GATE EE Solved Paper by RK Kanodia
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GATE 2014 Electrical Engineering Topicwise Solved Paper 2013 - 2000 RK Kanoida & Ashish Murolia
GATE MCQ Electrical Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia
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CONTENTS 1 2 3 4 5 6 7 8 9 10
Engineering Mathematics Electric Circuit and Fields Signals and Systems Electrical Machines Control Systems Power Systems Electrical & Electronics Measurement Analog and Digital Electronics Power Electronics General Aptitude
1 ENGINEERING MATHEMATICS
YEAR 2013 MCQ 1.1.1
MCQ 1.1.2
MCQ 1.1.3
ONE MARK
Given a vector field Fv = y2 xavx - yzavy - x2 avz , the line integral along a segment on the x -axis from x = 1 to x = 2 is (A) - 2.33 (B) 0 (C) 2.33 (D) 7
# Fv : dlv evaluated
2 - 2 x1 0 The equation > = > H has H > H 1 - 1 x2 0 (A) no solution
x1 0 (B) only one solution > H = > H x2 0
(C) non-zero unique solution
(D) multiple solutions
Square roots of - i , where i = - 1 , are (A) i , - i (B) cos d- p n + i sin d- p n, cos b 3p l + i sin b 3p l 4 4 4 4 (C) cos d p n + i sin b 3p l, cos b 3p l + i sin d p n 4 4 4 4 (D) cos b 3p l + i sin b- 3p l, cos b- 3p l + i sin b 3p l 4 4 4 4
MCQ 1.1.4
The curl of the gradient of the scalar field defined by V = 2x2 y + 3y2 z + 4z2 x is (A) 4xyavx + 6yzavy + 8zxavz (B) 4avx + 6avy + 8avz (C) ^4xy + 4z2h avx + ^2x2 + 6yz h avy + ^3y2 + 8zx h avz (D) 0
MCQ 1.1.5
A continuous random variable X has a probability density function f ^x h = e-x , 0 < x < 3. Then P "X > 1, is (A) 0.368 (B) 0.5 (C) 0.632 (D) 1.0 YEAR 2013
MCQ 1.1.6
MCQ 1.1.7
TWO MARKS
When the Newton-Raphson method is applied to solve the equation f ^x h = x3 + 2x - 1 = 0 , the solution at the end of the first iteration with the initial value as x 0 = 1.2 is (A) - 0.82 (B) 0.49 (C) 0.705 (D) 1.69 A function y = 5x2 + 10x is defined over an open interval x = ^1, 2h. Atleast at one point in this interval, dy/dx is exactly (A) 20 (B) 25
GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) 30 MCQ 1.1.8
# zz
, is
2 2
(D) 35
- 4 dz evaluated anticlockwise around the circle z - i = 2 , where i = +4
(A) - 4p (C) 2 + p MCQ 1.1.9
Page 4
-1
(B) 0 (D) 2 + 2i
1 A Matrix has eigenvalues - 1 and - 2 . The corresponding eigenvectors are > H -1 1 and > H respectively. The matrix is -2 1 1 (A) > - 1 - 2H
1 2 (B) > - 2 - 4H
-1 0 (C) > 0 - 2H
0 1 (D) > - 2 - 3H
YEAR 2012
ONE MARK
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MCQ 1.1.10
Two independent random variables X and Y are uniformly distributed in the interval 6- 1, 1@. The probability that max 6X, Y @ is less than 1/2 is (B) 9/16 (A) 3/4 (C) 1/4 (D) 2/3
MCQ 1.1.11
If x = - 1, then the value of xx is (A) e- p/2 (C) x
MCQ 1.1.12
(B) e p/2 (D) 1
1 - 2 . If C is a counter clockwise path in the z -plane z+1 z+3 such that z + 1 = 1, the value of 1 f (z) dz is 2p j C (A) - 2 (B) - 1 (C) 1 (D) 2 Given f (z) =
#
MCQ 1.1.13
With initial condition x (1) = 0.5 , the solution of the differential equation t dx + x = t , is dt (B) x = t 2 - 1 (A) x = t - 1 2 2 2
(C) x = t 2
(D) x = t 2
YEAR 2012 MCQ 1.1.14
MCQ 1.1.15
TWO MARKS
-5 -3 1 0 Given that A = > , the value of A3 is and I = > H 2 0 0 1H (A) 15A + 12I (B) 19A + 30I (C) 17A + 15I (D) 17A + 21I The maximum value of f (x) = x3 - 9x2 + 24x + 5 in the interval [1, 6] is (A) 21 (B) 25
GATE MCQ Electrical Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) 41
Page 5
(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 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
MCQ 1.1.18
Consider the differential equation d 2 y (t) dy (t) dy +2 + y (t) = d (t) with y (t) t = 0 =- 2 and 2 dt dt dt dy The numerical value of is dt t = 0 (A) - 2 (B) - 1 (C) 0 (D) 1 -
=0 t = 0-
+
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YEAR 2011
ONE MARK
MCQ 1.1.19
Roots of the algebraic equation x3 + x2 + x + 1 = 0 are (A) (+ 1, + j, - j) (B) (+ 1, - 1, + 1) (C) (0, 0, 0) (D) (- 1, + j, - j)
MCQ 1.1.20
With K as a constant, the possible solution for the first order differential equation dy = e-3x is dx (B) - 1 e3x + K (A) - 1 e-3x + K 3 3 (C) - 1 e-3x + K (D) - 3e-x + K 3
MCQ 1.1.21
A point Z has been plotted in the complex plane, as shown in figure below.
The plot of the complex number Y = 1 is Z
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
YEAR 2011 MCQ 1.1.22
Page 6
TWO MARKS
Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method. Equation (1) 10x2 sin x1 - 0.8 = 0 Equation (2) 10x 22 - 10x2 cos x1 - 0.6 = 0 Assuming the initial values are x1 = 0.0 and x2 = 1.0 , the jacobian matrix is 10 - 0.8 (A) > 0 - 0.6H
10 0 (B) > 0 10H
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0 - 0.8 (C) > 10 - 0.6H
10 0 (D) > 10 - 10H
MCQ 1.1.23
The function f (x) = 2x - x2 - x3 + 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 the signal is (B) a (A) a 3 2 (C) a 2 (D) a 3
MCQ 1.1.25
MCQ 1.1.26
2 1 The matrix [A] = > is decomposed into a product of a lower 4 - 1H triangular matrix [L] and an upper triangular matrix [U]. The properly decomposed [L] and [U] matrices respectively are 1 0 1 1 2 0 1 1 and > (B) > and > H (A) > H H H 4 -1 0 -2 4 -1 0 1 1 0 2 1 2 0 1 1.5 (C) > H and > (D) > and > H H 4 1 0 -1 4 -3 0 1H The two vectors [1,1,1] and [1, a, a2] where a = c- 1 + j 3 m, are 2 2 (A) Orthonormal (B) Orthogonal (C) Parallel (D) Collinear
GATE MCQ Electrical Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
YEAR 2010
ONE MARK 1
MCQ 1.1.27
The value of the quantity P , where P = # xex dx , is equal to 0 (A) 0 (B) 1 (C) e (D) 1/e
MCQ 1.1.28
Divergence of the three-dimensional radial vector field r is (A) 3 (B) 1/r t t t t) (C) i + j + k (D) 3 (ti + tj + k YEAR 2010
MCQ 1.1.29
MCQ 1.1.30
MCQ 1.1.31
MCQ 1.1.32
MCQ 1.1.33
TWO MARKS
A box contains 4 white balls and 3 red balls. In succession, two balls are randomly and removed form the box. Given that the first removed ball is white, the probability that the second removed ball is red is (A) 1/3 (B) 3/7 (C) 1/2 (D) 4/7
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At t = 0 , the function f (t) = sin t has t (A) a minimum (C) a point of inflection
(B) a discontinuity (D) a maximum
J1 1 0N K O An eigenvector of P = K0 2 2O is K0 0 3O (A) 8- 1 1 1BT L P T (C) 81 - 1 2B
(B) 81 2 1BT (D) 82 1 - 1BT
2 For the differential equation d x2 + 6 dx + 8x = 0 with initial conditions x (0) = 1 dt dt and dx = 0 , the solution is dt t = 0 (A) x (t) = 2e- 6t - e- 2t (B) x (t) = 2e- 2t - e- 4t (C) x (t) =- e- 6t + 2e- 4t (D) x (t) = e- 2t + 2e- 4t
For the set of equations, x1 + 2x2 + x 3 + 4x 4 = 2 and 3x1 + 6x2 + 3x 3 + 12x 4 = 6 . The following statement is true. (A) Only the trivial solution x1 = x2 = x 3 = x 4 = 0 exists (B) There are no solutions (C) A unique non-trivial solution exists (D) Multiple non-trivial solutions exist YEAR 2009
MCQ 1.1.34
Page 7
ONE MARK
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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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 8
YEAR 2009 MCQ 1.1.35
TWO MARKS
f (x, y) is a continuous function defined over (x, y) ! [0, 1] # [0, 1]. Given the two constraints, x > y2 and y > x2 , the volume under f (x, y) is y=1
x= y
(A)
#y = 0 #x = y
(C)
#y = 0 #x = 0
y=1
2
x=1
f (x, y) dxdy
f (x, y) dxdy
y=1
x=1
(B)
#y = x #x = y
(D)
#y = 0
2
y= x
2
f (x, y) dxdy
x= y
#x = 0
f (x, y) dxdy
MCQ 1.1.36
Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of at least two people in the room being born on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5 ? (A) 20 (B) 7 (C) 15 (D) 16
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 x2 - 117 = 0 . The iterative steps for the solution using Newton-Raphon’s method is given by (B) xk + 1 = xk - 117 (A) xk + 1 = 1 bxk + 117 l 2 xk xk (C) xk + 1 = xk - xk (D) xk + 1 = xk - 1 bxk + 117 l 2 xk 117
MCQ 1.1.39
F (x, y) = (x2 + xy) at x + (y2 + xy) at y . It’s (x, y) = (0, 2) to (x, y) = (2, 0) evaluates (A) - 8 (C) 8
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line integral over the straight line from to (B) 4 (D) 0
YEAR 2008
ONE MARKS
MCQ 1.1.40
X is a uniformly distributed random variable that takes values between 0 and 1. The value of E {X3} will be (A) 0 (B) 1/8 (C) 1/4 (D) 1/2
MCQ 1.1.41
The characteristic equation of a (3 # 3 ) matrix P is defined as a (l) = lI - P = l3 + l2 + 2l + 1 = 0 If I denotes identity matrix, then the inverse of matrix P will be (A) (P2 + P + 2I) (B) (P2 + P + I) (C) - (P2 + P + I)
MCQ 1.1.42
(D) - (P2 + P + 2I)
If the rank of a (5 # 6) matrix Q is 4, then which one of the following statement is correct ? (A) Q will have four linearly independent rows and four linearly independent columns (B) Q will have four linearly independent rows and five linearly independent
GATE MCQ Electrical Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia
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Page 9
columns T (C) QQ will be invertible (D) QT Q will be invertible YEAR 2008
TWO MARKS
MCQ 1.1.43
Consider function f (x) = (x2 - 4) 2 where x is a real number. Then the function has (A) only one minimum (B) only tow minima (C) three minima (D) three maxima
MCQ 1.1.44
Equation ex - 1 = 0 is required to be solved using Newton’s method with an initial guess x0 =- 1. Then, after one step of Newton’s method, estimate x1 of the solution will be given by (A) 0.71828 (B) 0.36784 (C) 0.20587 (D) 0.00000
MCQ 1.1.45
A is m # n full rank matrix with m > n and I is identity matrix. Let matrix A' = (AT A) - 1 AT , 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 - 2t u (t), has to be solved using trapezoidal rule of integration with a step size h = 0.01 s. Function u (t) indicates a unit step function. If x (0 -) = 0 , then value of x at t = 0.01 s will be given by (A) 0.00099 (B) 0.00495 (C) 0.0099 (D) 0.0198
MCQ 1.1.47
Let P be a 2 # 2 real orthogonal matrix and x is a real vector [x1, x2] T with length x = (x12 + x22) 1/2 . Then, which one of the following statements is correct ? (A) Px # x where at least one vector satisfies Px < x
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(B) Px # x for all vector x (C) Px $ x where at least one vector satisfies Px > x (D) No relationship can be established between x and Px YEAR 2007 MCQ 1.1.48
x = 8x1 x2 g xn 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 T
YEAR 2007 MCQ 1.1.49
ONE MARK
TWO MARKS
1-x The differential equation dx dt = t is discretised using Euler’s numerical integration method with a time step 3 T > 0 . What is the maximum permissible value of 3 T to ensure stability of the solution of the corresponding discrete time equation ? (A) 1 (B) t/2
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) t MCQ 1.1.50
(D) 2t
The value of C
(A) 2pi
where C # (1 dz + z2)
is the contour z - i/2 = 1 is (B) p (D) pi tan - 1 z
-1
(C) tan z MCQ 1.1.51
MCQ 1.1.52
Page 10
The integral 1 2p (A) sin t cos t (C) (1/2) cos t
2p
#0 sin (t - t) cos tdt equals (B) 0 (D) (1/2) sin t
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 of values 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
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MCQ 1.1.53
Let x and y be two vectors in a 3 dimensional space and < x, y > denote their dot product. Then the determinant < x, x > < x, y > det =< y, x > < y, y >G (A) is zero when x and y are linearly independent (B) is positive when x and y are linearly independent (C) is non-zero for all non-zero x and y (D) is zero only when either x or y is zero
MCQ 1.1.54
The linear operation L (x) is defined by the cross product L (x) = b # x , where T T b = 80 1 0B and x = 8x1 x2 x3 B are three dimensional vectors. The 3 # 3 matrix M of this operations satisfies R V Sx1 W L (x) = M Sx2 W SSx WW 3 T X Then the eigenvalues of M are (A) 0, + 1, - 1 (B) 1, - 1, 1 (C) i, - i, 1 (D) i, - i, 0
Statement for Linked Answer Question 46 and 47.
MCQ 1.1.55
Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix -3 2 A == - 2 0G A satisfies the relation
GATE MCQ Electrical Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
MCQ 1.1.56
(A) A + 3I + 2A - 1 = 0 (C) (A + I) (A + 2I)
(B) A2 + 2A + 2I = 0 (D) exp (A) = 0
A9 equals (A) 511A + 510I (C) 154A + 155I
(B) 309A + 104I (D) exp (9A)
YEAR 2006 MCQ 1.1.57
The expression V = (A)
(C) MCQ 1.1.59
#0
R
TWO MARKS
#0
H
pR2 (1 - h/H) 2 dh for the volume of a cone is equal to
pR2 (1 - h/H) 2 dr
(B)
#0
R
pR2 (1 - h/H) 2 dh
2 2prH`1 - r j dr R A surface S (x, y) = 2x + 5y - 3 is integrated once over a path consisting of the points that satisfy (x + 1) 2 + (y - 1) 2 = 2 . The integral evaluates to (A) 17 2 (B) 17 2
(C) MCQ 1.1.58
Page 11
H
#0 2prH (1 - r/R) dh
(D)
#0
R
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(D) 0
Two fair dice are rolled and the sum r of the numbers turned up is considered (A) Pr (r > 6) = 1 6 (B) Pr (r/3 is an integer) = 5 6 (C) Pr (r = 8 ; r/4 is an integer) = 5 9 (D) Pr (r = 6 ; r/5 is an integer) = 1 18
Statement for Linked Answer Question 60 and 61. VT R VT R VT R S 2 W S- 2 W S- 10 W P = S 1 W , Q = S- 5 W , R = S- 7 W are three vectors. SS 12 WW SS 9 WW SS 3 WW X T X T X T MCQ 1.1.60 An orthogonal set of vectors having a span that contains P, Q, Ris V R V R V R V R V R S- 6 W S 4 W S- 4 W S 5 W S 8 W (A) S- 3 W S- 2 W (B) S 2 W S 7 W S 2 W SS- 6 WW SS 3 WW SS 4 WW SS- 11WW SS- 3 WW TR VX R T V RX VT X RT VX RT VX R V 6 3 3 S W S W S W S 4 W S 1 W S5 W (C) S 7 W S 2 W S 9 W (D) S 3 W S31W S 3 W SS- 1WW SS- 2 WW SS- 4 WW SS11WW SS 3 WW SS 4 WW X T X vector T X is linearly dependent upon T X T X TtheX solution MCQ 1.1.61 The Tfollowing to the previous problem V R V R S8 W S -2 W (A) S 9 W (B) S- 17 W SS 3 WW SS 30 WW TR V X RT VX S4 W S 13 W (C) S 4 W (D) S 2 W SS 5 WW SS- 3 WW X T X T and For more GATE Resources, Mock Test Study material
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
YEAR 2005
Page 12
ONE MARK
MCQ 1.1.62
In the matrix equation Px = q , which of the following is a necessary condition for the existence of at least on solution for the unknown vector x (A) Augmented matrix [Pq] must have the same rank as matrix P (B) Vector q must have only non-zero elements (C) Matrix P must be singular (D) Matrix P must be square
MCQ 1.1.63
If P and Q are 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 P and Q are mutually exclusive, then they must be independent (D) Probability (P + Q) # Probability (P)
MCQ 1.1.64
If S =
MCQ 1.1.65
#1
3 -3
x dx , then S has the value
(B) 1 (A) - 1 3 4 (C) 1 (D) 1 2 The solution of the first order DE x' (t) =- 3x (t), x (0) = x0 is (B) x (t) = x0 e - 3 (A) x (t) = x0 e - 3t
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(C) x (t) = x0 e - 1/3
(D) x (t) = x0 e - 1
YEAR 2005 MCQ 1.1.66
MCQ 1.1.67
MCQ 1.1.68
TWO MARKS
V R S3 - 2 2 W For the matrix p = S0 - 2 1 W, one of the eigen values is equal to - 2 SS0 0 1 WW T is anXeigen vector ? Which of the following R V R V S 3 W S- 3 W W S (A) - 2 (B) S 2 W SS 1 WW SS- 1WW TR V X RT VX 1 S W S2 W (C) S- 2 W (D) S 5 W SS 3 WW SS 0 WW T RX T X V 1 0 1 W S If R = S2 1 - 1W, then top row of R - 1 is SS2 3 2 WW X T (A) 85 6 4B (B) 85 - 3 1B (C) 82 0 - 1B (D) 82 - 1 1/2B
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 (B) 1 (A) 1 8 2 (C) 3 (D) 3 8 4
GATE MCQ Electrical Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 1.1.69
MCQ 1.1.70
MCQ 1.1.71
Page 13
For the function f (x) = x2 e - x , the maximum occurs when x is equal to (A) 2 (B) 1 (C) 0 (D) - 1 2 y2 For the scalar field u = x + , magnitude of the gradient at the point (1, 3) is 2 3 13 9 (A) (B) 9 2 (C) 5 (D) 9 2 For the equation x'' (t) + 3x' (t) + 2x (t) = 5 ,the solution x (t) approaches which of the following values as t " 3 ? (A) 0 (B) 5 2
(C) 5
(D) 10 ***********
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
SOLUTION
SOL 1.1.1
Page 14
Option (B) is correct. Given, the field vector Fv = y2 xavx - yzavy - x2 avz For the line segment along x -axis, we have dlv = dxavx So, Fv : dlv = ^y2 x h^dx h Since, on x -axis y = 0 so, Fv : dlv = 0 or, Fv : dlv = 0
#
SOL 1.1.2
Option (D) is correct. Given the equations in matrix form as 2 - 2 x1 0 > H> H = > H 1 - 1 x2 0
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So, it is a homogenous set of linear equation. It has either a trivial solution ^x1 = x2 = 0h or an infinite no. of solution. Since, for the matrix 2 -2 H A => 1 -1 we have the determinant A =0 Hence, it will have multiple solutions SOL 1.1.3
Option (B) is correct. We know that (In phasor form) i = eip/2 -ip/2 or, -i = e 1/2 So, - i = !^e-ip/2h = ! e-ip/4 = !=cos d p n - i sin d p nG 4 4 = cos d p n - i sin d p n ; - cos d p n + i sin d p n 4 4 4 4 = 1 - i ; -1 + i 2 2 2 This is equivalent to the given option (B) only.
SOL 1.1.4
Option (D) is correct. Given the scalar field V = 2x2 y + 3y2 z + 4z2 x Its gradient is given by dV = ^4xy + 4z2h avx + _2x2 + 6yz i avy + ^3y2 + 8zx h avz So, the curl of the gradient is obtained as avx avy avz 2 2 2 d # ^dvh = rx ry rz 2 2 2 4xy + 4z 2z + 6yz 3y + 8zx
GATE MCQ Electrical Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia
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Page 15
= avx ^6y - 6y h - avy ^8z - 8z h + avz ^4x - 4x h =0 Note : From the properties of curl, we know that curl of gradient of any scalar field 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 x as f ^x h = e-x So, P ^x > 1h =
#e
3-x
1
0 H, so, we have -1 1 a b 1 >c d H>- 1H =- 1 >- 1H or, Similarly, for a >c or,
a - b =- 1 c-d = 1
....(1) ....(2)
1 eigen value - 2 with eigen vector > H, we obtain -2 1 b 1 =- 2 > H -2 d H>- 2H ....(3) a - 2b =- 2 ....(4) c - 2d = 4
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 17
Solving Eqs. (1) and (3), we obtain a = 0, b = 1 and solving Eqs. (2) and (4), we obtain c =- 2, d = 3 Thus, the required matrix is 0 1 a b >c d H = >- 2 - 3H SOL 1.1.10
Option (B) is correct. Probability density function of uniformly distributed variables X and Y is shown as
P &[max (x, y)] < 1 0 2 Since X and Y are independent. P &[max (x, y)] < 1 0 = P b X < 1 l P bY < 1 l 2 2 2 P b X < 1 l = shaded area = 3 2 4 Similarly for Y : P bY < 1 l = 3 2 4 So P &[max (x, y)] < 1 0 = 3 # 3 = 9 2 4 4 16
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Alternate method:
From the given data since random variables X and Y lies in the interval [- 1, 1] as from the figure X , Y lies in the region of the square ABCD . Probability for max 6X, Y @ < 1/2 : The points for max 6X, Y @ < 1/2 will be inside the region of square AEFG . So, P &max 6X, Y @ < 1 0 = Area of 4AEFG 2 Area of square ABCD
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 18
3 3 #2 2 = = 9 2#2 16 SOL 1.1.11
Option (A) is correct. - 1 = i = cos p + i sin p 2 2
x=
p
x = ei 2
So,
xx = ^ei 2 h & ^ei 2 h = e- 2 p i
p x
SOL 1.1.12
p
Option (C) is correct. f (z) = 1 - 2 z+1 z+3 1 f (z) dz = sum of the residues of the poles which lie inside the given 2p j C closed region.
#
C & z+1 = 1 Only pole z =- 1 inside the circle, so residue at z =- 1 is. (z + 1) (- z + 1) 2 -z + 1 f (z) = = lim = =1 2 z "- 1 (z + 1) (z + 3) (z + 1) (z + 3) 1 So f (z) dz = 1 2p j C Option (D) is correct. t dx + x = t dt dx + x = 1 t dt dx + Px = Q (General form) dt
nodia.co.in #
SOL 1.1.13
IF = e # = e = e lnt = t
Integrating factor, Solution has the form
Pdt
1 # dt t
# ^Q # IF hdt + C x # t = # (1) (t) dt + C
x # IF =
2 xt = t + C 2 Taking the initial condition
x (1) = 0.5 0.5 = 1 + C 2 C =0 2 xt = t & x = t 2 2
So, SOL 1.1.14
Option (B) is correct. Characteristic equation. A - lI = 0 -5 - l -3 =0 2 -l 5l + l2 + 6 = 0 l2 + 5l + 6 = 0
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 19
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 SOL 1.1.15
Option (B) is correct.
&
f (x) = x3 - 9x2 + 24x + 5 df (x) = 3x2 - 18x + 24 = 0 dx df (x) = x2 - 6x + 8 = 0 dx
x = 4, x = 2 d 2 f (x) = 6x - 18 dx 2 d 2 f (x) For x = 2, = 12 - 18 =- 6 < 0 dx2 So at x = 2, f (x) will be maximum
nodia.co.in f (x)
SOL 1.1.16
max
= (2) 3 - 9 (2) 2 + 24 (2) + 5 = 8 - 36 + 48 + 5 = 25
Option (C) is correct. 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, ........... 3 5 P = 1 + b 1 l + b 1 l + ..... 2 2 2 1 P = 2 =2 3 1- 1 4
Probability
SOL 1.1.17
Option (A) is correct. Divergence of A in spherical coordinates is given as d:A = 12 2 (r 2 Ar ) = 12 2 (krn + 2) r 2r r 2r = k2 (n + 2) rn + 1 r = k (n + 2) rn - 1 = 0 (given) n+2 = 0 n =- 2
SOL 1.1.18
Option (D) is correct. d 2 y (t) 2dy (t) + + y (t) = d (t) dt dt 2 By taking Laplace transform with initial conditions dy 2 ;s Y (s) - sy (0) - dt E + 2 [sy (s) - y (0)] + Y (s) = 1 t=0 &
2 6s Y (s) + 2s - 0@ + 2 6sY (s) + 2@ + Y (s) = 1
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 20
Y (s) [s2 + 2s + 1] = 1 - 2s - 4 Y (s) = 2- 2s - 3 s + 2s + 1 We know If,
y (t) dy (t) dt
then,
L
Y (s)
L
sY (s) - y (0)
(- 2s - 3) s +2 (s2 + 2s + 1) 2 2 = - 2s - 32 s + 2s + 4s + 2 (s + 2s + 1) + s 2 s+1 + 1 sY (s) - y (0) = 2 = 2 (s + 1) (s + 1) (s + 1) 2 1 = 1 + s + 1 (s + 1) 2 By taking inverse Laplace transform dy (t) = e-t u (t) + te-t u (t) dt dy At t = 0+ , = e0 + 0 = 1 dt t = 0 Option (D) is correct. So,
sY (s) - y (0) =
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SOL 1.1.19
or and SOL 1.1.20
SOL 1.1.21
x3+x2+x+1 x 2 (x + 1) + (x - 1) (x + 1) (x 2 + 1) x+1
=0
=0 =0 = 0 & x =- 1 x2 + 1 = 0 & x =- j, j x =- 1, - j, j
Option (A) is correct. dy = e-3x dx dy = e-3x dx by integrating, we get y =- 1 e-3x + K , where K is constant. 3 Option (D) is correct.
Z is Z = 0 where q is around 45c or so. Thus
Z = Z 45c where Z < 1 1 = 1 - 45c Y = 1 = Z Z Z 45c Y > 1 [a Z < 1]
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 21
So Y will be out of unity circle. SOL 1.1.22
Option (B) is correct. f 1 = 10x2 sin x1 - 0.8 f 2 = 10x 22 - 10x2 cos x1 - 0.6 Jacobian matrix is given by R V S2f 1 2f 1 W 10x2 cos x1 10 sin x1 2x 2x J = SS2f 21 2f 22 WW = > 10x2 sin x1 20x2 - 10 cos x1H S2x1 2x2 W T X 10 0 For x1 = 0, x2 = 1, J = > 0 10H
SOL 1.1.23
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Option (C) is correct.
f (x) = 2x - x 2 + 3 f l (x) = 2 - 2x = 0 x =1 f m (x) =- 2 f m (x) is negative for x = 1, so the function has a maxima at x = 1. SOL 1.1.24
Option (A) is correct. Let a signal p (x) is uniformly distributed between limits - a to + a .
Variance
sp =
#
a
-a
x 2 p (x) dx =
x2 : 1 dx 2a -a
#
a
3 2 3 a = 1 :x D = 2a = a 6 3 2a 3 -a It means square value is equal to its variance 2 2 = sp = a p rms 3 p rms = a 3
SOL 1.1.25
Option (D) is correct. We know that matrix A is equal to product of lower triangular matrix L and upper triangular matrix U . A = 6L@6U @ only option (D) satisfies the above relation.
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.1.26
Page 22
Option (B) is correct. Let the given two vectors are X1 = [1, 1, 1] X2 = [1, a, a 2] Dot product of the vectors
Where
R1V S W X 1 $ X 2 = X1 X 2T = 81 1 1BS a W = 1 + a + a2 SSa 2WW T X a =- 1 + j 3 = 1 - 2p/3 2 2
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so,
X1, X2 are orthogonal 1 + a + a2 = 0 Note: We can see that X1, X2 are not orthonormal as their magnitude is ! 1 SOL 1.1.27
Option (B) is correct.
P =
#0
1
xex dx
= 6x # e = 6xe
x
dx @0 - # 1 : d
@0 - #0
x 1
1
0
1
dx
(x) # ex dx D dx
(1) ex dx = (e1 - 0) - 6e
@0
x 1
= e1 - [e1 - e0] = 1 SOL 1.1.28
Option (A) is correct. t Radial vector r = xti + ytj + zk
SOL 1.1.29
SOL 1.1.30
Divergence = 4$ r t : _xti + ytj + zk ti = c 2 ti + 2 tj + 2 k 2x 2y 2z m 2y 2z = 2x + + = 1+1+1 = 3 2x 2y 2z Option (C) is correct. No of white balls = 4 , no of red balls = 3 If first removed ball is white then remaining no of balls = 6 (3 white, 3 red) we have 6 balls, one ball can be choose in 6 C1 ways, since there are three red balls so probability that the second ball is red is 6 P = 3 C1 = 3 = 1 6 2 C1 Option (D) is correct. Function f (t)= sin t = sin ct has a maxima at t = 0 as shown below t
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
SOL 1.1.31
Page 23
Option (B) is correct. Let eigen vector
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X = 8x1 x2 x 3BT Eigen vector corresponding to l1 = 1 8A R0 S S0 SS0 T
I BX = 0 Rx V R0V S 1W S W Sx2W = S0W SSx WW SS0WW 3 T X T X x2 = 0 x2 + 2x 3 = 0 & x 3 = 0 (not given in the option) Eigen vector corresponding to l2 = 2 8A - l2 I B X = 0 R- 1 1 0V Rx V R0V W S 1W S W S S 0 0 2W Sx2W = S0W SS 0 0 1WW SSx WW SS0WW 3 X T X T X T - x1 + x 2 = 0 2x 3 = 0 & x 3 = 0 (not given in options.) Eigen vector corresponding to l3 = 3 8A - l3 I B X = 0 R- 2 1 0V Rx V R0V W S 1W S W S S 0 - 1 2W Sx2W = S0W SS 0 0 0WW SSx WW SS0WW 3 X T X T X T - 2x1 + x2 = 0 - x2 + 2x 3 = 0 - l1 1 0VW 1 2W 0 2WW X
Put x1 = 1, x2 = 2 and x 3 = 1 So Eigen vector Rx V R1V S 1W S W X = Sx2W = S2W = 81 2 1BT SSx WW SS1WW 3 T X T X
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.1.32
Page 24
Option (B) is correct. d2 x + 6 dx + 8x = 0 dt dt2 Taking Laplace transform (with initial condition) on both sides s2 X (s) - sx (0) - x' (0) + 6 [sX (s) - x (0)] + 8X (s) = 0 s2 X (s) - s (1) - 0 + 6 [sX (s) - 1] + 8X (s) = 0 X (s) [s2 + 6s + 8] - s - 6 = 0 X (s) = By partial fraction
(s + 6) (s + 6s + 8) 2
2 - 1 s+2 s+4 Taking inverse Laplace transform X (s) =
x (t) = (2e- 2t - e- 4t) SOL 1.1.33
Option (C) is correct. Set of equations .....(1) x1 + 2x2 + x 3 + 4x 4 = 2 .....(2) 3x1 + 6x2 + 3x 3 + 12x 4 = 6 or 3 (x1 + 2x2 + x 3 + 4x 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.
SOL 1.1.34
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Option (C) is correct.
a b A => H c d Trace of a square matrix is sum of its diagonal entries Trace A = a + d =- 2 Determinent ad - bc =- 35 Eigenvalue A - lI = 0 a-l b =0 c d-l Let the matrix is
(a - l) (d - l) - bc = 0 l2 - (a + d) l + (ad - bc) = 0 l2 - (- 2) l + (- 35) = 0 l2 + 2l - 35 = 0 (l - 5) (l + 7) = 0 l1, l2 = 5, - 7 SOL 1.1.35
Option (A) is correct. Given constraints x > y2 and y > x2
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Limit of y : y = 0 to y = 1 Limit of x : x = y2 to x2 = y & x = So volume under f (x, y) V = SOL 1.1.36
y=1
x= y
#y = 0 #x = y
2
Page 25
y
f (x, y) dx dy
Option (B) is correct. No of events of at least two people in the room being born on same date = n C2 three people in the room being born on same date = n C 3 Similarly four for people = n C 4 n n n n Probability of the event, 0.5 $ C2 $ C 3 $ C 4 g Cn & N = 7 N
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SOL 1.1.37
Option ( ) is correct. Assume a Cubic polynomial with real Coefficients = a 0 x 3 + a1 x 3 + a 2 x + a 3 = 3a 0 x2 + 2a1 x + a2 = 6a 0 x + 2a1 = 6a 0 Piv (x) = 0
P (x) P' (x) P'' (x) P''' (x) SOL 1.1.38
Option (D) is correct. An iterative sequence in Newton-Raphson’s method is obtained by following expression f (xk ) xk + 1 = xk f' (xk )
So So SOL 1.1.39
a 0, a1, a2, a 3 are real
f (x) = x2 - 117 f' (x) = 2x f (xk ) = x k2 - 117 f' (xk ) = 2xk = 2 # 117 2 xk + 1 = xk - x k - 117 = xk - 1 :xk + 117 D 2 xk 2xk
Option (D) is correct. Equation of straight line y - 2 = 0 - 2 (x - 0) 2-0 y - 2 =- x F $ dl = [(x2 + xy) at x + (y2 + xy) at y] [dxat x + dyat y + dzat z]
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Page 26
= (x2 + xy) dx + (y2 + xy) dy Limit of x : 0 to 2 Limit of y : 2 to 0 2
# F $ dl Line So
SOL 1.1.40
= # (x2 + xy) dx + 0 y - 2 =- x dy =- dx
# F $ dl
#2
0
(y2 + xy) dy
=
#0
2
[x2 + x (2 - x)] dx +
=
#0
2
2xdx +
#2
0
#2
0 2
y + (2 - y) y dy
2y dy
2 2 y2 0 = 2 :x D + 2 ; E = 4 - 4 = 0 2 0 2 2 Option (C) is correct. X is uniformly distributed between 0 and 1 So probability density function 1, 0 < x < 1 fX (X) = ) 0, otherwise 1 1 3 So, E {X } = # X3 fX (X) dx = # X3 (1) dx
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0
4 1
SOL 1.1.41
= :X D = 1 4 0 4 Option (D) is correct. According to CAYLEY-HAMILTON Theorem every non-singular square matrix satisfies its own characteristic equation. Characteristic equation a (l) = lI - P = l3 + l2 + 2l + 1 = 0 Matrix P satisfies above equation P 3 + P 2 + 2P + I = 0 I =- (P3 + P2 + 2P) Multiply both sides by P- 1 P- 1 =- (P2 + P + 2I)
SOL 1.1.42
Option (A) is correct. Rank of a matrix is no. of linearly independent rows and columns of the matrix. Here Rank r (Q) = 4 So Q will have 4 linearly independent rows and flour independent columns.
SOL 1.1.43
Option (B) is correct. Given function f (x) = (x2 - 4) 2 f' (x) = 2 (x2 - 4) 2x To obtain minima and maxima f' (x) = 0 2
4x (x - 4) = 0 x = 0, x2 - 4 = 0 & x = ! 2 So, x = 0, + 2, - 2 f'' (x) = 4x (2x) + 4 (x2 - 4) = 12x2 - 16 For
x = 0, f'' (0) = 12 (0) 2 - 16 =- 16 < 0
(Maxima)
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
x =+ 2, f'' (2) = 12 (2) 2 - 16 = 32 > 0 x =- 2, f'' (- 2) = 12 (- 2) 2 - 16 = 32 > 0 So f (x) has only two minima SOL 1.1.44
Page 27
(Minima) (Minima)
Option (A) is correct. An iterative sequence in Newton-Raphson method can obtain by following expression f (xn) xn + 1 = xn f' (xn) We have to calculate x1 , so n = 0 x1 = x 0 -
f (x 0) , Given x 0 =- 1 f' (x 0)
f (x 0) = ex - 1 = e- 1 - 1 =- 0.63212 f' (x 0) = ex = e- 1 = 0.36787 0
0
x1 =- 1 -
So,
(- 0.63212) (0.36787)
=- 1 + 1.71832 = 0.71832
SOL 1.1.45
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Option (D) is correct.
A' = (AT A) - 1 AT = A- 1 (AT ) - 1 AT = A- 1 I Put A' = A- 1 I in all option. AA'A = A AA- 1 A = A A =A (AA') 2 = I (AA- 1 I) 2 = I (I) 2 = I A'A = I -1 A IA = I I =I AA'A = A' AA- 1 IA = A = Y A'
option (A)
option (B)
option (C)
option (D) SOL 1.1.46
(true)
(true)
(true) (false)
Option (C) is correct. dx = e- 2t u (t) dt x = #e
- 2t
u (t) dt =
1
# e- 2t dt 0
=
1
# f (t) dt , 0
t = .01 s From trapezoid rule t + nh f (t) dt = h 6f (0) + f (.01)@ #t 2 1 e0 + e- .02@, h = .01 #0 f (t) dt = .01 2 6 0
0
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 28
= .0099 SOL 1.1.47
Option (B) is correct. P is an orthogonal matrix So cos q Let assume P => sin q cos q PX = > sin q cos q => sin q PX PX
SOL 1.1.48
PPT = I - sin q cos q H - sin q x x T cos q H8 1 2B x1 cos q - x2 sin q - sin q x1 => H > H cos q x2 x1 sin q + x2 cos qH
=
(x1 cos q - x2 sin q) 2 + (x1 sin q + x2 cos q) 2
=
x 12 + x 22
= X
Option (D) is correct. x = 8x1 x2 g xnBT V = xxT R V R V Sx1W Sx1W Sx2W Sx2W =S W S W ShW ShW SxnW SxnW So rank of V is n . T X T X
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SOL 1.1.49
Option ( ) is correct.
SOL 1.1.50
Option (A) is correct. Given # dz 2 = # (z + idz ) (z - i) 1+z C C Contour z- i = 1 2 P(0, 1) lies inside the circle z - i = 1 and P (0, 1) does not lie. 2 So by Cauchy’s integral formula 1 (z - i) # dz 2 = 2pi lim z"i + ( z i ) (z - i) + 1 z C = 2pi lim 1 = 2pi # 1 = p 2i z"i z + i Option ( ) is correct.
SOL 1.1.51 SOL 1.1.52
Option (C) is correct. Probability of occurrence of values 1,5 and 6 on the three dice is P (1, 5, 6) = P (1) P (5) P (6) = 1#1#1 = 1 128 8 4 4 In option (A) P (3, 4, 5) = P (3) P (4) P (5) =1#1#1 = 1 512 8 8 8 In option (B) P (1, 2, 5) = P (1) P (2) P (5) = 1#1#1 = 1 256 8 8 4
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.1.53
Page 29
Option (D) is correct. x$x x$y det >y $ x y $ yH = (x : x) (y : y) - (x : y) (y : x) = 0 only when x or y is zero
SOL 1.1.54
Option ( ) is correct.
SOL 1.1.55
Option (C) is correct. For characteristic equation -3 - l 2 > - 1 0 - lH = 0 (- 3 - l) (- l) + 2 = 0 (l + 1) (l + 2) = 0 According to Cayley-Hamiliton theorem
or
(A + I) (A + 2I) = 0 SOL 1.1.56
Option (A) is correct. According to Cayley-Hamiliton theorem or or or
nodia.co.in (A + I) (A + 2I) = 0 A2 + 3A + 2I = 0
A2 =- (3A + 2I) A 4 = (3A + 2I) 2 = (9A2 + 12A + 4I) = 9 (- 3A - 2I) + 12A + 4I =- 15A - 14I
A8 = (- 15A - 14I) 2 = 225A2 + 420A + 196 = 225 (- 3A - 2I) + 420A + 196I =- 255A - 254I A9 =- 255A2 - 254A =- 255 (- 3A - 2I) - 254A = 511A + 510I SOL 1.1.57
Option (D) is correct. Volume of the cone V =
#0
H
2 pR2 b1 - h l dh H
Solving the above integral V = 1 pR 2 H 3 Solve all integrals given in option only for option (D) R #0 2prH a1 - Rr k2 dr = 13 pR2 H SOL 1.1.58
Option ( ) is correct.
SOL 1.1.59
Option (C) is correct. By throwing dice twice 6 # 6 = 36 possibilities will occur. Out of these sample space consist of sum 4, 8 and 12 because r/4 is 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 8 =5 9
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.1.60
Option ( ) is correct.
SOL 1.1.61
Option ( ) is correct.
SOL 1.1.62
Option (A) is correct. Matrix equation PX = q has a unique solution if Where
SOL 1.1.63
Page 30
r (P) = r (r) r (P) " rank of matrix P r (r) " rank of augmented matrix [P] r = 8P : qB
Option (D) is correct. for two random events conditional probability is given by probability (P + Q) = probability (P) probability (Q) probability (P + Q) probability (Q) = #1 probability (P) so
SOL 1.1.64
probability (P + Q) # probability (P)
Option (C) is correct.
SOL 1.1.65
-2 3 = :x D = 1 -2 1 2
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# 3 x- 3 dx 1
Option (A) is correct. We have xo(t) or xo(t) + 3x (t) A.E. D+3 Thus solution is x (t) From x (0) = x 0 we get C1 Thus x (t)
=- 3x (t) =0 =0
= C1 e- 3t = x0 = x 0 e- 3t
SOL 1.1.66
Option (D) is correct. For eigen value l =- 2 V Rx V R0V R3 - (- 2) -2 2 W S 1W S W S - 2 - (- 2) 0 1 W Sx2W = S0W S SS 0 0 1 - (- 2)WW SSx 3WW SS0WW T R5 - 2 2XV TRx XV TR0XV W S 1W S W S S0 0 1W Sx2W = S0W SS0 0 1WW SSx WW SS0WW 3 X T X T X T 5x1 - 2x2 + x 3 = 0
SOL 1.1.67
Option (B) is correct. C11 C21 C 31 R
SOL 1.1.68
= 2 - (- 3) = 5 =- (0 - (- 3)) =- 3 = (- (- 1)) = 1 = (1) C11 + 2C21 + 2C 31 = 5 - 6 + 2 = 1
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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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.1.69
Option (A) is correct. We have f (x) = x2 e- x or f' (x) = 2xe- x - x2 e- x = xe- x (2 - x) f'' (x) = (x2 - 4x + 2) e- x Now for maxima and minima, f' (x) = 0 xe- x (2 - x) = 0 or x = 0, 2 at x = 0 f'' (0) = 1 (+ ve) at x = 2 f'' (2) =- 2e- 2 (- ve) Now f'' (0) = 1 and f'' (2) =- 2e- 2 < 0 . Thus x = 2 is point of maxima
SOL 1.1.70
Option (C) is correct. 4 u = cti 2 + tj 2 m u = ti2u + tj2u = xti + 2 ytj 3 2x 2y 2x 2y 2 At (1, 3) magnitude is 4 u = x2 + b 2 y l = 1 + 4 = 5 3 Option (B) is correct. d2 x + 3dx + 2x (t) = 5 dt dt2 Taking Laplace transform on both sides of above equation. s2 X (s) + 3sX (s) + 2X (s) = 5 s 5 X (s) = 2 s (s + 3s + 2) From final value theorem lim x (t) = lim X (s) = lim s 2 5 =5 2 t"3 s"0 s " 0 s (s + 3s + 2)
SOL 1.1.71
Page 31
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2 ELECTRICAL CIRCUITS & FIELDS
YEAR 2013 MCQ 1.2.1
ONE MARK
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
MCQ 1.2.2
The flux density at a point in space is given by Bv = 4xavx + 2kyavy + 8avz Wb/m2 . The value of constant k must 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 current flowing from the load to the source is 10+ - 150cA and if the voltage at the load terminal is 100+60cV , 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
MCQ 1.2.5
A source vs ^ t h = V cos 100pt has an internal impedance of ^4 + j3h W . 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 The transfer function
(A) 0.5s + 1 s+1
V2 ^s h of the circuit shown below is V1 ^s h
(B) 3s + 6 s+2
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(C) s + 2 s+1
Page 33
(D) s + 1 s+2
YEAR 2013
TWO MARKS
MCQ 1.2.6
A dielectric slab with 500 mm # 500 mm cross-section is 0.4 m long. The slab is subjected to a uniform electric field of Ev = 6avx + 8avy kV/mm . The relative permittivity of the dielectric material is equal to 2. The value of constant e0 is 8.85 # 10-12 F/m . 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 capacitors C1 , C2 and C 3 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,
nodia.co.in (A) 2.8 and 36 (C) 2.8 and 32 MCQ 1.2.8
(B) 7 and 119 (D) 7 and 80
In the circuit shown below, if the source voltage VS = 100+53.13c V then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is
(A) 100+90c (C) 800+90c
(B) 800+0c (D) 100+60c
YEAR 2012 MCQ 1.2.9
ONE MARK
In the circuit shown below, the current through the inductor is
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(B) - 1 A 1+j
2 A 1+j (C) 1 A 1+j (A)
MCQ 1.2.10
Page 34
(D) 0 A
The impedance looking into nodes 1 and 2 in the given circuit is
nodia.co.in (A) 50 W (C) 5 kW MCQ 1.2.11
(B) 100 W (D) 10.1 kW
(s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) is excited by sin (wt). The steady-state output of the system is zero at (A) w = 1 rad/s (B) w = 2 rad/s (C) w = 3 rad/s (D) w = 4 rad/s A system with transfer function G (s) =
MCQ 1.2.12
The average power delivered to an impedance (4 - j3) W by a current 5 cos (100pt + 100) A is (A) 44.2 W (B) 50 W (C) 62.5 W (D) 125 W
MCQ 1.2.13
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
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YEAR 2012 MCQ 1.2.14
TWO MARKS
If VA - VB = 6 V then VC - VD is
(A) - 5 V (C) 3 V MCQ 1.2.15
Page 35
(B) 2 V (D) 6 V
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
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(A) 0.8 W (C) 2 W
(B) 1.4 W (D) 2.8 W
Common Data for Questions 16 and 17 : 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 3 A (ii) 2.5 W connected at port B draws a current of 2 A
MCQ 1.2.16
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) 1 A (D) 9/7 A
MCQ 1.2.17
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
Statement for Linked Answer Questions 18 and 19 :
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Page 36
In the circuit shown, the three voltmeter readings are V1 = 220 V, V2 = 122 V, V3 = 136 V .
MCQ 1.2.18
MCQ 1.2.19
The power factor of the load is (A) 0.45 (C) 0.55
(B) 0.50 (D) 0.60
If RL = 5 W , the approximate power consumption in the load is (A) 700 W (B) 750 W (C) 800 W (D) 850 W
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YEAR 2011 MCQ 1.2.20
ONE MARK
The r.m.s value of the current i (t) in the circuit shown below is (B) 1 A (A) 1 A 2 2 (C) 1 A
MCQ 1.2.21
(D)
The voltage applied to a circuit is 100 2 cos (100pt) volts and the circuit draws a current of 10 2 sin (100pt + p/4) amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is (B) 10 - p/4 (A) 10 2 - p/4 (C) 10 + p/4
MCQ 1.2.22
2A
(D) 10 2 + p/4
In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3 W is
(A) zero (C) 6 W
(B) 3 W (D) infinity
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YEAR 2011 MCQ 1.2.23
TWO MARKS
A lossy capacitor Cx , rated for operation at 5 kV, 50 Hz is represented by an equivalent circuit with an ideal capacitor C p in parallel with a resistor R p . The value C p is found to be 0.102 μF and value of R p = 1.25 MW . Then the power loss and tan d of 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
MCQ 1.2.24
Page 37
(D) 20 W and 0.04
A capacitor is made with a polymeric dielectric having an er of 2.26 and a dielectric 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 a length of 40 cm, then the maximum electric charge in the capacitor is (A) 2 μC (B) 4 μC (C) 8 μC (D) 10 μC
Common Data questions: 25 & 26
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The input voltage given to a converter is vi = 100 2 sin (100pt) V The current drawn by the converter is ii = 10 2 sin (100pt - p/3) + 5 2 sin (300pt + p/4) + 2 2 sin (500pt - p/6) A 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 R is (A) 0.5 W (B) 1 W (C)
MCQ 1.2.28
2W
The current IC in the figure above is
(D) 2 W
(A) - j2 A
(B) - j 1 A 2
(C) + j 1 A 2
(D) + j2A
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YEAR 2010 MCQ 1.2.29
ONE MARK
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 mF capacitor is
(A) 0 A (C) 1.25 A MCQ 1.2.30
Page 38
(B) 1 A (D) 5 A
As shown in the figure, a 1 W resistance is connected across a source that has a load line v + i = 100 . The current through the resistance is
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(A) 25 A (C) 100 A
(B) 50 A (C) 200 A
YEAR 2010 MCQ 1.2.31
TWO MARKS
If the 12 W resistor draws a current of 1 A as shown in the figure, the value of resistance R is
(A) 4 W (C) 8 W MCQ 1.2.32
(B) 6 W (D) 18 W
The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a,b) and (c,d) respectively. It has an impedance matrix Z with parameters denoted by Zij . A 1 W resistor is connected in series with the network at port 1 as shown in the figure. The impedance matrix of the modified two-port network (shown as a dashed box ) is
Z11 + 1 (A) e Z21 Z11 + 1 (C) e Z21
Z12 + 1 Z22 + 1o Z12 Z22 o
Z11 + 1 Z12 (B) e Z21 Z22 + 1o Z11 + 1 Z12 (D) e Z21 + 1 Z22 o
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Page 39
YEAR 2009 MCQ 1.2.33
The current through the 2 kW resistance in the circuit shown is
(A) 0 mA (C) 2 mA MCQ 1.2.34
ONE MARK
(B) 1 mA (D) 6 mA
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
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YEAR 2009 MCQ 1.2.35
In the figure shown, all elements used are ideal. For time t < 0, S1 remained closed and S2 open. At t = 0, S1 is opened and S2 is closed. If the voltage Vc2 across the capacitor C2 at t = 0 is zero, the voltage across the capacitor combination at t = 0+ will be
(A) 1 V (C) 1.5 V MCQ 1.2.36
(B) 2 V (D) 3 V
The equivalent capacitance of the input loop of the circuit shown is
(A) 2 mF (C) 200 mF MCQ 1.2.37
TWO MARKS
(B) 100 mF (D) 4 mF
For the circuit shown, find out the current flowing through the 2 W resistance. Also identify the changes to be made to double the current through the 2 W resistance.
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(A) (5 A; PutVS = 30 V) (C) (5 A; Put IS = 10 A)
Page 40
(B) (2 A; PutVS = 8 V) (D) (7 A; Put IS = 12 A)
Statement for Linked Answer Question 38 and 39 :
MCQ 1.2.38
For the circuit given above, the Thevenin’s resistance across the terminals A and B is (A) 0.5 kW (B) 0.2 kW (C) 1 kW (D) 0.11 kW
MCQ 1.2.39
For the circuit given above, the Thevenin’s voltage across the terminals A and B is (A) 1.25 V (B) 0.25 V (C) 1 V (D) 0.5 V
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YEAR 2008 MCQ 1.2.40
The number of chords in the graph of the given circuit will be
(A) 3 (C) 5 MCQ 1.2.41
ONE MARK
(B) 4 (D) 6
The Thevenin’s equivalent of a circuit operation at w = 5 rads/s, has Voc = 3.71+ - 15.9% V and Z0 = 2.38 - j0.667 W . At this frequency, the minimal realization of the Thevenin’s impedance will have a (A) resistor and a capacitor and an inductor (B) resistor and a capacitor (C) resistor and an inductor (D) capacitor and an inductor
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YEAR 2008 MCQ 1.2.42
(B) 1/4 s (D) 9 s
The resonant frequency for the given circuit will be
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(A) 1 rad/s (C) 3 rad/s MCQ 1.2.44
TWO MARKS
The time constant for the given circuit will be
(A) 1/9 s (C) 4 s MCQ 1.2.43
Page 41
(B) 2 rad/s (D) 4 rad/s
Assuming ideal elements in the circuit shown below, the voltage Vab will be
(A) - 3 V (C) 3 V
(B) 0 V (D) 5 V
Statement for Linked Answer Question 38 and 39. The current i (t) sketched in the figure flows through a initially uncharged 0.3 nF capacitor.
MCQ 1.2.45
The charge stored in the capacitor at t = 5 ms , will be
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(A) 8 nC (C) 13 nC
Page 42
(B) 10 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 the capacitor after 1 ms will approximately be (A) 18.8 V (B) 23.5 V (C) - 23.5 V (D) - 30.6 V
MCQ 1.2.47
In the circuit shown in the figure, the value of the current i will be given by
(A) 0.31 A (C) 1.75 A
(B) 1.25 A (D) 2.5 A
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MCQ 1.2.48
Two point charges Q1 = 10 mC and Q2 = 20 mC are placed at coordinates (1,1,0) and (- 1, - 1, 0) respectively. The total electric flux passing through a plane z = 20 will be (A) 7.5 mC (B) 13.5 mC (C) 15.0 mC (D) 22.5 mC
MCQ 1.2.49
A capacitor consists of two metal plates each 500 # 500 mm2 and spaced 6 mm apart. The space between the metal plates is filled with a glass plate of 4 mm thickness and a layer of paper of 2 mm thickness. The relative primitivities of the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the capacitance will be (Given that e0 = 8.85 # 10 - 12 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 coil corresponding to a magnetizing current of 3 A will be (Given that m0 = 4p # 10 - 7 H/m) (A) 37.68 mH (B) 113.04 mH (C) 3.768 mH (D) 1.1304 mH YEAR 2007
MCQ 1.2.51
ONE MARK
Divergence of the vector field V (x, y, z) =- (x cos xy + y) it + (y cos xy) jt + (sin z2 + x2 + y2) kt is (A) 2z cos z2 (B) sin xy + 2z cos z2 (C) x sin xy - cos z (D) None of these YEAR 2007
MCQ 1.2.52
TWO MARKS
The state equation for the current I1 in the network shown below in terms of the
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Page 43
voltage VX and the independent source V , is given by
MCQ 1.2.53
(A) dI1 =- 1.4VX - 3.75I1 + 5 V (B) dI1 = 1.4VX - 3.75I1 - 5 V dt 4 dt 4 (C) dI1 =- 1.4VX + 3.75I1 + 5 V (D) dI1 =- 1.4VX + 3.75I1 - 5 V dt 4 dt 4 The R-L-C series circuit shown in figure is supplied from a variable frequency voltage source. The admittance - locus of the R-L-C network at terminals AB for increasing frequency w is
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MCQ 1.2.54
In the circuit shown in figure. Switch SW1 is 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. SW2 is closed at t = 0 and SW1 is opened at t = 0 . The current through C and the voltage across L at (t = 0+) is
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(A) 55 A, 4.5 V (C) 45 A, 5.5 A MCQ 1.2.55
Page 44
(B) 5.5 A, 45 V (D) 4.5 A, 55 V
In the figure given below all phasors are with reference to the potential at point ''O'' . The locus of voltage phasor VYX as R is varied from zero to infinity is shown by
nodia.co.in MCQ 1.2.56
A 3 V DC supply with an internal resistance of 2 W supplies a passive non-linear 2 resistance characterized by the relation VNL = INL . 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 = [V1V2 .....V6]T denote the vector of branch voltages while I = [i1 i2 .....i6]T that of branch currents. The vector E = [e1 e2 e3 e4]T denotes the vector of node voltages relative to a common ground. R1 1 1 0 0 0 V S W S 0 -1 0 -1 1 0 W S- 1 0 0 0 - 1 - 1W S W S 0 0 -1 1 0 1 W T X Which of the following statement is true ? (A) The equations V1 - V2 + V3 = 0,V3 + V4 - V5 = 0 are KVL equations for the network for some loops (B) The equations V1 - V3 - V6 = 0,V4 + V5 - V6 = 0 are KVL equations for the network for some loops (C) E = AV (D) AV = 0 are KVI equations for the network
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Page 45
A solid sphere made of insulating material has a radius R and has a total charge Q distributed uniformly in its volume. What is the magnitude of the electric field intensity, E , at a distance r (0 < r < R) inside the sphere ? 1 Qr 4pe0 R3 Q (C) 1 2 4pe0 r
(A)
3 Qr 4pe0 R3 QR (D) 1 4pe0 r3
(B)
Statement for Linked Answer Question 59 and 60. An inductor designed with 400 turns coil wound on an iron core of 16 cm2 cross sectional area and with a cut of an air gap length of 1 mm. The coil is connected to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and leakage inductance, (m0 = 4p # 10 - 7 H/M) MCQ 1.2.59
MCQ 1.2.60
The current in the inductor is (A) 18.08 A (C) 4.56 A
(B) 9.04 A (D) 2.28 A
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The average force on the core to reduce the air gap will be (A) 832.29 N (B) 1666.22 N (C) 3332.47 N (D) 6664.84 N YEAR 2006
MCQ 1.2.61
ONE MARK
In the figure the current source is 1+0 A, R = 1 W , the impedances are ZC =- j W and ZL = 2j W . The Thevenin equivalent looking into the circuit across X-Y is
(A)
2 +0 V, (1 + 2j) W
(C) 2+45% V, (1 + j) W
(B) 2+45% V, (1 - 2j) W (D)
2 +45% V, (1 + j) W
YEAR 2006 MCQ 1.2.62
TWO MARKS
The parameters of the circuit shown in the figure are Ri = 1 MW R0 = 10 W, A = 106 V/V If vi = 1 mV , the output voltage, input impedance and output impedance respectively are
(A) 1 V, 3, 10 W
(B) 1 V, 0, 10 W
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(D) 10 V, 3, 10 W
(C) 1 V, 0, 3 MCQ 1.2.63
Page 46
In the circuit shown in the figure, the current source I = 1 A , the voltage source V = 5 V, R1 = R2 = R3 = 1 W, L1 = L2 = L3 = 1 H, C1 = C2 = 1 F
The currents (in A) through R3 and through the voltage source V respectively will 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 port parameters that describe the circuit shown are
nodia.co.in 0 (A) z parameters, = 0 0 (C) h parameters, = 0 MCQ 1.2.65
0 0G 0 0G
1 (B) h parameters, = 0 1 (D) z parameters, = 0
0 1G 0 1G
The circuit shown in the figure is energized by a sinusoidal voltage source V1 at a frequency 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
Page 47
An ideal capacitor is charged to a voltage V0 and connected at t = 0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let w0 = 1 , the voltage across the capacitor at time t > 0 is given by LC (A) V (B) V cos (w t) 0
(C) V0 sin (w0 t)
0
(D) V0 e
0
- w0t
cos (w0 t)
MCQ 1.2.67
An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. The heater 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
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YEAR 2005 MCQ 1.2.69
In the figure given below the value of R is
(A) 2.5 W (C) 7.5 W MCQ 1.2.70
(B) 5.0 W (D) 10.0 W
The RMS value of the voltage u (t)= 3 + 4 cos (3t) is (A) 17 V (B) 5 V (C) 7 V
MCQ 1.2.71
ONE MARK
(D) (3 + 2 2 ) V
For the two port network shown in the figure the Z -matrix is given by
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MCQ 1.2.72
Z1 Z1 + Z2 (A) = Z1 + Z2 Z2 G
Z1 Z1 (B) = Z1 + Z2 Z2 G
Z1 Z2 (C) = Z2 Z1 + Z2 G
Z1 Z1 (D) = Z1 Z1 + Z2 G
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
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(A) 20 V (C) 5 V MCQ 1.2.73
Page 48
(B) 10 V (D) 0 V
If Ev is the electric intensity, 4 (4 # Ev) is equal to (A) Ev (B) Ev (C) null vector (D) Zero YEAR 2005
TWO MARKS
Statement for Linked Answer Question 74 and 75. A coil of inductance 10 H and resistance 40 W is connected as shown in the figure. After the switch S has been in contact with point 1 for a very long time, it is moved 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 R is
(A) 0 W (C) 40 W
(B) 20 W (D) 60 W
MCQ 1.2.75
For the value as obtained in (a), the time taken for 95% of the stored energy to be 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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Page 49
sinusoidal voltage source Vin . At 100 Hz, the Rand L elements each have a voltage drop mRMS .If the frequency of the source is changed to 50 Hz, then new voltage drop across R is
5u (B) 8 RMS 8u (C) (D) 5 RMS For the three-phase circuit shown in the figure is given by (A)
MCQ 1.2.77
2u 3 RMS 3u 2 RMS the ratio of the currents IR: IY : IB
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3
(C) 1 : 1 : 0 MCQ 1.2.78
(D) 1 : 1 : 3/2
The circuit shown in the figure is in steady state, when the switch is closed at t = 0 .Assuming that the inductance is ideal, the current through the inductor at t = 0+ equals
(A) 0 A (C) 1 A MCQ 1.2.79
(B) 1 : 1 : 2
(B) 0.5 A (D) 2 A
In the given figure, the Thevenin’s equivalent pair (voltage, impedance), as seen at the terminals P-Q, is given by
(A) (2 V, 5 W) (C) (4 V, 5 W)
(B) (2 V, 7.5 W) (D) (4 V, 7.5 W)
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 1.2.80
Page 50
The charge distribution in a metal-dielectric-semiconductor specimen is shown in the figure. The negative charge density decreases linearly in the semiconductor as shown. The electric field distribution is as shown in
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ONE MARK
The value of Z in figure which is most appropriate to cause parallel resonance at 500 Hz is
(A) 125.00 mH (C) 2.0 mF MCQ 1.2.82
(B) 304.20 mF (D) 0.05 mF
A parallel plate capacitor is shown in figure. It is made two square metal plates of 400 mm side. The 14 mm space between the plates is filled with two layers of dielectrics of er = 4 , 6 mm thick and er = 2 , 8 mm thick. Neglecting fringing of fields at the edge the capacitance is
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(A) 1298 pF (C) 354 pF MCQ 1.2.83
(B) 944 pF (D) 257 pF
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
MCQ 1.2.84
Page 51
TWO MARKS
In figure, the value of the source voltage is
nodia.co.in (A) 12 V (C) 30 V MCQ 1.2.85
In figure, Ra , Rb and Rc are 20 W, 20 W and 10 W respectively. The resistances R1 , R2 and R 3 in W of an equivalent star-connection are
(A) 2.5, 5, 5 (C) 5, 5, 2.5 MCQ 1.2.86
(B) 24 V (D) 44 V
(B) 5, 2.5, 5 (D) 2.5, 5, 2.5
In figure, the admittance values of the elements in Siemens are YR = 0.5 + j0, YL = 0 - j1.5, YC = 0 + j0.3 respectively. The value of I as a phasor when the voltage E across the elements is 10+0% V
(A) 1.5 + j0.5
(B) 5 - j18
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(C) 0.5 + j1.8 MCQ 1.2.87
MCQ 1.2.90
(B) 20 (D) 40
In figure, the capacitor initially has a charge of 10 Coulomb. The current in the circuit one second after the switch S is closed will be
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(A) 14.7 A (C) 40.0 A MCQ 1.2.89
(D) 5 - j12
In figure, the value of resistance R in W is
(A) 10 (C) 30 MCQ 1.2.88
(B) 18.5 A (D) 50.0 A
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 0.9 0.2 The Z-matrix of a 2-port network as given by = 0.2 0.6G The element Y22 of 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
MCQ 1.2.91
ONE MARK
Figure Shows the waveform of the current passing through an inductor of resistance 1 W and inductance 2 H. The energy absorbed by the inductor in the first four seconds is
(A) 144 J (C) 132 J MCQ 1.2.92
Page 52
(B) 98 J (D) 168 J
A segment of a circuit is shown in figure vR = 5V, vc = 4 sin 2t .The voltage vL is
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given by
(A) 3 - 8 cos 2t (C) 16 sin 2t MCQ 1.2.93
(B) 32 sin 2t (D) 16 cos 2t
In the figure, Z1 = 10+ - 60%, Z2 = 10+60%, Z3 = 50+53.13% . Thevenin impedance seen form X-Y is
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MCQ 1.2.95
(A) 56.66+45%
(B) 60+30%
(C) 70+30%
(D) 34.4+65%
Two conductors are carrying forward and return current of +I and - I as shown in figure. The magnetic field intensity H at point P is
(A) I Y (B) I X pd pd (C) I Y (D) I X 2pd 2pd Two infinite strips of width w m in x -direction as shown in figure, are carrying forward and return currents of +I and - I in the z - direction. The strips are separated 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 snow reduced to x/2 m, the inductance per unit length of the configuration is
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(A) 2L H/m (C) L/2 H/m
(B) L/4 H/m (D) 4L H/m
YEAR 2003 MCQ 1.2.96
Page 54
TWO MARKS
In the circuit of figure, the magnitudes of VL and VC are twice that of VR . Given that f = 50 Hz , the inductance of the coil is
nodia.co.in (A) 2.14 mH (C) 31.8 mH MCQ 1.2.97
In figure, the potential difference between points P and Q is
(A) 12 V (C) - 6 V MCQ 1.2.98
(B) 5.30 H (D) 1.32 H
(B) 10 V (D) 8 V
Two ac sources feed a common variable resistive load as shown in figure. Under the maximum power transfer condition, the power absorbed by the load resistance RL is
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(A) 2200 W (C) 1000 W MCQ 1.2.99
(B) 1250 W (D) 625 W
In figure, the value of R is
(A) 10 W (C) 24 W MCQ 1.2.100
Page 55
(B) 18 W (D) 12 W
In the circuit shown in figure, the switch S is closed at time (t = 0). The voltage across the inductance at t = 0+ , is
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(A) 2 V (C) - 6 V MCQ 1.2.101
The h-parameters for a two-port network are defined by E1 h11 h12 I1 = I G = =h h G =E G 2 21 22 2 For the two-port network shown in figure, the value of h12 is given by
(A) 0.125 (C) 0.625 MCQ 1.2.102
(B) 4 V (D) 8 V
(B) 0.167 (D) 0.25
A point charge of +I nC is placed in a space with permittivity of 8.85 # 10 - 12 F/m as shown in figure. The potential difference VPQ between 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
Page 56
(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 with a permittivity of 8.85 # 10 - 12 F/m. The charge on the capacitor is 100 V. The stored 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 material with different thickness (t1 and t2 ) as shown in figure. The two different dielectric materials are separated by a conducting foil F. The voltage of the conducting foil is
(A) 52 V (C) 67 V
(B) 60 V (D) 33 V
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YEAR 2002 MCQ 1.2.105
MCQ 1.2.106
MCQ 1.2.107
ONE MARK
A current impulse, 5d (t), is forced through a capacitor C . The voltage , vc (t), across the capacitor is given by (A) 5t (B) 5u (t) - C 5u (t) (C) 5 t (D) C C The graph of an electrical network has N nodes and B branches. 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 - 2B - 1 Given a vector field F, the divergence theorem states that (A)
# F : dS = # 4: FdV S
(C)
# F # dS = # 4: FdV S
(B)
V
V
# F : dS = # 4# FdV S
(D)
V
# F # dS = # 4: FdV S
V
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 distance between their centres is 1 m. If this distance is doubled, then the inductance per unit 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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between the conductor and the ground. The maximum electric stress occurs at (A) The upper surface of the conductor (B) The lower surface of the conductor. (C) The ground surface. (D) midway between the conductor and ground. YEAR 2002 MCQ 1.2.110
TWO MARKS
A two port network shown in Figure, is described by the following equations I1 = Y11 E1 + Y12 E2 I1 = Y21 E1 + Y22 E2
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The admittance parameters, Y11, Y12, Y21 and Y22 for the network shown are (A) 0.5 mho, 1 mho, 2 mho and 1 mho respectively (B) 13 mho, - 16 mho, - 16 mho and 13 mho respectively
MCQ 1.2.111
(C) 0.5 mho, 0.5 mho, 1.5 mho and 2 mho respectively (D) - 2 mho, - 3 mho, 3 mho and 25 mho respectively 5 7 7 In the circuit shown in Figure, what value of C will cause a unity power factor at the ac source ?
