GATE EE Solved Paper by RK Kanodia

7/26/2019 GATE EE Solved Paper by RK Kanodia

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GATE 2014Electrical Engineering

Topicwise Solved Paper

2013 - 2000

GATE MCQ Electrical Engineering (Vol-1, 2 & 3)

by RK Kanodia & Ashish Murolia

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RK Kanoida & Ashish Murolia


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CONTENTS

1 Engineering Mathematics

2 Electric Circuit and Fields

3 Signals and Systems

4 Electrical Machines

5 Control Systems

6 Power Systems

7 Electrical & Electronics Measurement

8 Analog and Digital Electronics

9 Power Electronics10 General Aptitude


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1ENGINEERING MATHEMATICS

YEAR 2013 ONE MARK

MCQ 1.1.1 Given a vector field F y xa yza x a x y z2 2= - -v v v v , the line integral F dl:v v# evaluated

along a segment on the x-axis from x 1= to x 2= is

(A) .2 33- (B) 0

(C) .2 33 (D) 7

MCQ 1.1.2 The equationx

x

21

21

00

1

2

--

=> > >H H Hhas(A) no solution (B) only one solution

x

x

0

01

2=> >H H

(C) non-zero unique solution (D) multiple solutions

MCQ 1.1.3 Square roots of i- , where i 1= - , are

(A) i, i-

(B) cos sini4 4p p

- + -d dn n, cos sini43

43p p

+b bl l(C) cos sini

4 43p p+d bn l, cos sini43 4p p+b dl n

(D) cos sini43

43p p

+ -b bl l, cos sini43 43p p- +b bl lMCQ 1.1.4 The curl of the gradient of the scalar field defined by V x y y z z x 2 3 42 2 2= + + is

(A) 4 6 8xya yza zxa x y z+ +v v v

(B) 4 6 8a a ax y z+ +v v v

(C) xy z a x yz a y zx a 4 4 2 6 3 8x y z2 2 2+ + + + +v v v^ ^ ^h h h

(D) 0

MCQ 1.1.5 A continuous random variable Xhas a probability density function f x ex= -^ h ,x0< < 3. Then P X 1>" ,is(A) 0.368 (B) 0.5

(C) 0.632 (D) 1.0

YEAR 2013 TWO MARKS

MCQ 1.1.6 When the Newton-Raphson method is applied to solve the equationf x x x 2 1 03= + - =^ h , the solution at the end of the first iteration with the initialvalue as .x 1 20= is

(A) .0 82- (B) .0 49(C) .0 705 (D) .1 69

MCQ 1.1.7 A function y x x5 102= + is defined over an open interval ,x 1 2= ^ h. Atleast at onepoint in this interval, /dy dxis exactly(A) 20 (B) 25


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(C) 30 (D) 35

MCQ 1.1.8

z

zdz

4

42

2

+

-# evaluated anticlockwise around the circle z i 2- = , where i 1= -, is

(A) 4p- (B) 0

(C) 2 p+ (D) i2 2+

MCQ 1.1.9 A Matrix has eigenvalues 1- and 2- . The corresponding eigenvectors are1

1-> Hand

1

2-> Hrespectively. The matrix is(A)

1

1

1

2- -> H (B)1

2

2

4- -> H

(C)1

0

0

2

-

-> H (D) 02

1

3- -> H

YEAR 2012 ONE MARK

MCQ 1.1.10 Two independent random variables Xand Yare uniformly distributed in theinterval ,1 1-6 @. The probability that ,maxX Y6 @is less than /1 2is(A) /3 4 (B) /9 16

(C) /1 4 (D) /2 3

MCQ 1.1.11 If ,x 1= - then the value of xxis(A) e /2p- (B) e/2p

(C) x (D) 1

MCQ 1.1.12 Given ( )f zz z1

13

2=+

-+

. If Cis a counter clockwise path in the z-plane

such that z 1 1+ = , the value of ( )j

f z dz21

Cp # is

(A) 2- (B) 1-

(C) 1 (D) 2

MCQ 1.1.13 With initial condition ( ) .x1 0 5= , the solution of the differential equationtdtdx

x t+ = , is

(A) x t21= - (B) x t

212= -

(C) x t2

2

= (D) x t2

=

YEAR 2012 TWO MARKS

MCQ 1.1.14 Given that andA I5

2

3

0

1

0

0

1

=- -

=

> >H H, the value of A 3is

(A) 15 12A I+ (B) 19 30A I+

(C) 17 15A I+ (D) 17 21A I+

MCQ 1.1.15 The maximum value of ( )f x x x x 9 24 53 2= - + + in the interval [ , ]1 6 is(A) 21 (B) 25


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(C) 41 (D) 46

MCQ 1.1.16 A fair coin is tossed till a head appears for the first time. The probability that the

number of required tosses is odd, is(A) /1 3 (B) /1 2

(C) /2 3 (D) /3 4

MCQ 1.1.17 The direction of vector A is radially outward from the origin, with krA n= .where r x y z 2 2 2 2= + + and kis a constant. The value of nfor which A 0:d = is(A) 2- (B) 2

(C) 1 (D) 0

MCQ 1.1.18 Consider the differential equation

( ) ( )

( )dt

d y t

dt

dy t

y t22

2

+ + ( )td= with ( ) 2 0andy t dtdy

tt

00=- == =

--

The numerical value ofdt

dy

t 0= +is

(A) 2- (B) 1-

(C) 0 (D) 1

YEAR 2011 ONE MARK

MCQ 1.1.19 Roots of the algebraic equation x x x 1 03 2+ + + = are(A) ( , , )j j1+ + - (B) ( , , )1 1 1+ - +

(C) ( , , )0 0 0 (D) ( , , )j j1- + -

MCQ 1.1.20 With Kas a constant, the possible solution for the first order differential equation

dx

dye

x3= - is

(A) e K31 x3- +- (B) e K

31 x3- +

(C) e K31 x3- +- (D) e K3 x- +-

MCQ 1.1.21 A point Zhas been plotted in the complex plane, as shown in figure below.

The plot of the complex numberYZ1= is


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YEAR 2011 TWO MARKS

MCQ 1.1.22 Solution of the variables x1and x2for the following equations is to be obtained

by employing the Newton-Raphson iterative method.

Equation (1) .sinx x10 0 8 02 1 - =

Equation (2) .cosx x x10 10 0 6 022 2 1- - =Assuming the initial values are .x 0 01 = and .x 1 02= , the jacobian matrix is

(A).

.

10

0

0 8

0 6

-

-> H (B) 100

0

10> H

(C).

.

0

10

0 8

0 6

-

-> H (D) 1010 010-> H

MCQ 1.1.23 The function ( )f x x x x 2 32 3= - - + has(A) a maxima at x 1= and minimum at x 5=

(B) a maxima at x 1= and minimum at x 5=-(C) only maxima at x 1= and

(D) only a minimum at x 5=

MCQ 1.1.24 A zero mean random signal is uniformly distributed between limits a- and a+ and its mean square value is equal to its variance. Then the r.m.s value of thesignal is

(A) a3

(B) a2

(C) a 2 (D) a 3

MCQ 1.1.25 The matrix [ ]A24

11

=-> His decomposed into a product of a lower

triangular matrix [ ]L and an upper triangular matrix [ ]U. The properly decomposed[ ]L and [ ]Umatrices respectively are

(A)1

4

0

1-> Hand1

0

1

2-> H (B)2

4

0

1-> Hand1

0

1

1> H(C)

1

4

0

1> Hand2

0

1

1-> H (D)2

4

0

3-> Hand.1

0

1 5

1> HMCQ 1.1.26 The two vectors [1,1,1] and [ , , ]a a1 2 where a j

2

1

2

3= - +

c m, are

(A) Orthonormal (B) Orthogonal

(C) Parallel (D) Collinear


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YEAR 2010 ONE MARK

MCQ 1.1.27 The value of the quantity P, where P xe dx x0

1=# , is equal to

(A) 0 (B) 1

(C) e (D) 1/e

MCQ 1.1.28 Divergence of the three-dimensional radial vector field ris(A) 3 (B) /r1

(C) i j k+ +t t t (D) 3( )i j k+ +t t t

YEAR 2010 TWO MARKS

MCQ 1.1.29 A box contains 4 white balls and 3 red balls. In succession, two balls are randomlyand removed form the box. Given that the first removed ball is white, theprobability that the second removed ball is red is(A) 1/3 (B) 3/7

(C) 1/2 (D) 4/7

MCQ 1.1.30 At t 0= , the function ( ) sinf ttt

= has

(A) a minimum (B) a discontinuity

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

MCQ 1.1.31 An eigenvector of P

1

0

0

1

2

0

0

2

3

=

J

L

KKK

N

P

OOO

is

(A) 1 1 1 T-8 B (B) 1 2 1 T8 B(C) 1 1 2 T-8 B (D) 2 1 1 T-8 B

MCQ 1.1.32 For the differential equationdt

d xdtdx

x6 8 022

+ + = with initial conditions ( )x0 1=

anddtdx 0

t 0

==

, the solution is

(A) ( ) 2x t e e t t6 2= -- - (B) ( ) 2x t e e t t2 4= -- -

(C) ( ) 2x t e e t t6 4=- +- - (D) ( ) 2x t e e t t2 4= +- -

MCQ 1.1.33 For the set of equations, x x x x 2 4 21 2 3 4+ + + = and 3 6 3 12 6x x x x 1 2 3 4+ + + = .

