLISHING FOR GATE

For , '' ( ) ( )x f0 0 12 0 16 16 0<2= = − =− (Maxima) , '' ( ) ( )x f2 2 12 2 16 32 0>2=+ = − = (Minima) , '' ( ) ( )x f2 2 12 2 16 32 0>2=− − = − − = (Minima)So ( )f x has only two minimaHence (B) is correct option.

MCQ 1.35 Equation e 1 0x − = is required to be solved using Newton’s method with an initial guess x 10 =− . Then, after one step of Newton’s method, estimate x1 of the solution will be given by(A) 0.71828 (B) 0.36784

(C) 0.20587 (D) 0.00000

SOL 1.35 An iterative sequence in Newton-Raphson method can obtain by following expression

xn 1+ ' ( )( )

xf xf x

nn

n= −

We have to calculate x1, so n 0=

x1 ' ( )( )

xf xf x

00

0= − , Given x 10 =−

( )f x0 1e e1x 10= − = −−

.0 63212=− ' ( )f x0 e ex 10= = −

.0 36787=

So, x1 ( . )( . )

10 367870 63212=− − −

.1 1 71832=− + .0 71832=Hence (A) is correct option

MCQ 1.36 A is m n# full rank matrix with m n> and I is identity matrix. Let matrix ' ( )A A A A1T T= - , Then, which one of the following statement is FALSE ?

(A) 'AA A A= (B) ( ')AA 2

(C) 'A A I= (D) ' 'AA A A=

SOL 1.36 Hence (D) is correct option 'A ( )A A AT T1= −

( )A A AT T1 1= − −

A I1= −

Put 'A A I1= − in all option.

option (A) 'AA A A= AA A1− A=



PUBLISHING FOR GATE

A A= (true)

option (B) ( ')AA 2 I= ( )AA I1 2− I= ( )I 2 I= (true)

option (C) 'A A I= A IA1− I= I I= (true)

option (D) 'AA A 'A= AA IA1− 'A A= =Y (false)

MCQ 1.37 A differential equation / ( )dx dt e u tt2= - , has to be solved using trapezoidal rule of integration with a step size .h 0 01= s. Function ( )u t indicates a unit step function. If ( )x 0 0=- , then value of x at .t 0 01= s will be given by(A) 0.00099 (B) 0.00495

(C) 0.0099 (D) 0.0198

SOL 1.37 Hence (C) is correct option

dtdx ( )e u tt2= −

x ( )e u t dtt2= −#

x e dtt2

0

1

= −#

x ( )f t dt0

1

= # ,

t .01= s

From trapezoid rule

( )f t dtt

t nh

0

0+# ( ) (. )h f f2 0 01= +6 @

( )f t dt0

1# . e e2

01 .0 02= + −6 @, .h 01=

.0099=

MCQ 1.38 Let P be a 2 2# real orthogonal matrix and x is a real vector [ ]x ,x1 2T with length

( )x x x /12

22 1 2= + . Then, which one of the following statements is correct ?

(A) Px x# where at least one vector satisfies Px x<

(B) Px x# for all vector x

(C) Px x$ where at least one vector satisfies Px x>

(D) No relationship can be established between x and Px

SOL 1.38 P is an orthogonal matrix So PP IT =



PUBLISHING FOR GATE

Let assume P cossin

sincos

θθ

θθ=

−> H

PX cossin

sincos x x T

1 2

θθ

θθ=

−> 8H B

cossin

sincos

xx

1

2

θθ

θθ=

−> >H H

cos sinsin cos

x xx x

1 2

1 2

θ θθ θ=

−+> H

PX ( ) ( )cos sin sin cosx x x x1 22

1 22θ θ θ θ= − + +

x x12

22= +

PX X=Hence (B) is correct option.

MCQ 1.39 Let ( )x t trect 21= −^ h (where ( ) 1rect x = for x2

121# #− and zero otherwise. If

( )xsinc ( )sinx

x= ππ , then the Fourier Transform of ( ) ( )x t x t+ − will be given by

(A) 2

sincπ

ω` j (B) 2

2sinc

πω

` j

(C) 2 cos2 2

sincπ

ω ω` `j j (D) sin

2 2sinc

πω ω

` `j j

SOL 1.39 Given signal

( )x t rect t21= −` j

So, ( )x t 1, 0 1

, elsewhere

t t21

21

21

0

or# # # #=

− −*

Similarly

( )x t− rect t21= − −` j

( )x t− 1, 1 0

, elsewhere

t t21

21

21

0

or# # # #=

− − − −*

[ ( ) ( )]x t x tF + − ( ) ( )x t e dt x t e dtj t j t= + −3

3

3

3ω ω-

-

-

-# #

( ) ( )e dt e dt1 1j t j t

0

1

1

0= +ω ω- -

-# #

j

ej

ej t j t

0

1

1

0

ω ω=

−+

−ω ω- -

-; ;E E

(1 ) ( 1)j

ej

e1 1j j

ω ω= − + −ω ω-



PUBLISHING FOR GATE

( ) ( )j

e e ej

e e e/

/ //

/ /j

j jj

j j2

2 22

2 2

ω ω= − + −

ωω ω

ωω ω

-- -

( )( )

je e e e/ / / /j j j j2 2 2 2

ω= − +ω ω ω ω- -

2sin cos22 2

$ω

ω ω= ` `j j

2 cos2 2

sincωπ

ω= ` j

Hence (C) is correct option.