(A) 68.1 mF (C) 0.681 mF
(B) 165 mF (D) 6.81 mF
MCQ 1.2.112
A series R-L-C circuit has R = 50 W ; L = 100 mH and C = 1 mF . The lower half power frequency of the circuit is (A) 30.55 kHz (B) 3.055 kHz (C) 51.92 kHz (D) 1.92 kHz
MCQ 1.2.113
A 10 V pulse of 10 ms duration is applied to the circuit shown in Figure, assuming that the capacitor is completely discharged prior to applying the pulse, the peak value of the capacitor voltage is
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(A) 11 V (C) 6.32 V MCQ 1.2.114
Page 58
(B) 5.5 V (D) 0.96 V
In the circuit shown in Figure, it is found that the input voltage (vi ) and current i are in phase. The coupling coefficient is K = M , where M is the mutual L1 L2 inductance between the two coils. The value of K and the dot polarity of the coil P-Q are
(A) K = 0.25 and dot at P (C) K = 0.25 and dot at Q MCQ 1.2.115
(B) K = 0.5 and dot at P (C) K = 0.5 and dot at Q
Consider the circuit shown in Figure If the frequency of the source is 50 Hz, then a value of t0 which results in a transient free response is
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(A) 0 ms (C) 2.71 ms MCQ 1.2.116
(B) 1.78 ms (D) 2.91 ms
In the circuit shown in figure, the switch is closed at time t = 0 . The steady state value of the voltage vc is
(A) 0 V (C) 5 V
(B) 10 V (D) 2.5 V
Common data Question for Q. 117-118* : A constant current source is supplying 10 A current to a circuit shown in figure. The switch is initially closed for a sufficiently long time, is suddenly opened at t=0
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MCQ 1.2.117
MCQ 1.2.118
The inductor current iL (t) will be (A) 10 A (C) 10e- 2t A
Page 59
(B) 0 A (D) 10 (1 - e- 2t) A
What is the energy stored in L, a long time after the switch is opened (A) Zero (B) 250 J (C) 225 J (D) 2.5 J
Common Data Question for Q. 119-120* : An electrical network is fed by two ac sources, as shown in figure, Given that Z1 = (1 - j) W , Z2 = (1 + j) W and ZL = (1 + j0) W .
nodia.co.in MCQ 1.2.119
*Thevenin voltage and impedance across terminals X and Y respectively are (A) 0 V, (2 + 2j) W (B) 60 V, 1 W (C) 0 V, 1 W (D) 30 V, (1 + j) W
MCQ 1.2.120
*Current iL through load is (A) 0 A (C) 0.5 A
MCQ 1.2.121
(B) 1 A (D) 2 A
*In the resistor network shown in figure, all resistor values are 1 W. A current of 1 A passes from terminal a to terminal b as shown in figure, Voltage between terminal a and b is
(A) 1.4 Volt
(B) 1.5 Volt
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(C) 0 Volt
(D) 3 Volt
YEAR 2001 MCQ 1.2.122
Page 60
ONE MARK
In a series RLC circuit at resonance, the magnitude of the voltage developed across the capacitor (A) is always zero (B) can never be greater than the input voltage (C) can be greater than the input voltage, however it is 90% out of phase with the input voltage (D) can be greater than the input voltage, and is in phase with the input voltage.
MCQ 1.2.123
Two incandescent light bulbs of 40 W and 60 W rating are connected in series across the mains. Then (A) the bulbs together consume 100 W (B) the bulbs together consume 50 W (C) the 60 W bulb glows brighter (D) the 40 bulb glows brighter
MCQ 1.2.124
A unit step voltage is applied at t = 0 to a series RL circuit with zero initial conditions. (A) It is possible for the current to be oscillatory.
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(B) The voltage across the resistor at t = 0+ is zero. (C) The energy stored in the inductor in the steady state is zero. (D) The resistor current eventually falls to zero. MCQ 1.2.125
Given two coupled inductors L1 and L2 , their mutual inductance M satisfies (L + L2) (A) M = L21 + L22 (B) M > 1 2 (C) M >
MCQ 1.2.126
L1 L 2
(D) M #
L1 L 2
A passive 2-port network is in a steady-state. Compared to its input, the steady state output can never offer (A) higher voltage (B) lower impedance (C) greater power (D) better regulation YEAR 2001
MCQ 1.2.127
TWO MARKS
Consider the star network shown in Figure The resistance between terminals A and B with C open is 6 W, between terminals B and C with A open is 11 W, and between terminals C and A with B open is 9 W. Then
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(A) (B) (C) (D)
Page 61
RA = 4 W, RB = 2 W, RC = 5 W RA = 2 W, RB = 4 W, RC = 7 W RA = 3 W, RB = 3 W, RC = 4 W RA = 5 W, RB = 1 W, RC = 10 W
MCQ 1.2.128
A connected network of N > 2 nodes has at most one branch directly connecting any pair of nodes. The graph of the network (A) Must have at least N branches for one or more closed paths to exist (B) Can have an unlimited number of branches (C) can only have at most N branches (D) Can have a minimum number of branches not decided by N
MCQ 1.2.129
A 240 V single-phase ac source is connected to a load with an impedance of 10+60% W . A capacitor is connected in parallel with the load. If the capacitor suplies 1250 VAR, the real power supplied by the source is (A) 3600 W (B) 2880 W (C) 240 W (D) 1200 W
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Common Data Questions Q.130-131*:
For the circuit shown in figure given values are R = 10 W , C = 3 mF , L1 = 40 mH, L2 = 10 mH and M = 10 mH
MCQ 1.2.130
MCQ 1.2.131
The resonant frequency of the circuit is (B) 1 # 105 rad/sec A) 1 # 105 rad/sec 3 2 (D) 1 # 105 rad/sec (C) 1 # 105 rad/sec 9 21 The Q-factor of the circuit in Q.82 is (A) 10 (B) 350 (C) 101 (D) 15
MCQ 1.2.132
Given the potential function in free space to be V (x) = (50x2 + 50y2 + 50z2) volts, the magnitude (in volts/metre) and the direction of the electric field at a point (1,-1,1), where the dimensions are in metres, are (B) 100/ 3 ; (it - tj + kt) (A) 100; (it + tj + kt) (C) 100 3 ; [(- it + tj - kt) / 3 ] (D) 100 3 ; [(- it - tj - kt) / 3 ]
MCQ 1.2.133
The hysteresis loop of a magnetic material has an area of 5 cm2 with the scales given as 1 cm = 2 AT and 1 cm = 50 mWb. At 50 Hz, the total hysteresis loss is. (A) 15 W (B) 20 W (C) 25 W (D) 50 W
MCQ 1.2.134
The conductors of a 10 km long, single phase, two wire line are separated by a
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
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distance of 1.5 m. The diameter of each conductor is 1 cm. If the conductors are of copper, the inductance of the circuit is (A) 50.0 mH (B) 45.3 mH (C) 23.8 mH (D) 19.6 mH ***********
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SOLUTION
SOL 1.2.1
Page 63
Option (B) is correct. In the equivalent star connection, the resistance can be given as Rb Ra Ra + Rb + Rc Ra Rc RB = Ra + Rb + Rc Rb Rc RA = Ra + Rb + Rc RC =
So, if the delta connection components Ra , Rb and Rc are scaled by a factor k then ^k Rb h^k Rc h RAl = kRa + kRb + kRc
nodia.co.in 2 Rb Rc =k k Ra + Rb + Rc
= k RA Hence, it is also scaled by a factor k SOL 1.2.2
Option (A) is correct. Given, flux density Bv = 4x avx + 2ky avy + 8 avz Since, magnetic flux density is always divergence less. i.e., d$Bv = 0 So, for given vector flux density, we have d$Bv = 4 + 2k + 0 = 0 or, k =- 2
SOL 1.2.3
Option (B) is correct. Consider the voltage source and load shown in figure
We obtain the power delivered by load as Pdelivered = I L* VL = ^10 + 150ch^10 60ch = 100 210c = 1000 cos 210c + j1000 sin 210c =- 866.025 - j500 As both the reactive and average power (real power) are negative so, power is
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absorbed by load. i.e., load absorbs real as well as reactive power. SOL 1.2.4
Option (C) is correct. 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. i.e., RL = 42 + 32 = 5W
SOL 1.2.5
Option (D) is correct. For the given capacitance, C = 100mF in the circuit, we have the reactance. XC = 1 sc 1 = s # 100 # 10-6 4 = 10 s So, 10 4 + 10 4 V2 ^s h = 4 s V1 ^s h 10 + 10 4 + 10 4 s s = s+1 s+2 Option (B) is correct. Energy density stored in a dielectric medium is obtained as
SOL 1.2.6
nodia.co.in wE = 1 e E 2 J/m2 2
The electric field inside the dielectric will be same to given field in free space only if the field is tangential to the interface 2 So, wE = 1 2e0 ^ 62 + 82 h # 106 /mm2 2 Therefore, the total stored energy is WE =
#W v
E
dv
= e0 100 # 106 /mm2 # ^500 # 500h mm2 # ^0.4h = e0 # 100 # 106 # 0.4 # 25 # 10 4 = 8.85 # 10-12 # 1013 = 88.5 J SOL 1.2.7
Option (C) is correct.
Consider that the voltage across the three capacitors C1 , C2 and C 3 are V1 , V2 and V3 respectively. So, we can write V2 = C 3 ....(1) V3 C2
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Since, Voltage is inversely proportional to capacitance Now, given that C1 = 10 mF ; ^V1hmax = 10V C2 = 5 mF ; ^V2hmax = 5 V C 3 = 2 mF ; ^V3hmax = 2V So, from Eq (1) we have V2 = 2 5 V3 for ^V3hmax = 2 We obtain, V2 = 2 # 2 = 0.8 volt < 5 5 i.e., V2 < ^V2hmax 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 Ceq = C2 C 3 + C1 C2 + C3 = 5 # 2 + 10 5+2 80 = mF 7 Equivalent voltage is (max. value) Vmax = V1 = 2.8 So, charge stored in the effective capacitance is Q = Ceq Vmax
nodia.co.in = b 80 l # ^2.8h 7 = 32 mC
SOL 1.2.8
Option (C) is correct. 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, ^ j4h Vs VL1 = j4 + 3
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j40 100 53.13c j4 + 3 40 90c = 100 53.13c 5 53.13c = 800 90c
VTh = 10 VL1 =
SOL 1.2.9
Option (C) is correct.
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Applying nodal analysis at top node. V1 + 1 0c V1 + 1 0c = 1 0c + 1 j1 V1 (j 1 + 1) + j 1 + 1 0c = j1 V1 = - 1 1 + j1
SOL 1.2.10
1 V1 + 1 0c - 1 + j + 1 j Current = I1 = = = 1 A j1 j1 (1 + j) j 1 + j Option (A) is correct. We put a test source between terminal 1, 2 to obtain equivalent impedance
ZTh = Vtest Itest By applying KCL at top right node Vtest + Vtest - 99I = I b test 9 k + 1k 100 Vtest + Vtest - 99I = I b test 10 k 100 But Ib =- Vtest =-Vtest 9k + 1k 10k
...(i)
Substituting Ib into equation (i), we have Vtest + Vtest + 99Vtest = I test 10 k 100 10 k
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SOL 1.2.11
Page 67
100Vtest + Vtest = I test 10 # 103 100 2Vtest = I test 100 Vtest ZTh = Itest = 50 W Option (C) is correct. (s2 + 9) (s + 2) G (s) = (s + 1) (s + 3) (s + 4) (- w2 + 9) (jw + 2) G (jw) = (jw + 1) (jw + 3) (jw + 4) The steady state output will be zero if G (jw) = 0 -w 2 + 9 = 0 w = 3 rad/s
SOL 1.2.12
Option (B) is correct. In phasor form
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Z = 4 - j3 Z = 5 - 36.86cW I = 5 100c A Average power delivered. Pavg. = 1 I 2 Z cos q = 1 # 25 # 5 cos 36.86c = 50 W 2 2
Alternate method:
SOL 1.2.13
Z = (4 - j3) W I = 5 cos (100pt + 100) A 2 Pavg = 1 Re $ I Z . = 1 # Re "(5) 2 # (4 - j3), = 1 # 100 = 50 W 2 2 2 Option (D) is correct. The s -domain equivalent circuit is shown as below.
vc (0) /s v (0) = c 1 + 1 1 + 1 C1 s C 2 s C1 C 2 I (s) = b C1 C2 l (12 V) C1 + C 2 I (s) =
vC (0) = 12 V
I (s) = 12Ceq Taking inverse Laplace transform for the current in time domain,
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i (t) = 12Ceq d (t) SOL 1.2.14
Page 68
(Impulse)
Option (A) is correct. In the given circuit,
VA - VB = 6 V So current in the branch, IAB = 6 = 3 A 2 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
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The total current in the resistor 1 W will be I1 = 2 + IDC = 2+3 = 5A So, VCD = 1 # (- I1) =- 5 V SOL 1.2.15
(By writing KCL at node D )
Option (A) is correct. We obtain Thevenin equivalent of circuit B .
Thevenin Impedance :
ZTh = R
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Thevenin Voltage : VTh = 3 0c V Now, circuit becomes as
I1 = 10 - 3 2+R Power transfer from circuit A to B P = (I 12) 2 R + 3I1 Current in the circuit,
2 = :10 - 3D R + 3 :10 - 3D = 49R 2 + 21 2+R 2+R (2 + R) (2 + R) 49R + 21 (2 + R) = = 42 + 70R2 (2 + R) 2 (2 + R) 2 dP = (2 + R) 70 - (42 + 70R) 2 (2 + R) = 0 dR (2 + R) 4 (2 + R) [(2 + R) 70 - (42 + 70R) 2] = 0 140 + 70R - 84 - 140R = 0 56 = 70R R = 0.8 W
nodia.co.in SOL 1.2.16
Option (C) is correct. 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 .
IL =
VTh,10 V RTh + RL
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For RL = 1 W , IL = 3 A VTh,10 V RTh + 1 For RL = 2.5 W , IL = 2 A V 2 = Th,10 V RTh + 2.5 Dividing above two 3 = RTh + 2.5 2 RTh + 1 3=
...(i)
...(ii)
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
nodia.co.in
For RL = 7 W , SOL 1.2.17
IL =
VTh,10 V = 9 = 1A 2 + RL 2 + 7
Option (B) is correct. Now, when 6 V connected at port A let Thevenin voltage seen at port B is VTh,6 V . Here RL = 1 W and IL = 7 A 3
VTh, 6 V = RTh # 7 + 1 # 7 = 2 # 7 + 7 = 7 V 3 3 3 3 This is a linear network, so VTh at port B can be written as VTh = V1 a + b where V1 is the input applied at port A. We have V1 = 10 V , VTh,10 V = 9 V 9 = 10a + b When V1 = 6 V , VTh, 6 V = 9 V 7 = 6a + b Solving (i) and (ii) a = 0.5 , b = 4
...(i) ...(ii)
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Thus, with any voltage V1 applied at port A, Thevenin voltage or open circuit voltage at port B will be So, VTh, V = 0.5V1 + 4 For V1 = 8 V (open circuit voltage) VTh,8 V = 0.5 # 8 + 4 = 8 = Voc 1
SOL 1.2.18
Option (A) is correct. By taking V1, V2 and V3 all are phasor voltages. V1 = V2 + V3 Magnitude of V1, V2 and V3 are given as V1 = 220 V , V2 = 122 V , V3 = 136 V Since voltage across R is in same phase with V1 and the voltage V3 has a phase difference of q with voltage V1 , we write in polar form V1 = V2 0c + V3 q V1 = V2 + V3 cos q + jV3 sin q V1 = (V2 + V3 cos q) + jV3 sin q V1 =
(V2 + V3 cos q) 2 + (V2 sin q) 2
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220 = (122 + 136 cos q) 2 + (136 sin q) 2 By solving, power factor cos q = 0.45
SOL 1.2.19
SOL 1.2.20
Option (B) is correct. Voltage across load resistance
VRL = V3 cos q = 136 # 0.45 = 61.2 V Power absorbed in RL 2 (61.2) 2 PL = V RL = - 750 W 5 RL Option (B) is correct. The frequency domains equivalent circuit at w = 1 rad/ sec .
Since the capacitor and inductive reactances are equal in magnitude, the net impedance of that branch will become zero. Equivalent circuit
Current
i (t) = sin t = (1 sin t) A 1W
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rms value of current i rms = 1 A 2 SOL 1.2.21
Option (D) is correct. Voltage in time domain v (t) = 100 2 cos (100pt) Current in time domain i (t) = 10 2 sin (100pt + p/4) Applying the following trigonometric identity sin (f) = cos (f - 90c) i (t) = 10 2 cos (100pt + p/4 - p/2)
So,
= 10 2 cos (100pt - p/4) In phasor form, SOL 1.2.22
I = 10 2 - p/4 2
Option (A) is correct.
nodia.co.in Power transferred to the load 2 10 RL l Rth + RL For maximum power transfer Rth , should be minimum. Rth = 6R = 0 6+R
P = I 2 RL = b
R =0 Note: Since load resistance is constant so we choose a minimum value of Rth SOL 1.2.23
Option (C) is correct.
2 (5 103) 2 Power loss = V rated = # = 20 W Rp 1.25 # 106 For an parallel combination of resistance and capacitor 1 1 tan d = = 1 = 0.025 = 40 wC p R p 2p # 50 # 1.25 # 0.102
SOL 1.2.24
Option (C) is correct. Charge Q = CV = e0 er A V = (e0 er A)V d d
C = e0 er A d
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Q = Q max We have e0 = 8.85 # 10-14 F/cm , er = 2.26 , A = 20 # 40 cm2 V = 50 103 kV/cm # d Maximum electrical charge on the capacitor V = V when b d l = 50 kV/cm d max Thus, SOL 1.2.25
Q = 8.85 # 10-14 # 2.26 # 20 # 40 # 50 # 103 = 8 mC
Option (C) is correct. vi = 100 2 sin (100pt) V Fundamental component of current ii = 10 2 sin (100pt - p/3) A 1
Input power factor I1 (rms) cos f1 Irms I1 (rms) is rms values of fundamental component of current and Irms is the rms value of converter current. 10 pf = cos p/3 = 0.44 102 + 52 + 22 Option (B) is correct. Only the fundamental component of current contributes to the mean ac input power. The power due to the harmonic components of current is zero. So, Pin = Vrms I1rms cos f1 = 100 # 10 cos p/3 = 500 W pf =
SOL 1.2.26
SOL 1.2.27
nodia.co.in
Option (B) is correct. Power delivered by the source will be equal to power dissipated by the resistor. P = Vs Is cos p/4 = 1 #
SOL 1.2.28
2 cos p/4 = 1 W
Option (D) is correct. IC = Is - I RL = =
2 p /4 -
2 - p /4
2 $^cos p/4 + j sin p/4h - ^cos p/4 - j sin p/4h.
= 2 2 j sin p/4 = 2j SOL 1.2.29
Option (B) is correct. For t < 0 , the switch was closed for a long time so equivalent circuit is
Voltage across capacitor at t = 0 vc (0) = 5 = 4 V 4#1 Now switch is opened, so equivalent circuit is
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For capacitor at t = 0+ vc (0+) = vc (0) = 4 V current in 4 W resistor at t = 0+ , i1 =
vc (0+) =1A 4
so current in capacitor at t = 0+ , ic (0+) = i1 = 1 A SOL 1.2.30
Option (B) is correct. Thevenin equivalent across 1 X resistor can be obtain as following Open circuit voltage vth = 100 V (i = 0) Short circuit current (vth = 0 ) isc = 100 A So, Rth = vth = 100 = 1 W isc 100
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Equivalent circuit is
i = 100 = 50 A 1+1 SOL 1.2.31
Option (B) is correct. The circuit is
Current in R W resistor is i = 2-1 = 1 A Voltage across 12 W resistor is So, SOL 1.2.32
VA = 1 # 12 = 12 V i = VA - 6 = 12 - 6 = 6 W 1 R
Option (C) is correct.
V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 Here, I1 = I l1, I2 = I l2
V l1 = Zl11 I l1 + Zl12 I l2 V l2 = Zl21 I l1 + Zl22 I l2
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When R = 1 W is connected V l1 = V1 + I l1 # 1 = V1 + I1 V l1 = Z11 I1 + Z12 I2 + I1 V l1 = (Z11 + 1) I1 + Z12 I2 So, Zl11 = Z11 + 1 Zl12 = Z12 Similarly for output port V l2 = Zl21 I l1 + Zl22 I l2 = Zl21 I1 + Zl22 I2 So, Zl21 = Z21 , Zl22 = Z22 Z11 + 1 Z12 Z-matrix is Z => Z21 Z22H SOL 1.2.33
Option (A) is correct.
nodia.co.in In the bridge R1 R 4 = R 2 R 3 = 1 So it is a balanced bridge I = 0 mA SOL 1.2.34
Option (D) is correct. Resistance of the bulb rated 200 W/220 V is 2 (220) 2 R1 = V = = 242 W 200 P1 Resistance of 100 W/220 V lamp is 2 (220) 2 RT = V = = 484 W 100 P2 To connect in series RT = n # R1 484 = n # 242 n =2
SOL 1.2.35
Option (D) is correct. For t < 0 , S1 is closed and S2 is opened so the capacitor C1 will charged upto 3 volt. VC1 (0) = 3 Volt Now when switch positions are changed, by applying charge conservation Ceq VC (0+) = C1 VC (0+) + C2 VC (0+) (2 + 1) # 3 = 1 # 3 + 2 # VC (0+) 1
1
2
2
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9 = 3 + 2VC (0+) VC (0+) = 3 Volt 2
2
SOL 1.2.36
Option (A) is correct.
Applying KVL in the input loop v1 - i1 (1 + 1) # 103 - 1 (i1 + 49i1) = 0 jw C v1 = 2 # 103 i1 + 1 50i1 jw C 1 Input impedance, Z1 = v1 = 2 # 103 + i1 jw (C/50) 100 nF Equivalent capacitance, Ceq = C = = 2 nF 50 50
nodia.co.in
SOL 1.2.37
Option (B) is correct. Voltage across 2 X resistor, VS = 2 V Current, I2W = VS = 4 = 2 A 2 2 To make the current double we have to take VS = 8 V
SOL 1.2.38
Option (B) is correct. To obtain equivalent Thevenin circuit, put a test source between terminals AB
Applying KCL at super node VP - 5 + VP + VS = I S 2 2 1 VP - 5 + VP + 2VS = 2IS 2VP + 2VS = 2Is + 5 VP + VS = IS + 2.5 VP - VS = 3VS & VP = 4VS So, 4VS + VS = IS + 2.5 5VS = IS + 2.5 VS = 0.2IS + 0.5 For Thevenin equivalent circuit
...(1)
...(2)
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VS = IS Rth + Vth By comparing (2) and (3), Thevenin resistance Rth = 0.2 kW SOL 1.2.39
Option (D) is correct. From above Vth = 0.5 V
SOL 1.2.40
Option (A) is correct. No. of chords is given as
Page 77
...(3)
l = b-n+1 b " no. of branches n " no. of nodes l " no. of chords
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b = 6, n = 4 l = 6 - 4 + 1= 3 SOL 1.2.41
Option (A) is correct. Impedance Zo = 2.38 - j0.667 W Constant term in impedance indicates that there is a resistance in the circuit. Assume that only a resistance and capacitor are in the circuit, phase difference in Thevenin voltage is given as (Due to capacitor) q =- tan- 1 (wCR) j Zo = R wC 1 = 0.667 So, wC and
R = 2.38 W q =- tan- 1 b 1 # 2.38 l =- 74.34c =[ 15.9c 0.667
given Voc = 3.71+ - 15.9c So, there is an inductor also connected in the circuit SOL 1.2.42
Option (C) is correct. Time constant of the circuit can be calculated by simplifying the circuit as follows
Ceq = 2 F 3 Equivalent Resistance
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Req = 3 + 3 = 6 W t = Req Ceq = 6 # 2 = 4 sec 3 Option (C) is correct. Impedance of the circuit is 1 R 1 - jwCR R Z = jwL + 1jwC = jw L + # 1 + jwCR 1 - jwCR j wC + R Time constant SOL 1.2.43
R (1 - jwCR) jwL (1 + w2 C2 R2) + R - jwCR2 = 1 + w2 C2 R2 1 + w2 C2 R2 j [wL (1 + w2 C2 R2) - wCR2] R = + 1 + w2 C2 R2 1 + w2 C2 R2 For resonance Im (Z) = 0 So, wL (1 + w2 C2 R2) = wCR2 L = 0.1 H, C = 1 F, R = 1 W So, w # 0.1 [1 + w2 (1) (1)] = w (1) (1) 2 = jw L +
nodia.co.in 1 + w2 = 10
& SOL 1.2.44
w=
9 = 3 rad/sec
Option (A) is correct. By applying KVL in the circuit Vab - 2i + 5 = 0 i = 1 A, Vab = 2 # 1 - 5 =- 3 Volt
SOL 1.2.45
Option (C) is correct. Charge stored at t = 5 m sec 5
Q =
# i (t) dt
=area under the curve
0
Q =Area OABCDO =Area (OAD)+Area(AEB)+Area(EBCD) = 1#2#4+1#2#3+3#2 2 2
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= 4 + 3 + 6 = 13 nC SOL 1.2.46
Option (D) is correct. Initial voltage across capacitor Q V0 = o = 13 nC = 43.33 Volt 0.3 nF C When capacitor is connected across an inductor it will give sinusoidal esponse as where At t = 1 m sec ,
SOL 1.2.47
vc (t) = Vo cos wo t 1 wo = 1 = -9 LC 0.3 # 10 # 0.6 # 10- 3 = 2.35 # 106 rad/sec vc (t) = 43.33 cos (2.35 # 106 # 1 # 10- 6) = 43.33 # (- 0.70) =- 30.44 V
Option (B) is correct. By writing node equations at node A and B Va - 5 + Va - 0 = 0 1 1
nodia.co.in 2Va - 5 = 0 Va = 2.5 V
Similarly Vb - 4Vab ++Vb - 0 = 0 3 1 Vb - 4 (Va - Vb) + Vb = 0 3
Vb - 4 (2.5 - Vb) + 3Vb = 0 8Vb - 10 = 0 Vb = 1.25 V Current i = Vb = 1.25 A 1 SOL 1.2.48
Option ( ) is correct.
SOL 1.2.49
Option (B) is correct. Here two capacitance C1 and C2 are connected in series, so equivalent capacitance is Ceq = C1 C2 C1 + C 2
- 12 -6 # 500 # 10 C1 = e0 er1 A = 8.85 # 10 # 8 # 500 d1 4 # 10- 3
= 442.5 # 10- 11 F - 12 -6 # 500 # 10 C2 = e0 er2 A = 8.85 # 10 # 2 # 500 d2 2 # 10- 3 = 221.25 # 10- 11 F - 11 - 11 Ceq = 442.5 # 10 - 11 # 221.25 # 10- 11 = 147.6 # 10- 11 442.5 # 10 + 221.25 # 10
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- 1476 pF SOL 1.2.50
Option (B) is correct. Circumference no. of turns Cross sectional area
l = 300 mm n = 300 A = 300 mm2
Inductance of coil
L =
4p # 10- 7 # (300) 2 # 300 # 10- 6 m0 n2 A = l (300 # 10- 3)
= 113.04 mH SOL 1.2.51
Option (A) is correct. Divergence of a vector field is given as Divergence = 4: V In cartesian coordinates 4 = 2 it + 2 tj + 2 kt 2x 2y 2z So 4: V = 2 6- (x cos xy + y)@ + 2 6(y cos xy)@ + 2 6(sin z2 + x2 + y2)@ 2x 2y 2z
nodia.co.in =- x (- sin xy) y + y (- sin xy) x + 2z cos z2 = 2z cos z2
SOL 1.2.52
Option (A) is correct. Writing KVL for both the loops V - 3 (I1 + I2) - Vx - 0.5 dI1 = 0 dt V - 3I1 - 3I2 - Vx - 0.5 dI1 = 0 dt
...(1)
In second loop - 5I2 + 0.2Vx + 0.5 dI1 = 0 dt
SOL 1.2.53
...(2) I2 = 0.04Vx + 0.1 dI1 dt Put I2 from eq(2) into eq(2) V - 3I1 - 3 :0.04Vx + 0.1 dI1 D - Vx - 0.5 dI1 = 0 dt dt 0.8 dI1 =- 1.12Vx - 3I1 + V dt dI1 =- 1.4V - 3.75I + 5 V x 1 4 dt Option (A) is correct. Impedance of the given network is Z = R + j b wL - 1 l wC 1 AdmittanceY = 1 = Z R + j b wL - 1 l wC R - j b wL - 1 l R - j b wL - 1 l wC wC 1 = = # 2 1 1 2 R + j b wL R - j b wL R + b wL - 1 l l l wC wC wC j b wL - 1 l wC R = 2 2 R 2 + b wL - 1 l R 2 + b wL - 1 l wC wC
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= Re (Y) + Im (Y) Varying frequency for Re (Y) and Im (Y) we can obtain the admittance-locus.
SOL 1.2.54
Option (D) is correct. At t = 0+ , when switch positions are changed inductor current and capacitor voltage does not change simultaneously So at t = 0+ vc (0+) = vc (0-) = 10 V iL (0+) = iL (0-) = 10 A The equivalent circuit is
nodia.co.in
Applying KCL vL (0+) vL (0+) - vc (0+) + = iL (0+) = 10 10 10 2vL (0+) - 10 = 100 Voltage across inductor at t = 0+ vL (0+) = 100 + 10 = 55 V 2
SOL 1.2.55
So, current in capacitor at t = 0+ v (0+) - vc (0+) iC (0+) = L = 55 - 10 = 4.5 A 10 10 Option (B) is correct. In the circuit
VX = V+0c Vy - 2V+0c + (Vy) jwC = 0 R
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Vy (1 + jwCR) = 2V+0c Vy = 2V+0c 1 + jwCR VYX = VX - VY = V VYX = V - 2V =- V VYX = V - 0 = V
R " 0, R " 3, SOL 1.2.56
2V 1 + jwCR
Option (A) is correct. The circuit is
Applying KVL 2 = VNL 3 - 2 # I NL 2 2 3 - 2I NL = I NL 2 = 3 & INL = 1 A 3I NL VNL = (1) 2 = 1 V So power dissipated in the non-linear resistance P = VNL INL = 1 # 1 = 1 W
nodia.co.in
SOL 1.2.57
SOL 1.2.58
Option (C) is correct. In node incidence matrix b1 b 2 b 3 b 4 b 5 b 6 V R n1 S 1 1 1 0 0 0 W n2S 0 - 1 0 - 1 1 0 W n 3SS- 1 0 0 0 - 1 - 1WW n 4S 0 0 - 1 1 0 1 W X T In option (C) E = AV R V S1 1 1 0 0 0W S 0 -1 0 -1 1 0 W T T 8e1 e2 e 3 e 4B = S- 1 0 0 0 - 1 - 1W8V1 V2 -- V6B S W S 0 0 -1 1 0 1 W X R V TR V Se1W S V1 + V2 + V3 W Se2W S- V2 - V4 + V5W Se W = S- V - V - V W which is true. 5 6W S 3W S 1 Se 4W S- V3 + V4 + V6W T X T X Option (A) is correct. Assume a Gaussian surface inside the sphere (x < R)
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From gauss law y = Q enclosed =
# D : ds = Q enclosed
3 Q 4 pr3 = Qr # 3 4 3 R3 3 pR 3 # D : ds = Qr3 R Q r Qr3 D # 4pr2 = 3 = pe0 R3 4 R
Q enclosed =
So, or SOL 1.2.59
SOL 1.2.60
SOL 1.2.61
a D = e0 E
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Option (D) is correct. Inductance is given as
4p # 10- 7 # (400) 2 # (16 # 10- 4) m0 N2 A = L = = 321.6 mH (1 # 10- 3) l V = IXL = 230 ` XL = 2pfL 2pfL 230 = 2.28 A = 2 # 3.14 # 50 # 321.6 # 10- 3 Option (A) is correct. Energy stored is inductor E = 1 LI2 = 1 # 321.6 # 10- 3 # (2.28) 2 2 2 Force required to reduce the air gap of length 1 mm is F = E = 0.835- 3 = 835 N l 1 # 10 Option (D) is correct. Thevenin voltage:
Vth = I (R + ZL + ZC ) = 1+0c [1 + 2j - j] = 1 (1 + j) = Thevenin impedance:
2 +45% V
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Zth = R + ZL + ZC = 1 + 2j - j = (1 + j) W SOL 1.2.62
Option (A) is correct. In the given circuit
Output voltage vo = Avi = 106 # 1 mV = 1 V Input impedance Zi = vi = vi = 3 0 ii Output impedance Zo = vo = Avi = Ro = 10 W io io Option (D) is correct. All sources present in the circuit are DC sources, so all inductors behaves as short circuit and all capacitors as open circuit Equivalent circuit is
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SOL 1.2.63
Voltage across R 3 is 5 = I1 R 3 5 = I1 (1) I1 = 5 A By applying KCL, current through voltage source
(current through R 3 )
1 + I 2 = I1 I2 = 5 - 1 = 4 A SOL 1.2.64
Option () is correct. Given Two port network can be described in terms of h-parametrs only.
SOL 1.2.65
Option (A) is correct. At resonance reactance of the circuit would be zero and voltage across inductor and capacitor would be equal
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VL = VC At resonance impedance of the circuit Current
Z R = R1 + R 2 IR = V1 +0c R1 + R 2
V2 = IR R2 + j (VL - VC ) V2 = V1 +0c R2 R1 + R 2 Voltage across capacitor VC = 1 # IR = 1 # VR +0c = VR + - 90c R1 + R 2 jw C jw C wC (R1 + R2) So phasor diagram is Voltage
SOL 1.2.66
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Option (B) is correct. This is a second order LC circuit shown below
Capacitor current is given as dv (t) iC (t) = C c dt Taking Laplace transform IC (s) = CsV (s) - V (0), V (0) "initial voltage Current in inductor iL (t) = 1 # vc (t) dt L V (s) IL (s) = 1 L s for t > 0 , applying KCL(in s-domain) IC (s) + IL (s) = 0 V (s) =0 CsV (s) - V (0) + 1 L s 1 2 :s + LCs D V (s) = Vo V (s) = Vo 2 s 2 , s + w0 Taking inverse Laplace transformation v (t) = Vo cos wo t , t > 0 SOL 1.2.67
a w20 = 1 LC
Option (B) is correct. Power dissipated in heater when AC source is connected
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Page 86
2
P = 2.3 kW = V rms R 2.3 # 103 =
(230) 2 R
R = 23 W (Resistance of heater) Now it is connected with a square wave source of 400 V peak to peak Power dissipated is 2 P = V rms , Vp - p = 400 V & Vp = 200 V R
SOL 1.2.68
(200) 2 = = 1.739 kW 23 Option (D) is correct. From maxwell’s first equation
Vrms = Vp =200 (for square wave)
4: D = rv r 4: E = v e (Divergence of electric field intensity is non-Zero) Maxwell’s fourth equation
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4: B = 0 (Divergence of magnetic field intensity is zero) SOL 1.2.69
Option (C) is correct. Current in the circuit
I =
Or SOL 1.2.70
Option (A) is correct. Rms value is given as 32 +
(4) 2 = 2
9 + 8 = 17 V
Option (D) is correct. Writing KVL in input and output loops V1 - (i1 + i2) Z1 = 0 V1 = Z1 i1 + Z1 i2 Similarly
SOL 1.2.72
(given)
100 = 8 R+5 R = 60 = 7.5 W 8
mrms = SOL 1.2.71
100 =8 A R + (10 || 10)
...(1)
V2 - i2 Z2 - (i1 + i2) Z1 = 0 ...(2) V2 = Z1 i1 + (Z1 + Z2) i2 From equation (1) and (2) Z -matrix is given as Z1 Z1 Z => Z1 Z1 + Z2H Option (B) is correct. In final steady state the capacitor will be completely charged and behaves as an open circuit
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SOL 1.2.73
Page 87
Steady state voltage across capacitor vc (3) = 20 (10) = 10 V 10 + 10 Option (D) is correct. We know that divergence of the curl of any vector field is zero 4 (4 # E) = 0
SOL 1.2.74
Option (A) is correct. When the switch is at position 1, current in inductor is given as
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120 = 2 A 20 + 40 At t = 0 , when switch is moved to position 1,inductor current does not change simultaneously so iL (0-) =
iL (0+) = iL (0-)=2 A Voltage across inductor at t = 0+ vL (0+) = 120 V By applying KVL in loop 120 = 2 (40 + R + 20) 120 = 120 + R R = 0W SOL 1.2.75
Option (C) is correct. Let stored energy and dissipated energy are E1 and E2 respectively. Then Current i 22 = E2 = 0.95 E1 i 12 i2 = 0.95 i1 = 0.97i1 Current at any time t, when the switch is in position (2) is given by R
60
i (t) = i1 e- L t = 2e- 10 t = 2e- 6t After 95% of energy dissipated current remaining in the circuit is i = 2 - 2 # 0.97 = 0.05 A
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So, SOL 1.2.76
Page 88
0.05 = 2e- 6t t . 0.50 sec
Option (C) is correct. At f1 = 100 Hz, voltage drop across R and L is mRMS V (jw L) mRMS = Vin .R = in 1 R + jw1 L R + jw1 L So, R = w1 L At f2 = 50 Hz, voltage drop across R mlRMS = Vin .R R + jw2 L mRMS R + jw2 L = = R + jw1 L mlRMS =
w12 L2 + w22 L2 , w12 L2 + w12 L2
=
w12 + w22 = 2w12 8m 5 RMS
mlRMS =
SOL 1.2.77
R2 + w22 L2 R2 + w12 L2 R = w1 L f 12 + f 22 2f 12
=
(100) 2 + (50) 2 = 2 (100) 2
5 8
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Option (A) is correct. In the circuit
I B = IR +0c + Iy +120c
Since so,
I B2 = I R2 + I y2 + 2IR Iy cos b 120c l = I R2 + I y2 + IR Iy 2 I R = Iy I B2 = I R2 + I R2 + I R2 = 3I R2 I B = 3 I R = 3 Iy
IR: Iy: IB = 1: 1: 3 SOL 1.2.78
Option (C) is correct. Switch was opened before t = 0 , so current in inductor for t < 0
iL (0-) = 10 = 1 A 10 Inductor current does not change simultaneously so at t = 0 when switch is closed current remains same iL (0+) = iL (0-)=1 A SOL 1.2.79
Option (A) is correct. Thevenin voltage:
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Page 89
Nodal analysis at P Vth - 4 + Vth = 0 10 10 2Vth - 4 = 0 Vth = 2 V Thevenin resistance:
Rth = 10 W || 10 W = 5 W SOL 1.2.80
Option (A) is correct. Electric field inside a conductor (metal) is zero. In dielectric charge distribution os constant so electric field remains constant from x1 to x2 . In semiconductor electric field varies linearly with charge density.
SOL 1.2.81
Option (D) is correct. Resonance will occur only when Z is capacitive, in parallel resonance condition, suseptance of circuit should be zero. 1 + jw C = 0 jw L
nodia.co.in 1 - w2 LC = 0
1 (resonant frequency) LC 1 1 C = 2 = = 0.05 m F 2 wL 4 # p # (500) 2 # 2 Option (D) is correct. Here two capacitor C1 and C2 are connected in series so equivalent Capacitance is Ceq = C1 C2 C1 + C 2 w=
SOL 1.2.82
8.85 # 10- 12 # 4 (400 # 10- 3) 2 C1 = e0 er1 A = d1 6 # 10- 3 Similarly
- 12 16 # 10- 2 = 94.4 10- 11 F = 8.85 # 10 # 4 # # 6 # 10- 3
8.85 # 10- 12 # 2 # (400 # 10- 3) 2 C2 = e0 er2 A = d2 8 # 10- 3
SOL 1.2.83
- 12 16 # 10- 12 = 35.4 10- 11 F = 8.85 # 10 # 2 # # -3 8 # 10 - 11 - 11 Ceq = 94.4 # 10 # 35.4 #-10 = 25.74 # 10- 11 - 257 pF (94.4 + 35.4) # 10 11 Option (C) is correct. Inductance of the Solenoid is given as m N2 A L = 0 l
Where
A " are of Solenoid
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Page 90
l " length 4p # 10- 7 # (3000) 2 # p (30 # 10- 3) 2 L = = 31.94 # 10- 3 H (1000 # 10- 3) - 32 mH SOL 1.2.84
Option (C) is correct. In the circuit
VA = (2 + 1) # 6 = 18 Volt 2 = E - VA 6 2 = E - 18 6
Voltage So,
nodia.co.in E = 12 + 18 = 30 V
SOL 1.2.85
Option (A) is correct. Delta to star (T - Y) conversions is given as Rb Rc R1 = = 10 # 10 = 2.5 W 20 + 10 + 10 Ra + Rb + Rc Ra Rc R2 = = 20 # 10 = 5 W 20 + 10 + 10 Ra + Rb + Rc Ra Rb R3 = = 20 # 10 = 5 W 20 + 10 + 10 Ra + Rb + Rc
SOL 1.2.86
Option (D) is correct. For parallel circuit I = E = EYeq Zeq Yeq " Equivalent admittance of the circuit Yeq = YR + YL + YC = (0.5 + j0) + (0 - j1.5) + (0 + j0.3) = 0.5 - j1.2 So, current I = 10 (0.5 - j1.2) = (5 - j12) A
SOL 1.2.87
Option (B) is correct. In the circuit
100 10R 100 (10 || R) = # b 10 R 10 +Rl 10 + (10 || R) f 10 + p 10 + R 1000 R 50 R = = 100 + 20R 5+R Current in R W resistor Voltage
VA =
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Page 91
2 = VA R 2=
R = 20 W
or SOL 1.2.88
50R R (5 + R)
Option (A) is correct. Since capacitor initially has a charge of 10 coulomb, therefore vc (0) " initial voltage across capacitor Q 0 = Cvc (0) 10 = 0.5vc (0) vc (0) = 10 = 20 V 0.5 When switch S is closed, in steady state capacitor will be charged completely and capacitor voltage is vc (3) = 100 V At any time t transient response is t
vc (t) = vc (3) + [vc (0) - vc (3)] e- RC
nodia.co.in t
vc (t) = 100 + (20 - 100) e- 2 # 0.5 = 100 - 80e- t Current in the circuit i (t) = C dvc = C d [100 - 80e- t] dt dt = C # 80e- t = 0.5 # 80e- t = 40e- t
At t = 1 sec, i (t) = 40e- 1 = 14.71 A SOL 1.2.89
Option (D) is correct. Total current in the wire I = 10 + 20 sin wt 102 +
Irms = SOL 1.2.90
300 = 17.32 A
Option (D) is correct. From Z to Y parameter conversion Y11 Y12 Z11 Z12 - 1 >Y Y H = >Z Z H 21 22 21 22 So,
SOL 1.2.91
(20) 2 = 100 + 200 = 2
Y11 Y12 0.6 - 0.2 >Y Y H = 0.150 >- 0.2 0.9 H 12 22 Y22 = 0.9 = 1.8 0.50
Option (C) is correct. Energy absorbed by the inductor coil is given as t
EL =
# Pdt 0
Where power
P = VI = I bL dI l dt t
So,
EL =
dt # LIb dI dt l
0
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For0 # t # 4 sec 4
dt # Ib dI dt l
EL = 2
0
a dI = 3, 0 # t # 2 , * dt 0 2 = 0, 2 < t < 4 2 = 6 # I.dt =6(area under the curve i (t) - t ) 2
4
# I (3) dt + 2 # I (0) dt
=2
0
= 6 # 1 # 2 # 6 = 36 J 2 Energy absorbed by 1 W resistor is t
ER =
# I2 Rdt 0
=
4
2
# (3t)
2
# 1dt +
# (6) 2 dt 2
0
I = 3t, 0 # t # 2 ) = 6A 2 # t # 4
3 2
4 = 9 # :t D + 36[t]2 = 24 + 72 =96 J 3 0 Total energy absorbed in 4 sec E = EL + ER = 36 + 96 = 132 J
SOL 1.2.92
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Option (B) is correct. Applying KCL at center node
iL = iC + 1 + 2 iL = iC + 3 iC =- C dvc =- 1 d [4 sin 2t] dt dt
SOL 1.2.93
=- 8 cos 2t so (current through inductor) iL =- 8 cos 2t + 3 Voltage across inductor vL = L diL = 2 # d [3 - 8 cos 2t] = 32 sin 2t dt dt Option (A) is correct. Thevenin impedance can be obtain as following
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Page 93
Zth = Z 3 + (Z1 || Z2) Given that
Z1 = 10+ - 60c = 10 c
1-
2
3j
m = 5 (1 -
3 j)
1+
3j m = 5 (1 + 3 j) 2 3 + 4j Z 3 = 50+53.13c = 50 b = 10 (3 + 4j) 5 l Z2 = 10+60c = 10 c
So,
5 (1 - 3j) 5 (1 + 3 j) 5 (1 - 3 j) + 5 (1 + 3 j) 25 (1 + 3) = 30 + 40j + 10 = 40 + 40j = 10 (3 + 4j) + 10
Zth = 10 (3 + 4j) +
Zth = 40 2 +45c W SOL 1.2.94
Option (A) is correct. Due to the first conductor carrying + I current, magnetic field intensity at point P is H 1 = I Y (Direction is determined using right hand rule) 2pd Similarly due to second conductor carrying - I current, magnetic field intensity is H 2 = - I (- Y) = I Y 2pd 2pd Total magnetic field intensity at point P. H = H1 + H 2 = I Y + I Y = I Y 2pd 2pd pd
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SOL 1.2.95
Option ( ) is correct.
SOL 1.2.96
Option (C) is correct. Given that magnitudes of VL and VC are twice of VR VL = VC = 2VR (Circuit is at resonance) Voltage across inductor VL = iR # jwL Current iR at resonance % iR = 5+0 = 5 = 1 A 5 R so,
SOL 1.2.97
VL = wL = 2VR wL = 2 # 5 2 # p # 50 # L = 10 L = 10 = 31.8 mH 314
VR = 5 V, at resonance
Option (C) is correct. Applying nodal analysis in the circuit At node P 2 + VP - 10 + VP = 0 2 8 16 + 4VP - 40 + VP = 0 5VP - 24 = 0 VP = 24 Volt 5 At node Q
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2=
SOL 1.2.98
Page 94
VQ - 10 VQ - 0 + 6 4
24 = 3VQ - 30 + 2VQ 5VQ - 54 = 0 VQ = 54 V 5 Potential difference between P-Q VPQ = VP - VQ = 24 - 54 =- 6 V 5 5 Option (D) is correct. First obtain equivalent Thevenin circuit across load RL
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Thevenin voltage Vth - 110+0c + Vth - 90+0c 0 = 6 + 8j 6 + 8j
2Vth - 200+0c = 0 Vth = 100+0c V Thevenin impedance
Zth = (6 + 8j) W || (6 + 8j) W = (3 + 4j) W For maximum power transfer RL = Zth =
32 + 42 = 5 W
Power in load 2 P = ieff RL
SOL 1.2.99
2
2 (100) 100 5 = 625 Watt 5 = 80 # 3 + 4j + 5 # Option (D) is correct. By applying mesh analysis in the circuit
P =
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Page 95
I1 = 10 A, I2 =- 5 A Current in 2 W resistor I2W = I1 - (- I2) = 10 - (- 5) = 15 A So, voltage VA = 15 # 2 = 30 Volt Now we can easily find out current in all branches as following
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Current in resistor R is 5 A 5 = 100 - 40 R R = 60 = 12 W 5 SOL 1.2.100
Option (B) is correct. Before t = 0 , the switch was opened so current in inductor and voltage across capacitor for t < 0 is zero vc (0-) = 0 , iL (0-) = 0 at t = 0 , when the switch is closed, inductor current and capacitor voltage does not change simultaneously so vc (0+) = vc (0-) = 0 , iL (0+) = iL (0-) = 0 At t = 0+ the circuit is
Simplified circuit
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SOL 1.2.101
Page 96
Voltage across inductor (at t = 0+ ) vL (0+) = 10 # 2 = 4 Volt 3+2 Option (D) is correct. Given that E1 = h11 I1 + h12 E2 and I2 = h21 I1 + h22 E2 Parameter h12 is given as h12 = E1 E2 I = 0 (open circuit) 1
At node A E A - E1 + E A - E 2 + E A = 0 2 2 4
nodia.co.in 5EA = 2E1 + 2E2
...(1)
Similarly
E1 - E A + E1 = 0 2 2
2E1 = EA
...(2)
From (1) and (2) 5 (2E1) = 2E1 + 2E2 4 8E1 = 2E2 h12 = E1 = 1 4 E2 SOL 1.2.102
Option (B) is correct. VPQ = VP - VQ =
KQ KQ OP OQ
9 10- 9 - 9 # 109 # 1 # 10- 9 = 9 # 10 # 1 # -3 40 # 10 20 # 10- 3 = 9 # 103 : 1 - 1 D =- 225 Volt 40 20
SOL 1.2.103
SOL 1.2.104
Option (D) is correct. Energy stored in Capacitor is E = 1 CV2 2 - 12 -6 # 10 = 8.85 # 10- 12 F C = e0 A = 8.85 # 10 # 100 3 d 0.1 # 10 E = 1 # 8.85 # 10- 12 # (100) 2 = 44.3 nJ 2 Option (B) is correct. The figure is as shown below
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Page 97
The Capacitor shown in Figure is made up of two capacitor C1 and C2 connected in series. C1 = e0 er1 A , C2 = e0 er2 A t1 t2 Since C1 and C2 are in series charge on both capacitor is same. Q1 = Q 2 C1 (100 - V) = C2 V (Let V is the voltage of foil) e0 er1 A (100 - V) = e0 er2 A V t1 t2 3 (100 - V) = 4 V 0.5 1
nodia.co.in 300 - 3V = 2V 300 = 5V & V = 60 Volt
SOL 1.2.105
Option (D) is correct. Voltage across capacitor is given by vc (t) = 1 C
SOL 1.2.106
3
# i (t) dt -3
= 1 C
3
# 5d (t) dt -3
= 5 # u (t) C
Option (C) is correct. No. of links is given by L = N-B+1
SOL 1.2.107
Option (A) is correct. Divergence theorem states that the total outward flux of a vector field F through a closed surface is same as volume integral of the divergence of F
# F $ ds s
SOL 1.2.108
=
# (4: F) dv V
Option (C) is correct. The figure as shown below
Inductance of parallel wire combination is given as ml L = 0 ln b d l p r Where
l " Length of wires d " Distance between wires r " Radius
L \ ln d So when d is double, inductance increase but does not double.
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SOL 1.2.109
Option (B) is correct. Since distance from ground to lower surface is less than from ground to upper surface so electric stress is maximum at lower surface.
SOL 1.2.110
Option (B) is correct. Writing node equation for the circuit
I 1 = E1 - E A 2 I2 = E2 - EA 2
and At node A E A - E1 + E A + E A - E 2 = 0 2 2 2
nodia.co.in 3EA = E1 + E2
...(1)
From eqn(1)
(E + E2) I 1 = 1 E1 - 1 1 2 2 3 I 1 = 1 E1 - 1 E 2 3 6 (E + E2) Similarly I2 = 1 E2 - 1 1 2 2 3 I2 =- 1 E1 + 1 E2 6 3 From (2) and (3) admittance parameters are
...(2)
...(3)
[Y11 Y12 Y21 Y22] = [1/3 - 1/6 - 1/6 1/3] SOL 1.2.111
Option (A) is correct. Admittance of the given circuit Y (w) = jwC + 1 ZL So,
ZL = 30+40c = 23.1 + j19.2 W 23.1 - j19.2 1 Y (w) = j2p # 50 # C + 23.1 + j19.2 # 23.1 - j19.2 23.1 - j19.2 = j (100p) C + 902.25 = 23.1 + j :(100p) C - 19.2 D 902.25 902.25
For unity power factor Im [Y (w)] = 0 100 # 3.14 # C = 19.2 902.25 C - 68.1 mF SOL 1.2.112
Option (B) is correct. In series RLC circuit lower half power frequency is given by following relations
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Page 99
w1 L - 1 =- R w1 C 1 =- 50 2p # f1 (1 # 10- 6) f1 = 3.055 kHz
(2p # f1 # 100 # 10- 6) -
SOL 1.2.113
Option (C) is correct. Since initial charge across capacitor is zero, voltage across capacitor at any time t is given as t
vc (t) = 10 (1 - e- t ) Time constant t = Req C = (10 kW || 1 kW) # C = b 10 l kW # 11 nF = 10 # 10- 6 sec = 10 m sec 11 t
So, vc (t) = 10 (1 - e- 10 m sec ) Pulse duration is 10 msec, so voltage across capacitor will be maximum at t = 10 m sec
nodia.co.in 10 m sec
vc (t = 10 m sec) = 10 (1 - e- 10 m sec ) = 10 (1 - e- 1) = 6.32 Volt
SOL 1.2.114
Option (C) is correct. Since voltage and current are in phase so equivalent inductance is Leq L1 + L2 ! 2M 8 + 8 ! 2M 16 - 2M M Coupling Coefficient
= 12 H = 12 M " Mutual Inductance = 12 = 12 (Dot is at position Q) =2 H
K =
2 = 0.25 8#8
SOL 1.2.115
Option ( ) is correct.
SOL 1.2.116
Option (C) is correct. In steady state there is no voltage drop across inductor (i.e. it is short circuit) and no current flows through capacitors (i.e. it is open circuit) The equivalent circuit is
So, SOL 1.2.117
vc (3) = 10 # 1=5 Volt 1+1
Option (C) is correct. When the switch was closed before t = 0 , the circuit is
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Current in the inductor iL (0-) = 0 A When the switch was opened at t = 0 , equivalent circuit is
In steady state, inductor behaves as short circuit and 10 A current flows through it
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iL (3) = 10 A Inductor current at any time t is given by
iL (t) = iL (3) + 6iL (0) - iL (3)@ e- L t R
5
= 10 + (0 - 10) e- 10 t = 10 (1 - e- 2t) A SOL 1.2.118
Option (B) is correct. Energy stored in inductor is E = 1 Li2 = 1 # 5 # (10) 2 = 250 J 2 2
SOL 1.2.119
Option (C) is correct. To obtain Thevenin’s equivalent, open the terminals X and Y as shown below,
By writing node equation at X Vth - V1 + Vth - V2 = 0 Z1 Z2 V1 = 30+45c = 30 (1 + j) 2 V2 = 30+ - 45c = 30 (1 - j) 2 So, Vth - 30 (1 + j) Vth - 30 (1 - j) 2 2 + =0 1-j 1+j
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Page 101
2Vth - 30 (1 + j) 2 - 30 (1 - j) 2 = 0 2 2 Vth = 0 Volt Thevenin’s impedance
Zth = Z1 || Z2 = (1 - j) || (1 + j) = SOL 1.2.120
(1 - j) (1 + j) = 1W (1 - j) + (1 + j)
Option (A) is correct. Drawing Thevenin equivalent circuit across load :
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So, current SOL 1.2.121
iL = 0 A
Option (A) is correct. In the circuit we can observe that there are two wheatstone bridge connected in parallel. Since all resistor values are same, therefore both the bridge are balanced and no current flows through diagonal arm. So the equivalent circuit is
We can draw the circuit as
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Page 102
From T - Y conversion
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Now the circuit is
SOL 1.2.122
VAB = 1 # 14 = 1.4 Volt 10 Option (C) is correct. In a series RLC circuit, at resonance, current is given as i = Vs +0c , VS " source voltage R So, voltage across capacitor at resonance
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Page 103
1 # Vs +0c jw C R Vc = Vs + - 90c wCR Vc =
Voltage across capacitor can be greater than input voltage depending upon values of w, C and R but it is 90c out of phase with the input SOL 1.2.123
Option (D) is correct. Let resistance of 40 W and 60 W lamps are R1 and R2 respectively a P \ 12 R P1 = R2 P2 R1 R2 = 40 R1 60 R2 < R1 40 W bulb has high resistance than 60 W bulb, when connected in series power is P1 = I2 R1
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P2 = I2 R2 a R1 > R2 , So P1 > P2 Therefore, 40 W bulb glows brighter SOL 1.2.124
Option (B) is correct. Series RL circuit with unit step input is shown in following figure
1, t > 0 u (t) = ) 0, otherwise Initially inductor current is zero i (0+) = 0 When unit step is applied, inductor current does not change simultaneously and the source voltage would appear across inductor only so voltage across resistor at t = 0+ vR (0+) = 0 SOL 1.2.125
Option (D) is correct. For two coupled inductors M = K L1 L2 Where K " coupling coefficient 0 2 Which of the following is true ? (A) x (t) has finite energy because only finitely many coefficients are non-zero (B) x (t) has zero average value because it is periodic (C) The imaginary part of x (t) is constant (D) The real part of x (t) is even MCQ 1.3.27
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The z-transform of a signal x [n] is given by 4z - 3 + 3z - 1 + 2 - 6z2 + 2z3 It is applied to a system, with a transfer function H (z) = 3z - 1 - 2 Let the output be y [n]. Which of the following is true ? (A) y [n] is non causal with finite support (B) y [n] is causal with infinite support (C) y [n] = 0; n > 3 (D) Re [Y (z)] z = e =- Re [Y (z)] z = e ji
- ji
Im [Y (z)] z = e = Im [Y (z)] z = e ; - p # q < p ji
YEAR 2008
- ji
ONE MARK
MCQ 1.3.28
The impulse response of a causal linear time-invariant system is given as h (t). Now consider the following two statements : Statement (I): Principle of superposition holds Statement (II): h (t) = 0 for t < 0 Which one of the following statements is correct ? (A) Statements (I) is correct and statement (II) is wrong (B) Statements (II) is correct and statement (I) is wrong (C) Both Statement (I) and Statement (II) are wrong (D) Both Statement (I) and Statement (II) are correct
MCQ 1.3.29
A signal e - at sin (wt) is the input to a real Linear Time Invariant system. Given K and f are constants, the output of the system will be of the form Ke - bt sin (vt + f) where (A) b need not be equal to a but v equal to w (B) v need not be equal to w but b equal to a (C) b equal to a and v equal to w (D) b need not be equal to a and v need not be equal to w
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Page 112
YEAR 2008 MCQ 1.3.30
TWO MARKS
A system with x (t) and output y (t) is defined by the input-output relation : y (t) =
- 2t
#- 3x (t) dt
The system will be (A) Casual, time-invariant and unstable (B) Casual, time-invariant and stable (C) non-casual, time-invariant and unstable (D) non-casual, time-variant and unstable MCQ 1.3.31
A signal x (t) = sinc (at) where a is a real constant ^sinc (x) = px h is the input to a Linear Time Invariant system whose impulse response h (t) = sinc (bt), where b is a real constant. If min (a, b) denotes the minimum of a and b and similarly, max (a, b) denotes the maximum of a and b, and K is a constant, which one of the following statements is true about the output of the system ? (A) It will be of the form Ksinc (gt) where g = min (a, b) (B) It will be of the form Ksinc (gt) where g = max (a, b) (C) It will be of the form Ksinc (at) (D) It can not be a sinc type of signal sin (px)
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MCQ 1.3.32
Let x (t) be a periodic signal with time period T , Let y (t) = x (t - t0) + x (t + t0) for some t0 . The Fourier Series coefficients of y (t) are denoted by bk . If bk = 0 for all odd k , then t0 can be equal to (A) T/8 (B) T/4 (C) T/2 (D) 2T
MCQ 1.3.33
H (z) is a transfer function of a real system. When a signal x [n] = (1 + j) n is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of ^1 - 12 z - 1h H(z) is the entire Z-plane (except z = 0 ). It can then be inferred that H (z) can have a minimum of (A) one pole and one zero (B) one pole and two zeros (C) two poles and one zero D) two poles and two zeros
MCQ 1.3.34
z Given X (z) = with z > a , the residue of X (z) zn - 1 at z = a for n $ 0 2 (z - a) will be (B) an (A) an - 1 (D) nan - 1
(C) nan MCQ 1.3.35
Let x (t) = rect^t - 12 h (where rect (x) = 1 for - 12 # x #
1 2
and zero otherwise.
sin (px)
MCQ 1.3.36
If sinc (x) = px , then the FTof x (t) + x (- t) will be given by (B) 2 sinc` w j (A) sinc` w j 2p 2p (C) 2 sinc` w j cos ` w j (D) sinc` w j sin ` w j 2p 2 2p 2 Given a sequence x [n], to generate the sequence y [n] = x [3 - 4n], which one of the following procedures would be correct ?