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

(B) There are no solutions

(C) A unique non-trivial solution exists

(D) Multiple non-trivial solutions exist

YEAR 2009 ONE MARK

MCQ 1.1.34 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-


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YEAR 2009 TWO MARKS

MCQ 1.1.35 ( , )f x yis a continuous function defined over ( , ) [ , ] [ , ]x y 0 1 0 1#! . Given the twoconstraints, x y> 2and y x> 2, the volume under ( , )f x yis

(A) ( , )f x y dxdyx y

x y

y

y

0

1

2=

=

=

= ## (B) ( , )f x y dxdyx y

x

y x

y 11

22 =

=

=

= ##

(C) ( , )f x y dxdyx

x

y

y

0

1

0

1

=

=

=

= ## (D) ( , )f x y dxdyx

x y

y

y x

00 =

=

=

= ##

MCQ 1.1.36 Assume for simplicity that Npeople, all born in April (a month of 30 days), arecollected in a room. Consider the event of at least two people in the room beingborn on the same date of the month, even if in different years, e.g. 1980 and 1985.What is the smallest Nso that the probability of this event exceeds 0.5 ?

(A) 20 (B) 7

(C) 15 (D) 16

MCQ 1.1.37 A cubic polynomial with real coefficients(A) Can possibly have no extrema and no zero crossings

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

(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

MCQ 1.1.38 Let x 117 02 - = . The iterative steps for the solution using Newton-Raphons

method is given by

(A) x xx2

1 117k k

k

1 = ++

b l (B) x x

x

117k k

k

1 = -+

(C) x x x117k k

k1 = -+ (D) x x x x2

1 117k k k

k1 = - ++ b l

MCQ 1.1.39 ( , ) ( ) ( )x y x xy y xy F a ax y2 2= + + +t t . Its 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

YEAR 2008 ONE MARKS

MCQ 1.1.40 Xis 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

MCQ 1.1.41 The characteristic equation of a (3 3# ) matrix Pis defined as

( )a I P 2 1 03 2l l l l l= - = + + + =If Idenotes identity matrix, then the inverse of matrix Pwill be

(A) ( )P P I22 + + (B) ( )P P I2 + +

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

MCQ 1.1.42 If the rank of a ( )5 6# matrix Qis 4, then which one of the following statementis correct ?(A) Qwill have four linearly independent rows and four linearly independent

columns

(B) Qwill have four linearly independent rows and five linearly independent


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columns

(C) QQTwill be invertible

(D) Q QT will be invertible

YEAR 2008 TWO MARKS

MCQ 1.1.43 Consider function ( ) ( 4)f x x2 2= - where xis a real number. Then the functionhas

(A) only one minimum (B) only tow minima

(C) three minima (D) three maxima

MCQ 1.1.44 Equation e 1 0x- = is required to be solved using Newtons method with aninitial guess x 10= - . Then, after one step of Newtons method, estimate x1of the

solution will be given by(A) 0.71828 (B) 0.36784

(C) 0.20587 (D) 0.00000

MCQ 1.1.45 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=

MCQ 1.1.46 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 stepfunction. If ( )x0 0=- , then value of xat .t 0 01= s will be given by(A) 0.00099 (B) 0.00495

(C) 0.0099 (D) 0.0198

MCQ 1.1.47 Let Pbe a 2 2# real orthogonal matrix and xis a real vector [ ]x ,x1 2Twith 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

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

YEAR 2007 ONE MARK

MCQ 1.1.48 x x xx n1 2T

g= 8 B is an n-tuple nonzero vector. The n n# matrix

V xxT=

(A) has rank zero (B) has rank 1

(C) is orthogonal (D) has rank n

YEAR 2007 TWO MARKS

MCQ 1.1.49 The differential equationdtdx x1= t

- is discretised using Eulers numerical integration

method with a time step T 0>3 . What is the maximum permissible value of T3 to ensure stability of the solution of the corresponding discrete time equation ?(A) 1 (B) /2t


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(C) t (D) 2t

MCQ 1.1.50 The value of

( )z

dz

1C2+

# where Cis the contour /z i2 1- = is

(A) i2p (B) p

(C) tanz1- (D) tani z1p -

MCQ 1.1.51 The integral ( )sin cost d21

0

2

p t t t-

p

# equals

(A) sin cost t (B) 0

(C) ( / )cos t1 2 (D) (1/2)sin t

MCQ 1.1.52 A loaded dice has following probability distribution of occurrences

Dice Value 1 2 3 4 5 6

Probability 1/4 1/8 1/8 1/8 1/8 1/4

If three identical dice as the above are thrown, the probability of occurrence ofvalues 1, 5 and 6 on the three dice is(A) same as that of occurrence of 3, 4, 5

(B) same as that of occurrence of 1, 2, 5

(C) 1/128

(D) 5/8

MCQ 1.1.53 Let xand ybe two vectors in a 3 dimensional space and x,y< >denote their

dot product. Then the determinant

detx,x

y,x

x,y

y,y

< >

< >

< >

< >= G

(A) is zero when xand yare linearly independent

(B) is positive when xand yare linearly independent

(C) is non-zero for all non-zero xand y

(D) is zero only when either xor yis zero

MCQ 1.1.54 The linear operation ( )L x is defined by the cross product L(x) b x#= , where

b 0 1 0T

= 8 B and x x xx 1 2 3T

= 8 B are three dimensional vectors. The 3 3# matrixMof this operations satisfies

( ) M

x

x

x

L x1

2

3

=

R

T

SSSS

V

X

WWWW

Then the eigenvalues of Mare(A) , ,0 1 1+ - (B) , ,1 1 1-

(C) , ,i i1- (D) , ,i i0-

Statement for Linked Answer Question 46 and 47.Cayley-Hamilton Theorem states that a square matrix satisfies its own

characteristic equation. Consider a matrix

A3

2

2

0=

-

-= G

MCQ 1.1.55 A satisfies the relation


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(A) A I A3 2 01+ + =- (B) 2 2 0A A I2 + + =

(C) ( )( )A I A I 2+ + (D) ( ) 0expA =

MCQ 1.1.56 A 9equals(A) 511 510A I+ (B) 309 104A I+

(C) 154 155A I+ (D) ( )exp A9

YEAR 2006 TWO MARKS

MCQ 1.1.57 The expression ( / )V R h H dh 1H

2 2

0p= -# for the volume of a cone is equal to

(A) ( / )R h H dr 1R

2 2

0p -# (B) ( / )R h H dh 1

R2 2

0p -#

(C) ( / )rH r R dh 2 1H

0p -# (D) rH Rr dr2 1

R

20

p -` j#MCQ 1.1.58 A surface ( , )S x y x y 2 5 3= + - is integrated once over a path consisting of the

points that satisfy ( ) ( )x y1 2 1 2 2+ + - = . The integral evaluates to

(A) 17 2 (B) 17 2

(C) /2 17 (D) 0

MCQ 1.1.59 Two fair dice are rolled and the sum rof the numbers turned up is considered

(A) ( )Prr 661> =

(B) Pr ( /r3 is an integer)

6

5=

(C) Pr ( /r r8 4;= is an integer)95=

(D) ( 6 /Prr r5;= is an integer)181=

Statement for Linked Answer Question 60 and 61.

, ,P Q R

10

1

3

2

5

9

2

7

12

T T T

=

-

=

-

- = -

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

are three vectors.

