GATE Electronics and Communication Topicwise Solved Papers 1996-2013

GATE
ELECTRONICS & COMMUNICATION
Topicwise Solved Paper
Year 2013- 1996
By RK Kanodia & Ashish Murolia
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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia Page 1
d y t
dy t
2 y t 2
dy
=- =
dy
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UNIT 1
ENGINEERING MATHEMATICS
2013 ONE MARK
1.1 The maximum value of q until which the approximation sinq . q
holds to within 10% error is
(A) 10c (B) 18c
(C) 50c (D) 90c
1.2 The minimum eigen value of the following matrix is
3
5
2
5
12
7
2
7
5
R
SSSS
T
V
WWWW
X
(A) 0 (B) 1
(C) 2 (D) 3
4
1.3 A polynomial f (x) a4x a 3
x a x a x a
3
2
2
= + + + 1 - 0 with all coefficients
positive has
(A) no real roots
(B) no negative real root
(C) odd number of real roots
(D) at least one positive and one negative real root
2013 TWO MARKS
1.4 Let A be an m#n matrix and B an n#m matrix. It is given that
determinant ^Im + ABh = determinant ^In + BAh, where Ik is the
k#k identity matrix. Using the above property, the determinant of
the matrix given below is
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
R
SSSSS
S
T
V
WWWWW
W
X
(A) 2 (B) 5
(C) 8 (D) 16
2012 ONE MARK
1.5 With initial condition x(1) = 0.5, the solution of the differential
equation
t dt
dx + x = t , is
(A) x t 2
= - 1 (B) x = t 2- 1
2
= (D) x t2
(C) x t
2
2
=
= 1
2 + - + .
1.6 Given f (z) z 1 z
3
If C is a counter clockwise path in the z -plane such that
z + 1 = 1, the value of 1
2 j f (z)dz
C p # is
(A) -2 (B) -1
(C) 1 (D) 2
1.7 If x = -1, then the value of xx is
(A) e-p/2 (B) ep/2
(C) x (D) 1
2012 TWO MARKS
1.8 Consider the differential equation
( ) ( )
( )
dt
dt
2
+ + = d(t)
with y(t) 2 and dt 0
t
t
0
0
=
=
-
-
The numerical value of dt
t=0+
is
(A) -2 (B) -1
(C) 0 (D) 1
1.9 The direction of vector A is radially outward from the origin, with
A = krn . where r2 = x2+ y2+ z2 and k is a constant. The value of
n for which d:A = 0 is
(A) -2 (B) 2
(C) 1 (D) 0
1.10 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
1.11 The maximum value of f (x) = x3- 9x2+ 24x + 5 in the interval [1,6]
is
(A) 21 (B) 25
(C) 41 (D) 46
1.12 Given that
- -
> H = > H, the value of A3 is
5
2
3
0
A and I
1
0
0
= 1
(A) 15A+ 12I (B) 19A+ 30I
(C) 17A+ 15I (D) 17A+ 21I
2011 ONE MARK
1.13 Consider a closed surface S surrounding volume V . If rv is the
position vector of a point inside S , with nt the unit normal on S ,
the value of the integral 5r n dS
## v $ t is
S
(A) 3V (B) 5V
(C) 10V (D) 15V
dy = ky y = c is
1.14 The solution of the differential equation dx , (0)
(A) x = ce-ky (B) x = kecy
(C) y = cekx (D) y = ce-kx
1.15 The value of the integral
# - + where c is the circle
z dz
4 5
3 4
(z z )
c
2+ +
z = 1 is given by
(A) 0 (B) 1/10
(C) 4/5 (D) 1

+ + =
+ + =
+ + =
x y z
x y y
x y z
6
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by RK Kanodia
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v = t + 2 t , then A dl
# v $ v over the path shown in the figure
2
-
- 1 and 1 (B) , 2
- 1 and -1
1 and 2
3
dy x
- 1 3 s
+ 1
; E. If limf (t) 1
3 2 =
+ + -
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2011 TWO MARKS
1.16 A numerical solution of the equation f (x) + x - 3 = 0 can be
obtained using Newton- Raphson method. If the starting value is
x = 2 for the iteration, the value of x that is to be used in the next
step is
(A) 0.306 (B) 0.739
(C) 1.694 (D) 2.306
1.17 The system of equations
4 6 20
4
l m
has NO solution for values of l and μ given by
(A) l = 6, m = 20 (B) l = 6, m =Y 20
(C) l =Y 6, m = 20 (D) l =Y 6, m = 20
1.18 A fair dice is tossed two times. The probability that the second toss
results in a value that is higher than the first toss is
(A) 2/36 (B) 2/6
(C) 5/12 (D) 1/2
2010 ONE MARKS
1.19 The eigen values of a skew-symmetric matrix are
(A) always zero (B) always pure imaginary
(C) either zero or pure imaginary (D) always real
1.20 The trigonometric Fourier series for the waveform f (t) shown below
contains
(A) only cosine terms and zero values for the dc components
(B) only cosine terms and a positive value for the dc components
(C) only cosine terms and a negative value for the dc components
(D) only sine terms and a negative value for the dc components
1.21 A function n(x) satisfied the differential equation
d n x
n x
( ) ( )
dx
0 2
L
2
2 - =
where L is a constant. The boundary conditions are :n(0) = K
and n(3) = 0. The solution to this equation is
(A) n(x) = Kexp(x/L) (B) n(x) = Kexp(- x/ L)
(C) n(x) = K2exp(- x/L) (D) n(x) = Kexp(- x/L)
2010 TWO MARKS
1.22 If ey = x1/x , then y has a
(A) maximum at x = e (B) minimum at x = e
(C) maximum at x = e-1 (D) minimum at x = e-1
1.23 A fair coin is tossed independently four times. The probability of the
event “the number of time heads shown up is more than the number
of times tail shown up”
(A) 1/16 (B) 1/3
(C) 1/4 (D) 5/16
1.24 If A xyax x ay
C
is
(A) 0 (B)
3
(C) 1 (D) 2 3
1.25 The residues of a complex function
( )
( )( )
x z
z z z z
1 2
= 1 2 - -
at its poles are
(A) 1
2,
2
1
2
(C) 2, 1
-3 (D) , 2
1 - 1 and 2
1.26 Consider differential equation
( )
dx ( )
- y x = x , with the initial
condition y(0) = 0. Using Euler’s first order method with a step
size of 0.1, the value of y(0.3) is
(A) 0.01 (B) 0.031
(C) 0.0631 (D) 0.1
1.27 Given ( )
( )
f t L
s 4 s k 3
s
t
=
"3
, then the value
of k is
(A) 1 (B) 2
(C) 3 (D) 4
2009 ONE MARK
1.28 The order of the differential equation
2 3
+ + 4 = - c m
d y
dt
dy y et
dt
2
is
(A) 1 (B) 2
(C) 3 (D) 4
1.29 A fair coin is tossed 10 times. What is the probability that only the
first two tosses will yield heads?
(A) 2
10 1
c 1 2
m (B) C 2
2
2
b l
1 10
c m (D) C 2
(C) 2
10 1
2
10
b l

dx t
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= + -1, then
1.30 If f (z) c0 c1z
# + f z
is given by
( )
z
dz
1
unit circle
(A) 2pc1 (B) 2p(1 + c0)
(C) 2pjc1 (D) 2p(1 + c0)
2009 TWO MARKS
1.31 The Taylor series expansion of sin
x
x
- p at x = p is given by
(A)
+ ( - p )
2 + (B)
!
...
x
1
3
- - ( - p )
2 +
!
...
x
1
3
(C)
- ( - p )
2 + (D)
!
...
x
1
3
- + ( - p )
2 +
!
...
x
1
3
1.32 Match each differential equation in Group I to its family of solution
curves from Group II
Group I Group II
A.
dy
dx
= y 1. Circles
x
B.
dy
dx
=-y 2. Straight lines
x
C.
dy
dx
y x
= 3. Hyperbolas
D.
dy
dx
y x
=-
(A) A - 2, B - 3,C - 3,D - 1
(B) A - 1, B - 3,C - 2,D - 1
(C) A - 2, B - 1,C - 3,D - 3
(D) A - 3, B - 2,C - 1,D - 2
1.33 The Eigen values of following matrix are
1
3
0
3
1
0
5
6
3
R
-
- -
SSSS
T
V
WWWW
X
(A) 3, 3 + 5j, 6 - j (B) -6 + 5j, 3 + j, 3 - j
(C) 3 + j, 3 - j, 5 + j (D) 3, - 1 + 3j, - 1 - 3j
2008 ONE MARKS
1.34 All the four entries of the 2 # 2 matrix P
p
p
p
p
11
21
12
22
= = G are nonzero,
and one of its eigenvalue is zero. Which of the following statements
is true?
(A) p11p12- p12p21 = 1 (B) p11p22- p12p21 =- 1
(C) p11p22- p12p21 = 0 (D) p11p22+ p12p21 = 0
1.35 The system of linear equations
4x + 2y = 7
2x + y = 6 has
(A) a unique solution
(B) no solution
(C) an infinite number of solutions
(D) exactly two distinct solutions
1.36 The equation sin(z) = 10 has
(A) no real or complex solution
(B) exactly two distinct complex solutions
(C) a unique solution
(D) an infinite number of complex solutions
1.37 For real values of x , the minimum value of the function
f (x) = exp(x) + exp(- x) is
(A) 2 (B) 1
(C) 0.5 (D) 0
1.38 Which of the following functions would have only odd powers of x
in its Taylor series expansion about the point x = 0 ?
(A) sin(x3) (B) sin(x2)
(C) cos (x3) (D) cos (x2)
1.39 Which of the following is a solution to the differential equation
( )
( )
dt
+ 3x t = 0 ?
(A) x(t) = 3e-t (B) x(t) = 2e-3t
(C) x(t) =- 3 t 2 (D) x(t) = 3t2
2
2008 TWO MARKS
1.40 The recursion relation to solve x = e x - using Newton - Raphson
method is
= n +
(A) xn e
x
1
- (B) x x e n n
x
= - n +
1
-
-
x
= +
e
n n n 1 x
(C) x (1 x )
+ + -
1 n
e
(D)
- - -
x e ( 1 x )
1
x
+ -
x e
n
n
x
n
x
n
1
2
n
n
=
-
-
1.41 The residue of the function f (z)
1
2 2 =
+ -
(z 2) (z 2)
at z = 2 is
(A)
- 1 (B)
32
- 1
16
(C)
1 (D)
16
1
32
1.42 Consider the matrix P
0
2
1
= =- -3G. The value of ep is
(A)
2 1
2 1
e e
e e
e e
e e
2 3
2 2 5
1 2
2 1
-
-
-
-
- -
- -
- -
> - - H (B)
- -
-
1 1
1 2
+
-
e e
e e
- -
2 1
1 2
-
+
e e
2
3 2
> - - H
2 4 e e
(C)
e e
e e
e e
e
5
2 6
3
4 6
2 1
2 1
1 2
2 1
-
-
-
+
- -
- -
- -
> - - H (D)
1 -
2
1 2
e e
e e
1 2
1 2
e e
e e
2
2 2 2
- +
-
- +
- -
- -
- -
> - - H
1.43 In the Taylor series expansion of exp(x) + sin(x) about the point
x = p, the coefficient of (x - p)2 is
(A) exp(p) (B) 0.5 exp(p)
(C) exp(p) + 1 (D) exp(p) - 1
1.44 The value of the integral of the function g(x,y) = 4x3+ 10y4 along
the straight line segment from the point (0,0) to the point (1,2) in
the x - y plane is
(A) 33 (B) 35
(C) 40 (D) 56
1.45 Consider points P and Q in the x - y plane, with P = (1,0) and
Q
# + along the semicircle
Q = (0,1). The line integral 2 (xdx ydy)
P
with the line segment PQ as its diameter
(A) is -1
(B) is 0

by RK Kanodia
Now in 3 Volume
2 - # is
ds
1
1
(C) is 1
(D) depends on the direction (clockwise or anit-clockwise) of the
semicircle
2007 ONE MARK
1.46 The following plot shows a function which varies linearly with x .
The value of the integral I ydx
1
2
= # is
(A) 1.0 (B) 2.5
(C) 4.0 (D) 5.0
1.47 For x << 1, coth (x) can be approximated as
(A) x (B) x2
(C)
1 (D)
x
1
2
x
sin 2
0 q
1.48 lim
q
q"
b l
is
(A) 0.5 (B) 1
(C) 2 (D) not defined
1.49 Which one of following functions is strictly bounded?
(A) 1/x2 (B) ex
(C) x2 (D) e x- 2
1.50 For the function e x - , the linear approximation around x = 2 is
(A) (3 - x)e-2 (B) 1 - x
(C) 63 + 3 2 - (1 - 2)x @e-2 (D) e-2
2007 TWO MARKS
1.51 The solution of the differential equation k
2 d y y y
dx
2
2
= - 2 under the
boundary conditions
(i) y = y1 at x = 0 and
(ii) y = y2 at x = 3, where k,y1 and y2 are constants, is
x y = 1- 2 a- 2 k+ 2 (B) y (y y )exp k
(A) y (y y )exp
k
x y = 2- 1 a- k+ 1
x y = ^ 1- 2h a k+ 1 (D) y y y exp k
(C) y y y sinh k
x y = ^ 1- 2h a- k+ 2
1.52 The equation x3- x2+ 4x - 4 = 0 is to be solved using the Newton
- Raphson method. If x = 2 is taken as the initial approximation of
the solution, then next approximation using this method will be
(A) 2/3 (B) 4/3
(C) 1 (D) 3/2
1.53 Three functions f1(t), f2(t) and f3(t) which are zero outside the interval
[0, T] are shown in the figure. Which of the following statements is
correct?
(A) f1(t) and f2(t) are orthogonal
(B) f1(t) and f3(t) are orthogonal
(C) f2(t) and f3(t) are orthogonal
D) f1(t) and f2(t) are orthonormal
1.54 If the semi-circular control D of radius 2 is as shown in the figure,
then the value of the integral
(s )
D
(A) jp (B) -jp
(C) -p (D) p
1.55 It is given that , ... X X XM
1 2 at M non-zero, orthogonal vectors.
The dimension of the vector space spanned by the 2M vectors
X1,X2,...XM, - X1, - X2,... - XM is
(A) 2M (B) M+ 1
(C) M
(D) dependent on the choice of , ,... X X XM
1 2
1.56 Consider the function f (x) = x2- x - 2. The maximum value of f (x)
in the closed interval [- 4, 4] is
(A) 18 (B) 10
(C) -225 (D) indeterminate
1.57 An examination consists of two papers, Paper 1 and Paper 2. The
probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2.
Given that a student has failed in Paper 2, the probability of failing
in Paper 1 is 0.6. The probability of a student failing in both the
papers is
(A) 0.5 (B) 0.18
(C) 0.12 (D) 0.06