MCQ 1.40 Two perfectly matched silicon transistor are connected as shown in the figure assuming the β of the transistors to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current I is

(A) 0 mA (B) 3.6 mA

(C) 4.3 mA (D) 5.7 mA

SOL 1.40 Hence (C) is correct option.

This is a current mirror circuit. Since β is high so ,I IC C1 2= I IB B1 2= VB ( . )5 0 7= − + 4.3=− voltDiode D1 is forward biased.So, current I is, I I IC C2 1= =

( . )

4.31

0 4 3= − − = mA

MCQ 1.41 In the voltage doubler circuit shown in the figure, the switch ‘S ’ is closed at t 0=



PUBLISHING FOR GATE

. Assuming diodes D1 and D2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero. The steady state voltage across capacitor C1 and C2 will be

(A) 10 , 5V VV Vc c1 2= = (B) 10 , 5V VV Vc c1 2= =−

(C) 5 , 10V VV Vc c1 2= = (D) 5 , 10V VV Vc c1 2= =−

SOL 1.41 In positive half cycle of input, diode D1 is in forward bias and D2 is off, the equivalent circuit is

Capacitor C1 will charge upto 5+ volt. 5VC1 =+ voltIn negative halt cycle diode D1 is off and D2 is on.

Now capacitor VC2 will charge upto 10− volt in opposite direction.Hence (D) is correct option

MCQ 1.42 The block diagrams of two of half wave rectifiers are shown in the figure. The transfer characteristics of the rectifiers are also shown within the block.

It is desired to make full wave rectifier using above two half-wave rectifiers. The resultants circuit will be



PUBLISHING FOR GATE

SOL 1.42 Let input Vin is a sine wave shown below

According to given transfer characteristics of rectifiers output of rectifier P is.



PUBLISHING FOR GATE

Similarly output of rectifier Q is

Output of a full wave rectifier is given as

To get output V0

( )V V VK P Q0 = − + K − gain of op-ampSo, P should connected at inverting terminal of op-amp and Q with non-inverting terminalHence (D) is correct Option

MCQ 1.43 A 3-line to 8-line decoder, with active low outputs, is used to implement a 3-variable Boolean function as shown in the figure

The simplified form of Boolean function ( , , )F A B C implemented in ‘Product of Sum’ form will be(A) ( )( )( )X Z X Y Z Y Z+ + + +

(B) ( )( )( )X Z X Y Z Y Z+ + + +

(C) ( )( )( )( )X Y Z X Y Z X Y Z X Y Z+ + + + + + + +

(D) ( )( )( )( )X Y Z X Y Z X Y Z X Y Z+ + + + + + + +



PUBLISHING FOR GATE

SOL 1.43 In SOP form, F is written as F (1, 3, 5, 6)mΣ= X YZ X YZ XYZ XYZ= + + +Solving from K- map

F XZ YZ XYZ= + +In POS form F ( )( )( )Y Z X Z X Y Z= + + + +since all outputs are active low so each input in above expression is complemented F ( )( )( )Y Z X Z X Y Z= + + + +Hence (B) is correct option.

MCQ 1.44 The truth table of monoshot shown in the figure is given in the table below :

Two monoshots, one positive edge triggered and other negative edge triggered, are connected shown in the figure, The pulse widths of the two monoshot outputs Q1 and Q2 are TON1

and TON2 respectively.

The frequency and the duty cycle of the signal at Q1 will respectively be

(A) ,fT T

D151

ON ON1 2

=+

=

(B) ,fT T

DT T

T1ON ON ON ON

ON

1 2 1 2

2=+

=+

(C) ,fT

DT T

T1ON ON ON

ON

1 1 2

1= =+



PUBLISHING FOR GATE

(D) ,fT

DT T

T1ON ON ON

ON

2 1 2

1= =+

SOL 1.44 In this case

f T T1

ON ON1 2

= +

and, D T TT

ON ON

ON

1 2

2= +


MCQ 1.45 The content of some of the memory location in an 8085 accumulator based system are given below

Address Content

g g

26FE 00

26FF 01

2700 02

2701 03

2702 04

g g

The content of stack (SP), program counter (PC) and (H,L) are 2700 H, 2100 H and 0000 H respectively. When the following sequence of instruction are executed.2100 H: DAD SP2101 H: PCHLthe content of (SP) and (PC) at the end of execution will be(A) 2102 , 2700PC H SP H= = (B) 2700 , 2700PC H SP H= =

(C) 2800 , 26PC H SP FE H= = (D) 2 02 , 2702PC A H SP H= =

SOL 1.45 Given that SP 2700 H= PC 2100 H= HL 0000 H=Executing given instruction set in following steps, DAD SP &Add register pair (SP) to HL register HL HL SP= + HL 0000 H 2700 H= + HL 2700 H= PCHL & Load program counter with HL contents PC = HL = 2700 HSo after execution contents are, PC = 2700 H, HL = 2700 H



PUBLISHING FOR GATE


MCQ 1.46 A waveform generator circuit using OPAMPs is shown in the figure. It produces a triangular wave at point ‘P’ with a peak to peak voltage of 5 V for 0V Vi = .

If the voltage Vi is made .2 5+ V, the voltage waveform at point ‘P’ will become

SOL 1.46 Hence ( ) is correct option.