MCQ 1.1.60 An orthogonal set of vectors having a span that contains P,Q,Ris

(A)

6

3

6

4

2

3

-

-

-

-

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

(B)

4

2

4

5

7

11

8

2

3

-

- -

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

(C)

6

7

1

3

2

2

3

9

4-

-

- -

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

(D)

4

3

11

1

31

3

5

3

4

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

MCQ 1.1.61 The following vector is linearly dependent upon the solution to the previousproblem

(A)89

3

R

T

SSSS

V

X

WWWW

(B)217

30

--

R

T

SSSS

V

X

WWWW

(C)

4

4

5

R

T

SSSS

V

X

WWWW

(D)

13

2

3-

R

T

SSSS

V

X

WWWW


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YEAR 2005 ONE MARK

MCQ 1.1.62 In the matrix equation x qP = , which of the following is a necessary condition forthe existence of at least on solution for the unknown vector x

(A) Augmented matrix [ ]qP must have the same rank as matrix P

(B) Vector qmust have only non-zero elements

(C) Matrix Pmust be singular

(D) Matrix Pmust be square

MCQ 1.1.63 If Pand Qare two random events, then the following is TRUE(A) Independence of P and Q implies that probability ( )P Q 0+ =

(B) Probability ( )P Q, $Probability (P) + Probability (Q)

(C) If Pand Qare mutually exclusive, then they must be independent

(D) Probability ( )P Q+ #Probability (P)

MCQ 1.1.64 If S x dx 31

= 3 -# , then Shas the value

(A)31- (B)

41

(C)21 (D) 1

MCQ 1.1.65 The solution of the first order DE '( ) ( )x t x t 3=- , (0)x x0= is

(A) ( )x t x e t03= - (B) ( )x t x e 0

3= -

(C) ( )x t x e /01 3= - (D) ( )x t x e 0

1= -

YEAR 2005 TWO MARKS

MCQ 1.1.66 For the matrix p

3

0

0

2

2

0

2

1

1

=

-

-

R

T

SSSS

V

X

WWWW

, one of the eigen values is equal to 2-

Which of the following is an eigen vector ?

(A)

3

2

1

-

R

T

SSSS

V

X

WWWW

(B)

3

2

1

-

-

R

T

SSSS

V

X

WWWW

(C)

1

2

3

-

R

T

SSSS

V

X

WWWW

(D)

2

5

0

R

T

SSSS

V

X

WWWW

MCQ 1.1.67 If R

1

2

2

0

1

3

1

1

2

=

-

-

R

T

SSSS

V

X

WWWW

, then top row of R1- is

(A) 5 6 48 B (B) 5 3 1-8 B(C) 2 0 1-8 B (D) /2 1 1 2-8 B

MCQ 1.1.68 A fair coin is tossed three times in succession. If the first toss produces a head,then the probability of getting exactly two heads in three tosses is

(A)81 (B)

21

(C)83 (D)

43


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MCQ 1.1.69 For the function ( )f x x e x2= - , the maximum occurs when xis equal to(A) 2 (B) 1

(C) 0 (D) 1-

MCQ 1.1.70 For the scalar field u x y

2 3

2 2

= + , magnitude of the gradient at the point (1, 3) is

(A)913 (B)

29

(C) 5 (D)29

MCQ 1.1.71 For the equation '' ( ) ' ( ) ( )x t x t x t 3 2 5+ + = ,the solution ( )x tapproaches which of

the following values as t " 3 ?

(A) 0 (B)25

(C) 5 (D) 10***********


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SOLUTION

SOL 1.1.1 Option (B) is correct.

Given, the field vector

Fv y xa yza x a x y z2 2= - -v v v

For the line segment along x-axis, we have

dlv dxax= v

So, F dl:v v y x dx2= ^ ^h hSince, on x-axis y 0= so,

F dl:v v 0=

or, F dl:v v# 0=SOL 1.1.2 Option (D) is correct.

Given the equations in matrix form as

x

x

2

1

2

11

2

-

-> >H H 0

0= > H

So, it is a homogenous set of linear equation. It has either a trivial solution

x x 01 2= =^ hor an infinite no. of solution. Since, for the matrix A

2

1

2

1=-

-> Hwe have the determinant A 0=

Hence, it will have multiple solutions


We know that i e/i 2= p (In phasor form)or, i- e /i 2= p-

So, i- e e// /i i2 1 2 4

! != =p p- -

^ h cos sini4 4! p p= -d dn n= G cos sini

4 4p p= -d dn n; cos sini

4 4p p- +d dn n

i2

12

= - ; i12

- +

This is equivalent to the given option (B) only.

SOL 1.1.4 Option (D) is correct.

Given the scalar field

V x y y z z x 2 3 42 2 2= + +

Its gradient is given by Vd xy z a x yz a y zx a 4 4 2 6 3 8x y z

2 2 2= + + + + +v v v^ _ ^h i hSo, the curl of the gradient is obtained as

v#d d ha

xy z

a

z yz

a

y zx4 4 2 6 3 8

x

rx

y

ry

z

rz

2 2 2

=

+ + +

2 2 2

v v v


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8 8a y y a z z a x x 6 6 4 4x y z= - - - + -v v v^ ^ ^h h h 0=

Note : From the properties of curl, we know that curl of gradient of any scalarfield is always zero. So, there is no need to solve the curl and gradient.

SOL 1.1.5 Option (A) is correct.Given, the PdF of random variable xas

f x^ h ex= - x0< < 3So,

P x 1>^ h e dxx1

=3

-# e

1

x

1

=-

3-: D e1= -

.0 368=

SOL 1.1.6 Option (C) is correct.Given, the equation

f x^ h x x2 1 03= + - =as initial condition is x0 .1 2=

so, from N R- method we obtain

xn 1+ xf x

f xn

n

n

= -l h

h

here, x0 .1 2=

f x0^ h . 2 . 1 3.1281 2 1 23= + - =^ ^h hAlso, f xl h x3 22= +So, f x0l h . .3 1 2 2 6 322= + =^ hHence, 1stiterative value is

x1 xf x

f x0

0

0= -

l hh

...1 26 323 128= -

.0 705=

SOL 1.1.7 Option (B) is correct.Given the function

y 5 10f x x x 2= = +^ h in the internal ,x 1 2= ^ hSince, function yis continuous in the interval ,1 2^ has well as its is differentiableat each point so, from Lagranges mean value theorem there exist at least a pointwhere

f cl hb a

f b f a =

-

-^ ^h hHere, we have a 1= , b 2=

So, for x a 1= = , we obtain y 5f a f1 1 10 1 152= = = + =^ ^ ^ ^h h h hand for x b 2= = y 5f b f2 2 10 2 402= = = + =^ ^ ^ ^h h h h


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Therefore,

f cl h 2 140 15 25= -- =SOL 1.1.8 Option (A) is correct.

Given the contour integral

z

zdz

44

2

2

+-#

It has two poles given as z i2!=Now, the contour is defined by circle z i 2- = which is shown in the figure below

So, it can be observed that the given contour enclosed z i2= while z i2=- is outof the contour. So, we obtain the residue at z i2= only as

residuez iz

24

z i

2

2

=+-

=

i i

i

i i

2 22 4

48 2

2

=+

-=

-=^ h

Hence, contour integral is given as

z

zdz

44

2

2

+-

# 2ip= (sum of residues) 2i i2p= ^ h 4p=-

SOL 1.1.9 Option (D) is correct.We know that the characteristic equation is given by

A X6 6@ @ Xl= 6 @where A6 @ is the matrix as l is the scalar which gives eigen values. Now, weconsider the matrix

A6 @ ac

b

d= > H matrix2 2#^ h

For eigen value 1- as eigen vector is1

1-> H, so, we have

a

c

b

d

1

1-> >H H 1 11=- -> Hor, a b- 1=- ....(1) c d- 1= ....(2)

Similarly, for eigen value 2- with eigen vector 12-> H, we obtain

a

c

b

d

1

2-> >H H 21

2=-

-> H

or, a b2- 2=- ....(3) c d2- 4= ....(4)


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Solving Eqs. (1) and (3), we obtain

a ,b0 1= =and solving Eqs. (2) and (4), we obtain

c ,d2 3=- =Thus, the required matrix is

a

c

b

d> H 02 13= - -> H


Probability density function of uniformly distributed variables Xand Yis shown

as

[ ( , )]maxP x y21


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2 223

23

169

#

#= =

SOL 1.1.11 Option (A) is correct.

x i1= - = cos sini2 2p p= +

So, x ei2= p

xx ei x2=

p^ h & ei i2p^ h e2= p-SOL 1.1.12 Option (C) is correct.

( )f zz z1

13

2=+

-+

( )j

f z dz21

Cp # =sum of the residues of the poles which lie inside the given

closed region. C z 1 1& + =

Only pole z 1=- inside the circle, so residue at z 1=- is.

( )f z( )( )z z

z1 3

1=+ +

- + ( )( )

( )( )lim

z z

z z

1 31 1

22 1

z 1=

+ +

+ - += =

"-

So ( )j

f z dz21

Cp # 1=


tdtdx

x+ t=

dtdx

t

x

+ 1=

dtdx

Px+ Q= (General form)

Integrating factor, I F e e e t lntPdt tdt1

= = = =# #

Solution has the form

x IF# Q IF dt C #= +^ h# x t# ( )( )t dt C1= +# xt t C

2

2

= +

Taking the initial condition

( )x1 .0 5=

0.5 C21= +

C 0=

So, xt t2

2

= x t2

& =


Characteristic equation.