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2006 ONE MARK
1.58 The rank of the matrix
1
1
1
1
1
1
1
0
1
-
R
SSSS
T
V
WWWW
X
is
(A) 0 (B) 1
(C) 2 (D) 3
1.59 4#4# P, where P is a vector, is equal to
(A) P P 2P #4# -4
(B) 42P +4 (4# P)
(C) 42P +4# P
(D) 4(4$ P) -42P
1.60 ##(4# P) $ ds , where P is a vector, is equal to
(A) # P $ dl
(B) # 4#4# P $ dl
(C) # 4# P $ dl
(D) ### 4$ Pdv
1.61 A probability density function is of the form
p(x) = Ke-a x ,x ! (-3,3)
The value of K is
(A) 0.5 (B) 1
(C) 0.5a (D) a
1.62 A solution for the differential equation xo (t) + 2x(t) = d(t) with
initial condition x(0-) = 0 is
(A) e-2tu(t) (B) e2tu(t)
(C) e-tu(t) (D) etu(t)
2006 TWO MARKS
1.63 The eigenvalue and the corresponding eigenvector of 2 # 2 matrix
are given by
Eigenvalue Eigenvector
l1 = 8 v
1
1
1 == G
l2 = 4 v
1
1
2 =
= G
-
The matrix is
(A)
6
2
2
= 6G (B)
4
6
6
= 4G
(C)
2
4
4
= 2G (D)
4
8
8
= 4G
1.64 For the function of a complex variable W = lnZ (where, W = u + jv
and Z = x + jy, the u = constant lines get mapped in Z -plane as
(A) set of radial straight lines (B) set of concentric circles
(C) set of confocal hyperbolas (D) set of confocal ellipses
1.65 The value of the constant integral
1
# dz is positive sense is
z 4
z j
2
2
+
- =
p (B)
(A) j
2
2 p
-
- p (D)
(C) j
2
2 p
q q p # is given by
1.66 The integral sin3 d
0
(A)
1 (B)
2
2
3
(C)
3 4
(D)
8
3
1.67 Three companies X,Y and Z supply computers to a university. The
percentage of computers supplied by them and the probability of
those being defective are tabulated below
Company % of Computer Sup-plied
Probability of being
supplied defective
X 60% 0.01
Y 30% 0.02
Z 10% 0.03
Given that a computer is defective, the probability that was sup-plied
by Y is
(A) 0.1 (B) 0.2
(C) 0.3 (D) 0.4
1.68 For the matrix
4
2
2
= 4G the eigenvalue corresponding to the eigenvector
101
=101G is
(A) 2 (B) 4
(C) 6 (D) 8
1.69 For the differential equation
2
d y k y 0 2
dx
+ 2 = the boundary conditions
are
(i) y = 0 for x = 0 and (ii) y = 0 for x = a
The form of non-zero solutions of y (where m varies over all inte-gers)
are
=/ p (B) y A cos
(A) y A sin
m x
a
m
m
=/ p
m x
a
m
m
= / m
p (D) y Ame a
(C) y Amx a
m
= / - p
m x
m
1.70 As x increased from -3 to 3, the function f (x)
e
1 e
x
x
=
+
(A) monotonically increases
(B) monotonically decreases
(C) increases to a maximum value and then decreases
(D) decreases to a minimum value and then increases
2005 ONE MARK
1.71 The following differential equation has
2 3 2 c m+ c m + + = x
3 d y
4 dy y 2 2
dt
dt
(A) degree = 2, order = 1 (B) degree = 1, order = 2
(C) degree = 4, order = 3 (D) degree = 2, order = 3
1.72 A fair dice is rolled twice. The probability that an odd number will
follow an even number is
(A) 1/2 (B) 1/6
(C) 1/3 (D) 1/4

d y
5 dy 6y 0 2
by RK Kanodia
Now in 3 Volume
-
= G and A
1 2 1
1
2
# c m is
= - 3
SSSSS
WWWWW
1
1
SSSSSS
WWWWWW
SSSSSS
WWWWWW
SSSSS
WWWWW
SSSSSS V
1
1
1
1
WWWWWW
1.73 A solution of the following differential equation is given by
dx
dx
2
- + =
(A) y = e2x + e-3x
(B) y = e2x + e3x
(C) y = e-2x + 33x
(D) y = e-2x + e-3x
2005 TWO MARKS
1.74 In what range should Re(s) remain so that the Laplace transform of
the function e(a+2)t+5 exits.
(A) Re(s) > a + 2
(B) Re(s) > a + 7
(C) Re(s) < 2
(D) Re(s) > a + 5
1.75 The derivative of the symmetric function drawn in given figure will
look like
1.76 Match the following and choose the correct combination:
Group I Group 2
E. Newton-Raphson method 1. Solving nonlinear equations
F. Runge-kutta method 2. Solving linear simultaneous
equations
G. Simpson’s Rule 3. Solving ordinary differential
equations
H. Gauss elimination 4. Numerical integration
5. Interpolation
6. Calculation of Eigenvalues
(A) E - 6, F - 1,G - 5, H - 3 (B) E - 1, F - 6,G - 4, H - 3
(C) E - 1, F - 3,G - 4, H - 2 (D) E - 5, F - 3,G - 4, H - 1
1.77 Given the matrix
-
= G, the eigenvector is
4
4
2
3
(A)
3
=2G (B)
4
=3G
(C)
2
=-1G (D)
-
= G
1
2
1.78 Let,
.
A
2
0
0 1
= 3
a
0 b
- = = G. Then (a + b) =
(A) 7/20 (B) 3/20
(C) 19/60 (D) 11/20
1.79 The value of the integral I exp x dx
2
8
p 0
(A) 1 (B) p
(C) 2 (D) 2p
1.80 Given an orthogonal matrix
A
1
1
1
0
1
1
1
0
1
1
0
1
1
1
0
1
= -
- -
R
T
V
X
6AAT @-1 is
(A)
0
0
0
0
0
0
0
0
0
0
0
0
4
4
2 1
2 1
R
T
V
X
(B)
0
0
0
0
0
0
0
0
0
0
0
0
2 1
2 1
2 1
2 1
R
T
V
X
(C)
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
R
T
V
X
(D)
0
0
0
0
0
0
0
0
0
0
0
0
4
4
4
4
R
T
X

SSSSS
S
WWWWW
W
R
SSSSS
S
SSSSS
S
WWWWW
W
WWWWW
W
SSSSS
S
WWWWW
W
SSSSS
S
WWWWW
W
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SOLUTIONS
1.1 Option (B) is correct.
Here, as we know
Limsin
0
q
q"
. 0
but for 10% error, we can check option (B) first,
= = p =
q 18 18 .
c c# 0 314
180
c
sinq = sin18c = 0.309
= 0 314 - 0 309 #100 = 0 49
. . % . % 0 309
% error .
Now, we check it for q = 50c
= = p =
q 50 50 .
c c# 0 873
180
c
sinq = sin50c = 0.77
0.77 0.873 . % = 0 873 12 25 - =-
% error .
so, the error is more than 10%. Hence, for error less than 10%,
q = 18c can have the approximation
sin q . q
1.2 Option (A) is correct.
For, a given matrix A 6 @ the eigen value is calculated as
A - lI = 0
where l gives the eigen values of matrix. Here, the minimum eigen
value among the given options is
l = 0
We check the characteristic equation of matrix for this eigen value
A - lI = A (for l = 0)
3
5
2
5
12
7
2
7
5
=
= 3^60 - 49h- 5^25 - 14h+ 2^35 - 24h
= 33 - 55 + 22
= 0
Hence, it satisfied the characteristic equation and so, the minimum
eigen value is
l = 0
1.3 Option (D) is correct.
Given, the polynomial
f^xh a4x 4
a x 3
a x 2
3
2
a x a
= + + + 1 - 0
Since, all the coefficients are positive so, the roots of equation is
given by
f^xh = 0
It will have at least one pole in right hand plane as there will be
least one sign change from â1h to â0h in the Routh matrix 1st col-umn.
Also, there will be a corresponding pole in left hand plane
i.e.; at least one positive root (in R.H.P)
and at least one negative root (in L.H.P)
Rest of the roots will be either on imaginary axis or in L.H.P
Consider the given matrix be
Im + AB
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
=
R
T
V
X
where m = 4 so, we obtain
AB
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
= -
T
R
T
V
X
V
X
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
=
R
T
V
X
1
1
1
1
=
R
T
V
X
61 1 1 1@
Hence, we get
A
1
1
1
1
=
R
SSSSS
S
T
V
WWWWW
W
X
, B = 81 1 1 1B
R
Therefore, BA = 81 1 1 1B 1
SSSSS
WWWWW
1
1
1
S
T
V
W
X
= 4
From the given property
Det Îm + ABh = DetÎm + BAh
& Det
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
R
SSSSS
S
T
V
WWWWW
W
X
Det
R
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
SSSSS
WWWWW
]]
]]
= + 4
S
T
V
W
X
Z
[

_
bb
bb
`
a
= 1 + 4
= 5
Note : Determinant of identity matrix is always 1.
dx + x = t
t dt
dx
dt
t x
+ = 1
dx + Px = Q (General form)
dt
1dt = # = # = =
Integrating factor, IF e Pdt e t elnt t
Solution has the form,
x#IF = # ^Q#IFhdt + C
x#t = # (1) (t)dt + C

2
2
= x t2
1
1
C p # = sum of the residues of the poles which lie
- +
z
1 3
= 1 + +
z z
1 1
( )( )
= + +
by RK Kanodia
Now in 3 Volume
dy t
dy
1 r A
2 r
1 krn
2
= 2 = 2 ( +
)
k n 2 rn
2
= + +1
= k(n + 2)rn-1 = 0 (given)
1
1
3 1 5
= +b l +b l +
2
1
1
df x
d f x
d f x
xt = t2 +
C
Taking the initial condition,
x(1) = 0.5
0.5 2 C
= 1 + & C = 0
So, xt t
2
& =
1.6 Option (C) is correct.
f (z) z 1 z
3
= 2 + - +
2 j f (z)dz
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 )
( )( )
lim
z z
1 3
2 2
1
z 1
+ - + = =
"-
1
C p # = 1
So 2 j f (z)dz
= p + p
x = -1 = i cos 2 i sin 2
So, x = ei 2p
= 2p ^ h & ei i
xx ei x
^ h = e 2-p
2 p
d y t
dy t
( ) ( )
( )
dt
dt
y t
2
2
2
+ + = d(t)
By taking Laplace transform with initial conditions
0 dy 2 sy s y 0 Y s
- - + - +
; E = 1
s Y(s) sy( ) dt [ ( ) ( )] ( )
t
2
0
=
& 6s2Y(s) + 2s - 0@+ 26sY(s) + 2@+ Y(s) = 1
Y(s) [s2+ 2s + 1] = 1 - 2s - 4
Y(s)
- -
s
2 1
2 3
2 =
+ +
s s
We know that, If, y(t) Y(s) L
then,
dy t
( )
dt
sY(s) y(0) L -
So, sY(s) - y(0)
- - +
s s
2 1
( 2 3
)
s s
2 2 =
+ +
( )
2 2
- - + + +
s s s s
2 3 2 4 2
(s 2 s 1
)
2
=
+ +
sY(s) - y(0)
s
s
+ 1
+
+ =
(s 1 ) ( s
) ( s
)
2
1
1
1
2 2 2 =
+
+
+
1
s 1 (s 1
)
1
2 = + +
+
Taking inverse Laplace transform
( )
dt
= e-tu(t) + te-tu(t)
At t = 0+, dt
t=0+
= e0+ 0 = 1
Divergence of A in spherical coordinates is given as
d:A ( )
r r
2
2
r r
2
2
( )
r
n + 2 = 0 & n =- 2
Probability of appearing a head is 1/2. If the number of required
tosses is odd, we have following sequence of events.
H, TTH, TTTTH, ...........
Probability P 2 2
2
.....
P
1 3
4
2
=
-
=
f (x) = x3- 9x2+ 24x + 5
( )
dx
df x
= 3x2- 18x + 24 = 0
&
( )
dx
= x2- 6x + 8 = 0 x = 4, x = 2
( )
dx
2
2
= 6x - 18
For x = 2,
( )
dx
12 18 6 < 0 2
2
= - =-
So at x = 2, f (x) will be maximum
f (x)
max
= (2)3- 9(2)2+ 24(2) + 5
= 8 - 36 + 48 + 5 = 25
Characteristic equation.
A- lI = 0
5
l 3
2
l
- - -
- = 0
5l + l2+ 6 = 0
l2+ 5l + 6 = 0
Since characteristic equation satisfies its own matrix, so
A2+ 5A+ 6 = 0 & A2 =- 5A- 6I
Multiplying with A
A3+ 5A2+ 6A = 0
A3+ 5(- 5A- 6I) + 6A = 0
A3 = 19A+ 30I
From Divergence theorem, we have
= # v $ t
###4v $ Avdv A n ds
s