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Page 113
(A) First delay x (n) by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and than finally time reverse z2 [n] to obtain y [n]. (B) First advance x [n] by 3 samples to generate z1 [n], then pick every 4th sample of z1 [n] to generate z2 [n], and then finally time reverse z2 [n] to obtain y [n] (C) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally advance v2 [n] by 3 samples to obtain y [n] (D) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n] to obtain v2 [n], and finally delay v2 [n] by 3 samples to obtain y [n] YEAR 2007 MCQ 1.3.37
ONE MARK
Let a signal a1 sin (w1 t + f) be applied to a stable linear time variant system. Let the corresponding steady state output be represented as a2 F (w2 t + f2). Then which of the following statement is true? (A) F is not necessarily a “Sine” or “Cosine” function but must be periodic with w1 = w2 . (B) F must be a “Sine” or “Cosine” function with a1 = a2 (C) F must be a “Sine” function with w1 = w2 and f1 = f2 (D) F must be a “Sine” or “Cosine” function with w1 = w2
MCQ 1.3.38
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The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will be
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Page 114
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TWO MARKS
A signal x (t) is given by 1, - T/4 < t # 3T/4 x (t) = *- 1, 3T/4 < t # 7T/4 - x (t + T) Which among the following gives the fundamental fourier term of x (t) ? (B) p cos ` pt + p j (A) 4 cos ` pt - p j p T 4 4 2T 4 (C) 4 sin ` pt - p j (D) p sin ` pt + p j p T 4 4 2T 4
Statement for Linked Answer Question 41 and 41 : MCQ 1.3.40
A signal is processed by a causal filter with transfer function G (s) For a distortion free output signal wave form, G (s) must (A) provides zero phase shift for all frequency (B) provides constant phase shift for all frequency (C) provides linear phase shift that is proportional to frequency (D) provides a phase shift that is inversely proportional to frequency
MCQ 1.3.41
G (z) = az - 1 + bz - 3 is a low pass digital filter with a phase characteristics same as that of the above question if (A) a = b (B) a =- b (C) a = b(1/3)
(D) a = b(- 1/3)
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Page 115
Consider the discrete-time system shown in the figure where the impulse response of G (z) is g (0) = 0, g (1) = g (2) = 1, g (3) = g (4) = g = 0
This system is stable for range of values of K (A) [- 1, 12 ] (B) [- 1, 1] 1 (C) [- 2 , 1] (D) [- 12 , 2] MCQ 1.3.43
If u (t), r (t) denote the unit step and unit ramp functions respectively and u (t) * r (t) their convolution, then the function u (t + 1) * r (t - 2) is given by (A) 12 (t - 1) u (t - 1) (B) 12 (t - 1) u (t - 2) (C)
MCQ 1.3.44
1 2
(t - 1) 2 u (t - 1)
(D) None of the above
X (z) = 1 - 3z - 1, Y (z) = 1 + 2z - 2 are Z transforms of two signals x [n], y [n] respectively. A linear time invariant system has the impulse response h [n] defined by these two signals as h [n] = x [n - 1] * y [n] where * denotes discrete time convolution. Then the output of the system for the input d [n - 1] (A) has Z-transform z - 1 X (z) Y (z) (B) equals d [n - 2] - 3d [n - 3] + 2d [n - 4] - 6d [n - 5]
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(C) has Z-transform 1 - 3z - 1 + 2z - 2 - 6z - 3 (D) does not satisfy any of the above three YEAR 2006
ONE MARK
MCQ 1.3.45
The following is true (A) A finite signal is always bounded (B) A bounded signal always possesses finite energy (C) A bounded signal is always zero outside the interval [- t0, t0] for some t0 (D) A bounded signal is always finite
MCQ 1.3.46
x (t) is a real valued function of a real variable with period T . Its trigonometric Fourier Series expansion contains no terms of frequency w = 2p (2k) /T; k = 1, 2g Also, no sine terms are present. Then x (t) satisfies the equation (A) x (t) =- x (t - T) (B) x (t) = x (T - t) =- x (- t) (C) x (t) = x (T - t) =- x (t - T/2) (D) x (t) = x (t - T) = x (t - T/2)
MCQ 1.3.47
A discrete real all pass system has a pole at z = 2+30% : it, therefore (A) also has a pole at 12 +30% (B) has a constant phase response over the z -plane: arg H (z) = constant constant (C) is stable only if it is anti-causal (D) has a constant phase response over the unit circle: arg H (eiW) = constant
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YEAR 2006 MCQ 1.3.48
MCQ 1.3.49
Page 116
TWO MARKS
x [n] = 0; n < - 1, n > 0, x [- 1] =- 1, x [0] = 2 is the input and y [n] = 0; n < - 1, n > 2, y [- 1] =- 1 = y [1], y [0] = 3, y [2] =- 2 is the output of a discrete-time LTI system. The system impulse response h [n] will be (A) h [n] = 0; n < 0, n > 2, h [0] = 1, h [1] = h [2] =- 1 (B) h [n] = 0; n < - 1, n > 1, h [- 1] = 1, h [0] = h [1] = 2 (C) h [n] = 0; n < 0, n > 3, h [0] =- 1, h [1] = 2, h [2] = 1 (D) h [n] = 0; n < - 2, n > 1, h [- 2] = h [1] = h [- 1] =- h [0] = 3 n The discrete-time signal x [n] X (z) = n3= 0 3 z2n , where 2+n transform-pair relationship, is orthogonal to the signal n (A) y1 [n] ) Y1 (z) = n3= 0 ` 2 j z - n 3
/
denotes a
/ (B) y2 [n] ) Y2 (z) = /n3= 0 (5n - n) z - (2n + 1) (C) y3 [n] ) Y3 (z) = /n3=- 3 2 - n z - n (D) y4 [n] ) Y4 (z) = 2z - 4 + 3z - 2 + 1
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MCQ 1.3.50
A continuous-time system is described by y (t) = e - x (t) , where y (t) is the output and x (t) is the input. y (t) is bounded (A) only when x (t) is bounded (B) only when x (t) is non-negative (C) only for t # 0 if x (t) is bounded for t $ 0 (D) even when x (t) is not bounded
MCQ 1.3.51
The running integration, given by y (t) =
t
#- 3 x (t') dt'
(A) has no finite singularities in its double sided Laplace Transform Y (s) (B) produces a bounded output for every causal bounded input (C) produces a bounded output for every anticausal bounded input (D) has no finite zeroes in its double sided Laplace Transform Y (s) YEAR 2005 MCQ 1.3.52
For the triangular wave from shown in the figure, the RMS value of the voltage is equal to
(A) (C) 1 3 MCQ 1.3.53
TWO MARKS
1 6
(B) (D)
1 3 2 3
2 The Laplace transform of a function f (t) is F (s) = 5s 2+ 23s + 6 as t " 3, f (t) s (s + 2s + 2) approaches
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MCQ 1.3.54
MCQ 1.3.55
(A) 3 (B) 5 (D) 3 (C) 17 2 The Fourier series for the function f (x) = sin2 x is (A) sin x + sin 2x (B) 1 - cos 2x (C) sin 2x + cos 2x (D) 0.5 - 0.5 cos 2x If u (t) is the unit step and d (t) is the unit impulse function, the inverse z -transform of F (z) = z +1 1 for k > 0 is (A) (- 1) k d (k)
(B) d (k) - (- 1) k
(C) (- 1) k u (k)
(D) u (k) - (- 1) k
YEAR 2004 MCQ 1.3.56
MCQ 1.3.57
TWO MARKS
The rms value of the periodic waveform given in figure is
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(A) 2 6 A
(B) 6 2 A
(C)
(D) 1.5 A
4/3 A
The rms value of the resultant current in a wire which carries a dc current of 10 A and a sinusoidal alternating current of peak value 20 is (A) 14.1 A (B) 17.3 A (C) 22.4 A (D) 30.0 A YEAR 2002
MCQ 1.3.58
Page 117
ONE MARK
Fourier Series for the waveform, f (t) shown in Figure is
(A) 82 8sin (pt) + 1 sin (3pt) + 1 sin (5pt) + .....B 9 25 p (B) 82 8sin (pt) - 1 cos (3pt) + 1 sin (5pt) + .......B 9 25 p
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MCQ 1.3.59
(C) 82 8cos (pt) + 1 cos (3pt) + 1 cos (5pt) + .....B 9 25 p (D) 82 8cos (pt) - 1 sin (3pt) + 1 sin (5pt) + .......B 9 25 p Let s (t) be the step response of a linear system with zero initial conditions; then the response of this system to an an input u (t) is t t (B) d ; s (t - t) u (t) dt E (A) s (t - t) u (t) dt dt 0 0
#
(C) MCQ 1.3.60
#
#0 s (t - t); #0 u (t1) dt1Edt t
t
(D)
1
#0 [s (t - t)] 2 u (t) dt
Let Y (s) be the Laplace transformation of the function y (t), then the final value of the function is (A) LimY (s) (B) LimY (s) s"0
s"3
(C) Lim sY (s)
(D) Lim sY (s)
s"0
MCQ 1.3.61
Page 118
s"3
What is the rms value of the voltage waveform shown in Figure ?
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(B) (100/p) V (D) 100 V
YEAR 2001 MCQ 1.3.62
ONE MARK
Given the relationship between the input u (t) and the output y (t) to be y (t) =
t
#0 (2 + t - t) e- 3(t - t)u (t) dt ,
The transfer function Y (s) /U (s) is - 2s (A) 2e s+3 (C) 2s + 5 s+3
s+2 (s + 3) 2 (D) 2s + 72 (s + 3) (B)
Common data Questions Q.63-64* Consider the voltage waveform v as shown in figure
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MCQ 1.3.64
The DC component of v is (A) 0.4 (C) 0.8
Page 119
(B) 0.2 (D) 0.1
The amplitude of fundamental component of v is (A) 1.20 V (B) 2.40 V (C) 2 V (D) 1 V ***********
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SOLUTION
SOL 1.3.1
Page 120
Option (A) is correct. Given, the maximum frequency of the band-limited signal fm = 5 kHz According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as fN = 2fm = 2 # 5 = 10 kHz So, the sampling frequency fs must satisfy fs $ fN fs $ 10 kHz only the option (A) does not satisfy the condition therefore, 5 kHz is not a valid sampling frequency.
SOL 1.3.2
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Option (A) is correct. Given, the signal
v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p4 h So we have w1 = 100 rad/s w2 = 300 rad/s w3 = 500 rad/s Therefore, the respective time periods are T1 = 2p = 2p sec w1 100 T2 = 2p = 2p sec w2 300 T3 = 2p sec 500
SOL 1.3.3
So, the fundamental time period of the signal is LCM ^2p, 2p, 2ph L.C.M. ^T1, T2 T3h = HCF ^100, 300, 500h or, T0 = 2p 100 Thus, the fundamental frequency in rad/sec is w0 = 2p = 100 rad/s 10 Option (C) is correct. If the two systems with impulse response h1 ^ t h and h2 ^ t h are connected in cascaded configuration as shown in figure, then the overall response of the system is the convolution of the individual impulse responses.
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Page 121
SOL 1.3.4
Option (C) is correct. For a system to be casual, the R.O.C of system transfer function H ^s h which is rational should be in the right half plane and to the right of the right most pole. For the stability of LTI system. All poles of the system should lie in the left half of S -plane and no repeated pole should be on imaginary axis. Hence, options (A), (B), (D) satisfies both stability and causality an LTI system. But, Option (C) is not true for the stable system as, S = 1 have one pole in right hand plane also.
SOL 1.3.5
Option (C) is correct. Given, the input x ^ t h = u ^t - 1h It’s Laplace transform is -s X ^s h = e s
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The impulse response of system is given
h^t h = t u^t h Its Laplace transform is H ^s h = 12 s Hence, the overall response at the output is
SOL 1.3.6
Y ^s h = X ^s h H ^s h -s =e3 s its inverse Laplace transform is ^t - 1h2 y^t h = u ^t - 1h 2 Option (B) is correct. Given, the impulse response of continuous time system
h ^ t h = d ^t - 1h + d ^t - 3h From the convolution property, we know x ^ t h * d ^t - t 0h = x ^t - t 0h So, for the input x ^ t h = u ^ t h (Unit step fun n ) The output of the system is obtained as y^t h = u^t h * h^t h
= u ^ t h * 6d ^t - 1h + d ^t - 3h@ = u ^t - 1h + u ^t - 3h at t = 2
y ^2 h = u ^2 - 1h + u ^2 - 3h
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Page 122
=1 SOL 1.3.7
Option (C) is correct. n n x [n] = b 1 l - b 1 l u [n] 3 2 n -n n = b 1 l u [n] + b 1 l u [- n - 1] - b 1 l u (n) 3 3 2 Taking z -transform X 6z @ = =
3
/ n =- 3 3
/
n =- 3 3
1 n -n b 3 l z u [ n] +
1 -n b 2 l z u [ n] = n
3
3
/ n =- 3 3
/ b 13 l z
n=0
m
-
m=1
1 44 2 44 3 II
14 42 4 43 I
n
-n
+
n=0
/ b 31z l + / b 13 z l n
1 -n -n b 3 l z u [ - n - 1] -1
/ n =- 3
3
1 -n -n b3l z -
/ b 21z l
n
3
/ b 12 l z n
-n
n=0
Taking m =- n
n=0
14 42 4 43 III
1 < 1 or z > 1 3 3z Series II converges if 1 z < 1 or z < 3 3 Series III converges if 1 < 1 or z > 1 2z 2 Region of convergence of X (z) will be intersection of above three So, ROC : 1 < z < 3 2 Option (D) is correct. Using s -domain differentiation property of Laplace transform. Series I converges if
SOL 1.3.8
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If
f (t)
F (s)
dF (s) ds 2s + 1 So, L [tf (t)] = - d ; 2 1 = ds s + s + 1E (s2 + s + 1) 2 Option (A) is correct. Convolution sum is defined as tf (t)
SOL 1.3.9
L
L
-
y [n] = h [n] * g [n] = For causal sequence,
y [n] =
3
/ h [n] g [n - k] k =- 3
3
/ h [n] g [n - k] k=0
y [n] = h [n] g [n] + h [n] g [n - 1] + h [n] g [n - 2] + ..... For n = 0 ,
y [0] = h [0] g [0] + h [1] g [- 1] + ........... = h [ 0] g [ 0] g [- 1] = g [- 2] = ....0 ...(i) = h [ 0] g [ 0]
For n = 1,
y [1] = h [1] g [1] + h [1] g [0] + h [1] g [- 1] + .... = h [ 1] g [ 1] + h [ 1 ] g [ 0 ] 1 = 1 g [ 1] + 1 g [ 0 ] 2 2 2
1 h [1] = b 1 l = 1 2 2
1 = g [1] + g [0] g [1] = 1 - g [0]
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SOL 1.3.10
Page 123
y [0] 1 = =1 h [ 0] 1
From equation (i),
g [0] =
So,
g [1] = 1 - 1 = 0
Option (C) is correct. (2 cos w) (sin 2w) H (jw) = = sin 3w + sin w w w w We know that inverse Fourier transform of sin c function is a rectangular function.
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So, inverse Fourier transform of H (jw) h (t) = h1 (t) + h2 (t)
h (0) = h1 (0) + h2 (0) = 1 + 1 = 1 2 2 SOL 1.3.11
Option (D) is correct. y (t) =
t
# x (t) cos (3t) dt -3
Time invariance : Let, x (t) = d (t) y (t) =
t
# d (t) cos (3t) dt -3
= u (t) cos (0) = u (t)
For a delayed input (t - t 0) output is y (t, t 0) = Delayed output
t
# d (t - t ) cos (3t) dt -3
0
= u (t) cos (3t 0)
y (t - t 0) = u (t - t 0) y (t, t 0) ! y (t - t 0) System is not time invariant. Stability : Consider a bounded input x (t) = cos 3t y (t) =
#
t
-3
cos2 3t =
1 - cos 6t = 1 2 2 -3
#
t
# 1dt - 12 # cos 6t dt t
t
-3
-3
As t " 3, y (t) " 3 (unbounded) System is not stable. SOL 1.3.12
Option (D) is correct.
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f (t) = a 0 +
Page 124
3
/ (an cos wt + bn sin nwt)
n=1
The given function f (t) is an even function, therefore bn = 0 f (t) is a non zero average value function, so it will have a non-zero value of a 0 T/2 a 0 = 1 # f (t) dt (average value of f (t)) ^T/2h 0 • an is zero for all even values of n and non zero for odd n T an = 2 # f (t) cos (nwt) d (wt) T 0 • •
So, Fourier expansion of f (t) will have a 0 and an , n = 1, 3, 5f3 SOL 1.3.13
Option (A) is correct.
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x (t) = e-t Laplace transformation X (s) = 1 s+1
y (t) = e-2t Y (s) = 1 s+2 Convolution in time domain is equivalent to multiplication in frequency domain. z (t) = x (t) ) y (t) Z (s) = X (s) Y (s) = b 1 lb 1 l s+1 s+2 By partial fraction and taking inverse Laplace transformation, we get Z (s) = 1 - 1 s+1 s+2 z (t) = e-t - e-2t SOL 1.3.14
Option (D) is correct. f (t)
L
F1 (s)
L
e-st F1 (s) = F2 (s) F (s) F 1)(s) e-st F1 (s) F 1)(s) G (s) = 2 = F1 (s) 2 F1 (s) 2 e-sE F1 (s) 2 ) 2 = "a F1 (s) F 1 (s) = F1 (s) F1 (s) 2 = e-st Taking inverse Laplace transform f (t - t)
g (t) = L - 1 [e-st] = d (t - t) SOL 1.3.15
Option (C) is correct. h (t) = e-t + e-2t
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Page 125
Laplace transform of h (t) i.e. the transfer function H (s) = 1 + 1 s+1 s+2 For unit step input r (t) = m (t) or R (s) = 1 s Output, Y (s) = R (s) H (s) = 1 : 1 + 1 D s s+1 s+2 By partial fraction Y (s) = 3 - 1 - b 1 l 1 2s s + 1 s+2 2 Taking inverse Laplace e-2t u (t) y (t) = 3 u (t) - e-t u (t) 2 2 = u (t) 61.5 - e-t - 0.5e-2t@ SOL 1.3.16
Option (C) is correct. System is given as
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H (s) =
Step input Output
Y (s) = H (s) R (s) =
2 1 =2- 2 (s + 1) b s l s (s + 1)
Taking inverse Laplace transform y (t) = (2 - 2e- t) u (t) Final value of y (t), yss (t) = lim y (t) = 2 t"3
Let time taken for step response to reach 98% of its final value is ts . So, 2 - 2e- ts = 2 # 0.98 0.02 = e- ts ts = ln 50 = 3.91 sec. SOL 1.3.17
Option (D) is correct. Period of x (t), T = 2p = 2 p = 2.5 sec 0.8 p w
SOL 1.3.18
Option (B) is correct. Input output relationship y (t) =
5t
#- 3x (t) dt,
t>0
Causality : y (t) depends on x (5t), t > 0 system is non-causal. For example t = 2 y (2) depends on x (10) (future value of input) Linearity :
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Page 126
Output is integration of input which is a linear function, so system is linear. SOL 1.3.19
Option (A) is correct. Fourier series of given function x (t) = A0 +
3
/ an cos nw0 t + bn sin nw0 t
n=1
So,
a x (t) =- x (t) odd function A0 = 0 an = 0 T bn = 2 x (t) sin nw0 t dt T 0
#
T /2 T = 2= (1) sin nw0 t dt + (- 1) sin nw0 t dt G T 0 T /2 T /2 T = 2 =c cos nw0 t m - c cos nw0 t m G - nw0 0 - nw0 T/2 T = 2 6(1 - cos np) + (cos 2np - cos np)@ nw0 T = 2 61 - (- 1) n @ np 4 , n odd bn = * np 0 , n even So only odd harmonic will be present in x (t) For second harmonic component (n = 2) amplitude is zero.
#
#
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SOL 1.3.20
Option (D) is correct. By parsval’s theorem 1 3 X (w) 2 dw = 2p - 3
#
3
#- 3 SOL 1.3.21
3
#- 3 x2 (t) dt
X (w) 2 dw = 2p # 2 = 4p
Option (C) is correct. Given sequences x [n] = {1, - 1}, 0 # n # 1 y [n] = {1, 0, 0, 0, - 1}, 0 # n # 4 If impulse response is h [n] then y [ n] = h [ n] * x [ n] Length of convolution (y [n]) is 0 to 4, x [n] is of length 0 to 1 so length of h [n] will be 0 to 3. Let h [n] = {a, b, c, d} Convolution
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Page 127
y [n] = {a, - a + b, - b + c, - c + d, - d} By comparing
So, SOL 1.3.22
a -a + b -b + c -c + d h [ n]
=1 =0 &b=a=1 =0 &c=b=1 =0 &d=c=1 = {1, 1, 1, 1} -
Option (D) is correct. We can observe that if we scale f (t) by a factor of 1 and then shift, we will get 2 g (t). First scale f (t) by a factor of 1 2 g1 (t) = f (t/2)
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g (t) = g1 (t - 3) = f` t - 3 j 2
g (t) = f` t - 3 j 2 2 SOL 1.3.23
Option (C) is correct. g (t) can be expressed as g (t) = u (t - 3) - u (t - 5) By shifting property we can write Laplace transform of g (t) - 3s G (s) = 1 e - 3s - 1 e - 5s = e (1 - e - 2s) s s s
SOL 1.3.24
Option (D) is correct. L Let x (t) X (s) L y (t) Y (s) L h (t) H (s) So output of the system is given as Y (s) = X (s) H (s) Now for input So now output is
x (t - t)
L
e - st X (s)
h (t - t)
L
e- st H (s)
(shifting property)
Y' (s) = e - st X (s) $ e - ts H (s) = e - 2st X (s) H (s) = e - 2st Y (s)
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y' (t) = y (t - 2t) SOL 1.3.25
Option (B) is correct. Let three LTI systems having response H1 (z), H2 (z) and H 3 (z) are Cascaded as showing below
Assume H1 (z) = z2 + z1 + 1 (non-causal) H2 (z) = z3 + z2 + 1 (non-causal) Overall response of the system H (z) = H1 (z) H2 (z) H3 (z) H (z) = (z2 + z1 + 1) (z3 + z2 + 1) H3 (z) To make H (z) causal we have to take H3 (z) also causal. H3 (z) = z - 6 + z - 4 + 1
Let
= (z2 + z1 + 1) (z3 + z2 + 1) (z - 6 + z - 4 + 1) H (z) " causal
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Similarly to make H (z) unstable atleast one of the system should be unstable. SOL 1.3.26
Option (C) is correct. Given signal x (t) =
3
/ak e j2pkt/T
k =- 3
Let w0 is the fundamental frequency of signal x (t) x (t) =
3
a 2p = w0 T
/ak e jkw t 0
k =- 3
x (t) = a - 2 e - j2w t + a - 1 e - jw t + a0 + a1 e jw t + a2 e j2w t 0
0
0
0
= (2 - j) e - 2jw t + (0.5 + 0.2j) e - jw t + 2j + 0
0
+ (0.5 - 0.2) e jw t + (2 + j) e j2w t 0
= 2 6e - j2w t + e j2w t @ + j 6e j2w t - e - j2w t @ + 0
0
0
0
0
0.5 6e jw t + e - jw t @ - 0.2j 6e+ jw t - e - jw t @ + 2j = 2 (2 cos 2w0 t) + j (2j sin 2w0 t) + 0.5 (2 cos w0 t) 0.2j (2j sin w0 t) + 2j = 6 4 cos 2w0 t - 2 sin 2w0 t + cos w0 t + 0.4 sin w0 t @ + 2j Im [x (t)] = 2 (constant) 0
SOL 1.3.27
0
0
0
Option (A) is correct. Z-transform of x [n] is X (z) = 4z - 3 + 3z - 1 + 2 - 6z2 + 2z3 Transfer function of the system H (z) = 3z - 1 - 2 Output Y (z) = H (z) X (z) Y (z) = (3z - 1 - 2) (4z - 3 + 3z - 1 + 2 - 6z2 + 2z3) = 12z -4 + 9z -2 + 6z -1 - 18z + 6z2 - 8z -3 - 6z -1 - 4 + 12z2 - 4z3 = 12z - 4 - 8z - 3 + 9z - 2 - 4 - 18z + 18z2 - 4z3
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Page 129
Or sequence y [n] is y [n] = 12d [n - 4] - 8d [n - 3] + 9d [n - 2] - 4d [n] 18d [n + 1] + 18d [n + 2] - 4d [n + 3] y [n] = Y 0, n < 0 So y [n] is non-causal with finite support. SOL 1.3.28
Option (D) is correct. Since the given system is LTI, So principal of Superposition holds due to linearity. For causal system h (t) = 0 , t < 0 Both statement are correct.
SOL 1.3.29
Option (C) is correct. For an LTI system output is a constant multiplicative of input with same frequency. Here
input g (t) = e - at sin (wt)
output y (t) = Ke - bt sin (vt + f) Output will be in form of Ke - at sin (wt + f) So \= b, v = w SOL 1.3.30
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Option (D) is correct. Input-output relation y (t) =
- 2t
#- 3x (t) dt
Causality : Since y (t) depends on x (- 2t), So it is non-causal. Time-variance : y (t) =
- 2t
#- 3x (t - t0) dt =Y y (t - t0)
So this is time-variant. Stability : Output y (t) is unbounded for an bounded input. For example Let
x (t) = e - t (bounded) y (t) =
SOL 1.3.31
#- 3e- t dt = 8 -e 1 B- 3 $ Unbounded - 2t
- t - 2t
Option (A) is correct. Output y (t) of the given system is y (t) = x (t) ) h (t) Or Y (jw) = X (jw) H (jw) Given that, x (t) = sinc (at) and h (t) = sinc (bt) Fourier transform of x (t) and h (t) are X (jw) = F [x (t)] = p rect` w j, - a < w < a a 2a H (jw) = F [h (t)] = p rect` w j, - b < w < b b 2b 2 Y (jw) = p rect` w j rect` w j ab 2a 2b
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
So, Where And SOL 1.3.32
Page 130
Y (jw) = K rect ` w j 2g g = min (a, b) y (t) = K sinc (gt)
Option (B) is correct. Let ak is the Fourier series coefficient of signal x (t) Given y (t) = x (t - t0) + x (t + t0) Fourier series coefficient of y (t) bk = e - jkwt ak + e jkwt ak bk = 2ak cos kwt0 bk = 0 (for all odd k ) kwt0 = p , k " odd 2 k 2p t0 = p 2 T For k = 1, t0 = T 4 0
0
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SOL 1.3.33
Option ( ) is correct.
SOL 1.3.34
Option (D) is correct. z , z >a (z - a) 2 Residue of X (z) zn - 1 at z = a is = d (z - a) 2 X (z) zn - 1 z = a dz z = d (z - a) 2 zn - 1 2 dz (z - a) z=a n-1 n d z = = nz z = a = nan - 1 dz z = a Option (C) is correct. Given signal x (t) = rect `t - 1 j 2 1, - 1 # t - 1 # 1 or 0 # t # 1 2 2 2 So, x (t) = * 0, elsewhere Similarly x (- t) = rect`- t - 1 j 2 1, - 1 # - t - 1 # 1 or - 1 # t # 0 2 2 2 x (- t) = * 0, elsewhere Given that
SOL 1.3.35
X (z) =
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F [x (t) + x (- t)] = =
3
Page 131
3
#- 3 x (t) e- jwt dt + #- 3 x (- t) e- jwt dt 0
1
#0 (1) e- jwt dt + #- 1 (1) e- jwt dt 1
0
- jw t - jw t = ; e E + ; e E = 1 (1 - e - jw) + 1 (e jw - 1) - jw 0 - jw - 1 jw jw - j w /2 j w /2 =e (e jw/2 - e - jw/2) + e (e jw/2 - e - jw/2) jw jw
SOL 1.3.36
(e jw/2 - e - jw/2) (e - jw/2 + e jw/2) = jw = 2 sin ` w j $ 2 cos ` w j = 2 cos w sinc` w j w 2 2 2 2p Option (B) is correct. In option (A) z1 [n] = x [n - 3] z2 [n] = z1 [4n] = x [4n - 3] y [n] = z2 [- n] = x [- 4n - 3] = Y x [3 - 4n] In option (B)
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z1 [n] = x [n + 3] z2 [n] = z1 [4n] = x [4n + 3] y [n] = z2 [- n] = x [- 4n + 3] In option (C)
v1 [n] = x [4n] v2 [n] = v1 [- n] = x [- 4n] y [n] = v2 [n + 3] = x [- 4 (n + 3)] = Y x [3 - 4n] In option (D) v1 [n] = x [4n] v2 [n] = v1 [- n] = x [- 4n] y [n] = v2 [n - 3] = x [- 4 (n - 3)] = Y x [3 - 4n] SOL 1.3.37
Option ( ) is correct. The spectrum of sampled signal s (jw) contains replicas of U (jw) at frequencies ! nfs . Where n = 0, 1, 2....... 1 fs = 1 = = 1 kHz Ts 1 m sec
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SOL 1.3.38
Option (D) is correct. For an LTI system input and output have identical wave shape (i.e. frequency of input-output is same) within a multiplicative constant (i.e. Amplitude response is constant) So F must be a sine or cosine wave with w1 = w2
SOL 1.3.39
Option (C) is correct. Given signal has the following wave-form
nodia.co.in Function x(t) is periodic with period 2T and given that x (t) =- x (t + T) (Half-wave symmetric) So we can obtain the fourier series representation of given function. SOL 1.3.40
Option (C) is correct. Output is said to be distortion less if the input and output have identical wave shapes within a multiplicative constant. A delayed output that retains input waveform is also considered distortion less. Thus for distortion less output, input-output relationship is given as y (t) = Kg (t - td ) Taking Fourier transform. Y (w) = KG (w) e - jwt = G (w) H (w) H (w) & transfer function of the system d
So, H (w) = Ke - jwt Amplitude response H (w) = K Phase response, qn (w) =- wtd For distortion less output, phase response should be proportional to frequency. d
SOL 1.3.41
Option (A) is correct. G (z) z = e = ae- jw + be- 3jw jw
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for linear phase characteristic a = b . SOL 1.3.42
Option (A) is correct. System response is given as G (z) H (z) = 1 - KG (z) g [n] = d [n - 1] + d [n - 2] G (z) = z - 1 + z - 2 (z - 1 + z - 2) = 2 z+1 -1 -2 z - Kz - K 1 - K (z + z ) For system to be stable poles should lie inside unit circle. So
H (z) =
z #1 z = K!
K2 + 4K # 1 K ! 2
K2 + 4K # 2
K2 + 4K # 2 - K K2 + 4K # 4 - 4K + K2 8K # 4 K # 1/2
SOL 1.3.43
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Option (C) is correct. Given Convolution is,
h (t) = u (t + 1) ) r (t - 2) Taking Laplace transform on both sides,
H (s) = L [h (t)] = L [u (t + 1)] ) L [r (t - 2)] We know that, L [u (t)] = 1/s L [u (t + 1)] = es c 12 m s and L [r (t)] = 1/s2 L r (t - 2) = e - 2s c 12 m s s 1 So H (s) = ;e ` jE;e - 2s c 12 mE s s -s 1 H (s) = e c 3 m s Taking inverse Laplace transform h (t) = 1 (t - 1) 2 u (t - 1) 2 SOL 1.3.44
(Time-shifting property)
(Time-shifting property)
Option (C) is correct. Impulse response of given LTI system. h [ n ] = x [ n - 1] ) y [ n ] Taking z -transform on both sides. H (z) = z - 1 X (z) Y (z) We have X (z) = 1 - 3z - 1 and Y (z) = 1 + 2z - 2 So
a x [n - 1]
Z
z - 1 x (z)
H (z) = z - 1 (1 - 3z - 1) (1 + 2z - 2) Output of the system for input u [n] = d [n - 1] is , y (z) = H (z) U (z)
U [n]
Z
U (z) = z - 1
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Page 134
So Y (z) = z - 1 (1 - 3z - 1) (1 + 2z - 2) z - 1 = z - 2 (1 - 3z - 1 + 2z - 2 - 6z - 3) = z - 2 - 3z - 3 + 2z - 4 - 6z - 5 Taking inverse z-transform on both sides we have output. y [n] = d [n - 2] - 3d [n - 3] + 2d [n - 4] - 6d [n - 5] SOL 1.3.45
Option (B) is correct. A bounded signal always possesses some finite energy. E =
t0
g (t) 2 dt < 3
#- t
0
SOL 1.3.46
Option (C) is correct. Trigonometric Fourier series is given as 3
/an cos nw0 t + bn sin nw0 t
x (t) = A0 +
n=1
Since there are no sine terms, so bn = 0 T x (t) sin nw0 t dt bn = 2 T0 0 0
#
= 2= T0
#0
= 2; T0
#T
#T /2 x (t) sin nw0 t dt G
nodia.co.in T0 /2
x (t) sin nw0 t dt +
0
Where t = T - t & dt =- dt
= 2; T0
= 2; T0
T
T0 /2
x (T - t) sin nw0 (T - t) (- dt)+
0
#T /2 x (t) sin nw0 t dt E T
0
TO
#T /2 x (T - t) sin n` 2Tp T - t j dt ++ #T /2 x (t) sin nw0 t dt E T
O
0
#T /2 x (T - t) sin (2np - nw0) dt+ #T /2 x (t) sin nw0 t dt E
= 2 ;T0
T0
T0
0
0
#T /2 x (T - t) sin (nw0 t) dt + + #T /2 x (t) sin nw0 t dt E T0
T0
0
0
bn = 0 if x (t) = x (T - t) From half wave symmetry we know that if x (t) =- x`t ! T j 2 Then Fourier series of x (t) contains only odd harmonics. SOL 1.3.47
Option (C) is correct. Z -transform of a discrete all pass system is given as -1 ) H (z) = z - z-0 1 1 - z0 z It has a pole at z 0 and a zero at 1/z) 0. Given system has a pole at z = 2+30% = 2
( 3 + j) = ( 3 + j) 2
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Page 135
system is stable if z < 1 and for this it is anti-causal. SOL 1.3.48
Option (A) is correct. According to given data input and output Sequences are x [n] = {- 1, 2}, - 1 # n # 0 y [n] = {- 1, 3, - 1, - 2}, - 1 # n # 2 If impulse response of system is h [n] then output y [n] = h [ n] ) x [ n] Since length of convolution (y [n]) is - 1 to 2, x [n] is of length - 1 to 0 so length of h [n] is 0 to 2. Let h [n] = {a, b, c} Convolution
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So, a=1
y [n] = {- a, 2a - b, 2b - c, 2c} y [n] = {- 1, 3, - 1, - 2} -
SOL 1.3.49
2a - b = 3 & b =- 1 2a - c =- 1 & c =- 1 Impulse response h [n] = "1, - 1, - 1, Option ( ) is correct.
SOL 1.3.50
Option (D) is correct. Output y (t) = e - x (t) If x (t) is unbounded, x (t) " 3 y (t) = e - x (t) " 0 (bounded) So y (t) is bounded even when x (t) is not bounded.
SOL 1.3.51
Option (B) is correct. Given
y (t) =
t
# x (t') dt' -3
Laplace transform of y (t)
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Y (s) =
X (s) , has a singularity at s = 0 s t
# x (t') dt'
For a causal bounded input, y (t)= SOL 1.3.52
Page 136
is always bounded.
-3
Option (A) is correct. RMS value is given by 1 T
Vrms = Where
#0
T
V2 (t) dt
2 T `T j t, 0 # t # 2
V (t) = * 0, So
1 T
#0
T
V 2 (t) dt = 1 = T
#0
T /2
T /2
2t 2 ` T j dt +
T
#0
T /2
nodia.co.in Vrms
Option (A) is correct. By final value theorem
lim f (t) = lim s F (s) = lim s
t"3
SOL 1.3.54
#T/2 (0) dt G
3 t2 dt = 43 ; t E T 3 0 3 = 43 # T = 1 6 24 T = 1 V 6
= 1 $ 42 T T
SOL 1.3.53
T q H = 6B@-1 >P H 3 3 where 6B@ is obtained as R V 1 S 1 + 1 W X12 X23 X23 W S 6B@ = S - 1 1 + 1 W X23 X13 W S X23 T X 1 + 1 -1 2 -1 H=> H => -1 1 + 1 -1 2 Its inverse is obtained as -1 2 -1 -1 6B@ = >1 - 2 H 2 1 0. 1 H> H = 1> 3 + 1 2 - 0. 2 Therefore, q2 1 2 1 0. 1 >q H = 3 >+ 1 2H>- 0.2H 3 0 H = 1> 3 - 0.3 0 => - 0.1H
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i.e., and SOL 1.5.3
q2 = 0 q3 =- 0.1 rad
Option (C) is correct. From the above solution, we have P2 = 0.1 P3 =- 0.2 since, P1 + P2 + P3 = 0 (Where P1 is injection at bus 1) So, P1 - P2 - P3 =- 0.1 + 0.2 = 0.1 pu Now, the apparent power delivered to base is 3 2 2 ^100 # 10 h V S = = 100 R 6 = 100 # 10 VA
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Page 240
Therefore, the real power delivered by slack bus (bus 1) P = P1 S = ^0.1h # 100 # 106 = 10 # 106 watt = 10 MW SOL 1.5.4
Option (B) is correct. For bus admittance matrix, Y11 + (Y12 + yline) + Y13 = 0 - j13 + (j10 + yline) + j 5 = 0 yline =- j2 Magnitude of susceptance is + 2
SOL 1.5.5
Option (A) is correct. i1 (t) = Im sin (wt - f1) i2 (t) = Im cos (wt - f2) We know that, cos (q - 90c) = sin q So, i1 (t) can be written as i1 (t) = Im cos (wt - f1 - 90c) i2 (t) = Im cos (wt - f2) Now, in phasor form I1 = Im f1 + 90c I 2 = Im f 2 Current are balanced if I1 + I 2 = 0 Im f1 + 90c + Im f 2 = 0 Im cos ^f1 + 90ch + jIm sin ^f1 + 90ch + cos f 2 + j sin f 2 = 0 Im 8cos ^f1 + 90ch + j sin ^f1 + 90chB + Im 6cos f 2 + j sin f 2@ = 0 Im 8cos ^f1 + 90ch + cos f 2B + jIm 8sin f 2 + sin ^f1 + 90chB = 0 cos ^f1 + 90ch + cos f2 = 0 cos ^f1 + 90ch =- cos f2 = cos ^p + f2h f1 + 90c = p + f2 or, f1 = p + f2 2 Option (A) is correct. Let penalty factor of plant G , is L1 given as 1 L1 = 1 - 2PL 2PG
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SOL 1.5.6
1
PL = 0.5PG2 2PL = 0.5 (2P ) = P G G 2PG 1 So, L1 = 1 - PG Penalty factor of plant G2 is 1 L2 = =1 1 - 2PL 2PG For economic power generation 1
1
1
2
2
2PL ca 2PG = 0 m 2
2
C1 # L1 = C 2 # L 2
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where C1 and C2 are the incremental fuel cost of plant G1 and G2 . So, (10000) b 1 l = 12500 # 1 1 - PG 4 = 1-P G 5 PG = 1 pu 5 It is an 100 MVA, so PG = 1 # 100 = 20 MW 5 2 Loss PL = 0.5 b 1 l = 1 pu 5 50 or PL = 1 # 100 = 2 MW 50 2
2
2
2
PL = PG + PG - PL 40 = 20 + P2 - 2 PG = 22 MW
Total power,
1
2
2
SOL 1.5.7
Option (C) is correct. For double line-to-ground (LLG ) fault, relation between sequence current is
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I positive =-^I negative + I zeroh Gives values satisfy this relation, therefore the type of fault is LLG . SOL 1.5.8
Option (B) is correct. Complex power for generator
SG = SD1 + SD2 = 1 + 1 = 2 pu Power transferred from bus 1 to bus 2 is 1 pu, so V1 V2 sin (q1 - q2) 1= X 1
= 1 # 1 sin (q1 - q2) 0.5
(Line is lossless)
V1 = V2 = 1 pu X = 0.5 pu
0.5 = sin (q1 - q2) q1 - q2 = 30c q2 = q1 - 30c =- 30c (q1 = 0c) So, V1 = 1 0c V V2 = 1 - 30c V 1 0c - 1 30c Current, I12 = V1 - V2 = = (1 - j 0.288) pu j 0.5 Z ) Current in SD is I2 , SD = V2 I2 1 = 1 - 30c I2) I2 = 1 - 30c pu Current in QG , IG = I2 - I12 = 1 - 30c - (1 - j 0.288) = 0.268 - 120c VAR rating of capacitor, 2
2
2
QC = V2 VG = 1 # 0.268 = 0.268 pu SOL 1.5.9
Option (D) is correct.
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Total reactance, X = j1 + j 0.5 = j1.5 pu Critical angle is given as, dcr = cos-1 [(p - 2d0) sin d0 - cos d0] d0 " steady state torque angle. Steady state power is given as
...(i)
Pm = Pmax sin d0 E V X E V Pm = sin d0 X (1.5) (1) 0.5 = sin d0 1.5
Pmax =
where, So,
Pm = 0.5 pu
nodia.co.in sin d0 = 0.5 d0 = 30c
d0 = 30c # p = 0.523 180c Substituting d0 into equation (i) dcr = cos-1 [(p - 2 # 0.523) sin 30c - cos 30c] In radian,
= cos-1 [(2.095) (0.5) - 0.866] = cos-1 (0.1815) - 79.6c SOL 1.5.10
Option ( ) is correct
SOL 1.5.11
Option (A) is correct. Negative phase sequence relay is used for the protection of alternators against unbalanced loading that may arise due to phase-to-phase faults.
SOL 1.5.12
Option (C) is correct. Steady state stability or power transfer capability E V Pmax = X To improve steady state limit, reactance X should be reduced. The stability may be increased by using two parallel lines. Series capacitor can also be used to get a better regulation and to increase the stability limit by decreasing reactance. Hence (C) is correct option.
SOL 1.5.13
Option (A) is correct. We know that loss \ PG2 loss \ length Distance of load from G1 is 25 km Distance of load from G2 & G 3 is 75 km generally we supply load from nearest generator. So maximum of load should be supplied from G1 . But G2 & G 3 should be operated at same minimum generation.
SOL 1.5.14
Option (B) is correct.
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Power angle for salient pole alternator is given by V sin f + Ia Xq tan d = t Vt cos f + Ia Ra Since the alternator is delivering at rated kVA rated voltage Ia = 1 pu Vt = 1 pu f = 0c sin f = 0, cos f = 1 Xq = 1 pu, Xd = 1.2 pu 1 0 + 1 (1) tan d = # 1+0 =1 d = 45c SOL 1.5.15
Option (B) is correct. The admittance diagram is shown below
here
nodia.co.in y10 =- 10j, y12 =- 5j, y23 = 12.5j, y 30 =- 10j
Note: y23 is taken positive because it is capacitive.
SOL 1.5.16
Y11 = y10 + y12 =- 10j - 5j =- 15j Y12 = Y21 =- y21 = 5j Y13 = Y31 =- y13 = 0 Y22 = y20 + y21 + y23 = 0 + (- 5j) + (12.5j) = 7.5j Y23 = Y32 =- y23 =- 12.5j Y33 = y 30 + y13 + y23 =- 10j + 0 + 12.5j = 2.5j So the admittance matrix is RY Y Y V R- 15j 5j 0 VW S 11 12 13W S Y = SY21 Y22 Y33W = S 5j 7.5j - 12.5j W SSY Y Y WW SS 0 - 12.5j 2.5j WW 33 32 31 X T X T Option (A) is correct. For generator G1 X mG = 0.25 # 100 = 0.1 pu 250 For generator G2 X mG = 0.10 # 100 = 0.1 pu 100 1
1
XL = XL = 0.225 # 10 = 2.25 W For transmission lines L1 and L2 2
1
2 Z base = kV base = 15 # 15 = 2.25 W 100 MVA base
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X mL (pu) = 2.25 = 1 pu 2.25 X mL (pu) = 2.25 = 1 pu 2.25 So the equivalent pu reactance diagram 2
1
SOL 1.5.17
Option (D) is correct. We can see that at the bus 3, equivalent thevenin’s impedance is given by Xth = ^ j0.1 + j1.0h || ^ j0.1 + j1.0h = j1.1 || j1.1 = j0.55 pu
SOL 1.5.18
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Fault MVA = Base MVA = 100 = 181.82 MVA 0.55 Xth Option (C) is correct. Given that, a a I > 0 so
SOL 1.5.19
I >0 VAB > 0 since it is Rectifier O/P VCD > 0 since it is Inverter I/P VAB > VCD , Than current will flow in given direction.
Option (A) is correct. Given step voltage travel along lossless transmission line.
a Voltage wave terminated at reactor as.
By Applying KVL V + VL = 0 VL =- V VL =- 1 pu SOL 1.5.20
Option (A) is correct. Given two buses connected by an Impedance of (0 + j5) W The Bus ‘1’ voltage is 100+30c V and Bus ‘2’ voltage is 100+0c V We have to calculate real and reactive power supply by bus ‘1’
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P + jQ = VI) = 100+30c ;100+30c - 100+0cE 5j = 100+30c [20+ - 60c - 20+ - 90c] = 2000+ - 30c - 2000+ - 60c P + jQ = 1035+15c real power P = 1035 cos 15c = 1000 W reactive power Q = 1035 sin 15c = 268 VAR SOL 1.5.21
SOL 1.5.22
Option (C) is correct. Given 3-f, 33 kV oil circuit breaker. Rating 1200 A, 2000 MVA, 3 sec Symmetrical breaking current Ib = ? Ib = MVA kA = 2000 = 34.99 kA - 35 kA 3 kV 3 # 33 Option (C) is correct. Given a stator winding of an alternator with high internal resistance fault as shown in figure
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Current through operating coil I1 = 220 # 5 A, I2 = 250 # 5 A 400 400 Operating coil current = I2 - I1 = (250 - 220) # 5/400 = 0.375 Amp SOL 1.5.23
Option (C) is correct. Zero sequence circuit of 3-f transformer shown in figure is as following:
No option seems to be appropriate but (C) is the nearest. SOL 1.5.24
Option (D) is correct. Given that A 50 Hz Generator is initially connected to a long lossless transmission line which is open circuited as receiving end as shown in figure. Due to ferranti effect the magnitude of terminal voltage does not change, and the
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field current decreases. SOL 1.5.25
Option (B) is correct. Given : 3-f, 50 Hz, 11 kV distribution system, We have to find out e1, e2 = ? Equivalent circuit is as following
11 (6C) e1 = 3 = 11 # 6 = 3.46 kV 11 6C + 5C 3
SOL 1.5.26
e2 = 11 # 5 = 2.89 kV 11 3 Option (A) is correct. Given : 3-f, 50 Hz, 11 kV cable
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C1 = 0.2 mF C2 = 0.4 mF Charging current IC per phase = ? Capacitance Per Phase C = 3C1 + C2
C = 3 # 0.2 + 0.4 = 1 mF w = 2pf = 314 3 Changing current IC = V = V (wC) = 11 # 10 # 314 # 1 # 10- 6 XC 3 = 2 Amp SOL 1.5.27
Option (B) is correct. Generator G1 and G2 XG1 = XG2 = X old # New MVA # b Old kV l New kV Old MVA 2 = j0.9 # 200 # b 25 l = j0.18 100 25 2 Same as XT1 = j0.12 # 200 # b 25 l = j0.27 90 25 2 XT2 = j0.12 # 200 # b 25 l = j0.27 90 25 X Line = 150 # 220 2 = j0.62 (220) The Impedance diagram is being given by as
2
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.5.28
Option ( ) is correct.
SOL 1.5.29
Option (C) is correct. We know complex power
a
So SOL 1.5.30
S = P + jQ = VI (cos f + j sin f) = VIe jf I = S jf Ve Real Power loss = I2 R 2 2 1 PL = c S jf m R = S j2R # 2 V Ve e f PL \ 12 V
Page 247
2 a S j2R = Constant e f
Option (C) is correct. YBus matrix of Y-Bus system are given as R V S- 5 2 2.5 0 W S 2 - 10 2.5 0 W YBus = j S W S2.5 2.5 - 9 4 W 4 4 - 8W S0 T X element We have to find out the buses having shunt R V Sy11 y12 y13 y14W Sy21 y22 y23 y24W We know YBus = S W Sy 31 y 32 y 33 y 34W Sy 41 y 42 y 43 y 44W T X Here y11 = y10 + y12 + y13 + y14 =- 5j y22 = y20 + y21 + y23 + y24 =- 10j y 33 = y 30 + y 31 + y 32 + y 34 =- 9j y 44 = y 40 + y 41 + y 42 + y 43 =- 8j y12 = y21 =- y12 = 2j y13 = y 31 =- y13 = 2.5j y14 = y 41 =- y14 = 0j y23 = y 32 =- y23 = 2.5j y24 = y 42 =- y24 = 4j So y10 = y11 - y12 - y13 - y14 =- 5j + 2j + 2.5j + 0j =- 0.5j y20 = y22 - y12 - y23 - y24 =- 10j + 2j + 2.5j + 4j =- 1.5j y 30 = y 33 - y 31 - y 32 - y 34 =- 9j + 2.5j + 2.5j + 4j = 0 y 40 = y 44 - y 41 - y 42 - y 43 =- 8j - 0 + 4j + 4j = 0 Admittance diagram is being made by as
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From figure. it is cleared that branch (1) & (2) behaves like shunt element. SOL 1.5.31
Option (B) is correct. We know that • Shunt Capacitors are used for improving power factor. • • •
Series Reactors are used to reduce the current ripples. For increasing the power flow in line we use series capacitor. Shunt reactors are used to reduce the Ferranti effect.
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SOL 1.5.32
Option (C) is correct. We know that for different type of transmission line different type of distance relays are used which are as follows. Short Transmission line -Ohm reactance used Medium Transmission Line -Reactance relay is used Long Transmission line -Mho relay is used
SOL 1.5.33
Option (C) is correct. Given that three generators are feeding a load of 100 MW. For increased load power demand, Generator having better regulation share More power, so Generator -1 will share More power than Generator -2.
SOL 1.5.34
Option (A) is correct. Given Synchronous generator of 500 MW, 21 kV, 50 Hz, 3-f, 2-pole P.F = 0.9 , Moment of inertia M = 27.5 # 103 kg-m2 Inertia constant H = ? Generator rating in MVA G = P = 500 MW = 555.56 MVA 0.9 cos f N = 120 # f = 120 # 50 = 3000 rpm 2 pole 2 Stored K.E = 1 Mw2 = 1 M b 2pN l 2 2 60 = 1 # 27.5 # 103 # b 2p # 3000 l MJ 2 60 = 1357.07 MJ Stored K.E Inertia constant (H) = Rating of Generator (MVA) H = 1357.07 555.56
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= 2.44 sec SOL 1.5.35
Option (D) is correct. Given for X to F section of phase ‘a’ Va -Phase voltage and Ia -phase current. Impedance measured by ground distance, Bus voltage = Va W Relay at X = Ia Current from phase 'a'
SOL 1.5.36
Option (D) is correct. For EHV line given data is Length of line = 300 km and b = 0.00127 S rad/km wavelength l = 2p = 2p = 4947.39 km 0.00127 b l % = 300 So 100 = 0.06063 # 100 4947.39 # l l % = 6.063 l
SOL 1.5.37
Option (B) is correct. For three phase transmission line by solving the given equation VRI V RDV V R(X - X ) 0 0 m WS aW S aW S s We get, 0 (Xs - Xm) 0 WSIbW SDVbW = S SSDV WW SS 0 0 (Xs + 2Xm)WWSSIcWW c XT X X T T Zero sequence Impedance = Xs + 2Xm = 48 and Positive Sequence Impedance = Negative Sequence Impedance
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...(1)
...(2)
By solving equation (1) and (2) Zs or Xs = 26 and Zm or Xm = 11 SOL 1.5.38
Option ( ) is correct.
SOL 1.5.39
Option (B) is correct. SIL has no effect of compensation So SIL = 2280 MW
SOL 1.5.40
Option (C) is correct. Given PG1 + PG2 = 250 MW C1 (PG1) = PG1 + 0.055PG12 and 4 C2 (PG2) = 3PG2 + 0.03PG22 from equation (2) dC1 = 1 + 0.11P G1 dPG1 dC2 = 3 + 0.06P and G2 dPG2 Since the system is loss-less dC1 = dC2 Therefore dPG1 dPG2 So from equations (3a) and (3b) We have 0.11PG1 - 0.06PG2 = 2 Now solving equation (1) and (4), we get
...(1) ...(2)
...(3a) ...(3b)
...(4)
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PG1 = 100 MW PG2 = 150 MW SOL 1.5.41
Option (B) is correct. After connecting both the generators in parallel and scheduled to supply 0.5 Pu of power results the increase in the current. ` Critical clearing time will reduced from 0.28 s but will not be less than 0.14 s for transient stability purpose.
SOL 1.5.42
Option (D) is correct. Given that the each section has equal impedance. Let it be R or Z , then by using the formula line losses = / I2 R On removing (e1); losses = (1) 2 R + (1 + 2) 2 R + (1 + 2 + 5) 2 R = R + 9R + 64R = 74R Similarly, On removing e2 ;losses = 52 R + (5 + 2) 2 R + (5 + 2 + 1) 2 R = 138R lossess on removing e 3 = (1) 2 R + (2) 2 R + (5 + 2) 2 R = 1R + 4R + 49R = 54R on removing e 4 lossless = (2) 2 R + (2 + 1) 2 R + 52 R
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= 4R + 9R + 25R = 38R So, minimum losses are gained by removing e 4 branch. SOL 1.5.43
Option (A) is correct. Given : V (t) = Vm cos (wt) For symmetrical 3 - f fault, current after the fault 2 Vm cos (wt - a) Z At the instant of fault i.e t = t 0 , the total current i (t) = 0 i (t) = Ae- (R/L) t +
0 = Ae- (R/L) t + 0
`
2 Vm cos (wt - a) 0 Z
Ae- (R/L) t =- 2 Vm cos (wt 0 - a) Z Maximum value of the dc offset current 0
Ae- (R/L) t =- 2 Vm cos (wt 0 - a) Z For this to be negative max. 0
or
(wt 0 - a) = 0 t0 = a w
...(1)
Z = 0.004 + j0.04 Z = Z +a = 0.0401995+84.29c a = 84.29cor 1.471 rad.
and
From equation (1) t0 =
1.471 = 0.00468 sec (2p # 50) t 0 = 4.682 ms
SOL 1.5.44
Option (C) is correct.
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Since the fault ‘F’ is at mid point of the system, therefore impedance seen is same from both sides.
Z = 0.0201+84.29c 2 Z1 (Positive sequence) = Z = 0.0201+84.29c 2 also Z1 = Z2 = Z 0 (for 3-f fault) `
1+0c I f (pu) = 1+0c = Z1 0.0201+84.29c
So magnitude
If
(p.u.)
I f = 49.8 #
` Fault current SOL 1.5.45
= 49.8 100 = 7.18 kA 3 # 400
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Option (A) is correct. If fault is LG in phase ‘a’
Z1 = Z = 0.0201+84.29c 2 and Then
Z2 = Z1 = 0.0201+84.29c Z 0 = 3Z1 = 0.0603+84.29c Ia /3 = Ia1 = Ia2 = Ia0 1.0+0c Z1 + Z 2 + Z 0 1. 0 = = 9.95 pu (0.0201 + 0.0201 + 0.0603)
Ia1 (pu) = and
SOL 1.5.46
SOL 1.5.47
Ia1
Fault Current I f = Ia = 3Ia1 = 29.85 pu 100 So Fault current I f = 29.85 # = 4.97 kA 3 # 400 Option (A) is correct. a Equal Phase shift of point A & B with respect to source from both bus paths. So the type of transformer Y-Y with angle 0c. Option (C) is correct. Given incremental cost curve
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PA + PB = 700 MW For optimum generator PA = ? , PB = ? From curve, maximum incremental cost for generator A = 600 at 450 MW and maximum incremental cost for generator B = 800 at 400 MW minimum incremental cost for generator B = 650 at 150 MW
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Maximim incremental cost of generation A is less than the minimum incremental constant of generator B. So generator A operate at its maximum load = 450 MW for optimum generation. PA = 450 MW PB = (700 - 450) = 250 MW SOL 1.5.48
Option (C) is correct. Here power sharing between the AC line and HVDC link can be changed by controlling the HVDC converter alone because before changing only grid angle we can change the power sharing between the AC line and HVDC link.
SOL 1.5.49
Option (B) is correct. We have to find out maximum electric field intensity at various points. Electric field intensity is being given by as follows
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From figures it is cleared that at point Y there is minimum chances of cancelation. So maximum electric field intensity is at point Y. SOL 1.5.50
Option (D) is correct. To increase capacitive dc voltage slowly to a new steady state value first we have to make d =- ve than we have to reach its original value.
SOL 1.5.51
Option (B) is correct. Given that .045 2p # 50 1 1 Suspectance of Line = 1.2 pu & C = 2p # 50 # 1.2 Reactance of line
= 0.045 pu & L =
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Velocity of wave propagation = 3 # 105 Km/sec Length of line l = ? We know velocity of wave propagation VX = l LC .45 1 1 l = VX LC = 3 # 105 2p # 50 # 2p # 50 # 1.2 = 185 Km SOL 1.5.52
Option (C) is correct. Due to the fault ‘F’ at the mid point and the failure of circuit-breaker ‘4’ the sequence of circuit-breaker operation will be 5, 6, 7, 3, 1, 2 (as given in options) (due to the fault in the particular zone, relay of that particular zone must operate first to break the circuit, then the back-up protection applied if any failure occurs.)
SOL 1.5.53
Option (A) is correct.
SOL 1.5.54
Option (C) is correct.
R V 1 - 1 W S 0 3 3 W RSiaVW S 1 W Si W R = [Van Vbn Vcn] SS- 1 0 b 3 3 W SS WW i S 1 W c - 1 0 WT X S S 3 W 3 T X By solving we get R = ;Van (ib - ic) + Vbn (ic - ia) + Vc (ia - ib)E 3 3 3 (ib - ic) R = 3 (VI) , where = I and Van = V 3
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Here
Since `
Page 254
P1 " power before the tripping of one ckt P2 " Power after tripping of one ckt P = EV sin d X Pmax = EV X here, [X2 = (0.1 + X) (pu)] P2 max = EX , X2
To find maximum value of X for which system does not loose synchronism P2 = Pm (shown in above figure) EV sin d = P ` m 2 X2 as Pm = 1 pu, E = 1.0 pu,V = 1.0 pu 1.0 # 1.0 sin 130c = 1 X2
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& & & SOL 1.5.55
X2 = 0.77 (0.1 + X) = 0.77 X = 0.67
Option (B) is correct. Given that FP = KAFS
Rf V Rf V S aW S pW where, Phase component FP = SfbW, sequence component FS = SfnW SSf WW SSf WW c o X X T T R 1 1 1V W S and A = Sa2 a 1W SS a a2 1WW X T VP = KAVS ` 3 IP = KAIS and VS = Zl [IS ] R0.5 0 0 V S W where Zl = S 0 0.5 0 W SS 0 0 2.0WW T X We have to find out Z if VP = ZIP From equation (2) and (3) VP = KAZl [IS ] -1 VP = KAZlb A l I p K
...(1)
...(2) ...(3)
...(4)
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...(5) VP = AZlA- 1 I p R 1 1 1V S W A = Sa2 a 1W SS a a2 1WW T X Adj A -1 A = A R 2V S1 a a W Adj A = S1 a2 a W S W S1 1 1 W T X A =1 3 R 2V S1 a a W A- 1 = 1 S1 a2 a W 3S W S1 1 1 W T X From equation (5) R 1 1 1VR0.5 0 0VR1 a a2V R 1 0.5 0.5V S W S S W S W W ...(6) Vp = 1 Sa2 a 1WS 0 0.5 0WS1 a2 a W I p = S0.5 1 0.5W I p 3S S W SS0.5 0.5 1 WW S a a2 1WWSS 0 0 2WWS1 1 1 W T T X X Comparing of equation (5)XTand (6) XT R 1 0.5 0.5V W S Z = S0.5 1 0.5W SS0.5 0.5 1 WW X T Option (A) is correct. Given that the first two power system are not connected and separately loaded. Now these are connected by short transmission line. as P1 = P2 = Q1 = Q2 = 0 So here no energy transfer. The bus bar voltage and phase angle of each system should be same than angle difference is
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SOL 1.5.56
q = 30c - 20c = 10c SOL 1.5.57
Option (B) is correct. Given that, 230 V, 50 Hz, 3-f, 4-wire system P = Load = 4 kw at unity Power factor IN = 0 through the use of pure inductor and capacitor Than L = ?, C = ? a IN = 0 = IA + IB + IC Network and its Phasor is being as
...(1)
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Here the inductor is in phase B and capacitor is in Phase C. We know P = VI 3 So Ia = P = 4 # 10 = 17.39 Amp. 230 V From equation (1) IA =- (IB + IC ) `
IA =-c IB #
`
IA = IB
Now
XC
and
XC
&
C
a Ib - Ic
3 +I 3 C # 2 2 m
3 IB = 3 IC - IC = 17.39 - 10 Amp 3 V 230 = = - 23 W 10 IC = 1 2pfC 1 = 139.02 mF = 1 = 2p # 50 # 23 2pfXC = V = 230 - 23 W = 2pfL 10 IL 23 = XL = = 72.95 mH 2p # 100 2pf
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&
So SOL 1.5.58
L
L = 72.95 mH in phase B C = 139.02 mF in phase C
Option (A) is correct. Maximum continuous power limit of its prime mover with speed governor of 5% droop. Generator feeded to three loads of 4 MW each at 50 Hz. Now one load Permanently tripped ` f = 48 Hz If additional load of 3.5 MW is connected than f = ? a Change in Frequency w.r.t to power is given as drop out frequency Df = # Change in power rated power = 5 # 3.5 = 1.16% = 1.16 # 50 = 0.58 Hz 15 100 System frequency is = 50 - 0.58 = 49.42 Hz
SOL 1.5.59
Option (B) is correct. With the help of physical length of line, we can recognize line as short, medium and long line.
SOL 1.5.60
Option (A) is correct. For capacitor bank switching vacuum circuit breaker is best suited in view of cost and effectiveness.
SOL 1.5.61
Option (B) is correct. Ratio of operating coil current to restraining coil current is known as bias in biased differential relay.
SOL 1.5.62
Option (B) is correct. HVDC links consist of rectifier, inverter, transmission lines etc, where rectifier
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consumes reactive power from connected AC system and the inverter supplies power to connected AC system. SOL 1.5.63
Option (C) is correct. Given ABCD constant of 220 kV line A = D = 0.94+10c, B = 130+730c, C = 0.001+900c, VS = 240 kV % voltage regulation is being given as - (VR) Full load (V ) %V.R. = R No Load # 100 VR (Full load) At no load IR = 0 (VR) NL = VS /A , (VR) Full load = 220 kV 240 - 220 0 %V.R. = .94 # 100 220 %V.R. = 16
SOL 1.5.64
Option ( ) is correct.
SOL 1.5.65
Option (B) is correct. Given that, Vab1 = X+q1 , Vab2 = Y+q2 , Phase to neutral sequence volt = ? First we draw phasor of positive sequence and negative sequence.
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From figure we conclude that positive sequence line voltage leads phase voltage by 30c VAN1 = X+q1 - 30c VAN2 = 4+q2 + 30c SOL 1.5.66
SOL 1.5.67
SOL 1.5.68
Option (A) is correct. For system base value 10 MVA, 69 kV, Load in pu(Z new ) = ? (MVA) old kVnew 2 Z new = Z old # # b kVold l (MVA) new 2 Z new = 0.72 # 20 # b 69 l = 36 pu 10 13.8 Option (A) is correct. Unreliable convergence is the main disadvantage of gauss seidel load flow method. Option (C) is correct. Generator feeds power to infinite bus through double circuit line 3-f fault at middle of line. Infinite bus voltage(V ) = 1 pu Transient internal voltage of generator(E ) = 1.1 pu Equivalent transfer admittance during fault = 0.8 pu = 1/X delivering power(PS ) = 1.0 pu
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Perior to fault rotor Power angle d = 30c, f = 50 Hz Initial accelerating power(Pa ) = ?
SOL 1.5.69
SOL 1.5.70
Pa = PS - Pm2 sin d = 1 - EV sin 30c = 1 - 1.1 # 1 # 1 = 0.56 pu 2 X 1/0.8 Option (B) is correct. If initial acceleration power = X pu Initial acceleration = ? Inertia constant = ? X (pu) # S 180 # 50 # X # S = a = Pa = M S#S SH/180F a = 1800X deg / sec2 Inertia const. = 1 = 0.056 18 Option (D) is correct. The post fault voltage at bus 1 and 3 are. Pre fault voltage. RV V R1+0cV S 1W S W VBus = SV2W = S1+0cW SSV WW SS1+0cWW 3 T X T X At bus 2 solid fault occurs Z (f) = 0 , r = 2 Fault current I f = Vr c = V2 c Zrr + Z f Z22 + Z f Z f = 1+0c =- 4j j0.24
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= Vi c (0) - Zir I (f), Vi c = Prefault voltage = Vi c - Z12 I f = 1+0c - j0.08 (- j4) = 1 - 0.32 = 0.68 pu = V3 c - Z 32 I f = 1+0c - j0.16 (- j4) = 1 - 0.64 = 0.36 pu
SOL 1.5.71
Option ( ) is correct.
SOL 1.5.72
Option (D) is correct. Rating of D-connected capacitor bank for unity p.f. PL = S cos f = 12 3 # 0.8 = 16.627 kW reactive power QL = S sin f = 12 3 # 0.6 = 12.47 kW For setting of unity p.f. we have to set capacitor bank equal to reactive power = 12.47 kW real power
SOL 1.5.73
Option (D) is correct. Given that pu parameters of 500 MVA machine are as following
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M = 20 pu, X = 2 pu Now value of M and X at 100 MVA base are for inertia (M) (pu) new = (pu) old # old MVA new MVA (M pu) new = (M Pu) old # 500 = 20 # 5 = 100 pu 100 1 and for reactance (X ) (pu) new = (pu) old # new MVA old MVA (X pu) new = (X pu) old # 100 500 (X Pu) new = 2 # 1 = 0.4 pu 5 SOL 1.5.74
Option (D) is correct. 800 kV has Power transfer capacity = P At 400 kV Power transfer capacity = ? We know Power transfer capacity P = EV sin d X
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SOL 1.5.75
So if V is half than Power transfer capacity is 1 of previous value. 4 Option (B) is correct. In EHV lines the insulation strength of line is governed by the switching over voltages.
SOL 1.5.76
Option (A) is correct. For bulk power transmission over very long distance HVDC transmission preferably used.
SOL 1.5.77
Option (D) is correct. Parameters of transposed overhead transmission line XS = 0.4 W/km , Xm = 0.1 W/km + ve sequence reactance X1 = ? Zero sequence reactance X 0 = ? We know for transposed overhead transmission line. + ve sequence component X1 = XS - Xm = 0.4 - 0.1 = 0.3 W/km Zero sequence component X 0 = XS + 2Xm = 0.4 + 2 (0.1) = 0.6 W/km
SOL 1.5.78
Option (C) is correct. Industrial substation of 4 MW load = PL QC = 2 MVAR for load p.f. = 0.97 lagging If capacitor goes out of service than load p.f. = ? cos f tan f QL - QC PL QL - 2 4
= 0.97 = tan (cos- 1 0.97) = 0.25 = 0.25 = 0.25 & QL = 3 MVAR
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f = tan- 1 c
Page 260
QL = tan- 1 b 3 l = 36c 4 PL m
cos f = cos 36c = 0.8 lagging SOL 1.5.79
Option (D) is correct. Y22 = ? I1 = V1 Y11 + (V1 - V2) Y12 = 0.05V1 - j10 (V1 - V2) =- j9.95V1 + j10V2 I2 = (V2 - V1) Y21 + (V2 - V3) Y23 = j10V1 - j9.9V2 - j0.1V3 Y22 = Y11 + Y23 + Y2 =- j9.95 - j9.9 - 0.1j =- j19.95
SOL 1.5.80
Option (C) is correct. F1 = a + bP1 + cP 12 Rs/hour F2 = a + bP2 + 2cP 22 Rs/hour For most economical operation P1 + P2 = 300 MW then P1, P2 = ? We know for most economical operation 2F1 = 2F2 2P1 2P2
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2cP1 + b = 4cP2 + b P1 = 2P2 P1 + P2 = 300 from eq (1) and (2)
...(1) ...(2)
P1 = 200 MW , P2 = 100 MW SOL 1.5.81
Option (B) is correct. A B V2 V1 We know that ABCD parameters > H = > C DH >I1H I1 , C = I1 B = V1 I2 V = 0 V2 I = 0 V1 Z 1 + Z2 In figure C = = 1 V1 Z2 Z Z1 + Z 2 # 2 1 or Z2 = 1 = = 40+ - 45c C 0.025+45c 2
SOL 1.5.82
2
Option (D) is correct. Given
Steady state stability Power Limit = 6.25 pu If one of double circuit is tripped than Steady state stability power limit = ? Pm1 = EV = 1 # 1 = 6.25 X 0.12 + X 2
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Page 261
1 = 6.25 0.12 + 0.5X & X = 0.008 pu If one of double circuit tripped than 1 Pm2 = EV = 1 # 1 = 0.12 + 0.08 X 0.12 + X Pm2 = 1 = 5 pu 0.2 SOL 1.5.83
SOL 1.5.84
Option (D) is correct. Given data Substation Level = 220 kV 3-f fault level = 4000 MVA LG fault level = 5000 MVA Positive sequence reactance: 4000 Fault current I f = 3 # 220 X1 = Vph /I f 220 3 = = 220 # 220 = 12.1 W 4000 4000 3 # 220 Option (B) is correct. Zero sequence Reactance X 0 = ? 5000 If = 3 # 220 I 5000 Ia1 = Ia2 = Ia0 = f = 3 3 3 # 220 220 V 3 X1 + X2 + X 0 = ph = 5000 Ia1 220 # 3 3 X1 + X2 + X 0 = 220 # 220 = 29.04 W 3 # 5000
nodia.co.in X1 = X2 = 12.1 W X 0 = 29.04 - 12.1 - 12.1 = 4.84 W
SOL 1.5.85
Option (B) is correct. Instantaneous power supplied by 3-f ac supply to a balanced R-L load. P = Va Ia + Va Ib + Vc Ic = (Vm sin wt) Im sin (wt - f) + Vm sin (wt - 120c) Im sin (wt - 120c - f) + Vm sin (wt - 240c) Im sin (wt - 240c - f) = VI [cos f - cos (2wt - f) + cos f - cos (2wt - 240 - f) + cos f - cos (2wt + 240 - f)] ...(1) P = 3VI cos f equation (1) implies that total instantaneous power is being constant.