A Il- 0=

5

2

3l

l

- - -

- 0=

5 62l l+ + 0=

5 62l l+ + 0=


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Since characteristic equation satisfies its own matrix, so

5 6A A2 + + 0= 5 6A A I2& =- -

Multiplying with A

5 6A A A3 2+ + 0=

5( 5 6 ) 6A A I A3 + - - + 0=

A 3 19 30A I= +


( )f x x x x9 24 53 2= - + +

( )dx

df x x x3 18 24 02= - + =

&( )dx

df x 6 8 0x x2= - + =

x 4= , x 2=

( )

dx

d f x2

2

x6 18= -

For ,x 2= ( )

dx

d f x12 18 6 0


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( ) [ ]Y s s s 2 12 + + s1 2 4= - -

( )Y ss s

s

2 12 3

2= + +- -

We know If, ( )y t ( )Y sL

then,( )dt

dy t ( ) ( )sY s y 0

L-

So, ( ) ( )sY s y 0- ( )

( )

s s

s s

2 1

2 322= + +

- -+

( )s s

s s s s

2 12 3 2 4 2

2

2 2

=+ +

- - + + +

( ) ( )sY s y 0- ( ) ( ) ( )ss

s

s

s12

11

11

2 2 2= ++

=++

++

( )s s1

1

1

12

=+

++

By taking inverse Laplace transform

( )dt

dy t ( ) ( )e u t te u t t t= +- -

At t 0= +,dt

dy

t 0= + e 0 10= + =


x x x 13 2+ + + 0=

( ) ( )x x x1 12 + + - 0=

( )( )x x1 12+ + 0=

or x 1+ 0 1x&= =-

and 1x2 + 0 ,x j j&= =-

x 1, ,j j=- -


dx

dy e x3= -

dy e dxx3= -

by integrating, we get

y e K3

1 x3

=- +

-

, where Kis constant.SOL 1.1.21 Option (D) is correct.

Zis Z 0= where qis around 45cor so.

Thus Z Z 45c= where Z 1


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So Ywill be out of unity circle.


f1 .sinx x10 0 82 1= -

f2 10 10 0.6cosx x x22

2 1= - -

Jacobian matrix is given by

Jcos

sin

sin

cosxf

x

fxf

x

f

x x

x x

x

x x

10

10

10

20 101

1

1

2

2

1

2

2

2 1

2 1

1

2 1

22

2

222

2

2= = -

R

T

SSSSS

>V

X

WWWWW

H

For 0, 1x x1 2= = , J10

0

0

10= > H

SOL 1.1.23 Option (C) is correct.

( )f x x x2 32= - +

( )f xl x2 2 0= - =

x 1= ( )f xm 2=-

( )f xm is negative for x 1= , so the function has a maxima at x 1= .


Let a signal ( )p xis uniformly distributed between limits a- to a+ .

Variance ps ( )x p x dxa

a2=

-# x adx21a

a2

:=-

#

ax

21

3 a

a3

=-

: D a a62 33 2

= =

It means square value is equal to its variance

prms2 a

3p2

s= =

prmsa

3=


We know that matrix A is equal to product of lower triangular matrix L and

upper triangular matrix U.

A L U= 6 6@ @only option (D) satisfies the above relation.


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Let the given two vectors are

X1 [ , , ]1 1 1=

X2 [ , , ]a a12=

Dot product of the vectors

X X1 2$ 1a

a

a aX X 1 1 1

1T

1 22

2= = = + +

R

T

SSSS

8V

X

WWWW

BWhere a 1 /j

21

23 2 3p=- + = -

so,

a a1 2+ + 0= ,X X1 2are orthogonal

Note: We can see that ,X X1 2are not orthonormal as their magnitude is 1!


P xe dxx

0

1

=# ( )

dxd

x e dx dxx

0

1

0

1x e dxx

= - :6 D@# ## ( )e dx1 x

0

1

0

1xex

= -6 @ # ( 0)e10

1ex

= - -6 @ [ ]e e e1 1 0= - - 1=


Radial vector r x y zi j k= + +t t t

Divergence r4$=

x y z

x y zi j k i j k :2

22

22

2= + + + +t t t t t tc _m i

x

x

y

y

z

z

2

2

2

2

2

2= + + 1 1 1= + + 3=


No of white balls 4= , no of red balls 3=

If first removed ball is white then remaining no of balls 6(3 ,3white red)=

we have 6 balls, one ball can be choose in C6 1ways, since there are three red balls

so probability that the second ball is red is

PCC

31

61=

63=

21=


Function ( )f t sintt

= sin ct= has a maxima at 0t= as shown below


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nodia.co.inSOL 1.1.31 Option (B) is correct.

Let eigen vector

X x x x1 2 3T= 8 B

Eigen vector corresponding to 11l =

A IX1l-8 B 0=

x

x

x

0

0

0

1

1

0

0

2

2

1

2

3

R

T

SS

SS

R

T

SS

SS

V

X

WW

WW

V

X

WW

WW

0

0

0

=

R

T

SS

SS

V

X

WW

WW x2 0=

x x2 02 3+ = x 03& = (not given in the option)


A IX2l-8 B 0=

x

x

x

1

0

0

1

0

0

0

2

1

1

2

3

-R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

0

0

0

=

R

T

SSSS

V

X

WWWW

x x1 2- + 0=

2 0x3= x 03& = (not given in options.)


A IX3l-8 B 0=

x

x

x

2

0

0

1

1

0

0

2

0

1

2

3

-

-

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

0

0

0

=

R

T

SSSS

V

X

WWWW

x x2 1 2- + 0=

x x22 3- + 0=

Put ,x x1 21 2= = and x 13=So Eigen vector

X

x

x

x

1

2

3

=

R

T

SSSS

V

X

WWWW

1

2

1

1 2 1 T= =

R

T

SSSS

8V

X

WWWW

B


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8dt

d xdtdx

x622

+ + 0=

Taking Laplace transform (with initial condition) on both sides ( ) ( ) ' ( ) [ ( ) ( )] ( )s X s sx x sX s x X s 0 0 6 0 82 - - + - + 0=

( ) ( ) [ ( ) ] ( )s X s s sX s X s 1 0 6 1 82 - - + - + 0=

( )[ ]X s s s s 6 8 62 + + - - 0=

( )X s( )

( )

s s

s

6 8

62= + +

+

By partial fraction

( )X ss s2

24

1=+

-+

Taking inverse Laplace transform

( )x t ( )e e2 t t2 4= -- -


Set of equations

x x x x 2 41 2 3 4+ + + 2= .....(1)

x x x x 3 6 3 121 2 3 4+ + + 6= .....(2)

or ( )x x x x 3 2 41 2 3 4+ + + 3 2#=

Equation (2) is same as equation(1) except a constant multiplying factor of 3. So

infinite (multiple) no. of non-trivial solution exists.


Let the matrix is A a

c

b

d= > H

Trace of a square matrix is sum of its diagonal entries

Trace A a d= + 2=-

Determinent ad bc- 35=-

Eigenvalue A Il- 0=

a

c

b

d

l

l

-

- 0=

( )( )a d bc l l- - - 0=

( ) ( )a d ad bc 2l l- + + - 0=

( ) ( )2 352l l- - + - 0=

2 352l l+ - 0=

( )( )5 7l l- + 0=

,1 2l l ,5 7= -


Given constraints x y> 2and y x> 2


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Limit of y: y 0= to y 1=

Limit of x: x y2= to x y x y 2 &= =

So volume under ( , )f x y

V ( , )f x y dx dyx y

x y

y

y

0

1

2=

=

=

=

= ##SOL 1.1.36 Option (B) is correct.

No of events of at least two people in the room being born on same date Cn 2=

three people in the room being born on same date Cn 3=

Similarly four for people Cn 4=

Probability of the event, 0.5N

C C C C N 7

n n n nn2 3 4

&$ $ g$ =

SOL 1.1.37 Option ( ) is correct.

Assume a Cubic polynomial with real Coefficients

( )P x a x a x a x a 03

13

2 3= + + + , , ,a a a a0 1 2 3are real

'( )P x a x a x a 3 202

1 2= + +

'' ( )P x a x a6 20 1= +

''' ( )P x 6a0=

( )P xi v 0=


An iterative sequence in Newton-Raphsons method is obtained by following

expression

xk 1+ ' ( )( )

xf x

f xk

k

k= -

( )f x x 1172= -

' ( )f x x2=

So ( )f xk x 117k2= -

' ( )f xk x2 2 117k #= =

So xk 1+ 117

xx

x2k k

k2

= - - x x

x21 117

k kk

= - +: DSOL 1.1.39 Option (D) is correct.

Equation of straight line

y 2- ( )x2 00 2 0=

--

-

y 2- x=-

F dl$ [( ) ( ) ] [ ]x xy y xy dx dy dz a a a a ax y x y z2 2= + + + + +t t t t t


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( ) ( )x xy dx y xy dy 2 2= + + +

Limit of x: 0 to 2

Limit of y: 2 to 0

F dl$# ( ) ( )x xy dx y xy dy 2 22

0

0

2= + + +##

Line y 2- x=-

dy dx=-

So F dl$# [ ( )] ( )x x x dx y y y dy 2 220

2 2

2

0= + - + + -# #

xdx y dy 2 22

0

0

2= +##

x y

22

22

2

0

2 2

2

0

= +: ;D E 4 4= - 0=SOL 1.1.40 Option (C) is correct.

