GATE Electrical Engineering Chapter-Wise Solved Papers (2000-2020) 9788126571857, 9788126595358
289 61 72MB
English Pages [605]
Polecaj historie
![GATE 2024 Electronics & Communication Engineering - 37 Years Topic-wise Previous Solved Papers [2023 ed.]
9356812144, 9789356812147](https://dokumen.pub/img/200x200/gate-2024-electronics-amp-communication-engineering-37-years-topic-wise-previous-solved-papers-2023nbsped-9356812144-9789356812147.jpg)
![AIIMS Specialist-Physics(26 Years' Chapterwise Solved Papers) [2 ed.]
9789313197126](https://dokumen.pub/img/200x200/aiims-specialist-physics26-years-chapterwise-solved-papers-2nbsped-9789313197126.jpg)
Table of contents :
Front Cover
TCY's Unique Success Mantra
Title Page
Copyright
Note to the Aspirants
Contents
C H A P T E R 1 Engineering Mathematics
C H A P T E R 2 Electric Circuits
C H A P T E R 3 Electromagnetic Fields
C H A P T E R 4 Signals and Systems
C H A P T E R 5 Electrical Machines
C H A P T E R 6 Power Systems
C H A P T E R 7 Control Systems
C H A P T E R 8 Electrical and Electronic Measurements
C H A P T E R 9 Analog and Digital Electronics
C H A P T E R 10 Power Electronics
APPENDIX Solved GATE (EE) 2019
APPENDIX Solved GATE (EE) 2020
Back Cover
Citation preview
GATE The book comprises previous years’ (2000 to 2020) GATE questions in Electrical Engineering covered chapter-wise with detailed solutions. Each chapter begins with a detailed year-wise analysis of topics on which questions are based and is followed by listing of important formulas and concepts for that chapter. The book is designed to make the students well-versed with the pattern of examination, level of questions asked and the concept distribution of questions and thus bring greater focus to their preparation. It aims to be a must-have resource for an essential step in their preparation, that is, solving and practicing previous years’ papers.
Salient Features Chapter-wise coverage of GATE Electrical Engineering previous years’ questions (2000 to 2018). Chapter-wise 2019 and 2020 GATE papers available as appendix
l
Chapter on Engineering Mathematics included for complete coverage of the technical section of the GATE paper
l
Detailed topic-wise analysis of questions for each chapter
l
Important Formulas and Concepts summarized for easy recall in each chapter
l
All questions marked for level of difficulty (i.e. 1 Mark or 2 Marks)
l
Detailed solutions for all questions, tagged topic-wise
l
Packaged with unique scratch code for free 7-day subscription for online GATE tests
Customer Care +91 120 6291100 [email protected] www.wileyindia.com www.wiley.com
eISBN 978-81-265-9535-8
9 788126 595358
Chatterjee, Lather, Gupta
Wiley India Pvt. Ltd.
CHAPTER-WISE SOLVED PAPERS (2000-2020)
l
GATE
ELECTRICAL ENGINEERING
CHAPTER-WISE SOLVED PAPERS
(2000-2020)
Inside
The Graduate Aptitude Test in Engineering (GATE) is an All-India level examination for engineering graduates aspiring to pursue Master’s or Ph.D. programmes in India. The Public Sector Undertakings (PSU’s) also use GATE used as a recruiting examination. The examination is highly competitive, and the GATE score plays an important role in fulfilling the academic and professional aspirations of the students. This book is aimed at supporting the efforts of GATE aspirants in achieving a high GATE score.
ELECTRICAL ENGINEERING
About the Book
Dr. Debashis Chatterjee Dr. J. S. Lather Dr. Lalita Gupta
l l l
Detailed Exam Analysis Chapter-wise and Topic-wise. Questions from previous years’ (2000 – 2020) papers. Unique scratch code that provides access to w Free online test with analytics. w 7-Day free subscription for topic-wise GATE tests. w Instant correction report with remedial action.
To get access code for the free tests, please write to us at [email protected], with a copy of the Invoice to verify purchase.
GATE
ELECTRICAL ENGINEERING
CHAPTER-WISE SOLVED PAPERS
(2000-2020)
Dr. Debashis Chatterjee Professor, Department of Electrical Engineering Jadavpur University Kolkata
Dr. Lalita Gupta Assistant Professor, Department of Electronics and Communication Engineering, Maulana Azad National Institute of Technology, Bhopal
Dr. J. S. Lather Professor, Department of Electrical Engineering National Institute of Technology Kurukshetra
Engineering Mathematics Contributed by Dr Anil Kumar Maini Outstanding Scientist and former Director of Laser Science and Technology Centre DRDO, New Delhi
Varsha Agrawal Senior Scientist, Laser Science and Technology Centre DRDO, New Delhi
Nakul Maini Analyst Ericsson (India)
GATE ELECTRICAL ENGINEERING Chapter-wise Solved Papers 2000-2020 Copyright © 2020 by Wiley India Pvt. Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher. Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely, Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services. Other Wiley Editorial Offices: John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 1 Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1 Edition: 2020 ISBN: 978-81-265-7185-7 ISBN: 978-81-265-9535-8 (ebk) www.wileyindia.com Printed at:
Note to the Aspirants About the Examination Graduate Aptitude Test in Engineering (GATE) is an examination conducted jointly by the Indian Institute of Science (IISc), Bangalore and the seven Indian Institutes of Technology (at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee) on behalf of the National Coordination Board (NCB)GATE, Department of Higher Education, Ministry of Human Resource Development (MHRD), and Government of India. Qualifying in GATE is a mandatory requirement for seeking admission and/or financial assistance to:
• M aster’s programs and direct Doctoral programs in Engineering/Technology/Architecture. • Doctoral programs in relevant branches of Science, in the institutions supported by the MHRD and other Government agencies. Even in some colleges and institutions, which admit students without MHRD scholarship/assistantship, the GATE qualification is mandatory. • Many Public Sector Undertakings (PSUs) have been using the GATE score in their recruitment process.
Graduate Aptitude Test in Engineering (GATE) is basically an examination on the comprehensive understanding of the candidates in various undergraduate subjects in Engineering/Technology/ Architecture and post-graduate level subjects in Science. GATE is conducted for 24 subjects (also referred to as “papers”) are generally held at the end of January or early February of the year. The GATE examination centres are spread in different cities across India, as well as, in six cities outside India. The examination is purely a Computer Based Test (CBT). The GATE score reflects the relative performance level of the candidate in a particular subject, which is quantified based on the several years of examination data. Validity
Eligibility for GATE Electrical Engineering
Bachelor’s degree holders in Electrical Engineering (4 years after 10+2 or 3 years after B.Sc./Diploma in Electrical).
About the Book This book GATE Chapter-wise Electrical Engineering Solved Papers is designed as a must have resource for the students preparing for the M.E./M.Tech./M.S./Ph.D. in Electrical Engineering. It offers Chapter-wise solved previous years’ GATE Electrical Engineering questions for the years 2000–2020. This book will help students become well-versed with the pattern of examination, level of questions asked and concept distribution in questions. Key Features of the Book
• C omplete solutions provided for every question, tagged for the topic on which the question is based. • Chapter-wise analysis of GATE questions provided at the beginning of the book to make students familiar with chapter-wise marks distribution and weightage of each. • Topic-wise analysis of GATE Questions provided at the beginning of each chapter to make students familiar with important topics on which questions are commonly asked. • Unique scratch code in the book that provides access to Free online test with analytics. 7-Day free subscription for topic-wise GATE tests. Instant correction report with remedial action. These features will help students develop problem-solving skills and focus in their preparation on important chapters and topics. ■
■
■
The GATE score is valid for THREE YEARS from the date of announcement of the results.
FM.indd 3
7/28/2020 11:57:04 AM
FM.indd 4
7/28/2020 11:57:04 AM
GATE 2011
GATE 2012
GATE 2013
GATE 2014
GATE 2015
GATE 2016
GATE 2017
GATE 2018
GATE 2019
GATE 2020
1
3
1
3
2
6
Electromagnetic Fields
Signals and Systems
Electrical Machines
Power Systems
Control Systems
Electrical and Electronic Measurements
Analog and Digital Electronics
Power Electronics
3
4
5
6
7
8
9
10
2
7
2
Electric Circuits
2
2
Engineering Mathematics
2
2
1
3
4
4
5
0
2
6
2
3
3
3
3
3
1
0
3
3
4
4
1
3
4
3
3
1
4
4
3
3
3
1
4
1
3
0
2
0
4
4
1
4
2
3
3
0
6
2
0
3
2
2
3
2
6
1
1
4
6
4
2
3
3
2
1
1
1
3
3
9
7
6
8
9
9
4
6
14
8
14
5
9
13
13
9
3
8
11
2
10
4
4
3
7
3
4
6
11
7
7
2
8
6
11
5
3
5
4
4
5
0
3
6
5
8
4
8
10
8
4
2
7
7
7
3
3
9
10
6
4
3
4
6
4
5
4
5
8
5
8
1
9
8
10
4
2
6
7
1
4
1
3
2
3
1
1
4
5
2
4
1
3
3
5
3
1
3
5
2
2
2
3
1
4
2
1
1
7
5
3
3
3
3
5
0
2
1
5
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks
1
S.No. Chapter Name
GATE 2010
The table given below depicts the chapter-wise question and marks distribution of previous years’ (2010–2020) GATE Electrical Engineering papers. This will help students understand the relative weightage of each chapter in terms of number of questions asked and thus bring focus in their preparation.
Gate Electrical Engineering: Chapter-Wise Question Distribution Analysis 2010–2020
Contents
FM.indd 5
Note to the Aspirants
iii
Chapter 1: Engineering Mathematics Important Formulas Questions Answers with Explanation
1 1 32 46
Chapter 2: Electric Circuits Important Formulas Questions Answers with Explanation
69 69 76 99
Chapter 3: E lectromagnetic Fields Important Formulas Questions Answers with Explanation
131 131 133 140
Chapter 4: Signals and Systems Important Formulas Questions Answers with Explanation
149 149 152 165
Chapter 5: Electrical Machines Important Formulas Questions Answers with Explanation
183 183 186 214
Chapter 6: Power Systems Important Formulas
247 248
Questions Answers with Explanation
252 282
Chapter 7: Control Systems Important Formulas Questions Answers with Explanation
315 315 319 340
Chapter 8: Electrical and Electronic Measurements379 Important Formulas 379 Questions 381 Answers with Explanation 394 Chapter 9: Analog and Digital Electronics Important Formulas Questions Answers with Explanation
411 411 422 465
Chapter 10: Power Electronics Important Formulas Questions Answers with Explanation
503 503 508 533
Solved GATE (EE) 2019
A1
Solved GATE (EE) 2020
B1
7/28/2020 11:57:04 AM
FM.indd 6
7/28/2020 11:57:04 AM
Engineering Mathematics
CHAPTER1
Syllabus Linear Algebra: Matrix Algebra, Systems of linear equations, Eigenvalues, Eigenvectors. Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series, Vector identities, Directional derivatives, Line integral, Surface integral, Volume integral, Stokes’s theorem, Gauss’s theorem, Green’s theorem. Differential Equations: First order equations (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s equation, Euler’s equation, Initial and boundary value problems, Partial Differential Equations, Method of separation of variables. Complex Variables: Analytic functions, Cauchy’s integral theorem, Cauchy’s integral formula, Taylor series, Laurent series, Residue theorem, Solution integrals. Probability and Statistics: Sampling theorems, Conditional probability, Mean, Median, Mode, Standard Deviation, Random variables, Discrete and Continuous distributions, Poisson distribution, Normal distribution, Binomial distribution, Correlation analysis, Regression analysis. Numerical Methods: Solutions of nonlinear algebraic equations, Single and Multi-step methods for differential equations. Transform Theory: Fourier Transform, Laplace Transform, z-Transform (or Z-Transform).
CHAPTER ANALYSIS Topic
GATE 2009
GATE 2010
GATE 2011
GATE 2012
GATE 2013
GATE 2014
GATE 2015
GATE 2016
GATE 2017
GATE 2018
Linear Algebra
0
2
0
1
2
4
2
5
2
2
Calculus
1
2
2
1
2
5
5
4
7
5
Differential Equations
0
1
1
0
0
3
2
4
1
1
Complex Variables
0
0
1
0
2
2
1
2
1
2
Probability and Statistics
0
1
0
0
0
5
2
2
3
0
Numerical Methods
0
0
3
0
1
2
0
0
1
0
Transform Theory
0
2
0
0
0
4
3
3
0
0
IMPORTANT FORMULAS Linear Algebra 1.
Types of Matrices (a) Row matrix: A matrix having only one row is called a row matrix or a row vector. Therefore, for a row matrix, m = 1. (b) Column matrix: A matrix having only one column is called a column matrix or a column vector. Therefore, for a column matrix, n = 1. (c) Square matrix: A matrix in which the number of rows is equal to the number of columns, say n, is called a square matrix of order n.
Ch wise GATE_EE_Ch1-a_Imp formula.indd 1
(d) Diagonal matrix: A square matrix is called a diagonal matrix if all the elements except those in the leading diagonal are zero, that is aij = 0 for all i ≠ j.
(e) Scalar matrix: A matrix A = [aij ]n × n is called a scalar matrix if (i) aij = 0, for all i ≠ j . (ii) aii = c, for all i, where c ≠ 0.
(f) Identity or unit matrix: A square matrix A = [aij]n × n is called an identity or unit matrix if (i) aij = 0, for all i ≠ j. (ii) aij = 1, for all i.
11/21/2018 12:09:42 PM
2
GATE EE Chapter-wise Solved Papers
(g) Null matrix: A matrix whose all the elements are zero is called a null matrix or a zero matrix.
(h) Upper triangular matrix: A square matrix A = [aij] is called an upper triangular matrix if aij = 0 for i > j. (i) Lower triangular matrix: A square matrix A = [aij] is called a lower triangular matrix if aij = 0 for i < j.
B+A=C+A⇒B=C 5.
Types of a Square Matrix (a) Nilpotent matrix: A square matrix A is called a nilpotent matrix if there exists a positive integer n such that An = 0. If n is least positive integer such that An = 0, then n is called index of the nilpotent matrix A.
2.
(e) Cancellation laws: If A, B and C are matrices of the same order, then A+B=A+C⇒B=C Some important properties of matrix multiplication are: (a) Matrix multiplication is not commutative.
(b) Matrix multiplication is associative, that is (AB)C = A(BC).
(c) Matrix multiplication is distributive over matrix addition, that is A(B + C) = AB + AC.
(b) Symmetrical matrix: It is a square matrix in which aij = aji for all i and j. A symmetrical matrix is necessarily a square one. If A is symmetric, then AT = A.
(d) If A is an m × n matrix, then ImA = A = AIn. (e) The product of two matrices can be the null matrix while neither of them is the null matrix.
(c) Skew-symmetrical matrix: It is a square matrix in which aij = −aji for all i and j. In a skew-symmetrical matrix, all elements along the diagonal are zero.
6.
(d) Hermitian matrix: It is a square matrix A in which (i, j)th element is equal to complex conjugate of the (j, i)th element, i.e. aij = a ji for all i and j.
Some of the important properties of scalar multiplication are: (a) k(A + B) = kA + kB
(b) (k + l) ⋅ A = kA + lA
(c) (kl) ⋅ A = k(lA) = l(kA)
(d) (−k) ⋅ A = −(kA) = k(−A)
(e) Skew-Hermitian matrix: It is a square matrix A = [aij] in which aij = − aij for all i and j.
(f) Orthogonal matrix: A square matrix A is called orthogonal matrix if AAT = ATA = I.
3.
Equality of a Matrix Two matrices A = [aij]m × n and B = [bij]x × y are equal if (a) m = x, that is the number of rows in A equals the number of rows in B.
(b) n = y, that is the number of columns in A equals the number of columns in B. (c) aij = bij for i = 1, 2, 3, …, m and j = 1, 2, 3, …, n.
4.
Some of the important properties of matrix addition are: (a) Commutativity: If A and B are two m × n matrices, then A + B = B + A, that is matrix addition is commutative.
(b) Associativity: If A, B and C are three matrices of the same order, then (A + B) + C = A + (B + C) that is matrix addition is associative.
(c) Existence of identity: The null matrix is the identity element for matrix addition. Thus, A + O = A = O + A (d) Existence of inverse: For every matrix A = [aij]m × n, there exists a matrix [aij]m × n, denoted by −A, such that A + (−A) = O = ( −A) + A
Ch wise GATE_EE_Ch1-a_Imp formula.indd 2
(e) 1 ⋅ A = A (f) −1 ⋅ A = −A Here, A and B are two matrices of same order and k and l are constants. If A is a matrix and A2 = A, then A is called idempotent matrix. If A is a matrix and satisfies A2 = I, then A is called involuntary matrix. 7.
Some of the important properties of transpose of a matrix are: (a) For any matrix A, (AT)T = A
(b) For any two matrices A and B of the same order (A + B)T = AT + BT (c) If A is a matrix and k is a scalar, then (kA)T = k(AT) (d) If A and B are two matrices such that AB is defined, then (AB)T = BTAT 8.
Some of the important properties of inverse of a matrix are: (a) A−1 exists only when A is non-singular, that is |A| ≠ 0.
(b) The inverse of a matrix is unique.
(c) Reversal laws: If A and B are invertible matrices of the same order, then (AB)−1 = B−1A−1
11/21/2018 12:09:43 PM
3
Chapter 1 • Engineering Mathematics
(d) If A is an invertible square matrix, then (AT)−1 = (A−1)T
(e) The inverse of an invertible symmetric matrix is a symmetric matrix.
(f) Let A be a non-singular square matrix of order n. Then
|adj A| = |A|n −1
adj (AB) = (adj B)(adj A)
(h) If A is an invertible square matrix, then adj AT = (adj A)T
(i) If A is a non-singular square matrix, then adj (adj A) = |A|n − 2 A
(j) If A is a non-singular matrix, then 1 |A | = |A| , that is | A | = | A|
−1
−1
−1
(k) Let A, B and C be three square matrices of same type and A be a non-singular matrix. Then AB = AC ⇒ B = C
9.
(m) The rank of transpose of a matrix is equal to rank of the original matrix. rank (A) = rank (AT)
(n) The rank of a matrix does not change by pre-multiplication or post-multiplication with a non-singular matrix.
(o) If A − B, then rank (A) = rank (B).
(g) If A and B are non-singular square matrices of the same order, then
(l) If A is any n-rowed square matrix of rank, n − 1, then adj A ≠ 0
and
BA = CA ⇒ B = C
The rank of a matrix A is commonly denoted by rank (A). Some of the important properties of rank of a matrix are: (a) The rank of a matrix is unique.
(b) The rank of a null matrix is zero.
(c) Every matrix has a rank.
(d) If A is a matrix of order m × n, then rank (A) ≤ m × n (smaller of the two)
(e) If rank (A) = n, then every minor of order n + 1, n + 2, etc., is zero.
(f) If A is a matrix of order n × n, then A is non-singular and rank (A) = n.
(g) Rank of IA = n.
(h) A is a matrix of order m × n. If every kth order minor (k < m, k < n) is zero, then rank (A) < k
(i) A is a matrix of order m × n. If there is a minor of order (k < m, k < n) which is not zero, then rank (A) ≥ k
(j) If A is a non-zero column matrix and B is a non-zero row matrix, then rank (AB) = 1.
(k) The rank of a matrix is greater than or equal to the rank of every sub-matrix.
Ch wise GATE_EE_Ch1-a_Imp formula.indd 3
(p) The rank of a product of two matrices cannot exceed rank of either matrix. rank (A × B) ≤ rank A or rank (A × B) ≤ rank B
(q) The rank of sum of two matrices cannot exceed sum of their ranks. (r) Elementary transformations do not change the rank of a matrix.
10. Determinants Every square matrix can be associated to an expression or a number which is known as its determinant. If A = [aij] is a square matrix of order n, then the determinant of A is denoted by det A or |A|. If A = [a11] is a square matrix of order 1, then determinant of A is defined as |A| = a11 a12 ⎤ ⎡a If A = ⎢ 11 ⎥ is a square matrix of order 2, then ⎣ a21 a23 ⎦ determinant of A is defined as |A| = a11a23 − a12a21 ⎡ a11 If A = ⎢⎢ a21 ⎢⎣ a31
a12 a22 a32
a13 ⎤ a23 ⎥⎥ is a square matrix of order 3, a33 ⎥⎦
then determinant of A is defined as |A| = a11 (a22a33 − a23a32) − a21 (a12a33 − a13a32) + a21 (a12a23 − a13a22) or |A| = a11 (a22a33 − a23a32) − a12 (a21a33 − a23a31) + a13 (a21a32 − a22a31) 11. Minors The minor Mij of A = [aij] is the determinant of the square sub-matrix of order (n − 1) obtained by removing i th row and j th column of the matrix A. 12. Cofactors The cofactor Cij of A = [aij] is equal to (−1)i + j times the determinant of the sub-matrix of order (n − 1) obtained by leaving i th row and j th column of A.
11/21/2018 12:09:43 PM
4
GATE EE Chapter-wise Solved Papers
13. Some of the important properties of determinants are: (a) Sum of the product of elements of any row or column of a square matrix A = [aij] of order n with their cofactors is always equal to its determinant. n
∑a c i =1
ij ij
n
= | A| = ∑ aij cij j =1
(b) Sum of the product of elements of any row or column of a square matrix A = [aij] of order n with the cofactors of the corresponding elements of other row or column is zero. n
∑a c i =1
ij ik
(d) Consider a square matrix A = [aij] of order n ≥ 2 and B obtained from A by interchanging any two rows or columns of A, then |B| = −A.
(e) For a square matrix A = [aij] of order (n ≥ 2), if any two rows or columns are identical, then its determinant is zero, that is |A| = 0.
(f) If all the elements of a square matrix A = [aij] of order n are multiplied by a scalar k, then the determinant of new matrix is equal to k|A|. (g) Let A be a square matrix such that each element of a row or column of A is expressed as the sum of two or more terms. Then |A| can be expressed as the sum of the determinants of two or more matrices of the same order.
• If A ≠ 0, system is consistent and has a unique solution given by X = A−1 B. • If A = 0 and (adj A)B = 0, system is consistent and has infinite solutions. • If A = 0 and (adj A)B ≠ 0, system is inconsistent. 16. Cramer’s Rule Suppose we have the following system of linear equations: a1x + b1 y + c1z = k1
a2x + b2 y + c2z = k2
a3x + b3 y + c3z = k3
(h) Let A be a square matrix and B be a matrix obtained from A by adding to a row or column of A a scalar multiple of another row or column of A, then |B| = |A|. (i) Let A be a square matrix of order n (≥ 2) and also a null matrix, then |A| = 0.
(j) Consider A = [aij] as a diagonal matrix of order n (≥ 2), then |A| = a11 × a22 × a33 × … × ann
(k) Suppose A and B are square matrices of same order, then |AB| = |A| ⋅ |B|
14. There are two cases that arise for homogeneous systems: (a) Matrix A is non-singular or |A| ≠ 0. The solution of the homogeneous system in the above equation has a unique solution, X = 0, that is x1 = x2 = … = xj = 0.
(b) Matrix A is singular or |A| = 0, then it has infinite many solutions. To find the solution when |A| = 0, put z = k (where k is any real number) and solve any
Ch wise GATE_EE_Ch1-a_Imp formula.indd 4
(b) If AX = B is a system with linear equations equal to the number of unknowns, then three cases arise:
j =1
n
(c) For a square matrix A = [aij] of order n, |A| = |AT|.
15. The method to solve a non-homogeneous system of simultaneous linear equations. Please note the number of unknowns and the number of equations. (a) Given that A is a non-singular matrix, then a system of equations represented by AX = B has the unique solution which can be calculated by X = A−1 B.
= 0 = ∑ aij ckj
two equations for x and y using the matrix method. The values obtained for x and y with z = k is the solution of the system.
Now, if ⎡ a1 Δ = ⎢⎢ a2 ⎢⎣ a3
b1 b2 b3
c1 ⎤ c2 ⎥⎥ ≠ 0 c3 ⎥⎦
⎡ k1 Δ1 = ⎢⎢ k2 ⎢⎣ k3
b1 b2 b3
c1 ⎤ c2 ⎥⎥ ≠ 0 c3 ⎥⎦
⎡ a1 Δ 2 = ⎢⎢ a2 ⎢⎣ a3
k1 k2 k3
c1 ⎤ c2 ⎥⎥ ≠ 0 c3 ⎥⎦
⎡ a1 Δ 3 = ⎢⎢ a2 ⎢⎣ a3
b1 b2 b3
k1 ⎤ k2 ⎥⎥ ≠ 0 k3 ⎥⎦
Thus, the solution of the system of equations is given by Δ1 Δ Δ y= 2 Δ Δ3 z= Δ x=
11/21/2018 12:09:44 PM
5
Chapter 1 • Engineering Mathematics
17. Augmented Matrix Consider the following system of equations:
problem. To solve the problem, we need to determine the value of X’s and l’s to satisfy the above-mentioned vector. Note that the zero vector (that is X = 0) is not of our interest. A value of l for which the above equation has a solution X ≠ 0 is called an eigenvalue or characteristic value of the matrix A. The corresponding solutions X ≠ 0 of the equation are called the eigenvectors or characteristic vectors of A corresponding to that eigenvalue, l. The set of all the eigenvalues of A is called the spectrum of A. The largest of the absolute values of the eigenvalues of A is called the spectral radius of A. The sum of the elements of the principal diagonal of a matrix A is called the trace of A.
a11 x1 + a12 x2 + + a1n xn = b1 a21 x1 + a22 x2 + + a2 n xn = b2
am1 x1 + am 2 x2 + + amn xn = bm
This system can be represented as AX = B.
⎡ a11 ⎢a where A = ⎢ 21 ⎢ ⎢ ⎣ am1
a12 a22 am 2
a1n ⎤ ⎡ x1 ⎤ ⎢ ⎥ ⎥ x a2 n ⎥ , X = ⎢ 2 ⎥ and ⎢ ⎥ ⎥ ⎢ ⎥ ⎥ amn ⎦ ⎢⎣ xn ⎥⎦
⎡b1 ⎤ ⎢ ⎥ b B = ⎢ 2 ⎥. ⎢ ⎥ ⎢ ⎥ ⎢⎣bn ⎥⎦
⎡ a11 ⎢ a The matrix ⎡⎣ A B ⎤⎦ = ⎢ 21 ⎢ ⎢ ⎢⎣ am1
a12 a22 am 2
a1n b1 ⎤ ⎥ a2 n b2 ⎥ is called ⎥ ⎥ amn bm ⎥⎦
augmented matrix. 18. Cayley–Hamilton Theorem According to the Cayley–Hamilton theorem, every square matrix satisfies its own characteristic equations. Hence, if A − l I = ( −1) n ( l n + a1l n −1 + a2 l n − 2 + + an )
is the characteristic polynomial of a matrix A of order n, then the matrix equation X n + a1 X n −1 + a2 X n − 2 + + an I = 0
is satisfied by X = A.
19. Eigenvalues and Eigenvectors If A = [aij]n × n is a square matrix of order n, then the vector ⎡ x1 ⎤ ⎢x ⎥ ⎢ 2⎥ equation AX = lX, where X = ⎢⎢. ⎥⎥ is an unknown vector ⎢. ⎥ ⎢. ⎥ ⎢ ⎥ ⎢⎣ xn ⎥⎦ and l is an unknown scalar value, is called an eigenvalue
Ch wise GATE_EE_Ch1-a_Imp formula.indd 5
20. Properties of Eigenvalues and Eigenvectors Some of the main characteristics of eigenvalues and eigenvectors are discussed in the following points: (a) If l1, l2, l3, …, ln are the eigenvalues of A, then kl1, kl2, kl3, …, kln are eigenvalues of kA, where k is a constant scalar quantity.
(b) If l1, l2, l3, …, ln are the eigenvalues of A, then 1 1 1 1 , , , ..., are the eigenvalues of A−1. l1 l 2 l3 ln
(c) If l1, l2, l3, …, ln are the eigenvalues of A, then
l1k , l 2k , l3k , ..., l nk are the eigenvalues of Ak.
(d) If l1, l2, l3, …, ln are the eigenvalues of A, then A
,
A
,
A
l1 l 2 l3
, ...,
A
ln
are the eigenvalues of adj A.
(e) The eigenvalues of a matrix A are equal to the eigenvalues of AT.
(f) The maximum number of distinct eigenvalues is n, where n is the size of the matrix A.
(g) The trace of a matrix is equal to the sum of the eigenvalues of a matrix.
(h) The product of the eigenvalues of a matrix is equal to the determinant of that matrix.
(i) If A and B are similar matrices, that is A = IB, then A and B have the same eigenvalues.
(j) If A and B are two matrices of same order, then the matrices AB and BA have the same eigenvalues. (k) The eigenvalues of a triangular matrix are equal to the diagonal elements of the matrix.
Calculus 21. Rolle’s Mean Value Theorem Consider a real-valued function defined in the closed interval [a, b], such that
11/21/2018 12:09:46 PM
6
GATE EE Chapter-wise Solved Papers
(a) It is continuous on the closed interval [a, b].
(b) It is differentiable on the open interval (a, b). (c) f(a) = f(b). Then, according to Rolle’s theorem, there exists a real number c ∈( a, b) such that f ′(c) = 0.
22. Lagrange’s Mean Value Theorem Consider a function f(x) defined in the closed interval [a, b], such that (a) It is continuous on the closed interval [a, b]. (b) It is differentiable on the closed interval (a, b). Then, according to Lagrange’s mean value theorem, there exists a real number c ∈( a, b), such that f ′( c ) = 23.
f ( b) − f ( a) b−a
Cauchy’s Mean Value Theorem Consider two functions f(x) and g(x), such that (a) f (x) and g(x) both are continuous in [a, b]. (b) f ′(x) and g ′(x) both exist in (a, b). Then there exists a point c ∈( a, b) such that f ′( c ) f ( b) − f ( a) = g ′( c ) g ( b) − g ( a)
24. Taylor’s Theorem If f(x) is a continuous function such that f ′(x), f ′( x ), f ′′( x ), …, f n −1 ( x ) are all continuous in [a, a + h] and f n ( x ) exists in (a, a + h) where h = b − a, then according to Taylor’s theorem, f ( a + h) = f ( a) + hf ′( a) +
h2 hn −1 f ′′( a) + + f n −1 ( a) 2! (n − 1)!
hn n + f (a) n! 25. Maclaurin’s Theorem If the Taylor’s series obtained in Section 24 is centered at 0, then the series we obtain is called the Maclaurin’s series. According to Maclaurin’s theorem, f ( h) = f (0) + hf ′(0) + +
hn n f (0 ) n!
f (x) = f (y) for all x ∈ (a, b)
Ch wise GATE_EE_Ch1-a_Imp formula.indd 6
f (x) ≥ f (y) for all x ∈ (a, b)
Local maxima and local minima of any function can be calculated as: Consider that f(x) be defined in (a, b) and y ∈ (a, b). Now, (a) If f ′( y ) = 0 and f ′( x ) changes sign from positive to negative as ‘x’ increases through ‘y’, then x = y is a point of local maximum value of f(x). (b) If f ′( y ) = 0 and f ′( x ) changes sign from negative to positive as ‘x’ increases through ‘y’, then x = y is a point of local minimum value of f(x).
27. Some important properties of maximum and minima are given as follows: (a) If f(x) is continuous in its domain, then at least one maxima and minima lie between two equal values of x. (b) Maxima and minima occur alternately, that is no two maxima or minima can occur together. 28. Maximum and minimum values in a closed interval [a, b] can be calculated using the following steps:
(a) Calculate f ′( x ).
(b) Put f ′( x ) = 0 and find value(s) of x. Let c1, c2, ..., cn be values of x. (c) Take the maximum and minimum values out of the values f(a), f(c1), f(c2), …, f(cn), f(b). The maximum and minimum values obtained are the absolute maximum and absolute minimum values of the function, respectively.
29. Partial Derivatives Partial differentiation is used to find the partial derivative of a function of more than one independent variable. The partial derivatives of f(x, y) with respect to x and y are defined by ∂f f ( x + ay ) − f ( x, y ) = lim ∂x a → 0 a
h2 hn −1 f ′′(0) + + f n −1 (0) 2! (n − 1)!
26. Maxima and Minima Suppose f(x) is a real-valued function defined at an internal (a, b). Then f(x) is said to have maximum value, if there exists a point y in (a, b) such that
Suppose f(x) is a real-valued function defined at the interval (a, b). Then f(x) is said to have minimum value, if there exists a point y in (a, b) such that
∂f f ( x , y + b) − f ( x , y ) = lim ∂x b → 0 b
and the above limits exist. ∂f ∂x is simply the ordinary derivative of f with respect to x keeping y constant, while ∂f ∂x is the ordinary derivative of f with respect to y keeping x constant. Similarly, second-order partial derivatives can be calculated by
11/21/2018 12:09:48 PM
Chapter 1 • Engineering Mathematics
∂ ⎛ ∂ f ⎞ ∂ ⎛ ∂ f ⎞ ∂ ⎛ ∂f ⎞ ∂ ⎛ ∂f ⎞ and is, respec, ⎜ ⎟, ⎜ ⎟, ∂x ⎝ ∂x ⎠ ∂x ⎜⎝ ∂y ⎟⎠ ∂y ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ ∂2 f ∂2 f ∂2 f ∂2 f tively, denoted by , , , . ∂x 2 ∂x ∂y ∂y ∂x ∂y 2
30. Some of the important relation are given as follows: (a) If u = f(x, y) and y = f(x), then ∂u ∂u ∂u ∂y = + ⋅ ∂x ∂x ∂y ∂x
Integration
Result
∫ cos
2
xdx
x 1 + sin x cos x + C 2 2
∫ cos
3
xdx
1 sin x − sin 3 x + C 3
∫ cos
n
1 n −1 cos n −1 x sin x + n n
xdx
∫ cos
(b) If u = f(x, y) and x = f1(t1, t2) and y = f2(t1, t2), then
and
∂u ∂u ∂ x ∂ u ∂ y = ⋅ + ⋅ ∂t1 ∂x ∂t1 ∂y ∂t1
n ∫ cos xdx
∂u ∂u ∂x ∂u ∂y = ⋅ + ⋅ ∂t 2 ∂x ∂t 2 ∂y ∂t 2
∫x
31. Integration Using Table Some of the common integration problems can be solved by directly referring the tables and computing the results. Table 1 shows the result of some of the common integrals we use. Table 1 | Table of common integrals
Integration 1 ∫ ax + b dx 1
∫ (ax + b)
2
1
∫ (ax + b) ∫a
2
n
dx
f ′ (x)
∫ f ( x ) dx 2
∫ sin
3
∫ sin
n
1 ln ax + b + C where C is a a constant 1 − +C a ( ax + b )
dx
1 dx + x2
∫ sin
Result
−
1 a ( n − 1) ( ax + b )
n −1
+C
1 ⎛ x⎞ tan −1 ⎜ ⎟ + C ⎝ a⎠ a ln f ( x ) + C
xdx
x 1 − sin x cos x + C 2 2
xdx
1 − cos x + cos3 x + C 3
xdx
1 n −1 − sin n −1 x cos x + n n n−2 ∫ sin xdx + C
2
n−2
xdx + C
tan −1 x − tan n − 2 xdx + C n −1 ∫ 1 x−a ln +C 2a x + a
dx − a2 dx
∫
ln x + x 2 ± a 2 + C
x ± a2 2
∫ x sin nxdx
1 (sin nx − nx cos nx ) + C n2
∫ x cos nxdx
1 (cos nx + nx sin nx ) + C n2
∫e
ax
e ax ( a sin bx − b cos bx )
sin bxdx
a2 + b2 e ax ( a cos bx + b sin bx )
ax ∫ e cos bxdx
a2 + b2
∫x
+C +C
1 ( − n2 x 2 cos nx + 2 cos nx n3 + 2nx sin nx ) + C
2 ∫ x sin nxdx
2
7
1 2 2 ( n x sin nx − 2 sin nx n3 + 2nx cos nx ) + C
cos nxdx
32. Integration by Partial Fraction The formula which come handy while working with partial fractions are given as follows: 1
∫ x − a dx = ln ( x − a) + C ∫a
2
∫a
2
1 1 ⎛ x⎞ dx = tan −1 ⎜ ⎟ + C 2 ⎝ a⎠ a +x
(
)
x 1 dx = ln a 2 + x 2 + C 2 2 +x
(Continued)
Ch wise GATE_EE_Ch1-a_Imp formula.indd 7
11/21/2018 12:09:51 PM
8
GATE EE Chapter-wise Solved Papers
33. Integration Using Trigonometric Substitution Trigonometric substitution is used to simplify certain integrals containing radical expressions. Depending on the function we need to integrate, we substitute one of the following expressions to simplify the integration:
(a) For
a 2 − x 2 , use x = a sin q .
(b) For
a 2 + x 2 , use x = a tan q .
(c) For
x − a , use x = a sec q . 2
b
∫ a
b
f ( x ) ⋅ dx = ∫ f ( y ) ⋅ dy a
a
(c)
∫ 2a
∫
(d)
0
(e) (f)
a
c
a
a
0
0
f ( x ) ⋅ dx = ∫ f ( x ) ⋅ dx + ∫ f ( 2a − x ) ⋅ dx
a
a
0
0
∫
−a
f ( x ) ⋅ dx = 0, if the function is odd.
−a
na
(g)
∫ 0
a
f ( x ) ⋅ dx = n∫ f ( x ) ⋅ dx if f(x) = f(x + a) 0
35. Some of the important properties of double integrals are: (a) When x1, x2 are functions of y and y1, y2 are constants, then f(x, y) is integrated with respect to x keeping y constant within the limits x1, x2 and the resulting expression is integrated with respect to y between the limits y1, y2.
y2 x2
∫∫
Ch wise GATE_EE_Ch1-a_Imp formula.indd 8
∫∫
f ( x, y ) dydx
x1 y1
Hence, in a double integral, the order of integration does not change the final result provided the limits are changed accordingly. However, if the limits are variable, the change of order of integration changes the limits of integration.
37. The length of the arc of the polar curve r = f (q ) between the points where q = α , β is
β
∫ α
⎡ 2 ⎛ dr ⎞ 2 ⎤ ⎢ r + ⎜⎝ ⎟⎠ ⎥ dq . dq ⎦ ⎣
The length of the arc of the curve x = f (t ) , y = g (t ) between the points where t = a and t = b is b
∫ a
⎡⎛ dx ⎞ 2 ⎛ dy ⎞ 2 ⎤ ⎢⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ ⎥ dt . dt ⎦ ⎣ dt
38. Fourier Series Fourier series is a way to represent a wave-like function as a combination of sine and cosine waves. It decomposes any periodic function into the sum of a set of simple oscillating functions (sines and cosines). The Fourier series for the function f(x) in the interval α < x < α + 2p is given by f ( x) =
∞ a0 ∞ + ∑ an cos nx + ∑ bn sin nx 2 n =1 n =1
where 1 α + 2p f ( x )dx p ∫α 1 α + 2p an = ∫ f ( x ) cos nxdx p α 1 α + 2p bn = ∫ f ( x ) sin nxdx p α a0 =
y1 x1
(b) When y1, y2 are functions of x and x1, x2 are constants, f(x, y) is first integrated with respect to y, keeping x constant and between the limits y1, y2 and
f ( x, y ) dydx
x1 y1
x2 y 2
f ( x, y ) dxdy =
y1 x1
y2 x2
∫∫
36. Change of Order of Integration As already discussed, if limits are constant
∫∫ f ( x, y) dxdy = ∫ ∫ f ( x, y) dxdy Q
∫∫
x2 y2
f ( x, y ) dxdy =
y1 x1
a
a
∫
y2 x2
f ( x, y ) dxdy =
f ( x ) ⋅ dx = 2∫ f ( x ) ⋅ dx , if the function is even. 0
f ( x, y ) dydx
x1 y1
Q
∫ f ( x) ⋅ dx = ∫ f (a − x) ⋅ dx a
b
f ( x ) ⋅ dx = ∫ f ( x ) ⋅ dx + ∫ f ( x ) ⋅ dx if a < c < b
a
c
∫∫
(c) When x1, x2, y1 and y2 are constants, then
∫∫
b
b
∫ f ( x) ⋅ dx = − ∫ f ( x) ⋅ dx
x2 y2
f ( x, y ) dxdy =
Q
a
b
(b)
∫∫
2
34. Some of the important properties of definite integrals are given as follows: (a) The value of definite integrals remains the same with change of variables of integration provided the limits of integration remain the same.
the resulting expression is integrated with respect to x within the limits x1, x2.
The values of a0, an and bn are known as Euler’s formulae.
11/21/2018 12:09:54 PM
9
Chapter 1 • Engineering Mathematics
39. Conditions for Fourier Expansion np x 1 c np x However, since f ( x ) sin is even, bn = ∫ f ( x ) sin dx = 0. Fourier expansion can be performed on any function f(x) c c −c c np x if it fulfills the Dirichlet conditions. The Dirichlet condi- 1 c bn = ∫ f ( x ) sin dx = 0. c − tions are given as follows: c c (a) f(x) should be periodic, single-valued and finite. Thus, if a periodic function f(x) is even, its Fourier expansion contains only cosine terms, a0 and an. (b) f(x) should have a finite number of discontinuities in (b) When f(x) is an odd function, any one period.
(c) f(x) should have a finite number of maxima and minima.
40. Fourier Expansion of Discontinuous Function We can define the Euler’s formulae as follows: α + 2p 1⎡ c ϕ ( x )dx + ∫ ϕ 2 ( x )dx ⎤ c ⎦⎥ p ⎣⎢ ∫α 1 c α + 2 p 1 ϕ 2 ( x ) cos nxdx ⎤ an = ⎡ ∫ ϕ 1 ( x ) cos nxdx + ∫ c ⎦⎥ p ⎣⎢ α
a0 =
α + 2p 1 c ϕ 2 ( x ) sin nxdx ⎤ bn = ⎡ ∫ ϕ 1 ( x ) sin nxdx + ∫ ⎢ ⎥⎦ c α p⎣
At x = c, there is a finite jump in the graph of function. Both the limits, left-hand limit (i.e. f(c − 0)) and righthand limit (i.e. f(c + 0)), exist and are different. At such a point, Fourier series gives the value of f(x) as the arithmetic mean of these two limits. Hence, at x = c, f ( x) =
1 ⎡ f (c − 0 ) + f (c + 0 )⎤⎦ 2⎣
41. Change of Interval Till now, we have talked about functions having periods of 2p . However, often the period of the function required to be expanded is some other interval (say 2c). Then, the Fourier expansion is given as follows: f ( x) =
a0 ∞ np x np x ∞ + ∑ an cos + ∑ bn sin 2 n =1 c c n =1
where 1 α + 2c a0 = ∫ f ( x )dx c α 1 α + 2c np x an = ∫ f ( x ) cos dx c α c 1 α + 2c np x bn = ∫ f ( x ) sin dx α c c
a0 =
1 c f ( x )dx = 0 c ∫− c
an =
1 c np x f ( x ) cos dx = 0 ∫ c − c c
However, bn =
1 c 2 c np x np x f ( x ) sin dx = ∫ f ( x ) sin dx . ∫ c − 0 c c c c
T hus, if a periodic function f(x) is odd, its Fourier expansion contains only sine terms and bn. 43. Vectors Vector is any quantity that has magnitude as well as direction. If we have two points A and B, then vector between A and B is denoted by AB . Position vector is a vector of any points, A, with respect to the origin, O. If A is given by the coordinates x, y and z. AP = x 2 + y 2 + z 2 44. Zero vector is a vector whose initial and final points are same. Zero vectors are denoted by 0 . They are also called null vectors. 45. Unit vector is a vector whose magnitude is unity or one. It is in the direction of given vector A and is denoted by A . 46. Equal vectors are those which have the same magnitude and direction regardless of their initial points. 47. Addition of Vectors According to triangle law of vector addition, as shown in Fig. 1, C c b
42. Fourier Series Expansion of Even and Odd Functions (a) When f(x) is an even function, 1 c 2 c a0 = ∫ f ( x )dx = ∫ f ( x )dx c −c c 0 c 1 2 c np x np x dx an = ∫ f ( x ) cos dx = ∫ f ( x ) cos c c −c c c 0
Ch wise GATE_EE_Ch1-a_Imp formula.indd 9
A
B a
Figure 1 | Triangle law of vector addition.
c = a+b
11/21/2018 12:09:55 PM
10
GATE EE Chapter-wise Solved Papers
According to parallelogram law of vector addition, as shown in Fig. 2, B
C
c
b
O
A
a
Figure 2 | Parallelogram law of vector addition.
c = a+b
If we have two vectors represented by adjacent sides of parallelogram, then the sum of the two vectors in magnitude and direction is given by the diagonal of the parallelogram. This is known as parallelogram law of vector addition.
48. Some important properties of vector addition are: (a) If we have two vectors a and b a+b = b +a (b) If we have three vectors a , b and c a+b +c = a+ b +c
(
)
(
)
49. Multiplication of Vectors (a) Multiplication of a vector with scalar: Consider a vector a and a scalar quantity v. Then ka = k a (b) Multiplication of a vector with another vector using dot product: Dot product or scalar product of two vectors is given by a ⋅ b = a b cos q where a = magnitude of vector a , b = magni tude of vector b and q = angle between a and b, 0 ≤ q ≤ p.
where a = magnitude of vector a b = magnitude of vector b q = angle between a and b , 0 ≤ q ≤ p n = Unit vector perpendicular to both a and b. The result of a × b is always a vector.
52. Some important laws of cross product are: (a) A × B = −B × A
(b) A × (B + C) = A × B + A × C
(c) m(A × B) = (ma) × B = A × (mB)
(d) i × i = j × j = k × k = 0, i × j = k, j × k = i, k × i = j
(e) If A = A1i + A2j + A3k and B = B1i + B2j + B3k, then
Some important laws of dot product are: (a) A ⋅B = B ⋅A (b) A ⋅B + C = A ⋅B + A ⋅C (c) k(A ⋅B) = (kA)⋅B = A ⋅(kB) (d) i ⋅ i = j ⋅ j = k ⋅ k = 1, i ⋅ j = j ⋅ k = k ⋅ i = 0
51. Multiplication of Vectors Using Cross Product The cross or vector product of two vectors is given by a × b = a b sin q nˆ, iˆ
Ch wise GATE_EE_Ch1-a_Imp formula.indd 10
j
k
A × B = A1 B1
A2 B2
A3 B3
53. Derivatives of Vector Functions The derivative of vector A(x) is defined as dA A( x + Δx ) − A( x ) = lim dx Δx → 0 Δx if the above limits exists. If A(x) = A1(x)i + A2(x)j + A3(x)k, then dA dA dA dA1 = i+ 2 j+ 3 k dx dx dx dx If A(x, y, z) = A1(x, y, z)i + A2(x, y, z)j + A3(x, y, z)k, then
50.
i
dA =
dA dA dA dx + dy + dz dx dy dz
d dB dA ( A ⋅ B) = A + B dy dy dy d dB dA ( A × B) = A × + ×B dz dz dz
A unit vector perpendicular to two given vectors a and b is given by a×b c= |a×b |
54. Gradient of a Scalar Field If we have a scalar function a(x, y, z), then the gradient of this scalar function is a vector function which is defined by ∂f ∂f ∂f i+ j+ k grad a = ∇a = ∂x ∂y ∂z
11/21/2018 12:09:58 PM
Chapter 1 • Engineering Mathematics
55. Divergence of a Vector If we have a differentiable vector A (x, y, z), then diver gence of vector A is given by
where Ax, Ay and Az are the components of vector A.
56. Curl of a Vector The curl of a continuously differentiable vector A is given by ⎛ ∂ ∂ ∂⎞ curl A = ∇A = ⎜ i + j + k ⎟ × ( Ax i + Ay j + Az k ) ⎝ ∂x ∂y ∂z ⎠ i ∂ = ∂x Ax
j ∂ ∂y Ay
k ∂ ∂z Az
58. Directional Derivative The directional derivative of a scalar function, f ( x ) = f ( x1 , x2 , …, xn ) along a vector v = ( v1 , …, vn ) is a function defined by the limit
Ch wise GATE_EE_Ch1-a_Imp formula.indd 11
The directional derivative is also often written as follows: d ∂ ∂ ∂ = v ⋅ ∇ = vx + vy + vz dv ∂x ∂y ∂z
59. Scalar Triple Product The scalar product of three vectors a, b and c is defined as
57. Some important points of divergence and curl are: ∂2 A ∂2 A ∂2 A (a) ∇ ⋅ ∇ A = ∇2A = + + ∂x 2 ∂y 2 ∂z 2 (b) ∇ × ∇ A = 0 (c) ∇ ⋅ ∇ × A = 0 (d) ∇ × (∇ × A ) = ∇(∇ ⋅ A ) × ∇2 A (e) ∇ (∇ ⋅ A ) = ∇ × (∇ × A ) + ∇2 A (f) ∇ ( A + B ) = ∇ ⋅ A + ∇ ⋅ B (g) ∇ × ( A + B ) = ∇ × A + ∇ × B (h) ∇ ⋅ ( A × B ) = B ⋅ (∇ × A ) − A ⋅ (∇ × B ) (i) ∇ × ( A × B ) = (B ⋅ ∇) A − B (∇⋅A) − (A∇) B + A (∇B)
h→ 0
where ∇ on the right-hand side of the equation is the v gradient and v is the unit vector given by v = . v
[a, b, c] = a ⋅ (b × c ) = b ⋅ (c × a) = c ⋅ (a × b)
⎛ ∂Ay ∂Ax ⎞ ⎛ ∂A ∂Ay ⎞ ⎛ ∂Ax ∂Az ⎞ j+⎜ k i+⎜ − − =⎜ z − ⎟ ⎟ ∂y ⎟⎠ ∂x ⎠ ∂z ⎠ ⎝ ∂z ⎝ ∂x ⎝ ∂y where Ax, Ay and Az are the components of vector A.
∇ v f ( x ) = lim
If the function f is differentiable at x, then the directional derivative exists alone any vector v, ∇ v f ( x ) = ∇f ( x ) ⋅ v
∂A ∂Ay ∂Az + div A = ∇A = x + ∂x ∂y ∂z
11
(
)
f x + hv − f ( x ) h
The volume of a parallelepiped whose sides are given by the vectors a, b and c, as shown in Fig. 3, can be calculated by the absolute value of the scalar triple product. Vparallelepiped = a ⋅ (b × c ) a×b
c b a Figure 3 | Parallelepiped with sides given by the vectors a, b and c.
60. Vector Triple Product If we have three vectors a, b and c, then the vector triple product is given by a × (b × c ) = b ( a ⋅ c ) − c ( a ⋅ b ) 61. Line integrals Let us say that points (ak, bk) are chosen so that they lie on the curve between points (xk−1, yk−1) and (xk, yk). Now, consider the sum, n
∑ {P (α k =1
k
, βk ) Δxk + Q(α k , βk ) Δyk }
11/21/2018 12:10:03 PM
12
GATE EE Chapter-wise Solved Papers
The limit of the sum as n → ∞ is taken in such a way that all the quantities Δ xk and Δ yk approach zero, and if such limit exists, it is called a line integral along the curve C and is denoted by
∆Ck C
∫ [ P ( x, y)dx + Q( x, y)dy] c
The limit exists if P and Q are continuous at all points of C. To understand better, refer to Fig. 4.
y
∆Ak = ∆xk ∆yk
C′
(xk + 1, yk + 1)
B(a2, b2)
(αk + 1, βk + 1) Figure 5 | Surface integrals.
(xk, yk)
C A(α1, β1)
63. Stokes’ Theorem
∫ A ⋅ dr = ∫∫ (∇ × A) ⋅ nds
(x1, y1)
c
s
64. Green’s Theorem
a2
a1 Figure 4 | Line integral.
⎛ ∂f1
∂f 2 ⎞
∫∫ ⎜⎝ ∂x − ∂y ⎟⎠ dxdy = ∫ ( f dx + f dy)
x
2
s
65. Gauss Divergence Theorem
In the same way, a line integral along a curve C in the three-dimensional space is given by
∫∫∫ ∇ ⋅ AdV = ∫∫ A ⋅ ndS v
∫ [ A dx + A dy + A dz ] 1
2
where A1, A2 and A3 are functions of x, y and z, respectively.
62. Surface Integrals Suppose g(x, y, z) is single valued and continuous for all values of C. Now, consider the sum n
∑ g (α k =1
k
Differential Equations
where (ak, bk, gk) is any arbitrary point of ΔCk. If the limit of this sum, n→∞ is in such a way that each ΔCk → 0 exists, the resulting limit is called the surface integral of g(x, y, z) over C. The surface integral is denoted by C
Figure 5 graphically represents surface integrals.
Ch wise GATE_EE_Ch1-a_Imp formula.indd 12
66. Variable Separable If all functions of x and dx can be arranged on one side and y and dy on the other side, then the variables are separable. The solution of this equation is found by integrating the functions of x and y.
∫ f ( x)dx = ∫ g ( y)dy + C
, βk , γ k ) Δc p
∫∫ g ( x, y, z ) ⋅ ds
s
3
c
1
c
67. Homogeneous Equation
Homogeneous equations are of the form dy f ( x, y ) = dx g ( x, y )
where f(x, y) and g(x, y) are homogeneous functions of the same degree in x and y. Homogeneous functions are those in which all the terms are of nth degree. To solve homogeneous equation,
11/21/2018 12:10:04 PM
Chapter 1 • Engineering Mathematics
dy dy = v + x . dx dx
(a) Put y = vx , then
(b) Separate v and x and then integrate. dy = f ( y/ x ) dx y/ x = v dv dx = f (v) − v x
⇒ ⇒
dv
∫ f (v) − v = log x + C
⇒
68. Linear Equation of First Order If a differential equation has its dependent variables and its derivatives occur in the first degree and are not multiplied together, then the equation is said to be linear. The standard equation of a linear equation of first order is given as
70. Integrating Factor A differential equation which is not exact can be converted into an exact differential equation by multiplication of a suitable function. This function is called integrating factor. Some of the important formulae are: (a) If the given equation of the form M dx + N dy = 0 is homogeneous and Mx + Ny ≠ 0, then
where P and Q are functions of x.
Integrating factor = (I.F.) = e ∫ y ⋅ e∫
P ⋅ dx
= ∫ Q ⋅ e∫
P ⋅ dx
P ⋅ dx
69. Exact Differential Equation A differential equation of the form M(x, y) dx + N(x, y) dy = 0, if df = M dx + N dy where f(x, y) is a function, is called an exact differential equation. Solution of differential equation is f(x, y) = C. The necessary condition for the differential equation to be exact is
I.F. =
The exact differential equation can be solved using the following steps: (a) First, integrate M with respect to x keeping y constant. (b) Next, integrate the terms of N with respect to y which do not involve x. (c) Then, the sum of the two expressions is equated to an arbitrary constant to give the solution.
Ch wise GATE_EE_Ch1-a_Imp formula.indd 13
∫ M ⋅ dx + ∫ N ⋅ dy = C
1 Mx − Ny
(c) If the given equation is of the form Mdx + Ndy = 0 1 ⎛ ∂M ∂N ⎞ − = f ( x ) where f ( x ) is a funcN ⎜⎝ ∂y ∂x ⎟⎠ tion of x, then and
I.F. = e ∫
f ( x )⋅∂x
(d) If the given equation is of the form Mdx + Ndy = 0 1 ⎛ ∂N ∂M ⎞ − = f ( y ) where f ( y ) in a funcand M ⎜⎝ ∂x ∂y ⎟⎠ tion of y, then f ( y )⋅∂y I .F . = e ∫
71. Clairaut’s Equation
dy An equation of the form y = px + f ( p), where p = dx is called a Clairaut’s equation. We know that y = px + f ( p) (1) Differentiating above equation w.r.t. x, we get
∂M ∂N = ∂y ∂x
1 Mx + Ny
(b) If the given equation is of the form f ( x, y ) y∂x + g ( x, y ) x∂y = 0 f ( x, y ) y∂x + g ( x, y ) x∂y = 0 and Mx − Ny ≠ 0, then
dx + C
⇒ y(I.F.) = ∫ Q(I.F.) dx + C
I.F. =
dy + Py = Q dx
13
dp dp + f ′( p) dx dx dp ⇒ [ x + f ′( p)] = 0 dx p = p+ x
Therefore, [ x + f ′( p)] = 0 or
dp =0 dx
dp = 0, then p = c.(2) dx
Now, if
Thus, eliminating p from Eqs. (1) and (2), we get
y = cx + f (c)
11/21/2018 12:10:08 PM
14
GATE EE Chapter-wise Solved Papers
as the general solution of the given equation. Hence, the solution of Clairaut’s equation is obtained on replacing p by c.
72. Linear Differential Equation The general form of a differential equation of nth order is denoted by
n −1
n
n−2
d y d y d y + p1 n −1 p2 n − 2 + + pn y = X n dx dx dx where p1, p2, …, pn and X are functions of x. d If operator is denoted by D, then substituting it in dx above equation, we get D n y + p1 D n −1 y + p2 D n − 2 y + + pn y = X
where f(D) = D + p1D + + pn = 0. As already mentioned before, the equation can be generalized as f ( D) y = X
The general solution of the above equation can be given as y = (Complementary Function) + (Particular Integral)
The general form of the linear differential equation with constant coefficients is given as dn y d n −1 y d n−2 + k1 n −1 + k2 n − 2 + + kn y = X n dx dx ∂x where k1, k2, …, kn are constants and X is a function of x only. The equation
y = (c1 + c2 x )e m1 x + c3 c m3 x + Case III: If the roots of A.E. are complex, that is α + iβ , α − iβ , m3 ,..., mn , then y = eα x (C1 cos β x + C2 sin β x ) + c3 e m3 x + + cn e mn x
where C1 = c1 + c2 and C2 = i(c1 − c2 ).
If the complex roots are equal, then y = eα x [(c1 x + c2 ) cos β x + (c3 x + c4 ) sin β x ] + + cn e mn x
73. Particular Integrals Consider the equation given in the previous section ( D n + k1 D n −1 + k2 D n − 2 + + kn ) y = X
The particular integral of the equation is given by P.I. =
n -1
n
dn y d n −1 y d n−2 + k + k + + kn y = 0 2 1 dx n dx n −1 dx n − 2
⇒ f ( D) y = X
1 X D n + k1 D n −1 + k2 D n − 2 + + kn
The following cases arise for particular integrals: (a) When X = eax, then 1 e ax f ( D)
P.I. =
1 ax e , if f ( a) ≠ 0 f ( a)
=
If f ( a) = 0, then
P.I. =
If f ′( a) = 0, then P.I. =
(b) When X = sin ax, then
( D n + k1 D n −1 + k2 D n − 2 + + kn ) y = 0
n −1
y = c1e m1 x + c2 e m2 x + c3 e m3 x + Case II: If some roots of A.E. are real and some are equal, that is m1 = m2 ≠ m3, then
Ch wise GATE_EE_Ch1-a_Imp formula.indd 14
=
n−2
and the equation D + k1 D + k2 D + + kn = 0 is called the auxiliary equation (A.E.). Now, if m1, m2, …, mn are the roots, then the complementary functions can be calculated using the following cases: Case I: If the roots of A.E. are distinct and real, that is m1 ≠ m2 ≠ m3 ≠ ≠ mn, then
x2 e ax , if f ′′( a) ≠ 0 f ′′( a)
P.I. =
where k1, k2, …, kn are constants can also be denoted as
n
x e ax , if f ′( a) ≠ 0 f ′( a)
=
1 2 sin ax , if f ( − a ) ≠ 0 f ( −a2 )
(c) When X = cos ax, then P.I. =
1 sin ax f (D2 )
1 cos ax f (D2 ) 1 cos ax , if f ( − a 2 ) ≠ 0 f ( −a2 )
(d) When X = xm, then P.I. =
1 x m = [ f ( D )]−1 x m f ( D)
11/21/2018 12:10:12 PM
Chapter 1 • Engineering Mathematics
Expansion of [ f ( D )]−1 is to be carried up to the term Dm because (m + 1)th and higher derivatives of xm are zero.
(e) When X = e ax v( x ), then P.I. =
1 e ax v( x ) f ( D)
P.I. = e ax
1 v( x ) f ( D + a)
(f) When X = x v(x), then
1 P.I. = x v( x ) f ( D) ⎡ f ′( D ) ⎤ 1 = ⎢x − ⋅ v( x ) f ( D ) ⎥⎦ f ( D ) ⎣
Thus, the following steps should be used to calculate the complete solution: (i) Find the auxiliary equation (A.E.) from the differential equation. (ii) Calculate the complementary function using the cases given before. (iii) Calculate particular integral (P.I.) using the cases explained before. (iv) To find the complete solution, use the following expression: y = C.F. + P.I.
74. Homogeneous Linear Equation Cauchy’s homogeneous linear equation is given by
where k1, k2, …, kn are constants and X is a function of x.
dy + Py = Qy n, where P and Q are funcdx tions of x and can be reduced to Leibnitz’s linear equation and is called the Bernoulli’s equation. To solve the Bernoulli’s equation, divide both sides by y n , so that The equation
y−n
dy + Py1− n = Q (3) dx
Putting y1− n = z , we get
dy dz (1− n) y − n = dx dx Therefore, Eq. (3) becomes 1 dz + Pz = Q 1− n dx
Ch wise GATE_EE_Ch1-a_Imp formula.indd 15
ax 2 y ′′ + bxy ′ + cy = 0 (4) where a, b and c are constants. This equation has two linearly independent solutions, which depend on the following quadratic equation:
This quadratic equation is known as characteristic equation. Let us consider Eq. (4) for x > 0 as this equation is singular for x = 0.
(a) If the roots of the characteristic equation are real and distinct, say m1 and m2, two linearly independent solutions of the differential equations are xm1 and xm2.
(b) If the roots of characteristic equation are real and equal, that is, m1 = m2 = m; the linearly independent solutions of the differential equation are xm and xm ln x. The general solution for the first case is given by y = c1xm1 + c2xm2 The general solution for the second case is given by y = (c1 + c2 ln x) xm
(c) If the roots of characteristic equation are complex conjugate, that is, m1 = a + jb and m2 = a - jb, Linearly independent solutions of differential are xa cos (b ln x) and xa sin (b ln x).
The general solution is given by
y = xa [C1 cos (b ln x) + C2 sin (b ln x)]
75. Bernoulli’s Equation
which is Leibnitz’s linear equation in z and can be solved using the methods mentioned under Integrating Factor.
am2 + (b - a) m + c = 0
n−2
n −1
dz + P (1 − n) z = Q (1 − n) dx
76. Homogeneous Euler–Cauchy Equation An ordinary differential equation is called homogeneous Euler-Cauchy equation, if it is in the form
d y d y ∂ + k1 x n −1 n −1 + k2 x n − 2 n − 2 + + kn y = X n dx dx ∂x n
xn
⇒
15
77. Non-homogeneous Euler–Cauchy Equation The non-homogeneous Euler-Cauchy equation is of the form
ax 2 y ′′ + bxy ′ + cy = r( x )
Here, a, b and c are constants. The method of variation of parameters to find solution to non-homogeneous Euler-Cauchy equation is given. As a first step, divide equation by ax2 so as to make coefficient of y ′′ as unity.
y ′′ +
or
y ′′ +
b c r( x ) y′ + 2 y = ax ax ax 2 b c y ′ + 2 y = r( x) ax ax r( x ) r( x) = ax 2
11/21/2018 12:10:14 PM
16
GATE EE Chapter-wise Solved Papers
Particular solution is given by y p ( x ) = − y1 ( x )∫
y2 ( x ) r ( x ) y ( x )r( x ) dx + y2 ( x )∫ 1 dx W ( y1 , y2 ) W ( y1 , y2 )
General solution to non-homogeneous Euler-Cauchy equation is therefore given by the following equation:
81. Exponential Function of Complex Variables z = re i q . 82. Circular Function of Complex Variables The circular functions of the complex variable z can be defined by the equations:
y(x) = C1 y1(x) + C2 y2(x) + yp(x)
78. Variation of Parameters Method The method of variation of parameters is a general method that can be used to solve non-homogeneous linear ordinary differential equations of the form y ′′ + py ′ + qy = x
sin z =
e iz − e − iz 2i
cos z =
e iz + e − iz 2
tan z =
sin z cos z
where p, q and x are functions of x. The particular integral in this case is given by
Now, cos z + i sin z =
y X yX P.I. = − y1 ∫ 2 dx + y2 ∫ 1 dx W W Here, y1 and y2 are the solutions of y ′′ + py ′ + qy = 0 y and W = 1 y1′
y2 is called the Wronskian of y1 and y2. y2′
79. Separation of Variables Method (a) Partial differential equations representing engineering problems include the one-dimensional heat flow equation given by ∂u ∂2u = c2 2 dt dx
(b) wave equation representing vibrations of a stretched string given by
with cosec z, sec z and cot z as respective reciprocals of above equations.
z = x + iy.
e iz + e − iz e iz − e − iz +i = e iz , where 2 2i
Also, e iy = cos y + i sin y, where y is real. Hence, e i q = cos q + i sin q , where q is real or complex.
This is called Euler’s theorem. Now, according to De Moivre’s theorem for complex number,
(cos q + i sin q )n = (eiq )
n
= cos nq + i sin nq ,
whether q is real or complex. 83. Hyperbolic Functions of Complex Variables Hyperbolic functions can be given as sin hx =
ex − e− x 2
cos h x =
ex + e− x 2
tanh x =
∂2u ∂2u + =0 ∂x 2 ∂y 2
ex − e− x ex + e− x
cot h x =
transmission line equations, two-dimensional wave equation and so on.
ex + e− x ex − e− x
sec h x =
2 e + e− x
Complex Variables
cosec h x =
∂2 y ∂2 y = c2 2 2 ∂t ∂x
(c) two-dimensional heat flow equation that becomes two-dimensional Laplace equation in steady state and given by
80. A number of the form x + iy, where x and y are real numbers and i = ( −1) , is called a complex number.
x is called the real part of x + iy and is written as R(x + iy), whereas y is called the imaginary part and is written as I(x + iy).
Ch wise GATE_EE_Ch1-a_Imp formula.indd 16
x
2 ex − e− x
Hyperbolic and circular functions are related as follows: sin ix = i sin h x cos ix = cos h x tan ix = i tan h x
11/21/2018 12:10:17 PM
17
Chapter 1 • Engineering Mathematics
84. Logarithmic Function of Complex Variables If z = x + iy and w = u + iy are related such that e w = z , then w is said to be a logarithm of x to the base e and is written as w = log e z Also, e w + 2inp = e w ⋅ e 2inp = z [ e 2inp = 1 ] ⇒ log z = w + 2 sin p 85. Cauchy–Riemann Equations A necessary condition for w = u ( x, y ) + iv ( x, y ) to be analytic in a region R is that u and v satisfy the following equations: ∂u ∂v = ∂z ∂y
∂u ∂v = ∂y ∂x Above equations are called Cauchy–Riemann equations. If the partial derivatives in above equations are continuous in R, the equations are sufficient conditions that f ( z ) be analytic in R. The derivative of f ( z ) is then given by ⎛ ∂u ∂v ⎞ ∂u ∂v f ′ ( z ) = lim ⎜ + i ⎟ = +i = ux + iv x Δy → 0 ⎝ ∂x ∂x ⎠ ∂x ∂x
⎛ ∂u ∂v ⎞ f ′ ( z ) = lim ⎜ +i or Δx → 0 ⎝ i∂y i∂y ⎟⎠ f ′ (z) =
1 ∂u ∂v ∂v ∂u + = −i = v y − iu y i ∂y ∂y ∂y ∂y
The real and imaginary parts of an analytic function are called conjugate functions.
86. Cauchy’s Theorem
If f ( z ) is an analytic function and f ′ ( z ) is continuous at each point within and on a closed curve c, then according to Cauchy’s theorem, f ( z ) ⋅ dz = 0
∫
c 87. Some of the important results that can be concluded are: (a) The line integral of a function f ( z ) , which is analytic in the region R, is independent of the path joining any two points of R.
(b) Extension of Cauchy’s theorem: If f(z) is analytic in the region R between the two simple closed curves c and c1, then f ( z ) ⋅ dz = f ( z ) ⋅ dz
Ch wise GATE_EE_Ch1-a_Imp formula.indd 17
∫ c
∫
c1
(c) If c1 , c2 , c3 , … be any number of closed curves within c, then
∫ f ( z ) ⋅ dz = ∫ f ( z ) ⋅ dz + ∫ f ( z ) ⋅ dz c
c1
c2
+ ∫ f ( z ) ⋅ dz + c3
88. Cauchy’s Integral Formula If f(x) is analytic within and on a simple closed curve c and a is any point interior to c, then f (α ) =
f (z)
Consider
f ( z ) dz 1 ∫ 2p i c z − α
in above equation, which is analytic at z −α all points within c except at z = α . Now, with a as center and r as radius, draw a small circle c1 lying entirely within c. Generally, we can write that f (z) n! dz ∫ 2p i c ( z − α )n +1
f n (α ) =
89. Taylor’s Series of Complex Variables If f(z) is analytic inside a circle C with centre at a, then for z inside C, f ( z ) = f (α ) + f ′ (α ) ( z − α ) + + +
f n (α ) n!
f ′′ (α ) 2!
( z − α )2
( z − α )n
90. Laurent’s Series of Complex Variables If f(z) is analytic in the ring-shaped region R bounded by the two concentric circles C and C1 of radii r and r1 (such that r > r1 ) and with center at a, then for all z in R ∞
f ( z ) = ∑ α ( z − α ) = a0 + a1 ( z − α ) + a2 ( z − α ) + 2
n
−∞
+ a−1( z − α ) + a−2 ( z − α ) −1
where an =
−2
f (t ) 1 dt . ∫ 2p i (t − α )n +1
91. Zeros and Poles of an Analytic Function (a) A zero of an analytic function f(z) is the value of z for which f(z) = 0. (b) A singular point of a function f(z) is a value of z at which f(z) fails to be analytic.
11/21/2018 12:10:21 PM
18
GATE EE Chapter-wise Solved Papers
92. Residues −1 The coefficient of ( z − α ) in the expansion of f ( z ) around an isolated singularity is called the residue of f ( z ) at the point.
1 d n −1 ⎡ ( z − α )n f ( z )⎤⎦ z → α ( n − 1)! dz z −1 ⎣
a−1 = lim
where n is the order of the pole. The residue of f(z) at z = a can also be found by Res f (α ) =
1 ∫c f ( z ) 2p i
93. Residue Theorem If f(z) is analytic in a region R except for a pole of order n at z = a and let C be a simple closed curve in R containing z = a, then
∫ f ( z )dz = 2p i × (sum of the residue at the singular c
points within C )
94. Calculation of Residues (a) If f(z) has a simple pole at z = a, then Res f (α ) = lim ⎡⎣( z − α ) f ( z )⎤⎦ z →α
(b) If f ( z ) = ϕ ( z ) /ψ ( z ) , where
ψ ( z ) = ( z − α ) f ( z ) , f (α ) ≠ 0 , then Res f (α ) =
ϕ (α ) ψ (α )
(c) If f(z) has a pole of order n at z = ¥∞, then Res f (α ) =
⎫ 1 ⎧ d n −1 ⎡ n ⎨ n −1 ⎣( z − α ) f ( x )⎤⎦ ⎬ (n − 1)! ⎩ dz ⎭ z =α
Probability and Statistics 95. Types of Events (a) Each outcome of a random experiment is called an elementary event. (b) An event associated with a random experiment that always occurs whenever the experiment is performed is called a certain event.
It can be found from the formula
(c) An event associated with a random experiment that never occurs whenever the experiment is performed is called an impossible event. (d) If the occurrence of any one of two or more events, associated with a random experiment, presents the occurrence of all others, then the events are called mutually exclusive events.
Ch wise GATE_EE_Ch1-a_Imp formula.indd 18
(e) If the union of two or more events associated with a random experiment includes all possible outcomes, then the events are called exhaustive events. (f) If the occurrence or non-occurrence of one event does not affect the probability of the occurrence or non-occurrence of the other, then the events are independent. (g) Two events are equally likely events if the p robability of their occurrence is same. (h) An event which has a probability of occurrence equal to 1− P, where P is the probability of o ccurrence of an event A, is called the complementary event of A.
96. Axioms of Probability (a) The numerical value of probability lies between 0 and 1. Hence, for any event A of S, 0 ≤ P ( A) ≤ 1 . (b) The sum of probabilities of all sample events is unity. Hence, P(S) = 1. (c) Probability of an event made of two or more sample events is the sum of their probabilities. 97. Conditional Probability Let A and B be two events of a random experiment. The probability of occurrence of A if B has already occurred and P ( B ) ≠ 0 is known as conditional probability. This is denoted by P ( A/B). Also, conditional probability can be defined as the probability of occurrence of B if A has already occurred and P ( A) ≠ 0 . This is denoted by P ( B /A). 98. Geometric Probability Due to the nature of the problem or the solution or both, random events that take place in continuous sample space may invoke geometric imagery. Hence, geometric probabilities can be considered as non-negative quantities with maximum value of 1 being assigned to subregions of a given domain subject to certain rules. If P is an expression of this assignment defined on a domain S, then 0 < P ( A) ≤ 1, A ⊂ S and P ( S ) = 1
The subsets of S for which P is defined are the random events that form a particular sample spaces. P is defined by the ratio of the areas so that if σ ( A) is defined as the area of set A, then σ ( A) P ( A) = σ ( s)
99. Rules of Probability Some of the important rules of probability are given as follows: (a) Inclusion–Exclusion principle of probability: P ( A ∪ B) = P ( A) + P ( B) − P ( A ∩ B)
11/21/2018 12:10:23 PM
Chapter 1 • Engineering Mathematics
I f A and B are mutually exclusive events, P ( A ∩ B) = 0 and then formula reduces to P ( A ∪ B) = P ( A) + P ( B)
(b) Complementary probability:
100. Arithmetic Mean Arithmetic Mean for Raw data Suppose we have values x1 , x2 ,..., xn and n are the total number of values, then arithmetic mean is given by x=
P ( A) = 1 − P ( A ) c
where P ( Ac ) is the complementary probability of A.
(c) P ( A ∩ B) = P ( A) ∗ P ( B /A) = P ( B) ∗ P ( A/B)
where P ( A/B) represents the conditional prob ability of A given B and P ( B /A) represents the conditional probability of B given A.
⎛ A⎞ + + P ⎜ ⎟ P ( Bn ) ⎝ Bn ⎠
(d) Conditional probability rule:
P ( A ∩ B) = P ( B) ∗ P ( A/B) ⇒ P ( A/B ) =
P ( A ∩ B) P ( B)
P ( B /A) =
P ( A ∩ B) P ( A)
Or
(e) Bayes’ theorem: Suppose we have an event A corresponding to a number of exhaustive events B1 , B2 ,..., Bn .
If P ( Bi ) and P ( A/Bi ) are given, then P ( Bi /A) =
P ( Bi ) P ( A/Bi ) ∑ P ( Bi ) P ( A/Bi )
(f) Rule of total probability: Consider an event E which occurs via two different values A and B. Also, let A and B be mutually exhaustive and collectively exhaustive events. Now, the probability of E is given as P( E ) = P( A ∩ E ) + P( B ∩ E ) = P ( A) ∗ P ( E /A) + P ( B) ∗ P ( E /B)
This is called the rule of total probability.
Ch wise GATE_EE_Ch1-a_Imp formula.indd 19
1 n ∑ xi n i =1
where x = arithmetic mean.
Arithmetic Mean for Grouped Data (Frequency Distribution) Suppose f i is the frequency of xi , then the arithmetic mean from frequency distribution can be calculated as
If B1 , B2 ,..., Bn are pairwise disjoint events of positive probability, then ⎛ A⎞ ⎛ A⎞ P ( A) = P ⎜ ⎟ P ( B1 ) + P ⎜ ⎟ P ( B2 ) ⎝ B1 ⎠ ⎝ B2 ⎠
19
x= where
1 n ∑ f i xi N i =1 n
N = ∑ fi i =1
101. Median Median for Raw Data Suppose we have n numbers of ungrouped/raw values and let values be x1 , x2 ,..., xn . To calculate median, arrange all the values in ascending or descending order. th
⎡ ( n + 1) ⎤ Now, if n is odd, then median = ⎢ ⎥ value ⎣ 2 ⎦ If n is even, then median
th ⎡⎛ n ⎞ th ⎤ ⎛n ⎞ = ⎢⎜ ⎟ value + ⎜ + 1⎟ value ⎥ ⎝2 ⎠ ⎢⎣⎝ 2 ⎠ ⎥⎦ 2
Median for Grouped Data To calculate median of grouped values, identify the class containing the middle observation.
⎤ ⎡ ( N + 1) − ( F + 1) ⎥ ⎢ Median = L + ⎢ 2 ⎥×h fm ⎥ ⎢ ⎥⎦ ⎢⎣ where L = lower limit of median class
N = total number of data items = ∑ f
F = cumulative frequency of class immediately preceding the median class
= frequency of median class f m h = width of class
11/21/2018 12:10:26 PM
20
GATE EE Chapter-wise Solved Papers
(a)
(b)
102. Mode Mode of raw values of data is the value with the highest frequency or simply the value which occurs the most number of times.
Mode of Raw Data Mode of raw data is calculated by simply checking which value is repeated the most number of times. Mode of Grouped Data Mode of grouped values of data is calculated by first identifying the modal class, i.e. the class which has the target frequency. The mode can then be calculated using the following formula:
Mode = L +
f m − f1 ×h 2 f m − f1 − f 2
(c) Figure 6 | (a) Symmetrical frequency distribution, (b) positively skewed frequency distribution and (c) negatively skewed frequency distribution.
104. Geometric Mean Geometric Mean of Raw Data
where L = lower limit of modal class
f m = frequency of modal class f1 = frequency of class preceding modal class f 2 = frequency of class following modal class h = width of model class
⎛ n⎞ ⎜⎝ ip=1⎟⎠
103. Relation Between Mean, Median and Mode
When an approximate value of mode is required, the given empirical formula for mode may be used. There are three types of frequency distribution: (a) Symmetric distribution: It has lower half equal to upper half of distribution. For symmetric distribution, Mean = Median = Mode
n
x1 ⋅ x2 ⋅ x3 ⋅⋅⋅⋅⋅ xn
Geometric Mean of Grouped Data Geometric mean for a frequency distribution is given by n
1 N
∑f i =1
(c) Negatively skewed distribution: It has a long tail on the left than on the right. Mean ≤ Median ≤ Mode
Figure 6 shows the three types of frequency distribution.
log( xi )
i
n
i =1
105. Harmonic Mean
Harmonic Mean of Raw Data Harmonic mean of n numbers x1 , x2 , x3 ,..., xn is calculated as H.M. =
Mode ≤ Median ≤ Mean
=
where N = ∑ f i .
(b) Positively skewed distribution: It has a longer tail on the right than on the left.
1/ n
log G.M. =
Empirical mode = 3 Median − 2 Mean
Geometric mean of n numbers x1 , x2 ,..., xn is given by
n n
1
∑x i =1
i
Harmonic Mean of Grouped Data Harmonic mean for a frequency distribution is calculated as H.M. =
N n
∑ ( f /x ) i =1
i
i
n
where N = ∑ f i . i =1
(a)
Ch wise GATE_EE_Ch1-a_Imp formula.indd 20
(b)
106. Mean Deviation Mean Deviation of Raw Data Suppose we have a given set of n values x1 , x2 , x3 ,..., xn , then the mean deviation is given by
11/21/2018 12:10:28 PM
Chapter 1 • Engineering Mathematics
M.D. =
1 n ∑ xi − X n i =1
21
The following steps should be followed to calculate standard deviation for discrete data:
where x = mean.
(a) Calculate mean ( X ) for given observations.
The following steps should be followed to calculate mean deviation of raw data: (a) Compute the central value or average ‘A’ about which mean deviation is to be calculated. (b) Take mod of the deviations of the observations about the central value ‘A’, that is xi − x .
(b) Take deviations of observations from the mean, i.e. ( xi − X ) .
(c) Square the deviations obtained in the previous step and find
n
(c) Obtain the total of these deviations, that is ∑ x2 − X .
(d) Divide the total obtained in step 3 by the number of observations. Mean Deviation of Discrete Frequency Distribution For a frequency deviation, the mean deviation is given by
n
∑ (x i =1
i =1
M.D. =
1 N
n
∑f i =1
i
i
(d) Divide the sum by n to obtain the value of variance, that is
σ2 =
− X )2
1 n ∑ ( xi − X )2 n i =1
(e) Take out the square root of variance to obtain standard deviation,
xi − x
1 n ∑ ( xi − X )2 n i =1
σ2 =
n
where N = ∑ fi . i =1
The following steps should be followed to calculate mean deviation of discrete frequency deviation: (a) Calculate the central value or average ‘A’ of the given frequency distribution about which mean deviation is to be calculated. (b) Take mod of the deviations of the observations from the central value, that is xi − x . (c) Multiply these deviations by respective frequencies and obtain the total ∑ f i xi − x .
Standard Deviation of Discrete Frequency Distribution If we have a discrete frequency distribution of X, then
σ2 =
⎤ 1 ⎡ n ∑ ( xi − X )2 ⎥ N ⎢⎣ i =1 ⎦
σ=
⎤ 1 ⎡ n ( xi − X ) 2 ⎥ ∑ ⎢ N ⎣ i =1 ⎦
n
where N = ∑ f . i =1
(d) Divide the total obtained by the number of observa-
The following steps should be followed to calculate variance if discrete distribution of data is given: (a) Obtain the given frequency distribution.
n
i =1
(b) Calculate mean ( X ) for given frequency distribution.
(c) Take deviations of observations from the mean, that is ( xi − X ).
(d) Square the deviations obtained in the previous step and multiply the squared deviations by respective frequencies, that is f i ( xi − X ).
107. Standard Deviation Standard Deviation of Raw Data
(e) Calculate the total, that is ∑ f i ( xi − X ) 2 .
tions, that is N = ∑ f i to obtain the mean deviation.
Mean Deviation of Grouped Frequency Distribution For calculating the mean deviation of grouped frequency distribution, the procedure is same as for a discrete frequency distribution. However, the only difference is that we have to obtain the mid-points of the various classes and take the deviations of these mid-points from the given central value.
If we have n values x1 , x2 , ..., xn of X, then 1 n σ = ∑ ( xi − x ) 2 n i =1
Ch wise GATE_EE_Ch1-a_Imp formula.indd 21
1 n ∑ ( xi − x )2 n i =1
i =1
n
(f) Divide the sum ∑ f i ( xi − X ) 2 by N, where N = ∑ f i i =1
N = ∑ f i , to obtain the variance, σ 2.
2
⇒ σ=
n
(g) Take out the square root of the variance to obtain standard deviation, σ =
1⎡n ⎤ ∑ f i ( xi − X ) 2 ⎥ . ⎢ i = 1 n⎣ ⎦
11/21/2018 12:10:31 PM
22
GATE EE Chapter-wise Solved Papers
Standard Deviation of Grouped Frequency Distribution For calculating standard deviation of a grouped frequency distribution, the procedure is same as for a discrete frequency distribution. However, the only difference is that we have to obtain the mid-point of the various classes and take the deviations of these midpoints from the given central point.
108. Coefficient of Variation Coefficient of variation (C.V.) is a measure of variability which is independent of units and hence can be used to compare two data sets with different units.
σ × 100 X where σ represents standard deviation and X represents mean. C.V. =
109. Random Variable Random variable can be discrete and continuous. Discrete random variable is a variable that can take a value from a continuous range of values. Continuous random variable is a variable that can take a value from a continuous range of values. If a random variable X takes x1 , x2 ,..., xn with respective probabilities P1 , P2 ,..., Pn , then X : x1 P ( X ) : P1
x2 P2
112. General Discrete Distribution Suppose a discrete variable X is the outcome of some random experiment and the probability of X taking the value xi is Pi , then P ( X = xi ) = Pi for i = 1, 2,...
where, P ( xi ) ≥ 0 for all values of i and ∑ P ( xi ) = 1.
The set of values xi with their probabilities Pi of a discrete variable X is called a discrete probability distribution. For example, the discrete probability distribution of X, the number which is selected by picking a card from a well-shuffled deck is given by the following table: X = xi : Ace 1 Pi : 13 X = xi : 8 1 Pi : 13
(b) E ( x ) = ∑ xP ( x )
(c) V ( x ) = E ( x 2 ) − [ E ( x )] = ∑ x 2 P ( x ) − [ ∑ xP ( x )]
where E(x) denotes the expected value or average value of a random variable x and V(x) denotes the variable of a random variable x.
2
2
x
f ( x ) dx
−∞ ∞
xf ( x ) dx
−∞
(c) V ( x ) = E ( x 2 ) − [ E ( x )] = 2
∞
∫x −∞
2
f ( x )dx
⎤ ⎡∞ − ⎢ ∫ xf ( x )dx ⎥ ⎣ −∞ ⎦
Ch wise GATE_EE_Ch1-a_Imp formula.indd 22
where x is any integer. The mean value ( x ) of the probability distribution of a discrete variable X is known as its expectation and is denoted by E(x). If f(x) is the probability density function of X, then n
E ( x ) = ∑ xi f ( xi ) i =1
Variable of a distribution is given by n
(a) Cumulative distribution function is given by
∫
The distribution function F(x) of discrete variable X is defined by
σ 2 = ∑ ( xi − x ) 2 f ( xi )
111. Properties of Continuous Distribution
(b) E ( x ) =
3 4 5 6 7 1 1 1 1 1 13 13 13 13 13 10 Jack Queen King 1 1 1 1 13 13 13 13
i =1
∫
2 1 13 9 1 13
n
(a) ∑ P ( x ) = 1
a
F ( x ) = P ( X ≤ x ) = ∑ Pi
b
is known as the probability distribution of X.
F ( x) =
(d) P ( a < x < b) = P ( a ≤ x ≤ b) = P ( a < x ≤ b) = P ( a ≤ x ≤ b) = ∫ f ( x ) dx
... xn ... Pn
x3 P3
110. Properties of Discrete Distribution
2
i =1
113. Binomial Distribution Binomial distribution is concerned with trails of a respective nature whose outcome can be classified as either a success or a failure. Suppose we have to perform n independent trails, each of which results in a success with probability P and in a failure with probability X which is equal to 1 − P. If X represents the number of successes that occur in the n trails, then X is said to be binomial random variable with parameters (n, p).
11/21/2018 12:10:33 PM
23
Chapter 1 • Engineering Mathematics
The binomial distribution occurs when the experiment performed satisfies the following four assumptions of Bernoulli’s trails. (a) They are finite in number.
(b) There are exactly two outcomes: success or failure.
(c) The probability of success or failure remains same in each trail.
(d) They are independent of each other.
The probability of obtaining x successes from n trails is given by the binomial distribution formula,
116. Geometric Distribution Consider repeated trails of a Bernoulli experiment E with probability of success, P, and probability of failure, q = 1 − p . Let x denote the number of times E must be repeated until finally obtaining a success. The distribution is
P ( X ) = nC x P x (1 − p) n − x
A random variable X, taking on one of the values 0, 1, 2, …, n, is said to be a Poisson random variable with parameters m if for some m > 0, P( x) =
e−m mx x!
Mean = E ( x ) = m
Variance = V ( x ) = m
i =1
i =1
m
P( x) =
C nx C y − x
m+ n
Cx
Also, p( x ) = q x p, x = 0, 1, 2, ..., q = 1 − p ∞
∞
x=0
x=0
Ch wise GATE_EE_Ch1-a_Imp formula.indd 23
x
= p
1 =1 1− q
The mean of geometric distribution = q /p.
The variance of geometric distribution = q /p 2.
117. General Continuous Distribution When a random variable X takes all possible values in an interval, then the distribution is called continuous distribution of X. A continuous distribution of X can be defined by a probability density function f(x) which is given by p( −∞ ≤ x ≤ ∞) =
∞
∫
f ( x )dx = 1
−∞
The expectation for general continuous distribution is given by E (x) =
∞
∫
−∞
x f ( x ) dx
The variance for general continuous distribution is given by V (x) = σ 2 =
∞
∫ ( x − X ) f ( x ) dx −∞ 2
118. Uniform Distribution If density of a random variable X over the interval −∞ < a < b < ∞ is given by f (x) =
, x = 0, 1, ..., y, y ≤ m, n
This distribution is known as hypergeometric distribution.
C xnC y − x = m + nC y
m
∑ P( x) = P ∑ q
Therefore, the expected value and the variable of a Poisson random variable are both equal to its parameters m.
115. Hypergeometric Distribution If the probability changes from trail to trail, one of the assumptions of binomial distribution gets violated; hence, binomial distribution cannot be used. In such cases, hypergeometric distribution is used; say a bag contains m white and n black balls. If y balls are drawn one at a time (with replacement), then the probability that x of them will be white is
n
For Poisson distribution,
n
∑ p( x) = 1, since ∑
where P is the probability of success in any trail and (1 − p) is the probability of failure.
114. Poisson Distribution Poisson distribution is a distribution related to the probabilities of events which are extremely rare but which have a large number of independent opportunities for occurrence.
For hypergeometric distribution,
1 , a< x 0 is given by ⎧l e − l x f (x) = ⎨ ⎩ 0
nσ xσ y
where X is deviation from the mean ( x − x ) , Y is deviation from the mean ( y − y ) , σ x is standard deviation
b
∑ XY
x′
°
x
y′ Figure 7 | Line of regression.
11/21/2018 12:10:38 PM
25
Chapter 1 • Engineering Mathematics
Numerical Methods 123. Bisection Method We begin the iterative cycle by choosing two trial points x0 and x1 , which enclose the actual root. Then f ( x0 ) and f ( x1 ) are of opposite signs. The interval ( x0 , x1 ) is bisected and its midpoint x2 is obtained as
y
x2 = ( x1 + x0 ) / 2
iterations are carried over with the new set of x0 and x1. Another x2 is found by the intersection of the straight line joining the new f ( x0 ) and f ( x1 ) points with x-axis. Each new or successive interval is smaller than the previous interval, and it is guaranteed to converge to the root. The procedure is illustrated in Fig. 9.