= No of favourable outcomes
= = 5
.
No of total outcomes
3
= /
= + w + w
= # T0" fundamental period
= # w
= # w
15
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The position vector
rv = ^utx x + utyy + utz zh
Here, Av = 5rv, thus
4$ Av
= c u
t + t 2
+ t 2 m ^t + t + t h
x y z : ux x uy y uz z
2
x
u
y
u
z
2
2
2
= c dx
+ + dz m = 3#5 = 15
dy
dx dy
dz
5
## v $ t = ### 15dv = 15V
So, 5r nds
s
dy = ky
We have dx
# dy = # k dx + A
Integrating y
or lny = kx + A
Since y(0) = c thus lnc = A
So, we get, lny = kx + lnc
or lny = lnekx + lnc
or y = cekx
C R Integrals is
# - + where C is circle z = 1
z dz
4 5
3 4
z z
C
2+ +
f (z)dz
C #
= 0 if poles are outside C.
Now z 2+ 4z + 5 = 0
(z + 2)2+ 1 = 0
Thus z1,2 =- 2!j & z1,2 > 1
So poles are outside the unit circle.
We have f (x) = x + x - 3 = 0
fl(x)
x
= 1
+ 1
2
Substituting x0 = 2 we get
fl(x0) = 1.35355 and f (x0) = 2 + 2 - 3 = 0.414
Newton Raphson Method
x1
f x
( )
= - 0
( )
x
f x
0
0
l
Substituting all values we have
2 . 1 3535
= - 0 414 = 1.694
x 1 .
Writing A:B we have
:
:
:
1
1
1
1
4
4
1
6
6
20
WWWW Apply R3" R3- R2
l m
R
SSSS
T
V
X
:
:
: 20
1
1
0
1
4
0
1
6
6
6
20
l - m -
R
SSSS
T
V
WWWW
X
For equation to have solution, rank of A and A:B must be same.
Thus for no solution; l = 6, m ! 20
Total outcome are 36 out of which favorable outcomes are :
(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6);
(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) which are 15.
Thus P(E) .
36
12
Eigen value of a Skew-symmetric matrix are either zero or pure
imaginary in conjugate pairs.
For a function x(t) trigonometric fourier series is
x(t) Ao [An cosn t Bn sinn t]
n 1
Where, Ao T1 x(t)dt
0T
0
An T2 x(t)cosn t dt
0T
0
Bn T2 x(t) sinn t dt
0T
0
For an even function x(t),Bn = 0
Since given function is even function so coefficient Bn = 0, only cosine
and constant terms are present in its fourier series representation.
Constant term :
3 T
/
4 A0 =
T1 x(t)dt
/
4
- #
T
T1 T
Adt 2Adt
= /
+ 3 4 -
-: # # D
/
/
/
T
T
T
4
4
4
= : 2 - 2 2 D A2
1 TA AT
T
=-
Constant term is negative.
Given differential equation
d n x
n x
( ) ( )
dx
2
L
2
2 - = 0
Let n(x) = Aelx
So, A e
x Ae
2 l l -
L
x
2
l
= 0
l 2 - 1
= 0 & l =!
2 L1 L
L 1
Boundary condition, n ( 3 ) = 0 so take l =-
L n ( x ) = Ae x
-
n(0) = Ae0 = K & A = K
So, n(x) = Ke-(x/L)
1 =
Given that ey x x
1 =
or ln ey ln x x
or y = x1 lnx
Now dx
dy ln x x x x 1 1
= + ^- - 1
ln
x
2h 1
2 2 = -
x x
For maxima and minima :
dy (1 ln ) 0
dx
1 x
2 = - =
x
lnx = 1 " x = e 1

d y
2 2 1 1
=- 3 - b- 3 l- 2 b l
2 2 1
2 3 3 =- + -
d x
2 2 1 < 0 2 3 3 = - + -
4 1
1
5 = : + : : =
1 4 8
1
= t + 2 t
dvl = dxatx + dyaty
# v : v (xyax x ay) (dxa dya )
= # t + 2 t : t + t
by RK Kanodia
Now in 3 Volume
1
- s
+ ; E
3 1 L 1
3 2 =
+ + -
= =
+
s :
( 3 s
1
)
4 3
s 0 3 2
+ + -
+
s ( 3 s
1
)
4 3
s 0 2+ + -
1-
1
10
= + -1
f z
1 1 c0 c1z
= z 1 + c 0 + c
1
= z 1 + c 0 + c
1
" c m = c1
#
= # + f z
= 2pj [Residue at z = 0]
= 2pjc1
= y
+ p =- y = -1 (sin y
)
3 5
1 y y y
= - c - + - m
Now
dx
2
2
ln
x
x
x x x
ln
x x
x
x
dy
atx e
2
2
= 1
e e e
So, y has a maximum at x = e1
According to given condition head should comes 3 times or 4 times
P(Heads comes 3 times or 4 times) C 2 C
2
2
4
4
4
3
3
= b l + b l b l
1 16
1
2
16
A v
xyax x ay
A dl
C
C
x y
= # + 2
(xydx x dy)
C
xdx 3 4
3 xdx dy 1
3
dy = # 2 3 + # 1 3
+ # + #
/
/
/
/
2 3
1
3
3
1
1 3
1
3 4
1
3
3
= : - D+ : - D+ - + 1 1 - 3
= 1
2 3
2
[ ] [ ]
1
3 4
3 4
3 1 3
Given function
X(z)
-
( ) ( ) z z z z
1 2
= 1 2 - -
Poles are located at z = 0, z = 1, and z = 2
At Z = 0 residues is
= :
R0 z Z
zX ( ) =
0
- 2
= 1 2#0 - -
(0 1) (0 2)
= 1
= - :
at z Z
Z= 1 , R1 ( Z 1) X ( ) =
1
- =
= 1 2#1 -
1 1(1 2)
At z = 2, R2 (z 2) X(z)
z 2
= - :
=
= 1 2 # 2
3 2(2 -
1)
2
- =-
Taking step size h = 0.1, y(0) = 0
x y
dy = x + y y y hdx
dx
dy
i+1 = i +
0 0 0 y1 = 0 + 0.1(0) = 0
0.1 0 0.1 y2 = 0 + 0.1(0.1) = 0.01
0.2 0.01 0.21 y3 = 0.01 + 0.21#0.1 = 0.031
0.3 0.031
From table, at x = 0.3,y(x = 0.3) = 0.031
Given that
f (t)
s 4 s (K 3
)s
limf (t)
t"3
= 1
By final value theorem
limf (t)
t"3
limsF(s) 1
s "
0
or
( )
lim
s s K s
"
= 1
or
[ ( )]
lim
s s s K
"
= 1
K 3
= 1
or K = 4
The highest derivative terms present in DE is of 2nd order.
Number of elements in sample space is 210. Only one element
"H,H,T,T,T,T,T,T,T,T, is event. Thus probability is
2
We have
f (z) c0 c1z
f1(z)
( )
z
z
1
= + = + + -
( )
z
2
Since f1(z) has double pole at z = 0, the residue at z = 0 is
Res f1(z)z=0 limz .f (z)
z 0
2
= 1
"
.
( )
lim z
z
z 0
2
2
Hence
f1(z)dz
unit circle
[ ()]
z
dz
1
unit circle
We have f (x) sin
x
x
p
=
-
Substituting x - p = y ,we get
f (y + p)
sin( ) sin
y
y
y
! !
...
y
3 5
or f (y + p)
2 4
1 y y ...
=- + - +
3 5
! !
Substituting x - p = y we get
f (x)
=- + ( - p )
2 - - p 4 +
!
( )
!
...
x x
1
3 5
(A)
dy
dx
= y
x
or
# dy
y
= # dx
x
or log y = log x + log c

f x
( )
= - n
x= - - n
-
-
x e
1 n x
n = -
x e
1
+
n
1
-
d z a f z
d z
1 2
1
; E
d
1
; E
=- 2 32
2
-
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or y = cx Straight Line
Thus option (A) and (C) may be correct.
(B)
dy
dx
=-y
x
or
# dy
y
=- # dx
x
or log y =- log x + log c
or log y = log 1 + log
c
x
or y
x c
= Hyperbola
Sum of the principal diagonal element of matrix is equal to the sum
of Eigen values. Sum of the diagonal element is -1 - 1 + 3 = 1.In
only option (D), the sum of Eigen values is 1.
The product of Eigen value is equal to the determinant of the matrix.
Since one of the Eigen value is zero, the product of Eigen value is
zero, thus determinant of the matrix is zero.
Thus p11p22- p12p21 = 0
The given system is
x
y
4
2
2
= 1G= G
7
= =6G
We have A
4
2
2
= = 1G
and A
4
2
2
= 1 = 0 Rank of matrix r(A) < 2
Now C
4
2
2
1
7
= = 6G Rank of matrix r(C) = 2
Since r(A) ! r(C) there is no solution.
sinz can have value between -1 to +1. Thus no solution.
We have f (x) = ex + e-x
For x > 0, ex > 1 and 0 < e-x < 1
For x < 0, 0 < ex < 1 and e-x > 1
Thus f (x) have minimum values at x = 0 and that is e0+ e-0 = 2.
sin x
3 5
x x x ...
= + + +
3 5
! !
cos x
2 4
1 x x ...
= + + +
2 4
! !
Thus only sin(x3) will have odd power of x .
We have
dx t
( )
( )
dt
+ 3x t = 0
or (D + 3)x(t) = 0
Since m =- 3, x(t) = Ce-3t Thus only (B) may be
solution.
We have x = e-x
or f (x) = x - e-x
f'(x) = 1 + e-x
The Newton-Raphson iterative formula is
xn+1
'( )
x
f x
n
n
Now f (xn) xn e
f'(xn) 1 e x= + - n
Thus xn+1 x
e
n
x
n
+
-
( )
e
1
x
n
x
n
n
=
+
-
-
Res f (z)z=a
( )!
( ) ()
n 1
dz
1
n
n
1 z a
=
-
- -
= 6 @
Here we have n = 2 and a = 2
Thus R es f (z)z=2
( )!
( )
dz ( ) ( )
2 1 z z
2 2
z a
2
2 2 =
-
-
- + =
dz ( z 2
)
z a
2 =
+ =
(z 2)
z a
3 =
+
=
; E
64
=- 1
eP = L-1 (sI -A)-1 6 @
-
L
s
0 s
0 0
2
1
3
1
1
= - - -
-
e= G = Go
L
s
2 s
1
3
1
1
=
-
+
-
-
e= G o
s
3
+
L ( )( )
2
-
( )( )
1
( )( )
( )( )
s s
s s
s s
s
s s
1 1 2
1 2
1 2
1 2
= - + +
+ +
+ +
+ +
f> Hp
1 -
2
1 2
e e
e e
1 2
e e
e e
2
2 2 2
= 1 2
- +
-
- +
- -
- -
- -
= - - G
Taylor series is given as
x a
f a x a f a "( ) ...
f (x) ( )
!
'( )
( )
!
f a
1 2
2
= + - + - +
For x = p we have
Thus f (x) ( )
f x f "( )...
!
'( )
( )
!
x
f x
1 2
2
= p + - p p + - p
Now f (x) = ex + sinx
f'(x) = ex + cos x
f"(x) = ex - sinx
f"(p) = ep - sinp = ep
Thus the coefficient of (x - p)2 is
p
f"( )
2
!
The equation of straight line from (0,0) to (1,2) is y = 2x .
Now g(x,y) = 4x3+ 10y4
or, g(x,2x) = 4x3+ 160x4
Now g(x,2x)
1 # (4x3 160x4)dx
0
0
1
= # +