Xis uniformly distributed between 0 and 1

So probability density function

( )f XX 1, 0 1

0,

x

otherwise

< 2=+ = - = (Minima)

, '' ( ) ( )x f2 2 12 2 16 32 0>2=- - = - - = (Minima)

So ( )f xhas only two minima


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

expression

xn 1+ ' ( )( )

xf x

f xn

n

n= -

We have to calculate x1, so n 0=

x1 ' ( )( )

xf x

f x0

0

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=


'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=

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 I A1- 'A A= =Y (false)


dtdx ( )e u tt2= -

x ( )e u t dtt2= -# e dtt20

1

= -#

( )f t dt0

1

=# ,

t .01= s

From trapezoid rule

( )f t dtt

t nh

0

0+# ( ) (. )hf f2 0 01= +6 @ ( )f t dt

0

1# . e e201 .0 02= + -6 @, .h 01=


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.0099=


P is an orthogonal matrix So PP IT=

Let assume Pcos

sin

sin

cos

q

q

q

q=

-> H PX

cos

sin

sin

cos x x

T1 2

q

q

q

q=

-> 8H B

cos

sin

sin

cos

x

x

1

2

q

q

q

q=

-> >H H cos sinsin cosx x

x x

1 2

1 2

q q

q q=

-

+> H

PX ( ) ( )cos sin sin cosx x x x 1 22

1 22q q q q= - + +

x x12

22= +

PX X=


x x x xn1 2T

g= 8 B V xxT=

x

x

x

x

x

xn n

1

2

1

2

h h=

R

T

SSSSSS

R

T

SSSSSS

V

X

WWWWWW

V

X

WWWWWW

So rank of Vis n.



Givenz

dz

1C

2+#

( )( )z i z i dz

C

=+ -

#

Contour z i2

- 1=

P(0, 1) lies inside the circle 1z i2

- = and ( , )P0 1 does not lie.

So by Cauchys integral formula

z

dz

1C

2+# 2 ( )

( )( )limi z i

z i z i

1z i

p= -+ -"

limiz i

2 1z i

p=+"

ii

221#p= p=



Probability of occurrence of values 1,5 and 6 on the three dice is

( , , )P1 5 6 ( ) ( ) ( )P P P1 5 6=

41

81

41

# #= 1281=

In option (A)

( , , )P3 4 5 ( ) ( ) ( )P P P3 4 5=

81

81

81

# #= 5121=

In option (B)

( , , )P1 2 5 ( ) ( ) ( )P P P1 2 5=

41

81

81

# #= 2561

=


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detx x

y x

x y

y y

$

$

$

$> H ( )( ) ( )( )x x y y x y y x: : : := -

0= only when xor yis zero



For characteristic equation

3

1

2

0

l

l

- -

- -> H 0=

or ( )( )3 2l l- - - + 0=

( )( )1 2l l+ + 0=

According to Cayley-Hamiliton theorem ( )( )A I A I 2+ + 0=


According to Cayley-Hamiliton theorem

( )( )A I A I 2+ + 0=

or A A I3 22 + + 0=

or A 2 ( )A I3 2=- +

or A 4 ( ) ( )A I A A I 3 2 9 12 42 2= + = + +

9( 3 2 ) 12 4A I A I = - - + + 15 14A I=- -

A 8 ( )A I A A15 14 225 420 1962 2= - - = + + 225( 3 2 ) 420 196A I A I = - - + + 255 254A I=- -

A 9 255 254A A2=- -

255( 3 2 ) 254A I A=- - - - A I511 510= +


Volume of the cone

V RHh

dh1H 2

0

2

p= -b l# Solving the above integral

V R H312p=

Solve all integrals given in option only for option (D)

rHRr

dr2 1R

0

2p -a k# R H31 2p=



By throwing dice twice 6 6 36# = possibilities will occur. Out of these sample

space consist of sum 4, 8 and 12 because /r4is an integer. This can occur in

following way :

(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2) and (6, 6)

Sample Space 9=

Favourable space is coming out of 8 5=

Probability of coming out 895

=


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SOL 1.1.62 Option (A) is correct.Matrix equation PX q= has a unique solution if

( )Pr ( )rr=

Where ( )P "r rank of matrix P

( )r "r rank of augmented matrix [ ]P

r :P q= 8 BSOL 1.1.63 Option (D) is correct.

for two random events conditional probability is given by

( )probabilityP Q+ ( ) ( )probability probabilityP Q=

( )Qprobability ( )( )PP Q 1probabilityprobability +

#=

so ( )P Qprobability + ( )Pprobability#


S x dx31

= 3 -# x2

2

1

=-

3-: D 21=SOL 1.1.65 Option (A) is correct.

We have ( )x to ( )x t3=-

or ( ) 3 ( )x t x t +o 0=

A.E. D 3+ 0=Thus solution is ( )x t C e t1

3= -

From ( )x x0 0= we get C1 x0=

Thus ( )x t x e t03= -


For eigen value l 2=-

( )

( )

( )

x

x

x

3 2

0

0

2

2 2

0

2

1

1 2

1

2

3

- - -

- - -

- -

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

0

0

0

=

R

T

SSSS

V

X

WWWW

x

x

x

5

0

0

2

0

0

2

1

1

1

2

3

-R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

0

0

0

=

R

T

SSSS

V

X

WWWW

x x x5 21 2 3- + 0=


C11 ( )2 3 5= - - =

C21 ( ( ))0 3 3=- - - =-

C31 ( ( ))1 1= - - =

R ( )C C C1 2 211 21 31= + + 5 6 2= - + 1=SOL 1.1.68 Option (B) is correct.

If the toss produces head, then for exactly two head in three tosses three tosses

there must produce one head in next two tosses. The probability of one head in

two tosses will be 1/2.


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We have ( )f x x ex2= -

or '( )f x xe x e 2 x x2= -- -

( )xe x2x= --

'' ( )f x ( )x x e4 2 x2= - + -

Now for maxima and minima, ' ( )f x 0=

( )xe x2x -- 0=

or x ,0 2=

at x 0= '' ( )f 0 1( )ve= +

at x 2= '' ( )f 2 2 ( )e ve2=- --

Now ''( )f 0 1= and '' ( )f e2 2 0


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2ELECTRICAL CIRCUITS & FIELDS

YEAR 2013 ONE MARK

MCQ 1.2.1 Consider a delta connection of resistors and its equivalent star connection as

shown below. If all elements of the delta connection are scaled by a factor k, k 0>, the elements of the corresponding star equivalent will be scaled by a factor of

(A) k2 (B) k

(C) /k1 (D) k

MCQ 1.2.2 The flux density at a point in space is given by 4 2 8 /Wb mB xa kya a x y z2= + +v v v v .

The value of constant kmust be equal to(A) 2- (B) .0 5-

(C) .0 5+ (D) 2+

MCQ 1.2.3 A single-phase load is supplied by a single-phase voltage source. If the currentflowing from the load to the source is 10 150 Ac+ - and if the voltage at the load

terminal is 100 60 Vc+ , then the(A) load absorbs real power and delivers reactive power

(B) load absorbs real power and absorbs reactive power

(C) load delivers real power and delivers reactive power

(D) load delivers real power and absorbs reactive power

MCQ 1.2.4 A source cosv t V t 100s p=^ h has an internal impedance of j4 3 W+^ h . If a purelyresistive load connected to this source has to extract the maximum power out ofthe source, its value in Wshould be(A) 3 (B) 4

(C) 5 (D) 7

MCQ 1.2.5 The transfer functionV s

V s

1

2 ^ hh of the circuit shown below is

(A) .ss

10 5 1

++ (B)

ss

23 6

++


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(C)ss

12

++ (D)

ss

21

++

YEAR 2013 TWO MARKS

MCQ 1.2.6 A dielectric slab with mm mm500 500# cross-section is 0.4 mlong. The slab

is subjected to a uniform electric field of 6 8 /kV mmE a ax y= +v v v . The relative

permittivity of the dielectric material is equal to 2. The value of constant 0e is8.85 10 /F m12#

- . The energy stored in the dielectric in Joules is(A) .8 85 10 11#

- (B) .8 85 10 5# -

(C) .88 5 (D) 885

MCQ 1.2.7 Three capacitorsC1,C2andC3whose values are 10 Fm , 5 Fm , and 2 Fm respectively,have breakdown voltages of 10 V, 5 Vand 2 Vrespectively. For the interconnection

shown below, the maximum safe voltage in Volts that can be applied across thecombination, and the corresponding total charge in Cm stored in the effectivecapacitance across the terminals are respectively,