MCQ 1.47 A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise (forward) direction at a speed of 1425 rpm. If the rotor resistance at standstill is 7.8 Ω, then the effective rotor resistance in the backward branch of the equivalent circuit will be(A) 2 Ω (B) 4 Ω

(C) 78 Ω (D) 156 Ω

SOL 1.47 Given: 230 V, 50 Hz, 4-Pole, 1-φ induction motor is rotating inclock-wise(forward) direction Ns 1425= rpm Rotar resistance at stand still(R2) .7 8 Ω=So



PUBLISHING FOR GATE

Ns 4120 50 1500#= =

Slip(S ) .15001500 1425 0 05= − =

Resistance in backward branch rb SR

22= −

..

2 0 057 8= −

4 Ω=Hence (B) is correct option.

MCQ 1.48 A 400 V, 50 Hz 30 hp, three-phase induction motor is drawing50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be(A) 23.06 kW (B) 24.11 kW

(C) 25.01 kW (D) 26.21 kW

SOL 1.48 Given: a 400 V, 50 Hz, 30 hp, 3-φ induction motorCurrent 50 A= at 0.8 p.f. laggingStator and rotor copper losses are 1.5 kW and 900 W fraction and windage losses 1050 W= Core losses 1200 W 1.2 kW==So, Input power in stator .3 400 50 0 8# # #= 27.71 kW= Air gap power . . .27 71 1 5 1 2= − − 25.01 kW=Hence (C) is correct option.

MCQ 1.49 The core of a two-winding transformer is subjected to a magnetic flux variation as indicated in the figure.

The induced emf ( )ers in the secondary winding as a function of time will be of the form



PUBLISHING FOR GATE

SOL 1.49 Hence (A) is correct option.

Induced emf in secondary N dtd

2φ=−

During t0 1< <−

E1 ( ) 12dtd100 Vφ=− =−

E1 and E2 are in opposition E2 2 24E V1= =

During time t1 2< <

dtdφ 0= , then E E 01 2= =

During .t2 2 5< <

E1 (100) 24dtd Vφ=− =−

Then E2 0 48 V=− −

MCQ 1.50 Voltages phasors at the two terminals of a transmission line of length 70 km have a magnitude of 1.0 per unit but are 180 degree out of phase. Assuming that the maximum load current in the line is 1/5th

of minimum 3-phase fault current. Which one of the following transmission line protection schemes will not pick up for this condition ?(A) Distance protection using ohm relay with zoen-1 set to 80% of the line

impedance.



PUBLISHING FOR GATE

(B) Directional over current protection set to pick up at 1.25 times the maximum load current

(C) Pilot relaying system with directional comparison scheme

(D) Pilot relaying system with segregated phase comparison scheme


MCQ 1.51 A loss less transmission line having Surge Impedance Loading (SIL) of 2280 MW is provided with a uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated transmission line will be(A) 1835 MW (B) 2280 MW

(C) 2725 MW (D) 3257 MW

SOL 1.51 SIL has no effect of compensationSo SIL 2280= MWHence (B) is correct option.

MCQ 1.52 A loss less power system has to serve a load of 250 MW. There are tow generation (G1 and G2) in the system with cost curves C1 and C2 respectively defined as follows ;

( ) 0.055C P P P12

G1 G1 G1#= +( ) 3 0.03C P P P2

2G2 G2 G2#= +

Where PG1 and PG2 are the MW injections from generator G1 and G2respectively. Thus, the minimum cost dispatch will be(A) 250 ; 0P PMW MWG1 G2= = (B) 150 ; 100P PMW MWG1 G2= =

(C) 100 ; 150P PMW MWG1 G2= = (D) 0 ; 250P PMW MWG1 G2= =

SOL 1.52 Hence (C) is correct optionGiven P PG1 G2+ 250= MW ...(1)

and ( )

( )

C P

C P1

2

G1

G2

.

.

P P

P P

0 055

3 0 03G1 G1

2

G2 G22

= +

= +4 ...(2)

from equation (2)

dPdC1

G1 1 0.11PG1= + ...(3a)

and dPdC2

G2 3 0.06PG2= + ...(3b)

Since the system is loss-less

Therefore dPdC1

G1 dP

dC2

G2=

So from equations (3a) and (3b)We have 0.11 0.06P P 2G1 G− 2= ...(4)Now solving equation (1) and (4), we get PG1 100= MW



PUBLISHING FOR GATE

PG2 150= MW

MCQ 1.53 A loss less single machine infinite bus power system is shown below :

The synchronous generator transfers 1.0 per unit of power to the infinite bus. The critical clearing time of circuit breaker is 0.28 s. If another identical synchronous generator is connected in parallel to the existing generator and each generator is scheduled to supply 0.5 per unit of power, then the critical clearing time of the circuit breaker will(A) reduce to 0.14 s (B) reduce but will be more than 0.14 s

(C) remain constant at 0.28 s (D) increase beyond 0.28 s

SOL 1.53 After connecting both the generators in parallel and scheduled to supply 0.5 Pu of power results the increase in the current.

Critical clearing time will reduced from 0.28 s but will not be less than 0.14 s for transient stability purpose.Hence (B) is correct option.