SOL 1.5.86
Option (C) is correct. In 3-f Power system, the rated voltage is being given by RMS value of line to
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Page 262
line voltage. SOL 1.5.87
Option (B) is correct.
In this figure the sequence is being given as RBY SOL 1.5.88
Option (C) is correct. In thermal power plants, the pressure in the working fluid cycle is developed by the help to feed water pump.
SOL 1.5.89
Option (A) is correct. Kaplan turbines are used for harnessing low variable waterheads because of high percentage of reaction and runner adjustable vanes.
SOL 1.5.90
Option (B) is correct. MHO relay is the type of distance relay which is used to transmission line protection. MHO Relay has the property of being inherently directional.
SOL 1.5.91
Option (C) is correct. Surge impedance of line is being given by as
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L = 11 # 10- 3 = 306.88 W C 11.68 # 10- 9 Ideal power transfer capability 2 (800) 2 P =V = = 2085 MW 306.88 Z0 Z =
SOL 1.5.92
Option (D) is correct. Given that, Power cable voltage = 110 kV C = 125 nF/km Dielectric loss tangent = tan d = 2 # 10- 4 Dielectric power loss = ? dielectric power loss is given by P = 2V2 wC tan d = 2 (110 # 103) 2 # 2pf # 125 # 10- 9 # 2 # 10- 4 = 2 (121 # 108 # 2 # 3.14 # 50 # 250 # 10- 13) = 189 W/km
SOL 1.5.93
Option (A) is correct. Given data Lightening stroke discharge impulse current of I = 10 kA Transmission line voltage = 400 kV Impedance of line Z = 250 W Magnitude of transient over-voltage = ? The impulse current will be equally divided in both directions since there is equal distribution on both sides. Then magnitude of transient over-voltage is
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Page 263
V = IZ/2 = 10 # 103 # 250 2 = 1250 # 103 V = 1250 kV SOL 1.5.94
Option (C) is correct. The A, B, C, D parameters of line A = D = 0.936+0.98c B = 142+76.4c C = (- 5.18 + j914) 10- 6 W At receiving end PR = 50 MW , VR = 220 kV p.f = 0.9 lagging VS = ? Power at receiving end is being given by as follows VS VR A VR 2 PR = cos (b - d) cos (b - a) B B VS # 220 0.936 (220) 2 = cos (76.4c - d) cos 75.6c 142 142 ` VS cos (76.4 - d) = 50 # 142 + 0.936 # 220 # 0.2486 = 32.27 + 51.19 220
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VS cos (76.4 - d) = 83.46 Same as QR = PR tan f = PR tan (cos- 1 f) = 50 tan (cos- 1 0.9) = 24.21 MW VS VR A VR 2 QR = sin (b - d) sin (b - a) B B VS # 220 0.936 # (220) 2 = sin (76.4c - d) sin 75.6c 142 142 (24.21) 142 + 0.936 # 220 # 0.9685 = VS sin (76.4c - d) 220 from equation (1) & (2) VS
2
...(2)
= (215) 2 + (83.46) 2
VS = SOL 1.5.95
...(1)
53190.5716 = 230.63 kV
Option (B) is correct. A new generator of Eg = 1.4+30c pu XS = 1.0 pu, connected to bus of Vt Volt Existing Power system represented by thevenin’s equivalent as Eth = 0.9+0c, Zth = 0.25+90c, Vt = ?
From the circuit given E - Eth 1.212 + j7 - 0.9 I = g = 1.4+30c - 0.9+0c = Zth + XS j (1.25) j (1.25) 0.312 + j7 = = 0.56 - 0.2496j j (1.25)
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Page 264
Vt = Eg - IXS = 1.212 + j7 - (0.56 - 0.2496j) (j1) = 1.212 - 0.2496 + j (0.7 - 0.56) = 0.9624 + j0.14 Vt = 0.972+8.3c SOL 1.5.96
Option (C) is correct. Given that 3-f Generator rated at 110 MVA, 11 kV Xdm = 19% , Xdl= 26% XS = 130% , Operating at no load 3-f short circuit fault between breaker and transformer symmetrical Irms at breaker = ? We know short circuit current Isc = 1 = 1 =- j5.26 pu Xdm j0.19 rating MVA of generator Base current IB = 3 # kV of generator 6 IB = 110 # 10 3 3 # 11 # 10 IB = 5773.67 Amp Symmetrical RMS current = IB # Isc
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SOL 1.5.97
= 5773.67 # 5.26 = 30369.50 Amp = 30.37 kA
Option (A) is correct. + ve sequence current Ia = 1 [Ia + aIb + a2 Ic] 3 = 1 [10+0c + 1+120c # 10+180c + 0] 3 = 1 [10+0c + 10+300c] = 1 [10 + 5 - j8.66] 3 3 = 1 [15 - j8.66] = 17.32+ - 30c 3 3 = 5.78+ - 30c
SOL 1.5.98
Option (D) is correct. Given data 500 MVA , 50 Hz, 3 - f generator produces power at 22 kV Generator " Y connected with solid neutral Sequence reactance X1 = X2 = 0.15 , X 0 = 0.05 pu Sub transient line current = ? E 1 Ia1 = = = 1 =- 2.857j j0.15 + j0.15 + j0.05 0.35j Z1 + Z 2 + Z 0 Now sub transient Line current Ia = 3Ia1 Ia = 3 (- 2.857j) =- 8.57j
SOL 1.5.99
Option (B) is correct. Given: 50 Hz, 4-Pole, 500 MVA, 22 kV generator p.f. = 0.8 lagging Fault occurs which reduces output by 40%. Accelerating torque = ? Power = 500 # 0.8 = 400 MW
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After fault,
Where
Page 265
Power = 400 # 0.6 = 240 MW a Pa = Ta # w Ta = Pa w w = 2pfmechanical fmechanical = felectrical # 2 = felectrical # 2 4 P Pa = 400 - 240 = 160 MW 160 Ta = 2 # p # 50/2 Ta = 1.018 MN
SOL 1.5.100
Option (D) is correct. Turbine rate speed N = 250 rpm To produce power at f = 50 Hz. No. of Poles P =? a N = 120 f P P = 120 f = 120 # 50 = 24 250 N
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SOL 1.5.101
Option (C) is correct. In case of bundled conductors, We know that self GMD of conductor is increased and in a conductor critical disruptive voltage of line depends upon GMD of conductor. Since GMD of conductor is increased this causes critical disruptive voltage is being reduced and if critical disruptive voltage is reduced, the corona loss will also be reduced.
SOL 1.5.102
Option (B) is correct. Given that no. of buses n = 300 Generator bus = 20 Reactive power support buses = 25 Fixed buses with Shunt Capacitor = 15 Slack buses (ns ) = 20 + 25 - 15 = 30 a Size of Jacobian Matrix is given as = 2 (n - ns) # 2 (n - ns) = 2 (300 - 30) # 2 (300 - 30) = 540 # 540
SOL 1.5.103
Option (B) is correct. Auxiliary component in HVDC transmission system are DC line inductor and reactive power sources.
SOL 1.5.104
Option (C) is correct. a Exchanged electrical power is being given as follows P = EV 6sin (d1 - d2)@ Xd Given that
...(1)
P " Power supply by generator = 0.5 pu
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E " Voltage for rotar generator = 2.0 pu V " Voltage of motor rotor = 1.3 pu Xd = X eq = Reactance of generator + Reactance of motor + Recatance of connecting line Xd = 1.1 + 1.2 + 0.5 = 2.8 d1 - d2 = Rotor angle difference = ? from eq(1), 0.5 = 2 # 1.3 sin (d1 - d2) 2.8 & d1 - d2 = sin- 1 b 2.8 # 0.5 l 2. 6 & d1 - d2 = 32.58 SOL 1.5.105
Option (B) is correct. Time period between energization of trip circuit and the arc extinction on an opening operation is known as the interrupting time of Circuit breaker.
SOL 1.5.106
Option (B) is correct. Given that ABCD parameters of line as A = D = 0.9+0c, B = 200+90% W , C = 0.95 # 10 - 3 +90% S . at no-load condition,
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Receiving end voltage (VR) = sending end voltage (VS ) ohmic value of reactor = ? We know VS = AVR + BIR VS = VR VR = AVR + BIR VR (1 - A) = BIR VR = B = 200+90c IR 1-A 1 - 0.9+0c VR = 2000+90c IR The ohmic value of reactor = 2000 W SOL 1.5.107
Option (A) is correct. Surge impedance of cable Z1 =
L; C
L = 0.4 mH/km, C = 0.5 mF /km
0.4 # 10- 3 = 28.284 0.5 # 10- 6 surge impedance of overhead transmission line Z2 = Z 3 = L ; L = 1.5 mm/km, C = 0.015 mF/km C =
1.5 # 10- 5 = 316.23 0.015 # 10- 6 Now the magnitude of voltage at junction due to surge is being given by as Vl = 2 # V # Z2 V = 20 kV Z 2 + Z1 Z2 = Z 3 =
3 = 2 # 20 # 10 # 316.23 316 + 28.284
= 36.72 kV
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Page 267
Option (D) is correct. Let that current in line is I amp than from figure current in line section PR is (I - 10) amp current in line section RS is (I - 10 - 20) = (I - 30) amp current in SQ Section is (I - 30 - 30) = (I - 60) amp Given that VP and VQ are such that VP - VQ = 3 V by applying KVL through whole line VP - VQ = (I - 10) 0.1 + (I - 30) 0.15 + (I - 60) # 0.2 & 3 = 0.45I - 17.5 I = 20.5 = 45.55 amp 0.45 Now the line drop is being given as = (I - 10) 0.1 + (I - 30) 0.15 + (I - 60) 0.2 = (33.55) 0.1 + (15.55) 0.15 + (14.45) 0.2 = 8.58 V The value of VP for minimum voltage of 220 V at any feeder is = 220 + Line voltage = 220 + 8.58 = 228.58 V
SOL 1.5.109
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Option (D) is correct. Given Load Power = 100 MW VS = VR = 11 kV j0.2 # (11) 2 p.u. # (kV) 2 Impedance of line ZL = = = j0.242 W 100 MV VS VR sin d We know PL = X 3 3 100 # 106 = 11 # 10 # 11 # 10 sin d 0.242 100 # 0.242 = sin d 121
d = sin- 1 (0.2) = 11.537c Reactive Power is being given by VS VR VR 2 QL = cos d X X 3
3
= 11 # 10 # 11 # 10 cos (11.537c) 0.242
(11 # 103) 2 0.242
6
SOL 1.5.110
= 121 # 10 [cos (11.537c) - 1] =- 10.1 MVAR 0.242 Option (B) is correct. Given the bus Impedance Matrix of a 4-bus Power System R V Sj0.3435 j0.2860 j0.2723 j0.2277W Sj0.2860 j0.3408 j0.2586 j0.2414W Z bus = S W Sj0.2723 j0.2586 j0.2791 j0.2209W Sj0.2277 j0.2414 j0.2209 j0.2791W T X bus 2 and reference Now a branch os j0.2 W is connected between
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RZ V S ij W 1 ZB(New) = ZB (Old) Sh W Z g Z jnB Zij + Zb S W8 ji SZnjW X in jth and reference bus New element Zb = j0.2 W isT connected j = 2 , n = 4 so R V SZ12W SZ22W 1 Z Z Z Z Zij + Zb SSZ23WW 8 21 22 23 24B SZ24W R V T X Sj0.2860W Sj0.3408W 1 = S W8j0.2860 j0.3408 j0.2586 j0.2414B ...(1) 6j (0.3408) + j0.2@ Sj0.2586W Sj0.2414W T X Given that we are required to change only Z22, Z23 j2 (0.3408) 2 So in equation (1) = j0.2147 Zl22 = j (0.5408) j2 (0.3408) (0.2586) = j0.16296 Zl23 = 0.5408
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Z22(New) = Z22(Old) - Zl22 = j0.3408 - j0.2147 = j0.1260 Z23(New) = Z23 (Old) - Zl23 = j0.2586 - j0.16296 = j0.0956 SOL 1.5.111
Option (D) is correct. Total zero sequence impedance, + ve sequence impedance and - ve sequence impedances Z0 Z1 Z2 Zn for L-G fault
= (Z 0) Line + (Z 0) Generator = j0.04 + j0.3 = j0.34 pu = (Z1) Line + (Z1) Generator = j0.1 + j0.1 = j0.2 pu = (Z2) Line + (Z2) Generator = j0.1 + j0.1 = j0.2 pu = j0.05 pu
Ia1 =
Ea 0.1 = j0.2 + j0.2 + j0.34 + j0.15 Z 0 + Z 1 + Z 2 + 3Z n
=- j1.12 pu generator MVA = IB = 3 generator kV Fault current
20 # 106 = 1750 Amp 3 # 6.6 # 103
I f = (3Ia) IB = 3 (- j1.12) (1750) =- j5897.6 Amp Neutral Voltage and
Vn = I f Zn Zn = ZB # Z pu (6.6) 2 0.05 = 0.1089 W = 20 # Vn = 5897.6 # 0.1089 = 642.2 V
SOL 1.5.112
Option (A) is correct. We know that Optimal Generation IC1 = IC2 , and P3 = 300 MW (maximum load) IC 3 = 30
(Independent of load)
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20 + 0.3P1 = 30 + 0.4 P2 0.3P1 - 0.4P2 = 10 P1 + P2 + P3 = 700 P1 + P2 + 300 = 700 P1 + P2 = 400 From equation (1) and (2)
Page 269
...(1)
...(2)
P1 = 242.8 MW P2 = 157.14 MW SOL 1.5.113
Option (A) is correct. For transmission line protection-distance relay For alternator protection-under frequency relay For bus bar protection-differential relay For transformer protection-Buchholz relay
SOL 1.5.114
Option (C) is correct.
nodia.co.in We know by equal area criteria PS (dm - d0) =
#d
dm
Pmax sin ddd
C
Pmax sin d0 (dm - d0) = Pmax [cos d0 - cos dm] Pmax = 2 P0 = Pmax sin d0 = 1 d0 = 30c dmax = 110c (given) Now from equation (1) 2 sin 30c (110 - 30) p = 2 [cos dc - cos 110c] 180 0.5 # 80p = cos dc + 0.342 180
...(1)
cos dc = 0.698 - 0.342 dc = 69.138c SOL 1.5.115
Option (D) is correct. a Both sides are granted So, Ia = Ea = 10+0c = 5+ - 90c 2j Za Ib = Eb = 10+ - 90c = 3.33+ - 180c 3j Zb Ic = Ec = 10+120c = 2.5+30c 4j Zc We know Ia = 1 [Ia + aIb + a2 Ic] 3 1
where a = 1+120c & a2 = 1+240c
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Page 270
Ia1 = 1 [5+ - 90c + 3.33+ ^- 180c + 120ch + 2.5+ ^240c + 30ch] 3 Ia1 = 1 [5+ - 90c + 3.33+ - 60c + 2.5+270c] 3 = 1 [- 5j + 1.665 - j2.883 - 2.5j] 3 = 1 [1.665 - j10.383] = 3.5+ - 80.89c 3 SOL 1.5.116
Option (B) is correct. Given data A balanced delta connected load = 8 + 6j = 2 V2 = 400 volt Improved Power Factor cos f2 = 0.9 f1 = tan- 1 ^6/8h = 36.85c f2 = cos- 1 (0.9) = 25.84c 400 = 40+ - 36.86c I = V = 400 = 8 + 6j 10+36.86c Z
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= 32 - j24 Since Power factor is Improved by connecting a Y-connected capacitor bank like as
Phasor diagram is being given by as follows
In figure
oa = I l cos f2 = I cos f1 I l cos 25.84c = 32 I l # 0.9 = 32 Il = 35.55 ac = 24 Amp. ab = I l sin f2 = 35.55 sin 25.84c
(ac = I sin f1)
ab = 15.49 Amp Ic = bc = ac - ab = 24 - 15.49 = 8.51 Amp KVAR of Capacitor bank = 3 # V # IC = 3 # 400 # 8.51 1000 1000 = 10.2 KVAR SOL 1.5.117
Option (B) is correct. Given power system with these identical generators
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Page 271
G1 has Speed governor G2 and G3 has dr0op of 5% When load increased, in steady state generation of G1 is only increased while generation of G2 and G3 are unchanged. SOL 1.5.118
Option (A) is correct. R1 , R2 -Distance Relay Zone-1 and Zone-2 setting for both the relays Correct setting for Zone-2 of relay R1 and R2 are given as TZ2 R = 0.6 sec, TZR = 0.3 sec a Fault at Zone-2, therefore firstly operated relay is R2 , so time setting of R2 is 0.3 sec and R1 is working as back up relay for zone-2, so time setting for R1 is 0.6 sec. 1
2
SOL 1.5.119
Option (B) is correct. The reactive power absorbed by the rectifier is maximum when the firing angle a = 30c.
SOL 1.5.120
Option (D) is correct. Given a power system consisting of two areas as shown connected by single tieline
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For load flow study when entering the network data, the tie line data inadvertently left out. If load flow programme is run with this incomplete data than load flow will not converge if only one slack bus is specified. SOL 1.5.121
Option (D) is correct. Given that XS = 0.2 pu Mid point voltage of transmission line = 0.98 pu VS = VR = 1 Steady state power transfer limit P = VS VR sin d = 1.1 sin 90c= 5 pu 0.2 XS
SOL 1.5.122
Option (B) is correct. We have to find out the thevenin’s equivalent zero sequence impedance Z 0 at point B.The zero sequence network of system can be drawn as follows
equivalent zero sequence impedance is being given as follows Z 0 = 0.1j + 0.05j + 0.07j + (3 # 0.25) Z 0 = 0.75 + j0.22 SOL 1.5.123
*Given data : ZC = 400 W (Characteristics Impedance)
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b = 1.2 # 10- 3 rad/km (Propagation constant) l = 100 km (length of line) Pmax = ? If VS = 230 kV VS = VR cos (bl) + jZC sin (bl) IR VS = AVR + BIR A = cos bl = cos (1.2 # 10- 3 # 100) = 0.9928+0c B = jZC sin (bl) = j400 sin (1.2 # 10- 3 # 100) = j47.88 = 47.88+90c VS = 230 kV, l = 100 km Since it is a short line, so VS - VR = 230 kV again we know for transmission line the equation (Pr - Pr0) 2 + (Qr - Qr0) = Pr2 Where
...(1) 2
Pr0 =- AVR cos (b - a) MW B
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Qr0 =- AVR sin (b - a) MW B Pr = VS VR MVA B
and maximum power transferred is being given by as Prm = Pr - Pr0 Pr = VS VR = 230 # 230 47.88 B Pr = 1104.84 MVA 2 Pr0 =- AVR cos (b - a) MW B =-
0.9928 # (230) 2 # cos (90c - 0) 47.88
Pr0 = 0 MW So maximum Power transferred Prm = Pr - Pr0 = 1104.84 MW SOL 1.5.124
*Given: two transposed 3-f line run parallel to each other. The equation for voltage drop in both side are given as R R V VR V S0.15 0.05 0.05 0.04 0.04 0.04WSIa1W SDVa1W S0.05 0.15 0.05 0.04 0.04 0.04WSIb1W SDVb1W S SDV W WS W S c1W = j S0.05 0.05 0.15 0.04 0.04 0.04WSIc1W S0.04 0.04 0.04 0.15 0.05 0.05WSIa2W SDVa2W S0.04 0.04 0.04 0.05 0.15 0.05WSIb2W SDVb2W S S W WS W S0.04 0.04 0.04 0.05 0.05 0.15WSIc2W SDVc2W T self and mutual zero sequence X XT impedance X We haveT to compute of the system i.e. Z 011, Z 012, Z 021, Z 022 in the following equation. DV01 = Z 011 I 01 + Z 021 I 02 DV02 = Z 021 I 01 + Z 022 I 02
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Page 273
We know that + ve , - ve and zero sequence Impedance can be calculated as respectively. Z1 = j (XS - Xm) Z2 = j (XS - Xm) Z 0 = j (XS + 2Xm) So zero sequence Impedance calculated as Z 011 = j (XS + 2Xm) Z 011 = j [0.15 # 2 (0.05)] = 0.25j Z 012 = Z 021 = j (XS + 2Xm) Z 012 = Z 021 = j [0.15 + 2 (0.04)] = 0.23j Z 022 = j (XS + 2Xm) = j [0.15 + 2 (0.01)] = 0.25j SOL 1.5.125
XS = 0.15 , Xm = 0.05 XS = 0.15 , Xm = 0.04 XS = 0.15 , Xm = 0.05
*Given
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X = 0.2 pu For generator X' = 0.1 pu , El = 1.0 pu, H = 5 MJ/MVA Mechanical Power Pm = 0.0 pu, wB = 2p # 50 rad/sec Initially generator running on open circuit, at switch closure d = 0 wB = dd = winit dt
maximum winit = ? , so that generator pulls into synchronism We know that swing equation H d2 d = (P - P ) pu ......(1) m e pf dt2 E V sin d = 1.1 sin d = 3.33 sin d Pe = 0. 3 X From equation (1) 5 d2 d = 0 - 3.33 sin d 3.14 # 50 dt2 d2 d =- 104.72 sin d dt2 integrating on both side. dd = 104.72 cos d + d 0 dt d0 = 0 (given) w = dd dt For (winit) max = b dd l dt max dd b dt l
when cos d = 1
max
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SOL 1.5.126
SOL 1.5.127
(winit) max = b dd l = 104.72 rad/sec dt max Option (C) is correct. A lossless radial transmission line with surge impedance loading has flat voltage profile and unity power factor at all points along it. Option (B) is correct. Given that 3-f transformer, 20 MVA, 220 kV(Y) - 33 kV(D) Xl = leakage Reactance = 12% X = reffered to LV in each phase = ? (LV side voltage) 2 = 3# Reactance of Leakage MVA Rating # = 3#
SOL 1.5.128
Page 274
(33 kV) 2 0.12 = 19.6 W 20 MVA #
Option (D) is correct. Given 75 MVA, 10 kV synchronous generator Xd = 0.4 pu We have to find out (Xd ) new at 100 MVA, 11 kV
nodia.co.in (Xd ) new = (X d) old # >
^kVhold 2 ^MVAhnew # H > H ^kVhnew ^MVAhold
(Xd ) new = 0.4 # b 10 l # 100 = 0.44 pu 11 75 Option (A) is correct. Given Y-alternator: 440 V, 50 Hz Per phase Xs = 10 W , Capacitive Load current I = 20 A For zero voltage regulation load p.f = ? Let Load Z = R + jX Zero voltage regulation is given so 2
SOL 1.5.129
E Ph - IXs - I (R + jX) = 0 440 - 20 (j10) - 20 (R + jX) = 0 ...(1) 3 separating real and imaginary part of equation (1) 20R = 440 3 R = 22 3 and 20 (X + 10) = 440 3 22 X = - 10 = 4.68 3 3 4.68/ 3 - 1 4.68 q = tan- 1 X = tan- 1 f p = tan b 22 l R 22/ 3 and power factor cos q = cos b tan- 1 4.68 l 22 cos q = 0.82 SOL 1.5.130
Option (B) is correct. Given 240 V, 1-f AC source, Load Impedance Z = 10+60c W Capacitor is in parallel with load and supplies 1250 VAR
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Page 275
The real power P by source = ?
from figure current through load IL = I + IC I = V = 240 = 24+ - 60c Z 10+60c IC = VAR = 1250 = 5.20j 240 V IL = 24+ - 60c + 5.20j = 12 - 15.60j a apparent power S = VI = P + jQ = 240 (12 + 15.60j) = 2880 + 3744j = P + jQ Where P = Real Power , Q = Reactive Power P = 2880 W
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SOL 1.5.131
Option ( ) is correct.
SOL 1.5.132
Option (C) is correct. We have to find out maximum voltage location on line by applying KVL in the circuit VS - VR = 0.05j , where VS = 1 voltage at voltage at voltage at
VR P1 P2 P3
= 1 - 0.05j = VS = 1 pu . = 1 - 0.1j (by applying KVL) = 1 - 0.1j + j0.15 (by applying KVL)
..(1) ...(2)
...(3) = 1 + 0.05j From equation (1), (2) and (3) it is cleared that voltage at P3 is maximum. SOL 1.5.133
Option (B) is correct. Given: two generators P1 = 50 (50 - f) P2 = 100 (51 - f) total load = 400 MW than f = ? P1 + P2 = 400 50 (50 - f) + 100 (51 - f) = 400 50 + 102 - 8 = 3f f = 48 Hz
SOL 1.5.134
*Given 132 kV transmission line connected to cable as shown in figure
Characteristics impedance of line and cable are 400 Wand 80 W 250 kV surge travels from A to B than (a) We have to calculate voltage surge at C.
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Page 276
(b) Reflected component of surge when reaches A. (c) Surge current in cable BC Vi = 250 kV , ZC1 = 400 W , ZC2 = 80 W (a) Voltage surge at C Vt = Z # ZC2 # Vi = 2 # 80 # 250 = 83.34 kV 400 + 80 ZC1 + ZC2 (b) Reflected voltage at A Vr = b ZC2 - ZC1 l Vi = 80 - 400 # 250 =- 166.67 kV 400 + 80 ZC2 + ZC1 (c) Surge current in cable BC It = Ii + Ir = Ii - aIi
SOL 1.5.135
= (1 - a) Ii , Here a = ZC2 - ZC1 ZC2 + ZC1 It = b1 - ZC2 - ZC1 l Vi = b1 + 320 l 250 480 400 ZC2 + ZC1 ZC1 = b1 + 4 l 25 = 1.04 kAmp 6 40 *We have to draw reactance diagram for given YBus matrix R V S- 6 2 2.5 0 W S 2 - 10 2.5 4 W YBus = j S W S2.5 2.5 - 9 4 W 4 4 - 8W S0 T X as a It is 4 # 4 matrix (admittance matrix) R V Sy11 y12 y13 y14W Sy21 y22 y23 y24W YBus = S W Sy 31 y 32 y 33 y 34W Sy 41 y 42 y 43 y 44W T X Here diagonal elements
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...(1) y11 = y10 + y12 + y13 + y14 =- 6j ...(2) y22 = y20 + y21 + y23 + y24 =- 10j ...(3) y 33 = y 30 + y 31 + y 32 + y 34 =- 9j ...(4) y 44 = y 40 + y 41 + y 42 + y 43 =- 9j and diagonal elements y12 = y21 =- y12 = 2j _ b y13 = y 31 =- y13 = 2.5j b b y14 = y 41 =- y14 = 0j b .....(5) ` y23 = y 32 =- y23 = 2.5j b y24 = y 42 =- y24 = 4j b bb y 34 = y 34 = 4j a from equation (1) y10 = y11 - y12 - y13 - y14 =- 6j + 2j + 2.5j + 0j =- 1.5j Same as from equation (2) y20 = y22 - y21 - y23 - y24 =- 10j + 2j + 2.5j + 4j =- 1.5j from equation (3) y 30 = y 33 - y 31 - y 32 - y 34 =- 9j + 2.5j + 2.5j + 4j = 0 from equation (4) y 40 = y 44 - y 41 - y 42 - y 43 =- 8j + 0 + 4j + 4j = 0 Now we have to draw the reactance diagram as follows
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SOL 1.5.136
Page 277
*Given synchronous generator is connected to infinite bus through loss less double circuit line Pd = 1+30c pu sudden fault reduces the peak power transmitted to 0.5 pu after clearance of fault, peak power = 1.5 pu Critical clearing angle ( dcr ) = ?
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d0 = 30c = 0.52 rad From equal area criteria
#d
dcr
0
Where
(PL11 - Pmax 11 sin d) dd =
#d
dmax
cr
(Pmax 111 sin d - Pm) dd
...(1)
dmax = p - sin- 1 b Pm l Pmax 111
dmax = p - 0.8729 = 2.41 rad By integrating equation (1) 6Pm d + Pmax 11 cos d@ddcrmax - Pmax 111 (cos dmax - cos dcr ) = 0 & Pm (dcr - ds) + Pmax 11 (cos dcr - cos d0) + Pm (dmax - dcr ) + Pmax 111 (cos dmax - cos dcr ) = 0 P (d - d0) - Pmax 11 cos d0 + Pmax 111 cos dmax & cos dcr = m max Pmax 111 - Pmax 11 1 (2.41 - 0.52) - 0.5 cos (0.52) + 1.5 cos (2.41) = 1.5 - 0.5 cos dcr = 0.35 dcr = cos- 1 0.35 = 1.21 rad SOL 1.5.137
*Given: L - G fault on unloaded generator Z 0 = j0.15 , Z1 = j0.25 , Z2 = j0.25 pu, Zn = j0.05 pu Vprefault = 1 pu
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Page 278
If = ? Fault Current I f = 3Ia1 =
3Vprefault 3#1 = Z1 + Z2 + Z 0 + 3Zn (j0.25 + j0.25 + j0.15) + 3 (j.05)
3 =- 3.75j 0.80j Sequence network is being drawn as follows =
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SOL 1.5.138
*Given power system has two generator 2 Generator - 1; C1 = 0.006P G1 + 8PG1 + 350 2 Generator - 2; C2 = 0.009P G2 + 7PG2 + 400 Generator Limits are 100 MW # PG1 # 650 MW 50 MW # PG2 # 500 MW PG1 + PG2 = 600 MW , PG1, PG2 = ? For optimal generation We know for optimal Generation 2C1 = 2C2 2PG1 2PG2 2C1 = 0.012P + 8 G1 2PG1 2C2 = 0.018P + 7 G2 2PG2 from equation (1) 0.012PG1 + 8 = 0.018PG2 + 7 0.012PG1 - 0.018PG2 =- 1 PG1 + PG2 = 600 From equation (2)
...(1)
...(2) ...(3)
0.012PG1 - 0.018 (600 - PG1) =- 1 & 0.03PG1 = 9.8 & PG1 = 326.67 MW PG2 = 600 - PG1 = 600 - 326.67 = 273.33 MW ***********
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6 CONTROL SYSTEMS
YEAR 2013 MCQ 1.6.1
ONE MARK
The Bode plot of a transfer function G ^s h is shown in the figure below.
The gain _20 log G ^s h i is 32 dB and - 8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all w. Then G ^s h is (B) 392.8 (A) 39.8 s s (C) 32 (D) 322 s s
MCQ 1.6.2
Assuming zero initial condition, the response y ^ t h of the system given below to a unit step input u ^ t h is
(A) u ^ t h 2 (C) t u ^ t h 2 YEAR 2013 MCQ 1.6.3
(B) tu ^ t h (D) e-t u ^ t h TWO MARKS
Y ^s h The signal flow graph for a system is given below. The transfer function for U ^s h this system is
GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
s+1 5s2 + 6s + 2 (C) 2 s + 1 s + 4s + 2 (A)
MCQ 1.6.4
Page 280
s+1 s 2 + 6s + 2 (D) 2 1 5s + 6s + 2 (B)
w ^s h The open-loop transfer function of a dc motor is given as = 10 . When V ^s h 1 + 10s connected in feedback as shown below, the approximate avalue of Ka that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is
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(A) 1 (C) 10
(B) 5 (D) 100
Common Data Questions: 5 & 6 The state variable formulation of a system is given as 1 - 2 0 x1 x1 xo1 > H=> H> H + > H u , x1 ^0 h = 0 , x2 ^0 h = 0 and y = 61 0@> H 0 - 1 x2 1 x2 xo2 MCQ 1.6.5
The response y ^ t h to the unit step input is (B) 1 - 1 e-2t - 1 e-t (A) 1 - 1 e-2t 2 2 2 2 (C) e-2t - e-t
MCQ 1.6.6
The system is (A) controllable but not observable (B) not controllable but observable (C) both controllable and observable (D) both not controllable and not observable YEAR 2012
MCQ 1.6.7
(D) 1 - e-t
TWO MARKS
The state variable description of an LTI system is given by Jxo1N J 0 a1 0NJx1N J0N K O K OK O K O Kxo2O = K 0 0 a2OKx2O + K0O u Kxo O Ka 0 0OKx 3O K 1O 3 3 L P L PL P L P Jx1N K O y = _1 0 0iKx2O Kx 3O GATE MCQ Electrical L P Engineering (Vol-1, 2 & 3) by RK Kanodia & Ashish Murolia
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Page 281
where y is the output and u is the input. The system is controllable for (A) a1 ! 0, a2 = 0, a 3 ! 0 (B) a1 = 0, a2 ! 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0 (D) a1 ! 0, a2 ! 0, a 3 = 0 MCQ 1.6.8
The feedback system shown below oscillates at 2 rad/s when
(A) K = 2 and a = 0.75 (C) K = 4 and a = 0.5
(B) K = 3 and a = 0.75 (D) K = 2 and a = 0.5
Statement for Linked Answer Questions 9 and 10 :
MCQ 1.6.9
MCQ 1.6.10
The transfer function of a compensator is given as Gc (s) = s + a s+b Gc (s) is a lead compensator if (A) a = 1, b = 2 (B) a = 3, b = 2 (C) a =- 3, b =- 1 (D) a = 3, b = 1
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The phase of the above lead compensator is maximum at (A) 2 rad/s (B) 3 rad/s (C)
(D) 1/ 3 rad/s
6 rad/s
YEAR 2011 MCQ 1.6.11
The frequency response of a linear system G (jw) is provided in the tubular form below G (jw)
1.3
+G (jw) - 130c (A) 6 dB and 30c (C) - 6 dB and 30c MCQ 1.6.12
ONE MARK
1.2
1.0
0.8
0.5
0.3
- 140c
- 150c
- 160c
- 180c
- 200c
(B) 6 dB and - 30c (D) - 6 dB and - 30c
The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r (t) having a magnitude of 10 and a duration of one second, as shown in the figure is
(A) 0 (C) 1
(B) 0.1 (D) 10
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Page 282
An open loop system represented by the transfer function (s - 1) is G (s) = (s + 2) (s + 3) (A) Stable and of the minimum phase type (B) Stable and of the non–minimum phase type (C) Unstable and of the minimum phase type (D) Unstable and of non–minimum phase type YEAR 2011
TWO MARKS
MCQ 1.6.14
The open loop transfer function G (s) of a unity feedback control system is given as K bs + 2 l 3 G (s) = 2 s (s + 2) From the root locus, at can be inferred that when K tends to positive infinity, (A) Three roots with nearly equal real parts exist on the left half of the s -plane (B) One real root is found on the right half of the s -plane (C) The root loci cross the jw axis for a finite value of K; K ! 0 (D) Three real roots are found on the right half of the s -plane
MCQ 1.6.15
A two loop position control system is shown below
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The gain K of the Tacho-generator influences mainly the (A) Peak overshoot (B) Natural frequency of oscillation (C) Phase shift of the closed loop transfer function at very low frequencies (w " 0) (D) Phase shift of the closed loop transfer function at very high frequencies (w " 3) YEAR 2010 MCQ 1.6.16
TWO MARKS
1 plotted in the s (s + 1) (s + 2) complex G (jw) plane (for 0 < w < 3) is The frequency response of G (s) =
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MCQ 1.6.17
o = AX + Bu with A = >- 1 2H, B = >0H is The system X 0 2 1
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(A) Stable and controllable (C) Unstable but controllable MCQ 1.6.18
(B) Stable but uncontrollable (D) Unstable and uncontrollable
The characteristic equation of a closed-loop system is s (s + 1) (s + 3) k (s + 2) = 0, k > 0 .Which of the following statements is true ? (A) Its root are always real (B) It cannot have a breakaway point in the range - 1 < Re [s] < 0 (C) Two of its roots tend to infinity along the asymptotes Re [s] =- 1 (D) It may have complex roots in the right half plane. YEAR 2009
MCQ 1.6.19
MCQ 1.6.20
Page 283
ONE MARK
The measurement system shown in the figure uses three sub-systems in cascade whose gains are specified as G1, G2, 1/G3 . The relative small errors associated with each respective subsystem G1, G2 and G3 are e1, e2 and e3 . The error associated with the output is :
(A) e1 + e2 + 1 e3
(B) e1 e2 e3
(C) e1 + e2 - e3
(D) e1 + e2 + e3
The polar plot of an open loop stable system is shown below. The closed loop system is
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Page 284
(A) always stable (B) marginally stable (C) un-stable with one pole on the RH s -plane (D) un-stable with two poles on the RH s -plane MCQ 1.6.21
The first two rows of Routh’s tabulation of a third order equation are as follows. s3 2 2 s2 4 4 This means there are (A) Two roots at s = ! j and one root in right half s -plane (B) Two roots at s = ! j2 and one root in left half s -plane (C) Two roots at s = ! j2 and one root in right half s -plane (D) Two roots at s = ! j and one root in left half s -plane
MCQ 1.6.22
The asymptotic approximation of the log-magnitude v/s frequency plot of a system containing only real poles and zeros is shown. Its transfer function is
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(A)
10 (s + 5) s (s + 2) (s + 25)
(C)
100 (s + 5) s (s + 2) (s + 25)
1000 (s + 5) s (s + 2) (s + 25) 80 (s + 5) (D) 2 s (s + 2) (s + 25)
(B)
2
YEAR 2009 MCQ 1.6.23
TWO MARKS
The unit-step response of a unity feed back system with open loop transfer function G (s) = K/ ((s + 1) (s + 2)) is shown in the figure. The value of K is
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(A) 0.5 (C) 4 MCQ 1.6.24
Page 285
(B) 2 (D) 6
The open loop transfer function of a unity feed back system is given by G (s) = (e - 0.1s) /s . The gain margin of the is system is (A) 11.95 dB (B) 17.67 dB (C) 21.33 dB (D) 23.9 dB
Common Data for Question 25 and 26 : A system is described by the following state and output equations dx1 (t) =- 3x1 (t) + x2 (t) + 2u (t) dt dx2 (t) =- 2x2 (t) + u (t) dt y (t) = x1 (t) when u (t) is the input and y (t) is the output MCQ 1.6.25
MCQ 1.6.26
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The system transfer function is (A) 2 s + 2 (B) 2 s + 3 s + 5s - 6 s + 5s + 6 (C) 2 2s + 5 (D) 2 2s - 5 s + 5s + 6 s + 5s - 6 The state-transition matrix of the above system is e - 3t 0 e - 3t e - 2t - e - 3t (B) (A) = - 2t G G = e + e - 3t e - 2t 0 e - 2t e - 3t e - 2t + e - 3t (C) = G 0 e - 2t
e 3t e - 2t - e - 3t (D) = G 0 e - 2t
YEAR 2008 MCQ 1.6.27
ONE MARK
A function y (t) satisfies the following differential equation : dy (t) + y (t) = d (t) dt where d (t) is the delta function. Assuming zero initial condition, and denoting the unit step function by u (t), y (t) can be of the form (B) e - t (A) et (C) et u (t)
(D) e - t u (t)
YEAR 2008
TWO MARK
MCQ 1.6.28
The transfer function of a linear time invariant system is given as G (s) = 2 1 s + 3s + 2 The steady state value of the output of the system for a unit impulse input applied at time instant t = 1 will be (A) 0 (B) 0.5 (C) 1 (D) 2
MCQ 1.6.29
The transfer functions of two compensators are given below :
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Page 286
10 (s + 1) , C2 = s + 10 (s + 10) 10 (s + 1) Which one of the following statements is correct ? (A) C1 is lead compensator and C2 is a lag compensator (B) C1 is a lag compensator and C2 is a lead compensator (C) Both C1 and C2 are lead compensator (D) Both C1 and C2 are lag compensator
C1 =
MCQ 1.6.30
The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the figure :
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This transfer function has (A) Three poles and one zero (C) Two poles and two zero MCQ 1.6.31
(B) Two poles and one zero (D) One pole and two zeros
Figure shows a feedback system where K > 0
The range of K for which the system is stable will be given by (A) 0 < K < 30 (B) 0 < K < 39 (C) 0 < K < 390 (D) K > 390 MCQ 1.6.32
The transfer function of a system is given as 100 2 s + 20s + 100 The system is (A) An over damped system (B) An under damped system (C) A critically damped system (D) An unstable system
Statement for Linked Answer Question 27 and 28.
MCQ 1.6.33
The state space equation of a system is described by Xo = AX + Bu,Y = CX where X is state vector, u is input, Y is output and 0 1 0 A == , B = = G, C = [1 0] G 0 -2 1 The transfer function G(s) of this system will be s (A) (B) s + 1 (s + 2) s (s - 2)
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Page 287
s 1 (D) (s - 2) s (s + 2) A unity feedback is provided to the above system G (s) to make it a closed loop system as shown in figure. (C)
MCQ 1.6.34
For a unit step input r (t), the steady state error in the input will be (A) 0 (B) 1 (C) 2 (D) 3 YEAR 2007 MCQ 1.6.35
ONE MARK
The system shown in the figure is
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If x = Re [G (jw)], and y = Im [G (jw)] then for w " 0+ , the Nyquist plot for G (s) = 1/s (s + 1) (s + 2) is (A) x = 0 (B) x =- 3/4 (C) x = y - 1/6
MCQ 1.6.37
TWO MARKS
(D) x = y/ 3
The system 900/s (s + 1) (s + 9) is to be such that its gain-crossover frequency becomes same as its uncompensated phase crossover frequency and provides a 45c phase margin. To achieve this, one may use (A) a lag compensator that provides an attenuation of 20 dB and a phase lag of 45c at the frequency of 3 3 rad/s (B) a lead compensator that provides an amplification of 20 dB and a phase lead of 45c at the frequency of 3 rad/s (C) a lag-lead compensator that provides an amplification of 20 dB and a phase lag of 45c at the frequency of 3 rad/s (D) a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45c at the frequency of 3 rad/s
MCQ 1.6.38
If the loop gain K of a negative feed back system having a loop transfer function
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Page 288
K (s + 3) / (s + 8) 2 is to be adjusted to induce a sustained oscillation then (A) The frequency of this oscillation must be 4 3 rad/s (B) The frequency of this oscillation must be 4 rad/s (C) The frequency of this oscillation must be 4 or 4 3 rad/s (D) Such a K does not exist MCQ 1.6.39
The system shown in figure below
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can be reduced to the form
with (A) X = c0 s + c1, Y = 1/ (s2 + a0 s + a1), Z = b0 s + b1 (B) X = 1, Y = (c0 s + c1) / (s2 + a0 s + a1), Z = b0 s + b1 (C) X = c1 s + c0, Y = (b1 s + b0) / (s2 + a1 s + a0), Z = 1 (D) X = c1 s + c0, Y = 1/ (s2 + a1 s + a), Z = b1 s + b0 MCQ 1.6.40
Consider the feedback system shown below which is subjected to a unit step input. The system is stable and has following parameters Kp = 4, Ki = 10, w = 500 and x = 0.7 .The steady state value of Z is
(A) 1 (C) 0.1
(B) 0.25 (D) 0
Data for Q.41 and Q.42 are given below. Solve the problems and choose the correct answers. R-L-C circuit shown in figure
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Page 289
MCQ 1.6.41
For a step-input ei , the overshoot in the output e0 will be (A) 0, since the system is not under damped (B) 5 % (C) 16 % (D) 48 %
MCQ 1.6.42
If the above step response is to be observed on a non-storage CRO, then it would be best have the ei as a (A) Step function (B) Square wave of 50 Hz (C) Square wave of 300 Hz (D) Square wave of 2.0 KHz
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YEAR 2006 MCQ 1.6.43
For a system with the transfer function 3 (s - 2) , H (s) = 2 4s - 2s + 1 the matrix A in the state space form Xo = AX + Bu is equal to V V R R S1 0 0 W S0 1 0 W (A) S 0 1 0 W (B) S 0 0 1 W SS- 1 2 - 4 WW SS- 1 2 - 4 WW XV TR V X RT 0 1 0 1 0 0 W W S S (C) S3 - 2 1 W (D) S 0 0 1 W SS1 - 2 4 WW SS- 1 2 - 4 WW X X T T YEAR 2006
MCQ 1.6.44
ONE MARK
TWO MARKS
Consider the following Nyquist plots of loop transfer functions over w = 0 to w = 3 . Which of these plots represent a stable closed loop system ?
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(A) (1) only (C) all, except (3) MCQ 1.6.45
The Bode magnitude plot H (jw) =
Page 290
(B) all, except (1) (D) (1) and (2) only 10 4 (1 + jw) is (10 + jw) (100 + jw) 2
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MCQ 1.6.46
A closed-loop system has the characteristic function (s2 - 4) (s + 1) + K (s - 1) = 0 . Its root locus plot against K is
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Page 291
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ONE MARK
MCQ 1.6.47
A system with zero initial conditions has the closed loop transfer function. s2 + 4 T (s) = (s + 1) (s + 4) The system output is zero at the frequency (A) 0.5 rad/sec (B) 1 rad/sec (C) 2 rad/sec (D) 4 rad/sec
MCQ 1.6.48
Figure shows the root locus plot (location of poles not given) of a third order system whose open loop transfer function is
(A) K3 s
MCQ 1.6.49
K s2 (s + 1) K K (C) (D) s (s2 + 1) s (s2 - 1) The gain margin of a unity feed back control system with the open loop transfer (B)
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function G (s) =
(s + 1) is s2
(A) 0 (C)
1 2 (D) 3
(B) 2
YEAR 2005 MCQ 1.6.50
Page 292
TWO MARKS
A unity feedback system, having an open loop gain K (1 - s) , G (s) H (s) = (1 + s) becomes stable when (A) K > 1 (C) K < 1
MCQ 1.6.51
(B) K > 1 (D) K < - 1
When subject to a unit step input, the closed loop control system shown in the figure will have a steady state error of
nodia.co.in (A) - 1.0 (C) 0 MCQ 1.6.52
(B) - 0.5 (D) 0.5
In the G (s) H (s)-plane, the Nyquist plot of the loop transfer function G (s) H (s) = pes passes through the negative real axis at the point (A) (- 0.25, j0) (B) (- 0.5, j0) (C) 0 (D) 0.5 -0.25s
MCQ 1.6.53
If the compensated system shown in the figure has a phase margin of 60c at the crossover frequency of 1 rad/sec, then value of the gain K is
(A) 0.366 (C) 1.366
(B) 0.732 (D) 2.738
Data for Q.54 and Q.55 are given below. Solve the problem and choose the correct answer. 0 1 1 X (t) + = Gu (t) with the initial G 0 -3 0 T condition X (0) = [- 1, 3] and the unit step input u (t) has
A state variable system Xo (t) = = MCQ 1.6.54
The state transition matrix
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MCQ 1.6.55
1 (A) = 0
1 3
(1 - e- 3t) G e- 3t
1 (B) > 0
1 3
1 (C) > 0
1 3
(e3 - t - e- 3t) H e- 3t
1 (D) > 0
(1 - e- t) H e- t
Page 293
(e- t - e- 3t) H e- t
The state transition equation t - e-t (A) X (t) = = - t G e
1 - e-t (B) X (t) = = - 3t G 3e
t - e 3t (C) X (t) = = - 3t G 3e
t - e - 3t (D) X (t) = = - t G e
YEAR 2004
ONE MARK
MCQ 1.6.56
The Nyquist plot of loop transfer function G (s) H (s) of a closed loop control system passes through the point (- 1, j 0) in the G (s) H (s)plane. The phase margin of the system is (A) 0c (B) 45c (C) 90c (D) 180c
MCQ 1.6.57
Consider the function, 5 F (s) = 2 s (s + 3s + 2) where F (s) is the Laplace transform of the of the function f (t). The initial value of f (t) is equal to (A) 5 (B) 25
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(C) MCQ 1.6.58
5 3
(D) 0
For a tachometer, if q (t) is the rotor displacement in radians, e (t) is the output voltage and Kt is the tachometer constant in V/rad/sec, then the transfer function, E (s) will be Q (s) (A) Kt s2 (C) Kt s
(B) Kt s (D) Kt
YEAR 2004
TWO MARKS
MCQ 1.6.59
For the equation, s3 - 4s2 + s + 6 = 0 the number of roots in the left half of s -plane will be (A) Zero (B) One (C) Two (D) Three
MCQ 1.6.60
For the block diagram shown, the transfer function
2 (A) s +2 1 s
C (s) is equal to R (s)
2 (B) s + s2 + 1 s
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MCQ 1.6.61
MCQ 1.6.62
Page 294
1 s2 + s + 1 o = AX where The state variable description of a linear autonomous system is, X X is the two dimensional state vector and A is the system matrix given by 0 2 . The roots of the characteristic equation are A == 2 0G (A) - 2 and + 2 (B) - j2 and + j2 (C) - 2 and - 2 (D) + 2 and + 2 (D)
The block diagram of a closed loop control system is given by figure. The values of K and P such that the system has a damping ratio of 0.7 and an undamped natural frequency wn of 5 rad/sec, are respectively equal to
(A) 20 and 0.3 (C) 25 and 0.3
(B) 20 and 0.2 (D) 25 and 0.2
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MCQ 1.6.63
The unit impulse response of a second order under-damped system starting from rest is given by c (t) = 12.5e - 6t sin 8t, t $ 0 . The steady-state value of the unit step response of the system is equal to (A) 0 (B) 0.25 (C) 0.5 (D) 1.0
MCQ 1.6.64
In the system shown in figure, the input x (t) = sin t . In the steady-state, the response y (t) will be
(A)
1 sin (t - 45c) 2
(C) sin (t - 45c) MCQ 1.6.65
(B)
1 sin (t + 45c) 2
(D) sin (t + 45c)
The open loop transfer function of a unity feedback control system is given as 1. G (s) = as + s2 The value of ‘a ’ to give a phase margin of 45c is equal to (A) 0.141 (B) 0.441 (C) 0.841 (D) 1.141 YEAR 2003
MCQ 1.6.66
ONE MARK
A control system is defined by the following mathematical relationship d2 x + 6 dx + 5x = 12 (1 - e - 2t) dt dt2 The response of the system as t " 3 is (A) x = 6 (B) x = 2 (C) x = 2.4 (D) x =- 2
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MCQ 1.6.68
A lead compensator used for a closed loop controller has the following transfer function K (1 + as ) (1 + bs ) For such a lead compensator (A) a < b (B) b < a (C) a > Kb (D) a < Kb 2 A second order system starts with an initial condition of = G without any external 3 e - 2t 0 input. The state transition matrix for the system is given by = G. The state 0 e-t of the system at the end of 1 second is given by 0.271 0.135 (A) = (B) = G 1.100 0.368G 0.271 (C) = 0.736G
0.135 (D) = 1.100 G
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YEAR 2003 MCQ 1.6.69
Page 295
TWO MARKS
A control system with certain excitation is governed by the following mathematical equation d2 x + 1 dx + 1 x = 10 + 5e- 4t + 2e- 5t 2 dt 18 dt2 The natural time constant of the response of the system are (A) 2 sec and 5 sec (B) 3 sec and 6 sec (C) 4 sec and 5 sec (D) 1/3 sec and 1/6 sec
MCQ 1.6.70
The block diagram shown in figure gives a unity feedback closed loop control system. The steady state error in the response of the above system to unit step input is
(A) 25% (C) 6% MCQ 1.6.71
The roots of the closed loop characteristic equation of the system shown above (Q-5.55)
(A) - 1 and - 15 (C) - 4 and - 15 MCQ 1.6.72
(B) 0.75 % (D) 33%
(B) 6 and 10 (D)- 6 and - 10
The following equation defines a separately excited dc motor in the form of a differential equation
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Page 296
d2 w + B dw + K2 w = K V dt J dt LJ LJ a The above equation may be organized in the state-space form as follows R 2 V Sd w W dw S dt2 W = P dt + QV > H a S dw W w S dt W Where Tthe PX matrix is given by K - B - LJ - K - BJ (B) = LJ (A) = J G G 0 1 1 0 2
0 1 (C) =- K - B G LJ J 2
2
1 0 (D) =- B - K G J LJ 2
MCQ 1.6.73
The loop gain GH of a closed loop system is given by the following expression K s (s + 2) (s + 4) The value of K for which the system just becomes unstable is (A) K = 6 (B) K = 8 (C) K = 48 (D) K = 96
MCQ 1.6.74
The asymptotic Bode plot of the transfer function K/ [1 + (s/a)] is given in figure. The error in phase angle and dB gain at a frequency of w = 0.5a are respectively
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(A) 4.9c, 0.97 dB (C) 4.9c, 3 dB MCQ 1.6.75
(B) 5.7c, 3 dB (D) 5.7c, 0.97 dB
The block diagram of a control system is shown in figure. The transfer function G (s) = Y (s) /U (s) of the system is
(A)
1 s 18`1 + j`1 + s j 12 3
(B)
1 s 27`1 + j`1 + s j 6 9
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(C)
1 s 27`1 + j`1 + s j 12 9
(D)
1 s 27`1 + j`1 + s j 9 3
YEAR 2002 MCQ 1.6.76
Page 297
ONE MARK
The state transition matrix for the system Xo = AX with initial state X (0) is (A) (sI - A) - 1 (B) eAt X (0) (C) Laplace inverse of [(sI - A) - 1] (D) Laplace inverse of [(sI - A) - 1 X (0)] YEAR 2002
MCQ 1.6.77
MCQ 1.6.78
MCQ 1.6.79
MCQ 1.6.80
TWO MARKS
2 3 1 X + = Gu , which of the following statements is true ? G 0 5 0 (A) The system is controllable but unstable (B) The system is uncontrollable and unstable (C) The system is controllable and stable (D) The system is uncontrollable and stable For the system Xo = =
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A unity feedback system has an open loop transfer function, G (s) = K2 . The root s locus plot is
The transfer function of the system described by d2 y dy + = du + 2u 2 dt dt dt with u as input and y as output is (s + 2) (s + 1) (A) 2 (B) 2 (s + s) (s + s) (C) 2 2 (D) 22s (s + s) (s + s) For the system 2 0 1 Xo = = X + = Gu ; Y = 84 0B X, 0 4G 1 with u as unit impulse and with zero initial state, the output y , becomes (A) 2e2t (B) 4e2t (C) 2e 4t
(D) 4e 4t
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MCQ 1.6.81
The eigen values of the system represented by R0 1 0 0 V S W 0 0 1 0W S Xo = S X are 0 0 0 1W W (A) 0, 0, 0, 0SS (B) 1, 1, 1, 1 0 0 0 1W T 1 X (C) 0, 0, 0, (D) 1, 0, 0, 0
MCQ 1.6.82
*A single input single output system with y as output and u as input, is described by d2 y dy du 2 + 2 dt + 10y = 5 dt - 3u dt for an input u (t) with zero initial conditions the above system produces the same output as with no input and with initial conditions dy (0-) =- 4 , y (0-) = 1 dt input u (t) is (B) 1 d (t) - 7 e- 3t u (t) (A) 1 d (t) - 7 e(3/5)t u (t) 5 25 5 25 (C) - 7 e- (3/5)t u (t) (D) None of these 25
MCQ 1.6.83
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*A system is described by the following differential equation d2 y dy + - 2y = u (t) e- t dt dt2 dy the state variables are given as x1 = y and x2 = b - y l et , the state dt variable representation of the system is 1 1 x1 1 1 xo1 xo1 1 e- t x 1 (B) > o H = > H> H + > H u (t) (A) > o H = > - tH> H + > H u (t) 0 1 x2 0 0 x2 x2 0 e x2 1 xo1 1 e- t x 1 (C) > o H = > >x H + >0H u (t) H x2 0 -1 2
(D) none of these
Common Data Question Q.84-86*.
MCQ 1.6.84
MCQ 1.6.85
MCQ 1.6.86
The open loop transfer function of a unity feedback system is given by 2 (s + a) G (s) = s (s + 2) (s + 10) Angles of asymptotes are (A) 60c, 120c, 300c (B) 60c, 180c, 300c (C) 90c, 270c, 360c (D) 90c, 180c, 270c Intercepts of asymptotes at the real axis is (A) - 6
(B) - 10 3
(C) - 4
(D) - 8
Break away points are (A) - 1.056 , - 3.471 (C) - 1.056, - 6.9433
(B) - 2.112, - 6.9433 (D) 1.056, - 6.9433
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YEAR 2001 MCQ 1.6.87
ONE MARK
The polar plot of a type-1, 3-pole, open-loop system is shown in Figure The closed-loop system is
(A) always stable (B) marginally stable (C) unstable with one pole on the right half s -plane (D) unstable with two poles on the right half s -plane. MCQ 1.6.88
-3 1 Given the homogeneous state-space equation xo = = x the steady state 0 - 2G T value of xss = lim x (t), given the initial state value of x (0) = 810 - 10B is
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0 (A) xss = = G 0
-3 (B) xss = = G -2
- 10 (C) xss = = 10 G
3 (D) xss = = G 3
YEAR 2001 MCQ 1.6.89
TWO MARKS
The asymptotic approximation of the log-magnitude versus frequency plot of a minimum phase system with real poles and one zero is shown in Figure. Its transfer functions is
(A)
20 (s + 5) s (s + 2) (s + 25)
(C)
20 (s + 5) s (s + 2) (s + 25) 2
10 (s + 5) (s + 2) 2 (s + 25) 50 (s + 5) (D) 2 s (s + 2) (s + 25) (B)
Common Data Question Q.90-93*. A unity feedback system has an open-loop transfer function of G (s) = 10000 2 s (s + 10)
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MCQ 1.6.90
Determine the magnitude of G (jw) in dB at an angular frequency of w = 20 rad/ sec. (A) 1 dB (B) 0 dB (C) - 2 dB (D) 10 dB
MCQ 1.6.91
The phase margin in degrees is (A) 90c (C) - 36.86c
(B) 36.86c (D) - 90c
The gain margin in dB is (A) 13.97 dB (C) - 13.97 dB
(B) 6.02 dB (D) None of these
The system is (A) Stable (C) Marginally stable
(B) Un-stable (D) can not determined
MCQ 1.6.92
MCQ 1.6.93
MCQ 1.6.94
*For the given characteristic equation s3 + s2 + Ks + K = 0 The root locus of the system as K varies from zero to infinity is
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Page 301
SOLUTION
SOL 1.6.1
Option (B) is correct. From the given plot, we obtain the slope as Slope =
20 log G2 - 20 log G1 log w2 - log w1
From the figure 20 log G2 20 log G1 and w1 w2 So, the slope is
=- 8 dB = 32 dB = 1 rad/s = 10 rad/s
Slope = - 8 - 32 log 10 - log 1 =- 40 dB/decade Therefore, the transfer function can be given as G ^s h = k2 S at w = 1 G ^ jwh = k 2 = k w In decibel, 20 log G ^ jwh = 20 log k = 32 or, k = 10 = 39.8 Hence, the Transfer function is G ^s h = k2 = 392.8 s s
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SOL 1.6.2
20
Option (B) is correct. The Laplace transform of unit step fun n is U ^s h = 1 s So, the O/P of the system is given as Y ^s h = b 1 lb 1 l s s = 12 s For zero initial condition, we check dy ^ t h u^t h = dt & U ^s h = SY ^s h - y ^0 h & U ^s h = s c 12 m - y ^0 h s 1 or, U ^s h = s Hence, the O/P is correct which is Y ^s h = 12 s
^y ^0 h = 0h
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Page 302
its inverse Laplace transform is given by y ^ t h = tu ^ t h SOL 1.6.3
Option (A) is correct. For the given SFG, we have two forward paths Pk1 = ^1 h^s-1h^s-1h^1 h = s-2 Pk2 = ^1 h^s-1h^1 h^1 h = s-1 since, all the loops are touching to both the paths Pk1 and Pk2 so, Dk 1 = Dk 2 = 1 Now, we have D = 1 - (sum of individual loops) + (sum of product of nontouching loops) Here, the loops are L1 = ^- 4h^1 h =- 4
L2 = ^- 4h^s-1h = 4s-1 L 3 = ^- 2h^s-1h^s-1h =- 2s-2 L 4 = ^- 2h^s-1h^1 h =- 2s-1 As all the loop L1, L2, L 3 and L 4 are touching to each other so,
nodia.co.in D = 1 - ^L1 + L2 + L 3 + L 4h
= 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h = 5 + 6s-1 + 2s-2 From Mason’s gain formulae Y ^s h = SPk Dk D U ^s h
s-2 + s-1 5 + 6s-1 + 2s-2 = 2s+1 5s + 6s + 2 Option (C) is correct. Given, open loop transfer function G ^s h = 10Ka = Ka 1 1 + 10s s + 10 By taking inverse Laplace transform, we have g ^ t h = e- t =
SOL 1.6.4
1 10
Comparing with standard form of transfer function, Ae-t/t , we get the open loop time constant, tol = 10 Now, we obtain the closed loop transfer function for the given system as G ^s h 10Ka H ^s h = = 1 + 10 s + 10Ka 1 + G ^s h Ka = s + ^Ka + 101 h By taking inverse laplace transform, we get h ^ t h = ka .e-^k + ht So, the time constant of closed loop system is obtained as tcl = 1 1 ka + 10 a
1 10
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(approximately) tcl = 1 ka Now, given that ka reduces open loop time constant by a factor of 100. i.e., tcl = tol 100 1 = 10 or, 100 ka or, ka = 10 or,
SOL 1.6.5
Option (A) is correct. Given, the state variable formulation, - 2 0 x1 1 xo1 > o H = > 0 - 1H>x H + >1H u x2 2 x1 and y = 61 0@> H x2
....(1) ....(2)
From Eq. (1) we get xo1 = 2x1 + u Taking Laplace transform So,
nodia.co.in ^s + 2h X1 = s1
^x1 ^0 h = 0h
1 s ^s + 2h Now, from Eq. (2) we have y = x1 Taking Laplace transform both the sides, Y = XL 1 or, Y = s ^s + 2h or, Y = 1 ;1 - 1 E 2 s s+2 Taking inverse Laplace transform y = 1 8u ^ t h - e-2t u ^ t hB 2 1 = - 1 e-2t 2 2 Option (A) is correct. From the given state variable system, we have -2 0 A => 0 1H or,
SOL 1.6.6
sX1 - x1 ^0 h =- 2X1 + 1 (Here, X1 denotes Laplace transform of x1 ) s
X1 =
....(3)
(from eq. (3))
^for t > 0h
1 B = > H; C = 61 0@ 1 Now, we obtain the controllability matrix CM = 6B : AB@ and
1 -2 => 2 1H and the observability matrix is obtained as C OM = > H CA
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Page 304
1 0 H => -2 0 So, we get Rank of the controllability matrix " Rank ^CM h = 2 Rank of the observability matrix " Rank ^OM h = 1 Since, the order of state variable is 2 ^x1 and x2h. Therefore, we have Rank ^CM h = order of state variables but, Rank (OM ) < order of state variables Thus, system is controllable but not observable SOL 1.6.7
Option (D) is correct. General form of state equations are given as xo = Ax + Bu yo = Cx + Du For the given problem R0V R 0 a 0V 1 S W W S A = S 0 0 a2W, B = S0W SS1WW SSa 0 0WW 3 XVR V R VT X TR 0 a 0 1 WS0W S 0W S AB = S 0 0 a2WS0W = Sa2W SSa 0 0WWSS1WW SS 0WW 3 XT X VTR XV R TR 0 0 a1 a2WS0W Sa1 a2VW S A2 B = Sa2 a 3 0 0WS0W = S 0W SS 0 a a 0WWSS1WW SS 0WW 3 1 X XT X T T For controllability it is necessary that following matrix has a tank of n = 3 . R0 0 a a V 1 2W S 2 U = 6B : AB : A B@ = S0 a2 0W SS1 0 0WW So, a2 ! 0 X T a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not.
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SOL 1.6.8
Option (A) is correct. K (s + 1) Y (s) = 3 [R (s) - Y (s)] s + as2 + 2s + 1 K (s + 1) K (s + 1) R (s) = 3 E 2 s + as2 + 2s + 1 s + as + 2s + 1 Y (s) [s3 + as2 + s (2 + k) + (1 + k)] = K (s + 1) R (s) Transfer Function, K (s + 1) Y (s) = 3 H (s) = 2 R (s) s + as + s (2 + k) + (1 + k) Routh Table : Y (s) ;1 +
3
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SOL 1.6.9
Page 305
For oscillation, a (2 + K) - (1 + K) =0 a a = K+1 K+2 Auxiliary equation A (s) = as2 + (k + 1) = 0 s2 =- k + 1 a k + 1 (k + 2) =- (k + 2) 2 s = (k + 1) s = j k+2 jw = j k + 2 (Oscillation frequency) w = k+2 = 2 k =2 and a = 2 + 1 = 3 = 0.75 2+2 4 Option (A) is correct. jw + a GC (s) = s + a = s+b jw + b Phase lead angle, f = tan-1 a w k - tan-1 a w k a b Jw - wN -1 K a bO = tan 2 KK OO w 1+ ab L P w (b - a) = tan-1 c ab + w 2 m For phase-lead compensation f > 0 b-a > 0 b >a Note: For phase lead compensator zero is nearer to the origin as compared to pole, so option (C) can not be true.