A (x0, f(x0))
If f ( x2 ) = 0 , then x2 itself is the root. If not, the root lies either between x0 and x2 or between x1 and x2 . That is, if f ( x0 ) and f ( x1 ) are of opposite signs, then root lies between x0 and x2 ; otherwise if f (x1) and f(x2) are of opposite signs, the root lies between x1 and x2 . The new interval for searching the root is therefore ( x0 , x1 ) in the first case and ( x1 , x2 ) in the second case, which is much less compared to the first interval ( x0 , x1 ) . This process is illustrated in Fig. 8. This new interval will be the beginning interval for the next iteration, and the processes of bisection and finding another such x2 are repeated.
x3 0
x0
x2 x1
x
P(x) B (x1, f(x1))
Figure 9 | Illustration of Regula–Falsi method.
y or f(x) f(x)
Actual root
From Fig. 9, it can be seen (from the equation of the straight line) that y − f ( x0 ) =
x0 x′
x =(x0+x1) x1 2
x
Third iteration Second iteration Starting interval Figure 8 | Illustration of bisection method.
124. Regula–Falsi Method (Method of False Position Method) The iterative procedure is started by choosing two values x0 and x1 such that f ( x0 ) and f ( x1 ) are of opposite signs. Then the two points [ x0 , f ( x0 )] and [ x1 , f ( x1 )] are joined by a straight line. The intersection of this line with the x-axis gives x2. If f ( x2 ) and f ( x0 ) are of opposite signs, then replace x1 by x2; otherwise, replace x0 by x2. This yields a new set of values for x0 and x1 . The present range is much smaller than the range or interval between the first chosen set of x0 and x1. The convergence is thus established, and the
Ch wise GATE_EE_Ch1-a_Imp formula.indd 25
where x2 = x0 −
f ( x1 ) − f ( x0 ) ( x − x0 ) x1 − x0
x0 − x1 f ( x0 ) f ( x1 ) − f ( x0 )
which is an approximation to the root.
125. Newton–Raphson Method Newton’s method, also known as Newton–Raphson method, is another iteration method for solving e quations f ( x ) = 0 , where f is assumed to have a continuous derivative f ′. The underlying idea is that we approximate the graph of f by suitable tangents. The Newton–Raphson method has second-order convergence.
Using an approximate value x0 obtained from the graph of f, we let x1 be the point of intersection of the x-axis and the tangent to the curve of f at x0 (Fig. 10). Then tan β = f ′( x0 ) =
Hence,
x1 = x0 −
f ( x0 ) x0 − x1
f ( x0 ) f ′( x0 )
11/21/2018 12:10:41 PM
26
GATE EE Chapter-wise Solved Papers
In the second step, we compute x2 =
x1 − f ( x1 ) f ′( x1 )
Consider Fig. 11. The slope of line from points (x0, f(x0)) to (x1, f(x1)) can be given by the following equation. y=
In the third step x3 from x2 again by the same formula, and so on.
f ( x1 ) − f ( x0 ) x1 − x0
We now calculate the solution of x by putting y = 0. x = x1 − f ( x1 )
y
y =f(x)
x2
x1
x0
Figure 10 | Newton–Raphson method.
126. Secant Method The secant method can be thought of as a finite difference approximation of Newton–Raphson method.
x2 = x1 − f ( x1 )
x1 − x0 f ( x1 ) − f ( x0 )
x3 = x2 − f ( x2 )
x2 − x1 f ( x2 ) − f ( x1 )
The generalized solution for the secant method as xn = xn −1 − f ( xn −1 )
x
x1 − x0 f ( x1 ) − f ( x0 )
We use the new value of x as x2 and repeat the same process using x1 and x2 instead of x0 and x1. This process is continued in the same manner unless we obtain xn = xn −1 .
f(x0)
b
( x − x1 ) + f ( x1 )
xn −1 − xn − 2 f ( xn −1 ) − f ( xn − 2 )
127. Jacobian If u and v are functions of two independent variables x ∂u / ∂x ∂u / ∂y and y, then the determinant is called ∂v / ∂x ∂v / ∂y the Jacobian of u, v with respect to x, y and is written as ∂ (u, v )
⎛ u, v ⎞ . or J ⎜ ⎝ x, y ⎟⎠ ∂ ( x, y )
Transform Theory 128. Laplace transform X(s) of a continuous time signal x(t) is given by
x3
x0
x2
X (s) =
x1
∞
∫ x (t ) e
− st
dt
−∞
Variable s is generally complex and is given by s = ( s + jω ) . Laplace transform defined by above equation is also called bilateral Laplace transform. A unilateral Laplace transform is defined as ∞
X ( s ) = ∫ x (t ) e − st dt 0
f(x)
Figure 11 | Secant method.
Ch wise GATE_EE_Ch1-a_Imp formula.indd 26
The two Laplace transforms are equivalent only when x(t ) = 0 for t < 0. Region of convergence (ROC) is defined with respect to Laplace transforms. The range of values of complex variables (s) for which Laplace transform converges is called the region of convergence.
11/21/2018 12:10:42 PM
27
Chapter 1 • Engineering Mathematics
129. Laplace Transform of Common Signals Laplace transform of a unit step function u(t) as shown in Fig. 12 is given by
1 (g) L ⎡⎣ −te − at u( −t ) ⎤⎦ = Re( s) < − Re( a) ( s + a) 2
u(t)
(h) L [cos ω 0 t u(t )] =
s Re( s) > 0 ( s 2 + ω 02 )
(i) L [sin ω 0 t u(t )] =
ω0 Re( s) > 0 ( s 2 + ω 02 )
1
0
(j) L ⎡⎣e − at cos ω 0 t u( −t ) ⎤⎦
t
Figure 12 | Unit step function.
= L [ u (t ) ] =
∞
∫
u (t ) e − st dt =
−∞
1 = − e − st s
∞
∫e
0 ∞
0
+
− st
dt
( s + a) Re( s) > − Re( a) ( s + a) 2 + ω 02
(k) L ⎡⎣e − at sin ω 0 t u(t ) ⎤⎦
+
1 = , Re( s) > 0 s
=
where 0 + = lim(0 + ε )
ω0 Re( s) > − Re( a) ( s + a) 2 + ω 02
132. Properties of Laplace Transform
ε →0
130. Laplace transform of a unit impulse function
Laplace transforms exhibit the following properties:
(a) Linearity: If X 1 ( s) and X 2 ( s) are, respectively, Laplace transforms of x1 (t ) and x2 (t ) , then
Laplace transform of a unit impulse function as shown in Fig. 13 is given by L [δ (t )] =
∞
∫ d (t ) e
− st
L [ a1 x1 (t ) + a2 x2 (t )] = a1 X 1 ( s) + a2 X 2 ( s)
dt = 1 for all s
−∞
If the regions of convergence of X 1 ( s) and X 2 ( s) are R1 and R2 , respectively, then the region of convergence of resultant transform R is equal to intersection of R1 and R2 .
δ(t)
1
t
L [ x(t − t0 )] = e − st0 X ( s)
0 Figure 13 | Unit impulse function.
131. Laplace Transform of Some Other Common Signals
Both transforms have same region of convergence.
1 (a) L [ −u( −t )] = Re( s) < 0 s (b) L [tu(t )] =
(b) Time shifting: If Laplace transform of x(t ) is X ( s), Laplace transform of x(t − t0 ) is given by
1 Re( s) > 0 s2
(c) Time scaling: If Laplace transform of x(t ) is X ( s) with a region of convergence equal to R , then Laplace transform of x( at ) is given by L[ x( at )] =
1 ⎛ s⎞ ×⎜ ⎟ | a | ⎝ a⎠
k! (c) L ⎡⎣t k u(t ) ⎤⎦ = ( k +1) Re( s) > 0 s
The region of convergence of resultant transform would be aR.
1 Re( s) > − Re( a) (d) L ⎡⎣e − at u(t ) ⎤⎦ = ( s + a)
1 Re( s) < − Re( a) (e) L ⎡⎣ −e − at u( −t ) ⎤⎦ = ( s + a) 1 Re( s) > − Re( a) (f) L ⎡⎣te u(t ) ⎤⎦ = ( s + a) 2 − at
Ch wise GATE_EE_Ch1-a_Imp formula.indd 27
(d) Shifting in s-domain: If Laplace transform of x(t ) is X ( s) having a region of convergence R, then L ⎡⎣e s0 t x(t ) ⎤⎦ = X ( s − s0 )
Region of convergence of resultant transform =
R + Re( so )
11/21/2018 12:10:46 PM
28
GATE EE Chapter-wise Solved Papers
(e) Time reversal: If Laplace transform of x(t) is X(s) with region of convergence R, then L[ x( −t )] = X ( − s)
134. Proper rational function (m < n): If the poles of X(s) are simple, then partial fraction expansion of X(s) can be written as X ( s) =
The transform has region of convergence −R.
(f) Differentiation in time domain: If Laplace transform of x(t) is X(s), then ⎡ dx(t ) ⎤ L⎢ ⎥ = sX ( s) ⎣ dt ⎦
(g) Differentiation in s-domain: If Laplace of x(t) is X(s) with a region of convergence R, then dX ( s) L[ −tx(t )] = ds
(h) Integration in time domain: If Laplace transform of x(t) is X(s), then t
lr − k = where
X ( s) =
If R is region of convergence for X(s), region of convergence R ′ of the resultant transform is expressed by
(i) Convolution: If X 1 ( s) and X 2 ( s) are Laplace transforms of x1 (t ) and x2 (t ) , respectively, then L[ x1 (t ) ∗ x2 (t )] = X 1 ( s) ⋅ X 2 ( s)
If R1 and R2 are regions of convergence of X 1 ( s) and X 2 ( s) , respectively, then region of convergence of resultant transform is expressed by R ′ ⊃ R1 ∩ R2. Convolution in time domain is multiplication in s-domain. 133. Inverse Laplace Transform
Inverse Laplace transform of X(s) is given by c + jω
L−1 X ( s) = x(t ) =
1 X ( s)e st ds 2p j c −∫jω
1 dk [( s − pi ) r X ( s)] s = pi k ! ds k
135. Improper rational function (m ≥ n): When X(s) is an improper rational function, that is, m ≥ n , X(s) can be written in the following form:
⎤ X ( s) ⎡t L ⎢ ∫ x(τ ) dτ ⎥ = s ⎣ −∞ ⎦
R ′ = R ∩ {Re( s) > 0}
If the denominator D(s) has multiple roots, then expansion of X(s) consists of terms of the form:
l1 l1 lr + ++ 2 ( s − pi ) ( s − pi ) ( s − pi ) r
The resultant transform also has region of convergence R.
where coefficients (Ck ) are given by Ck = ( s − pk ) X ( s)| s = pk
If R is region of convergence of X(s), then region of convergence R ′ of the resultant transform is expressed by R ′ ⊃ R.
cn c1 ++ ( s − pn ) ( s − p1 )
X(s) is a polynomial in s with degree (m − n). R(s) is a polynomial in s with degree strictly less than n. The inverse Laplace transform of X(s) can be determined from inverse Laplace transform of Q(s) and R(s)/ D(s). The inverse Laplace transforms of Q(s) can be determined by using transform below: L
R( s) N ( s) = Q ( s) + D ( s) D ( s)
d k δ (t ) = s k v , k = 1, 2, 3, … dt k
Laplace transform of R(s)/D(s) can be computed by partial fraction method explained earlier. Note that R(s)/ D(s) is a proper rational function.
136. The z-transform X(z) of a discrete time signal or sequence is defined by X ( z) =
∞
∑ x[n]z
−n
n = −∞
Variable z is generally complex valued and is expressed in polar form by Z = re jΩ where r and Ω , respectively, are magnitude and angle of z. The z-transform defined by above equation is a bilateral or two-sided transforms unilateral or one-sided z-transform is defined as follows: ∞
X ′( z ) = ∑ x[n]z − n n= 0
Ch wise GATE_EE_Ch1-a_Imp formula.indd 28
11/21/2018 12:10:48 PM
Chapter 1 • Engineering Mathematics
x[n] and X(z) are said to form a transform pair denoted as x[n] ↔ X ( z )
(k) ( r n sin Ω0 n)u[n] ↔
(l) ⎡ a n , for 0 ≤ n ≤ N − 1⎤ ⎢ ⎥↔ 0 , otherwise ⎣ ⎦ N −N 1− a z for | z | > 0 1 − az −1
∞
X ( z ) = ∑ δ [n]z − n = z − n = 1 for all z n= 0
δ [n] ↔ 1 for all z.
138. z-Transform of a Unit Step Sequence u[n] is given by X ( z) =
∞
∑ u[n] z
−n
n = −∞
u[n] ↔ Therefore,
=
1 z = ,| z | > 1 1 − z −1 z − 1
z for all z. z −1
139. Some Other Common z-Transform Pairs (a) −u[ − n − 1] ↔
z for | z | < 1 z −1
(b) δ [n − m] ↔ z − m for all z except z = 0 (for m > 0) and ∞(for m < 0) (c) a n u[n] ↔ −
z for | z | > | a | z−a
(d) − a n u[ − n − 1] ↔ (e) na n u[n] ↔
z for | z | < | a | z−a
az for | z | > | a | ( z − a) 2 az for | z | < | a | ( z − a) 2
(f) − na n u[ − n − 1] ↔
2
⎡ z ⎤ (g) ( n + 1)a n u[n] ↔ ⎢ ⎥ for | z | > | a | ⎣ ( z − a) ⎦ (h) (cos Ω0 n)u[n] ↔
z 2 − (cos Ω0 ) z z − ( 2 cos Ω0 ) z + 1 2
for | z | > |1| (i) (sin Ω0 n)u[n] ↔
(sin Ω0 ) z z 2 − ( 2 cos Ω0 ) z + 1
for | z | > |1| (j) ( r n cos Ω0 n)u[n] ↔ z 2 − ( r cos Ω0 ) z for | z | z 2 − ( 2r cos Ω0 ) z + r 2
140. Properties of z -Transform (a) z-Transform exhibits the property of linearity. If X 1 ( z ) and X 2 ( z ) , respectively, are z-transforms of X 1 [n] and X 2 [n] , then a1 x1 [n] + a2 x2 [n ] ↔ a1 X 1 ( z ) + a2 X 2 ( z )
If R1 and R2 are ROC of X 1 ( z ) and X 2 ( z ) , respectively, then ROC of a1 X 1 ( z ) + a2 X 2 ( z ) is given by R ′ ⊃ R1 ∩ R2 . ( a1 ) and ( a2 ) are arbitrary constants. (b) If z-transform of x[n] is X(z), then z-transform of x[n − n0 ] is given by z − n 0 X ( z ) That is,
x[n − n0 ] ↔ z n 0 X ( z )
The region of convergence is given by R ′ = R ∩ {0 < | z | < ∞} , where R is region of convergence for x(z). This is the time shifting property. As an example, z-transform of x[n - 1] is given by z −1 X ( z ) with region of convergence given by R ′ = R ∩ {| z | < ∞}. If X(z) is z-transform of x[n] with ROC (R), then z-transform of z0 n x[n] is given by
⎛ z⎞ z0n x[n] ↔ x ⎜ ⎟ with ROC R = | z0 | R ⎝ z0 ⎠
(c) A pole (or zero) at z = zk in X(z) moves to z = z0 zk after multiplication of x[n] by z0 n . The region of convergence also gets multiplied by | zo |. As a special case of this property, following transform holds: e jΩ0 n x[n] ↔ X (e − jΩ0 z ) with R ′ = R In this special case, all poles and zeros are rotated by angle Ω0 and the region of convergence remains unchanged. (d) According to the time reversal property of z-transform, if X(z) is z-transform of x[n] with ROC = R , then
Ch wise GATE_EE_Ch1-a_Imp formula.indd 29
( r sin Ω0 ) z for | z | > r z 2 − ( 2r cos Ω0 ) z + r 2
137. z-Transform of a unit impulse sequence δ [n] is given by
Therefore,
29
1 ⎛ 1⎞ x[ − n] ↔ X ⎜ ⎟ with ROC R ′ = ⎝ z⎠ R
11/21/2018 12:10:52 PM
30
GATE EE Chapter-wise Solved Papers
⎛ 1⎞ This implies that a pole or zero at z = zk moves to ⎜ ⎟ ⎝ zk ⎠ after time reversal. As far as the ROC is concerned, time reversal means that a right-sided sequence becomes a left-sided sequence and vice versa.
(e) In the case of multiplication of discrete sequence x[n] by n, the following transform pair holds:
143. Partial Fraction Expansion Partial fraction expansion method can be used to determine inverse z-transform when X(z) is a rational function of z. If X(z) is expressed by X ( z) =
dX ( z ) nx[n] ↔ − z dz The region of convergence remains unchanged, that is, R ′ = R.
(f) If X(z) is z-transform of x[n] with ROC = R, following transform pair holds: n
1
∑ x[k ] ↔ 1 − z
k = −∞
−1
X ( z) =
n
∑ x[k ] is the discrete time counterpart of integra-
=
n
X ( z ) = c0 + ∑ ck k =1
X ( z) =
If X 1 ( z ) is z-transform of x1 [n] with ROC = R1 x1 [n] ∗ x2 [n] ↔ X 1 ( z ) ⋅ X 2 ( z )
And R ′ ⊃ R1 ∩ R2
∫ X ( z ) z
n −1
142. z-Transform as Power Series Expansion z-Transform given by above equation earlier can also be represented by a power series where the sequence values of x[n] are nothing but the coefficients of z − n. That is, X ( z) =
∞
∑ x[n]z
−n
n = −∞
= ...... + x[ −2]z 2 + x[ −1]z + x(0) + x(1) z −1 + x[2]z −2 + ...
This approach is particularly useful for finite-length sequences.
Ch wise GATE_EE_Ch1-a_Imp formula.indd 30
where
dz
c where C is the counterclockwise contour of integration enclosing origin.
z z − pk
m−n
n
q=0
k =1
∑ bqz q + ∑ ck
z z − pk
In the case of multiple order poles, the partial fracX ( z) tion expansion of will consists of terms of z the form (assuming pole ( pi ) has multiplicity of r),
141. Inverse z-Transform Inverse z-transform of x(z) is given by 1 x[n] = 2p j
X ( z) |z = pk z
(b) If m > n, then a polynomial of z must be added to the right-hand side of X(z) expression the order of which is (m - n). In that case, partial fraction expansion has the following form:
(g) According to the convolution property of z-transform,
Then
Equation written as
k = −∞
X 2 ( z ) is z-transform of x2 [n] with ROC = R2
n c0 c +∑ k z k =1 z − pk
where c0 = X ( z ) |z = 0 and ck = ( z − pk )
tion in the continuous time domain.
(a) If n ≥ m and all poles are simple, then cn c c2 X ( z ) c0 = + 1 + ++ z z z − p1 z − p2 z − pn
z X ( z) ( z − 1)
Region of convergence R ′ is given by R ′ = R ⊃ {| z | > 1}
( z − z1 )( z − z2 ) ( z − zm ) N ( z) = k⋅ ( z − p1 )( z − p2 ) ( z − pn ) D( z )
l1 l2 lr + ++ ( z − pi ) ( z − pi ) 2 ( z − pi ) r lr − k
1 dk = k ! dz k
⎡ r X ( z) ⎤ ⎢( z − pi ) z ⎥ ⎣ ⎦ z = pi
144. Relationship between z-Transform and Laplace Transform z-Transform is related to Laplace transform by following expressions.
1 ln z T (b) X(s) = X(z)|z = e+Ts where T is the sampling period. (a) s =
Fourier Transform 145. Fourier transform transforms a time domain signal into its frequency domain counterpart. If X(w) is the Fourier transform of x(t), then
11/21/2018 12:10:54 PM
31
Chapter 1 • Engineering Mathematics ∞
X (ω ) =
∫
x(t )e − jωt dt
(d) Scaling in time domain causes inverse scaling in frequency domain.
−∞
x( at ) ↔
146. The inverse Fourier transform is defined by x (t ) =
1 2p
∞
∫ X (ω )e
jω t
dω
−∞ 147. x(t) and X (ω ) form a Fourier transform pair written as
(e) Inverse scaling of both frequency variable ( ω s ) and
x(t ) ↔ X (ω ) 148. X (ω ) is, in general, complex and is expressed as X (ω ) = X (ω ) e jϕ (ω )
∞
∫ x (t )e
jω t
dt
(f) According to property of duality or symmetry X (t ) ↔ 2 p x( −ω )
(g) Differentiation in time domain causes multiplication is frequency domain is dx(t ) ↔ jω X (ω ) dt
(h) Differentiation in frequency domain causes multiplication in time domain is dX (ω ) ( − jt ) x(t ) ↔ dω
(i) Fourier transform of integration in time domain is given by
−∞
And X (−ω ) = X (ω ) *
150. Convergence of Fourier Transforms Dirichlet conditions for convergence of Fourier transforms are defined as follows:
t
∫ x(τ ) dτ ↔ p X (0)δ (ω ) +
∞
(a) x(t) is absolutely integrable, that is
∫ | x(t )| dt < ∞.
−∞
(b) x(t) has a finite number of maxima and minima within any finite integer. (c) x(t) has a finite number of discontinuities within any finite interval. Each of the discontinuities is finite.
151. Properties of Fourier Transform Important properties of Fourier transforms, many of which are similar to those of Laplace transforms are briefly described as under. (a) According to linearity property, if X 1 (ω ) and X 2 (ω ) are Fourier transform of x1 (t ) and x2 (t ) , respectively, then
−∞
e
Ch wise GATE_EE_Ch1-a_Imp formula.indd 31
jω 0 t
x(t ) ↔ X (ω − ω 0 )
(j) Convolution in time domain is multiplication in frequency domain x1 (t ) ∗ x2 (t ) ↔ X 1 (ω ) ⋅ X 2 (ω ) he above equation is also called time convolution T theorem. (k) Convolution in frequency domain is multiplication in time domain this is also known as frequency convolution theorem. 1 x1 (t ) ⋅ x2 (t ) ↔ X 1 (ω ) ∗ X 2 (ω ) 2p (l) If x(t) is real and given by xe (t ) + xo (t ) where xe (t ) and xo (t ) are even and odd components of x(t) and if x(t ) ↔ X (ω ) = A(ω ) + jB(ω )
( a1 ) and ( a2 ) are arbitrary constants. (b) According to the time shifting property, a shift in time causes addition of a linear phase shift in the Fourier spectrum.
(c) Complex modulation in time domain causes frequency shifting in frequency domain, that is,
1 X (ω ) jω
a1 x1 (t ) + a2 x2 (t ) ↔ a1 X 1 (ω ) + a2 X 2 (ω )
x(t − t0 ) ↔ e − jωt0 X (ω )
1 ω amplitude scaling of X ⎛⎜ ⎞⎟ by takes place. ⎝ a⎠ |a| According to time reversal property, x( −t ) ↔ X ( −ω )
149. |X (ω ) | is known as the magnitude spectrum and ϕ (ω ) is called the phase spectrum of X(ω ). If X(t) is a real signal, we can write X ( −ω ) =
1 ⎛ω⎞ ×⎜ ⎟ |a| ⎝ a ⎠
and
X ( −ω ) = X * (ω )
then
xe (t ) ↔ A(ω )
and
xo (t ) ↔ j B(ω )
This implies that Fourier transform of an even signal is a real function of ω and Fourier transform of an odd signal is a pure imaginary function of ω .
(m) Bilateral Laplace transform of x(t) can be interpreted as the Fourier transform of x(t ) e −σ t .
11/21/2018 12:10:57 PM
32
GATE EE Chapter-wise Solved Papers
QUESTIONS Linear Algebra 0 0 ⎡ 1 ⎢100 1 0 1. The determinant of the matrix ⎢⎢ 100 200 1 ⎢ ⎣100 200 300
Common Data for Questions 6 and 7 0⎤ 0 ⎥⎥ is: 0⎥ ⎥ 1⎦
(a) 100 (c) 1
(b) 200 (d) 300 (GATE 2002: 1 Mark)
2. The eigen ⎡0 1 ⎢0 0 X = ⎢ ⎢0 0 ⎢ ⎣0 0
values of the system represented by 0 0⎤ 1 0 ⎥⎥ X are 0 1⎥ ⎥ 0 1⎦
(a) 0, 0, 0, 0 (c) 0, 0, 0, −1 3.
(b) 1, 1, 1, 1 (d) 1, 0, 0, 0 (GATE 2002: 2 Marks)
In the matrix equation Px = q, which one of the following options is a necessary condition for the existence of at least one 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 (GATE 2005: 1 Mark)
⎡ 1 0 -1⎤ 4. If R = ⎢⎢ 2 1 -1⎥⎥ , then top row of R−1 is ⎢⎣ 2 3 2⎥⎦
T
6.
5.
(c) ⎡ 1⎤ (d) ⎡ 2⎤ ⎢ -2⎥ ⎢ 5⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 3⎥⎦ ⎢⎣0 ⎥⎦ (GATE 2005: 2 Marks)
Ch wise GATE_EE_Ch1-b_Qns.indd 32
T
An orthogonal set of vectors having a span that contains P, Q and R is (a) ⎡ -6 ⎤ ⎢ -3⎥ ⎢ ⎥ ⎢⎣ 6 ⎥⎦
⎡ 4⎤ ⎢ -2⎥ ⎢ ⎥ ⎢⎣ 3⎥⎦
(c) ⎡ 6 ⎤ ⎢ 7⎥ ⎢ ⎥ ⎢⎣ -1⎥⎦
⎡ -3⎤ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ -2⎥⎦
(b) ⎡ -4 ⎤ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ 4 ⎥⎦ ⎡ 3⎤ (d) ⎡ 4 ⎤ ⎢ 3⎥ ⎢ 9⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ -4 ⎥⎦ ⎢⎣11⎥⎦
⎡ 5⎤ ⎢ 7⎥ ⎢ ⎥ ⎢⎣ -11⎥⎦ ⎡ 1⎤ ⎢31⎥ ⎢ ⎥ ⎢⎣ 3⎥⎦
⎡ 8⎤ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ -3⎥⎦ ⎡ 5⎤ ⎢ 3⎥ ⎢ ⎥ ⎢⎣ 4 ⎥⎦
(GATE 2006: 2 Marks) 7. The following vector is linearly dependent upon the solution to the previous problem: (a) ⎡ 8⎤ ⎢ 9⎥ ⎢ ⎥ ⎢⎣ 3⎥⎦
(b) ⎡ -2⎤ ⎢ -17⎥ ⎢ ⎥ ⎢⎣ 30 ⎥⎦
(c) ⎡ 4 ⎤ ⎢4⎥ ⎢ ⎥ ⎢⎣ 5⎥⎦
(d) ⎡ 13⎤ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ -3⎥⎦ (GATE 2006: 2 Marks)
8.
X = [ x1 , x2 ,..., xn ]T is an n-tuple non-zero vector. The n × n matrix V = XX T
(a) [5 6 4] (b) [5 −3 1] (c) [2 0 −1] (d) [2 −1 1/2] (GATE 2005: 2 Marks) ⎡ 3 -2 2⎤ ⎢ For the matrix A = ⎢0 -2 1⎥⎥ , one of the eigenvalues ⎢⎣0 0 1⎥⎦ is equal to −2. Which one of the following options is an eigenvector? (a) ⎡ 3⎤ (b) ⎡ -3⎤ ⎢ 2⎥ ⎢ -2⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ -1⎥⎦ ⎢⎣ 1⎥⎦
T
⎡ -10 ⎤ ⎡ -2⎤ ⎡ 2⎤ P = ⎢⎢ -1⎥⎥ , Q = ⎢⎢ -5⎥⎥ and R = ⎢⎢ -7⎥⎥ are three vectors. ⎢⎣ 3⎥⎦ ⎢⎣ 9⎥⎦ ⎢⎣ 12⎥⎦
(a) has rank zero (c) is orthogonal 9.
(b) has rank 1 (d) has rank n (GATE 2007: 1 Mark)
The linear operation L(x) is defined by the cross product L(x) = b × X, where b = [0 1 0]T and X = [x1 x2 x3]T are three-dimensional vectors. The 3 × 3 matrix M of this operation satisfies ⎡ x1 ⎤ L( x ) = M ⎢⎢ x2 ⎥⎥ ⎢⎣ x3 ⎥⎦
Then the eigenvalues of M are (a) 0, + 1, −1 (b) 1, −1, 1 (c) i, −i, 1 (d) i, −i, 0 (GATE 2007: 2 Marks)
11/21/2018 12:10:03 PM
33
Chapter 1 • Engineering Mathematics
Common Data for Questions 10 and 11: Cayley–Hamilton theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix
(a) Only the trivial solution x1 = x2 = x3 = x4 = 0 exists (b) There is no solution (c) A unique non-trivial solution exists (d) Multiple non-trivial solutions exist (GATE 2010: 2 Marks)
⎡ -3 2⎤ A= ⎢ ⎥ ⎣ -1 0 ⎦ 10. A satisfies the relation (a) A + 3I + 2 A−1 = 0 (b) A2 + 2A + 2I = 0 (c) (A + 1) (A + 2) (d) exp (A) = 0
⎡ 1 1 0⎤ 16. An eigenvector of P = ⎢⎢0 2 2⎥⎥ is ⎢⎣0 0 3⎥⎦ [1 2 1]T (a) [ -1 1 1]T (b)
(GATE 2007: 2 Marks) 11. A9 equals (a) 511 A + 510 I (c) 154 A + 155 I
[2 1 - 1]T (c) [1 - 1 2]T (d) (b) 309 A + 104 I (d) exp (9A) (GATE 2007: 2 Marks)
(GATE 2010: 2 Marks) 17. Given that ⎡ 1 0⎤ ⎡ -5 -3⎤ A= ⎢ and I = ⎢ ⎥ ⎥, ⎣ 2 0⎦ ⎣0 1⎦
12. The characteristic equation of a (3 × 3) matrix P is defined as a( λ ) = λ I - P = λ 3 + λ 2 + 2λ + 1 = 0
If I denotes identity matrix, then the inverse of matrix P will be (a ) ( P 2 + P + 2 I )
( b) ( P 2 + P + 1)
(c) ( - P 2 + P + 1)
(d ) - ( P 2 + P + 2 I ) (GATE 2008: 1 Mark)
13. If the rank of a (5 × 6) matrix Q is 4, then which one of the following statements is correct? (a) Q will have four linearly independent rows and four linearly independent columns (b) Q will have four linearly independent rows and five linearly independent columns (c) QQ T will be invertible
(a) no solution ⎡ x1 ⎤ ⎡0 ⎤ (b) only one solution ⎢ x ⎥ = ⎢0 ⎥ ⎣ 2⎦ ⎣ ⎦ (c) non-zero unique solution (d) multiple solutions (GATE 2013: 1 Mark) 19. A matrix has eigenvalues −1 and −2. The correspond⎡ 1⎤ ⎡ 1⎤ ing eigenvectors are ⎢ ⎥ and ⎢ ⎥ , respectively. The 1 ⎣ ⎦ ⎣ -2⎦ matrix is 1⎤ (a) ⎡ 1 (b) ⎡ 1 2⎤ ⎢ -1 -2⎥ ⎢ -2 -4 ⎥ ⎣ ⎦ ⎣ ⎦ 1⎤ (c) ⎡ -1 0 ⎤ (d) ⎡ 0 ⎢ 0 -2⎥ ⎢ -2 -3⎥ ⎣ ⎦ ⎣ ⎦
(GATE 2008: 1 Mark) 14. A is m × n full rank matrix with m > n and I is the identity matrix. Let matrix A′ = ( AT A) -1 AT , then which one of the following statements is true? ( AA′ ) 2 AA′ A = A (b)
(c)
AA′ A = 1 (d) AA′ A = A′ (GATE 2008: 2 Marks)
15. For the set of equations x1 + 2 x2 + x3 + 4 x4 = 2 3 x1 + 6 x2 + 3 x3 + 12 x4 = 6
the following statement is true:
Ch wise GATE_EE_Ch1-b_Qns.indd 33
(b) 19A + 30I (d) 17A + 15I (GATE 2012: 2 Marks)
⎡ 2 -2⎤ ⎡ x1 ⎤ ⎡0 ⎤ 18. The equation ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ has ⎣ 1 -1⎦ ⎣ x2 ⎦ ⎣0 ⎦
(d) Q T Q will be invertible
(a)
the value A3 is (a) 15A + 12I (c) 17A + 15I
(GATE 2013: 2 Marks) 20. Given a system of equations: x + 2y + 2z = b1
5x + y + 3z = b2 Which of the following is true regarding its solutions?
11/21/2018 12:10:07 PM
34
GATE EE Chapter-wise Solved Papers
(a) The system has a unique solution for any given b1 and b2 (b) The system will have infinitely many solutions for any given b1 and b2 (c) Whether or not a solution exists depends on the given b1 and b2 (d) The system would have no solution for any values of b1 and b2 (GATE 2014: 1 Mark) 2l. Which one of the following statements is true for all real symmetric matrices? (a) All the eigenvalues are real. (b) All the eigenvalues are positive. (c) All the eigenvalues are distinct. (d) Sum of all the eigenvalues is zero. (GATE 2014: 1 Mark) 22. Two matrices A and B are given below: ⎡ p q⎤ A= ⎢ ⎥; ⎣ r s⎦
⎡ p2 + q2 B=⎢ ⎣ pr + qs
pr + qs ⎤ ⎥ r 2 + s2 ⎦
If the rank of matrix A is N, then the rank of matrix B is (a) N/2 (b) N-1 (c) N (d) 2 N (GATE 2014: 1 Mark)
23. A system matrix is given as follows: 1 -1⎤ ⎡ 0 A = ⎢⎢ -6 -11 6 ⎥⎥ ⎢⎣ -6 -11 5 ⎥⎦
The absolute value of the ratio of the maximum eigenvalue to the minimum eigenvalue is . (GATE 2014: 2 Marks)
24. If the sum of the diagonal elements of a 2 × 2 matrix is −6, then the maximum possible value of determinant of the matrix is . (GATE 2015: 1 Mark) 25. We have a set of three linear equations in three unknowns. ‘X ≡ Y ’ means X and Y are equivalent statements and ‘X ≡ Y ’ means X and Y are not equivalent statements. P: There is a unique solution. Q: The equations are linearly independent. R: All eigenvalues of the coefficient matrix are non-zero. S: The determinant of the coefficient matrix is nonzero. Which one of the following is TRUE? (a) P ≡ Q ≡ R ≡ S (b) P ≡ R ≡ Q ≡ S (c) P ≡ Q ≡ R ≡ S
(d) P ≡ Q ≡ R ≡ S (GATE 2015: 1 Mark)
Ch wise GATE_EE_Ch1-b_Qns.indd 34
26. Consider the function f(z) = z + z* where z is a complex variable and z* denotes its complex conjugate. Which one of the following is TRUE? (a) f(z) is both continuous and analytic. (b) f(z) is continuous but not analytic. (c) f(z) is not continuous but is analytic. (d) f(z) is neither continuous nor analytic. (GATE 2016: 1 Mark) 27. Consider a 3 × 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ________. (GATE 2016: 1 Mark) 28. Let the eigenvalues of a 2 × 2 matrix A be 1, −2 with eigenvectors x1 and x2, respectively. Then the eigenvalues and eigenvectors of the matrix A2 - 3A + 4I would, respectively, be (a) 2, 14; x1, x2 (b) 2, 14; x1 + x2, x1 - x2 (c) 2, 0; x1, x2 (d) 2, 0; x1 + x2, x1 - x2 (GATE 2016: 2 Marks) 29. Let A be a 4 × 3 real matrix with rank 2. Which one of the following statements is TRUE? (a) Rank of ATA is less than 2. (b) Rank of ATA is equal to 2. (c) Rank of ATA is greater than 2. (d) Rank of ATA can be any number between 1 and 3. (GATE 2016: 2 Marks) ⎡3 1 ⎤ ⎛ x⎞ 30. Let P = ⎢ . Consider the set S of all vectors ⎜ ⎟ ⎥ ⎝ y⎠ ⎣1 3⎦ ⎛ x⎞ ⎛ a⎞ such that a2 + b2 = 1 where ⎜ ⎟ = P ⎜ ⎟ . Then S is ⎝ y⎠ ⎝ b⎠ (a) a circle of radius (b) a circle of radius
10 1 10
⎛1⎞ (c) an ellipse with major axis along ⎜ ⎟ ⎝1⎠ ⎛1⎞ (d) an ellipse with minor axis along ⎜ ⎟ ⎝1⎠ (GATE 2016: 2 Marks) 1⎤ ⎡3 ⎢2 0 2⎥ ⎢ ⎥ 31. The matrix A = ⎢0 -1 0 ⎥ has three distinct eigen⎢1 3⎥ ⎢ ⎥ 0 ⎡1 ⎤ 2⎦ ⎣2 values and one of its eigenvectors is ⎢⎢0 ⎥⎥ . Which one of ⎢⎣1 ⎥⎦ the following can be another eigenvector of A?
11/21/2018 12:10:09 PM
Chapter 1 • Engineering Mathematics
⎡ 0⎤ (a) ⎢⎢ 0 ⎥⎥ (b) ⎢⎣ -1⎥⎦
⎡ -1⎤ ⎢ 0 ⎥ (c) ⎢ ⎥ ⎢⎣ 0 ⎥⎦
⎡1 ⎤ ⎢0 ⎥ (d) ⎢ ⎥ ⎢⎣ -1⎥⎦
⎡ 1⎤ ⎢ -1⎥ ⎢ ⎥ ⎢⎣ 1⎥⎦
(c) (d)
(GATE 2017: 1 Mark) 32. The eigenvalues of the matrix given below are ⎡0 1 0 ⎤ ⎢0 0 1 ⎥ ⎢ ⎥ ⎢⎣0 -3 -4 ⎥⎦ (a) (0, −1, −3) (c) (0, 2, 3)
(b) (0, −2, −3) (d) (0, 1, 3) (GATE 2017: 2 Marks)
33. Consider a non-singular 2 ×´ 2 square matrix A. If trace (A) = 4 and trace (A2) = 5, the determinant of the matrix A is (up to 1 decimal place). (GATE 2018: 1 Mark) ⎡ 1 0 -1⎤ ⎢ ⎥ 34. Let A = ⎢ -1 2 0 ⎥ and B = A3 - A2 - 4A + 5I, where ⎢⎣ 0 0 -2⎥⎦ I is the 3 ×´ 3 identify matrix. The determinant of B is (up to 1 decimal place). (GATE 2018: 2 Marks)
Calculus 35. If S =
∞
∫1
x −3 dx , then S has the value
(a) −1/3 (c) 1/2
(b) 1/4 (d) 1 (GATE 2005: 1 Mark)
x2 y2 + , magnitude of the 2 3 gradient at the point (1, 3) is
36. For the scalar field u =
13 9 (b) 9 2 9 5 (d) 2
(a) (c)
(GATE 2005: 2 Marks) H
37. The expression V = ∫ π R 2 (1 - h /H ) 2 dh for the vol0
ume of a cone is equal to (a)
∫
R
(b)
∫
R
0
0
Ch wise GATE_EE_Ch1-b_Qns.indd 35
π R 2 (1 - h /H ) 2 dr π R 2 (1 - h /H ) 2 dh
∫
H
∫
R
0
0
35
2π rH (1 - r /R)dh 2
r⎞ ⎛ π rH ⎜1 - ⎟ dr ⎝ R⎠ (GATE 2006: 2 Marks)
38. Let x and y be two vectors in a three-dimensional space and denotes their dot product. Then the ⎡ < x, x > < x, y > ⎤ determinant of ⎢ ⎥ is ⎣< y, x > < y, y > ⎦ (a) (b) (c) (d)
zero when x and y are linearly independent positive when x and y are linearly independent non-zero for all non-zero x and y zero only when either x or y is zero (GATE 2007: 2 Marks)
39. 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 (a)
y =1 x = y
∫ ∫
f ( x, y )dx dy
y = 0 x = y2
(b)
y =1 x =1
∫ ∫
f ( x, y )dx dy
y = x2 x = y2
(c)
y =1 x =1
∫ ∫
f ( x, y )dx dy
y=0 x=0
(d)
y= x x= y
∫ ∫
y=0
f ( x, y )dx dy
x=0
(GATE 2009: 2 Marks) 1
40. The value of the quantity P, where P = ∫ xe x dx, is equal 0 to (a) 0 (b) 1 (c) e (d) 1/e (GATE 2010: 1 Mark) 41. Divergence of the three-dimensional radial vector field r is (a) 3 (b) 1/r
(
)
3 i + j + k (c) i + j + k (d) (GATE 2010: 1 Mark) 42. The function f (x) = 2x − x + 3 has (a) a maxima at x = 1 and a minima at x = 5 (b) a maxima at x = 1 and a minima at x = −5 (c) only a maxima at x = 1 (d) only a minima at x = 1 (GATE 2011: 2 Marks) 2
11/21/2018 12:10:12 PM
36
GATE EE Chapter-wise Solved Papers
43. The two vectors [1, 1, 1] and [1, a, a2], where a = ⎛ 1 3⎞ are a = ⎜- + j 2 ⎟⎠ ⎝ 2 (a) orthonormal (c) parallel
(b) orthogonal (d) collinear (GATE 2011: 2 Marks)
51. The minimum value of the function f(x) = x3 - 3x2 - 24x + 100 in the interval [-3, 3] is (a) 20 (b) 28 (c) 16 (d) 32 (GATE 2014: 2 Marks) 1 52. Consider a function f = 2 rˆ, where r is the distance r from the origin and rˆ is the unit vector in the radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is (a) 0 (b) 2p (c) 4p (d) Rp (GATE 2015: 1 Mark) rˆ
iˆ
44. The direction of vector A is radically outward from the origin, with A = kr n where r 2 = x 2 + y 2 + z 2 and k is a constant. The value of n for which ∇ ⋅ A = 0 is (a) −2 (b) 2 (c) 1 (d) 0 (GATE 2012: 2 Marks)
(a) 4 xy ax + 6 yz a y + 8 zx az
53. If a continuous function f (x) does not have a root in the interval [a, b], then which one of the following statements is TRUE? (a) f(a)⋅f(b) = 0 (b) f(a)⋅f(b) < 0 (c) f(a)⋅f(b) > 0 (d) f(a)/f(b) ≤ 0 (GATE 2015: 1 Mark)
(b) 4 ax + 6 a y + 8az
54. The maximum value of ‘a’ such that the matrix
45. The curl of the gradient of the scalar field defined by V = 2 x 2 y + 3 y 2 z + 4 z 2 x is
⎡ -3 0 -2⎤ ⎢ 1 -1 0 ⎥ ⎢ ⎥ ⎢⎣ 0 a -2⎥⎦
(c) ( 4 xy + 4 z 2 )ax + ( 2 x 2 + 6 yz )a y + (3 y 2 + 8 zx )az (d) 0 (GATE 2013: 1 Mark) 46. A vector field is given, F = y 2 xa x - yza y - x 2 a z , the line integral ∫ F ⋅ dl evaluated along a segment on the x-axis from x = 1 to x = 2 is
(a) −2.33 (c) 2.33
(b) 0 (d) 7 (GATE 2013: 1 Mark)
47. Let f (x) = xe-x. The maximum value of the function in the interval (0, ∞) is (a) e-1 (b) e (c) 1 − e-1 (d) 1 + e-1 (GATE 2014: 1 Mark) 48. Minimum of the real-valued function f (x) = (x − l)2/3 occurs at x equal to (a) -∞
(b) 0 (c) 1 (d) ∞ (GATE 2014: 1 Mark)
49. Let ∇ · (f v) = x2y + y2z + z2x, where f and v are scalar and vector fields, respectively. If v = yiˆ + zjˆ + xkˆ, then v · ∇f is (a) x2y + y2z + z2x (b) 2xy + 2yz + 2zx (c) x + y + z (d) 0 (GATE 2014: 1 Mark) 50. The line integral of function F = yzi, in the counterclockwise direction, along the circle x2 + y2 = 1 at z = 1 is (a) -2p
Ch wise GATE_EE_Ch1-b_Qns.indd 36
(b) -p
(c) p (d) 2p (GATE 2014: 2 Marks)
has three linearly independent real eigenvectors is (a) (c)
2 3 3
1 (b) 3 3
1+ 3 (d) 3 3 3 3 (GATE 2015: 1 Mark)
1+ 2 3
55. Match the following: P. Stokes’ theorem Q. Gauss’s theorem R. Divergence theorem S. Cauchy’s integral theorem 1.
∫∫ D ⋅ ds = Q
2.
∫ f ( z )dz = 0
3.
∫∫∫ (∇ ⋅ A)dv = ∫∫ A ⋅ ds
4.
∫∫ (∇ × A) ⋅ ds = ∫ A ⋅ dl P
Q
R
S
(a)
2
1
4
3
(b)
4
1
3
2
(c)
4
3
1
2
(d)
3
4
2
1
(GATE 2015: 1 Mark)
11/21/2018 12:10:15 PM
37
Chapter 1 • Engineering Mathematics
56. The volume enclosed by the surface f (x, y) = ex over the triangle bounded by the lines x = y; x = 0; y = 1 in the xy plane is . (GATE 2015: 1 Mark) 57. The maximum value attained by the function f(x) = x(x - 1)(x - 2) in the interval [1, 2] is ________. (GATE 2016: 1 Mark) 58. The value of the line integral
∫ (2 xy dx + 2 x
2
2
ydy + dz )
63. Let x and y be integers satisfying the following equations. 2x2 + y2 = 34 x + 2y = 11 The value of (x + y) is _______. (GATE 2017: 1 Mark) 64. Let y2 - 2y + 1 = x and x + y = 5 . The value of x + y equals _______. (Give the answer up to three decimal places.) (GATE 2017: 1 Mark)
along a path joining the origin (0, 0, 0) and the point 65. A function f (x) is defined as (1, 1, 1) is ⎧ ex x 0 Consider the composition of f and g, i.e. (fog)(x) = f(g(x)). The number of discontinuities in (fog)(x) present in the interval ( -∞, 0) is (a) 0 (b) 1 (c) 2 (d) 4 (GATE 2017: 2 Marks)
11/21/2018 12:10:17 PM
38
GATE EE Chapter-wise Solved Papers
68. Let f be a real-valued function of a real variable defined as f(x) = x2 for x ³ 0, and f(x) = -x2 for x < 0. Which one of the following statements is true? (a) f(x) is discontinuous at x = 0. (b) f(x) is continuous but not differentiable at x = 0. (c) f(x) is differentiable but its first derivative is not continuous at x = 0. (d) f(x) is differentiable but its first derivative is not differentiable at x = 0. (GATE 2018: 1 Mark) 69. The value of the directional derivative of the function ϕ ( x, y, z ) = xy 2 + yz 2 + zx 2 at the point (2, -1, 1) in the direction of the vector p = iˆ + 2 ˆj + 2kˆ is (a) 1 (c) 0.93
(b) 0.95 (d) 0.9 (GATE 2018: 1 Mark)
x(t ) = 2e -2t - e -4 t (a) x(t ) = 2e -6 t - e -2t (b) (c) x(t ) = 2e -6 t + e -4 t (d) x(t ) = 2e -2t + e -4 t (GATE 2010: 2 Marks) 75. With K as a constant, the possible solution for the dy first-order differential equation = e -3 x is dx 1 1 (a) - e -3 x + K (b) - e3 x + K 3 3 (c) -3e -3 x + K (d) -3e - x + K (GATE 2011: 1 Mark) 76. The solution for the differential equation d2x = -9 x, with initial conditions x (0) = 1 and dt 2 dx dt
z +1 70. The value of the integral ∫C z 2 - 4 dz in counter clockwise direction around a circle C of radius 1 with center at the point z = -2 is
πi 2 πi (c) - 2
(a)
(b) sin 3t +
71. Let f be a real-valued function of a real variable defined as f(x) = x - [x], where [x] denotes the largest integer less
1 sin 3t + cos 3t 3 (d) cos 3t + t (GATE 2014: 1 Mark) d2 y dy + x - y = 0. 2 dx dx Which of the following is a solution to this differential equation for x > 0?
77. Consider the differential equation x 2
1.25
to 2 decimal places).
f ( x )dx is
1 2 cos 3t + 3 3
(c)
(d) -2pi (GATE 2018: 1 Mark)
∫
= 1, is
(a) t2 + t + 1
(b) 2pi
than or equal to x. The value of
t =0
(up
0.25
(a) ex
(GATE 2018: 1 Mark) 72. Let f(x) = 3x3 - 7x2 + 5x + 6. The maximum value of f(x) over the interval [0, 2] is (up to 1 decimal place). (GATE 2018: 2 Marks)
(b) x2
(c) 1/x
(d) ln x
(GATE 2014: 1 Mark) 78. A particle, starting from origin at t = 0 s, is traveling along x-axis with velocity v=
Differential Equations
π ⎛π ⎞ cos ⎜ t ⎟ m/s ⎝2 ⎠ 2
(a) x(t ) = x0 = x0 e -3t (b) x(t ) = x0 e -3
At t = 3 s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is . (GATE 2014: 1 Mark)
(c) x(t ) = x0 e -1/ 3 (d) x(t ) = x0 e -1
79. A solution of the ordinary differential equation
73. The solution of the first-order differential equation x ′(t ) = -3 x(t ), x(0) = x0 is
(GATE 2005: 1 Mark) d2x dx 74. For the differential equation + 6 + 8 x = 0 with 2 dt dt dx initial conditions x(0) = 1 and = 0 , the solution is dt t = 0
Ch wise GATE_EE_Ch1-b_Qns.indd 38
d2 y dy + 5 + 6y = 0 dt dt 2
is such that y(0) = 2 and
dy d2 y dy (0) is + 5 + 6 y = 0 . The value of 2 dt dt dt
.
(GATE 2015: 2 Marks)
11/21/2018 12:10:21 PM
Chapter 1 • Engineering Mathematics
(a) x1 f < x2 f < ∞ (b) x2 f < x1 f < ∞
di - 0.2i = 0 is applicable over dt −10 < t < 10. If i(4) = 10, then i(−5) is . (GATE 2015: 2 Marks)
80. A differential equation
(c) x1 f = x2 f < ∞ (d) x1 f = x2 f = ∞ (GATE 2018: 2 Marks)
81. A function y(t), such that y(0) = 1 and y(1) = 3e-1, is a solution of the differential equation d2 y dy + 2 + y = 0. Then y(2) is 2 dt dt (a) 5e-1 (c) 7e-1
Complex Variables 87. The value of z - i /2 = 1 , is
(b) 5e-2 (d) 7e-2 (GATE 2016: 1 Mark)
82. The solution of the differential equation, for t > 0, y ′′(t ) + 2 y ′(t ) + y(t ) = 0 with initial conditions y(0) = 0 and y ′(0) = 1, is [u(t) denotes the unit step function], (a) te-t u(t) (b) (e-t - te-t)u(t) -t -t (c) (-e + te )u(t) (d) e-tu(t) (GATE 2016: 1 Mark)
39
dz
∫ (1 + z ) , 2
where C is the contour
C
(a) 2pi (b) p (c) tan -1 z (d) π i tan -1 z (GATE 2007: 2 Marks) 88. A point z has been plotted in the complex plane, as shown in the figure below: Im
Unit circle z
∞
Re
83. Let S = ∑ nα n where ½a½ < 1. The value of a in the n= 0
range 0 < a < 1, such that S = 2a is ________. (GATE 2016: 2 Marks) 84. Let y(x) be the solution of the differential equation d2 y dy - 4 + 4 y = 0 with initial conditions y(0) = 0 and dx dx 2 dy = 1. Then the value of y(1) is ________. dx x = 0 (GATE 2016: 2 Marks)
The plot of the complex number y = (a)
Im
1 is? z
Unit circle
Re y
85. Consider the differential equation dy (t 2 - 81) + 5t y = sin(t ) with y(1) = 2p. There exists dt a unique solution for this differential equation when t belongs to the interval (a) (−2, 2) (b) (−10, 10) (c) (−10, 2) (d) (0, 10) (GATE 2017: 2 Marks)
(b)
Im
Unit circle
Re y
86. Consider a system governed by the following equations dx1 (t ) = x2 (t ) - x1 (t ) dt dx2 (t ) = x1 (t ) - x2 (t ) dt
The initial conditions are such that x1 (0) < x2 (0) < ∞.
(c)
Im
Unit circle
y Re
Let x1 f = lim x1 (t ) and x2 f = lim x2 (t ). Which one of the t →∞
t →∞
following is true?
Ch wise GATE_EE_Ch1-b_Qns.indd 39
11/21/2018 12:10:24 PM
40
GATE EE Chapter-wise Solved Papers
(d)
Im
(d) If f(z) is differentiable at z0, then so are its real and imaginary parts. (GATE 2015: 1 Mark)
Unit circle
94. The value of the integral Re C
(GATE 2011: 1 Mark) 89. Square roots of −i, where i = -1 , are (a) i, −i ⎛ π⎞ ⎛ π⎞ ⎛ 3π ⎞ ⎛ 3π ⎞ (b) cos ⎜ - ⎟ + i sin ⎜ - ⎟ , cos ⎜ ⎟ + i sin ⎜ ⎟ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎛π⎞ ⎛ 3π ⎞ ⎛ 3π ⎞ ⎛π⎞ (c) cos ⎜ ⎟ + i sin ⎜ ⎟ , cos ⎜ ⎟ + i sin ⎜ ⎟ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎛ 3π ⎞ ⎛ 3π ⎞ ⎛ 3π ⎞ ⎛ 3π ⎞ (d) cos ⎜ ⎟ + i sin ⎜ - ⎟ , cos ⎜ - ⎟ + i sin ⎜ ⎟ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ 90.
∫
(GATE 2013: 1 Mark) z2 - 4 dz evaluated anticlockwise around the circle z2 + 4
z - 1 = 2 , where i = -1 , is (a) −4p (b) 0 (c) 2 + p (d) 2 + 2i (GATE 2013: 2 Marks) 91. All the values of the multi-valued complex function 1i, where i = -1 , are (a) purely imaginary (b) real and non-negative (c) on the unit circle (d) equal in real and imaginary parts (GATE 2014: 1 Mark) 2
z , in z2 -1 the counterclockwise direction, around z - 1 = 1, is
92. Integration of the complex function f ( z ) =
(a) -pi
(b) 0
(c) pi (d) 2pi (GATE 2014: 2 Marks)
93. Given f(z) = g(z) + h(z), where f, g, h are complex valued functions of a complex variable z. Which one of the following statements is TRUE? (a) If f(z) is differentiable at z0, then g(z) and h(z) are also differentiable at z0. (b) If g(z) and h(z) are differentiable at z0, then f(z) is also differentiable at z0. (c) If f(z) is continuous at z0, then it is differentiable at z0.
Ch wise GATE_EE_Ch1-b_Qns.indd 40
2z + 5 dz 1⎞ 2 ⎜⎝ z - ⎟⎠ ( z - 4 z + 5) 2
∫ ⎛
y
over the contour |z| = 1, taken in the anti-clockwise direction, would be 24π i 48π i (b) (a) 13 13 24 12 (c) (d) 13 13 (GATE 2016: 1 Mark)
95. A 3 × 3 matrix P is such that, P3 = P. Then the e igenvalues of P are (a) 1, 1, -1 (b) 1, 0.5 + j0.866, 0.5 - j0.866 (c) 1, -0.5 + j0.866, -0.5 - j0.866 (d) 0, 1, -1 (GATE 2016: 1 Mark) 96. For a complex number z, lim z →i
(a) −2i
(b) −i
z2 + 1 is z 3 + 2 z - i ( z 2 + 2) (c) i (d) 2i (GATE 2017: 1 Mark)
97. If C is a circle |z| = 4 and f ( z ) =
∫ f ( z )dz
is
z2 , then ( z 2 - 3 z + 2) 2
C
(a) 1 (b) 0 (c) -1 (d) -2 (GATE 2018: 2 Marks) 98. As shown in the figure, C is the arc from the point (3, 0) to the point (0, 3) on the circle x2 + y2 = 9. The value of the integral ∫ ( y 2 + 2 yx )dx + ( 2 xy + x 2 )dy is
(up to
C
2 decimal places). y
(0, 3) C
x (3, 0) (GATE 2018: 2 Marks)
11/21/2018 12:10:26 PM
Chapter 1 • Engineering Mathematics
Probability and Statistics 99. 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
41
104. Consider a dice with the property that the probability of a face with n dots showing up is proportional to n. The probability of the face with three dots showing . up is (GATE 2014: 1 Mark)
105. Lifetime of an electric bulb is a random variable with (b) Probability ( P ∪ Q ) ≥ Probability ( P ) + ProbabilProbability (Q ) density f (x) = kx2, where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 ity (Q) years, respectively, then the value of k is . (c) If P and Q are mutually exclusive, then they must (GATE 2014: 1 Mark) be independent (d) Probability ( P ∩ Q ) ≤ Probability ( P ) (GATE 2005: 1 Mark)
100. A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is (a) 1/8 (b) 1/2 (c) 3/8 (d) 3/4 (GATE 2005: 2 Marks) 101. Two fair dice are rolled and the sum r of the numbers turned up is considered: (a) P(r > 6) = (1/6) (b) P(r/3 is an integer) = (5/6) (c) P(r = 8 | r/4 is an integer) = (5/9) (d) P(r = 6 | r/5 is an integer) = (1/18) (GATE 2006: 2 Marks) 102. A loaded dice has the 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 (GATE 2007: 2 Marks)
103. A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is (a) 2/315 (b) 1/630 (c) 1/1260 (d) 1/2520 (GATE 2010: 2 Marks)
Ch wise GATE_EE_Ch1-b_Qns.indd 41
106. A fair coin is tossed n times. The probability that the difference between the number of heads and tails is (n-3) is (a) 2-n (b) 0 (c) nCn-32-n (d) 2-n + 3 (GATE 2014: 2 Marks) 107. Let X be a random variable with probability density function ⎧0.2, ⎪ f ( x ) = ⎨0.1, ⎪0, ⎩
for
x ≤1
for 1 < x ≤ 4 otherwise
. The probability P(0.5 < x < 5) is (GATE 2014: 2 Marks)
108. The mean thickness and variance of silicon steel laminations are 0.2 mm and 0.02, respectively. The varnish insulation is applied on both die sides of the laminations. The mean thickness of one side insulation and its variance are 0.1 mm and 0.01, respectively. If the transformer core is made using 100 such varnish coated laminations, the mean thickness and variance of the core, respectively, are (a) 30 mm and 0.22 (b) 30 mm and 2.44 (c) 40 mm and 2.44 (d) 40 mm and 0.24 (GATE 2014: 2 Marks) 109. A random variable X has the probability density function f(x) as given below: ⎧a + bx for 0 < x < 1 f ( x) = ⎨ otherwise ⎩ 0 If the expected value E[X] = 2/3, then Pr[X < 0.5] is . (GATE 2015: 1 Mark) 110. Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that A wins the game is (a) 5/11 (b) 1/2 (c) 7/13 (d) 6/11 (GATE 2015: 1 Mark)
11/21/2018 12:10:27 PM
42
GATE EE Chapter-wise Solved Papers
111. Candidates were asked to come to an interview with 3 pens each. Black, blue, green and red were the permitted pen colours that the candidate could bring. The probability that a candidate comes with all 3 pens having the same colour is ________. (GATE 2016: 2 Marks) 112. Let the probability density function of a random variable, X, be given as 3 -3 x e u( x ) + ae 4 x u( - x ) 2 where u(x) is the unit step function. Then the value of ‘a’ and Prob{X ≤ 0}, respectively, are f X ( x) =
(a) 2,
1 1 (b) 4, 2 2
1 1 (c) 2, (d) 4, 4 4 (GATE 2016: 2 Marks) 113. An urn contains 5 red balls and 5 black balls. In the first draw, one ball is picked at random and discarded without noticing its colour. The probability to get a red ball in the second draw is (a)
5 1 4 6 (b) (c) (d) 9 2 9 9 (GATE 2017: 1 Mark)
114. Two resistors with nominal resistance values R1 and R2 have additive uncertainties ΔR1 and ΔR2, respectively. When these resistances are connected in parallel, the standard deviation of the error in the equivalent resistance R is 2
2
2
2
⎫ ⎧ ∂R ⎫ ⎧ ∂R (a) ± ⎨ ΔR1 ⎬ + ⎨ ΔR2 ⎬ ⎩ ∂R1 ⎭ ⎩ ∂R2 ⎭ ⎫ ⎧ ∂R ⎫ ⎧ ∂R (b) ± ⎨ ΔR1 ⎬ + ⎨ ΔR2 ⎬ ⎩ ∂R2 ⎭ ⎩ ∂R1 ⎭ 2
2
⎧ ∂R ⎫ ⎧ ∂R ⎫ (c) ± ⎨ ⎬ ΔR2 + ⎨ ⎬ ΔR1 ⎩ ∂R1 ⎭ ⎩ ∂R2 ⎭ 2
2
⎧ ∂R ⎫ ⎧ ∂R ⎫ (d) ± ⎨ ⎬ ΔR1 + ⎨ ⎬ ΔR2 R ∂ ⎩ 1⎭ ⎩ ∂R2 ⎭ (GATE 2017: 1 Mark) 115. A person decides to toss a fair coin repeatedly until he gets a head. He will make at most 3 tosses. Let the random variable Y denote the number of heads. The value of var(Y ), where var {•} denotes the variance, equals
Ch wise GATE_EE_Ch1-b_Qns.indd 42
(a)
7 49 7 105 (b) (c) (d) 8 64 64 64 (GATE 2017: 2 Marks)
Numerical Methods 116. The differential equation ( dx /dt ) = ⎡⎣(1 - x ) /τ ⎤⎦ is discretised using Euler’s numerical integration method with a time step ΔT > 0 . What is the maximum permissible value of ΔT to ensure stability of the solution of the corresponding discrete time equation? (a) 1 (b) t/2 (c) t (d) 2t (GATE 2007: 2 Marks) 117. Equation e x - 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 (GATE 2008: 2 Marks) 118. Roots of the algebraic equation x 3 + x 2 + x + 1 = 0 are (a) (c)
(+1, + j, - j ) (b) (+1, -1, +1) (0, 0, 0) (d) ( -1, + j, - j ) (GATE 20011: 1 Mark)
⎡2 1 ⎤ 119. The matrix [ A] = ⎢ ⎥ is decomposed into a product ⎣ 4 -1⎦ of a lower triangular matrix x [ L] and an upper triangular matrix [U ] . The properly decomposed [ L] and [U ] matrices, respectively, are ⎡1 0 ⎤ ⎡1 1 ⎤ and ⎢ (a) ⎢ ⎥ ⎥ ⎣ 4 -1⎦ ⎣0 -2⎦ ⎡1 1⎤ ⎡2 0 ⎤ (b) ⎢ ⎥ and ⎢0 1⎥ 4 1 ⎣ ⎦ ⎣ ⎦ ⎡2 1 ⎤ ⎡1 0⎤ (c) ⎢ and ⎢ ⎥ ⎥ ⎣0 -1⎦ ⎣4 1⎦ ⎡1 0.5⎤ ⎡2 0 ⎤ (d) ⎢ and ⎢ ⎥ ⎥ ⎣ 4 -3⎦ ⎣0 1 ⎦ (GATE 2011: 2 Marks) 120. Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton– Raphson iterative method:
11/21/2018 12:10:31 PM
43
Chapter 1 • Engineering Mathematics
Transform Theory
Equation (i) 10 x2 sin x1 - 0.8 = 0
Equation (ii) 10 x22 - 10 x2 cos x1 - 0.6 = 0
Assuming the initial values x1 = 0.0 and x2 = 1.0, the J acobian matrix is
Common Data Questions 125 and 126: Given f (t ) and g (t ) as shown below:
⎡10 0 ⎤ ⎡10 -0.8⎤ (a) ⎢ (b) ⎢ 0 10 ⎥ ⎥ ⎦ ⎣ ⎣ 0 -0.6 ⎦
f(t)
g(t)
1
1
⎡ 0 -0.8⎤ ⎡10 0 ⎤ (c) ⎢ (d) ⎥ ⎢10 -10 ⎥ ⎣10 -0.6 ⎦ ⎣ ⎦ (GATE 2011: 2 Marks) 121. When the Newton–Raphson method is applied to solve the equation f ( x ) = x 3 + 2 x - 1 = 0, the solution at the end of the first iteration with the initial guess values as x0 = 1.2 is (a) −0.82 (b) 0.49 (c) 0.705 (d) 1.69 (GATE 2013: 2 Marks)
t 0
(a) g (t ) = f ( 2t - 3)
(c) g (t ) = f ( 2t - 3/ 2)
(b) (c) (d)
0
0
4
1
0
0
4
1
0
0
4
2
0
0
(a)
g (t ) = f (t /2 - 3)
(d)
g (t ) = f (t / 2 - 3/ 2)
(c)
(
)
(
1 3s 1 -5 s e - e 5 s (b) e - e -3 s s s
)
e -3 s 1 5s e - e3s 1 - e -2 s (d) s s (GATE 2010: 2 Marks)
(
)
(
)
3s + 5 be the Laplace transform of a s + 10 s + 21 signal x(t). Then, x(0+) is (a) 0 (b) 3 (c) 5 (d) 21 (GATE 2014: 1 Mark)
127. Let X ( s ) =
( 2 u du) dv ∫ (∫ 2 u du ) dv ∫ (∫ u du ) dv ∫ (∫ u du ) dv ∫ ∫
(b)
126. The Laplace transform of g (t ) is
y ⎛ 2x - y ⎞ and v = . dy, we make the substitution u = ⎜ ⎟ ⎝ 2 ⎠ 2 The integral will reduce to (a)
5
(GATE 2010: 2 Marks)
⎛ ( y / 2 ) +1 ⎛ 2 x - y ⎞ ⎞ 122. To evaluate the double integral ∫ ⎜ ∫ ⎜⎝ ⎟ dx dy, 0 ⎝ y/2 2 ⎠ ⎟⎠
2
3
0
125. g (t ) can be expressed as
8
4
t
1
2
128. A function f(t) is shown in the figure. f(t) 1/2
(GATE 2014: 2 Marks)
T/2 -T/2
123. The function f (x) = ex - 1 is to be solved using N ewton– Raphson method. If the initial value of xo is taken as 1.0, then the absolute error observed at 2nd iteration is . (GATE 2014: 2 Marks) 124. Only one of the real roots of f (x) = x6 - x - 1 lies in the interval 1 ≤ x ≤ 2 and bisection method is used to find its value. For achieving an accuracy of 0.001, the required minimum number of iterations is _______. (Give the answer up to two decimal places.) (GATE 2017: 2 Marks)
Ch wise GATE_EE_Ch1-b_Qns.indd 43
0
t
-1/2
The Fourier transform F(w) of f (t) is (a) (b) (c) (d)
real and even function of w real and odd function of w imaginary and odd function of w imaginary and even function of w (GATE 2014: 1 Mark)
11/21/2018 12:10:34 PM
44
GATE EE Chapter-wise Solved Papers
129. Let g: [0, ∞) → [0, ∞) be a function defined by g(x) = x − [x], where [x] represents the integer part of x. (That is, it is the largest integer which is less than or equal to x). The value of the constant term in the Fourier series expansion of g(x) is . (GATE 2014: 2 Marks) 130. A differentiable non-constant even function x(t) has a derivative y(t), and their respective Fourier transforms are X(w) and Y(w). Which of the following statements is TRUE? (a) X(w) and Y(w) are both real (b) X(w) is real and is Y(w) imaginary (c) X(w) and Y(w) are both imaginary (d) X(w) is imaginary and Y(w) is real (GATE 2014: 2 Marks)
134. The Laplace transform of f (t) = e2t sin (5t) u(t) is (a)
5 5 (b) 2 s - 4 s + 29 s +5
(c)
s-2 5 (d) s+5 s 2 + 4 s + 29
2
(GATE 2016: 1 Mark) 135. Let f (x) be a real, periodic function satisfying f (-x) = -f (x). The general form of its Fourier series representation would be
131. The Laplace transform of f (t ) = 2 t /π is s−3/2. The Laplace transform of g (t ) = 1/π t is
(a)
f ( x ) = a0 + ∑ k =1 ak cos( kx )
(b)
f ( x ) = ∑ k =1 bk sin( kx )
(c)
f ( x ) = a0 + ∑ k =1 a2 k cos( kx )
(d)
f ( x ) = ∑ k =1 a2 k +1 sin( 2k + 1) x
∞
∞
∞
∞
(GATE 2016: 1 Mark) (a) 3s−5/2/2 (b) s−1/2 136. Suppose x1(t) and x2(t) have the Fourier transforms as (c) s1/2 (d) s3/2 shown below. (GATE 2015: 1 Mark)
X1(jw)
132. Consider a signal defined by ⎧⎪e j 10 t x (t ) = ⎨ ⎩⎪0
for t ≤ 1 for t > 1
1 0.5
Its Fourier transform is
0.3
(a)
2 sin(ω - 10) (b) 2e ω - 10
(c)
2 sin ω j10 ω 2 sin ω ω (d) e ω - 10
sin( ω -10 ) j10 ω -10
−1
0
2 w
1
X2(jw)
(GATE 2015: 2 Marks)
1 133. The z-transform of a sequence x[n] is given as X(z) = 2z + 4 - 4/z + 3/z2. If y[n] is the first difference of x[n], then Y(x) is given by
0.5 0.3 −2
8 7 3 (a) 2 z + 2 - + 2 - 2 z z z
6 1 3 (b) -2 z + 2 - + 2 - 3 z z z (c) -2 z - 2 + (d) 4 z - 2 -
8 7 3 + + z z 2 z3
0
1 w
Which one of the following statements is TRUE? (a) x1(t) and x2(t) are complex and x1(t)x2(t) is also complex with non-zero imaginary part. (b) x1(t) and x2(t) are real and x1(t)x2(t) is also real. (c) x1(t) and x2(t) are complex but x1(t)x2(t) is real.
8 1 3 - + z z 2 z3
(d) x1(t) and x2(t) are imaginary but x1(t)x2(t) is real. (GATE 2015: 2 Marks)
Ch wise GATE_EE_Ch1-b_Qns.indd 44
−1
(GATE 2016: 2 Marks)
11/21/2018 12:10:37 PM
45
Chapter 1 • Engineering Mathematics
ANSWER KEY Linear Algebra 1. (c)
2. (d)
3. (a)
4. (b)
5. (d) 6. (a)
7. (b)
8. (b)
9. (d)
10. (a)
11. (a)
12. (d)
13. (a)
14. (a)
15. (d) 16. (b)
17. (b)
18. (d)
19. (d)
20. (b)
21. (a)
22. (c)
23. (0.33) 24. (9)
25. (a) 26. (b)
27. (3)
28. (a)
29. (b)
30. (d)
31. (c)
32. (a)
33. (5.5)
34. (1)
Calculus 35. (c)
36. (c)
37. (d)
38. (b)
39. (a) 40. (b)
41. (a)
42. (c)
43. (b)
44. (a)
45. (d)
46. (b)
47. (a)
48. (c)
49. (a) 50. (b)
51. (b)
52. (a)
53. (c)
54. (b)
55. (b)
56. (e − 2) 57. (0)
58. (b)
59. (d) 60. (4.41) 61. (0.99) 62. (40)
63. (7)
64. (5.732)
65. (a)
66. (b)
68. (d)
69. (a) 70. (a)
67. (a)
71. (0.5) 72. (12)
Differential Equations 73. (a)
74. (b)
83. (0.29) 84. (7.38)
75. (a)
76. (c)
85. (a)
86. (c)
77. (c)
78. (2)
79. (–3)
80. (1.65) 81. (b)
82. (a)
94. (c)
96. (d)
Complex Variables 87. (b)
88. (d)
97. (b)
98. (0)
89. (b)
90. (a)
91. (b) 92. (c)
93. (b)
95. (d)
Probability and Statistics 99. (d)
100. (b)
109. (0.25) 110. (d)
101. (c)
102. (c)
111. (0.167) 112. (a)
103. (c) 104. (0.14) 105. (0.43) 106. (b) 113. (a) 114. (a)
107. (0.4) 108. (*)
115. (c)
Numerical Methods 116. (d)
117. (a)
118. (d)
119. (d)
120. (b) 121. (c)
122. (b)
123. (0.06) 124. (10)
Transform Theory 125. (d)
126. (c)
135. (b)
136. (c)
127. (b)
128. (c)
129. (0.5) 130. (b)
131. (b)
132. (a)
133. (a)
134. (a)
* None of the options is correct
Ch wise GATE_EE_Ch1-b_Qns.indd 45
11/21/2018 12:10:37 PM
46
GATE EE Chapter-wise Solved Papers
ANSWERS WITH EXPLANATION Linear Algebra
Corresponding to eigenvalue l = −2, let us find the eigenvector using the following equation:
1. (c) The given matrix is a lower triangular matrix. Since for any triangular matrix, the determinant |A| = product of its principle diagonal elements, hence, |A| = 1 × 1 × 1 × 1 = 1
[ A − λ I ] X =0
2. (d) For the given 4 × 4 matrix [A]. Eigen values are calculated by A − λI = 0
1 −λ 0 0
0 1 −λ 0
0 0 =0 1 −λ
which gives the equations 5 x − 2 x2 + 2 x3 = 0 (1) 1 x3 = 0 (2)
It gives λ =1, 0, 0, 0 3. (a) Rank [Pq] = Rank [P] is necessary for existence of at least one solution to Px = q. 4. (b) We have
3x3 = 0
(3)
Putting x2 = k and x3 = 0 in Eq. (1), we get 5 x1 − 2( k ) + 2 × 0 = 0
⎡ 1 0 −1⎤ R = ⎢⎢ 2 1 −1⎥⎥ ⎢⎣ 2 3 2⎥⎦ adj( R) R −1 = | R|
⇒ x1 = 2/5k Therefore, eigenvectors are of the form
1 0 −1 ⇒ | R| = 2 1 −1 = 1( 2 + 3) − 0 ( 4 + 2) − 1(6 − 2) 2 3 2 = 5−4 =1 As we need only the top row of R−1, we need to find only the first row of adj (R). adj (1, 1) = +
1 −1 = 2+3= 5 3 2
adj (1, 2) = +
0 −1 = −3 1 −1
adj (1, 3) = +
0 −1 = +11 1 −1
⎡ x1 ⎤ ⎡ 2 / 5k ⎤ ⎢x ⎥ = ⎢ k⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ x3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⇒ x1 : x2 : x3 = 2/5k : k : 0 = 2/5 : 1 : 0 = 2 : 5 : 0. ⎡ x1 ⎤ ⎡ 2⎤ ⎢ ⎥ ⎢ ⎥ Therefore, ⎢ x2 ⎥ = ⎢ 5⎥ is an eigenvector of matrix A. ⎢⎣ x3 ⎥⎦ ⎢⎣0 ⎥⎦ 6. (a) We are looking for orthogonal vectors having a span that contains P, Q and R. ⎡ 4⎤ ⎡ −6 ⎤ ⎢ −2⎥ ⎢ −3⎥ Considering option (a), ⎢ ⎥ and ⎢ ⎥ ⎢⎣ 3⎥⎦ ⎢⎣ 6 ⎥⎦
Dividing by |R| = 1 gives the top row of R = [5 −3 1]. −1
5. (d) As matrix is triangular, the eigenvalues are the diagonal elements themselves, namely l = 3, −2 and 1.
Ch wise GATE_EE_Ch1-c_Exp.indd 46
Putting l = −2 in the above equation, we get ⎡ 5 −2 2⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢0 0 1⎥⎥ ⎢⎢ x2 ⎥⎥ = ⎢⎢0 ⎥⎥ ⎢ ⎢⎣0 0 3⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣0 ⎥⎦
that is −λ 0 0 0
−2 2⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡3 − λ ⎢ 0 −2 − λ 1⎥⎥ ⎢⎢ x2 ⎥⎥ = ⎢⎢0 ⎥⎥ ⎢ ⎢⎣ 0 0 1 − λ ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣0 ⎥⎦
Firstly, these are orthogonal because their dot product is given by ( −6 × 4) + ( −3 × −2) + (6 × 3) = 0 The space spanned by these two vectors is
11/21/2018 12:11:31 PM
Chapter 1 • Engineering Mathematics
⎡ 4⎤ ⎡ −6 ⎤ k1 ⎢⎢ −3⎥⎥ + k2 ⎢⎢ −2⎥⎥ (1) ⎢⎣ 6 ⎥⎦ ⎢⎣ 3⎥⎦
⎡ −6 ⎤ ⎡ 4⎤ ⎢ −3⎥ and ⎢ −2⎥ The span of ⎢ ⎥ ⎢ ⎥ contains P, Q and R. ⎢⎣ 6 ⎥⎦ ⎢⎣ 3⎥⎦ We can show this by successively setting Eq. (1) to P, Q and R one by one and solving for k1 and k2 uniquely. Also, the options (b), (c) and (d) are wrong because none of them are orthogonal as can be seen by taking pairwise dot products. 7. (b) We find the linear dependency by finding the determinant of the vectors taken as a matrix. Considering option (a) and finding the determinant of ⎡ −6 −3 6 ⎤ ⎢ 4 −2 3⎥ , ⎢ ⎥ we get ⎢⎣ 8 9 3⎥⎦
−6 −3 6 4 −2 3 = −6( −6 − 27) + 3(12 − 24) 8 9 3
9. (d) We have b = [0 1 0]T and X = [x1 x2 x3]T. The cross product of b and X can be written as i b× x = 0 x1
−6 −3 6 4 −2 3 = −6( −60 + 51) + 3(120 + 6) −2 −17 30 + 6( −68 − 4) = 0
Hence, it is linearly dependent. 8. (b) If X = [ x1 , x2 ,..., xn ]T
⎡ c1 ⎢c A = Let ⎢ 4 ⎢⎣ c7
⎡ c1 ⎢c ⎢ 4 ⎢⎣ c7
However, as both X and XT are non-zero vectors, neither of their ranks can be zero. Therefore, XXT has a rank 1.
Ch wise GATE_EE_Ch1-c_Exp.indd 47
c2 c5 c8
c3 ⎤ ⎡ x1 ⎤ ⎡ x3 ⎤ ⎢ ⎥ c6 ⎥⎥ ⎢⎢ x2 ⎥⎥ = ⎢0 ⎥ c9 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ − x1 ⎥⎦
Comparing L.H.S. and R.H.S., we get ⎡ 0 0 1⎤ ⎡ x1 ⎤ ⎡ x3 ⎤ ⎢ 0 0 0⎥ ⎢ x ⎥ = ⎢ 0⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −1 0 0 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ − x1 ⎥⎦ ⎡ 0 0 1⎤ ⎢ ⎥ So, A = ⎢ 0 0 0 ⎥ ⎢⎣ −1 0 0 ⎥⎦ Now, eigenvalues of A are calculated using characteristic equation |A − λI| = 0 −λ 0 1 0 =0 ⇒ 0 −λ −1 0 −λ ⇒ − λ ( λ 2 − 0) − 1(0 + λ ) = 0 ⇒ λ3 + λ = 0 ⇒ λ (λ 2 + λ ) = 0
Now, rank ( X T ) ≤ ( rank X , rank X T )
So, XX has a rank of either 0 or 1.
0 x1 ]
⎡ x1 ⎤ c3 ⎤ ⎢ ⎥ ⎥ c6 ⎥ . Now, A ⎢ x2 ⎥ = b × x ⎢⎣ x3 ⎥⎦ c9 ⎥⎦
c2 c5 c8
rank (XT) = rank (X) = 1
T
x2
k 0 = x3 i + 0 j − x1 k x3
⎡ x1 ⎤ Now, let L( x ) = b × x = A ⎢⎢ x2 ⎥⎥ , where A is a 3 × 3 ⎢⎣ x3 ⎥⎦ matrix.
Now, it is clear that rank (X) = 1.
⇒ rank( XX T ) ≤ 1
j 1
= [ x3
+ 6(36 + 16) ≠ 0
Considering option (b) and finding the determinant of ⎡ −6 −3 6 ⎤ ⎢ 4 −2 3⎥ , ⎢ ⎥ we get ⎣⎢ −2 −17 30 ⎥⎦
47
⇒ λ = 0, λ = ±1 So, the eigenvalues of A are i, −i and 0. 10. (a) We have ⎡ −3 2⎤ A= ⎢ ⎥ ⎣ −1 0 ⎦
11/21/2018 12:11:33 PM
48
GATE EE Chapter-wise Solved Papers
linearly independent columns, since rank = row rank = column rank.
Now, A − λI = 0 2 −3 − l =0 −1 0 − λ
⇒
⇒ ( −3 − λ )( − λ ) + 2 = 0 ⇒ λ + 3λ + 2 = 0 2
A will satisfy this equation according to Cayley–Hamilton theorem, that is, A2 + 3 A + 2 I = 0 Multiplying by A−1 on both sides, we get A−1 A2 + 3 A−1 A + 2 A−1 I = 0 ⇒ A + 3I + 2 A−1 = 0 11. (a) In the previous question, we derived A2 + 3 A + 2 I = 0.
= 9 A2 + 12 A + 4 I = 9( −3 A − 2 I ) + 12 A + 4 I = −15 A − 14 I A8 = A4 × A4 = ( −15 A − 141)( −15 A − 141) = 225 A2 + 420 A + 156 I = 225( −3 A − 21) + 420 A + 156 I = −225 A − 254 I A9 = A × A8 = A( −255 A − 254 I ) = −255 A2 − 254 A = −255( −3 A − 2 I ) − 254 A = 511A + 510I Hence, A9 = 511A + 510 I .
λ 3 + λ 2 + 2λ + 1 = 0 By Cayley–Hamilton theorem, P 3 + P 2 + 2P + I = 0 I = − P 3 − P 2 − 2P Multiplying by P−1 on both sides, we get = − P − P − 2I = − ( P + P + 2I ) 2
13. (a) If rank of (5 × 6) matrix is 4, then surely it must have exactly 4 linearly independent rows as well as 4
Ch wise GATE_EE_Ch1-c_Exp.indd 48
Then A[ X −1 X ] = A ⋅ I = A Hence, the correct option is (a). 15. (d) The given equations are x1 + 2 x2 + x3 + 4 x4 = 2 3 x1 + 6 x2 + 3 x3 + 12 x4 = 6 The augmented matrix is given by ⎡ 1 2 1 4 2⎤ ⎢ ⎥ ⎣3 6 3 12 6 ⎦
rank( A) = rank( A | B) = 1 Therefore, the system is consistent. As the system has rank( A) = rank( A | B) = 1 which is less than the number of variables, only infinite (multiple) non-trivial solutions exist. ⎡ 1 1 0⎤ 16. (b) Given, P = ⎢⎢0 2 2⎥⎥ ⎢⎣0 0 3⎥⎦ Since P is triangular, the eigenvalues are the diagonal elements themselves. Eigenvalues are therefore λ 1 = 1, λ 2 = 2 and λ 3 = 3. Now, the eigenvalue problem is [ A − λ ] X =0
12. (d) If characteristic equation is
P
Let AT A = X
⎡ 1 2 1 4 2⎤ R2 − 3 R1 ⎡ 1 2 1 4 2⎤ ⎢ ⎥ ⎯⎯⎯→ ⎢ ⎥ ⎣3 6 3 12 6 ⎦ ⎣0 0 0 0 0⎦
A4 = A2 × A2 = ( −3 A − 2 I )( −3 A − 2 I )
2
AA′ A = A[( AT A) −1 AT ] A = A ⎣⎡( AT A) −1 AT A⎤⎦
Performing Gauss elimination on this, we get
⇒ A2 = − 3 A − 21
−1
14. (a) Since
1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡1 − λ ⎢ 0 2−λ 2⎥⎥ ⎢⎢ x2 ⎥⎥ = ⎢⎢0 ⎥⎥ ⎢ ⎢⎣ 0 0 3 − λ ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣0 ⎥⎦ Putting l1 = 1, we get the eigenvector corresponding to this eigenvalue ⎡0 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ 0 1 2⎥ ⎢ x ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣0 0 2⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣0 ⎥⎦ which gives the equations
11/21/2018 12:11:35 PM
49
Chapter 1 • Engineering Mathematics
x2 = 0 x2 + 2 x3 = 0 2 x3 = 0
⎡ 1 0⎤ ⎡ −5 −3⎤ 17. (b) We have A = ⎢ ⎥ and I = ⎢0 1⎥ 2 0 ⎦ ⎦ ⎣ ⎣ Characteristic equation of A is given by
The solution is x2 = 0, x3 = 0, and x1 = k. ⎡k ⎤ ⎢ 0⎥ . X = So, one eigenvector is ⎢ ⎥ ⎢⎣ 0 ⎥⎦ Hence, x1 : x2 : x3 = k : 0 : 0. We need to find the other eigenvectors corresponding to the other eigenvalues because none of the eigenvectors given in the options matches with this ratio. Now, corresponding to l2 = 2, we get the following set of equations by substituting l = 2 in the eigenvalue problem, ⎡ −1 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ 0 0 2⎥ ⎢ x ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎣⎢ 0 0 1⎦⎥ ⎢⎣ x3 ⎥⎦ ⎣⎢0 ⎦⎥ which gives
−5 − λ −3 =0 2 0−λ ( −5 − λ )( − λ ) + 6 = 0 ⇒ λ 2 + 5λ + 6 = 0 So, A2 + 5 A + 6 I = 0 (by Cayley–Hamilton theorem) ⇒ A2 = −5 A − 6 I Multiplying by A on both sides, we have A3 = −5 A2 − 6 A ⇒ A3 = −5( −5 A − 6 I ) − 6 A = 19 A + 30 I 18. (d) The system of equations given is ⎡ 2 −2⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ 1 −1⎥ ⎢ x ⎥ = ⎢0 ⎥ ⎦⎣ 2⎦ ⎣ ⎦ ⎣ 2 x1 − 2 x2 = 0
− x1 + x2 = 0
x1 − x2 = 0
2 x3 = 0
⇒ x1 = x2
x3 = 0 Hence, the solution is x3 = 0, x1 = k , x2 = k . ⎡k ⎤ ⎢ ⎥ Therefore, X 2 = ⎢ k ⎥ ⇒ x1 : x2 : x3 = 1 : 1 : 0. ⎢⎣ 0 ⎥⎦ As none of the eigenvectors given in the options is of this ratio, we need the third eigenvector also. By putting l = 3 in the eigenvalue problem, we get 1 0 ⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎡ −2 ⎢ 0 −1 2⎥ ⎢ x ⎥ = ⎢0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣0 ⎥⎦ −2 x1 + x2 = 0 − x2 + 2 x3 = 0 Putting x1 = k, we get x2 = 2k and x3 = x2/2 = k ⎡ k⎤ X 3 = ⎢⎢ 2k ⎥⎥ ⇒ x1 : x2 : x3 = 1 : 2 : 1. Therefore, ⎢⎣ k ⎥⎦ ⎡ 1⎤ ⎢ ⎥ Only the eigenvector given in option (b) ⎢ 2⎥ is in this ⎢⎣ 1⎥⎦ ratio.
Ch wise GATE_EE_Ch1-c_Exp.indd 49
Hence, x1 and x2 are having infinite number of solutions or multiple solutions. 19. (d) We know that AX = λ X ⎡ a b ⎤ ⎡ 1⎤ ⎡ 1⎤ ⎢ c d ⎥ ⎢ −1⎥ = ( −1) ⎢ −1⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ a − b = −1 (1) c − d = 1 (2) ⎡ 1⎤ ⎡ a b ⎤ ⎡ 1⎤ ⎢ c d ⎥ ⎢ −2⎥ = ( −2) ⎢ −2⎥ ⎣ ⎦ ⎦⎣ ⎦ ⎣ a − 2b = −2 (3) c − 2d = 4 (4) Subtracting Eq. (1) from Eq. (3), we get b=1 Substituting value of b in Eq. (1), we get a − 1 = −1 ⇒ a = 0 Subtracting Eq. (2) from Eq. (4), we get − d = 3 ⇒ d = −3
11/21/2018 12:11:37 PM
50
GATE EE Chapter-wise Solved Papers
Substituting value of d in Eq. (2), we get c − ( −3) = 1 ⇒ c = 4
Maximum possible ( −3) × ( −3) = 9.
value
of
1⎤ ⎡a b⎤ ⎡ 0 =⎢ Therefore, A = ⎢ ⎥ ⎥ ⎣ c d ⎦ ⎣ −2 −3⎦
25. (a) The correct answer is option (a).
20. (b) We are given a set of two equations
Z * = A − iB Therefore, f(z) = 2A (continuous) u = 2A v = 0
x + 2 y + 2 z = b1 5 x + y + 3 z = b2 ⎡1 2 2 b1 ⎤ ⎥ ⎣ 1 3 b2 ⎦
[ A/B] = ⎢5
⎡1 2 2 b1 ⎤ ⎢ ⎥ ⎣0 −9 −7 b2 − 5b1 ⎦ Thus, rank(A) = rank(A/B) < number of unknowns, for all values of b1 and b2. Therefore, the equations have infinitely many solutions, for any given b1 and b2.
23. (0.33) The characteristic equation is A − λI = 0 −λ ⇒ −6 −6
1 −1 −11 − λ 6 =0 −11 5−λ
⇒ λ 3 + 6 λ 2 + 11λ + 6 = 0
du du = 2 =0 dA dB dv dv = 0 =0 dA dB
Therefore, it is not analytic. 27. (3) Det [A - lI]
1− λ 1 1 = 1 1− λ 1 1 1 1− λ l = 1 × order of matrix
2l. (a) Eigenvalues of a real symmetric matrix are all real. 22. (c) Rank of a matrix is unaltered by the elementary transformations, that is row/column operators. (Here, B is obtained from A by applying row/column operations on A). Since rank of A is N. Therefore, rank of B is also N.