MCQ 1.54 Single line diagram of a 4-bus single source distribution system is shown below. Branches , ,e e e1 2 3 and e4 have equal impedances. The load current values indicated in the figure are in per unit.

Distribution company’s policy requires radial system operation with minimum loss. This can be achieved by opening of the branch(A) e1 (B) e2

(C) e3 (D) e4



PUBLISHING FOR GATE

SOL 1.54 Given that the each section has equal impedance.Let it be R or Z , then by using the formula line losses I R2= / On removing ( );e1 losses ( ) (1 2) (1 2 5)R R R1 2 2 2= + + + + + R R R R9 64 74= + + =Similarly, On removing e2;losses ( ) ( )R R R5 5 2 5 2 12 2 2= + + + + + R138= lossess on removing e3 (1) ( ) ( )R R R2 5 22 2 2= + + + R R R1 4 49= + + R54= on removing e4 lossless ( ) ( )R R R2 2 1 52 2 2= + + + R R R4 9 25= + + R38=So, minimum losses are gained by removing e4 branch.Hence (D) is correct option

MCQ 1.55 A single phase fully controlled bridge converter supplies a load drawing constant and ripple free load current, if the triggering angle is 30c, the input power factor will be(A) 0.65 (B) 0.78

(C) 0.85 (D) 0.866

SOL 1.55 Given 30cα = , in a 1-φ fully bridge converterwe know that, Power factor cosDistortion factor# α= D.f. (Distortion factor) / 0.9I Is s(fundamental)= = power factor . cos0 9 30# c= .0 78=Hence (B) is correct option

MCQ 1.56 A single-phase half controlled converter shown in the figure feeding power to highly inductive load. The converter is operating at a firing angle of 60c.

If the firing pulses are suddenly removed, the steady state voltage ( )V0 waveform of the converter will become



PUBLISHING FOR GATE

SOL 1.56 Output of this

Here the inductor makes T1 and T3 in ON because current passing through T1 and



PUBLISHING FOR GATE

T3 is more than the holding current.Hence (A) is correct option

MCQ 1.57 A 220 20 ,1000V A rpm, separately excited dc motor has an armature resistance of 2.5 Ω. The motor is controlled by a step down chopper with a frequency of 1 kHz. The input dc voltage to the chopper is 250 V. The duty cycle of the chopper for the motor to operate at a speed of 600 rpm delivering the rated torque will be(A) 0.518 (B) 0.608(C) 0.852 (D) 0.902


MCQ 1.58 A 220 ,1400 , 40V rpm A separately excited dc motor has an armature resistance of 0.4 Ω. The motor is fed from a single phase circulating current dual converter with an input ac line voltage of 220 V (rms). The approximate firing angles of the dual converter for motoring operating at 50% of rated torque and 1000 rpm will be(A) ,43 137c c (B) ,43 47c c

(C) ,39 141c c (D) ,39 51c c


MCQ 1.59 A single phase source inverter is feeding a purely inductive load as shown in the figureThe inverter is operated at 50 Hz in 180c square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i0 will be

(A) 6.37 A (B) 10 A

(C) 20 A (D) 40 A

SOL 1.59 Input is given as

Here load current does not have any dc component



PUBLISHING FOR GATE

Peak current occur at ( / )π ω

Vs Ldtdi=

200 . dtdi0 1#=

Here di 2 501

1001

ππ= =a bk l

So di( )max .200 1001

0 11

# #= 20 A=


MCQ 1.60 A 400 V, 50 Hz, 4-pole, 1400 rpm, star connected squirrel cage induction motor has the following parameters referred to the stator:

1.0 , ' 1.5R X X'r s rΩ Ω= = =

Neglect stator resistance and core and rotational losses of the motor. The motor is controlled from a 3-phase voltage source inverter with constant /V f control. The stator line-to-line voltage(rms) and frequency to obtain the maximum torque at starting will be :(A) 20.6 V, 2.7 Hz (B) 133.3 V, 16.7 Hz

(C) 266.6 V, 33.3 Hz (D) 323.3 V, 40.3 Hz

SOL 1.60 Given 400 V, 50 Hz, 4-Pole, 1400 rpm star connected squirrel cage induction motor. R 1.00 , 1.5X Xs rΩ Ω= = =l

So,for max. torque slip

Sm X X

Rsm rm

r=+ ll

for starting torque S 1m =Then X Xsm rm+ l R r= l

.f L f L2 0 2m s m rπ π+ l 1=Frequency at max. torque

fm 2 ( )L L

1s rπ

=+ l

Ls .X

2 50 2 501 5s

# #π π= =

L rl .2 50

1 5#π=

fm . .501 5

501 5

1=+

350=

fm 16.7 Hz=In const /V f control method



PUBLISHING FOR GATE

fV

1

1 50400 8= =

a fV

1

2 8=So V2 f 82 #= .16 7 8#= V2 133.3 V=Hence (B) is correct option.

MCQ 1.61 A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure.

Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of 10I0 = A will be(A) 44c (B) 51c

(C) 129c (D) 136c

SOL 1.61 Here for continuous conduction mode, by Kirchoff’s voltage law, average load current

V I2 150a− + 0=

Ia V

2150= +

10I1` = A, So V 130=− V

cosV2 m

π α 130=−

cos2 2 230# #π α 130c=−



PUBLISHING FOR GATE

α 129c=Hence (C) is correct option.