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SOL 1.6.10
Option (A) is correct. f = tan-1 a w k - tan-1 a w k a b 1/a 1/b df = =0 2 2 dw 1 +awk 1 +awk a b 2 2 1 + w = 1+1w a ab2 b b a2 1 - 1 = w2 1 - 1 a b ab b a b l w = ab = 1 # 2 =
SOL 1.6.11
2 rad/ sec
Option (A) is correct. Gain margin is simply equal to the gain at phase cross over frequency ( wp ). Phase cross over frequency is the frequency at which phase angle is equal to - 180c. From the table we can see that +G (jwp) =- 180c, at which gain is 0.5. 1 GM = 20 log 10 e = 20 log b 1 l = 6 dB 0.5 G (jwp) o
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Phase Margin is equal to 180c plus the phase angle fg at the gain cross over frequency ( wg ). Gain cross over frequency is the frequency at which gain is unity. From the table it is clear that G (jwg) = 1, at which phase angle is - 150c fPM = 180c + +G (jwg) = 180 - 150 = 30c SOL 1.6.12
Option (A) is correct. We know that steady state error is given by sR (s) ess = lim s " 0 1 + G (s) where
R (s) " input G (s) " open loop transfer function For unit step input R (s) = 1 s sb 1 l s So ess = lim = 0.1 s " 0 1 + G (s) 1 + G (0) = 10 G (0) = 9 Given inputr (t)
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R (s) = 10 :1 - 1 e-sD = 10 :1 - e D s s s So steady state error (1 - e-s) s # 10 10 (1 - e0) s el = =0 ss = lim 1+9 s"0 1 + G (s) Option (B) is correct. Transfer function having at least one zero or pole in RHS of s -plane is called nonminimum phase transfer function. s-1 G (s) = (s + 2) (s + 3) • In the given transfer function one zero is located at s = 1 (RHS), so this is a non-minimum phase system. • Poles - 2, - 3 , are in left side of the complex plane, So the system is stable or
SOL 1.6.13
SOL 1.6.14
= 10 [m (t) - m (t - 1)]
-s
Option (A) is correct. K bs + 2 l 3 G (s) = 2 s (s + 2) Steps for plotting the root-locus (1) Root loci starts at s = 0, s = 0 and s =- 2 (2) n > m , therefore, number of branches of root locus b = 3 (3) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n-m (2 # 0 + 1) 180c (I) = 90c (3 - 1) (2 # 1 + 1) 180c (II) = 270c (3 - 1) (4) The two asymptotes intersect on real axis at centroid
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- 2 - b- 2 l 3 x = =- 2 n-m 3-1 3 (5) Between two open-loop poles s = 0 and s =- 2 there exist a break away point. s2 (s + 2) K =2 bs + 3 l dK = 0 ds
/ Poles - / Zeroes =
s =0 Root locus is shown in the figure
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Three roots with nearly equal parts exist on the left half of s -plane. SOL 1.6.15
Option (A) is correct. The system may be reduced as shown below
1 s (s + 1 + K ) Y (s) 1 = 2 = 1 R (s) 1 + s + s (1 + K ) + 1 s (s + 1 + K ) This is a second order system transfer function, characteristic equation is s2 + s (1 + K) + 1 = 0 Comparing with standard form s2 + 2xwn s + wn2 = 0 We get x = 1+K 2 Peak overshoot M p = e- px/
1 - x2
So the Peak overshoot is effected by K . SOL 1.6.16
Option (A) is correct. Given
G (s) =
1 s (s + 1) (s + 2)
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1 jw (jw + 1) (jw + 2) 1 G (jw) = 2 w w + 1 w2+ 4 +G (jw) =- 90c - tan- 1 (w) - tan- 1 (w/2) In nyquist plot For w = 0, G (jw) = 3 G (jw) =
+G (jw) =- 90c For w = 3, G (jw) = 0 +G (jw) =- 90c - 90c - 90c =- 270c Intersection at real axis 1 1 G (jw) = = jw (jw + 1) (jw + 2) jw (- w2 + j3w + 2) - 3w2 - jw (2 - w2) 1 = # - 3w2 + jw (2 - w2) - 3w2 - jw (2 - w2) - 3w2 - jw (2 - w2) = 9w4 + w2 (2 - w2) 2 2 jw (2 - w2) = 4 -23w 9w + w (2 - w2) 2 9w4 + w2 (2 - w2) 2 At real axis
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Im [G (jw)] = 0 w (2 - w2) So, =0 9w4 + w2 (2 - w2) 2 - w2 = 0 & w = 2 rad/sec At w = 2 rad/sec, magnitude response is 1 G (jw) at w = 2 = =1 -> A - lI = > H H 0 2 - lH 0 2 0 l A - lI = (- 1 - l) (2 - l) - 2 # 0 = 0 & l1, l2 =- 1, 2 Since eigen values of the system are of opposite signs, so it is unstable Controllability : 0 -1 2 , B=> H A => H 1 0 2 2 AB = > H 2 0 2 [B: AB] = > H 1 2 Y 0 6B: AB@ = So it is controllable.
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Page 309
Option (C) is correct. Given characteristic equation s (s + 1) (s + 3) + K (s + 2) = 0 ; s (s2 + 4s + 3) + K (s + 2) = 0 s3 + 4s2 + (3 + K) s + 2K = 0 From Routh’s tabulation method
K>0
s3
1
3+K
s2
4
2K
s1
4 (3 + K) - 2K (1) 12 + 2K = >0 4 4
s0
2K
There is no sign change in the first column of routh table, so no root is lying in right half of s -plane. For plotting root locus, the equation can be written as K (s + 2) =0 1+ s (s + 1) (s + 3)
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Open loop transfer function
G (s) =
K (s + 2) s (s + 1) (s + 3)
Root locus is obtained in following steps: 1. No. of poles n = 3 , at s = 0, s =- 1 and s =- 3 2. No. of Zeroes m = 1, at s =- 2 3. The root locus on real axis lies between s = 0 and s =- 1, between s =- 3 and s =- 2 . 4. Breakaway point lies between open loop poles of the system. Here breakaway point lies in the range - 1 < Re [s] < 0 . 5. Asymptotes meet on real axis at a point C , given by C =
/ real part of poles - / real parts of zeroes
n-m (0 - 1 - 3) - (- 2) = 3-1
=- 1 As no. of poles is 3, so two root loci branches terminates at infinity along asymptotes Re (s) =- 1 SOL 1.6.19
SOL 1.6.20
Option (D) is correct. Overall gain of the system is written as G = G1 G 2 1 G3 We know that for a quantity that is product of two or more quantities total percentage error is some of the percentage error in each quantity. so error in overall gain G is 3 G = e1 + e2 + 1 e3 Option (D) is correct.
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From Nyquist stability criteria, no. of closed loop poles in right half of s -plane is given as Z = P-N P " No. of open loop poles in right half s -plane N " No. of encirclement of (- 1, j0)
Here N =- 2 (` encirclement is in clockwise direction) P = 0 (` system is stable) So, Z = 0 - (- 2) Z = 2 , System is unstable with 2-poles on RH of s -plane. SOL 1.6.21
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Option (D) is correct. Given Routh’s tabulation. s3
2
2
s2
4
4
s1
0
0
So the auxiliary equation is given by, 4s 2 + 4 = 0 s2 =- 1 s =! j From table we have characteristic equation as 2s3 + 2s + 4s2 + 4 = 0 s3 + s + 2s2 + 2 = 0 s (s2 + 1) + 2 (s2 + 1) = 0 (s + 2) (s2 + 1) = 0 s =- 2 , s = ! j SOL 1.6.22
Option (B) is correct. Since initial slope of the bode plot is - 40 dB/decade, so no. of poles at origin is 2. Transfer function can be written in following steps: 1. Slope changes from - 40 dB/dec. to - 60 dB/dec. at w1 = 2 rad/sec., so at w1 there is a pole in the transfer function. 2. Slope changes from - 60 dB/dec to - 40 dB/dec at w2 = 5 rad/sec., so at this frequency there is a zero lying in the system function. 3. The slope changes from - 40 dB/dec to - 60 dB/dec at w3 = 25 rad/sec, so there is a pole in the system at this frequency. Transfer function T (s) =
K (s + 5) s (s + 2) (s + 25) 2
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Constant term can be obtained as. T (jw) at w = 0.1 = 80 So,
80 = 20 log
K (5) (0.1) 2 # 50
K = 1000 therefore, the transfer function is 1000 (s + 5) T (s) = 2 s (s + 2) (s + 25) SOL 1.6.23
Option (D) is correct. From the figure we can see that steady state error for given system is ess = 1 - 0.75 = 0.25 Steady state error for unity feed back system is given by sR (s) ess = lim = G s " 0 1 + G (s) s ^ 1s h ; R (s) = 1 (unit step input) = lim s s"0> K H 1+ (s + 1) (s + 2) = 1K = 2 2+K 1+ 2 So, ess = 2 = 0.25 2+K
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SOL 1.6.24
Option (D) is correct. Open loop transfer function of the figure is given by, - 0.1s G (s) = e s - j0.1w G (jw) = e jw Phase cross over frequency can be calculated as,
+G (jwp) =- 180c 180 b- 0.1wp # p l - 90c =- 180c 0.1wp # 180c = 90c p 0.1wp = 90c # p 180c wp = 15.7 rad/sec So the gain margin (dB) 1 = 20 log e = 20 log G (jwp) o >
1 1 b 15.7 l H
= 20 log 15.7 = 23.9 dB SOL 1.6.25
Option (C) is correct. Given system equations dx1 (t) =- 3x1 (t) + x2 (t) + 2u (t) dt
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dx2 (t) =- 2x2 (t) + u (t) dt y (t) = x1 (t) Taking Laplace transform on both sides of equations. sX1 (s) =- 3X1 (s) + X2 (s) + 2U (s) (s + 3) X1 (s) = X2 (s) + 2U (s) Similarly sX2 (s) =- 2X2 (s) + U (s) (s + 2) X2 (s) = U (s) From equation (1) & (2) U (s) + 2U (s) (s + 3) X1 (s) = s+2 U (s) 1 + 2 (s + 2) X1 (s) = E s + 3; s + 2 (2s + 5) = U (s) (s + 2) (s + 3) From output equation,
...(1) ...(2)
Y (s) = X1 (s)
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So,
SOL 1.6.26
Y (s) = U (s)
(2s + 5) (s + 2) (s + 3)
System transfer function (2s + 5) Y (s) (2s + 5) = 2 T.F = = U (s) (s + 2) (s + 3) s + 5s + 6 Option (B) is correct. Given state equations in matrix form can be written as, - 3 1 x1 2 xo1 > o H = > 0 - 2H>x H + >1H u (t) x2 2 dX (t) = AX (t) + Bu (t) dt State transition matrix is given by f (t) = L- 1 6F (s)@ F (s) = (sI - A) - 1 s 0 -3 1 (sI - A) = > H - > 0 s 0 - 2H s + 3 -1 (sI - A) = > 0 s + 2H s+2 1 1 > 0 s + 3H (s + 3) (s + 2) R V 1 S 1 W (s + 3) (s + 3) (s + 2)W S -1 So F (s) = (sI - A) = S W 1 S 0 (s + 2) W T - 3t - 2t X e e - e- 3t -1 f (t) = L [F (s)] = > H 0 e- 2t (sI - A) - 1 =
SOL 1.6.27
Option (D) is correct. Given differential equation for the function dy (t) + y (t) = d (t) dt
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Taking Laplace on both the sides we have, sY (s) + Y (s) = 1 (s + 1) Y (s) = 1 Y (s) =
1 s+1
Taking inverse Laplace of Y (s) y (t) = e- t u (t), t > 0 SOL 1.6.28
Option (A) is correct. Given transfer function G (s) =
1 s + 3s + 2 2
r (t) = d (t - 1) R (s) = L [d (t - 1)] = e- s Output is given by -s Y (s) = R (s) G (s) = 2 e s + 3s + 2 Steady state value of output -s =0 lim y (t) = lim sY (s) = lim 2 se t"3 s"0 s " 0 s + 3s + 2 Input
SOL 1.6.29
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Option (A) is correct. For C1 Phase is given by
qC = tan- 1 (w) - tan- 1 a w k 10 Jw - w N -1 K 10 O = tan 2 KK O w 1+ O 10 P L = tan- 1 c 9w 2 m > 0 (Phase lead) 10 + w Similarly for C2 , phase is qC2 = tan- 1 a w k - tan- 1 (w) 10 J w - wN - 1 K 10 O = tan 2 KK O w 1+ O 10 P L = tan- 1 c - 9w 2 m < 0 (Phase lag) 10 + w Option (C) is correct. From the given bode plot we can analyze that: 1. Slope - 40 dB/decade"2 poles 2. Slope - 20 dB/decade (Slope changes by + 20 dB/decade)"1 Zero 3. Slope 0 dB/decade (Slope changes by + 20 dB/decade)"1 Zero 1
SOL 1.6.30
So there are 2 poles and 2 zeroes in the transfer function. SOL 1.6.31
Option (C) is correct. Characteristic equation for the system K =0 1+ s (s + 3) (s + 10)
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s (s + 3) (s + 10) + K = 0 s3 + 13s2 + 30s + K = 0 Applying Routh’s stability criteria s3
1
30
s2
13
K
s1
(13 # 30) - K 13 K
s0
For stability there should be no sign change in first column So, 390 - K > 0 & K < 390 K >0 0 < K < 90 SOL 1.6.32
Option (C) is correct. Given transfer function is 100 s2 + 20s + 100 Characteristic equation of the system is given by H (s)) =
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or
s2 + 20s + 100 = 0 wn2 = 100 & wn = 10 rad/sec. 2xwn = 20 x = 20 = 1 2 # 10
(x = 1) so system is critically damped. SOL 1.6.33
Option (D) is correct. State space equation of the system is given by, o = AX + Bu X Y = CX Taking Laplace transform on both sides of the equations.
So
sX (s) (sI - A) X (s) X (s) ` Y (s) Y (s) T.F =
= AX (s) + BU (s) = BU (s) = (sI - A) - 1 BU (s) = CX (s) = C (sI - A) - 1 BU (s)
Y (s) = C (sI - A) - 1 B U (s)
s 0 0 1 s -1 => (sI - A) = > H - > H 0 s 0 -2 0 s + 2H R V 1 W S1 1 >s + 2 1H = Ss s (s + 2)W (sI - A) - 1 = S0 1 W s (s + 2) 0 s S (s + 2) W T X Transfer function
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V V R R 1 W S1 S 1 W s s (s + 2)W 0 Ss (s + 2)W = 1 0 G (s) = C [sI - A] - 1 B = 81 0BSS > H 8 B S 1 W 1 W1 S0 (s + 2) W S (s + 2) W X X T T 1 = s (s + 2) SOL 1.6.34
Option (A) is correct. Steady state error is given by,
Here
SOL 1.6.35
sR (s) ess = lim = G s " 0 1 + G (s) H (s) R (s) = L [r (t)] = 1 (Unit step input) s 1 G (s) = s (s + 2)
H (s) = 1 (Unity feed back) V R sb 1 l W S s W So, ess = lim S 1 s"0S W + 1 S s (s + 2) W X T s (s + 2) = lim = G =0 s " 0 s (s + 2) + 1 Option (D) is correct. For input u1 , the system is (u2 = 0)
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System response is (s - 1) (s - 1) (s + 2) H1 (s) = = (s - 1) 1 (s + 3) 1+ (s + 2) (s - 1) Poles of the system is lying at s =- 3 (negative s -plane) so this is stable. For input u2 the system is (u1 = 0)
System response is 1 (s - 1) (s + 2) H2 (s) = = 1 s ( ) ( s 1) (s + 3) 1+ 1 (s - 1) (s + 2) One pole of the system is lying in right half of s -plane, so the system is unstable.
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Page 316
Option (B) is correct. Given function is. 1 s (s + 1) (s + 2) 1 G (jw) = jw (1 + jw) (2 + jw) G (s) =
By simplifying - jw 1 - jw 2 - jw 1 1 G (jw) = c 1 # jw - jw mc 1 + jw # 1 - jw mc 2 + jw # 2 - jw m = c-
- jw (2 - w2 - j3w) jw 1 - j w 2 - j w = w2 mc 1 + w2 mc 4 + w2 m w2 (1 + w2) (4 + w2)
jw (w2 - 2) - 3w2 + w2 (1 + w2) (4 + w2) w2 (1 + w2) (4 + w2) G (jw) = x + iy x = Re [G (jw)] w " 0 = - 3 =- 3 1#4 4 Option (D) is correct. Let response of the un-compensated system is 900 H UC (s) = s (s + 1) (s + 9) Response of compensated system. 900 HC (s) = G (s) s (s + 1) (s + 9) C =
+
SOL 1.6.37
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Where GC (s) " Response of compensator Given that gain-crossover frequency of compensated system is same as phase crossover frequency of un-compensated system So, (wg) compensated = (wp) uncompensated - 180c = +H UC (jwp) - 180c =- 90c - tan- 1 (wp) - tan- 1 a
wp 9k
J w + wp N p 9 O 90c = tan KK 2 O K 1 - wp O 9 P L 2 w 1- p = 0 9 -1
wp = 3 rad/sec. So, (wg) compensated = 3 rad/sec. At this frequency phase margin of compensated system is fPM = 180c + +HC (jwg) 45c = 180c - 90c - tan- 1 (wg) - tan- 1 (wg /9) + +GC (jwg) 45c = 180c - 90c - tan- 1 (3) - tan- 1 (1/3) + +GC (jwg) R 1 V + 3 S 3 WW + +GC (jwg) 45c = 90c - tan- 1 S SS1 - 3 b 1 lWW 3 X T
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45c = 90c - 90c + +GC (jwg) +GC (jwg) = 45c The gain cross over frequency of compensated system is lower than un-compensated system, so we may use lag-lead compensator. At gain cross over frequency gain of compensated system is unity so. HC (jwg) = 1 900 GC (jwg) wg
wg2 + 1 wg2 + 81
=1
GC (jwg) = 3 9 + 1 9 + 81 = 3 # 30 = 1 10 900 900 in dB GC (wg) = 20 log b 1 l 10 =- 20 dB (attenuation) SOL 1.6.38
Option (B) is correct. Characteristic equation for the given system, K (s + 3) =0 1+ (s + 8) 2
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(s + 8) 2 + K (s + 3) = 0 s2 + (16 + K) s + (64 + 3K) = 0 By applying Routh’s criteria. 64 + 3K s2 1 s1
16 + K
s0
64 + 3K
0
For system to be oscillatory 16 + K = 0 & K =- 16 Auxiliary equation A (s) = s2 + (64 + 3K) = 0 &
SOL 1.6.39
s2 + 64 + 3 # (- 16) = 0 s2 + 64 - 48 = 0 s2 =- 16 & jw = 4j w = 4 rad/sec
Option (D) is correct. From the given block diagram we can obtain signal flow graph of the system. Transfer function from the signal flow graph is written as c 0 P + c1 P s s2 T.F = a Pb a 0 1 + 1 + 2 - 2 0 - Pb1 s s s s (c 0 + c1 s) P (s + a1 s + a 0) - P (b 0 + sb1) (c 0 + c1 s) P 2 s ^ + a1 s + a 0 h = P (b + sb1) 1- 2 0 s + a1 s + a 0 from the given reduced form transfer function is given by =
2
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XYP 1 - YPZ by comparing above two we have T.F =
X = (c 0 + c1 s) 1 Y = 2 s + a1 s + a 0 Z = (b 0 + sb1) SOL 1.6.40
Option (A) is correct. For the given system Z is given by Z = E (s) Ki s Where E (s) " steady state error of the system Here sR (s) E (s) = lim s " 0 1 + G (s) H (s) Input
R (s) = 1 (Unit step) s
nodia.co.in w2 G (s) = b Ki + K p le 2 s s + 2xws + w2 o H (s) = 1 (Unity feed back)
So,
SOL 1.6.41
R V sb 1 l S W s Wb Ki l Z = lim S 2 s"0S Ki w W s S1 + b s + K p l (s2 + 2xws + w2) W T X Ki = lim = Ki = 1 2 s"0 >s + (Ki + K p s) 2 w H Ki 2 (s + 2xws + w )
Option (C) is correct. System response of the given circuit can be obtained as. 1 bCs l e 0 (s) H (s) = = 1 ei (s) bR + Ls + Cs l 1 b LC l 1 = H (s) = LCs2 + RCs + 1 s2 + R s + 1 L LC Characteristic equation is given by, s2 + R s + 1 = 0 L LC Here natural frequency wn = 1 LC 2xwn = R L Damping ratio x = R LC = R C 2 L 2L Here
x = 10 2
1 # 10- 3 = 0.5 (under damped) 10 # 10- 6
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Page 319
So peak overshoot is given by SOL 1.6.42 SOL 1.6.43
% peak overshoot = e Option ( ) is correct.
- px 1 - x2
# 100 = e
- p # 0.5 1 - (0.5) 2
# 100 = 16%
Option (B) is correct. In standard form for a characteristic equation give as sn + an - 1 sn - 1 + ... + a1 s + a 0 = 0 in its state variable representation matrix A is given as R V 1 0 g 0 W S 0 S 0 0 1 g 0 W A =S W Sh h h h h W S- a 0 - a1 - a2 g - an - 1W T X Characteristic equation of the system is
SOL 1.6.44
SOL 1.6.45
4s2 - 2s + 1 = 0 So, a2 = 4, a1 =- 2, a 0 = 1 R 0 1 0 VW RS 0 S A =S 0 0 1 W=S 0 SS- a - a - a WW SS- 1 0 1 2 T X T Option (A) is correct. In the given options only in option (A) circle (- 1, j0), So this is stable.
1 0 VW 0 1W 2 - 4WW X
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the nyquist plot does not enclose the unit
Option (A) is correct. Given function is, 10 4 (1 + jw) H (jw) = (10 + jw) (100 + jw) 2 Function can be rewritten as, 10 4 (1 + jw) H (jw) = 2 10 91 + j w C 10 4 91 + j w C 10 100 =
0.1 (1 + jw) w w 2 a1 + j 10 ka1 + j 100 k
The system is type 0, So, initial slope of the bode plot is 0 dB/decade. Corner frequencies are w1 = 1 rad/sec w 2 = 10 rad/sec w 3 = 100 rad/sec As the initial slope of bode plot is 0 dB/decade and corner frequency w1 = 1 rad/ sec, the Slope after w = 1 rad/sec or log w = 0 is(0 + 20) =+ 20 dB/dec. After corner frequency w2 = 10 rad/sec or log w2 = 1, the Slope is (+ 20 - 20) = 0 dB/dec. Similarly after w3 = 100 rad/sec or log w = 2 , the slope of plot is (0 - 20 # 2) =- 40 dB/dec. Hence (A) is correct option. SOL 1.6.46
Option (B) is correct.
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Given characteristic equation. (s2 - 4) (s + 1) + K (s - 1) = 0 K (s - 1) or =0 1+ 2 (s - 4) (s + 1) So, the open loop transfer function for the system. K (s - 1) , no. of poles n = 3 G (s) = (s - 2) (s + 2) (s + 1) no of zeroes m = 1 Steps for plotting the root-locus (1) Root loci starts at s = 2, s =- 1, s =- 2 (2) n > m , therefore, number of branches of root locus b = 3 (3) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n-m (I)
(2 # 0 + 1) 180c = 90c (3 - 1)
(II)
(2 # 1 + 1) 180c = 270c (3 - 1)
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(4) The two asymptotes intersect on real axis at / Poles - / Zeroes = (- 1 - 2 + 2) - (1) =- 1 x = 3-1 n-m
(5) Between two open-loop poles s =- 1 and s =- 2 there exist a break away point. (s2 - 4) (s + 1) K =(s - 1) dK = 0 ds s =- 1.5 SOL 1.6.47
Option (C) is correct. Closed loop transfer function of the given system is, s2 + 4 T (s) = (s + 1) (s + 4) T (jw) =
(jw) 2 + 4 (jw + 1) (jw + 4)
If system output is zero 4 - w2 =0 T (jw) = ^ jw + 1h (jw + 4) 4 - w2 = 0 w2 = 4 & w = 2 rad/sec SOL 1.6.48
Option (A) is correct. From the given plot we can see that centroid C (point of intersection) where asymptotes intersect on real axis) is 0 So for option (a) G (s) = K3 s
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Centroid = SOL 1.6.49
Page 321
/ Poles - / Zeros = 0 - 0 = 0 n-m
3-0
Option (A) is correct. Open loop transfer function is. (s + 1) G (s) = s2 jw + 1 G (jw) = - w2 Phase crossover frequency can be calculated as. +G (jwp) =- 180c tan- 1 (wp) =- 180c wp = 0 Gain margin of the system is. G.M =
SOL 1.6.50
1 G (jwp)
=
1 = 2 wp + 1 w2p
w2p =0 w2p + 1
Option (C) is correct. Characteristic equation for the given system
nodia.co.in 1 + G (s) H (s) = 0 (1 - s) =0 1+K (1 + s)
(1 + s) + K (1 - s) = 0 s (1 - K) + (1 + K) = 0 For the system to be stable, coefficient of characteristic equation should be of same sign. 1 - K > 0, K + 1 > 0 K < 1, K > - 1 -1 < K < 1 K H - > H = > = s2 - 4 = 0 0 s 2 0 -2 s H s1, s2 = ! 2
SOL 1.6.62
Option (D) is correct. For the given system, characteristic equation can be written as, 1 + K (1 + sP) = 0 s (s + 2) s (s + 2) + K (1 + sP) = 0 s2 + s (2 + KP) + K = 0 From the equation. wn = K = 5 rad/sec (given) So, K = 25 and 2xwn = 2 + KP 2 # 0.7 # 5 = 2 + 25P or P = 0.2 so K = 25 , P = 0.2
SOL 1.6.63
Option (D) is correct. Unit - impulse response of the system is given as, c (t) = 12.5e- 6t sin 8t , t $ 0 So transfer function of the system. H (s) = L [c (t)]
=
12.5 # 8 (s + 6) 2 + (8) 2
100 s2 + 12s + 100 Steady state value of output for unit step input, H (s) =
lim y (t) = lim sY (s) = lim sH (s) R (s)
t"3
s"0
s"0
100 1 = 1.0 = lim s ; 2 E s s"0 s + 12s + 100
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.6.64
Page 326
Option (A) is correct. System response is. H (s) = s s+1 jw H (jw) = jw + 1 Amplitude response w w+1 Given input frequency w = 1 rad/sec. 1 So H (jw) w = 1 rad/sec = = 1 1+1 2 Phase response H (jw) =
SOL 1.6.65
qh (w) = 90c - tan- 1 (w) qh (w) w = 1 = 90c - tan- 1 (1) = 45c So the output of the system is y (t) = H (jw) x (t - qh) = 1 sin (t - 45c) 2 Option (C) is correct. Given open loop transfer function jaw + 1 G (jw) = (jw) 2 Gain crossover frequency (wg) for the system.
nodia.co.in G (jwg) = 1 a2 wg2 + 1 =1 - wg2
a2 wg2 + 1 = wg4 wg4 - a2 wg2 - 1 = 0 Phase margin of the system is
...(1)
fPM = 45c = 180c + +G (jwg) 45c = 180c + tan- 1 (wg a) - 180c
SOL 1.6.66
tan- 1 (wg a) = 45c wg a = 1 From equation (1) and (2) 1 -1-1 = 0 a4 a 4 = 1 & a = 0.841 2 Option (C) is correct. Given system equation is. d 2 x + 6 dx + 5x = 12 (1 - e- 2t) dt dt 2 Taking Laplace transform on both side. s2 X (s) + 6sX (s) + 5X (s) = 12 :1 - 1 D s s+2
(2)
(s2 + 6s + 5) X (s) = 12 ; 2 E s (s + 2)
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Page 327
System transfer function is X (s) =
24 s (s + 2) (s + 5) (s + 1)
Response of the system as t " 3 is given by lim f (t) = lim sF (s) (final value theorem) t"3
s"0
24 = lim s ; s"0 s (s + 2) (s + 5) (s + 1)E = SOL 1.6.67
24 = 2.4 2#5
Option (A) is correct. Transfer function of lead compensator is given by. K a1 + s k a H (s) = s a1 + b k R w V S1 + j a a kW H (jw) = K S W SS1 + j a w kWW b T X So, phase response of the compensator is. qh (w) = tan- 1 a w k - tan- 1 a w k a b
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Jw - wN w (b - a) = tan K a b2 O = tan- 1 ; KK O ab + w2 E 1+ w O ab P L qh should be positive for phase lead compensation w (b - a) So, qh (w) = tan- 1 ; >0 ab + w2 E -1
b >a SOL 1.6.68
Option (A) is correct. Since there is no external input, so state is given by X (t) = f (t) X (0) f (t) "state transition matrix X [0] "initial condition e- 2t 0 2 So x (t) = > H> H 0 e- t 3 2e- 2t x (t) = > - t H 3e At t = 1, state of the system 0.271 2e- 2 x (t) t = 1 = > - 1H = > 1.100H 2e
SOL 1.6.69
Option (B) is correct. Given equation d2 x + 1 dx + 1 x = 10 + 5e- 4t + 2e- 5t dt2 2 dt 18 Taking Laplace on both sides we have
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Page 328
s2 X (s) + 1 sX (s) + 1 X (s) = 10 + 5 + 2 2 18 s s+4 s+5 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) (s2 + 1 s + 1 ) X (s) = 2 18 s (s + 4) (s + 5) System response is, 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) X (s) = s (s + 4) (s + 5) bs2 + 1 s + 1 l 2 18 10 (s + 4) (s + 5) + 5s (s + 5) + 2s (s + 4) s (s + 4) (s + 5) bs + 1 lbs + 1 l 3 6 We know that for a system having many poles, nearness of the poles towards imaginary axis in s -plane dominates the nature of time response. So here time constant given by two poles which are nearest to imaginary axis. Poles nearest to imaginary axis s1 =- 1 , s2 =- 1 3 6 =
nodia.co.in So, time constants )
SOL 1.6.70
t1 = 3 sec t2 = 6 sec
Option (A) is correct. Steady state error for a system is given by sR (s) ess = lim s " 0 1 + G (s) H (s) Where input R (s) = 1 (unit step) s G (s) = b 3 lb 15 l s + 15 s + 1 H (s) = 1 So
SOL 1.6.71
(unity feedback) sb 1 l s ess = lim = 15 = 15 60 45 15 + 45 s"0 1+ (s + 15) (s + 1)
%ess = 15 # 100 = 25% 60 Option (C) is correct. Characteristic equation is given by Here
So,
1 + G (s) H (s) = 0 H (s) = 1
(unity feedback) G (s) = b 3 lb 15 l s + 15 s + 1 1 + b 3 lb 15 l = 0 s + 15 s + 1
(s + 15) (s + 1) + 45 = 0 s2 + 16s + 60 = 0 (s + 6) (s + 10) = 0 s =- 6, - 10 SOL 1.6.72
Option (A) is correct.
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Page 329
Given equation can be written as, d 2 w =- b dw - K 2 w + K V J dt LJ LJ a dt 2 Here state variables are defined as, dw = x 1 dt w = x2 So state equation is 2 xo1 =- B x1 - K x2 + K Va J LJ LJ xo2 = dw = x1 dt In matrix form K/LJ xo1 - B/J - K 2 /LJ x1 >o H = > >x H + > 0 H Va H x2 1 0 2 R 2 V Sd w W S dt2 W = P >dwH + QVa dt S dw W S dt W T X So matrix P is
nodia.co.in - B/J - K 2 /LJ > 1 H 0
SOL 1.6.73
Option (C) is correct. Characteristic equation of the system is given by 1 + GH = 0 K =0 1+ s (s + 2) (s + 4) s (s + 2) (s + 4) + K = 0 s3 + 6s2 + 8s + K = 0 Applying routh’s criteria for stability s3
1
8
s2
6 K - 48 6
K
s
1
s0
SOL 1.6.74
K
System becomes unstable if K - 48 = 0 & K = 48 6 Option (A) is correct. The maximum error between the exact and asymptotic plot occurs at corner frequency. Here exact gain(dB) at w = 0.5a is given by 2
1 + w2 a (0.5a) 2 1/2 = 20 log K - 20 log ;1 + E a2 = 20 log K - 0.96 Gain(dB) calculated from asymptotic plot at w = 0.5a is gain(dB) w = 0.5a = 20 log K - 20 log
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= 20 log K Error in gain (dB) = 20 log K - (20 log K - 0.96) dB = 0.96 dB Similarly exact phase angle at w = 0.5a is. qh (w) w = 0.5a =- tan- 1 a w k =- tan- 1 b 0.5a l =- 26.56c a a Phase angle calculated from asymptotic plot at (w = 0.5a) is - 22.5c Error in phase angle =- 22.5 - (- 26.56c) = 4.9c SOL 1.6.75
Option (B) is correct. Given block diagram
Given block diagram can be reduced as
nodia.co.in 1 bs l
1 3 = s+3 1 1 +b l s 1 bs l G2 = = 1 s + 12 1 1 + b l 12 s Further reducing the block diagram. Where
G1 =
2G1 G2 1 + (2G1 G2) 9 (2) b 1 lb 1 l s + 3 s + 12 = 1 + (2) b 1 lb 1 l (9) s + 3 s + 12
Y (s) =
2 2 = 2 (s + 3) (s + 12) + 18 s + 15s + 54 1 2 = = s (s + 9) (s + 6) 27 a1 + ka1 + s k 9 6 Option (C) is correct. Given state equation is, o = AX X =
SOL 1.6.76
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Taking Laplace transform on both sides of the equation, sX (s) - X (0) = AX (s) (sI - A) X (s) = X (0) X (s) = (sI - A) - 1 X (0) = F (s) X (0) Where f (t) = L- 1 [F (s)] = L- 1 [(sI - A) - 1] is defined as state transition matrix SOL 1.6.77
Option (B) is correct. State equation of the system is given as, o = >2 3H X + >1H u X 0 5 0 Here
2 3 1 A = > H, B = > H 0 5 0
Check for controllability: 2 2 3 1 AB = > H> H = > H 0 0 5 0 1 2 U = [B : AB] = > H 0 0
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U = (1 # 0 - 2 # 0) = 0 Matrix U is singular, so the system is uncontrollable. Check for Stability: Characteristic equation of the system is obtained as, sI - A = 0 s 0 2 3 (sI - A) = > H - > H 0 s 0 5 s - 2 -3 => 0 s - 5H
sI - A = (s - 2) (s - 5) = 0 s = 2, s = 5 There are two R.H.S Poles in the system so it is unstable. SOL 1.6.78
Option (B) is correct. Given open loop transfer function, no of poles = 2 G (s) = K2 , s no of zeroes = 0 For plotting root locus: (1) Poles lie at s1, s2 = 0 (2) So the root loci starts (K = 0) from s = 0 and s = 0 (3) As there is no open-loop zero, root loci terminates (K = 3) at infinity. (4) Angle of asymptotes is given by (2q + 1) 180c , q = 0, 1 n-m So the two asymptotes are at an angle of (i)
(2 # 0 + 1) 180c = 90c 2
(2 # 1 + 1) 180c (ii) = 270c 2 (5) The asymptotes intersect on real axis at a point given by
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x=
/ Poles - / zeros n-m
Page 332
= 0-0 = 0 2
(6) Break away points 1 + K2 = 0 s K =- s2 dK =- 2s = 0 & s = 0 ds So the root locus plot is.
SOL 1.6.79
Option (A) is correct. System is described as. d2 y dy = du + 2u + dt dt2 dt
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Taking Laplace transform on both sides. s2 Y (s) + sY (s) = sU (s) + 2U (s) (s2 + s) Y (s) = (s + 2) U (s) So, the transfer function is Y (s) (s + 2) T.F = = 2 U (s) (s + s) SOL 1.6.80
Option (A) is correct. Here, we have 1 2 0 A = > H, B = > H, C = [4, 0] 1 0 4 We know that transfer function of the system is given by. Y (s) G (s) = = C (sI - A) - 1 B U (s) s 0 2 0 s-2 0 [sI - A ] = > H - > H = > 0 s 0 4 0 s - 4H (s - 4) 0 1 > 0 (s - 2)H (s - 2) (s - 4) R V S 1 0 W (s - 2) W = SS 1 W S 0 (s - 4)W T X R V R V S 1 W S 1 0 W1 (s - 2) S(s - 2)W W = [4 0] SS 1 W>1H = [4 0] S 1 W S(s - 4)W S 0 (s - 4)W T X X T 4 = (s - 2)
(sI - A) - 1 =
So,
Y (s) U (s) Y (s) U (s)
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Page 333
Here input is unit impulse so U (s) = 1 and output Y (s) = 4 (s - 2) Taking inverse Laplace transfer we get output y (t) = 4e2t SOL 1.6.81
Option (D) is correct. Given state equation R V S0 1 0 0W S W o = S0 0 1 0W X X S0 0 0 1W S0 0 0 1W RT VX S0 1 0 0W S0 0 1 0W Here A =S W S0 0 0 1W S0 0 0 1W T obtained X as Eigen value can be
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or SOL 1.6.82
A - lI = 0 R V R V S0 1 0 0W Sl 0 0 0W S0 0 1 0W S0 l 0 0W (A - lI) = S W-S W S0 0 0 1W S0 0 l 0W S0 0 0 1W S0 0 0 lW T X T X R V 0 W S- l 1 0 S 0 -l 1 0 W =S W S 0 0 -l 1 W S 0 0 0 1 - lW T X A - lI = l3 (1 - l) = 0 l1, l2, l3 = 0 , l4 = 1
Option (A) is correct. Input-output relationship is given as d 2y dy du 2 + 2 dt + 10y = 5 dt - 3u dt Taking Laplace transform on both sides with zero initial condition. s 2 Y (s) + 2sY (s) + 10Y (s) = 5sU (s) - 3U (s) (s2 + 2s + 10) Y (s) = (5s - 3) U (s) (5s - 3) Output Y (s) = 2 U (s) (s + 2s + 10) With no input and with given initial conditions, output is obtained as d 2y dy 2 + 2 dt + 10y = 0 dt Taking Laplace transform (with initial conditions) [s2 Y (s) - sy (0) - y' (0)] + 2 [sY (s) - y (0)] + 10Y (s) = 0 Given that y' (0) =- 4 , y (0) = 1 [s2 Y (s) - s - (- 4)] + 2 (s - 1) + 10Y (s) = 0
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Page 334
Y (s) [s2 + 2s + 10] = (s - 2) (s - 2) Y (s) = 2 (s + 2s + 10) Output in both cases are same so (s - 2) (5s - 3) U (s) = 2 (s + 2s + 10) (s2 + 2s + 10) U (s) =
(s - 2) (5s - 10) =1 5 (5s - 3) (5s - 3)
(5s - 3) 7 = 1= 5 5s - 3 (5s - 3)G 7 U (s) = 1 ;1 5 (5s - 3)E Taking inverse Laplace transform, input is u (t) = 1 :d (t) - 5 e3/5t u (t)D 5 5 = 1 d (t) - 7 e3/5t u (t) 5 25 SOL 1.6.83
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Option (C) is correct. d 2 y dy + - 2y = u (t) e- t dt 2 dt State variable representation is given as o = AX + Bu X Or Here
...(1)
x1 xo1 > o H = A >x H + Bu x2 2 x1 = y , x 2 = b
dy - y l et dt
dx1 = dy = x e- t + y = x e- t + x 2 2 1 dt dt dx1 = x + x e- t + (0) u (t) or 1 2 dt Similarly 2 dx2 = d y et + dy et - et dy - yet dt dt dt dt 2 d 2y Put from equation (1) dt 2 dx2 = u (t) e- t - dy + 2y et - yet So, : D dt dt = u (t) -
...(2)
dy t e + 2yet - yet dt
= u (t) - [x2 e- t + y] et + yet = u (t) - x2 dx2 = 0 - x + u (t) 2 dt
...(3)
From equation (2) and (3) state variable representation is 0 xo1 1 e- t x 1 >o H = > >x H + >1H u (t) H x2 0 -1 2
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.6.84
Page 335
Option (B) is correct. Characteristic equation of the system 1 + G (s) = 0 2 (s + a) =0 1+ s (s + 2) (s + 10) s (s + 2) (s + 10) + 2 (s + a) = 0 s3 + 12s2 + 20s + 2s + 2a = 0 s3 + 12s2 + 22s + 2a = 0 2a 1+ 3 =0 s + 12s2 + 22s No of poles n = 3 No. of zeros m = 0 Angle of asymptotes (2q + 1) 180c fA = , q = 0, 1, 2 n-m fA =
(2q + 1) 180c = (2q + 1) 60c 3
nodia.co.in fA = 60c, 180c, 300c
SOL 1.6.85
Option (A) is correct. Asymptotes intercepts at real axis at the point / real Parts of Poles - / real Parts of zeros C = n-m Poles at
s1 = 0 s2 =- 2 s 3 =- 10
C = 0 - 2 - 10 - 0 =- 4 3-0 Option (C) is correct. Break away points da = 0 ds So
SOL 1.6.86
a =- 1 [s3 + 12s2 + 22s] 2 da =- 1 [3s2 + 24s + 22] = 0 2 ds s1, s2 =- 1.056, - 6.9433 SOL 1.6.87
Option ( ) is correct.
SOL 1.6.88
Option (A) is correct. Given state equation o = >- 3 1 H X X 0 -2 Or
o = AX , where A = >- 3 1 H X 0 -2
Taking Laplace transform on both sides. sX (s) - X (0) = AX (s)
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X (s) (sI - A) = X (0) X (s) = (sI - A) - 1 X (0) Steady state value of X is given by xss = lim sX (s) = lim s (sI - A) - 1 X (0) s"0
s"0
s 0 -3 1 s + 3 -1 => (sI - A) = > H - > H 0 s 0 -2 0 s + 2H s+2 1 1 > 0 s + 3H (s + 3) (s + 2) R V 1 S 1 W (s + 3) (s + 2) (s + 3)W S =S W 1 S 0 (s + 2) W T value X So the steady state R V 1 S 1 W (s + 3) (s + 2) (s + 3)W 10 S xss = lim s S W>- 10H 1 s"0 S 0 (s + 2) W TR XV 10 10 S W 0 (s + 3) (s + 2) (s + 3)W S => H = lim s S W 0 s"0 - 10 S W (s + 2) T X Option (D) is correct. Initial slope of the bode plot is - 40 dB/dec. So no. of poles at origin is 2. Then slope increased by - 20 dB/dec. at w = 2 rad/sec, so one poles lies at this frequency. At w = 5 rad/sec slope changes by + 20 dB/dec, so there is one zero lying at this frequency. Further slope decrease by - 20 dB/dec at w = 25 so one pole of the system is lying at this frequency. Transfer function K (s + 5) H (s) = 2 s (s + 2) (s + 25) At w = 0.1, gain is 54 dB, so 5K 54 = 20 log (0.1) 2 (2) (25) (sI - A- 1) =
SOL 1.6.89
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K = 50 H (s) = SOL 1.6.90
50 (s + 5) s (s + 2) (s + 25) 2
Option (B) is correct. Open loop transfer function of the system is 10 4 G (s) = s (s + 10) 2 10 4 10 4 = jw (jw + 10) 2 jw (100 - w2 + j20w) 10 4 G (jw) = w (100 - w2) 2 + 400w2
G (jw) = Magnitude
At w = 20 rad/sec
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Page 337
10 4 20 9 # 10 4 + 16 # 10 4 10 4 =1 = 20 # 5 # 102 Magnitude in dB = 20 log 10 G (j20) = 20 log 10 1 = 0 dB G (j20) =
SOL 1.6.91
Option (C) is correct. Since G (j w) = 1 at w = 20 rad/sec, So this is the gain cross-over frequency wg = 20 rad/sec Phase margin fPM = 180c + +G (jwg) 20 wg +G (jwg) =- 90c - tan- 1 = 100 - wg2 G fPM = 180 - 90c - tan- 1 ; 20 # 20 2 E 100 - (20) =- 36.86c
SOL 1.6.92
Option (C) is correct. To calculate the gain margin, first we have to obtain phase cross over frequency (wp). At phase cross over frequency
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+G (jwp) =- 180c 20wp =- 180c - 90c - tan- 1 = 100 - w2p G 20wp = 90c tan- 1 = 100 - w2p G 100 - w2p = 0 & wp = 10 rad/sec. 1 Gain margin in dB = 20 log 10 e G (jwp) o 10 4 G (jwp) = G (j10) = 10 (100 - 100) 2 + 400 (10) 2
10 4 =5 10 # 2 # 102 G.M. = 20 log 10 b 1 l =- 13.97 dB 5 Option (B) is correct. Since gain margin and phase margin are negative, so the system is unstable. =
SOL 1.6.93
SOL 1.6.94
Option (C) is correct. Given characteristic equation s3 + s2 + Ks + K = 0 K (s + 1) =0 1+ 3 s + s2 K (s + 1) =0 1+ 2 s (s + 2) so open loop transfer function is K (s + 1) G (s) = 2 s (s + 1) root-locus is obtained in following steps: 1. Root-loci starts(K = 0 ) at s = 0 , s = 0 and s =- 2
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2. 3. 4.
Page 338
There is one zero at s =- 1, so one of root-loci terminates at s =- 1 and other two terminates at infinity No. of poles n = 3 , no of zeros ,m = 1 Break - Away points
dK = 0 ds Asymptotes meets on real axis at a point C / poles - / zeros C = n-m (0 + 0 - 2) - (- 1) = =- 0.5 3-1 ***********
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7 ELECTRICAL & ELECTRONIC MEASUREMENTS
YEAR 2013 MCQ 1.7.1
Three moving iron type voltmeters are connected as shown below. Voltmeter readings are V , V1 and V2 as indicated. The correct relation among the voltmeter readings is
(A) V = V1 + V2 2 2 (C) V = V1 V2 MCQ 1.7.2
ONE MARK
(B) V = V1 + V2 (D) V = V2 - V1
The input impedance of the permanent magnet moving coil (PMMC) voltmeter is infinite. Assuming that the diode shown in the figure below is ideal, the reading of the voltmeter in Volts is
(A) 4.46 (C) 2.23 YEAR 2013
(B) 3.15 (D) 0 TWO MARKS
MCQ 1.7.3
Two magnetically uncoupled inductive coils have Q factors q1 and q2 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) ^q1 R1 + q2 R2h / ^R1 + R2h (D) ^q1 R2 + q2 R1h / ^R1 + R2h
MCQ 1.7.4
A strain gauge forms one arm of the bridge shown in the figure below and has a nominal resistance without any load as Rs = 300 W . Other bridge resistances are R1 = R2 = R 3 = 300 W . The maximum permissible current through the strain gauge is 20 mA. During certain measurement when the bridge is excited by
GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 340
maximum permissible voltage and the strain gauge resistance is increased by 1% over the nominal value, the output voltage V0 in mV is
(A) 56.02 (C) 29.85
(B) 40.83 (D) 10.02
YEAR 2012 MCQ 1.7.5
ONE MARK
A periodic voltage waveform observed on an oscilloscope across a load is shown. A permanent magnet moving coil (PMMC) meter connected across the same load reads
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(A) 4 V (C) 8 V
(B) 5 V (D) 10 V
MCQ 1.7.6
The bridge method commonly used for finding mutual inductance is (A) Heaviside Campbell bridge (B) Schering bridge (C) De Sauty bridge (D) Wien bridge
MCQ 1.7.7
For the circuit shown in the figure, the voltage and current expressions are v (t) = E1 sin (wt) + E 3 sin (3wt) and i (t) = I1 sin (wt - f1) + I 3 sin (3wt - f3) + I5 sin (5wt) The average power measured by the wattmeter is
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Page 341
(A) 1 E1 I1 cos f1 2 (B) 1 [E1 I1 cos f1 + E1 I 3 cos f3 + E1 I5] 2 (C) 1 [E1 I1 cos f1 + E 3 I 3 cos f3] 2 (D) 1 [E1 I1 cos f1 + E 3 I1 cos f1] 2 YEAR 2012 MCQ 1.7.8
TWO MARKS
An analog voltmeter uses external multiplier settings. With a multiplier setting of 20 kW, it reads 440 V and with a multiplier setting of 80 kW, it reads 352 V. For a multiplier setting of 40 kW, the voltmeter reads (A) 371 V (B) 383 V (C) 394 V (D) 406 V YEAR 2011
MCQ 1.7.9
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Consider the following statement (1) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. (2) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit. (A) (1) is true but (2) is false (B) (1) is false but (2) is true (C) both (1) and (2) are true
MCQ 1.7.10
(D) both (1) and (2) are false
The bridge circuit shown in the figure below is used for the measurement of an unknown element ZX . The bridge circuit is best suited when ZX is a
(A) low resistance (C) low Q inductor MCQ 1.7.11
ONE MARK
(B) high resistance (D) lossy capacitor
A dual trace oscilloscope is set to operate in the ALTernate mode. The control input of the multiplexer used in the y -circuit is fed with a signal having a frequency equal to (A) the highest frequency that the multiplexer can operate properly (B) twice the frequency of the time base (sweep) oscillator (C) the frequency of the time base (sweep) oscillator (D) haif the frequency of the time base (sweep) oscillator
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YEAR 2011 MCQ 1.7.12
TWO MARKS
A 4 12 digit DMM has the error specification as: 0.2% of reading + 10 counts. If a dc voltage of 100 V is read on its 200 V full scale, the maximum error that can be expected in the reading is (A) ! 0.1% (B) ! 0.2% (C) ! 0.3% (D) ! 0.4% YEAR 2010
MCQ 1.7.13
Page 342
ONE MARK
A wattmeter is connected as shown in figure. The wattmeter reads.
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(A) Zero always (B) Total power consumed by Z1 and Z 2 (C) Power consumed by Z1 (D) Power consumed by Z2 MCQ 1.7.14
An ammeter has a current range of 0-5 A, and its internal resistance is 0.2 W. In order to change the range to 0-25 A, we need to add a resistance of (A) 0.8 W in series with the meter (B) 1.0 W in series with the meter (C) 0.04 W in parallel with the meter (D) 0.05 W in parallel with the meter
MCQ 1.7.15
As shown in the figure, a negative feedback system has an amplifier of gain 100 with ! 10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately :
(A) 10 ! 1% (C) 10 ! 5%
(B) 10 ! 2% (D) 10 ! 10%
YEAR 2010 MCQ 1.7.16
TWO MARKS
The Maxwell’s bridge shown in the figure is at balance. The parameters of the inductive coil are.
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(A) (B) (C) (D)
Page 343
R = R2 R 3 /R 4, L = C 4 R2 R 3 L = R2 R 3 /R 4, R = C 4 R2 R 3 R = R 4 /R2 R 3, L = 1/ (C 4 R2 R 3) L = R 4 /R2 R 3, R = 1/ (C 4 R2 R 3)
YEAR 2009
ONE MARK
MCQ 1.7.17
The pressure coil of a dynamometer type wattmeter is (A) Highly inductive (B) Highly resistive (C) Purely resistive (D) Purely inductive
MCQ 1.7.18
The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. The following inference can be made from this. (A) The signals are not sinusoidal (B) The amplitudes of the signals are very close but not equal (C) The signals are sinusoidal with their frequencies very close but not equal (D) There is a constant but small phase difference between the signals
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YEAR 2009 MCQ 1.7.19
The figure shows a three-phase delta connected load supplied from a 400V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be
(A) 0 (C) 800 Watt MCQ 1.7.20
TWO MARKS
(B) 1600 Watt (D) 400 Watt
An average-reading digital multi-meter reads 10 V when fed with a triangular wave, symmetric about the time-axis. For the same input an rms-reading meter will read (B) 10 (A) 20 3 3
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(C) 20 3
(D) 10 3
YEAR 2008 MCQ 1.7.21
ONE MARK
Two 8-bit ADCs, one of single slope integrating type and other of successive approximate type, take TA and TB times to convert 5 V analog input signal to equivalent digital output. If the input analog signal is reduced to 2.5 V, the approximate time taken by the two ADCs will respectively, be (A) TA, TB (B) TA /2, TB (C) TA, TB /2 (D) TA /2, TB /2 YEAR 2008
MCQ 1.7.22
Page 344
TWO MARKS
Two sinusoidal signals p (w1, t) = A sin w1 t and q (w2 t) are applied to X and Y inputs of a dual channel CRO. The Lissajous figure displayed on the screen shown below : The signal q (w2 t) will be represented as
nodia.co.in (A) q (w2 t) = A sin w2 t, w2 = 2w1 (C) q (w2 t) = A cos w2 t, w2 = 2w1 MCQ 1.7.23
(B) q (w2 t) = A sin w2 t, w2 = w1 /2 (D) q (w2 t) = A cos w2 t, w2 = w1 /2
The ac bridge shown in the figure is used to measure the impedance Z .
If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance Z will be (A) (260 + j0) W (B) (0 + j200) W (C) (260 - j200) W (D) (260 + j200) W YEAR 2007 MCQ 1.7.24
ONE MARK
The probes of a non-isolated, two channel oscillocope are clipped to points A, B and C in the circuit of the adjacent figure. Vin is a square wave of a suitable low frequency. The display on Ch1 and Ch2 are as shown on the right. Then the “Signal” and “Ground” probes S1, G1 and S2, G2 of Ch1 and Ch2 respectively are connected to points :
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(A) A, B, C, A (C) C, B, A, B
(B) A, B, C, B (D) B, A, B, C
YEAR 2007 MCQ 1.7.25
Page 345
TWO MARKS
A bridge circuit is shown in the figure below. Which one of the sequence given below is most suitable for balancing the bridge ?
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(A) First adjust (B) First adjust (C) First adjust (D) First adjust YEAR 2006 MCQ 1.7.26
R4 , and then adjust R1 R2 , and then adjust R3 R2 , and then adjust R4 R4 , and then adjust R2 ONE MARK
The time/div and voltage/div axes of an oscilloscope have been erased. A student connects a 1 kHz, 5 V p-p square wave calibration pulse to channel-1 of the scope and observes the screen to be as shown in the upper trace of the figure. An unknown signal is connected to channel-2(lower trace) of the scope. It the time/ div and V/div on both channels are the same, the amplitude (p-p) and period of the unknown signal are respectively
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(A) 5 V, 1 ms (C) 7.5 V, 2 ms MCQ 1.7.27
Page 346
(B) 5 V, 2 ms (D) 10 V, 1 ms
A sampling wattmeter (that computes power from simultaneously sampled values of voltage and current) is used to measure the average power of a load. The peak to peak voltage of the square wave is 10 V and the current is a triangular wave of 5 A p-p as shown in the figure. The period is 20 ms. The reading in W will be
(A) 0 W (C) 50 W
(B) 25 W (D) 100 W
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YEAR 2006
TWO MARKS
MCQ 1.7.28
A current of - 8 + 6 2 (sin wt + 30%) A is passed through three meters. They are a centre zero PMMC meter, a true rms meter and a moving iron instrument. The respective reading (in A) will be (A) 8, 6, 10 (B) 8, 6, 8 (D) - 8 ,2,2 (C) - 8 ,10,10
MCQ 1.7.29
A variable w is related to three other variables x ,y ,z as w = xy/z . The variables are measured with meters of accuracy ! 0.5% reading, ! 1% of full scale value and ! 1.5% reading. The actual readings of the three meters are 80, 20 and 50 with 100 being the full scale value for all three. The maximum uncertainty in the measurement of w will be (A) ! 0.5% rdg (B) ! 5.5% rdg (C) ! 6.7 rdg (D) ! 7.0 rdg
MCQ 1.7.30
A 200/1 Current transformer (CT) is wound with 200 turns on the secondary on a toroidal core. When it carries a current of 160 A on the primary, the ratio and phase errors of the CT are found to be - 0.5% and 30 minutes respectively. If the number of secondary turns is reduced by 1 new ratio-error(%) and phaseerror(min) will be respectively (A) 0.0, 30 (B) - 0.5, 35 (C) - 1.0, 30 (D) - 1.0, 25
MCQ 1.7.31
R1 and R4 are the opposite arms of a Wheatstone bridge as are R3 and R2 . The source voltage is applied across R1 and R3 . Under balanced conditions which one of the following is true (A) R1 = R3 R4 /R2 (B) R1 = R2 R3 /R4 (C) R1 = R2 R4 /R3 (D) R1 = R2 + R3 + R4
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Page 347
YEAR 2005 MCQ 1.7.32
MCQ 1.7.33
The Q-meter works on the principle of (A) mutual inductance (C) series resonance
(B) self inductance (D) parallel resonance
A PMMC voltmeter is connected across a series combination of DC voltage source V1 = 2 V and AC voltage source V2 (t) = 3 sin (4t) V. The meter reads (A) 2 V (B) 5 V (C) (2 +
MCQ 1.7.34
ONE MARK
3 /2) V
(D) ( 17 /2) V
A digital-to-analog converter with a full-scale output voltage of 3.5 V has a resolution close to 14 mV. Its bit size is (A) 4 (B) 8 (C) 16 (D) 32 YEAR 2005
TWO MARKS
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MCQ 1.7.35
The simultaneous application of signals x (t) and y (t) to the horizontal and vertical plates, respectively, of an oscilloscope, produces a vertical figure-of-8 display. If P and Q are constants and x (t) = P sin (4t + 30c), then y (t) is equal to (B) Q sin (2t + 15c) (A) Q sin (4t - 30c) (C) Q sin (8t + 60c) (D) Q sin (4t + 30c)
MCQ 1.7.36
A DC ammeter has a resistance of 0.1 W and its current range is 0-100 A. If the range is to be extended to 0-500 A, then meter required the following shunt resistance (A) 0.010 W (B) 0.011 W (C) 0.025 W (D) 1.0 W
MCQ 1.7.37
The set-up in the figure is used to measure resistance R .The ammeter and voltmeter resistances are 0.01W and 2000 W, respectively. Their readings are 2 A and 180 V, respectively, giving a measured resistances of 90 W The percentage error in the measurement is
(A) 2.25% (C) 4.5% MCQ 1.7.38
(B) 2.35% (D) 4.71%
A 1000 V DC supply has two 1-core cables as its positive and negative leads : their insulation resistances to earth are 4 MW and 6 MW, respectively, as shown in the figure. A voltmeter with resistance 50 kW is used to measure the insulation of the cable. When connected between the positive core and earth, then voltmeter reads
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(A) 8 V (C) 24 V MCQ 1.7.39
Page 348
(B) 16 V (D) 40 V
Two wattmeters, which are connected to measure the total power on a three-phase system supplying a balanced load, read 10.5 kW and - 2.5 kW, respectively. The total power and the power factor, respectively, are (A) 13.0 kW, 0.334 (B) 13.0 kW, 0.684 (C) 8.0 kW, 0.52 (D) 8.0 kW, 0.334 YEAR 2004
ONE MARK
MCQ 1.7.40
A dc potentiometer is designed to measure up to about 2 V with a slide wire of 800 mm. A standard cell of emf 1.18 V obtains balance at 600 mm. A test cell is seen to obtain balance at 680 mm. The emf of the test cell is (A) 1.00 V (B) 1.34 V (C) 1.50 V (D) 1.70 V
MCQ 1.7.41
The circuit in figure is used to measure the power consumed by the load. The current coil and the voltage coil of the wattmeter have 0.02 W and 1000W resistances respectively. The measured power compared to the load power will be
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(A) 0.4 % less (C) 0.2% more MCQ 1.7.42
(B) 0.2% less (D) 0.4% more
A galvanometer with a full scale current of 10 mA has a resistance of 1000 W. The multiplying power (the ratio of measured current to galvanometer current) of 100 W shunt with this galvanometer is (A) 110 (B) 100 (C) 11 (D) 10 YEAR 2004
MCQ 1.7.43
TWO MARKS
A CRO probe has an impedance of 500 kW in parallel with a capacitance of 10 pF. The probe is used to measure the voltage between P and Q as shown in figure. The measured voltage will be
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(A) 3.53 V (C) 4.54 V
Page 349
(B) 4.37 V (D) 5.00 V
MCQ 1.7.44
A moving coil of a meter has 100 turns, and a length and depth of 10 mm and 20 mm respectively. It is positioned in a uniform radial flux density of 200 mT. The coil carries a current of 50 mA. The torque on the coil is (A) 200 mNm (B) 100 mNm (C) 2 mNm (D) 1 mNm
MCQ 1.7.45
A dc A-h meter is rated for 15 A, 250 V. The meter constant is 14.4 A-sec/rev. The meter constant at rated voltage may be expressed as (A) 3750 rev/kWh (B) 3600 rev/kWh (C) 1000 rev/kWh (D) 960 rev/kWh
MCQ 1.7.46
A moving iron ammeter produces a full scale torque of 240 mNm with a deflection of 120c at a current of 10 A . The rate of change of self induction (mH/radian) of the instrument at full scale is (A) 2.0 mH/radian (B) 4.8 mH/radian (C) 12.0 mH/radian (D) 114.6 mH/radian
MCQ 1.7.47
A single-phase load is connected between R and Y terminals of a 415 V, symmetrical, 3-phase, 4-wire system with phase sequence RYB. A wattmeter is connected in the system as shown in figure. The power factor of the load is 0.8 lagging. The wattmeter will read
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(A) - 795 W (C) + 597 W
(B) - 597 W (D) + 795 W
MCQ 1.7.48
A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1 W. The magnetizing ampere-turns is 200. The phase angle between the primary and second current is (A) 4.6c (B) 85.4c (C) 94.6c (D) 175.4c
MCQ 1.7.49
The core flux in the CT of Prob Q.44, under the given operating conditions is (A) 0 (B) 45.0 mWb (C) 22.5 mWb (D) 100.0 mWb
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Page 350
YEAR 2003
ONE MARK
MCQ 1.7.50
A Manganin swap resistance is connected in series with a moving coil ammeter consisting of a milli-ammeter and a suitable shunt in order to (A) minimise the effect of temperature variation (B) obtain large deflecting torque (C) reduce the size of the meter (D) minimise the effect of stray magnetic fields
MCQ 1.7.51
The effect of stray magnetic field on the actuating torque of a portable instrument is maximum when the operating field of the instrument and the stray fields are (A) perpendicular (B) parallel (C) inclined at 60%
MCQ 1.7.52
(D) inclined at 30%
A reading of 120 is obtained when standard inductor was connected in the circuit of a Q-meter and the variable capacitor is adjusted to value of 300 pF. A lossless capacitor of unknown value Cx is then connected in parallel with the variable capacitor and the same reading was obtained when the variable capacitor is readjusted to a value of 200 pF. The value of Cx in pF is (A) 100 (B) 200 (C) 300 (D) 500
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YEAR 2003 MCQ 1.7.53
TWO MARKS
The simplified block diagram of a 10-bit A/D converter of dual slope integrator type is shown in figure. The 10-bit counter at the output is clocked by a 1 MHz clock. Assuming negligible timing overhead for the control logic, the maximum frequency of the analog signal that can be converted using this A/D converter is approximately
(A) 2 kHz (C) 500 Hz MCQ 1.7.54
(B) 1 kHz (D) 250 Hz
The items in Group-I represent the various types of measurements to be made with a reasonable accuracy using a suitable bridge. The items in Group-II represent the various bridges available for this purpose. Select the correct choice of the item in Group-II for the corresponding item in Group-I from the following List-I
List-II
P.
Resistance in the milli-ohm 1. range
Wheatstone Bridge
Q.
Low values of Capacitance
R.
Comparison of resistance which 3. are nearly equal
2.
Kelvin Double Bridge Schering Bridge
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S.
Inductance of a coil with a 4. large time-constant
Codes : (A) P=2, Q=3, R=6, S=5 (C) P=2, Q= 3, R=5, S=4 MCQ 1.7.55
Page 351
Wien’s Bridge
5.
Hay’s Bridge
6.
Carey-Foster Bridge (B) P=2, Q=6, R=4, S=5 (D) P=1, Q=3, R=2, S=6
A rectifier type ac voltmeter of a series resistance Rs , an ideal full-wave rectifier bridge and a PMMC instrument as shown in figure. The internal. resistance of the instrument is 100 W and a full scale deflection is produced by a dc current of 1 mA. The value of Rs required to obtain full scale deflection with an ac voltage of 100 V (rms) applied to the input terminals is
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(A) 63.56 W (C) 89.93 W
(B) 69.93 W (D) 141.3 kW
MCQ 1.7.56
A wattmeter reads 400 W when its current coil is connected in the R-phase and its pressure coil is connected between this phase and the neutral of a symmetrical 3-phase system supplying a balanced star connected 0.8 p.f. inductive load. This phase sequence is RYB. What will be the reading of this wattmeter if its pressure coil alone is reconnected between the B and Y phases, all other connections remaining as before ? (A) 400.0 (B) 519.6 (C) 300.0 (D) 692.8
MCQ 1.7.57
The inductance of a certain moving-iron ammeter is expressed as L = 10 + 3q - (q2 /4) mH , where q is the deflection in radians from the zero position. The control spring torque is 25 # 10 - 6 Nm/radian. The deflection of the pointer in radian when the meter carries a current of 5 A, is (A) 2.4 (B) 2.0 (C) 1.2 (D) 1.0
MCQ 1.7.58
A 500A/5A, 50 Hz transformer has a bar primary. The secondary burden is a pure resistance of 1 W and it draws a current of 5 A. If the magnetic core requires 250 AT for magnetization, the percentage ratio error is (A) 10.56 (B) - 10.56 (C) 11.80 (D) - 11.80
MCQ 1.7.59
The voltage-flux adjustment of a certain 1-phase 220 V induction watt-hour meter is altered so that the phase angle between the applied voltage and the flux due to it is 85c(instead of 90c). The errors introduced in the reading of this meter when the current is 5 A at power factor of unity and 0.5 lagging are respectively
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(A) 3.8 mW, 77.4 mW (C) - 4.2 W, - 85.1 W MCQ 1.7.60
Page 352
(B) - 3.8 mW, - 77.4 mW (D) 4.2 W, 85.1 W
Group-II represents the figures obtained on a CRO screen when the voltage signals Vx = Vxm sin wt and Vy = Vym sin (wt + F) are given to its X and Y plates respectively and F is changed. Choose the correct value of F from Group-I to match with the corresponding figure of Group-II. Group-I Group-II P. F = 0
Q. F = p/2
R. p < F < 3p/2
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S. F = 3p/2
Codes : (A) P=1, Q= 3, R=6, S=5 (C) P=2, Q= 3, R=5, S=4
(B) P=2, Q= 6, R=4, S=5 (D) P=1, Q=5, R=6, S=4
YEAR 2002
ONE MARK
MCQ 1.7.61
Two in-phase, 50 Hz sinusoidal waveforms of unit amplitude are fed into channel-1 and channel-2 respectively of an oscilloscope. Assuming that the voltage scale, time scale and other settings are exactly the same for both the channels, what would be observed if the oscilloscope is operated in X-Y mode ? (A) A circle of unit radius (B) An ellipse (C) A parabola (D) A straight line inclined at 45c with respect to the x-axis.