(A) 2.8 and 36 (B) 7 and 119(C) 2.8 and 32 (D) 7 and 80

MCQ 1.2.8 In the circuit shown below, if the source voltage 100 53.13 VVS c+= then theThevenins equivalent voltage in Volts as seen by the load resistance RL is

(A) 100 90c+ (B) 800 0c+

(C) 800 90c+ (D) 100 60c+

YEAR 2012 ONE MARK

MCQ 1.2.9 In the circuit shown below, the current through the inductor is


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(A) Aj1

2+

(B) Aj11

+-

(C) Aj1

1+

(D) 0 A

MCQ 1.2.10 The impedance looking into nodes 1 and 2 in the given circuit is

(A) 05 W (B) 100 W

(C) 5 kW (D) 10.1kW

MCQ 1.2.11 A system with transfer function ( )( )( )( )

( )( )G s

s s s

s s

1 3 49 22

=+ + +

+ +

is excited by ( )sin tw . The steady-state output of the system is zero at

(A) 1 /rad sw= (B) /rad s2w=

(C) /rad s3w= (D) /rad s4w=

MCQ 1.2.12 The average power delivered to an impedance (4 3)j W- by a current

5 (100 100)cos Atp + is(A) 44.2 W (B) 50 W

(C) 62.5W (D) 125 W

MCQ 1.2.13 In the following figure, C1and C2are ideal capacitors. C1has been charged to 12

V before the ideal switch Sis closed at .t 0= The current ( )i tfor all tis

(A) zero (B) a step function

(C) an exponentially decaying function (D) an impulse function


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YEAR 2012 TWO MARKS

MCQ 1.2.14 If 6 VV VA B- = then V VC D- is

(A) 5 V- (B) V2

(C) V3 (D) V6

MCQ 1.2.15 Assuming both the voltage sources are in phase, the value of Rfor which maximum

power is transferred from circuit A to circuit Bis

(A) 0.8 W (B) 1.4 W

(C) 2 W (D) 2.8 W

Common Data for Questions 16 and 17 :

With 10 Vdc connected at port A in the linear nonreciprocal two-port networkshown below, the following were observed :(i) 1 Wconnected at port Bdraws a current of 3 A

(ii) 2.5 Wconnected at port Bdraws a current of 2 A

MCQ 1.2.16 With 10 Vdc connected at port A , the current drawn by 7 Wconnected at portBis(A) 3/7 A (B) 5/7 A

(C) 1 A (D) 9/7 A

MCQ 1.2.17 For the same network, with 6 Vdc connected at port A , 1 Wconnected at portBdraws 7/3 .A If 8 Vdc is connected to port A , the open circuit voltage at portBis(A) 6 V (B) 7 V

(C) 8 V (D) 9 V

Statement for Linked Answer Questions 18 and 19 :


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In the circuit shown, the three voltmeter readings are 220 ,VV1 = 122 ,VV2=

136 VV3= .

MCQ 1.2.18 The power factor of the load is(A) 0.45 (B) 0.50

(C) 0.55 (D) 0.60

MCQ 1.2.19 If 5RL W= , the approximate power consumption in the load is(A) 700 W (B) 750 W

(C) 800 W (D) 850 W

YEAR 2011 ONE MARK

MCQ 1.2.20 The r.m.s value of the current ( )i tin the circuit shown below is

(A) A21 (B) A

21

(C) 1 A (D) A2

MCQ 1.2.21 The voltage applied to a circuit is ( )cos t100 2 100p volts and the circuit draws

a current of ( / )sin t

10 2 100 4p p+ amperes. Taking the voltage as the referencephasor, the phasor representation of the current in amperes is

(A) /10 2 4p- (B) /10 4p-

(C) /10 4p+ (D) /10 2 4p+

MCQ 1.2.22 In the circuit given below, the value of Rrequired for the transfer of maximumpower to the load having a resistance of 3 Wis

(A) zero (B) 3 W

(C) 6 W (D) infinity


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YEAR 2011 TWO MARKS

MCQ 1.2.23 A lossy capacitor Cx, rated for operation at 5 kV, 50 Hz is represented by anequivalent circuit with an ideal capacitor Cpin parallel with a resistor Rp. The

value Cpis found to be 0.102 F and value of 1.25MRp W= . Then the power lossand tan dof the lossy capacitor operating at the rated voltage, respectively, are(A) 10 W and 0.0002 (B) 10 W and 0.0025

(C) 20 W and 0.025 (D) 20 W and 0.04

MCQ 1.2.24 A capacitor is made with a polymeric dielectric having an re of 2.26 and adielectric breakdown strength of 50 kV/cm. The permittivity of free space is 8.85

pF/m. If the rectangular plates of the capacitor have a width of 20 cm and alength of 40 cm, then the maximum electric charge in the capacitor is(A) 2 C (B) 4 C

(C) 8 C (D) 10 C

Common Data questions: 25 & 26

The input voltage given to a converter is vi 100 (100 )sin Vt2 p=The current drawn by the converter is

10 (100 /3) 5 (300 /4)sin sini t t2 2i p p p p= - + + 2 (500 /6)sin At2 p p+ -

MCQ 1.2.25 The input power factor of the converter is(A) 0.31 (B) 0.44

(C) 0.5 (D) 0.71

MCQ 1.2.26 The active power drawn by the converter is(A) 181 W (B) 500 W

(C) 707 W (D) 887 W

Common Data questions: 27 & 28

An RLC circuit with relevant data is given below.

MCQ 1.2.27 The power dissipated in the resistor Ris(A) 0.5 W (B) 1 W

(C) W2 (D) 2 W

MCQ 1.2.28 The current ICin the figure above is(A) 2 Aj- (B) Aj

21-

(C) Aj2

1+ (D) 2Aj+


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YEAR 2010 ONE MARK

MCQ 1.2.29 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 Fm capacitor is

(A) 0 A (B) 1 A

(C) 1.25 A (D) 5 A

MCQ 1.2.30 As shown in the figure, a 1 Wresistance is connected across a source that has a

load line v i 100+ = . The current through the resistance is

(A) 25 A (B) 50 A

(C) 100 A (C) 200 A

YEAR 2010 TWO MARKS

MCQ 1.2.31 If the 12 Wresistor draws a current of 1 A as shown in the figure, the value ofresistance R is

(A) 4 W (B) 6 W

(C) 8 W (D) 18 W

MCQ 1.2.32 The two-port network P shown in the figure has ports 1 and 2, denoted byterminals (a,b) and (c,d) respectively. It has an impedance matrix Z withparameters denoted by Zi j. A 1Wresistor is connected in series with the network

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

(A)Z

Z

Z

Z

1 1

111

21

12

22

+ +

+e o (B)

Z

Z

Z

Z

1

111

21

12

22

+

+e o

(C)Z

Z

Z

Z

111

21

12

22

+e o (D)

Z

Z

Z

Z

1

111

21

12

22

+

+e o


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YEAR 2009 ONE MARK

MCQ 1.2.33 The current through the 2 kWresistance in the circuit shown is

(A) 0 mA (B) 1 mA

(C) 2 mA (D) 6 mAMCQ 1.2.34 How many 200 W/220 V incandescent lamps connected in series would consume

the same total power as a single 100 W/220 V incandescent lamp ?(A) not possible (B) 4

(C) 3 (D) 2

YEAR 2009 TWO MARKS

MCQ 1.2.35 In the figure shown, all elements used are ideal. For time ,t S0< 1remained closed

and S2open. At ,t S0 1= is opened and S2is closed. If the voltage Vc2across the

capacitor C2 at t 0= is zero, the voltage across the capacitor combination att 0= +will be

(A) 1 V (B) 2 V

(C) 1.5 V (D) 3 V

MCQ 1.2.36 The equivalent capacitance of the input loop of the circuit shown is

(A) 2 Fm (B) 100 Fm

(C) 200 Fm (D) 4 Fm

MCQ 1.2.37 For the circuit shown, find out the current flowing through the 2 Wresistance.Also identify the changes to be made to double the current through the 2 Wresistance.


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(A) (5 ; 30 )VA Put VS= (B) (2 ; 8 )VA Put VS=

(C) (5 ; 10 )IA Put AS= (D) (7 ; 12 )IA Put AS=

Statement for Linked Answer Question 38 and 39 :

MCQ 1.2.38 For the circuit given above, the Thevenins resistance across the terminals A and

B is(A) 0.5 kW (B) 0.2 kW

(C) 1kW (D) 0.11 kW

MCQ 1.2.39 For the circuit given above, the Thevenins voltage across the terminals A and Bis(A) 1.25 V (B) 0.25 V

(C) 1 V (D) 0.5 V

YEAR 2008 ONE MARK

MCQ 1.2.40 The number of chords in the graph of the given circuit will be

(A) 3 (B) 4

(C) 5 (D) 6

MCQ 1.2.41 The Thevenins equivalent of a circuit operation at 5w= rads/s, has

3.71 15.9V Voc += - % and 2.38 0.667Z j0 W= - . At this frequency, the minimal

realization of the Thevenins 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


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YEAR 2008 TWO MARKS

MCQ 1.2.42 The time constant for the given circuit will be

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

(C) 4 s (D) 9 s

MCQ 1.2.43 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

MCQ 1.2.44 Assuming ideal elements in the circuit shown below, the voltage Vabwill be

(A) 3 V- (B) 0 V

(C) 3 V (D) 5 V


The current ( )i tsketched in the figure flows through a initially uncharged 0.3 nFcapacitor.