MCQ 1.62 A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30c. The approximate Total harmonic Distortion (%THD) and the rms value of fundamental component of input current will respectively be(A) 31% and 6.8 A (B) 31% and 7.8 A

(C) 66% and 6.8 A (D) 66% and 7.8 A

SOL 1.62 Hence (B) is correct option.

Total rms current Ia 10 8.1632 A#= =

Fundamental current Ia1 0.78 10 7.8 A#= =

THD 1 1DF2= −

where

DF .. 0.955I

I0 816 100 78 10

a

a1

##= = =

THD . 3 %0 9551 1 1

2= − =b l

MCQ 1.63 In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous.

The average voltage across the load and the average current through the diode will respectively be(A) 10 V, 2 A (B) 10 V, 8 A

(C) 40 V 2 A (D) 40 V, 8 A

SOL 1.63 Hence (C) is correct option.



PUBLISHING FOR GATE

In the given diagramwhen switch S is open 4 20I I VA, VL s0 = = =when switch S is closed 0, 0I V VD 0= =

Duty cycle .0 5= so average voltage is V1

s

δ−

Average current 220 4= + = amp

Average voltage . 401 0 520 V= − =

MCQ 1.64 The transfer function of a linear time invariant system is given as

( )G ss s3 2

12=

+ +The steady state value of the output of the system for a unit impulse input applied at time instant t 1= will be(A) 0 (B) 0.5

(C) 1 (D) 2

SOL 1.64 Given transfer function

( )G s s s3 2

12=+ +

Input ( )r t ( )t 1δ= − ( )R s [ ( )]t e1L sδ= − = −

Output is given by ( )Y s ( ) ( )R s G s=

( )Y s s s

e3 2

s

2=+ +

−

Steady state value of output ( )limy t

t"3 ( )lim sY s

s 0=

"

lims s

se3 2s

s

0 2=+ +"

− 0=

Hence (A) is correct option.

MCQ 1.65 The transfer functions of two compensators are given below :

( )( )

,( )

Cs

sC

ss

1010 1

10 110

1 2=+

+ =+

+

Which one of the following statements is correct ?(A) C1 is lead compensator and C2 is a lag compensator

(B) C1 is a lag compensator and C2 is a lead compensator

(C) Both C1 and C2 are lead compensator

(D) Both C1 and C2 are lag compensator

SOL 1.65 For C1 Phase is given by



PUBLISHING FOR GATE

C1θ ( )tan tan 101 1ω ω= −− −

a k

C1θ tan1 10

1012ω

ω ω=

+

−−

J

L

KKK

N

P

OOO

C1θ tan10

9 0>12ω

ω=+

−c m (Phase lead)

Similarly for C2, phase is

C2θ ( )tan tan101 1ω ω= −− −a k

tan1 10

1012ω

ω ω=

+

−−

J

L

KKK

N

P

OOO

tan10

9 0<12ω

ω=+

−−c m (Phase lag)

Hence (A) is correct option

MCQ 1.66 The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the figure :

This transfer function has(A) Three poles and one zero (B) Two poles and one zero

(C) Two poles and two zero (D) One pole and two zeros

SOL 1.66 From the given bode plot we can analyze that:Slope 40− dB/decade"2 polesSlope 20− dB/decade (Slope changes by 20+ dB/decade)"1 ZeroSlope 0 dB/decade (Slope changes by 20+ dB/decade)"1 ZeroSo there are 2 poles and 2 zeroes in the transfer function.Hence (C) is correct option.

MCQ 1.67 Figure shows a feedback system where K 0>



PUBLISHING FOR GATE

The range of K for which the system is stable will be given by(A) 0 30K< < (B) 0 39K< <

(C) 0 390K< < (D) 390K >

SOL 1.67 Characteristic equation for the system

1( )( )s s s

K3 10

+ + + 0=

( 3)( 10)s s s K+ + + 0= 13 30s s s K3 2+ + + 0=

Applying Routh’s stability criteria

s3 1 30

s2 13 K

s1 ( ) K13

13 30# −

s0 K

For stability there should be no sign change in first columnSo, K390 − 0> K 390<&

K 0> 0 K 90< <Hence (C) is correct option

MCQ 1.68 The transfer function of a system is given as

s s20 100

1002 + +

The system is(A) An over damped system (B) An under damped system

(C) A critically damped system (D) An unstable system

SOL 1.68 Given transfer function is

( )H s ) s s20 100

1002=+ +

Characteristic equation of the system is given by s s20 1002 + + 0= 100n

2ω = 10n& ω = rad/sec. 2 nξω 20=

or ξ 2 1020 1#

= =



PUBLISHING FOR GATE

( )1ξ = so system is critically damped.Hence (C) is correct option.

MCQ 1.69 Two sinusoidal signals ( , ) sinp t A t1 1ω ω= and ( )q t2ω are applied to X and Y inputs of a dual channel CRO. The Lissajous figure displayed on the screen shown below :The signal ( )q t2ω will be represented as

(A) ( ) , 2sinq t A t2 2 2 1ω ω ω ω= = (B) ( ) , /sinq t A t 22 2 2 1ω ω ω ω= =

(C) ( ) ,cosq t A t 22 2 2 1ω ω ω ω= = (D) ( ) , /cosq t A t 22 2 2 1ω ω ω ω= =


Frequency ratio ffX

Y meeting points of vertical tangentsmeeting points of horizontal tangents=

ffX

Y 42=

fY ( )f21

X=

2ω /21ω=Since the Lissajous figures are ellipse, so there is a phase difference of 90c exists between vertical and horizontal inputs.So ( )q t2ω ,cosA t2ω= /22 1ω ω=

MCQ 1.70 The ac bridge shown in the figure is used to measure the impedance Z .