=1×3 =3 (Eigenvalues are 0, 0, 3.) 28. (a) Eigenvalues of A2 - 3A + 4I are (1)2 = 3(1) + 4 and (-2)2 - 3(-2) + 4 = 2, 14 29. (b) Rank (ATA) = Rank (A) 30. (d) Given that ⎡ 3 1⎤ ⎡ x ⎤ ⎡ a ⎤ ⎢1 3⎥ ⎢ y ⎥ = ⎢ b ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦
Then 3x + y = a
So, λ = −1, − 2, and − 3 are the eigenvalues of A.
x + 3y = b
Now, λ max = −1 and λ min = −3 . Hence,
a2 + b2 = 1
λ max −1 1 = = = 0.33 λ min −3 3 24. (9) Sum of the diagonals elements is −6 for 2 × 2 matrix. The possible eigenvalues are −1, −5 −5, −1 −8, 2 −2, −3 −4, −2 −9, 3 − − − −3, −1 −3, −3 −10, 4
Ch wise GATE_EE_Ch1-c_Exp.indd 50
is
26. (b) Let z = A + iB. Therefore,
Applying Row transformation, R2 → R2 − 5 R1
determinant
Therefore, (3 x + y ) 2 + ( x + 3 y ) 2 = 1 10 x 2 + 10 y 2 + 12 xy = 1 ⇒ a = 10, b = 10 and h = 6
This represents an ellipse. Then length of semi-axis is
( ab − h2 )r 4 − ( a + b)r 2 + 1 = 0 1 1 ⇒ r = or 2 4
11/21/2018 12:11:39 PM
Chapter 1 • Engineering Mathematics
⎛ 1⎞ 1 Length of minor axis = 2r = 2 ⎜ ⎟ = ⎝ 4⎠ 2 Equation of minor axis is 1⎞ ⎛ ⎜⎝ a − 2 ⎟⎠ x + hy = 0 r ⇒ (10 − 16) x + 6 y = 0 ⇒ y−x=0 Similarly, major axis equation is
34. (1) We have
⎡ 1 0 −1⎤ A = ⎢⎢ −1 2 0 ⎥⎥ ⎢⎣ 0 0 −2⎥⎦
A − λI = 0
y+x=0
31. (c) This matrix A is symmetric and all its eigenvalues are distinct. Therefore, all the eigenvalues are orthogonal. We know, ⎡1 ⎤ ⎢ X 1 = ⎢0 ⎥⎥ ⎢⎣1 ⎥⎦ The corresponding orthogonal vector matches option (c), ⎡1⎤ that is, X 2 = ⎢⎢ 0 ⎥⎥ as X 1T ⋅ X 2 = 0 . ⎢⎣ −1⎥⎦ 32. (a) |A - lI | = 0 0 1 =0 −4 − λ
0 2−λ 0
⇒ A3 - A2 - 4A + 4I = 0
⇒ A3 - A2 - 4A + 5I = I
⇒ B = I (according to the question)
⇒ B = I =1
Calculus 35. (c) We have S=
∞
Tr(A ) = 5
Þ⇒ λ12 + λ 22 = 5
Þ⇒ ( λ1 + λ 2 ) 2 = λ12 + λ 22 + 2λ1λ 2
Þ⇒ 16 = 5 + 2l1l2
⇒ 2l1l2 = 11
2
11 2 11 ⇒ A = = 5.5 2
∞
36. (c) We have the scalar field, u=
At (1, 3),
Tr(A) = 4 ⇒Þ l1 + l2 = 4
x −3 dx
⎡ x −2 ⎤ ⎡ 1 ⎤ ⎡ 1 1⎤ 1 S=⎢ ⎥ = −⎢ 2 ⎥ = −⎢ − ⎥ = 2 − 2 x ⎣ ⎦1 ⎣∞ 2⎦ 2 ⎦1 ⎣
⇒ l = 0 or l = −1, −3
∞
∫1
Now, integrating, we get
Now, grad u = i
Þ⇒ λ1λ 2 =
−1 0 =0 −2 − λ
⇒ −( λ 3 + 4 λ + 3) = 0
33. (5.5) A is 2 ´× 2 matrix
Ch wise GATE_EE_Ch1-c_Exp.indd 51
1− λ ⇒ −1 0
By Cayley-Hamilton theorem
⎡1⎤ So, ⎢ ⎥ lies on the minor axis. ⎣1⎦
0−λ 1 ⇒ 0 0−λ 0 −3
51
x2 y2 + 2 3
∂u ∂u 2 +j = xi + yj ∂x ∂y 3
⎛2 ⎞ grad u = (1) j + ⎜ ⋅ 3⎟ j = i + 2 j ⎝3 ⎠ grad u = 12 + 22 = 5 37. (d) We consider options (a) and (d) only, because these contain variable r, as variable of integration. On integrating option (d), we get 1/ 3π a 2 H , which is volume of cone. 38. (b) Let D =
x⋅x y⋅x
x⋅ y y⋅ y
Let x = x1i + x2 j and y = y1i + y2 j.
11/21/2018 12:11:42 PM
52
GATE EE Chapter-wise Solved Papers
Then,
Now,
x ⋅ x = x12 + x22
∫ udv = ux − ∫ vdu x x x ∫ xe dx = xe − ∫ e dx
Therefore,
y ⋅ y = y12 + y22
= x ex – ex + c
x ⋅ y = x1 y1 + x2 y2 Therefore, D =
x12 + x2 2 x1 y1 + x2 y2
x1 y1 + x2 y2 y12 + y2 2
= ( x12 + x2 2 )( y12 + y2 2 ) − ( x1 y1 + x2 y2 ) 2 = x2 2 y12 + x12 y2 2 − 2 x1 y1 x2 y2 Now, D = 0. Thus, x2 y1 − x1 y2 = 0 x1 y = 1 x2 y2
Hence, vectors x1i + x2 j and y1i + y2 j are linearly dependent. Therefore, Linear dependence ⇒ D = 0 So, linear independence ⇒ D ≠ 0 That is, D is negative or positive. However, since D = ( x2 y1 − x1 y2 ) 2, it cannot be negative. Hence, D is positive when x and y are linearly independent. 39. (a) x2 = y y2 = x
1
0
y =1
x= y
y=0
x = y2
∫ ∫
f ( x, y )dx dy
40. (b) We have 1
P=
∫ xe
x
dx
Integrating by parts, we get Let u = x , v = ex 0
dv = e x dx du = dx
Ch wise GATE_EE_Ch1-c_Exp.indd 52
v = ∫ e x dx = e x
x
x
− e x ]10
0
= (1⋅ e1 − e1 ) − (0 ⋅ e 0 − e 0 ) = 0 − ( −1) =1 41. (a) We have r = xi + y j + z k Now, divergence of vector is given by div r = ∇ ⋅ r
= ( x2 y1 − x1 y2 ) 2
⇒
1
∫ xe dx = [ xe
(
⎛ ∂ ∂ ∂⎞ = ⎜ i + j + k ⎟ ⋅ xi + y j + z k ⎝ ∂x ∂y ∂z ⎠
)
∂x ∂y ∂z + + ∂x ∂y ∂z = 1+1+1 = 3 =
42. (c) We have
f ( x) = 2 x − x 2 + 3 f ′( x ) = 2 − 2 x = 0
Now, x = 1 isNow, the stationary x = 1 is thepoint. stationary point. f ′′( x ) = −2 f ′′( x ) = −2 f ′′(1) = −2 1 or ⎪⎭ ∀x in (1, 1 + h)
53. (c) We know that according to intermediate value theorem, if f ( a) ⋅ f (b) < 0 then f(x) has at least one root in (a, b). f(x) does not have roots in (a, b) means f ( a) ⋅ f (b) > 0.
Thus, h is positive and small. Therefore, f has local minima at x = 1 and the minimum value is ‘0’. 49. (a) We have,
Now, ∇ ⋅ v = 0 + 0 + 0 = 0 Thus, Eq. (1) becomes
(1)
54. (b) The characteristic equation of A is A − XI = 0 ⇒ f ( x ) = x 3 + 6 x 2 + 11x + 6 + 2a = ( x + 1)( x + 2)( x + 3) + 2a = 0 f(x) cannot have all 3 roots (if any) equal
x y + y z + z x = f ( 0 ) + v ∇f 2
f ( −3) = 118, f (3) = 28
and f ( −2) = 128, f ( 4 ) = 44
Differentiating again, we get
∇ ⋅ ( f v ) = f ( ∇ ⋅ v ) + ∇f ⋅ v
2π
2π
1 ⎡ sin 2θ ⎛ cos 2θ − 1⎞ ⎤ ∫0 ⎜⎝ 2 ⎟⎠ dθ = 2 ⎢⎣ 2 − θ ⎥⎦ 0 = −π =
Differentiating w.r.t. x, we get
f ′ ( x ) = 0 ⇒ e − x (1 − x ) = 0 ⇒ x = 1
2
Therefore, v ⋅ ∇f = x 2 y + y 2 z + z 2 x .
Ch wise GATE_EE_Ch1-c_Exp.indd 53
0
f ( x ) = x 3 − 3 x 2 − 24 x + 100
f ( x ) = xe − x For maximum value of the function, we calculate
2
∫ (sin θ )(1) ( − sin θ dθ )
51. (b) We are given
47. (a) We are given,
f ( x ) = ( x − 1) Differentiating, we get
2
2π
2
Putting y = 0, z = 0, dy = 0 and dz = 0 , we get F ⋅ dl =0 ∫
C
[C is circle x + y = 1 at z = 1 ⇒ x = cosq, y = sinq and θ = 0 to 2π ] 2
45. (d) The curl of the gradient of a scalar field is always zero. ∇ × ∇V = 0
Now,
53
since if f ( x ) = ( x − k )3 , then comparing coefficients, we get 6 = −3k , 3k 2 = 11
11/21/2018 12:11:47 PM
54
GATE EE Chapter-wise Solved Papers
No such k exists.
Hence,
(a) Thus f(x) = 0 has repeated (2) roots (say) α , α , β
F ⋅ di =
(1,1,1)
∫
or
( 0,0,0 )
59. (d) We know that
f ′( x ) = 0 ⇒ x1 = x2 =
fore,
−6 − 3 ≈ −2.577a 3
∫
dϕ = x 2 y 2 + 2
( 0,0,0 )
(b) f(x) = 0 has real roots (say) α , β , γ . Now,
(1,1,1)
1,1,1 0,0,0
=2
sin 2π t is an even function. Thereπt
∞
∞
sin( 2π t ) sin( 2π t ) dt = 2 × 2∫ dt π t πt −∞ 0
2∫
−6 + 3 ≈ −1.422 3
∞
At x1, f (x) has relative maxima and at x2, f(x) has relative minima.
sin( 2π t ) 1 dt = π t 2 0
∫ Therefore,
In all possible cases, we have
∞
sin( 2π t ) 1 = 4× = 2 π t 2 −∞
4∫
⎛ 3⎞ 1 ≤0⇒a≤ f ( x2 ) ≤ 0 ⇒ 2 ⎜ a − ⎟ 9 ⎠ 3 3 ⎝
60. (4.41) Given that E = 5 xzi + (3 x 2 + 2 y ) j + x 2 zk = ∫ F ⋅ dr
55. (b) The correct answer is option (b). 56. (e - 2) The triangle is bounded by x = y, x = 0, y = 1 in xy plane.
C 1
Required volume =
∫∫
1
f ( x, y )dxdy =
1
∫
1
e x ( y )1x dx =
x=0
∫
∫ ∫
e x dxdy
0
x=0 y= x
OAB
=
= ∫ 5t 2 dt + (3t 2 + 2t 2 )2t dt + t 3 dt
1
= 4.41
1
e x (1 − x )dx =
x=0
∫ (e
x
− xe x )dx
x=0
61. (0.99) I = c ∫∫ xy 2 dxdy
= (e x )10 − (e x ( x − 1))10 = (e1 − 1) − [0 − ( −1)] = e − 2
5
f(x) = x(x - 1) (x - 2)
5
xy 2 dydx
1
f ′( x ) = 3 x − 6 x + 2 2
=
Therefore, there are no critical points in the given interval. Checking value at end points, that is, 1 and 2. f(2) = 0
5
= 1
8c (625 − 0.2) 3
62. (40) ∂f = ( x 2 + y 2 − 2 z 2 )(0) + ( y 2 + z 2 )( 2 x ) ∂z
58. (b) F = xy 2 i + 2 x 2 y j + k As ∇ × F = 0
⇒
⇒ F = ∇ϕ Let ϕ x = 2 xy , ϕ y = 2 xy , ϕ z = 1 2
Therefore,
ϕ = x y + z+c 2
∂f ∂x
= (12 + 32 )( 2 × 2) = 40 x = 2 , y =1, z = 3
63. (7) Given that
2
8c x 5 3 5
0
5
8 dx = c ∫ x 4 dx 3 1
8 = (6 × 10 −4 )(625 − 0.2) = 0.99968 3
f(1) = 0 Hence, maximum value of f(x) in the interval [1, 2] = 0.
2x
xy 3 3
= c∫
= x(x2 - 3x + 2) = x3 - 3x2 + 2x
Ch wise GATE_EE_Ch1-c_Exp.indd 54
∫
x =1 y = 0
57. (0) Given that
2
2x
=c∫
2x2 + y2 = 34
(1)
x + 2y = 11
(2)
11/21/2018 12:11:50 PM
Chapter 1 • Engineering Mathematics
From Eq. (2), we have x + 2y = 11 ⇒ x = 11 - 2y(3)
Substituting the value of x in Eq. (1), we have 2(11 - 2y)2 + y2 = 34 ⇒ 2(121 + 4y2 - 44y) + y2 = 34 ⇒ 242 + 8y2 - 88y + y2 = 34 ⇒ 9y2 - 88y + 208 = 0 ⇒ y = 5.77 or 4 From Eq. (3), we have x = 3 at y = 4 Therefore, x+y=3+4=7 64. (5.732) Given that y2 - 2y + 1 = x(1)
and
⇒ dy = dx
I = ∫ ( x 2 + iy 2 )dz = ∫ ( x 2 + iy 2 )( dx + idy ) C
C
= ∫ ( x 2 + ix 2 )( dx + idx ) C 1
= ∫ x 2 dx + ix 2 dx + ix 2 dx − x 2 dx 0
=
2i 3
67. (a) fog(x) is continuous at x < 0. Therefore, there are no discontinuities. 68. (d) Given f(x) = x2, x ³ 0 and f(x) = -x2, x < 0. First derivative: ⎧ 2x f ′( x ) = ⎨ ⎩ −2 x
From Eq. (1),
y2 - 2y + 1 = x
⇒ (y - 1) = x 2
⇒ y −1 =
x (3)
Since, lim+ f ( x ) ≠ lim− f ( x ) x→0
x→0
⎧ 2, x ≥ 0 f ′′( x ) = ⎨ ⎩ −2, x < 0
y - 1 + y = 5 ⇒ y=3
From Eq. (3), 3 −1 = x ⇒ x=4
Therefore,
x + y = 4 + 3 = 4 + 1.732 = 5.732
So, f(x) is differentiable but its 1st derivative is not differentiable at x = 0. 69. (a) Given ϕ = xy 2 + yz 2 + zx 2 ∂ϕ + ∇ϕ = iˆ ∂x
ˆj ∂ϕ + kˆ ∂ϕ ∂y ∂z
= iˆ( y 2 + 2 xz ) + ˆj ( 2 xy + z 2 ) + kˆ( 2 yz + x 2 )
65. (a) Given that ⎧e x , x 0, we have a
a + b2 2
Therefore, solution, x2 (t ) = C1 + C2 e −2t x2 f = lim x2 (t ) = C1
−b
a + b2 2
t →∞
⇒ x1 f = x2 f < ∞
Complex Variables
So,
2
1 1 = z + 1 ( z − i )( z + 1)
=
Hence, we have poles at i and −i, that is at (0, 1) and (0, −1).
Since
From the figure of z − i / 2 = 1, we see that pole (0, 1), that is i is inside C, whereas pole (0, −1), that is −i is outside C.
1 a + b2 2
1 a + b2 0 < a2 + b2 < 1 1 a + b2
>1
1 is outside the unit circle in the fourth quadrant. z
89. (b) We know that
(0,1) (0,1/2) C
⎛ π⎞ ⎛ π⎞ ⎛ 3π ⎞ ⎛ 3π ⎞ −i = cos ⎜ − ⎟ + i sin ⎜ − ⎟ , cos ⎜ ⎟ + i sin ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⇒ ( −i )
1/ 2
(0,−1/2)
Ch wise GATE_EE_Ch1-c_Exp.indd 59
2
2
2
So,
0
1 is in 4th quadrant. z
87. (b) We know that
z −i/2 = 1
4th Quadrant
rant, a and b both are +ve and 0 < a 2 + b 2 < 1 .
⇒ D = 0, −2
59
1/ 2
⎡ ⎛ π⎞ ⎛ π⎞⎤ = ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥ ⎝ 2⎠⎦ ⎣ ⎝ 2⎠
⎛ π⎞ ⎛ π⎞ ⎛ 3π ⎞ ⎛ 3π ⎞ = cos ⎜ − ⎟ + i sin ⎜ − ⎟ , cos ⎜ ⎟ + i sin ⎜ ⎟ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠
11/21/2018 12:12:02 PM
60
GATE EE Chapter-wise Solved Papers
90. (a) We have
94. (c) For the given integral z −4 z −4 = 2 z + 4 ( z + 2i ) ( z − 2i ) 2
∫
2
Hence, we have poles at 2i and −2i, that is (0, 2i) and (0, −2i) Imaginary
Poles: z =
2z + 5 1⎞ 2 ⎛ ⎜⎝ z − ⎟⎠ ( z − 4 z + 5) 2
1 , 2±i 2
⎛ 1⎞ Of these, only z = ⎜ ⎟ lies inside | z| = 1. ⎝ 2⎠
3i 2i
Using residue theorem,
∫ = 2π i( R
i
1/ 2
)
C
Real 0 −i −2i From the figure of z − i = 2 , we see that the pole is inside C while the pole, −2i, is outside C. Therefore,
∫
z2 − 4 dz = 2π i × Re s F ( z ) z2 + 4
⎡ ⎛ ⎤ 1⎞ ⎢ ⎜⎝ z − 2 ⎟⎠ ( 2 z + 5)dz ⎥ 24 1 ⎥= = lim ⎢ Residue at z →1/ 2 ⎢ ⎛ 1⎞ 2 ⎥ 13 2 ⎢ ⎜⎝ z − 2 ⎟⎠ ( z − 4 z + 5) ⎥ ⎣ ⎦ 95. (d) Using Cayley Hamilton theorem, we get l3 = l
( z − 2i )( z 2 − 4) = 2π i ⋅ ( z + 2i )( z − 2i )
l = 0, 1, -1 z = 2i
⎡( 2i ) 2 − 4 ⎤⎦ = 2π i ⎣ = −4π ( 2i + 2i ) 91. (b) We know that 1 = cos ( 2k π ) + i sin ( 2k π ) , where k is an integer = e i (2 kπ )
96. (d) lim z →i
z2 + 1 (% form) z 3 + 2 z − i ( z 2 + 2)
On differentiating, we get
lim z →i
2z 2i = = 2i 3 z 2 + 2 − i( 2 z ) 3i 2 + 2 − i( 2i )
97. (b)
Therefore, 1i = e − (2 kπ )
f(z) =
Hence, all values are real and non-negative. 92. (c) z = −1, 1 are the simple poles of f(z) and z = 1 lies inside C : z − 1 = 1. Therefore,
z2 ∫ ( z 2 − 3z + 2) dz
=∫
z2 dz ( z − 1) ( z − 2) 2 2
∫ f ( z ) dz = 2π i × ⎡⎣Re s f ( z )⎤⎦ = 2π i × ⎡⎣lim ( z − 1) ⋅ f ( z )⎤⎦ z =1
C
z →1
⎡ z2 ⎤ = 2π i × ⎢ lt ⎥ = πi ⎣ z→1 z + 1⎦ 93. (b) Given f ( z ) = g ( z ) + h( z ) f ( z ), g ( z ) and h( z ) are complex variable functions.
-4
1
2
4
Option (c) is not correct, since every continuous function need not be differentiable. Let g ( z ) = xh( z ) = iy ⇒ g ( z ) = x + i(0) h( z ) = 0 + iy
Ch wise GATE_EE_Ch1-c_Exp.indd 60
11/21/2018 12:12:04 PM
Chapter 1 • Engineering Mathematics
z =1
Therefore, p ( P ∪ Q ) ≤ p ( P ) + p (Q )
⎛ 2 z ( z − 2) 2 − 2 z 2 ( z − 2) ⎞ = lim ⎜ ⎟⎠ z →1 ⎝ ( z − 2) 4
Option (c) is false because independence and mutually exclusion are unrelated properties. Now, as P ∩ Q ⊆ P
⎛ 2 z ( z − 2) − 2 z 2 ⎞ = lim ⎜ ⎟⎠ z →1 ⎝ ( z − 2) 3 =
z→2
z=2
⇒ n (P ∩ Q) ≤ n (P )
−4 =4 −1
Re s. f ( z ) = lim
Hence, option (d) is true.
⎞ 1 d ⎛ z ( z − 2) 2 ⋅ 2 2⎟ ⎜ 1! dz ⎝ ( z − 1) ( z − 2) ⎠ 2
100. (b) We know that P ( A | B) =
⎛ ( z − 1) 2 z − z 2( z − 1) ⎞ = lim ⎜ ⎟⎠ z→2 ⎝ ( z − 1) 4 2
⎛ 2 z ( z − 1) − 2 z ⎞ = lim ⎜ ⎟⎠ z→2 ⎝ ( z − 1)3
=
4 −8 = −4 1
q varies from 0 to p/2. So,
π /2
∫ (9 sin
2
θ + 18 sin θ cos θ )( −3 sin θ dθ )
0
+ (18 sin θ cos θ + 9 cos 2 θ )(3 cos θ )dθ
=
π /2
∫ ( −27 sin 0
θ − 54 sin 2 θ cos θ + 54 sin cos 2 θ
101. (c) If two fair dice are rolled, the probability distribution of r where r is the sum of the numbers on each dice is given by gg
2
P( gg )) P(
1 2 3 4 5 6 5 4 3 2 1 36 36 36 36 36 36 36 36 36 36 36
+ 27 cos3 θ )dθ
3
=0
Probability and Statistics 99. (d) Option (a) is false because P and Q are independent p( P ∩ Q ) = p( P ) ∗ p(Q ) which need not be zero. Option (b) is false because
Ch wise GATE_EE_Ch1-c_Exp.indd 61
P (first toss is head )
1 1 Therefore, required probability = 4 = 1 2 2
dx = -3 sin q d q and dy = 3 cos q d q
=
P ( 2 heads in 3 tosses and first toss is head )
P(2 head in 3 tosses and first toss is head) = P(HHT) + 1 1 1 1 1 1 1 P(HTH) = × × + × × = 2 2 2 2 2 2 4
98. (0) Given x2 + y2 = 9. Since the arc is from the point (3, 0) to the point (0, 3), we have x = 3 cos q and y = 3 sin q.
+ 2 xy )dx + ( 2 xy + x 2 )dy
P ( B)
P(first toss is head) = 1/2
By residue theorem, I = 2pi (4 - 4) = 0
2
P ( A ∩ B)
Therefore, P(2 heads in 3 tosses | first toss is head)
2
∫(y
⇒ pr ( P ∩ Q ) ≤ pr ( P )
2
=
p( P ∪ Q ) = p ( P ) + p(Q ) − p( P ∩ Q )
⎞ 1 d ⎛ z2 2 1 ( z − ) ⋅ 2 2⎟ z →1 1! dz ⎜ ( z − 1) ( z − 2) ⎠ ⎝
Re s. f ( z ) = lim
61
3
4
5
6
7
8
9
10 11 12
The above table has been obtained by taking all different ways of obtaining a particular sum. For example, a sum of 5 can be obtained by (1, 4), (2, 3), (3, 2) and (4, 1). P(x = 5) = 4/36 Now, let us consider option (a), P(r > 6) = P(r > 7)
=
6 5 4 3 2 1 21 7 + + + + + = = 36 36 36 36 36 36 36 12
Therefore, option (a) is wrong.
11/21/2018 12:12:06 PM
62
GATE EE Chapter-wise Solved Papers
Consider option (b),
P(2 washers, then 3 nuts, then 4 bolts)
P(r/3 is an integer) = P(r = 3) + P(r = 6) + P(r = 9) + 2 5 4 1 12 1 P(r = 12) = + + + = = 36 36 36 36 36 3
1 ⎛ 2 1 ⎞ ⎛ 3 2 1 ⎞ ⎛ 4 3 2 1⎞ = ⎜ × ⎟ ×⎜ × × ⎟ ×⎜ × × × ⎟ = ⎝ 9 8 ⎠ ⎝ 7 6 5 ⎠ ⎝ 4 3 2 1⎠ 1260 104. (0.14) P(n) = K ⋅ n where n = 1 to 6. We know that,
Therefore, option (b) is wrong.
∑ P ( x ) = 1 ⇒ K [1 + 2 + 3 + 4 + 5 + 6] = 1
Consider option (c),
n
P(r = 8 | r/4 is an integer) =
⇒K=
1 36
Now, P(r/4 is an integer) = P(r = 4) + P(r = 8) + 5 1 9 1 3 P(r = 12) = + + = = 36 36 36 36 4 P(r = 8 and r/4 is an integer) = P(r = 8) = 5/36
105. (0.43) Density of the random variable is f(x) = kx2. Since f(x) is a PDF
Therefore, 2
∫
Hence, option (c) is correct.
1
102. (c) Dice value
1
2
3
4
5
6
Probability
1 4
1 8
1 8
1 8
1 8
1 4
As the dice are independent, 1 1 1 1 P (1, 5, 6 ) = × × = 4 8 4 128 1 1 1 1 P (3, 4, 5) = × × = 8 8 8 512 1 1 1 1 P (1, 2, 5) = × × = 4 8 8 256 1 is correct. 128
103. (c) Box contains 2 washers, 3 nuts and 4 bolts. 2 1 Probability of drawing 2 washers first = ⎛⎜ × ⎞⎟ ⎝ 9 8⎠ Probability of drawing 3 nuts after drawing 2 washers = ⎛ 3 2 1⎞ ⎜⎝ × × ⎟⎠ 7 6 5 Probability of drawing 4 bolts after drawing 4 bolts = ⎛ 4 3 2 1⎞ ⎜⎝ × × × ⎟⎠ 4 3 2 1
Ch wise GATE_EE_Ch1-c_Exp.indd 62
1 = 0.14. 7
Therefore, required probability is P(3) = 3K =
⎧kx 2 , 1 < x < 2 f (x) = ⎨ ⎩ 0, otherwise
5 20 5 36 P(r = 8 | r/4 is an integer) = = = 1 36 9 4
Therefore, option (c), P(1, 5 and 6) =
1 21
2
⎡ x3 ⎤ 3 f ( x ) dx = 1 ⇒ k ⎢ ⎥ = 1 ⇒ k = = 0.428 0.43 7 3 ⎣ ⎦1
106. (b) Let X be the difference between number of heads and tails. Take n = 2 ⇒ S = {HH , HT , TH , TT } and X = -2, 0, 2. Here, n − 3 = −1 is not possible. Take n = 3 ⇒ S = {HHH , HHT , HTH , HTT , THH , THT , TTH , TTT } and X = -3, -1, 1, 3. Here, n − 3 = 0 is not possible. Similarly, if a coin is tossed n times, then the difference between heads and tails is n − 3, which is possible. Thus, required probability is 0. 107. (0.4) The probability density function of the random variable is ⎧0.2, for x ≤ 1 ⎪ f ( x ) = ⎨ 0.1, for 1 < x ≤ 4 ⎪ 0, otherwise ⎩ P (0.5 < x < 5) =
5
∫ f ( x ) dx
0.5 1
=
∫
0.5
4
5
f ( x ) dx + ∫ f ( x ) dx + ∫ f ( x ) dx 1
4
= (0.2)( x )0.5 + (0.1)( x )1 + 0 1
4
= 0.1 + 0.3 = 0.4
11/21/2018 12:12:08 PM
Chapter 1 • Engineering Mathematics
108. (*) Mean thickness of laminations = 0.2 mm
110. (d) Probability of getting 6 =
Mean thickness of one side varnish = 0.1 mm
6 1 = 36 6
Mean thickness of varnish including both sides = 0.2 mm
That is, probability of A wins the game =
Therefore, mean thickness of one lamination varnished on both sides = 0.4 mm
Probability of A not wins the game = 1 −
Thickness of the stack of 100 such laminations = 0.4 × 100 = 40 mm
Probability of B wins the game =
If there are 100 laminations with mean thickness = 0.2 mm and variance = 0.1 mm, then we can write the following expressions:
Probability of B not winning the game =
(d1 − 0.2)2 + (d2 − 0.2)2 + … + (d100 − 0.2)2/100 = 0.02 (x1 − 0.1)2 + (x2 − 0.1)2 + … + (x100 − 0.1)2 = 0.01
63
1 6
1 5 = 6 6
1 6 5 6
If A starts the game, probability that A wins the game = P ( A) + P ( A) P ( B ) P ( A) + P ( A) P ( B ) P ( A) P ( B ) P ( A) +
Assuming all thicknesses to be equal to d and each side lamination varnish thickness to be equal to x, the two equations reduce to:
=
1 551 55551 + + + 6 666 66666
d = 0.3414 and x = 0.2
=
2 4 ⎤ 1⎡ 55 5555 ⎤ 1 ⎡ ⎛ 5⎞ ⎛ 5⎞ 1 1 + + + = + + ⎢ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ + ⎥ ⎢ ⎥ 6⎣ 66 6666 6 6 ⎦ 6 ⎢⎣ ⎥⎦
Each lamination thickness therefore including steel and two-side varnish coating thicknesses = 0.3414 + 0.4 = 0.7414 mm. Variance of overall core is therefore given by 100 × (0.7414 − 0.4)2/100 = 0.1166 Therefore, none of the given answers matches the correct answer. 109. (0.25) We have ⎧a + bx for 0 < x < 1 f ( x) = ⎨ otherwise ⎩ 0 ∞
∫
f ( x )dx = 1
1
So, ∫ (a + bx)dx = 1 0
Given that,
b = 1 ⇒ 2a + b = 2 2 1
2 a b = + ⇒ 3a + 2b = 4 (2) 3 2 3
From Eqs. (1) and (2), we get a = 0 and b = 2 0.5
Pr[ X < 0.5] =
∫ 0
Ch wise GATE_EE_Ch1-c_Exp.indd 63
0.5
f ( x )dx = 2 ∫ xdx = 0.25 0
aaa bbb ccc ddd
Let first pen be a. (1)
2 E[ X ] = = ∫ x[a + bx ]dx 3 0
111. (0.167) Let Black = a Blue = b Green = c Red = d For all three pens to be of the same colour, four cases are possible:
−∞
a+
⎤ ⎡ ⎢ ⎥ 1 1 ⎥ = 1 × 36 = 6 = ⎢ 2 6 ⎢ ⎛ 5 ⎞ ⎥ 6 11 11 ⎢1 − ⎜ ⎟ ⎥ ⎣ ⎝ 6⎠ ⎦
Next, choose 1 from 4 = 4C1 Next, choose 1 from 4 = 4C1 = a × 4C1 × 4C1 = 16 cases Similarly,
b ×4C1×4C1 = 16 cases
c ×4C1 ×4C1 = 16 cases d ×4C1 ×4C1 = 16 cases Total 64 cases. Therefore, probability 4 1 = = or 0.167 64 6
11/21/2018 12:12:10 PM
64
GATE EE Chapter-wise Solved Papers
112. (a) For the given function:
E (Y 2 ) =
⎧ ae , x < 0 ⎪ f ( x ) = ⎨ 3 −3 x ⎪ e , x≥0 ⎩2 4x
+∞
∫
2
Var{Y } =
f x ( x) = 1
Therefore,
∫ ae
−∞
ae 4 x 4
dx 1 − x 116. (d) Here, = dt τ
∞
4x
0
−∞
3 dx + ∫ e −3 x dx = 1 2 0
3 ⎛ −1⎞ + ⎜ ⎟ e −3 x 2⎝ 3 ⎠
7 ⎛ 7⎞ 7 −⎜ ⎟ = 8 ⎝ 8⎠ 64
Numerical Methods
∞
0
7 8
∞ 0
f ( x, y ) =
Here,
=1
1− x τ
The Euler’s method equation is given by xi +1 = xi + h f ( xi , yi )
a 1 + = 1⇒ a = 2 4 2
⇒
⎛ 1 − xi ⎞ xi +1 = xi + h ⎜ ⎝ τ ⎟⎠
⇒
h ⎛ h⎞ xi +1 = ⎜1 − ⎟ xi + ⎝ τ⎠ τ
113. (a) Let the first ball be red. Then P(red) in the second draw = 4/9.
For stability 1 −
h 9.96 ⇒ n = 10
Transform Theory
Applying Fourier transform, Y (ω ) = jω X (ω )
125. (d) Let us take g(t) = f (t/2 − 3/2) This
g (3) = f (0 ) ,
gives
⇒ if X (ω ) is real, Y (ω ) is imaginary.
which
is
true
and
g (5) = f (5/ 2 − 3/ 2) = f (1) , which is true.
131. (b) Given the Laplace transform of
126. (c) Laplace transform of 3
5
∞
0
3
5
g (t ) = ∫ e − st ⋅ 0 dt + ∫ e − st ⋅1 dt + ∫ e − st ⋅ 0 dt
Given that g ( f ) =
=−
e
(e
s
−2 s
)
− 13 =
e
s
−2 s
127. (b) We have to find the Laplace transform of X (s) =
3s + 5 s + 10 s + 21
s →∞
Ch wise GATE_EE_Ch1-c_Exp.indd 66
(s
s ⋅ 3s + 5 2
∞
1 −3/ 2 ⎧ f (t ) ⎫ 1 L{g (t )} = L ⎨ ⎬ = ∫ { f (t )}ds = ∫ s ds 2s ⎩ 2t ⎭ 2 s ∞
=
1 ⎛ s −3/ 2 +1 ⎞ 1 = ( −2)[0 − s −1/ 2 ] = s −1/ 2 2 ⎜⎝ −3 / 2 + 1⎟⎠ s 2
132. (a) We are given
2
x[0+] = lim
2 t /π f (t ) = 2t 2t 0
(1 − e )
t π
1 πt
⇒ g (t ) =
5
⎡ e −5 s − e −3 s ⎤ ⎡ e − st ⎤ = =⎢ ⎥ ⎥ ⎢ s ⎦ ⎣ s ⎦3 ⎣ −3 s
f (t ) = 2
is s −3/ 2 .
No other answer satisfies these conditions.
−3 s
d x (t ) dt
)
+ 10 s + 21
[using initial value theorem]
j 10 t ⎪⎧e x (t ) = ⎨ ⎩⎪0
for t ≤ 1 for t > 1
11/21/2018 12:12:19 PM
Chapter 1 • Engineering Mathematics
Applying Fourier transform, we get 1
1
=
b 5 = ( s − a) 2 + b 2 ( s − 2) 2 + 52
2 sin (ω − 10 ) e = j (10 − ω ) −1 (ω − 10)
=
5 5 = s 2 + 4 − 4 s + 25 s 2 − 4 s + 29
=
−1
j (10 − ω )t
1
133. (a) Say, y[n] is first difference of x[n]. So g[n] = x[n] − x[n − 1] ⇒ Y ( z ) = X (1 − z −1 ) = X ( z ) − z −1 Z ( z ) Y ( z ) = [ 2 z + 4 − 4 z −1 + 3 z −2 ] − [2 + 4 z −1 − 4 z −2 ]
135. (b) f(-x) = -f(x) implies odd function. Odd functions have only sine terms in their Fourier expression. Therefore, correct option is (b). 136. (c) X1(jω) and X2(jω) are not conjugate symmetric. Therefore, x1(t) × x2(t) are not real. Fourier transform of
= 2 z + 4 − 4 z −1 + 3 z −2 − 2 − 4 z −1 + 4 −2 − 3 z −3 = 2 z + 2 − 8 z −1 + 7 z −2 − 3 z −3 134. (a) For the given function e 25 sin(5t )u(t ), the standard form of Laplace transform is
Ch wise GATE_EE_Ch1-c_Exp.indd 67
e at sin (bt )u(t ) ⎯LT ⎯→
X (ω ) = ∫ e j10⋅t ⋅ e − jωt dt = ∫ e j (10 −ω )t dt −1
67
x1 (t ) ⋅ x2 (t ) →
1 X 1 ( jω ) * X 2 ( jω ) 2π
X1(jω) * X2(jω) is conjugate symmetric, so x1(t) ⋅ x2(t) will be real.
11/21/2018 12:12:19 PM
Ch wise GATE_EE_Ch1-c_Exp.indd 68
11/21/2018 12:12:19 PM
Electric Circuits
CHAPTER2
Syllabus Network graph, KCL, KVL, Node and Mesh analysis, Transient response of dc and ac networks, Sinusoidal steady-state analysis, Resonance, Passive filters, Ideal current and voltage sources, Thevenin’s theorem, Norton’s theorem, Superposition theorem, Maximum power transfer theorem, Two-port networks, Three phase circuits, Power and power factor in ac circuits.
CHAPTER ANALYSIS Topic
GATE 2009
GATE 2010
GATE 2011
GATE 2012
GATE 2013
GATE 2014
GATE 2015
GATE 2016
GATE 2017
GATE 2018
Network Graph
1
KCL
2
2
KVL
1
Node and Mesh Analysis
2
1
1
2
1
2
1
1
Transient Response of DC and AC Networks
2
1
1
2
3
3
2
1
3
3
2
1
1
Sinusoidal Steady-State Analysis
1
Resonance Passive Filters
1
Ideal Current and Voltage Sources Thevenin’s Theorem
2
1
Norton’s Theorem
1
1
Superposition Theorem Maximum Power Transfer Theorem
1
Two-Port Networks
2
1
1
2
1
Three Phase Circuits Power and Power Factor in AC Circuits
1 1
2
2
2 1
1 3
2
3
2
2
1
2
2
2
3
IMPORTANT FORMULAS I. Circuit Concepts and Laws 1. Current density i J = A/m 2 A 2. Power dissipation
⎛ W⎞ J ⋅ E = (σ E ) ⋅ E = sE 2 ⎜ 3 ⎟ ⎝m ⎠
Ch wise GATE_EE_Ch2.indd 69
3. Ohm’s law (a) J = σ E (in network theory) where s = conductivity of material. dq (b) V = Ri Volts ⇒ V = R ⋅ (in circuit theory) dt (c) i = G ⋅V Ampere where G is the conductance (expressed in Siemens or 1 mho) = R
11/21/2018 12:05:13 PM
70
GATE EE Chapter-wise Solved Papers
4. Resistance R=
10. Voltage and current relationship in an inductor
l Ω σ⋅A
V = L⋅
5. Resistivity
di dt
11. Voltage and current relationship in a capacitor
1 Ωm σ 6. Power
12. In practical voltage sources
V2 Watts R 7. Energy associated with resistor
V = VS - IR S 13. In practical current sources
ρ=
i =C⋅
P=
E R = ∫ i 2 R ⋅ dt Joules
dV dt
I = IS -
V RS
8. Inductor (a) Total flux
ψ = Nϕ where N is the number of turns of coil and f is flux per turn ψ = Li Webers
14. Parallel impedance and current division
II. Circuit Analysis Techniques
(b) Voltage across the inductor
dψ di = L ⋅ Volts dt dt (c) Current across the inductor is VL =
t
iL =
1 ∫ V ⋅ dt Amperes L -∞
(d) Power P = Li ⋅
di d ⎛ 1 2 ⎞ = ⎜ Li ⎟ Watts ⎠ dt dt ⎝ 2
(e) Energy in the inductor 1 L ⋅ i 2 Joules 2
WL =
9. Capacitor (a) Current through capacitor
i C = C.
dV dt
(b) Voltage across the capacitor 1 VC = C
(c) Power
P = CV
t
∫
i dt Volts
-∞
dV Watts dt
(d) Energy of the capacitor
Ch wise GATE_EE_Ch2.indd 70
W =
1 CV 2 Joules 2
I R2 I R1 Z = R ⇒ I1 = , I2 = R1 + R2 R1 + R2 I ⋅ C2 I ⋅ C1 1 Z= ⇒ I1 = , I2 = C1 + C2 C1 + C2 jw C I ⋅ L2 I ⋅ L1 Z = jw L ⇒ I1 = L + L , I 2 = L + L 1 2 1 2 15. Series impedance and voltage division VR1 VR2 Z = R ⇒ V1 = , V2 = R1 + R2 R1 + R2 VL1 VL2 Z = jw L ⇒ V1 = L + L , V2 = L + L 1 2 1 2 VC2 VC1 1 Z= ⇒ V1 = , V2 = C1 + C2 C1 + C2 jw C 16. Mesh analysis: Steps to follow
(i) Identify the meshes.
(ii) Assign the mesh current.
(iii) Write the mesh equation using KVL + Ohm's law.
17. Nodal analysis: Steps to follow
(i) Identify the nodes.
(ii) Assign the node voltage and ground node.
(iii) Write nodal equations using KCL + Ohm's law.
III. Topology 18. For an n-node graph, the number of tree branches, is n − 1.
11/21/2018 12:05:16 PM
Chapter 2 • Electric Circuits
Number of possible trees = det[Ar][Ar]T where A is the reduced incidence matrix and [A]T is its transpose.
30. Tellengens theorem
19. Order of incidence matrix is n × b
where b is the number of branches.
20. Number of branches of a co-tree is (b − n + 1)
31. Millman’s theorem
21. Number of independent KCL equation = Number of twigs.
b
∑v i k =1
V′ =
22. The rank of complete incidence matrix is (n − 1). 23. If a network contains b branches and n nodes, then the number of mesh currents is b − (n − 1). 24. Number of independent KVL equation = Number of links. 25. Inter-relationships between different matrices are as follows:
Bt = [IAt-1Al]T
Ql = At-1Al
BQT = QBT = 0
26. Superposition theorem R2 RR V+ 1 2 I t = tV + tI = R1 + R2 R 1 + R2
27. Thevenin’s theorem I L =
VTh R Th + R L
28. Maximum power theorem (a) Under variable load conditions, RS and RL Pmax =
2 S
V Watts 4RL
(b) For XL variable
1 Rn
32. Delta to Wye Conversion ZA =
Z1 Z3 Z1 + Z 2 + Z3
ZC =
Z1 Z3 Z1 + Z 2 + Z3
ZB =
Z1 Z 2 Z1 + Z 2 + Z3
Z1 = Z A + Z B +
ZA ZB ZC
Z3 = Z C + Z A +
ZC Z A ZB
Z2 = ZB + ZC +
ZB ZC ZA
V. Laplace Transform 34. Laplace transform of a signal (bilateral Laplace transform) ∞
L[ f (t )] =
∫
f (t )e - st dt
-∞
35. Laplace transform of causal signal (unilateral Laplace transform): ∞
VS RL Pmax= ( R S + R L ) 2 + X S2 VS2 VS RL 4 R Ssimultaneously (c) Both RL and XL are varied ( R S + R L ) 2 + X S2 Pmax =
VS2 4 RS
29. Reciprocity theorem V1 V2 = I1 I 2
Ch wise GATE_EE_Ch2.indd 71
1 G1 + G2 + + Gn
33. Wye to Delta Conversion
IV. Network Theorems
where Gn =
=0
V1G1 + V2G2 + V3G3 + + VnGn G1 + G2 + + Gn
R′ =
k k
71
L[ f (t )] = ∫ f (t )e - st dt 0
36. Differentiating property L[tf (t )] =
-d F ( s) ds
37. Integrating property ∞
⎡ f (t ) ⎤ L⎢ ⎥ = ∫ F ( s) ds ⎣ t ⎦ s 38. Shifting theorem f (t - a ) u (t - a ) ↔ e - as F ( s )
11/21/2018 12:05:18 PM
72
GATE EE Chapter-wise Solved Papers
39. (a) Initial value theorem
(c) Source free parallel RLC circuit
lim f (t ) = lim[ sF ( s)] t →0
Cs 2 +
s →∞
(b) Final value theorem
(d) Source free series RLC circuit
lim f (t ) = lim[ sF ( s)] t →∞
s→0
40. Any function convolving with the impulse signal results in the same function.
Ls 2 + Rs +
i (t ) =
41. Laplace transform for the periodic function
(General formula for calculating current through inductor.)
(f) Source driven RC circuits
42. Initial and steady state (a) The inductor current at t = 0− and t = 0+
i (t ) =
i L (0 + ) = iL (0 - )
E L( 0 + ) = E L ( 0 - ) (c) If VL(t) = d(t), then
(General formula for calculating voltage through capacitor.)
(g) Source driven RLC circuit in parallel
1 L (d) The capacitor voltage at t = 0− and t = 0+ i L (t ) = i L ( 0 - ) +
V (t ) = L
EL(0+) = EC(0−)
(h) Source driven RLC circuit in series
(e) For all excitations except impulse, VC (0 + ) = VC (0 - ) +
1 C
i (t ) = C
44. AC Transients
i(t ) = I 0 e - ( R / L )t for t ≥ 0 i(t ) = I 0 e - t / t where t =
(a) Phase notations (i) Voltage excitation
L seconds R
V (t ) = I m cos(w t + θ )
(b) Source free RC circuit (for t ≥ 0)
where, t = RC seconds
(iii) Impedance phasor
V I (iv) Admittance phasor
(for t ≥ 0) di (t ) VL (t ) = L L , (for t ≥ 0) dt dVC (t ) dt
(ii) Current phasor I = I m e jϕ = I m ∠ϕ = I m (cos θ + j sin θ )
VC (t ) = VC (∞) + [VC (0) - VL (∞)]e - t / t
i L (t ) = C
dV (t ) (capacitor current) dt
V d 2V (t ) R dV (t ) 1 + + V (t ) = 0 2 L dt LC LC dt
43. DC transients (a) Source free RL circuit
-t /t
di(t ) (inductor voltage) dt
I d 2 i (t ) 1 di(t ) i(t ) + + = 0 2 RC dt LC LC dt
VC(0+) = VC(0−)
V (t ) = V0 e
V0 - t / RC e R
VC (t ) = [VC (0 + ) - VC (∞)]e - t / RC + VC (∞)
(b) For all excitations except impulse,
Ch wise GATE_EE_Ch2.indd 72
V0 (1 - e - tR / L ) R
i(t ) = [i(0 + ) - i(∞) e - tR / L + i(∞)]
1 F1 ( s) 1 - e -Ts
VI. Transients
1 =0 C
(e) Source driven RL circuit
f (t) ∗ δ(t − t0) → f (t − t0)
F ( s) =
1 1 s+ = 0 R L
Z=
Y =
I V
11/21/2018 12:05:21 PM
Chapter 2 • Electric Circuits
VII. Sinusoidal Steady-State Analysis using Phasors
(b) RL circuit with AC excitation: (i) Current response
45. Phasor relationship for circuit elements
i(t) = itr(t) + iss(t) = ke i(t ) = ke
R - t L
R - t L
(a) When current and voltage are in the same phase
+ iss (t )
V = RI (b) When current lags the voltage by 90°
⎛ ⎛ wL⎞ ⎞ sin ⎜ w t + ϕ - tan -1 ⎜ ⎝ R ⎟⎠ ⎟⎠ ⎝ R + (w L) Vm
+
2
V = jw LI
2
(c) When the current leads the voltage by 90°
(by Laplace transform) ⎛ ⎛ wL⎞ ⎞ .sin ⎜ w t + ϕ - tan -1 ⎜ ⎝ R ⎟⎠ ⎟⎠ ⎝ R + (w L)
2
(ii) For voltage excitation V (t ) = Vm cos(w t + ϕ ) i(t ) = ke
R - t L
+
⎛ ⎛ L⎞⎞ cos ⎜ w t + ϕ - tan -1 ⎜ ⎟ ⎟ ⎝ R⎠ ⎠ ⎝ R + (w L) 2
⎛ wL⎞ π ϕ - tan -1 ⎜ = ⎝ R ⎟⎠ 2
Voltage V = VR2 + VL2
⎛V ⎞ Impedance angle ϕ = tan -1 ⎜ L ⎟ ⎝ VR ⎠
Power factor = cosf (lagging)
⎛ wL⎞ π ϕ = tan -1 ⎜ + (at t = 0) ⎝ R ⎟⎠ 2 ⎛ wL⎞ π + (at t = t0) w t0 + ϕ = tan -1 ⎜ ⎝ R ⎟⎠ 2 (c) RC circuits with AC excitations (i) Capacitor voltage
(b) Series RC circuit
Voltage V = VC2 + VR2
⎛V ⎞ Impedance angle ϕ = tan -1 ⎜ C ⎟ ⎝ V2 ⎠
Power factor = cos f (leading)
(c) Parameters for Series RLC circuit conditions
(d) Parallel RL Circuit
VC(t) = VC(tr)(t) + VC(ss)(t) Vm sin(w t + ϕ - tan -1 (w CR)) 2 1 + (w CR)
VC (t ) = VC( tr ) (t ) +
(ii) For voltage excitation V (t ) = Vm cos(w t + ϕ ) VC (t ) = ke
46. Phasor diagrams (a) Series RL circuit
Vm
2
(iii) Conditions for transient-free response
-
t RC
+
Vm 1 + (wCR) 2
.cos(w t + ϕ - tan (wCR)) -1
(iii) Conditions for transient-free response π ϕ = tan -1 (w CR) + 2
Ch wise GATE_EE_Ch2.indd 73
I jw C
2
(by phasor method)
I = jw CV and V =
Vm
iss (t ) =
73
ϕ = tan -1 (w CR) +
π (at t = 0) 2
ϕ + w 0 t0 = tan -1 (w CR) +
π (at t = t0) 2
I = I R2 + I L2
⎛I ⎞ ϕ = tan -1 ⎜ L ⎟ ⎝ IR ⎠
Power factor = cos ϕ (lagging)
(e) Parallel RC circuit
I = I R2 + I C2
⎛I ⎞ ϕ = tan -1 ⎜ C ⎟ ⎝ IR ⎠
Power factor = cos ϕ =(leading) (f) Parameters for parallel RLC circuit conditions
11/21/2018 12:05:23 PM
74
GATE EE Chapter-wise Solved Papers
VIII. Magnetically Coupled Circuits
(b) Capacitor current
47. Mutual inductance V2 (t ) = M 21
IC =
di1 (t ) dt
(c) Frequency response
di (t ) V1 (t ) = M12 2 dt
V = IRjw 0C = w 0 RC I ∠90° = QI ∠90° ZC
V =
M12 = M21 = M
48. For two inductors connected in series (a) For current entering dotted terminals in both the inductors
ϕ = L1i1 + Mi2 + L2 i2 + Mi1
i1 = i2 = i L = L1 + L2 + 2 M
(b) For current entering on dotted and second undotted terminal
ϕ = L1i1 - Mi2 + L2 i2 - Mi1 i1 = i2 = i
49. For two inductors connected in parallel (a) Leq =
Bandwidth =
L1 L2 - M 2 L1 + L2 - 2 M
53. High-pass filter Vo ( s) RCs = (first-order) Vi ( s) 1 + RCs 54. Band-pass filters: Necessary condition to obtain from combination of LPF and HPF fc (LPF) > fc (HPF)
55. Band elimination (band stop) filters: Necessary condition to obtain from combination of HPF and LPF fc (HPF) > fc (LPF)
IX. Resonance
50. Series resonance (Series RLC circuit) (a) Quality factor
XI. Three Phase Circuits 56. Balanced three phase circuits
w0 L 1 1 L = = R w 0 RC R C
(b) Capacitor voltage VC = QV ∠( -90°)
V = Z
V 1 ⎞ ⎛ R2 + ⎜ w L ⎟ ⎝ w C⎠
fH - fL =
2
f0 Q
51. Parallel resonance (Parallel RLC circuit) (a) Quality factor R Q= = w 0CR w0 L
Ch wise GATE_EE_Ch2.indd 74
In Y connection, I ph = I L and Vph =
In ∆ connection, I ph =
(c) Frequency response I =
f0 Q
Vo ( s) 1 = (first-order) Vi ( s) 1 + RCs
L L - M2 (b) Leq = 1 2 L1 + L2 + 2 M
Q=
2
X. Filters
L = L1 + L2 - 2 M
1 ⎛ 1 ⎞ + ⎜w C ⎟ wL⎠ R2 ⎝
52. Low-pass filter
I
IL 3
VL 3
and Vph = VL
XII. AC Power Analysis 57. Average power (a) Average of the instantaneous power over all period T
P=
1 p(t )dt T ∫0
(b) Average power for periodic waveform P=
1 T
t1 + T
∫
p(t )dt
t1
11/21/2018 12:05:25 PM
Chapter 2 • Electric Circuits
(c) Average power in the sinusoidal steady state 1 P = Vm I m cos(w t - ϕ ) 2 (d) Average power absorbed by an ideal resistor PR =
Vm2 2R
(e) Average power absorbed by purely reactive element PX = 0 (f) Average power for non-periodic functions P=
1 2 ( I m1 + I m2 2 + ...) R 2
58. Effective or rms values (a) For any function 1 2 x (t ) dt T ∫0
(b) Power 2 P = I eff R T
1 2 i dt T ∫0
where I = eff
T
1 2 I m cos 2 (w t + ϕ ) dt = T ∫0
Im 2
(d) Average power from effective values of current and voltage P = Veff I eff cos(θ - ϕ ) =
Veff2 R
(e) Effective values with multiple frequency circuits
(
P= I
2 eff
+I
2 2 eff
)
+ R
2 I eff = I1eff + I 22eff +
(f) Power factor =
Average power Apparant power =
I1 = Y11V1 + Y12V2
I2 = Y21V1 + Y22V2
Symmetric condition Y11 = Y22
Reciprocal condition Y12 = Y21
61. T or Transmission (A, B, C, D) parameters
V1 = AV2 − BI2
I1 = CV2 − DI2
Symmetric condition A = D
Reciprocal condition |AD − BC | = 1
V2 = A′V1 − B ′I1
I2 = C ′V1 − D ′I1
Symmetric condition A′ = D ′
Reciprocal condition |A′D ′ − B ′C ′| = 1
63. h or Hybrid parameters
(c) Effective (rms) value of current in sinusoidal wave I eff =
60. Y-parameters or short circuit admittance parameters
62. Inverse transmission (A′B ′C ′D ′) parameters T
X rms =
P = cos(θ - ϕ ) Veff I eff
V1 = h11I1 + h12V2
I2 = h21I1 + h22V2
Symmetric condition |h| = 1 ⇒ |h11h22 − h21h12| = 1
Reciprocal condition h12 = −h21
64. g-Parameters
I1 = g11V1 + g12I2
V2 = g21V1 + g22I2
Symmetric condition |g| = 1 ⇒ |g11g22 − g12g21| = 1 Reciprocal condition g12 = −g21
65. Relation among the two-port network parameters: Y = Z −1
T ′ = T −1
XIII. Two-Port Network
g = h−1
59. Z-parameters or open circuit impedance parameters V1 = Z11I1 + Z12I2
66. Interconnections of two-port network
V2 = Z21I1 + Z22I2
Symmetric condition Z11 = Z22
Reciprocal condition Z12 = Z21
Ch wise GATE_EE_Ch2.indd 75
75
(a) Series connection: [ Z ] = [ Z A ] + [ Z B ] (b) Parallel connection: [Y ] = [YA ] + [YB ] (c) Cascade connection: [T ] = [TA ].[TB ]
11/21/2018 12:05:27 PM
76
GATE EE Chapter-wise Solved Papers
QUESTIONS 1. A two-port device is defined by the following pair of equations: i1 = 2v1 + v2 and i2 = v1 + v2 Its impedance parameters (Z11, Z12, Z21, Z22) are given by (a) (2, 1, 1, 1) (b) (1, −1, −1, 2) (c) (1, 1, 1, 2) (d) (2, −1, −1, 1) (GATE 2000: 2 Marks) 2. The circuit shown in the following figure is equivalent to a load of 2W
I 4W
(a)
4 ohms 3
+ −
(b)
(c) 4 ohms
2I
8 ohms 3
(d) 2 ohms (GATE 2000: 2 Marks)
3. The impedance seen by the source in the circuit in the following figure, is given by
4Ω
−j2 Ω
1:4 Z L = 10 Ð 30°
+ −
5. 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 W bulb glows brighter (GATE 2001: 1 Mark) 6. 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. (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. (GATE 2001: 1 Mark) 7. 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 (GATE 2001: 1 Mark) 8. A 50 Hz balanced three-phase, Y-connected supply is connected to a balanced three phase Y-connected load. If the instantaneous phase-α of the supply voltage is V cos(w t ) and the phase-α of the load current is I cos(w t - ϕ ), the instantaneous three-phase power is (a) a constant with a magnitude of VI cos ϕ (b) a constant with a magnitude of (3/ 2)VI cos ϕ (c) time-varying
(a) (b) (c) (d)
with an average value (3/ 2)VI cos ϕ and a frequency of 100 Hz
(0.54 + j0.313) ohms (4 – j2) ohms (4.54 – j1.69) ohms (4 + j2) ohms
(d) time-varying with an average value of VI cos ϕ and a frequency of 50 Hz (GATE 2001: 1 Mark) (GATE 2000: 2 Marks)
4. 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. (GATE 2001: 1 Mark)
Ch wise GATE_EE_Ch2.indd 76
of
9. 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 (GATE 2001: 2 Marks)
11/21/2018 12:05:28 PM
77
Chapter 2 • Electric Circuits
10. Consider the star network shown in in the following figure. The resistance between terminals A and B with C open is 6 Ω, between terminals B and C with A open is 11 Ω, and between terminals C and A with B open is 9 Ω. Then
I1
I2
2Ω E1
2Ω
2Ω E2
A The admittance parameters Y11, Y12, Y21 Y22 for the network shown are
RA RB
RC
(a) 0.5 mho, 1 mho, 2 mho and 1 mho respectively (b)
B C
(c) 0.5 mho, 0.5 mho, 1.5 mho and 2 mho respectively
(a) RA = 4 Ω, RB = 2 Ω, RC = 5 Ω (b) RA = 2 Ω, RB = 4 Ω, RC = 7 Ω
3 3 2 2 mho, - mho, mho and mho respec5 7 7 5 tively (GATE 2002: 2 Marks)
(d) -
(c) RA = 3 Ω, RB = 3 Ω, RC = 4 Ω (d) RA = 5 Ω, RB = 1 Ω, RC = 10 Ω (GATE 2001: 2 Marks) 11. A 240 V single-phase AC source is connected to a load with an impedance of 10 ∠60°Ω. A capacitor is connected in parallel with the load. If the capacitor supplies 1250-VAR, the real power supplied by the source is (a) 3600 W (b) 2880 W (c) 2400 W (d) 1200 W (GATE 2001: 2 Marks) 12. A current impulse, 5δ (t), is forced through a capacitor C. The voltage, VC(t), across the capacitor is given by (a) 5t (b) 5u(t) – C (c)
5 5u (t ) t (d) C C (GATE 2002: 1 Mark)
13. 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 (GATE 2002: 1 Mark) 14. A two port network, as shown in the following figure is described by the following equations I1 = Y11E1 + Y12E2 I1 = Y21E1 + Y22E2
Ch wise GATE_EE_Ch2.indd 77
1 1 1 1 mho, - mho, - mho and mho respec3 6 6 3 tively
15. In the circuit as shown in the following figure, what value of C will cause a unity power factor at the AC source?
230 V C 50 Hz
(a) 68.1 μF (c) 0.681 μF
ZL = 30 Ð 40°
(b) 165 μF (d) 6.81 μF (GATE 2002: 2 Marks)
16. A first order, low pass filter is given with R = 50 Ω and C = 5 μF. 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 (c) 2.46 kHz (d) 24.6 kHz (GATE 2002: 2 Marks) 17. A series RLC circuit has R = 50 Ω, L = 100 μH and C = 1 μF. 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 (GATE 2002: 2 Marks) 18. In the circuit shown in the following figure, it is found that the input AC voltage (v) and current i are in phase. M The coupling coefficient is K = , where M is the L1 L2
11/21/2018 12:05:29 PM
78
GATE EE Chapter-wise Solved Papers
mutual inductance between the two coils. The value of K and the dot polarity of the coil PQ are
6A
K t Q
P −j12 W
10 W
L1
V1
(a) (b) (c) (d)
j8 W
0
j8 W −
2s
4s
(a) 144 J (c) 132 J
(b) 98 J (d) 168 J (GATE 2003: 1 Mark)
L2
22. A segment of a circuit is shown in the figure given below. vR = 5 V, vC = 4 sin 2t. The voltage v L is given by
K = 0.25 and dot at P K = 0.5 and dot at P K = 0.25 and dot at Q K = 0.5 and dot at Q
Q
(GATE 2002: 2 Marks) P 19. Consider the circuit shown in the following figure. If the frequency of the source is 50 Hz, then a value of t0 which results in a transient free response is 5W
+ 5 Ω vR − 1F 1A + − + vC 2A 2 H vL − S
0.01 H (a) 3 - 8 cos 2t (c) 16 sin 2t
t = t0 sin(w t)
R
(b) 32 sin 2t (d) 16 cos 2t (GATE 2003: 1 Mark)
23. In the following figure, Z1 = 10 ∠( - 60°), Z 2 = 10 ∠60°, Z3 = 50 ∠53.13° (b) 1.78 ms . Z1 =2.91 10 ∠ms ( - 60°), Z 2 = 10 ∠ (d) 60°, Z3 = 50 ∠53.13° (GATE 2002: 2 Marks) Thevenin impedance seen from X-Y is
(a) 0 ms (c) 2.71 ms
20. In the circuit shown in the following figure, the switch is closed at time t = 0. The steady state value of the voltage VC is
+
Z1
X
Z3
100∠0°
Z2 Y
t = to + 10 V −
1 mH 1Ω
(a) 56.6∠45° (c) 70∠30°
1 mH
1Ω
+ −
VC
(b) 60∠30° (d) 34.4∠65° (GATE 2003: 1 Mark)
24. In the circuit of figure given below, the magnitudes of VL and VC are twice that of VR. The inductance of the coil is VR
(a) 0 V (c) 5 V
(b) 10 V (d) 2.5 V (GATE 2002: 2 Marks)
5Ω
VC C
5∠0° 21. Figure given below shows the waveform of the c urrent passing through an inductor of resistance 1 Ω and inductance 2 H. The energy absorbed by the inductor in the first four seconds is
Ch wise GATE_EE_Ch2.indd 78
L
VL
11/21/2018 12:05:31 PM
79
Chapter 2 • Electric Circuits
(a) 2.14 mH (c) 31.8 mH
(b) 5.30 H (d) 1.32 H (GATE 2003: 2 Marks)
28. In the circuit shown in the following figure, the switch S is closed at time t = 0. The voltage across the inductance at t = 0+, is 3Ω
25. In the figure given below, the potential difference between points P and Q is
S 4F
2A 10 V 2Ω
+ −
4Ω
P
4Ω
4Ω
4H
Q + (a) 2 V (c) -6 V
10 V
(b) 4 V (d) 8 V (GATE 2003: 2 Marks)
8Ω
6Ω (b) 10 V (d) 8 V (GATE 2003: 2 Marks)
29. The h-parameters for a two-port network are defined by
26. Two AC sources feed a common variable resistive load as shown in the following figure. Under the maximum power transfer condition, the power absorbed by the load resistance RL is
For the two-port network shown in the figure given below, the value of h12 is given by
(a) 12 V (c) -6 V
⎡ E1 ⎤ ⎡ h 11 ⎢ ⎥ = ⎢h ⎣ I 2 ⎦ ⎣ 21
h 12 ⎤ ⎡ I1 ⎤ h 22 ⎥⎦ ⎢⎣ E2 ⎥⎦
4Ω
2Ω
2Ω
I1 6Ω
j8 Ω
6Ω
I2
j8 Ω E1
2Ω
4Ω
E2
RL 110∠0°
90∠0°
(a) 2200 W (c) 1000 W
(b) 1250 W (d) 625 W (GATE 2003: 2 Marks)
RΩ
+ 100 V −
(a) 10 Ω (c) 24 Ω
Ch wise GATE_EE_Ch2.indd 79
(GATE 2003: 2 Marks)
1Ω
10 A
(b) 0.167 (d) 0.25
30. The value of Z in the following figure which is most appropriate to cause parallel resonance at 500 Hz is
27. In the given figure, the value of R is
14 Ω
(a) 0.125 (c) 0.625
5Ω 5A
2Ω
2H
Z
+ − 40 V
(b) 18 Ω (d) 12 Ω (GATE 2003: 2 Marks)
(a) 125.00 mH (c) 2.0 mF
(b) 304.20 mF (d) 0.05 mF (GATE 2004: 1 Mark)
11/21/2018 12:05:32 PM
80
GATE EE Chapter-wise Solved Papers
31. The rms value of the periodic waveform given in the following figure is I
YR
YL
YC
E = 10∠10° V
6A t T/2 −6 A
35. In the figure given below, the value of resistance R in Ω is
(a) 2 6A (b) 6 2A (c)
4 / 3A
(b) 5 - 18 j (d) 5 - 12 j (GATE 2004: 2 Marks)
(a) 1.5 + 0.5 j (c) 0.5 + 1.8 j
T
V1
(d) 1.5 A (GATE 2004: 2 Marks)
10 Ω
32. In the figure given below, the value of the source voltage is
100 V
2A
10 Ω
+ −
R
P 2A 10 Ω
6Ω 6Ω
1A
(a) 10 (c) 30
+ − E
36. In the following figure, the capacitor initially has a charge of 10 Coulombs. The current in the circuit one second after the switch S is closed will be
Q (a) 12 V (c) 30 V
(b) 20 (d) 40 (GATE 2004: 2 Marks)
(b) 24 V (d) 44 V (GATE 2004: 2 Marks)
S
2Ω +
33. In the following figure, Ra, Rb and Rc are 20 Ω, 10 Ω and 10 Ω, respectively. The resistances R1, R2 and R3 in Ω of an equivalent star-connection are
100 V
+ −
−
0.5 F
a (a) 14.7 A (c) 40.0 A
a R1
Rb
c
Rc
Ra (a) 2.5, 5, 5 (c) 5, 5, 2.5
R3 b c
R2 b
(b) 5, 2.5, 5 (d) 2.5, 5, 2.5 (GATE 2004: 2 Marks)
34. In the figure given below, the admittance values of the elements in Siemens are YR = 0.5 + 0j, YL = 0 - 1.5 j, YC = 0 + 0.3 j, respectively. The value of I as a phasor when the voltage E across the elements is 10 ∠10° V is
Ch wise GATE_EE_Ch2.indd 80
(b) 18.5 A (d) 50.0 A (GATE 2004: 2 Marks)
37. 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 A is (a) 14.1 A (b) 17.3 A (c) 22.4 A (d) 30.0 A (GATE 2004: 2 Marks) 38. The Z matrix of a 2-port network as given by
⎡0.9 0.2⎤ ⎢0.2 0.6 ⎥ ⎦ ⎣ The element Y22 of the corresponding Y matrix of the same network is given by
11/21/2018 12:05:33 PM
81
Chapter 2 • Electric Circuits
(a) 1.2 (c) -0.4
(b) 0.4 (d) 1.8 (GATE 2004: 2 Marks)
R
L
39. In the figure given below, the value of R is 8A
R
10 Ω
100 V
(a) 2.5 Ω (c) 7.5 Ω
10 Ω
(b) 5.0 Ω (d) 10.0 Ω (GATE 2005: 1 Mark)
(a)
5 2 urms urms 8 3 (b)
(c)
8 3 urms urms 5 2 (d) (GATE 2005: 2 Marks)
43. For the three-phase circuit shown in the figure given below ratio of the currents IR : IY : IB is given by IR
40. RMS value of the voltage u(t) = 3 + 4 cos (3t) is (a)
17 V
R
(b) 5 V (d) (3 + 2 2 )V (GATE 2005: 1 Mark)
(c) 7 V
41. For the two-port network shown in the figure given below, the Z-matrix is given by
R1 Balanced three-phase voltage source
IB B R1 IY Y
Z2
i1
v1
Z1
⎡ Z1 (a) ⎢ ⎣ Z1 + Z 2
Z1 + Z 2 ⎤ Z 2 ⎥⎦
⎡ Z1 (b) ⎢ ⎣ Z1 + Z 2
Z1 ⎤ Z 2 ⎥⎦
i2
v2
(a) 1 : 1 : 3
(b) 1 : 1 : 2
(c) 1 : 1 : 0
(d) 1 : 1 :
3 2 (GATE 2005: 2 Marks)
44. For the triangular wave form shown in the figure given below, rms value of the voltage is equal to V(t) 1
⎡ Z1 (c) ⎢ ⎣ Z2
Z2 ⎤ Z1 + Z 2 ⎥⎦
⎡ Z1 (d) ⎢ ⎣ Z1
Z1 ⎤ Z1 + Z 2 ⎥⎦
T/2 (a)
(GATE 2005: 1 Mark) 42. The RL circuit of the figure is fed from a constant magnitude, variable frequency sinusoidal voltage source vin . At 100 Hz, the R and L elements each have a voltage drop urms. If the frequency of the source is changed to 50 Hz, then new voltage drop across R is
Ch wise GATE_EE_Ch2.indd 81
(c)
T
3T/2
2T
t
1 1 (b) 6 3 2 1 (d) 3 3 (GATE 2005: 2 Marks)
45. 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
11/21/2018 12:05:35 PM
82
GATE EE Chapter-wise Solved Papers
X
10 Ω 10 V
10 mH
(a) 0 A (c) 1 A
t=0
(b) 0.5 A (d) 2 A (GATE 2005: 2 Marks)
46. In the given figure, the Thevenin’s equivalent pair (voltage and impedance), as seen at the terminals PQ, is given by 10 Ω
Y (a)
(b) 2∠ 45° V, (1 + 2 j ) Ω (c) 2∠ 45° V, (1 + j ) Ω (d)
P
2∠0 V, (1 + 2 j ) Ω
2∠ 45°V , (1 + j ) Ω (GATE 2006: 1 Mark)
20 Ω
10 Ω
4V
Unknown network
Q (a) 2 V, 5 Ω (c) 4 V, 5 Ω
(b) 2 V, 7.5 Ω (d) 4 V, 7.5 Ω (GATE 2005: 2 Marks)
50. The three-limbed non-ideal core shown in the figure given below has three windings with nominal inductances L each when measured individually with a single phase AC source. The inductance of the windings as connected will be (a) very low (b) L/3 (c) 3L (d) very high R
Y
B
Linked Answer Questions 47 and 48: A coil of inductance 10 H and resistance 40 Ω 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. 47. If, at t = 0+, the voltage across the coil is 120 V, the value of resistance R is 20 Ω
1 2
S
10 H
120 V R
(a) 0 Ω (c) 40 Ω
51. The parameters of the circuit shown in the figure given below are Ri = 1 MΩ, Ro = 10 Ω, A = 106 V/V. If Vi = 1 mV, then output voltage, input impedance and output impedance, respectively, are
40 Ω
49. In the figure given below, the current source is I < 0 A, R = 1 Ω, the impedances are ZC = -j Ω and ZL = 2 j Ω. The Thevenin equivalent looking into the circuit across XY is
Ro
Ri +
(b) 20 Ω (d) 60 Ω (GATE 2005: 2 Marks)
48. For the value of obtained in the above question, the time taken for 95% of the stored energy to be dissipated is close to (a) 0.10 s (b) 0.15 s (c) 0.50 s (d) 1.0 s (GATE 2005: 2 Marks)
Ch wise GATE_EE_Ch2.indd 82
(GATE 2006: 1 Mark)
Vi
+ −
AVi
− (a) 1 V, ∞, 10 Ω (c) l V, 0, ∞
(b) l V, 0, 10 Ω (d) 10 V, ∞, 10 Ω (GATE 2006: 2 Marks)
52. In the circuit shown in the figure given below, current source I = 1 A, voltage source V = 5 V, R1 = R2 = R3 = 1 Ω, L1 = L2 = L3 = 1 H, C1 = C2 = 1 F. The currents (in A) through R3 and the voltage source V, respectively, will be
11/21/2018 12:05:36 PM
83
Chapter 2 • Electric Circuits
(b) R1
R2
L1
V2 I
L3 + −
C1 C2
L2
V1 V
R3
I VC (a) 1, 4 (c) 5, 2
(b) 5, 1 (d) 5, 4 (GATE 2006: 2 Marks)
(c) VC
V1
53. The parameter type and the matrix representation of the relevant two-port parameters that describe the circuit shown are I1
I2
+
+
V1
V2
−
−
V2 I (d) V1 V2
⎡0 0 ⎤ (a) Z parameters, ⎢ ⎥ ⎣0 0⎦
I VC
⎡ 1 0⎤ (b) h parameters, ⎢ ⎥ ⎣0 1⎦
(GATE 2006: 2 Marks)
⎡0 0 ⎤ (c) h parameters, ⎢ ⎥ ⎣0 0⎦ ⎡ 1 0⎤ (d) Z parameters, ⎢ ⎥ ⎣0 1⎦
(GATE 2006: 2 Marks)
54. The circuit shown in the following figure is energised by a sinusoidal voltage source V1 at a frequency which causes resonance with a current of I.
55. 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 1 , the voltage across the capacitor at time let w 0 = LC t > 0 is given by (a) V0 (b) V0cos(w0t) V0 e - w0 t cos (w 0 t ) (c) V0sin (w0t) (d) (GATE 2006: 2 Marks) 56. The state equation for the current I1 shown in the network given below in terms of the voltage Vx and the independent source V, is given by
I
3 Ω 0.2 H
V2 V1 V
VC
+ −
+ Vx
− I1
5Ω I2 0.5 H
− + 0.2 Vx
The phasor diagram which is applicable to this c ircuit is (a) I VC
Ch wise GATE_EE_Ch2.indd 83
V2
(a)
dI1 5 = -1.4Vx - 3.75 I1 + V dt 4
(b)
dI1 5 = -1.4Vx - 3.75 I1 - V dt 4
V1
11/21/2018 12:05:39 PM
84
GATE EE Chapter-wise Solved Papers
(c)
dI1 5 = -1.4Vx + 3.75 I1 + V dt 4
(d)
dI1 5 = -1.4Vx - 3.75 I1 - V dt 4
59. The matrix A given below is the node incidence matrix of a network. The columns correspond to branches of the network while the rows correspond to nodes.
(GATE 2007: 2 Marks) 57. In the circuit shown in the following figure switch SW1 is initially closed and SW2 is open. The inductor L carries a current of 10 A and the capacitor is charged to 10 V with polarities as indicated. SW2 is initially closed at t = 0- and SW1 is opened at t = 0. The current through C and the voltage across L at t = 0+ is SW2 SW1 R1 10 Ω
L
R2 10 Ω 10 A C
Let V = [v1v2 v6 ]T denote the vector of branch volt ages 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. Which of the following statements is true?
+ 10 V −
(a) 55 A, 4.5 V (c) 45 A, 5.5 V
⎡ 1 1 1 0 0 0⎤ ⎢ 0 -1 0 -1 1 0 ⎥ ⎥ A= ⎢ ⎢ -1 0 0 0 -1 -1⎥ ⎢ ⎥ 1⎦ ⎣ 0 0 -1 1 0
(b) 5.5 A, 45 V (d) 4.5 A, 55 V (GATE 2007: 2 Marks)
58. In the figure given below, all phasors are with r eference to the potential at point “O”. The locus of voltage phasor VYX as R is varied from zero to infinity is shown by
R V∠0°
(a) The equations v1 - v2 + v3 = 0, v3 + v4 - v5 = 0 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 = A V (d) A V = 0 are KVL equations for the network (GATE 2007: 2 Marks) 60. The number of chords in the graph of the given circuit will be
VYX X Y + −
V∠0°
O (a) O
(a) 3 (c) 5
2 V (b) VYX VYX
Locus of VYX (c) O
2V O Locus of VYX
2 V (d) VYX
Locus of VYX
VYX 2V O Locus of VYX (GATE 2007: 2 Marks)
Ch wise GATE_EE_Ch2.indd 84
(b) 4 (d) 6 (GATE 2008: 1 Mark)
61. The Thevenin’s equivalent of a circuit operating at w = 5 rad/s, has V∞ = 3.71∠ -15.9° V and Z0 = 2.38 - j 0.667 Ω. At this frequency, the minimal realization of the Thevenin’s impedance will have a (a) resistor and a capacitor and an inductor (b) resistor and a capacitor (c) resistor and an inductor (d) capacitor and an inductor (GATE 2008: 1 Mark)
11/21/2018 12:05:40 PM
85
Chapter 2 • Electric Circuits
1Ω
62. The time constant for the given circuit will be 1F
ab
3Ω
−b
3W
4Vab 1Ω
+ 5V − 1F
a+V
I
3W
1F
1Ω
+ −
3A (a) 0.31 A (c) 1.75A
(b) 1.25A (d) 2.5A (GATE 2008: 2 Marks)
1 (a) 1 s (b) s 9 4 Linked Answer Questions 67 and 68: The current I(t) sketched (c) 4 s (d) 9 s (GATE 2008: 2 Marks) in the figure flows through an initially uncharged 0.3 nF capacitor. 63. The resonant frequency for the given circuit will be 0.1 H I(t) mA
1F
1W
6 5 4 3 2 1
(a) 1 rad/s (c) 3 rad/s
(b) 2 rad/s (d) 4 rad/s (GATE 2008: 2 Marks)
64. Assuming ideal elements in the circuit shown below, the voltage Vab will be a + 1A
Vab
2Ω
I
− + 5V
− b (a) -3 V (c) 3 V
(b) 0 V (d) 5 V (GATE 2008: 2 Marks)
65. 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 permittivities 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) 983.33 pF (b) 1475 pF (c) 6637.5 pF (d) 9956.25 pF (GATE 2008: 2 Marks) 66. In the circuit shown in the figure given below, the value of the current i will be given by
Ch wise GATE_EE_Ch2.indd 85
0 1 2 3 4 5 6 7 8 9 t(ms) 67. The charge stored in the capacitor at t = 5 ms, will be (a) 8 nC (b) 10 nC (c) 13 nC (d) 16 nC (GATE 2008: 2 Marks) 68. The capacitor charged up to 5 ms, as per the c urrent profile given in the figure, is connected across an inductor of 0.6 mH. Then 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 (GATE 2008: 1 Mark) 69. The current through the 2 kΩ resistance in the circuit shown below is 1 kΩ C 1 kΩ A
B 2 kΩ 1 kΩ D
1 kΩ
6V (a) 0 mA (c) 2 mA
(b) 1 mA (d) 6 mA (GATE 2009: 1 Mark)
11/21/2018 12:05:41 PM
86
GATE EE Chapter-wise Solved Papers
70. 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 (GATE 2009: 1 Mark)
Linked Answer Questions 74 and 75: 2 kΩ
+−
5V + −
3 VAB A
2 kΩ
1 kΩ B
71. In the figure shown below, 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 S1
S2
C1
3V
1F
(a) 1 V (c) 1.5 V
C2
2F
(b) 2 V (d) 3 V (GATE 2009: 1 Mark)
74. For the circuit given above, Thevenin’s resistance across the terminals A and B is (a) 0.5 kΩ (b) 0.2 kΩ (c) 1 kΩ (d) 0.11 kΩ (GATE 2009: 2 Marks) 75. For the circuit given above, 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 (GATE 2009: 2 Marks) 76. 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 1Ω
72. The equivalent capacitance of the input loop of the circuit shown below is
t=0 i1 1 kΩ
1 kΩ
1 µF
5V
4Ω
1 kΩ Input loop
49i1
100 µF
100 µF (a) 2 mF (c) 200 mF
(b) 100 mF (d) 4 mF (GATE 2009: 1 Mark)
(a) 0 A (c) 1.25 A
(b) 1 A (d) 5 A (GATE 2010: 1 Mark)
77. As shown in the figure given below, a 1 Ω resistance is connected across a source that has a load line v + i = 100. The current through the resistance is
73. For the circuit shown below, find out the current flowing through the 2 Ω resistance. Also, identify the changes to be made to double the current through the 2 Ω resistance.
i + Source
(a) (b) (c) (d)
2Ω
(5 A; Put VS = 20 V) (2 A; Put VS = 8 V) (5 A; Put VS = 10 A) (7 A; Put VS = 12 A) (GATE 2009: 2 Marks)
Ch wise GATE_EE_Ch2.indd 86
1W
−
IS = 5 A VS = 4 V + −
v
(a) 25 A (c) 100 A
(b) 50 A (d) 200 A (GATE 2010: 1 Mark)
78. If 12 Ω resistor draws a current of 1 A as shown in the figure given below, the value of resistance R is (a) 4 Ω (b) 6 Ω (c) 8 Ω (d) 18 Ω
11/21/2018 12:05:43 PM
Chapter 2 • Electric Circuits
1Ω
2A
R
R
1A
87
12 Ω
6V
6Ω
+
3Ω
10 V
Load
(GATE 2010: 2 Marks) 79. 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 Ω 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
e
1Ω
c
a P b
d
Z12 ⎞ ⎛ Z11 + 1 Z12 + 1⎞ ⎛ Z11 + 1 (b) (a) ⎜ ⎜⎝ Z ⎟ Z 22 + 1⎟⎠ Z ⎝ Z 21 21 22 + 1⎠
(GATE 2011: 1 Mark) 82. 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) a band stop (band-reject) filter. (d) a band-pass filter. (GATE 2011: 1 Mark) Common Data Questions 83 and 84: An RLC circuit with relevant data is given below.
⎛ Z11 + 1 Z12 ⎞ ⎛ Z11 + 1 Z12 ⎞ (d) (c) ⎜ ⎟ ⎜⎝ Z + 1 Z ⎟⎠ Z 22 ⎠ ⎝ Z 21 21 22
IS +
C
I RL = 2∠ - π / 4 A.
1Ω
83. The power dissipated in the resistor R is (a) 0.5 W (b) 1 W
1Ω + (1.0 sin t) V
1 (a) 1 A (b) A 2 2 (d)
2A
(GATE 2011: 1 Mark) 81. In the circuit given below, the value of R required for the transfer of maximum power to the load having resistance of 3 Ω is
Ch wise GATE_EE_Ch2.indd 87
R
We have VS = 1∠ 0 V , I S = 2∠ π / 4 A,
1H
(c) 1 A
IC
L
80. The rms value of the current i(t) in the circuit shown below is
i(t)
IRL
VS
(GATE 2010: 2 Marks)
1F
(b) 3 Ω (d) infinity
(a) zero (c) 6 Ω
(c)
2W
(d) 2 W (GATE 2011: 2 Marks)
84. The current IC in the figure above is 1 (a) -j 2A (b) -j A 2 1 (c) + jA (d) + j 2A 2 (GATE 2011: 2 Marks) 85. In the circuit shown below, the current through the inductor is
11/21/2018 12:05:44 PM
88
GATE EE Chapter-wise Solved Papers
88. If VA-VB = 6 V; then VC -VD is j1 Ω
1Ω
−
1∠0A 1∠0V 1∠0V + + 1∠0A
(a)
(d) 0 A (GATE 2012: 1 Mark)
86. The impedance looking into nodes 1 and 2 in the given circuit is ib 1 kΩ
99ib 1
9 kΩ
+ 1V −
100 Ω 2 (a) 50 Ω (c) 5 kΩ
(b) 100 Ω (d) 10.1 kΩ (GATE 2012: 1 Mark)
87. 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 S
C1
i(t)
t=0
C2
(a) Zero (b) A step function (c) An exponentially decaying function (d) An impulse function (GATE 2012: 1 Mark)
Ch wise GATE_EE_Ch2.indd 88
2Ω
R
VB
− 10 V +
R
R
R
1Ω
R R
2 -1 A A (b) 1+ j 1+ j
1 (c) A 1+ j
VA
R − 1Ω
−j1 Ω
R
+ − 5V (a) -5 V (c) 3 V
VC
VD 2A (b) 2 V (d) 6 V (GATE 2012: 2 Marks)
89. 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 2Ω
R
+ 10 V −
−j1 Ω
Circuit A
Circuit B
(a) 0.8 Ω (c) 2 Ω
+ 3V −
(b) 1.4 Ω (d) 2.8 Ω (GATE 2012: 2 Marks)
Common Data Questions 90 and 91: With 10 V DC connected at port A in the linear non-reciprocal two-port network shown below, the following were observed: (i) 1 Ω connected at port B draws a current of 3 A. (ii) 2.5 Ω, connected at port B draws a current of 2 A. + A
B −
90. For the same network, with 6 V DC connected at port A, 1 Ω 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 (GATE 2012: 2 Marks) 91. With 10 V DC connected at port A, the current drawn by 7 Ω connected at port B is 5 (a) 3 A (b) A 7 7 (c) 1 A (d) 9 A 7 (GATE 2012: 2 Marks)
11/21/2018 12:05:46 PM
89
Chapter 2 • Electric Circuits
Linked Answer Questions 92 and 93: In the circuit shown, the three voltmeter readings are V1 = 220 V, V2 = 122 V, V3 = 136 V. R
96. The three circuit elements shown in the figure are part of an electric circuit. The total power absorbed by the three . circuit elements in watts is
I
10 A
8A
V2 RL V1
100 V
Load
V3
80 V
X
15 V 92. The power factor of the load is (a) 0.45 (b) 0.50 (c) 0.55 (d) 0.60 (GATE 2012: 2 Marks) 93. If RL = 5 Ω, the approximate power consumption in the load is (a) 700 W (b) 750 W (c) 800 W (d) 850 W (GATE 2012: 2 Marks) 94. 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
(GATE 2014: 1 Mark) 97. A combinations of 1 mF capacitor with an initial voltage VC (0) = -2 V in series with a 100 Ω resistor is connected to a 20 mA ideal DC current source by operating both switches at t = 0 s as shown. Which of the following graphs shown in the options approximates the voltage VS across the current source over the next few seconds?
VC +
−
+ t=0 VS
Ra
RC
Rb
Rc
RB
−
t=0
RA VS VS
(a)
VS VS
VS VS
(a) k2 (b) k (c) 1/k
(d)
−2 −2
k
(GATE 2013: 1 Mark) 95. In the circuit shown below, if the source voltage VS = 100∠53.13° V, then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is 3Ω
j4 Ω + − VL1 j40I2 I1
(a) 100∠90° (c) 800∠90°
j6 Ω
−2 −2
t
t
t
t −2 −2
(a) (a)
−2 −2
(a) (a)
(c)
+ − 10 VL1
RL = 10 Ω I2
(b) 800∠0° (d) 100∠60°
−2 −2
t
t
t
t
t
( )( )
(d) VS VS
VS VS
5Ω
t
( )( )
VS VS
V S VS t
+ −
(b) VS VS
t
t t t −2 −2 (d) 1 Mark) (c) (c) (d) (GATE 2014: −2 −2 −2 −2 (c)SW 98. The switch shown in the circuit is(d) kept (c) (d)at position ‘1’ for a long duration. At t = 0+, the switch is moved to position ‘2’. Assuming |V02| > |V01|, the voltage vC(t) across the capacitor is t
(GATE 2013: 2 Marks)
Ch wise GATE_EE_Ch2.indd 89
11/21/2018 12:05:47 PM
90
GATE EE Chapter-wise Solved Papers
R
`2’ SW
I 300 V
`1’ V02
R
R
V01 C
(GATE 2014: 2 Marks)
vC
103. The reading of the voltmeter (rms) in Volts, for the circuit shown in the figure is .
(a) vC(t) = -V02(1 - e-t/2RC) -V01
R = 0.5 W
(b) vC(t) = V02(1 - e-t/2RC) + V01 (c) vC(t) = -(V02 + V01)(1 - e-t/2RC) - V01 (d) vC(t) = (V02 - V01)(1 - e-t/2RC) + V01 (GATE 2014: 1 Mark)
1j W
99. The line A to neutral voltage is 10 ∠15°V for a balanced three phase star-connected load with phase sequence ABC. The voltage of line B with respect to line C is given by 10 ∠105°V (a) 10 3∠105°V (b)
1/j W
100sin wt
V 1/j W
1j W
(c) 10 3∠ - 75°V (d) -10 3∠90°V (GATE 2014: 1 Mark) 100. A non-ideal voltage source VS has an internal impedance of ZS. If a purely resistive load is to be chosen that maximises the power transferred to the load, its value must be (a) 0 (b) real part of ZS (c) magnitude of ZS (d) complex conjugate of ZS (GATE 2014: 1 Mark)
(GATE 2014: 2 Marks) 104. The voltage across the capacitor, as shown in the figure, is expressed as VC(t) = A1 sin(w1t - q1) + A2 sin(w2t - q2) 1Ω
20 sin10t 101. The driving point impedance Z(s) for the following circuit is 1H
Z(s)
(a) (c)
VC(t)
1F
10 sin5t
1H
1F
1F
s 4 + 3s 2 + 1 (b) s 4 + 2s 2 + 4 s3 + 2 s s2 + 2 s 2 + 1 (d) s3 + 1 4 2 4 s + s +1 s + s2 + 1 (GATE 2014: 1 Mark)
102. In the following figure, the value of resistor R is (25 + I/2) ohms, where I is the current in amperes. The current I is .
Ch wise GATE_EE_Ch2.indd 90
1H
The values of A1 and A2, respectively, are (a) 2.0 and 1.98 (b) 2.0 and 4.20 (c) 2.5 and 3.50 (d) 5.0 and 6.40 (GATE 2014: 2 Marks) 105. The total power dissipated in the circuit, shown in the figure, is 1 kW. 10 A 2 A 1 Ω
XC1 XL
AC source
R Load
XC2
V
200 V
11/21/2018 12:05:48 PM
Chapter 2 • Electric Circuits
91
The voltmeter, across the load, reads 200 V. The value of XL is . (GATE 2014: 2 Marks)
35 kΩ
106. The Norton’s equivalent source in amperes as seen into the terminals X and Y is .
V sin(w t)
C1 G 0.1 µF
105 kΩ
2.5 V X 2.5 Ω
(GATE 2014: 2 Marks) 110. A diode circuit feeds an ideal inductor as shown in the following figure. Given vs = 100sin(wt)V, where w = 100p rad/s, and L = 31.83 mH. The initial value of inductor current is zero. Switch S is closed at t = 2.5 ms. The peak value of inductor current iL (in A) in the first cycle is .