MCQ 1.7.62
The line-to-line input voltage to the 3-phase, 50 Hz, ac circuit shown in Figure is 100 V rms. Assuming that the phase sequence is RYB, the wattmeters would read.
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(A) W1 =886 W and W2 =886 W (C) W1 =0 W and W2 =1000 W
Page 353
(B) W1 =500 W and W2 =500 W (D) W1 =250 W and W2 =750 W
YEAR 2001
ONE MARK
MCQ 1.7.63
If an energy meter disc makes 10 revolutions in 100 seconds when a load of 450 W is connected to it, the meter constant (in rev/kWh) is (A) 1000 (B) 500 (C) 1600 (D) 800
MCQ 1.7.64
The minimum number of wattmeter(s) required to measure 3-phase, 2-wire balanced or unbalanced power is (A) 1 (B) 2 (C) 3 (D) 4
MCQ 1.7.65
A 100 mA ammeter has an internal resistance of 100 W. For extending its range to measure 500 mA , the shunt required is of resistance (in W) (A) 20.0 (B) 22.22 (C) 25.0 (D) 50.0
MCQ 1.7.66
Resistance R1 and R2 have, respectively, nominal values of 10 W and 5 W, and tolerance of ! 5% and ! 10% . The range of values for the parallel combination of R1 and R2 is (A) 3.077 W to 3.636 W (B) 2.805 W to 3.371 W (C) 3.237 W to 3.678 W (D) 3.192 W to 3.435 W
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SOLUTIONS
SOL 1.7.1
Page 354
Option (B) is correct. For an ideal voltmeter interval resistance is always zero. So we can apply the KVL along the two voltmeters as or
V - V1 - V2 = 0 V = V1 + V2
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SOL 1.7.2
Option (A) is correct. For the + ve half cycle of I/p voltage, diode will be forward biased (Vg = 0 , ideal diode) Therefore, the voltmeter will be short circuited and reads (for + ve half cycle) V1 = 0 volt Now, for - ve half cycle, diode will be reverse biased and treated as open circuit. So, the voltmeter reads the voltage across 100 kW. Which is given by 14.14 0c V2 = 100 # 100 + 1 So, V2,rms = 14 volt 2 Therefore, the average voltage for the whole time period is obtained as V + V2, rms 0 + ^14/ 2 h Vave = 1 = = 14 2 2 2 2 = 4.94 . 4.46 volt
SOL 1.7.3
Option (C) is correct. The quality factor of the inductances are given by q 1 = wL 1 R1 and q 2 = wL 2 R2 So, in series circuit, the effective quality factor is given by XLeq = wL 1 + wL 2 Q = Req R1 + R 2 w L 1 + wL 2 R = 1 R 2 R1 R 2 1 + 1 R 2 R1 q1 q + 2 R R 2 2 = 1 + 1 R 2 R1
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=
Page 355
q1 R1 + q 2 R 2 R1 + R 2
SOL 1.7.4
Option (C) is correct.
SOL 1.7.5
Option (A) is correct. PMMC instrument reads average (dc) value. 20 T 1 v (t) dt Vavg = 1 v (t) dt = 3 T 0 20 # 10 0
#
#
10 = 1 ; tdt + 20 0
#
20
12
# (- 5) dt + # 5dtE 10
12
2 10
20 = 1 c :t D - 5 6t @12 + 5 6t @12 m 10 20 2 0
SOL 1.7.6
SOL 1.7.7
= 1 [50 - 5 (2) + 5 (8)] = 80 = 4 V 20 20 Option (A) is correct. Heaviside mutual inductance bridge measures mutual inductance is terms of a known self-inductance.
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Option (C) is correct. Let wt = q , we have instaneous voltage and current as follows. v (t) = E1 sin q + E 3 sin 3q i (t) = I1 sin (q - f1) + I 3 sin (3q - f3) + I5 sin (5q) We know that wattmeter reads the average power, which is gives as 2p ...(i) P = 1 v (t) i (t) dq 2p 0 We can solve this integration using following results. 2p (i) 1 A sin (q + a):B sin (q + b) dq = 1 AB cos (a - b) 2p 0 2
#
#
SOL 1.7.8
2p
(ii) 1 2p
#
(iii) 1 2p
# A sin (mq + a):B cos (nq + b) dq = 0
(iv) 1 2p
# A sin (mq + a):B cos (nq + b) dq = 0
0
A sin (q + a):B cos (q + a) dq = 1 AB sin (a - b) 2
2p
0
2p
0
Result (iii) and (iv) implies that power is transferred between same harmonics of voltages and currents. Thus integration of equation (i) gives. P = 1 E1 I1 cos f + 1 E 3 I 3 cos f3 2 2 Option (D) is correct. A voltmeter with a multiplier is shown in figure below.
Here
Im = Fully scale deflection current of meter.
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Rm Rs V Vm V V Vm Here when, So,
Page 356
= Internal resistance of meter = Voltage across the meter = Full range voltage of instrument = Im Rm = Im (Rm + Rs) = R m + Rs = 1 + Rs Rm Rm
Rs1 = 20 kW , Vm1 = 440 V V = 1 + 20k 440 Rm
...(i)
Rs2 = 80 kW , Vm2 = 352 V V = 1 + 80 k ...(ii) So, 352 Rm Solving equation (i) and (ii), we get V = 480 V , Rm = 220 kW So when Rs3 = 40 kW , Vm3 = ? 480 = 1 + 40 k & V - 406 V m2 Vm3 220 k Option (A) is correct. The compensating coil compensation the effect of impedance of current coil. When,
SOL 1.7.9
SOL 1.7.10
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Option (C) is correct. Let Z1 = R1 || jwC1 so admittance Y1 = 1 = 1 + jwC1 Z1 R 1
Z2 = R2 and Z 4 = R 4 Let ZX = RX + jwLX (Unknown impedance) For current balance condition of the Bridge Z 2 Z 4 = Z X Z1 = Z X Y1 Let
ZX = Z2 Z 4 Y1
R X + jw L X = R 2 R 4 b 1 + jw C 1 l R1 Equating imaginary and real parts RX = R2 R 4 and LX = R2 R 4 C1 R1 Quality factor of inductance which is being measured Q = wL X = wR 1 C 1 RX From above equation we can see that for measuring high values of Q we need a large value of resistance R 4 which is not suitable. This bridge is used for measuring low Q coils. Note: We can observe directly that this is a maxwell’s bridge which is suitable for low values of Q (i.e. Q < 10 ) SOL 1.7.11
Option (C) is correct. In the alternate mode it switches between channel A and channel B, letting each through for one cycle of horizontal sweep as shown in the figure.
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Page 357
nodia.co.in SOL 1.7.12
Option (C) is correct. 4 1 digit display will read from 000.00 to 199.99 So error of 10 counts is equal to 2 = ! 0.10 V For 100 V, the maximum error is e = ! (100 # 0.002 + 0.1) = ! 0.3 V Percentage error
SOL 1.7.13
= ! 0.3 # 100 % = ! 0.3 % of reading 100 Option (D) is correct. Since potential coil is applied across Z2 as shown below
Wattmeter read power consumed by Z2 SOL 1.7.14
Option (D) is correct. Given that full scale current is 5 A
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SOL 1.7.15
Page 358
Current in shunt Il = IR - I fs = 25 - 5 = 20 A 20 # Rsh = 5 # 0.2 Rsh = 1 = .05 W 20 Option (A) is correct. Overall gain of the system is 100 g = = 10 (zero error) 1 + 100 b 9 l 100 Gain with error 110 100 + 10% g = = 10.091 = 110 9 #9 1+ 1 + (100 + 10%) b l 100 100
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error 3 g = 10.091 - 10 - 0.1 Similarly 90 100 - 10% g = = 9 1 + (100 - 10%) 1 + 90 # 9 100 100 = 9.89 error 3 g = 9.89 - 10 -- 0.1 So gain g = 10 ! 0.1 = 10 ! 1% SOL 1.7.16
Option (A) is correct. At balance condition -j = R2 R 3 wC 4 m - jR 4 wC 4 (R + jwL) = R2 R 3 j c R 4 - wC 4 m - jRR 4 wLR 4 + = R2 R 3 R 4 wC 4 wC 4 - jRR 4 LR 4 + = R2 R 3 R 4 wC 4 C4 Comparing real & imaginary parts. RR 4 = R2 R 3 wC 4 wC 4 R = R2 R 3 R4 Similarly, LR 4 = R R R 2 3 4 C4 (R + jwL) c R 4 <
jR2 R 3 wC 4 jR2 R 3 wC 4
L = R2 R3 C 4 SOL 1.7.17
Option (B) is correct. Since Potential coil is connected across the load terminal, so it should be highly
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Page 359
resistive, so that all the voltage appears across load. SOL 1.7.18
Option (D) is correct. A circle is produced when there is a 90c phase difference between vertical and horizontal inputs.
SOL 1.7.19
Option (C) is correct. Wattmeter reading P = VPC ICC VPC " Voltage across potential coil. ICC " Current in current coil. VPC = Vbc = 400+ - 120c ICC = Iac = 400+120c = 4+120c 100 P = 400+ - 120c # 4+120c = 1600+240c = 1600 # 1 = 800 Watt 2 Option (D) is correct. Average value of a triangular wave Vav = Vm 3 rms value Vms = Vm 3 Given that Vav = Vm = 10 V 3 So Vrms = Vm = 3 Vav = 10 3 V 3 Power
SOL 1.7.20
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SOL 1.7.21
Option (A) is correct. Conversion time does not depend on input voltage so it remains same for both type of ADCs.
SOL 1.7.22
Option (D) is correct.
Frequency ratio
meeting points of horizontal tangents fY = fX meeting points of vertical tangents fY =2 4 fX fY = 1 (fX ) 2
w2 = w1 /2 Since the Lissajous figures are ellipse, so there is a phase difference of 90c exists between vertical and horizontal inputs. So q (w2 t) = A cos w2 t, w2 = w1 /2
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.7.23
Page 360
Option (A) is correct. Impedance of different branches is given as ZAB = 500 W 1 + 300 W j # 2p # 2 # 103 # 0.398 mF - (- 200j + 300) W
ZBC =
ZAD = j # 2p # 2 # 103 # 15.91 mH + 300 W - (200j + 300) W To balance the bridge ZAB ZCD 500Z 500Z Z
= ZAD ZBC = (200j + 300) (- 200j + 300) = 130000 = (260 + j0) W
SOL 1.7.24
Option (B) is correct. Since both the waveform appeared across resistor and inductor are same so the common point is B. Signal Probe S1 is connecte with A, S2 is connected with C and both the grount probes G1 and G2 are connected with common point B.
SOL 1.7.25
Option (A) is correct. To balance the bridge
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(R1 + jX1) (R 4 - jX 4) = R2 R 3 (R1 R 4 + X1 X 4) + j (X1 R 4 - R1 X 4) = R2 R 3 comparing real and imaginary parts on both sides of equations
...(1) R1 R 4 + X1 X 4 = R 2 R 3 ...(2) X1 R 4 - R1 X 4 = 0 & X1 = R 1 X4 R4 from eq(1) and (2) it is clear that for balancing the bridge first balance R 4 and then R1 . SOL 1.7.26
Option (C) is correct. From the Calibration pulse we can obtain Voltage (3 V) = 5 = 2.5 V 2 Division Time (3 T) = 1 ms = 1 msec 4 4 Division So amplitude (p-p) of unknown signal is VP - P = 3 V # 5 = 2.5 # 5 = 7.5 V Time period T = 3 T # 8 = 1 # 8 = 2 ms 4
SOL 1.7.27
Option (A) is correct. Reading of wattmeter (Power) in the circuit T Pav = 1 # VIdt = Common are between V - I T 0
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Page 361
total common area = 0 (Positive and negative area are equal) So Pav = 0 SOL 1.7.28
Option (C) is correct. PMMC instrument reads only dc value so I PMMC =- 8 A rms meter reads rms value so (- 8) 2 +
Irms =
(6 2 ) 2 2
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= 64 + 36 = 10 A Moving iron instrument also reads rms value of current So I MI = 10 mA Reading are (I PMMC, Irms, I MI) = (- 8 A, 10 A, 10 A) SOL 1.7.29
Option (D) is correct.
Given that w =
xy z
SOL 1.7.30
log w = log x + log y - log z Maximum error in w dy ! dz % dw = ! dx ! x y z w dx = ! 0.5% reading x dy = ! 1% full scale y = ! 1 # 100 = ! 1 100 dy = ! 1 # 100 = ! 5% reading y 20 dz = 1.5% reading z So % dw = ! 0.5% ! 5% ! 1.5% = ! 7% w Option ( ) is correct.
SOL 1.7.31
Option (B) is correct.
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In balanced condition there is no current in CD arm so VC = VD Writing node equation at C and D VC - V + VC = 0 & V = V R3 C b R1 + R 3 l R1 R3 V0 - V + VD = 0 & V = V R4 D b R2 + R 4 l R2 R4 So Vb R3 l = Vb R 4 l R1 + R 3 R2 + R 4 R 2 R 3 + R 3 R 4 = R1 R 4 + R 3 R 4 R1 = R2 R 3 /R 4 SOL 1.7.32
Option (C) is correct. Q-meter works on the principle of series resonance.
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At resonance VC = VL and I = V R Quality factor Q = wL = 1 = wL # I = VL = VC R R#I E E wCR Thus, we can obtain Q SOL 1.7.33
Option (A) is correct. PMMC instruments reads DC value only so it reads 2 V.
SOL 1.7.34
Option (B) is correct. Vfs 2 -1 V = 3.5 2n - 1 3.5 = 14 # 10- 3 = 250 = 251 = 8 bit
Resolution of n-bit DAC = So
14 mv 2n - 1 2n - 1 2n n
SOL 1.7.35
n
Option (B) is correct. We can obtain the frequency ratio as following
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Page 363
meeting points of horizontal tangents fY = fX meeting points of vertical tangents fY =2 4 fX fY = 1 fX 2 There should exist a phase difference(15c) also to produce exact figure of-8. SOL 1.7.36
Option (C) is correct. The configuration is shown below
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It is given that Im = 100 A Range is to be extended to 0 - 500 A, I = 500 A So,
SOL 1.7.37
Im Rm = (I - Im) Rsh 100 # 0.1 = (500 - 100) Rsh Rsh = 100 # 0.1 = 0.025 W 400 Option (D) is correct. The configuration is shown below
Current in voltmeter is given by IV = E = 180 = .09 A 2000 2000 So
Ideally
I + IV = 2 amp I = 2 - .09 = 1.91 V R = E = 180 = 94.24 W 1.91 I R 0 = 180 = 90 W 2 % error = R - R 0 # 100 = 94.24 - 90 # 100 = 4.71% 90 R0
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.7.38
Page 364
Option (A) is correct. The measurement system is shown below
Voltmeter reading 1000 (50 kW z 4 MW) 6 MW + 50 kW z 4 MW l = 1000 # .049 = 8.10 V 6 + .049 Option (D) is correct. Total power P = P1 + P2 = 10.5 - 2.5 = 8 kW Power factor = cos q Where q = tan- 1 ; 3 b P2 - P1 lE = tan- 1 : 3 # - 13D 8 P2 + P1 V =b
SOL 1.7.39
nodia.co.in =- 70.43c
Power factor = cos q = 0.334 SOL 1.7.40 CHECK
SOL 1.7.41
Option (B) is correct. for the dc potentiometer E \ l E1 = l 1 so, E2 l2 l E2 = E1 d 1 n = (1.18) # 680 = 1.34 V 600 l2 Option (C) is correct. Let the actual voltage and current are I1 and V1 respectively, then
Current in CC is 20 A 20 = I1 b
1000 1000 + 0.02 l
I1 = 20.0004 A - 20 A 200 = V1 - .02 # 20 = 200.40 Power measured Pm = V1 I1 = 20 (200.40) = 4008 W Load power PL = 20 # 200 = 4000 W % Change = Pm - PL = 4008 - 4000 # 100 4000 PL = 0.2% more SOL 1.7.42
Option (C) is correct.
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Page 365
We have to obtain n = I I1
I1 = Rsh = 100 = 1 Rm 1000 10 I2 I1 + I 2 = I I1 + 10I1 = I 11I1 = I n = I = 11 I1 SOL 1.7.43
Option (B) is correct. In the following configuration
nodia.co.in 1 = 1 jw C 2p # 100 # 103 # 10 # 10- 12 writing node equation at P VP - 10 + V 1 + 1 - j = 0 Pb 100 100 500 159 l Rectance Xc =
SOL 1.7.44
10 - VP = VP (1.2 - j0.628) 10 = (2.2 - j0.628) VP VP 10 = 4.38 V 2.28 Option (A) is correct. The torque on the coil is given by t = NIBA N " no. of turns, I " current, B " magnetic field, So,
SOL 1.7.45
N I B A
= 100 = 50 mA = 200 mTA " Area, = 10 mm # 20 mm
t = 100 # 50 # 10- 3 # 200 # 10- 3 # 200 # 10- 3 # 10- 3 = 200 # 10- 6 Nm
Option (C) is correct. Meter constant (A-sec/rev) is given by 14.4 = I speed I 14.4 = K # Power
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SOL 1.7.46
SOL 1.7.47
Page 366
Where ‘K ’ is the meter constant in rev/kWh. I 14.4 = K # VI 15 14.4 = K # 15 # 250 1 K = 250 # 14.4 1 K = = 1000 # 3600 = 1000 rev/kWh. 3600 250 # 14.4 b 1000 # 3600 l Option (B) is correct. For moving iron ameter full scale torque is given by tC = 1 I2 dL 2 dq 240 # 10- 6 = 1 (10) 2 dL 2 dq Change in inductance dL = 4.8 mH/radian dq
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Option (B) is correct. In the figure
VRY = 415+30c VBN = 415 +120c 3 Current in current coil IC = VRY = 415+30c Z 100+36.87c = 4.15+ - 6.87
` power factor = 0.8 cos f = 0.8 & f = 36.87c
Power = VI) = 415 +120c # 4.15+6.87c 3 = 994.3+126.87c Reading of wattmeter P = 994.3 ^cos 126.87ch = 994.3 (- 0.60) =- 597 W SOL 1.7.48
Option (A) is correct. For small values of phase angle IP = nf , f " Phase angle (radians) IS n " turns ratio Magnetizing ampere-turns = 200 So primary current IP = 200 # 1 = 200 amp Turns ratio n = 500 Secondary current IS = 5 amp 200 = 500f So 5 f (in degrees) = b 180 lb 200 l p 5 # 500 - 4.58c
SOL 1.7.49
Option (B) is correct.
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Page 367
Voltage appeared at secondary winding ES = IS # ZL = 5 # 1 = 5 Volts Voltage induced is given by ES = 5= f=
2 pfNf , f " flux 2 # 3.14 # 50 # 500 # f 5 = 45 # 10- 6 wb 2 # 3.14 # 25 # 103
SOL 1.7.50
Option (A) is correct. In PMCC instruments, as temperature increases the coil resistance increases. Swamp resistors are connected in series with the moving coil to provide temperature compensation. Swamping resistors is made of magnin, which has a zero-temperature coefficient.
SOL 1.7.51
Option () is correct. Effect of stray magnetic field is maximum when the operating field and stray fields are parallel.
SOL 1.7.52
Option (A) is correct. Let C1 = 300 pF
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1 wC 1 R Now when Cx is connected in parallel with variable resistor C1 ' = 200 pF 1 Q = 120 = w (C1 ' + Cx ) R Q = 120 =
So
SOL 1.7.53
Option (B) is correct. Maximum frequency of input in dual slop A/D converter is given as where
so
SOL 1.7.54
C1 = C1 ' + C x 300 = 200 + Cx Cx = 100 pF
Tm = 2n TC fm = 1 " maximum frquency of input Tm fC = 1 " clock frequency TC f fm = Cn , n = 10 2 6 = 10 = 1 kHz (approax) 1024
Option (A) is correct. Kelvin Double bridge is used for measuring low values of resistances. (P " 2) Low values of capacitances is precisely measured by schering bridge (Q " 3)
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Page 368
Inductance of a coil with large time constant or high quality factor is measured by hay’s bridge (R " 5) SOL 1.7.55
Option (C) is correct. Full scale deflection is produced by a dc current of 1 mA (Idc) fs = 1 mA For full wave reactifier (Idc) fs = 2Im , Im "peak value of ac current p 1 mA = 2Im 3.14 Im = 1.57 mA Full scale ac current (Irms) fs = 1.57 = 1.11 mA 2
nodia.co.in V = (Rs + Rm) (Irms) fs 100 = (Rs + 100) (1.11 mA)
100 = Rs + 100 (1.11 mA) 100 # 900 = Rs + 100 Rs = 89.9 kW SOL 1.7.56
Option (B) is correct. First the current coil is connected in R-phase and pressure coil is connected between this phase and the neutral as shown below
reading of wattmeter W1 = IP VP cos q1 , cos q1 = 0.8 & q1 = 36.86c 400 = IL VL cos q1 3 400 = IL VL # 0.8 3
...(1)
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Page 369
Now when pressure coil is connected between B and Y-phases, the circuit is
phasor diagram
nodia.co.in angle q2 = 23.14c + 30c = 54.14c2 now wattmeter reading W2 = VYB IL cos q2 from equation (1) so
VL IL = 400 # 3 0.8 W2 = 400 # 3 # cos 53.14c 0.8 = 519.5 W
SOL 1.7.57
Option (C) is correct. In a moving-iron ammeter control torque is given as tc = Kq = 1 I2 dL 2 dq Where K " control spring constant q " deflection 2 Given that L = 10 + 3q - q 4 dL = 3 - q mH/rad b 2l dq
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So, tc = (25 # 10- 6) q = 1 (5) 2 b 3 - q l # 10- 6 2 2 2q = 3 - q 2 5q = 3 & q = 6 = 1.2 rad. 2 5 SOL 1.7.58
Option (B) is correct. Magnetizing current Im = 250 = 250 amp 1 Primary current I p = 500 amp Secondary current Is = 5 amp I Turn ratio n = p = 500 = 100 5 Is Total primary current (IT ) =
[primary current (I p)] 2 + [magnetising current (I m)] 2
nodia.co.in IT = =
I p2 + I m2
(500) 2 + (250) 2 = 559.01 amp
Turn ratio n' = IT = 559.01 = 111.80 5 Is Percentage ratio error 3 n = n - nl # 100 nl 100 - 111.80 100 =- 10.55% = # 111.80 SOL 1.7.59
Option (C) is correct. Power read by meter Pm = VI sin (3 - f) Where 3 "Phase angle between supply voltage and pressure coil flux. f "Phase angle of load Here 3 = 85c, f = 60c "a cos f = 0.5 So measured power Pm = 200 # 5 sin (85c - 60c) = 1100 sin 25c = 464.88 W PO = VI cos f = 220 # 5 # 0.5 = 550 W Error in measurement = Pm - PO = 464.88 - 550 =- 85.12 W For unity power factor cos f = 1
Actual power
f = 0c Pm = 220 # 5 sin (85c - 0c) = 1095.81 W PO = 220 # 5 cos 0c = 1100 Error in Measurement = 1095.81 - 1100 =- 4.19 W
So
SOL 1.7.60
Option (A) is correct.
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Page 371
We can obtain the Lissaju pattern (in X-Y mode) by following method. For f = 0c, Vx = Vxm sin wt Vy = Vym sin (wt + 0c) = sin wt Draw Vx and Vy as shown below
nodia.co.in Divide both Vy and Vx equal parts and match the corresponding points on the screen. Similarly for f = 90c Vx = Vxm sin wt Vy = Vym sin (wt + 90c)
Similarly for f = 3p 2
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SOL 1.7.61
Page 372
we can also obtain for 0 < f < 3p 2 Option (D) is correct. We can obtain the Lissaju pattern (in X-Y made) by following method.
nodia.co.in
Divide the wave forms appearing an channel X and channel Y in equal parts, match the corresponding points on the screen. We would get a straight line in X - Y mode. SOL 1.7.62
Option (C) is correct. In two wattmeters method angle between phase voltage and phase current is given by f = tan- 1 b 3 W2 - W1 l W2 + W1 here f =- 60c readings in option (C) only satisfies this equation. f = tan- 1 b 3 0 - 1000 l =- 60c 0 + 1000
SOL 1.7.63
Option (D) is correct.
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Speed (rev/sec) of the energy meter is given. S = K # power K " meter constant S = 10 rev = K # 450 100 sec 10 rev K = = 10 # 1000 # 3600 100 # 450 kWh 100 # 450 b 1000 # 3600 l = 800 rev/kWh SOL 1.7.64
Option (B) is correct. Power in a 3-phase three wire system, with balanced load can be measured by using two wattmeters. The load may be star or delta connected.
SOL 1.7.65
Option (C) is correct. Ameter configuration is given below
Here
SOL 1.7.66
nodia.co.in IR 500 Isn Im Ish 100 400 Rsh
= Im + Ish = 100 + Ish = 400 mA = Rsh Rm = Rsh 100 = 25 W
Option (A) is correct. Equivalent resistance when connected in parallel is R = R1 R 2 R1 + R 2 Let So
R1 + R2 = R sum R = R1 R2 = 10 # 5 = 3.33 W 15 R sum % error in R = DR1 (%) + DR2 (%) - DR sum (%) DR sum = (10 ! 5%) + (5 ! 10%) = (10 ! 0.5) + (5 ! 0.5) = 15 ! 0.1 DR sum (%) = 15 ! 1 # 100% = 15 ! 6.66% 15 % error in R = 5% + 10% - 6.66% = 8.33% value of R = 3.33 ! 8.33% = 3.05 W to 3.61 W ***********
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8 ANALOG & DIGITAL ELECTRONICS
YEAR 2013
ONE MARK
MCQ 1.8.1
A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles (A) and AND gate (B) an OR gate (C) an XOR gate (D) a NAND gate
MCQ 1.8.2
In a voltage-voltage feedback as shown below, which one of the following statements is TRUE if the gain k is increased?
(A) (B) (C) (D) MCQ 1.8.3
The The The The
input input input input
increases and output impedance decreases increases and output impedance also increases decreases and output impedance also decreases decreases and output impedance increases
In the circuit shown below what is the output voltage ^Vouth if a silicon transistor Q and an ideal op-amp are used?
(A) - 15 V (C) + 0.7 V YEAR 2013 MCQ 1.8.4
impedance impedance impedance impedance
(B) - 0.7 V (D) + 15 V TWO MARKS
In the circuit shown below the op-amps are ideal. Then, Vout in Volts is
GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
(A) 4 (C) 8 MCQ 1.8.5
Page 375
(B) 6 (D) 10
In the circuit shown below, Q1 has negligible collector-to-emitter saturation voltage and the diode drops negligible voltage across it under forward bias. If Vcc is + 5 V , X and Y are digital signals with 0 V as logic 0 and Vcc as logic 1, then the Boolean expression for Z is
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(A) XY (C) XY MCQ 1.8.6
The clock frequency applied to the digital circuit shown in the figure below is 1 kHz. If the initial state of the output of the flip-flop is 0, then the frequency of the output waveform Q in kHz is
(A) 0.25 (C) 1 MCQ 1.8.7
(B) XY (D) XY
(B) 0.5 (D) 2
A voltage 1000 sin wt Volts is applied across YZ . Assuming ideal diodes, the voltage measured across WX in Volts, is
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(A) sin wt (C) ^sin wt - sin wt h /2 MCQ 1.8.8
Page 376
(B) _sin wt + sin wt i /2 (D) 0 for all t
In the circuit shown below, the knee current of the ideal Zener dioide is 10 mA . To maintain 5 V across RL , the minimum value of RL in W and the minimum power rating of the Zener diode in mW, respectively, are
nodia.co.in (A) 125 and 125 (C) 250 and 125
(B) 125 and 250 (D) 250 and 250
YEAR 2012 MCQ 1.8.9
ONE MARK
/
In the sum of products function f (X, Y, Z) = (2, 3, 4, 5), the prime implicants are (A) XY, XY (B) XY, X Y Z , XY Z (C) XY Z , XYZ, XY
MCQ 1.8.10
(D) XY Z , XYZ, XY Z , XY Z
The i -v characteristics of the diode in the circuit given below are v - 0.7 A, v $ 0.7 V i = * 500 0A v < 0. 7 V
The current in the circuit is (A) 10 mA (C) 6.67 mA
(B) 9.3 mA (D) 6.2 mA
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MCQ 1.8.11
The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B . The number of combinations for which the output is logic 1, is (A) 4 (B) 6 (C) 8 (D) 10
MCQ 1.8.12
Consider the given circuit
In this circuit, the race around (A) does not occur (B) occur when CLK = 0 (C) occur when CLK = 1 and A = B = 1 (D) occur when CLK = 1 and A = B = 0
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YEAR 2012 MCQ 1.8.13
The voltage gain Av of the circuit shown below is
(A) Av . 200 (C) Av . 20 MCQ 1.8.14
TWO MARKS
(B) Av . 100 (D) Av . 10
The state transition diagram for the logic circuit shown is
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MCQ 1.8.15
Page 378
The circuit shown is a
1 rad/s (R1 + R2) C (B) high pass filter with f3dB = 1 rad/s R1 C (C) low pass filter with f3dB = 1 rad/s R1 C 1 (D) high pass filter with f3dB = rad/s (R1 + R2) C (A) low pass filter with f3dB =
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YEAR 2011 MCQ 1.8.16
ONE MARK
A low-pass filter with a cut-off frequency of 30 Hz is cascaded with a high pass filter with a cut-off frequency of 20 Hz. The resultant system of filters will function as (A) an all – pass filter (B) an all – stop filter (C) an band stop (band-reject) filter (D) a band – pass filter
MCQ 1.8.17
The CORRECT transfer characteristic is
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MCQ 1.8.18
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The output Y of the logic circuit given below is
(A) 1 (C) X
(B) 0 (D) X
YEAR 2011 MCQ 1.8.19
Page 379
TWO MARKS
A portion of the main program to call a subroutine SUB in an 8085 environment is given below. h LXI D, DISP LP : CALL SUB LP+3 h It is desired that control be returned to LP+DISP+3 when the RET instruction is executed in the subroutine. The set of instructions that precede the RET instruction in the subroutine are
POP D (A) DAD H PUSH D
POP DAD INX (B) INX INX PUSH
POP H (C) DAD D PUSH H
XTHL INX D (D) INX D INX D Test and Study XTHL
H D H H H H
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MCQ 1.8.20
Page 380
The transistor used in the circuit shown below has a b of 30 and ICBO is negligible
If the forward voltage drop of diode is 0.7 V, then the current through collector will be (A) 168 mA (B) 108 mA (C) 20.54 mA (D) 5.36 mA MCQ 1.8.21
A two bit counter circuit is shown below
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It the state QA QB of the counter at the clock time tn is ‘10’ then the state QA QB of the counter at tn + 3 (after three clock cycles) will be (A) 00 (B) 01 (C) 10 (D) 11 MCQ 1.8.22
A clipper circuit is shown below.
Assuming forward voltage drops of the diodes to be 0.7 V, the input-output transfer characteristics of the circuit is
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YEAR 2010 MCQ 1.8.23
(B) 6 V (D) 12.12 V
Assuming that the diodes in the given circuit are ideal, the voltage V0 is
(A) 4 V (C) 7.5 V YEAR 2010 MCQ 1.8.25
ONE MARK
Given that the op-amp is ideal, the output voltage vo is
(A) 4 V (C) 7.5 V MCQ 1.8.24
Page 381
(B) 5 V (D) 12.12 V TWO MARKS
The transistor circuit shown uses a silicon transistor with VBE = 0.7, IC . IE and
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a dc current gain of 100. The value of V0 is
(A) 4.65 V (C) 6.3 V MCQ 1.8.26
(B) 5 V (D) 7.23 V
The TTL circuit shown in the figure is fed with the waveform X (also shown). All gates have equal propagation delay of 10 ns. The output Y of the circuit is
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MCQ 1.8.27
When a “CALL Addr” instruction is executed, the CPU carries out the following sequential operations internally : Note: (R) means content of register R ((R)) means content of memory location pointed to by R. PC means Program Counter SP means Stack Pointer (A) (SP) incremented (B) (PC)!Addr (PC)!Addr ((SP))!(PC) ((SP))!(PC) (SP) incremented (C) (PC)!Addr (D) ((SP))!(PC) (SP) incremented (SP) incremented ((SP))!(PC) (PC)!Addr
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Statement For Linked Answer Questions: 28 & 29 The following Karnaugh map represents a function F .
MCQ 1.8.28
MCQ 1.8.29
A minimized form of the function F is (A) F = X Y + YZ (C) F = X Y + Y Z
(B) F = X Y + YZ (D) F = X Y + Y Z
Which of the following circuits is a realization of the above function F ?
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YEAR 2009 MCQ 1.8.30
ONE MARK
The following circuit has a source voltage VS as shown in the graph. The current through the circuit is also shown.
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Page 384
The element connected between a and b could be
MCQ 1.8.31
The increasing order of speed of data access for the following device is (I) Cache Memory (II) CD-ROM (III) Dynamic RAM (IV) Processor Registers (V) Magnetic Tape (A) (V), (II), (III), (IV), (I) (B) (V), (II), (III), (I), (IV) (C) (II), (I), (III), (IV), (V) (D) (V), (II), (I), (III), (IV)
MCQ 1.8.32
The nature of feedback in the op-amp circuit shown is
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(A) Current-Current feedback (C) Current-Voltage feedback MCQ 1.8.33
(B) Voltage-Voltage feedback (D) Voltage-Current feedback
The complete set of only those Logic Gates designated as Universal Gates is (A) NOT, OR and AND Gates (B) XNOR, NOR and NAND Gates (C) NOR and NAND Gates (D) XOR, NOR and NAND Gates YEAR 2009
MCQ 1.8.34
TWO MARKS
The following circuit has R = 10 kW, C = 10 mF . The input voltage is a sinusoidal at 50 Hz with an rms value of 10 V. Under ideal conditions, the current Is from
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Page 385
the source is
MCQ 1.8.35
(A) 10p mA leading by 90%
(B) 20p mA leading by 90%
(C) 10p mA leading by 90%
(D) 10p mA lagging by 90%
Transformer and emitter follower can both be used for impedance matching at the output of an audio amplifier. The basic relationship between the input power Pin and output power Pout in both the cases is (A) Pin = Pout for both transformer and emitter follower (B) Pin > Pout for both transformer and emitter follower (C) Pin < Pout for transformer and Pin = Pout for emitter follower (D) Pin = Pout for transformer and Pin < Pout for emitter follower
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MCQ 1.8.36
In an 8085 microprocessor, the contents of the Accumulator, after the following instructions are executed will become XRA A MVI B, F0 H SUB B (A) 01 H (B) 0F H (C) F0 H (D) 10 H
MCQ 1.8.37
An ideal op-amp circuit and its input wave form as shown in the figures. The output waveform of this circuit will be
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YEAR 2008 MCQ 1.8.38
Page 386
ONE MARK
The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure. (I) (II)
If such a diode is used in clipper circuit of figure given above, the output voltage V0 of the circuit will be
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YEAR 2008 MCQ 1.8.39
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In the voltage doubler circuit shown in the figure, the switch ‘S ’ is closed at t = 0 . Assuming diodes D1 and D2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero. The steady state voltage across capacitor C1 and C2 will be
(A) Vc1 = 10 V,Vc2 = 5 V (C) Vc1 = 5 V,Vc2 = 10 V MCQ 1.8.41
TWO MARKS
Two perfectly matched silicon transistor are connected as shown in the figure assuming the b of the transistors to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current I is
(A) 0 mA (C) 4.3 mA MCQ 1.8.40
Page 387
(B) Vc1 = 10 V,Vc2 =- 5 V (D) Vc1 = 5 V,Vc2 =- 10 V
The block diagrams of two of half wave rectifiers are shown in the figure. The transfer characteristics of the rectifiers are also shown within the block.
It is desired to make full wave rectifier using above two half-wave rectifiers. The resultants circuit will be
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A waveform generator circuit using OPAMPs is shown in the figure. It produces a triangular wave at point ‘P’ with a peak to peak voltage of 5 V for Vi = 0 V .
If the voltage Vi is made + 2.5 V, the voltage waveform at point ‘P’ will become
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Page 389
Statement for Linked Answer Questions 21 and 22. A general filter circuit is shown in the figure :
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MCQ 1.8.43
If R1 = R2 = RA and R3 = R4 = RB , the circuit acts as a (A) all pass filter (B) band pass filter (C) high pass filter (D) low pass filter
MCQ 1.8.44
The output of the filter in Q.21 is given to the circuit in figure : The gain v/s frequency characteristic of the output (vo) will be
MCQ 1.8.45
A 3-line to 8-line decoder, with active low outputs, is used to implement a 3-variable Boolean function as shown in the figure
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The simplified form of Boolean function F (A, B, C) implemented in ‘Product of Sum’ form will be (A) (X + Z) (X + Y + Z ) (Y + Z) (B) (X + Z ) (X + Y + Z) (Y + Z ) (C) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z ) (D) (X + Y + Z) (X + Y + Z ) (X + Y + Z) (X + Y + Z ) MCQ 1.8.46
The content of some of the memory location in an 8085 accumulator based system are given below Address
Content
g
g
26FE
00
26FF
01
2700
02
2701
03
2702
04
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g g The content of stack (SP), program counter (PC) and (H,L) are 2700 H, 2100 H and 0000 H respectively. When the following sequence of instruction are executed. 2100 H: DAD SP 2101 H: PCHL the content of (SP) and (PC) at the end of execution will be (A) PC = 2102 H, SP = 2700 H (B) PC = 2700 H, SP = 2700 H (C) PC = 2800 H, SP = 26FE H (D) PC = 2A02 H, SP = 2702 H YEAR 2007 MCQ 1.8.47
ONE MARK
The common emitter forward current gain of the transistor shown is bF = 100
The transistor is operating in (A) Saturation region (C) Reverse active region MCQ 1.8.48
(B) Cutoff region (D) Forward active region
The three-terminal linear voltage regulator is connected to a 10 W load resistor as
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Page 391
shown in the figure. If Vin is 10 V, what is the power dissipated in the transistor ?
(A) 0.6 W (C) 4.2 W MCQ 1.8.49
(B) 2.4 W (D) 5.4 W
The circuit shown in the figure is
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rV R1 < R2 r < R2 (B) a voltage source with voltage V R1 r < R2 V (C) a current source with current c R1 + R2 m r (D) a current source with current c R2 mV R1 + R2 r A, B, C and D are input, and Y is the output bit in the XOR gate circuit of the figure below. Which of the following statements about the sum S of A, B, C, D and Y is correct ? (A) a voltage source with voltage
MCQ 1.8.50
(A) S is always with zero or odd (B) S is always either zero or even (C) S = 1 only if the sum of A, B, C and D is even (D) S = 1 only if the sum of A, B, C and D is odd YEAR 2007 MCQ 1.8.51
TWO MARKS
The input signal Vin shown in the figure is a 1 kHz square wave voltage that alternates between + 7 V and - 7 V with a 50% duty cycle. Both transistor have the same current gain which is large. The circuit delivers power to the load resistor RL . What is the efficiency of this circuit for the given input ? choose the
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closest answer.
(A) 46% (C) 63% MCQ 1.8.52
(B) 55% (D) 92%
The switch S in the circuit of the figure is initially closed, it is opened at time t = 0 . You may neglect the zener diode forward voltage drops. What is the behavior of vout for t > 0 ?
nodia.co.in (A) It makes a transition from - 5 V (B) It makes a transition from - 5 V (C) It makes a transition from + 5 V (D) It makes a transition from + 5 V
to + 5 V at t = 12.98 ms to + 5 V at t = 2.57 ms to - 5 V at t = 12.98 ms to - 5 V at t = 2.57 ms
MCQ 1.8.53
The Octal equivalent of HEX and number AB.CD is (A) 253.314 (B) 253.632 (C) 526.314 (D) 526.632
MCQ 1.8.54
IC 555 in the adjacent figure is configured as an astable multi-vibrator. It is enabled to to oscillate at t = 0 by applying a high input to pin 4. The pin description is : 1 and 8-supply; 2-trigger; 4-reset; 6-threshold 7-discharge. The waveform appearing across the capacitor starting from t = 0 , as observed on a storage CRO is
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What are the states of the three ideal diodes of the circuit shown in figure ?
(A) D1 ON, D2 OFF, D3 OFF (C) D1 ON, D2 OFF, D3 ON MCQ 1.8.56
ONE MARK
(B) D1 OFF, D2 ON, D3 OFF (D) D1 OFF, D2 ON, D3 ON
For a given sinusoidal input voltage, the voltage waveform at point P of the clamper circuit shown in figure will be
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TWO MARKS
Assuming the diodes D1 and D2 of the circuit shown in figure to be ideal ones, the transfer characteristics of the circuit will be
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MCQ 1.8.58
Page 395
Consider the circuit shown in figure. If the b of the transistor is 30 and ICBO is 20 mA and the input voltage is + 5 V , the transistor would be operating in
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(A) saturation region (C) breakdown region
(B) active region (D) cut-off region
MCQ 1.8.59
A relaxation oscillator is made using OPAMP as shown in figure. The supply voltages of the OPAMP are ! 12 V . The voltage waveform at point P will be
MCQ 1.8.60
A TTL NOT gate circuit is shown in figure. Assuming VBE = 0.7 V of both the transistors, if Vi = 3.0 V, then the states of the two transistors will be
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(A) Q1 (B) Q1 (C) Q1 (D) Q1 MCQ 1.8.61
Page 396
ON and Q2 OFF reverse ON and Q2 OFF reverse ON and Q2 ON OFF and Q2 reverse ON
A student has made a 3-bit binary down counter and connected to the R-2R ladder type DAC, [Gain = (- 1 kW/2R)] as shown in figure to generate a staircase waveform. The output achieved is different as shown in figure. What could be the possible cause of this error ?
nodia.co.in (A) The resistance values are incorrect option. (B) The counter is not working properly (C) The connection from the counter of DAC is not proper (D) The R and 2R resistance are interchanged MCQ 1.8.62
A 4 # 1 MUX is used to implement a 3-input Boolean function as shown in figure. The Boolean function F (A, B, C) implemented is
(A) F (A, B, C) = S (1, 2, 4, 6) (C) F (A, B, C) = S (2, 4, 5, 6)
(B) F (A, B, C) = S (1, 2, 6) (D) F (A, B, C) = S (1, 5, 6)
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MCQ 1.8.63
A software delay subroutine is written as given below : DELAY : MVI H, 255D MVI L, 255D LOOP : DCR L JNZ LOOP DCR H JNZ LOOP How many times DCR L instruction will be executed ? (A) 255 (B) 510 (C) 65025 (D) 65279
MCQ 1.8.64
In an 8085 A microprocessor based system, it is desired to increment the contents of memory location whose address is available in (D,E) register pair and store the result in same location. The sequence of instruction is (A) XCHG (B) XCHG INR M INX H (C) INX D (D) INR M XCHG XCHG
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YEAR 2005 MCQ 1.8.65
ONE MARK
Assume that D1 and D2 in figure are ideal diodes. The value of current is
(A) 0 mA (C) 1 mA
(B) 0.5 mA (D) 2 mA
MCQ 1.8.66
The 8085 assembly language instruction that stores the content of H and L register into the memory locations 2050H and 2051H , respectively is (A) SPHL 2050H (B) SPHL 2051H (C) SHLD 2050H (D) STAX 2050H
MCQ 1.8.67
Assume that the N-channel MOSFET shown in the figure is ideal, and that its threshold voltage is + 1.0 V the voltage Vab between nodes a and b is
(A) 5 V (C) 1 V
(B) 2 V (D) 0 V
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 1.8.68
The digital circuit shown in the figure works as
(A) JK flip-flop (C) T flip-flop
(B) Clocked RS flip-flop (D) Ring counter
YEAR 2005 MCQ 1.8.69
Page 398
TWO MARKS
The common emitter amplifier shown in the figure is biased using a 1 mA ideal current source. The approximate base current value is
nodia.co.in (A) 0 mA (C) 100 mA MCQ 1.8.70
Consider the inverting amplifier, using an ideal operational amplifier shown in the figure. The designer wishes to realize the input resistance seen by the small-signal source to be as large as possible, while keeping the voltage gain between - 10 and - 25 . The upper limit on RF is 1 MW. The value of R1 should be
(A) Infinity (C) 100 kW MCQ 1.8.71
(B) 10 mA (D) 1000 mA
(B) 1 MW (D) 40 kW
The typical frequency response of a two-stage direct coupled voltage amplifier is as shown in figure
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MCQ 1.8.72
Page 399
In the given figure, if the input is a sinusoidal signal, the output will appear as shown
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MCQ 1.8.73
Select the circuit which will produce the given output Q for the input signals X1 and X2 given in the figure
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MCQ 1.8.74
Page 400
If X1 and X2 are the inputs to the circuit shown in the figure, the output Q is
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(A) X1 + X2 (C) X1 : X2 MCQ 1.8.75
(B) X1 : X2 (D) X1 : X2
In the figure, as long as X1 = 1 and X2 = 1, the output Q remains
(A) at 1 (C) at its initial value
(B) at 0 (D) unstable
Data for Q. 76 and Q. 77 are given below. Solve the problems and choose the correct option. Assume that the threshold voltage of the N-channel MOSFET shown in figure is + 0.75 V. The output characteristics of the MOSFET are also shown
MCQ 1.8.76
The transconductance of the MOSFET is (A) 0.75 ms (B) 1 ms
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MCQ 1.8.77
(C) 2 ms
(D) 10 ms
The voltage gain of the amplifier is (A) + 5 (C) + 10
(B) - 7.5 (D) - 10
YEAR 2004 MCQ 1.8.78
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The feedback used in the circuit shown in figure can be classified as
(A) shunt-series feedback (C) series-shunt feedback MCQ 1.8.81
(B) 3.3 mA (D) 0 mA
Two perfectly matched silicon transistor are connected as shown in figure. The value of the current I is
(A) 0 mA (C) 4.3 mA MCQ 1.8.80
ONE MARK
The current through the Zener diode in figure is
(A) 33 mA (C) 2 mA MCQ 1.8.79
Page 401
(B) shunt-shunt feedback (D) series-series feedback
The digital circuit using two inverters shown in figure will act as
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(A) a bistable multi-vibrator (B) an astable multi-vibrator (C) a monostable multi-vibrator (D) an oscillator MCQ 1.8.82
The voltage comparator shown in figure can be used in the analog-to-digital conversion as
(A) a 1-bit quantizer (B) a 2-bit quantizer (C) a 4-bit quantizer (D) a 8-bit quantizer
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YEAR 2004 MCQ 1.8.83
Assuming that the diodes are ideal in figure, the current in diode D1 is
(A) 9 mA (C) 0 mA MCQ 1.8.84
TWO MARKS
(B) 5 mA (D) - 3 mA
The trans-conductance gm of the transistor shown in figure is 10 mS. The value of the input resistance Rin is
(A) 10.0 kW
(B) 8.3 kW
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(C) 5.0 kW MCQ 1.8.85
(D) 2.5 kW
The value of R for which the PMOS transistor in figure will be biased in linear region is
(A) 220 W (C) 680 W MCQ 1.8.86
Page 403
(B) 470 W (D) 1200 W
In the active filter circuit shown in figure, if Q = 1, a pair of poles will be realized with w0 equal to
nodia.co.in (A) 1000 rad/s (C) 10 rad/s MCQ 1.8.87
The input resistance Rin = vx /ix of the circuit in figure is
(A) + 100 kW (C) + 1 MW MCQ 1.8.88
(B) - 100 kW (D) - 1 MW
The simplified form of the Boolean expression Y = (A $ BC + D) (A $ D + B $ C ) can be written as (A) A $ D + B $ C $ D (C) (A + D) (B $ C + D )
MCQ 1.8.89
(B) 100 rad/s (D) 1 rad/s
(B) AD + B $ C $ D (D) A $ D + BC $ D
A digit circuit which compares two numbers A3 A2 A1 A0 and B 3 B2 B1 B 0 is shown in figure. To get output Y = 0 , choose one pair of correct input numbers.
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(A) 1010, 1010 (C) 0010, 0010 MCQ 1.8.90
Page 404
(B) 0101, 0101 (D) 1010, 1011
The digital circuit shown in figure generates a modified clock pulse at the output. Choose the correct output waveform from the options given below.
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MCQ 1.8.91
If the following program is executed in a microprocessor, the number of instruction cycle it will take from START to HALT is START MVI A, 14H ; Move 14 H to register A SHIFT RLC ; Rotate left without carry JNZ SHIFT ; Jump on non-zero to SHIFT HALT (A) 4 (B) 8 (C) 13 (D) 16
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 1.8.92
In the Schmitt trigger circuit shown in figure, if VCE (sat) = 0.1 V , the output logic low level (VOL) is
(A) 1.25 V (C) 2.50 V
(B) 1.35 V (D) 5.00 V
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YEAR 2003 MCQ 1.8.93
Page 405
ONE MARK
The variation of drain current with gate-to-source voltage (ID - VGS characteristic) of a MOSFET is shown in figure. The MOSFET is
(A) an n-channel depletion mode device (B) an n-channel enhancement mode device (C) an p-channel depletion mode device (D) an p-channel enhancement mode device MCQ 1.8.94
In the circuit of figure, assume that the transistor has hfe = 99 and VBE = 0.7 V. The value of collector current IC of the transistor is approximately
(A) [3.3/3.3] mA (C) [3.3/.33] mA MCQ 1.8.95
(B) [3.3/(3.3+3.3)] mA (D) [3.3(33+3.3)] mA
For the circuit of figure with an ideal operational amplifier, the maximum phase shift of the output vout with reference to the input vin is
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(B) - 90c (D) ! 180c
(A) 0c (C) + 90c MCQ 1.8.96
Page 406
Figure shows a 4 to 1 MUX to be used to implement the sum S of a 1-bit full adder with input bits P and Q and the carry input Cin . Which of the following combinations of inputs to I0, I1, I2 and I3 of the MUX will realize the sum S ?
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(A) I0 = I1 = Cin; I2 = I3 = Cin (C) I0 = I3 = Cin; I1 = I2 = Cin MCQ 1.8.97
(B) I0 = I1 = C in; I2 = I3 = Cin (D) I0 = I3 = C in; I1 = I2 = Cin
When a program is being executed in an 8085 microprocessor, its Program Counter contains (A) the number of instructions in the current program that have already been executed (B) the total number of instructions in the program being executed. (C) the memory address of the instruction that is being currently executed (D) the memory address of the instruction that is to be executed next YEAR 2003
MCQ 1.8.98
For the n-channel enhancement MOSFET shown in figure, the threshold voltage Vth = 2 V. The drain current ID of the MOSFET is 4 mA when the drain resistance RD is 1 kW.If the value of RD is increased to 4 kW, drain current ID will become
(A) 2.8 mA (C) 1.4 mA MCQ 1.8.99
TWO MARKS
(B) 2.0 mA (D) 1.0 mA
Assuming the operational amplifier to be ideal, the gain vout /vin for the circuit shown in figure is
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(A) - 1 (C) - 100 MCQ 1.8.100
Page 407
(B) - 20 (D) - 120
A voltage signal 10 sin wt is applied to the circuit with ideal diodes, as shown in figure, The maximum, and minimum values of the output waveform Vout of the circuit are respectively
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(A) + 10 V and - 10 V (C) + 7 V and - 4 V MCQ 1.8.101
(B) + 4 V and - 4 V (D) + 4 V and - 7 V
The circuit of figure shows a 555 Timer IC connected as an astable multi-vibrator. The value of the capacitor C is 10 nF. The values of the resistors RA and RB for a frequency of 10 kHz and a duty cycle of 0.75 for the output voltage waveform are
(A) RA = 3.62 kW, RB = 3.62 kW (B) RA = 3.62 kW, RB = 7.25 kW (C) RA = 7.25 kW, RB = 3.62 kW (D) RA = 7.25 kW, RB = 7.25 kW MCQ 1.8.102
The boolean expression X Y Z + XY Z + XYZ + XY Z + XYZ can be simplified to (A) XZ + X Z + YZ (B) XY + Y Z + YZ (C) XY + YZ + XZ (D) XY + YZ + X Z
MCQ 1.8.103
The shift register shown in figure is initially loaded with the bit pattern 1010. Subsequently the shift register is clocked, and with each clock pulse the pattern gets shifted by one bit position to the right. With each shift, the bit at the serial
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Page 408
input is pushed to the left most position (msb). After how many clock pulses will the content of the shift register become 1010 again ?
(A) 3 (C) 11 MCQ 1.8.104
(B) 7 (D) 15
An X-Y flip-flop, whose Characteristic Table is given below is to be implemented using a J-K flip flop
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(A) J = X, K = Y (C) J = Y, K = X
(B) J = X, K = Y (D) J = Y , K = X
MCQ 1.8.105
A memory system has a total of 8 memory chips each with 12 address lines and 4 data lines, The total size of the memory system is (A) 16 kbytes (B) 32 kbytes (C) 48 kbytes (D) 64 kbytes
MCQ 1.8.106
The following program is written for an 8085 microprocessor to add two bytes located at memory addresses 1FFE and 1FFF LXI H, 1FFE MOV B, M INR L MOV A, M ADD B INR L MOV M, A XOR A On completion of the execution of the program, the result of addition is found (A) in the register A (B) at the memory address 1000 (C) at the memory address 1F00 (D) at the memory address 2000 YEAR 2002
MCQ 1.8.107
ONE MARK
The frequency of the clock signal applied to the rising edge triggered D-flip-flop shown in Figure is 10 kHz. The frequency of the signal available at Q is.
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(A) 10 kHz (C) 20 kHz MCQ 1.8.108
MCQ 1.8.109
Page 409
(B) 2.5 kHz (D) 5 kHz
The forward resistance of the diode shown in Figure is 5 W and the remaining parameters are same at those of an ideal diode. The dc component of the source current is
(A) Vm (B) Vm 50p 50p 2 Vm (C) (D) 2Vm 50p 100p 2 The cut-in voltage of both zener diode DZ and diode D shown in Figure is 0.7 V, while break-down voltage of DZ is 3.3 V and reverse break-down voltage of D is 50 V. The other parameters can be assumed to be the same as those of an ideal diode. The values of the peak output voltage (Vo) are
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(A) 3.3 V in the positive half cycle and 1.4 V in the negative half cycle. (B) 4 V in the positive half cycle and 5 V in the negative half cycle. (C) 3.3 V in both positive and negative half cycles. (D) 4 V in both positive and negative half cycle MCQ 1.8.110
The logic circuit used to generate the active low chip select (CS ) by an 8085 microprocessor to address a peripheral is shown in Figure. The peripheral will respond to addresses in the range.
(A) E000-EFFF (C) 1000-FFFF YEAR 2002 MCQ 1.8.111
(B) 000E-FFFE (D) 0001-FFF1 TWO MARKS
A first order, low pass filter is given with R = 50 W and C = 5 mF . What is the frequency at which the gain of the voltage transfer function of the filter is 0.25 ? (A) 4.92 kHz (B) 0.49 kHz
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(C) 2.46 kHz MCQ 1.8.112
Page 410
(D) 24.6 kHz
The output voltage (vo) of the Schmitt trigger shown in Figure swings between + 15 V and - 15 V . Assume that the operational amplifier is ideal. The output will change from + 15 V to - 15 V when the instantaneous value of the input sine wave is
(A) 5 V in the positive slope only (B) 5 V in the negative slope only (C) 5 V in the positive and negative slopes (D) 3 V in the positive and negative slopes. MCQ 1.8.113
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For the circuit shown in Figure, the boolean expression for the output Y in terms of inputs P, Q, R and S is
(A) P + Q + R + S (C) (P + Q ) (R + S )
(B) P + Q + R + S (D) (P + Q) (R + S)
Common Data Questions Q.114-116* For the circuit shown in Figure, IE = 1 mA, b = 99 and VBE = 0.7 V
MCQ 1.8.114
The current through RC is (A) 0.99 mA
(B) 1.1 mA
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MCQ 1.8.115
MCQ 1.8.116
(C) 1.20 mA
(D) 1 mA
Output voltage V0 will be (A) 16.1 Volt (C) 13.9 Volt
(B) 14 Volt (D) None of these
Value of resistance RF is (A) 110.9 kW (C) 130.90 kW
(B) 124.5 kW (D) None of these
Page 411
Common data question Q.117-119*. The following network is used as a feedback circuit in an oscillator shown in figure to generate sinusoidal oscillations. Assuming that the operation amplifier is ideal. given that R = 10 kW and C = 100 pF
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The transfer function
Vy of the first network is Vx
jwCR (1 - w R2 C 2) + j3wCR jwCR (C) 1 + j3wCR
(A)
MCQ 1.8.118
MCQ 1.8.119
MCQ 1.8.120
2
The frequency of oscillation will be (A) 1 RC (C) 1 4RC Value of RF is (A) 1 kW (C) 2 kW
jwCR (1 - w R2 C 2) + j2wCR jwCR (D) 1 + j2wCR (B)
(B)
2
1 2RC
(D) None of these (B) 4 kW (D) 8 kW
*The ripple counter shown in figure is made up of negative edge triggered J-K flip-flops. The signal levels at J and K inputs of all the flip flops are maintained at logic 1. Assume all the outputs are cleared just prior to applying the clock
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Page 412
signal. module no. of the counter is:
(A) 7 (C) 4 MCQ 1.8.121
(B) 5 (D) 8
*In Figure , the ideal switch S is switched on and off with a switching frequency f = 10 kHz . The switching time period is T = tON + tOFF ms. The circuit is operated in steady state at the boundary of continuous and discontinuous conduction, so that the inductor current i is as shown in Figure. Values of the on-time tON of the switch and peak current ip . are
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(A) 63.33 msec , 63.33 A (C) 66.66 msec , 66.66 mA
(B) 63.33 msec , 63.33 mA (D) none of these
Common Data Questions Q.122-123* In the circuit shown in Figure, the source I is a dc current source.The switch S is operated with a time period T and a duty ratio D . You may assume that the capacitance C has a finite value which is large enough so that the voltage. VC has negligible ripple, calculate the following under steady state conditions, in terms of D , I and R
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MCQ 1.8.122
The voltage Vc, with the polarity shown in Figure, (B) I (1 - DT) (A) I C C (C) I (1 - D) T (D) - I T C C
MCQ 1.8.123
The average output voltage V0, with the polarity shown in figure (B) - I D2 T (A) - I T C 2C (C) I (1 - DT) (D) I (1 - D) T 2C 2C YEAR 2001
MCQ 1.8.124
ONE MARK
In the single-stage transistor amplifier circuit shown in Figure, the capacitor CE is removed. Then, the ac small-signal mid-band voltage gain of the amplifier
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(A) increase (C) is unaffected
(B) decreases (D) drops to zero
MCQ 1.8.125
Among the following four, the slowest ADC (analog-to-digital converter) is (A) parallel-comparator (i.e. flash) type (B) successive approximation type (C) integrating type (D) counting type
MCQ 1.8.126
The output of a logic gate is “1” when all its inputs are at logic “0”. The gate is either (A) a NAND or an EX-OR gate (B) a NOR or an EX-OR gate (C) an AND or an EX-NOR gate (D) a NOR or an EX-NOR gate
MCQ 1.8.127
The output f of the 4-to-1 MUX shown in Figure is
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(A) xy + x (C) x + y MCQ 1.8.128
(B) x + y (D) xy + x
An op-amp has an open-loop gain of 105 and an open-loop upper cut-off frequency of 10 Hz. If this op-amp is connected as an amplifier with a closed-loop gain of 100, then the new upper cut-off frequency is (A) 10 Hz (B) 100 Hz (C) 10 kHz (D) 100 kHz YEAR 2001
MCQ 1.8.129
MCQ 1.8.130
MCQ 1.8.131
Page 414
TWO MARKS
For the oscillator circuit shown in Figure, the expression for the time period of oscillation can be given by (where t = RC )
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(A) t ln 3 (C) t ln 2
(B) 2t ln 3 (D) 2t ln 2
An Intel 8085 processor is executing the MVI A, 10 H MVI B, 10 H BACK: NOP ADD B RLC INC BACK HLT The number of times that the operation (A) 1 (C) 3
program given below.
NOP will be executed is equal to (B) 2 (D) 4
A sample-and-hold (S/H) circuit, having a holding capacitor of 0.1 nF, is used at the input of an ADC (analog-to-digital converter). The conversion time of the ADC is 1 m sec, and during this time, the capacitor should not loose more than 0.5% of the charge put across it during the sampling time. The maximum value of the input signal to the S/H circuit is 5 V. The leakage current of the S/H circuit should be less than
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(A) 2.5 mA (C) 25.0 mA
Page 415
(B) 0.25 mA (D) 2.5 mA
MCQ 1.8.132
An op-amp, having a slew rate of 62.8 V/ m sec , is connected in a voltage follower configuration. If the maximum amplitude of the input sinusoidal is 10 V, then the minimum frequency at which the slew rate limited distortion would set in at the output is (A) 1.0 MHz (B) 6.28 MHz (C) 10.0 MHz (D) 62.8 MHz
MCQ 1.8.133
An n-channel JFET, having a pinch off voltage (Vp ) of - 5 V , shows a transconductance (gm) of 1 mA/V when the applied gate -to-source voltage (VGS ) is - 3 V . Its maximum transconductance (in mA/V) is (A) 1.5 (B) 2.0 (C) 2.5 (D) 3.0
MCQ 1.8.134
*The circuit shown in the figure is a MOD-N ring counter. Value of N is (assume initial state of the counter is 1110 i.e. Q 3 Q2 Q1 Q 0 = 1110 ).
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(A) 4 (C) 7 MCQ 1.8.135
(B) 15 (D) 6
*For the op-amp circuit shown in Figure, determine the output voltage vo . Assume that the op-amps are ideal.
(A) - 8 V 7
(B) - 20 V 7
(C) - 10 V
(D) None of these
Common Data Questions Q.136-137*. The transistor in the amplifier circuit shown in Figure is biased at IC = 1 mA Use VT = kT/q = 26 mV, b0 = 200, r b = 0, and r 0 " 3
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MCQ 1.8.136
Small-signal mid-band voltage gain vo /vi is (A) - 8 (B) 38.46 (C) - 6.62 (D) - 1
MCQ 1.8.137
What is the required value of CE for the circuit to have a lower cut-off frequency of 10 Hz (A) 0.15 mF (B) 1.59 mF (C) 5 mF (D) 10 mF
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Common Data Questions Q.138-139* For the circuit shown in figure
MCQ 1.8.138
MCQ 1.8.139
The circuit shown is a (A) Low pass filter (C) Band Reject filter
(B) Band pass filter (D) High pass filter
If the above filter has a 3 dB frequency of 1 kHz, a high frequency input resistance of 100 kW and a high frequency gain of magnitude 10. Then values of R1, R2 and C respectively are :(A) 100 kW, 1000 kW, 15.9 nF (B) 10 kW, 100 kW, 0.11 mF (C) 100 kW, 1000 kW, 15.9 nF (D) none of these ************
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SOLUTION
SOL 1.8.1
Option (C) is correct. Let A denotes the position of switch at ground floor and B denotes the position of switch at upper floor. The switch can be either in up position or down position. Following are the truth table given for different combinations of A and B A
B
Y(Bulb)
up(1)
up(1)
OFF(0)
Down(0)
Down(0)
OFF(0)
up(1)
Down(0)
ON(1)
Down(0) up(1) ON(1) When the switches A and B are both up or both down, output will be zero (i.e. Bulb will be OFF). Any of the switch changes its position leads to the ON state of bulb. Hence, from the truth table, we get
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Y = A5B i.e., the XOR gate SOL 1.8.2
Option (A) is correct. The i/p voltage of the system is given as
Vin = V1 + Vf = V1 + k Vout = V1 + k A 0 V1 ^Vout = A 0 V1h = V1 ^1 + k A 0h Therefore, if k is increased then input voltage is also increased so, the input impedance increases. Now, we have Vout = A 0 V1 Vin = A0 ^1 + k A 0h A 0 Vin ^1 + k A 0h Since, Vin is independent of k when seen from output mode, the output voltage decreases with increase in k that leads to the decrease of output impedance. Thus, input impedance increases and output impedance decreases. =
SOL 1.8.3
Option (B) is correct. For the given ideal op-amp, negative terminal will be also ground (at zero voltage) and so, the collector terminal of the BJT will be at zero voltage. i.e., VC = 0 volt The current in 1 kW resistor is given by I = 5 - 0 = 5 mA 1 kW This current will flow completely through the BJT since, no current will flow into the ideal op-amp ( I/P resistance of ideal op-amp is infinity). So, for BJT we have
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Page 418
VC = 0 VB = 0 IC = 5 mA i.e.,the base collector junction is reverse biased (zero voltage) therefore, the collector current (IC ) can have a value only if base-emitter is forward biased. Hence, VBE = 0.7 volts & VB - VE = 0.7 & 0 - Vout = 0.7 or, Vout =- 0.7 volt SOL 1.8.4
Option (C) is correct.
nodia.co.in For the given ideal op-Amps we can assume V 2- = V 2+ = V2 (ideal) V 1+ = V 1- = V1 (ideal) So, by voltage division V1 = Vout # 1 2 Vout = 2V1 and, as the I/P current in Op-amp is always zero therefore, there will be no voltage drop across 1 KW in II op-amp i.e., V2 = 1 V Therefore, V1 - V2 = V2 - ^- 2h 1 1 or, V1 - 1 = 1 + 2 or, V1 = 4 So, Vout = 2V1 = 8 volt SOL 1.8.5
Option (B) is correct. For the given circuit, we can make the truth table as below X Y Z 0 0 0 0 1 1 1 0 0 1 1 0
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Page 419
Logic 0 means voltage is v = 0 volt and logic 1 means voltage is 5 volt For x = 0 , y = 0 , Transistor is at cut off mode and diode is forward biased. Since, there is no drop across forward biased diode. So, Z =Y=0 For x = 0 , y = 1, Again Transistor is in cutoff mode, and diode is forward biased. with no current flowing through resistor. So, Z =Y=1 For x = 1, y = 0 , Transistor is in saturation mode and so, z directly connected to ground irrespective of any value of Y . i.e., Z = 0 (ground) Similarly for X = Y = 1 Z = 0 (ground) Hence, from the obtained truth table, we get Z =XY SOL 1.8.6
Option (B) is correct. From the given logic diagram, we obtain
X = ^Q 5 Q h $ ^Q 5 Q h =0=1 So, the input is always ‘1’ at T , since, clock is - ve edge trigged therefore, at the negative edge Q changes its state as shown in waveform below
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SOL 1.8.7
Hence, as obtained from the waveform, time period of Q is double to that of CLK I/p and so, frequency is 12 of clock frequency Thus, fQ = FCLK = 1 = 0.5 kHz 2 2 Option (D) is correct. Given, the input voltage VYZ = 100 sin wt
For + ve half cycle VYZ > 0
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i.e., VY is a higher voltage than VZ So, the diode will be in cutoff region. Therefore, there will no voltage difference between X and W node. i.e., VWX = 0 Now, for - ve half cycle all the four diodes will active and so, X and W terminal is short circuited i.e., VWX = 0 Hence, VWX = 0 for all t SOL 1.8.8
Option (B) is correct.