= + 1 =
Q
Q
Q
= # + = 2 # xdx + 2 #
ydy
= 9
- 4 = 2 5
= cosh
x
by RK Kanodia
Now in 3 Volume
2
y y
= - x
+ x +
x = - - +
f x
( )
= - 0
3 2
- + -
2 2 42 4
( )
2
+ # = 0
2 - # = 2pj [sum of residues]
-
1
# = 2 p j ` 1 j = p
j
2 - For more GATE Resources, Mock Test and
[x4 32x5] 33
0
I 2 (xdx ydy)
P
P
P
2 xdx 2 ydy 0
0
1
1
0
= # + # =
The given plot is straight line whose equation is
x y
-1 1
+ = 1
or y = x + 1
Now I ydx
1
2
= # (x 1)dx
1
2
= # +
(x )
2
1 2 2
= +
; E .
2
2
coth x
sinh
x
as x << 1, coshx . 1 and sinhx . x
Thus coth x
. 1
x
sin
0
lim
^ q h
2
q" q
q
^
sin
0 2 2
^
h
= 2
lim
q" q
h
= 2
lim
sin
1
2
q" q
0 2
q
^
^
h
h
= 1 = 0 .
5
2
We have, lim
1
x"0 2
x
= 3
lim x
x
2
"3
= 3
lim e
x
x
"3
- = 3
lim e
x
x2
"3
- = 0
lim e
x
x
0
2
"
- = 1 Thus e x- 2 is strictly bounded.
We have f (x) = e-x = e-(x - 2) - 2 = e-(x - 2)e-2
x ...
( )
( )
!
x
1 2 e
2
2 2
= - - + - -2 ; E
= 1 - (x - 2) e-2 6 @ Neglecting higher powers
= (3 - x)e-2
We have k
2 d y
dx
2
2
= y - y2
or
d y
dx
y
k
2
2
2 -
y
2
=- 2
k
A.E. D
2 1
2 - = 0
k
or D
=! 1
k
= - x
+
x C.F. C1e C2e k
k
P.I.
D
k
k
1
1
2
2
2
2
= 2
-
- = c m
Thus solution is
y C1e k
C2e k
y2 From y(0) = y1 we get
C1+ C2 = y1- y2
From y(3) = y2 we get that C1 must be zero.
Thus C2 = y1- y2
y (y1 y2)e y2 k
We have
f (x) = x3- x2+ 4x - 4
f'(x) = 3x2- 2x + 4
Taking x0 = 2 in Newton-Raphosn method
x1
'( )
x
f x
0
0
( ) ( )
2
3 2 2 2 4
= -
- +
3 4
=
For two orthogonal signal f (x) and g(x)
f (x)g(x)dx
3
3
-
i.e. common area between f (x) and g(x) is zero.
We know that
s
ds
1
1
D
Singular points are at s =! 1 but only s =+ 1 lies inside the given
contour, Thus Residue at s =+ 1 is
lim(s 1) f (s)
s "
1
lim(s )
s
1
1
1
2
s 1 2 = -
-
=
"
s
ds
1
1
D
2
For two orthogonal vectors, we require two dimensions to define
them and similarly for three orthogonal vector we require three
dimensions to define them. 2M vectors are basically M orthogonal
vector and we require M dimensions to define them.
We have
f (x) = x2- x + 2
1 " =
f'(x) = 2x - 1 = 0 x
2
f"(x) = 2
Since f"(x) = 2 > 0, thus x
= 1 is minimum point. The maximum
2
value in closed interval 6-4,4@ will be at x =- 4 or x = 4
Now maximum value
= max[f (- 4), f (4)]
= max(18,10) = 18
Probability of failing in paper 1 is P(A) = 0.3
Possibility of failing in Paper 2 is P(B) = 0.2

2 + - = #
1
1
= p -
" + -
= p p
# =
= q q p #
= # p 3 q - ` 3
q j q
sin3q = 3 sinq - 4 sin3q
= - q = w q
p p
3
: D : D
1
i
2
1
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Probability of failing in paper 1, when
student has failed in paper 2 is P 0.6 B
^A h =
We know that
A b l
P B
P + B =
( )
P ( B
)
= bAl = 0.6 # 0.2 = 0.12
or P(A + B) P(B)P B
We have
A
R
1
1
1
1
1
1
1
0
1
1
1
0
1
1
0
1
0
0
SSSS
SSSS
WWWW
= - + -
T
R
T
V
X
V
WWWW
X
R3- R1
Since one full row is zero, r(A) < 3
Now
1
1
1
-1 =- 2 ! 0, thus r(A) = 2
The vector Triple Product is
A# (B #C) = B(A$C) -C(A$ B)
Thus 4#4# P =4 (4$ P) - P(4$4)
=4 (4$ P) -42P
The Stokes theorem is
##(4# F) $ ds = #A $ dl
3
-# = 1
We know p(x)dx
3
3 -a
-# = 1
Thus Ke x dx
3
- # # = 1
or Ke x dx Ke x dx
0
0
+ 3
3
a -a
or
K e x 0 k e x
a a 0 3+ -
6 a @ 6 - a @ 3
= 1
( )
-
or K K
a + a = 1
or K
2 a
=
We have xo (t) + 2x(t) = s(t)
Taking Laplace transform both sides
sX(s) - x(0) + 2X(s) = 1
or sX(s) + 2X(s) = 1 Since x(0-) = 0
X(s)
= 1
+
s 2
Now taking inverse Laplace transform we have
x(t) = e-2tu(t)
Sum of the Eigen values must be equal to the sum of element of
principal diagonal of matrix.
Only matrix
6
2
2
= 6G satisfy this condition.
We have W = lnz
u + jv = ln(x + jy)
or eu+jv = x + jy
or eu ejv = x + jy
eu (cos v + j sinv) = x + jy
Now x = eu cos v and y = eu sinv
Thus x2+ y2 = e2u Equation of circle
We have
z
dz
4
1
z j
2
(z i) (z i)
dz
2 2
z j 2
=
+ -
- = #
P(0,2) lies inside the circle z - j = 2 and P(0, - 2) does not lie.
Thus By cauchy’s integral formula
I ( )
( )( )
i lim z i
z i z i
2 2
2 2
z 2i
i i
2 2
2
C
+
I sin3 d
0
sin sin d
4
0
3
s
4
cos 12
0 0
4 3
4 3
12
12
3 4
= 8 + B-8 + B =
Let d " defective and y " supply by Y
y
a k
p d
P y+d =
( )
P ( d
)
P(y + d) = 0.3 # 0.02 = 0.006
P(d) = 0.6 # 0.1 + 0.3 # 0.02 + 0.1 # 0.03 = 0.015
P y
a d
k
. .
0 015
= 0 006 = 0 4
.
We have A
4
2
2
= = 4G
Now 6A - lI @[X] = 0
or
-
= - G= G
4
2
2
4
101
101
l
l
0
= =0G
or (101)(4 - l) + 2(101) = 0
or l = 6
We have
2
d y k y 2
dx
+ 2 = 0
or D2y + k2y = 0
The AE is m2+ k2 = 0
The solution of AE is m =! ik
Thus y = Asinkx + Bcos kx
From x = 0, y = 0 we get B = 0 and x = a,y = 0 we get
Asinka = 0
or sinka = 0
k
= mpx
a
=/ ` p j
Thus y A sin
m x
a
m
m

e
1 x
# = 1 .
d y
5 dy 6y 2
by RK Kanodia
Now in 3 Volume
- l = 0
- -
= - G= G
-
= G and A
1 2 1
-
= G
= 1 and a
= 1
= 1
+ 1
= 7
1 (x )
-m # for -3 # x # 3
3 -
-
3
-# = 1
1 x
-# 3 -
2
= 1
1
2
1
2
We have f (x)
e
x
=
+
For x " 3, the value of f (x) monotonically increases.
Order is the highest derivative term present in the equation and
degree is the power of highest derivative term.
Order = 2, degree = 1
Probability of coming odd number is 2 1
and the probability of
coming even number is 2 1
. Both the events are independent to each
other, thus probability of coming odd number after an even number
is 2 1
2 1
4
We have
dx
dx
2
- + = 0
The A.E. is m2- 5m + 6 = 0
m = 3,2
The CF is yc = C 3
x 2
x
1
e + C 2
e Since Q = 0, thus y = C e x + C e 2
x
1
3
2
Thus only (B) may be correct.
We have f (t) = e(a+2)t+5 = e5.e(a+2)t
Taking Laplace transform we get
F(s)
= 5 1
; - + E Thus Re(s) > (a +
2) ( )
e
s a 2
For x > 0 the slope of given curve is negative. Only (C) satisfy this
condition.
Newton - Raphson " Method-Solving nonlinear eq.
Runge - kutta Method " Solving ordinary differential eq.
Simpson’s Rule " Numerical Integration
Gauss elimination " Solving linear simultaneous eq.
We have A
-
= G
4
4
2
= 3
Characteristic equation is
A - lI = 0
or
4
4
2
3
l
l
-
- = 0
or (- 4 - l)(3 - l) - 8 = 0
or -12 + l + l2- 8 = 0
or l2+ l - 20 = 0
or l =- 5,4 Eigen values
Eigen vector for l =- 5
( ) A I Xi
( ) x
x
1 5
4
2
8 4
1
2
0
= =0G
x
x
1
0
2
0
1
2
= =G G
0
= =0G R2- 4R1
x1+ 2x2 = 0
Let -x1 = 2 & x2 =- 1,
Thus X
2
= =-1G Eigen vector
We have
A
2 .
0
0 1
= 3
a
0 b
- = = G
Now AA-1 = I
or
. a
b
2
0
0 1
3 0 2 1
-
= =G G
1
0
0
= = 1G
or
a . b
b
1
0
2 01
3
1
0
0
= = 1G
or 2a - 0.1 = 0 and 3b = 1
Thus solving above we have b
3
60
Therefore a + b
3
60
20
Gaussian PDF is
f (x) e dx
2
2 2
2
p s
=
3
s
and f (x)dx
3
Substituting m = 0 and s = 2 in above we get
e dx
2 2
8
p 3
or 2 e dx
2 2
0
x
8
p
# 3 - = 1
or e dx
2
0
x
8
p
# 3 - = 1
From orthogonal matrix
[AAT] = I
Since the inverse of I is I , thus
[AAT]-1 = I-1 = I

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UNIT 2
NETWORKS
2013 ONE MARK
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) 1/k (D) k
2.2 The transfer function
^
h
V s
1
2
^
V s
h
of the circuit shown below is
(A) .
s
+ (B) s
s
1
0 5 1
+
s
+
3 6
+
2
s
(C) s
+ 2
+
1
(D) s
s
1
+
2
+
2.3 A source vs ^t h = Vcos 100pt has an internal impedance of ^4 + j3hW
. If a purely resistive load connected to this source has to extract the
maximum power out of the source, its value in W should be
(A) 3 (B) 4
(C) 5 (D) 7
2013 TWO MARKS
2.4 In the circuit shown below, if the source voltage VS = 100+53.13cV
then the Thevenin’s equivalent voltage in Volts as seen by the load
resistance RL is
(A) 100+90c (B) 800+0c
(C) 800+90c (D) 100+60c
2.5 The following arrangement consists of an ideal transformer and
an attenuator which attenuates by a factor of 0.8. An ac voltage
VWX1 = 100 V is applied across WX to get an open circuit voltage
VYZ1 across YZ. Next, an ac voltage VYZ2 = 100 V is applied across
YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1/VWX1
, VWX2/VYZ2 are respectively,
(A) 125/100 and 80/100 (B) 100/100 and 80/100
(C) 100/100 and 100/100 (D) 80/100 and 80/100
2.6 Three capacitors C1, C2 and C3 whose values are 10 mF, 5 mF, and
2 mF respectively, have breakdown voltages of 10 V, 5 V and 2 V
respectively. For the interconnection shown below, the maximum
safe voltage in Volts that can be applied across the combination,
and the corresponding total charge in mC stored in the effective
capacitance across the terminals are respectively,
(A) 2.8 and 36 (B) 7 and 119
(C) 2.8 and 32 (D) 7 and 80
Common Data For Q. 8 and 9:
Consider the following figure
2.7 The current IS in Amps in the voltage source, and voltage VS in
Volts across the current source respectively, are
(A) 13, - 20 (B) 8, - 10
(C) -8,20 (D) -13,20
2.8 The current in the 1W resistor in Amps is
(A) 2 (B) 3.33
(C) 10 (D) 12
2.9 Two magnetically uncoupled inductive coils have Q factors q1 and q2

by RK Kanodia
Now in 3 Volume
at the chosen operating frequency. Their respective resistances are
R1 and R2. When connected in series, their effective Q factor at the
same operating frequency is
(A) q1+ q2 (B) ^1/q1h+^1/q2h
(C) ^q1R1+ q2R2h/^R1+ R2h (D) ^q1R2+ q2R1h/^R1+ R2h
2012 ONE MARK
2.10 In the following figure, C1 and C2 are ideal capacitors. C1 has been
charged to 12 V before the ideal switch S is closed at t = 0. The
current i (t) for all t is
(A) zero (B) a step function
(C) an exponentially decaying function (D) an impulse function
2.11 The average power delivered to an impedance (4 - j3)W by a current
5 cos (100pt + 100)A is
(A) 44.2W (B) 50W
(C) 62.5W (D) 125W
2.12 In the circuit shown below, the current through the inductor is
2+
(A) 1 j A
-
1
+
(B) 1 j A
1+
(C) 1 j A
(D) 0 A
2012 TWO MARKS
2.13 Assuming both the voltage sources are in phase, the value of R for
which maximum power is transferred from circuit A to circuit B is
(A) 0.8 W (B) 1.4 W
(C) 2 W (D) 2.8 W
2.14 If VA- VB = 6 V then VC - VD is
(A) -5 V (B) 2 V
(C) 3 V (D) 6 V
Common Data For Q. 48 and 49 :
With 10 V dc connected at port A in the linear nonreciprocal two-port
network shown below, the following were observed :
(i) 1 W connected at port B draws a current of 3A
(ii) 2.5 W connected at port B draws a current of 2A
2.15 With 10 V dc connected at port A, the current drawn by 7 W
connected at port B is
(A) 3/7 A (B) 5/7 A
(C) 1A (D) 9/7 A
2.16 For the same network, with 6 V dc connected at port A, 1 W
connected at port B draws 7/3 A. If 8 V dc is connected to port A
, the open circuit voltage at port B is
(A) 6 V (B) 7 V
(C) 8 V (D) 9 V
2011 ONE MARK
2.17 In the circuit shown below, the Norton equivalent current in amperes
with respect to the terminals P and Q is
(A) 6.4 - j 4.8 (B) 6.56 - j 7.87
(C) 10 + j0 (D) 16 + j0
2.18 In the circuit shown below, the value of RL such that the power