PUBLISHING FOR GATE

If the bridge is balanced for oscillator frequency 2f = kHz, then the impedance Z will be(A) (260 0)j Ω+ (B) (0 200)j Ω+

(C) (260 200)j Ω− (D) (260 200)j Ω+

SOL 1.70 Impedance of different branches is given as ZAB 500= Ω

ZBC .

300j F2 2 10 0 398

13

# # # #π μ= + Ω

( )j200 300- − + Ω ZAD 2 2 10 15.91 300j mH3

# # # #π Ω= +

( )j200 300- + Ω

To balance the bridge Z ZAB CD Z ZAD BC= 500Z ( )( )j j200 300 200 300= + − + 500Z 130000= Z (260 0)j= + ΩHence (A) is correct option.

Common Data for Questions 71, 72 and 73:Consider a power system shown below:

Given that: 1 0V V j p.us1 s2= = + ;The positive sequence impedance are

0.001 0.01Z Z j p.us1 s2= = + and 0.006 0.06Z j p.uL = +3-phase Base 100MVA =voltage base 400= kV(Line to Line)Nominal system frequency 50= Hz.



PUBLISHING FOR GATE

The reference voltage for phase ‘a’ is defined as ( ) ( )cosV t V tm ω= .A symmetrical three phase fault occurs at centre of the line, i.e. point ‘F’ at time ‘t0

’. The positive sequence impedance from source S1 to point ‘F’ equals 0.004 0.04j+ p.u. The wave form corresponding to phase ‘a’ fault current from bus X reveals that decaying d.c. offset current is negative and in magnitude at its maximum initial value, Assume that the negative sequence impedances are equal to postive sequence impedance and the zero sequence impedances are three times positive sequence impedances.

MCQ 1.71 The instant ( )t0 of the fault will be(A) 4.682 ms (B) 9.667 ms

(C) 14.667 ms (D) 19.667 ms

SOL 1.71 Given ( )V t ( )cosV tm ω=For symmetrical 3 φ− fault, current after the fault

( )i t ( )cosAe ZV t2( / )R L t m ω α= + −−

At the instant of fault i.e t t0= , the total current ( )i t 0=

0 ( )cosAe ZV t2( / )R L t m

00 ω α= + −−

Ae ( / )R L t0− ( )cosZV t2 m

0ω α=− −

Maximum value of the dc offset current

Ae ( / )R L t0− ( )cosZV t2 m

0ω α=− −

For this to be negative max. ( )t0ω α− 0=

or t0 ωα= ...(1)

and Z . .j0 004 0 04= + Z 0.0401995 84.29Z c+ +α= =

α 84.29 1.471orc= rad.From equation (1)

t0 ( ). .

2 501 471 0 00468#π= = sec

t0 .4 682= msHence (A) is correct option.

MCQ 1.72 The rms value of the component of fault current ( )If will be(A) 3.59 kA (B) 5.07 kA

(C) 7.18 kA (D) 10.15 kA



PUBLISHING FOR GATE

SOL 1.72 Since the fault ‘F’ is at mid point of the system, therefore impedance seen is same from both sides.

Z2 . .0 0201 84 29c+=

Z1(Positive sequence) Z2= . .0 0201 84 29c+=

also Z Z Z1 2 0= = (for 3-φ fault)

( )I puf . .Z

1 00 0201 84 29

1 01

cc

c+++= =

So magnitude If(p.u.)

.49 8=

Fault current If 49.83 400100

##

=

7.18= kAHence (C) is correct option.

MCQ 1.73 Instead of the three phase fault, if a single line to ground fault occurs on phase ‘a’ at point ‘F’ with zero fault impedance, then the rms of the ac component of fault current ( )Ix for phase ‘a’ will be(A) 4.97 p.u (B) 7.0 p.u

(C) 14.93 p.u (D) 29.85 p.u

SOL 1.73 If fault is LG in phase ‘a’

Z1 . .Z2 0 0201 84 29c+= =

Z2 . .Z 0 0201 84 291 c+= =and Z0 . .Z3 0 0603 84 291 c+= =



PUBLISHING FOR GATE

Then /I 3a I I Ia a a1 2 0= = =

( )I pua1 .Z Z Z

1 0 01 2 0

c+= + +

and Ia1 ( . . . )

. .0 0201 0 0201 0 0603

1 0 9 95= + + = pu

Fault Current If .I I3 29 85a a1= = = pu

So Fault current If .29 853 400100

##

=

.4 97= kAHence (A) is correct option.

Common Data for Questions 74 and 75:A 3-phase, 440 V, 50 Hz, 4-pole slip ring induction motor is feed from the rotor side through an auto-transformer and the stator is connected to a variable resistance as shown in the figure.

The motor is coupled to a 220 V, separately excited d.c generator feeding power to fixed resistance of 10 Ω. Two-wattmeter method is used to measure the input power to induction motor. The variable resistance is adjusted such the motor runs at 1410 rpm and the following readings were recorded 1800W1 = W, 200W2 =− W.