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From the circuit, we have
Is = I Z + I L or, (1) I Z = Is - I L Since, voltage across zener diode is 5 V so, current through 100 W resistor is obtained as Is = 10 - 5 = 0.05 A 100 Therefore, the load current is given by IL = 5 RL Since, for proper operation, we must have IZ $ Iknes So, from Eq. (1), we write 0.05 A - 5 $ 10 mA RL 50 mA - 5 $ 10 mA RL 40 mA $ 5 RL 40 # 10-3 $ 5 RL RL 1 -3 # 5 40 # 10 5 # RL 40 # 10-3 or, 125 W # RL Therefore, minimum value of RL = 125 W Now, we know that power rating of Zener diode is given by PR = VZ IZ^maxh IZ^maxh is maximum current through zener diode in reverse bias. Maximum currrent through zener diode flows when load current is zero. i.e., IZ^maxh = Is = 10 - 5 = 0.05 100
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PR = 5 # 0.05 W = 250 mW
Therefore, SOL 1.8.9
SOL 1.8.10
SOL 1.8.11
Page 421
Option (A) is correct. Prime implicants are the terms that we get by solving K-map
F = XY + XY 1prime 44 2 44 3 implicants
Option (D) is correct. Let v > 0.7 V and diode is forward biased. Applying Kirchoff’s voltage law 10 - i # 1k - v = 0 10 - :v - 0.7 D (1000) - v = 0 500 10 - (v - 0.7) # 2 - v = 0 v = 11.4 = 3.8 V > 0.7 3 So, i = v - 0.7 = 3.8 - 0.7 = 6.2 mA 500 500 Option (B) is correct.
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(Assumption is true)
Y = 1, when A > B A = a1 a 0, B = b1 b 0 a1
a0
b1
b0
Y
0
1
0
0
1
1
0
0
0
1
1
0
0
1
1
1
1
0
0
1
1
1
0
1
1
1
1
1
0
1
Total combination = 6 SOL 1.8.12
Option (A) is correct. The given circuit is
Condition for the race-around
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It occurs when the output of the circuit (Y1, Y2) oscillates between ‘0’ and ‘1’ checking it from the options. 1. Option (A): When CLK = 0 Output of the NAND gate will be A1 = B1 = 0 = 1. Due to these input to the next NAND gate, Y2 = Y1 : 1 = Y1 and Y1 = Y2 : 1 = Y2 . If Y1 = 0 , Y2 = Y1 = 1 and it will remain the same and doesn’t oscillate. If Y2 = 0 , Y1 = Y2 = 1 and it will also remain the same for the clock period. So, it won’t oscillate for CLK = 0 . So, here race around doesn’t occur for the condition CLK = 0 . 2. Option (C): When CLK = 1, A = B = 1 A1 = B1 = 0 and so Y1 = Y2 = 1 And it will remain same for the clock period. So race around doesn’t occur for the condition. 3. Option (D): When CLK = 1, A = B = 0 So, A1 = B1 = 1 And again as described for Option (B) race around doesn’t occur for the condition. So, Option (A) will be correct. SOL 1.8.13
Option (D) is correct. DC Analysis :
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Using KVL in input loop, VC - 100IB - 0.7 = 0 VC = 100IB + 0.7 IC - IE = 13.7 - VC = (b + 1) IB 12k 13.7 - VC = 100I B 12 # 103 Solving equation (i) and (ii),
...(i)
...(ii)
IB = 0.01 mA Small Signal Analysis : Transforming given input voltage source into equivalent current source.
This is a shunt-shunt feedback amplifier.
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Given parameters, rp = VT = 25 mV = 2.5 kW IB 0.01 mA b 100 gm = = = 0.04 s rp 2.5 # 1000 Writing KCL at output node v0 + g v + v0 - vp = 0 m p RC RF v 0 : 1 + 1 D + v p :gm - 1 D = 0 RC RF RF Substituting RC = 12 kW, RF = 100 kW, gm = 0.04 s v 0 (9.33 # 10-5) + v p (0.04) = 0 v 0 =- 428.72Vp Writing KCL at input node vi = v p + v p + v p - vo = v 1 + 1 + 1 - v 0 p: Rs Rs rp RF Rs rp RF D RF = v p (5.1 # 10-4) - v 0 RF Substituting Vp from equation (i)
...(i)
nodia.co.in vi = - 5.1 # 10-4 v - v 0 0 428.72 Rs RF
SOL 1.8.14
SOL 1.8.15
vi (source resistance) =- 1.16 # 10-6 v 0 - 1 # 10-5 v 0 Rs = 10 kW 10 # 103 vi =- 1.116 # 10-5 10 # 103 1 Av = v 0 = - 8.96 3 vi 10 # 10 # 1.116 # 10-5 Option (D) is correct. Let Qn + 1 is next state and Qn is the present state. From the given below figure. D = Y = AX 0 + AX1 Qn + 1 = D = AX 0 + AX1 Qn + 1 = A Qn + AQn X 0 = Q , X1 = Q If A = 0, (toggle of previous state) Qn + 1 = Qn If A = 1, Qn + 1 = Qn So state diagram is
Option (B) is correct. First we obtain the transfer function.
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Page 424
0 - Vi (jw) 0 - Vo (jw) =0 + 1 +R R2 1 jw C Vo (jw) - Vi (jw) = 1 +R R2 1 jw C Vo (jw) =-
Vi (jw) R2 R1 - j 1 wC
1 " 3, so V = 0 o wC 1 " 0, so V (jw) =- R2 V (jw) At w " 3 (higher frequencies), o R1 i wC The filter passes high frequencies so it is a high pass filter. H (jw) = Vo = - R2 Vi R1 - j 1 wC At w " 0 (Low frequencies),
nodia.co.in H (3) = - R2 = R2 R1 R1
At 3 dB frequency, gain will be
So,
2 times of maximum gain 6H (3)@
H ^ jw0h = 1 H (3) 2 R2 = 1 b R2 l 2 R1 R 12 + 21 2 w0 C 2R 12 = R 12 + w0 =
SOL 1.8.16
Option (D) is correct.
1 & R2 = 1 1 w C2 w 2C 2 2 0
1 R1 C
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Page 425
nodia.co.in So, it will act as a Band pass filter. SOL 1.8.17
Option (D) is correct.
The first half of the circuit is a differential amplifier (negative feedback) Va =- (Vi) Second op-amp has a positive feedback, so it acts as an schmitt trigger. Since Va =- Vi this is a non-inverting schmitt trigger. Threshold value VTH = 12 = 6 V 2 VTL =- 6 V SOL 1.8.18
Option (A) is correct.
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Page 426
Y = X5X = X X + XX = XX + X X = X+X = 1 SOL 1.8.19
Option (C) is correct. LXI D, DISP LP : CALL SUB LP + 3 When CALL SUB is executed LP+3 value is pushed(inserted) in the stack. POP H & HL = LP + 3 DAD D
& HL = HL + DE = LP + 3 + DE
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PUSH H SOL 1.8.20
& The last two value of the stack will be HL value i.e, LP + DISP + 3
Option (D) is correct. Zener Diode is used as stabilizer. The circuit is assumed to be as
We can see that both BE and BC Junction are forwarded biased. So the BJT is operating in saturation. Collector current IC = 12 - 0.2 = 5.36 mA 2.2k Y bIB Note:- In saturation mode IC SOL 1.8.21
Option (C) is correct. The characteristics equation of the JK flip-flop is Q n + 1 = JQ n + KQn From figure it is clear that
Qn + 1 is the next state
J = QB ; K = QB The output of JK flip flop QA (n + 1) = QB QA + QB QA = QB (QA + QA) = QB Output of T flip-flop QB (n + 1) = Q A
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SOL 1.8.22
Page 427
Clock pulse
QA
QB
QA (n + 1)
QB (n + 1)
Initially(tn )
1
0
1
0
tn + 1
1
0
1
0
tn + 2
1
0
1
0
tn + 3
1
0
1
0
Option (C) is correct. We can obtain three operating regions depending on whether the Zener and PN diodes are forward biased or reversed biased. 1. vi #- 0.7 V , zener diode becomes forward biased and diode D will be off so the equivalent circuit looks like
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The output vo =- 0.7 V 2. When - 0.7 1 vi # 5.7 , both zener and diode D will be off. The circuit is
Output follows input i.e vo = vi Note that zener goes in reverse breakdown(i.e acts as a constant battery) only when difference between its p-n junction voltages exceeds 10 V. 3. When vi > 5.7 V , the diode D will be forward biased and zener remains off, the equivalent circuit is
vo = 5 + 0.7 = 5.7 V SOL 1.8.23
Option (B) is correct. Since the op-amp is ideal v+ = v- =+ 2 volt By writing node equation v- - 0 + v- - vo = 0 R 2R
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Page 428
2 + (2 - vo) = 0 R 2R 4 + 2 - vo = 0 vo = 6 volt SOL 1.8.24
Option (B) is correct. Given circuit is,
We can observe that diode D2 is always off, whether D1 ,is on or off. So equivalent circuit is.
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10 10 10 + 10 #
= 5 volt SOL 1.8.25
Option (A) is correct. By writing KVL equation for input loop (Base emitter loop) 10 - (10 kW) IB - VBE - V0 = 0 Emitter current IE = V0 100
...(1)
IC - IE = bIB V0 = 100I B 100 V0 IB = 10 # 103
So,
Put IB into equation (1) 10 - (10 # 103)
& SOL 1.8.26
V0 - 0.7 - V0 = 0 10 # 103 9.3 - 2V0 = 0 V0 = 9.3 = 4.65 A 2
Option (A) is correct. The circuit is
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Page 429
Output Y is written as Y = X5B Since each gate has a propagation delay of 10 ns.
nodia.co.in SOL 1.8.27
Option (D) is correct. CALL, Address performs two operations (1) PUSH PC & Save the contents of PC (Program Counter) into stack. SP = SP - 2 (decrement) ((SP)) ! (PC) (2) Addr stored in PC. (PC) ! Addr
SOL 1.8.28
Option (B) is correct. Function F can be minimized by grouping of all 1’s in K-map as following.
F = X Y + YZ SOL 1.8.29
Option (D) is correct. Since F = X Y + YZ In option (D)
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Page 430
SOL 1.8.30
Option (A) is correct. Figure shows current characteristic of diode during switching.
SOL 1.8.31
Option (B) is correct. The increasing order of speed is as following Magnetic tape> CD-ROM> Dynamic RAM>Cache Memory>Processor register
SOL 1.8.32
Option (B) is correct. Equivalent circuit of given amplifier
nodia.co.in Feedback samples output voltage and adds a negative feedback voltage (vfb) to input. So, it is a voltage-voltage feedback. SOL 1.8.33
Option () is correct. NOR and NAND gates considered as universal gates.
SOL 1.8.34
Option (A) is correct. Let voltages at positive and negative terminals of op-amp are V+ and V- respectively, then V+ = V- = Vs (ideal op-amp) In the circuit we have, V- - 0 + V- - V0 (s) = 0 1 R ` Cs j (RCs) V- + V- - V0 (s) = 0 (1 + RCs) Vs = V0 (s) Similarly current Is is, Is = Vs - V0 R Is = RCs Vs R Is = jwCVs Is = wCVs + + 90% Is = 2pf # 10 # 10 - 6 # 10 Is = 2 # p # 50 # 10 # 10 - 6 # 10
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Page 431
Is = 10p mA, leading by 90% SOL 1.8.35
Option (D) is correct. Input and output power of a transformer is same Pin = Pout for emitter follower, voltage gain (A v) = 1 current gain (Ai) > 1 Power (Pout) = Av Ai Pin Since emitter follower has a high current gain so Pout > Pin
SOL 1.8.36
Option (D) is correct. For the given instruction set, XRA A & XOR A with A & A = 0 MVI B, F0 H&B = F0 H SUB B &A = A - B A B 2’s complement of (- B) A + (- B) = A - B
SOL 1.8.37
= 00000000 = 1111 0 0 0 0 = 0 0 010 0 0 0 = 0 0 010 0 0 0 = 10 H
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Option (D) is correct. This is a schmitt trigger circuit, output can takes two states only. VOH =+ 6 volt VOL =- 3 volt Threshold voltages at non-inverting terminals of op-amp is given as VTH - 6 + VTH - 0 = 0 2 1 3VTH - 6 = 0 VTH = 2 V (Upper threshold) Similarly VTL - (- 3) VTL =0 + 2 1 3VTL + 3 = 0 VTL =- 1 V (Lower threshold) For Vin < 2 Volt, V0 =+ 6 Volt Vin > 2 Volt, V0 =- 3 Volt Vin < - 1 Volt V0 =+ 6 Volt Vin > - 1 Volt V0 =- 3 Volt Output waveform
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SOL 1.8.38
Page 432
Option (A) is correct. Assume the diode is in reverse bias so equivalent circuit is
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V0 = 10 sin wt # 10 = 5 sin wt 10 + 10 Due to resistor divider, voltage across diode VD < 0 (always). So it in reverse bias for given input. Output, V0 = 5 sin wt Output voltage
SOL 1.8.39
Option (C) is correct.
This is a current mirror circuit. Since b is high so IC1 = IC2, IB1 = IB2 VB = (- 5 + 0.7) =- 4.3 volt Diode D1 is forward biased. So, current I is, I = IC2 = IC1 0 - (- 4.3) = = 4.3 mA 1 SOL 1.8.40
Option (D) is correct. In positive half cycle of input, diode D1 is in forward bias and D2 is off, the equivalent circuit is
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Page 433
Capacitor C1 will charge upto + 5 volt. VC1 =+ 5 volt In negative halt cycle diode D1 is off and D2 is on.
Now capacitor VC2 will charge upto - 10 volt in opposite direction. SOL 1.8.41
Option () is correct. Let input Vin is a sine wave shown below
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According to given transfer characteristics of rectifiers output of rectifier P is.
Similarly output of rectifier Q is
Output of a full wave rectifier is given as
To get output V0
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Page 434
K - gain of op-amp V0 = K (- VP + VQ) So, P should connected at inverting terminal of op-amp and Q with non-inverting terminal. SOL 1.8.42
Option () is correct.
SOL 1.8.43
Option (C) is correct. For low frequencies, w " 0 , so 1 " 3 wC Equivalent circuit is,
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Applying node equation at positive and negative input terminals of op-amp. vA - vi + vA - vo = 0 R1 R2 2vA = vi + vo ,
a R1 = R 2 = R A
Similarly, vA - vi + vA - 0 = 0 R3 R4 2vA = vin ,
a R 3 = R 4 = RB
So, vo = 0 It will stop low frequency signals. For high frequencies, w " 3 , then 1 " 0 wC Equivalent circuit is,
Output, vo = vi So it will pass high frequency signal. This is a high pass filter. SOL 1.8.44
Option (D) is correct. In Q.7.21 cutoff frequency of high pass filter is given by, 1 wh = 2pRA C
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Page 435
Here given circuit is a low pass filter with cutoff frequency, 1 2 wL = = p R 2 R A AC 2p C 2 wL = 2wh When both the circuits are connected together, equivalent circuit is,
So this is is Band pass filter, amplitude response is given by.
SOL 1.8.45
Option (B) is correct. In SOP form, F is written as F = Sm (1, 3, 5, 6) = X Y Z + X YZ + XY Z + XYZ Solving from K- map
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F = X Z + Y Z + XYZ In POS form F = (Y + Z) (X + Z) (X + Y + Z ) Since all outputs are active low so each input in above expression is complemented F = (Y + Z ) (X + Z ) (X + Y + Z) SOL 1.8.46
Option (B) is correct. Given that SP = 2700 H PC = 2100 H HL = 0000 H Executing given instruction set in following steps, DAD SP & Add register pair (SP) to HL register HL = HL + SP HL = 0000 H + 2700 H HL = 2700 H PCHL & Load program counter with HL contents PC = HL = 2700 H So after execution contents are, PC = 2700 H, HL = 2700 H
SOL 1.8.47
Option (D) is correct. If transistor is in normal active region, base current can be calculated as following, By applying KVL for input loop, 10 - IC (1 # 103) - 0.7 - 270 # 103 IB = 0
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Page 436
bIB + 270 IB = 9.3 mA, ` IC = bIB IB (b + 270) = 9.3 mA IB = 9.3 mA = 0.025 mA 270 + 100 In saturation, base current is given by, 10 - IC (1) - VCE - IE (1) = 0 10 = I C (sat) 2
IC - IE VCE - 0
IC (sat) = 5 mA IC (sat) = 5 = .050 mA b 100 IB 1 IB(sat), so transistor is in forward active region. IB(sat) =
SOL 1.8.48
Option (B) is correct. In the circuit
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We can analyze that the transistor is operating in active region. VBE(ON) = 0.6 volt VB - VE = 0.6 6.6 - VE = 0.6 VE = 6.6 - 0.6 = 6 volt At emitter (by applying KCL), IE = IB + IL IE = 6 - 6.6 + 6 - 0.6 amp 1 kW 10 W VCE = VC - VE = 10 - 6 = 4 volt Power dissipated in transistor is given by. PT = VCE # IC = 4 # 0.6 = 2.4 W SOL 1.8.49
` IC - IE = 0.6 amp
Option (D) is correct. This is a voltage-to-current converter circuit. Output current depends on input voltage.
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Page 437
Since op-amp is ideal v+ = v- = v1 Writing node equation. v1 - v + v1 - 0 = 0 R1 R2 v1 c R1 + R2 m = V R1 R1 R2
R2 R1 + R2 m Since the op-amp is ideal therefore iL = i1 = v1 = V c R2 m r r R1 + R2 Option (D) is correct. In the circuit output Y is given as v1 = V c
SOL 1.8.50
Y = [A 5 B] 5 [C 5 D] Output Y will be 1 if no. of 1’s in the input is odd. SOL 1.8.51
Option () is correct. This is a class-B amplifier whose efficiency is given as h = p VP 4 VCC
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where VP " peak value of input signal VCC " supply voltage here VP = 7 volt, VCC = 10 volt so, SOL 1.8.52
h = p # 7 # 100 = 54.95% - 55% 10 4
Option (B) is correct. In the circuit the capacitor starts charging from 0 V (as switch was initially closed) towards a steady state value of 20 V. for t < 0 (initial) for t " 3 (steady state)
So at any time t , voltage across capacitor (i.e. at inverting terminal of op-amp) is given by vc (t) = vc (3) + [vc (0) - vc (3)] e
-t RC
-t RC
vc (t) = 20 (1 - e ) Voltage at positive terminal of op-amp v+ - vout v+ - 0 =0 + 10 100 v+ = 10 vout 11 Due to zener diodes, - 5 # vout # + 5 So, v+ = 10 (5) V 11
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Page 438
Transistor form - 5 V to + 5 V occurs when capacitor charges upto v+ . So 20 (1 - e - t/RC ) = 10 # 5 11 1 - e - t/RC = 5 22 17 = e - t/RC 22 t = RC ln ` 22 j = 1 # 103 # .01 # 10 - 6 # 0.257 = 2.57 msec 17 Voltage waveforms in the circuit is shown below
SOL 1.8.53
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Option (B) is correct. First convert the given number from hexadecimal to its binary equivalent, then binary to octal. Hexadecimal no. AB. CD 1 0 10 S 1 0 1 1 $ 1A BB 1 0B0C S 11 0 1 Binary equivalent S C A B D To convert in octal group three binary digits together as shown 0 1 0 1 0 1 0 11 $ 11 0 0 11 0 1 0 SSSSSS 5 2 3 6 3 2 So,
SOL 1.8.54
(AB.CD) H = (253.632) 8
Option (B) is correct. In a 555 astable multi vibrator circuit, charging of capacitor occurs through resistor (RA + RB) and discharging through resistor RB only. Time for charging and discharging is given as. TC = 0.693 (RA + RB) C = 0.693 RB C But in the given circuit the diode will go in the forward bias during charging, so the capacitor will charge through resistor RA only and discharge through RB only. a So
SOL 1.8.55
RA = RB TC = TD
Option (A) is correct. First we can check for diode D2 . Let diode D2 is OFF then the circuit is
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Page 439
In the above circuit diode D1 must be ON, as it is connected with 10 V battery now the circuit is
Because we assumed diode D2 OFF so voltage across it VD2 # 0 and it is possible only when D3 is off.
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So, all assumptions are true. SOL 1.8.56
Option (D) is correct. In the positive half cycle of input, Diode D1 will be reverse biased and equivalent circuit is.
Since there is no feed back to the op-amp and op-amp has a high open loop gain so it goes in saturation. Input is applied at inverting terminal so. VP =- VCC =- 12 V In negative half cycle of input, diode D1 is in forward bias and equivalent circuit is shown below.
Output VP = Vg + VOp-amp is at virtual ground so V+ = V- = 0 and VP = Vg = 0.7 V Voltage wave form at point P is
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SOL 1.8.57
Page 440
Option (A) is correct. In the circuit when Vi < 10 V, both D1 and D2 are off. So equivalent circuit is,
Output,
Vo = 10 volt
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When Vi > 10 V (D1 is in forward bias and D2 is off So the equivalent circuit is,
Output, Vo = Vi Transfer characteristic of the circuit is
SOL 1.8.58
Option (B) is correct. Assume that BJT is in active region, thevenin equivalent of input circuit is obtained as
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Page 441
Vth - Vi + Vth - (- 12) = 0 15 100 20Vth - 20Vi + 3Vth + 36 23Vth Vth Thevenin resistance Rth
=0 = 20 # 5 - 36 , Vi = 5 V = 2.78 V = 15 KW || 100 KW = 13.04 KW
So the circuit is
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Writing KVL for input loop
2.78 - Rth IB - 0.7 = 0 IB = 0.157 mA Current in saturation is given as, I IB(sat) = C(sat) b IC(sat) = 12.2 = 5.4 mA 2.2 So, IB(sat) = 5.45 mA = 0.181 mA 30 Since IB (sat) > IB , therefore assumption is true. SOL 1.8.59
Option (C) is correct. Here output of the multi vibrator is V0 = ! 12 volt Threshold voltage at positive terminal of op-amp can be obtained as following When output V0 =+ 12 V, equivalent circuit is,
writing node equation at positive terminal of op-amp Vth - 12 + Vth - 0 = 0 10 10
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Page 442
Vth = 6 volt (Positive threshold) So, the capacitor will charge upto 6 volt. When output V0 =- 12 V, the equivalent circuit is.
node equation Vth + 12 + Vth - 0 = 0 2 10
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5 Vth + 60 + Vth = 0 Vth =- 10 volt (negative threshold) So the capacitor will discharge upto - 10 volt. At terminal P voltage waveform is.
SOL 1.8.60
Option () is correct.
SOL 1.8.61
Option () is correct.
SOL 1.8.62
Option (A) is correct. Function F can be obtain as, F = I0 S1 S0 + I1 S1 S0 + I2 S1 S0 + I3 S1 S0 = AB C + A B C + 1 $ BC + 0 $ BC = AB C + A BC + BC = AB C + A BC + BC (A + A) = AB C + A BC + ABC + A BC = S (1, 2, 4, 6)
SOL 1.8.63
Option (A) is correct. MVI H and MVI L stores the value 255 in H and L registers. DCR L decrements L by 1 and JNZ checks whether the value of L is zero or not. So DCR L executed 255 times till value of L becomes ‘0’. Then DCR H will be executed and it goes to ‘Loop’ again, since L is of 8 bit so no more decrement possible and it terminates.
SOL 1.8.64
Option (A) is correct. XCHG & Exchange the contain of DE register pair with HL pair So now addresses of memory locations are stored in HL pair.
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INR M & pair. SOL 1.8.65
Page 443
Increment the contents of memory whose address is stored in HL
Option (A) is correct. From the circuit we can observe that Diode D1 must be in forward bias (since current is flowing through diode). Let assume that D2 is in reverse bias, so equivalent circuit is.
Voltage Vn is given by Vn = 1 # 2 = 2 Volt Vp = 0 Vn > Vp (so diode is in reverse bias, assumption is true) Current through D2 is ID2 = 0 SOL 1.8.66
SOL 1.8.67
SOL 1.8.68
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Option (C) is correct. SHLD transfers contain of HL pair to memory location. SHLD 2050 & L " M [2050H] H " M [2051H] Option (D) is correct. This is a N-channel MOSFET with VGS = 2 V VTH =+ 1 V VDS(sat) = VGS - VTH VDS(sat) = 2 - 1 = 1 V Due to 10 V source VDS > VDS(sat) so the NMOS goes in saturation, channel conductivity is high and a high current flows through drain to source and it acts as a short circuit. So, Vab = 0 Option (C) is correct. Let the present state is Q(t), so input to D-flip flop is given by, D = Q (t) 5 X Next state can be obtained as,
and
Q (t + 1) = D = Q (t) 5 X = Q (t) X + Q (t) X = Q (t), if X = 1 Q (t + 1) = Q (t), if X = 0
So the circuit behaves as a T flip flop. SOL 1.8.69
Option (B) is correct. Since the transistor is operating in active region. IE . bIB
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Page 444
IB = IE = 1 mA = 10 mA 100 b SOL 1.8.70
Option (C) is correct. Gain of the inverting amplifier is given by, 6 Av =- RF =- 1 # 10 , RF = 1 MW R1 R1 6
R1 =- 1 # 10 Av Av =- 10 to - 25 so value of R1 6 R1 = 10 = 100 kW 10
for Av =- 10
6 R1' = 10 = 40 kW 25
for Av =- 25
R1 should be as large as possible so R1 = 100 kW SOL 1.8.71
Option (B) is correct. Direct coupled amplifiers or DC-coupled amplifiers provides gain at dc or very low frequency also.
SOL 1.8.72
Option (C) is correct. Since there is no feedback in the circuit and ideally op-amp has a very high value of open loop gain, so it goes into saturation (ouput is either + V or - V ) for small values of input. The input is applied to negative terminal of op-amp, so in positive half cycle it saturates to - V and in negative half cycle it goes to + V .
SOL 1.8.73
Option (B) is correct. From the given input output waveforms truth table for the circuit is drawn as X1 X2 Q 1 0 1 0 0 1 0 1 0 In option (A), for X1 = 1, Q = 0 so it is eliminated. In option (C), for X1 = 0, Q = 0 (always), so it is also eliminated. In option (D), for X1 = 0, Q = 1, which does not match the truth table. Only option (B) satisfies the truth table.
CHECK
SOL 1.8.74
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Option (D) is correct. In the given circuit NMOS Q1 and Q3 makes an inverter circuit. Q4 and Q5 are in parallel works as an OR circuit and Q2 is an output inverter. So output is Q = X1 + X2 = X1 .X2
SOL 1.8.75
Option (D) is correct. Let Q (t) is the present state then from the circuit,
So, the next state is given by
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Page 445
Q (t + 1) = Q (t) (unstable) SOL 1.8.76
Option (B) is correct. Trans-conductance of MOSFET is given by gm = 2iD 2VGS (2 - 1) mA = = 1 mS (2 - 1) V
SOL 1.8.77
Option (D) is correct. Voltage gain can be obtain by small signal equivalent circuit of given amplifier.
So,
vo =- gm vgs RD vgs = vin vo =- gm RD vin Av = vo =- gm RD vi
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Voltage gain
=- (1 mS) (10 kW) =- 10
SOL 1.8.78
Option (C) is correct. Given circuit,
In the circuit
SOL 1.8.79
V1 = 3.5 V (given) Current in zener is. IZ = V1 - VZ = 3.5 - 3.33 = 2 mA RZ 0.1 # 10 Option (C) is correct. This is a current mirror circuit. Since VBE is the same in both devices, and transistors are perfectly matched, then IB1 = IB2 and IC1 = IC2 From the circuit we have,
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IR = IC1 + IB1 + IB2 = IC1 + 2IB2 = IC2 + 2IC2 b IR = IC2 c1 + 2 m b IR IC2 = I = 2 c1 + b m IR can be calculate as
Page 446
a IB1 = IB2 a IC1 = IC2, IC2 = bIB2
nodia.co.in IR = - 5 + 03.7 =- 4.3 mA 1 # 10
So, SOL 1.8.80
I =
4. 3 - 4.3 mA 2 1 + ` 100 j
Option (B) is correct. The small signal equivalent circuit of given amplifier
Here the feedback circuit samples the output voltage and produces a feed back current Ifb which is in shunt with input signal. So this is a shunt-shunt feedback configuration. SOL 1.8.81
Option (A) is correct. In the given circuit output is stable for both 1 or 0. So it is a bistable multivibrator.
SOL 1.8.82
Option (A) is correct. Since there are two levels (+ VCC or - VCC ) of output in the given comparator circuit. For an n -bit Quantizer 2n = No. of levels 2n = 2 n =1
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Page 447
Option (C) is correct. From the circuit, we can see the that diode D2 must be in forward Bias. For D1 let assume it is in reverse bias. Voltages at p and n terminal of D1 is given by Vp and Vn Vp < Vn (D1 is reverse biased)
Applying node equation Vp - 5 Vp + 8 =0 + 1 1 2Vp =- 3 Vp =- 1.5 Vn = 0 Vp < Vn (so the assumption is true and D1 is in reverse bias) and current in D1 ID1 = 0 mA SOL 1.8.84
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Option (D) is correct. The small signal ac equivalent circuit of given amplifier is as following.
RB = (10 kW < 10 kW) = 5 kW gm = 10 ms 50 a gm rp = b & rp = = 5 kW 10 # 10 - 3 Input resistance Here
Rin = RB < rp = 5 kW < 5 kW = 2.5 kW SOL 1.8.85
Option (D) is correct. For PMOS to be biased in non-saturation region. VSD < VSD(sat) and
So,
VSD(sat) = VSG + VT VSD(sat) = 4 - 1 = 3 Volt VSD < 3 VS - VD < 3 4 - ID R < 3 1 < ID R
"a VSG = 4 - 0 = 4 volt
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ID R > 1, R > 1000 W
Page 448
ID = 1 mA
SOL 1.8.86
Option () is correct.
SOL 1.8.87
Option (B is correct. If op-amp is ideal, no current will enter in op-amp. So current ix is v - vy ...(1) ix = x 1 # 106 (ideal op-amp) v+ = v- = vx vx - vy + vx - 0 3 = 0 3 100 # 10 10 # 10 vx - vy + 10vx = 0 11vx = vy For equation (1) & (2) ix = vx - 11v6 x =- 10v6x 1 # 10 10 Input impedance of the circuit. 6 Rin = vx =- 10 =- 100 kW 10 ix
SOL 1.8.88
...(2)
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Option (A) is correct. Given Boolean expression,
Y = (A $ BC + D) (A $ D + B $ C ) = (A $ BCD) + (ABC $ B $ C ) + (AD) + B C D = A BCD + AD + B C D = AD (BC + 1) + B C D = AD + B C D SOL 1.8.89
Option (D) is correct. In the given circuit, output is given as. Y = (A0 5 B0) 9 (A1 5 B1) 9 (A2 5 B2) 9 (A3 5 B3) For option (A) Y = (1 5 1) 9 (0 5 0) 9 (1 5 1) 9 (0 5 0) = 0909090 = 1 For option (B) Y = (0 5 0) 9 (1 5 1) 9 (0 5 0) 9 (1 5 1) = 0909090 = 1 For option (C) Y = (0 5 0) 9 (0 5 0) 9 (1 5 1) 9 (0 5 0) = 0909090 = 1 For option (D) Y = (1 5 1) 9 (0 5 0) 9 (1 5 1) 9 (0 5 1) = 0909091 = 0
SOL 1.8.90
Option (B) is correct. In the given circuit, waveforms are given as,
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SOL 1.8.91
Page 449
Option (C) is correct. The program is executed in following steps. START MVI A, 14H " one instruction cycle. RLC & rotate accumulator left without carry RLC is executed 6 times till value of accumulator becomes zero. JNZ, JNZ checks whether accumulator value is zero or not, it is executed 5 times. HALT " 1-instruction cycle.
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So total no. of instruction cycles are n = 1+6+5+1 = 13 SOL 1.8.92
Option (B) is correct. In the given circuit Vi = 0 V So, transistor Q1 is in cut-off region and Q2 is in saturation. 5 - IC RC - VCE(sat) - 1.25 = 0
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5 - IC RC - 0.1 - 1.25 = 0 5 - IC RC = 1.35 V0 = 1.35
Page 450
"a V0 = 5 - IC RC
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SOL 1.8.93
Option (C) is correct. Since there exists a drain current for zero gate voltage (VGS = 0), so it is a depletion mode device. ID increases for negative values of gate voltages so it is a p-type depletion mode device.
SOL 1.8.94
Option (B) is correct. Applying KVL in input loop, 4 - (33 # 103) IB - VBE - (3.3 # 103) IE = 0 4 - (33 # 103) IB - 0.7 - (3.3 # 103) (hfe + 1) IB = 0
a IE = (hfe + 1) IB
3.3 = 6(33 # 103) + (3.3 # 103) (99 + 1)@ IB 3.3 IB = 33 # 103 + 3.3 # 103 # 100 IC = hfe IB 99 # 3.3 3.3 = mA = mA [0.33 + 3.3] # 100 0.33 + 3.3
SOL 1.8.95
Option (D) is correct. Let the voltages at positive and negative terminals of op-amp are v+ and vrespectively. Then by applying nodal equations. v- - vin + v- - vout = 0 R1 R1 2 v-- = vin + vout
..(1)
Similarly, v+ - vin v -0 =0 + + R 1 c jwC m v+ - vin + v+ (jwCR) = 0 v+ (1 + jwCR) = Vin By equation (1) & (2)
..(2)
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2vin = vin + vout 1 + jwCR 2 - 1E = vout vin ; 1 + jwCR (1 - jwCR) vout = vin 1 + jwCR
Page 451
"a v+ = v- (ideal op-amp)
Phase shift in output is given by q = tan - 1 (- wCR) - tan - 1 (wCR) = p - tan - 1 (wCR) - tan - 1 (wCR) Maximum phase shift SOL 1.8.96
= p - 2 tan - 1 (wCR) q =p
Option (C) is correct. In given circuit MUX implements a 1-bit full adder, so output of MUX is given by. F = Sum = A 5 Q 5 Cin Truth table can be obtain as. Cin P Q Sum
nodia.co.in 0
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
0
0
1
1
0
1
0
1
1
0
0
1
1
1
1
Sum = P Q Cin + PQ Cin + P Q Cin + P Q Cin Output of MUX can be written as F = P Q $ I0 + PQ $ I1 + PQ $ I2 + PQ $ I3 Inputs are, I0 = Cin, I1 = Cin, I2 = Cin, I3 = Cin SOL 1.8.97
Option (D) is correct. Program counter contains address of the instruction that is to be executed next.
SOL 1.8.98
Option (A) is correct. For a n -channel enhancement mode MOSFET transition point is given by, a VTH = 2 volt VDS (sat) = VGS - VTH VDS (sat) = VGS - 2 From the circuit, VDS = VGS So VDS (sat) = VDS - 2 & VDS = VDS (sat) + 2 VDS > VDS (sat) Therefore transistor is in saturation region and current equation is given by. ID = K (VGS - VTH ) 2 4 = K (VGS - 2) 2
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Page 452
VGS is given by VGS = VDS = 10 - ID RD = 10 - 4 # 1 = 6 Volt 4 = K (6 - 2) 2 K =1 4 ' ' Now RD is increased to 4 kW, Let current is ID' and voltages are VDS = VGS Applying current equation. So,
' ID' = K (VGS - VTH ) 2 ' ID' = 1 (VGS - 2) 2 4 ' ' = VDS VGS = 10 - ID' # RD' = 10 - 4ID'
So, 4ID' = (10 - 4ID' - 2) 2 = (8 - 4ID' ) 2 = 16 (2 - ID' ) 2 ID' = 4 (4 + I'D2 - 4ID' ) 4I'D2 - 17 + 16 = 0
nodia.co.in I'D2 = 2.84 mA
SOL 1.8.99
Option (D) is correct. Let the voltages at input terminals of op-amp are v- and v+ respectively. So, v+ = v- = 0 (ideal op-amp)
Applying node equation at negative terminal of op-amp, 0 - vin + 0 - vx = 0 1 10 At node x vx - 0 + vx - vout + vx - 0 = 0 10 10 1
From equation (1),
SOL 1.8.100
...(1)
vx + vx - vout + 10vx = 0 12 vx = vout vx = vout 12 vin + vx = 0 1 10 vin =- vout 120 vout =- 120 vin
Option (D) is correct. In the positive half cycle (when Vin > 4 V ) diode D2 conducts and D1 will be off so the equivalent circuit is,
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Page 453
Vout = + 4 Volt In the negative half cycle diode D1 conducts and D2 will be off so the circuit is,
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Applying KVL
Vin - 10I + 4 - 10I = 0 Vin + 4 = I 20 Vin =- 10 V (Maximum value in negative half cycle) So, I = - 10 + 4 =- 3 mA 20 10 Vin - Vout = I 10 - 10 - Vout =- 3 10 10 Vout =- (10 - 3) Vout =- 7 volt SOL 1.8.101
Option (C) is correct. In the circuit, the capacitor charges through resistor (RA + RB) and discharges through RB . Charging and discharging time is given as. TC = 0.693 (RA + RB) C TD = 0.693 RB C 1 1 f= 1 = = T TD + TC 0.693 (RA + 2RB) C 1 = 10 # 103 -9 0.693 (RA + 2RB) # 10 # 10
Frequency
14.4 # 103 = RA + 2RB duty cycle = TC = 0.75 T 0.693 (RA + RB) C =3 0.693 (RA + 2RB) C 4
...(1)
4RA + 4RB = 3RA + 6RB
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RA = 2RB
Page 454
...(2)
From (1) and (2) 2RA = 14.4 # 103 RA = 7.21 kW RB = 3.60 kW
and SOL 1.8.102
Option (B) is correct. Given boolean expression can be written as, F = XYZ + X Y Z + XY Z + XYZ + XYZ = X YZ + Y Z (X + X ) + XY (Z + Z) = XYZ + Y Z + XY = Y Z + Y (X + X Z ) a A + BC = (A + B) (A + C) = Y Z + Y (X + X ) (X + Z ) = Y Z + Y (X + Z ) = Y Z + YX + YZ
SOL 1.8.103
Option (B) is correct.
nodia.co.in X = X1 5 X 0 , Y = X 2 Serial Input Z = X 5 Y = [X1 5 X0] 5 X2 Truth table for the circuit can be obtain as. Clock pulse
Serial Input
Shift register
Initially
1
1010
1
0
1101
2
0
0110
3
0
0011
4
1
0001
5
0
1000
6
1
0100
7 1 1010 So after 7 clock pulses contents of the shift register is 1010 again. SOL 1.8.104
Option (D) is correct. Characteristic table of the X-Y flip flop is obtained as. X
Y
Qn
Qn+1
0
0
0
1
0
0
1
1
0
1
0
0
0
1
1
1
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1
0
0
1
1
0
1
0
1
1
0
0
1
1
0
0
Page 455
Solving from k-map
Characteristic equation of X-Y flip flop is Qn + 1 = Y Qn + XQn Characteristic equation of a J-K flip-flop is given by Qn + 1 = KQn + J Qn by comparing above two characteristic equations J =Y, K=X SOL 1.8.105
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Option (A) is correct. Total size of the memory system is given by. = (212 # 4) # 8 bits = 214 # 8 bits = 214 Bytes = 16 K bytes
SOL 1.8.106
Option (C) is correct. Executing all the instructions one by one. LXI H, 1FFE & H = (1F) H, L = (FE) H MOV B, M & B = Memory [HL] = Memory [1FFE] INR L & L = L + (1) H = (FF) H MOV A, M & A = Memory [HL] = Memory [1FFF] ADD B & A = A + B INR L & L = L + (1) H = (FF) H + (1) H = 00 MOV M, A & Memory [HL] = A Memory [1F00] = A XOR A & A = A XOR A = 0 So the result of addition is stored at memory address 1F00.
SOL 1.8.107
Option (D) is correct. Let the initial state Q(t) = 0, So D = Q = 1, the output waveform is.
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SOL 1.8.108
Page 456
So frequency of the output is, f f out = in = 10 = 5 kHz 2 2 Option (A) is correct. This is a half-wave rectifier circuit, so the DC voltage is given by Vdc = Vm p Equivalent circuit with forward resistance is
DC current in the circuit Idc
nodia.co.in Idc
SOL 1.8.109
Vm (Vm /p) = p = rf + R (5 + 45) = Vm 50p
Option (B) is correct. In the positive half cycle zener diode (Dz ) will be in reverse bias (behaves as a constant voltage source) and diode (D) is in forward bias. So equivalent circuit for positive half cycle is.
Vo = VD + Vz = 0.7 + 3.3 = 4 Volt In the negative halt cycle, zener diode (Dz ) is in forward bias and diode (D) is in reverse bias mode. So equivalent circuit is. Output
So the peak output is, Vo =
10 # 1 (1 + 1)
Vo = 5 Volt SOL 1.8.110
Option (A) is correct.
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Page 457
For active low chip select CS = 0 , so the address range can be obtain as, A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 A0 1110 0000 0000 0000 h h h h 1110
1111
1111
1111
So address range is E000-EFFF SOL 1.8.111
Option (C) is correct. A first order low pass filter is shown in following figure.
Transfer function V0 (jw) 1 1 = # 1 = j w cR +1 V1 (jw) 1 j w C R+ jw C H (jw1) = 0.25
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Given that
1
=1 4 1 16 = w12 R2 C2 + 1
w12 C2 R2 +
w12 R2 C2 = 15 4p2 f12 (50) 2 (5 # 10 - 6) 2 = 15 f 1 = 2.46 kHz SOL 1.8.112
SOL 1.8.113
Option (A) is correct. In the circuit, voltage at positive terminal of op-amp is given by v+ - vo v+ - 2 =0 + 10 3 3 (v+ - vo) + 10 (v+ - 2) = 0 13v+ = 20 + 3vo Output changes from + 15 V to - 15 V ,when v- > v+ 20 + (3 # 15) v+ = = 5 Volt (for positive half cycle) 13 Option (B) is correct. Output for each stage can be obtain as,
So final output Y is.
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Y = P Q $ R S = (P + Q) $ (R + S) = P+Q+R+S SOL 1.8.114
Page 458
a AB = A + B
Option (B) is correct. We can analyze that the transistor is in active region. b IC = I (b + 1) E 99 (1 mA) = 0.99 mA = (99 + 1) In the circuit
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In the circuit
VBE = 0.7 V VE = IE # 1 kW
=1V
VB - VE = 0.7 VB = 0.7 + 1 = 1.7 volt Current throughR1 IR = VB = 1.7 = 100 mA 17 kW 17 kW IB = IE = 1 mA = 10 mA b+1 (99 + 1) Current through RF , by writing KCL at Base 1
IRF = IB + IR1 = 10 + 100 = 110 mA Current through RC I1 = IC + IRF = 0.99 mA + 110 mA = 1.1 mA SOL 1.8.115
Option (D) is correct. Output voltage V0 = 15 - I1 RC = 15 - (1.1 mA) (1 kW) = 13.9 V
SOL 1.8.116
Option (A) is correct. Current in RF IRF = V0 - VB RF 0.11 mA = 13.9 - 1.7 kW RF
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Page 459
RF = 110.9 kW SOL 1.8.117
Option (A) is correct. By writing node equations in the circuit
Va - Vx + V Cs + (V - V ) Cs = 0 a a y R Va (1 + 2RCs) - Vx - sCRVy = 0
or
(Vy - Va) Cs +
or
Vy =0 R
or Vy (1 + sCR) - Va sCR = 0 From equation (1) & (2) 1 + sCR c sCR m (1 + 2sCR) Vy - Vx - sCRVy = 0 (1 + sCR) (1 + 2sCR) Vy ; - sCR E = Vx sCR
nodia.co.in Vy
...(1)
...(2)
(1 + 3sCR + 2s2 C2 R2 - s2 C2 R2) = Vx sCR
Transfer function Vy sCR = Vx 1 + 3sCR + s2 C2 R2 jwCR jwCR T (jw) = = 2 2 2 2 2 2 1 + j3wCR - C R w (1 - C R w ) + 3jwCR
T (s) =
SOL 1.8.118
Option (A) is correct. Applying Barkhausen criterion of oscillation phase shift will be zero. +T (jw0) = 0 w0 " frequency of oscillation. 1 - C2 R2 w20 = 0 1 R C2 w0 = 1 RC w20 =
SOL 1.8.119
2
Option (C) is correct. In figure V0 R RF + R V jw0 CR = y = 2 V0 1 - w0 C2 R2 + j3w0 CR = 1 RC j = =1 3j 3
Vy = T (jw) w So,
Vy V0
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Page 460
R =1 RF + R 3 RF = 2R = 2 # 1 = 2 kW SOL 1.8.120
Option (C) is correct. By writing truth table for the circuit Q2 Q1 CLK
Q0
Initially
0
0
0
1
0
0
1
2
0
1
0
3
0
1
1
4
1
0
0
1 0 1 All flip flops are reset. When it goes to state 101, output of NAND gate becomes 0 or CLR = 0, so all FFs are reset. Thus it is modulo 4 counter. SOL 1.8.121
Option (A) is correct. When the switch is closed (i.e. during TON ) the equivalent circuit is
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Diode is off during TON .writing KVL in the circuit. 100 - (100 # 10- 6) di = 0 dt di = 106 dt i = # 106 dt = 106 t + i (0) Since initial current is zero i (0) = 0 So, i = 106 t After a duration of TON the current will be maximum given as i Peak = 106 TON When the switch is opened (i.e. during Toff ) the equivalent circuit is
Diode is ON during Toff , writing KVL again 500 =- (100 # 10- 6) di dt i =- 5 # 106 t + i (0) i (0) = i p = 106 TON So, i =- 5 # 106 t + 106 TON After a duration of Toff , current i = 0
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So, & Given that
Page 461
0 =- 5 # 106 t Toff + 106 TON TON = 5 Toff
TON + Toff = 100 m sec TON + TON = 100 m sec 5 TON = 100 = 63.33 m sec 1.2 Peak current SOL 1.8.122
i p = 106 # TON = 63.33 # 10- 6 # 106 = 63.33 A
Option (C) is correct. When the switch is opened, current flows through capacitor and diode is ON in this condition. so the equivalent circuit during TOFF is
nodia.co.in & Initially
At
Duty cycle
I = C dVc dt Vc = I t + Vc (0) C
Vc (0) = 0 Vc = I t C t = Toff Vc = I Toff C TON D = = TON TON + TOFF T
TON = DT TOFF = T - TON = T - DT So, Vc = I (T - DT) C = I (1 - D) T C During TOFF , output voltage V0 = 0 volt . SOL 1.8.123
Option (B) is correct. When the switch is closed, diode is off and the circuit is
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Page 462
In steady state condition C dVc = I2 dt a dVc = I dt C
I2 = C I C V0 =- Vc = - I t C Average output voltage DT = T I V0 = 1 ; # b-C t l dt + T 0 ON
TOFF
#0
0 dtE
2 DT 2 2 2 =- 1 . I :t D =- 1 . I . D T =- I D .T 2 T C 2 0 T C C 2
SOL 1.8.124
Option (B) is correct. Equivalent hybrid circuit of given transistor amplifier when RE is by passed is shown below.
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In the circuit
ib = vs hie
...(1)
vo = hfe ib .RC = hfe . vs .RC hie h R Voltage gain Av = vo = fe C vi hie Equivalent hybrid circuit when RE is not bypassed by the capacitor. 1
In the circuit vs = ib hie + (ib + hfe ib) RE vs = ib [hie + (1 + hfe) RE ] v0 = hfe ib .RC from equation (2) and (3)
...(2) ...(3)
vs hie + (1 + hfe) RE hfe RC Voltage gain, Av2 = v0 = vs hie + (1 + hfe) RE Av1 = hie + (1 + hfe) RE = 1 + (1 + hfe) RE So hie hie Av2 v0 = hfe .RC
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Av < Av 2
SOL 1.8.125
Page 463
1
Option (C) is correct. Conversion time for different type of ADC is given as Counting type TT " Conversion time TT = 2n TC TC " Clock period Integrating type TT = 2n + 1 TC Successive Approximation type TT = nTC Parallel (flash) type " fastest Conversion time is highest for integrating type ADC. So it is slowest.
SOL 1.8.126
Option (D) is correct. F = A + B (NOR) Output is 1 when A = B = 0 OR, F = A 9 B (Ex-NOR) Output is 1 when A = B = 0
SOL 1.8.127
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Option (B) is correct. Output of the multiplexer is written as
f = I0 S1 S0 + I1 S1 S0 + I2 S1 S0 + I3 S1 S0 I0 = 0 , I1 = I2 = I 3 = 1
f = 0 + xy + xy + xy = xy + xy + xy = xy + x (y + y ) = xy + x a y+y = 1 = (x + x) (x + y) A + BC = (A + B) (A + C) = x+y a x+x = 1
So,
SOL 1.8.128
Option (C) is correct. Since gain-bandwidth product remains constant Therefore
SOL 1.8.129
105 # 10 = 100 # fCL fCL = 10 kHz
Option (B) is correct. Given circuit is an astable multi vibrator circuit, time period is given as 1+b , t = RC T = 2t ln c 1 - bm b " feedback factor
b=
v+ =1 vo 2
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SOL 1.8.130
Page 464
J1 + 1 N 2 O = 2t ln 3 So, T = 2t ln KK 1 K1 - OO 2P L Option (C) is correct. MVI A, 10 H & MOV (10) H in accumulator A =(10)H MVI B, 10 H & MOV (10) H in register B B = (10) H BACK : NOP ADDB & Adds contents of register B to accumulator and result stores in accumulator A = A + B = (10) H + (10) H 000 10000 ADD 0 0 0 1 0 0 0 0 A=001 00000 = (20) H RLC & Rotate accumulator left without carry
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JNC BACK & JUMP TO Back if CY = 0 NOP ADD B &A = A + B = (40) H + (10) H 0100 0000 ADD 0 0 0 1 0 0 0 0 A=0101 0000 = (60) H
A = (A0) H JNC BACK NOP ADDB & A = A + B = (A0) H + (10) H 1010 0000 ADD 0 0 0 1 0 0 0 0 A=1011 0 0 0 0 A = (B0) H
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Page 465
CY = 1 So it goes to HLT. therefore NOP will be executed 3 times. SOL 1.8.131
SOL 1.8.132
Option (D) is correct. Leakage current is given by 1 0.5 # 1 # CV Ql 0.5 # 100 # Q 100 I Leakage = = = t t t -2 10- 9 # 5 = 0.5 # 10 # 0.1-# 6 1 # 10 - 13 = 25 # -10 = 2.5 # 10- 6 = 2.5 mA 10 6 Option (A) is correct. Slew rate is defined as the maximum rate of change in output voltage per unit time. Slew rate = dv0 dt
nodia.co.in v0 = vin Slew rate = dvin , vin = 10 sin wt dt = d (10 sin wt) = 10w cos wt dt
For voltage follower, So,
= 10w = 62.8 volt/msec (given) 10 # 2pf = 62.8 # 106 6 f = 62.8 # 10 = 1 MHz 62.8
SOL 1.8.133
Option (C) is correct. Trans conductance of an n-channel JFET, is given by. gm = 2IDS = - 2IDSS c1 - VGS m 2VGS VP VP Trans conductance (gm) is maximum when gate - to - source voltage VGS = 0 (gm) max = - 2IDSS VP gm = (gm) max c1 - VGS m VP (- 3) Here 1 = (gm) max ;1 = (gm) max # 2 (- 5) E 5 (gm) max = 5 = 2.5 2 Option () is correct. The circuit is a synchronous counter. Where input to the flip flops are So,
SOL 1.8.134
D3 = Q3 + Q2 + Q1 D2 = Q3 , D1 = Q2 , D 0 = Q1 Truth table of the circuit can be drawn as
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Page 466
CLK
Q3
Q2
Q1
Q0
Initial state
1
1
1
0
1
0
1
1
1
2
0
0
1
1
3
0
0
0
1
4
1
0
0
0
5
0
1
0
0
6
0
0
1
0
7
0
0
0
1
8 1 0 0 0 From the truth table we can see that counter states at N = 4 and N = 8 are same. So mod number is 4. SOL 1.8.135
Option (B) is correct. In the circuit
nodia.co.in Writing node equation in the circuit at the negative terminal of op amp-1 v1 - 1 + v1 - v2 = 0 1 2 3v1 - v2 = 2 Similarly, at the positive terminal of op amp-1 v1 - vo + v1 - 0 = 0 3 1
...(1)
4v1 - vo = 0 At the negative terminals of op-amp-2 - 1 - v2 + - 1 - vo = 0 m c m c 4 8
...(2)
- 2 - 2v2 - 1 - vo = 0 vo + 2v2 =- 3 From equation (1) and (2) 3 vo - 2v2 = 1 4 From equation (3) 3 v - 2 (- 3 - v ) = 1 o 4 o 3 v + v =- 5 o 4 o 7 v =- 5 4 o
...(3)
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SOL 1.8.136
Page 467
vo =- 20 volt 7 Option (C) is correct. Small signal circuit is (mid-band frequency range)
CE " 0 , for mid-band frequencies vo =- gm vp RC In the input loop vi rp RB + rp - gm RC rp vi So, vo = RB + rp - gm rp RC Gain Av = vo = vi RB + rp Trans-conductance (1 mA) gm = IC = = 1 A/V VT (26 mV) 26 b gm rp = b0 & rp = 0 = 200 # 26 = 5.2 kX gm - 200 # (1 kW) So gain Av = =- 6.62 (25 kW + 5.2 kW) vp =
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SOL 1.8.137
Option (B) is correct. Cut off frequency due to CE is obtained as
f0 =
1 2pReq CE
Req " Equivalent resistance seen through capacitor CE
Req = RE < RB + rp =
RE (RB + rp) RE + RB + rp
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SOL 1.8.138
So
f0 =
1 (RE + RB + rp) = 10 Hz (given) 2pRE (RB + rp) CE
So,
CE =
(0.1 + 25 + 5.2) # 103 = 1.59 mF 2p # 0.1 (25 + 5.2) # 106
Page 468
Option (D) is correct. We can approximately analyze the circuit at low and high frequencies as following. For low frequencies w " 0 & 1 " 3 (i.e. capacitor is open) wc Equivalent circuit is
So, it does not pass the low frequencies. For high frequencies w " 3 & 1 " 0 (i.e. capacitor is short) wc Equivalent circuit is
nodia.co.in vo =- R2 vi R1
So it does pass the high frequencies. This is a high pass filter. SOL 1.8.139
At high frequency w " 3 & 1 " 0 , capacitor behaves as short circuit wc and gain of the filter is given as Av = - R2 = 10 R1 R2 = 10 R1 Input resistance of the circuit Rin = R1 = 100 kW So, R2 = 10 # 100 kW = 1 MW Transfer function of the circuit Vo (jw) - jwR2 C = 1 + jw R1 C Vi (jw) High frequency gain Av3 = 10 At cutoff frequency gain is - jwc R2 C Av = 10 = 1 + jwc R1 C 2 10 = 2
wc R2 C 1 + wc2 R 12 C2
100 + 100wc2 R 12 C2 = 2wc2 R 22 C2 100 + 100 # wc2 # 1010 # C2 = 2 # wc2 # 1012 # C2 100 = wc2 C2 # 1012
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100 wc2 # 1012 1 C = 2pfc # 10 4
C2 =
=
1 2 # 3.14 # 10 3 # 10 4
= 15.92 nF ***********
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9 POWER ELECTRONICS
YEAR 2013 MCQ 1.9.1
Thyristor T in the figure below is initially off and is triggered with a single pulse of width 10 ms . It is given that L = b 100 l mH and C = b 100 l mF . Assuming p p latching and holding currents of the thyristor are both zero and the initial charge on C is zero, T conducts for
(A) 10 ms (C) 100 ms MCQ 1.9.2
TWO MARKS
(B) 50 ms (D) 200 ms
The separately excited dc motor in the figure below has a rated armature current of 20 A and a rated armature voltage of 150 V. An ideal chopper switching at 5 kHz is used to control the armature voltage. If La = 0.1 mH , Ra = 1 W , neglecting armature reaction, the duty ratio of the chopper to obtain 50% of the rated torque at the rated speed and the rated field current is
(A) 0.4 (C) 0.6
(B) 0.5 (D) 0.7
Common Data Questions: 3 & 4 In the figure shown below, the chopper feeds a resistive load from a battery source. MOSFET Q is switched at 250 kHz, with duty ratio of 0.4. All elements of the circuit are assumed to be ideal
GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia MCQ 1.9.3
The Peak to Peak source current ripple in amps is (A) 0.96 (B) 0.144 (C) 0.192 (D) 0.228
MCQ 1.9.4
The average source current in Amps in steady-state is (A) 3/2 (B) 5/3 (C) 5/2 (D) 15/4
Page 471
Statement for Linked Answer Questions: 5 & 6 The Voltage Source Inverter (VSI) shown in the figure below is switched to provide a 50 Hz, square wave ac output voltage Vo across an RL load. Reference polarity of Vo and reference direction of the output current io are indicated in the figure. It is given that R = 3 ohms, L = 9.55 mH .
nodia.co.in MCQ 1.9.5
MCQ 1.9.6
In the interval when V0 < 0 and i 0 > 0 the pair of devices which conducts the load current is (A) Q1, Q2 (B) Q 3, Q 4 (C) D1, D2 (D) D 3, D 4 Appropriate transition i.e., Zero Voltage Switching ^ZVS h/Zero Current Switching ^ZCS h of the IGBTs during turn-on/turn-off is (A) ZVS during turn off (B) ZVS during turn-on (C) ZCS during turn off (D) ZCS during turn-on YEAR 2012
ONE MARK
MCQ 1.9.7
A half-controlled single-phase bridge rectifier is supplying an R-L load. It is operated at a firing angle a and the load current is continuous. The fraction of cycle that the freewheeling diode conducts is (B) (1 - a/p) (A) 1/2 (C) a/2p (D) a/p
MCQ 1.9.8
The typical ratio of latching current to holding current in a 20 A thyristor is (A) 5.0 (B) 2.0 (C) 1.0 (D) 0.5
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YEAR 2012 MCQ 1.9.9
Page 472
TWO MARKS
In the circuit shown, an ideal switch S is operated at 100 kHz with a duty ratio of 50%. Given that Dic is 1.6 A peak-to-peak and I 0 is 5 A dc, the peak current in S , is
(A) 6.6 A (C) 5.8 A
(B) 5.0 A (D) 4.2 A
Common Data for Questions 10 and 11 In the 3-phase inverter circuit shown, the load is balanced and the gating scheme is 180c conduction mode. All the switching devices are ideal.
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MCQ 1.9.10
MCQ 1.9.11
The rms value of load phase voltage is (A) 106.1 V (C) 212.2 V
(B) 141.4 V (D) 282.8 V
If the dc bus voltage Vd = 300 V, the power consumed by 3-phase load is (A) 1.5 kW (B) 2.0 kW (C) 2.5 kW (D) 3.0 kW YEAR 2011
MCQ 1.9.12
ONE MARK
A three phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown below. Switches S1 to S6 are identical switches.
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Page 473
The proper configuration for realizing switches S1 to S6 is
MCQ 1.9.13
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Circuit turn-off time of an SCR is defined as the time (A) taken by the SCR turn to be off (B) required for the SCR current to become zero (C) for which the SCR is reverse biased by the commutation circuit (D) for which the SCR is reverse biased to reduce its current below the holding current YEAR 2011
MCQ 1.9.14
TWO MARKS
A voltage commutated chopper circuit, operated at 500 Hz, is shown below.
If the maximum value of load current is 10 A, then the maximum current through the main (M) and auxiliary (A) thyristors will be (A) iM max = 12 A and iA max = 10 A (B) iM max = 12 A and iA max = 2 A (C) iM max = 10 A and iA max = 12 A (D) iM max = 10 A and iA max = 8 A
Statement for Linked Answer Questions: 9 & 10 A solar energy installation utilize a three – phase bridge converter to feed energy
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Page 474
into power system through a transformer of 400 V/400 V, as shown below.
The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance 10 W. MCQ 1.9.15
The maximum current through the battery will be (A) 14 A (B) 40 A (C) 80 A (D) 94 A
MCQ 1.9.16
The kVA rating of the input transformer is (A) 53.2 kVA (B) 46.0 kVA (C) 22.6 kVA (D) 7.5 kVA YEAR 2010
MCQ 1.9.17
ONE MARK
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The power electronic converter shown in the figure has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a
(A) step down chopper (buck converter) (B) half-wave rectifier (C) step-up chopper (boost converter) (D) full-wave rectifier MCQ 1.9.18
Figure shows a composite switch consisting of a power transistor (BJT) in series with a diode. Assuming that the transistor switch and the diode are ideal, the I -V characteristic of the composite switch is
MCQ 1.9.19
The fully controlled thyristor converter in the figure is fed from a single-phase
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Page 475
source. When the firing angle is 0c, the dc output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60c, assuming continuous conduction
(A) 150 V (C) 300 V YEAR 2009 MCQ 1.9.20
(B) 210 V (D) 100p V ONE MARK
An SCR is considered to be a semi-controlled device because (A) It can be turned OFF but not ON with a gate pulse. (B) It conducts only during one half-cycle of an alternating current wave. (C) It can be turned ON but not OFF with a gate pulse. (D) It can be turned ON only during one half-cycle of an alternating voltage wave.
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YEAR 2009 MCQ 1.9.21
TWO MARKS
The circuit shows an ideal diode connected to a pure inductor and is connected to a purely sinusoidal 50 Hz voltage source. Under ideal conditions the current waveform through the inductor will look like.
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MCQ 1.9.22
Page 476
The Current Source Inverter shown in figure is operated by alternately turning on thyristor pairs (T1, T2) and (T3, T4). If the load is purely resistive, the theoretical maximum output frequency obtainable will be
nodia.co.in (A) 125 kHz (C) 500 kHz MCQ 1.9.23
In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio of 0.8 which is much larger the commutation interval. If the maximum allowable reapplied dv/dt on TM is 50 V/ ms , what should be the theoretical minimum value of C1 ? Assume current ripple through L 0 to be negligible.
(A) 0.2 mF (C) 2 mF MCQ 1.9.24
(B) 250 kHz (D) 50 kHz
(B) 0.02 mF (D) 20 mF
Match the switch arrangements on the top row to the steady-state V -I characteristics on the lower row. The steady state operating points are shown by large black dots.
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Page 477
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(B) P-II, Q-IV, R-I, S-III (D) P-IV, Q-III, R-II, S-I
YEAR 2008 MCQ 1.9.25
In the single phase voltage controller circuit shown in the figure, for what range of triggering angle (a), the input voltage (V0) is not controllable ?
(A) 0c < a < 45c (C) 90c < a < 180c MCQ 1.9.26
ONE MARK
(B) 45c < a < 135c (D) 135c < a < 180c
A 3-phase voltage source inverter is operated in 180c conduction mode. Which one of the following statements is true ? (A) Both pole-voltage and line-voltage will have 3rd harmonic components (B) Pole-voltage will have 3rd harmonic component but line-voltage will be free from 3rd harmonic (C) Line-voltage will have 3rd harmonic component but pole-voltage will be free from 3rd harmonic (D) Both pole-voltage and line-voltage will be free from 3rd harmonic components
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Page 478
YEAR 2008 MCQ 1.9.27
TWO MARKS
The truth table of monoshot shown in the figure is given in the table below :
Two monoshots, one positive edge triggered and other negative edge triggered, are connected shown in the figure, The pulse widths of the two monoshot outputs Q1 and Q2 are TON and TON respectively. 1
2
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The frequency and the duty cycle of the signal at Q1 will respectively be TON 1 1 (A) f = (B) f = , D= , D= 1 TON + TON TON + TON TON + TON 5 TON TON (C) f = 1 , D = (D) f = 1 , D = TON TON + TON TON TON + TON A single phase fully controlled bridge converter supplies a load drawing constant and ripple free load current, if the triggering angle is 30c, the input power factor will be (A) 0.65 (B) 0.78 (C) 0.85 (D) 0.866 1
1
2
1
1
MCQ 1.9.28
MCQ 1.9.29
1
2
2
1
2
1
2
2
1
2
A single-phase half controlled converter shown in the figure feeding power to highly inductive load. The converter is operating at a firing angle of 60c.
If the firing pulses are suddenly removed, the steady state voltage (V0) waveform of the converter will become
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MCQ 1.9.30
A single phase source inverter is feeding a purely inductive load as shown in the figure. The inverter is operated at 50 Hz in 180c square wave mode. Assume that the load current does not have any dc component. The peak value of the inductor current i0 will be
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(A) 6.37 A (C) 20 A MCQ 1.9.31
Page 479
(B) 10 A (D) 40 A
A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure.
Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of I0 = 10 A will be (A) 44c (B) 51c (C) 129c (D) 136c MCQ 1.9.32
A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30c. The approximate Total harmonic Distortion (%THD) and the rms value of fundamental component of input current will respectively be (A) 31% and 6.8 A (B) 31% and 7.8 A (C) 66% and 6.8 A (D) 66% and 7.8 A
MCQ 1.9.33
In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous.