2
-
> H (B) .
-
> H
4
2
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transferred to RL is maximum is
(A) 5 W (B) 10 W
(C) 15 W (D) 20 W
2.19 The circuit shown below is driven by a sinusoidal input
vi = Vp cos (t/RC). The steady state output vo is
(A) (Vp/3)cos (t/RC) (B) (Vp/3) sin(t/RC)
(C) (Vp/2)cos (t/RC) (D) (Vp/2) sin(t/RC)
2011 TWO MARKS
2.20 In the circuit shown below, the current I is equal to
(A) 1.4+0cA (B) 2.0+0cA
(C) 2.8+0cA (D) 3.2+0cA
2.21 In the circuit shown below, the network N is described by the
following Y matrix:
0 .
1
0 .
01
.
.
S
S
S
0 01
-
> H. the voltage gain V
= 0 1
Y S
V
1
2 is
(A) 1/90 (B) –1/90
(C) –1/99 (D) –1/11
2.22 In the circuit shown below, the initial charge on the capacitor is 2.5
mC, with the voltage polarity as indicated. The switch is closed at
time t = 0. The current i (t) at a time t after the switch is closed is
(A) i (t) 15 exp( 2 103t)A = - #
(B) i (t) 5 exp( 2 103t)A = - #
(C) i (t) 10 exp( 2 103t)A = - #
(D) i (t) 5 exp( 2 103t)A =- - #
2010 ONE MARK
2.23 For the two-port network shown below, the short-circuit admittance
parameter matrix is
(A) S
- 4
.
S
1
0 5
0 5
- 1
(C) .
.
S
1
0 5
0 5
4
2
2
> 1 H (D) S
> 4H
2.24 For parallel RLC circuit, which one of the following statements is
NOT correct ?
(A) The bandwidth of the circuit decreases if R is increased
(B) The bandwidth of the circuit remains same if L is increased
(C) At resonance, input impedance is a real quantity
(D) At resonance, the magnitude of input impedance attains its
minimum value.
2010 TWO MARKS
2.25 In the circuit shown, the switch S is open for a long time and is
closed at t = 0. The current i (t) for t $ 0+ is
(A) i (t) = 0.5 - 0.125e-1000tA (B) i (t) = 1.5 - 0.125e-1000tA
(C) i (t) = 0.5 - 0.5e-1000tA (D) i (t) = 0.375e-1000t A
2.26 The current I in the circuit shown is
(A) -j1A (B) j1A

by RK Kanodia
Now in 3 Volume
(C) 0 A (D) 20 A
2.27 In the circuit shown, the power supplied by the voltage source is
(A) 0 W (B) 5 W
(C) 10 W (D) 100 W
GATE 2009 ONE MARK
2.28 In the interconnection of ideal sources shown in the figure, it is
known that the 60 V source is absorbing power.
Which of the following can be the value of the current source I ?
(A) 10 A (B) 13 A
(C) 15 A (D) 18 A
2.29 If the transfer function of the following network is
V s
i
o
( )
V ( s
)
= 1
+
2 sCR
The value of the load resistance RL is
(A) R4
(B) R2
(C) R (D) 2R
2.30 A fully charged mobile phone with a 12 V battery is good for a 10
minute talk-time. Assume that, during the talk-time the battery
delivers a constant current of 2 A and its voltage drops linearly from
12 V to 10 V as shown in the figure. How much energy does the
battery deliver during this talk-time?
(A) 220 J (B) 12 kJ
(C) 13.2 kJ (D) 14.4 J
GATE 2009 TWO MARK
2.31 An AC source of RMS voltage 20 V with internal impedance
Zs = (1 + 2j)W feeds a load of impedance ZL = (7 + 4j)W in the
figure below. The reactive power consumed by the load is
(A) 8 VAR (B) 16 VAR
(C) 28 VAR (D) 32 VAR
2.32 The switch in the circuit shown was on position a for a long time,
and is move to position b at time t = 0. The current i (t) for t > 0
is given by
(A) 0.2e-125tu(t) mA (B) 20e-1250tu(t) mA
(C) 0.2e-1250tu(t) mA (D) 20e-1000tu(t) mA
2.33 In the circuit shown, what value of RL maximizes the power delivered
to RL?

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8 W
(A) 2.4 W (B) 3
(C) 4 W (D) 6 W
2.34 The time domain behavior of an RL circuit is represented by
di + Ri V (1 Be Rt/L sint)u(t)
L
dt
= 0 + - .
0 = V0 , the steady state value of the
For an initial current of i ( )
R
current is given by
(A) i (t)
" V0 (B) i (t)
R
" 2V0
R
" V0 1 + B (D) i (t) ( )
(C) i (t) ( )
R
" 2V0 1 + B
R
GATE 2008 ONE MARK
2.35 In the following graph, the number of trees (P) and the number of
cut-set (Q) are
(A) P = 2,Q = 2 (B) P = 2,Q = 6
(C) P = 4,Q = 6 (D) P = 4,Q = 10
2.36 In the following circuit, the switch S is closed at t = 0. The rate of
di 0+ is given by
change of current ( )
dt
(A) 0 (B)
Rs Is
L
(C)
R + Rs Is
( )
L
(D) 3
GATE 2008 TWO MARKS
2.37 The Thevenin equivalent impedance Zth between the nodes P and
Q in the following circuit is
1 + + 1
(A) 1 (B) s
s
2 + + 1 (D)
(C) s
s
2
+ +
s s
s s
1
2 1
2
+ +
2.38 The driving point impedance of the following network is given by
Z(s)
.
0 2
s .
s
s
0 1 2
2 =
+ +
The component values are
(A) L = 5 H,R = 0.5 W,C = 0.1 F
(B) L = 0.1H,R = 0.5 W,C = 5 F
(C) L = 5 H,R = 2 W,C = 0.1 F
(D) L = 0.1H,R = 2 W,C = 5 F
2.39 The circuit shown in the figure is used to charge the capacitor C
alternately from two current sources as indicated. The switches S1
and S2 are mechanically coupled and connected as follows:
For 2nT # t # (2n + 1)T, (n = 0,1,2,..) S1 to P1 and S2 to P2
For (2n + 1)T # t # (2n + 2)T, (n = 0,1,2,...) S1 to Q1 and S2 to
Q2
Assume that the capacitor has zero initial charge. Given that u(t)
is a unit step function , the voltage vc (t) across the capacitor is
given by
(A) ( 1)n tu(t nT)
n 1
- -
3
= /
3
= /
(B) u(t) + 2 ( - 1)nu(t -
nT)
n 1
3
= /
(C) tu(t) + 2 ( - 1)nu(t - nT) (t -
nT)
n 1
/ 6 @
(D) 0.5 e (t nT) 0.5e (t nT) T
n
2 2
1
- + -
3
- - - -
=
Common Data For Q. 2.23 & 2.24 :
The following series RLC circuit with zero conditions is excited by
a unit impulse functions d(t).

2 t t 2
2 2
3 - - ^ h (B) te t
2
2 t 3
2
2 t 3
1 2 t t
1 _ - - i
t 3
1
3
2 1 - -
= c m c mG
2
2 t 3
2
2 t 3
by RK Kanodia
Now in 3 Volume
2.40 For t > 0, the output voltage vC ^t h is
(A) e e
3
1
2
3
-1
(C) e cos t
3
2
-1
c m (D) e sin t
3
2
-1
c m
2.41 For t > 0, the voltage across the resistor is
(A) e e
3
3
2
(B) e cos t sin t
2
3
2
(C) e sin t
3
2
-1
c m
(D) e cos t
3
2
-1
c m
Statement for linked Answers Questions 2.25 & 2.26:
A two-port network shown below is excited by external DC source.
The voltage and the current are measured with voltmeters V1,V2
and ammeters. A1,A2 (all assumed to be ideal), as indicated
Under following conditions, the readings obtained are:
(1) S1 -open, S2 - closed A1 = 0,V1 = 4.5 V,V2 = 1.5 V,A2 = 1 A
(2) S1 -open, S2 - closed A1 = 4 A,V1 = 6 V,V2 = 6 V,A2 = 0
2.42 The z -parameter matrix for this network is
(A)
.
.
.
.
1 5
4 5
1 5
= 1 5G (B)
.
.
.
.
1 5
1 5
4 5
= 4 5G
(C)
.
.
.
.
1 5
1 5
4 5
= 1 5G (D)
.
.
.
.
4 5
1 5
1 5
= 4 5G
2.43 The h-parameter matrix for this network is
-
3
3
=- 1
0 67
G (B) .
(A) .
- -
= G
3
3
1
0 67
3
1
3
= 0 67G (D) .
(C) .
3
3
1
=- -0 67G
GATE 2007 ONE MARK
2.44 An independent voltage source in series with an impedance
Zs = Rs + jXs delivers a maximum average power to a load impedance
ZL when
(A) ZL = Rs + jXs (B) ZL = Rs
(C) ZL = jXs (D) ZL = Rs - jXs
2.45 The RC circuit shown in the figure is
(A) a low-pass filter (B) a high-pass filter
(C) a band-pass filter (D) a band-reject filter
GATE 2007 TWO MARKS
2.46 Two series resonant filters are as shown in the figure. Let the 3-dB
bandwidth of Filter 1 be B1 and that of Filter 2 be B2. the value
B
B
2
1 is
(A) 4 (B) 1
(C) 1/2 (D) 1/4
2.47 For the circuit shown in the figure, the Thevenin voltage and
resistance looking into X - Y are
(A) V, 2 3 4
W (B) 4 V, 3 2
W
(C) V, 3 4
3 2
W (D) 4 V, 2 W
2.48 In the circuit shown, vC is 0 volts at t = 0 sec. For t > 0, the capacitor
current iC (t), where t is in seconds is given by
(A) 0.50 exp(- 25t) mA (B) 0.25 exp(- 25t) mA
(C) 0.50 exp(- 12.5t) mA (D) 0.25 exp(- 6.25t) mA
2.49 In the ac network shown in the figure, the phasor voltage VAB (in
Volts) is
(A) 0 (B) 5+30c
(C) 12.5+30c (D) 17+30c

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GATE 2006 TWO MARKS
2.50 A two-port network is represented by ABCD parameters given by
V
I
1
1
= G
A
C
B
D
V
I
2
2
= = G=- G
If port-2 is terminated by RL, the input impedance seen at port-1
is given by
(A)
A + BR
(B)
C DR
L
L
+
AR C
L
+
BR D
L
+
(C)
DR A
L
+
BR C
L
+ (D)
+
B AR
L
D CR
L
+
2.51 In the two port network shown in the figure below, Z12 and Z21 and
respectively
(A) re and br0 (B) 0 and -br0
(C) 0 and bro (D) re and -br0
2.52 The first and the last critical frequencies (singularities) of a driving
point impedance function of a passive network having two kinds of
elements, are a pole and a zero respectively. The above property
will be satisfied by
(A) RL network only (B) RC network only
(C) LC network only (D) RC as well as RL networks
2.53 A 2 mH inductor with some initial current can be represented as
shown below, where s is the Laplace Transform variable. The value
of initial current is
(A) 0.5 A (B) 2.0 A
(C) 1.0 A (D) 0.0 A
2.54 In the figure shown below, assume that all the capacitors are initially
uncharged. If vi (t) = 10u(t) Volts, vo (t) is given by
(A) 8e-t/0.004 Volts (B) 8(1 - e-t/0.004) Volts
(C) 8u(t) Volts (D) 8 Volts
2.55 A negative resistance Rneg is connected to a passive network N
having driving point impedance as shown below. For Z2(s) to be
positive real,
(A) Rneg # ReZ1(jw),6w (B) Rneg # Z1(jw) ,6w
(C) Rneg # ImZ1(jw),6w (D) Rneg # +Z1(jw),6w
GATE 2005 ONE MARK
2.56 The condition on R,L and C such that the step response y(t) in the
figure has no oscillations, is
(A) R
L
$ 1 (B) R C
C
2
$ L
$ 2 L (D) R
(C) R C
= 1
LC
2.57 The ABCD parameters of an ideal n:1 transformer shown in the
figure are
n
0 x
0
> H
The value of x will be
(A) n (B)
1
n
(C) n2 (D)
1
2
n
= 1 mF The
2.58 In a series RLC circuit, R = 2 kW, L = 1 H, and C 400
resonant frequency is
(A) 2 104 # Hz (B) 1 104
p # Hz
(C) 104 Hz (D) 2p # 104 Hz
2.59 The maximum power that can be transferred to the load resistor RL
from the voltage source in the figure is