MCQ 1.74 The speed of rotation of stator magnetic field with respect to rotor structure will be(A) 90 rpm in the direction of rotation

(B) 90 rpm in the opposite direction of rotation

(C) 1500 rpm in the direction of rotation

(D) 1500 rpm in the opposite direction of rotation

SOL 1.74 Given 3-φ, 440 V, 50 Hz, 4-Pole slip ring motor



PUBLISHING FOR GATE

Motor is coupled to 220 V N 1410 , 1800 , 200W Wrpm W W1 2= = =So,

Ns Pf120=

15004120 50 rpm#= =

Relative speed 1500 1410= − 90 rpm= in the direction of rotation.Hence (A) is correct option.

MCQ 1.75 Neglecting all losses of both the machines, the dc generator power output and the current through resistance ( )Rex will respectively be(A) 96 W, 3.10 A (B) 120 W, 3.46 A

(C) 1504 W, 12.26 A (D) 1880 W, 13.71 A

SOL 1.75 Neglecting losses of both machines

Slip(S ) NN N

s

s= −

15001500 1410= − .0 06=

total power input to induction motor is Pin 1800 200= − 1600 W=Output power of induction motor Pout (1 )S Pin= − ( . )1 0 06 1600= − 1504 W=Losses are neglected so dc generator input power output power= 1504 W=So, I R2 1504=

I 12.26101504 A= =

Hence (C) is correct option.

Linked Answer Questions: Q.76 to Q.85 carry two marks each.

Statement for Linked Answer Questions 76 and 77:The current ( )i t sketched in the figure flows through a initially uncharged 0.3 nF



PUBLISHING FOR GATE

capacitor.

MCQ 1.76 The charge stored in the capacitor at 5t sμ= , will be(A) 8 nC (B) 10 nC

(C) 13 nC (D) 16 nC

SOL 1.76 Charge stored at 5t μ= sec

Q ( )i t dt0

5

= #

=area under the curve

Q =Area OABCDO =Area (OAD)+Area(AEB)+Area(EBCD)

2 4 2 3 3 221

21

# # # # #= + +

4 3 6= + + 13= nCHence (C) is correct option



PUBLISHING FOR GATE

MCQ 1.77 The capacitor charged upto 5 ms, as per the current profile given in the figure, is connected across an inductor of 0.6 mH. Then the value of voltage across the capacitor after 1 sμ will approximately be(A) 18.8 V (B) 23.5 V

(C) 23.5 V− (D) 30.6 V−

SOL 1.77 Initial voltage across capacitor

V0 CQo= . nF

nC0 313=

43.33= VoltWhen capacitor is connected across an inductor it will give sinusoidal esponse as ( )v tc cosV to oω=

where oω LC1=

. .0 3 10 0 6 10

19 3

# # #=

− −

2.35 106#= rad/sec

at 1 sect μ= So, ( )v tc 43.33 (2.35 10 1 10 )cos 6 6

# # #= −

43.33 ( 0.70)#= − 30.44=− VHence (D) is correct option.

Statement for Linked Answer Question 78 and 79.The state space equation of a system is described by ,A B CX X u Y X= + =o where X is state vector, u is input, Y is output and

, , [1 0]A B C00

12

01= − = == =G G

MCQ 1.78 The transfer function G(s) of this system will be

(A) ( )s

s2+

(B) ( )s ss

21

−+

(C) ( )s

s2−

(D) ( )s s 2

1+

SOL 1.78 State space equation of the system is given by, Xo A BX u= + Y CX=Taking Laplace transform on both sides of the equations. ( )s sX ( ) ( )A s B sX U= + ( ) ( )sI A sX− ( )B sU= ( )sX ( ) ( )sI A B sU1= − −

( )sY` ( )C sX=



PUBLISHING FOR GATE

So ( )sY ( ) ( )C sI A B sU1= − −

( )( )ss

UY

T.F = ( )C sI A B1= − −

( )sI A− s

s00 0

012= − −> >H H

ss0

12=

−+> H

( )sI A 1− − ( )s s

ss2

1 20= ++ 1

> H

( )

( )

s s s

s

1

0

21

21=+

+

R

T

SSSSS

V

X

WWWWW

Transfer function

( ) [ ]G s C sI A B1= − − ( )

( )

s s s

s

1 0

1

0

21

21

01=

+

+

R

T

SSSSS

8 >

V

X

WWWWW

B H ( )

( )

s s

s

1 02

1

21=+

+

R

T

SSSSS

8

V

X

WWWWW

B

( )s s 2

1= +


MCQ 1.79 A unity feedback is provided to the above system ( )G s to make it a closed loop system as shown in figure.

For a unit step input ( )r t , the steady state error in the input will be(A) 0 (B) 1

(C) 2 (D) 3

SOL 1.79 Steady state error is given by,

ess 1 ( ) ( )( )

limG s H ssR s

s 0= +"

= G

Here ( )R s [ ( )]r t s1L= = (Unit step input)

( )G s ( )s s 2

1= +

( )H s 1= (Unity feed back)

So, ess 1

( 2)1

1

lim

s s

s ss 0

=+ +

"

b lR

T

SSSS

V

X

WWWW

( 2) 1

( 2)lim

s ss s

s 0= + +

+"

= G



PUBLISHING FOR GATE

0=Hence (A) is correct option.