MCQ 1.2.45 The charge stored in the capacitor at 5t sm= , will be


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(A) 8 nC (B) 10 nC

(C) 13 nC (D) 16 nC

MCQ 1.2.46 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 thecapacitor after 1 sm will approximately be

(A) 18.8 V (B) 23.5 V

(C) 23.5 V- (D) 30.6V-

MCQ 1.2.47 In the circuit shown in the figure, the value of the current iwill be given by

(A) 0.31 A (B) 1.25 A

(C) 1.75 A (D) 2.5 A

MCQ 1.2.48 Two point charges 10Q C1 m= and 20Q2= mC are placed at coordinates (1,1,0)

and ( , , )1 1 0- - respectively. The total electric flux passing through a plane0z 2= will be

(A) 7.5 Cm (B) 13.5 Cm

(C) 15.0 Cm (D) 22.5 Cm

MCQ 1.2.49 A capacitor consists of two metal plates each 500 500# mm2and spaced 6 mmapart. The space between the metal plates is filled with a glass plate of 4 mmthickness 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 10012

#e =- F/m )

(A) 983.3 pF (B) 1475 pF

(C) 637.7 pF (D) 9956.25 pF

MCQ 1.2.50 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 coilcorresponding to a magnetizing current of 3 A will be

(Given that 4 1007#m p=

- H/m)(A) 37.68 Hm (B) 113.04 Hm

(C) 3.768 Hm (D) 1.1304 Hm

YEAR 2007 ONE MARK

MCQ 1.2.51 Divergence of the vector field

( , , ) ( ) ( ) ( )cos cos sinV x y z x xy y i y xy j z x y k 2 2 2=- + + + + +t t tis

(A) cosz z22

(B) sin cosxy z z 22

+(C) sin cosx xy z - (D) None of these

YEAR 2007 TWO MARKS

MCQ 1.2.52 The state equation for the current I1in the network shown below in terms of the


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voltage VXand the independent source V, is given by

(A) 1.4 3.75dt

dIV I V

45

X1

1=- - + (B) 1.4 3.75dt

dII V

45VX1 1= - -

(C) 1.4 3.75dt

dIV I V

45

X1

1=- + + (D) 1.4 3.75dt

dII V

45VX1 1=- + -

MCQ 1.2.53 The R-L-C series circuit shown in figure is supplied from a variable frequencyvoltage source. The admittance - locus of the R-L-C network at terminals AB forincreasing frequency wis

MCQ 1.2.54 In the circuit shown in figure. Switch SW1is initially closed and SW2 is open.The inductor L carries a current of 10 A and the capacitor charged to 10 V with

polarities as indicated. SW2is closed at t 0= and SW1is opened at t 0= . The

current through Cand the voltage across L at ( )t 0= + is


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(A) 55 A, 4.5 V (B) 5.5 A, 45 V

(C) 45 A, 5.5 A (D) 4.5 A, 55 V

MCQ 1.2.55 In the figure given below all phasors are with reference to the potential at point'' ''O. The locus of voltage phasorVYXas Ris varied from zero to infinity is shownby

MCQ 1.2.56 A 3 V DC supply with an internal resistance of 2 Wsupplies a passive non-linearresistance characterized by the relation V INL NL

2= . The power dissipated in the

non linear resistance is(A) 1.0 W (B) 1.5 W

(C) 2.5 W (D) 3.0 W

MCQ 1.2.57 The matrix A given below in the node incidence matrix of a network. The columns

correspond to branches of the network while the rows correspond to nodes. Let[ ..... ]V V V V T1 2 6= denote the vector of branch voltages while [ ..... ]I i i i T

1 2 6= that

of branch currents. The vector [ ]E e e e eT1 2 3 4= denotes the vector of node voltagesrelative to a common ground.

1

0

1

0

1

1

0

0

1

0

0

1

0

1

0

1

0

1

1

0

0

0

1

1

-

-

-

-

- -

R

T

SSSSS

V

X

WWWWW

Which of the following statement is true ?

(A) The equations ,V V V V V V 0 01 2 3 3 4 5- + = + - = are KVL equations for the

network for some loops

(B) The equations ,V V V V V V 0 01 3 6 4 5 6- - = + - = are KVL equations for thenetwork for some loops

(C) E AV=

(D) AV 0= are KVI equations for the network


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(C) 1 , 0,V 3 (D) 10 , , 10V 3 W

MCQ 1.2.63 In the circuit shown in the figure, the current source 1I A= , the voltage source

5 , 1 , 1 , 1V R R R L L L C C V H F1 2 3 1 2 3 1 2W= = = = = = = = =

The currents (in A) through R3and through the voltage source Vrespectivelywill be(A) 1, 4 (B) 5, 1

(C) 5, 2 (D) 5, 4

MCQ 1.2.64 The parameter type and the matrix representation of the relevant two portparameters that describe the circuit shown are

(A) z parameters,00

00= G

(B) h parameters,10

01= G

(C) h parameters,0

0

0

0= G(D) z parameters,

1

0

0

1= G

MCQ 1.2.65 The circuit shown in the figure is energized by a sinusoidal voltage sourceV1at afrequency which causes resonance with a current of I.

The phasor diagram which is applicable to this circuit is


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MCQ 1.2.66 An ideal capacitor is charged to a voltage V0and connected at t 0= across anideal inductor L. (The circuit now consists of a capacitor and inductor alone). If

we letLC

10w = , the voltage across the capacitor at time t 0> is given by

(A) V0 (B) ( )cosV t0 0w

(C) ( )sin tV0 0w (D) ( )cosV e tt0 00 ww-

MCQ 1.2.67 An energy meter connected to an immersion heater (resistive) operating on anAC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. Theheater is removed from the supply and now connected to a 400 V peak square

wave source of 150 Hz. The power in kW dissipated by the heater will be(A) 3.478 (B) 1.739

(C) 1.540 (D) 0.870

MCQ 1.2.68 Which of the following statement holds for the divergence of electric and magnetic

flux densities ?(A) Both are zero

(B) These are zero for static densities but non zero for time varying densities.

(C) It is zero for the electric flux density

(D) It is zero for the magnetic flux density

YEAR 2005 ONE MARK

MCQ 1.2.69 In the figure given below the value of Ris

(A) 2.5 W (B) 5.0 W

(C) 7.5 W (D) 10.0 W

MCQ 1.2.70 The RMS value of the voltage ( )u t ( )cos t3 4 3= + is

(A) 17 V (B) 5 V

(C) 7 V (D) (3 2 )2 V+

MCQ 1.2.71 For the two port network shown in the figure the Z-matrix is given by


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(A)Z

Z Z

Z Z

Z

1

1 2

1 2

2+

+= G (B)

Z

Z Z

Z

Z

1

1 2

1

2+= G

(C)Z

Z

Z

Z Z

1

2

2

1 2+= G (D)

Z

Z

Z

Z Z

1

1

1

1 2+= G

MCQ 1.2.72 In the figure given, for the initial capacitor voltage is zero. The switch is closed

at t 0= . The final steady-state voltage across the capacitor is

(A) 20 V (B) 10 V

(C) 5 V (D) 0 V

MCQ 1.2.73 If Evis the electric intensity, ( )E4 4 #v is equal to

(A) Ev (B) Ev

(C) null vector (D) Zero

YEAR 2005 TWO MARKS


A coil of inductance 10 H and resistance 40 Wis connected as shown in the figure.

After the switch S has been in contact with point 1 for a very long time, it ismoved to point 2 at, t 0= .