5Ω 5Ω
5Ω 5V
Y
t = 2.5 ms iL
(GATE 2014: 2 Marks) 107. The power delivered by the current source, in the following figure is .
S + Vs
L −
1V +− 1Ω
1V
+ −
(GATE 2014: 2 Marks)
1Ω
2A
1Ω
(GATE 2014: 2 Marks) 108. A series RLC circuit is observed at two frequencies. At w1 = 1 krad/s, we note that source voltage V1 = 100 ∠0° V results in a current I1 = 0.03∠31° A. At w2 = 2 krad/s, the source voltage V2 = 100 ∠0° V results in a current I 2 = 2∠0° A. The closest values for R, L, C out of the following options are (a) R = 50 Ω; L = 25 mH; C = 10 μF (b) R = 50 Ω; L = 10 mH; C = 25 μF (c) R = 50 Ω; L = 50 mH; C = 5 μF (d) R = 50 Ω; L = 5 mH; C = 50 μF (GATE 2014: 2 Marks) 109. In the bridge circuit shown, the capacitors are loss free. At balance, the value of capacitance C1 in microfarad is .
Ch wise GATE_EE_Ch2.indd 91
111. The voltages developed across the 3 Ω and 2 Ω resistors shown in the figure are 6 V and 2 V, respectively, with the polarity as marked. What is the power (in Watt) delivered by the 5 V voltage source? − 6V + Network N1
3Ω 2Ω − + 2V −
Network N2
+ 5V
(a) 5 (c) 10
(b) 7 (d) 14 (GATE 2015: 1 Mark)
112. For the given circuit, the Thevenin equivalent is to be determined. The Thevenin voltage, VTh (in volt), seen from terminal AB is 20 i 1Ω − + A + 2V
−
1Ω
i
2Ω B (GATE 2015: 1 Mark)
11/21/2018 12:05:50 PM
92
GATE EE Chapter-wise Solved Papers
113. An inductor is connected in parallel with a capacitor as shown in the following figure.
C1
1 µF ± 10%
C2
9 µF ± 10% Vout
Vbus
L i C
Z (GATE 2015: 1 Mark)
As the frequency of current i is increased, the impedance (Z) of the network varies as (a)
Z
Inductive
115. A series RL circuit is excited at t = 0 by closing a switch as shown in the figure. Assuming zero initial conditions, 2 the value of d i at t = 0+ is dt 2 R
f L
V
Capacitive (a) (b)
Inductive
V -V (b) L R
Z
(d) -RV L2
(c) 0 f
(GATE 2015: 1 Mark) Capacitive (c) Z Inductive
116. The current i (in Ampere) in the 2 W resistor of the given network is –––. 1Ω
Capacitive f
+ 5V −
1Ω
1Ω
i 2Ω
(d) Z
1Ω
(GATE 2015: 1 Mark) Inductive f Capacitive
(GATE 2015: 1 Mark) 114. A capacitive voltage divider is used to measure the bus voltage Vbus in a high voltage 50 Hz AC system as shown in the figure. The measurement capacitors C1 and C2 have tolerances of ±10% on their nominal capacitance values. If the bus voltage Vbus is 100 kV rms, the maximum rms output voltage Vout (in kV), considering the capacitor tolerances, is ________.
Ch wise GATE_EE_Ch2.indd 92
117. The circuit shown in the following figure has two sources connected in series. The instantaneous voltage of the AC source (in Volt) is given by v(t) = 12 sin t. If the circuit is in steady state, then the rms value of the current (in Ampere) flowing in the circuit is ————————. 1Ω v(t) + 8V
−
1H
(GATE 2015: 2 Marks)
11/21/2018 12:05:51 PM
93
Chapter 2 • Electric Circuits
118. In a linear two-port network, when 10 V is applied to Port 1, a current of 4 A flows through Port 2 when it is short circuited. When 5 V is applied to Port 1, a current of 1.25 A flows through a 1 W resistance connected across Port 2. When 3 V is applied to Port 1, the current (in Ampere) through a 2 W resistance connected across Port 2 is ————————. (GATE 2015: 2 Marks)
122. RA and RB are the input resistances of circuits as shown in the following figures. The circuits extend infinitely in the direction shown. Which one of the following statements is TRUE? 2Ω RA
V
5Ω
2Ω RB
+V − 0 10 Ω + 4V −
5A
2Ω
1Ω
1Ω
2Ω
2Ω
1Ω
1Ω
1Ω
119. In the given circuit, the parameter k is positive, and the power dissipated in the 2 W resistor is 12.5 W. The value of k is ————————. 2Ω
2Ω
1Ω
1Ω
(a) RA = RB (b) RA = RB = 0 RA (c) RA < RB (d) RB = (1 + RA )
kV0
(GATE 2015: 2 Marks) 120. In the given network V1 = 100 ∠0° V, V2 = 100 ∠ - 120° V, V3 = 100 ∠ + 120° V. The phasor current i (in Ampere) is
(GATE 2016: 1 Mark) 123. In the portion of a circuit shown, if the heat generated in 5 Ω resistance is 10 calories per second, then heat generated by the 4 Ω resistance, in calories per second, is ——————. 4Ω
6Ω
−j1 V1 5Ω V2
j1 (GATE 2016: 1 Mark)
V3
124. In the given circuit, the current supplied by the battery, in ampere, is ————————.
i
(a) 173.2∠( - 60°) (b) 173.2∠120° (c) 100.0 ∠( - 60°) (d) 100.0 ∠120° (GATE 2015: 2 Marks) 121. A symmetrical square wave of 50% duty cycle has amplitude of ±15 V and time period of 0.4p ms. This square wave is applied across a series RLC circuit with R = 5 W, L = 10 mH, and C = 4 mF. The amplitude of the 5000 rad/s component of the capacitor voltage (in Volt) is ——————. L
I1
1Ω
1 Ω I2 1Ω
1V I2 (GATE 2016: 1 Mark) 125. A resistance and a coil are connected in series and supplied from a single phase, 100 V, 50 Hz AC source as shown in the figure below. The rms values of plausible voltages across the resistance (VR) and coil (VC) respectively, in Volts, are
C
VR + −
R
VS
VC
(GATE 2015: 2 Marks)
Ch wise GATE_EE_Ch2.indd 93
11/21/2018 12:05:53 PM
94
GATE EE Chapter-wise Solved Papers
(a) 65, 35 (c) 60, 90
(b) 50, 50 (d) 60, 80 (GATE 2016: 1 Mark)
126. The voltage (V ) and current (A) across a load are as follows. v(t) = 100 sin(w t) i(t) = 10 sin(w t − 60°) + 2 sin(3w t) + 5 sin(5w t). The average power consumed by the load, in W, is ————————. (GATE 2016: 1 Mark)
131. In the circuit shown below, the supply voltage is 10 sin(1000 t) Volts. The peak value of the steady state current through the 1 Ω resistor, in Amperes, is ————————.
4Ω 250 µF
127. In the circuit shown below, the voltage and current sources are ideal. The voltage (Vout) across the current source, in volts, is 2Ω + 10 V −
2 µF 1Ω
4 mH 10 sin(1000t) (GATE 2016: 2 Marks)
+
5A
Vout −
(a) 0 (b) 5 (c) 10 (d) 20 (GATE 2016: 1 Mark)
132. The circuit below is excited by a sinusoidal source. The value of R, in Ω, for which the admittance of the circuit becomes a pure conductance at all frequencies, is ———————— .
128. The graph associated with an electrical network has 7 branches and 5 nodes. The number of independent KCL equations and the number of independent KVL equations, respectively, are (a) 2 and 5 (b) 5 and 2 (c) 3 and 4 (d) 4 and 3 (GATE 2016: 1 Mark) 129. For the network shown in the figure below, the frequency (in rad/s) at which the maximum phase lag occurs is, ________.
100 mF
R
0.02 H
R
(GATE 2016: 2 Marks) 133. In the circuit shown below, the node voltage VA is —————— V.
9Ω
A I1 1Ω
5Ω 5Ω
1F
130. In the circuit shown, switch S2 has been closed for a long time. At time t = 0 switch S1 is closed. At t = 0+, the rate of change of current through the inductor, in amperes per second, is ——————. 1Ω
(GATE 2016: 2 Marks) 134. The driving point input impedance seen from the source VS of the circuit shown below, in W, is ——————.
S2
+ V1 − IS 2Ω
1H 3V
(GATE 2016: 2 Marks)
Ch wise GATE_EE_Ch2.indd 94
5Ω
− 10I1 + 10 V − +
5A
(GATE 2016: 1 Mark)
3V
5Ω
Vo
Vin
S1
500 mH 5Ω
+ VS −
2Ω
2Ω 3Ω
4 V1 4 Ω
(GATE 2016: 2 Marks)
11/21/2018 12:05:55 PM
Chapter 2 • Electric Circuits
135. The Z-parameters of the two port network shown in the figure are Z11 = 40 W, Z12 = 60 W, Z21 = 80 W and Z22 = 100 W. The average power delivered to RL = 20 W, in watts, is ——————. 10 Ω I1
+ 20 V −
139. The power supplied by the 25 V source in the figure shown below is W. I
I2
+
V2
[Z]
−
R1 − + 17 V
+ 25 V −
+
V1
+
14 A −
(GATE 2016: 2 Marks) 136. In the balanced three-phase, 50 Hz, circuit shown below, the value of inductance (L), is 10 mH. The value of the capacitance (C) for which all the line currents are zero, in millifarads, is —————— .
0.4I
(GATE 2017: 1 Mark) 140. The equivalent resistance between the terminals A and B is Ω. 1Ω
2Ω
A
1Ω
6Ω 3Ω
C
R2
RL
−
L
95
1Ω
6Ω 3Ω
L B
0.8 Ω
C LC
(GATE 2017: 1 Mark) (GATE 2016: 2 Marks) 137. In the circuit shown below, the initial capacitor voltage is 4 V. Switch S1 is closed at t = 0. The charge (in μC) lost by the capacitor from t = 25 μs to t = 100 μs is —————— .
141. The following measurements are obtained on a single phase load: V = 200 V ± 1%, I = 50 A ± 1% and W = 555 W ± 2%. If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is . (Give answer up to one decimal place.)
S1
4V 5 µF
(GATE 2017: 1 Mark)
5Ω
142. For the given 2-port network, the value of transfer impedance Z21 in ohms is . 2Ω
(GATE 2016: 2 Marks) 1 138. Three single-phase transformers are connected to form a delta-star three-phase transformer of 110 kV/11 kV. The transformer supplies at 11 kV a load of 8 MW at 0.8 pf lagging to a nearby plant. Neglect the transformer losses. The ratio of phase currents in delta side to star side is (a) 1 : 10 3 (b) 10 3 : 1 (c) 1:10
(d) 3 : 10 (GATE 2016: 2 Marks)
Ch wise GATE_EE_Ch2.indd 95
1′
4Ω
2Ω
2
2Ω 2′ (GATE 2017: 1 Mark)
143. The initial charge in the 1 F capacitor present in the circuit shown is zero. The energy in joules transferred from the DC source until steady state condition is reached equals . (Give the answer up to one decimal place.)
11/21/2018 12:05:56 PM
GATE EE Chapter-wise Solved Papers
5Ω
5Ω 10 V
8Ω
8Ω
+ − 50 V
+ −
Ω
6Ω
1F
32 Ω
32
96
2H
5Ω
5Ω
5Ω (a) 2.5 e-4t
(b) 5 e-4t
(c) 2.5 e-0.25t
(d) 5 e-0.25t
(GATE 2017: 1 Mark) 144. Two passive two-port networks are connected in cascade as shown in figure. A voltage source is connected at port 1.
(GATE 2017: 2 Marks) 146. In the circuit shown below, the maximum power transferred to the resistor R is W. 3Ω
I1
I2 I3 + + Two-port Two-port V1 V2 V3 network 2 network 1 − − − Port 1 Port 2 Port 3 +
Given, V1 = A1V2 + B1I2
5Ω
+− 6V
+ 5V −
5Ω
R
2A
I1 = C1V2 + D1I2 V2 = A2V3 + B2I3
(GATE 2017: 2 Marks)
I2 = C2V3 + D2I3 A1, B1, C1, D1, A2, B2, C2 and D2 are the generalized circuit constants. If the Thevenin equivalent circuit at port 3 consists of a voltage source VT and an impedance ZT, connected in series, then (a) VT =
V1 A B + B1 D2 , ZT = 1 2 A1 A2 A1 A2 + B1C2
V1 A B + B1 D2 (b) VT = , ZT = 1 2 A1 A2 + B1C2 A1 A2 V1 A B + B1 D2 (c) VT = , ZT = 1 2 A1 + A2 A1 + A2 V1 A B + B1 D2 (d) VT = , ZT = 1 2 A1 A2 + B1C2 A1 A2 + B1C2 (GATE 2017: 2 Marks) 145. The switch in the figure below was closed for a long time. It is opened at t = 0. The current in the inductor of 2 H for t ≥ 0, is
Ch wise GATE_EE_Ch2.indd 96
147. For the balanced Y-Y connected 3-phase circuit shown in the figure below, the line–line voltage is 208 V rms and the total power absorbed by the load is 432 W at a power factor of 0.6 leading. a
A Van
Z N
n Vcn
Z
Vbn c
b
Z
C
B
The approximate value of the impedance Z is (a) 33∠ ( -53.1°) Ω (b) 60 ∠53.1° Ω (c) 60 ∠ ( -53.1°) Ω (d) 180 ∠ ( -53.1°) Ω (GATE 2017: 2 Marks) 148. In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is
11/21/2018 12:05:58 PM
Chapter 2 • Electric Circuits
97
(a) P1 = P2 = P3 = 0
R2 = 0.5Ω
(b) P1 < 0, P2 > 0, P3 > 0 (c) P1 < 0, P2 > 0, P3 < 0
5 mH
(d) P1 > 0, P2 < 0, P3 > 0
V(t) = 10 sin(100t) 1Ω
153. In the two-port network shown, the h11 parameter ⎛ ⎞ V1 . (up to ⎜⎝ where, h11 = I , when V2 = 0⎟⎠ in ohms is
Load (b) 1 µF (d) 10 mF
(a) 1 nF (c) 1 mF
(GATE 2018: 1 Mark)
C
1
2 decimal places).
(GATE 2017: 2 Marks)
2I1
149. For the network given in figure below, the Thevenin’s voltage Vab is 10 Ω
1W
1W +
+
a
5Ω
6A
10 Ω
10 Ω
+ 16 V −
b (a) −1.5 V (c) 0.5 V
(b) −0.5 V (d) 1.5 V
I1 1W
V1
V2
-
-
(GATE 2017: 2 Marks)
(GATE 2018: 1 Mark)
150. A single-phase 100 kVA, 1000 V/100 V, 50 Hz transformer has a voltage drop of 5% across its series impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at full load with 0.8 lagging power factor is (a) 4.8 (b) 6.8 (c) 8.8 (d) 10.8 (GATE 2018: 1 Mark)
154. The equivalent impedance Zeq for the infinite ladder circuit shown in the figure is
151. The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph is (a) 11 (b) 12 (c) 13 (d) 14 (GATE 2018: 1 Mark) 152. In the figure, the voltages are v1(t) = 100 cos (wt), v2(t) = 100 cos (wt + p/18) and v3(t) = 100 cos (wt + p/36). The circuit is in sinusoidal steady state, and R 0) are 2 and 5, respectively. Region 1 has uniform electric field E = 3aˆ x + 4 aˆ y + 2aˆ z where a x , a y , and a z are unit vectors along the x, y and z axes, respectively. The electric field in region 2 is
(a) 2 2k
(b) -2k
(a) 3a x + 1.6 a y + 2 a z
(c) 2k
(d) -2 2k
(b) 1.2 a x + 4 a y + 2 a z
(GATE 2014: 2 Marks)
(c) 12 a x + 4 a y + 0.8 a z
35. Consider a one turn rectangular loop of wire placed in a uniform magnetic field as shown in the figure. The plane of the loop is perpendicular to the field lines. The resistance of the loop is 0.4 Ω, and its inductance is negligible. The magnetic flux density (in Tesla) is a function of time, and is given by B(t) = 0.25 sinwt, where w = 2p × 50 radian/second. The power absorbed (in Watt) by the loop from the magnetic field is ————————.
(d) 3a x + 10 a y + 0.8 a z
Ch wise GATE_EE_Ch3.indd 137
5 cm
(GATE 2015: 1 Mark) 38. A circular turn of radius 1 m revolves at 60 rpm about its diameter aligned with the x-axis as shown in the figure. −7 The value of m0 is 4p × 10 in SI7 unit. If a uniform magnetic field of intensity H = 10 z A/m is applied, then the peak value of the induced voltage, Vturn (in Volts), is ________.
11/21/2018 12:00:35 PM
138
GATE EE Chapter-wise Solved Papers
(a) increases with r (c) is 3
z H
Vturn
x
(GATE 2015: 1 Mark) 39. Two semi-infinite conducting sheets are placed at right angles to each other as shown in the following figure. A point charge of +Q is placed at a distance of d from both Q2 K sheets. The net force on the charge is where K 4pε 0 d 2 is given by
(b) is 0 (d) decreases with z (GATE 2016: 1 Mark)
43. A soft-iron toroid is concentric with a long straight conductor carrying a direct current I. If the relative permeability mr of soft-iron is 100, the ratio of the magnetic flux densities at two adjacent points located just inside and just outside the toroid, is ——————. (GATE 2016: 1 Mark) 44. A parallel plate capacitor filled with two dielectrics is shown in the figure below. If the electric field in the region A is 4 kV/cm, the electric field in the region B, in kV/cm, is A
y
er = 1
B er = 4 2 cm
d +Q
(a) 1 (c) 4
d x
(a) 0
1 1 (b) - i - j 4 4
1 1 (c) - i - j 8 8
(d)
1- 2 2 1- 2 2 i+ j 8 2 8 2 (GATE 2015: 2 Marks)
40. A parallel plate capacitor is partially filled with glass of dielectric constant 4.0 as shown below. The dielectric strengths of air and glass are 30 kV/cm and 300 kV/cm, respectively. The maximum voltage (in kilovolts), which can be applied across the capacitor without any breakdown, is ————————. Air, er = 1.0
5 mm
10 mm Glass, er = 4.0 (GATE 2015: 2 Marks) 41. Two identical coils each having inductance L are placed together on the same core. If an overall inductance of a L is obtained by interconnecting these two coils, the minimum value of a is ________. (GATE 2015: 2 Marks) 42. In cylindrical coordinate system, the potential produced by a uniform ring charge is given by f = f (r, z), where f is a continuous function of r and z. Let E be the resulting electric field. Then the magnitude of ∇ × E
Ch wise GATE_EE_Ch3.indd 138
(b) 2 (d) 16 (GATE 2016: 1 Mark)
45. Two electrodes, whose cross-sectional view is shown in the figure below, are at the same potential. The maximum electric field will be at the point
A
D
C
B (a) A (c) C
(b) B (d) D (GATE 2016: 1 Mark)
46. Two electric charges q and −2q are placed at (0, 0) and (6, 0) on the x-y plane. The equation of the zero equipotential curve in the x-y plane is (a) x = −2 (b) y=2 (d) (x + 2)2 + y2 = 16 (c) x2 + y2 = 2 (GATE 2016: 2 Marks) 47. The flux linkage (l) and current (i) relation for an electromagnetic system is λ = ( i ) /g . When i = 2 A and g (air-gap length) = 10 cm, the magnitude of mechanical force on the moving part, in N, is ________. (GATE 2016: 2 Marks) 48. A rotating conductor of 1 m length is placed in a radially outward (about the z-axis) magnetic flux density (B) of 1 Tesla as shown in figure below. Conductor is parallel to and at 1 m distance from the z-axis. The speed of the conductor in rpm required to induce a voltage of 1 V across it, should be ________.
11/21/2018 12:00:37 PM
Chapter 3 • Electromagnetic Fields
(a) ∇ ⋅ X (b) ∇ ⋅ X (c) ∇ ⋅ X (d) ∇ ⋅ X
z
B
1m 1m
(GATE 2016: 2 Marks) 49. A solid iron cylinder is placed in a region containing a uniform magnetic field such that the cylinder axis is parallel to the magnetic field direction. The magnetic field lines inside the cylinder will (a) bend closer to the cylinder axis (b) bend farther away from the axis (c) remain uniform as before (d) cease to exist inside the cylinder (GATE 2017: 1 Mark) 50. Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the centre of the line joining the electron and proton. At t = 0, the particles are released but are constrained to move along the same straight line. Which of these will collide first? (a) The particles will never collide. (b) All will collide together. (c) Proton and neutron. (d) Electron and neutron. (GATE 2017: 1 Mark) 51. Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million electrons are added to this sphere, these electrons will be distributed (a) uniformly over the entire volume of the sphere (b) uniformly over the outer surface of the sphere (c) concentrated around the centre of the sphere (d) along a straight line passing through the centre of the sphere (GATE 2017: 1 Mark) 52. The figures show diagrammatic representations of vec tor fields X , Y and Z , respectively. Which one of the following choices is true? X Y Z
Ch wise GATE_EE_Ch3.indd 139
139
= 0, ∇ × Y ≠ 0, ∇ × Z = 0 ≠ 0, ∇ × Y = 0, ∇ × Z ≠ 0 ≠ 0, ∇ × Y ≠ 0, ∇ × Z ≠ 0 = 0, ∇ × Y = 0, ∇ × Z = 0 (GATE 2017: 1 Mark)
53. The magnitude of magnetic flux density (B) in micro Tesla (µT) at the centre of a loop of wire wound as a regular hexagon of side length 1 m carrying a current (I = 1 A) and placed in vacuum as shown in the figure is . (Give the answer up to two decimal places.)
I (GATE 2017: 2 Marks) 54. A thin soap bubble of radius R = 1 cm and thickness a = 3.3 mm (a a z < a
(d) Inverse Z-transform by power series expansion X ( z) =
31. Z-transform (a) Expression for discrete-time signal ∞
Ai 1 - g i z -1
∞
∑ x(n) z
-n
n = -∞
= ... + x( -3) z 3 + x( -2) z 2 + x( -1) z + x(0) + x(1) z -1 + x( 2) z -2 + ...
-n
n = -∞
(e) Common Z-transform pairs.
(b) All the properties of Z-transform 32. Inverse Z-transform (a) Expression X ( z) =
∞
∑ x( k ) z
-k
k = -∞
QUESTIONS 1. If an AC voltage wave is corrupted with an arbitrary number of harmonics, then the overall voltage waveform differs from its fundamental frequency component in terms of (a) only the peak values (b) only the rms values (c) only the average values (d) all the three measures (peak, rms and average values) (GATE 2000: 2 Marks) 2. Fourier series for the waveform, f(t) as shown in the following figure is
1 0
8 π2
(b)
8 ⎡ 1 1 ⎤ sin(π t ) - cos(3π t ) + sin(5π t ) + ⎥ 9 25 π 2 ⎢⎣ ⎦
(c)
8 ⎡ 1 1 ⎤ cos(π t ) + cos(3π t ) + cos(5π t ) + ⎥ 9 25 π 2 ⎢⎣ ⎦
(d)
8 ⎡ 1 1 ⎤ cos( π t ) - sin(3π t ) + sin(5 π t ) + ⎥ 9 25 π 2 ⎢⎣ ⎦ (GATE 2002: 1 Mark)
3. Let Y(s) be the Laplace transformation of the function y(t), then the final value of the function is (a) lim Y ( s)
f(t)
−1
s →0
1
2
3
(b) lim Y ( s) s →∞
(c) lim sY ( s) −1
s →0
(d) lim sY ( s) s →∞
Ch wise GATE_EE_Ch4.indd 152
1 1 ⎡ ⎤ ⎢sin( π t ) + 9 sin(3π t ) + 25 sin(5 π t ) + ⎥ ⎣ ⎦
(a)
(GATE 2002: 1 Mark)
11/21/2018 11:55:42 AM
Chapter 4 • Signals and Systems
4. What is the rms value of the voltage waveform as shown in the following figure?
100 V
(a) (b) (c) (d)
153
h[n] = 0; n2, h[0] = l, h[l] = h[2] = -l h[n] = 0; nl, h[-l] = l, h[0] = h[l] = 2 h[n] = 0; n3, h[0] = -l, h[l] = 2, h[2] = l h[n] = 0; nl, h[-2] = h[1] = h[-l] =-h[0] = 3 (GATE 2006: 2 Marks)
q
p
x[n] ↔ X ( z ) = ∑ n = 0 ∞
−100 V p/3 2p/3 4p/3 5p/3 2p
200 (a) V π
n
∞ ⎛ 2⎞ y1 [n] ↔ Y1 ( z ) = ∑ n = 0 ⎜ ⎟ z - n ⎝ 3⎠
(
)
(b) y2 [n] ↔ Y2 ( z ) = ∑ 5n - n z - ( 2 n +1) n= 0 ∞
100 V π (c) 200 V (d) 100 V
(c) (GATE 2002: 1 Mark)
5. If u(t) is the unit step and d (t) is the unit impulse func1 tion, the inverse Z-transform of F ( z ) = for k > 0 is z +1 (a) ( -1) k d( k ) (b) d( k ) - ( -1) k (d) u( k ) - ( -1) k (GATE 2005: 2 Marks)
6. Which of 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. (GATE 2006: 1 Mark) 7. x (t) is a real-valued function of a real variable with period T. Its trigonometric Fourier series expansion contains no terms of frequency, that is w = 2p (2k)/T; k = l, 2,.... 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) (GATE 2006: 1 Mark) 8. x[n] = 0; n0, x[- 1] = - 1, x[0] = 2 is the input and y[n] = 0; n2, 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
Ch wise GATE_EE_Ch4.indd 153
3n 2 n z 2+n
where ↔ denotes a transform-pair relationship, is orthogonal to the signal (a)
(b)
(c) ( -1) k u( k )
9. The discrete-time signal
y3 [n] ↔ Y3 ( z ) = ∑ n = -∞ 2 - n z - n ∞
(d) y4 [n] ↔ Y4 ( z ) = 2 z -4 + 3 z -2 + 1 (GATE 2006: 2 Marks) 10. 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. (GATE 2006: 2 Marks) 11. y[n] denotes output and x [n] denotes input of a d iscretetime system given by the difference equation y[n] - 0.8y[n - 1] = x[n] + 1.25x[n + 1]. Its right-sided impulse response is (a) causal. (b) unbounded. (c) periodic. (d) non-negative. (GATE 2006: 2 Marks) 12. Let a signal a1sin(w 1t + f 1) be applied to a stable linear time-invariant system. Let the corresponding steady state output be represented as a 2F (w 2t + f2). Then which of the following statements is true? (a) F is not necessarily a “sine” or “cosine” function but must be periodic with w 1 = w 2. (b) F must be a “sine” or “cosine” function with a 1 = a 2. (c) F must be a “sine” function with w 1 = w 2 and f 1 = f 2. (d) F must be a “sine” or “cosine” function with w 1 = w 2. (GATE 2007: 1 Mark)
11/21/2018 11:55:44 AM
154
GATE EE Chapter-wise Solved Papers
13. The frequency spectrum of a signal is shown in the following figure. If this signal is ideally sampled at intervals of 1 ms, then frequency spectrum of the sampled signal will be
(a)
1 (t - 1)(t - 2) 2
(b)
(c)
1 (t - 1) 2 u(t - 1) 2
(d) None of these
U(jw)
w U ∗(jw)
(a)
(GATE 2007: 2 Marks)
16. X(z); = 1 - 3z-1, Y(z) = 1 + 2z-2 are Z-transforms of two signals x[n] and 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]
1 kHz
1 (t - 1)(t - 2) 2
where * denotes discrete time convolution. Then the output of the system for the input d [n - 1] w
(b)
w
(c)
w
(d)
w (GATE 2007: 1 Mark)
14. A signal x(t) is given by - T 4 < t ≤ 3T 4 ⎧1, ⎪ x(t ) = ⎨ -1, 3T 4 < t ≤ 7T 4 ⎪ - x (t + T ) ⎩
Which among the following gives the fundamental Fourier term of x(t)? (a)
4 ⎛πt π⎞ cos ⎜ - ⎟ ⎝ T 4⎠ π
4 ⎛ πt π⎞ (b) cos ⎜ + ⎝ 2T 4 ⎟⎠ π
(c)
4 ⎛πt π⎞ sin ⎜ - ⎟ ⎝ T 4⎠ π
4 ⎛ πt π⎞ (d) sin ⎜ + ⎝ 2T 4 ⎟⎠ π (GATE 2007: 2 Marks)
15. 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
Ch wise GATE_EE_Ch4.indd 154
(a) has Z-transform z-1 X(z) Y(z) (b) equals d [n - 2] - 3d [n - 3] + 2d [n - 4] - 6d [n - 5] (c) has Z-transform 1 - 3z -1 + 2z -2 - 6z-3 (d) does not satisfy any of the above three. (GATE 2007: 2 Marks) 17. A signal e-atsin(w t) is the input to a linear-time invariant system. Given K and f are constants, the output of the system will be of the form Ke-bt sin(nt + f), where (a) b need not be equal to a but n equal to w (b) n need not be equal to w but b equal to a (c) b equal to a and n equal to w (d) b need not be equal to a and n need not to be equal to w (GATE 2008: 1 Mark) 18. 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) (b) (c) (d)
Statement (I) is correct and Statement (II) is wrong. Statement (II) is correct and Statement (I) is wrong. Both Statement (I) and Statement (II) are wrong. Both Statement (I) and Statement (II) are correct. (GATE 2008: 1 Mark)
19. Given a sequence x[n], to generate the sequence y[n] = x[3 - 4n]. Which one of the following procedures would be correct? (a) First delay x[n] by 3 sample to generate z1[n], then pick every fourth sample of z1[n] to generate z2[n], and then finally time reverse z2 [n] to obtain y[n].
11/21/2018 11:55:45 AM
Chapter 4 • Signals and Systems
(b) First advance x[n] by 3 samples to generate z[n], then pick every fourth 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 ν11[n], time-reverse ν1 [n] to obtain ν2 [n] and finally advance ν2 [n] by 3 samples to obtain y[n]. (d) First pick every fourth sample of x[n] to generate ν1 [n], time-reverse ν1 [n] to obtain ν2 [n], and finally delay ν2 [n] by 3 samples to obtain y[n]. (GATE 2008: 2 Marks) 20. A system with input x(t) and output y(t) is defined by the input-output relation: y (t ) =
(a) (b) (c) (d)
one pole and one zero. one pole and two zeros. two poles and one zero. two poles and two zeros. (GATE 2008: 2 Marks)
24. If X ( z ) =
z with |z|>a, then residue of X[z]zn - 1 at ( z - a) 2
z = a for n ≥ 0 will be (a) an - 1 (b) an n (c) na (d) nan - 1 (GATE 2008: 2 Marks)
-2 t
∫ x(t )dt
-∞
The system will be (a) causal, time-invariant and unstable (b) causal, time-invariant and stable (c) non-causal, time-invariant and unstable (d) non-causal, time-variant and unstable (GATE 2008: 2 Marks) 21. A signal x(t) = sinc (a t) where a is a real constant. ⎛ sin(π x ) ⎞ ⎜⎝ sin c( x ) = π x ⎟⎠ 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 K sin(g t) where g = min (a, b ) (b) It will be of the form K sin(g t) where g = max (a, b ) (c) It will be of the form K sinc(a t) (d) It cannot be a sinc type of signal (GATE 2008: 2 Marks)
155
1 1 ⎛ 1⎞ 25. Let x(t) = rect ⎜ t - ⎟ (where rect (x) = 1 for - ≤ x ≤ ⎝ 2⎠ 2 2 and zero otherwise). sin(π x ) , then Fourier transform of πx x(t) + x(-t) will be given by
Then if sinc ( x ) =
⎛w⎞ ⎛w⎞ (a) sin c ⎜ ⎟ (b) 2 sin c ⎜ ⎟ ⎝ 2π ⎠ ⎝ 2π ⎠ ⎛w⎞ ⎛w⎞ (c) 2 sin c ⎜ ⎟ cos ⎜ ⎟ ⎝ 2π ⎠ ⎝ 2⎠
⎛w⎞ ⎛w⎞ (d) sin c ⎜ ⎟ sin ⎜ ⎟ ⎝ 2π ⎠ ⎝ 2⎠ (GATE 2008: 2 Marks)
26. A linear time invariant system with an impulse response h(t) produces output y(t) when input x(t) is applied. When the input x(t - t ) is applied to a system with impulse response h(t - t ), then output will be (a) y(t) (b) y(2(t - t )) (c) y(t - t ) (d) y(t - 2t ) (GATE 2009: 1 Mark)
22. 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
27. A cascade of three linear-time invariant systems is causal and unstable. From this, we conclude that (a) each system in the cascade is individually causal and unstable. (b) at least one system is unstable and at least one T T (a) (b) system is causal. 8 4 (c) at least one system is causal and all systems are T (c) (d) 2T unstable. 2 (d) the majority are unstable and the majority are (GATE 2008: 2 Marks) causal. 23. H(z) is a transfer function of a real system. When a sig(GATE 2009: 2 Marks) nal x[n] = (1 + j)n is the input to such a system, the output is zero. Further, the region of convergence (ROC) 28. The Fourier series coefficients of a periodic signal x(t), expressed as ⎛ 1 ⎞ of ⎜1 - z -1 ⎟ H(z) is the entire z-plane (except z = 0). ⎝ 2 ⎠ ∞ x(t ) = ∑ k = -∞ ak e j 2π kt / T It can then be inferred that H(z) can have a minimum of
Ch wise GATE_EE_Ch4.indd 155
11/21/2018 11:55:46 AM
156
GATE EE Chapter-wise Solved Papers
are given by a-2 = 2 -j1; a-1 = 0.5 + j 0.2; a0 = j2; a1 = 0.5 - j 0.2; a2 = 2+ j1; and ak = 0; for |k| > 2.
(a) 0 2 π
(c)
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. (GATE 2009: 2 Marks) 29. The Z-transform of a signal x[n] is given by 4 z - 3 + 3 z -1 + 2 - 6 z 2 + 2 z 3 It is applied to a system, with a transfer function
(b) 1 (d)
(GATE 2010: 1 Mark)
33. At t = 0, the function f (t ) = (a) a minimum. (c) a point of inflection.
sin t has t (b) a discontinuity. (d) a maximum. (GATE 2010: 2 Marks)
34. x(t) is a positive rectangular pulse from t = -1 to t = +1 with unit height as shown in the following figure. The ∞
value of
∫
2 X (w ) dw {where X(w) is the Fourier trans-
-∞
H ( z ) = 3 z -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
form of x(t)} is x(t) 1
t
−1
(d) Re[Y ( z )]z = e jθ = - Re[Y ( z )]z = e jθ ; Im[Y ( z )]z = e jθ = Im[Y ( z )]z = e jθ ; π ≤ θ < π (GATE 2009: 2 Marks)
π⎞ ⎛ 30. The period of the signal x(t) = 8 sin ⎜ 0.8π t+ ⎟ is ⎝ 4⎠ (a) 0.4 p s (c) 1.25 s
5
(b) 0.8 p s (d) 2.5 s (GATE 2010: 1 Mark)
1
0
(a) 2 (c) 4
(b) 2p (d) 4p (GATE 2010: 2 Marks)
35. Given the finite length input x[n] and the corresponding finite length output y[n] of an LTI system as shown in the following figure. The impulse response h[n] of the system is
31. The system represented by the input-output relationship 5t
y (t ) =
∫ x(t )dt , t > 0 , is
↑
-∞
(a) (b) (c) (d)
h[n]
[n] = {1, -1}
linear and causal linear but not causal causal but not linear neither linear nor causal
↑
(a) h[n] = {1, 0, 0,1} ↑
(c) h[n] = {1,1,1,1} ↑
(GATE 2010: 1 Mark) 32. The second harmonic component of the periodic waveform given in the figure has an amplitude of
y[n] = {1, 0, 0 , -1} (b) h[n] = {1, 0,1} ↑
(d) h[n] = {1,1,1} ↑
(GATE 2010: 2 Marks) Common Data for Questions 36 and 37: Given f(t) and g(t) as shown in the following figure: f(t)
g(t)
1
1
+1 1 0 −1
Ch wise GATE_EE_Ch4.indd 156
T/2
T 0
1
t
0
3
5
t
11/21/2018 11:55:48 AM
Chapter 4 • Signals and Systems
version f(t − t) be F2(s). Let F1*(s) be the complex conjugate of F1(s) with the Laplace variable set as s = s + jw. If
36. g(t) can be expressed as (a) g(t) = f(2t - 3)
⎛t ⎞ (b) g (t ) = f ⎜ - 3⎟ ⎝3 ⎠
3⎞ ⎛ (c) g (t ) = f ⎜ 2t - ⎟ ⎝ 2⎠
⎛ t 3⎞ (d) g (t ) = f ⎜ - ⎟ ⎝ 2 2⎠
G ( s) =
(GATE 2010: 2 Marks) 37. The Laplace transform, of g(t) is 1 3s 1 -5 s (a) (e - e 5 s ) (b) (e - e -3 s ) s s (c)
e -3 s 1 5s (1 - e -2 s ) (d) (e - e3s ) s s (GATE 2010: 2 Marks)
38. The Fourier series expansion
157
F2 ( s) ⋅ F1* ( s) F1 ( s)
2
then the inverse Laplace transform of G(s) is (a) an ideal impulse d (t). (b) an ideal delayed impulse d (t −t). (c) an ideal step function u(t). (d) an ideal delayed step function impulse u(t − t). (GATE 2011: 2 Marks) 42. The unilateral Laplace transform of f(t) is The unilateral Laplace transform of t f(t) is
∞
f (t ) = a0 + ∑ an cos nw t + bn sin n w t n =1
of the periodic signal shown below will contain the following non-zero terms
(a) (c)
f(t)
1 . s + s +1 2
s 2s + 1 (b) - 2 2 ( s + s + 1) ( s + s + 1) 2 2
s ( s 2 + s + 1) 2
(d)
2s + 1 ( s 2 + s + 1) 2
(GATE 2012: 1 Mark) n
t 0 (a) (b) (c) (d)
a0 and bn, n = l, 3, 5, ……. ∞ a0 and an, n = 1, 2, 3, …. ∞ a0, an and bn, n = 1, 2, 3, … ∞ a0 and an, n = 1, 3, 5, … ∞ (GATE 2011: 1 Mark)
39. Given two continuous time signals x(t) = e-t and y(t) = e-2t which exist for t > 0, the convolution z(t) = x(t) * y(t) is (a) e-t - e-2t (b) e-3t +t (c) e (d) e-t + e-2t (GATE 2011: 1 Mark) 40. 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 rms value of the signal is α α (a) (b) 3 2
α 3 (c) α 2 (d) (GATE 2011: 2 Marks) 41. Let the Laplace transform of a function f(t) which exists for t >0 be F1(s) and the Laplace transform of its delayed
Ch wise GATE_EE_Ch4.indd 157
43. If x[n] = (1 3)|n| - (1 2) u[n] , then the region of convergence (ROC) of its Z-transform in the z-plane will be 1 < z 1
sin( w -10 ) j10 w -10 (b) 2e
j10w 2 sinw w (d) e (GATE 2015: 2 Marks)
77. For linear time invariant systems that are Bounded Input Bounded Output stable, which one of the following statements is TRUE? (a) The impulse response will be integrable, but may not be absolutely integrable. (b) The unit impulse response will have finite support. (c) The unit step response will be absolutely integrable. (d) The unit step response will be bounded. (GATE 2015: 2 Marks) 78. The Z-transform of a sequence x[n] is given as X(z) = 2z + 4 − 4/z + 3/z2. If y[n] is the first difference of x[n], then Y(z) is given by 8 7 3 (a) 2 z + 2 - + 2 - 3 z z z 6 1 3 (b) -2 z + 2 - + 2 - 3 z z z (c) -2 z - 2 + (d) 4 z - 2 -
8 7 3 + + z z 2 z3
8 1 3 - 2+ 3 z z z (GATE 2015: 2 Marks)
79. Consider a continuous-time system with input x(t) and output y(t) given by y(t) = x(t)cos(t) This system is (a) Linear and time-invariant (b) Non-linear and time-invariant
Ch wise GATE_EE_Ch4.indd 161
(c) Linear and time-varying (d) Non-linear and time-varying (GATE 2016: 1 Mark) 80. The value of
∫
+∞ -∞
e - t d ( 2t - 2)dt , where d (t ) is the Dirac
delta function, is (a)
1 2 (b) 2e e
(c)
1 (d) 1 e2 2e 2 (GATE 2016: 1 Mark)
81. Consider a causal LTI system characterised by differendy(t ) 1 tial equation + y(t ) = 3 x(t ). The response of the dt 6 t system to the input x(t ) = 3e 3 u(t ), where u(t) denotes the unit step function, is -
t
t
(a) 9e 3 u(t ). (b) 9e 6 u(t ). -
t
-
t
-
t
-
t
(c) 9e 3 u(t ) - 6e 6 u(t ). (d) 54e 6 u(t ) - 54e 3 u(t ). (GATE 2016: 1 Mark) 82. Suppose the maximum frequency in a band-limited signal x(t) is 5 kHz. Then, the maximum frequency in x(t) cos(2000p t), in kHz, is ________. (GATE 2016: 1 Mark) 83. Consider the function f(z) = z + z* where z is a complex variable and z* denotes its complex conjugate. Which one of the following is TRUE? (a) f(z) is both continuous and analytic (b) f(z) is continuous but not analytic (c) f(z) is not continuous but is analytic (d) f(z) is neither continuous nor analytic (GATE 2016: 1 Mark) 84. The solution of the differential equation, for t > 0, y ′′(t ) + 2 y ′(t ) + y(t ) = 0 with initial conditions y(0) = 0 and y ′(0) = 1, is [u(t) denotes the unit step function], (a) te−t u(t)
(b) (e−t − te−t) u(t)
(c) (−e−t + te−t) u(t) (d) e−t u(t) (GATE 2016: 1 Mark) 85. Let f(x) be a real, periodic function satisfying f(−x) = −f(x). The general form of its Fourier series representation would be
11/21/2018 11:55:56 AM
162
GATE EE Chapter-wise Solved Papers
88. Let x1 (t ) ↔ X 1 (w ) and x2 (t ) ↔ X 2 (w ) be two signals whose Fourier transforms are as shown in the figure below. In the figure, h(t) = e−2|t | denotes the impulse response.
(a)
f ( x ) = a0 + ∑ k =1 ak cos( kx )
(b)
f ( x ) = ∑ k =1 bk sin( kx )
(c)
f ( x ) = a0 + ∑ k =1 a2 k cos( kx )
(d)
f ( x ) = ∑ k =1 a2 k +1 sin( 2k + 1) x
∞
∞
∞
X1(w) X2(w)
∞
(GATE 2016: 1 Mark) 86. Suppose x1(t) and x2(t) have the Fourier transforms as shown below.
−B1−B1 2
B1 B1 2
w
−B2
w B2
x1(t) X1(jw) h(t) = e−2t
y(t)
1 x2(t) 0.5 0.3 −1
0
2 w
1
X2(jw) 1 0.5 0.3 −2
−1
0
1 w
Which one of the following statements is TRUE? (a) x1(t) and x2(t) are complex and x1(t) x2(t) is also complex with non-zero imaginary part (b) x1(t) and x2(t) are real and x1(t) x2(t) is also real (c) x1(t) and x2(t) are complex but x1(t) x2(t) is real (d) x1(t) and x2(t) are imaginary but x1(t) x2(t) is real (GATE 2016: 2 Marks) 87. The output of a continuous-time, linear time-invariant system is denoted by T{x(t)} where x(t) is the input signal. A signal z(t) is called eigen-signal of the system T, when T{z(t)} = g z(t), where g is a complex number, in general, and is called an eigenvalue of T. Suppose the impulse response of the system T is real and even. Which of the following statements is TRUE? (a) cos(t) is an eigen-signal but sin(t) is not (b) cos(t) and sin(t) are both eigen-signals but with different eigenvalues (c) sin(t) is an eigen-signal but cos(t) is not (d) cos(t) and sin(t) are both eigen-signals with identical eigenvalues (GATE 2016: 2 Marks)
Ch wise GATE_EE_Ch4.indd 162
For the given system, the minimum sampling rate required to sample y(t), so that y(t) can be uniquely reconstructed from its samples, is (a) 2B1 (b) 2(B1 + B2) (c) 4(B1 + B2) (d) ∞ (GATE 2016: 2 Marks) 89. Let z(t) = x(t) * y(t), where “*” denotes convolution. Let c be a positive real-valued constant. Choose the correct expression for z(ct). x(ct ) * y(ct ) (a) c ⋅ x(ct ) * y(ct ) (b)
c ⋅ x(ct ) * y(t ) (c) c ⋅ x(t ) * y(ct ) (d) (GATE 2017: 1 Mark) 90. Consider the system with the following input–output relation y[n] = (1 + (-1)n) x[n] where x[n] is the input and y[n] is the output. The system is (a) invertible and time invariant (b) invertible and time varying (c) non-invertible and time invariant (d) non-invertible and time varying (GATE 2017: 1 Mark) ⎧t - ⎢⎣t ⎥⎦ , t≥0 91. Consider g (t ) = ⎨ , where t ∈ . ⎩t - ⎡⎢t ⎤⎥ , otherwise Here ⎢⎣t ⎥⎦ represents the largest integer less than or equal to t and ⎡⎢t ⎤⎥ denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g(t) is . (GATE 2017: 1 Mark)
11/21/2018 11:55:58 AM
Chapter 4 • Signals and Systems
92. The pole—zero plots of three discrete time systems P, Q and R on the z-plane are shown below.
(c) -7e -2t u(t ) - 2e -5t u(t ) e -2t u(t ) - 2e -5t u(t ) (d) -7Im(z) R
Im(z)
Im(z) P
163
Q
(GATE 2017: 2 Marks)
2 Poles 95. Let the signal
0.5 Re(z) —0.5 Unit circle
Re(z)
Re(z) Unit circle
Unit circle
Im(z)
Im(z)
+∞
x (t ) =
k
k ⎞
k = -∞
be passed through an LTI system with frequency response H(w), as given in the figure below
R
Q
⎛
∑ ( -1) d ⎜⎝ t - 2000 ⎟⎠
H(w) 1
0.5
Re(z)
Re(z)
Re(z) —0.5
−5000 p
Unit circle
Unit circle
Which one of the following is TRUE about the frequency selectively of these systems? (a) All three are high-pass filters. (b) All three are band-pass filters. (c) All three are low-pass filters. (d) P is a low-pass filter, Q is a band-pass filter and R is a high-pass filter. (GATE 2017: 1 Mark) 93. The mean square value of the given periodic waveform f(t) is .
(GATE 2017: 2 Marks) 96. The output y(t) of the following system is to be sampled, so as to reconstruct it from its samples uniquely. The required minimum sampling rate is X(w)
4 —1000p 2.7 —1.3 —0.3 —2
w
The Fourier series representation of the output is given as (a) 4000 + 4000 cos (2000 pt) + 4000 cos (4000 pt) (b) 2000 + 2000 cos (2000 pt) + 2000 cos (4000 pt) (c) 4000 cos (2000 pt) (d) 2000 cos (2000 pt)
f(t)
—3.3
5000 p
0.7
w 1000p sin(1500p t) y(t)
t 3.7 4.7
h(t) = X(w)
x(t)
pt
cos (1000p t) (GATE 2017: 1 Mark) 94. Let a causal LTI system be characterized by the following differential equation, with initial rest condition d2 y dy dx(t ) + 7 + 10 y(t ) = 4 x(t ) + 5 2 dt dt dt
(a) 1000 samples/s (c) 2000 samples/s
(b) 1500 samples/s (d) 3000 samples/s (GATE 2017: 2 Marks)
97. A cascade system having the impulse responses h1 ( n) = {1- 1} and h2 ( n) = {1- 1} is shown in the figure ↑
↑
where x(t) and y(t) are the input and output, respectively. The impulse response of the system is (u(t) is the unit step function)
below, where symbol ↑ denotes the time origin.
(a) 2e -2t u(t ) - 7e -5t u(t )
The input sequence x(n) for which the cascade system produces an output sequence y( n) = {1, 2, 1, -1, - 2 , -1} ↑ is
(b) -2e -2t u(t ) + 7e -5t u(t )
Ch wise GATE_EE_Ch4.indd 163
x( n) → h1 ( n) → h2 ( n) → y( n)
11/21/2018 11:56:00 AM
164
GATE EE Chapter-wise Solved Papers
(a) x( n) = {1, 2, 1, 1} (b) x( n) = {1, 1, 2, 2} ↑
↑
(c) x( n) = {1, 1, 1, 1} (d) x( n) = {1, 2, 2, 1} ↑
↑
(GATE 2017: 2 Marks) 98. Match the transfer functions of the second-order systems with the nature of the systems given below. Transfer functions 15 s 2 + 5s + 15 25 Q: 2 s + 10 s + 25 35 R: 2 s + 18s + 35 P:
Nature of system
These signals are sampled with a sampling period of T = 0.25 seconds to obtain discrete-time signals x1[n] and x2[n], respectively. Which one of the following statements is true? (a) The energy of x1[n] is greater than the energy of x2[n]. (b) The energy of x2[n] is greater than the energy of x1[n]. (c) x1[n] and x2[n] have equal energies. (d) Neither x1[n] nor x2[n] is a finite-energy signal.
I. Overdamped
(GATE 2018: 2 Marks) II. Critically damped 100. The signal energy of the continuous-time signal III. Under damped
x(t) = [(t - 1)u(t - 1)] - [(t - 2)u(t - 2)] - [(t - 3)u(t - 3)] + [(t - 4)u(t - 4)] is
(a) P-I, Q-II, R-III
(a) 11/3 (c) 1/3
(b) P-II, Q-I, R-III (c) P-III, Q-II, R-I
(b) 7/3 (d) 5/3 (GATE 2018: 2 Marks)
(d) P-III, Q-I, R-II (GATE 2018: 1 Mark) 99. Consider the two continuous-time signals defined below:
⎧ t , -1 ≤ t ≤ 1 x1 (t ) = ⎨ ⎩ 0, otherwise
⎧1 - t , -1 ≤ t ≤ 1 x2 (t ) = ⎨ 0, otherwise ⎩
101. The Fourier transform of a continuous-time signal x(t) is 1 given by x(w ) = , -∞ 2, n < 0
⎤ ⎡ T0 ⎢ - ∫ x(T - t ) sin ( nw0 t )dt ⎥ ⎥ 2 ⎢ T 2 = ⎢ 0T ⎥ T0 ⎢ ⎥ + ∫ x(t ) sin n w0 t dt ⎢ ⎥ ⎣ T0 2 ⎦
{
Left hand limit of y[n] is -1, By using y[n] = x[n] + h[n], we get,
}
Let h( n) = a, b, c
(a) H ( z ) = 1 +
By convolution
So, the system is casual.
↑
a
b
c
−1
−a
−b
−c
2
2a
2b
2c
y ( n) = {- a, ( 2a - b ) , ( 2b - c ) , 2c} a = 1 ; 2a - b = 3 ⇒ b = -1 2a - c = -1 ⇒ c = -1
{
}
When a = -1, h ( n) = 1, - 1, - 1 ↑
9. Topic: Shifting and Scaling Operations (a) According to the question,
( ) y[n] = ( -1, 3, -1, -2)
x[n] = 0, 0, 0, -1, 2, 0, 0, 0 ↑
↑
Ch wise GATE_EE_Ch4.indd 166
1.25 Z 1 - 0.8Z -1
12. Topic: Linear Time Invariant and Causal Systems (d) Given that A signal = α1 sin(w1t + ϕ1) Steady-state = α2 F(w2t + ϕ2), then F must be a “sine” or “cosine” function with w1 = w2. 13. Topic: Fourier Series Representation of Continuous Periodic Signals (d) Given that frequency of the signal w = 1 kHz. The spectrum of the sampled signal s(jw) will contain signals that replicate the given signal for U(jw) at frequencies ± nfs where n = 0, 1, 2, …. The sampling frequency fs =
1 1 = = 1 kHz Ts 1 × 10 -3
The signal sampled at intervals of 1 ms is given by
11/21/2018 11:56:05 AM
Chapter 4 • Signals and Systems
167
⎛ 1⎞ LT[r (t - 2)] = e -2 s ⎜ 2 ⎟ ( by time shifting property ) ⎝s ⎠
S(f)
⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ H ( s ) = e s ⎜ ⎟ e -2 s ⎜ 2 ⎟ = e - s ⎜ 3 ⎟ ⎝ s⎠ ⎝s ⎠ ⎝s ⎠ h(t ) = L-1 [ H ( s)] =
−3 −2 −1
0
1
2
f(kHz)
3
The frequency spectrum of the sampled signal is given by S(f)
1 (t - 1) 2 u (tt -1) 2
16. Topic: Laplace Transform and Z-Transform (b) Impulse response h( n) = x( n -1) * y( n) Taking Z-transform, we get H ( z ) = z -1 X ( z )Y ( z ) Given that X ( z ) = 1 - 3 z -1 , Y ( z ) = 1 + 2 z -2 , so H ( z ) = z -1 (1 - 3 z -1 )(1 + 2 z -2 )
f(kHz) 14. Topic: Fourier Series Representation of Continuous Periodic Signals (c) Function x(t) is periodic with a time period 2T and also given that x(t ) = - x(t + T ) , so half-wave symmetry
Y ( z ) = H ( z )U ( z ) Y ( z ) = z -1 (1 - 3 z -1 )(1 + 2 z -2 ) z -1 Y ( z ) = z -2 - 3 z -3 + 2 z -4 - 6 z -5
+1 3T/4 7T/4 t
−T/4 0
Taking inverse Z-transform; y( n) = d ( n - 2) - 3d ( n - 3) + 2d ( n - 4) - 6d ( n - 5) 17. Topic: Linear Time Invariant and Causal Systems (a) The signal input is
−1 Hence, the fundamental Fourier term is given by 4 ⎛π t π⎞ sin ⎜ - ⎟ ⎝ T 4⎠ π 15. Topic: Laplace Transform and Z-Transform (c) Convolution sum is given by, h (t ) = u (t + 1) * r (t - 2) Taking Laplace transform H ( s) = L[h(t )] = L[u(t + 1)]* L[r (t - 2)] 1 LT[u(t )] = u( s) = s ⎛ 1⎞ LT[u(t + 1)] = e s ⎜ ⎟ ( by time shifting property ) ⎝s ⎠
Ch wise GATE_EE_Ch4.indd 167
Z u ( n) ←⎯ → U ( z ) = z -1
= z -2 (1 - 3 z -1 + 2 z -2 - 6zz -3 )
x(t)
LT[r (t )] = R( s) =
The output of the system, for u( n) = d ( n - 1) is,
1 s2
g (t ) = e - α t sin(w t ) and the output is given by y(t ) = Ke - β t sin(ν t +ϕ ) For linear time invariant systems, the frequency of the output must be equal to the input frequency, so n is equal to w. The degree of attenuation will b will depend on the system parameters, so need not be equal to a. 18. Topic: Linear Time Invariant and Causal Systems (d) For a linear time invariant system, principle of superposition holds good. Therefore, for causal system, h(t ) = 0 for t < 0 . 19. Topic: Sampling Theorem (b) Let us consider the procedures given in the options As per option (a) z1 ( n) = x ( n - 3)
z2 ( n) = z1 ( 4 n) = x ( 4 n - 3)
y ( n) = z2 ( - n) = x ( -4 n - 3) ≠ x (3 - 4 n)
11/21/2018 11:56:07 AM
168
GATE EE Chapter-wise Solved Papers
As per option (b) z1 ( n) = x ( n + 3)
z2 ( n) = z1 ( 4 n) = x ( 4 n + 3)
y ( n) = z2 ( - n) = x ( -4 n + 3)
X ( jw ) = F [ x(t )] =
π ⎛ w⎞ rect ⎜ ⎟ ; for - α < w < -α ⎝ 2α ⎠ α
H ( jw ) = F [h(t )] =
⎛w⎞ π rect ⎜ ⎟ ; for - β < w < -β β ⎝ 2β ⎠
Y ( jw ) =
As per option (c)
ν1 ( n) = x( 4n)
⎛w⎞ π2 ⎛ w⎞ rect ⎜ ⎟ rect ⎜ ⎟ ⎝ 2α ⎠ αβ ⎝ 2β ⎠
⎛w⎞ Y ( jw ) = K rect ⎜ ⎟ ⎝ 2g ⎠
ν2 ( n) = ν1 ( - n) = x( -4 n) y( n) = ν2 ( n + 3) = x[ -4( n + 3)] ≠ x(3 - 4 n)
where g is min(a, b ). So y (t ) = K sin c (g t)
As per option (d)
ν1 ( n) = x( 4n)
22. Topic: Fourier Series Representation of Continuous Periodic Signals (b) Given that y ( t ) = x ( t - t 0 ) + x (t + t 0 ) .
ν2 ( n) = ν1 ( - n) = x( -4 n) y( n) = ν2 ( n - 3) = x[ -4( n - 3)] ≠ x(3 - 4 n) Hence, correct option is (b).
Let ak be Fourier series coefficient of x(t)
20. Topic: Linear Time Invariant and Causal Systems (d) The input-output relation is given by y (t ) =
Then Fourier coefficients of y(t) are bk = e - jkw t0 ak + e jkw t0 ak
-2 t
∫ x(t ) dt
bk = 2ak cos k w t0
-∞
Causality: y (t ) depends on x( -2t ) , therefore system is non-causal. Non-linear time-variance: y (t ) =
For all odd values of k, bk = 0, so for odd k,
π 2 2π π k⋅ t0 = T 2 k w t0 =
-2 t
∫ x (t - t ) dt ≠ y (t - t ) 0
0
-∞
Therefore, the system is time variant. Stability: y(t) is unbounded for bounded input, that is
For k = 1,
x(t ) = e -t (bounded) y (t ) =
-2 t
t0 =
T 4
23. Topic: Laplace Transform and Z-Transform (b) Given that x[n] = (1 + j ) n and
-2 t
⎡ e -t ⎤ -t ∫-∞ e dt = ⎢⎣ -1 ⎥⎦ (unbounded) ∞
⎛ 1 ⎞ ROC = ⎜1 - z -1 ⎟ H ( z ) ⎝ 2 ⎠
21. Topic: Applications of Fourier Transform (a) The output is given by
⎛ 2 z - 1⎞ =⎜ H ( z) ⎝ 2 z ⎟⎠
y ( t ) = x ( t ) * h (t )
So, H(z) is of the form
Y ( jw ) = X ( jw ) H ( jw )
⎛ zn ⎞ ⎜⎝ ( 2 z - 1) ⎟⎠
Given that x(t ) = sin c (α t)
(1)
h(t ) = sin c (β t )
(2)
Taking Fourier transform of x (t ) and h (t ) in Eqs. (1) and (2), we get
Ch wise GATE_EE_Ch4.indd 168
So, the correct answer is one pole and two zeros. 24. Topic: Laplace Transform and Z-Transform (d) Given that X ( z) =
z ; with z > a ( z - a) 2
11/21/2018 11:56:09 AM
169
Chapter 4 • Signals and Systems
The output is
n -1 Residue of X ( z ) z at z = a,
d ( z - a) 2 × ( z ) z n -1 dz =
The input is LT[ x(t - t )] = e - st X ( s)
z=a
d z z n -1 ( z - a) 2 × dz ( z - a) 2
d n = z dz
Y ( s) = X ( s) H ( s)
LT[h(t - t )] = e - st H ( s) Y ′( s) = e - st X ( s) ⋅ e - st H ( s)
z=a
= e -2 st X ( s) H ( s) = e -2 st Y ( s)
= nz n -1 |z = a = na n -1
y ′ (t ) = y (t - 2t )
z=a
25. Topic: Applications of Fourier Transform (c) Given that
27. Topic: Linear Time Invariant and Causal Systems (b) For a cascaded system
⎛ 1⎞ x (t ) = rect ⎜ t - ⎟ ⎝ 2⎠
H1(z)
Input
⎪⎧1, -0.5 ≤ (t - 0.5) ≤ 0.5 or 0 ≤ t ≤ 1 ⇒ x (t ) = ⎨ ⎪⎩0, otherwise
H 1 ( z ) = z 2 + z1 + 1 Therefore,
⎪⎧1, -0.5 ≤ ( -t - 0.5) ≤ 0.5 (or ) - 1 ≤ t ≤ 0 x ( -t ) = ⎨ ⎩⎪0, otherwise F [ x (t ) + x ( -t )] =
∞
∫
-∞
0
jw t jw t 1
0 0
∞
x (t ) e - jwt dt + ∫ x ( -t ) e - jwt dt -∞
dt + ∫ (1) e dt + -∫1 (1) e -1
)) ((
dt dt
= ( z 2 + z + 1)( z 3 + z 2 + 1) H 3 ( z ) (1) As, H(z) is causal and H3(z) is causal, so H 3 ( z ) = z -6 + z -4 + 1 Substituting in Eq. (1), we have H(z) is causal
Hence, H(z) will be unstable, when any one of the system is unstable.
0
28. Topic: Fourier Series Representation of Continuous Periodic Signals (c) Given that ⎡⎣e jjww 22 - e -- jjww ⎡⎣e -e
))
2 2
⎤⎦ ⎤⎦
26. Topic: Laplace Transform and Z-Transform (d) Let LT[ x(t )] = X ( s); LT[ y(t )] = Y ( s); LT[h(t )] = H ( s)
Ch wise GATE_EE_Ch4.indd 169
H ( z ) = H1 ( z ) H 2 ( z ) H 3 ( z )
H ( z ) = ( z 2 + z + 1) ( z 3 + z 2 + 1) ( z -6 + z -4 + 1) - jw t - jw t
⎡ e - jwt ⎤1 ⎡ e - jwt ⎤ 0 = ⎢⎡ e - jwt ⎥⎤ + ⎢⎡ e - jwt ⎥⎤ = ⎣⎢ - jw ⎦⎥ 0 + ⎣⎢ - jw ⎥⎦ -1 ⎣ - jw ⎦ 0 ⎣ - jw ⎦ -1 1 1 = 1 ⎡⎣1 - e -- jjww ⎤⎦ + 1 ⎡⎣e jjww - 1⎤⎦ = jw ⎡⎣1 - e ⎤⎦ + jw ⎡⎣e - 1⎤⎦ jw jw e -- jjww 22 jw 2 e jw 2 - jw 2 ⎡⎣e jw 2 - e - jw 2 ⎤⎦ + e jw 2 =e ⎤⎦ + jw = jw ⎡⎣e -e jw jw jw 2 - jw 2 - jw 2 e jw 2 - e - jw 2 e - jw 2 + e jjww 22 -e +e e = e = jw jw ⎛w⎞ ⎛w⎞ 2 sin ⎛⎜ w ⎞⎟ 2 cos ⎜⎛ w ⎟⎞ 2 sin ⎜⎝ 2 ⎠⎟ 2 cos ⎝⎜ 2 ⎠⎟ ⎝ 2⎠ ⎝ 2⎠ = = w w ⎛w⎞ ⎛w⎞ = 2 cos ⎛⎜ w ⎟⎞ sinc ⎛⎜ w ⎞⎟ = 2 cos ⎝⎜ 2 ⎠⎟ sinc ⎝⎜ 2π ⎠⎟ ⎝ 2⎠ ⎝ 2π ⎠
((
H(z)
H 2 ( z) = z3 + z 2 + 1
1⎞ ⎛ x ( -t ) = rect ⎜ -t - ⎟ ⎝ 2⎠
= ∫ (1) e = ∫0 (1) e
H3(z)
For non-causal system, let
Also
1 1
H2(z)
x (t ) =
∞
∑ae
k = -∞
k
j 2π kt T
Fundamental frequency of signal w 0 = 2π T . Therefore, x (t ) = ( a-2 )e - j 2w0 t + ( a-1 )e - jw0 t + a0 + a1e jw0 t + a2 e j 2w0 t On substituting the constant values given, we get x(t ) = ( 2 - j )e -2 jw0 t + (0.5 + 0.2 j )e - jw0 t + 2 j + (0.5 - 0.2 j )e jw0 t + ( 2 + j )e j 2w0 t = 2[e - j 2w0 t + e j 2w0 t ] + j[e j 2w0 t - e - j 2w0 t ] + 0.5[e jw0 t + e - jw0 t ] - 0.2 j[e jw0 t - e - jw0 t ] + 2 j = 2( 2 cos 2w 0 t ) + j ( 2 j sin 2w 0 t ) + 0.5( 2 cos 2w 0 t ) - 0.2 j ( 2 j sin w 0 t ) + 2 j = 4 cos 2w 0 t - 2 sin 2w 0 t + cos w 0 t + 0.4 sin w 0 t + 2 j Thus imaginary part is constant.
11/21/2018 11:56:10 AM
T
bn = 170
T 2 T ⎤ 2⎡ ⎢ ∫ (1) sin n w 0 t dt + ∫ ( -1) sin n w 0 t dt ⎥ T ⎢⎣ 0 ⎥⎦ T 2 T 2 T ⎤ ⎛ cos nw 0 t ⎞ 2 ⎡⎛ cos nw 0 t ⎞ ⎥ + = ⎢⎜ ⎟ ⎟ ⎜ T ⎢⎝ - nw 0 ⎠ 0 ⎝ nw 0 ⎠ T 2 ⎥⎦ ⎣ 2 = [(1 - cos nπ ) + (cos 2nπ - cos nπ )] nw 0T
GATE EE Chapter-wise Solved Papers
=
29. Topic: Laplace Transform and Z-Transform (a) Given that X [ z ] = 4 z -3 + 3 z -1 + 2 - 6 z 2 + 2 z 3 H ( z ) = 3 z -1 - 2 Y ( z) = H ( z) X ( z) = (3 z -1 - 2)( 4 z -3 + 3 z -1 + 2 - 6 z 2 + 2 z 3 ) -4
-2
-1
= 12 z + 9 z + 6 z - 18 z + 6 z - 8 z 2
=
-3
-6 z -1 - 4 + 12 z 2 - 4 z 3 -4
-3
2 x (t ) sin nw 0 t ⋅ dt T ∫0
2 n [1 - ( -1) ] nπ
Therefore,
-2
= 12 z - 8 z + 9 z - 4 - 18 z + 18 z - 4 z The output is given by 2
3
⎧ 4 ⎪ , n odd bn = ⎨ nπ ⎪⎩0, n even
y ( n) = 12d ( n - 4 ) - 8d ( n - 3) + 9d ( n - 2) - 4d ( n) -18d ( n + 1) + 18d ( n + 2) - 4d ( n + 3)
Therefore, y ( n) ≠ 0; n < 0 , so y(n) is non-causal with finite support. 30. Topic: Representation of Continuous and DiscreteTime Signals
Thus, only odd harmonics will be present in x(t) for second harmonic component, and the amplitude is zero. 33. Topic: Linear Time Invariant and Causal Systems (d) Given that at t = 0; f (t ) =
π⎞ ⎛ (d) Given that x(t ) = 8 sin ⎜ 0.8π t + ⎟ ⎝ 4⎠
By L’hospital rule, lim
2π 2π Also, T = = = 2.5 s w 0.8π
x→0
31. Topic: Linear Time Invariant and Causal Systems (b) Given that 5t
y (t ) =
∫ x(t ) dt ,
t>0
-∞
Causality: y (t ) depends on x(5t ), t > 0 So, the system is non-causal, as it depends on future value of input. Linearity: The given system is linear as output is the integration of the input (linear function). 32. Topic: Applications of Fourier Transform (a) Fourier series expression is given as, ∞
x(t ) = A0 + ∑ an cos nw 0 t + bn sin nw 0 t n =1
For odd function, x(t ) = - x(t ). Also, A0 = 0 and an = 0
T 2 T ⎤ 2⎡ ⎢ ∫ (1) sin n w 0 t dt + ∫ ( -1) sin n w 0 t dt ⎥ T ⎢⎣ 0 T 2 ⎦⎥ T 2 T ⎤ ⎛ cos nw 0 t ⎞ 2 ⎡⎛ cos nw 0 t ⎞ ⎥ +⎜ = ⎢⎜ ⎟ ⎟ T ⎢⎝ - nw 0 ⎠ 0 ⎝ nw 0 ⎠ T 2 ⎥⎦ ⎣ 2 = [(1 - cos nπ ) + (cos 2nπ - cos nπ )] nw 0T
=
2
sin t t
sin x =1 x
As f(t) is sinc function, so it is maximum at t = 0 . 34. Topic: Applications of Fourier Transform (d) By Parseval’s identity, 1 2π
∞
∫
X (w ) dw = 2
-∞
∞
∫ x (t ) dt 2
-∞
∞
∫ X (w )
2
dw = 2π × 2 = 4π
-∞
35. Topic: Linear Time Invariant and Causal Systems (c) Given that, finite length input is x[n] = {1, - 1}
0 ≤ n ≤ 1 (1)
And output is y[n] = {-1, 0, 0, 0, -1}
0 ≤ n ≤ 4 (2)
Impulse response, y[n] = h[n] * x[n]
T
2 bn = ∫ x (t ) sin nw 0 t ⋅ dt T 0
Ch wise GATE_EE_Ch4.indd 170
[∵ w 0 = π T ]
n
We have, length of y[n] = 0 → 4; length of x[n] = 0 → 1; length of h[n] = 0 → 3 Let h[n] = {a, b, c, d}
11/21/2018 11:56:13 AM
Chapter 4 • Signals and Systems
By convolution
On taking Laplace transform (shifting properties) a
b
c
d
a
b
c
d
1 1 e -3 s G ( s) = e -3 s - e -5 s = [1 - e -2 s ] s s s
1
−a
−b
−c
−d
−1 y[n] = {a, - a + b, - b + c, - c + d , - d} On comparing with output given in Eq. (2), we get a =1 -a + b = 0 ; b = a = 1 -b + c = 0 ; c = b = 1 -c + d = 0 ; d = c = 1
38. Topic: Applications of Fourier Transform (d) The given waveform satisfies the condition of half wave symmetry. For half wave symmetry, the signal contains only odd harmonics, there will not be any even harmonics. As it satisfies the condition for even symmetry, so bn = 0. Therefore, two possible options are (b) and (d). However, from half wave symmetry (odd harmonic), we get n = 1, 3, 5, 7 ∞ , therefore a0 and an, n = 1, 3, 5, …∞ is the correct answer. 39. Topic: Laplace Transform and Z-Transform (a) Given two continuous time signals that exist for t >0 x (t ) = e - t
Therefore,
y (t ) = e -2t
h ( n) = {1, 1, 1,1}
Convolution is z (t ) = x (t ) * y (t )
↑
36. Topic: Shifting and Scaling Operations (d) First scale f (t) by a factor of 1/2. Then
On taking Laplace-transform
⎛t⎞ g1 (t ) = f ⎜ ⎟ (1) ⎝ 2⎠
g1(t)
Z ( s) = X ( s).Y ( s) =
1 1 ⋅ s +1 s + 2
By partial fraction method, 1 A B = + s + 1 s + 2 s 1 s 2 + + ( )( )
1 = A ( s + 2) + B ( s + 1)
1 2
s = -1; s = -2; A = 1 B = -1 1 1 z ( s) = s +1 s + 2 LT -1 z ( s) = z (t ) = e - t - e -2t
t
Next shift g1(t) by 3, then ⎛ t - 3⎞ g (t ) = g1 (t - 3) = f ⎜ (2) ⎝ 2 ⎟⎠ g1(t)
1
40. Topic: Representation of Continuous and DiscreteTime Signals 1/ 2 (a) We know that rms value = [ variance ] Variance =
3
5
t
From Eqs. (1) and (2), we get ⎛ t 3⎞ g (t ) = f ⎜ - ⎟ ⎝ 2 2⎠ 37. Topic: Laplace Transform and Z-Transform (c) g (t ) can be expressed as, g (t ) = u(t - 3) - u(t - 5)
Ch wise GATE_EE_Ch4.indd 171
171
[α - ( -α )]2 4α 2 α 2 = = 12 12 3
Therefore, rms value is
α2 α = 3 3 41. Topic: Laplace Transform and Z-Transform (b) Given that LT [f (t)] = F1 (s) for t > 0 F2 (s) = LT {F (t − t)} = e −st F1 (s)
11/21/2018 11:56:15 AM
172
GATE EE Chapter-wise Solved Papers
Let m = -n, then
(Delayed version of F1 (s)). Given,
n
G ( s) = = =
F2 ( s) ⋅ F1 *( s) F1 ( s)
2 ∞
e - st F1 ( s) ⋅ F1 *( s) F1 ( s) e - st F1 ( s) F1 ( s)
2
2
2
1 1 < 1 or z > 3z 3
1 ⎛1 ⎞ ∑ ⎜⎝ z ⎟⎠ ; converges if z < 1 or z < 3 3 3 m =1 ∞
⎛ 1⎞
∑ ⎜⎝ 2 z ⎟⎠
n
; converges if
n= 0
1 s + s +1 2
1 1 < 1 or z > 2z 2
Therefore, region of convergence =
1 < z 1
=
1 = 0.5 2
x(t) 63. Topic: Shifting and Scaling Operations (a) The given function can be considered as a combination of
1
−1
1
t
t=2 ⎛t⎞ ⎛t⎞ x(t ) = rect ⎜ ⎟ = rect ⎜ ⎟ (as t = 2) ⎝ 2⎠ ⎝t⎠ Therefore, ⎛t⎞ ⎛ w C⎞ rect ⎜ ⎟ ⎯FT ⎯→ t sa ⎜ ⎝t⎠ ⎝ 2 ⎟⎠
1
t T
2T
2 sin w ⎛t⎞ X ( F ) = rect ⎜ ⎟ ⎯FT ⎯→ 2 sa(w ) = ⎝t⎠ w
Ch wise GATE_EE_Ch4.indd 175
11/21/2018 11:56:23 AM
176
GATE EE Chapter-wise Solved Papers
65. Topic: Fourier Series Representation of Continuous Periodic Signals (b) f (t) is a continuous time signal. So, F(w) = FT(f (t))
(i) 1
∞
∫
F (w ) =
f (t )e - jw t ⋅ dt (1)
-∞ ∞
∫ F (u)e
Function g(t) is defined as
t T
- jut
⋅ du
-∞
This is unit step function u(t).
By doing inverse Fourier transform on Eq. (1), we get
(ii)
f (t ) = 1
1 2π
∞
∫ F (w )e
jw t
⋅ dw
-∞
For the defined function g(t), replace w by u
f (t ) =
1 2π
∞
∫ F (u)e
jut
⋅ du (2)
-∞
t 0
T
- u (t - T ) +
2T
Given that g (t ) =
∞
∫ F (u)e
(t - T ) u (t - T ) T
Replacing t by -t in Eq. (2), we get f ( -t ) =
1 2π
∞
∫ F (u)e
f ( -t ) ≥
⋅ du
2T
(t - 2T ) u(t - 2T ) T
So, overall function can be represented as (t - T ) (t - 2T ) u (t - T ) u(t - 2T ) T T
64. Topic: Laplace Transform and Z-Transform 1 is the causal signal. Now, we 1 - z -3 have to find x(2) and x(3). On expanding X(z), we get
(b) Given X ( z ) =
66. Topic: Laplace Transform and Z-Transform (a) We know that X = Ax + Bu From the given difference equation, a - 1⎤ ⎡1 0 ⎤ ⎡ a A= ⎢ for a = 1; A = ⎢ ⎥ ⎥ ⎣a + 1 a ⎦ ⎣2 1⎦ ⎡ s 0⎤ ⎡1 0⎤ ⎡ s - 1 0 ⎤ sI - A = ⎢ ⎥-⎢ ⎥=⎢ ⎥ ⎣0 s ⎦ ⎣ 2 1 ⎦ ⎣ -2 s - 1⎦
X(z) = 1 + z-3 +z-6 +z-9
sI - A = ( s - 1) 2 - 0 = 0
where X(0) corresponds to constant term = 1
(s - 1)2 = 0 ⇒ s = 1 ± j 0
x(1) corresponds to the coefficient of z -1 = 0 x(2) corresponds to the coefficient of z -2 = 0 -3
x(3) corresponds to the coefficient of z = 1 So, x(2) and x(3) = 0 and 1.
1 g (t ) 2π
If f(t) = f(-t) then it is an even function, therefore, g(t) = 2p f(t) g (t ) ∝ f (t ) So, (as 2p is constant) Thus, g (t) is proportional to f (t) if f (t) is an even function.
t
Ch wise GATE_EE_Ch4.indd 176
- jut
-∞
Substituting for g(t) from Eq. (3), we get
1
u (t ) - u (t - T ) +
⋅ du (3)
-∞
(iii)
-
- jut
67. Topic: Sampling Theorem (c) Given that
x(t ) = 2 + 5 sin(100πt ) (1)
11/21/2018 11:56:25 AM
Chapter 4 • Signals and Systems
f s = 400 Hz =
1 Ts
where Ts is the sampling period. Y ( z) 1 ⎛ 1 - z - N ⎞ (2) = ⎜ X ( z ) N ⎝ 1 - z -1 ⎟⎠
So, we replace t = nTs in Eq. (1) and get x(nTs) = 2 + 5 sin(100p.nTs) 1 ⎞ ⎛ = 2 + 5 sin ⎜100π n ⎟ ⎝ 400 ⎠ ⎛π ⎞ = 2 + 5 sin ⎜ n⎟ ⎝4 ⎠
⎛ m⎞ N = 2π ⎜ ⎟ ⎝w⎠
π , we have 4
⎛ m ⎞ N = 2π ⎜ ⎝ π / 4 ⎟⎠
where m should be the smallest integer such that N be a smallest integer. If m = 1, we have N = 8. Substituting z = e jw , we have Y (e ) 1 ⎡1 - e = ⎢ X ( e jw ) N ⎣ 1 - e - jw jw
f
−fx
fx
(50 + fx)
jw N
⎤ jw ⎥ = H (e ) ⎦
Given that LPF = 25 Hz, output = 20 Hz. Therefore, the lower component, that is (fs - fx) passes through the filter 50 - fx = 20 ⇒fx = 30 Hz 70. Topic: Fourier Series Representation of Continuous Periodic Signals d x (t ) dt
On Fourier transformation, Y(w) = jwX(jw). Therefore, X(w) is real and Y(w) is imaginary. 71. Topic: Linear Time Invariant and Causal systems (90) Given that input signal is sinusoidal signal. This implies u(t) = sin(wt) To find, w
w = 2π f = 2π ×
1 (given f = 1/2T) 2T
wT = p
or
x(n) = x1(n) + x2(n) When r = 0;
(50 + fx) 50 − fx
(b) Given that y(t ) =
To find the fundamental period,
as w =
177
t
The given expression is y(t ) =
y1 ( n) = H (e jw ) r = 0 = x1 ( n) and y1 ( n) = 2.
1 u( t) d t T t -∫T
t
Now, y2 ( n) = H (e jw ) w = π / 4 = 0
=
1 sin( wt )dt T t -∫T
=
1 ⎡ - cos wt ⎤ 1 ⎥ = wT [ - cos w t + cos w (t - T )] T ⎢⎣ w ⎦ t -T
=
1 [ - cos w t + cos w t cos w t + sin w t sin w t ] wT
=
1 [ - cos w t + cos w t cos π + sin w t sin π ] π
=
1 [ - cos w t - cos w t + 0] π
=
-2 2 cos w t = sin(90° + w t ) π π
Therefore, y(n) = y1(n) + y2(n) = 2 + 0 = 2. 68. Topic: Representation of Continuous and DiscreteTime Signals (c) Given the following statements: (i) Even symmetric square wave (ii) Centre frequency 30 kHz From (i), the frequency component present will be 10 kHz, 30 kHz, 50 kHz, 70 kHz. From (ii), the 30 kHz component will pass through filter. Thus, the filter output will be a near cosine wave (due to even signal) at 30 kHz. 69. Topic: Sampling Theorem (c) We have fs = 50 Hz. Let the frequency of signal at x(t) be fx. Then
Ch wise GATE_EE_Ch4.indd 177
t
As x(t) = sin wt, therefore, ϕ = 90° .
11/21/2018 11:56:28 AM
178
GATE EE Chapter-wise Solved Papers
72. Topic: Linear Time Invariant and Causal Systems (d) The continuous time signal is g (t ) = u(t ) - u(t -1) . Therefore,
cos(t) t
g (t ) × g (t ) = [u(t ) - u(t - 1)] × [u(t ) - u(t - 1)] = u(t ) × u(t ) - u(t ) × u(t - 1)) - u(t - 1) sgncos(t)
× u(t ) + u(t - 1) × u (t - 1) = r (t ) - r (t - 1) - r (t - 1) + r (t - 2) So, g (t )eq = r (t ) - 2r (t - 1) + r (t - 2) The continuous time signal is depicted in the following figure. h(t)
75. Topic: Laplace Transform and Z-Transform (c) The given equation is
1 m=1 m = −1 t 1 Overall impulse response h(f ) = g(f ) * g(f ). So, the maximum value of geq = 1. 73. Topic: Laplace Transform and Z-Transform (b) From the given question L[f(t)] = s−3/2 and L(g(t)) = ?
j 10 t ⎪⎧e x (t ) = ⎨ ⎩⎪0
for t ≤ 1 for t > 1
t
1
-t
-1
- jw t j10 t - jw t ∫ x(t ) e dt = ∫ e e dt
∞
⎡ ⎤ 1 -3/ 2 1 ⎢ s -3/ 2 +1 ⎥ 1 -1/ 2 = ∫ s ⋅ ds = ⎢ = ⎥ =s 3 2s 2⎢ s - + 1⎥ ⎣ 2 ⎦s 74. Topic: Fourier Series Representation of Continuous Periodic Signals (d) Given that
1
⎡ e j (10 -w )t ⎤ = ∫ e j (10 -w )t dt = ⎢ ⎥ ⎣ j (10 - w ) ⎦ -1 -1 2 sin(w - 10) = (w - 10) 1
∞
Ch wise GATE_EE_Ch4.indd 178
76. Topic: Applications of Fourier Transform (a) We have
x(w ) =
⎡ f (t ) ⎤ Now, L[ g (t ) = L ⎢ ⎥ ⎣ 2t ⎦
⎧ x ⎪ ; x≠0 sgn( x ) = ⎨ | x | ⎪ 0; x = 0 ⎩
x(n) = (−0.25)n u(n) + (0.5)n u(−n − 1). So, the region of convergence: for (−0.25)n u(n) is |z | > |−0.25| for (0.5)n u(−n − 1) is |z | < |0.5| Therefore, ROC is 0.25 < |z | < 0.5
for which the Fourier transform is,
or, L[2 t / π ] = s -3/ 2 and L( 1 / π t ) = ?
cost > 0 ⇒ sgn(cost) = 1 cost < 0 ⇒ sgn(cost) = −1
sgn(cost) is a square wave with even and half wave symmetry function. Thus, it includes cosine terms for all odd harmonics.
77. Topic: Linear Time Invariant and Causal Systems (a) For linear time invariant systems that are Bounded Input Bounded Output stable, the unit step response will be bounded. 78. Topic: Laplace Transform and Z-Transform (a) We have y(n) = x(n) - x(n - 1) Taking Z-transform, we get Y ( z ) = X ( z ) - z -1 X ( z ) X(z) = 2z + 4 - 4z−1 + 3z−2 z−1X(z) = 2 + 4z−1 - 4z−2 + 3z−3
11/21/2018 11:56:29 AM
Chapter 4 • Signals and Systems
Y(z) = (2z + 4 - 4z−1 + 3z−2) - 2 − 4z−1 + 4z−2 − 3z−3
=
= 2z + 2 − 8z−1 + 7z−2 − 3z−3 79. Topic: Representation of Continuous and DiscreteTime Signals (c) Let there be two inputs Ax1 (t) and Bx2(t). Apply x(t) = Ax1(t) + Bx2(t) y(t) = [Ax1(t) + Bx2(t)]cos(t) = Ax1(t)cos(t) + Bx2(t)cos(t) = y1(t) + y2(t) Thus the function is linear. Let input be time shifted by t0. Then output y(t) = x(t - t0)cos(t) y(t) does not shift t0, hence it is time-varying. 80. Topic: Linear Time Invariant and Causal Systems (a) Given function +∞
∫e
-t
d ( 2t - 2)
⎛ 1⎞ -⎜ ⎟ t ⎝ 6⎠
83. Topic: Laplace Transform and Z-Transform (b) Let z = A + iB. Therefore, Z * = A - iB Therefore, f(z) = 2A (continuous) u = 2A; v=0 du = 2; dA
du =0 dB
dv = 0; dA
dv =0 dB
=
1 -t e |t =1 2
Therefore, it is not analytic.
=
1 2e
84. Topic: Laplace Transform and Z-Transform (a) Using Laplace transform
dy(t ) 1 + y (t ) = 3 x (t ) dt 6
Therefore, Y ( s) =
3 9 . 1⎞ ⎛ 1⎞ ⎛ ⎜⎝ s + ⎟⎠ ⎜⎝ s + ⎟⎠ 6 3
) ⋅ u (t )
Therefore, maximum frequency of x(t)cos(2000 p t) will be convolution of spectrums in frequency domain, that is, 6 kHz.
3 X ( s) 9 H ( s) = and X ( s) = 1⎞ 1⎞ ⎛ ⎛ ⎜⎝ s + ⎟⎠ ⎜⎝ s + ⎟⎠ 6 3
⎛ 1⎞ -⎜ ⎟ t ⎝ 3⎠
Maximum frequency of x(t) = 5 kHz. Maximum fre2000 π quency of cos(2000pt) = = 1000 Hz = 1 kHz 2π
1 -t ∫ e d (t - 1) dt 2 -∞
1 sY ( s) + Y ( s) = 3 X ( s) 6
- 54e
82. Topic: Sampling Theorem (6) Given that
=
81. Topic: Laplace Transform and Z-Transform (d) Using Laplace transform
Ch wise GATE_EE_Ch4.indd 179
54 54 1⎞ ⎛ 1⎞ ⎛ ⎜⎝ s + ⎟⎠ ⎜⎝ s + ⎟⎠ 6 3
y(t ) = (54e
-∞
∞
179
d 2 y 2dy + +y=0 dt dt 2 [ s 2Y ( s) - sy(0) - y ′(0)] + 2[ sY ( s) - y(0)] + Y ( s) = 0 Y ( s) =
sY (0) + y ′(0) + 2 y(0) s 2 + 2s + 1
y ′(0) = 1, y(0) = 0 Y ( s) =
1 ⇒ y (t ) = t ⋅ e - t u (t ) ( s + 1) 2
85. Topic: Fourier Series Representation of Continuous Periodic Signals (b) f(−x) = −f(x) implies odd function. Odd functions have only sine terms in their Fourier expression. Therefore, the correct option is (b).
11/21/2018 11:56:31 AM
180
GATE EE Chapter-wise Solved Papers
86. Topic: Applications of Fourier Transform (c) X1(jw) and X2(jw) are not conjugate symmetric. Therefore, x1(t) . x2(t) are not real. Fourier transform of x1 (t ) ⋅ x2 (t ) ↔
These two are not equivalent. Hence, system is time varying. If x[n] = d [n], y[n] = [1 + (-1)0] d [n] = 2d [n]
1 X 1 ( jw ) * X 2 ( jw ) 2π
If x[n] = 2d [n],
X1(jw)*X2(jw) is conjugate symmetric, so, x1(t) . x2(t) will be real. 87. Topic: Linear Time Invariant and Causal Systems (d) Since impulse response is real and even, H(jw) will also be real and even. Therefore,
y[n] = [1 + (-1)0]2d n = 4d [n] This suggests that the system is non-invertible. 91. Topic: Applications of Fourier Transform (0.318) Given that
H(j w0) = H(−j w0)
⎧t - ⎢⎣t ⎥⎦ , t≥0 g (t ) = ⎨ t t , otherwise ⎡ ⎤ ⎩ ⎢⎥
e jt + e - jt cos(t) input = input 2 Therefore, output =
g(t)
H ( j1)e jt + H ( - j1)e - jt = H ( j1) cos(t ) 2
Similarly for sin(t), output = H(j1)sint Output sin(t) and cos(t) are eigen signals with same eigenvalues.
1
t −2
−1
88. Topic: Sampling Theorem (b) For the given system
1
g ′ (t ) = 1 +
∞
∑ - d(t - k )
Fourier series coefficient of g ′(t ) is ⎧ 1, k = 0 Gk′ = ⎨ ⎩ -1, k ≠ 0
Bandwidth of X1 = B1 and bandwidth of X2 = B2, therefore, bandwidth of X 1 * X 2 = B1 + B2
Gk =
Therefore, bandwidth of Y ( jw ) = B1 + B2 Hence, sampling frequency ≥ 2(B1 + B2)
x(t) * y(t) = z(t) Then 1 z (ct ) c
⇒ z(ct) = c[x(ct) * y(ct)] 90. Topic: Representation of Continuous and DiscreteTime Signals (d) y(n - 1) = [1 + (-1)n -1] x(n - 1) y[x(n - 1)] = [1 + (-1)n ] x(n - 1)
1 Gk′ 1 1 ; G0 = ∫ g (t )dt = jw 0 k 10 2
Gk is imaginary and thus the series has only sine terms. ⎡ -1 -1 ⎤ -2π bk = j ⎢ ⎥= k ⎣ jπ k jπ k ⎦
89. Topic: Applications of Fourier Transform (a) Using time scaling, we get
Ch wise GATE_EE_Ch4.indd 180
3
k = -∞
Y ( jw ) = [ X 1 ( jw ) * X 2 ( jw )]H ( jw )
x (t ) * y (t ) =
2
⇒ b2 =
-2π = 0.318 2
92. Topic: Representation of Continuous and DiscreteTime Signals (b) First figure: H ( z) =
k ( z 2 - 1) z2
At low frequency, z = 1 ⇒|H(1)| = 0
11/21/2018 11:56:33 AM
181
Chapter 4 • Signals and Systems
At high frequency, z = −1
Now,
⇒|H(−1)| = 0 Therefore, the first system is a band-pass filter.