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Page 480
The average voltage across the load and the average current through the diode will respectively be (A) 10 V, 2 A (B) 10 V, 8 A (C) 40 V 2 A (D) 40 V, 8 A YEAR 2007
ONE MARK
MCQ 1.9.34
A single-phase fully controlled thyristor bridge ac-dc converter is operating at a firing angle of 25c and an overlap angle of 10c with constant dc output current of 20 A. The fundamental power factor (displacement factor) at input ac mains is (A) 0.78 (B) 0.827 (C) 0.866 (D) 0.9
MCQ 1.9.35
A three-phase, fully controlled thyristor bridge converter is used as line commutated inverter to feed 50 kW power 420 V dc to a three-phase, 415 V(line), 50 Hz ac mains. Consider dc link current to be constant. The rms current of the thyristor is (A) 119.05 A (B) 79.37 A (C) 68.73 A (D) 39.68 A
MCQ 1.9.36
A single phase full-wave half-controlled bridge converter feeds an inductive load. The two SCRs in the converter are connected to a common DC bus. The converter has to have a freewheeling diode. (A) because the converter inherently does not provide for free-wheeling (B) because the converter does not provide for free-wheeling for high values of triggering angles (C) or else the free-wheeling action of the converter will cause shorting of the AC supply (D) or else if a gate pulse to one of the SCRs is missed, it will subsequently cause a high load current in the other SCR.
MCQ 1.9.37
“Six MOSFETs connected in a bridge configuration (having no other power device) must be operated as a Voltage Source Inverter (VSI)”. This statement is (A) True, because being majority carrier devices MOSFETs are voltage driven. (B) True, because MOSFETs hav inherently anti-parallel diodes (C) False, because it can be operated both as Current Source Inverter (CSI) or a VSI (D) False, because MOSFETs can be operated as excellent constant current sources in the saturation region.
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YEAR 2007 MCQ 1.9.38
TWO MARKS
A single-phase voltages source inverter is controlled in a single pulse-width modulated mode with a pulse width of 150c in each half cycle. Total
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Page 481
harmonic distortion is defined as 2 - V 12 V rms # 100 V1 where V1 is the rms value of the fundamental component of the output voltage. The THD of output ac voltage waveform is (A) 65.65% (B) 48.42% (C) 31.83% (D) 30.49%
THD =
MCQ 1.9.39
A three-phase, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V dc, 15 kW, 1500 rpm separately excited dc motor with a ripple free continuos current in the dc link under all operating conditions, Neglecting the losses, the power factor of the ac mains at half the rated speed is (A) 0.354 (B) 0.372 (C) 0.90 (D) 0.955
MCQ 1.9.40
A single-phase, 230 V, 50 Hz ac mains fed step down transformer (4:1) is supplying power to a half-wave uncontrolled ac-dc converter used for charging a battery (12 V dc) with the series current limiting resistor being 19.04 W. The charging current is (A) 2.43 A (B) 1.65 A (C) 1.22 A (D) 1.0 A
MCQ 1.9.41
In the circuit of adjacent figure the diode connects the ac source to a pure inductance L.
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The diode conducts for (A) 90c (C) 270c MCQ 1.9.42
(B) 180c (D) 360c
The circuit in the figure is a current commutated dc-dc chopper where, Th M is the main SCR and Th AUX is the auxiliary SCR. The load current is constant at 10 A. Th M is ON. Th AUX is trigged at t = 0 . Th M is turned OFF between.
(A) 0 ms < t # 25 ms (C) 50 ms < t # 75 ms
(B) 25 ms < t # 50 ms (D) 75 ms < t # 100 ms
Common Data for Question 37 and 38.
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Page 482
A 1:1 Pulse Transformer (PT) is used to trigger the SCR in the adjacent figure. The SCR is rated at 1.5 kV, 250 A with IL = 250 mA, IH = 150 mA, and IG max = 150 mA, IG min = 100 mA. The SCR is connected to an inductive load, where L = 150 mH in series with a small resistance and the supply voltage is 200 V dc. The forward drops of all transistors/diodes and gate-cathode junction during ON state are 1.0 V
MCQ 1.9.43
MCQ 1.9.44
The resistance R should be (A) 4.7 kW (C) 47 W
(B) 470 kW (D) 4.7 W
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The minimum approximate volt-second rating of pulse transformer suitable for triggering the SCR should be : (volt-second rating is the maximum of product of the voltage and the width of the pulse that may applied) (B) 200 mV-s (A) 2000 mV-s (C) 20 mV-s (D) 2 mV-s YEAR 2006
ONE MARK
MCQ 1.9.45
The speed of a 3-phase, 440 V, 50 Hz induction motor is to be controlled over a wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage source inverter. It is desired to keep the flux in the machine constant in the constant torque region by controlling the terminal voltage as the frequency changes. The inverter output voltage vs frequency characteristic should be
MCQ 1.9.46
A single-phase half wave uncontrolled converter circuit is shown in figure. A 2-winding transformer is used at the input for isolation. Assuming the load current to be constant and V = Vm sin wt , the current waveform through diode D2 will be
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YEAR 2006 MCQ 1.9.47
Page 483
TWO MARKS
A single-phase inverter is operated in PWM mode generating a single-pulse of width 2d in the centre of each half cycle as shown in figure. It is found that the output voltage is free from 5th harmonic for pulse width 144c. What will be percentage of 3rd harmonic present in the output voltage (Vo3 /Vo1 max) ?
(A) 0.0% (C) 31.7%
(B) 19.6% (D) 53.9%
MCQ 1.9.48
A 3-phase fully controlled bridge converter with free wheeling diode is fed from 400 V, 50 Hz AC source and is operating at a firing angle of 60c. The load current is assumed constant at 10 A due to high load inductance. The input displacement factor (IDF) and the input power factor (IPF) of the converter will be (A) IDF = 0.867; IPF = 0.828 (B) IDF = 0.867; IPF = 0.552 (C) IDF = 0.5; IPF = 0.478 (D) IDF = 0.5; IPF = 0.318
MCQ 1.9.49
A voltage commutation circuit is shown in figure. If the turn-off time of the SCR is 50 msec and a safety margin of 2 is considered, then what will be the approximate minimum value of capacitor required for proper commutation ?
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(A) 2.88 mF (C) 0.91 mF
Page 484
(B) 1.44 mF (D) 0.72 mF
MCQ 1.9.50
A solar cell of 350 V is feeding power to an ac supply of 440 V, 50 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at 20 A. If the solar cell resistance is 0.5 W ,then each thyristor will be reverse biased for a period of (A) 125c (B) 120c (C) 60c (D) 55c
MCQ 1.9.51
A single-phase bridge converter is used to charge a battery of 200 V having an internal resistance of 0.2 W as shown in figure. The SCRs are triggered by a constant dc signal. If SCR 2 gets open circuited, what will be the average charging current ?
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(A) 23.8 A (C) 11.9 A MCQ 1.9.52
(B) 15 A (D) 3.54 A
An SCR having a turn ON times of 5 msec, latching current of 50 A and holding current of 40 mA is triggered by a short duration pulse and is used in the circuit shown in figure. The minimum pulse width required to turn the SCR ON will be
(A) 251 msec (C) 100 msec
(B) 150 msec (D) 5 msec
Data for Q. 53 and Q. 54 are given below. Solve the problems and choose the correct answers. A voltage commutated chopper operating at 1 kHz is used to control the speed of dc as shown in figure. The load current is assumed to be constant at 10 A
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MCQ 1.9.53
The minimum time in msec for which the SCR M should be ON is. (A) 280 (B) 140 (C) 70 (D) 0
MCQ 1.9.54
The average output voltage of the chopper will be (A) 70 V (B) 47.5 V (C) 35 V (D) 0 V YEAR 2005
ONE MARK
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MCQ 1.9.55
The conduction loss versus device current characteristic of a power MOSFET is best approximated by (A) a parabola (B) a straight line (C) a rectangular hyperbola (D) an exponentially decaying function
MCQ 1.9.56
A three-phase diode bridge rectifier is fed from a 400 V RMS, 50 Hz, three-phase AC source. If the load is purely resistive, then peak instantaneous output voltage is equal to (A) 400 V (B) 400 2 V (C) 400 2 V (D) 400 V 3 3 The output voltage waveform of a three-phase square-wave inverter contains (A) only even harmonics (B) both odd and even harmonic (C) only odd harmonics (D) only triple harmonics
MCQ 1.9.57
YEAR 2005 MCQ 1.9.58
TWO MARKS
The figure shows the voltage across a power semiconductor device and the current through the device during a switching transitions. If the transition a turn ON transition or a turn OFF transition ? What is the energy lost during the transition?
(A) Turn ON, VI (t1 + t2) 2 (C) Turn ON, VI (t1 + t2)
(B) Turn OFF, VI (t1 + t2) (D) Turn OFF, VI (t1 + t2) 2
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Page 486
An electronics switch S is required to block voltage of either polarity during its OFF state as shown in the figure (a). This switch is required to conduct in only one direction its ON state as shown in the figure (b)
Which of the following are valid realizations of the switch S?
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(A) Only P (C) P and R MCQ 1.9.60
(B) P and Q (D) R and S
The given figure shows a step-down chopper switched at 1 kHz with a duty ratio D = 0.5 . The peak-peak ripple in the load current is close to
(A) 10 A (C) 0.125 A
(B) 0.5 A (D) 0.25 A
MCQ 1.9.61
An electric motor, developing a starting torque of 15 Nm, starts with a load torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/sec2 , the moment of inertia of the system must be (neglecting viscous and coulomb friction) (A) 0.25 kg-m2 (B) 0.25 Nm2 (C) 4 kg-m2 (D) 4 Nm2
MCQ 1.9.62
Consider a phase-controlled converter shown in the figure. The thyristor is fired at an angle a in every positive half cycle of the input voltage. If the peak value of the instantaneous output voltage equals 230 V, the firing angle a is close to
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(A) 45c (C) 90c
Page 487
(B) 135c (D) 83.6c
YEAR 2004
ONE MARK
MCQ 1.9.63
A bipolar junction transistor (BJT) is used as a power control switch by biasing it in the cut-off region (OFF state) or in the saturation region (ON state). In the ON state, for the BJT (A) both the base-emitter and base-collector junctions are reverse biased (B) the base-emitter junction is reverse biased, and the base-collector junction is forward biased (C) the base-emitter junction is forward biased, and the base-collector junction is reverse biased (D) both the base-emitter and base-collector junctions are forward biased
MCQ 1.9.64
The circuit in figure shows a full-wave rectifier. The input voltage is 230 V (rms) single-phase ac. The peak reverse voltage across the diodes D 1 and D 2 is
MCQ 1.9.65
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(A) 100 2 V
(B) 100 V
(C) 50 2 V
(D) 50 V
The triggering circuit of a thyristor is shown in figure. The thyristor requires a gate current of 10 mA, for guaranteed turn-on. The value of R required for the thyristor to turn on reliably under all conditions of Vb variation is
(A) 10000 W (C) 1200 W MCQ 1.9.66
(B) 1600 W (D) 800 W
The circuit in figure shows a 3-phase half-wave rectifier. The source is a symmetrical, 3-phase four-wire system. The line-to-line voltage of the source is
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100 V. The supply frequency is 400 Hz. The ripple frequency at the output is
(A) 400 Hz (C) 1200 Hz
(B) 800 Hz (D) 2400 Hz
YEAR 2004 MCQ 1.9.67
TWO MARKS
A MOSFET rated for 15 A, carries a periodic current as shown in figure. The ON state resistance of the MOSFET is 0.15 W. The average ON state loss in the MOSFET is
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(A) 33.8 W (C) 7.5 W MCQ 1.9.68
The triac circuit shown in figure controls the ac output power to the resistive load. The peak power dissipation in the load is
(A) 3968 W (C) 7935 W MCQ 1.9.69
(B) 5290 W (D) 10580 W
Figure shows a chopper operating from a 100 V dc input. The duty ratio of the main switch S is 0.8. The load is sufficiently inductive so that the load current is ripple free. The average current through the diode D under steady state is
(A) 1.6 A (B) 8.0 A MCQ 1.9.70
(B) 15.0 W (D) 3.8 W
(B) 6.4 A (D) 10.0 A
Figure shows a chopper. The device S 1 is the main switching device. S 2 is the auxiliary commutation device. S 1 is rated for 400 V, 60 A. S 2 is rated for 400 V,
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Page 489
30 A. The load current is 20 A. The main device operates with a duty ratio of 0.5. The peak current through S 1 is
(A) 10 A (C) 30 A
(B) 20 A (D) 40 A
MCQ 1.9.71
A single-phase half-controlled rectifier is driving a separately excited dc motor. The dc motor has a back emf constant of 0.5 V/rpm. The armature current is 5 A without any ripple. The armature resistance is 2 W. The converter is working from a 230 V, single-phase ac source with a firing angle of 30c. Under this operating condition, the speed of the motor will be (A) 339 rpm (B) 359 rpm (C) 366 rpm (D) 386 rpm
MCQ 1.9.72
A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. Figure shows the reversing torque and the speed during the transient. The moment of inertia of the drive is
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(A) 0.048 kg-m2 (C) 0.096 kg-m2 YEAR 2003 MCQ 1.9.73
(B) 0.064 km-m2 (D) 0.128 kg-m2 ONE MARK
Figure shows a thyristor with the standard terminations of anode (A), cathode (K), gate (G) and the different junctions named J1, J2 and J3. When the thyristor is turned on and conducting
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Page 490
(A) J1 and J2 are forward biased and J3 is reverse biased (B) J1 and J3 are forward biased and J2 is reverse biased (C) J1 is forward biased and J2 and J3 are reverse biased (D) J1, J2 and J3 are all forward biased MCQ 1.9.74
Figure shows a MOSFET with an integral body diode. It is employed as a power switching device in the ON and OFF states through appropriate control. The ON and OFF states of the switch are given on the VDS - IS plane by
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The speed/torque regimes in a dc motor and the control methods suitable for the same are given respectively in List-II and List-I List-I P.
Field Control
1.
Below base speed
Q.
Armature Control
2.
Above base speed
3.
Above base torque
4.
Below base torque
Codes: (A) P-1, Q-3 (C) P-2, Q-3 MCQ 1.9.76
List-II
(B) P-2, Q-1 (D) P-1, Q-4
A fully controlled natural commutated 3-phase bridge rectifier is operating with a firing angle a = 30c, The peak to peak voltage ripple expressed as a ratio of the peak output dc voltage at the output of the converter bridge is
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(A) 0.5
(B)
3 /2
(C) c1 - 3 m 2
(D)
3 -1
YEAR 2003 MCQ 1.9.77
Page 491
TWO MARKS
A phase-controlled half-controlled single-phase converter is shown in figure. The control angle a = 30c
The output dc voltage wave shape will be as shown in
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A chopper is employed to charge a battery as shown in figure. The charging current is 5 A. The duty ratio is 0.2. The chopper output voltage is also shown in the figure. The peak to peak ripple current in the charging current is
(A) 0.48 A (C) 2.4 A MCQ 1.9.79
(B) 1.2 A (D) 1 A
An inverter has a periodic output voltage with the output wave form as shown in figure
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Page 492
When the conduction angle a = 120c, the rms fundamental component of the output voltage is (A) 0.78 V (B) 1.10 V (C) 0.90 V (D) 1.27 V MCQ 1.9.80
With reference to the output wave form given in above figure , the output of the converter will be free from 5 th harmonic when (A) a = 72c (B) a = 36c (C) a = 150c (D) a = 120c
MCQ 1.9.81
An ac induction motor is used for a speed control application. It is driven from an inverter with a constant V/f control. The motor name-plate details are as follows (no. of poles = 2) V: 415 V VPh: 3 V f: 50 Hz N: 2850 rpm The motor runs with the inverter output frequency set at 40 Hz, and with half the rated slip. The running speed of the motor is (A) 2400 rpm (B) 2280 rpm (C) 2340 rpm (D) 2790 rpm
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YEAR 2002
ONE MARK
MCQ 1.9.82
A six pulse thyristor rectifier bridge is connected to a balanced 50 Hz three phase ac source. Assuming that the dc output current of the rectifier is constant, the lowest frequency harmonic component in the ac source line current is (A) 100 Hz (B) 150 Hz (C) 250 Hz (D) 300 Hz
MCQ 1.9.83
A step-down chopper is operated in the continuous conduction mode is steady state with a constant duty ratio D . If V0 is the magnitude of the dc output voltage and if Vs is the magnitude of the dc input voltage, the ratio V0 /Vs is given by (A) D (B) 1 - D (C) 1 (D) D 1-D 1-D YEAR 2002
MCQ 1.9.84
TWO MARKS
In the chopper circuit shown in figure, the input dc voltage has a constant value Vs . The output voltage V0 is assumed ripple-free. The switch S is operated with a switching time period T and a duty ratio D . What is the value of D at the boundary of continuous and discontinuous conduction of the inductor current iL ?
(A) D = 1 - Vs V0
(B) D = 2L RT
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(C) D = 1 - 2L RT MCQ 1.9.85
Vs f p 2
f 2p
f p
(D) Vs p
In the single phase diode bridge rectifier shown in figure, the load resistor is R = 50 W . The source voltage is V = 200 sin (wt), where w = 2p # 50 radians per second. The power dissipated in the load resistor R is
(A) 3200 W p
(B) 400 W p
(C) 400 W
(D) 800 W
*The semiconductor switch S in the circuit of figure is operated at a frequency of 20 kHz and a duty ratio D = 0.5 . The circuit operates in the steady state. Calculate the power transferred from the dc voltage source V2 .
YEAR 2001 MCQ 1.9.88
(B) Vs
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(C) Vs
MCQ 1.9.87
(D) D = RT L
Figure(a) shows an inverter circuit with a dc source voltage Vs . The semiconductor switches of the inverter are operated in such a manner that the pole voltage V10 and V20 are as shown in figure(b). What is the rms value of the pole-to-pole voltage V12 ?
(A)
MCQ 1.9.86
Page 493
ONE MARK
The main reason for connecting a pulse transformer at the output stage of thyristor triggering circuit is to (A) amplify the power of the triggering pulse
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(B) provide electrical isolation (C) reduce the turn on time of thyristor (D) avoid spurious triggering of the thyristor due to noise MCQ 1.9.89
AC-to-DC circulating current dual converters are operated with the following relationship between their triggering angles( a1 and a2 ) (A) a1 + a2 = 180c (B) a1 + a2 = 360c (C) a1 - a2 = 180c (D) a1 + a2 = 90c YEAR 2001
MCQ 1.9.90
A half-wave thyristor converter supplies a purely inductive load as shown in figure. If the triggering angle of the thyristor is 120c, the extinction angle will be
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(A) 240c (C) 200c MCQ 1.9.91
(B) 180c (D) 120c
A single-phase full bridge voltage source inverter feeds a purely inductive load as shown in figure, where T1 , T2 , T3 , T4 are power transistors and D 1 , D 2 , D 3 , D 4 are feedback diodes. The inverter is operated in square-wave mode with a frequency of 50 Hz. If the average load current is zero, what is the time duration of conduction of each feedback diode in a cycle?
(A) 5 msec (C) 20 msec MCQ 1.9.92
TWO MARKS
(B) 10 msec (D) 2.5 msec
*A voltage commutated thyristor chopper circuit is shown in figure. The chopper is operated at 500 Hz with 50% duty ratio. The load takes a constant current of 20 A. (a) Evaluate the circuit turn off time for the main thyristor Th 1 . (b) Calculate the value of inductor L, if the peak current through the main thyristor Th 1 is limited to 180% of the load current. (c) Calculate the maximum instantaneous output voltage of chopper.
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MCQ 1.9.93
Page 495
*A separately excited dc motor is controlled by varying its armature voltage using a single-phase full-converter bridge as shown in figure. The field current is kept constant at the rated value. The motor has an armature resistance of 0.2 W, and the motor voltage constant is 2.5 V/(rad/sec). The motor is driving a mechanical load having a constant torque of 140 Nm. The triggering angle of converter is 60c . The armature current can be assumed to be continuous and ripple free.
nodia.co.in (a) Calculate the motor armature constant. (b) Evaluate the motor speed in rad/sec. (c) Calculate the rms value of the fundamental component of the input current to the bridge. ***********
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SOLUTION
SOL 1.9.1
Page 496
Option (C) is correct. Given, L = 100 mH p C = 100 mF p When the circuit is triggered by 10 ms pulse, the thyristor is short circuited and so, we consider IC = Im sin wt Therefore, voltage stored across capacitor is VC = 1 IC dt C = Vm ^1 - cos wt h where w is angular frequency obtained as 1 w = 1 = = p # 10 4 100 -6 LC b p l # 10 So, T = 1 = 2p = 200 ms w f
#
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SOL 1.9.2
As IC = Im sin wt oscillates between - ve and - ve half cycle so, circuit is conducting for only half of cycle and thyristor is open after half cycle. i.e., the conduction period = T = 100 ms 2 Option (D) is correct. Given, the rated armature current Ia^rated h = 20 A as rated armature voltage Va^rated h = 150 volt Also, for the armature, we have La = 0.1 mH , Ra = 1 W and T = 50% of Trated ^T " Torqueh So, we get I = 6Ia^rotatedh@^0.5h = 10 A N = Nrated , I f = I f rated " rated field current At the rated conditions, E = V - Ia^ratedh Ra = 150 - 20 ^1 h = 130 volt For given torque, V = E + Ia Ra = 130 + ^10h^1 h = 140 V Therefore, chopper output = 140 V or, D ^200h = 140 (D " duty cycle) or, D = 140 = 0.7 200
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Page 497
Option (C) is correct. Here, as the current from source of 12 V is the same as that pass through inductor. So, the peak to peak current ripple will be equal to peak to peak inductor current. Now, the peak to peak inductor current can be obtained as IL (Peak to Peak) = Vs D Ts L where,
Vs " source voltage = 12 volt , L " inductance = 100mH = 10-4 H , D " Duty ration = 0.4 ,
TS " switching time period of MOSFET = 1 fS and fs " switching frequency = 250 kHz Therefore, we get 1 IL^Peak to Peakh = 12-4 # 0.4 # 250 # 103 10 = 0.192 A This is the peak to peak source current ripple. SOL 1.9.4
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Option (B) is correct. Here, the average current through the capacitor will be zero. (since, it is a boost converter). We consider the two cases : Case I : When MOSFET is ON (i 0 is output current) ic =- i 0 (since, diode will be in cut off mode) Case II : When MOSFET is OFF Diode will be forward biased and so (Is is source current) ic = Is - i 0 Therefore, average current through capacitor i + Ic Ic, avg = c 2 DTs ^- io h + ^1 - D h Ts ^Is - io h (D is duty ratio) & 0= 2 Solving the equation, we get i0 ....(1) Is = ^1 - D h Since, the output load current can be given as V/ 12/0.6 i 0 = V0 = s 1 - D = = 1A 20 R R Hence, from Eq. (1) Is = i 0 = 1 = 5 A 0.6 3 1-D 1
1
1
SOL 1.9.5
2
Option (D) is correct. We consider the following two cases : Case I : When Q1, Q2 ON In this case the + ve terminal of V0 will be at higher voltage. i.e. V0 > 0 and so i 0 > 0 (i.e., it will be + ve ). Now, when the Q1 , Q2 goes to OFF condition we consider the second case. Case II : When Q 3 , Q 4 ON and Q , Q2 OFF : In this condition, - ve terminal of applied voltage V0 will be at higher potential i.e., V0 < 0 and since, inductor opposes the change in current so, although the
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Page 498
polarity of voltage V0 is inversed, current remains same in inductor i.e. I 0 > 0 . This is the condition when conduction have been asked. In this condition ^V0 > 0, I 0 > 0h since, IGBT’s can’t conduct reverse currents therefore current will flow through D 3, D 4 until ID becomes negative. Thus, D 3 and D 4 conducts. SOL 1.9.6
Option (D) is correct. When Q 3, Q 4 is switched ON, initially due to the reverse current it remain in OFF state and current passes through diode. In this condition the voltage across Q 3 and Q 4 are zero as diodes conduct. Hence, it shows zero voltage switching during turn-ON
SOL 1.9.7
Option (D) is correct. The circuit of a single-phase half controlled bridge rectifier with RL load and free wheel diode is shown as below.
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The voltage current wave forms are shown in figure below.
We note that, for continuous load current, the flywheel diode conducts from p
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Page 499
to p + a in a cycle. Thus, fraction of cycle that freewheel diode conducts is a/p. Thus fraction of cycle that freewheel diode conducts is a/p. SOL 1.9.8
Option (B) is correct. The latching current is higher than the holding current. Usually, latching current is taken two to three times the holding currents.
SOL 1.9.9
Option (C) is correct.
SOL 1.9.10
IS = I 0 + Tic = 5 + 0.8 = 5.8 A 2 Option (B) is correct. For a three-phase bridge inverter, rms value of output line voltage is VL =
2V = 2 300 3 dc 3 #
Vdc = 300 V
= 141.4 V SOL 1.9.11
SOL 1.9.12
Option (D) is correct. 2 (141.4) 2 - 3 kW P = 3 # VL = 3 # 20 R Option (C) is correct. Only option C allow bi direction power flow from source to the drive
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SOL 1.9.13
Option (C) is correct. Once the SCR start conducting by an forward current, the gate has no control on it and the device can be brought back to the blocking state only by reducing the forward current to a level below that of holding current. This process of turn-off is called commutation. This time is known as the circuit turn-off time of an SCR.
SOL 1.9.14
Option (A) is correct. Maximum current through main thyristor C = 10 + 200 L Maximum current through auxiliary thyristor IM (max) = I 0 + Vs
0.1 # 10-6 = 12 A 1 # 103
IA (max) = I 0 = 10 A SOL 1.9.15
Option (A) is correct. Output voltage of 3-phase bridge converter V0 = 3 3 Vph cos a p Maximum output (V0) max = 3 3 Vph cos a = 1 p = 3 3 # 400 # 2 p 3
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Page 500
= 540.6 V Resistance of filter choke is 10 W, So (V0) max = E + IR chock 540.6 = 400 + I (10) I - 14 A SOL 1.9.16
Option (D) is correct. kVA rating =
3 VL IL =
3 # 400 # 6 # 14 p
= 7.5 kVA SOL 1.9.17
Option (A) is correct. The figure shows a step down chopper circuit. a Vout = DVin where, D = Duty cycle and D < 1
SOL 1.9.18
Option (C) is correct. Given figure as
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The I -V characteristic are as
Since diode connected in series so I can never be negative. When current flows voltage across switch is zero and when current is zero than there may be any voltage across switch. SOL 1.9.19
Option (A) is correct. Given fully-controlled thyristor converter, when firing angle a = 0 , dc output voltage Vdc = 300 V If a = 60c, then Vdc = ? For fully-controlled converter 0
Vdc = 0
2 2 Vdc cos a p 1
a a = 0 , Vdc = 300 V 0
2 2 Vdc cos 0c p = 300p 2 2
300 = Vdc
1
1
At a = 60c, Vdc = ? 2
Vdc = 2 2 # 300p cos 60c = 300 # 1 = 150 V p 2 2 2 Option (C) is correct. SCR has the property that it can be turned ON but not OFF with a gate pulse, So SCR is being considered to be a semi-controlled device. 2
SOL 1.9.20
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Page 501
Option (D) is correct.
Current wave form for iL vL = LdiL dt iL = 1 # vL dt 2 vL = vin = 10 sin wt = diL dt iL = 1 # vL dt =- cos 100pt + C 2
for 0 < wt +p,
iL = 0 , C = 0 iL =- 100 cos pt iL (peak) = 1 Amp
at 100pt = p/2 ,
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for p < wt vL = vin = 0
Option (C) is correct. In CSI let T3 and T4 already conducting at t = 0 At triggering T1 and T2 , T3 and T4 are force cumulated. Again at t = T , T1 and T2 are force cumulated. This completes a cycle. 2
Time constant t = RC = 4 # 0.5 = 2 m sec 1 Frequency f = 1 = = 500 kHz t 2 # 10- 6 SOL 1.9.23
Option (A) is correct. duty ratio TM = 0.8 Maximum dv on TM = 50 V/msec dt Minimum value of C1
=?
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Page 502
Given that current ripple through L 0 is negligible. Current through TM = Im = duty ratio # current a
SOL 1.9.24
= 0.8 # 12.5 = 10 A Im = C1 dv dt 10 = C1 # 50- 6 10 C1 = 50 # 10- 6 = 0.2 mF 10
Option (C) is correct. Characteristics are as
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Option (A) is correct.
`
R + jXL = 50 + 50j tan f = wL = 50 = 1 50 R
f = 45c so, firing angle ‘a’ must be higher the 45c, Thus for 0 < a < 45c, V0 is uncontrollable. SOL 1.9.26
Option (D) is correct. A 3-f voltage source inverter is operated in 180c mode in that case third harmonics
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Page 503
are absent in pole voltage and line voltage due to the factor cos (np/6). so both are free from 3rd harmonic components. SOL 1.9.27
Option (B) is correct. In this case and,
SOL 1.9.28
1 TON1 + TON 2 TON 2 D = TON1 + TON 2 f =
Option (B) is correct. Given a = 30c, in a 1-f fully bridge converter we know that, Power factor = Distortion factor # cos a D.f. (Distortion factor) = Is(fundamental) /Is = 0.9 power factor = 0.9 # cos 30c = 0.78
SOL 1.9.29
Option (A) is correct. Output of this
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Here the inductor makes T1 and T3 in ON because current passing through T1 and T3 is more than the holding current. SOL 1.9.30
SOL 1.9.31
Option (C) is correct. Input is given as
Here load current does not have any dc component Peak current occur at (p/w) ` ` Vs = L di dt 200 = 0.1 # di dt Here di = a p kb 1 l = 1 2p 50 100 So di(max) = 200 # 1 # 1 = 20 A 100 0.1 Option (C) is correct. Here for continuous conduction mode, by Kirchoff’s voltage law, average load
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Page 504
current
V - 2Ia + 150 = 0 Ia = V + 150 2 ` I1 = 10 A, So V =- 130 V 2Vm cos a =- 130 p 2#
2 # 230 cos a =- 130c p a = 129c
SOL 1.9.32
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Option (B) is correct.
Total rms current Ia =
2 10 = 8.16 A 3#
Fundamental current Ia1 = 0.78 # 10 = 7.8 A THD =
where
1 -1 DF2
DF = Ia1 = 0.78 # 10 = 0.955 0.816 # 10 Ia THD =
` SOL 1.9.33
1 2 b 0.955 l - 1 = 31%
Option (C) is correct.
In the given diagram when switch S is open I 0 = IL = 4 A, Vs = 20 V when switch S is closed ID = 0, V0 = 0 V Duty cycle = 0.5 so average voltage is Vs 1-d Average current = 0 + 4 = 2 amp 2 Average voltage = 20 = 40 V 1 - 0.5 SOL 1.9.34
Option (A) is correct. Firing angle a = 25c Overlap angle m = 10c
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so,
I 0 = Vm [cos a - cos (a + m)] wLs
`
20 =
Page 505
230 2 [cos 25c - cos (25c + 10c)] 2p # 50Ls
Ls = 0.0045 H
`
V0 = 2Vm cos a - wLsI 0 p p -3 = 2 # 230 2 cos 25c - 2 # 3.14 # 50 # 4.5 # 10 # 20 3.14 3.14
= 187.73 - 9 = 178.74c Displacement factor = V0 I 0 = 178.25 # 20 = 0.78 230 # 20 Vs Is SOL 1.9.35
SOL 1.9.36
Option (C) is correct. Given that P = 50 # 1000 W Vd = 420 So P = Vd # Id Id = 50 # 1000 = 119.05 420 RMS value of thyristor current = 119.05 = 68.73 3 Option (B) is correct. Single phase full wave half controlled bridge converter feeds an Inductive load. The two SCRs in the converter are connected to a common dc bus. The converter has to have free wheeling diode because the converter does not provide for free wheeling for high values of triggering angles.
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SOL 1.9.37
Option (D) is correct. If we connect the MOSFET with the VSI, but the six MOSFETs are connected in bridge configuration, in that case they also operated as constant current sources in the saturation region so this statement is false.
SOL 1.9.38
Option (C) is correct. Given that, total harmonic distortion THD =
Vrms - V 12 # 100 V1 2
Pulse width is 150c Here
SOL 1.9.39
150 V = 0.91V s 180 l s V1 = Vrms(fundamental) = 0.4Vs sin 75c = 0.8696Vs p# 2 2 (0.91Vs) - (0.87Vs) 2 THD = = 31.9% (0.87Vs) 2 Vrms = b
Option (A) is correct. When losses are neglected, 750 # 2p 2 # 440 cos a = K m# 60 p Here back emf e with f is constant 3#
e = V0 = Km wm
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Page 506
440 = Km # 1500 # 2p 60 Km = 2.8 cos a = 0.37 at this firing angle Vt = 3 2 # 440 # (0.37) = 219.85 V p Ia = 1500 = 34.090 440 Isr = Ia 2/3 = 27.83 p.f. = SOL 1.9.40
Vt Is = 0.354 3 Vs Isr
Option (D) is correct. Vs = 230 = 57.5 4 Here charging current = I Vm sin q = 12 q1 = 8.486 = 0.148 radian Vm = 81.317 V
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e = 12 V There is no power consumption in battery due to ac current, so average value of charging current. 1 Iav(charging) = [2Vm cos q1 - e (p - 2q1)] 2p # 19.04 1 = [2 V cos q1 - 12 (p - 2q1)] 2p # 19.04 # m # = 1.059 W/A SOL 1.9.41
Option (C) is correct. Conduction angle for diode is 270c as shown in fig.
SOL 1.9.42
Option ( ) is correct.
SOL 1.9.43
Option (C) is correct. Here, Vm = maximum pulse voltage that can be applied so = 10 - 1 - 1 - 1 = 7 V Here 1 V drop is in primary transistor side, so that we get 9V pulse on the secondary side. Again there are 1 V drop in diode and in gate cathode junction each. I g max = 150 mA
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7 R = Vm = = 46.67 W Ig max 150 mA
So SOL 1.9.44
Page 507
Option (A) is correct. We know that the pulse width required is equal to the time taken by ia to rise upto iL so, Vs = L di + Ri (VT . 0) dt ia = 200 [1 - e- t/0.15] 1 Here also
t = T, 0.25 = 200 [1 - e- T/0.5]
ia = iL = 0.25
T = 1.876 # 10- 4 = 187.6 ms Width of pulse = 187.6 ms Magnitude of voltage = 10 V Vsec rating of P.T. = 10 # 187.6 ms = 1867 mV-s is approx to 2000 mV-s
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SOL 1.9.45
Option (D) is correct. If we varying the frequency for speed control, V/f should be kept as constant so that, minimum flux density (Bm ) also remains constant So, V = 4.44NBm Af
SOL 1.9.46
Option (D) is correct. In first half cycle D 1 will conduct and D 2 will not and at q = 0 there is zero voltage. So current wave form is as following
SOL 1.9.47
Option (B) is correct. In the PWM inverter V0 = output voltage of inverter 3 V0 = / 4Vs sin nd sin nwt sin np/2 n = 1 np So the pulse width = 2d = 144c V01 = 4Vs sin 72c sin wt p V03 = 4Vs sin ^3 # 72ch sin 3wt 3p 4Vs sin (3 # 72c) V 3p # 03 so, = 19.61% bV01 max l = 4Vs sin 72c p
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.9.48
Option (C) is correct. Given that 400 V, 50 Hz AC source, a = 60c, IL = 10 A so, Input displacement factor = cos a = 0.5 and, input power factor = D.F. # cos a
so, SOL 1.9.49
SOL 1.9.50
Page 508
4 # 10 sin 60c Is(fundamental) distortion factor = = p# 2 Is 10 # 2/3 = 0.955 input power factor = 0.955 # 0.5 = 0.478
Option (A) is correct. We know that T = RC ln 2 100 T So = 2.88 mF C = = 50 # 0.693 R # 0.693 Option (A) is correct. Let we have R solar = 0.5 W , I 0 = 20 A so Vs = 350 - 20 # 0.5 = 340 V
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`
340 = 3 # 440 # p
2 cos a
cos a = 55c So each thyristor will reverse biased for 180c - 55c = 125c. SOL 1.9.51
Option (C) is correct. In this circuitry if SCR gets open circuited, than circuit behaves like a half wave rectifier.
So I avg = Average value of current p-q = 1 # (Vm sin wt - E) dq 2pR q I 0(avg) = 1 62Vm cos q - E (p - 2q1)@ 2p R 1
1
a
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Page 509
1 [2 (230 # 2 ) cos q - 200 (p - 2q1)] 2p # 2 # 200 q1 = sin- 1 b E l = sin- 1 c = 38c = 0.66 Rad Vm 230 # 2 m 1 [2 2 I 0 (avg) = # 230 cos 38c - 200 (p - 2 # 0.66)] 2p # 2 =
`
= 11.9 A SOL 1.9.52
Option (B) is correct.
In this given circuit minimum gate pulse width time= Time required by ia rise up to iL i2 = 100 3 = 20 mA 5 # 10 i1 = 100 [1 - e- 40t] 20 `
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anode current I = I1 + I2 = 0.02 + 5 [1 - e- 40t] 0.05 = 0.05 + 5 [1 - e- 40t] 1 - e- 40t = 0.03 5 T = 150 ms
SOL 1.9.53
Option (B) is correct. Given IL = 10 A . So in the + ve half cycle, it will charge the capacitor, minimum time will be half the time for one cycle. so min time required for charging = p = p LC w0 = 3.14 #
SOL 1.9.54
2 # 10- 3 # 10- 6 = 140 m sec
Option (C) is correct. Given Ton = 140 m sec Average output = Ton # V Ttotal Ttotal = 1/f = 1 = 1 msec 103 -6
so average output = 140 # 10-3 # 250 = 35 V 1 # 10 SOL 1.9.55
Option (A) is correct. The conduction loss v/s MOSFET current characteristics of a power MOSFET is best approximated by a parabola.
SOL 1.9.56
Option (B) is correct. In a 3-f bridge rectifier Vrms = 400 V , f = 50 Hz This is purely resistive then
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instantaneous voltage
V0 =
Page 510
2 Vrms = 400 2 V
SOL 1.9.57
Option (C) is correct. A 3-f square wave (symmetrical) inverter contains only odd harmonics.
SOL 1.9.58
Option (A) is correct. In Ideal condition we take voltage across the device is zero. average power loss during switching = VI (t1 + t2) (turn ON) 2 Option (C) is correct. So in P thyristor blocks voltage in both polarities until gate is triggered and also in R transistor along with diode can do same process.
SOL 1.9.59
SOL 1.9.60
Option (C) is correct. Duty ratio a = 0.5 1 -3 sec - 3 = 10 1 # 10 Ta = L = 200 mH = 40 msec 5 R T =
here
(1 - e- aT/Ts) (1 - e- (1 - a) T/Ta) Ripple = Vs = G R 1 - e- T/Ts 100 (TI) max = Vs = 4fL 4 # 103 # 200 # 10- 3 = 0.125 A
SOL 1.9.61
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Option (C) is correct. Tst TL a T T
SOL 1.9.62
= 15 Nm = 7 Nm
= 2 rad/sec2 = Ia so = Tst - TL = 8 Nm I = 8 = 4 kgm2 2 Option (B) is correct. We know that Vrms = 230 V so, If whether Then
Vm = 230 # 2 V a 1 90c Vpeak = Vm sin a = 230
230 2 sin a = 230 sin a = 1 2 angle a = 135c SOL 1.9.63
Option (D) is correct. When we use BJT as a power control switch by biasing it in cut-off region or in the saturation region. In the on state both the base emitter and base-collector junction are forward biased.
SOL 1.9.64
Option (A) is correct. Peak Inverse Voltage (PIV) across full wave rectifier is 2Vm Vm = 50 2 V
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so, SOL 1.9.65
Page 511
PIV = 100 2 V
Option (D) is correct. Vb = 12 ! 4 V Vb max = 16 V Vb min = 8 V V (min) 8 Required value of R = b = = 800 W Ig 10 # 10- 3
SOL 1.9.66
Option (C) is correct.
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Ripple frequency = 3f = 3 # 400 = 1200 Hz So from V0 ripple frequency = 1200 Hz SOL 1.9.67
Option (C) is correct. Given that
R = 0.15 W I = 15 A 1 # p/w I 2 Rdt So average power losses = (2p/w) 0 = w # 102 # 0.15 # p/w 2p = 7.5 W SOL 1.9.68
Option (D) is correct. Output dc voltage across load is given as following 1
Vdc = =
SOL 1.9.69
SOL 1.9.70
2 V ; 1 &(2p - a) + sin 2a 0E2 ap 2 1
1
sin p/2 2 p 'a2p - 4 k + b 2 l1H p
2 # 230 2 >p # 4 = 317.8 V 2 (317.8) 2 losses = V dc = = 10100 W 100 R Option (C) is correct. Vs = 100 V , duty ratio = 0.8 , R = 10 W
So average current through diode = aVs R = 0.8 # 100 = 8 A 10 Option (D) is correct. Peak current through S 1 I = I 0 + VS C/L = 20 + 200
2 # 10- 6 = 40 A 200 # 10- 6
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GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia SOL 1.9.71
Option ( ) is correct.
SOL 1.9.72
Option (C) is correct.
Page 512
so a =; and
500 - (- 1500) 2p 2 E # 60 = 418.67 rad/sec 0.5
T = 40 Nm T = Ia I = T # 40 = 0.096 kgm2 a 418.67
SOL 1.9.73
Option (D) is correct. When thyristor turned on at that time J2 junction will break. So J1, J2, J3 all are in forward bias.
SOL 1.9.74
Option (D) is correct. The ON-OFF state of switch is given on VDS - IS plane as following
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When VDS =+ ve , diode conducts and IS = 0 VDS =- ve , diode opens, but IS = 0 , D "- ve potential. SOL 1.9.75
Option (B) is correct. P. Field control-Above base speed Q. Armature control-below base torque
SOL 1.9.76
Option (A) is correct. As we know in fully controlled rectifier. or or
VPP = Vm - Vm cos (p/6 + a) VPP = Vm [1 - cos (p/6 + 30c)] VPP = 0.5 Vm
a a = 30c
SOL 1.9.77
Option ( ) is correct.
SOL 1.9.78
Option (A) is correct. In the chopper during turn on of chopper V -t area across L is, T T di dt = # imax Ldi #0 onVL dt = #0 on L b dt l i min = L (i max - i min) = L ^DI h
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Page 513
V -t are applied to ‘L’ is = (60 - 12) Ton = 48Ton So now volt area -3 DI = 48Ton = 48 # 0.2 #-10 = 0.48 A L 20 # 10 3 SOL 1.9.79
Option (A) is correct.
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4VS b np l^sin nd h^sin nwt h^sin np/2h n = 1, 3, 5 ` RMS value of fundamental component Vrms(fundamental) = 4VS sin d # 1 2p a = 120c, 2d = 120c & d = 60c Output voltage V0 =
3
/
Vrms(fundamental) = 4VS # sin 60c 2p = 0.78VS = 0.78 V SOL 1.9.80
Option (A) is correct. After removing 5 th harmonic 5d = 0, p, 2p `
Pulse width = 2d = a = 0, 2p , 4p 5 5 = 0c, 72c, 144c
SOL 1.9.81
Option (C) is correct. NSa = 3000 rpm Na = 2850 rpm SFL = 3000 - 2850 = 0.05 3000
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Page 514
where by (V/f) control
SOL 1.9.82
SOL 1.9.83
Nsb = 3000 b 40 l = 2400 rpm 50 ` N2 = new running speed of motor = 2400 b1 - 0.05 l = 2340 rpm 2 Option (C) is correct. For six pulse thyristor rectifier bridge the lowest frequency component in AC source line current is of 250 Hz. Option (A) is correct. Given a step down chopper is operated in continuous conduction mode in steady state with a constant duty Ratio D . V0 " dc output voltage . Vs " dc input voltage V0 = D = duty ratio Vs
SOL 1.9.84
Option ( ) is correct.
SOL 1.9.85
Option (B) is correct. From figure
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1/2
= Vs # f = Vs p
SOL 1.9.86
f p
Option (C) is correct. Given that, V = 200 sin wt f = 50 Hz Power dispatched in the load resistor R = ? First we have to calculate output of rectifier. 1/2 p (V0) rms = : 1 # (200 sin wt) 2 dwtD p0 1/2 p = 200 ; # b 1 - cos 2wt l dwtE 2 p 0 p 1/2 = 200 ;1 b wt - sin 2wt l E 2 p 2 0 1/2 200 1 200 = : # pD = p 2 2 Power dissipiated to resistor 2 ^V0h2rms e 200 o = 400 W = = 50 R 2
PR SOL 1.9.87
* Given
f = 20 kHz D = 0.5 Power transferred from source V1 to V2 = ? 1 Time period t = 1 = = 50 m sec f 20 # 10- 3 D = 0.5
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Page 515
tON = 25 m sec , t off = 25 m sec at tON , energy will stored in inductor circuit v = L di dt 100 = 100 # 10- 6 di dt di = 106 dt i = 106 t + i (0) i = 106 t
a i (0) = 0 ...(1)
E = 1 Li2 2 E = 1 # 100 # 10- 6 # 1012 # 25 # 25 # 10- 12 2
SOL 1.9.88
SOL 1.9.89
E = 3.1250 # 10- 2 J Now power transferred during t off -2 Pt = 3.1250 # 10 = 12.5 # 102 W -6 25 # 10 Option (B) is correct. For providing electrical isolation it is necessary to connect a pulse transformer at the output stage of a thyristor triggering circuit.
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Option (A) is correct. In ac to dc circulating current dual converters if triggering angles are a1 and a 2 , than it is necessary that a1 + a2 = 180c
SOL 1.9.90
Option (D) is correct. Given a half wave Thyristor converter supplies a purely inductive load Triggering angle a = 120c than extinction angle b = ?
First we have to draw its output characteristics as shown below
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output is given by i 0 = Vm sin (wt - f) - Vm sin (a - f) exp b - R - a l Z wL Z We know at extinction angle i.e. wt = b , i 0 = 0 from equation (1), at (wt = b) 0 = Vm sin (b - f) - Vm sin (a - f) ec Z Z
Page 516
...(1)
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or sin (b - f) = sin (a - f) or b-f = a-f or b = a = 120c SOL 1.9.91
Option (D) is correct. f = 50 Hz So total time = 1 = 1 = 20 msec 50 f Conduction time for each feedback diode in a cycle is being given by t conduction = 20 = 2.5 msec 8
a
SOL 1.9.92
* Given a voltage commulated thyristor chopper circuit in figure which is operated at 500 Hz, with 50% duty ratio. IL = 20 A (constant) We have to evaluate (a) Toff for thyristor Th 1 (b) L = ? if peak current through Th 1 is 180% limited (c) Maximum instantaneous output voltage -6 Turn off time Toff = CVs = 6 # 10 # 100 = 30 m sec 20 IL Peak current through Th 1 i Th = I 0 + Vdc C L 1
a
i Th = 1.8IL = 1.8 # 20 = 36 A 1
36 = 20 + 100
6 # 10- 6 L
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6 # 10- 6 L
0.16 =
or
Page 517
-6 L = 6 # 10 2 = 2.34 # 10- 4 H (0.16) Maximum instantaneous output voltage
Vm = 2Vdc = 200 V SOL 1.9.93
* Given in figure separately excited dc motor is controlled by varying its armature voltage using 1-f full converter bridge. Motor voltage constant Kv = 2.5 V/rad/sec Motor Torque T = 140 Nm , a = 60c armature current continuous and ripple free. (a) Ia = ? (b) Nm = ? (c) rms of fundamental component of input current.
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(a) a Motor Torque Than
and Eb = Kv w
Kv wIa = Tw
Ia = T = 140 = 50 Amp 25 Kv
(b) In dc motor we know Ia = V0 - Eb Ra Eb = V0 - Ia Ra
V0 = 2Vm cos a p
Eb = 500 2 # 2 - 20 (0.2) p
= 2 # 250 2 cos 60c p
Eb = 215.2 V w = Ea Ia = 215.2 # 20 = 30.74 rad/sec T 140 (c) Rms value of fundamental component of input current Ior Isr = 1/2 2 ; 1 b(p - a) + 1 sin 2a lE 2 p Ior = 56 Amp , a = 60c 56 Isr = 1/2 p 1 2 : ap - k + 1 sin 120cD p 3 2 Isr =
39.6 = 61.34 Amp 2 - 1 1/2 b3 4l ***********
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10 GENERAL APTITUDE
2012
ONE MARK
MCQ 1.10.1
If (1.001) 1259 = 3.52 and (1.001) 2062 = 7.85, then (1.001) 3321 (A) 2.23 (B) 4.33 (C) 11.37 (D) 27.64
MCQ 1.10.2
Choose the most appropriate alternate from the options given below to complete the following sentence : If the tired soldier wanted to lie down, he..................the mattress out on the balcony. (A) should take (B) shall take (C) should have taken (D) will have taken
MCQ 1.10.3
Choose the most appropriate word from the options given below to complete the following sentence : Give the seriousness of the situation that he had to face, his........was impressive. (A) beggary (B) nomenclature (C) jealousy (D) nonchalance
MCQ 1.10.4
Which one of the following options is the closest in meaning to the word given below ? Latitude (A) Eligibility (B) Freedom (C) Coercion (D) Meticulousness
MCQ 1.10.5
One of the parts (A, B, C, D) in the sentence given below contains an ERROR. Which one of the following is INCORRECT ? I requested that he should be given the driving test today instead of tomorrow. (A) requested that (B) should be given (C) the driving test (D) instead of tomorrow 2012
MCQ 1.10.6
TWO MARKS
One of the legacies of the Roman legions was discipline. In the legious, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them. Which one of the following statements best sums up the meaning of the above passage ? (A) Through regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances. (B) The legions were treated inhumanly as if the men were animals (C) Disciplines was the armies inheritance from their seniors
GATE Electrical Engineering Topicwise Solved Paper by RK Kanodia & Ashish Murolia
Page 519
(D) The harsh discipline to which the legions were subjected to led to the odds and conditions being against them. MCQ 1.10.7
Raju has 14 currency notes in his pocket consisting of only Rs. 20 notes and Rs. 10 notes. The total money values of the notes is Rs. 230. The number of Rs. 10 notes that Raju has is (A) 5 (B) 6 (C) 9 (D) 10
MCQ 1.10.8
There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is (A) 2 (B) 3 (C) 4 (D) 8
MCQ 1.10.9
The data given in the following table summarizes the monthly budget of an average household. Category
Amount (Rs.)
Food
4000
Clothing
1200
Rent
2000
Savings
1500
Other expenses
1800
nodia.co.in The approximate percentages of the monthly budget NOT spent on savings is (A) 10% (B) 14% (C) 81% (D) 86% MCQ 1.10.10
A and B are friends. They decide to meet between 1 PM and 2 PM on a given day. There is a conditions that whoever arrives first will not wait for the other for more than 15 minutes. The probability that they will meet on that days is (A) 1/4 (B) 1/16 (C) 7/16 (D) 9/16 2011
ONE MARK
MCQ 1.10.11
There are two candidates P and Q in an election. During the campaign, 40% of voter promised to vote for P , and rest for Q . However, on the day of election 15% of the voters went back on their promise to vote for P and instead voted for Q . 25% of the voter went back on their promise to vote for Q and instead voted for P . Suppose, P lost by 2 votes, then what was the total number of voters ? (A) 100 (B) 110 (C) 90 (D) 95
MCQ 1.10.12
The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relations in the original pair : Gladiator : Arena (A) dancer : stage (B) commuter : train (C) teacher : classroom (D) lawyer : courtroom
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Page 520
MCQ 1.10.13
Choose the most appropriate word from the options given below to complete the following sentence : Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which................... treatments are unsatisfactory. (A) similar (B) most (C) uncommon (D) available
MCQ 1.10.14
Choose the word from the from the options given below that is most opposite in meaning to the given word : Frequency (A) periodicity (B) rarity (C) gradualness (D) persistency
MCQ 1.10.15
Choose the most appropriate word from the options given below to complete the following sentence : It was her view that the country’s had been ............. by foreign techno-crafts, so that to invite them to come back would be counter-productive. (A) identified (B) ascertained (C) exacerbated (D) analysed
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2011 MCQ 1.10.16
TWO MARKS
The fuel consumed by a motor cycle during a journey while travelling at various speed is indicated in the graph below.
The distance covered during four laps of the journey are listed in the table below Lap
Distance (Kilometres)
Average speed (kilometres per hour)
P
15
15
Q
75
45
R
40
75
S 10 10 From the given data, we can conclude that the fuel consumed per kilometre was least during the lap (A) P (B) Q (C) R (D) S MCQ 1.10.17
The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of disease until their blood build up
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Page 521
immunities. Then a serum was made from their blood. Serums to fight with diphteria and tetanus were developed this way. It can be inferred from the passage, that horses were (A) given immunity to diseases (B) generally quite immune to diseases (C) given medicines to fight toxins (D) given diphtheria and tetanus serums MCQ 1.10.18
The sum of n terms of the series 4 + 44 + 444 + ........ (A) (4/81) [10n + 1 - 9n - 1] (B) (4/81) [10n - 1 - 9n - 1] (C) (4/81) [10n + 1 - 9n - 10] (D) (4/81) [10n - 9n - 10]
MCQ 1.10.19
Given that f (y) = y /y, and q is any non-zero real number, the value of f (q) - f (- q) is (A) 0 (B) - 1 (C) 1 (D) 2
MCQ 1.10.20
Three friends R, S and T shared toffee from a bowl. R took 1/3 rd of the toffees, but returned four to the bowl. S took 1/4 th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left, how many toffees were originally there in the bowl ? (A) 38 (B) 31 (C) 48 (D) 41
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ONE MARK
MCQ 1.10.21
Which of the following options is the closest in meaning to the word below ? Circuitous (A) Cyclic (B) Indirect (C) Confusing (D) Crooked
MCQ 1.10.22
The question below consist of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed : Worker (A) Fallow : Land (B) Unaware : Sleeper (C) Wit : Jester (D) Renovated : House
MCQ 1.10.23
Choose the most appropriate word from the options given below to complete the following sentence : If we manage to ........ our natural resources, we would leave a better planet for our children. (A) unhold (B) restrain (C) cherish (D) conserve
MCQ 1.10.24
Choose the most appropriate word from the options given below to complete the following sentence : His rather casual remarks on politics..................his lack of seriousness about the subject. (A) masked (B) belied
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(C) betrayed MCQ 1.10.25
Page 522
(D) suppressed
25 persons are in a room 15 of them play hockey, 17 of them play football and 10 of them play hockey and football. Then the number of persons playing neither hockey nor football is (A) 2 (B) 17 (C) 13 (D) 3 2010
TWO MARKS
MCQ 1.10.26
Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare ; and regretfully, their exist people in military establishments who think that chemical agents are useful fools for their cause. Which of the following statements best sums up the meaning of the above passage ? (A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in ware fare would be undesirable. (D) People in military establishments like to use chemical agents in war.
MCQ 1.10.27
If 137 + 276 = 435 how much is 731 + 672 ? (A) 534 (B) 1403 (C) 1623 (D) 1531
MCQ 1.10.28
5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall ? (A) 20 days (B) 18 days (C) 16 days (D) 15 days
MCQ 1.10.29
Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how much distinct 4 digit numbers greater than 3000 can be formed ? (A) 50 (B) 51 (C) 52 (D) 54
MCQ 1.10.30
Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and sisters.) All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts : 1. Hari’s age + Gita’s age > Irfan’s age + Saira’s age. 2. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest. 3. There are no twins. In what order were they born (oldest first) ? (A) HSIG (B) SGHI (C) IGSH (D) IHSG
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SOLUTIONS
SOL 1.10.1
Page 523
Option (D) is correct. Let 1.001 = x So in given data :
Again
x1259 = 3.52 x2062 = 7.85 x3321 = x1259 + 2062 = x1259 x2062 = 3.52 # 7.85 = 27.64
SOL 1.10.2
Option (C) is correct.
SOL 1.10.3
Option (D) is correct.
SOL 1.10.4
Option (B) is correct.
SOL 1.10.5
Option (B) is correct.
SOL 1.10.6
Option (A) is correct.
SOL 1.10.7
Option (A) is correct. Let no. of notes of Rs.20 be x and no. of notes of Rs. 10 be y . Then from the given data.
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x + y = 14 20x + 10y = 230 Solving the above two equations we get x = 9, y = 5 So, the no. of notes of Rs. 10 is 5. SOL 1.10.8
Option (A) is correct. We will categorize the 8 bags in three groups as : (i) A1 A2 A 3 , (ii) B1 B2 B 3 , (iii) C1 C2 Weighting will be done as bellow : 1st weighting " A1 A2 A 3 will be on one side of balance and B1 B2 B 3 on the other. It may have three results as described in the following cases. Case 1 : A1 A 2 A 3 = B1 B 2 B 3 This results out that either C1 or C2 will heavier for which we will have to perform weighting again. 2 nd weighting " C1 is kept on the one side and C2 on the other. if C1 > C2 then C1 is heavier. then C2 is heavier. C1 < C 2 Case 2 : A1 A 2 A 3 > B1 B 2 B 3 it means one of the A1 A2 A 3 will be heavier So we will perform next weighting as: 2 nd weighting " A1 is kept on one side of the balance and A2 on the other. it means A 3 will be heavier if A1 = A2
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Page 524
then A1 will be heavier A1 > A 2 then A2 will be heavier A1 < A 2 Case 3 : A1 A 2 A 3 < B1 B 2 B 3 This time one of the B1 B2 B 3 will be heavier, So again as the above case weighting will be done. 2 nd weighting " B1 is kept one side and B2 on the other if B1 = B2 B 3 will be heavier B1 > B 2 B1 will be heavier B1 < B 2 B2 will be heavier So, as described above, in all the three cases weighting is done only two times to give out the result so minimum no. of weighting required = 2. SOL 1.10.9
Option (D) is correct. Total budget = 4000 + 1200 + 2000 + 1500 + 1800 = 10, 500 The amount spent on saving = 1500 So, the amount not spent on saving = 10, 500 - 1500 = 9000 So, percentage of the amount = 9000 # 100% = 86% 10500
SOL 1.10.10
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Option (S) is correct. The graphical representation of their arriving time so that they met is given as below in the figure by shaded region.
So, the area of shaded region is given by Area of 4PQRS - (Area of TEFQ + Area of TGSH ) = 60 # 60 - 2 b 1 # 45 # 45 l 2 = 1575 SOL 1.10.11
So, the required probability = 1575 = 7 3600 16 Option (A) is correct. Let us assume total voters are 100. Thus 40 voter (i.e. 40 %) promised to vote for P and 60 (rest 60 % ) promised to vote fore Q. Now, 15% changed from P to Q (15 % out of 40)
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Page 525
15 40 = 6 100 #
Changed voter from P to Q
Now Voter for P 40 - 6 = 34 Also, 25% changed form Q to P (out of 60%) 25 Changed voter from Q to P 60 = 15 100 # Now Voter for P 34 + 15 = 49 Thus P P got 49 votes and Q got 51 votes, and P lost by 2 votes, which is given. Therefore 100 voter is true value. SOL 1.10.12
Option (A) is correct. A gladiator performs in an arena. Commutators use trains. Lawyers performs, but do not entertain like a gladiator. Similarly, teachers educate. Only dancers performs on a stage.
SOL 1.10.13
Option (D) is correct. Available is appropriate because manipulation of genes will be done when other treatments are not useful.
SOL 1.10.14
Option (B) is correct. Periodicity is almost similar to frequency. Gradualness means something happening with time. Persistency is endurance. Rarity is opposite to frequency.
SOL 1.10.15
Option (C) is correct. The sentence implies that technocrats are counterproductive (negative). Only (C) can bring the same meaning.
SOL 1.10.16
Option (B) is correct. Since fuel consumption/litre is asked and not total fuel consumed, only average speed is relevant. Maximum efficiency comes at 45 km/hr, So least fuel consumer per litre in lap Q
SOL 1.10.17
Option (B) is correct. Option B fits the sentence, as they built up immunities which helped humans create serums from their blood.
SOL 1.10.18
Option (C) is correct.
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4 + 44 + 444 + .............. 4 (1 + 11 + 111 + .......) = 4 (9 + 99 + 999 + ............) 9 = 4 [(10 - 1) + (100 - 1) + ........] 9 = 4 [10 (1 + 10 + 102 + 103) - n] 9 n = 4 :10 # 10 - 1 - nD 9 10 - 1 = 4 610n + 1 - 10 - 9n@ 81
SOL 1.10.19
Option (D) is correct. y y -y f (- y) = =- f (y) y f (y) =
Now
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or SOL 1.10.20
Page 526
f (q) - f (- q) = 2f (q) = 2
Option (C) is correct. Let total no of toffees be x . The following table shows the all calculations. Friend
Bowl Status
R
= x -4 3
= 2x + 4 3
S
= 1 :2x + 4D - 3 4 3 = x +1-3 = x -2 6 6
= 2x + 4 - x + 2 3 6 = x +6 2
T
= 1 a x + 6k - 2 2 2 = x +1 4
= x +6-x -1 2 4 = x +5 4
Now,
x + 5 = 17 4 x = 17 - 5 = 12 4
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or
x = 12 # 4 = 48
SOL 1.10.21
Option (B) is correct. Circuitous means round about or not direct. Indirect is closest in meaning to this circuitous (A) Cyclic : Recurring in nature (B) Indirect : Not direct (C) Confusing : lacking clarity of meaning (D) Crooked : set at an angle; not straight
SOL 1.10.22
Option (B) is correct. A worker may by unemployed. Like in same relation a sleeper may be unaware.
SOL 1.10.23
Option (D) is correct. Here conserve is most appropriate word.
SOL 1.10.24
Option (C) is correct. Betrayed means reveal unintentionally that is most appropriate.
SOL 1.10.25
Option (D) is correct. Number of people who play hockey n (A) = 15 Number of people who play football n (B) = 17 Persons who play both hockey and football n (A + B) = 10 Persons who play either hockey or football or both : n (A , B) = n (A) + n (B) - n (A + B) = 15 + 17 - 10 = 22 Thus people who play neither hockey nor football = 25 - 22 = 3
SOL 1.10.26
Option (D) is correct.
SOL 1.10.27
Option (C) is correct. Since 7 + 6 = 13 but unit digit is 5 so base may be 8 as 5 is the remainder when 13 is divided by 8. Let us check.
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137 8 276 8 435 SOL 1.10.28
Thus here base is 8. Now
Page 527
731 8 672 8 1623
Option (D) is correct. Let W be the total work. =W 20 Per day work of one skill worker = W =W 5 # 20 100 Similarly per day work of 1 semi-skilled workers = W = W 8 # 25 200 Similarly per day work of one semi-skill worker = W = W 10 # 30 300 Thus total per day work of 2 skilled, 6 semi-skilled and 5 unskilled workers is = 2W + 6W + 5W = 12W + 18W + 10W = W 100 200 300 600 15 Therefore time to complete the work is 15 days. Per day work of 5 skilled workers
SOL 1.10.29
Option (B) is correct. As the number must be greater than 3000, it must be start with 3 or 4. Thus we have two case: Case (1) If left most digit is 3 an other three digits are any of 2, 2, 3, 3, 4, 4, 4, 4. (1) Using 2, 2, 3 we have 3223, 3232, 3322 i.e. 3! = 3 no. 2! (2) Using 2, 2, 4 we have 3224, 3242, 3422 i.e. 3! = 3 no. 2! (3) Using 2, 3, 3 we have 3233, 3323, 3332 i.e. 3! = 3 no. 2!
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(4) Using 2, 3, 4 we have 3! = 6 no. (5) Using 2, 4, 4 we have 3244, 3424, 3442 i.e. 3! = 3 no. 2! (6) Using 3, 3, 4 we have 3334, 3343, 3433 i.e. 3! = 3 no. 2! (7) Using 3, 4, 4 we have 3344, 3434, 3443 i.e. 3! = 3 no. 2! (8) Using 4, 4, 4 we have 3444 i.e. 3! = 1 no. 3! Total 4 digit numbers in this case is 1 + 3 + 3 + 3 + 6 + 3 + 3 + 3 + 1 = 25 Case 2 : If left most is 4 and other three digits are any of 2, 2, 3, 3, 3, 4, 4, 4. (1) Using 2, 2, 3 we have 4223, 4232, 4322 i.e. . 3! = 3 no 2! (2) Using 2, 2, 4 we have 4224, 4242, 4422 i.e. . 3! = 3 no 2! (3) Using 2, 3, 3 we have 4233, 4323, 4332 i.e. . 3! = 3 no 2! (4) Using 2, 3, 4 we have i.e. . 3! = 6 no
(5) Using 2, 4, 4 we have 4244, 4424, 4442 i.e. . 3! = 3 no 2! (6) Using 3, 3, 3 we have 4333 i.e 3! = 1. no. 3!
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Page 528
(7) Using 3, 3, 4 we have 4334, 4343, 4433 i.e. . 3! = 3 no 2! (8) Using 3, 4, 4 we have 4344, 4434, 4443 i.e. . 3! = 3 no 2! (9) Using 4, 4, 4 we have 4444 i.e. 3! = 1. no 3! Total 4 digit numbers in 2nd case = 3 + 3 + 3 + 6 + 3 + 3 + 1 + 3 + 1 = 26 Thus total 4 digit numbers using case (1) and case (2) is = 25 + 26 = 51 SOL 1.10.30
Option (B) is correct. Let H , G , S and I be ages of Hari, Gita, Saira and Irfan respectively. Now from statement (1) we have H + G > I + S Form statement (2) we get that G - S = 1 or S - G = 1 As G can’t be oldest and S can’t be youngest thus either GS or SG possible. From statement (3) we get that there are no twins (A) HSIG : There is I between S and G which is not possible (B) SGHI : SG order is also here and S > G > H > I and G + H > S + I which is possible. (C) IGSH : This gives I > G and S > H and adding these both inequalities we have I + S > H + G which is not possible.
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(D) IHSG : This gives I > H and S > G and adding these both inequalities we have I + S > H + G which is not possible. **********
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