by RK Kanodia
Now in 3 Volume
10
1
-
1
0 05
10
1
1
(A) 1 W (B) 10 W
(C) 0.25 W (D) 0.5 W
2.60 The first and the last critical frequency of an RC -driving point
impedance function must respectively be
(A) a zero and a pole (B) a zero and a zero
(C) a pole and a pole (D) a pole and a zero
GATE 2005 TWO MARKS
2.61 For the circuit shown in the figure, the instantaneous current i1(t) is
10 3 c A (B)
(A) 90
2
10 3 -90c A
2
(C) 5 60c A (D) 5 -60c A
2.62 Impedance Z as shown in the given figure is
(A) j29 W (B) j9 W
(C) j19 W (D) j39 W
2.63 For the circuit shown in the figure, Thevenin’s voltage and Thevenin’s
equivalent resistance at terminals a - b is
(A) 5 V and 2W (B) 7.5 V and 2.5W
(C) 4 V and 2W (D) 3 V and 2.5W
2.64 If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in
the figure, then the reading in the ideal voltmeter connected between
a and b is
(A) 0.238 V (B) 0.138 V
(C) -0.238 V (D) 1 V
2.65 The h parameters of the circuit shown in the figure are
(A)
.
.
.
.
0 1
0 1
0 1
=- 0 3G (B) .
= G
(C)
30
20
20
= 20G (D) .
=- 0 05G
2.66 A square pulse of 3 volts amplitude is applied to C - R circuit
shown in the figure. The capacitor is initially uncharged. The output
voltage V2 at time t = 2 sec is
(A) 3 V (B) -3 V
(C) 4 V (D) -4 V
GATE 2004 ONE MARK
2.67 Consider the network graph shown in the figure. Which one of the
following is NOT a ‘tree’ of this graph ?

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(A) a (B) b
(C) c (D) d
2.68 The equivalent inductance measured between the terminals 1 and 2
for the circuit shown in the figure is
(A) L1+ L2+M (B) L1+ L2-M
(C) L1+ L2+ 2M (D)L1+ L2- 2M
= 1
W = 1H and C = 3 F
2.69 The circuit shown in the figure, with R 3 , L
4
has input voltage v(t) = sin2t . The resulting current i (t) is
(A) 5 sin(2t + 53.1c)
(B) 5 sin(2t - 53.1c)
(C) 25 sin(2t + 53.1c)
(D) 25 sin(2t - 53.1c)
2.70 For the circuit shown in the figure, the time constant RC = 1 ms.
= 3 . The output voltage v (t) o is
The input voltage is vi (t) 2 sin10 t
equal to
(A) sin(103t - 45c) (B) sin(103t + 45c)
(C) sin(103t - 53c) (D) sin(103t + 53c)
2.71 For the R - L circuit shown in the figure, the input voltage
vi (t) = u(t). The current i (t) is
GATE 2004 TWO MARKS
2.72 For the lattice shown in the figure, Za = j2 W and Zb = 2 W. The
values of the open circuit impedance parameters z
z
z
z
z
11
21
12
22
6 @ = = G are
(A)
j
j
j
j
1
1
1
1
-
+
+
= + G (B)
j
j
j
j
1
1
1
1
-
- +
+
= - G
(C)
j
j
j
j
1
1
1
1
+
-
+
= - G (D)
j
j
j
j
1
1
1
1
+
- +
- +
= + G
2.73 The circuit shown in the figure has initial current iL (0-) = 1 A
through the inductor and an initial voltage vC (0-) =- 1 V across
the capacitor. For input v(t) = u(t), the Laplace transform of the
current i (t) for t $ 0 is
(A)
s
2+ + 1
s s
(B)
s
+
s s
1
2
2+ +
(C)
s
- (D)
s s
1
2
2+ +
1
2+ +
s s 1
2.74 The transfer function ( )
V s
i
( )
= o of an RLC circuit is given by
( )
H s
V s
H(s)
10
6
2 6
s 20s 10
=
+ +
The Quality factor (Q-factor) of this circuit is
(A) 25 (B) 50

V s
i
( )
= c is
1
2+ 6 + 6
10
6
2 3 6
+ +
10
3
2 3 6
+ +
10
6
2 6 6
+ +
by RK Kanodia
Now in 3 Volume
2
+ + = (B) ( ) cos
2 d i
2 di i t t 2
2
+ + =
d i
2 di 2i t t 2
2
+ + = (D) ( ) sin
2 d i
2 di i t t 2
2
+ + =
d i
2 di 2i t t 2
3 W
(C) 100 (D) 5000
2.75 For the circuit shown in the figure, the initial conditions are zero. Its
transfer function ( )
( )
H s
V s
(A)
s 10 s 10
(B)
s 10 s 10
(C)
s 10 s 10
(D)
s 10 s 10
2.76 Consider the following statements S1 and S2
S1 : At the resonant frequency the impedance of a series RLC
circuit is zero.
S2 : In a parallel GLC circuit, increasing the conductance G re-sults
in increase in its Q factor.
Which one of the following is correct?
(A) S1 is FALSE and S2 is TRUE
(B) Both S1 and S2 are TRUE
(C) S1 is TRUE and S2 is FALSE
(D) Both S1 and S2 are FALSE
GATE 2003 ONE MARK
2.77 The minimum number of equations required to analyze the circuit
shown in the figure is
(A) 3 (B) 4
(C) 6 (D) 7
2.78 A source of angular frequency 1 rad/sec has a source impedance
consisting of 1 W resistance in series with 1 H inductance. The load
that will obtain the maximum power transfer is
(A) 1 W resistance
(B) 1 W resistance in parallel with 1 H inductance
(C) 1 W resistance in series with 1 F capacitor
(D) 1 W resistance in parallel with 1 F capacitor
2.79 A series RLC circuit has a resonance frequency of 1 kHz and a
quality factor Q = 100. If each of R,L and C is doubled from its
original value, the new Q of the circuit is
(A) 25 (B) 50
(C) 100 (D) 200
2.80 The differential equation for the current i (t) in the circuit of the
figure is
(A) ( ) sin
dt
dt
dt
dt
(C) ( ) cos
dt
dt
dt
dt
GATE 2003 TWO MARKS
2.81 Twelve 1 W resistance are used as edges to form a cube. The resistance
6 between two 5
diagonally opposite corners of the cube is
(A) W (B) 1 W
(C) 5 6
W (D) 2
2.82 The current flowing through the resistance R in the circuit in the
figure has the form Pcos 4t where P is
(A) (0.18 + j0.72) (B) (0.46 + j1.90)
(C) -(0.18 + j1.90) (D) -(0.192 + j0.144)
The circuit for Q. 2.66 & 2.67 is given below.
Assume that the switch S is in position 1 for a long time and
thrown to position 2 at t = 0.
2.83 At t = 0+, the current i1 is
(A)
V
2 -
R
(B)
-V
R
(C)
- (D) zero
V
4
R
2.84 I1(s) and I2(s) are the Laplace transforms of i1(t) and i2(t) respectively.
The equations for the loop currents I1(s) and I2(s) for the circuit
shown in the figure, after the switch is brought from position 1 to
position 2 at t = 0, are
(A)
( )
( )
R Ls
Ls
Ls
R
I s
I s 0
Cs
Cs
s V
1
1
1
2
+ +
-
-
> + H= G = = G
(B)
( )
( )
R Ls
Ls
Ls
R
I s
I s 0
Cs
Cs
s V
1
1
1
2
+ +
-
-
+ =
-
> H= G = G

40 W
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(C)
( )
( )
R Ls
Ls
Ls
R Ls
I s
I s 0
Cs
Cs
s V
1
1
1
2
+ +
-
-
+ + =
-
> H= G = G
(D)
( )
( )
R Ls
Ls
Cs
R Ls
I s
I s 0
Cs
Cs
s V
1
1
1
2
+ +
-
-
> + + H= G = = G
2.85 The driving point impedance Z(s) of a network has the pole-zero
locations as shown in the figure. If Z(0) = 3, then Z(s) is
(A)
+
s
2 3
3 3
2+ +
( )
s s
(B)
+
s
2 2
2 3
2+ +
( )
s s
(C)
+
s
2 2
3 3
2+ +
( )
s s
(D)
-
s
2 3
2 3
2- -
( )
s s
2.86 An input voltage v(t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c)
V is applied to a series combination of resistance R = 1W and
an inductance L = 1 H. The resulting steady-state current i (t) in
ampere is
(A) 10 cos (t + 55c) + 10 cos (2t + 10c + tan-12)
(B) 10 cos (t + 55 c ) + 10 3 cos(2t + 55 2
c
) (C) 10 cos (t - 35c) + 10 cos (2t + 10c - tan-12)
(D) 10 cos (t - 35 c ) + 3 cos(2t - 35 ) 2
c
2.87 The impedance parameters z11 and z12 of the two-port network in
the figure are
(A) z11 = 2.75 W and z12 = 0.25 W
(B) z11 = 3 W and z12 = 0.5 W
(C) z11 = 3 W and z12 = 0.25 W
(D) z11 = 2.25 W and z12 = 0.5 W
GATE 2002 ONE MARK
2.88 The dependent current source shown in the figure
(A) delivers 80 W (B) absorbs 80 W
(C) delivers 40 W (D) absorbs 40 W
2.89 In the figure, the switch was closed for a long time before opening
at t = 0. The voltage vx at t = 0+ is
(A) 25 V (B) 50 V
(C) -50 V (D) 0 V
GATE 2002 TWO MARKS
2.90 In the network of the fig, the maximum power is delivered to RL if
its value is
(A) 16 W (B) 3
(C) 60 W (D) 20 W
2.91 If the 3-phase balanced source in the figure delivers 1500 W at
a leading power factor 0.844 then the value of ZL (in ohm) is
approximately
(A) 90+32.44c (B) 80+32.44c
(C) 80+ - 32.44c (D) 90+ - 32.44c
GATE 2001 ONE MARK
2.92 The Voltage e0 in the figure is
(A) 2 V (B) 4/3 V
(C) 4 V (D) 8 V
2.93 If each branch of Delta circuit has impedance 3 Z , then each
branch of the equivalent Wye circuit has impedance
(A) Z
3
(B) 3Z
(C) 3 3 Z (D) Z
3
2.94 The admittance parameter Y12 in the 2-port network in Figure is

by RK Kanodia
Now in 3 Volume
6
16
11 = W 21 = W (B) z 11 ; z
4
11 = W 21 = W
6
16
11 = W 21 =- W (D) z 11 ; z
4
11 = W 21 = W
(A) -0.02 mho (B) 0.1 mho
(C) -0.05 mho (D) 0.05 mho
GATE 2001 TWO MARKS
2.95 The voltage e0 in the figure is
(A) 48 V (B) 24 V
(C) 36 V (D) 28 V
2.96 When the angular frequency w in the figure is varied 0 to 3, the
locus of the current phasor I2 is given by
2.97 In the figure, the value of the load resistor RL which maximizes the
power delivered to it is
(A) 14.14 W (B) 10 W
(C) 200 W (D) 28.28 W
2.98 The z parameters z11 and z21 for the 2-port network in the figure are
(A) z 11 ; z
11
6
11
(C) z 11 ; z
11
4
11
GATE 2000 ONE MARK
2.99 The circuit of the figure represents a
(A) Low pass filter (B) High pass filter
(C) band pass filter (D) band reject filter
2.100 In the circuit of the figure, the voltage v(t) is
(A) eat - ebt (B) eat + ebt
(C) aeat - bebt (D) aeat + bebt
2.101 In the circuit of the figure, the value of the voltage source E is
(A) -16 V (B) 4 V
(C) -6 V (D) 16 V
GATE 2000 TWO MARKS
2.102 Use the data of the figure (a). The current i in the circuit of the
figure (b)