T0 = 2T =
2π 1 = 2000π rad ; w0 = T 1000
The series contains only odd harmonics. Therefore,
Second figure: H ( z) =
k ( z 2 - 1) z 2 + 0.25
a2 =
2 T
At low frequency, z = 1 =
⇒|H(1)| = 0 At high frequency, z = −1
∫
x(t ) cos 2w 0 t dt
T0 / 2 T0 / 2
∫
d (t ) cos 2 w 0 t dt = 4000
0
Hence,
⇒|H(−1)| = 0
y(t) = 4000 cos (2000 p t)
Therefore, the second system is a band-pass filter. Third figure: H ( z) =
4 T0
+ T0 / 2
z2 -1 z2 + 1
96. Topic: Sampling Theorem (b) z(t) = x(t) cos 1000 p t Z (w ) =
At low frequency, z = 1
1 [ X (w + 1000π ) + x(w - 1000π )] 2 Z(w)
⇒|H(1)| = 0 At high frequency, z = −1 ⇒|H(−1)| = 0
w
Therefore, the third system is a band-pass filter.
—2000p
2000p
93. Topic: Applications of Fourier Transform (6) Mean square value =
1 T0
H(w)
∫
2
f (t ) dt
T0
1 24 = [4 2 × 1 + 22 × 2] = =6 4 4
94. Topic: Laplace Transform and Z-Transform (b) Taking Laplace transform, equation becomes (s2 + 7s + 10)˙y(s) = (4 + 5s) X(s) ⇒
Y ( s) 4 + 5s = X ( s) 52 + 7 s + 10
1500p
Y(w) has the frequency = 1500p rad/s ⇒ Minimum fs = 1500 samples/s 97. Topic: Applications of Fourier Transform (d) Overall, h[n] = h1[n] * h2[n] ↑
H(z) = 1 - z−2 y(z) = 1 + 2z−1 + z−2 - z−3 - 2z−4 − z−5
⎧ Y ( s) ⎫ Impulse response = L-1 ⎨ ⎬ ⎩ X ( s) ⎭ 7 ⎫ ⎧ 2 = L-1 ⎨ + ⎬ ⎩ s + 2 s + 5⎭ = -2e-2tu(t) + 7e-5tu(t) 95. Topic: Applications of Fourier Transform x (t ) =
+∞
⎛
k ⎞
∑ ( -1) d ⎜⎝ t - 2000 ⎟⎠
k = -∞
Ch wise GATE_EE_Ch4.indd 181
w —1500p
= [1, -1] * [1,1] = [1, 0, -1]
Therefore,
(c)
1
k
X ( 2) =
1 + 2 z -1 + z -2 - z -3 - 2 z -4 - z -5 1 - z -2
= 1 + 2z−1 + 2z−2 + z−3 = [1, 2, 2, 1] ↑
98. Topic: Laplace Transform and Z-Transform (c) We start with P =
15 . s + 5s + 15 2
11/21/2018 11:56:35 AM
182
GATE EE Chapter-wise Solved Papers
⎫ ⎧ ⎪ ⎪ x2 [n] = ⎨0, 0.25, 0.5, 0.75,1, 0.75, 0.5, 0.25, 0⎬ ⎪ ⎪ ↑ ⎩ ⎭
w n 2 = 15 ⇒ w n = 15 = 3.872 rad/sec. 2x ´× 3.872 = 5 ⇒ Þξ =
5 = 0.64 < 1, so underdamped 2 × 3.872
We can see clearly, x1 [n] has greater energy than x2 [n]
Now, Q=
x1 [n] = 2 × 12 + 2 × (0.75) 2 + 2 × (0.5) 2 + 2 × (0.25) 2 + 0 2
energy in x1 [n] = 2 × 12 + 2 × (0.75) 2 + 2 × (0.5) 2 + 2 × (0.25) 2 + 0 2
25 s 2 + 10 s + 25
Energy in x2 [n] = 2 × 0 2 + 2 × (0.75) 2 + 2 × (0.25) 2 + 2 × (0.5) 2 + 0 2
w n = 25 ⇒ w n = 25 = 5 rad/sec. x2 [n] = 2 × 0 2 + 2 × (0.75) 2 + 2 × (0.25) 2 + 2 × (0.5) 2 + 0 2 2x ´× 5 = 10 Þ⇒ x = 1, so critically damped 2
100. Topic: Representation of Continuous and DiscreteTime Signals (d) x(t) = r(t - 1) - r(t - 2) - r(t - 3) + r(t - 4)
35 s + 18s + 35 = 35 ⇒ w n = 5.91
R=
2
wn2
18 = 1.522 > 1, so overdamped 2 × 5.91
ξ=
x(t) (t − 1)
99. Topic: Representation of Continuous and DiscreteTime Signals (a)
(4 − t) t
1 E x (t ) =
⎧ t , -1 ≤ t ≤ 1 x1 (t ) = ⎨ ⎩ 0, otherwise
2
3
4
∞
∫
x(t ) 2 dt
-∞ 2
3
4
1
2
3
= ∫ (t - 1) 2 dt + ∫ 12 dt + ∫ ( 4 - t ) 2 dt
x1(t)
= 1 t
-1
1
1 ⎧1 - t , -1 ≤ t ≤ 1 x2 (t ) = ⎨ 0, otherwise ⎩
1 1 2 5 +1+ = +1 = 3 3 3 3
101. Topic: Applications of Fourier Transform (10) X (w ) =
1 (10 + jw ) 2
Inverse Fourier,
x2(t)
X ( s) =
1
1 ( s + a) 2
x(t ) = te - at u(t ) ⇒Þ x(t ) = te -10 t u(t )
-1
t 1
⎧ ⎫ ⎪ ⎪ x1 [n] = ⎨+1, +0.75, +0.5, 0.25, 0, 0.25, 0.5, 0.75,1⎬ ⎪ ⎪ ↑ ⎩ ⎭
Ch wise GATE_EE_Ch4.indd 182
{ln x(t ) } = ln t -10t ln e e At t = 1, we have |ln (x(t))| = |ln (1) - 10 ´× 1| = |-10| = +10
11/21/2018 11:56:37 AM
Electrical Machines
CHAPTER5
Syllabus Single phase transformer: equivalent circuit, phasor diagram, open circuit and short circuit tests, regulation and efficiency; Three phase transformers: connections, parallel operation; Auto-transformer, Electromechanical energy conversion principles, DC machines: separately excited, series and shunt, motoring and generating mode of operation and their characteristics, starting and speed control of dc motors; Three phase induction motors: principle of operation, types, performance, torque-speed characteristics, no-load and blocked rotor tests, equivalent circuit, starting and speed control; Operating principle of single phase induction motors; Synchronous machines: cylindrical and salient pole machines, performance, regulation and parallel operation of generators, starting of synchronous motor, characteristics; Types of losses and efficiency calculations of electric machines.
CHAPTER ANALYSIS Topic
GATE 2009
Single Phase Transformer
1
Three Phase Transformers
2
Auto-Transformer
2
GATE 2010
GATE 2011
GATE 2012
GATE 2013
GATE 2014
1
1
1
3
GATE 2015
GATE 2016
GATE 2017
GATE 2018
2
1
2
3
2
4
1
1
1
1
2
4
6
1
4
1
4
1
2
2
1
2
1
1
1
5
3
Electromechanical Energy Conversion Principles DC Machines
1
Three Phase Induction Motors
1
Operating Principle of Single Phase Induction Motors
2
2
Synchronous Machines Types of Losses and Efficiency Calculations of Electric Machines
2
1
1
1
2 2
1
3
1
4
2
2
4
1
2
IMPORTANT FORMULAS I. Ideal Transformer 1.
Instantaneous values of the applied voltage and induced emf e1 based on Kirchhoff’s voltage law: v1 - e1 = 0
2.
RMS value of induced emf E2 = 4.44 fN 2ϕ 2 m V
3.
Voltage ratio between the primary and secondary windings E1 N1 = E2 N 2
Ch wise GATE_EE_Ch5.indd 183
RMS value E1 I1 = E2 I 2 4.
Equivalent circuit parameters,
and
Rc =
V1 I 0 cos ϕ0
Xm =
V1 I 0 sin ϕ0
where, I c = I 0 cos ϕ 0 and I m = I 0 sin ϕ 0
11/21/2018 11:53:19 AM
184 5.
GATE EE Chapter-wise Solved Papers
The equivalent circuit parameters for short circuit test Psc = I Re 2 sc
Re =
14. Per unit (pu) quantities Actual value of the quantity in any unit Base value of the qunatity in the same unit
Psc I sc2
Base current,
Xe = Z - R 2 e
6.
2 e
Voltage regulation
3 × Vbase
Base impedance,
Vnl - Vl
2
Vnl R = ε r cos q 2 - ε x sin q 2 7.
Pbase
I base =
⎛V ⎞ P Z 2 = Z1 × ⎜ base_old ⎟ base_new ⎝ Vbase_new ⎠ Pbase_old IV. DC Machines
Efficiency (per unit) xV1 cos ϕ xV1 cos ϕ + x 2 Pcu + Pcore
15. Reactance voltage, Vc = 1.11L
Condition for maximum efficiency x= 8.
9.
where TC is the commutation time.
Pcore Pcu
16. The total induced emf is,
xV2 I 2 cos ϕ × 100% xV2 I 2 cos ϕ + Pcore + x 2 Pcu
All day efficiency:
II. Auto-transformer V2 N 2 = = a (also represented by K ) V1 N1 Transformed power (VA) (V1 - V2 ) I1 = V1 I1 Input power (VA) = 1-
V2 = 1- a V1
Conducted power (VA) V2 I1 V2 = = =a Input power (VA) V1 I1 V1
III. Three Phase Transformers 13. Power delivered by all three transformers in open-delta connection PV = 3 × VL × I ph
Power delivered for all three transformers in delta connection PD = 3 × VL × I ph Therefore, PV = 0.577 or 57.7% PD
Ch wise GATE_EE_Ch5.indd 184
17. Induced emf in armature E =
pϕw Z V 2π a
E = K a ϕw where Ka = emf constant =
pZ V/Wb/rad/s 2π a
18. Electromagnetic torque Te = K a ϕ I a Nm
10. Turns or transformation ratio
12.
Z V a
General form,
Total output for 24 hours ×100% Total output for 24 hour + Total losses for 24 hours
11.
E = Bavg ⋅ L ⋅ v
Percentage transformer efficiency %η =
2I TC
19. Characteristics of DC generators Terminal voltage, V = E - I a Ra - Vb where E is the induced emf, Vb is brush contact drop (in Volts) and Ia is the armature current (in Amperes) 20. Characteristics of DC shunt motors Torque-current relationship, Te = K a ϕ I a Voltage equation, V = E + I a Ra Speed-current characteristics
w=
(V - I a Ra ) Kaϕ
Speed-torque characteristics
w=
TR V - e a K a ϕ ( K a ϕ )2
21. Characteristics of DC series motors Torque-current relationship Te = K a ϕ I a
11/21/2018 11:53:21 AM
Chapter 5 • Electrical Machines
Voltage equation V = E + I a ( Ra + Rf )
29. Series parameters Pbr = Vbr I br cos ϕ br
Speed-current characteristic
w=
and
V K a′ Te
-
Z br =
Ra Ka and
V - I a Ra rpm ϕ
23. Efficiency of DC machines For generator,
ηg =
X br = Z br sin ϕ br 30. Power flow diagram Pem = Pg - sPg = Pg (1 - s) 31. Per phase rotor current Ir =
Output VI = Output +losses VI + ( I + I sh ) 2 Ra + Pc
For motor, Input - losses VI - ( I - I sh ) 2 Ra + Pc = ηm = VI Input
Pg = I r =
Pc = VI 0 - ( I 0 - I sh ) Ra 2
V. Three Phase Induction Motors
Te =
ia = I m cos 2π ft ic = I m cos( 2π ft - 240°)
Fr (t ) = Fa cos w t cos q + Fb cos(w t - 2π / 3) cos(q - 2π / 3) + Fc cos(w t + 2π /3) cos(q + 2π / 3) or Fr (t ) = 3Fa / 2 cos(w t - q ) where Fa= Fb = Fc = NIm.
2
Rr′ s Vs 2 2
R r′ ⎞ ⎛ 2 ⎜⎝ Rs + s ⎟⎠ + j ( X ls + X lr′ )
Rr′ s
Vs2 Rr′ 3 2 s ws ⎛ Rr′ ⎞ ⎜⎝ Rs + s ⎟⎠ + ( X ls + X lr′ )
34. Maximum torque
where Im is the maximum value of the stator current in each phase. 26. Resultant flux of the three phases,
⎛ R′ ⎞ Rs + ⎜ r ⎟ + j ( X ls + X lr′ ) ⎝ s⎠
33. Electromagnetic torque
25. Three phase currents are given by, ib = I m cos( 2π ft - 120°)
Vs
32. Air-gap power
24. Testing of DC Machines In Swinburne’s test constant losses,
Tmax = ±
Vs 2 3 w s 2 R ± R2 + X 2 s s e
(
)
The positive sign is for motor and the negative sign is for the generator. So Tmax( m ) < Tmax( g ) where X e = X ls + X lr′
35. Methods of starting three-phase motors
27. Rotor current
Direct-on-line starting Ir =
E1 Rr2 + ( X lr2 ) s2
28. Shunt parameters Pnl = Vnl I nl cos ϕ nl V Rc = nl cos ϕ nl I nl
2
Tst ⎛ I sc ⎞ = × sfl Tfl ⎜⎝ I fl ⎟⎠ Autotransformer starting 2
⎛I ⎞ Tst = K 2 ⎜ sc ⎟ × sfl Tfl ⎝ I fl ⎠ Start-delta starting
and Xm =
Ch wise GATE_EE_Ch5.indd 185
Vbr I br
R br = Z br cos ϕ br
22. Speed of DC motors N=K
185
Vnl sin ϕ nl I nl
2
Tst 1 ⎛ I sc ⎞ = × sfl Tfl 3 ⎜⎝ I fl ⎟⎠
11/21/2018 11:53:24 AM
186
GATE EE Chapter-wise Solved Papers
VI. Single Phase Induction Motors 36. Slip of the rotor with respect to the forward-rotating field s+ = s =
Ns - N r N = 1- r Ns Ns
Vt = E0 + I a Ra + jI d X d + jI q X q where, I a = I d + jI q and Vt = Vd + jVq
37. Slip of the rotor with respect to the backward-rotating field s- = s =
41. The terminal voltage considering the two reaction theory for synchronous motor
42. Synchronous motor power equation P = VI a cos ϕ = -
⎛ N ⎞ Ns - (- N r ) = 2 - ⎜1 - r ⎟ Ns ⎝ Ns ⎠
⎛ VE ⎞ P = 3 × VI a cos ϕ = 3 × ⎜ - ⎟ sin d ⎝ Xs ⎠
VII. Synchronous Machines
(for three phase motor)
38. Induced emf
43. Synchronising power
E0 (rms) = 4.44 f ϕ Nk p k d Volts
Psy = E1 I sy cos ϕ 1 = E1 I sy cos(90° - q ) = E1 I sy sin q
No load emf E = (Vt cos ϕ + IR) + (Vt sin ϕ + IX s ) 2 0
2
Psy =
2
39. Zero power characteristics Regulation % =
VE sin d (for single phase motor) Xs
αE2 2Xs
44. Synchronising current,
Eph - Vph Vph
× 100%
I sy =
40. Generator voltage regulation
Er αE = 2Xs 2Xs
45. In slip test
( E0 - Vt ) × 100% Vt
Xq =
V Vmin and X d = max I min I max
QUESTIONS 1. A 3-phase, 4-pole squirrel cage induction motor has 36 stator and 28 rotor slots. The number of phases in the rotor is: (a) 3 (b) 9 (c) 7 (d) 8 (GATE 2000: 1 Mark) 2. The compensating winding in a DC machine (a) is located in armature slots for compensation of the armature reaction (b) is located on commutating poles for improving the commutation (c) is located on poles shoes for avoiding the flashover at the commutation surface (d) is located on pole shoes to avoid the sparking at the brushes (GATE 2000: 1 Mark) 3. In a constant voltage transformer (CVT), the output voltage remains constant due to
Ch wise GATE_EE_Ch5.indd 186
(a) capacitor (c) saturation
(b) input inductor (d) tapped windings (GATE 2000: 1 Mark)
4. The phase sequence of a three-phase alternator will reverse if (a) the field current is reversed keeping the direction of rotation same (b) the field current remains the same but the direction of rotation is reversed (c) the field current is reversed and the number of poles is doubled (d) the number of poles is doubled without reversing the current (GATE 2000: 1 Mark) 5.
A 1.8° step, 4-phase stepper motor has a total of 40 teeth on 8 poles of stator. The number of rotor teeth for this motor will be
11/21/2018 11:53:25 AM
Chapter 5 • Electrical Machines
(a) 40 (c) 100
(b) 50 (d) 80 (GATE 2000: 1 Mark)
6. Ratio of the rotor reactance X to the rotor resistance R for a two-phase servomotor (a) is equal to that of a normal induction motor (b) is less than that of a normal induction motor (c) is greater than that of a normal induction motor (d) may be less or greater than that of a normal induction motor (GATE 2000: 1 Mark) 7. A 3-phase delta/star transformer is supplied at 6000 V on the delta-connected side. The terminal voltage on the secondary side when supplying full load as 0.8 lagging power-factor is 415 V. The equivalent resistance and reactance drops for the transformer are 1% and 5% respectively. The turn’s ratio of the transformer is: (a) 14 (b) 24 (c) 42 (d) 20 (GATE 2000: 2 Marks) 8. A 240 V DC series motor takes 40 A when giving its rated output at 1500 rpm. Its resistance is 0.3 ohms. The value of resistance which must be added to obtain rated torque at 1000 rpm is: (a) 6 ohms (b) 5.7 ohms (c) 2.2 ohms (d) 1.9 ohms (GATE 2000: 2 Marks) 9. The power input to a 415 V, 50 Hz, 6 pole, 3-phase induction motor running at 975 rpm is 40 kW. The stator losses are 1 kW and friction and windage losses total 2 kW. The efficiency of the motor is (a) 92.5% (b) 90% (c) 91% (d) 88% (GATE 2000: 2 Marks) 10. A single-phase, 2000 V alternator has armature resistance and reactance of 0.8 ohms and 4.94 ohms respectively. The voltage regulation of the alternator at 100A load at 0.8 leading power-factor is: (a) 7% (b) −8.9% (c) 14% (d) 0% (GATE 2000: 2 Marks) 11. A permanent magnet DC commutator motor has a no load speed of 6000 rpm when connected to a 120V DC supply. The armature resistance is 2.5 ohms and other losses may be neglected. The speed of the motor with supply voltage of 60 V developing a torque 0.5 Nm, is: (a) 3000 rpm (b) 2673 rpm (c) 2836 rpm (d) 5346 rpm (GATE 2000: 2 Marks)
Ch wise GATE_EE_Ch5.indd 187
187
12. A three phase, wound rotor induction motor is to be operated with slip energy recovery in the constant torque mode, when it delivers an output power Po at slope s. Then theoretically, the maximum power that is available for recovery at the rotor terminals, is equal to (a) Po (b) Po·S (c)
Po Po ⋅ s (d) (1- s) (1 - s) (GATE 2000: 2 Marks)
13. A single-phase transformer is to be switched to the supply to have minimum inrush current. The switch should be closed at (a) maximum supply voltage (b) zero supply voltage 1 (c) maximum supply voltage 2 1 maximum supply voltage (d) 2 (GATE 2001: 1 Mark) 14. It is desirable to eliminate 5th harmonic voltage from the phase voltage of an alternator. The coils should be shortpitched by an electrical angle of (a) 30° (b) 36° (c) 72° (d) 18° (GATE 2001: 1 Mark) 15. The following figure shows the magnetization curves of an alternator at rated armature current, unity power factor and also at no load. The magnetization curve for rated armature current, 0.8 power factor leading is given by Rated armature current unit pf
No load
A
C B
D
Exciting current (a) curve A (c) curve C
(b) curve B (d) curve D (GATE 2001: 1 Mark)
16. The core flux of a practical transformer with a resistive load (a) is strictly constant with load changes (b) increases linearly with load (c) increases as the square root of the load (d) decreases with increased load (GATE 2001: 1 Mark)
11/21/2018 11:53:25 AM
GATE EE Chapter-wise Solved Papers
17. X d , X d′ and X d′′ are steady state d-axis synchronous reactance, transient d-axis reactance and sub-transient d-axis reactance of a synchronous machine respectively. Which of the following statements is true? (a) X d > X d′ > X d′′ (b) X d′′ > X d′ > X d (c) X d′ > X d′′ > X d (d) X d > X d′′ > X d′ (GATE 2001: 1 Mark) 18. In case of an armature controlled separately excited DC motor drive with closed-loop speed control, an inner current loop is useful because it (a) limits the speed of the motor to a safe value (b) helps in improving the drive energy efficiency (c) limits the peak current of the motor to the permissible value (d) reduces the steady state speed error (GATE 2001: 1 Mark) 19. 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 (GATE 2001: 2 Marks) 20. An electric motor with ‘constant output power’ will have a torque speed characteristic in the form of a (a) straight line through the origin (b) straight line parallel to the speed axis (c) circle about the origin (d) rectangular hyperbola (GATE 2001: 2 Marks) 21. A 3-phase transformer has rating of 20 MVA, 220 kV (star) − 33 kV (delta ) with leakage reactance of 12%. The transformer reactance (in ohms) referred to each phase of the L.V. delta-connected side is (a) 23.5 (b) 19.6 (c) 18.5 (d) 8.7 (GATE 2001: 2 Marks) 22. If a 400 V, 50 Hz, star connected, 3 phase squirrel cage induction motor is operated from a 400 V, 75 H supply, the torque that the motor can now provide while drawing rated current from the supply? (a) reduces (b) increases (c) remains the same (d) increase or reduces depending upon the rotor resistance (GATE 2002: 1 Mark)
Ch wise GATE_EE_Ch5.indd 188
23. A DC series motor fed from rated supply voltage is overloaded and its magnetic circuit is saturated. The torquespeed characteristic of this motor will be approximately represented by which curve of as shown in the following figure?
A
Speed
188
B D
C
Torque (a) Curve A (c) Curve C
(b) Curve B (d) Curve D (GATE 2002: 1 Mark)
24. A 1 kVA, 230 V/100 V, single phase, 50 Hz transformer having negligible winding resistance and leakage inductance is operating under saturation, while 250 V, 50 Hz sinusoidal supply is connected to the high voltage winding. A resistive load is connected to the low voltage winding which draws rated current. Which one of the following quantities will not be sinusoidal? (a) (b) (c) (d)
Voltage induced across the low voltage winding Core flux Load current Current drawn from the source (GATE 2002: 1 Mark)
25. A 400 V/200 V/200 V, 50 Hz three winding transformer is connected as shown in the following figure. The reading of the voltmeter, V, will be
V
400 V 50 Hz 400:200:200 (a) 0 V (c) 600 V
(b) 400 V (d) 800 V (GATE 2002: 1 Mark)
26. A 200 V, 2000 rpm, 10 A, separately excited DC motor has an armature resistance of 2 Ω. Rated DC voltage is applied to both the armature and field winding of the motor. If the armature drawn 5 A from the source, the torque developed by the motor is
11/21/2018 11:53:26 AM
Chapter 5 • Electrical Machines
(a) 4.30 Nm (c) 0.45 Nm
(b) 4.77 Nm (d) 0.50 Nm (GATE 2002: 2 Marks)
27. The rotor of a three phase, 5 kW, 400 V, 50 Hz, slip ring induction motor is wound for 6 poles while its stator is wound for 4 poles. The approximate average no load steady state speed when this motor is connected to 400 V, 50 Hz supply is (a) 1500 rpm (b) 500 rpm (c) 0 rpm (d) 1000 rpm (GATE 2002: 2 Marks) 28. The flux per pole in a synchronous motor with the field circuit ON and the stator disconnected from the supply is found to be 25 mWb. When the stator is connected to the rated supply with the field excitation unchanged, the flux per pole in the machine is found to be 20 mWb while the motor is running on no load. Assuming no load losses to be zero, the no load current down by the motor from the supply (a) lags the supply voltage (b) leads the supply voltage (c) is in phase with the supply voltage (d) is zero (GATE 2002: 2 Marks) 29. A single-phase transformer has a maximum efficiency of 90% at full load and unity power factor. Efficiency at half load at the same power factor is (a) 86.7% (b) 88.26% (c) 88.9% (d) 87.8% (GATE 2003: 1 Mark) 30. Group I lists different applications and Group II lists the motors for these applications. Match the application with the most suitable motor and choose the right combination among the choices given thereafter Group I
Group II
P. Food mixer
1. Permanent magnet DC motor
Q. Cassette tape
2. Single-phase induction recorder motor
R. Domestic water pump
3. Universal motor
S. Escalator
4. Three-phase induction motor 5. DC series motor 6. Stepper motor
Ch wise GATE_EE_Ch5.indd 189
189
Code: P Q R S (a) 3 6 4 5 (b) 1 3 2 4 (c) 3 1 2 4 (d) 3 2 1 4 (GATE 2003: 1 Mark) 31. A stand-alone engine-driven synchronous generator is feeding a partly inductive load. A capacitor is now connected across the load to completely nullify the inductive current. For this operating condition (a) field current and fuel input have to be reduced (b) field current and fuel input have to be increased (c) field current has to be increased and fuel input left unaltered (d) field current has to be reduced and fuel input left unaltered (GATE 2003: 1 Mark) 32. Curves X and Y in the figure denote open circuit and full-load zero power factor (ZPF) characteristics of a synchronous generator. Q is a point on the ZPF characteristics at 1.0 per unit (pu) voltage. The vertical distance PQ in in the figure gives the voltage drop across (a) synchronous reactance (b) magnetising reactance (c) potier reactance (d) leakage reactance Voltage (pu)
X P Y
1.0
Q
Field current (GATE 2003: 1 Mark) 33. No-load test on a three-phase induction motor was conducted at different supply voltages and a plot of input power versus voltage was drawn. This curve was extrapolated to intersect the y-axis. This intersection point yields (a) core loss (b) stator copper loss (c) stray load loss (d) friction and windage loss (GATE 2003: 1 Mark)
11/21/2018 11:53:26 AM
190
GATE EE Chapter-wise Solved Papers
34. The speed-torque regimes in a DC motor and the control methods suitable for the same are given, respectively, in Group II and Group I Group II
P. Field control
1. Below base speed
Q. Armature control
2. Above base speed
E2
E2
Code: P Q (a) 1 3 (b) 2 1 (c) 2 3 (d) 1 4 (GATE 2003: 1 Mark) 35. The given figure shows an ideal single-phase transformer. The primary and secondary coils are wound on the core as shown. Turns ratio (N1/N2) = 2. The correct phasors of voltages E1, E2, currents I1, I2 and core flux f are as shown in the following figure f I1 I2 N2
f
I2
The match between the control method and the speed− torque regime is as follows:
N1
I1
f
3. Above base torque
E1
E1
I1
Group I
4. Below base torque
E1
E2
R
I2
(C)
(D)
(a) Figure A (c) Figure C
(b) Figure B (d) Figure D (GATE 2003: 2 Marks)
36. To conduct load test on a DC shunt motor, it is coupled to a generator which is identical to the motor. The field of the generator is also connected to the same supply source as the motor. The armature of the generator is connected to a load resistance. The armature resistance is 0.02 pu. Armature reaction and mechanical losses can be neglected. With rated voltage across the motor, the load resistance across the generator is adjusted to obtain rated armature current in both motor and generator. The pu value of this load resistance is (a) 1.0 (b) 0.98 (c) 0.96 (d) 0.94 (GATE 2003: 2 Marks) 37. The given figure shows an ideal three-winding transformer. The three windings 1, 2, 3 of the transformer are wound on the same core as shown. The turns ratio N1:N2:N3 is 4:2:1. A resistor of 10 Ω is connected across winding-2. A capacitor of reactance 2.5 Ω is connected across winding-3. Winding-1 is connected across a 400 V AC supply. If the supply voltage phasor V1 = 400 ∠0° , the supply current phasor I1 is given by I1
E1
E1
I2 E2
N1
N2
2
R = 10 Ω
N3 E2
3
I1 f
Ch wise GATE_EE_Ch5.indd 190
1
I2
I1
V1
(A)
f
Xc = 2.5 Ω (B)
(a) (-10 + j10) A (c) (10 + j10) A
(b) (-10 - j10) A (d) (10 - j10) A (GATE 2003: 2 Marks)
11/21/2018 11:53:27 AM
191
Chapter 5 • Electrical Machines
38. Following are some of the properties of rotating electrical machines P. Stator winding current is dc, rotor-winding current is ac Q. Stator winding current is AC, rotor-winding current is DC R. Stator winding current is AC, rotor-winding current is ac S. Stator has salient poles and rotor has commutator T. Rotor has salient poles and slip rings and stator is cylindrical U. Both stator and rotor have poly-phase windings DC machines, synchronous machines and induction machines exhibit some of the above properties as given in the following table. Indicate the correct combination from this table DC machines
Synchronous machines
Induction machines
Stator surface CDA forms
Rotor surface abc forms
Rotor surface cda forms
Torque
(a) North pole
South pole
North pole
South pole
Clockwise
(b) South pole
North pole
North pole
South pole
Counter clockwise
(c) North pole
South pole
South pole
North pole
Counter clockwise
(d) South pole
North pole
South pole
North pole
Clockwise
(GATE 2003: 2 Marks) 40. A 4-pole, three-phase, double-layer winding is housed in a 36-slot stator for an AC machine with 60° phase spread. Coil span is 7 slot pitches. Number of slots in which top and bottom layers belong to different phases is
(a)
P.S.
Q.T.
R.U.
(b)
Q.U.
P.T.
R.S.
(c)
P.S.
R.U.
Q.T.
(GATE 2003: 2 Marks)
(d)
R.S.
Q.U.
P.T.
41. A three-phase induction motor is driving a constant torque load at rated voltage and frequency. If both voltage and frequency are halved, following statements relate to the new condition if stator resistance, leakage reactance and core loss are ignored
(GATE 2003: 2 Marks) 39. When stator and rotor windings of a 2-pole rotating electrical machine are excited, each would produce a sinusoidal mmf distribution in the air gap with peak values Fs and Fr, respectively. The rotor mmf lags stator mmf by a space angle d at any instant as shown in the figure. Thus, half of stator and rotor surfaces will form one pole with the other half forming the second pole. Further, the direction of torque acting on the rotor can be clockwise or counter clockwise Stator Air gap Rotor
C c b B d d
a
D Fs
Stator mmf axis
Fr Rotor mmf axis
A
Stator surface ABC forms
The following table gives four sets of statements as regards poles and torque. Select the correct set corresponding to the mmf axes as shown in the given figure.
Ch wise GATE_EE_Ch5.indd 191
(a) 24
(b) 18
(c) 12
(d) 0
P. The difference between synchronous speed and actual speed remains same Q. The air gap flux remains same R. The stator current remains same S. The pu slip remains same Among the above, correct statements are (a) All (c) Q, R and S
(b) P, Q and R (d) P and S (GATE 2003: 2 Marks)
42. A single-phase induction motor with only the main winding excited would exhibit which of the following response at synchronous speed? (a) Rotor current is zero. (b) Rotor current is non-zero and is at slip frequency. (c) Forward and backward totalling fields are equal. (d) Forward rotating field is more than the backward rotating field. (GATE 2003: 2 Marks)
11/21/2018 11:53:27 AM
192
GATE EE Chapter-wise Solved Papers
43. A DC series motor driving an electric train faces a constant power load. It is running at rated speed and rated voltage. If the speed has to be brought down to 0.25 pu, the supply voltage has to be approximately brought down to (a) 0.75 pu (b) 0.5 pu (c) 0.25 pu (d) 0.125 pu (GATE 2003: 2 Marks) 44. 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: V = 415 V Ph = 3 f = 50 Hz N = 2850 rpm The motor is run 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 (GATE 2003: 2 Marks) 45. A 500 kVA, three-phase transformer has iron losses of 300 W and full load copper losses of 600 W. The percentage load at which the transformer is expected to have maximum efficiency is (a) 50.0% (b) 70.7% (c) 141.4% (d) 200.0% (GATE 2004: 1 Mark) 46. For a given stepper motor, which of the following torque has the highest numerical value? (a) Detent torque (b) Pull-in torque (c) Pull-out torque (d) Holding torque (GATE 2004: 1 Mark) 47. Which of the following motor definitely has a permanent magnet rotor? (a) DC commutator motor (b) Brushless DC motor (c) Stepper motor (d) Reluctance motor (GATE 2004: 1 Mark) 48. The type of single-phase induction motor having the highest power factor at full load is (a) shaded pole type (b) split-phase type (c) capacitor-start type (d) capacitor-run type (GATE 2004: 1 Mark) 49. The direction of rotation of a three-phase induction motor is clockwise when it is supplied with three-phase
Ch wise GATE_EE_Ch5.indd 192
sinusoidal voltage having phase sequence A-B-C. For counter clockwise rotation of the motor, the phase sequence of the power supply should be (a) B-C-A (b) C-A-B (c) A-C-B (d) B-C-A or C-A-B (GATE 2004: 1 Mark) 50. The synchronous speed for the seventh space harmonic mmf wave of a three-phase, 8 pole, 50 Hz induction machine is (a) 107.14 rpm in forward direction (b) 107.14 rpm in reverse direction (c) 5250 rpm in forward direction (d) 5250 rpm in reverse direction (GATE 2004: 2 Marks) 51. A rotating electrical machine having its self-inductances of both the stator and the rotor windings independent of the rotor position will definitely not develop (a) starting torque (b) synchronising torque (c) hysteresis torque (d) reluctance torque (GATE 2004: 2 Marks) 52. The armature resistance of a permanent magnet DC motor is 0.8 Ω. At no load, the motor draws 1.5 A from a supply voltage of 25 V and runs at 1500 rpm. The efficiency of the motor while it is operating on load at 1500 rpm drawing a current of 3.5 A from the same source will be (a) 48.0% (b) 57.1% (c) 59.2% (d) 88.8% (GATE 2004: 2 Marks) 53. A 50 kVA, 3300/230 V single-phase transformer is connected as an autotransformer shown in the figure. The nominal rating of the autotransformer will be
N2 Vout N1 Vin = 3300 V
(a) 50.0 kVA (c) 717.4 kVA
(b) 53.5 kVA (d) 767.4 kVA (GATE 2004: 2 Marks)
11/21/2018 11:53:28 AM
193
Chapter 5 • Electrical Machines
54. The resistance and reactance of a 100 kVA 11000/400 Δ−Y distribution transformer are 0.02 and 0.07 pu, respectively. The phase impedance of the transformer referred to the primary is (a) (0.02 + j 0.07) Ω (b) (0.55 + j 925) Ω (c) (15.125 + j 52.94) Ω (d) (72.6 + j 254.1) Ω (GATE 2004: 2 Marks) 55. A single-phase, 230 V, 50 Hz, 4 pole, capacitor-start induction motor has the following stand-still impedances Main winding Zm = 6.0 + j 4.0 Ω Auxiliary winding Za = 8.0 + j 6.0 Ω The value of the starting capacitor required to produce 90° phase difference between the currents in the main and auxiliary windings will be (a) 176.84 µF (b) 187.24 µF (c) 265.26 µF (d) 280.86 µF (GATE 2004: 2 Marks) 56. Two three-phase, Y-connected alternators are to be paralleled to a set of common bus bars. The armature has a per phase synchronous reactance of 1.7 Ω and negligible armature resistance. The line voltage of the first machine is adjusted to 3300 V and that of the second machine is adjusted to 3200 V. The machine voltages are in phase at the instant they are paralleled. Under this condition, the synchronising current per phase will be (a) 16.98 A (b) 29.41 A (c) 33.96 A (d) 58.82 A (GATE 2004: 2 Marks)
60. A 400 V, 50 kVA, 0.8 pf leading Δ-connected, 50 Hz synchronous machine has a synchronous reactance of 2 Ω and negligible armature resistance. The friction and windage losses are 2 kW and the core loss is 0.8 kW. The shaft is supplying 9 kW load at a power factor of 0.8 leading. The line current drawn is (a) 12.29 A (c) 21.29 A
(b) 16.24 A (d) 36.88 A (GATE 2004: 2 Marks)
61. A 500 MW three-phase Y-connected synchronous generator has a rated voltage of 21.5 kV at 0.85 pf. The line current when operating at full load rated c onditions will be (a) 13.43 kA (c) 23.25 kA
(b) 15.79 kA (d) 27.36 kA (GATE 2004: 2 Marks)
62. A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. The given figure shows the reversing torque and the speed during the transient. The moment of inertia of the drive is
+20 Nm
t Torque 0.5 s
57. A 400 V, 15 kW, 4 pole, 50 Hz, Y-connected induction motor has full load slip of 4%. The output torque of the machine at full load is (a) 1.66 Nm (b) 95.50 Nm (c) 99.47 Nm (d) 624.73 Nm (GATE 2004: 2 Marks) 58. For a 1.8°, two-phase bipolar stepper motor, the stepping rate is 100 steps/s. The rotational speed of the motor in rpm is (a) 15 (b) 30 (c) 60 (d) 90 (GATE 2004: 2 Marks) 59. A 8-pole, DC generator has a simplex wave-wound armature containing 32 coils of 6 turns each. Its flux per pole is 0.06 Wb. The machine is running at 250 rpm. The induced armature voltage is (a) 96 V (b) 192 V (c) 384 V (d) 768 V (GATE 2004: 2 Marks)
Ch wise GATE_EE_Ch5.indd 193
+500 rpm Speed
t
−1500 rpm
(a) 0.048 kgm2 (c) 0.096 kgm2
(b) 0.064 kgm2 (d) 0.128 kgm2 (GATE 2004: 2 Marks)
63. The equivalent circuit of a transformer has leakage reactances X1, X2 ′ and magnetising reactance XM. Their magnitudes satisfy
11/21/2018 11:53:28 AM
194
GATE EE Chapter-wise Solved Papers
(a) X1 >> X2 ′ >> XM (b) X1 f 2 (b) f 1 < 50 Hz and f 2 > 50 Hz (c) f 1, f 2 < 50 Hz and f 2 > f 1 (d) f 1 > 50 Hz and f 2 < 50 Hz (GATE 2014: 1 Mark) 153. A single phase induction motor draws 12 MW power at 0.6 lagging power. A capacitor is connected in parallel to the motor to improve the power factor of the combination of motor and capacitor to 0.8 lagging. Assuming that the real and reactive power drawn by the motor remains same as before, the reactive power delivered by the capacitor in MVAR is . (GATE 2014: 1 Mark) 154. In a synchronous machine, hunting is predominantly damped by (a) mechanical losses in the rotor (b) iron losses in the rotor (c) copper losses in the stator (d) copper losses in the rotor (GATE 2014: 1 Mark)
Torque
0.7 Ns Ns Speed (d) Torque
0.7 Ns Ns Speed (GATE 2014: 1 Mark) 156. The no-load speed of a 230 V separately excited DC motor is 1400 rpm. The armature resistance drop and the brush drop are neglected. The field current is kept constant at rated value. The torque of the motor in Nm for an armature current of 8 A is . (GATE 2014: 1 Mark) 157. The torque-speed characteristics of motor (TM) and load (TL) for two cases are shown in the figures (a) and (b). The load torque is equal to motor torque at points P, Q, R and S (a) Speed Torque TM
P 155. A single phase induction motor is provided with capacitor and centrifugal switch in series with auxiliary winding. The switch is expected to operate at a speed of 0.7 Ns, but due to malfunctioning the switch fails to operate. The torque-speed characteristics of the motor is represented by
Ch wise GATE_EE_Ch5.indd 205
Q TL
11/21/2018 11:53:33 AM
206
GATE EE Chapter-wise Solved Papers
(b)
TM S
Speed
TL
R
161. A 3-phase, 50 Hz, six pole induction motor has a rotor resistance of 0.1 Ω and reactance of 0.92 Ω. Neglect the voltage drop in stator and assume that the rotor resistance is constant. Given that the full load slip is 3%, the ratio of maximum torque to full load torque is (a) 1.567 (b) 1.712 (c) 1.948 (d) 2.134 (GATE 2014: 2 Marks)
Torque The stable operating points are (a) P and R (b) P and S (c) Q and R (d) Q and S (GATE 2014: 1 Mark) 158. Assuming an ideal transformer, the Thevenin’s equivalent voltage and impedance as seen from the terminals x and y for the circuit in figure are as follows: 1Ω X
162. A 250 V DC shunt machine has armature circuit resistance of 0.6 Ω and field circuit resistance of 125 Ω. The machine is connected to 250 V supply mains. The motor is operated as a generator and then as a motor separately. The line current of the machine in both the cases is 50 A. The ratio of the speed as a generator to the . speed as a motor is (GATE 2014: 2 Marks) 163. A three-phase slip-ring induction motor, provided with a commutator winding, is shown in the figure. The motor rotates in clockwise direction when the rotor windings are closed.
sin(wt) 3-phase AC, f Hz Y 1:2 (a) (b) (c) (d)
2 sin(wt), 4 Ω l sin(wt), 1 Ω 1 sin(wt), 2 Ω 2 sin(wt), 0.5 Ω
f2 Prime mover fr (GATE 2014: 1 Mark)
159. The core loss of a single phase, 230/115 V, 50 Hz power transformer is measured from 230 V side by feeding the primary (230 V side) from a variable voltage variable frequency source while keeping the secondary open circuited. The core loss is measured to be 1050 W for 230 V, 50 Hz input. The core loss is again measured to be 500 W for 138 V, 30 Hz input. The hysteresis and eddy current losses of the transformer for 230 V, 50 Hz input are respectively, (a) 508 W and 542 W (b) 468 W and 582 W (c) 498 W and 552 W (d) 488 W and 562 W (GATE 2014: 2 Marks) 160. A 15 kW, 230 V DC shunt motor has armature circuit resistance of 0.4 Ω and field circuit resistance of 230 Ω. At no load and rated voltage, the motor runs at 1400 rpm and the line current drawn by the motor is 5 A. At full load, the motor draws a line current of 70 A. Neglect armature reaction. The full load speed of the motor in . rpm is (GATE 2014: 2 Marks)
Ch wise GATE_EE_Ch5.indd 206
Slip ring induction motor f1
If the rotor winding is open circuited and the system is made to run at rotational speed fr with the help of primemover in anti-clockwise direction, then the frequency of voltage across slip rings is f1 and frequency of voltage across commutator brushes is f2. The values of f1 and f2 respectively are (a) f + fr and f (b) f − fr and f (c) f − fr and f + f (d) f + fr and f − f (GATE 2014: 2 Marks)
164. A 20-pole alternator is having 180 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are connected to realise single-phase winding, the generated voltage is V1. If the coils are reconnected to realise three-phase star-connected winding, the generated phase voltage is V2. Assuming full pitch, single-layer winding, the ratio V1/V2 is (a)
1 (b) 1 (c) 3 (d) 2 2 3 (GATE 2014: 2 Marks)
11/21/2018 11:53:34 AM
Chapter 5 • Electrical Machines
165. For a single phase, two winding transformer, the supply frequency and voltage are both increased by 10%. The percentage changes in the hysteresis loss and eddy current loss, respectively, are (b) −10 and 21 (d) −21 and 10 (GATE 2014: 2 Marks)
(a) 10 and 21 (c) 21 and 10
166. An open circuit test is performed on 50 Hz transformer, using variable frequency source and keeping V/f ratio constant, to separate its eddy current and hysteresis losses. The variation of core loss/frequency as function of frequency is shown in the figure below.
15 Pc/f 10 (W/Hz) 5
25
50 f(Hz)
The hysteresis and eddy current losses of the transformer at 25 Hz, respectively, are (a) 250 W and 2.5 W (b) 250 W and 62.5 W (c) 312.5 W and 62.5 W (d) 312.5 W and 250 W (GATE 2014: 2 Marks) 167. A non-salient pole synchronous generator having synchronous reactance of 0.8 pu is supplying 1 pu power to a unity power factor load at a terminal voltage of 1.1 pu. Neglecting the armature resistance, the angle of the voltage behind the synchronous reactance with respect to the angle of the terminal voltage in degrees is . (GATE 2014: 2 Marks) 168. A separately excited 300 V DC shunt motor under no load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5 Ω and leakage inductance is 0.01 H. When loaded, the armature current . is 15 A. Then the speed in rpm is (GATE 2014: 2 Marks) 169. The load shown in the figure absorbs 4 kW at a power factor of 0.89 lagging. 1Ω 1:2 50 Hz AC source
Ch wise GATE_EE_Ch5.indd 207
X
110 V
ZL
207
Assuming the transformer to be ideal, the value of the reactance X to improve the input power factor to unity is . (GATE 2014: 2 Marks)
170. The parameters measured for a 220 V/110 V, 50 Hz, single-phase transformer are: Self-inductance of primary winding = 45 mH Self-inductance of secondary winding = 30 mH Mutual inductance between primary and secondary windings = 20 mH Using the above parameters, the leakage (Ll1, Ll2) and magnetising (Lm) inductances as referred to primary side in the equivalent circuit respectively, are (a) 5 mH, 20 mH and 40 mH (b) 5 mH, 80 mH and 40 mH (c) 25 mH, 10 mH and 20 mH (d) 45 mH, 30 mH and 20 mH (GATE 2014: 2 Marks) 171. A separately excited DC generator has an armature resistance of 0.1 W and negligible armature inductance. At rated field current and rated rotor speed, its open circuit voltage is 200 V. When this generator is operated at half the rated speed, with half the rated field current, an uncharged 1000 mF capacitor is suddenly connected across the armature terminals. Assume that the speed remains unchanged during the transient. At what time (in microsecond) after the capacitor is connected will the voltage across it reach 25 V? (a) 62.25 (b) 69.3 (c) 73.25 (d) 77.3 (GATE 2015: 1 Mark) 172. The self-inductance of the primary winding of a single phase, 50 Hz transformer is 800 mH, and that of the secondary winding is 600 mH. The mutual inductance between these two windings is 480 mH. The secondary winding of the transformer is short circuited and primary winding is connected to a 50 Hz, single phase, sinusoidal voltage source. The current flowing in both the windings is less than their respective rated currents. The resistance of both windings can be neglected. In this condition, what is the effective inductance (in mH) seen by the source? (a) 416 (b) 440 (c) 200 (d) 920 (GATE 2015: 1 Mark) 173. The primary mmf is least affected by the secondary terminal conditions in a (a) power transformer (b) potential transformer (c) current transformer (d) distribution transformer (GATE 2015: 1 Mark)
11/21/2018 11:53:35 AM
208
GATE EE Chapter-wise Solved Papers
174. Find the transformer ratios a and b such that the impedance (Zin) is resistive and equals 2.5 W when the network is excited with a sine wave voltage of angular frequency of 5000 rad/s. C = 10 µF
L = 1 mH R = 2.5 Ω
Zin 1:b
1:a
(a) a = 0.5, b = 2.0 (c) a = 1.0, b = 1.0
(b) a = 2.0, b = 0.5 (d) a = 4.0, b = 0.5 (GATE 2015: 1 Mark)
175. A shunt-connected DC motor operates at its rated terminal voltage. Its no-load speed is 200 radian/second. At its rated torque of 500 Nm, its speed is 180 radian/second. The motor is used to directly drive a load whose load torque TL depends on its rotational speed wr (in radian/second), such that TL = 2.78 × wr. Neglecting rotational losses, the steady-state speed (in radian/second) of the motor, when it drives this load, is . (GATE 2015: 1 Mark) 176. The figure shows the per-phase equivalent circuit of a two-pole three-phase induction motor operating at 50 Hz. The “air-gap” voltage, Vg across the magnetising inductance, is 210Vrms, and the slip, is 0.05. The torque (in Nm) produced by the motor is ––––––––––––. 0.04 Ω j0.22 Ω
Vs
chronous speed, and producing 1 pu of real power. The initial value of the rotor angle d is 5°, when a bolted three phase to ground short circuit fault occurs at the terminal of the generator. Assuming the input mechanical power to remain at 1 pu, the value of d in degrees, 0.02 second after the fault is ————————. (GATE 2015: 2 Marks)
j6.28 Ω
j0.22 Ω
Vg
0.05 Ω s
(GATE 2015: 1 Mark) 177. A 4-pole, separately excited, wave wound DC machine with negligible armature resistance is rated for 230 V and 5 kW at a speed of 1200 rpm. If the same armature coils are reconnected to form a lap winding, what is the rated voltage (in volts) and power (in kW), respectively, at 1200 rpm of the reconnected machine if the field circuit is left unchanged? (a) 230 and 5 (b) 115 and 5 (c) 115 and 2.5 (d) 230 and 2.5 (GATE 2015: 1 Mark)
179. A separately excited DC motor runs at 1000 rpm on no load when its armature terminals are connected to a 200 V DC source and the rated voltage is applied to the field winding. The armature resistance of this motor is 1 Ω. The no load armature current is negligible. With the motor developing its full load torque, the armature voltage is set so that the rotor speed is 500 rpm. When the load torque is reduced to 50% of the full load value under the same armature voltage conditions, the speed rises to 520 rpm. Neglecting the rotational losses, the full load armature current (in Ampere) is ——————. (GATE 2015: 2 Marks) 180. A DC motor has the following specifications: 10 hp, 37.5 A, 230 V; flux/pole = 0.01 Wb, number of poles = 4, number of conductors = 666, number of parallel paths = 2. Armature resistance = 0.267 W. The armature reaction is negligible and rotational losses are 600 W. The motor operates from a 230 V DC supply. If the moor runs at 1000 rpm, the output torque produced (in Nm) is ———————. (GATE 2015: 2 Marks) 181. A 200/400 V, 50 Hz, two winding transformer is rated at 20 kVA. Its windings are connected as an auto-transformer of rating 200/600 V. A resistive load of 12 W is connected to the high voltage (600 V) side of the auto-transformer. The value of equivalent load resistance (in ohm) as seen from low voltage side is ——————. (GATE 2015: 2 Marks) 182. A 3-phase transformer rated for 33 kV/11kV is connected in delta/star as shown in figure. The current transformers (CTs) on low and high voltage sides have a ratio of 500/5. Find the currents i1 and i2, if the fault current is 300 A as shown in figure.
a
b 300 A c i1
i2
178. A 50 Hz generating unit has H-constant of 2 MJ/MVA. The machine is initially operating in steady state at syn-
Ch wise GATE_EE_Ch5.indd 208
11/21/2018 11:53:35 AM
Chapter 5 • Electrical Machines
(a) i1 =
1 3
(a) 1∠90° (b) 1∠270°
A, i2 = 0 A
(c) 4 ∠90° (d) 4 ∠270°
(b) i1 = 0 A, i2 = 0 A 1
(c) i1 = 0 A, i2 = (d) i1 =
1 3
3
A, i2 =
209
(GATE 2015: 2 Marks) A 1 3
A (GATE 2015: 2 Marks)
183. A balanced (positive sequence) three-phase AC voltage source is connected to a balanced, star connected load through a star–delta transformer as shown in the figure. The line-to-line voltage rating is 230 V on the star side, and 115 V on the delta side. If the magnetising current is → → neglected and Is I s = 100 ∠0° A , then what is the value → of Ip in Ampere? ®
®
Ip
Is a
A R B
185. With an armature voltage of 100 V and rated field winding voltage, the speed of a separately excited DC motor driving a fan is 1000 rpm, and its armature current is 10 A. The armature resistance is 1 W. The load torque of the fan load is proportional to the square of the rotor speed. Neglecting rotational losses, the value of the armature voltage (in Volt) which will reduce the rotor speed to 500 rpm is . (GATE 2015: 2 Marks) 186. A three-phase, 11 kV, 50 Hz, 2-pole, star connected, cylindrical rotor synchronous motor is connected to an 11 kV, 50 Hz source. Its synchronous reactance is 50 W per phase, and its stator resistance is negligible. The motor has a constant field excitation. At a particular load torque, its stator current is 100 A at unity power factor. If the load torque is increased so that the stator current is 120 A, then the load angle (in degrees) at this load is . (GATE 2015: 2 Marks)
R
R
b C c
(a) 50 ∠30° (b) 50 ∠ - 30°
187. A 220 V, 3-phase, 4-pole, 50 Hz inductor motor of wound rotor type is supplied at rated voltage and frequency. The stator resistance, magnetising reactance, and core loss are negligible. The maximum torque produced by the rotor is 225% of full load torque and it occurs at 15% slip. The actual rotor resistance is 0.03 W/phase. The value of external resistance (in ohm) which must be inserted in a rotor phase if the maximum torque is to occur at start is . (GATE 2015: 2 Marks)
(c) 50 3∠30° (d) 200 ∠30° (GATE 2015: 2 Marks) 184. A three-winding transformer is connected to an AC voltage source as shown in the figure. The number of turns are as follows: N1 = 100, N2 = 50, N3 = 50. If the magnetising current is neglected, and the currents in two windings are I 2 = 2∠30° A and I 3 = 2∠150° A, then what is the value of the current I1 in Ampere?
A2 a1
V1
B2
b2 B1
®
I1
a2 A1
b1
C2
N2
®
188. Two three-phase transformers are realized using single-phase transformers are shown in the figure.
c2
C1
I2
c1 A2 A1
®
I3
b2
B2 B1
N3
Ch wise GATE_EE_Ch5.indd 209
V2
b1 c1
C2
N1
a2 a1
c2
C1
11/21/2018 11:53:36 AM
210
GATE EE Chapter-wise Solved Papers
The phase difference (in degree) between voltages V1 and V2 is . (GATE 2015: 2 Marks)
nals, the ratio of initial and final values of the sinusoidal component of the short circuit current is _______. (GATE 2016: 1 Mark)
189. In a constant v/f induction motor drive, the slip at the maximum torque
193. The direction of rotation of a single-phase capacitor run induction motor is reversed by (a) interchanging the terminals of the AC supply. (b) interchanging the terminals of the capacitor. (c) interchanging the terminals of the auxiliary winding. (d) interchanging the terminals of both the windings. (GATE 2016: 1 Mark)
(a) is directly proportional to the synchronous speed. (b) remains constant with respect to the synchronous speed. (c) has an inverse relation with the synchronous speed. (d) has no relation with the synchronous speed. (GATE 2016: 1 Mark) 190. A 4-pole, lap-connected, separately excited DC motor is drawing a steady current of 40 A while running at 600 rpm. A good approximation for the waveshape of the current in an armature conductor of the motor is given by (a)
194. If an ideal transformer has an inductive load element at port 2 as shown in the figure below, the equivalent inductance at port 1 is n:1 L
(b) 40 A
I 10 A
I Port 1
Port 2
(a) nL (b) n2L t (c)
n2 n (d) L L (GATE 2016: 1 Mark)
t
(c)
(d) I
195. If the star side of the star-delta transformer shown in the figure is excited by a negative sequence voltage, then
I 10 A 10 A
T = 25 ms
T = 25 ms
t
A
(GATE 2016: 1 Mark)
B
191. The following figure shows the connection of an ideal transformer with primary to secondary turns ratio of 1:100. The applied primary voltage is 100 V (rms), 50 Hz, AC. The rms value of the current I, in ampere, is________.
C
t T = 25 ms
−10 A
XL = 10 Ω
T = 25 ms −10 A
XC = 40 kΩ
(GATE 2016: 1 Mark) 192. A 50 MVA, 10 kV, 50 Hz, star-connected, unloaded three-phase alternator has a synchronous reactance of 1 pu and a sub-transient reactance of 0.2 pu. If a threephase short circuit occurs close to the generator termi-
Ch wise GATE_EE_Ch5.indd 210
(a) (b) (c) (d)
N c b
VAB leads Vab by 60° VAB lags Vab by 60° VAB leads Vab by 30° VAB lags Vab by 30° (GATE 2016: 2 Marks)
1:100 R = 80 kΩ
I 100 V
a
196. A single-phase 400 V, 50 Hz transformer has an iron loss of 5000 W at the rated condition. When operated at 200 V, 25 Hz, the iron loss is 2000 W. When operated at 416 V, 52 Hz, the value of the hysteresis loss divided by the eddy current loss is ______. (GATE 2016: 2 Marks) 197. A DC shunt generator delivers 45 A at a terminal voltage of 220 V. The armature and the shunt field resistances are 0.01 Ω and 44 Ω respectively. The stray losses are
11/21/2018 11:53:37 AM
Chapter 5 • Electrical Machines
375 W. The percentage efficiency of the DC generator is ____________.
211
(GATE 2016: 2 Marks)
(c) less than the terminal voltage Vt (d) dependent upon supply voltage Vt (GATE 2017: 1 Mark)
198. A three-phase, 50 Hz salient-pole synchronous motor has a per-phase direct-axis reactance (Xd) of 0.8 pu and a per-phase quadrature-axis reactance (Xq) of 0.6 pu. Resistance of the machine is negligible. It is drawing full-load current at 0.8 pf (leading). When the terminal voltage is 1 pu, per-phase induced voltage, in pu, is ________.
205. A 3-phase, 4-pole, 400 V, 50 Hz squirrel-cage induction motor is operating at a slip of 0.02. The speed of the rotor flux in mechanical rad/s, sensed by a stationary observer, is closest to (a) 1500 (b) 1470 (c) 157 (d) 154 (GATE 2017: 1 Mark)
(GATE 2016: 2 Marks)
206. A 375 W, 230 V, 50 Hz capacitor start single-phase induction motor has the following constants for the main the auxiliary windings (at starting); Zm = (12.50 + j15.75)Ω (main winding), Za = (24.50 + j12.75)Ω (auxiliary winding). Neglecting the magnetizing branch, the value of the capacitance (in mF) to be added in series with the auxiliary winding to obtain maximum torque at starting is . (GATE 2017: 2 Marks)
199. A single-phase, 22 kVA, 2200 V/ 220 V, 50 Hz, distribution transformer is to be connected as an auto-transformer to get an output voltage of 2420 V. Its maximum kVA rating as an autotransformer is (a) 22 (c) 242
(b) 24.2 (d) 2420 (GATE 2016: 2 Marks)
200. The starting line current of a 415 V, three-phase, delta connected induction motor is 120 A, when the rated voltage is applied to its stator winding. The starting line current at a reduced voltage of 110 V, in ampere, is ________. (GATE 2016: 2 Marks) 201. A single-phase, 2 kVA, 100/200 V transformer is reconnected as an auto-transformer such that its kVA rating is maximum. The new rating, in kVA, is ________. (GATE 2016: 2 Marks) 202. A 4-pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor is RPM is (a) 1350 (b) 1650 (c) 1950 (d) 2250 (GATE 2017: 1 Mark) 203. A three-phase, 50 Hz, star-connected cylindrical-rotor synchronous machine is running as a motor. The machine is operated from a 6.6 kV grid and draws current at unity power factor (UPF). The synchronous reactance of the motor is 30 Ω per phase. The load angle is 30°. . The power delivered to the motor in kW is (Give the answer up to one decimal place.) (GATE 2017: 1 Mark) 204. If a synchronous motor is running at a leading power factor, its excitation induced voltage (Ef) is (a) equal to the terminal voltage Vt (b) higher than the terminal voltage Vt
Ch wise GATE_EE_Ch5.indd 211
207. Two parallel connected, three-phase, 50 Hz, 11 kV, star-connected synchronous machines A and B are operating as synchronous condensers. They together supply 50 MVAR to a 11 kV grid. Current supplied by both the machines are equal. Synchronous reactances of machine A and machine B are 1 Ω and 3 Ω, respectively. Assuming the magnetic circuit to be linear, the ratio of excitation current of machine A to that of machine B is . (Give the answer up to two decimal places.) (GATE 2017: 2 Marks) 208. A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistance are 0.4 Ω and 0.1 Ω, respectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50%, the resistance in ohms that should be added in series with the armature is . (Give the answer up to two decimal places.) (GATE 2017: 2 Marks) 209. A three-phase, three winding D/D/Y (1.1 kV/ 6.6 kV/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The rms line current in ampere drawn by the 1.1 kV winding from the mains is . (Give the answer up to one decimal place.) (GATE 2017: 2 Marks)
11/21/2018 11:53:38 AM
212
GATE EE Chapter-wise Solved Papers
210. A separately excited DC generator supplies 150 A to a 145 V DC grid. The generator is running at 800 rpm. The armature resistance is 0.1 Ω. If the speed of the generator is increased to 1000 rpm, the current in amperes supplied by the generator to the DC grid is . (Give the answer up to one decimal place.) (GATE 2017: 2 Marks) 211. A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra = 0.02 W. When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply is (a) 34.2 A (b) 30 A (c) 22 A (d) 4.84 A (GATE 2017: 2 Marks) 212. A 120 V DC shunt motor takes 2 A at no load. It takes 7 A on full load while running at 1200 rpm. The armature resistance is 0.8 W, and the shunt field resistance is 240 W. The no load speed, in rpm, is . (GATE 2017: 2 Marks) 213. A star-connected, 12.5 kW, 208 V (line), 3-phase, 60 Hz squirrel cage induction motor has following equivalent circuit parameters per phase referred to the stator: R1 = 0.3 W, R2 = 0.3 W, X1 = 0.41 W, X2 = 0.41 W. Neglect shunt branch in the equivalent circuit. The starting current (in Ampere) for this motor when connected to an 80 V (line), 20 Hz, 3-phase AC source is . (GATE 2017: 2 Marks) 214. A 25 kVA, 400 V, Δ-connected, 3-phase, cylindrical rotor synchronous generator requires a field current of 5 A to maintain the rated armature current under short-circuit condition. For the same field current, the open-circuit voltage is 360 V. Neglecting the armature resistance and magnetic saturation, its voltage regulation (in % with respect to terminal voltage), when the generator delivers the rated load at 0.8 pf leading, at rated terminal voltage is . (GATE 2017: 2 Marks) 215. If the primary line voltage rating is 3.3 kV (Y side) of a 25 kVA, Y−Δ transformer (the per phase turns ratio is 5:1), then the line current rating of the secondary side (in Ampere) is . (GATE 2017: 2 Marks) 216. In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is (a) 0 (b) 45 (c) 60 (d) 90 (GATE 2018: 1 Mark)
Ch wise GATE_EE_Ch5.indd 212
217. The positive, negative and zero sequence impedances of a 125 MVA, three-phase, 15.5 kV, star-grounded, 50 Hz generator are j 0.1 pu, j 0.05 pu and j 0.01 pu, respectively, on the machine rating base. The machine is unloaded and working at the rated terminal voltage. If the grounding impedance of the generator is j 0.01 pu, then the magnitude of fault current for a b-phase to ground fault (in kA) is (up to 2 decimal places). (GATE 2018: 1 Mark) 218. A separately excited dc motor has an armature r esistance Ra = 0.05 W. The field excitation is kept constant. At an armature voltage of 100 V, the motor produces a torque of 500 Nm at zero speed. Neglecting all mechanical losses, the no-load speed of the motor (in radian/s) for an armature voltage of 150 V is (up to 2 decimal places). (GATE 2018: 1 Mark) 219. A transformer with toroidal core of permeability m is shown in the figure. Assuming uniform flux density across the circular core cross-section of radius r X d′ > X d′′
Ch wise GATE_EE_Ch5.indd 216
18. Topic: DC Machines: Starting and Speed Control of DC Motors (c) In drives, the inner loop is used to limit the peak current of the machine to the permissible value. 19. Topic: Types of Losses and Efficiency Calculations of Electric Machines (c) Given that: area = 5 cm2, scale on x-axis, 1 cm = 2 AT, scale on y-axis, 1cm = 50 mWb and frequency = 50 Hz. If hysteresis loop is given, then the hysteresis loss Wn = Area of the loop × Frequency Wn = (5 × 2 AT × 50 × 10−3 Wb) × 50 = 25 W 20. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (d) Given that the output of the motor is constant, that is, EaIa = constant Also, power output = Tw and output power is constant. Therefore, Torque, T w = constant Then, if power is constant, 1 w So, torque-speed characteristic is a rectangular h yperbola. T∝
21. Topic: Three Phase Transformers (b) For a three-phase system, base impedance, (Vbase ) 2 3 (VAbase )
Z base = Base impedance per phase Z base/phase =
(Vbase ) 2 (33 kV ) 2 = (VAbase/phase ) ⎛ 20 ⎞ ⎜⎝ ⎟⎠ MVA 3
Z b = 163.35 Ω Thus,
Zactual = Zb × Zpu
12 = 0.12 100 Therefore, Z actual = 163.35 × 0.12 = 19.602 Ω = 19.6 Ω Z pu =
22. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (a) For a three-phase induction machine, when the machine is drawing rated current, the overall torque for 3 2 Rr three phases is given by T = I ws r s where w s is the synchronous speed of the machine in mechanical radians per second and s is the slip. Also,
11/21/2018 11:53:44 AM
217
Chapter 5 • Electrical Machines
ws = And N s =
Therefore, from the voltage equation we have Eb = V - I a Ra = 200 - 10 × 2 = 180 V
2π N s 60
Therefore, the rated torque
120 f p
where f is the frequency and p is the number of poles. Since, motor is drawing the same rated current, from above relations we have 1 1 1 T∝ ∝ ∝ ws N s f Thus, on increasing frequency, torque will reduce. 23. Topic: DC Machines: Series and Shunt (b) Since the magnetic circuit is saturated, torquespeed characteristic will be linear. Curve B fit best. 24. Topic: Single Phase Transformer: Regulation and Efficiency (d) In the single-phase transformer under loaded condition, the flux is sinusoidal in the core of the transformer, so induced emf is always sinusoidal. To produce sinusoidal flux during saturation condition of transformer, the current drawn from supply should be a peaky waveform, not sinusoidal. 25. Topic: Single Phase Transformer: Regulation and Efficiency (a) Redrawing the given circuit as shown below
Also for a separately excited DC motor T ∝ Ia (Therefore ϕ = constant ) T2 I a 2 = T1 I a1
Therefore, T2 = 8.598 ×
5 = 4.299 Nm ≈ 4.30 Nm 10
27. Topic: Three Phase Induction Motors: Starting and Speed Control (c) Since number of poles on rotor and stator are different. The induction motor will not work. 28. Topic: Synchronous Machines: Starting of Synchronous Motor (b) Due to armature reaction, main flux is deteriorated. Therefore, the motor must be opening with leading power factor. 29. Topic: Types of Losses and Efficiency Calculations of Electric Machines (d) Efficiency of the transformer is given by,
400 V 50 Hz 400 V
Eb I a w 180 × 10 = = 8.598 Nm 2000 2π 60
T=
200 V
η=
200 V
where Pi and Pcu are iron and copper losses, respectively.
V Clearly, both 200 V windings are additive hence, voltmeter reads V = 400 – (200 + 200) = 0 V 26. Topic: DC Machine: Separately Excited (a) For the separately excited dc motor, given that V = 200 V, N = 200 rpm, Ia= 10 A and Ra= 2 W ,
Eb Ra = 2 W
V2 I 2 cos ϕ 2 V2 I 2 cos ϕ 2 + Pi + Pcu
200 V
At full load, the load factor (LF) =
Pi = V2 I 2 = 1 Pcu
Let output = 1 pu, then we have
1×1 90% = 11+× 12 Pi 90% = 1 2 Pi mVA Pi = 0+.0556 Pi = 0.0556 mVA At half load, LF = 0.5, therefore
η=
(as Pi = Pcu ) (as Pi = Pcu )
0.5 × 1 × 100 = 87.8% 0.5 × 1 + 0.0556 + 0.0556
30. Topic: Operating Principle of Single Phase Induction Motors (c) P → 3; Q → 1; R → 2; S → 4.
Ch wise GATE_EE_Ch5.indd 217
11/21/2018 11:53:45 AM
218
GATE EE Chapter-wise Solved Papers
31. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (d) Given that (i) Synchronous generator feeds a partly inductive load. (ii) Capacitor connected across the load nullifies the inductive current. Therefore, field current has to be reduced and fuel input should be left unaltered. 32. Topic: Synchronous Machines: Characteristics (a) Given that, on zero power factor curve, at point Q the voltage is 1.0 pu. Therefore, voltage drop between points P and Q is across synchronous reactance. 33. Topic: Types of Losses and Efficiency Calculations of Electric Machines (d) The intersection point gives friction and windage loss.
E1 I2 E2 I1 f 36. Topic: Types of Losses and Efficiency Calculations of Electric Machines (c) Given that, armature resistance is Ra = 0.02. Mechanical losses and armature reaction are neglected, therefore back emf in the motor Eb = 1 - 0.02 × 1 = 0.98 pu As the motor is coupled to the generator, emf generated in the motor Eg = E b Therefore, terminal voltage of the generator Vg = 0.98 - 1 × 0.02 = 0.96 pu The value of load resistance is
Po
RL = PF & W
34. Topic: DC Machines: Starting and Speed Control of DC Motors (b) For above base speed we use field control method and for below base speed we use armature control method. 35. Topic: Single Phase Transformer: Phasor Diagram (a) Let flux be ϕ = ϕ m sin w t
I1 =
Ch wise GATE_EE_Ch5.indd 218
0.96 = 0.96 pu 1
We know that V1 N1 4 2 2 = = ⇒ V2 = V1 = × 400 V2 N 2 2 4 4 Also, V1 N1 4 = = V3 N 3 1 1 1 V3 = V1 = × 400 4 4 Therefore, V2 = 200 V; V3 = 100 V Considering secondary winding V2 = I 2 R
E1 E = 1 ∠( - 90°) jX m X m
Hence, I1 lags behind E1 by 90° Since, we know that I1 ∝ f1, so, f1 also lags behind E1 by 90°. From the above statements we can conclude the phasor diagram as:
=
R = 10 Ω; V1 = 400 V and N1 : N 2 : N 3 = 4 : 2 :1
E2 = -
Hence, we can see that f lags behind E1 by 90° I2 is in phase with E2 (Load is purely resistive) Now, magnetising current
Ia
37. Topic: Three Phase Transformers: Connections, Parallel Operation (d) Given that
Therefore, emf induced in secondary dϕ N 2 = -ϕ mw cosw t ⋅ N 2 dt = ϕ mw sin (w t - 90°) ⋅ N 2
Vg
I2 =
V2 200 = = 20 A R 10
On connecting the capacitor, I3 =
V3 100 = = 40 j ∠0 - j × C - j × 2.5
Secondary current I2, referred to primary be I1′
11/21/2018 11:53:47 AM
Chapter 5 • Electrical Machines
I1′ N 2 2 = = I 2 N1 4 I1′ = 20 ×
I1′′ N 3 1 = = - I 3 N1 4
(as I 3 flows opposite to I1 )
I1′′= - j × 10 Therefore, I1 = I1′ + I1′′= 10 - 10 j A 38. Topic: Synchronous Machines: Characteristics (a) In DC motor: (i) Stator winding is connected to dc supply and rotor-winding flow current is ac (P). (ii) Stator is made of salient pole and rotor will have commutator to carry ac power (S) In induction machines: (i) The AC supply is connected to stator winding (R) (ii) Rotor and stator are made of poly-phase winding (U). In synchronous machines: (i) Stator is fed with ac supply, but rotor will be excited by dc supply (Q). (ii) Stator is cylindrical, rotor is made of both salient pole and slip rings (T). Hence, the correct combination is represented by option (a). 39. Topic: Three Phase Induction Motors: No-Load and Blocked Rotor Tests (c) Given that: Fs is the peak value of stator mmf; Fr is the peak value of rotar mmf and d is the space angle. Direction of rotation of rotor is clockwise or counter clockwise. Thus, the rotor tends to move when the opposite poles are produced in same half of the stator and rotor. Therefore, stator surface ABC forms north pole; rotor surface abc forms South pole; rotor surface cda forms North pole and stator surface CDA forms South pole with torque in counter clockwise direction. 40. Topic: Three Phase Induction Motors: Principle of Operation (a) Given that: number of poles = 4; number of phases = 3, number of slots = 36, phase spread = 36° and coil span = 7 (short-pitched).
Ch wise GATE_EE_Ch5.indd 219
Slot 36 = =9 Pole 4 9 Slot/pole/phase = = 3 3 Pole pitch =
2 = 10 A 4
Let third winding current I3 referred to primary be I 2′′
219
Therefore, out of 3, 2 will have different phases which implies 36 out of 24 will have different phases. Hence, number of slots in which top and bottom layers belong to different phases is 24. 41. Topic: Types of Losses and Efficiency Calculations of Electric Machines (b) Given that the induction motor drives a constant torque load at rated voltage and frequency. Rs, XL and core loss are ignored. Also V=
X F ,F= 2 2
We know that Ns =
120 f = actual speed p
(i) When leakage reactance are ignored, air-gap flux remains the same. (ii) Rs ignored implies that Is remains the same. 42. Topic: Operating Principle of Single Phase Induction Motors (d) Slip corresponding to the forward field: sf =
Ns - N r Ns
And, slip corresponding to the backward field: sb =
N s - (- N r ) Ns
Now, at rotor speed Nr = Ns, sf = 0 sb = 2 Hence, their no relative motion between forward field and rotor winding, but there is a relative motion between backward field and rotor winding. Therefore, forward rotating field is more than the backward rotating field. 43. Topic: DC Machines: Series and Shunt (b) In DC motor Eb = V - I a Ra = V - I a ( Ra + Rse ) ϕ ZNp Eb = 60 a Eb = K ϕ N N=
[For series motor I a = I se ]
[as Z , p, a are constant , so K is a constant = Z p / 60 a]
Eb Kϕ
11/21/2018 11:53:48 AM
220
GATE EE Chapter-wise Solved Papers
In series motor f ∝ Ise N=
V - I ( Ra + Rse ) KI se
At constant power, Eb I a = T × W = constant
(1)
T ∝ ϕ Ia
( 2)
T ∝I
2 a
( For series motor )
To maintain constant power, when W is decreased, torque is to be increased W =
1 , then T = 4 4
Therefore, Ia is increased two times and voltage is reduced to 0.5 pu. 44. Topic: Three Phase Induction Motors: Starting and Speed Control (c) Given that: V = 415 V, Ph = 3, f = 50 Hz and N = 2850 rpm. We have 120 f N1 = p 120 × 50 =2 p= 2850 When the output frequency is set at 40 Hz, 120 f1 P 120 × 40 = = 2400 2
NT =
When the slip is reduced to half, N = 2340 rpm. 45. Topic: Types of Losses and Efficiency Calculations of Electric Machines (b) Given rating = 500 kVA Iron losses = 300 W Full load copper losses = 600 W Maximum efficiency occurs at Wi = X 2Wcu X =
Wi = Wcu
300 = 0.707 600
% Efficiency = 0.707 × 100 = 70.7% 46. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (c) (i) Holding torque is the amount of torque that the motor produces when it has rated current flowing through the windings but the motor is at rest.
Ch wise GATE_EE_Ch5.indd 220
(ii) Detent torque is the amount of torque that the motor produces when it is not energised. No current is flowing through the windings. (iii) Pull-in torque curve shows the maximum value of torque at given speeds that the motor can start, stop or reverse in synchronism with the input pulses. The motor cannot start at a speed that is beyond this curve. (iv) Pull-out torque curve shows the maximum value of torque at given speeds that the motor can generate while running in synchronism. If the motor is run outside of this curve, it will stall. Therefore, the higher value of torque is pull-out torque and less torque when the torque is pull-in torque. 47. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (b) Brushless DC motor has a permanent magnet motor. 48. Topic: Operating Principle of Single Phase Induction Motors (d) Single-phase induction motor is not self-starting. Capacitor-run type motor will have high power factor in which capacitor will be connected in running condition. 49. Topic: Three Phase Induction Motors: Starting and Speed Control (c) Clockwise rotation is A-B-C Counter clockwise direction will be A-C-B (any two phases can be reversed). 50. Topic: Synchronous Machines: Characteristics (a) Given that number of poles p = 8 and frequency f = 50 Hz. Ns at 7th harmonic = Ns =
Ns 7
120 f 120 × 50 = = 750 rpm p 8
N s at 7th harmonic =
750 = 107.14 rpm 7
51. Topic: Synchronous Machines: Characteristics (b) Rotating electrical machines having its self-inductance of stator and rotor windings is independent of the rotor position of synchronising torque. 52. Topic: Types of Losses and Efficiency Calculations of Electric Machines (a) Given that Ra = 0.8 Ω, Vnl = 25 V, N = 1500 rpm, Inl = 1.5 A and Ifl = 3.5 A.
11/21/2018 11:53:49 AM
221
Chapter 5 • Electrical Machines
Under no load condition, we have P =V ×I = (V - I a Ra ) I
= ( 25 - 1.5 × 0.8)1.5 = 35.7 W
Under load condition: Iron loss = I 2 R = (3.5) 2 × 0.8 = 9.8 W Total loss = No load losses + Iron losses = 35.7 + 9.8 = 45.5 W Total power, P = VI = 25 × 3.5 = 87.5 W Efficiency =
55. Topic: Operating Principle of Single Phase Induction Motors (a) Given that V = 230 V, f = 50 Hz and number of poles p = 4 For capacitor-start induction motor , Z m = Rm + X m = 6 + j 4 Ω (Main winding) Z a = Ra + X a = 8 + j 6 Ω ∠I m = ∠ - Z m = - ∠6 + 4 j = - 33.7° 1 (as capacitor is connected in jw C series) j = ∠8 + 6 j wC α = ∠I a - ∠I m
∠I a = ∠8 + j 6 +
1 ⎛ ⎞ 6⎜ ⎟ w C 90 = - tan ⎜ - ( -33.7)⎟ 8 ⎜ ⎟ ⎝ ⎠
Pout × 100 Pin
-1
Total power - losses Total power 87.5 - 45.5 = 48% = 87.5 =
1 = 18 wC 1 (∵ w = 2π f ) 2π f × 18 1 = 176.8 mF = 2 × 3.14 × 50 × 18
C=
53. Topic: Auto-Transformer (d) Given rating = 50 kVA Vin = 3300 V Vout = 3300 + 230 = 3530 V Output current I 2 =
In kVA rating P2 = Vout × I 2 = 3530 × 217.4 = 767.42 kVA 54. Topic: Three Phase Transformers: Connections, Parallel Operation (d) Given that Z pu = 0.02 + j 0.07, V2 I 2 = 100 kVA Z Base =
Vp 2 V2 I 2 3
=
(11 × 103 ) 2 = 3630 Ω 100 × 103 3
Phase impedance referred to primary, Z1 = Z pu × Z Base = (0.02 + j 0.07)(3630) = (72.6 + j 254.1) Ω
Ch wise GATE_EE_Ch5.indd 221
56. Topic: Three Phase Induction Motors: Performance (a) We have
Power Output voltage
50 × 103 = = 217.4 A 230
(Auxiliary winding)
Ef 1 =
3300 3
Synchronizing current =
;
Ef 2 =
3200 3
Ec X s1 + X s 2
Reactance of both the alternator will be the same, therefore, synchronizing current will be =
Ef 1 - Ef 2 1 ⎛ 3300 - 3200 ⎞ = ⎜ ⎟ = 16.98 A X s1 + X s 2 3 ⎝ 1.7 + 1.7 ⎠
57. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (c) Given that V = 400 V P = 15 kW and number of poles p = 4 120 f Ns = p 120 × 50 = = 1500 rpm 4 Actual speed = Synchronised speed − Slip ⎛ 4 ⎞ = 1500 - ⎜ × 1500⎟ = 1440 rpm ⎝ 100 ⎠
11/21/2018 11:53:50 AM
222
GATE EE Chapter-wise Solved Papers
The output torque T= =
2π N ⎞ ⎛ ⎜⎝ Therefore w s (1 - s) = ⎟ 60 ⎠
P w s (1 - s)
15 × 10 = 99.47 Nm 2π × 1440 60 3
(b) Number of steps per revolution =
360 = 200 steps 1.8
Time required for one revolution = 2 seconds Revolutions/s = 0.5 rps Revolutions/min = 0.5 × 60 = 30 rpm 59. Topic: DC Machines: Motoring and Generating Mode of Operation and their Characteristics (c) Given that number of poles p = 8, N = 250 rpm and flux f = 0.06. The voltage of the generator is given by
ϕ ZNp 60 a
where, a = 2 (wave winding) and Z = 2 × 32 × 6 = 384. Therefore, Eg =
0.06 × 250 × 8 × 384 = 384 V 60 × 2
60. Topic: Types of Losses and Efficiency Calculations of Electric Machines (c) Given that V = 400 V, f = 50 Hz and pf = 0.8. Total input power = 9 + 2 + 0.8 kW = 11.8 kW
11.8 kW 3 × 400 × 0.8
= 21.29 A
61. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (b) Input power 3 VL I L = 500 MW IL =
500 × 106
3 × 21.5 × 103 × 0.85 I L = 15.79 kA
= 15.79 × 103
62. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (c) As we know by formula: T = Iα
Ch wise GATE_EE_Ch5.indd 222
500 - ( -1500) 2π = 418.67 rad/s2 0.5 60
I=
40 418.67
Moment of inertia = 0.096 kgm2 63. Topic: Single Phase Transformer: Equivalent Circuit X 2′X ′ (d) As X 1X= = 1 2 X 2′X ′ > 1Xand and Therefore,X MX > X M
1
2
X 1 ≈ X 2′ X M 64. Topic: Three Phase Transformers: Connections, Parallel Operation (b) Only in Y- Δ and Δ -Y , phase shift of ±30° can be introduced. Therefore, connection is star−delta. 65. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (a) When speed of the motor is in the forward direction, slip varies from 0 to 1. When speed of the motor is in the reverse direction, then slip > 1. Therefore, slip > 1 at point W. 66. Topic: Operating Principle of Single Phase Induction Motors (b) Gross power = 1- s Air-gap power
Input power 3 V2 I 2 = 11.8 kW I2 =
α=
Therefore, I will be:
58. Topic: Three Phase Induction Motors: Torque-Speed Characteristics
Eg =
Torque is given as 40 Nm. Now, a is equal to:
67. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (a) At c1 emf will be induced upwards, whereas at c2 and c′ 2 no emf is induced. 68. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (b) We know that P ∝N where N is the rated speed. At half the rated speed, Pnew =
50 = 25 kW 2
At 1.5 times the rated speed, P = constant. Therefore, Pnew = 50 kW
11/21/2018 11:53:52 AM
Chapter 5 • Electrical Machines
69. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (d) In a DC motor, (i) E = K ϕw n (ii) T =
pϕ ZI a 2π a
(iii) Thus, from the above expressions power P depends on f, w and Ia. Therefore, P – 2 Q – 3, R – 1 70. Topic: Synchronous Machines: Characteristics (c) (i) In a synchronous machine, when the armature terminals are shorted, the If should first be reduced to zero and then started as alternator. (ii) In open-circuit, the synchronous machine runs at rated Ns. (iii) The field current is gradually increased in steps. (iv) The short-circuit ratio is the ratio of If required to produce the rated voltage on open circuit to the rated armature current. 71. Topic: Three Phase Induction Motors: No-Load and Blocked Rotor Tests (b) Under no load condition, when the applied voltage to an induction motor is reduced from the rated voltage to half the rated value, both speed and stator current decrease. 72. Topic: Three Phase Induction Motors: Starting and Speed Control (b) Given that starting current Ist is six times the full load current Ifl I = 6 I fl ⇒
I st =6 I fl
Full-load slip = 4%, therefore, the required ratio is 2
Tst ⎛ I st ⎞ = × Sfl Tfl ⎜⎝ I fl ⎟⎠ = 6 2 × 0.04 = 1.44 73. Topic: Operating Principle of Single Phase Induction Motors (b) The increased rotor resistance will increase starting torque and hence it will have high acceleration. 74. Topic: Three Phase Induction Motors: Starting and Speed Control (a) In an induction motor E = 4.44 kw1 f ϕ T1
Ch wise GATE_EE_Ch5.indd 223
223
In V/f control, the maximum developed torque remains the same. (i) To avoid saturation and to minimise losses, motor is operated at rated air gap flux by varying terminal voltage with frequency. (ii) To maintain V/f as constant, the magnetic flux should be maintained constant at the rated value which keeps maximum torque constant. 75. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (b) Given that P = 1000 kVA, V = 6.6 kV and reactance = 20 Ω. P = 3VL I L IL =
1000 × 103
= 87.47 A 3 × 6.6 × 103 XI L = 87.47 × 20 = 1.75 kV 2
⎛ 6.6 ⎞ 2 + (1.75) Eph = ⎜ ⎝ 3 ⎟⎠ = 4.2 kV EL = 3Eph = 3 × 4.2 = 7.26 kV 76. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (c) Torque angle is given by ⎛X ⎞ α z = tan -1 ⎜ s ⎟ Ra ⎞⎠ ⎛⎝ X α z = tan -1 ⎜ s ⎟ Ra3⎠ × 1.75 ⎞ ⎝⎛have Substituting values, we = 24.6° α z = tan -1 ⎜ 6×.61.75 ⎟⎞⎠ ⎝ ⎛ 3 α z = tan -1 ⎜ ⎟ = 24.6° ⎝ 6.6 ⎠ 77. Topic: Types of Losses and Efficiency Calculations of Electric Machines (d) In open-circuit test, current is drawn at low power factor while in short-circuit test, current is drawn at high power factor. 78. Topic: Operating Principle of Single Phase Induction Motors (b) Main and auxiliary windings are displaced by 90° in space. The direction of rotation can be changed by reversing the main winding terminals. 79. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (b)
11/21/2018 11:53:53 AM
224
GATE EE Chapter-wise Solved Papers
80. Topic: Three Phase Induction Motors: Principle of Operation (d) Real power, P = S cos ϕ = 12 × 3 × 0.8 = 16.6 kW Reactive power, Q = S sin ϕ = 12 × 3 × 0.6 = 12.47 kW
84. Topic: Three Phase Induction Motors: Starting and Speed Control (c) Given that: number of poles, p = 4, f1 = 50 Hz, V1 = 400 V, N1 = 1440 rpm and T1 = T2. Since V ∝ f , therefore, V1 f = 1 V2 f2
For unity power factor, we have to set capacitor bank to Q = 12.47 kW. 81. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (a) For the generator, given that: V = 220 V, Ia = 20 A and Ra = 0.2 Ω Therefore, Eg = V + I a Ra = 200 + ( 20)(0.2) = 204 V Now, Eb = V - I a Ra
When the flux is increased by 10%, the ratio of motor to generator speed can be calculated as Eb
=
Ng Nm
×
=
-1
×
1 kVA
500 kVA × 0.05 = 250 kVA × x
Ch wise GATE_EE_Ch5.indd 224
2
s2 ⎛ V1 ⎞ f T = × 2× 2 s1 ⎜⎝ V2 ⎟⎠ f1 T1 30 ⎛ 400 ⎞ s2 = (0.04) ⎜ × ⎝ 240 ⎟⎠ 50 = 0.066
(as T1 = T2 )
We know that Ns =
120 f p
120 × 30 (1 - 0.066) 4 = 840 rpm
Nr =
1 1.1
196 = 0.87 204 × 1.1
500 × 0.05 = 0.1 pu 250
Therefore,
ϕm
83. Topic: Three Phase Transformers: Connections, Parallel Operation (b) Given T1 = 500 kVA, leakage impedance = 0.05 pu and T2 = 250 kVA. Let pu leakage impedance be x.
x=
⎛V 2 ⎞ T ∝⎜ ⎟ s ⎝ f ⎠
N r = N s (1 - s);
82. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (b) The distributor of current and fluxes with phase π angle ϕ = + . 2 For lagging load, the current will be shifted in space by an angle p/2 from maximum emf which coincides with centre of pole. The field created by the armature reaction mmf will be in direction opposite to main field flux and will therefore have complete demagnetising effect.