MCQ 1.2.74

If, at t = 0+

, the voltage across the coil is 120 V, the value of resistance Ris

(A) 0 W (B) 20 W

(C) 40 W (D) 60 W

MCQ 1.2.75 For the value as obtained in (a), the time taken for 95% of the stored energy tobe dissipated is close to(A) 0.10 sec (B) 0.15 sec

(C) 0.50 sec (D) 1.0 sec

MCQ 1.2.76 The RL circuit of the figure is fed from a constant magnitude, variable frequency


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sinusoidal voltage sourceVin. At 100 Hz, the Rand L elements each have a voltage

drop RM Sm .If the frequency of the source is changed to 50 Hz, then new voltagedrop across Ris

(A)85 uRMS (B) 3

2 uRMS

(C)58 uRMS (D) 2

3 uRMS

MCQ 1.2.77

For the three-phase circuit shown in the figure the ratio of the currents : :I I IR Y Bis given by

(A) : :1 1 3 (B) : :1 1 2

(C) : :1 1 0 (D) : : /1 1 3 2

MCQ 1.2.78 The circuit shown in the figure is in steady state, when the switch is closed att 0= .Assuming that the inductance is ideal, the current through the inductor at

t 0= +equals

(A) 0 A (B) 0.5 A

(C) 1 A (D) 2 A

MCQ 1.2.79 In the given figure, the Thevenins equivalent pair (voltage, impedance), as seenat the terminals P-Q, is given by

(A) (2 5 )V, W (B) (2 , 7.5 )V W

(C) (4 , 5 )V W (D) (4 , 7.5 )V W


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MCQ 1.2.80 The charge distribution in a metal-dielectric-semiconductor specimen is shown in

the figure. The negative charge density decreases linearly in the semiconductor asshown. The electric field distribution is as shown in

YEAR 2004 ONE MARK

MCQ 1.2.81 The value of Z in figure which is most appropriate to cause parallel resonance at

500 Hz is

(A) 125.00 mH (B) 304.20 Fm

(C) 2.0 Fm (D) 0.05 Fm

MCQ 1.2.82 A parallel plate capacitor is shown in figure. It is made two square metal platesof 400 mm side. The 14 mm space between the plates is filled with two layers of

dielectrics of 4re = , 6 mm thick and 2re = , 8 mm thick. Neglecting fringing offields at the edge the capacitance is


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(A) 1298 pF (B) 944 pF

(C) 354 pF (D) 257 pF

MCQ 1.2.83 The inductance of a long solenoid of length 1000 mm wound uniformly with 3000

turns on a cylindrical paper tube of 60 mm diameter is(A) 3.2 mH (B) 3.2 mH

(C) 32.0 mH (D) 3.2 H

YEAR 2004 TWO MARKS

MCQ 1.2.84 In figure, the value of the source voltage is

(A) 12 V (B) 24 V(C) 30 V (D) 44 V

MCQ 1.2.85 In figure, Ra, Rband Rcare 20 W, 20 Wand 10 Wrespectively. The resistances R1, R2and R3in Wof an equivalent star-connection are

(A) 2.5, 5, 5 (B) 5, 2.5, 5

(C) 5, 5, 2.5 (D) 2.5, 5, 2.5

MCQ 1.2.86 In figure, the admittance values of the elements in Siemens are0.5 0, 0 1.5, 0 0.3Y j Y j Y j R L C= + = - = + respectively. The value of Ias a phasor

when the voltage Eacross the elements is 10 0 V+ %

(A) 1.5 0.5j+ (B) j5 18-


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(C) .5 .j0 1 8+ (D) 5 j12-

MCQ 1.2.87 In figure, the value of resistance Rin Wis

(A) 10 (B) 20

(C) 30 (D) 40

MCQ 1.2.88 In figure, the capacitor initially has a charge of 10 Coulomb. The current in thecircuit one second after the switch S is closed will be

(A) 14.7 A (B) 18.5 A

(C) 40.0 A (D) 50.0 A

MCQ 1.2.89 The rms value of the current in a wire which carries a d.c. current of 10 A and a

sinusoidal alternating current of peak value 20 A is(A) 10 A (B) 14.14 A

(C) 15 A (D) 17.32 A

MCQ 1.2.90 The Z-matrix of a 2-port network as given by.

.

.

.

0 9

0 2

0 2

0 6= G

The element Y22of the corresponding Y-matrix of the same network is given by

(A) 1.2 (B) 0.4

(C) .0 4- (D) 1.8

YEAR 2003 ONE MARK

MCQ 1.2.91 Figure Shows the waveform of the current passing through an inductor ofresistance 1 Wand inductance 2 H. The energy absorbed by the inductor in the

first four seconds is

(A) 144 J (B) 98 J

(C) 132 J (D) 168 J

MCQ 1.2.92 A segment of a circuit is shown in figure 5 , 4 2sinv V v t R c= = .The voltage vL is


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given by

(A) 3 8 cost2- (B) 2sint32

(C) sint16 2 (D) 16 2cost

MCQ 1.2.93 In the figure, 10 60 , 10 60 , 50 53.13 .Z Z Z1 2 3+ + += - = =% % %

Thevenin impedance seen form X-Y is

(A) .56 66 45+

%

(B) 60 30+

%

(C) 70 30+ % (D) .34 4 65+ %

MCQ 1.2.94 Two conductors are carrying forward and return current of +I and I- as shown

in figure. The magnetic field intensity Hat point P is

(A)dIY

p (B)

dI X

p

(C)d

I Y2p

(D)d

I X2p

MCQ 1.2.95 Two infinite strips of width w m in x-direction as shown in figure, are carryingforward and return currents of +I and I- in the z- direction. The strips areseparated by distance of x m. The inductance per unit length of the configuration

is measured to be L H/m. If the distance of separation between the strips in snowreduced to x/2 m, the inductance per unit length of the configuration is


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(A) L2 H/m (B) /L 4 H/m

(C) /L 2 H/m (D) L4 H/m

YEAR 2003 TWO MARKS

MCQ 1.2.96 In the circuit of figure, the magnitudes of VL and VCare twice that of VR. Given

that 50f Hz= , the inductance of the coil is

(A) 2.14 mH (B) 5.30 H(C) 31.8 mH (D) 1.32 H

MCQ 1.2.97 In figure, the potential difference between points P and Q is

(A) 12 V (B) 10 V

(C) 6 V- (D) 8 V

MCQ 1.2.98 Two ac sources feed a common variable resistive load as shown in figure. Underthe maximum power transfer condition, the power absorbed by the load resistance

RL is


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(A) 2200 W (B) 1250 W

(C) 1000 W (D) 625 W

MCQ 1.2.99 In figure, the value of Ris

(A) 10 W (B) 18 W(C) 24 W (D) 12 W

MCQ 1.2.100 In the circuit shown in figure, the switch S is closed at time (t = 0). The voltage

across the inductance at t 0= +, is

(A) 2 V (B) 4 V

(C) 6- V (D) 8 V

MCQ 1.2.101 The h-parameters for a two-port network are defined by

E

I

h

h

h

h

I

E

1

2

11

21

12

22

1

2== = =G G G

For the two-port network shown in figure, the value of h12is given by

(A) 0.125 (B) 0.167

(C) 0.625 (D) 0.25

MCQ 1.2.102 A point charge of +I nC is placed in a space with permittivity of .8 85 10 12# - F/mas shown in figure. The potential difference VPQbetween two points P and Q at

distance of 40 mm and 20 mm respectively fr0m the point charge is

(A) 0.22 kV (B) 225- V


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(C) .2 24- kV (D) 15 V

MCQ 1.2.103 A parallel plate capacitor has an electrode area of 100 mm2, with spacing of

0.1 mm between the electrodes. The dielectric between the plates is air witha permittivity of .8 85 10 12# - F/m. The charge on the capacitor is 100 V. Thestored energy in the capacitor is

(A) 8.85 pJ (B) 440 pJ

(C) 22.1 nJ (D) 44.3 nJ

MCQ 1.2.104 A composite parallel plate capacitor is made up of two different dielectric materialwith different thickness (t1and t2) as shown in figure. The two different dielectric

materials are separated by a conducting foil F. The voltage of the conducting foilis

(A) 52 V (B) 60 V

(C) 67 V (D) 33 V

YEAR 2002 ONE MARK

MCQ 1.2.105

A current impulse, ( )t

5d

, is forced through a capacitorC

. The voltage , ( )v tc ,across the capacitor is given by

(A) 5t (B) ( )u t C5 -

(C)Ct5 (D)

( )C

u t5

MCQ 1.2.106 The graph of an electrical network has Nnodes and Bbranches. The number of

links L , with respect to the choice of a tree, is given by(A) B N 1- + (B) B N+

(C) N B 1- + (D) N B2 1- -

MCQ 1.2.107 Given a vector field F, the divergence theorem states that

(A) d dVF S FS V

: 4:=# # (B) d dVF S FVS

: 4#= ##

(C) d dVF S FS V

# 4:=# # (D) d dVF S FS V

# 4:=# #

MCQ 1.2.108 Consider a long, two-wire line composed of solid round conductors. The radius

of both conductors. The radius of both conductors is 0.25 cm and the distancebetween their centres is 1 m. If this distance is doubled, then the inductance perunit length(A) doubles

(B) halves(C) increases but does not double

(D) decreases but does not halve

MCQ 1.2.109 A long wire composed of a smooth round conductor runs above and parallel

to the ground (assumed to be a large conducting plane). A high voltage exists


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