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(A) -2 A (B) 2 A
(C) -4 A (D) 4 A
GATE 1999 ONE MARK
2.103 Identify which of the following is NOT a tree of the graph shown in
the given figure is
(A) begh (B) defg
(C) abfg (D) aegh
2.104 A 2-port network is shown in the given figure. The parameter h21 for
this network can be given by
(A) -1/2 (B) +1/2
(C) -3/2 (D) +3/2
GATE 1999 TWO MARK
2.105 The Thevenin equivalent voltage VTH appearing between the
terminals A and B of the network shown in the given figure is given
by
(A) j16(3 - j4) (B) j16(3 + j4)
(C) 16(3 + j4) (D) 16(3 - j4)
2.106 The value of R (in ohms) required for maximum power transfer in
the network shown in the given figure is
(A) 2 (B) 4
(C) 8 (D) 16
2.107 A Delta-connected network with its Wye-equivalent is shown in the
given figure. The resistance R1,R2 and R3 (in ohms) are respectively
(A) 1.5, 3 and 9 (B) 3, 9 and 1.5
(C) 9, 3 and 1.5 (D) 3, 1.5 and 9
GATE 1998 ONE MARK
2.108 A network has 7 nodes and 5 independent loops. The number of
branches in the network is
(A) 13 (B) 12
(C) 11 (D) 10
2.109 The nodal method of circuit analysis is based on
(A) KVL and Ohm’s law (B) KCL and Ohm’s law
(C) KCL and KVL (D) KCL, KVL and Ohm’s law
2.110 Superposition theorem is NOT applicable to networks containing
(A) nonlinear elements (B) dependent voltage sources
(C) dependent current sources (D) transformers
2.111 The parallel RLC circuit shown in the figure is in resonance. In this
circuit
(A) IR < 1 mA (B) IR + IL > 1 mA
(C) IR + IC < 1 mA (D) IR + IC > 1 mA
0 /
1 2
2.112 The short-circuit admittance matrix a two-port network is /
1 2
0
-
> H
The two-port network is
(A) non-reciprocal and passive (B) non-reciprocal and active
(C) reciprocal and passive (D) reciprocal and active
2.113 The voltage across the terminals a and b in the figure is
(A) 0.5 V (B) 3.0 V
(C) 3.5 V (D) 4.0 V
2.114 A high-Q quartz crystal exhibits series resonance at the frequency

by RK Kanodia
Now in 3 Volume
b16 lW (B) 3
b 8 + 12 lW (D) None of the above
ws and parallel resonance at the frequency wp . Then
(A) ws is very close to, but less than wp
(B) ws << wp
(C) ws is very close to, but greater than wp
(D) ws >> wp
GATE 1997 ONE MARK
2.115 The current i4 in the circuit of the figure is equal to
(A) 12 A (B) -12 A
(C) 4 A (D) None or these
2.116 The voltage V in the figure equal to
(A) 3 V (B) -3 V
(C) 5 V (D) None of these
2.117 The voltage V in the figure is always equal to
(A) 9 V (B) 5 V
(C) 1 V (D) None of the above
2.118 The voltage V in the figure is
(A) 10 V (B) 15 V
(C) 5 V (D) None of the above
2.119 In the circuit of the figure is the energy absorbed by the 4 W resistor
in the time interval (0,3) is
(A) 36 Joules (B) 16 Joules
(C) 256 Joules (D) None of the above
2.120 In the circuit of the figure the equivalent impedance seen across
terminals a,b, is
(A) 3
b 8 lW
(C) 3 j
GATE 1996 ONE MARK
2.121 In the given figure, A1,A2 and A3 are ideal ammeters. If A2 and A3
read 3 A and 4 A respectively, then A1 should read
(A) 1 A (B) 5 A
(C) 7 A (D) None of these
2.122 The number of independent loops for a network with n nodes and
b branches is
(A) n - 1
(B) b - n
(C) b - n + 1
(D) independent of the number of nodes
GATE 1996 TWO MARKS
2.123 The voltages VC1,VC2, and VC3 across the capacitors in the circuit in

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the given figure, under steady state, are respectively.
(A) 80 V, 32 V, 48 V (B) 80 V, 48 V, 32 V
(C) 20 V, 8 V, 12 V (D) 20 V, 12 V, 8 V

R R
= b a + +
R R
= a c + +
R R
= b c + +
^ ^h h
k R k R
a b c
b c = + +
k
b c
R R R
by RK Kanodia
Now in 3 Volume
^ h
j V
4 3
4 s = +
j
4 3
40 = + c
40 90
c
= c
= 800 90c
V
WX
= 1 25
VYZ = 0.8 (attenuation factor)
V
WX
YZ = ^0.8h^1.25h = 1
100
YZ
1 =
V
100
WX
2 =
= w 1
L
1
= w 2
L
2
X
= Leq
= w L 1 +
w L
2 +
w L
1
w 2
R R
R R
1 2
1 2
+
R q
R q
1
2
2
+
1 1
+
q R q R
= 1 1 2 2 +
SOLUTIONS
In the equivalent star connection, the resistance can be given as
RC R R R
a b c
RB R R R
a b c
RA R R R
a b c
So, if the delta connection components Ra , Rb and Rc are scaled
by a factor k then
RAl kR kR kR
k
R R
a b c
2
= + +
= k RA
hence, it is also scaled by a factor k
For the given capacitance, C = 100mF in the circuit, we have the
reactance.
= 1
XC sc
1
= - s
6 # #
s 100 10
104
=
So,
^
h
V s
1
2
^
V s
h
4
10 10
s
4
+
=
4
4 10 + 10 4
+
10
s s
s
= 1 s
+
2
+
For the purely resistive load, maximum average power is transferred
when
= 2 + 2
RL RTh XTh
where RTh + jXTh is the equivalent thevinin (input) impedance of
the circuit. Hence, we obtain
RL = 42+ 32
5 W
For evaluating the equivalent thevenin voltage seen by the load RL
, we open the circuit across it (also if it consist dependent source).
The equivalent circuit is shown below
As the circuit open across RL so
I2 = 0
or, j40I2 = 0
i.e., the dependent source in loop 1 is short circuited. Therefore,
VL1 j
VTh= 10VL1 j 100 53.13
.
100 53.13
5 5313
c
For the given transformer, we have
V
.
1
Since, V
So, V
or, VYZ = VWX
at VWX1 = 100 V; V
V
100
WX
1
at VWZ2 = 100 V; V
100
YZ
2
The quality factor of the inductances are given by
q1 R
and q2 R
So, in series circuit, the effective quality factor is given by
Q R
R R
eq
1 2
R R
L
1 1
2 1
=
+
R R
2
2 1
=
+
R R
1 2
Consider that the voltage across the three capacitors C1, C2 and C3
are V1, V2 and V3 respectively. So, we can write
V
3
= 3
2 C
V
C
2
....(1)

v s
v
0
c c 0
C C
1 2
= 1 2 b + l = 12Ceq vC (0) = 12 V
= 1 2 q 25 5 cos 36.86 2
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Since, Voltage is inversely proportional to capacitance
Now, given that
C1 = 10 mF ; V 10V ^ 1hmax =
C2 = 5 mF ; V 5 V ^ 2hmax =
C3 = 2 mF ; V 2V ^ 3hmax =
So, from Eq (1) we have
V
3
2 5 2
V
=
for V ^ 3hmax = 2
We obtain,
= 2#2 = <5
V2 5 0.8 volt
i.e., V2 < V ^ 2hmax
Hence, this is the voltage at C2. Therefore,
V3 = 2 volt
V2 = 0.8 volt
and V1 = V2+ V3 = 2.8 volt
Now, equivalent capacitance across the terminal is
C C C
2 3
= + + 1 5 2
Ceq 2 3
C C
= 5#2 + + 10 = 80 7 m
F
Equivalent voltage is (max. value)
Vmax = V1 = 2.8
So, charge stored in the effective capacitance is
= b 80 l#^2 8h = 32 mC
Q = CeqVmax 7 .
At the node 1, voltage is given as
V1 = 10 volt
Applying KCL at node 1
I V V
S 2 1 2 0
+ 1 + 1 - =
I 2 2
S+ 10
+ 10 1
- =
0 IS =- 13 A
Also, from the circuit,
VS - 5#2 = V1
VS = 10 + V1
= 20 volt
Again from the shown circuit, the current in 1 W resistor is
I V 10 A
= 1 = 10 =
1 1
The s-domain equivalent circuit is shown as below.
I (s)
( )/ ( )
C s C s
C C
1 1
1 1
1 2 1 2
=
+
=
+
I (s) C C (12 V)
Taking inverse Laplace transform for the current in time domain,
i (t) = 12Ceq d(t) (Impulse)
In phasor form, Z = 4 - j3 = 5 -36.86cW
I = 5 100c A
Average power delivered.
Pavg. 2 I Zcos
1 = # # c = 50W
Alternate method:
Z = (4 - j3)W, I = 5 cos (100pt + 100)A
1 2 = $ . 2 Re ( ) ( j )
Pavg 2 Re I Z
1 5 2 4 3 = # " # - ,
1 = # =
2 100 50W
2.12 Option (C) is correct
Applying nodal analysis at top node.
V 1+ 1 0
c 1
+ V
1+ 1
0c j
1
= 1 0c
V1(j1 + 1) + j1 + 1 0c = j1
V1 = -
1 1 +
j1
V j
Current I1 1
j 1
j
0
1
1
1 1
1 c = + =
- + +
j
1 1 j
= 1 + = +
( )
A
j j
We obtain Thevenin equivalent of circuit B.

by RK Kanodia
Now in 3 Volume
= 6 =
Thevenin Impedance :
ZTh = R
Thevenin Voltage :
VTh = 3 0c V
Now, circuit becomes as
-
= 10 3 +
Current in the circuit, I1 2 R
Power transfer from circuit A to B
2 2
= + 1
P (I1) R 3I
- : D : D
- + +
2
P 10 3 10 3
= +
2 R R 3 2 R
P
R
49
2 R
( R) ( 2
)
21
2 =
+
+ +
P
+ +
R R
2
49 21 2
( )
R
2 =
+
( )
P
R
+
42 70
=
( 2 +
R
)
2 dP
dR
+ - + +=
R R R
( 2 ) 70 ( 42 70 ) 22
( )
R
2
( )
0 4
2
=
+
(2 + R) [(2 + R)70 - (42 + 70R)2] = 0
140 + 70R - 84 - 140R = 0
56 = 70R
R = 0.8 W
In the given circuit
VA- VB = 6 V
So current in the branch will be
IAB 2 3 A
We can see, that the circuit is a one port circuit looking from
terminal BD as shown below
For a one port network current entering one terminal, equals the
current leaving the second terminal. Thus the outgoing current from
A to B will be equal to the incoming current from D to C as shown
i.e. IDC = IAB = 3 A
The total current in the resistor 1 W will be
I1 = 2 + IDC (By writing KCL at node D)
= 2 + 3 = 5 A
So, VCD = 1#(- I1) =- 5 V
When 10 V is connected at port A the network is
Now, we obtain Thevenin equivalent for the circuit seen at load
terminal, let Thevenin voltage is VTh,10 V with 10 V applied at port A
and Thevenin resistance is RTh .
V ,10 V
Th L
= Th+
IL R R
For RL = 1W, IL = 3 A
V
= Th + ...(i)
3 R
,10 V
1
Th
For RL = 2.5 W, IL = 2 A
V
= Th+ ,10 V
R 2 .
5
...(ii)
Th

= 10#10 10 + + = 15 W
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Dividing above two
R
+
3 .
2
= Th R
+
2 5
1
Th
3RTh + 3 = 2RTh + 5
RTh = 2 W
Substituting RTh into equation (i)
VTh,10 V = 3(2 + 1) = 9 V
Note that it is a non reciprocal two port network. Thevenin voltage
seen at port B depends on the voltage connected at port A. Therefore
we took subscript VTh,10 V. This is Thevenin voltage only when 10 V
source is connected at input port A. If the voltage connected to port
A is different, then Thevenin voltage will be different. However,
Thevenin’s resistance remains same.
Now, the circuit is as shown below :
V2
,10 V 9
= Th + = + =
For RL = 7 W, IL R 2 7
1A
L
Now, when 6 V connected at port A let Thevenin voltage seen at
port B is VTh,6 V. Here RL = 1W and I L =
7
3 A
7 1 3
VTh,6 V R 3
7
= 7
# + 7 =
= Th# + # 2 3 3
7 V
This is a linear network, so VTh at port B can be written as
VTh = V1a + b
where V1 is the input applied at port A.
We have V1 = 10 V, VTh,10 V = 9 V
` 9 = 10a + b ...(i)
When V1 = 6 V, VTh, 6 V = 9 V
` 7 = 6a + b ...(ii)
Solving (i) and (ii)
a = 0.5, b = 4
Thus, with any voltage V1 applied at port A, Thevenin voltage or
open circuit voltage at port B will be
So, VTh,V1 = 0.5V1+ 4
For V1 = 8 V
VTh,8 V = 0.5#8 + 4 = 8 = Voc (open circuit voltage)
Replacing P - Q by short circuit as shown below we have
Using current divider rule the current Isc is
ISC = 25 + 1255+ j30 (16 0) = (6.4 - j4.8)A
Power transferred to RL will be maximum when RL is equal to the
Thevenin resistance. We determine Thevenin resistance by killing
all source as follows :
RTH 10 10
The given circuit is shown below
For parallel combination of R and C equivalent impedance is
Zp
1
R j C
w
= w
R j C
R
j RC
1
1
$
w
+
= +
Transfer function can be written as
V
out Z Z
V
in
Z
s p
= p +
= w
1
R
w w
R
j RC
1
+ + +
R j C 1
j RC
+
= w
j RC
w + 1 +
w 2
j RC ( j RC)
=
j
j + ( 1 +
j)
2 w = 1
Here RC
V
out
V
in
j
1 3
( j) j
1
2 =
+ +
=
V
3 t RC
= b p l
Thus vout cos ( / )
From star delta conversion we have