Per unit impedance ∝
Also,
ϕg
N m ⎛ 204 ⎞ =⎜ ⎟ N g ⎝ 196 ⎠
30 = 240 V 50
2
= 200 - ( 20)(0.12) = 196 V
Eg
V2 = 400 ×
85. Topic: Three Phase Induction Motors: Starting and Speed Control (a) Given that number of poles, p = 4, V = 400 V, Ω, r2′ = 0.5 Ω , x1 = x2′ = 1.2 Ω f = 50 Hz, r1 = 1 x1 = x2′ = 1.2 Ω and xm = 35Ω . Therefore, Ns =
120 f 120 × 50 = = 1500 rpm p 4
Tst =
V 2 r2′ 180 × 2π N s ( r1 + r2′) 2 + ( x1 + x2′ ) 2 2
⎛ 400 ⎞ ⎜⎝ ⎟ × 0.5 180 3⎠ = × 2 × 3.14 × 1500 (1.5) 2 + ( 2.4) 2 = 63.58 Nm 86. Topic: Types of Losses and Efficiency Calculations of Electric Machines (b) Given that P = 10 kW, V = 400 V, p = 4, f = 50 Hz, Ifl = 20 A
11/21/2018 11:53:54 AM
Chapter 5 • Electrical Machines 2
⎛ 20 ⎞ Full load copper loss = ⎜ ⎟ × 762 ⎝ 15 ⎠
Slot angle =
= 1354.67
87. Topic: Types of Losses and Efficiency Calculations of Electric Machines (a) Given that number of phase = 3, V = 400 V, P = 5 kW, internal reactance = 10 Ω, and power factor = unity power factor. Full load current is given by P 3V × 1
=
5 × 103 3 × 400 × 1
= 7.22 A 1
E = ⎡⎣(V cos q - I a Ra ) 2 + (V sin q - I a X s ) 2 ⎤⎦ 2 1
⎡⎛ 400 ⎞ 2 ⎤2 2 = ⎢⎜ + × . 10 3 6 ⎥ ) ( ⎟ ⎢⎣⎝ 3 ⎠ ⎥⎦ E = 233.73
Eph = 4.44 × 50 × 0.957 × 0.951 × 160 × 0.025 = 808 V EL = 3 × 808 = 1399.5 V ≈ 1400 V 89. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (a) For a two-phase winding, Slots per pole per phase =
48 =6=m 4×2
480 = 240 2 180 × 4 Slot angle = = 15° = β 48
1 = 236 100
Using these values, we have 1
2 2 ⎛ ⎛ 400 ⎞ ⎞ I a x = E22 - V 2 = ⎜ 236 2 - ⎜ ⎟ = 48.9 ⎝ 3 ⎟⎠ ⎠ ⎝
48.9 48.9 Ia = = = 4.8 x 10 4.8 Load in % = = 67% 7.2 88. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (c) Given that number of poles, p = 4, f = 50 Hz, number of slots = 48, number of turns = 10, a = 36° (electrical) and d = 0.05 Wb. We know that Eph = 4.44 f ϕ Nk p kd Volts (i) (1) where, f = flux per pole, f = frequency of generation, N = number of turns, kp = pitch factor and kd = distribution factor. 48 The number of slots per pole per phase m = =4 4×3 48 Slots/pole = = 12 4
Ch wise GATE_EE_Ch5.indd 225
Therefore, substituting values in Eq. (1), we have
Tph =
When excitation is increased by 1%, E2 = 2833.73 ×
sin( mβ / 2) sin ( 4 × 15 / 2) = = 0.957 4 sin(15 / 2) m sin β / 2 α 36 K p = cos = cos = cos 180° = 0.951 2 2 48 × 10 Tph = = 160 3 Kd =
Output 10 × 103 = × 100 = 81% Input 10 × 103 + 2356.67
I fl =
180 = 15o = β 12
Using these values, we have
Total losses = 1354.67 + 1002 = 2356.67 The efficiency of the motor at full load is
η=
225
Kd =
sin (6 × 15/ 2) 6 sin(15/ 2)
= 0.903
⎛ 36 ⎞ K p = cos ⎜ ⎟ = 0.951 ⎝ 2⎠ Eph = 4.444 × 0.025 × 50 × 240 × 0.951 × 0.903 = 1143.8 V 90. Topic: DC Machines: Starting and Speed Control of DC Motors (a) For the fifth harmonic component of phase emf, Angle = 180/5 = 36° ⎛ nα ⎞ Maximum value, cos ⎜ ⎟ = Em cos(5 × 18) = Em cos( 90°) = 0 ⎝ 2 ⎠ ⎛ nα ⎞ cos ⎜ ⎟ = Em cos(5 × 18) = Em cos( 90°) = 0 ⎝ 2 ⎠ Therefore, phase emf of 5th harmonic is zero. 91. Topic: Auto-Transformer (a) There exist equal phase shift at points A and B with respect to source from both the buses. Therefore, type of transformer is star–star with phase shift of 0°.
11/21/2018 11:53:55 AM
226
GATE EE Chapter-wise Solved Papers
92. Topic: Single Phase Transformer: Regulation and Efficiency (c) In a transformer, zero voltage regulation at full load gives leading power factor.
When the transformer is reconfigured at 500 V/ 750 V, we have 150 η= = 98.29% 150 + 2.6
93. Topic: DC Machines: Starting and Speed Control of DC Motors (d) Differential compound motor provides zero speed regulation at full load without any controller.
99. Topic: Types of Losses and Efficiency Calculations of Electric Machines (d) Given that Tst = 150% and Tmax = 300%
94. Topic: DC Machines: Starting and Speed Control of DC Motors (c) From the characteristics given in the options points A and D are stable. 95. Topic: Synchronous Machines: Characteristics (b) The given three-phase synchronous motor, is running at full load with unity power factor. On reducing the shaft load by half, If remains constant; this implies leading power factor. 96. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (a) Given that for the star-connected synchronous machine P = 100 kVA, Voc = 415 V, If = 15 A Line synchronous impedance = =
Tst = 1.5Tfl
(1)
Tmax = 3Tfl
(2)
Dividing Eq. (1) by Eq. (2), we have Tst 1.5Tfl 1 = = Tmax 3Tfl 2
3 × short circuit phase current 415 ⎛ 100 × 1000 ⎞ 3×⎜ ⎝ 3 × 415 ⎟⎠
97. Topic: Three Phase Induction Motors: Starting and Speed Control (b) Given, number of poles p = 20 and number of phases = 3. Therefore, step angle
β=
360 = 6° 2 × 20
98. Topic: Types of Losses and Efficiency Calculations of Electric Machines (c) Given that P = 50 kVA, V = 250 V/500 V and h = 95 at unity power factor. We know that efficiency is given by Output power Output power + Iron loss + Copper loss
Therefore,
η=
50 × 1 × 1 = 95% 50 + (Wc + Wi )
⇒ Wcu + Wi = 2.6
(3)
For squirrel cage induction motor, Tst 2s = 2 mT 2 Tmax smT + 1
(4)
Equating eqs. (3) and (4), we get 2 smT = 0.5. 2 + 12 smT 2 + 0.5 - 2 smT = 0 0.5smT
smT = 26.7%
Voc
= 1..722 ≈ 1.731
η=
Let Tfl be full load torque. Then,
100. Topic: Auto-Transformer (c) Given that Ist = 7Ifl and sfl = 5%. Also, the ratio of starting torque to full-load torque is 2
Tst ⎛ I st ⎞ = × x 2 × sfl Tfl ⎜⎝ I fl ⎟⎠ 1.5 = 72 × x 2 × 0.05 ⇒ x = 0.782 Therefore ratio is 78.2%. 101. Topic: Three Phase Induction Motor: Principle of Operation (b) On using star−delta starter, 2
Tst 1 ⎛ I st ⎞ = × sfl Tfl 3 ⎜⎝ I fl ⎟⎠ 1 2 × 7 × 0.05 3 = 0.816
=
102. Topic: Three Phase Induction Motor: Principle of Operation (c) Given Tst = 0.5 pu We have 2
Tst ⎛ I sc ⎞ = × Sfl Tfl ⎜⎝ I fl ⎟⎠ 2
⎛I ⎞ 0.5 = ⎜ sc ⎟ × 0.05 ⎝I ⎠ fl
Ch wise GATE_EE_Ch5.indd 226
11/21/2018 11:53:57 AM
2
Tst ⎛ I sc ⎞ = × Sfl Tfl ⎜⎝ I fl ⎟⎠ 2
⎛I ⎞ 0.5 = ⎜ sc ⎟ × 0.05 ⎝ I fl ⎠ ⇒
I sc = I fl
0.5 = 3.162 0.05
103. Topic: Single Phase Transformer: Regulation and Efficiency (d) Since the secondary winding is at 5 V with opposite polarity, primary shows -5 V (due to the same number of turns). The same -5 V will appear across the secondary which is shorted by a 10 Ω resistor. -5 Current through the resistance = = 0.5 A 10 104. Topic: Synchronous Machines: Characteristics (d) Distributed winding and short chording in AC machines leads to reduction in emf and harmonics. 105. Topic: Single Phase Transformer: Regulation and Efficiency (b)
Chapter 5 • Electrical Machines
Pin = 3VL I L cos ϕ = 3 × 400 × 50 × 0.8 = 27.7 kW Therefore, air-gap power is Pg = 27.71 - (1.5 + 1.2) = 25.01 kW 110. Topic: Single Phase Transformer: Regulation and Efficiency (a) Given N1 = 100 and N2 = 200 and
ϕ m = 0.12 =
dϕ dt
We know that E1 ∝ N1 Therefore, E2 ∝ 2 E1 E = -N
dϕ dt
In the interval, 0 < t < 1, dϕ = -100 × 0.12 = -12 V dt E2 = 2 E1 = 2 × (12) = 24 V E1 = -100
106. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (d) Detent torque is defined as the maximum load torque that can be applied to the shaft of an unexcited motor, without causing continuous rotation.
(E1 and E2 are in opposition.) In the interval, 1 < t2 < 2
107. Topic: Single Phase Transformer: Regulation and Efficiency (d)
dϕ = 0, E1 = E2 = 0 dt In the region, 2 < t < 2.5,
108. Topic: Operating Principle of Single Phase Induction Motors (b) Given that: V = 230 V, p = 4, R2 = 7.8 Ω (resistance at standstill), f = 50 Hz and Nr = 1425 rpm. Then the standstill speed can be calculated as Ns = s=
120 f 120 × 50 = = 1500 rpm p 4
N s - N r 1500 - 1425 = = 0.05 Ns 1500
Resistance in the background branch =
R2 7.8 = = 4Ω 2 - s 2 - 0.05
109. Topic: Types of Losses and Efficiency Calculations of Electric Machines (c) Given that: V = 400 V, stator current loss = 1.5 kW, f = 50 Hz, rotor current loss = 900 W, P = 30 hp, friction and windage loss PF & W = 1050 W I = 50 A, core loss = 1200 W and power factor = 0.8 lag. We have
227
E1 = -100
dϕ = - 24 V dt
E2 = -48 V Therefore, option (a) fit best. 111. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (b) Given that V = 400 V, f = 50 Hz, p = 4, N = 1400 rpm, R = 100 Ω and X s = X r′ = 1.5 Ω We have Sm =
Rr′ X sm + X rm ′
At starting, Sm = 1, therefore, X sm + X rm ′ = R ′r 2π f m Ls + 0.2π f m Lr′ = 1 Frequency at maximum torque, 1 fm = 2π ( Ls + Lr′) Xs 1.5 = 2π f 2 p × 50 1.5 Lr′ = 2π × 50 Ls =
Ch wise GATE_EE_Ch5.indd 227
11/21/2018 11:53:58 AM
228
GATE EE Chapter-wise 1Solved Papers
fm =
2π ( Ls + Lr′)
Xs 1.5 = 2π f 2 p × 50 1.5 Lr′ = 2π × 50 Ls =
We have V V II L L
1
1.5 ⎤ ⎡ 1.5 2π ⎢ + 2 50 × × 50 ⎥⎦ π 2 π ⎣ 1 = = 16.7 Hz 3/ 50
a
Therefore,
120 f 120 × 50 = = 1500 rpm p 4
Ia = 12 ×
(c)
s=
Ns - N r 90 = = 0.06 Ns 1500
Pin = W1 + W2 = 1800 - 200 = 1600 W Po = (1 - s) Pin = (11 - 0.06)1600 = 1504 W Pin = Po [ neglecting losses] I 2 R = 1504 W I=
1504 = 12.26 A 10
114. Topic: DC Machines: Series and Shunt (d) Given that: V = 240 V, Ra = 0.5 Ω, IL = 15 A, Rsh = 80 Ω and N = 80 rad/s.
Ch wise GATE_EE_Ch5.indd 228
V Ra + Rext
125 474 = 100 0.5 + Rext
⇒ Rext = 31.1 Ω 116. Topic: Synchronous Machines: Characteristics (d) Given that: V = 1∠0° pu and Ia = 0.6∠0° pu We have Zs = R a + jX s = 0 + j = 1∠90° pu Also, V = E ∠d + I a Zs E ∠d = 1∠0° - (0.6 ∠0° × 1∠90°)
Relative speed = Ns−Nr = 1500 − 1410 = 90 rpm (in the direction of rotation) 113. Topic: Auto-Transformer
sh
115. Topic: DC Machines: Series and Shunt (a) On adding external resistance Rext to limit Ia to 125%
= 8 × 16.7 = 133.6 V
Ns =
L
= = 15 15 - 33 = = 12 12 A A
= 240 + 234 = 474 V
V2 = 8 × f 2
112. Topic: Auto-Transformer (a) Given that: number of phases = 3, N = 1410 rpm, V = 440 V, W1 = 1800 W, f = 50 Hz, W2 = -200 W p = 4 and the motor coupled is to 220 V. We have
sh
Vplugging = V + Eb
V2 =8 f2
Hence stator line-to-line voltage is 133.6 V and frequency required to obtain maximum torque is 16.7 Hz.
a
Eb = V - I a Ra = 240 - (12 × 0.5) = 234 V
By constant V/f method, V1 400 = = 8 = constant f1 50
E Ebb + + II aa R Raa II a + I + I sh
V V = 240 240 3 A II sh = = 80 = = 3A sh = R sh Rsh 80 II a = = II L - II sh
Therefore, fm =
= = = =
= 1.166 ∠( -30.96°) pu Therefore, magnitude of excitation voltage = 1.17 pu and load angle d = 30.96° (lagging) 117. Topic: Synchronous Machines: Characteristics (d)
I a = 1.2 × 0.6 = 0.72 pu E ( -d ) = V ∠0° - I a ∠ϕ ° × Z s 1 - 0.6 j = 1∠0° - 0.72∠ϕ ° × 1∠90° 1 - 0.6∠90° = 1 - 0.72∠ϕ ° × 1∠90° 0.6 ∠90° = 0.72∠ϕ ° × ∠90° 0.6 ∠ϕ ° = = 0.833 0.72 180 ∠ϕ ° = 0.833 × = 47.74 rad π
Therefore, cosf = 0.6724 (leading). The nearest option is 0.848 leading
11/21/2018 11:54:00 AM
229
Chapter 5 • Electrical Machines
118. Topic: Operating Principle of Single Phase Induction Motors (d) Given that I f = 20 A, I a = 400 A, Vsc = 200 V, I oc = 200 A. Therefore, internal resistance Rint = Internal voltage drop Vint = 5 × 200 = 1000 V 119. Topic: Single Phase Transformer: Regulation and Efficiency NBA 1 Inductance is directly proportional to cross- sectional area. Therefore, when area of cross-section is halved, inductance becomes half.
(c) Inductance, L =
120. Topic: Three Phase Induction Motors: Torque Speed Characteristics (a) At point A: As speed increases, the torque increases and as speed decreases, the torque decreases. At point B: As speed increases, the torque decreases. Therefore, A is stable and B is unstable. 121. Topic: Operating Principle of Single Phase Induction Motors (c) Z aux = ra + jw La 10 π = 1 + j1000 1000 ∠90° = 1 + j ( 2π × 50) ×
Z m = rm + jw Lm
I aux
wt 0 0.2p
Vsc 2000 = =5Ω Ia 400
= 0.1 + j ( 2π × 50) ×
VA′B′
0.1 π
= 0.1 + j10 = ∠90° V = = I a ∠( -90°) Z Aw0
0.4p
0.6p
0.8p
p
123. Topic: Three Phase Transformers: Connections, Parallel Operation (b) When the switches S1 and S2 are open, the star connection consists of third harmonics in line current and attains saturation (due to hysteresis). 124. Topic: Three Phase Transformers: Connections, Parallel Operation (a) When S1 is open and S2 is closed, the connection is open delta. The output will be sinusoidal at fundamental frequency. 125. Topic: Types of Losses and Efficiency Calculations of Electric Machines (d) When S is open, peak voltage across A and B is given by di = M × (slope of I - t curve) dt 400 ⎡ 10 ⎤ = × 10 -3 × ⎢ -3 ⎥ π ⎣ 5 × 10 ⎦ 800 V = π
V=M
126. Topic: Types of Losses and Efficiency Calculations of Electric Machines (a) When S is closed, peak voltage across A and B is given by d (10 sin 100π t ) di 400 = × 10 -3 V=M π dt dt 400 = × 10 -3 × 10 × 100π cos 100π t π = 400 cos 100π t Therefore, peak voltage is 400 V.
127. Topic: Auto-Transformer (a) Given that N1 = 4000, N2 = 6000, I = 25 A, f = 50 Hz, V = 400 V. Therefore, starting torque = kI a I m sin( I a , I m ) = k ′ sin(q ) = 0 Coils are to be connected to obtain single phase, 400/1000 V autotransformer to drive a load 10 kVA. This = kI a I m sin( I a , I m ) = k ′ sin(q ) = 0 can be realised, if we connect A and D; common to B. So, motor does not rotate at starting. Therefore, single IM =
V = I M ∠( -90°) ZM
phase motor is not self-starting. 122. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (a) The voltage waveform VA′B′ resembles
Ch wise GATE_EE_Ch5.indd 229
128. Topic: Auto-Transformer (d) Given that load = 10 kVA. Therefore, primary current 10 × 103 = 25A 400 10 × 103 I2 = = 10 A 1000 I1 =
11/21/2018 11:54:01 AM
230
3 GATE EE Chapter-wise Solved Papers 10 × 10
I1 =
= 25A 400 10 × 103 = 10 A Secondary current I 2 = 1000
We know that N ∝ Eb, therefore N1 Eb1 = N 2 Eb 2
Thus, current through Coil-1 = I1 – I2 = 15 A Current through Coil-2 = I2 = 10 A
Eb 2 =
C 10 A
=
25 A A
15 A 1000 V
N2 ⋅ Eb1 N1 1400 × 200 = 186.67 V 1500
T = Eb
I a 186.67 × 60 = Ia w 2π × 1400
Given T = 5 Nm, therefore
B
D Ia =
129. Topic: Types of Losses and Efficiency Calculations of Electric Machines
V = Eb + I a Ra
(c) Magnetising current
Ra =
I 0 = 1A I 2 = 1A Secondary current referred to primary I3 =
2 ×1 = 2A 1
Primary current I p = I 02 + I 32 = 12 + 22 = 5 = 2.24 A 130. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (c) Given that f = 50 Hz, p = 4, N = 1600 rpm We have Ns = s=
120 f 120 × 50 = = 1500 rpm p 4 N s - N r 1500 - 1600 -1 = = Ns 1500 15
= -0.066 (negative) Hence, induction machine acts as induction generator, DC machine acts as DC motor. 131. Topic: Types of Losses and Efficiency Calculations of Electric Machines (b) Given N0 = 1500 rpm, Va = 200 V, T = 5 Nm, N = 1400 rpm So, under no load Eb 0 = Va - I a 0 Ra I a0 = 0 Eb 0 = Va = 200 V
Ch wise GATE_EE_Ch5.indd 230
5 × 2π × 1400 = 3.926 A 186.67 × 60 Va - Eb 200 - 186.67 = = 3.4 Ω Ia 3.926
132. Topic: Types of Losses and Efficiency Calculations of Electric Machines (b) Given that T = 2.5 Nm at 1400 rpm. We know that T=
Eb I b 186.6 × I a × 60 ⇒ 2.5 = w 2π × 1400
Ia =
2.5 × 2π × 1400 = 1.963 A 186.6 × 60
V = Eb + I a Ra = 186.6 + (1.963)(3.4) = 193.34 V 133. Topic: DC Machines: Starting and Speed Control of DC Motors 110 (d) The phase voltage in the primary side = 3 110 Thus the turns ratio = 1.732 × 220 The delta equivalent load 4 × 4 + 4 × 4 + 4 × 4 / 4 = 12 ohm ) (
impedance
=
Therefore the impedance referred to primary side = 2 1 ⎛ ⎞ ⎜⎝ ⎟ × 12 = (1 + j 0) 2 × 1.732 ⎠ 134. Topic: DC Machines: Starting and Speed Control of DC Motors (a) A DC shunt motor should only rotate under rated speed because if the speed is more than the rated speed, motor will get damaged. To control the speed of a motor, a rheostat is connected to the field side of the motor and by varying the resistance, we can control the speed of the motor.
11/21/2018 11:54:03 AM
Chapter 5 • Electrical Machines
135. Topic: Synchronous Machines: Starting of Synchronous Motor (b) In a salient pole synchronous motor, under normal excitation with no-load condition, when the field side excitation is reduced, power generated will be high because of increase in armature current. Also, field excitation and armature current will have inversely proportional relation. Therefore, armature current first increases and then decreases steeply. 136. Topic: Single Phase Transformer: Regulation and Efficiency (c) The main property of air-core transformer is that there will be no saturation effect. Hence, output will be pure sinusoidal waveform. 137. Topic: DC Machines: Starting and Speed Control of DC Motors (a) Given that Va = 220 V , N = 1440 rpm , Ra =10 Ω, I a =10 A If the excitation is reduced by 10%, flux will be reduced by 10%, therefore 10 ⎞ ⎛ ϕ new = ⎜1 ϕ = 0.9ϕ ⎝ 100 ⎟⎠ The torque remains constant. I a ϕ = I a(new)ϕ (new) I a(new) =
I a ϕ 10ϕ = = 11.11 A ϕ new 0.9ϕ
In general, N∝
Eb , therefore ϕ
N r = N s - sN s 5 (1000) 100 = 950 rpm = 1000 -
Stator and rotor magnetic field rotate with the same speed = 1000 rpm, therefore Relative speed = 1000 - 1000 = 0 Nr with respect to stator magnetic field = 950 − 1000 = −50 rpm 139. Topic: Synchronous Machines: Characteristics (b) Given that xd = 1.2 pu , xq = 1 pu , ra = 0 , PF angle = 0, I a = 1 pu I a = 1 pu , Vt = 1 pu Power angle, tan d =
220 - (10)(1) × 0.9 1= 220 - 11.11(1 + R) 1 + R = 2.79 R = 1.79 Ω 138. Topic: Three Phase Induction Motors: Starting and Speed Control (*) Given that Vs = 440 V, p = 6, f = 50 Hz, s = 5% = 0.05 Ns =
Ch wise GATE_EE_Ch5.indd 231
120 f 120 × 50 = = 1000 rpm p 6
Vt + I a ( xq sin q - ra cos q )
140. Topic: Synchronous Machines: Characteristics (d) We know that slip is given by s=
Ns - N r Ns
Therefore, slip depends on synchronous speed (Ns) and rotor speed (Nr). Torque increases as slip increases up to a maximum value and then decreases. Slip does not depend on core-iron component. 141. Topic: DC Machines: Series and Shunt (d) Given that V = 220 V, Ra = 0.25 Ω, N1 = 1000 rpm, I L1 = 68 A, I L 2 = 52.8 A,
Using Eb = V − IaRa, we have 220 - I a Ra 0.9ϕ 1 × ϕ1 220 - I a(new) ( Ra + R)
I a ( xq cos q + ra sin q )
tan d = 1 ⇒ d = tan -1 1 = 45°
ϕ Eb N = × new N new Ebnew ϕ
1=
231
I F1 = 2.2 A, I F 2 = 1.8 A Case (i): We have I L1 = I a1 + I F1 I a1 = 68 - 2.2 = 65.8 A Therefore, E1 = V - I a1 Ra = 220 - (65.8)(0.25) = 203.55 V Case (ii): We have I a 2 = I L2 - I F2 = 52.8 - 1.8 = 51 A E2 = 220 - (51)(0.25) = 207.25 V
11/21/2018 11:54:05 AM
X M 2 = 2 X M1 232
GATE EE Chapter-wise Solved Papers
IM =
We know that
I M1 V1 X M 2 = . I M 2 V2 X M1
E N∝ ϕ or, E ∝ N ϕ
or
I M1 ⎛ V1 ⎞ ⎛ 2 X M1 ⎞ = I M 2 ⎜⎝ 2V1 ⎟⎠ ⎜⎝ X M1 ⎟⎠
Therefore,
⇒ I M 2 = 2 I M1
E1 N1 ϕ 1 = . E2 N 2 ϕ 2 203.55 1000 ⎛ ϕ 1 ⎞ = 207.25 1600 ⎜⎝ ϕ 2 ⎟⎠
ϕ 2 = 0.637ϕ 1 ϕ -ϕ2 % reduction in flux = 1 × 100 ϕ1 =
ϕ 1 - 0.637ϕ 1 × 100 = 36.3% (decrease) ϕ1
142. Topic: Three Phase Induction Motors: No-Load and Blocked Rotor Tests (b) Locked rotor line current I2 =
E2 = Z2
I M 2 = 2 × 0.5 = 0.707A V2 = 2V1 and I 2 = 2 I1 Therefore, power P2 = 2 2 × 55 = 155.6 W 144. Topic: Types of Losses and Efficiency Calculations of Electric Machines (c) Given that Core loss, Wcore = 64 W Copper loss, Wcu = 81 W (at 90% load) Therefore, at x% load, ⎛I ⎞ Wcu ( x ) = Wc × ⎜ x ⎟ ⎝ I fl ⎠
E2 R + X 22 2 2
⇒ I 2′ = 45 A 143. Topic: Single Phase Transformer: Regulation and Efficiency (b) Given that area a2 = 2a1 , length l2 = 2l1 We know that magnetizing inductance, L=
N 2 ma a ⇒L∝ l l
L1 a1 l2 = . L2 a2 l1 L1 a 2l1 = 1 . L2 2a1 l1 ⇒ L2 = 2 L 1 To find magnetizing current IM2 X M 2 = 2 X M1 IM =
1 x2 1 = 81 × = 100 W (0.91) 2
E2 E E = 2 ∝ 2 X 2 w L2 f
50 ⎛ 230 ⎞ ⎛ 57 ⎞ =⎜ ⎟⎜ ⎟ I 2′ ⎝ 236 ⎠ ⎝ 50 ⎠
V XM
I M1 V1 X M 2 = . I M 2 V2 X M1
2
Wcu(x ) = Wc × x 2
Given R2 = 0, therefore, I2 =
V XM
Wcu = Wcu ( x )
Under maximum efficiency, X =
Wi 64 = = 0.8 = 80% 100 Wcu
145. Topic: Operating Principle of Single Phase Induction Motors (d) Leakage flux of induction motor is the flux that links the stator winding or rotor winding, but not both. 146. Topic: Three Phase Induction Motors: Starting and Speed Control (b) Given that p = 4 , f = 50 Hz, N r = 1440 rpm f = 50 Hz, N r = 1440 rpm 120 × f 120 × 50 = = 1500 rpm 4 p N - N r 1500 - 1440 60 = = 0.004 s= s = 1500 1500 Ns
Ns =
Therefore, electrical frequency (Hz) of induced negative sequence current in rotor = ( 2 - s) f s = ( 2 - 0.04 ) 50 = 98 Hz
I M1 ⎛ V1 ⎞ ⎛ 2 X M1 ⎞ = I M 2 ⎜⎝ 2V1 ⎟⎠ ⎜⎝ X M1 ⎟⎠ Ch wise GATE_EE_Ch5.indd 232
⇒I
= 2I
11/21/2018 11:54:07 AM
Chapter 5 • Electrical Machines
147. Topic: Single Phase Transformer: Regulation and Efficiency (c) For the given transformer V 1.25 = VWX 1 Attenuation factor Attenuation factor VYZ = 0.8 V Therefore,
VYZ = (0.8) × 1.25 = 1 VWX ⇒ VYZ = VWX
When Vwx1 = 100 V ⇒ When Vwx2 = 100 V ⇒
VYZ1 VWX1
100 = 100
VWX 2 100 = VYZ 2 100
148. Topic: Three Phase Induction Motors: Starting and Speed Control (3.33) Given that p = 8, f = 50 Hz, N = 700 rpm, number of phases = 3. The rotor current frequency is given by fr = s ⋅f, where s is the slip. s=
Therefore, kVA rating of transformer = 1100 × 500 = 550 kVA 152. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (c) We have that f1 is the frequency of generating power to load and f2 is the new frequency after reducing load. When load is decreased, speed increases, thus increasing 120 f frequency as given by N = p Therefore, f2 > f1. N=
153. Topic: Operating Principle of Single Phase Induction Motors (7) Let P1 be power drawn without capacitor, P2 be the power drawn after adding parallel capacitor, then P1 = 12 MW, cos ϕ 1 = 0.6 , cos ϕ 2 = 0.8 and QC is the reactive power delivered by capacitor. To calculate VA,
Thus the per phase XS (assuming star connection) = ( 260.87 / 1.732) /10 ohm = 15.06 ohm/phase . 151. Topic: Auto-Transformer (550) Given that rating is 50 kVA and primary to secondary voltage ratio, 1000/100 V. 50 × 103 Secondary current, I 2 = = 500 A 100
Ch wise GATE_EE_Ch5.indd 233
P1 S1
cos ϕ 1 = S1 =
12 × 106 = 20 MVA 0.6
Q1 = 20 2 - 122 =16 MVAR
Therefore, fr = 0.067 × 50 = 3.33 Hz.
150. Topic: Synchronous Machines: Characteristics (15.06) The machine develops 400 V(L-L) at 2.3A field current. Thus assuming linear magnetic circuit, the emf developed for 1.5A field current during short circuit test is (1.5 / 2.3) × 400 V = 260.87V( L- L )
120 f 120 × f ⇒ 1500 = ⇒ f = 50 p 4
Thus, f1, f2 < 50 Hz.
N s - N 750 - 700 = = 0.067 Ns 750
149. Topic: Single Phase Transformer: Regulation and Efficiency (c) Radius of the core is reduced by half, which implies reluctance becomes double. Therefore, to maintain the no-load current and the flux, primary turns should be doubled. Therefore, the factor is 2.
233
cos ϕ 2 =
P1 (since real power drawn is the same) S2
12 × 106 = 15 MVA 0.8 Therefore, reactive power, S2 =
Q2 = S22 - P12 = 152 - 122 = 81 = 9 MVAR To draw the same reactive power, QC = Q1 - Q2 = 16 - 9 = 7 MVAR. 154. Topic: Types of Losses and Efficiency Calculations of Electric Machines (d) 155. Topic: Operating Principle of Single Phase Induction Motors (c) When switch fails to close, there will be no increase in the torque. Thus, there would not be any increase in torque after 0.7 Ns and hence no discontinuity in the curve. Therefore, option (c) curve fit the best.
11/21/2018 11:54:09 AM
234
GATE EE Chapter-wise Solved Papers
156. Topic: DC Machines: Starting and Speed Control of DC Motors (12.5) Given that V = 230 V, N = 1400 rpm, Vbrush = 0, IaRa = 0, Ia = 8 A Torque, T =
VI a 230 × 8 = = 12.5 Nm 2 π × 1400 w 60
157. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (b) The speed of motor increases with increase in torque. So to restore the speed to original and make system stable, the slope of torque-speed characteristic curve should be negative. Thus, from the given characteristics, points P and S represent stable points. 158. Topic: Single Phase Transformer: Regulation and Efficiency (a) Given that R = 1 Ω, Voc = sinwt and N2/N1 = 2/1. Therefore,
To obtain Wh and We at 50 Hz, we have Wh = 10.16 × 50 = 508 W We = 0.216 × 50 2 = 542 W 160. Topic: DC Machines: Series and Shunt (1240.63) Given that Va = 230 V, N1 = 1400 rpm, Ra = 0.4 Ω, Ra = 230 Ω. Under no load condition: 1A 4A 230 Ω
230 =1 A 230 Ia = 5 -1 = 4 A If =
2
159. Topic: Types of Losses and Efficiency Calculations of Electric Machines (a) Given that V1 = 230 V (primary), V2 = 115 V (secondary), f = 50 Hz and secondary (115 V) side is open circuited. Also, For V1 = 230 V, f = 50 Hz, the core loss = 1050 W (1) For V1 = 138 V; f = 30 Hz, the core loss = 500 W (2) Core loss = Hysteresis loss + Eddy current loss Wc = Wh + We 230 V / f for first measurement = = 4.6 50 V / f for first measurement =
138 = 4.6 30
Since (V / f) ratio is constant for Eqs. (1) and (2), let us assume Wc = A f + B f 2 For f = 50 Hz,
1050 = 50 A + 2500 B(3)
For f = 30 Hz, 500 = 30 A + 900 B(4) Solving Eqs. (3) and (4), we get A = 10.16, B = 0.216
Ch wise GATE_EE_Ch5.indd 234
230 V
Current through 230 Ω resistor is
2 Vxy = Voc = 2 sin w t 1 ⎛ 2⎞ Rxy = 1 × ⎜ ⎟ = 4 Ω ⎝ 1⎠
5A
Eb 0 = Va - I a Ra = 230 - ( 4)(0.4) = 228.4 V Under full load condition: 70 A 1A
69 A
230 Ω
230 V
Current through Rj is If =
230 = 1 A and I a = 69 A 230
E bf = 230 - (69)(0.4) = 202.4 V N 2 Eb 2 ϕ 1 = × N1 Eb1 ϕ 2 For constant flux machine
N 2 = 1400 ×
N 2 Eb 2 = N1 Eb1
202.4 = 1240.63 rpm 228.4
11/21/2018 11:54:10 AM
Chapter 5 • Electrical Machines
161. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (c) Given that number of phases, ϕ = 3 , f = 50 Hz, p = 6, Rr = 0.1 Ω (constant), Xr = 0.92 Ω sfl = 3% =
3 = 0.03 100
163. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (a) Frequency of voltage across slip rings under open circuit condition and rotating in anticlockwise is f1 =
Ratio of maximum to full load torque is given by Tmax sm2 + sfl2 = Tfl 2 sm sfl
Therefore, Tfl 2 2 = = = 1 / 1.954 Tmax ⎛ Sfl Sm ⎞ ⎛ 0.03 0.109 ⎞ + ⎜⎝ S + S ⎟⎠ ⎜⎝ 0.109 0.03 ⎟⎠ m fl Tmax = 1.954 Tfl
162. Topic: DC Machines: Series and Shunt (1.27) Consider Case (i): Machine as a motor 2A
50 A 48 A
Rsh = 125 Ω
250 V Ra = 0.6 Ω
Here, Eb = V − IaRa = 250 − (48) (0.6) = 221.2 V Case (ii): Machine as a generator 50 A 2A Rsh = 125 Ω
52 A 250 V Ra = 0.6 Ω
Eg = V + IaRa = 250 + (52) (0.6) = 281.2 V Therefore, N (generator ) Eg 281.2 = = = 1.27 N ( motor ) Eb 221.2
Ch wise GATE_EE_Ch5.indd 235
(Ns + N r ) p 120
Ns p + N r p 120 f1 = f + fr Frequency of voltage across commutator f1 =
The slip at which max torque occurs is given by, R 0.1 Sm = r = = 0.109 X r 0.92
or
235
f2 =
Ns p = f 120
164. Topic: Synchronous Machines: Characteristics (d) Given that p = 20, number of slots = 180, Z = 6/slot. Therefore, Ztotal = 6 × 180 = 1080. For single phase winding, slot span r = 20° and number of coils in a phase group = 180/(20 × 1) = 9 For three phase winding, slot span r = 20° and number of coils in a phase group = 180/(20 × 3) = 3 20 ⎞ ⎛ ⎛ nr ⎞ sin ⎜ ⎟ sin ⎜ 9 × ⎟ ⎝ ⎝ 2⎠ 2⎠ = 0.6398; = kd1 = ⎛ 20 ⎞ ⎛ r⎞ n sin ⎜ ⎟ 9 × sin ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ 1080 = 540 2 ⎛ 3 × 20 ⎞ ⎛ nr ⎞ sin ⎜ ⎟ sin ⎜ ⎝ 2⎠ ⎝ 2 ⎟⎠ = = 0.9597; kd 2 = ⎛ r⎞ ⎛ 20 ⎞ n sin ⎜ ⎟ 3 × sin ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ N1 =
N2 =
1080 1 × =180 2 3
V1 4.44 kd1ϕ fN1 kd1 × N1 0.6398 × 540 = = = ≈2 V2 4.44 kd 2ϕ fN 2 kd 2 × N 2 0.9597 × 180 165. Topic: Types of Losses and Efficiency Calculations of Electric Machines (a) For the given single-phase transformer, V/f is constant. Hysteresis loss, Wn ∝ f. Therefore, when f is increased by 10%, Wn increases by 10%. Eddy current loss, We ∝ f 2, so We increases by 21%. 166. Topic: Types of Losses and Efficiency Calculations of Electric Machines (b) From the given frequency function
11/21/2018 11:54:11 AM
236
GATE EE Chapter-wise Solved Papers
Load power, P2 = V2I2 × cos f2 15 I2 =
Pc/f 10
q
4 × 103 = 40.86 A 110 × 0.89
I1 = KI 2 ; where K = 5 f 25
I1 =
50
We have Wh = Af and We = B × f 2 where A = 10 and 5 = 0.1 50 Therefore, at 25 Hz, Wh = 10 × 25 = 250 W We = 0.1 × 252 = 62.5 W
N2 1 × I 2 = × 40.6 = 20.436 A 2 N1
To improve power factor to unity, the reactive power Q=
B = tan q =
167. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (33.46) Given that P = 1 pu, VT = 1.1 pu, Xs = 0.8 pu, Ra = 0, pf = 1. So, we have P=
EV sin d Xs
E = [(V cos ϕ + I a Ra ) 2 + (V sin ϕ + I a X s ) 2 ]1/ 2 We know that P = VI cos ϕ ⇒ I a =
1 = 0.91 1.1 × 1
As cos f = 1 ⇒ f = 0 ⇒ sin f = 0. E = [{(1.1 × 1) + (0.91 × 0)} + {(1.1 × 0) + (0.91 × 0.8) 2 }2 ] 2
1/ 2
= ⎡⎣(1.1) 2 + 0.7282 ⎤⎦ P=
= 1.319 V
1.319 × 1.1 sin d ⇒ d = 33.46° 0.8
168. Topic: DC Machines: Series and Shunt (880) Given that V = 300 V, Ra = 0.5 Ω, N0 = 900 rpm, I a 0 = 2A, I a1 = 15A . Under no load condition: Eb 0 = V - I a 0 Ra = 300 - ( 2)(0.5) = 299 V Under load condition: Eb1 = 300 - I a1 Ra = 300 - (15)(0.5) = 292.5 V We know that N1 Eb1 292.5 = ⇒ N1 = × 900 = 880 rpm N 0 Eb 0 299 169. Topic: Single Phase Transformer: Regulation and Efficiency (23.72) Given that Secondary side: V2 = 110 V, P2 = 4 kW, cos f2 = 0.89
Ch wise GATE_EE_Ch5.indd 236
N2 N1
V12 X
Therefore, V12 V12 ( 220) 2 = = Q VI sin ϕ 220 × 20.43 × sin( 27) = 23.72 Ω
X =
170. Topic: Single Phase Transformer: Regulation and Efficiency (b) Given that V1 = 220 V, V2 = 110 V, f = 50 Hz, L1 = 45 mH, L2 = 30 mH, M = 20 mH Leakage inductance, Ll1 = L1 - 2 M = 45 - 20( 2) = 5 mH Ll 2 = L2 - 2 M = 30 - 2( 20) = 10 mH Ll2 as referred to primary side is Ll′2 =
30 = 2( 20) = 120 - 40 = 80 mH (1/ 2) 2
Magnetising inductance with reference to primary is given by Lm = kL1 = 40 mH where k is the coupling coefficient. 171. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (b) Given that Ra = 0.1 W, Eb2 = 200 V, C = 1000 mF Also, since the generator is rotated at half the rated 1 speed and half field current N2 = N = 0.5 N1, and 2 1 f2 = 0.5f1 We know that Eb2 N 2ϕ 2 0.5 N1 × 0.5ϕ 1 1 = = 2 Eb1 N1ϕ 1 N1ϕ 1 So, Eb2 = 0.25 Eb1 or Eb2 = 50 V. Now, t = RC = 0.1 × 1000 mF = 100 mF From the relation, t = 69.3 ms
50 = 2000e - t /100 mF
we have
11/21/2018 11:54:14 AM
Chapter 5 • Electrical Machines
172. Topic: Single Phase Transformer: Regulation and Efficiency (a) Given that L1 = 800 mH, L2 = 600 mH, M = 480 mH. We know that input impedance seen by the source is V w2M 2 Z in = 1 = ( R1 + jX 1 ) + I1 R2 + jX 2 + Z L X1 = 2pf L1 = 2 × 3.14 × 50 × 0.8 = 251.32 X2 = 2 × 3.14 × 50 × 0.6 = 188.4 Z in = jX 1 +
⎛ w 2 m2 w 2 m2 ⎞ = j ⎜ X1 jX 2 x2 ⎟⎠ ⎝
⎡ 314 2 × 0.482 ⎤ = j ⎢ 251.32 ⎥ 188.4 ⎣ ⎦ = j[251.32 − 120.58] = j130.74 Now, j130.74 = jw Leff Leff =
130.14 = 0.416 or 416 mH 314
173. Topic: Single Phase Transformer: Regulation and Efficiency (c) 174. Topic: Single Phase Transformer: Regulation and Efficiency (b) Given that XC = −20j W, XL = j5 W, Zin = 2.5 W, Z in =
1 ⎡ 2.5 ⎤ + 5 j ⎥ - 20 j 2 ⎢ 2 b ⎣a ⎦
2.5 =
2.5 5 j 20 j + - 2 a2b2 b2 b
Equating real and imaginary parts gives
At steady state, Te = TL, that is 25 ( 200 - w r ) = 2.78w r or, w r = 179.98 rad/s 176. Topic: Three Phase Induction Motors: Principle of Operation (401.67) Given that Vg = 210 V and s = 0.05. So, we have The per phase rotor current is Ir =
210 (0.22 + (0.05 / 0.05) 2 ) 2
5 = 20 ⇒ b = 0.5 b2 175. Topic: DC Machines: Series and Shunt (179.98) The motor torque speed characteristics can be derived as:
(w r - 200) = (Te - 0) (200 - 180) (0 - 500) or, Te = 25 ( 200 - w r )
= 205.1 A
⎛ 3 ⎞ × 205.12 × 1 = 401.67 Nm Torque = ⎜ ⎝ 314.59 ⎟⎠ 177. Topic: DC Machines: Series and Shunt (b) We have 1 E∝ Number of parallel paths ( A) For wave wound, A = 2 and lap wound, A = p = 4, therefore 230 4 = ⇒ E = 115 V E 2 Thus, the voltage gets reduced to half as number of parallel paths is doubled, but the rated power remains the same P = 5 kW 178. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (5.9) Given that inertia constant, H = 2 MJ/MVA, initial rotor angle = 5°, Pm = 1 pu 2 GH = π f 180 × 50 1 pu = 4500
Mechanical momentum, m =
2.5 = 2.5 ⇒ a = 2 a 2 (0.5) 2 5 Also, -20 + 2 = 0 b
237
A bolted three phase to ground short circuit fault occurs at the terminal of the generator, so Pm - 0 m 1 = = 4500 deg/s 1/ 4500
mα = Pm - Pe ⇒ α =
The change in angle d is 1 1 Δd = × t 2 = × 4500 × (0.02) 2 2 2 = 4500 × 0.01 × 0.02 = 0.9° Now, 0.02 s after the fault, the angle is
d ′ = d + Δd = 5 + 0.9 = 5.9°
Ch wise GATE_EE_Ch5.indd 237
11/21/2018 11:54:15 AM
238
GATE EE Chapter-wise Solved Papers
Alternately Using swing equation, Pm - Pe =
2 H w pu ⎛ d 2d ⎞ 2 × 2 ⎛ d 2d ⎞ ⇒1= 2 ⎟ ⎜ 314.16 ⎜⎝ dt 2 ⎟⎠ w s ⎝ dt ⎠
78.554t 2 ⇒ d (t ) = 2 + d0 At t = 0.02 s, d(t) = 0.103 rad = 5.9° 179. Topic: DC Machines: Starting and Speed Control of DC Motors (8) Given that N0 = 1000 rpm, V = 200 V, E2 = 200 V, Ra = 1 Ω, N2 = 520 rpm and N1 = 500 rpm and we need to determine E1. The motor develops E = 200 V at 1000 rpm. Thus, at 500 rpm 200 × 500 = 100 V 1000 = V - I a Ra ⇒ 100 V
E1 =
= V - I a × 1 = 100 Also, at 520 rpm 200 × 520 = 104 V = V - 0.5 I a Ra 1000 ⇒ 104 V = V - 0.5 I a × 1
For the auto transformer K = RL = 12 × (0.33) 2 = 1.33 Ω
182. Topic: Three Phase Transformers: Connections, Parallel Operation (a) As the 3-phase transformer 33/11 kV is connected in delta-star connection, the fault is in star connected side, so all the current flows through fault and no current will pass through circuit of secondary side, hence i2 = 0. Now, primary kVA = secondary kVA ( 3 ⋅VL I L )1 = ( 3 ⋅VL I L ) 2 As the current flowing through the fault circuit, i2 = 0 3 × 33 × 103 × I L = 3 × 11 × 103 × 300 × or
⎛ 230 ⎞ ⎛ 100 ⎞ ∠30 ⎟ I p = (115∠60 ) ⎜ ⎜⎝ ⎠ ⎝ 3 ⎟⎠ 3 Ip =
0.5 I a = 4 or, I a = 8A
ϕ ZNp 0.01 × 666 × 1000 × 4 = = 222 V 60 A 60 × 2
We have E = V − IaRa 222 = 230 − Ia (0.267) ⇒ Ia = 29.96 A P = EIa = 6651.6 W Pout = 6651.6 − 600 = 6051.68 P 6051.68 = 57.8 N m T = out = 2π N 2 × 3.14 × 1000 60 60 181. Topic: Auto-Transformer (1.33) We have V1 = K , where K = 200/400 = 0.5 V2
Ch wise GATE_EE_Ch5.indd 238
1 1 A I ph ⇒ i1 = 3 3
183. Topic: Three Phase Transformers: Connections, Parallel Operation (a) For a constant kVA/phase
Solving,
E=
5 500
IL = 1 A
We know that I L =
E2 =
180. Topic: DC Machines: Starting and Speed Control of DC Motors (57.8)
200 = 0.33 600
115∠60° × 100 = 50 ∠30° 230 ∠30°
184. Topic: Three Phase Transformers: Connections, Parallel Operation →
(a) Given that N1 = 100, I 2 = 2∠30° A , N2 = 50, →
I 3 = 2∠150° A , N3 = 50. →
→
→
I 1 N1 = I 2 N 2 + I 3 N 3 I1 × 100 = 2∠30° × 50 + 2∠150° × 5 100 I1 = 100 ∠30° + 100 ∠150° I1 = 1∠30° + 1∠150° = 1∠90° 185. Topic: DC Machines: Starting and Speed Control of DC Motors (47.5) Given that N1 = 1000, V1 = 100 V, N2 = 500, V = 50 V2. T1 ⎛ I a1 ⎞ N12 = = T2 ⎜⎝ I a 2 ⎟⎠ N 22 Thus I a 2 = 2.5A Eb1 = 100 - (10 × 1) = 90 V = K fi N1
11/21/2018 11:54:18 AM
Chapter 5 • Electrical Machines
We know that synchronous speed
Eb1 N1 = Eb 2 N 2
Thus
Ns =
90 or, Eb 2 = = 45V 2
Therefore, smT ∝
Thus, applied voltage V = Eb 2 + I a 2 Ra = 45 + (1 × 2.5) = 47.5 V 186. Topic: Synchronous Machines: Starting of Synchronous Motor (−48.3) Given that Vt = 11 kV, Xs = 50 W/Ph, f = 50 Hz, Ia = 100 A and p = 2. 11 Ef = Vt - I a X s = × 103 - j100 × 50 3 = 6350 - 5000 j
⇒
120 f = 600 4
( I a X s ) 2 = E f2 + Vt 2 - 2 Ef Vt cos d
f = 4×
(120 × 50) = (8080) + (6350) − 2(8082) (6350) cosd d = −48.3° 2
2
187. Topic: Types of Losses and Efficiency Calculations of Electric Machines (0.17) Given wound rotor induction motor with 220 V, 3-phase, 4-pole, 50 Hz. At full load torque, R 0.03 s = r ⇒ 0.15 = ⇒ X2 = 0.2 W X2 X2
R ′r = X 2 = 0.2 Ω
188. Topic: Three Phase Transformers: Connections, Parallel Operation (30) From the given figure, we have Secondary of upper transformer is ∆ connected. Secondary of lower transformer is Y connected. Phase angle between ∆ voltage and Y voltage is 30°. 189. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (c) Given that v/f = constant, the slip at maximum torque 1 smT ∝ f
600 = 20 Hz 120
1 1 = = 0.05 s = 50 ms f 20
Therefore, T = 25 ms 191. Topic: Single Phase Transformer: Regulation and Efficiency (10) Referring entire circuit to primary 80 × 103 10 Ω (100)2 1:100
R R′ = r ⇒ 1 = r ⇒ R ′r = X 2 = 0.2 Ω X2 X2
External resistance to be added, 0.2 − 0.03 = 0.17 W
Ch wise GATE_EE_Ch5.indd 239
Time period =
Tst 2 = =1 1 Tmax + smT smT
Therefore, smT = 1, so smT
1 Ns
120 f = 600 p
N=
Now,
120 f p
190. Topic: DC Machines: Starting and Speed Control of DC Motors (c) Current in armature is AC, it is rectified by commutator segments. Lap connection implies four parallel paths, so in each path, maximum magnitude = 40/4 = 10 A. We know that
| E f | = 6350 2 - 5000 2 = 8082 V
2
239
40 × 103 (100)2
100 V
I=
100 ∠0° 8 + j (10 - 4)
I =
100 82 + 6 2
= 10 A
192. Topic: Synchronous Machines: Characteristics (5) From the given question I ′′ = I′ =
Eg X d′′ Eg Xd
=
1 pu = 5 pu 0.2 pu
=
1 pu = 1 pu 1 pu
11/21/2018 11:54:20 AM
240
GATE EE Chapter-wise Solved Papers
Therefore, the required ratio is
PE PH + = 2000 2 4
I ′′ 5 = =5 I′ 1
At 52 Hz,
193. Topic: Operating Principle of Single Phase Induction Motors (c)
PE′′= (1.04) 2 PE and PH′′ → (1.04) PH Solving Eqs. (1) and (2), we get PH = 3000 W and PE = 2000 W PE′′= 2163.2 W and PH′′ = 3120 W
194. Topic: Single Phase Transformer: Regulation and Efficiency (b) Let impedance due to L = wL = XL Referred to primary, it becomes L ⎛ 1⎞ ⎜⎝ ⎟⎠ n
2
(2)
PH′′ = 1.44 PE′′
197. Topic: Types of Losses and Efficiency Calculations of Electric Machines (86.84) For the given DC shunt generator
= n2 L
0.01 W
195. Topic: Three Phase Transformers: Connections, Parallel Operation (d) The star – delta transformers show a lag of 30° on star side or the phase of Vab is same as phase of VA. The phase sequence of input is AB. The voltage phasor diagram of primary side is as follows:
45 A 44 W
If =
VB
220 V
220 V =5A 44 Ω
Therefore, Ia = 50 A I2R losses = (5)2 × 44 + (50)2 × 0.01 = 1100 W + 25 W
VA 30°
lag of 30°
η= VC
VAB
=
−VB
196. Topic: Types of Losses and Efficiency Calculations of Electric Machines (1.44) We know that hysteresis and eddy current losses are given by PH = K1 Bm1.6 f
Output 220 × 45 = Output + Losses 220 × 45 + 1100 + 25 + 375 9900 9900 = 9900 + 1100 + 25 + 375 11400
= 0.8684 = 86.84% 198. Topic: Synchronous Machines: Starting of Synchronous Motor (1.602) I A X A cos q tan (d ) = Vt + I a X q sin q
(
PE = K 2 Bm2 f 2 t 2 V/f = constant for all there given cases, so Bm = constant. Therefore, PE ∝ f 2 and PH ∝ f At 50 Hz, PE + PH = 5000 At 25 Hz,
Ch wise GATE_EE_Ch5.indd 240
P P PE′ → E and PH′ → H 4 2
(1)
)
=
1.0 × 0.6 × 0.8 ⎡⎣1 + (1.0 × 0.6 × 0.6 )⎤⎦
=
0.48 1.36
or, d = 19.44° Thus I d = 1.0 × sin (19.44 + 36.97) = 0.832 A
E0 = Vt cos d + I d X d = 1.602
11/21/2018 11:54:22 AM
Chapter 5 • Electrical Machines
199. Topic: Auto-Transformer (c) kVA of autotransformer
Therefore, 3 × 3810.51 × 4400.127 × 0.5 30 = 838336.4 = 838.336 kW
P=
= (kVA)old (1 + K) ⎡ 2200 ⎤ = 242 kVA = 22 × ⎢1 + 220 ⎥⎦ ⎣
30Ω Ia
applied
d
201. Topic: Auto-Transformer (6) (kVA)auto = (1 + k) (kVA)old ⎛ 200 ⎞ = ⎜1 + 2 kVA = 6 kVA ⎝ 100 ⎟⎠ 202. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (c) In generator mode, slip is negative. Also, rotor frequency = sf ⇒ 5 = s(60) ⇒ s =
Ns =
cosf = 1 204. Topic: Synchronous Machines: Starting of Synchronous Motor (b) For a synchronous motor at a leading power factor, Ef > Vt 205. Topic: Three Phase Induction Motors: Torque-Speed Characteristics (c) Rotor flux speed is same as stator flux speed. Ns =
1 12
120 × 60 = 1800 rpm 4
To get inductance generator mode, rotor speed should be greater than synchronous speed, S=
N r - N s N r - 1800 1 = = Ns 1800 12
=
Z a′ = 24.5 + j12.75 - jX C Z m = 12.5 + j15.75 = 20.1∠51.56° ⇒ Im lags by 51.56°
and
Ch wise GATE_EE_Ch5.indd 241
ϕa = 90° - 51.56° = 38.44° (for maximum torque) For auxiliary winding,
3EV sin d Xs
tan ϕa =
3
= 3810.51 V
⇒ tan(38.44°) =
3810.51 V E= = = 4400.127 V cos d cos 30°
2π × 1500 = 157 rad/s 60
206. Topic: Operating Principle of Single Phase Induction Motors (98.93) Za = 24.5 + j12.75 Ω Let XC = capacitor in auxiliary winding. Then
203. Topic: Synchronous Machines: Starting of Synchronous Motor (838.3) We know
6.6 × 1000
120 × 50 = 1500 rpm 4
Therefore, the speed of the rotor flux in mechanical rad/s
⇒ Nr = 1950 rpm
P=
Ebcos d
Eb ∠(−30°)
Now,
V=
Ia
I 2 110 = I1 415
110 I2 = × 120 = 31.8 A 415
Now,
V = Ia Xs E b
Vt = 6.6 kV
200. Topic: Three Phase Induction Motors: Starting and Speed Control (31.8) I ∝ V start
241
Xa R X 24.5
⇒ X = 19.44 Ω Now, XC − 12.75 = 19.44 ⇒ XC = 32.19 Ω
11/21/2018 11:54:24 AM
242
GATE EE Chapter-wise Solved Papers
Therefore,
Since magnetic circuit is linear, we have C=
1 1 = 2π fX C 2 × π × 50 × 32.19
I fA EA 7663.05 = = = 0.745 I fB EB 10287.45
= 98.93 mF 207. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (0.745) Since, the current is same at same terminal voltage, the reactive power supplied by both the motors are same, i.e. QA = QB = 25 MVAR Assuming they are at no load sin f = 1 Now, QA
IA =
3VA sin ϕ QB
IB =
2VB sin ϕ
25 × 10
=
6
2 × 11 × 103
= 1312.2A
208. Topic: DC Machines: Starting and Speed Control of DC Motors (10.75) Given that τ ∝ N 2 , ϕ ∝ I a , τ ∝ I a2 ⇒ N 2 ∝ I a2 or N ∝ I a N 2 Ia2 Now, = N1 I a1 I ⇒ 0.5 = a 2 ⇒ I a 2 = 15A 30 For a series motor, Eb 2 I a 2 N 2 = Eb1 I a1 N1
= 1312.2A
EA = V - jI A X s =
=
11 × 103
- j1312.2 × 1∠90° 3 = 7663.05 V
EB = V - jI B X s =
11 × 10
N 2 Eb 2 I a1 = N1 Eb1 I a 2
⇒ 0.5 =
3
- j1312.2 × 3∠90° 3 = 10287.45 V
⇒
N 22 1 = N12 4
Eb 2 30 × Eb1 15
Eb 2 1 = Eb1 4
(1)
Also,
11 kV
Eb1 = 220 − 30(0.4 + 0.1) = 205 V 50 MVAR 1Ω
3Ω
Also,
50 × 106 IL = Ib
Ia
A
3 × 11 × 103
IaXs E
Vt
d 90°
Ch wise GATE_EE_Ch5.indd 242
= 2624.31
Eb2 = V − Ia2 (Ra + Rs + Rext) ⇒ 51.25 = 220 − 15 (0.4 + 0.1 + Rext) ⇒ Rext = 10.75 Ω
B
Ia
Therefore, from Eq. (1), we get 1 Eb 2 = × 205 = 51.25 V 4
209. Topic: Three Phase Transformers: Connections, Parallel Operation (623 to 627) Load on 6.6 kV winding, 900 × 10 I2 = = 78.73∠ - 36.86°A 3 × 6.6 × 103 I2 =
1.1 1 = 6.6 6
I 2′ =
I2 = 6 × 78.73∠ - 36.86° k
11/21/2018 11:54:26 AM
Chapter 5 • Electrical Machines
= 472.4 ∠ - 36.86°A
Thus,
(220 - I a × 0.02) I a = 70 × 94.24
Load on 400 V winding, I3 = k= I 3′ =
300 × 103 1.1 × 103 400 / 3
I a = 30.16 A ≈ 30 A
= 433.025∠ - 53.13°A
3 × 400
212. Topic: DC Machines: Starting and Speed Control of DC Motors (1241.81) The circuit diagram of shunt motor is as follows.
= 4.76
I 3 433.025∠ - 53.13° = k 4.76
0.5 A
7A
= 157.56 ∠ - 53.13°A
6.5 A
Hence,
240 Ω
I1 = I 2 + I 3 = 472.4 ∠ - 36.86° + 157.56 ∠ - 53.13°
120 V 0.8 Ω
= 625.19∠ - 40.9°A 210. Topic: DC Machines: Motoring and Generating Mode of Operation and Their Characteristics (550)
150 A
I=
Eb1 = 120 − 65(0.8) = 114.8 V Eb2 = 120 − 1.5(0.8) = 118.8 V
Grid
Grid Ra
120 = 0.5A 240
and
150 A
0.1 Ω
Here,
145 V − +
145 V − +
Ra
243
0.1 Ω
N1 = 1200 rpm (given) Therefore, N 2 Eb 2 118.8 = = N1 Eb1 114.8
Eg2
Eg1
800 rpm From the first figure,
Eg1 = 145 + 150 × 0.1 = 160 V Now, Eg 2 Eg1
=
⇒ N 2 = 118.8 × 1200 = 1241.81 rpm 114.8
1000 rpm
213. Topic: Three Phase Induction Motors: Equivalent Circuit (70.05) Circuit during starting is
N 2 1000 = N1 800
0.3
1000 × 160 = 200 V 800 ( 200 - 145) = 550 A = 0.1 = 550 A
⇒ Eg 2 = Thus
Ia2 ⇒ Ia2
211. Topic: DC Machines: Starting and Speed Control of DC Motors (b) The torque is given by, Eb I a = 70 900 ⎞ ⎛ ⎜⎝ 2π × ⎟ 60 ⎠
Ch wise GATE_EE_Ch5.indd 243
0.3 j0.41
j0.41
80√3
Reactance at 60 Hz = j 0.82 Reactance at 20 Hz =
20 j 0.82 = j 0.273 Ω 60
Therefore,
( neglecting losses) I start =
80 3 = 70.05A (0.16) 2 + (0.273) 2
11/21/2018 11:54:28 AM
244
GATE EE Chapter-wise Solved Papers
214. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (-14.56) I rated = I phase = Xs =
25000 3 × 400 36.085 3 VOC I rated
= 36.085A
=
= 15.789 pu I base =
= 20.83A
I f = constant
=
3 ×1 0.1 + 0.05 + 0.01 + 3(0.01) P 3VL
=
125 × 106 3 × 15.5 × 103
= 4656.05 A
I f = I fPu × I base = 15.789 × 4656.05
360 = 17.28 Ω phase 20.83
E = (V cos ϕ + I a Ra ) 2 + (V sin ϕ - I a X s ) 2 = ( 400 × 0.8) 2 + ( 400 × 0.6 - 20.83(17.28)) 2
= 73516.538 = 73.52 kA 218. Topic: DC Machines: Starting and Speed Control of DC Motors (600) Given, separately initiated dc motor. Field excitation is constant. Produces a torque of 500 Nm.
= 341.76 V %VR =
341.76 - 400 × 100 = -14.56% 400
0.05 Ω
100 V 215. Topic: Three Phase Transformers: Connections, Parallel Operation (37.88) Using turns ratio 5:1, V2 = 381 V (phase) = Vrms 3V2 I 2 = 25000 ⇒ I2 =
25000 3 × 381
At zero speed, N = 0, Eb = 0. = 37.88A
Eb = V - IaRa
⇒ Þ 0 = 100 - Ia1(0.05)
216. Topic: Synchronous Machines: Starting of Synchronous Motor (b) Salient pole synchronous motor power and torque relations per phase: P=
T=
EbV V 2 ⎛ Xd - Xq ⎞ ⋅ sin d + ⎜ ⎟ sin( 2d ) Watts Xt 2 ⎝ Xd Xq ⎠ 60 2π N S
⎤ ⎡E V V 2 ⎛ Xd - Xq ⎞ ⎢ b sin d + ⎜ ⎟ sin 2d ⎥ N-m 2 ⎝ Xd Xq ⎠ ⎢⎣ X d ⎥⎦
100 = 2000 A 0.05 60 60 ϕ ZNP T= Eb I a = ⋅ Ia 2π N 2π N 60 A
⇒ I a1 =
Let T1 =
1 ZP 1 ZPϕ ϕ I a1 Nm = kI a1 where k = 2π A 2π A
Now, T1 = 500 Nm, Ia1 = 2000 A, k =
500 = 0.25 2000
The second term is reluctance power or reluctance torque, which is directly proportional to sin 2d. Therefore reluctance torque will be maximum, when d = 45°. This is true because
When motor runs on no-load given all mechanical losses neglected, no-load current is negligible and the voltage drop at no-load can be negligible. Therefore
sin 2(45°) ⇒Þ sin 90° = 1 (Maximum)
Eb V = 150 V
217. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (73.52) For LG fault I fPu =
Ch wise GATE_EE_Ch5.indd 244
3VTh z1 + z2 + z0 + 3 zn
Eb =
ϕ ZNP 60A
Now
k=
1 ⎛ ZPϕ ⎞ ⎜ ⎟ 2π ⎝ A ⎠
11/21/2018 11:54:30 AM
Chapter 5 • Electrical Machines
⇒
ZPϕ = 2π ( k ) A
If =
N ⇒ Eb = 2π ( k ) ⋅ 60
245
3E g Z1 + Z 2
Case II: With Zf
⇒ Eb = k ⋅ w (w : No-load speed in rad/s)
⇒w=
Eb k
⇒w=
150 = 600 rad/s 0.25
219. Topic: Single Phase Transformers: Regulation and Efficiency (d) The inductance of primary is L = N P2 /S , where S is reluctance of magnetic path. S=
∼
Z2
If1 =
V = IX L =
3 Eg Z1 + Z 2 + Z f
Given, If1 = kIf . So
w N P2 m r 2 XL = wL = 2R
I w N P2 m r 2 2R
⎛ 3 Eg ⎞ =⎜ ⎟k Z1 + Z 2 + Z f ⎝ Z1 + Z 2 ⎠ 3 Eg
⇒ Z1 + Z2 = k(Z1 + Z2 + Zf) ⇒ Z f =
m I r 2 N P2 w R= 2V 220. Topic: Synchronous Machines: Regulation and Parallel Operation of Generators (a) For line to line fault Case I: Without Zf
Z1
Zf
2π R 2R = = 2 2 m A m[π r ] m r
N 2 mr 2 N2 ∴L = P = P 2R 2R mr 2
Therefore
Eg
Z1
Z2
( Z1 + Z 2 )(1 - k ) k
221. Topic: Three Phase Transformers: Connections, Parallel Operation (97.36) 3 aph = = 3; 3/ 3
I HV = I1
900 = 100 A 3⋅ 3
R1eq = R1 + a 2 R2 = 0.3 + 9 × (0.02) = 0.48 Ω Cu loss (3 - ϕ ) = 3I12 R1eq
Eg
∼
= 3 × 100 2 × 0.48 = 14400 W = 14.4 kW
Ch wise GATE_EE_Ch5.indd 245
11/21/2018 11:54:32 AM
246
GATE EE Chapter-wise Solved Papers 12 Ω
Core loss = 10 kW %η =
j12 Ω
900 = 97.36% 900 + 10 + 14.4 j120 Ω
222. Topic: DC Machines: Series and Shunt (825) Rse
12 2(0) j 12 2
10 A
12 2(2 – 0)
j120 Ω
j 12 2
200 V On no-load, N; Ns slip, s ≅ 0
From circuit diagram, we have
12 Ω
Eb1 = V - Ia(Ra + Rse)
j12 Ω
= 200 - 10(1) = 190 V N1 = 1000 rpm
j120 Ω 3Ω
Load torque increased by 44% (T ∝ I a2 ) . So T2 ⎛ I a 2 ⎞ = T1 ⎜⎝ I a1 ⎟⎠ ⇒
j120 Ω
2
1.44T1 ⎛ I a 2 ⎞ =⎜ ⎟ ⎝ 10 ⎠ T1
j6 Ω
2
⇒ I a22 = 144 ⇒ I a 2 = 12 A
On no-load, N; Ns slip, s ≅ 0
Simplifying the above circuit into a simple R-L circuit, we have
At same voltage,
I
Eb 2 = V - I a 2 ( Ra + Rse ) R
= 200 - 12(1) = 188 V For series motor,
Zeq = R + jX Z∠θ
V∠0 N∝
Eb (ϕ ∝ I a ) ϕ
jX
So,
N 2 Eb 2 I a1 = × N1 Eb1 I a 2 ⇒
N2 188 10 = × 1000 190 12
N 2 = 824.56 rpm 825 rpm
223. Topic: Operating Principle of Single Phase Induction Motors (0.106)
Ch wise GATE_EE_Ch5.indd 246
Z eq =
(3 + j 6)( j120) + [12 + j132] (3 + j126)
= 138.563∠83.9 Ω Current drawn by motor is
I=
V ∠0 Z ∠q
(Therefore q = impedance angle = pf angle)
Therefore, no-load lagging pf of motor is (cos q ) cos (83.9) = 0.106 lagging power factor.
11/21/2018 11:54:34 AM
Power Systems
CHAPTER6
Syllabus Power generation concepts, AC and DC transmission concepts, Models and performance of transmission lines and cables, Series and shunt compensation, Electric field distribution and insulators, Distribution systems, Per-unit quantities, Bus admittance matrix, Gauss-Seidel and Newton-Raphson load flow methods, Voltage and Frequency control, Power factor correction, Symmetrical components, Symmetrical and unsymmetrical fault analysis, Principles of over-current, differential and distance protection; Circuit breakers, System stability concepts, Equal area criterion.
CHAPTER ANALYSIS Topic Power Generation Concepts
GATE 2009
GATE 2010
GATE 2011
GATE 2012
GATE 2013
2
GATE 2014
GATE 2015
GATE 2016
GATE 2017
1
4 2
1
3
2
1
2
1
1
1
1
3
2
AC and DC Transmission Concepts Models and Performance of Transmission Lines and Cables
2
Series and Shunt Compensation
1
Electric Field Distribution and Insulators Distribution Systems
4
2 1
Per-Unit Quantities Bus Admittance Matrix
2
2
1
1
2
1
1
1 1
1
2
1
3
2
3
2 1
Gauss-Seidel and NewtonRaphson Load Flow Methods
2
Voltage and Frequency Control Power Factor Correction
2 1
GATE 2018
1
Symmetrical Components
2
2
1
3
2
1
1
1
2
4
1
Symmetrical and Unsymmetrical Fault Analysis
2
Principles of Over-Current
2
Differential and Distance Protection
1
Circuit Breakers
1
1
1 1
System Stability Concepts Equal Area Criterion
Ch wise GATE_EE_Ch6.indd 247
11/21/2018 11:50:45 AM
248
GATE EE Chapter-wise Solved Papers
IMPORTANT FORMULAS 1. Some terms related to power generation (a)
(b)
(c)
Demand factor = Load factor =
Maximum demand Connected load
Average demand Maximum demand
Diversity factor Sum of individual maximum demands = Maximum demand of the system
3. Capacitance calculations of: (a) Two wire line 0.0121 μF/km CAB = log D / ra rb (b) Three phase line of equal spacing and of same area CAN =
Plant capacity factor
Average demand = m installed capacity) Plant capacity (maximum
CAN =
μ0 ⎡ ⎛ D⎞ ⎤ 0.25 + ln ⎜ ⎟ ⎥ H ⎢ ⎝ r ⎠⎦ π ⎣ (b) Group of conductors in a transmission line:
{I1 ln(1/r1 ) + I 2 ln(1/D12 ) + I 3 ln(1/D13 ) + ... + I n ln(1/D1n )}
ψ 1 = 2 × 10
−7
(c) Three asymmetrically conductors spaced in a transmission line
⎡ 1 1 3 b⎤ La = 2 × 10 −7 ⎢ln − ln −j ln ⎥ 2 c⎦ bc ⎣ r1′
⎡ 1 1 3 a⎤ Lc = 2 × 10 −7 ⎢ln − ln −j ln ⎥ 2 b⎦ ba ⎣ r1′ ⎡ 1 1 3 c⎤ Lb = 2 × 10 −7 ⎢ln − ln −j ln ⎥ 2 a⎦ ac ⎣ r1′
(d) Composite conductors:
⎛D ⎞ LA = 2 × 10 −7 ln ⎜ m ⎟ H/m ⎝ Ds ⎠
where Dm is the geometric mean distance (GMD) Dm = mn ( D11′ D12′ ...D1′n )( D21′ D22′ ...D2′n )...( Dm′ 1 Dm′ 2 ...Dmn ′ ) and Ds is the geometric mean radius (GMR) Ds = m ( r ′D12 D13... D1m )( r ′D21 D23... D2 m )...( r ′Dm1 Dm 2 ... Dmm )
qA = VAN
2πε 0.0242 = μF/km ⎛ Deq ⎞ ⎛ Deq ⎞ ln ⎜ log ⎜ ⎝ r ⎟⎠ ⎝ r ⎟⎠
Deq = DAB DBC DCA 4. Effective resistance of a conductor
2. Inductance of: (a) Two-wire transmission line (per unit length) L=
qA 2πε 0.0242 μF/km = = VAN ln D / r log( D / r )
(c) Three phase line with unsymmetrical spacing
Station output in kWh (d) Utilisation factor = Plant capacity × hours of use (e)
)
(
R=
Ploss I2
5. Performance of transmission lines (a) Percentage efficiency of transmission lines =
Power received at the receiving end × 100 Power delivered at sendding end
(b) Percentage voltage regulation
⎛ No load receiving − Full load receiving ⎞ ⎜⎝ end voltage end voltage ⎟⎠ × 100 = Full load receiving end voltage 6. Short transmission lines (a) Relation between the delivering end and the receiving end parameters Vs cos ϕ s = Vr cos ϕ r + I r R Vs sin ϕ s = Vr sin ϕ r + I r X where Vr, Ir and fr are the receiving end voltage, current and phase angle and Vs and fs are the sending end voltage and phase angle, respectively and X is reactance (b) Regulation (in pu) =
Ir R I X cos ϕ r ± r sin ϕ r Vr Vr
(c) ABCD parameters of short transmission line Vs = AVr + BI r I s = CVr + DI r
2
Ch wise GATE_EE_Ch6.indd 248
A = 1, B = 2, C = 0 and D = 1
11/21/2018 11:50:47 AM
Chapter 6 • Power Systems
(d) Percentage efficiency %η =
(c) Equivalent T representation (i) Y and Z parameters
P × 100% P + 3I r2 R
where γ = α + jβ (ii) ABCD parameters:
P × 100% P + 1.5( I s2 + I r2 ) R
(ii) ABCD parameters: A = D = 1 + YZ/2 B=Z
where Z is the series impedance per unit length and Y is the shunt admittance per unit length. (b) Nominal Π representation (i) Current and no load receiving end voltage
Vr′ =
ωC ωC +j 2 2
sinh γ l and γl
Y ′/ 2 =
Y tanh(γ l/ 2) 2 (γ l/ 2)
(ii) ABCD parameters A = D = cosh( γ l)
R + jX − 2 j/ω C
B = Z c sinh(γ l) C=
P × 100% P + 3I s2 R
(iii) Percentage voltage regulation
(iv) ABCD parameters YZ A = D = 1+ 2 B=Z
C = Y (1 + YZ/4) 8. Long transmission lines (a) Current and voltage at point x from the receiving end: V ( x ) = cosh( γ x)Vr + Z c sinh( γ x) I r 1 I ( x) = sinh( γ x)Vr + cosh( γ x) I r Zc
(b) Current and voltage at the sending end (l is the length of the line) Vs = cosh( γ l)Vr + Z c sinh( γ l) I r Ir =
1 sinh( γ l)Vr + cosh( γ l) I r Zc
sinh(γ l) Zc
where Zc is the characterstic impedance. 9. Combination of networks (a) Cascaded ⎡ A B ⎤ ⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤ ⎢C D ⎥ = ⎢C D ⎥ + ⎢C D ⎥ ⎦ ⎣ 1 1⎦ ⎣ 2 2 ⎦ ⎣
Vr′− Vr × 100% Vr
Ch wise GATE_EE_Ch6.indd 249
Z′ = Z
B=Z
(ii) Efficiency
YZ 2
Vs ( −2 j/ω C )
%η =
A = D = 1+
C = Y (1 + YZ /4) (d) Equivalent Π representation (i) Y and Z parameters
and C = Y (1 + YZ/4)
I s = I r (cos ϕ r − j sin ϕ r ) + jVr
sinh γ l Z tanh(γ l/2) and Z ′ = γl 2 (γ l/2)
Y′ =Y
7. Medium transmission lines (a) Nominal T representation (i) Percentage efficiency %η =
249
(b) Parallel A= D = B=
A1 B2 + A2 B1 B1 + B2
B1 B2 B1 + B2
and AD − BC = 1 10. Surge impedance Zc =
Z Y
where, Z = R + jwL is the series impedance and Y = G + jwC is the shunt admittance of the line. (a) Surge impedance loading SIL =
VLL 2 Zc
where VLL is the line-to-line voltage
11/21/2018 11:50:50 AM
250
GATE EE Chapter-wise Solved Papers
11. AC and DC transmission (a) Power transmitted with DC and AC links Pd ⎧1.414 at pf = 1 2 = =⎨ Pa cos ϕ ⎩1.768 at pf = 0.8 (b) Power loss in DC transmission Ploss = (P/Vm) Rd = (P/Vm) .(rI/Ad) 2
2
13. Voltage distribution in suspension insulator String efficiency =
Voltage across the string n × voltage across the disc which is nearest to the string
14. Per unit (pu) system Per unit value of any quantity
(c) Power loss in AC transmission
=
2
Ploss = ⎡ 2 P/(V cos ϕ) ⎤ R m ⎣ ⎦ a = 2( P /Vm ) 2 ⋅ ( ρ I /Aa cos 2 ϕ ) (d) Ratio of areas: Ad cos 2 ϕ ⎧0.5 at pf = 1 = =⎨ Aa 2 ⎩0.2 at pf = 0.8
Quantity Base value of that quantity
15. Gauss–Seidel method of load flow analysis (a) Complex power at i bus in a power system n
(b) Pi , inj − jQi , inj = Vi * ∑ YikVk k =1
= Vi *[Yi1V1 + Yi 2V2 + …
12. Cables (a) Classification
+ YiiVi + …+ YinVn ]
Low-tension (LT) cables
up to 1000 V
High-tension (HT) cables
up to 11,000 V
Super-tension (ST) cables
from 22 kV to 33 kV
Extra high-tension (EHT) from 33 kV to 66 kV cables Extra super voltage cables
beyond 132 kV
(b) Dielectric stress in cables: Potential gradient
g min
ΔPi = Pi , inj − Pi , calc = PGi − PLi − Pi , calc and
⎡ d1 d2 D⎤ ⎢ d ln + d1 ln + d2 ln ⎥ d d d ⎣ 1 2⎦
g1 max = g 2 max = g3 max = g max =
V2 d1 d2 ln 2 d1
=
= Vi , acc ( k −1) + λ ⎡⎣Vi ( k ) − Vi , acc ( k −1) ⎤⎦
16. Newton–Raphson method for load flow analysis 17. Shunt compensator (a) Compensator current
Maximum stress between core and inner sheaths
Ch wise GATE_EE_Ch6.indd 250
2V = V/m D ln D/d
(d) Grading in cables: For capacitance grading, total potential difference between the core and earthed outer sheath
d d1 ln 2 d
(d) Error between the actual and injected real and reactive powers
Vi , acc ( k ) = (i − λ )Vi , acc ( k −1) + λVi ( k )
D = e = 2.718 d
V1
1 ⎡ Pi , inj − jQi , inj − Yi1V1 ⎢ Yii ⎣ Vi * − Yi 2V2 …− YinVn ]
ΔQi = Qi , inj − Qi , calc = QGi − QLi − Qi , calc (e) Voltage of ith bus using acceleration constant l
(c) Most economical size of conductor
g V = max 2
Vi =
V V/m x ln D/d 2V = V/m ln D/d
g= g max
(c) Voltage of the ith bus using the procedure
4V [1 − cos(δ / 2)] ∠(δ / 2) X
(b) Apparent power supplied by the source Ps + jQs = Vs I s*
=
2V 2 sin(δ / 2) 2V 2 [1 − cos(δ / 2)] +j X X
(c) Apparent power delivered at the receiving end Pr + jQr = Vr I r*
V3 d2 D ln 2 d2
IQ = j
=
2V 2 sin(δ / 2) 2V 2 [cos(δ / 2) − 1] +j X X
11/21/2018 11:50:51 AM
Chapter 6 • Power Systems
(d) Real power transmitted over the line PE = Ps + Pr =
2V 2 sin(δ / 2) X
(e) Reactive power consumed by the line
QE = Qs + QQ − Qr =
V = λ Is Q (b) Line current
Is =
22. Normalised inertia constant (H ) and swing equation H= =
8V 2 [1 − cos(δ / 2)] X
18. Series compensator (a) Injected voltage
Stored kinetic energy at synchronous speed Generator MVA rating J ω s2 MJ/MVA 2Srated 2 H d 2δ = Pm = Pe = Pa ω s dt 2
23. Equal area criterion (a) Area of acceleration
∓ j 90°
δc
A1 = ∫ ( Pm − Pe )dδ
V ∠δ − V j( X ∓ λ )
δ0
(b) Area of deceleration
(c) Capacitive mode of operation
VQ = λ I s e
− j 90
V ∠δ − V ⇒ Is = j( X − λ )
δm
A2 =
PE = Ps + Pr =
δm
δ0
δ cr
⎡Va0 ⎤ ⎡1 ⎢V ⎥ = 1 ⎢1 ⎢ a1 ⎥ 3 ⎢ ⎢⎣Va2 ⎥⎦ ⎢⎣1
I f(a0) =
Ch wise GATE_EE_Ch6.indd 251
m
)dδ
1 1 ⎤ ⎡Va ⎤ a a 2⎥⎥ ⎢⎢Vb ⎥⎥ a 2 a ⎥⎦ ⎢⎣Vc ⎥⎦
Z kk0
Vf + Z kk1 + Z kk 2 + 3Z f
(ii) Current for line to line fault:
I f(a1) = =
Vf Z kk1 + Z kk 2 + Z f
(iii) Positive, negative and zero sequence current for LLG fault:
(b) Reactive power supplied by the compensator
e
I f(a1) = − I f(a2) =
V2 sin δ X ∓λ
2λV 2 QQ = − j (1 − cos δ ) ( X − λ )2
∫ (P − P
25. Unsymmetrical fault analysis (a) Single-line-to-ground fault (LG) (i) Current for single line to ground fault:
ΔPref , i = K i ∫ ACE dt
21. Power angle characteristics (a) Real power transmitted over the line is given by
)dδ
24. Symmetrical component transformation matrix
20. Frequency control (a) Equation for the area control error (ACE)
(b) Change in the reference of the power setting DPref,i, for any area i
m
δ cr
∫ ( Pm − Pe )dδ =
Vs1Vr − Vr 2 N r 2 = RT PR + X T Qr
ACE = ( Ptie − Psch ) + Bf Δf = ΔPtie + Bf Δf where Ptie and Psch are tie-line and scheduled power through tie-line respectively; Bf is frequency bias constant and Df is the frequency deviation.
e
(c) From equal area criterion
V ∠δ − V j( X + λ )
19. Voltage control using tap changing transformers
∫ (P − P
δc
(d) Inductive mode of operation VQ = λ I s e + j 90 ⇒ I s =
251
Vf Z kk1 + Z kk 2 ( Z kk 0 + 3Z f ) Vf Z kk 2 ( Z kk 0 + 3Z f ) Z kk1 + Z kk 2 + Z kk 0 + 3Z f
11/21/2018 11:50:53 AM
252
GATE EE Chapter-wise Solved Papers
⎞ ⎛ Z kk 2 I f(a0) = − I f(a1) ⎜ ⎟ 3 Z + Z + Z ⎝ kk 2 kk 0 f ⎠
⎞ ⎛ Z kk 0 + 3Z f I f(a2) = − I f(a1) ⎜ ⎟ Z + Z + Z 3 ⎝ kk 2 kk 0 f ⎠
where I ″, I ′ and Iss are sub-transient, transient and steady state current, respectively. 27. Open conductor faults (a) One phase conductor open (i) Current I a(1) = I ij
26. Symmetrical fault analysis (a) Bus voltage at the faulted bus
Vk(f) = Vk (0) − Z kk I k(f)
Vi (1) ( f ) = Vi (1) (0) + ΔVi (1) I k(f) =
Vi ( 2 ) ( f ) = ΔVi (1)
Vk (0) Z kk
Vi ( 0 ) ( f ) = ΔVi ( 2 ) (b) Two phase conductors open
(c) Fault current for non-zero fault impedance I k(f) =
Vk (0) Z kk + Z f
1 Ia 3
(ii) Sequence voltages
(
Vi(f) = Vi (0) − Z ik I k(f) ∀i = 1, 2, … n, i ≠ k
)
(0) ( 2) Vkk(1)′ = I a(1) Z kk ′ + Z kk ′ = I ij
(e) Voltage at the ith bus
( 2) Vkk(2′) = − I a(2) Z kk ′ = − I ij
Z ik = Vi (0) − Vk (0) Z kk + Z f
(f) Current flowing between the ith and jth bus (post fault) I ij(f) =
(i) Current I a( 0 ) = I a(1) = I a( 2 ) =
(d) B us voltage for the unfaulted buses (after the event of fault)
(0) Vkk(0′) = − I a(0) Z kk ′ = − I ij
(
(1) (0) ( 2) Z kk ′ Z kk ′ + Z kk ′
Z
(0) kk ′
(1) kk ′
Z
(0) kk ′
+Z
(1) kk ′
+Z
)
( 2) kk ′
( 2) kk ′
Z Z (1) ( 2) + Z kk ′ + Z kk ′
(0) Z k(1k)′ Z kk ′ (0) (1) ( 2) Z kk ′ + Z kk ′ + Z kk ′
28. Economics of power generation (a) E = Rs. (a + b kW + c kWh),
Vi(f) − Vj(f) Z ij
where Z ij is the impedance of line between ith and jth bus. (g) F or synchronous generator, the variation of AC fault current (rms value) with time (t)
( 2) ( 0) ( 0 ) (1) (1) ( 2 ) Z kk ′ Z kk ′ + Z kk Z kk ′ + Z kk ′ Z kk ′
(ii) Bus voltages after fault
(b) Fault current
Vi(f)
(1) (0) ⎡ ( 2) ⎤ Z kk ′ ⎣ Z kk ′ + Z kk ′ ⎦
where, a is the fixed charge for power generation. b is the semi fixed charge and c is the running or variable charge for power generation. (b) E = Rs. (d kW + e kWh)
I (t ) = ( I ′′ − I ′ )e − t / T ′′ + ( I ′ − I ss )e − t / T ′ + I ss
where a fixed sum per kW of maximum demand represented by d and a running charge per kWh of energy represented by e.
QUESTIONS 1. In a thermal power plant the feed water coming to the economizer is heated using (a) H.P. steam (b) L.P. steam (c) direct heat in the furnace (d) flue gases (GATE 2000: 1 Mark)
Ch wise GATE_EE_Ch6.indd 252
2. For given base voltage and base volt-amperes, the per unit impedance value of an element is x. What will be the per unit impedance value of this element when the voltage and volt-ampere bases are both doubled? (a) 0.5x (b) 2x (c) 4x (d) x (GATE 2000: 1 Mark)
11/21/2018 11:50:54 AM
Chapter 6 • Power Systems
3. In an inverse definite minimum time, electromagnetic type over-current relay the minimum time feature is achieved because of (a) saturation of the magnetic circuit (b) proper mechanical design (c) appropriate time delay element (d) electromagnetic damping (GATE 2000: 1 Mark) 4. Out of the considerations (a) to (d) listed below: (i) No distance limitation related to steady state stability (ii) No reactive power requirement from the system at the two terminals (iii) No substantial effect on fault level of the two systems at the terminals inspite of the interconnection (iv) No corona problems The considerations, which constitute advantages of HVDC transmission, are (a) all of the above (b) (i) and (iii) (c) (iii) and (iv) (d) (ii) and (iv) (GATE 2000: 1 Mark) 5. In a 3-step distances protection, the reach of the three zones of the relay at the beginning of the first line typically extends up to (a) 100% of the first line, 50% of the second line and 20% of the third line (b) 80% of the first line, 50% of the second line and 20% of the third line (c) 80% of the first line, 20% of the second line and 10% of the third line (d) 50% of the first line, 50% of second line and 20% of the third line (GATE 2000: 1 Mark) 6. A transmission line has equal voltages at the two ends, maintained constant by two sources. A third source is to be provided to maintain constant voltage (equal to end voltages) at either the midpoint of the line or at 755 of the distance from the sending end. Then the maximum power transfer capabilities of the line in the original case and the other two cases respectively will be in the following ratios. 1 (a) 1:1:1 (b) 1:2: 0.75 (c) 1:2:4
(d) 1:4:16 (GATE 2000: 2 Marks)
7. The plug setting of a negative sequence relay is 0.2 A. The current transformer ratio is 5:1. The minimum value of line-to-line fault current for the operation of the relay is
Ch wise GATE_EE_Ch6.indd 253
(a) 1 A
(b)
1 A 1.732
(c) 1.732 A
(d)
0.2 A 1.732
253
(GATE 2000: 2 Marks) 8. The incremental cost characteristics of two generators delivering 200 MW are as follows dF1 dF = 2.0 + 0.01P1 , 2 = 1.6 + 0.2 P2 dP1 dP2 For economic operation, the generations P1 and P2 should be (a) P1 = P2 = 100 MW (b) P1 = 80 MW, P2 = 120 MW (c) P1 = 200 MW, P2 = 0 MW (d) P1 = 120 MW, P2 = 80 MW (GATE 2000: 2 Marks) 9. The corona loss on a particular system at 50 Hz is 1 kW/km per phase. The corona loss at 60 Hz would be (a) 1 kW/km per phase (b) 0.83 kW/km per phase (c) 1.2 kW/km per phase (d) 1.13 kW/km per phase (GATE 2000: 2 Marks) 10. The severity of line-to-ground and three phase faults at the terminals of an unloaded synchronous generator is to be same. If the terminal voltage is 1.0 p.u. and z1 = z2 = j0.1 p.u., z0 = j0.05 p.u. for the alternator, then the required inductive reactance for neutral grounding is: (a) 0.0166 p.u. (b) 0.05 p.u. (c) 0.1 p.u. (d) 0.15 p.u. (GATE 2000: 2 Marks) 11. In the protection of transformers, harmonic restraint is used to guard against (a) magnetizing inrush current (b) unbalanced operation (c) lightning (d) switching over-voltages (GATE 2001: 1 Mark) 12. A lossless radial transmission line with surge impedance loading (a) takes negative VAR at sending end and zero VAR at receiving end (b) takes positive VAR at sending end and zero VAR at receiving end
11/21/2018 11:50:55 AM
254
GATE EE Chapter-wise Solved Papers
(c) has flat voltage profile and unity power factor at all points along it (d) has sending end voltage higher than receiving end voltage and unity power factor at sending end (GATE 2001: 1 Mark) 13. A 75 NVA, 10 kV synchronous generator has Xd = 0.4 p.u. The Xd value (in p.u.) is a base of 100 MVA, 11 kV is (a) 0.578 (b) 0.279 (c) 0.412 (d) 0.44 (GATE 2001: 2 Marks) 14. A star-connected 440 V, 50 Hz alternators has per phase synchronous reactance of 10 Ω. It supplies a balanced capacitive load current of 20 A, as shown in the per phase equivalent circuit given below. It is desirable to have zero voltage regulation. The load power factor should be j10 W
Eph =
20 A
440 V Ö3 Load
(a) 0.82 (c) 0.39
(b) 0.47 (d) 0.92 (GATE 2001: 2 Marks)
15. A 50 Hz alternator is rated 500 MVA, 20 kV, with X d = 1.0 per unit and X d′′ = 0.2 per unit. It supplies a purely resistive load of 400 MW at 20 kV. The load is connected directly cross the generator terminals when a symmetrical fault-occurs at the load terminals. The initial rms current in the generator in per unit is (a) 7.22 (b) 6.4 (c) 3.22 (d) 2.2 (GATE 2001: 2 Marks) 16. Consider the model shown in the following figure of a transmission line with a series capacitor at its mid-point. The maximum voltage on the line is at the location P1
j0.1 pu
P2 −j0.1 pu
P3
j0.1 pu
P4 I = 1 pu pf = 1 Vr
Ch wise GATE_EE_Ch6.indd 254
(a) P1 (c) P3
(b) P2 (d) P4 (GATE 2001: 2 Marks)
17. A power system has two synchronous generators. The Governor-turbine characteristics corresponding to the generators are P1 = 50(50 – f ), P2 = 100(51 – f ) where f denotes the system frequency in Hz, and P1 and P2 are, respectively, the power outputs (in MW) of turbines 1 and 2. Assuming the generators and transmission network to be lossless, the system frequency for a total load of 400 MW is (a) 47.5 Hz (b) 48.0 Hz (c) 48.5 Hz (d) 49.0 Hz (GATE 2001: 2 Marks) 18. The conductors of a 10 km long, single phase, two wire line are separated by a 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 (GATE 2001: 2 Marks) 19. Consider a long, two-wire line composed of solid round conductors. The radius of both conductors is 0.25 cm and the distance between their centers is 1m. 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 (GATE 2002: 1 Mark) 20. Consider a power system with three identical generators. The transmission losses are negligible. One generator (G1) has a speed governor which maintains its speed constant at the rated value, while the other generators (G2 and G3) have governors with a droop of 5%. If the load of the system is increased, then in steady state. (a) generation of G2 and G3 is increased equally while generation of G1 is unchanged. (b) generation of G1 alone is increased while generation of G2 and G3 is unchanged. (c) generation of G1, G2 and G3 is increased equally. (d) generation of G1, G2 and G3 is increased in the ratio 0.5:0.25:0.25. (GATE 2002: 1 Mark) 21. 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 between
11/21/2018 11:50:55 AM
Chapter 6 • Power Systems
255
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 (GATE 2002: 1 Mark)
voltage of the transmission line is always maintained at 0.98 pu. If voltage at both ends of the line are maintained at 1.0 pu, then the steady state power transfer limit of the transmission line is (a) 9.8 pu (b) 4.9 pu (c) 19.6 pu (d) 5 pu (GATE 2002: 2 Marks)
22. Consider the problem of relay co-ordination for the distance relays R1 and R2 on adjacent lines of a transmission system as shown in the following figure. The Zone 1 and Zone 2 settings for both the relays are indicated on the diagram. Which of the following indicates the correct time setting for the Zone 2 of relays R1 and R2.