Statement for Linked Answer Questions 80 and 81.A general filter circuit is shown in the figure :

MCQ 1.80 If R R RA1 2= = and R R RB3 4= = , the circuit acts as a(A) all pass filter (B) band pass filter

(C) high pass filter (D) low pass filter

SOL 1.80 For low frequencies,

0"ω , so C1

" 3ω

Equivalent circuit is,

By applying node equation at positive and negative input terminals of op-amp.

R

v vR

v vA i A o

1 2

− + − 0=

2vA v vi o= + , R R RA1 2a = =

Similarly,

R

v vR

v 0A i A

3 4

− + − 0=



PUBLISHING FOR GATE

2vA vin= , R R RB3 4a = =

So, 0vo =It will stop low frequency signals.For high frequencies,

" 3ω , then 0C1

"ω

Equivalent circuit is,

Output, v vo i=So it will pass high frequency signal.This is a high pass filter.Hence (C) is correct option

MCQ 1.81 The output of the filter in Q.80 is given to the circuit in figure :The gain v/s frequency characteristic of the output ( )vo will be

SOL 1.81 In Q.80 cutoff frequency of high pass filter is given by,

R C21

hA

ωπ

=



PUBLISHING FOR GATE

Here given circuit is a low pass filter with cutoff frequency,

R C R C22

12

2L

A Aω

π π= =

2L hω ω=When both the circuits are connected together, equivalent circuit is,

So this is is Band pass filter, amplitude response is given by.


Statement for Linked Answer Question 82 and 83.

MCQ 1.82 A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of 80 rad/s. The armature resistance is 0.5 Ω and the field winding resistance is 80 Ω.The net voltage across the armature resistance at the time of plugging will be(A) 6 V (B) 234 V

(C) 240 V (D) 474 V

SOL 1.82 Given: 240V V= , dc shunt motor I 15 A= Rated load at a speed 80 rad/s= Armature Resistance .0 5 Ω= Field winding Resistance 80 Ω=So, E 240 12 0.5#= − E 234= Vplugging V E= + 240 234= + 474 V=Hence (D) is correct option.

MCQ 1.83 The external resistance to be added in the armature circuit to limit the armature current to 125% of its rated value is(A) 31.1 Ω (B) 31.9 Ω

(C) 15.1 Ω (D) 15.9 Ω]



PUBLISHING FOR GATE

SOL 1.83 External Resistance to be added in the armature circuit to limit the armature current to 125%.

So Ia . R R12 1 25 474a external

#= = +

R Ra external+ .31 6= Rexternal .31 1 Ω=Hence (A) is correct option

Statement for Linked Answer Question 84 and 85.A synchronous motor is connected to an infinite bus at 1.0 pu voltage and draws 0.6 pu current at unity power factor. Its synchronous reactance is 1.0 pu resistance is negligible.

MCQ 1.84 The excitation voltage (E ) and load angle ( )δ will respectively be(A) 0.8 pu and 36.86c lag (B) 0.8 pu and 36.86c lead

(C) 1.17 pu and 30.96c lead (D) 1.17 pu and 30.96c lag

SOL 1.84 A synchronous motor is connected to an infinite bus at 1.0 p.u. voltage and 0.6 p.u. current at unity power factor. Reactance is 1.0 p.u. and resistance is negligible.So, V 1 0c+= p.u. Ia 0.6 0c+= p.u. Zs 0 1 1 90R jX ja s c+= + = + = p.u. V 1 0 0.6 0 1 90E I Za s #c c c+ + + +δ= + = − E+δ 1.166 30.96c+= − p.u. Excitation voltage .1 17= p.u. Load angle ( )δ .30 96c= (lagging)Hence (D) is correct option.

MCQ 1.85 Keeping the excitation voltage same, the load on the motor is increased such that the motor current increases by 20%. The operating power factor will become(A) 0.995 lagging (B) 0.995 leading

(C) 0.791 lagging (D) 0.848 leading




PUBLISHING FOR GATE

Answer Sheet

1. (A) 19. (D) 37. (C) 55. (B) 73. (A)2. (A) 20. (D) 38. (B) 56. (A) 74. (A)3. (D) 21. (C) 39. (C) 57. (*) 75. (C)4. (C) 22. (C) 40. (C) 58. (*) 76. (C)5. (D) 23. (A) 41. (D) 59. (C) 77. (D)6. (A) 24. (B) 42. (D) 60. (B) 78. (D)7. (D) 25. (B) 43. (B) 61. (C) 79. (A)8. (A) 26. (B) 44. (B) 62. (B) 80. (C)9. (A) 27. (*) 45. (B) 63. (C) 81. (D)10. (*) 28. (B) 46. (*) 64. (A) 82. (D)11. (D) 29. (D) 47. (B) 65. (A) 83. (A)12. (B) 30. (A) 48. (C) 66. (C) 84. (D)13. (D) 31. (B) 49. (A) 67. (C) 85. (D)14. (D) 32. (*) 50. (*) 68. (C)15. (D) 33. (D) 51. (B) 69. (D)16. (B) 34. (B) 52. (C) 70. (A)17. (A) 35. (A) 53. (B) 71. (A)18. (D) 36. (D) 54. (D) 72. (C)

**********

Engineering

Gate ee 2008 with solutions