R R
6 6
= a b W + + = + + =
+ j - j
+ =
(2 4)(2 4)
= + - 2 7 W
j j
14 0c c
= + = +
by RK Kanodia
Now in 3 Volume
= 1 + 1- 2 4V 2V = 1- 2 ...(1)
V V V
0 5 0 5
= 2 + 2- 1 2V 4V =- 1+ 2 ...(2)
V V V
0 5 0 5
-
> H
1
w w
1
w = wC
= 1 = (maximum at resonance)
Thus R1 . R R R 6 6 6
2
a b c
Here R1 = R2 = R3 = 2 W
Replacing in circuit we have the circuit shown below :
Now the total impedance of circuit is
Z
(2 4)(2 4)
Current I 7 2 0
From given admittance matrix we get
I1 = 0.1V1- 0.01V2 and ...(1)
I2 = 0.01V1+ 0.1V2 ...(2)
Now, applying KVL in outer loop;
V2 =- 100I2
or I2 =- 0.01V2 ...(3)
From eq (2) and eq (3) we have
-0.01V2 = 0.01V1+ 0.1V2
-0.11V2 = 0.01V1
V
1
2 11
V
= -1
Here we take the current flow direction as positive.
At t = 0- voltage across capacitor is
- C
VC (0 )
=-Q 2 . 5 10
50 V
=- # =- -
50 10
3
6
#
-
+ =- 50 V
Thus VC (0 )
In steady state capacitor behave as open circuit thus
V(3) = 100 V
= 3 + + - 3 -t RC
Now, VC (t) V ( ) (V (0 ) V ( ))e /
C C C
-
t
- = 100 + (- 50 - 100)e
100 150e (2 103t) = - - #
10 50 10 6 # #
= dV
50 10 6 150 2 103e 2 103t A
Now ic (t) C dt
= # - # # # - #
15e 2 103t = - #
ic (t) 15 exp( 2 103t)A = - #
Given circuit is as shown below
By writing node equation at input port
I1 . .
By writing node equation at output port
I2 . .
From (1) and (2), we have admittance matrix
Y
4
2
2
= - 4
A parallel RLC circuit is shown below :
Input impedance Zin
R j L j C 1 1
=
+ +
At resonance L
So, Zin
/R
R
1
Thus (D) is not true.
Furthermore bandwidth is B w i.e R 1
wB and is independent of L
,
Hence statements A, B, C, are true.
Let the current i (t) = A + Be-t/t t " Time constant
When the switch S is open for a long time before t < 0, the circuit is
At t = 0, inductor current does not change simultaneously, So the
circuit is

A- + + A- 0 = 0
= 10 - 4 =
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Current is resistor (AB)
= 0 75 =
i (0) .2 0.375 A
Similarly for steady state the circuit is as shown below
= 15 =
i (3) 3 0.5 A
L
eq
=
t R
-
3
15 10 10
= # 3 + =
( || )
sec
10 10 10
-
i (t) A Be
t
- - = A + Be-100t
= + 1#10 3
Now i (0) = A + B = 0.375
and i (3) = A = 0.5
So, B = 0.375 - 0.5 =- 0.125
Hence i (t) = 0.5 - 0.125e-1000 t A
Circuit is redrawn as shown below
Where, Z1 j L j 103 20 10 3 20j = w = # # # - =
Z2 = R || XC
XC j C j
1 20 3 6 w # # # = = =- -
1 j
10 50 10
Z2
j
-
( )
= 1 20 -
j
1 20
R = 1W
Voltage across Z2
Z 0
1 2
j
j
20
-
-
1 20
= 2 : + 20
VZ2 Z Z 20
= :
j
20
- -
j j
20 1 20
c
c
m
m
- j
c m
=- ( 20)
j j 20
20 400 20 j = + - :
Current in resistor R is
=VZ2 j j A =-1 =-
I R
The circuit can be redrawn as
Applying nodal analysis
V V
2
10 1 2
2VA- 10 + 2 = 0 = V4 = 4 V
Current, I1 2 3 A
Current from voltage source is
I2 = I1- 3 = 0
Since current through voltage source is zero, therefore power
delivered is zero.
Circuit is as shown below
Since 60 V source is absorbing power. So, in 60 V source current
flows from + to - ve direction
So, I + I1 = 12
I = 12 - I1
I is always less then 12 A So, only option (A) satisfies this condi-tions.
For given network we have
V0
R X V
( )
R R X
L C i =
( )
L C
+
V s
i
0
( )
V ( s
)
R
sR C
L
R
sR C
R
1
1
L
L
L
=
+
+
+
R
L L
= L
+ +
R RRsC R
R
L L
= L
+ +
R RRsC R
1 R RsC
R
1
L
=
+ +
But we have been given
T.F.
V s
i
( )
V ( s
)
= 0
= 1
+
2 sCR
Comparing, we get
1 R
+ = 2 & RL = R
R
L

t t
= # = # = I # Area
2 ( ) .
= # 1 10 + 12 # 600 = 13 2 kJ
V
s L
= 20 + 0
c 20+0c
= 1 8 - 6
20 0c
+
f
= = 2+ - f where tan
20+0c 2
= # + = (7 + 4j) = 28 + 16j
by RK Kanodia
Now in 3 Volume
= 100 = 12 5 A
= + x = 100 - 50 = 12 5 A
100
= oc = = W
= V0
The energy delivered in 10 minutes is
E VIdt I Vdt
0 0
2
From given circuit the load current is
IL
Z Z
=
+
(1 2j) (7 4j) 8 6
j
+ + +
=
+
( j)
5
10
+
4 3
f = -1
The voltage across load is
VL = ILZL
The reactive power consumed by load is
Pr V I*
L = L L = I *
L Z # I 2
L L =
Z L I (7 4j) 8 6j
Thus average power is 28 and reactive power is 16.
At t = 0-, the circuit is as shown in fig below :
V(0-) = 100 V
Thus V(0+) = 100 V
At t = 0+, the circuit is as shown below
+ = 100 = 20 mA
I ( )
k
0
5
At steady state i.e. at t = 3 is I (3)= 0
Now i (t) = I (0 + )e - t
RC eq
u(t)
Ceq
= m m m m
+ = F
( . . ) .
0 5 0 3 0 2
m + m +
m
. . .
0.16
0 5 0 3 0 2
1
eq
RC
= = -
3 6 # # #
.
1250
1
5 10 0 16 10
i (t) = 20e-1250tu(t) mA
For Pmax the load resistance RL must be equal to thevenin resistance
Req i.e. RL = Req . The open circuit and short circuit is as shown
below
The open circuit voltage is
Voc = 100 V
From fig I1 .
8
Vx =- 4 # 12.5 =- 50 V
I2 V .
4
100
4
Isc = I1+ I2 = 25 A
Rth 4
I V
25
sc
Thus for maximum power transfer RL = Req = 4 W
Steady state all transient effect die out and inductor act as short
circuits and forced response acts only. It doesn’t depend on initial
current state. From the given time domain behavior we get that
circuit has only R and L in series with V0. Thus at steady state
i (t) " i (3)
R
The given graph is
There can be four possible tree of this graph which are as follows:
There can be 6 different possible cut-set.
Initially i (0-) = 0 therefore due to inductor i (0+) = 0. Thus all
current Is will flow in resistor R and voltage across resistor will be
IsRs . The voltage across inductor will be equal to voltage across Rs
as no current flow through R.

t
= # = -
V s
s
( )
V s
1
=
s
^ 1
+ + h
1
2 =
+ +
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+ I R = s s
Thus vL (0 )
+
but vL (0 )
= di 0
( )
L dt
+
Thus
di 0+
( )
( )
dt
= L 0 = IsRs
L
v
L
+
Killing all current source and voltage sources we have,
= + 1 +
Zth (1 s) (s 1)
+ s
+ =
+ + +
( 1 )( 1
) [ ]
s
( ) ( )
s
s
1 1
1 1
1 1
s
s
s
s
1
1
1
1
=
+ + +
+ + +
or Zth = 1
Alternative :
Here at DC source capacitor act as open circuit and inductor act
as short circuit. Thus we can directly calculate thevenin Imped-ance
as 1 W
Z(s) R
sC
sL
s
1
s
RC
LC
C s
2 1 = =
+ +
We have been given
Z(s)
0 2
.
s .
s
s
0 1 2
2 =
+ +
Comparing with given we get
1 = 0.2 or C = 5 F
C
1 = 0.1 or R = 2 W
RC
1 = 2 or L = 0.1 H
LC
Voltage across capacitor is
Vc
1 idt t
= #
C
0
Here C = 1 F and i = 1 A. Therefore
t
= #
Vc dt
0
For 0 < t < T, capacitor will be charged from 0 V
t
= # =
Vc dt t
0
At t = T,Vc = T Volts
For T < t < 2T, capacitor will be discharged from T volts as
Vc T dt 2T t
T
t
= - # = -
= = volts
At 2 , 0 t T Vc
For 2T < t < 3T, capacitor will be charged from 0 V
t
= # = -
Vc dt t 2T
T
2
At t = 3T,Vc = T Volts
For 3T < t < 4T, capacitor will be discharged from T Volts
t
= - # = -
Vc T dt 4T t
T
3
= = Volts
At 4 , 0 t T Vc
For 4T < t < 5T, capacitor will be charged from 0 V
Vc dt t 4T
T
4
At t = 5T,Vc = T Volts
Thus the output waveform is
Only option C satisfy this waveform.
Writing in transform domain we have
( )
c
s s 1
(s s 1)
Since Vs (t) = d(t) " Vs (s) = 1 and
Vc (s)
1
2 =
+ +
(s s 1)
or Vc (s)
2
2 1
3
3 (s )
2
3
4
2
=
+ + = G
Taking inverse Laplace transform we have
2
t 3
= -2 c m
Vt e sin t
3
2
Let voltage across resistor be vR
V s
S
( )
V s
( )
R
s
1 2 =
+ +
=
( s 1
) (s s )
1
+ +
s 1
Since vs = d(t) " Vs (s) = 1 we get
VR(s)
s
3 =
(s s ) ( s
)
s
2 1
2 1
2
4
+ +
=
+ +
s
+
( )
s 3
( )
s 2 1
( )
2
4
2 1
2 1
2
3
4
2 1
=
+ +
-
+ +
3
1
2
1
3 2
or vR(t) e cos 2 t 2
e sin t
3
2
1
2
= - - # -
3
1
3 t2
= - - = G
e cos 2 t sin t
3
2
From the problem statement we have
i v
6 1 5
= 1
= = W
z11 .
4
1 i
=
0 2
i v
z12 . .
4 5 4 5
= = = W
1
1
2 i
=
0 1
i v
6 1 5
= 2
= = W
z21 .
4
1 i
=
0 2
i v
z22 . .
1 5 1 5
= = = W
1
2
2 i
=
0 2
Thus z -parameter matrix is

4 .
5 3
v
= 1
= =
1 0 67
= 2
= =
i
2
1
= - z 21
= - 1 .
5 =- 1
=
h
21 1
by RK Kanodia
Now in 3 Volume
R
R
R
R
= = =
B
1
1 =
th + th + th - = 2
= th
= th
th + th + th - th = 2
i V
4
= th = = W
z
z
z
z
11
21
12
22
= G
.
.
.
.
1 5
1 5
4 5
= = 1 5G
From the problem statement we have
h12
.
v
1 5
2 i
=
0 1
h22
.
.
v i
1 5
2 i
=
0 1
From z matrix, we have
v1 = z11i1+ z12i2
v2 = z21i1+ z22i2
If v2 = 0
Then
i
.
z
1 5
22
or i2 =- i1
Putting in equation for v1, we get
v1 = (z11- z12) i1
i v
1 v
0 2=
= h11 = z11- z12 = 1.5 - 4.5 =- 3
Hence h -parameter will be
h
h
h
h
11
21
12
22
-
=- G
3
1
3
= G .
= 0 67
According to maximum Power Transform Theorem
ZL = Z* s = (R s - jX s
)
At w " 3, capacitor acts as short circuited and circuit acts as shown
in fig below
Here we get
V 0 =
0
V
i
At w " 0, capacitor acts as open circuited and circuit look like as
shown in fig below
Here we get also
V 0 =
0
V
i
So frequency response of the circuit is as shown in fig and circuit is
a Band pass filter.
We know that bandwidth of series RLC circuit is
L R
. Therefore
Bandwidth of filter 1 is B
L
1
1
=
Bandwidth of filter 2 is
/
B
L
L
L
4
4
2
2 1 1
Dividing above equation
B
4
2
Here Vth is voltage across node also. Applying nodal analysis we get
V V V i
2 1 1 2
But from circuit i V V
1
Therefore
V V V V
2 1 1 2
or Vth = 4 volt
From the figure shown below it may be easily seen that the short
circuit current at terminal XY is isc = 2 A because i = 0 due to
short circuit of 1 W resistor and all current will pass through short
circuit.
Therefore Rth 2
2
sc
The voltage across capacitor is
At t = 0+, Vc (0+) = 0
At t = 3, VC (3) = 5 V
The equivalent resistance seen by capacitor as shown in fig is
Req = 20 20 = 10kW
Time constant of the circuit is
t = ReqC = 10k # 4m = 0.04 s
Using direct formula
= 3 - 3 - -t t
V ( ) (1 e / ) V (0)e /
Vc (t) V ( ) [V ( ) V (0)]e /
C c c
= 3 - - t + -t t = 5(1 - e-t/0.04)
C
t
C
or Vc (t) = 5(1 - e-25t)

GATE Electronics and Communication Topicwise Solved Papers 1996-2013

GATE Electronics and Communication Topicwise Solved Papers 1996-2013

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