25. A generator is connected to a transformer which feeds another transformer through a short feeder (as shown in the following figure). The zero sequence impedance values are expressed in pu on a common base and are indicated in the following figure. The Thevenin equivalent zero sequence impedance at point B is
Zone 2 (R2) Zone 1 (R2)
R1
X0 = 0.03
X0 = 0.1
X0 = 0.05
X0 = 0.07
R2 0.25
0.25
Zone 1 (R1) Zone 2 (R1) (a) TZ2R1 = 0.6 s, TZ2R2 = 0.3 s
(a) 0.8 + j0.6
(b) 0.75 + j0.22
(b) TZ2R1 = 0.3 s, TZ2R2 = 0.6 s
(c) 0.75 + j0.25
(d) 1.5 + j0.25 (GATE 2002: 2 Marks)
(c) TZ2R1 = 0.3 s, TZ2R2 = 0.3 s (d) TZ2R1 = 0.1 s, TZ2R2 = 0.3 s (GATE 2002: 1 Mark) 23. A power system consists of 2 areas (Area 1 and Area 2) connected by a single tie line (as shown in the following figure). It is required to carry out a load flow study on this system. While entering the network data, the tie-line data (connectivity and parameters) is inadvertently left out. If the load flow program is run with this incomplete data Tie line
Area 1
Area 2
(a) The load-flow will converge only if the slack bus is specified in Area 1 (b) The load-flow will converge only if the slack bus is specified in Area 2 (c) The load-flow will converge if the slack bus is specified in either Area 1 or Area 2 (d) The load-flow will not converge if only one slack bus is specified. (GATE 2002: 2 Marks) 24. A transmission line has a total series reactance of 0.2 pu. Reactive power compensation is applied at the midpoint of the line and it is controlled such that the midpoint
Ch wise GATE_EE_Ch6.indd 255
26. Bundled conductors are mainly used in high voltage overhead transmission lines to (a) reduce transmission line losses. (b) increase mechanical strength of the line. (c) reduce corona. (d) reduce sag. (GATE 2003: 1 Mark) 27. A power system consists of 300 buses out of which 20 buses are generator buses, 25 buses are the ones with reactive power support and 15 buses are the ones with fixed shunt capacitors. All the other buses are load buses. It is proposed to perform a load flow analysis for the system using Newton-Raphson method. The size of the Newton-Raphson Jacobian matrix is (a) 553 × 553 (b) 540 × 540 (c) 555 × 555 (d) 554 × 554 (GATE 2003: 1 Mark) 28. Choose two appropriate auxiliary components of a HVDC transmission system from the following: P. DC line inductor Q. AC line inductor R. Reactive power sources S. Distance relays on DC line T. Series capacitance on AC line
11/21/2018 11:50:56 AM
256
GATE EE Chapter-wise Solved Papers
(a) P and Q (c) Q and S
(b) P and R (d) S and T (GATE 2003: 1 Mark)
(a) 240 V (c) 415 V
(b) 480 V (d) 0 V (GATE 2003: 2 Marks)
29. A round rotor generator with internal voltage E1 = 2.0 pu and X = 1.1 pu is connected to a round rotor synchronous motor with internal voltage E2 = 1.3 pu and X = 1.2 pu. The reactance of the line connecting the generator to the motor is 0.5 pu. When the generator supplies 0.5 pu power, the rotor and the difference between the machines will be (a) 57.42° (b) 1° (c) 32.58° (d) 122.58° (GATE 2003: 1 Mark)
33. The ABCD parameters of a three-phase overhead transmission line are A = D = 0.9∠0°, B = 200∠90° Ω and C = 0.95 × 10−3∠90°. At no-load condition, a shunt inductive reactor is connected at the receiving end of the line to limit the receiving-end voltage to be equal to the sending-end voltage. The ohmic value of the reactor is (a) ∞ Ω (b) 2000 Ω (c) 105.26 Ω (d) 1052.6 Ω (GATE 2003: 2 Marks)
30. The interrupting time of a circuit breaker is the period between the instant of (a) initiation of short circuit and the arc extinction on an opening operation. (b) energising of the trip circuit and the arc extinction on an opening operation. (c) initiation of short circuit and the parting of primary arc contacts. (d) energising of the trip circuit and the parting of primary arc contacts. (GATE 2003: 1 Mark)
34. A surge of 20 kV magnitude travels along a lossless cable towards its junction with two identical lossless overhead transmission lines. The inductance and the capacitance of the cable are 0.4 mH and 0.5 mF per km. The inductance and capacitance of the overhead transmission lines are 1.5 mH and 0.015 mF per km. The magnitude of the voltage at the junction due to surge is (a) 36.72 kV (b) 18.36 kV (c) 6.07 kV (d) 33.93 kV (GATE 2003: 2 Marks)
31. A balanced delta connected load of (8 + j 6) Ω per phase is connected to a 400 V, 50 Hz, three-phase supply lines. If the input power factor is to be improved to 0.9 by connecting a bank of star connected capacitors, the required kVAR of the bank is (a) 42.7 (b) 10.2 (c) 28.8 (d) 38.4 (GATE 2003: 2 Marks)
35. A DC distribution system is shown in the following figure with load currents as marked. The two ends of the feeder are fed by voltage sources such that VP − VQ = 3 V. The value of the voltage VP for a minimum voltage of 220 V at any point along the feeder is VP
VQ 0.1 Ω
R 0.15 Ω S
32. The following figure shows a ΔY connected three-phase distribution transformer used to step down the voltage from 11,000 V to 415 V line-to-line. It has two switches S1 and S2. Under normal conditions S1 is closed and S2 is open. Under certain special conditions S1 is open and S2 is closed. In such a case, the magnitude of the voltage across the LV terminals a and c is ∆
HV
Y
LV
A
a
B
b S2 c
C S1
Ch wise GATE_EE_Ch6.indd 256
10 A (a) 225.89 V (c) 220.0 V
0.2 Ω Q
P
20 A
30 A
15 A
(b) 222.89 V (d) 228.58V (GATE 2003: 2 Marks)
36. A three-phase 11 kV generator feeds power to a constant power unity power factor load of 100 MW through a three‑phase transmission line. The line-to-line voltage at the terminals of the machine is maintained constant at 11 kV. The per unit positive sequence impedance of the line based on 100 MVA and 11 kV is j 0.2. The line‑to‑line voltage at the load terminals is measured to be less than 11 kV. The total reactive power to be injected at the terminals of the load to increase the lineto-line voltage at the load terminals to 11 kV is (a) 100 MVAR (b) 10.1 MVAR (c) −100 MVAR (d) −10.1 MVAR (GATE 2003: 2 Marks)
11/21/2018 11:50:56 AM
257
Chapter 6 • Power Systems
37. The bus impedance matrix of a 4-bus power system is given by
Z bus
⎡ j 0.3435 ⎢ j 0.2860 =⎢ ⎢ j 0.2723 ⎢ ⎣ j 0.2277
j 0.2860 j 0.3408 j 0.2586 j 0.2414
j 0.2723 j 0.2586 j 0.2791 j 0.2209
j 0.2277 ⎤ j 0.2414 ⎥⎥ j 0.2209 ⎥ ⎥ j 0.2791 ⎦
A branch having an impedance of j 0.2Ω is connected between bus 2 and the reference. Then the values of Z22,new and Z23,new of the bus impedance matrix of the modified network are, respectively, (a) (b) (c) (d)
j 0.5408 Ω and j 0.4586 Ω j 0.1260 Ω and j 0.0956 Ω j 0.5408 Ω and j 0.0956 Ω j 0.1260 Ω and j 0.1630 Ω
(c) P1 = 300.0 MW; P2 = 300.0 MW; and P3 = 100 MW (d) P1 = 233.3 MW; P2 = 233.3 MW; and P3 = 233.4 MW (GATE 2003: 2 Marks) 40. A list of relays and the power system components protected by the relays are given in Group I and Group II, respectively. Choose the correct match from the codes given below. Group I
Group II
P. Distance relay
1. Transformers
Q. Under frequency relay
2. Turbines
R. Differential relay
3. Busbars
S. Buchholz relay
4. Shunt capacitors 5. Alternators
(GATE 2003: 2 Marks) 38. A 20 MVA, 6.6 kV, three-phase alternator is connected to a three-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j 0.1, j 0.l and j 0.04, respectively. The neutral of the alternator is connected to ground through an inductive reactor of j 0.05 pu. The per unit positive-, negative- and zero-sequence impedances of the transmission line are j 0.1, j 0.1 and j 0.3, respectively. All per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage of the alternator neutral with respect to ground during the fault is (a) 513.8 V (b) 889.9 V (c) 1112 V (d) 642.2 V (GATE 2003: 2 Marks) 39. Incremental fuel costs (in some appropriate unit) for a power plant consisting of three generating units are IC1 = 20 + 0.3P1 IC2 = 30 + 0.4P2 IC3 = 30 where Pi is the power in MW generated by unit i, for i = 1, 2 and 3. Assume that all the three units are operating all the time. Minimum and maximum loads on each unit are 50 MW and 300 MW, respectively. If the plant is operating on economic load dispatch to supply the total power demand of 700 MW, the power generated by each unit is (a) P1 = 242.86 MW; P2 = 157.14 MW; and P3 = 300 MW (b) P1 = 157.14 MW; P2 = 242.86 MW; and P3 = 300 MW
Ch wise GATE_EE_Ch6.indd 257
6. Transmission lines Code: P Q R S (a) 6 5 3 1 (b) 4 3 2 1 (c) 5 2 1 6 (d) 6 4 5 3 (GATE 2003: 2 Marks) 41. A generator delivers power of 1.0 pu to an infinite bus through a purely reactive network. The maximum power that could be delivered by the generator is 2.0 pu A three-phase fault occurs at the terminals of the generator which reduces the generator output to zero. The fault is cleared after tc seconds. The original network is then restored. The maximum swing of the rotor angle is found to be dmax = 110 electrical degree. Then the rotor angle in electrical degrees at t = tc is (a) 55 (b) 70 (c) 69.14 (d) 72.4 (GATE 2003: 2 Marks) 42. A three-phase alternator generating unbalanced voltages is connected to an unbalanced load through a three-phase transmission line as shown in the following figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltages of the alternator are Ea = 10 ∠ 0° V, E b = 10 ∠ −90° V, Ec = 10 ∠ 120° V. The positive-sequence component of the load current is Ea
j1.0 Ω
j1.0 Ω
Eb
j1.0 Ω
j2.0 Ω
Ec
j1.0 Ω
j3.0 Ω
11/21/2018 11:50:56 AM
258
GATE EE Chapter-wise Solved Papers
(a) 1.310 ∠ −107° A (c) 0.996 ∠ −120° A
(b) 0.332 ∠ −120° A (d) 3.510 ∠ −81° A (GATE 2003: 2 Marks)
43. The rated voltage of a three-phase power system is given as (a) rms phase voltage (b) peak phase voltage (c) rms line to line voltage (d) peak line to line voltage (GATE 2004: 1 Mark) 44. The phase sequence of the three-phase system shown in the following figure is R
B (b) RBY (d) YBR (GATE 2004: 1 Mark)
45. In thermal power plants, the pressure in the working fluid cycle is developed by (a) condenser (b) super heater (c) feed water pump (d) turbine (GATE 2004: 1 Mark) 46. For harnessing low variable water heads, the suitable check for coverage in text with high percentage of reaction and runner adjustable vanes is (a) Kaplan (b) Francis (c) Pelton (d) Impeller (GATE 2004: 1 Mark) 47. The transmission line distance protection relay having the property of being inherently directional is (a) impedance relay (b) mho relay (c) ohm relay (d) reactance relay (GATE 2004: 1 Mark) 48. A 800 kV transmission line is having per phase line inductance of 1.1 mH/km and per phase line capacitance of 11.68 nF/km. Ignoring the length of the line, its ideal power transfer capability in MW is (a) 1204 MW (b) 1504 MW (c) 2085 MW (d) 2606 MW (GATE 2004: 2 Marks)
Ch wise GATE_EE_Ch6.indd 258
50. A lightning stroke discharges impulse current of 10 kA (peak) on a 400 kV transmission line having surge impedance of 250 Ω. The magnitude of transient over-voltage travelling waves in either direction assuming equal distribution from the point of lightning strike will be (a) 1250 kV (b) 1650 kV (c) 2500 kV (d) 2900 kV (GATE 2004: 2 Marks) 51. The generalised circuit constants of a three-phase, 220 kV rated voltage, medium length transmission line are
Y
(a) RYB (c) BRY
49. A 110 kV, single core coaxial, XLPE insulated power cable delivering power at 50 Hz, has a capacitance of 125 nF/km. If the dielectric loss tangent of XLPE is 2 × 10−4, then dielectric power loss in this cable in W/km is (a) 5.0 (b) 31.7 (c) 37.8 (d) 189.0 (GATE 2004: 2 Marks)
A = D = 0.936 + j 0.016 = 0.936 ∠0.98° B = 33.5 + j138 = 142.0 ∠76.4°Ω C = ( −5.18 + j 914) × 10 −6 Ω If load at the receiving end is 50 MW at 220 kV with a power factor of 0.9 lagging, then magnitude of line to line sending end voltage should be (a) 133.23 kV (b) 220.00 kV (c) 230.78 kV (d) 246.30 kV (GATE 2004: 2 Marks) 52. A new generator having Eg = 1.4 ∠30° pu [equivalent to (1.212 + j 0.70) pu] and synchronous reactance Xs of 1.0 pu on the system base, is to be connected to a bus having voltage Vt in the existing power system. This existing power system can be represented by Thevenin’s voltage ETh = 0.9∠0° pu in series with Thevenin’s impedance Z Th = 0.25∠90° pu. The magnitude of the bus voltage Vt of the system in pu will be (a) 0.990 V (b) 0.973 V (c) 0.963 V (d) 0.900 V (GATE 2004: 2 Marks) 53. A three-phase generator rated at 110 MVA, 11 kV is connected through circuit breakers to a transformer. The generator is having direct axis sub-transient reactance Xd ″ = 19%, transient reactance Xd ′ = 26% and synchronous reactance = 130%. The generator is operating at no load and rated voltage when a three phase short circuit fault occurs between the breakers and the transformer. The magnitude of initial symmetrical RMS current in the breakers will be
11/21/2018 11:50:57 AM
Chapter 6 • Power Systems
(a) 4.44 kA (c) 30.39 kA
(b) 22.20 kA (d) 38.45 kA (GATE 2004: 2 Marks)
54. A three-phase transmission line supplies Δ-connected load Z. The conductor c of the line develops an open circuit fault as shown in the following figure. The currents in the lines are as shown in the diagram. The positive sequence current component in line a will be Ia = 10–0°A a Z Ib = 10–180°A b Ic = 0 c
Z
Z
5.78∠90°A (a) 5.78∠ −30° A (b) 10.00 ∠ −30° A (c) 6.33∠90°A (d) (GATE 2004: 2 Marks) 55. A 500 MVA, 50 Hz, three-phase turbo-generator produces power at 22 kV. Generator is Y-connected and its neutral is solidly grounded. Its sequence reactances are X1 = X2 = 0.15 and X0 = 0.05 pu. It is operating at rated voltage and disconnected from the rest of the system (no load). The magnitude of the sub-transient line current for single line to ground fault at the generator terminal in pu will be (a) 2.851 (b) 3.333 (c) 6.667 (d) 8.553 (GATE 2004: 2 Marks) 56. A 50 Hz, 4-pole, 500 MVA, 22 kV turbo-generator is delivering rated megavolt-amperes at 0.8 power factor. Suddenly, a fault occurs reducing electric power output by 40%. Neglect losses and assume constant power input to the shaft. The accelerating torque in the generator in MNm at the time of the fault will be (a) 1.528 (b) 1.018 (c) 0.848 (d) 0.509 (GATE 2004: 2 Marks) 57. A hydraulic turbine having rated speed of 250 rpm is connected to a synchronous generator. In order to produce power at 50 Hz, the number of poles required in the generator are (a) 6 (b) 12 (c) 16 (d) 24 (GATE 2004: 2 Marks) 58. The pu parameters for a 500 MVA machine on its own base are inertia, M = 20 pu; reactance, X = 2 pu.
Ch wise GATE_EE_Ch6.indd 259
259
The pu values of inertia and reactance on 100 MVA common base, respectively, are (a) 4, 0.4 (c) 4, 10
(b) 100, 10 (d) 100, 0.4 (GATE 2005: 1 Mark)
59. A 800 kV transmission line has a maximum power transfer capacity of P. If it is operated at 400 kV with the series reactance unchanged, then new maximum power transfer capacity is approximately (a) P (b) 2P (c) P/2 (d) P/4 (GATE 2005: 1 Mark) 60. The insulation strength of an EHV transmission line is mainly governed by (a) load power factor (b) switching over-voltages (c) harmonics (d) corona (GATE 2005: 1 Mark) 61. High voltage DC (HVDC) transmission is mainly used for (a) bulk power transmission over very long distances. (b) inter-connecting two systems with the same nominal frequency. (c) eliminating reactive power requirement in the operation. (d) minimizing harmonics at the converter stations. (GATE 2005: 1 Mark) 62. The parameters of a transposed overhead transmission line are given as: Self reactance, Xs = 0.4 Ω/km Mutual reactance, Xm =0.1 Ω/km The positive sequence reactance X1 and zero sequence reactance X0, respectively, in Ω/km are (a) 0.3, 0.2 (b) 0.5, 0.2 (c) 0.5, 0.6 (d) 0.3, 0.6 (GATE 2005: 2 Marks) 63. At an industrial sub-station with a 4 MW load, a capacitor of 2 MVAR is installed to maintain the load power factor at 0.97 lagging. If the capacitor goes out of service, the load power factor becomes (a) 0.85 lag (b) 1.00 lag (c) 0.80 lag (d) 0.90 lag (GATE 2005: 2 Marks) 64. The network shown in the given figure has impedances in pu as indicated. The diagonal element Y22 of the bus admittance matrix Ybus of the network is
11/21/2018 11:50:57 AM
260
GATE EE Chapter-wise Solved Papers
1
2
+j0.1
−j20.0
3
+j0.1
−j20.0
power limit of the system is 6.25 pu. If one of the double-circuit is tripped, then resulting steady state stability power limit in pu will be
−j10.0
1
2
3 x x
(a) −j 19.8 (b) + j 0.2
(b) + j 20.0 (d) −j 19.95 (GATE 2005: 2 Marks)
65. A load centre is at an equidistant from the two thermal generating stations G1 and G2 as shown in the given figure. The fuel cost characteristics of the generating stations are given by F1= a + bP1+ cP1 2 Rs/hour F2 = a + bP2 + 2cP2 2 Rs/hour, where P1and P2 are the generation in MW of G1 and G2, respectively.
G1
G2 Load
For most economic generation to meet 300 MW of load P1 and P2, respectively, are (a) 150, 150 (b) 100, 200 (c) 200, 100 (d) 175, 125 (GATE 2005: 2 Marks) 66. Two networks are connected in cascade as shown in the following figure. With the usual notations the equivalent A, B, C and D constants are obtained. Given that, C = 0.025 ∠45°, the value of Z2 is Z1 = 10–30°Ω
Z2
(a) 10∠30°Ω (c) 1 Ω
(b) 40∠−45°Ω (d) 0 Ω (GATE 2005: 2 Marks)
67. A generator with constant 1.0 pu terminal voltage supplies power through a step-up transformer of 0.12 pu reactance and a double-circuit line to an infinite bus bar as shown in the following figure. The infinite bus voltage is maintained at 1.0 pu. Neglecting the resistances and susceptances of the system, the steady state stability
Ch wise GATE_EE_Ch6.indd 260
(a) 12.5 pu (c) 10.0 pu
(b) 3.125 pu (d) 5.0 pu (GATE 2005: 2 Marks)
68. 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/s2, then moment of inertia of the systems must be (neglecting viscous and Coulomb friction) (a) 0.25 kgm2 (b) 0.25 Nm2 2 (c) 4 kgm (d) 4 Nm2 (GATE 2005: 2 Marks) Common Data for Questions 69 and 70: At a 220 kV substation of a power system, it is given that the three-phase fault level is 4000 MVA and single-line to ground fault level is 5000 MVA. Neglecting the resistance and the shunt susceptances of the system, 69. The positive sequence driving point reactance at the bus is (a) 2.5 Ω (b) 4.033 Ω (c) 5.5 Ω (d) 12.1 Ω (GATE 2005: 2 Marks) 70. The zero sequence driving point reactance at the bus is (a) 2.2 Ω (b) 4.84 Ω (c) 18.18 Ω (d) 22.72 Ω (GATE 2005: 2 Marks) 71. The concept of an electrically short, medium, and long line is primarily based on the (a) nominal voltage of the line (b) physical length of the line (c) wavelength of the line (d) power transmitted over the line (GATE 2006: 1 Mark) 72. Keeping in view the cost and overall effectiveness, the following circuit breaker is best suited for capacitor bank switching (a) vacuum (b) air blast (c) SF6 (d) oil (GATE 2006: 1 Mark) 73. In a biased differential relay, the bias is defined as a ratio of (a) number of turns of restraining and operating coil (b) operating coil current and restraining coil current
11/21/2018 11:50:58 AM
Chapter 6 • Power Systems
(c) fault current and operating coil current (d) fault current and restraining coil current (GATE 2006: 1 Mark) 74. An HVDC link consists of rectifier, inverter transmission line and other equipment. Which one of the following is true for this link? (a) The transmission line produces/supplies reactive power. (b) The rectifier consumes reactive power and the inverter supplies reactive power from/to the respective connected AC systems. (c) Rectifier supplies reactive power and the inverter consumes reactive power to/from the respective connected AC systems. (d) Both the converters (rectifier and inverter) consume reactive power from the respective connected AC systems. (GATE 2006: 1 Mark) 75. A 400 V, 50 Hz, three-phase balanced source supplies power to a star-connected load whose rating is 12 3 kVA , 0.8 pf (lag). The rating (in kVAR) of the delta-connected (capacitive) reactive power bank necessary to bring the pf to unity is (a) 28.78 (c) 16.60
(b) 21.60 (d) 12.47 (GATE 2006: 2 Marks)
76. The A, B, C and D constants of a 220 kV line are: A = D = 0.94 ∠1°; B = 130 ∠73°; C = 0.001 ∠ 90° If sending end voltage of the line for a given load delivered at nominal voltage is 240 kV, then % voltage regulation of the line is (a) 5 (b) 9 (c) 16 (d) 21 (GATE 2006: 2 Marks) 77. A single-phase transmission line and a telephone line are both symmetrically strung one below the other, in horizontal configurations, on a common tower. The shortest and longest distances between the phase and telephone conductors are 2.5 m and 3 m, respectively. The voltage (volt/km) induced in the telephone circuit, due to 50 Hz current of 100 A in the power circuit is (a) 4.81 (b) 3.56 (c) 2.29 (d) 1.27 (GATE 2006: 2 Marks) 78. Three identical star connected resistors of 1.0 pu are connected to an unbalanced three-phase supply. The load neutral is isolated. The symmetrical components of the line voltages in pu are:
Ch wise GATE_EE_Ch6.indd 261
261
Vab1 = X∠q1, Vab2 = Y∠q2 If all the pu calculations are with the respective base values, the phase to neutral sequence voltages are: (a) Vanl = X∠(q1 + 30o); Van2 =Y∠(q2 − 30o) (b) Vanl = X∠(q1 − 30o); Van2 =Y∠(q2 + 30o) (c) Van1= (d) Van1=
1 3 1 3
X ∠ (θ 2 − 30°); Van2 =
1
X ∠ (θ1 − 60°); Van2 =
1
3 3
Y ∠ (θ 2 − 30°) Y ∠ (θ 2 − 60°)
(GATE 2006: 2 Marks) 79. A generator is connected through a 20 MVA, 13.8/138 kV step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 10 MVA, 138/69 kV rating. A 0.72 pu load, evaluated on load side transformer ratings as base values, is supplied from the above system. For system base values of 10 MVA and 69 kV in load circuit, the value of the load (in per unit) in generator circuit will be (a) 36 (b) 1.44 (c) 0.72 (d) 0.18 (GATE 2006: 2 Marks) 80. The Gauss Seidel load flow method has the following disadvantages. Tick the incorrect statement. (a) Unreliable convergence (b) Slow convergence (c) Choice of slack bus affects convergence (d) A good initial guess for voltages is essential for convergence (GATE 2006: 2 Marks) Common Data for Questions 81 and 82: A generator feeds power to an infinite bus through a double circuit transmission line. A three phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of 5 MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of 30°. The system frequency is 50 Hz. 81. The initial accelerating power (in pu) will be (a) 1.0 (b) 0.6 (c) 0.56 (d) 0.4 (GATE 2006: 2 Marks) 82. If the initial accelerating power is X pu, the initial acceleration in elect deg/s2, and the inertia constant in MJ-s/elect degree respectively, will be (a) 31.4 X, 18 (b) 1800 X, 0.056 (c) X/1800, 0.056 (d) X/31.4, 18 (GATE 2006: 2 Marks)
11/21/2018 11:50:58 AM
262
GATE EE Chapter-wise Solved Papers
Linked Answer Questions 83 and 84: For a power system the admittance and impedance matrices for the fault studies are as follows.
Ybus
Z bus
⎡ j 0.16 ⎢ = ⎢ j 0.08 ⎢⎣ j 0.12
(GATE 2007: 1 Mark) 86. Two regional systems, each having several synchronous generators and loads are interconnected by an AC line and a HVDC link as shown in the following figure. Which of the following statements is true in the steady state?
j 0.12 ⎤ j 0.16 ⎥⎥ j 0.34 ⎥⎦
j 0.08 j 0.24 j 0.16
Generator A: 400 MW, Generator B: 300 MW Generator A: 350 MW, Generator B: 350 MW Generator A: 450 MW, Generator B: 250 MW Generator A: 425 MW, Generator B: 275 MW
PDC
The pre-fault voltages are 1.0 pu at all the buses. The system was unloaded prior to the fault. A solid three phase fault takes place at bus 2. 83. The post fault voltages at buses 1 and 3 in per unit, respectively, are (a) 0.24, 0.63 (b) 0.31, 0.76 (c) 0.33, 0.67 (d) 0.67, 0.33 (GATE 2006: 2 Marks) 84. The per unit fault feeds from generators connected to buses 1 and 2, respectively, are (a) 1.20, 2.51 (b) 1.55, 2.61 (c) 1.66, 2.50 (d) 5.00, 2.50 (GATE 2006: 2 Marks) 85. The incremental cost curves in Rs/ MW hour for two generators supplying a common load of 700 MW are shown in the following figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is:
HVDC line Region 1
Region 2 AC line PAC
(a) Both regions need not have the same frequency. (b) The total power flow between the regions (PAC + PDC) can be changed by controlling the HVDC converters alone. (c) The power sharing between the AC line and the HVDC link can be changed by controlling the HVDC converters alone. (d) The direction of power flow in the HVDC link (PDC) cannot be reversed. (GATE 2007: 1 Mark)
87. Consider a bundled conductor of an overhead line, conGenerator B sisting of three identical sub-conductors placed at the corners of an equilateral triangle as shown in the figure. 800 If we neglect the charges on the other phase conductors 650 and ground, and assume that spacing between sub-conductors is much larger than their radius, the maximum electric field intensity is experienced at
Generator A
600 450
YP
P 200 MW
450 MW
150 MW
400 MW
Incremental cost Rs/MWh X
Generator A
0 MW
j 2.50 ⎤ ⎡ − j8.75 j1.25 ⎢ = ⎢ − j1.25 − 6.25 j 2.50 ⎥⎥ ⎢⎣ − j 2.50 − j 2.50 − j 5.00 ⎥⎦
(a) (b) (c) (d)
Generator B Z W 800 650
(a) Point X (c) Point Z P 450 MW
P 150 MW
400 MW
(b) Point Y (d) Point W (GATE 2007: 1 Mark)
mental cost Rs/MWh
Ch wise GATE_EE_Ch6.indd 262
11/21/2018 11:50:58 AM
Chapter 6 • Power Systems
263
88. The following figure shows a three phase self- 91. A three-phase balanced star connected voltage source commutated voltage source converter connected to with frequency w rad/s is connected to a star connected a power system. The converter’s DC bus capacitor is balanced load which is purely inductive. The instantamarked as C in the figure. The circuit is initially operneous line currents and phase to neutral voltages are ating in steady state with d = 0 and the capacitor DC denoted by (ia, ib, ic) and (van, vbn, vcn) respectively and voltage is equal to VDC0. You may neg1ect all losses and their RMS values are denoted by V and I. If harmonics. What action should be taken to increase the ⎡ 1 1 ⎤ capacitor DC voltage slowly to a new steady state value? − ⎢0 ⎥ 3 3⎥ ⎢ ⎡ia ⎤ ⎢ 1 Three phase 1 ⎥⎢ ⎥ R = v v v − 0 [ an bn cn ] ⎢ ⎥ ib C voltage source 3 ⎥⎢ ⎥ ⎢ 3 ⎢i ⎥ converter ⎢ 1 ⎥⎣ c⎦ 1 E–d E–0 − 0 ⎢ ⎥ 3 ⎣ 3 ⎦ (a) Make d positive and maintain it at a positive value (b) Make d positive and return it to its original value then the magnitude of R is (a) 3VI (b) VI (c) Make d negative and maintain it at a negative value (c) 0.7 VI (d) 0 (d) Make d negative and return it to its original value (GATE 2007: 2 Marks) (GATE 2007: 2 Marks) 89. The total reactance and total susceptance of a lossless overhead EHV line, operating at 50 Hz, are given by 0.045 pu and 1.2 pu, respectively. If the velocity of wave propagation is 3 × 105 km/s, then approximate length of the line is (a) 122 km (b) 185 km (c) 222 km (d) 272 km (GATE 2007: 2 Marks) 90. Consider the protection system shown in the following figure. The circuit breakers, numbered from 1 to 7 are of identical type. A single line to ground fault with zero fault impedance occurs at the midpoint of the line (at point F), but circuit breaker 4 fails to operate (“stuck breaker”). If the relays are coordinated correctly, a valid sequence of circuit breaker operations is 1
92. Consider a synchronous generator connected to an infinite bus by two identical parallel transmission lines. The transient reactance x of the generator is 0.1 pu and the mechanical power input to it is constant at 1.0 pu. Due to some previous disturbance, the rotor angle (d ) is undergoing an undamped oscillation, with the maximum value of d (t) equal to 130°. One of the parallel lines trips due to relay mal-operation at an instant when d (t) = 130° as shown in the following figure. The maximum value of the per unit line reactance, x, such that the system does not lose synchronism subsequent to this tripping is 1.0–0 x x¢ = 0.1 pu x
3 Transmission line
1.0–d Bus C
2
4
δ
6 Stuck breaker F
Bus A
One line trips
130° 7
5 Transmission line Bus B
(a) 1, 2, 6, 7, 3, 5 (c) 5, 6, 7, 3, 1, 2
t (b) 1, 2, 5, 6, 7, 3 (d) 5, 1, 2, 3, 6, 7 (GATE 2007: 2 Marks)
Ch wise GATE_EE_Ch6.indd 263
(a) 0.87 (c) 0.67
(b) 0.74 (d) 0.54 (GATE 2007: 2 Marks)
11/21/2018 11:50:59 AM
264
GATE EE Chapter-wise Solved Papers
93. Suppose we define a sequence transformation between “a-b-c’’ and “p-n-o’’ variables as follows: 1 1⎤ ⎡ f p ⎤ ⎡ fa ⎤ ⎡1 ⎢ f ⎥ = k ⎢α 2 α 1⎥ ⎢ f ⎥ ⎢ b⎥ ⎢ ⎥⎢ n⎥ ⎢⎣ f c ⎥⎦ ⎢⎣α α 2 1⎥⎦ ⎢⎣ f o ⎥⎦ where, α = e
j
2π 3
1.02–0° 1.02–-20°
1.0–-5° 1.02– 0°
X
Y
1.02–10°
1.02–15°
and k = constant.
Now, if it is given that: (a)
⎡Vp ⎤ ⎡0.5 0 0 ⎤ ⎡ip ⎤ ⎢ ⎥ ⎢ ⎢ ⎥ 0.5 0 ⎥⎥ ⎢in ⎥ and ⎢Vn ⎥ = ⎢0 ⎢V ⎥ ⎢⎣0 0 2.0 ⎥⎦ ⎢⎣io ⎥⎦ ⎣ o⎦
P1, Q1 P2, Q2
⎡Va ⎤ ⎡ia ⎤ ⎢V ⎥ = Z ⎢i ⎥ ⎢ b⎥ ⎢ b⎥ ⎢⎣Vc ⎥⎦ ⎢⎣ic ⎥⎦ X
then 0.5 0.75⎤ ⎡1.0 ⎢ 0.5 ⎥⎥ (a) Z = ⎢0.75 1.0 ⎢⎣ 0.5 0.75 1.0 ⎥⎦
(b)
⎡1.0 0.5 0.5⎤ ⎢ ⎥ (b) Z = ⎢0.5 1.0 0.5⎥ ⎢⎣ 0.5 0.5 1.0 ⎥⎦ 0.75 0.5 ⎤ ⎡1.0 ⎢ 0.75⎥⎥ (c) Z = 3k ⎢0.5 1.0 ⎢⎣ 0.75 0.5 1.0 ⎥⎦ 2
⎡ 1.0 −0.5 −0.5⎤ k2 ⎢ 1.0 −0.5⎥⎥ (d) Z = 3 ⎢ −0.5 ⎢⎣ −0.5 −0.5 1.0 ⎥⎦ (GATE 2007: 2 Marks) 94. Consider the two power systems shown in the following Figure (a), which are initially not interconnected, and are operating in steady state at the same frequency. Separate load flow solutions are computed individually for the two systems, corresponding to this scenario. The bus voltage phasors so obtained are indicated on Figure (a). These two isolated systems are now interconnected by a short transmission line as shown in Figure (b), and it is found that P1 = P2 = Q1 = Q2 = 0
Ch wise GATE_EE_Ch6.indd 264
Y
The bus voltage phase angular difference between generator bus X and generator bus Y after the interconnection is (a) 10° (b) 25° (c) −30° (d) 30° (GATE 2007: 2 Marks) 95. A 230 V (phase), 50 Hz, three-phase, 4-wire system has a phase sequence ABC. A unity power-factor load of 4 kW is connected between phase A and neutral N. It is desired to achieve zero neutral current through the use of a pure inductor and a pure capacitor in the other two phases. The value of inductor and capacitor is (a) 72.95 mH in phase C and 139.02 mF in phase B (b) 72.95 mH in phase B and 139.02 mF in phase C (c) 42.12 mH in phase C and 240.79 mF in phase B (d) 42.12 mH in phase B and 240.79 mF in phase C (GATE 2007: 2 Marks) 96. An isolated 50 Hz synchronous generator is rated at 15 MW which is also the maximum continuous power limit of its prime mover. It is equipped with a speed governor with 5% droop. Initially, the generator is feeding three loads of 4 MW each at 50 Hz. One of these loads is programmed to trip permanently if the frequency falls
11/21/2018 11:51:00 AM
Chapter 6 • Power Systems
below 48 Hz. If an additional load of 3.5 MW is connected, then frequency will settle down to (a) 49.417 Hz (b) 49.917 Hz (c) 50.083 Hz (d) 50.583 Hz (GATE 2007: 2 Marks) 97. A two machine power system is shown in the following figure. Transmission line XY has positive sequence impedance of Z1 Ω and zero sequence impedance of Z0 Ω. X
⎡ ΔVa ⎤ ⎡ Zs ⎢ ΔV ⎥ = ⎢ Z ⎢ b⎥ ⎢ m ⎢⎣ ΔVc ⎥⎦ ⎢⎣ Z m
Zm Zs Zm
265
Zm ⎤ ⎡ I a ⎤ Z m ⎥⎥ ⎢⎢ I b ⎥⎥ Zs ⎥⎦ ⎢⎣ I c ⎥⎦
Shunt capacitance of the line can be neglected. If the line has positive sequence impedance of 15 Ω and zero sequence impedance of 48 Ω, then the values of Zs and Zm will be (a) Zs = 31.5 Ω; Zm = 16.5 Ω (b) Zs = 26 Ω; Zm = 11 Ω (c) Zs = 16.5 Ω; Zm = 31.5 Ω (d) Zs = 11 Ω; Zm = 26 Ω
Y
(GATE 2008: 1 Mark) F
(GATE 2008: 1 Mark)
100. Voltage phasors at the two terminals of a transmission line of length 70 km have a magnitude of 1.0 per unit but are 180° out of phase. Assuming that the maximum load current in the line is 1/5 th of minimum three-phase fault current, which one of the following transmission line protection schemes will NOT pick up for this condition? (a) Distance protection using mho relays with Zone-1 set to 80% of the line impedance (b) Directional over current protection set to pick up at 1.25 times the maximum load current (c) Pilot relaying system with directional comparison scheme (d) Pilot relaying system with segregated phase comparison scheme (GATE 2008: 2 Marks)
98. An extra high voltage transmission line of length 300 km can be approximated by a lossless line having propagation constant. Then percentage ratio of line length to wavelength will be given by (a) 24.24% (b) 12.12% (c) 19.05% (d) 6.06% (GATE 2008: 1 Mark)
101. A lossless transmission line having surge impedance loading (SIL) of 2280 MW is provided with a uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated transmission line will be (a) 1835 MW (b) 2280 MW (c) 2725 MW (d) 3257 MW (GATE 2008: 2 Marks)
99. A 3-phase transmission line is shown in the following figure.
102. A lossless power system has to serve a load of 250 MW. There are two generators (G1 and G2) in the system with cost curves C1 and C2, respectively, defined as follows:
An ‘a’ phase to ground fault with zero fault impedance occurs at the centre of the transmission line. Bus voltage at X and line current from X to F for the phase ‘a’, are given by Va Volts and Ia Amperes, respectively. Then, the impedance measured by the ground distance relay located at the terminal X of line XY will be given by (a)
Z1 Z0 Ω (b) Ω 2 2
(c)
Va ( Z 0 + Z1 ) Ω Ω (d) Ia 2
∆ Va
C1 ( PG1 ) = PG1 + 0.055 × P G2 1 Ia
∆ Vb Ib ∆ Vc Ic Voltage drop across the transmission line is given by the following equation:
Ch wise GATE_EE_Ch6.indd 265
C2 ( PG 2 ) = 3PG 2 + 0.03 × P G2 2 where PG1 and PG2 are the MW injections from generators G1 and G2, respectively. Then the minimum cost dispatch will be (a) PG1 = 250 MW; PG2 = 0 MW (b) PG1 = 150 MW; PG2 = 100 MW (c) PG1 = 100 MW; PG2 = 150 MW (d) PG1 = 0 MW; PG2 = 250 MW (GATE 2008: 2 Marks)
11/21/2018 11:51:01 AM
266
GATE EE Chapter-wise Solved Papers
103. A lossless single machine infinite bus power system is shown in the following figure 1.0–d pu
1.0–0 pu 1.0 pu
The synchronous generator transfers 1.0 per unit of power to the infinite bus. The critical clearing time of circuit breaker is 0.28 s. If another identical synchronous generator is connected in parallel to the existing generator and each generator is scheduled to supply 0.5 per unit of power. Then the critical clearing time of the circuit breaker will (a) reduce to 0.14 s (b) reduce but will be more than 0.14 s (c) remain constant at 0.28 s (d) increase beyond 0.28 s (GATE 2008: 2 Marks) 104. Single line diagram of a 4-bus single source distribution system is shown in the following figure. Branches e1, e2, e3 and e4 have equal impedances. The load current values indicated in the figure are in per unit.
e1
e2
e3
e4
2+j0 Distribution company’s policy requires radial system operation with minimum loss. This can be achieved by opening of the branch (a) e1 (b) e2 (c) e3 (d) e4 (GATE 2008: 2 Marks) Common Data for Questions 105 and 106: Consider a power system shown in the following figure. X
Y
Vs1
IX
Ch wise GATE_EE_Ch6.indd 266
Vs2
ZL
Zs1
F
Zs2 If
A symmetrical three phase fault occurs at centre of the line, that is, point F at time t0. The positive sequence impedance from source S1 to point F equals 0.004 + j 0.04 pu. The waveform corresponding to phase a fault current from bus X reveals that decaying DC offset current is negative and in magnitude at its maximum initial value. Assume that the negative sequence impedances are equal to positive sequence impedances, and the zero sequence impedances are three times positive sequence impedances. 105. The instant (t0) of the fault will be (a) 4.682 ms (b) 9.667 ms (c) 14.667 ms (d) 19.667 ms (GATE 2008: 2 Marks) 106. The rms value of the AC component of fault current (IX) will be (a) 3.59 kA (b) 5.07 kA (c) 7.18 kA (d) 10.15 kA (GATE 2008: 2 Marks)
5+j0
1+j0
Given that: Vs1 = Vs2 = 1.0 + j 0.0 pu The positive sequence impedance are: Zs1 = Za2 = 0.001 + j 0.01 pu and ZL = 0.006 + j 0.06 pu Three-phase base MVA = 100 Voltage base = 400 kV (Line to line) Nominal system frequency = 50 Hz The reference voltage for phase a is defined as v(t) = Vmcos(w t).
107. Instead of the three phase fault, if a single line to ground fault occurs on phase a at point F with zero fault impedance, then the rms value of the AC component of fault current (IX ) for phase a will be (a) 4.97 pu (b) 7.0 pu (c) 14.93 pu (d) 29.85 pu (GATE 2008: 2 Marks) 108. Out of the following plant categories (i) Nuclear (ii) Run-of-river (iii) Pump Storage (iv) Diesel the base load power plants are (a) (i) and (ii) (b) (ii) and (iii) (c) (i), (ii) and (iv) (d) (i), (ii) and (iv) (GATE 2009: 1 Mark) 109. For a fixed value of complex power flow in a transmission line having a sending end voltage V, the real power loss will be proportional to
11/21/2018 11:51:01 AM
267
Chapter 6 • Power Systems
(a) V (c)
Codes: A B C (a) 2 3 1 (b) 3 2 1 (c) 1 2 3 (d) 1 3 2
(b) V 2
1 1 (d) V V2 (GATE 2009: 1 Mark)
110. For the Y-bus matrix of a 4-bus system given in per unit, the buses having shunt elements are
Ybus
2 2.5 0 ⎤ ⎡ −5 ⎢ 2 −10 2.5 4 ⎥ ⎥ = j⎢ ⎢ 2.5 2.5 −9 4 ⎥ ⎢ ⎥ 4 4 −8⎦ ⎣ 0
(a) 3 and 4 (c) 1 and 2
(GATE 2009: 2 Marks) 113. Three generators are feeding a load of 100 MW. The details of the generators are given below. Rating (MW)
(b) 2 and 3 (d) 1, 2 and 4 (GATE 2009: 2 Marks)
111. Match the items in List I with the items in List II and select the correct answer using the codes given below the lists. List I (To)
1. Shunt reactor
B. Reduce the current ripples
2. Shunt capacitor
C. Increase the power
3. Series capacitor flow in line
D. Reduce the Ferranti effect
4. Series reactor
Codes: A B C D (a) 2 3 4 1 (b) 2 4 3 1 (c) 4 3 1 2 (d) 4 1 3 2 (GATE 2009: 2 Marks) 112. Match the items in List I with the items in List II and select the correct answer using the codes given below the lists.
Generator-1
100
20
0.02
Generator-2
100
30
0.04
Generator-3
100
40
0.03
114. A 500 MW, 21 kV, 50 Hz, three-phase, 2-pole synchronous generator having a rated pf = 0.9 has a moment of inertia of 27.5 × 103 kgm2. The inertia constant (H) will be (a) 2.44 s (b) 2.71 s (c) 4.88 s (d) 5.42 s (GATE 2009: 2 Marks) 115. Power is transferred from system A to system B by an HVDC link as shown in the figure given below. If the voltages VAB and VCD are as indicated in the figure, and I > 0, then AC system
Power flow A
A
C VAB
i
AC system B
VCD
List I (Type of transmission line)
List II (Type of distance relay preferred)
A. Short line
1. Ohm relay
B. Medium line
2. Reactance relay
(c) VAB > 0, VCD > 0, VAB > VCD
C. Long
3. Mho relay
(d) VAB > 0, VCD < 0
Ch wise GATE_EE_Ch6.indd 267
Regulation (pu) on 100 MVA base
In the event of increased load power demand, which of the following will happen? (a) All the generators will share equal power (b) Generator-3 will share more power compared to Generator-1 (c) Generator-1 will share more power compared to Generator-2 (d) Generator-2 will share more power compared to Generator-3 (GATE 2009: 2 Marks)
List II (Use)
A. Improve power factor
Efficiency (%)
B
D
Rectifier
Inverter
(a) VAB < 0, VCD < 0, VAB > VCD (b) VAB > 0, VCD > 0, VAB < VCD
(GATE 2010: 1 Mark)
11/21/2018 11:51:01 AM
268
GATE EE Chapter-wise Solved Papers
116. A balanced three-phase voltage is applied to a star-connected induction motor, the phase to neutral voltage being V. The stator resistance, rotor resistance referred to the stator, stator leakage reactance, rotor leakage reactance referred to the stator, and the magnetising reactance are denoted by rs, rr, xs, xr and Xm, respectively. The magnitude of the starting current of the motor is given by (a)
(b)
(c)
(d)
120. Consider a stator winding of an alternator with an internal high-resistance ground fault. The currents under the fault condition are as shown in the following figure. The winding is protected using a differential current scheme with current transformers of ratio 400/5 A as shown in the following figure. The current through the operating coil is CT ratio 400/5
V
CT ratio 400/5
(200 + j0)A
(250 + j0)A
( rs + rr ) + ( xs + xr ) 2 2
V ( rs + ( xs + X m )) 2 2
Operating coil
V ( rs + rr ) + ( X m + xr ) 2
(a) 0.1875 A (c) 0.375 A
2
V
(b) 0.2 A (d) 60 kA (GATE 2010: 1 Mark)
121. The zero-sequence circuit of the three phase transformer shown in the given figure is
( rs + ( X m + xr )) 2 2
(GATE 2010: 1 Mark)
R r
117. Consider a step voltage wave of magnitude 1 pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant the travelling wave reaches the reactor is Y B
A
Reactor
(a) −l pu (c) 2 pu
(b) 1 pu (d) 3 pu (GATE 2010: 1 Mark)
118. Consider two buses connected by an impedance of (0 +j 5) Ω. The bus 1 voltage is 100 ∠30° V, and bus 2 voltage is 100 ∠0° V. The real and reactive power supplied by bus 1, respectively, are (a) 1000 W, 268 VAR (b) −1000 W, −134 VAR (c) 276.9 W, −56.7 VAR (d) −276.9 W, 56.7 VAR (GATE 2010: 1 Mark) 119. A three-phase, 33 kV oil circuit breaker is rated 1200 A, 2000 MVA, 3s. The symmetrical breaking current is (a) 1200 A (b) 3600 A (c) 35 kA (d) 104.8 kA (GATE 2010: 1 Mark)
Ch wise GATE_EE_Ch6.indd 268
b
y
(a) R
r
G (b) R
r
G (c) R
r
G (d) R
r
G (GATE 2010: 1 Mark) 122. A 50 Hz synchronous generator is initially connected to a long lossless transmission line which is open circuited at the receiving end. With the field voltage held constant,
11/21/2018 11:51:02 AM
269
Chapter 6 • Power Systems
the generator is disconnected from the transmission line. Which of the following may be said about the steady state terminal voltage and field current of the generator? Long transmission line Receiving end (a) The magnitude of terminal voltage decreases, and the field current does not change. (b) The magnitude of terminal voltage increases, and the field current does not change. (c) The magnitude of terminal voltage increases, and the field current increases. (d) The magnitude of terminal voltage does not change, and the field current decreases. (GATE 2010: 2 Marks)
(a) 2.0 A (c) 2.7 A
(b) 2.4 A (d) 3.5 A (GATE 2010: 2 Marks)
125. For the power system shown in the following figure, specifications of the components are following: G1: 25 kV, 100 MVA, X = 9% G2: 25 kV, 100 MVA, X = 9% T1: 25 kV/ 220 kV, 90 MVA X = 12% T2: 25 kV/25 kV, 90 MVA, X = 12% Line 1:220 kV, X = 150 ohms T1
T2 Line 1 Bus 1
123. Consider a three-phase, 50 Hz, 11 kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self- capacitance of the insulator is five times the shunt capacitance between the link and the ground, as shown in the following figure. The voltage across the two insulators are
C
G2
Choose 25 kV as the base voltage at the generator G1, and 200 MVA as the MVA base. The impedance diagram is (a)
j0.27
j0.42
j0.27
j0.18
j0.18
5C e2 5C e1
G1
G2
(b)
Conductor (a) (b) (c) (d)
Bus 2
G1
e1 = 3.74 kV, e2 = 2.61 kV e1 = 3.46 kV, e2 = 2.89 kV e1 = 6.0 kV, e2 = 4.23 kV e1 = 5.5 kV, e2 = 5.5 kV
j0.27
j0.62
j0.27
j0.18
j0.18
G1
G2
(GATE 2010: 2 Marks) 124. Consider a three-core, three-phase, 50 Hz, 11 kV cable whose conductors are denoted as R, Y and B in the given figure. The inter-phase capacitance (C1) between each pair of conductors is 0.2 μF and the capacitance (C2) between each line conductor and the sheath is 0.4 μF. The per-phase charging current is
(c)
j0.27
j0.42
j0.27
j0.21
j0.21
G1
G2
C2 (d) C1
R
B
j0.3
j0.42
j0.3
C1 j0.21
j0.21
G1
G2
Y C1
C2
C2 Outer sheath
Ch wise GATE_EE_Ch6.indd 269
(GATE 2010: 2 Marks)
11/21/2018 11:51:03 AM
270
GATE EE Chapter-wise Solved Papers
126. A nuclear power station of 500 MW capacity is located at 300 km away from a load centre. Select the most suitable power evacuation transmission configuration among the following options: (a)
132 kV, 300 km double circuit Load centre
(b) Load centre 132 kV, 300 km single circuit with 40% series capacitor compensation
(a) P1 = 80 MW + losses P2 = 20 MW P3 = 20 MW (b) P1 = 60 MW P2 = 30 MW + losses P3 = 30 MW (c) P1 = 40 MW P2 = 40 MW P3 = 40 MW + losses (d) P1 = 30 MW + losses P2 = 45 MW P3 = 45 MW (GATE 2011: 2 Marks) 130. A three-bus network is shown in the following figure indicating the pu impedances of each element.
(c) Load centre
2
1
3
400 kV, 300 km single circuit −j0.08
j0.2
(d)
j0.1
400 kV, 300 km double circuit
j0.1
Load centre (GATE 2011: 1 Mark) 127. A negative sequence relay is commonly used to protect (a) an alternator (b) a transformer (c) a transmission line (d) a bus bar (GATE 2011: 1 Mark) 128. For enhancing the power transmission in a long EHV transmission line, the most preferred method is to connect a (a) series inductive compensator in the line. (b) shunt inductive compensator at the receiving end. (c) series capacitive compensator in the line. (d) shunt capacitive compensator at the sending end. (GATE 2011: 1 Mark) 129. A load centre of 120 MW derives power from two power stations connected by 220 kV transmission lines of 25 km and 75 km as shown in the following figure. The three generators G1, G2 and G3 are to 100 MW capacity each and have identical fuel cost characteristics. The minimum loss generation schedule for supplying the 120 MW load is
The bus admittance matrix, Y-bus, of the network is (a)
0⎤ ⎡ 0.3 −0.2 ⎢ j ⎢ −0.2 0.12 0.08⎥⎥ ⎢⎣ 0 0.08 0.02⎥⎦
(b)
5 0⎤ ⎡ −15 j ⎢⎢ 5 7.5 −12.5⎥⎥ ⎢⎣ 0 −12.5 2.5⎥⎦
(c)
0⎤ ⎡ 0.1 0.2 ⎢ j ⎢0.2 0.12 −0.08⎥⎥ ⎢⎣ 0 −0.08 0.10 ⎥⎦
(d)
5 0⎤ ⎡ −10 ⎢ j ⎢ 5 7.5 12.5⎥⎥ ⎢⎣ 0 12.5 −10 ⎥⎦ (GATE 2011: 2 Marks)
Common Data for Questions 131 and 132: Two generator units G1 and G2 are connected by 15 kV line with a bus at the midpoint as shown below.
G2 25 km
G2
75 km
25 km
G1
75 km
G1 G3 120 MW
Ch wise GATE_EE_Ch6.indd 270
G3 120 MW
11/21/2018 11:51:05 AM
271
Chapter 6 • Power Systems
G1 = 250 MVA, 15 kV, positive sequence reactance X = 25% on its own base G2 = 100 MVA, 15 kV, positive sequence reactance X = 10% on its own base L1 and L2 = 10 km, positive sequence reactance X = 0.225 Ω/km
133. The bus admittance matrix of a three-bus three-line system is
131. For the above system, the positive sequence diagram with the pu values on the 100 MVA common base is
If each transmission line between the two buses is represented by an equivalent p-network, the magnitude of the shunt susceptance of the line connecting bus 1 and 2 is (a) 4 (b) 2 (c) 1 (d) 0 (GATE 2012: 1 Mark)
(a)
3
1 j1.0
2 j1.0 j0.10
j0.10
5⎤ ⎡ −13 10 Y = j ⎢⎢ 10 −18 10 ⎥⎥ ⎢⎣ 5 10 −13⎥⎦
134. A two-phase load draws the following phase currents: i1 (t ) = I m sin(ω t − ϕ 1 ), i2 (t ) = I m cos(ω t − ϕ 2)
(b)
3
1
These currents are balanced if f1 is equal to (a) −f2 (b) f2
2
j1.0
(c)
j1.0 j0.25
j0.25
⎛π ⎞ ⎜⎝ − ϕ 2⎟⎠ 2
(d)
⎛π ⎞ ⎜⎝ + ϕ 2⎟⎠ 2
(GATE 2012: 1 Mark)
135. The figure shows a two-generator system supplying a load of PD = 40 MW, connected at bus 2. Bus 1 (c)
1
3
Bus 2
2 G2 G1
j2.25
PG2
j2.25
j0.10
j0.10 PG1
PD = 40 MW (d)
3
1 j2.25 j0.25
The fuel cost of generators G1 and G2 are C1(PG1) = 10000 Rs/MWh and C2(PG2) = 12500 Rs/MWh
2 j2.25 j0.10
2 , where the and the loss in the line is Ploss(pu) = 0.5 PG1(pu) loss coefficient is specified in pu on a 100 MVA base. The most economic power generation schedule in MW is (a) PG1 = 20, PG2 = 22 (b) PG1 = 22, PG2 = 20
(c) PG1 = 20, PG2 = 20
(d) PG1 = 0, PG2 = 40 (GATE 2012: 1 Mark)
(GATE 2011: 2 Marks) 132. In the above system, the three-phase fault MVA at the bus 3 is (a) 82.55 MVA (b) 85.11 MVA (c) 170.91 MVA (d) 181.82 MVA (GATE 2011: 2 Marks)
Ch wise GATE_EE_Ch6.indd 271
136. The sequence components of the fault current are as follows: Ipositive = j 1.5 pu, Inegative = −j 0.5 pu, Izero = −j 1 pu. The type of fault in the system is (a) LG (b) LL (c) LLG (d) LLLG (GATE 2012: 1 Mark)
11/21/2018 11:51:06 AM
272
GATE EE Chapter-wise Solved Papers
137. For the system shown below, SD1 and SD2 are complex power demands at bus 1 and bus 2, respectively. If |V2| = 1 pu, the VAR rating of the capacitor (QG2) connected at bus 2 is Bus 1 V1 = 1–0 pu
Bus 2 V2
Z = j 0.5 pu
SG1
QG2
SD2 = 1 pu (a) 0.2 pu (c) 0.312 pu
SD2 = 1 pu (b) 0.268 pu (d) 0.4 pu (GATE 2012: 2 Marks)
138. A cylindrical rotor generator delivers 0.5 pu power in the steady-state to an infinite bus through a transmission line of reactance 0.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator is 5 MW-s/MVA and the generator reactance is 1 pu. The critical clearing angle, in degrees, for a three-phase dead short circuit fault at the generator terminal is (a) 53.5 (b) 60.2 (c) 70.8 (d) 79.6 (GATE 2012: 2 Marks) 139. A source vs (t ) = V cos 100π t has an internal impedance of (4 +j 3) Ω. If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in Ω should be (a) 3 (b) 4 (c) 5 (d) 7 (GATE 2013: 1 Mark) 140. A single-phase load is supplied by a single-phase voltage source. If the current flowing from the load to the source is 10∠ − 150°A and if the voltage at the load terminals is 100 ∠ 60° V, 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. (GATE 2013: 1 Mark)
Ch wise GATE_EE_Ch6.indd 272
141. The angle d in the swing equation of a synchronous generator is the (a) angle between stator voltage and current (b) angular displacement of the rotor with respect to the stator (c) angular displacement of the stator mmf with respect to a synchronously rotating axis (d) angular displacement of an axis fixed to the rotor with respect to a synchronously rotating axis (GATE 2013: 1 Mark) 142. For a power system network with n nodes, Z33 of its bus impedance matrix is j 0.5 per unit. The voltage at node 3 is 1.3∠—10° per unit. If a capacitor having reactance of −j 3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is (a) 0.325 ∠−100° (b) 0.325 ∠80° (c) 0.371∠−100° (d) 0.433 ∠80° (GATE 2013: 2 Marks) Statement for Linked Answer Questions 143 and 144: In the following network, the voltage magnitudes at all buses are equal to 1 pu, the voltage phase angles are very small, and the line resistances are negligible. All the line reactances are equal to j1 Ω. Bus 1 (slack)
j1 Ω
Bus 2
P2 = 0.1 pu j1 Ω
j1 Ω
Bus 3
P3 = 0.2 pu
143. The voltage phase angles in rad at buses 2 and 3 are (a) q2 = −0.1, q3 = −0.2 (b) q2 = 0, q3 = −0.1 (c) q2 = 0.1, q3 = 0.1 (d) q2 = 0.1, q3 = 0.2 (GATE 2013: 2 Marks) 144. If the base impedance and the line-to-line base voltage are 100 Ω and 100 kV, respectively, then the real power in MW delivered by the generator connected at the slack bus is (a) −10 (b) 0 (c) 10 (d) 20 (GATE 2013: 2 Marks)
11/21/2018 11:51:06 AM
Chapter 6 • Power Systems
145. The undesirable property of an electrical insulating material is (a) high dielectric strength (b) high relative permittivity (c) high thermal conductivity (d) high insulation resistivity (GATE 2014: 1 Mark) 146. Three-phase to ground fault takes place at locations F1 and F2 in the system shown in the figure IF1 IF2 F1
EA∠d
A
B
F2
EB∠0
VF1 VF2 If the fault takes place at location F1, then the voltage and the current at bus A are VF1 and IF1 respectively. If the fault takes place at location F2, then the voltage a nd the current at bus A are VF2 and IF2 respectively. The correct statement about voltages and currents during faults at F1 and F2 is (a) (b) (c) (d)
VF1 leads IF1 and VF2 leads IF2 VF1 leads IF1 and VF2 lags IF2 VF1 lags IF1 and VF2 leads IF2 VF1 lags IF1 and VF2 lags IF2 (GATE 2014: 1 Mark)
147. A 2-bus system and corresponding zero sequence network are shown in the figure. Bus 1
Bus 2
XGn
XMn T1
T2 (a)
X0G X0T 1
X0L
X0T 2
X0M
3XGn
3XMn (b)
The transformers T1 and T2 are connected as (a)
and
∆
(b)
and
∆
∆
(c)
(d)
∆
and
∆
and (GATE 2014: 1 Mark)
Ch wise GATE_EE_Ch6.indd 273
273
148. A three phase star-connected load is drawing power at a voltage of 0.9 pu and 0.8 power factor lagging. The three phase base power and base current are 100 MVA and 437.38 A respectively. The line-to-line load voltage in kV is . (GATE 2014: 1 Mark) 149. Shunt reactors are sometimes used in high voltage transmission systems to (a) limit the short circuit current through the line. (b) compensate for the series reactance of the line under heavily loaded condition. (c) limit over-voltages at the load side under lightly loaded condition. (d) compensate for the voltage drop in the line under heavily loaded condition. (GATE 2014: 1 Mark) 150. In a long transmission line with r, l, g and c as the resistance, inductance, shunt conductance and capacitance per unit length, respectively, the condition for distortionless transmission is (a) rc = lg (b) r = l /c (c) rg = lc
(d) g = c /l (GATE 2014: 1 Mark)
151. For a fully transposed transmission line (a) positive, negative and zero sequence impedances are equal. (b) positive and negative sequence impedances are equal. (c) zero and positive sequence impedances are equal. (d) negative and zero sequence impedances are equal. (GATE 2014: 1 Mark) 152. A 183-bus power system has 150 PQ buses and 32 PV buses. In the general case, to obtain the load flow solution using Newton–Raphson method in polar coordinates, the minimum number of simultaneous equations to be solved is . (GATE 2014: 1 Mark) 153. In an unbalanced three phase system, phase current Ia = 1∠(- 90°) pu, negative sequence current Ib2 = 4 ∠ (- 150°) pu, zero sequence current Ic0 = 3∠90° pu. The magnitude of phase current Ib in pu is (a) 1.00 (b) 7.81 (c) 11.53 (d) 13.00 (GATE 2014: 2 Marks) 154. A synchronous generator is connected to an infinite bus with excitation voltage Ef = 1.3 pu. The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6 pu to the bus. Assume the infinite bus
11/21/2018 11:51:07 AM
274
GATE EE Chapter-wise Solved Papers
voltage to be 1.0 pu. Neglect stator resistance. The reactive power (Q) in pu supplied by the generator to the bus under this condition is . (GATE 2014: 2 Marks)
155. A three phase synchronous generator is to be connected to the infinite bus. The lamps are connected as shown in the figure for the synchronisation. The phase sequence of bus voltage is R-Y-B and that of incoming generator voltage is R′-Y′-B′.
(a) 75 A and 25 A (c) 25 A and 75 A
(b) 50 A and 50 A (d) 0 A and 100 A (GATE 2014: 2 Marks)
157. A two bus power system shown in the figure supplies load of 1.0 + j 0.5 pu. G1
Bus 1 V1 ∠ 0°
Bus 2 1 ∠ δ2 j0.1 1.0 + j0.5
RY B j2 R¢ Y¢
The values of V1 in pu and d2, respectively, are
B¢
(b) 1.05 and - 5.44° (d) 1.1 and - 27.12° (GATE 2014: 2 Marks)
(a) 0.95 and 6.00° (c) 1.1 and - 6.00°
La Infinite bus Lb
Incoming generator
Lc
Plant P2: C2 = 0.10 Pg 22 + 3 APg 2 + 2 B
It was found that the lamps are becoming dark in the sequence La-Lb-Lc. It means that the phase sequence of incoming generator is (a) opposite to infinite bus and its frequency is more than infinite bus. (b) opposite to infinite bus but its frequency is less than infinite bus. (c) same as infinite bus and its frequency is more than infinite bus. (d) same as infinite bus and its frequency is less than infinite bus. (GATE 2014: 2 Marks) 156. A distribution feeder of 1 km length having resistance, but negligible reactance, is fed from both the ends by 400 V, 50 Hz balanced sources. Both voltage sources S1 and S2 are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure. S1
S2 400 m 200 m 200 m 200 m
400 V 50 Hz
200 A 100 A 200 A
400 V 50 Hz
The contributions of S1 and S2 in 100 A current supplied at location P respectively, are
Ch wise GATE_EE_Ch6.indd 274
158. The fuel cost functions of two power plants are Plant P1: C1 = 0.05 Pg12 + APg1 + B where Pg1 and Pg2 are the generated powers of two plants, and A and B are the constants. If the two plants optimally share 1000 MW load at incremental fuel cost of 100 Rs./ MWh, the ratio of load shared by plants P1 and P2 is (a) 1:4 (b) 2:3 (c) 3:2 (d) 4:1 (GATE 2014: 2 Marks) 159. The overcurrent relays for the line protection and loads connected at the buses are shown in the f ollowing figure. A
B
RA 300 A
C
RB 200 A
100 A
The relays are IDMT in nature having the characteristic top =
0.14 × Time multiplier setting (Plug setting multiplier)0.02 − 1
The maximum and minimum fault currents at bus B are 2000 A and 500 A, respectively. Assuming the time multiplier setting and plug setting for relay RB to be 0.1 and 5 A, respectively, the operating time of RB (in seconds) . is (GATE 2014: 2 Marks)
11/21/2018 11:51:07 AM
Chapter 6 • Power Systems
160. There are two generators in a power system. No-load frequencies of the generators are 51.5 Hz and 51 Hz, respectively, and both are having droop constant of 1 Hz/MW. Total load in the system is 2.5 MW. Assuming that the generators are operating under their respective droop characteristics, the frequency of the power system in Hz in steady state is . (GATE 2014: 2 Marks) 161. A three phase, 100 MVA, 25 kV generator has solidly grounded neutral. The positive, negative, and the zero sequence reactances of the generator are 0.2 pu, 0.2 pu, and 0.05 pu, respectively, at the machine base quantities. If a bolted single phase to ground fault occurs at the terminal of the unloaded generator, the fault current in amperes immediately after the fault is . (GATE 2014: 2 Marks) 162. The mean thickness and variance of silicon steel laminations are 0.2 mm and 0.02 respectively. The varnish insulation is applied on both the sides of the laminations. The mean thickness of one side insulation and its variance are 0.1 mm and 0.01 respectively. If the transformer core is made using 100 such varnish coated laminations, the mean thickness and variance of the core respectively are (a) 30 mm and 0.22 (b) 30 mm and 2.44 (c) 40 mm and 2.44 (d) 40 mm and 0.24 (GATE 2014: 2 Marks) 163. For a 400 km long transmission line, the series impedance is (0.0 + j 0.5) Ω/km and the shunt admittance is (0.0 + j 5.0) μmho/km. The magnitude of the series impedance (in Ω) of the equivalent p circuit of the transmission line is . (GATE 2014: 2 Marks) 164. The complex power consumed by a constant-voltage load is given by (P1 + jQ1), where, 1 kW ≤ P1 ≤ 1.5 kW and 0.5 kVAR ≤ Q1 ≤ 1 kVAR. A compensating shunt capacitor is chosen such that |Q| ≤ 0.25 kVAR where Q is the net reactive power consumed by the capacitor-load combination. The reactive power (in kVAR) supplied by the capacitor is . (GATE 2014: 2 Marks)
275
The inertia constant of the synchronous generator H = 5 MW-s/MVA. Frequency is 50 Hz. Mechanical power is 1 pu. The system is operating at the stable equilibrium point with rotor angle d equal to 30°. A threephase short circuit fault occurs at a certain location on one of the circuits of the double circuit transmission line. During fault, electrical power in pu is Pmax sin d. If the values of d and dd/dt at the instant of fault clearing are 45° and 3.762 radian/s respectively, then Pmax (in pu) is . (GATE 2014: 2 Marks) 166. Base load power plants are P. Wind farms Q. Run-of-river plants R. Nuclear power plants S. Diesel power plants (a) (b) (c) (d)
P, Q and S only P, R and S only P, Q and R only Q and R only (GATE 2015: 1 Mark)
167. The synchronous generator shown in the following figure is supplying active power to an infinite bus via two short, lossless transmission lines, and is initially in steady state. The mechanical power input to the generator and the voltage magnitude E are constant. If one line is tripped at time t1 by opening the circuit breakers at the two ends (although there is no fault), then it is seen that the generator undergoes a stable transient. Which one of the following waveforms of the rotor angle d shows the transient correctly? Synchronous generator XS
Line 1 Infinite Bus 1∠0
E∠d Line 2 (a)
d
165. The figure shows the single line diagram of a single machine infinite bus system.
Infinite bus
Ch wise GATE_EE_Ch6.indd 275
d=0
time t1
11/21/2018 11:51:08 AM
276
GATE EE Chapter-wise Solved Papers
(b)
d
169. Consider the economic dispatch problem for a power plant having two generating units. The fuel costs in Rs./MWh along with the generation limits for the two units are given below: C1 ( P1 ) = 0.01P12 + 30 P1 + 10; 100 MW ≤ P1 ≤ 150 MW C2 ( P2 ) = 0.05 P22 + 10 P2 + 10;
d=0
time t1 d
(c)
t1
d=0
time
d
(d) d=0
t1 time
100 MW ≤ P2 ≤ 180 MW The incremental cost (in Rs./MWh) of the power plant when it supplies 200 MW is ————————. (GATE 2015: 2 Marks) 170. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion (A): Fast decoupled load flow method gives approximate load flow solution because it uses several assumptions. Reason (R): Accuracy depends on the power mismatch vector tolerance. (a) Both (A) and (R) are true and (R) is the correct reason for (A) (b) Both (A) and (R) are true but (R) is not the correct reason for (A) (c) Both (A) and (R) are false (d) (A) is false and (R) is true (GATE 2015: 2 Marks) 171. A sustained three-phase fault occurs in the power system shown in the following figure. The current and voltage phasors during the fault (on a common reference), after the natural transients have died down, are also shown. Where is the fault located? → →
(GATE 2015: 1 Mark) L 168. A 3-bus power system network consists of 3 transmission lines. The bus admittance matrix of the uncompensated system is
→
I1
Vi
Ch wise GATE_EE_Ch6.indd 276
→
V2
S
Q P → R Transmission
I2
j4 ⎤ ⎡ − j6 j3 ⎢ j 3 − j 7 j 5 ⎥ pu ⎢ ⎥ ⎢⎣ j 4 j 5 − j8⎥⎦ If the shunt capacitance of all transmission lines is 50% compensated, the imaginary part of the 3rd row 3rd column element (in pu) of the bus admittance matrix after compensation is (a) −j 7.0 (b) −j 8.5 (c) −j 7.5 (d) −j 9.0 (GATE 2015: 1 Mark)
I3
Transmission line
line
→
I4 →
V2 →
V1 →
→
I3
I2
→
I1
→
I4 (a) Location P (c) Location R
(b) Location Q (d) Location S (GATE 2015: 2 Marks)
11/21/2018 11:51:09 AM
277
Chapter 6 • Power Systems
172. Two single phase transformers T1 and T2 each rated at 500 kVA are operated in parallel. Percentage impedances of T1 and T2 are (1 + j 6) and (0.8 + j 4.8), respectively. To share a load of 1000 kVA at 0.8 lagging power factor, the contribution of T2 (in kVA) is ————————. (GATE 2015: 2 Marks) 173. The incremental costs (in Rupees/MWh) of operating two generating units are functions of their respective powers P1 and P2 in MW, and are given by dC = 0.2 P1 + 50 dP1 dC2 = 0.24 P2 + 40 dP2 where 20 MW ≤ P1 ≤ 150 MW and 20 MW ≤ P2 ≤ 150 MW. For a certain load demand, P1 and P2 have been chosen such that dC1/dP1 = 76 Rs./MWh and dC2/dP2 = 68.8 Rs./MWh. If the generations are rescheduled to minimise the total cost, then P2 is ________. (GATE 2015: 2 Marks) 174. A composite conductor consists of three conductors of radius R each. The conductors are arranged as shown in the following figure. The geometric mean radius (GMR) (in cm) of the composite conductor is kR. The value of k is ________. 3R R
(d) The number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes decreases by two. (GATE 2016: 1 Mark) 176. The magnitude of three-phase fault currents at buses A and B of a power system are 10 pu and 8 pu, respectively. Neglect all resistances in the system and consider the pre-fault systems to be unloaded. The pre-fault voltage at all buses in the system is 1.0 pu. The voltage magnitude at bus B during a three-phase fault at bus A is 0.8 pu. The voltage magnitude at bus A during a three-phase fault at bus B, in pu, is ________. (GATE 2016: 1 Mark) 177. Consider a system consisting of a synchronous generator working at a lagging power factor, a synchronous motor working at an overexcited condition and a directly grid-connected induction generator. Consider capacitive VAr to be a source and inductive VAr to be a sink of reactive power. Which one of the following statements is TRUE? (a) Synchronous motor and synchronous generator are sources and induction generator is a sink of reactive power. (b) Synchronous motor and induction generator are sources and synchronous generator is a sink of reactive power. (c) Synchronous motor is a source and induction generator and synchronous generator are sinks of reactive power. (d) All are sources of reactive power. (GATE 2016: 1 Mark)
60° 178. A power system with two generators is shown in the figure below. The system (generators, buses and transmission lines) is protected by six overcurrent relays R1 to R6. Assuming a mix of directional and non-directional relays at appropriate locations, the remote backup relays for R4 are
60°
(GATE 2015: 2 Marks) 175. In a 100 bus power system, there are 10 generators. In a particular iteration of Newton Raphson load flow technique (in polar coordinates), two of the PV buses are converted to PQ type. In this iteration, (a) The number of unknown voltage angles increases by two and the number of unknown voltage magnitudes increases by two. (b) The number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes increases by two. (c) The number of unknown voltage angles increases by two and the number of unknown voltage magnitudes decreases by two.
Ch wise GATE_EE_Ch6.indd 277
S1
R1
R2
S2 R5
R3
R6
R4
(a) R1, R2
(b) R2, R6
(c) R2, R5
(d) R1, R6 (GATE 2016: 1 Mark)
179. A power system has 100 buses including 10 generator buses. For the load flow analysis using Newton-Raphson method in polar coordinates, the size of the Jacobian is (a) 189 × 189 (b) 100 × 100 (c) 90 × 90 (d) 180 × 180 (GATE 2016: 1 Mark)
11/21/2018 11:51:09 AM
278
GATE EE Chapter-wise Solved Papers
180. The inductance and capacitance of a 400 kV, three-phase, 50 Hz lossless transmission line are 1.6 mH/km/phase and 10 nF/km/phase, respectively. The sending end voltage is maintained at 400 kV. To maintain a voltage of 400 kV at the receiving end, when the line is delivering 300 MW load, the shunt compensation required is (a) capacitive (b) inductive (c) resistive (d) zero (GATE 2016: 1 Mark) 181. A three-phase cable is supplying 800 kW and 600 kVAr to an inductive load. It is intended to supply an additional resistive load of 100 kW through the same cable without increasing the heat dissipation in the cable, by providing a three-phase bank of capacitors connected in star across the load. Given the line voltage is 3.3 kV, 50 Hz, the capacitance per phase of the bank, expressed in microfarads, is ________. (GATE 2016: 2 Marks) 182. A 30 MVA, 3-phase, 50 Hz, 13.8 kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%, 15% and 5% respectively. A reactance (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ‘b’ and ‘c’, with a fault impedance of j 0.1 pu. The value of Xn (in pu) that will limit the positive sequence generator current to 4270 A is _________. (GATE 2016: 2 Marks) 183. A single-phase transmission line has two conductors each of 10 mm radius. These are fixed at a centre-to-centre distance of 1 m in a horizontal plane. This is now converted to a three-phase transmission line by introducing a third conductor of the same radius. This conductor is fixed at an equal distance D from the two single-phase conductors. The three-phase line is fully transposed. The positive sequence inductance per phase of the three-phase system is to be 5% more than that of the inductance per conductor of the single-phase system. The distance D, in meters, is _______. (GATE 2016: 2 Marks) 184. The single line diagram of a balanced power system is shown in the figure. The voltage magnitude at the generator internal bus is constant and 1.0 pu. The pu. reactances of different components in the system are also shown in the figure. The infinite bus voltage magnitude is 1.0 pu. A three phase fault occurs at the middle of line 2. The ratio of the maximum real power that can be transferred during the pre-fault condition to the maximum real power that can be transferred under the faulted condition is ________.
Ch wise GATE_EE_Ch6.indd 278
Infinite bus
Generator internal bus j0.1
j0.5 Line 1
j0.2
j0.5 Line 2 j0.1 (GATE 2016: 2 Marks) 185. At no load condition, a three-phase, 50 Hz, lossless power transmission line has sending-end and receiving-end voltages of 400 kV and 420 kV respectively. Assuming the velocity of travelling wave to be the velocity of light, the length of the line, in km, is ________. (GATE 2016: 2 Marks) 186. The power consumption of an industry is 500 kVA, at 0.8 pf lagging. A synchronous motor is added to raise the power factor of the industry to unity. If the power intake of the motor is 100 kW, the pf of the motor is ________. (GATE 2016: 2 Marks) 187. Two identical unloaded generators are connected in parallel as shown in the figure. Both the generators are having positive, negative and zero sequence impedances of j 0.4 pu, j 0.3 pu and j 0.15 pu, respectively. If the pre-fault voltage is 1 pu, for a line-to-ground (L-G) fault at the terminals of the generators, the fault current, in pu, is ________.
(GATE 2016: 2 Marks) 188. A source is supplying a load through a 2-phase, 3-wire transmission system as shown in the figure below. The instantaneous voltage and current in phase-a are van = 220 sin(100 p t) V and ia = 10 sin(100 p t) A, respectively. Similarly for phase-b, the instantaneous voltage and current are vbn = 220 cos(100 p t) V and ib = 10 cos(100 p t) A, respectively. a
ia
a′
b
ib
b′
n
+ nbn −
Source
Load n′
11/21/2018 11:51:10 AM
Chapter 6 • Power Systems
The total instantaneous power flowing from the source to the load is (a) 2200 W (b) 2200 sin2 (100 p t) W (c) 4400 W (d) 2200 sin (100 p t) cos(100 p t) W (GATE 2017: 1 Mark) 189. A 3-bus power system is shown in the figure below, where the diagonal elements of Y-bus matrix are: Y11 = -j 12 pu, Y22 = -j 15 pu and Y33 = -j 7 pu. Bus-1
Bus-2
jq
279
For the given circuit YBUS and ZBUS are bus admittance matrix and bus impedance matrix, respectively, each of size 2 × 2. Which one of the following statements is true? (a) Both YBUS and ZBUS are symmetric. (b) YBUS is symmetric and ZBUS is unsymmetric. (c) YBUS is unsymmetric and ZBUS is symmetric. (d) Both YBUS and ZBUS are unsymmetric. (GATE 2017: 1 Mark) 192. The nominal-p circuit of a transmission line is shown in the figure. Z
jp
jr
X
X
Bus-3 The per unit values of the line reactances p, q and r shown in the figure are (a) p = -0.2, q = -0.1, r = -0.5 (b) p = 0.2, q = 0.1, r = 0.5 (c) p = -5, q = -10, r = -2 (d) p = 5, q = 10, r = 2 (GATE 2017: 1 Mark) 190. A 10-bus power system consists of four generator buses indexed at G1, G2, G3, G4 and six load buses indexed as L1, L2, L3, L4, L5, L6. The generator-bus G1 is considered as slack bus, and the load buses L3 and L4 are voltage controlled buses. The generator at bus G2 cannot supply the required reactive power demand, and hence it is operating at its maximum reactive power limit. The number of non-linear equations required for solving the load flow problem using Newton–Raphson method in polar form is . (GATE 2017: 1 Mark) 191. The figure shows the per-phase representation of a phase-shifting transformer connected between buses 1 and 2, where a is a complex number with non-zero real and imaginary parts.
Z
1 a
(GATE 2017: 1 Mark) 193. In a load flow problem solved by Newton–Raphson method with polar coordinates, the size of the Jacobian is 100 × 100. If there are 20 PV buses in addition to PQ buses and a slack bus, the total number of buses in the system is . (GATE 2017: 1 Mark) 194. A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that 1 kW ≤ P ≤ kW and 1 kVAR ≤ Q ≤ 2 kVAR. A capacitor connected across the load for power correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is (a) 0.447 lag (b) 0.707 lag (c) 0.894 lag (d) 1 (GATE 2017: 2 Marks) 195. The bus admittance matrix for a power system network is
Ideal transformer Bus 1
Impedance Z = 100 ∠80°Ω and reactance X = 3300 W. The magnitude of the characteristic impedance of the transmission line, in W, is . (Give the answer up to one decimal place.)
Bus 2
j 20 j 20 ⎤ ⎡ − j 39.9 ⎢ j 20 − j 39.9 j 20 ⎥⎥ pu ⎢ ⎢⎣ j 20 j 20 − j 39.9⎥⎦ There is a transmission line, connected between buses 1 and 3, which is represented by the circuit shown in figure.
Ch wise GATE_EE_Ch6.indd 279
11/21/2018 11:51:10 AM
280
GATE EE Chapter-wise Solved Papers
Reactance is 0.05 pu
Susceptance is 0.05 pu
Susceptance is 0.05 pu
198. Consider an overhead transmission line with 3-phase, 50 Hz balanced system with conductors located at the vertices of an equilateral triangle of length Dab = Dbc = Dca = 1 m as shown in figure below. The resistances of the conductors are neglected. The geometric mean radius (GMR) of each conductor is 0.01 m. Neglecting the effect of ground, the magnitude of positive sequences reactances in W/km (rounded off to three decimal places) is .
If this transmission line is removed from service, what is the modified bus admittance matrix?
c
j 20 0 ⎤ ⎡ − j19.9 ⎢ j 20 − j 39.9 j 20 ⎥⎥ pu (a) ⎢ ⎢ 0 j 20 − j19.9⎥⎦ ⎣ j 20 0 ⎤ ⎡ − j 39.95 ⎢ − j 39.9 j 20 ⎥⎥ pu (b) ⎢ j 20 ⎢⎣ 0 j 20 − j 39.95⎥⎦ j 20 0 ⎤ ⎡ − j19.95 ⎢ j 20 − j 39.9 j 20 ⎥⎥ pu (c) ⎢ ⎢ 0 j 20 − j19.95⎥⎦ ⎣ j 20 j 20 ⎤ ⎡ − j19.95 ⎢ j 20 − j 39.9 j 20 ⎥⎥ pu (d) ⎢ ⎢ j 20 j 20 − j19.95⎥⎦ ⎣ (GATE 2017: 2 Marks) 196. The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu and 0.1 pu, respectively. The generator is on open circuit with a terminal voltages of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs . (Assume fault impedance at the terminals is to be zero. Give the answer up to one decimal place.) (GATE 2017: 2 Marks) 197. The figure shows the single-line diagram of a power system with a double circuit transmission line. The expression for electrical power is 1.5 sin d, where d is the rotor angle. The system is operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the transmission line circuits is removed, the maximum value of d, as the rotor swings is 1.221 radian. If the expression for electrical power with one transmission line circuit removed is Pmax sin d, the value of Pmax, in pu, is . (Give the answer up to one decimal place.)
Dca a
Ch wise GATE_EE_Ch6.indd 280
b Dab (GATE 2017: 2 Marks)
199. A 3-phase, 50 Hz generator supplies power of 3 MW at 17.32 kV to a balanced 3-phase inductive load through an overhead line. The per phase line resistance and reactance are 0.25 W and 3.925 W, respectively. If the voltage at the generator terminal is 17.87 kV, the power factor of the load is . (GATE 2017: 2 Marks) 200. Two generating units rated 300 MW and 400 MW have governor speed regulation of 6% and 4%, respectively from no load to full load. Both the generating units are operating in parallel to share a load of 600 MW. Assuming free governor action, the load shared by the larger units is MW. (GATE 2017: 2 Marks) 201. A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivery 60 MW power angle at a power of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is electrical degrees. (GATE 2017: 2 Marks) 202. Consider a lossy transmission line with V1 and V2 as the sending and receiving end voltages, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be (a) greater than
(c) equal to (GATE 2017: 2 Marks)
Dbc
V1V2 X
V1V2 X
(b) less than
V1V2 X
(d) equal to
V1V2 Z
(GATE 2018: 1 Mark)
11/21/2018 11:51:11 AM
Chapter 6 • Power Systems
203. A 1000´ × 1000 bus admittance matrix for an electric power system has 8000 non-zero elements. The minimum number of branches (transmission lines and transformers) in this system are (up to 2 decimal places). (GATE 2018: 1 Mark) 204. Consider the two bus power system network with given loads as shown in the figure. All the values shown in the figure are in per unit. The reactive power supplied by generator G1 and G2 are QG1 and QG2 respectively. The per unit values of QG1, QG2, and line reactive power loss (Qloss) respectively are 1.0∠δ
1.0∠0
j0.1
G1
the value of d for P = 0.8 pu. If the initial guess is 30°, then its value (in degree) at the end of the first iteration is (a) 15° (b) 28.48° (c) 28.74° (d) 31.20° (GATE 2018: 2 Marks) 206. A three-phase load is connected to a three-phase balanced supply as shown in the figure. If Van = 100 ∠0° V, Vbn = 100 ∠ − 120° V and Vcn = 100 ∠ − 240° V (angles are considered positive in the anti-clockwise direction), the value of R for zero current in the neutral wire is W (upto 2 decimal places).
G2 Qloss
20 + jQG1
a
15 + jQG2
15 + j5 (a) 5.00, 12.68, 2.68 (c) 6.34, 11.34, 2.68
281
20 + j10
n
(b) 6.34, 10.00, 1.34 (d) 5.00, 11.34, 1.34
c
R
j10 -j10
(GATE 2018: 2 Marks) 205. The per-unit power output of a salient-pole generator which is connected to an infinite bus, is given by the expression, P = 1.4 sin d + 0.15 sin 2d, where d is the load angle. Newton-Raphson method is used to c alculate
b (GATE 2018: 2 Marks)
ANSWER KEY 1. (d)
2. (a)
3. (a)
4. (b)
5. (b)
6. (b)
7. (c)
8. (d)
9. (d)
10. (a)
11. (a)
12. (c)
13. (d)
14. (d)
15. (∗)
16. (c)
17. (b)
18. (c)
19. (c)
20. (b)
21. (b)
22. (a)
23. (d)
24. (a)
25. (b)
26. (c)
27. (b)
28. (b)
29. (c)
30. (b)
31. (b)
32. (d)
33. (b)
34. (a)
35. (d)
36. (d)
37. (b)
38. (d)
39. (a)
40. (a)
41. (c)
42. (d)
43. (c)
44. (b)
45. (c)
46. (a)
47. (b)
48. (c)
49. (d)
50. (a)
51. (c)
52. (b)
53. (c)
54. (a)
55. (d)
56. (b)
57. (d)
58. (d)
59. (d)
60. (b)
61. (a)
62. (d)
63. (c)
64. (d)
65. (c)
66. (b)
67. (d)
68. (c)
69. (d)
70. (b)
71. (b)
72. (a)
73. (b)
74. (b)
75. (d)
76. (c)
77. (c)
78. (b)
79. (a)
80. (a)
81. (c)
82. (b)
83. (d)
84. (c)
85. (c)
86. (c)
87. (b)
88. (d)
89. (b)
90. (c)
91. (a)
92. (c)
93. (b)
94. (a)
95. (b)
96. (a)
97. (d)
98. (d)
99. (b)
100. (a)
101. (b)
102. (c)
103. (b)
104. (d)
105. (a)
106. (c)
107. (a)
108. (a)
109. (c)
110. (c)
111. (b)
112. (c)
113. (c)
114. (a)
115. (c)
116. (a)
117. (a)
118. (a)
119. (c)
120. (c)
121. (c)
122. (d)
123. (b)
124. (a)
125. (b)
126. (a)
127. (a)
128. (c)
129. (a)
130. (b)
131. (a)
132. (d)
133. (b)
134. (d)
135. (a)
136. (c)
137. (b)
138. (d)
139. (c)
140. (b)
141. (d)
142. (d)
143. (b)
144. (c)
145. (b)
146. (c)
147. (b)
148. (118.8) 149. (c)
150. (a)
151. (b)
152. (332)
153. (c)
154. (0.1)
161. (15396) 162. (d)
163. (186.66) 164. (0.75)
155. (a)
156. (d)
157. (b)
158. (d)
159. (0.023) 160. (50)
165. (0.24)
166. (c)
167. (a)
168. (b)
169. (20)
176. (0.84)
171. (b)
172. (555)
173. (136.36) 174. (1.914) 175. (a)
181. (47.96)
182. (1.07)
183. (1.427)
191. (d) 201. (12.7)
170. (d)
177. (c)
178. (d)
179. (a)
180. (b)
184. (2.4)
185. (294.59) 186. (0.316) 187. (6)
188. (a)
189. (b)
190. (14)
192. (406.2) 193. (61)
194. (b)
195. (c)
196. (0.1)
202. (b)
204. (c)
205. (c)
206. (5.77)
203. (3500)
197. (1.22) 198. (0.289) 199. (0.8)
200. (333)
* None of the given options are correct.
Ch wise GATE_EE_Ch6.indd 281
11/21/2018 11:51:12 AM
282
GATE EE Chapter-wise Solved Papers
ANSWERS WITH EXPLANATION 1. Topic: Power Generation Concepts (d) In thermal plants, the waste heat from fuel gases gets stored in the economizer and further used in preheating the feed water of boilers. 2. Topic: Per-Unit Quantities (a) The pu impedance value of the element on two base values is related as:
6. Topic: Models and Performance of Transmission Lines and Cables (b) Maximum power transfer capability of a line
Pmax =
V2 X
Now, with compensation at mid point,
Therefore, Pmid =
2
⎛ KV ⎞ ⎛ KVA2 ⎞ Z base2 = Z base1 × ⎜ 1 ⎟ × ⎜ ⎝ KV2 ⎠ ⎝ KVA1 ⎟⎠
X
Since, V1 = V2 = V (say)
2
Z base2 ⎛ KV1 ⎞ ⎛ KVA2 ⎞ = × Z base1 ⎜⎝ KV2 ⎟⎠ ⎜⎝ KVA1 ⎟⎠
V1 V2
Pmax =
V2 ⎛δ⎞ sin ⎜ ⎟ X /2 ⎝ 2 ⎠
Therefore,
2
2 ⎛ 1⎞ = x × ⎜ ⎟ × = 0.5 x ⎝ 2⎠ 1 3. Topic: Principles of Over-Current (a) In an inverse definite minimum time electromagnetic type over-current relay, when the electric current in the system increases, the secondary electric current of the current transformer (CT) is increased proportionally. The secondary electric current is fed to the relay electric current coil. But when the CT becomes saturated, there is no further proportional increase of CT secondary electric current with increased system current. As the relay electric current is not increased further, there would not be any further reduction in time of operation in the relay. This time is referred as minimum time of operation. 4. Topic: System Stability Concepts (b) From the given options advantages of HVDC transmission are: (i) No distance limitation related to steady state stability, therefore used for very long distance of transmission. (ii) No substantial effect on fault level of the two systems at the terminals inspite of the interconnection. 5. Topic: Models and Performance of Transmission Lines and Cables (b) Distance protection function used in protection of transmission lines, where the response is dependent upon the electrical distance between the relay location and the fault. Distance protection using mho relays with Zone 1 set to 80% of the line impedance, 50% of the second line and 20% of the third line.
Ch wise GATE_EE_Ch6.indd 282
( Pmid ) max =
V2 X /2
(Pmid)max = 2Pmax And, with compensator connected at 75% distance ( P75% ) max =
=
V2 3 4 X 1 Pmax 0.75
Thus, the ratios will be, 1: 2 :
1 0.75
7. Topic: Circuit Breakers (c) As per the given plug setting, Primary current of CT = 5 × 0.2 = 1 A Thus, line to line current will be I L-L = 3 × 1= 1.732 A 8. Topic: Power Generation Concepts (d) Given that, P1 + P2 = 200 MW (1) For economic operations, dF1 dF2 = dP1 dP2 20 + 0.1 P1 = 16 + 0.2P2
2 P2 − P1 = 40 MW (2)
11/21/2018 11:51:13 AM
Chapter 6 • Power Systems
Solving Eqs. (1) and (2), we get P1 = 120 MW P2 = 80 MW
283
13. Topic: Per-Unit Quantities (d) According per unit system per unit value of any quantity is given by Quantity Base value of that quantity
9. Topic: Power Factor Correction (d) Corona loss ∝ ( f + 25) Therefore,
Therefore, PC1 f + 25 50 + 25 = 1 = PC 2 f 2 + 25 60 + 25
X d2
= 0.4 × 85 = 1.13 kW/km 75
10. Topic: Symmetrical and Unsymmetrical Fault Analysis (a) The fault current in the alternator during line-toground fault is given by IF =
I 3ϕ =
EF z1
100 ⎛ 10 ⎞ × ⎜ ⎟ = 0.4407 p.u. 75 ⎝ 11 ⎠
14. Topic: Voltage and Frequency Control (d) For zero voltage regulation, we know that |E| = |V| Therefore,
3EF z0 + z1 + z2 + 3 zn
Also, during a three-phase fault
cos(θ + ϕ ) =
3 z1 = z1 + z2 + z0 + 3 zn
=
2 z1 − z2 − z0 3 2(0.1) − 0.1 − 0.05 3
2V 440 3
and θ = 90° .
Therefore, cos(θ + ϕ ) =
I F = I 3ϕ 3EF E = F z1 + z2 + z0 + 3 zn z1
− I a Zs
Here, I a = 20 A, Zs = 10 Ω, V =
From the given condition,
zn =
2
2
Give that PC1 = 1 kW/km, PC 2 = 1 kW/km ×
( MVA) 2 ⎛ KV1 ⎞ = X d1 × × ( MVA)1 ⎜⎝ KV2 ⎟⎠
Therefore,
−20 × 10 = − 0.3936 440 2× 3
ϕ + θ = 113.178° ϕ = 113.178 − 90 = 23.178°
Therefore, power factor of the load, cos ϕ = 0.9192 15. Topic: Symmetrical and Unsymmetrical Fault Analysis (*) The given circuit is X″d = 0.2 pu
znn = 0.167 pu 11. Topic: Differential and Distance Protection (a) In transformers, the magnetizing inrush current can damage the transformer. This inrush current consists of large magnitude and low frequency harmonics (both even and odd). So, harmonic restrain to guard the transformer against magnetizing inrush current. 12. Topic: Models and Performance of Transmission Lines and Cables (c) A lossless radial transmission line with surge impedance loading shows zero voltage regulation, that is, flat voltage profile. Also, the power factor with surge impedance loading becomes unity at all points along it.
Ch wise GATE_EE_Ch6.indd 283
E ″g
RL If
Also, Base MVA = 500 MVA Base voltage = 20 kV Actual power = 400 MW
11/21/2018 11:51:15 AM
284
GATE EE Chapter-wise Solved Papers
Therefore, Power per unit,
P1 P − 2 = −1 50 100 Ppu =
400 = 0.8 pu 500
Also,
2 P1 − P2 = −100 (4) Solving Eqs. (3) and (4), we get P1 = 100 MW
Ppu = VPi I Pu cos ϕ I pu =
Therefore,
0.8 = 0.8∠0° 1.0 ∠0°
[∵Vpu = 1∠0°, cos ϕ = 1 for RL ]
Thus,
Eg′′ = 1∠0° + j 0.2(0.8∠0°) = 1.01∠9.09° pu
P2 = 300 MW Therefore, from Eq (1), we get f = 50 −
100 = 48 Hz 50
18. Topic: Models and Performance of Transmission Lines and Cables (c) For the given transmission line 1.5 m
Eg′′
1.01 I f′′= = = 5.05 pu X d′′ 0.2 16. Topic: Models and Performance of Transmission Lines and Cables (c) Given that VS = 1∠0° pu and I = 1 pu Power factor (pf ) = 1; ϕ = 0° Now, writing KVL at different points, VP1 = 1 pu VP2 = 1 pu − 1 pu × j 0.1 pu VP3 = 1 pu + 1 pu × j 0.05 pu VP4 = 1 pu − 1 pu × j 0.05 pu Hence, VP3 is maximum. 17. Topic: Power Generation Concepts (b) Given that
r = 1 cm Then inductance is given by ⎛ D⎞ L = 2 × 10 −7 ln ⎜ ⎟ mH/km ⎝ r′ ⎠ 1.5 ⎛ ⎞ = 2 × 10 −7 ln ⎜ mH/km ⎝ 0.7788 × 0.5 × 10 −2 ⎟⎠ For 10 km long line, 1.5 ⎛ ⎞ mH/km L = 2 × 10 −7 ln ⎜ ⎝ 0.7788 × 0.5 × 10 −2 ⎟⎠ = 23.815 mH/km 19. Topic: Models and Performance of Transmission Lines and Cables (c) For the given two-wire line composed of solid round conductors
P1 = 50(50 – f )
⇒ f = 50 −
P1 (1) 50
and P2 = 100 (51 – f) Also,
⇒ f = 51 −
P2 (2) 100
P1 + P2 = 400 (3)
Equating Eqs. (1) and (2), we have 50 −
Ch wise GATE_EE_Ch6.indd 284
1m
P1 P = 51 − 2 50 100
r = 0.25 cm We know that inductance, ⎛ D⎞ L ∝ ln ⎜ ⎟ ⎝ r′ ⎠ Therefore, for the same radius, if the distance between the wires is doubled ln (2D) > ln D
11/21/2018 11:51:17 AM
285
Chapter 6 • Power Systems
20. Topic: Power Generation Concepts (b) Since the generator G1 has a governor attached to maintain its speed constant, that is, no droop. Therefore, when the load increases to maintain the speed as constant, the power generation of G1 will increase. 21. Topic: Models and Performance of Transmission Lines and Cables (b) Since the high voltage exists between the conductor and ground, the lower surface of conductor will suffer the maximum electric stress. 22. Topic: Models and Performance of Transmission Lines and Cables (a) From the diagram of the given transmission, it is clear that the time setting of relays R1 should be greater than that of relay R2. And only option (a) is giving this relation. 23. Topic: Gauss-Seidel and Newton-Raphson Load Flow Methods (d) Since the lie-line data is left out, the two areas will now be considered as two separate areas.
Therefore,
Z th = 3 × 0.25 + j (0.1 + 0.05 + 0.07) = 0.75 + j 0.22
26. Topic: Models and Performance of Transmission Lines and Cables (c) Bundled conductors are used to reduce corona loss. When geometric mean distance (GMD) of the conductor is increased, critical disruptive voltage will be reduced, thus, reducing the corona loss. 27. Topic: Gauss Seidel and Newton-Raphson Load Flow Method (b) Given that: n = 300 Generator bus = 20 Reactive power support bus = 25 Fixed bus with shunt capacitor = 15 Therefore, number of slack buses, ns = 20 + 25 − 15 = 30 Size of Jacobian matrix = 2( n − ns ) × 2( n − ns ) = 2(300 − 30) × 2(300 − 30)
And hence two slack buses (that is one for each area) is required. 24. Topic: Models and Performance of Transmission Lines and Cables (a) The given system can be represented as follows. 0.1 pu 0.98 pu
0.1 pu
1 pu
1 pu Compensation
28. Topic: Models and Performance of Transmission Lines and Cables (b) Auxiliary components in HVDC transmission are: DC line inductor and reactive power sources. That is P and R. 29. Topic: Per Unit Quantities (c) Given that: P = 0.5 pu, E = 2 pu and V = 1.3 pu. The reactance of motor + generator + connecting lines:
To determine the rotor angle difference, we know that
1 × 0.98 P= = 9.8 pu 0.1
P=
25. Topic: Symmetrical and Unsymmetrical Fault Analysis (b) The zero sequence impedance diagram is as shown below 3 × 0.25
EV [sin(δ1 − δ 2 )] Xd
2 × 1.3 sin(δ1 − δ 2 ) 2.8 δ1 − δ 2 = 32.58°
0.5 =
30. Topic: Circuit Breakers (b) The interrupting time of CB is the time period between energising of trip circuit and the arc extinction on an opening operation.
j0.05 j0.07
j0.03 3 × 0.25
= 540 × 540
X d = X eq = 1.1 + 1.2 + 0.5 = 2.8
Thus, the steady state power transfer limit
j0.1
= 2( 270) × 2( 270)
Zth
31. Topic: AC and DC Transmission Concepts (b) A balanced delta connected load = 8 + 6 j. Given V2 = 400 V and cos ϕ 2 = 0.9.
Ch wise GATE_EE_Ch6.indd 285
11/21/2018 11:51:19 AM
286
GATE EE Chapter-wise Solved Papers
33. Topic: Series and Shunt Compensation (b) Given that
⎛ 6⎞ ϕ 1 = tan −1 ⎜ ⎟ = 36.86° ⎝ 8⎠
ϕ 2 = cos −1 0.9 = 25.84°
A = D = 0.9∠0° B = 200 ∠90°Ω
Therefore,
C = 0.95 × 10 −3 ∠90°S
V 400 400 I= = = = 40 ∠ − 36 − 86° Z 8 + 6 j 10 ∠36.86 I = 32 − 24 j The power factor is improved by connecting a delta connected capacitor bank. The phasor diagram is given as follows: IC
b
c
a f2 f1
I1
O Therefore,
Receiving end voltage = sending end voltage. We know that Vs = AVr + BI r Also,
s
= Vr ]
Vr (1 − A) = BI r Vr 200 ∠90° B = = = 2000 ∠90° I r 1 − A 1 − 0.9 ∠0° 34. Topic: Models and Performance of Transmission Lines and Cables (a) For the cable: L1 = 0.4 mH/km C1 = 0.5 μ F/km
oa = I ′ cos ϕ2 = I cos ϕ1 I ′ cos 25.84 = 32 32 I′ = = 35.55 0.9 ac = 24 A ab = I ′ sin ϕ 2 = 35.5 sin 25.84 = 15.5 A I C = ac − ab = 24 − 15.5 = 8.5 A 3VI C 3 × 400 × 8.5 = = = 10.2 kVAR 1000 1000 32. Topic: Distribution Systems (d) Under the given condition S1 → ON and S2 → OFF 11000 V → 415 V Under the special condition S1 → OFF and S2 → ON Two-phase supply is connected to the transformer V1 N1 11000 = = = 26.5 V2 N 2 415 As output terminals a and c are in opposite phase and cancel each other, terminal voltage = 0 V.
V = 20 kV
For the overhead transmission line L2 = 1.5 mH/km C2 = 0.5 μ F/km Then, for the cable Z1 =
L1 = C1
0.4 × 10 −3 = 28.284 0.5 × 10 −6
For the overhead transmission line Z 2 = Z3 =
L2 = C2
1.5 × 10 −5 = 316.23 0.015 × 10 −6
Therefore, voltage at junction due to surge is given by V=
Step down voltage output is obtained as
Ch wise GATE_EE_Ch6.indd 286
[∵ V
Vr = AVr + BI r
=
2 × V × Z2 Z1 + Z 2 2 × 20 × 103 × 316.23 = 36.72 kV 316 + 28.284
35. Topic: Distribution Systems (d) If the total current = I A, then Current through section PR = (I − 10) A Current through section RS = I − 10 − 20 = (I − 30) A Current through section SQ = I − 30 − 30 = (I − 60) A VP − VQ = 3 V
11/21/2018 11:51:20 AM
Chapter 6 • Power Systems
By Kirchhoff’s voltage law (KVL),
Substituting values, we have
VP − VQ = ( I − 10) 0.1 + ( I − 30) 0.15 + ( I − 60) × 0.2
⎡ j 0.2860 ⎤ ⎢ j 0.3408⎥ 1 ⎥× = ×⎢ j 0.3408 + j 0.2 ⎢ j 0.2586 ⎥ ⎥ ⎢ ⎣ j 0.2414 ⎦
3 = 0.45 I − 17.5 20.5 = 45.6 A I= 0.45 Line drop, VL = 8.58 V
[ j 0.2860 j 0.3408 j 0.2586 j 0.2414]
[∵ I = 45.6 A ]
The required change is in Z 22 and Z 23
Hence, VP = 220 + VL = 220 + 8.58 = 228.58 V 36. Topic: Per-Unit Quantities (d) Given that, PL = 100 MW, ZL(pu) = 0.2j, Vs = Vr = 11 kV (kV) 2 (11) 2 = j 0.2 × = j 0.242 Ω MV 100 V s V r sin δ 3
3
j 2 (0.3408)(0.2586) = j 0.16296 j (00.5408)
Vs V r cos δ
38. Topic: Models and Performance of Transmission Lines and Cables (d) Zero sequence impedance for generator and transmission line,
2
−
Vr
Z L( 0 ) + ZG ( 0 ) = Z0
ZL
11 × 10 × 11 × 10 (11 × 10 ) cos(11.54) − 0.242 0.242 6 121 × 10 = [cos(11.54) − 1] = −10.1 MVAR 0.242 3
3
= j 0.1260 = Z 23(old) − Z ′23 = j 0.2586 − j 0.16296 = j 0.0956
Reactive power is given by
=
Z ′23 =
Z 23(new)
11 × 10 × 11 × 10 sin δ 0.242 or, sin δ = 0.2 ⇒ δ = 11.54
ZL
j 2 (0.3408) 2 = j 0.2147 j (0.5408)
Z 22(new) = Z 22(old) − Z ′22 = j 0.3408 − j 0.2417
ZL
100 × 106 =
Qr =
Z ′22 =
Therefore,
Z L = Z L(base) × PL =
287
Z 0 = j 0.04 + j 0.3 = j 0.34 pu
3 2
Positive sequence impedance, Z1 = Z L (1) + Z G (1) = j 0.1 + j 0.1 = j 0.2 pu Negative sequence impedance,
37. Topic: Bus Admittance Matrix (b) When a branch of 0.2j Ω is connected between bus 2 and reference Z B(new) = Z B(old)
⇒
⎡ Z ij ⎤ 1 ⎢ ⎥ − × ⎢ ⎥ × ⎡⎣ Z ji … Z jn ⎤⎦ Z ij + Z b ⎢ ⎥ ⎣ Z nj ⎦
⎡ Z12 ⎤ ⎢Z ⎥ 1 × ⎢ 22 ⎥ × [ Z 21 Z 22 Z 23 Z 24 ] Z ij + Z b ⎢ Z 23 ⎥ ⎢ ⎥ ⎣ Z 24 ⎦
Z = j 0.2 Ω , reference bus j = 2, n = 4
Ch wise GATE_EE_Ch6.indd 287
Z 2 = Z L(2) + Z G ( 2 ) = j 0.1 + j 0.1 = j 0.2 pu Also, given that ZN = j 0.05 pu For line to ground fault, the current is Ia =
Ea Z 0 + Z1 + Z 2 + 3Z N
0.1 j 0.2 + j 0.2 + j 0.34 + j 0.15 = − j 1.12 pu =
The base current is given by IB =
Generator MVA 3 Generator KV
=
20 × 106 3 × 6.6 × 103
= 1750 A
11/21/2018 11:51:22 AM
288
GATE EE Chapter-wise Solved Papers
Therefore, the fault current is
By equal area criterion, we get
I f = 3 I a I B = 3( − j1.12) (1750) = − j 5897.6 A
Ps (δ max − δ 0 ) = ∫
Neutral voltage is given by
= Pmax [cos δ 0 − cos δ max ]
where
P0 = Pmax sin δ 0 = 1 2 sin δ 0 = 1 ⇒ δ = 30°
VN = 5897.6 × 0.1089 = 642.2 V 39. Topic: Power Generation Concepts (a) For optimal power generation, P3 = maximum load = 300 MW and IC1 = IC2
Substituting the value of d in Eq. (1) and using given value d max = 110°, we get 2 sin 30°(110 − 30) 0.5 ×
Therefore, 0.3 P1 − 0.4 P2 = 10 (1)
δ c = 69.138° ≈ 69.14 42. Topic: Models and Performance of Transmission Lines and Cables (d) The three phase currents are
P1 + P2 + P3 = 700 P1 + P2 + 300 = 700 P1 + P2 = 400 (2)
Solving Eqs. (1) and (2), we get P1 = 242.8 MW and P2 = 157.14 MW 40. Topic: Differential and Distance Protection (a) Group I
Group II
Distance relay
Transmission lines
Under frequency relay
Alternators
Differential relay
Busbars
Buchholz relay
Transformers
1 A1
Ch wise GATE_EE_Ch6.indd 288
Eb 10 ∠ − 90° = = 3.33∠ −180° Zb 3j
Ic =
Ec 10 ∠120° = 2.5∠30° = Zc 4j
I a1 =
1 [I a + α I b + α 2 Ic ] 3
I a1 =
1 ⎡5∠ − 90° + 3.33∠( −180° + 120°) ⎤ ⎥ 3 ⎢⎣ +2.5∠( 240° + 30°) ⎦
1 [5∠ − 90° + 3.33∠ − 60° + 2.5∠270°] 3 1 = [ − 5 j + 1.665 − j 2.883 − 2.5 j ] 3 1 = [1.665 − j10.383] 3 = 3.5∠ −80. 89°
A2
dc
Ib =
=
2
30°
Ea 10 ∠0° = = 5∠ − 90° Za 2j
2 [∵ α = 1 ∠120°α = 1 ∠240°]
P = 2 sind
P1
Ia =
We know that
41. Topic: Per-Unit Quantities (c) From the given data, we have
Pmax
π = 2 (cos δ c − cos 110°) 180
80π = cos δ c + 0.342 180 cos δ c = 0.698 − 0.342
20 + 0.3 P1 = 30 + 0.4 P2
(1)
Given that Pmax = 2 pu, therefore
(6.6) 2 × 0.05 = 0.1089 Ω 20
Therefore,
Also,
Pmax sin δ ⋅ dδ
Pmax sin [δ 0 (δ max − δ 0 )]
VN = I f Z N
Z N = Z B Z pu =
δ max
δc
d max
d
43. Topic: Models and Performance of Transmission Lines and Cables (c)
11/21/2018 11:51:23 AM
Chapter 6 • Power Systems
44. Topic: Models and Performance of Transmission Lines and Cables (b) The phase sequence is RBY
This is because impulse current will be equally divided in both directions. Therefore, substituting values, we have V=
R Y
10 × 103 × 250 = 1250 kV 2
R 51. Topic: Models and Performance of Transmission Lines and Cables (c) Given the circuit parameter constants A, B, C and D and
Y B B 45. Topic: Power Generation Concepts (c)
PR = 50 MW
α = 0.98o
Vr = 220 kV
β = 76.4o
46. Topic: Power Generation Concepts (a) Kaplan turbines are used for harnessing low variable water heads. This is because of high percentage of reaction and runner adjustable vanes.
pf = 0.9 lag
47. Topic: Differential and Distance Protection (b) One of the types of distance relay is mho. This has the property of being inherently directional which is used for transmission line protection. 48. Topic: Distribution Systems (c) Given that V = 800 kV, L = 1.1 mH/km and C = 11.68 nF/km. Then, Surge impedance, 1.1 × 10 −3 Z = L/C = = 306.88 Ω 11.68 × 10 −9 Power ( P ) =
V 2 (800) 2 = = 2085 MW Z 306.88
Pr = 50 =
Vs Vr B
cos( β − δ ) −
Vs × 220
A Vr B
cos( 76.4° − δ ) −
142 Vs [cos( 76.4 − δ )] = 83.46
2
cos(β − α ) (0..936) ( 220) 2 cos( 75.6°) 142 (1)
Also, for reactive power we have Qr = PR tan ϕ = PR tan(cos −1 ϕ ) = 50 tan(cos −1 0.9) = 24.2 MW Vs sin(76.4 − δ ) = 215 (2) From Eqs. (1) and (2), we get 2
V = 110 kV; C = 125 nF/km tan δ = 2 × 10 −4 Dielectric power loss is given by
Vs = ( 215) 2 + (83.46) 2 = 230.63 kV 52. Topic: Per-Unit Quantities (b) Given that Eg = 1.4 ∠30° pu; ETh = 0.90 ∠0°;
PL = 2V 2ω C tan δ = 2 (110 × 103 ) 2 2 π f × 125 × 10 −9 × 2 × 10 −4 = 2 (110 × 103 ) 2 2π × 50 × 125 × 10 −9 × 2 × 10 −4 = 189 W/km 50. Topic: Principles of Over-Current (a) Given that I = 10 kA V = 400 kV and Z = 250 Ω. Magnitude of transient over voltage V =
We know that
Vs = ( 215) 2 + (83.46) 2
49. Topic: Power Factor Correction (d) Given for the cross-linked polyethylene power cable:
Ch wise GATE_EE_Ch6.indd 289
289
IZ 2
X s = 1.0 pu and Z Th = 0.25∠90 ° Thevenin’s equivalent circuit is given by ZTh
Xs I
+ −
ETh
Vt
+ E g −
11/21/2018 11:51:25 AM
290
GATE EE Chapter-wise Solved Papers
Therefore,
Then, I = =
Eg − ETh Z Th + X s
=
1.4 ∠30° − 0.9∠0° j1.25
I a0 =
1.212 + j 7 − 0.9 j 1.25
= 0.56 − 0.2496 j Vt = Eg − IX s = 1.212 + j 7 − (0.56 − 0.2496 j ) (1 j ) = 0.972∠8.3° 53. Topic: Circuit Breakers (c) Given that, rating = 110 MVA, voltage = 11 kV, sub-transient reactance Xd ″ = 19%, transient reactance Xd′ = 26% and synchronous reactance = 130%. Short circuit current I sc 1 1 = = − j 5.26 pu X d′′ j 0.19
1∠0° j 0.15 + j 0.15 + j 0.05
=
1 = −2.857 j 0.35 j
I f = 3I a ( 0 )
(∵ I a ( 0 ) = I a (1) = I a ( 2 ) )
= 3( − 2.857 j ) = 8.57 A ≈ 8.533 A 56. Topic: Symmetrical and Unsymmetrical Fault Analysis (b) Given for the 500 MVA and 22 kV generator, that f = 50 Hz, number of poles, p = 4 and power factor lag = 0.8. Also, the output is reduced by 40% when fault occurs. Power (without fault) = 500 × 0.8 = 400 MW Power (after fault) = 400 × 0.6 = 240 MW
Base current IB =
=
Sub-transient line current for a single line to ground fault is given by
= 1.212 − 0.249 + j (0 0.7 − 0.56)
I sc =
Ea X 0 + X1 + X 2
Generator MVA
3 generator KV = 5773.67A
=
110 × 10
6
3 × 11 × 10
3
I rms = I B × I sc = 5773.67 × 5.26 = 30369.5 A = 30.37 kA
Then accelerating torque at the time of fault can be determined as Pa = Ta × ω Ta =
Pa ω
( where ω = 2π f mech)
2 2 50 = 50 × = = 25 p 4 2 54. Topic: Series and Shunt Compensation − 240 = 160 MW Pa = I400 (a) From the figure, we have I a = 10 ∠0° A, I b = 10 ∠180°A and c = 0 I a = 10 ∠0° A, I b = 10 ∠180°A and I c = 0 160 Ta = = 1.0019 MNm 2π × 25 Then positive sequence current component in line a will be f mech = f elec ×
1 I a1 = [ I a +α I b +α 2 I c] 3 1 = [10∠0° + (1∠120° × 10 ∠180°) + 0] 3 1 = [10 ∠0° + 10 ∠300°] 3 1 15 − j 8.66 = [10 + 5 − j 8.66] = 3 2 17.32 ∠ −30° = 5.78 ∠ − 30° = 3 55. Topic: Symmetrical and Unsymmetrical Fault Analysis (d) Given 500 MVA three-phase turbo generator, with f = 50 Hz, V = 22 kV, X 1 = X 2 = 0.15 and X 0 = 0.05 pu
Ch wise GATE_EE_Ch6.indd 290
57. Topic: Power Generation Concepts (d) Given that for the turbine, rated speed N = 250 rpm and f = 50 Hz. We know that, N=
120 f 120 f ⇒ p= p N
where p is the number of poles. Substituting values, we get p=
120 × 50 = 24 250
58. Topic: Per-Unit Quantities (d) Given that, at 500 MVA, M = 20 pu and X = 2 pu. We know that the relation between old and new pu values is given by pu (new) = pu (old)
11/21/2018 11:51:26 AM
The pu values of inertia (M) can be determined as: M (new) (pu) = M (old) (pu) × = 20 ×
6 • Power Systems QL Chapter − QC = 0.25 PL SubstitutingSubstituting values, we get values
QL − 2 = 0.25 ⇒ QL = 3 MVAR 4
MVA (old) MVA (new)
⎛Q ⎞ ⎛ 3⎞ ϕ = tan −1 ⎜ L ⎟ = tan −1 ⎜ ⎟ = 36° ⎝ 4⎠ ⎝ PL ⎠
500 = 100 pu 100
Similarly, reactance (X) can be determined as X (new) (pu) = X (old) (pu) = 2×
MVA (new) MVA (old)
100 = 0.4 pu 500
cos ϕ = cos 36° = 0.8 lag 64. Topic: Bus Admittance Matrix (d) For the given network I1 = V1Y11 + (V1 − V2 )Y12 = 0.05 V1 − j10 (V1 − V2 )
59. Topic: Series and Shunt Compensation (d) Given that P is the maximum power transfer capacity of 800 kV transmission line. Then, the maximum power transfer capacity of 400 kV can be determined from the following relation P=
EV sin δ ⇒ P ∝ V 2 X
= − j 9.95 V1 + j10V2 I 2 = (V2 − V1 )Y21 + (V2 − V3 )Y23 = j10V1 − j 9.9 V2 − j 0.1V3 Then, Y22 = Y11 + Y23 + Y2 = − j 9.95 − j 9.9 − j 0.1 = − j19.95
Therefore, when V is halved; power transfer capacity becomes 1/4 of the pervious value. 60. Topic: Models and Performance of Transmission Lines and Cables
291
65. Topic: Power Generation Concepts (c) Given that F1 = a + bP1 + CP12 Rs/h
(b) 61. Topic: Models and Performance of Transmission Lines and Cables (a) 62. Topic: Models and Performance of Transmission Lines and Cables (d) Given that X s = 0.4 Ω / km
F2 = a + bP2 + 2CP22 Rs/h
X 1 = X s − X m = 0.4 − 0.1 = 0.3 Ω /km
P1 = 300 − P2
For economical operation ∂F1 ∂F2 = ∂P1 ∂P2
(2)
0 + b + 2CP1 = 0 + b + 4CP2
X m = 0.1 Ω / km
Then positive sequence reactance is
(1)
P1 + P2 = 300 MW
P1 = 2 P2
From Eqs. (1) and (2),we get (300 − P2 ) = 2 P2
Zero sequence reactance X 0 = X s + 2 X m = 0.4 + 2(0.1) = 0.6 Ω /km 63. Topic: Power Factor Correction (c) Given that PL = 4 MW load, QC = 2 MVAR for load and power factor pf = 0.97 lag. Then, cos ϕ = 0.97 ⇒ ϕ = cos −1 0.97 −1
tan ϕ = tan(cos 0.97) = 0.25 QL − QC = 0.25 PL
300 = 3P2 ⇒ P2 = 100 MW; P1 = 200 MW 66. Topic: Bus Admittance Matrix (b) We know that ⎡V1 ⎤ ⎡ A B ⎤ ⎡V2 ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎣ I1 ⎦ ⎣C D ⎦ ⎣ I1 ⎦ B=
V1 I2
C= V2 = 0
I1 Y2
I2 = 0
Substituting values QL − 2 = 0.25 ⇒ QL = 3 MVAR 4 Ch wise GATE_EE_Ch6.indd 291
11/21/2018 11:51:28 AM
292
GATE EE Chapter-wise Solved Papers
The reactance is given by
From the given figure,
C=
⎛ V1 ⎞ ⎜⎝ Z + Z ⎟⎠ 1 2
X =
⎛ V1 ⎞ ⎜⎝ Z + Z ⎟⎠ × Z 2 1 2
=
1 Z2
Vph If
If =
1 1 = = 40 ∠ − 45° C 0.025∠45°
I a(1) = I a(2) = I a(0) =
67. Topic: Per-Unit Quantities (d) The given circuit can be represented as 1.0 pu
1
2
0.12 V
X 0 + X1 + X 2 = =
x T1
1.0 pu
Steady state stability power limit Pm is given by EV Pm = X
X 2
X = 0.08 pu In one of the double circuit is tripped Pm ( new ) =
1 EV 1×1 = = = 5 pu X 0.12 + 0.08 0.2
68. Topic: Electric Field Distribution and Insulators (c) Given that Tst = 15 Nm, TL = 7 Nm and a = 2 rad/s2. We know that T = Tst − TL = 15 − 7 = 8 Nm T=I α 8 I = = 4 kgm 2 2 69. Topic: Symmetrical and Unsymmetrical Fault Analysis (d) Given that substation level = 220 kV; three-phase fault level = 4000 MVA and ground fault level = 5000 MVA. The fault current is If =
Ch wise GATE_EE_Ch6.indd 292
)
= 12.1 Ω
5000 3 × 220 5000
If = 3 3 × 3 × 220
Vph I a(1)
220 / 3 5000 / 3 3 × 220
=
127 = 29.036 4.37
Since, X 1 = X 2 = −12.1 Ω Therefore,
71. Topic: Power Generation Concepts (b)
1×1 0.12 +
3 × 220
X 0 = 29.036 − 12.1 − 12.1 = 4.836 Ω ≈ 4.84 Ω
Substituting values, we have 6.25 =
(
3 x
G
4000 /
70. Topic: Symmetrical and Unsymmetrical Fault Analysis (b) The fault current is
Therefore, Z2 =
220 / 3
=
4, 000 3 × 220
72. Topic: Circuit Breakers (a) 73. Topic: Differential and Distance Protection (b) 74. Topic: Models and Performance of Transmission Lines and Cables (b) 75. Topic: Per-Unit Quantities (d) We know that Real power, P = S cos ϕ = 12 × 3 × 0.8 = 16.6 kW Reactive power, Q = S sin ϕ = 12 × 3 × 0.6 = 12.5 kW where S is the generator ratings. For unity power factor, we have to set capacitor bank to Q = 12.5 kW. 76. Topic: Distribution Systems (c) Given that A = D = 0.94 ∠1°; B = 130 ∠73°; C = 0.001 ∠90°; Vs = 240 kV
11/21/2018 11:51:29 AM
293
Chapter 6 • Power Systems
We know that % voltage regulation =
Vr(no load) − Vr(full load) VR(full load)
Ib × 100
Ir(no load) = 0. Therefore, Vr(no load) =
Vs 240 = = 255.32 A 0.94
So, % voltage regulation 255.32 − 220 × 100 = 16% 220
77. Topic: Symmetrical and Unsymmetrical Fault Analysis (c) Let P be the power line and T be telephone line. Let points P1T1 and P2T2 denote the points of shortest (Ds) and maximum (Dm) distance between the two lines. Then inductance is given by ⎛D ⎞ LP = 2 × 10 ln ⎜ m ⎟ H/m ⎝D ⎠ −7
s
⎛D ⎞ LT = 2 × 10 −7 ln ⎜ m ⎟ H/m ⎝ Ds ⎠ Total inductance due to conductor P and T, L = LP + LT , that is, ⎛D ⎞ L = 4 × 10 −7 ln ⎜ P2T2 ⎟ H/m ⎝ DP1T1 ⎠ ⎛ 3 ⎞ = 4 × 10 −7 ln ⎜ = 0.73 × 10 −7 H/m ⎝ 2.5 ⎟⎠ Hence, voltage induced by power line into telephone line is VT = ω LI = 2π fLI = 2 × 3.14 × 50 × (0.73 × 10 −7 ) × 100 = 2.29 V/km 78. Topic: Per-Unit Quantities (b) Given that Vab1 = X∠q1, Vab2 = Y∠q2. The sequence for given star connected resistors can be represented as shown in the following figure. We know that in positive sequence, line voltage leads phase voltage by 30° and converse is true for negative sequence. Therefore, Van1 = X ∠θ1 − 30° and Van2 = Y ∠θ 2 + 30°
Ch wise GATE_EE_Ch6.indd 293
Vbn n
Van
Ia Za
Zb k Zc
Vcn Ic c 79. Topic: Per-Unit Quantities (a) We know that
Vr(full load) = 220 kV
=
a b
⎛ MVA (old) ⎞ ⎛ KV(new) ⎞ Z new = Z old × ⎜ ⎟ ⎜ ⎟ ⎝ MVA (new) ⎠ ⎝ KV(old) ⎠
2
2
= 0.72 ×
20 ⎛ 69 ⎞ ×⎜ ⎟ = 36 pu 10 ⎝ 13.8 ⎠
Z new = 36 pu 80. Topic: Gauss Seidel and Newton-Raphson Load Flow Method (a) 81. Topic: Per-Unit Quantities (c) Given that V = 1 pu; E = 1.1 pu, Ps = 1.0 pu; 1/X = 0.8 pu; power angle d = 30° and f = 50 Hz. We know that accelerating power is given by Pa = Ps − Pe = 1−
EV 1.1 × 1 sin δ = 1 − sin 30° X 1/ 0.8
= 0.56 pu 82. Topic: Per-Unit Quantities (b) Let Pa be the initial acceleration power, then acceleration is given by a=
Pa X ( pu ) × S = M SH /180 f
where H is normalised inertia constant and S is generator MVA rating. Substituting values, we have 180 × 50 × X × S S ×5 = 1800 X deg /s 2
a=
Intertial constant =
1 = 0.056 18
11/21/2018 11:51:30 AM
294
GATE EE Chapter-wise Solved Papers
83. Topic: Symmetrical and Unsymmetrical Fault Analysis (d) The pre‑fault voltage is
Vbus
⎡V1 ⎤ ⎡1∠0°⎤ ⎢ = ⎢V2 ⎥⎥ = ⎢⎢1∠0°⎥⎥ ⎢⎣V3 ⎥⎦ ⎢⎣1∠0°⎥⎦
The fault occurs at bus 2, so r = 2. For solid fault Zf = 0. Then the fault current is Vr ( 0 ) V2( 0 ) = Z rr + Z f Z 22 + Z f
If = =
Minimum cost for Generator B = 650 at 150 MW. Generator A = 450 at 200 MW. The maximum cost of generator A is less than minimum cost of generator B, so PA = 450 MW PB = 700 − 450 = 250 MW. 86. Topic: AC and DC Transmission Concepts (c) 87. Topic: Electric Field Distribution and Insulators (b) The electric field intensity at the given points is
1∠0° = −4 j j 0.24
E1
Let the pre‑fault voltage be Vi (0) , then post fault voltage
E3
E2
E3
E2
X Vi( f) = Vi (0) − Z ir ⋅ I f E1
Using values from the given impedance matrix V1(f) = V1 (0) − Z12 I f = 1∠0° − j 0.08 ( − j 4)
X
Y
= 0.68 pu E1
V3(f) = V3 (0) − Z32 I f = 1∠0° − j 0.16 ( − j 4)
E2
E2
E3
= 0.36 pu
E3
E1 84. Topic: Symmetrical and Unsymmetrical Fault Analysis (c) The post fault voltage current is given by I g1(f) = [ Eg1 − V1(f) ]Y10
(1)
I g2(f) = [ Eg2 − V2(f) ]Y20
(2)
From the given admittance matrix Y10 = Y11 + Y12 + Y13 = − j8.75 + j1.25 + j 2.5 = − j 5 Y20 = Y21 + Y22 + Y23 = − j1.25 + j 6.25 + j 2.5 = − j 2.5 Substituting values in Eqs. (1) and (2), we get I g1(f) = [1∠0° − 0.68]( − j 5) = − j1.65 I g2(f) = [1∠0° − 0)]( − j 2.5)) = − j2.5
Z
W
Thus, the maximum electric field intensity will be observed at point Y as there will be minimum cancellation of the electric fields generated due to the three points. 88. Topic: AC and DC Transmission Concepts (d) The value of the capacitor DC voltage can be increased to a new steady state value by making d negative and returning to its original value. 89. Topic: Models and Performance of Transmission Lines and Cables (b) Given that X = 0.045 pu, then inductance is L=
0.045 0.045 = 2π × 50 2π f
Given that susceptance = 1.2 pu, then capacitance is C=
85. Topic: AC and DC Transmission Concepts (c) Given that PA + PB = 700 MW Maximum incremental cost for, Generator A = 600 at 450 MW. Generator B = 800 at 400 MW.
Ch wise GATE_EE_Ch6.indd 294
1 1 × 2π × 50 1.2
The velocity of wave propagation v = 3 × 105 km/s. We now that v=
l LC
⇒ l = v LC
11/21/2018 11:51:32 AM
295
Chapter 6 • Power Systems
Therefore, substituting values, length is
90. Topic: Circuit Breakers (c) Given that the fault F occurs at the midpoint and failure occurs at circuit breaker 4. So the correct sequence of operation is 5-6-7-3-2-1. This is based on the back-up mechanism for protection, wherein relays are so coordinated that relay in the section where fault occurs operate first and then the relays in the adjoining sections, in case the first relay fails. 91. Topic: Bus Admittance Matrix (a) We have from the given matrix van 3
(ib − ic ) +
vbn 3
(ic − ia ) +
vcn 3
(ia − ib )
Setting i b − ic i b − ic = I and van = V = I3and van = V 3 We have, R = 3VIR .= 3VI 92. Topic: Per-Unit Quantities (c) Let P1e be the electric power before tripping of circuit and P2e be the electric power after tripping. We know that. P=
⎡ fp ⎤ ⎡ fa ⎤ ⎡1 ⎢ ⎥ ⎥ ⎢ f P = ⎢ f b ⎥ , fS = ⎢ f n ⎥ and A = ⎢⎢α 2 ⎢f ⎥ ⎢⎣ f c ⎥⎦ ⎢⎣ α ⎣ o⎦
0.045 = 185 km 2π × 50 × 2π × 50 × 1.2
l = 3 × 105
R=
Therefore, from the given matrix equation
Similarly, VP = kAVS and I P = kAI S
Also,
⎡0.5 0 0 ⎤ Z = ⎢⎢ 0 0.5 0 ⎥⎥ ⎢⎣ 0 0 2⎥⎦ ⎡ 1 α α2 ⎤ 1 ⎢ ⎥ A−1 = ⎢ 1 α 2 α ⎥ 3 ⎢1 1 1 ⎥⎦ ⎣ ⎡ 1 α α2 ⎤ 1⎢ ⎥ VP = ⎢ 1 α 2 α ⎥ 3 ⎢1 1 1 ⎥⎦ ⎣ ′
X 2 = sin 130° = 0.766
α α2 2
1⎤ 1⎥⎥ I P 1⎥⎦
⎡ A−1 ⎤ −1 = kA Z ′ ⎢ ⎥ I P = AZ ′A I P k ⎣ ⎦ ⎡ 1 0.5 0.5⎤ VP = ⎢⎢0.5 1 0.5⎥⎥ I P ⎢⎣ 0.5 0.5 1 ⎥⎦ So, ⎡ 1 0.5 0.5⎤ ⎢ Z = ⎢0.5 1 0.5⎥⎥ ⎢⎣0.5 0.5 1 ⎥⎦ 94. Topic: Bus Admittance Matrix (a) The two power systems are connected by short transmission lines as P1 = P2 = Q1 = Q2 = 0 The phase angle difference for no energy transfer is
θ = 30° − 20° = 10°
We have X 2 = 0.1 + X X = 0.766 − 0.1 = 0.666 ≈ 0.67 93. Topic: Bus Admittance Matrix (b) We have f P = kAfS
Ch wise GATE_EE_Ch6.indd 295
1
VP = kA Z ′ [ I S ]
P2e = Pm
1.0 × 1.0 sin 130° = 1 X2
⎡0.5 0 0 ⎤ ⎡ 1 ⎢ 0 0.5 0 ⎥ ⎢α 2 ⎢ ⎥ ⎢ ⎢⎣ 0 0 2⎥⎦ ⎢⎣ α
Therefore,
where X 2 = (0.1 + X ) pu.
EV sin δ 2 = Pm X2
VS = Z ′ [ I S ]
Then from the given matrix equation, we have
EV EV EV sin δ ; Pmax = ; P2 max = X X X2
Since the system does not lose synchronism subsequent to tripping,
1 1⎤ α 1⎥⎥ α 2 1⎥⎦
95. Topic: Power Factor Correction (b) Given that V = 230 V, P = 4 kW (load), f = 50 Hz, IN = 0 Since, IN = 0, we have
I A + I B + I C = 0 (1)
11/21/2018 11:51:33 AM
296
GATE EE Chapter-wise Solved Papers
98. Topic: Models and Performance of Transmission Lines and Cables (d) Given that length l = 300 km, b = 0.00127 rad/km. We know that
We know that P = VI and therefore, IA =
P 4 × 10 = = 17.39A V 230 3
From Eq. (1),
λ=
IA = −IB − IC
l 300 %= × 100 = 6.06% λ 4947.4
⎛ 3 3⎞ IA = − ⎜ IB. + IC . ⎟ 2 2 ⎠ ⎝
[∵ I
I A = 3I B = 3I C IB =
17.39
XC =
3
= 10 A ; X C =
B
IC ]
99. Topic: Series and Shunt Compensation (b) We know that, 0 0 ⎤ ⎡ Ia ⎤ ⎡ ΔVa ⎤ ⎡( Zs − Z m ) ⎢ ΔV ⎥ = ⎢ ⎥ ⎢ ⎥ 0 0 ( Zs − Z m ) ⎢ b⎥ ⎢ ⎥ ⎢Ib ⎥ ⎢⎣ ΔVc ⎥⎦ ⎢⎣ 0 0 ( Zs + 2 Z m ) ⎥⎦ ⎢⎣ I c ⎥⎦
V 230 = = 23 Ω IC 10
1 1 1 ;C = = = 139 μF 2π f C 2π fX c 2π × 50 × 23
Zero sequence impedance
Thus, X L (or 2π fL) = L=
V 230 = = 23 Ω IL 10
23 23 = = 72.95 mH 2π f 2π × 50
Therefore, L = 72.95 mH in phase B and C =139 mF in phase C. 96. Topic: Symmetrical Components (a) Change in frequency w.r.t to power, Drop out frequency × Change in power Rated power 5 = × 3.5 = 1.16 15
Δf =
The change in terms of actual frequency Δf = 1.16 ×
50 = 0.58 Hz 100
Therefore, the overall system frequency = 50 - 0.58 = 49.42 Hz. 97. Topic: Symmetrical and Unsymmetrical Fault Analysis (d) Let phase voltage be Va and phase current be Ia. V Bus voltage = aΩ Current from phase a I a
Ch wise GATE_EE_Ch6.indd 296
Zs + 2 Z m = 48 (1)
Positive sequence impedance = Negative sequence
I m = Zs − Z m = 15 (2)
Solving Eqs. (1) and (2), we get Zs = 26 and Z m = 11 100. Topic: Per-Unit Quantities (a) Distance protection using mho relays with Zone‑1 set to 80% of the line impedance. 101. Topic: Distribution Systems (b) SIL of the compensated transmission line will be 2280 MW as there will be no effect of capacitive compensation on the active power of transmission line. 102. Topic: Power Generation Concepts (c) Given that C1 ( PG1 ) = PG1 + 0.055 × PG2
1
C2 ( PG2 ) = 3PG2 + 0.03P
2 G2
and
PG1 + PG2 = 250 MW (1) Then, dC1 = 1 + 0.11 PG1 and dPG1
Impedance measured from ground distance =
2π 2π = = 4947.4 km β 0.00217
dC2 = 3 + 0.06 PG2 dPG2
(2)
11/21/2018 11:51:35 AM
297
Chapter 6 • Power Systems
Maximum value of DC offset current,
dC1 dC2 = dPG1 dPG2
For lossless system,
Ae − ( R / L )t0 = −
Therefore, from Eq. (2) 1 + 0.11 PG1 = 3 + 0.06 PG2 0.11 PG1 − 0.06 PG2 = 2
For negative maximum, ω t0 − α = 0 ⇒ t0 = α /ω (3)
Solving Eqs. (1) and (3), we get PG = 100 MW and 1 PG = 150 MW
It is given that, Z = 0.004 + j 0.04 = 0.0402∠84.29° Therefore,
α = 84.29° or 1.471 rad
2
103. Topic: Circuit Breakers (b) When the generators are connected in parallel and supplied 0.5 pu of power, there will be increase in the current. Hence, the clearance time will be reduced, not less than 0.14 seconds, so as to maintain transient stability. 104. Topic: Distribution Systems (d) We know that line loss is given by
Thus, t0 =
Z1 =
Line loss = Σ I R where R is the impedance.
1.471 = 0.00468 s = 4.68 ms 2π × f 0
106. Topic: AC and DC Transmission Concepts (c) Positive sequence impedance,
2
Z1 = Z 2 = Z 0
(i) Without e1 ; losses
I f(pu) =
= (1) R + (1 + 2) R + (1 + 2 + 5) R = 74 R 2
2
(ii) Without e2 ; loss
= 52 R + (5 + 2) 2 R + (5 + 2 + 1) 2 R = 138 R
IX = If ×
= (1) 2 R + 22 R + (5 + 2) 2 R = 54 R (iv) Without e4 ; loss = 2 R + ( 2 + 1) R + 5 R = 38 R 2
2
So, minimal loss can be achieved by removing e4
For symmetrical three phase fault i(t ) = Ae − ( R / L )t +
2Vm cos(ω t − α ) Z
Therefore i (t) = 0 and so
Ch wise GATE_EE_Ch6.indd 297
3 × 400
= 49.8 ×
100 3 × 400
= 7.18 kA
Z1 = Z 2 = 0.0201∠84.29° Z 0 = 3Z1 = 0.0603∠84.29° Ia 3 1.0 = = 9.95 pu 0.0201 + 0.0201 + 0.0603
I a0 = I a1 = I a2 = I a1
Fault current, I f = I a = 3I a1 = 29.85 pu
Let fault occur at t = t0
Ae − ( R / L )t0 +
100
107. Topic: Symmetrical and Unsymmetrical Fault Analysis (a) Line to ground fault is at phase a, therefore,
105. Topic: Symmetrical and Unsymmetrical Fault Analysis (a) Given that v(t ) = Vm cos ωt
1∠0° V = = 49.8 Z1 0.0201∠84.29°
Therefore, rms value of the AC component of fault current is
(iii) Without e3 ; loss
2
Z = 0.0201∠84.29° 2
For three phase fault
Consider the case for removal for each of the branch:
2
2Vm cos(ω t0 − α ) Z
2Vm cos(ω t0 − α ) = 0 Z
Therefore, the rms value of the AC component of fault current for phase ‘a’ is I X = 29.85 ×
100 3 × 400
= 4.9 kA
11/21/2018 11:51:37 AM
298
GATE EE Chapter-wise Solved Papers
108. Topic: Power Generation Concepts (a) 109. Topic: Models and Performance of Transmission Lines and Cables (c) We know that
S1
Y10 1
2 Y12 Y24
S = P + jQ = VI [cos ϕ + j sin ϕ ] = VIe jϕ ⇒ I=
S2
Y20
Y23
S Ie jϕ
Y13
Also, power loss 2
Ploss
S2R ⎛ S ⎞ = I R = ⎜ jϕ ⎟ R = 2 jϕ 2 ⎝ Ve ⎠ V ⋅ (e )
4
3
2
Y43 S3
S4
Thus, PL ∝
1 V2
Therefore, branches (1) and (2) behave like shunt element. 111. Topic: Series and Shunt Compensation (b)
2
where S 2 R e jϕ is constant. 110. Topic: Bus Admittance Matrix (c) In general,
Ybus
⎡Y11 ⎢Y = ⎢ 21 ⎢Y31 ⎢ ⎣Y41
Y12 Y22 Y32 Y42
Y13 Y14 ⎤ Y23 Y24 ⎥⎥ Y33 Y34 ⎥ ⎥ Y43 Y44 ⎦
Y11 = Y10 + Y12 + Y13 + Y14 = −5 j Y22 = Y20 + Y21 + Y23 + Y24 = −10 j Y33 = Y30 + Y31 + Y32 + Y34 = −9 j
Elements
Uses
Shunt capacitor
Improving PF
Series reactors
Reduce the current ripples
Series capacitor
Increase power flow in line
Shunt reactors
Reduce the Ferranti effect
112. Topic: Models and Performance of Transmission Lines and Cables (c)
Y12 = Y21 = −Y12 = 2 j
Transmission line
Distance relay
Y13 = Y31 = −Y13 = 2.5 j
Short line
Ohm reactance
Y14 = Y41 = −Y41 = 0 j
Medium line
Reactance relay
Y24 = Y42 = −Y24 = 4 j
Long
Mho relay
Therefore, Y10 = Y11 − Y12 − Y13 − Y14 = −5 j + 2 j + 2.5 j + 0 j = −0.5 j Y20 = Y22 − Y12 − Y23 − Y24 = −10 j + 2 j + 2.5 j + 4 j = −1.5 j
113. Topic: Power Generation Concepts (c) Generator having better regulation share more power, for increased load power. Therefore, G-1 shares more power than G-2 114. Topic: Distribution Systems (a) Inertia constant
Similarly, Y30 = 0 , Y40 = 0
Ch wise GATE_EE_Ch6.indd 298
H=
Kinetic energy storted MVA rating of generator
(1) (i)
11/21/2018 11:51:39 AM
Chapter 6 • Power Systems
Therefore, complex power is given by
Rating G=
100 ∠30° − 100 ∠0° 5j = 2000∠30° − 2000∠60° = 1035∠15°°
500 MW P = = 555.56 MVA 0.9 cos ϕ
P + jQ = VI * = 100 ∠30°
Therefore, kinetic energy stored KE =
299
1 1 ⎛ 2π N ⎞ mω 2 = M ⎜ ⎟ 2 2 ⎝ 60 ⎠
2
(ii) (2)
Reactive power, Q = 1035 sin 15° = 268 VAR
Here, N is given in terms of frequency and number of poles as N=
120 f 120 × 50 = = 3000 rpm p 2
Substituting values in Eq. (2), we get 1 ⎛ 2π × 3000 ⎞ KE = × 27.5 × 103 × ⎜ ⎟⎠ ⎝ 2 60 = 1357.07 MJ Therefore, from Eq. (1), inertia constant is H=
Therefore, real power, P = 1035 cos 15° = 1000 W
1357.07 = 2.44s 555.56
119. Topic: Circuit Breakers (c) Symmetrical breaking current is given by, Ib =
MVA 3 kV
kA =
2000 3 × 33
= 34.9 kA ≈ 35 kA
120. Topic: Differential and Distance Protection (c) Current through operating coil 5 I OC1 = 220 × = 2.75 400 5 I OC2 = 250 × = 3.125 400 Net operating coil current = 3.125 − 2.750 = 0.375 A
115. Topic: Models and Performance of Transmission Lines and Cables (c) From the given diagram, we understand that I > 0; rectifier output VAB > 0 and inverter input VCD > 0. Therefore, I > 0 and VAB > VCD. 116. Topic: Electric Field Distribution and Insulators (a)
121. Topic: Models and Performance of Transmission Lines and Cables (c) Three phase star connected transformer with neutral grounded has infinite impedance to zero sequence current. A delta connected circuit, provided path for the zero sequence circuit to flow through the line and it circulates within the winding. r
R 117. Topic: Per-Unit Quantities (a) From the circuit, we have V A G + −
VL
V + VL = 0 Therefore,
Therefore, VL = −V = −1 pu
118. Topic: Bus Admittance Matrix (a) Given that impedance = 5j Ω; Vbus1 = 100 ∠30° V and Vbus 2 =100 ∠0° V.
Ch wise GATE_EE_Ch6.indd 299
122. Topic: Electric Field Distribution and Insulators (d) Due to Ferranti effect, the magnitude of terminal voltage does not change and the field current increases. 123. Topic: Distribution Systems (b) For the given distribution system
6C e2 5C e1 11 kV 3
11/21/2018 11:51:40 AM
300
GATE EE Chapter-wise Solved Papers
126. Topic: Models and Performance of Transmission Lines and Cables (a) Double circuit transmission enables the transfer of more power over a particular distance and is hence more preferred at lower voltage. Hence, circuit shown in option (a) is most suitable.
11 (6C ) 11 6 3 = × = 3.46 kV e1 = 6C + 5C 3 11 11 5 × = 2.89 kV e2 = 3 11
127. Topic: Differential and Distance Protection (a)
124. Topic: Distribution Systems (a) Given that C1 = 0.2 μF and C2 = 0.4 μF Capacitance per phase is
128. Topic: Models and Performance of Transmission Lines and Cables (c) For EHV transmission line,
C = 3C1 + C2 = 3 × 0.2 + 0.4 = 1 μF Charging current IC =
V = V ωC XC
P∝
= V 2π f =
11 × 103 3
1 ( XL − XC )
Therefore, the most preferred method is to connect a series capacitive compensator in the line.
× 2 × 3.14 × 50 × 10 −6
129. Topic: Distribution Systems (a) We know that
=2A 125. Topic: Power Factor Correction (b) For the given power system, using 25 kV as the base voltage at generator G1 and 200 MVA as the MVA base, we have impedances for generators G1 and G2 as X G1 = X G2 = X old ×
MVA(new) ⎛ kV(old) ⎞ × MVA(old) ⎜⎝ kV(new) ⎟⎠
2
Power loss ∝ length of transmission line Power loss ∝ P 2 On comparing all the given options, based on this, we have that minimum loss generation schedule is as given in option (a). Alternately
200 ⎛ 25 ⎞ × ⎜ ⎟ = j 0.18 100 ⎝ 25 ⎠
All the three generators are identical (each 100 MW) and have the same fuel characteristics.
Similarly, impedances for the transformers T1 and T2 are
The generators G1 and G2 can be represented by equivalent generator G23 and load by P23, so
2
= j 0..9 ×
The total load is P = P1 + P2 + P3.
2
X T1 = j 0.12 ×
200 ⎛ 25 ⎞ × ⎜ ⎟ = j 0.27 90 ⎝ 25 ⎠
P = P1 + P23. The length of the lines are L1 and 3L1, therefore if resistance R12 = R, then R23 = 3R = 3R.
2
X T2 = j 0.12 ×
200 ⎛ 25 ⎞ × ⎜ ⎟ = j 0.27 90 ⎝ 25 ⎠
Power delivered by the generator is
The impedance for the transmission line is X line(pu) = X (Ω) ×
P = 3VI cos ϕ ⇒ I =
Sbase 220 150 × = j 0.62 ( kV ) 2 ( 220) 2
Therefore, P ∝ I = kI (1) (i )
So, the impedance diagram is j0.27 j0.18
G1
XG1
j0.62
where
j0.27 XG2
j0.18
G2
1 3V cos ϕ
is constant k .
Assuming that all generators are operating at same power factor, total transmission loss can be expressed as PL = I12 R12 + I 2 2 R23 = R( I12 + 3I 2 2 ) = k 2 R( P12 + 3P2 2 )
Ch wise GATE_EE_Ch6.indd 300
P 3 cos ϕV
[from Eq. (1)]
11/21/2018 11:51:41 AM
301
Chapter 6 • Power Systems
Using this relation power losses can be worked out for each of the given options and minimum transmission loss is with operating conditions given in option (a). 130. Topic: Bus Admittance Matrix (b) Number of bus = 3.
133. Topic: Bus Admittance Matrix (b) From the bus admittance matrix, Y11 + (Y12 + Yline ) + Y13 = 0 − j13 + ( j10 + Yline ) + j 5 = 0 Yline = − j 2
⎡Y11 Y12 Y13 ⎤ Y = ⎢⎢Y21 Y22 Y23 ⎥⎥ ⎢⎣Y31 Y32 Y33 ⎥⎦ 1 1 + = −10 j − 5 j = −15 j Y11 = j 0.1 j 0.2
The magnitude of Yline = +2 134. Topic: Electric Field Distribution and Insulators (d) Given that i1 (t ) = I m sin(ω t − ϕ 1) i2 (t ) = I m cos(ω t − ϕ 2)
Therefore, option (b) is correct match. 131. Topic: Models and Performance of Transmission Lines and Cables (a) From the given values, we have
Also cos(ϕ − 90°) = sin ϕ Therefore,
X L1 = 0.225 × 10 = 2.25 Ω X L 2 = 0.225 × 10 = 2.25 Ω Taking 100 MVA, 15 kV as base, we get the reactance of generator G1 as
i1 (t ) = I m cos(ω t − ϕ 1 − 90°) i2 (t ) = I m cos(ω t − ϕ 2 )
In the phasor form
2
X G1 = 0.25 ×
100 ⎛ 15 ⎞ × ⎜ ⎟ = 0.1 pu 250 ⎝ 15 ⎠
The reactance for the line is X L1
Sbase =X× ( KVbase ) 2 = 2.25 ×
100 = 1 pu = X L 2 (15) 2
(1)
I1 = I m ∠ϕ 1 + 90°; I 2 = I m ∠ϕ 2; Under balanced condition, I1 + I2 = 0, so I m ∠ϕ 1 + 90° + I m ∠ϕ 2 = 0 I m [cos(ϕ 1 + 90°) + cos ϕ 2)]
+ j I m [(sin ϕ 2 + sin(ϕ 1 + 90°)]
Equating real part to zero, we get
Also, Z base X G2
cos(ϕ 1 + 90°) + cos ϕ 2 = 0
152 = = 2.25 Ω 100 100 = 0.1 × = 0.1 pu 100
cos( ϕ 2ϕ=2 0 cos(ϕ ϕ 11 + + 90 90°°)) += cos − cos cos( ϕ 11 + − cosπϕ+2 ϕ ) cos(ϕ + 90 90°°)) = = cos( 2
ϕ 1 +° =90π°)+=ϕcos( π + ϕ 2) ϕcos( 1 + 90 2
Hence, option (a) matched correct. 132. Topic: Models and Performance of Transmission Lines and Cables (d) For parallel connection, Thevenin’s fault impedance is given by
Ch wise GATE_EE_Ch6.indd 301
MVA (base) X Th
=
Alternately
1.1 × 1.1 = 0.55 1.1 + 1.1
We know that for balanced current the angle difference between two currents should be 180° or −180°. Therefore, from Eq. (1), we have
100 = 181.82 MVA 0.55
−ϕ 1 + ϕ 2 − 90° = −180° ⇒ ϕ 1 = 90° + ϕ 2
X Th = Fault =
ϕ 1 + 90 π ° = π +ϕ2 ϕ1 = + ϕ2 π2 ϕ1 = + ϕ2 2
11/21/2018 11:51:43 AM
302
GATE EE Chapter-wise Solved Papers
135. Topic: Power Factor Correction (a) Penalty factor, L1 =
137. Topic: Bus Admittance Matrix (b) From given the figure, we have
1 ( Plant G1 ) ∂P 1− L ∂PG1
(Plant G1)
V1 = 1∠0° pu V2 = 1 pu SD1 = 1 pu SD2 = 1 pu Total complex power
SG1 = SD1 + SD 2 = 1 + 1 = 2 pu
We have PL = 0.5 P , so 2 G1
Power from bus 1 to bus 2
∂PL = 0.5( 2 PG1 ) = PG1 ∂PG2
P=
Therefore, =
1 L1 = 1 − PG2 Penalty factor L2 =
X 1×1 sin(ϕ 1 − ϕ 2) 0.5
sin(ϕ 1 − ϕ 2) = 0.5
( Plant G 2 )
ϕ 1 − ϕ 2 = sin −1 0.5 = 30°
∂PL = 0 ⇒ L2 = 1 ∂PG2
ϕ 2 = ϕ 1 − 30 o = 0 − 30° = −30° Therefore,
For power generation to be economical, C1 L1 = C2 L2 ⎛ 1 ⎞ (10000) ⎜ = 12500 × 1 ⎝ 1 − PG 2 ⎟⎠ 1 = pu 5
V1 = 1∠0° I12 =
V2 = 1∠ − 30°
V1 − V2 1∠0° − 1∠ − 30° = = 1 − j 0.288 pu Z j 0.5
We know that complex power demand is given by SD 2 = V2 I 2∗
1 For 100 MVA, PG1 = × 100 = 20 MW 5 2
1 ⎛ 1⎞ PL = 0.5( PG2 ) 2 = 0.5 ⎜ ⎟ = pu ⎝ 5⎠ 50 1 PL = × 100 = 2 MW 50 Therefore, total power PLtotal = PG1 + PG2 − PL 40 = 20 + PG2 − 2 PG 2 = 22 MW 136. Topic: Symmetrical and Unsymmetrical Fault Analysis (c) For double line ground fault (LLG), I positive = −( I negative + I 0 ) The given values satisfy this condition.
Ch wise GATE_EE_Ch6.indd 302
V1 V2 sin(ϕ 1 − ϕ 2)
We have
1 ∂P 1− L ∂PG2
PG 2
Z = j 0.5 pu
= 1∠ − 30° I 2∗ ⇒ I 2 = 1∠ − 30° pu I G = I 2 − I12 = 1∠ − 30° − (1 − j 0.288) = 0.268 ∠ − 120° Rating of the capacitor, QG1 = V2 VG = 1 × 0.268 = 0.268 pu 138. Topic: Symmetrical and Unsymmetrical Fault Analysis (d) Representing the generator on a single line diagram, we have E = 1.5 pu
3f fault
V = 1 pu ∞ j0.5 pu
j1 pu X = j1 + j 0.5 = j1.5 pu
11/21/2018 11:51:45 AM
303
Chapter 6 • Power Systems
Critical angle,
δ cr = cos −1 [(π − 2δ 0 ) sin δ 0 − cos δ 0 ] To find steady state torque angle δ 0 : PM = Pmax sin δ 0 Pmax =
Since,
E V
E V X
=
1.3∠ − 10° = 0.433∠80° 3 − ∠ − 90°
(b) From the figure we have P2 = 0.1 pu and P3 = 0.2 pu.
Therefore,
δ cr = cos −1 ⎡⎣[π − ( 2 × 0.523)] sin 30° − cos 30°⎤⎦ = cos −1 [( 2.095)(0.5) − 0.866] = cos −1 (0.1815) = 79.6° 139. Topic: Power Factor Correction (c) Given that impedance = (4 + j 3) Ω For maximum power transfer from load to source, RL = R s + X s = 4 + 3 = 16 + 9 2
2
2
2
= 25 = 5 Ω 140. Topic: Power Factor Correction (b) Given that I L = 10 ∠ − 150° and VL = 100 ∠60°. We know that complex power is given by S = VI ∗ = (100 ∠60°) × (10∠ − 150°)∗ = 1000 ∠ 210° = −866.022 − j 500
Active power = −866.02 W Reactive power = − j 500 VAR Since, both are negative, therefore load absorbs real power and reactive power.
Ch wise GATE_EE_Ch6.indd 303
1.3∠ − 10° 1.3∠ − 10° 1.3∠ − 10° = = Z33 + Z new j 0.5 + ( − j 3.5) − j3
143. Topic: Bus Admittance Matrix
⇒ δ 0 = 30° or δ 0 = 0.523 in radian
V Z
= sin δ 0
(1.5)(1) 0.5 = sin δ 0 1.5 sin δ 0 = 0.5
142. Topic: Per-Unit Quantities (d) We know that I=
X
We have PM =
141. Topic: Symmetrical Components (d) In the swing equation, the angle d is the angular displacement of an axis fixed to the rotor with respect to a synchronous rotation axis.
For a load flow problem P ⎡θ 2 ⎤ −1 ⎡ 2 ⎤ ⎢θ ⎥ = [ B ] ⎢ P ⎥ ⎣ 3⎦ ⎣ 3⎦ 1 ⎡ 1 ⎢X + X 12 13 B=⎢ ⎢ −1 ⎢ ⎣ X 23
−1 X 23 1 X 23 +
⎡1 + 1 −1 ⎤ ⎡ 2 =⎢ ⎥=⎢ ⎣ −1 1 + 1⎦ ⎣ −1
⎤ ⎥ ⎥ ⎥ ⎥ X 13 ⎦ −1⎤ 2 ⎥⎦
⎡θ 2 ⎤ ⎡ 2 −1⎤ ⎡ 0.1⎤ ⎡ 0 ⎤ ⎢θ ⎥ = ⎢ −1 2 ⎥ ⎢ 0.2⎥ = ⎢ 0.1⎥ ⎦ ⎣− ⎦ ⎣− ⎦ ⎣ 3⎦ ⎣ 144. Topic: Bus Admittance Matrix (c) In the given network P1 + P2 + P3 = 0. P1 = − P2 − P3 P1 = −0.1 + 0.2 = 0.1 pu VA (base) =
(100 × 103 ) 2 = 100 × 106 100
MVA (base) = 100 Real power P = 100 × P2 = 100 × 0.1 = 10 MW
11/21/2018 11:51:47 AM
304
GATE EE Chapter-wise Solved Papers
145. Topic: Electric Field Distribution and Insulators (b) 146. Topic: Symmetrical and Unsymmetrical Fault Analysis (c) Consider the following two cases for the given system
150. Topic: Models and Performance of Transmission Lines and Cables (a) For distortionless transmission, R G = L C Therefore, required condition is
Case I: If fault occurs at F1
rc = lg A
F1 then current is feeding into Bus A. It functions as generator, that is, delivers power to Bus A. Case II: If fault occurs at F2 B F2 Then the current is received by load, that is, F2 acts as load and absorbs power from the generator. Therefore, VF1 lags IF1 and VF2 leads IF2. 147. Topic: Bus Admittance Matrix (b) From the given Figures, we have Transformer connection
151. Topic: Models and Performance of Transmission Lines and Cables (b) The impedance offered by a fully transposed transmission line to positive and negative sequence currents is equal because such lines are completely symmetrical. As a result, the impedance offered is independent of the phase sequence of the balanced set of current. 152. Topic: Bus Admittance Matrix (332) Number of load (PQ) buses = 150; number of generator (PV) buses = 32. Therefore, Number of simultaneous equations is = (2 × PQ) + PV = (2 × 150) + 32 = 332 (minimum) 153. Topic: Per-Unit Quantities (c) Given that I a = 1∠( − 90°) pu (phase current) I b2 = 4 ∠( −150°) pu (negative sequence current) I c0 = 3∠90° pu (zero sequence current)
Zero sequence network
α = 1∠120°
Xot1
I a = I a1 + I a 2 + I a 0 (1)
D
148. Topic: Power Factor Correction (118.8) Given that P = 100 MVA, IL = 437.38, Vpu = 0.9. Now, in three-phase star connected load, power is given by P = 3VL I L Base value of VL =
Thus, Vpu =
100 × 10
6
3 × 431.38
Ia2 =
149. Topic: Series and Shunt Compensation (c)
4 ∠( −150) = 4 ∠( − 270) (2) 1 ∠120
Given I a 0 = I b 0 = I c 0 = 3∠90° (3)
Substituting the values of I a , I and I in Eq. (1), we get a0 a2 I a = I a1 + I a2 + I a0
Vactual ⇒ Vactual Vbase
= 0.9 × 132 kV = 118.8 kV
Ch wise GATE_EE_Ch6.indd 304
4 ∠ −150° = 1 ∠120 × I a 2
= 132 kV (base)
I b2 = α I a2
1 ∠( − 90°) = I a1 + 4 ∠( − 270°) + 3 ∠90° I a1 = 8 ∠ − 90° As, I b1 = α 2 I a1 = (1 ∠120°) 2 ⋅ (8 ∠( − 90°)) = (1 ∠240°) × (8 ∠ − 90°) = 8 ∠150°
11/21/2018 11:51:48 AM
Chapter 6 • Power Systems
For phase B, we have similar equation as (1) I b = I b0 + I b1 + I b2 = 3 ∠90° + 8 ∠150° + 4 ∠( − 150°) = 11.53 ∠154° pu 154. Topic: Per-Unit Quantities (0.1) Given that Ef = 1.3 pu, Xs = 1.1 pu, P = 0.6 pu, V = 1.0 pu, Q = ? EV P= sin δ Xs 1.3 × 1 0.6 = sin δ ⇒ sind = 0.507 1.1
400rI + 200rI - 40000r + 200rI - 60000 + 200rI 100000r = 0 Solving, we get I = 200 A. Current through the second branch (B) = I - 200 = 200 - 200 = 0 A. Hence, source S1 does not supply current to point P. Source S2 supplies current of 100 A. 157. Topic: Power Factor Correction (b) G1
B1 V1 Ð 0°
B2 1 Ð d2 j0.1
So, d = 30.5°. Therefore, Q= =
1.0 + j0.5 j2
Vs [ E cos δ − V ] Xs
Load
1 [1.3 cos(30.5) − 1] = 0.1092 ≈ 0.1 1.1
IS
155. Topic:
Models and Performance of Transmission Lines and Cables (a) Given that the phase sequence of bus voltage is R-B-Y. The phase sequence of generator voltage is R′-Y′-B′. According to the given lamp connection, they become dark in the sequence La → Lb → Lc
⎡1 0.1 ∠ 90°⎤ ⎢0 ⎥ 1 ⎣ ⎦
BIr = Vs - AVr Ir =
200 m
S1
S2 I 200 A
100 A
[ j = ∠ 90°]
Vs = AVr + BIr
156. Topic: Distribution Systems (d) From the circuit for distribution feeder 200 m
200 A
Vs AVr − B B
Given that A = 1∠0°, B = 0.1∠90°, Vs = V1∠0°, Vr = 1∠δ 2 .
Let I is current from source S1, r is resistance per length, X is negligible.
Ir =
Current through first line (400 m) = I, current through second line (200 m) = (I - 200) A, current through third line (200 m) = I - 200 - 100 = (I - 300) A, current through fourth line (200 m) = I - 300 - 200 = (I - 500) A.
I r* = 10V1 ∠90° − 10 ∠(90° − δ 2 )
Supply voltage = Sum of all receiving side drops 400 = ( 400 r ) I + ( 200 r )( I − 200) + ( 200 r )( I − 300) + ( 200 r )( I − 500) + 400
+ B2 − load
⎡Vs ⎤ ⎡ A B ⎤ ⎡Vr ⎤ ⎢ I ⎥ = ⎢C D ⎥ ⎢ I ⎥ ⎦⎣ r⎦ ⎣ s⎦ ⎣
(ii) Frequency will be more than infinite bus.
200 m B
Vr
⎡ A B ⎤ ⎡1 0.1 j ⎤ ⎡1 2⎤ = ⎢C D ⎥ = ⎢0 1 ⎥⎦ ⎢⎣0 1 ⎥⎦ ⎦ ⎣ ⎣ ↓
(i) With the connection, lamp becomes dark, hence the phase sequence are different.
A
Ir Z
B1 + source −Vs
Thus,
400 m
305
V1 ∠0° 1 ∠ 0° − ⋅1 ∠δ 2 0.1 ∠90° 0.1 ∠90° = 10 V1 ∠ − 90° − 10 ∠(δ 2 − 90°)
Sr = Vr I r* or Pr + jQr = 1 ∠δ 2 [10V1 ∠90° − 10 ∠(90 − δ 2 )] = 10V1 ∠(90° + δ 2 ) − 10 ∠(90° − δ 2 + δ 2 ) = 10V1 ∠(90° + δ 2 ) − 10 ∠90° = 10V1 (cos(90° + δ 2 ) + j sin(90° + δ 2 ) − 10( j )) [∠90° → j ] = 10V1 ( − sin d2 + j cos d2 ) − 10 j
Ch wise GATE_EE_Ch6.indd 305
11/21/2018 11:51:50 AM
Sr = Vr I r* or Pr + jQr = 1 ∠δ 2 [10V1 ∠90° − 10 ∠(90 − δ 2 )] 306
GATE EE = 10Chapter-wise V1 ∠(90° + δ 2 ) −Solved 10 ∠(90°Papers − δ2 + δ2 )
= 10V1 ∠(90° + δ 2 ) − 10 ∠90° = 10V1 (cos(90° + δ 2 ) + j sin(90° + δ 2 ) − 10( j )) [∠90° → j ] = 10V1 ( − sin d2 + j cos d2 ) − 10 j Given load Sr = 1 + j 0.5(1) Sr = -10V1sind2 + j 10V1cosd2 - 10j = -10V1sind 2 + j (10V1cosd2 - 10)
(2)
Equating real and imaginary parts of Eqs. (1) and (2), we get (3) -10V1sind2 = 1 10V1 cos d2 - 10 = 0⋅5 10V1 cos d2 = 10⋅5(4) Dividing Eq. (3) by Eq. (4), we get 10V1 sin δ 2 −1 = 10V1 cos δ 2 10.5 tan δ 2 = −
1 ⇒ δ 2 = −5.44 10.5
Substituting d2 in Eq. (3), we get −10V1 sin( −5.44) = 1 ⇒ V1 = 1.054 158. Topic: Power Generation Concepts (d) Given that Plant P1 : C1 = 0.05 Pg12 + APg1 + B (1) Plant P2 : C2 = 0.10 Pg 22 + 3 APg 2 + B (2) where Pg1, Pg2 are the generated powers and A and B are constants. Pg1 + Pg2 = 1000 MW,
dC1 dC2 = = 100 dPg1 dPg2
Differentiating Eq. (1) w.r.t. Pg1, we get
dC1 = 2 × 0.05 Pg1 + A = 100 (3) dPg1
⇒ 0.1Pg1 + A = 100 Differentiating Eq. (2) w.r.t Pg2, we get
dC2 = 2 × 0.10 Pg2 + 3 A = 100 (4) dPg2 0.2 Pg2 + 3A = 100
From Eq. (3), we have A = 100 - 0.1 Pg1 Substituting in Eq. (4), we get 0.2 Pg2 + 3(100 - 0.1 Pg1) = 100
Ch wise GATE_EE_Ch6.indd 306
0.2Pg2 + 300 - 0.3Pg1 = 100 - 0.3Pg1 + 0.2Pg2 = -200 0.3 Pg1 + 0.2 Pg2 = 200 Pg1 + Pg2 = 1000
(5) (6)
From Eqs. (5) and (6), we get Pg1 = 800 MW, Pg2 = 200 MW Pg1 Pg2
=
800 4 = 200 1
159. Topic: Differential and Distance Protection (0.023) Given that top =
0.14 × Time multiplier setting (TMS) (Plug setting multipiler, PSM)0.02 − 1
IF(max) = 2000 A and IF(min) = 500 A, TMS = 0.1; PSM = 5 A relay B, I1(max) at C = 100 A. For a setting current of 1 A, plug setting is 100% PSM =
2000 Maximum fault current = 20 = CT ratio Current setting 100 × 1 top =
0.14 × 0.1 = 0.023 s ( 20)0.02 − 1
160. Topic: Voltage and Frequency Control (50) Given that f01 = 51.5 Hz, total load = 2.5 MW, f02 = 51 Hz. So, x1 + x2 = 2.5 (1) For generator 1, f = −x1 + f01 = −x1 + 51.5 For generator 2, f = −x2 + f02 = −x2 + 50 With respect to their droop characteristics, −x1 + 51.5 = −x2 + 51 51.5 − 51 = −x2 + x1 x1 − x2 = 0.5 Solving Eqs. (1) and (3), we get x1 =
(2)
(3)
3 = 1.5 2
Substituting in Eq. (2), we get f = −1.5 + 51.5 = 50 Hz 161. Topic: Per-Unit Quantities (15396) Given a rating, P = 100 MVA, 3 phase, 25 kV generator with positive sequence reactance, Z1 = 0.2 pu; negative sequence reactance Z2 = 0.2 pu; and zero sequence reactance Z0 = 0.05 pu.
11/21/2018 11:51:51 AM
Chapter 6 • Power Systems
Fault current single line to ground fault IF = 3Ia1. We know that I a1 =
Ea 1 + j0 = = − j 2.22 pu Z1 + Z 2 + Z 0 j 0.2 + j 0.2 + j 0.05
Therefore, IF = 3(−j 2.22) = −j 6.66 pu. So, Base current =
100 × 10
6
3 × 25 × 103
= 2309 pu
yz ⎡ ⎢ 1+ 2 A B ⎤ ⎢ ⎡ ⎢C D ⎥ = ⎢ yz ⎞ ⎛ ⎦ ⎣ ⎢ y ⎜1 + ⎟ 6⎠ ⎣ ⎝
307
yz ⎞ ⎤ ⎛ z ⎜1 + ⎟ ⎥ ⎝ 6 ⎠⎥ yz ⎥ 1+ ⎥ 2 ⎦
Total series impedance, Z = j 0.5 Ω × 400 = j 200 Ω Y = j 5 × 10−6/km = j 5 × 10−6 × 400 = −2j × 10−3 The p-equivalent circuit is:
Therefore, fault current in fault circuit = Base current × I a1 = 15396 A
Z Y2
Y1 162. Topic: Electric Field Distribution and Insulators (d) For silicon steel lamination, mean thickness, T = 0.2 mm and variance, V = 0.02. For vanish insulation lamination, one side, T1 = 0.1 mm, V1 = 0.01. Mean thickness of two-side insulation T2 = 2 × T1 So, the mean thickness of one lamination = T + T2 = 0.2 + 0.2 = 0.4 mm For 100 laminations, mean thickness = 0.4 × 100 = 40 mm ⎡ Σ( x − x ) 2 ⎤ Variance = ⎢ ⎥ n ⎣ ⎦ ⇒
( d1 − 0.2) 2 + ( d2 − 0.2) 2 + + ( d100 − 0.2) 2 = 0.02 100
As all have same thickness, for each lamination, 100 × ( d − 0.2) 2 = 0.02 100 So, ( d − 0.2) = 0.02 ⇒ d = 0.3414 mm To find each side insulation thickness, 100 × ( x − 0.1) 2 = 0.01 ⇒ x = 0.2 mm 100 Thickness of each lamination with Si steel and insulation = 0.3414 + 0.4 = 0.7414 mm Variance of overall core = 2×
100(0.7414 − 0.4) 2 = 2 × 0.1166 0.24 100
Hence, mean thickness = 40 mm; variance ≅ 0.24 163. Topic: Bus Admittance Matrix (186.66) For nominal p-circuit of long transmission lines, the ABCD parameters are given by
Ch wise GATE_EE_Ch6.indd 307
z ⎤ 1 + y2 z ⎡ A B⎤ ⎡ ⎢C D ⎥ = ⎢ y + y + y y z 1 + y z ⎥ ⎣ ⎦ ⎣ 1 2 1 2 1 ⎦ Therefore, ⎡ ( j 200) × ( j 2 × 10 −3 ) ⎤ ⎛ 1 + YZ ⎞ Z = B= Z⎜ = j 200 ⎢1 + ⎥ ⎟ ⎝ 6 ⎠ 6 ⎦ ⎣ ⎡1 − 0.4 ⎤ = j 200 ⎢ ⎥ = j186.66 ⎣ 6 ⎦ 164. Topic: Power Factor Correction (0.75) Reactive power consumed by load = 0.25 kVAR Reactive power required for load = 1 kVAR Reactive power supplied by capacitor = 1 - 0.25 = 0.75 kVAR 165. Topic: Voltage and Frequency Control (0.24) Given that H = 5 MW = s/MVA, d0 = 30°, dδ f = 50 Hz, d = 45°, = 3.762 rad/s (at fault clearing). dt We have d 2δ 1 = [ Pm − Pe ] 2 M dt dδ ⎧⎪ 2 =⎨ dt ⎪⎩ M ⎪⎧ 2 3.762 = ⎨ ⎩⎪ H /π f
1/ 2
⎫⎪ P P ds ( − ) ⎬ m e ∫ δ0 ⎪⎭ δ
1/ 2
⎪⎫ ∫30 (1 − Pmax sin δ ) ds⎪⎬ ⎭ 45
45 ⎛ 2 × π × 50 3.762 = ⎜ [δ + Pmax cos δ ]30 ⎞⎟⎠ ⎝ 5
1/ 2
Squaring both sides 14.1376 = 20π ⎡⎣( 45° − 30°) + Pmax (cos 45° − cos 30°)⎤⎦ Pmax = 0.24 pu
11/21/2018 11:51:53 AM
308
GATE EE Chapter-wise Solved Papers
166. Topic: Power Generation Concepts (c) Base load power plants are wind, coal, gas, nuclear and hydroelectric power plants. 167. Topic: Circuit Breakers (a) When operating as generator, the rotor angle d will be positive. Also, when any one line is tripped at time t due to unstable operation of generator, the rotor angle d starts increasing. So, option (a) curve fit on thin condition. 168. Topic: Bus Admittance Matrix (b) The y-bus matrix is given as follows ⎡ y10 + y12 + y13 ybus = ⎢⎢ − y12 ⎢⎣ − y13
− y12 y20 + y21 + y23 − y23
− y13 ⎤ ⎥ − y23 ⎥ y30 + y31 + y32 ⎥⎦
From the given matrix, y31 = y13 = −j 4 y23 = y32 = −j 5 Also, y10 + y12 + y13 = −j 6 y20 + y21 + y23 = −j 7 y30 + y31 + y32 = −j 8 y30 − 4j − 5j = j 8 ⇒ y30 = j After 50% compensation, 50 y30 = ⋅ j = 0.5 j 100 Therefore, [y30 + y31 + y32 ]New = j 0.5 − 4j − 5j = −8.5j 169. Topic: Power Generation Concepts (20) The given fuel costs are: C1 ( P1 ) = 0.01P + 30 P1 + 10 where generation limit is 100 MW ≤ P1 ≤ 150 MW 2 1
C2 ( P2 ) = 0.01P22 + 30 P2 + 10 where generation limit is 100 MW ≤ P2 ≤ 180 MW Differentiating, we get
dC1 = 0.02 P1 + 30 dP1
dC2 = 0.1P2 + 10 dP2
0.02P1 + 30 − 0.1P2 − 10 = 0 0.02P1 − 0.1P2 = 20 P1 + P2 = 200 where P1 = 200 and P2 = 0. Thus, the increment cost is Rs. 20/MWh.
Ch wise GATE_EE_Ch6.indd 308
170. Topic: Power Factor Correction (d) Fast decoupled load flow method can be made to give any special accuracy. So assertion is false. The power mismatch vector is re-estimated until a specified tolerance is reached, which decides the accuracy of the solution. 171. Topic: Symmetrical and Unsymmetrical Fault Analysis (b) From the given phasor diagram, we have that all the current except I2 are lagging. If, the current is leading, I2 will be opposite in direction to that shown in the figure. So, we get the actual current direction flow diagram as shown in the question figure. This implies that the fault should be at point Q only. Also V1 < V2, indicating that fault is at location Q. In the other case, if the fault is located at point S, then current I2 and I4 will flow in the direction R-S-Q. 172. Topic: Power Factor Correction (555) Given that T1 and T2 rating is 500 kVA, Z 2 = 0.8 + 4.8 j = 4.8∠80.5 Z1 = 1 + 6 j = 6.08∠80.5 , and S = 1000 kVA. Now, S2 = S1 ×
Z1 6.08∠80.5 = 100 × Z1 + Z 2 (6.08∠80.5) + ( 4.8∠80.5) ⎛ 6.08∠80.5 ⎞ = 1000 ⎜ = 555 kVA ⎝ 10.88∠80.5 ⎟⎠
173. Topic: Power Generation Concepts (136.36) Given that dC1 dC2 = 0.2 P1 + 50 = 76; = 0.24 P2 + 40 = 68.8 dP1 dP2 Now, P1 = 130 and P2 = 120. So, P1 + P2 = 250 For minimization of total cost dC1 dC2 ⇒ 0.2P1 + 50 − 0.24P2 − 40 = 0 = dP1 dP2 0.2[250 − P2] + 50 − 0.24P2 − 40 = 0 Therefore, P2 = 136.36. 174. Topic: Power Generation Concepts (1.914) We know that, GMR = [0.7788R × 3R × 3R]1/3 kR = 1.914 R k = 1.914 175. Topic: Gauss-Seidel and Newton-Raphson Load Flow Methods (a) For a PQ bus, both active and reactive power is known. But we do not know voltage magnitude or angles.
11/21/2018 11:51:54 AM
Chapter 6 • Power Systems
176. Topic: Symmetrical and Unsymmetrical Fault Analysis (0.84) Given that voltage magnitude at bus B, during a three-phase fault at bus A is VB = 0.8 pu We know that, the voltage at the ith bus when the fault occurs at kth bus is ⎛ Zik ⎞ Vi = E ⎜1 − ⎝ Z kk + Z f ⎟⎠ The fault current is If =
Vprefault X (pu)
Prefault current at A is 10 pu, therefore 1
10 =
X A(pu)
⇒ X A = 0.1 pu
309
Assume one of the 10 generator buses to be the slack bus. Size = (2 × 100 − 10 − 1) × (2 × 100 − 10 - 1) = 189 × 189 180. Topic: Series and Shunt Compensation (b) We have 1.6 × 10 −3 L = = 400 Ω C 10 × 10 −9
ZC =
Then surge impedance loading is SIL =
EV 400 × 103 × 400 × 103 = = 400 MW 400 ZC
Now, required load is 300 MW, therefore, ZC should increase. For ZC to increase, either L should increase or C should reduce. Therefore, compensation is inductive.
Prefault current at B is 8 pu, therefore, 8=
1 X B(pu)
⇒ X B = 0.125 pu
181. Topic: Voltage and Frequency Control (47.96) In the first case (kVA) old = 800 2 + 600 2 = 1000 kVA
Now, ⎛ Z ⎞ VB = E ⎜1 − AB ⎟ ⎝ Z ⎠
In the second case P2 = 800 + 100 = 900 kW, therefore,
BB
Substituting values, ⎛ Z ⎞ 0.8 = 1 ⎜1 − AB ⎟ ⇒ Z AB = 0.2 ⎝ 0.1 ⎠
Q2 = 1000 2 − 900 2 = 435.889 kVAr Reactive power provided by phase bank = 600 - 435.889 = 164.11 kVAr Per phase, Qbank =
Similarly, ⎛ Z ⎞ VA = E ⎜1 − AB ⎟ ⎝ Z AA ⎠ Substituting values,
Voltage / phase =
178. Topic: AC and DC Transmission Concepts (d) 179. Topic: Gauss-Seidel and Newton-Raphson Load Flow Methods (a) Size of the Jacobian = (2m − g − 1) × (2n − g − 1)
Ch wise GATE_EE_Ch6.indd 309
3.3 3
= 1.9 kV
From capacitor/phase, Q =
0.2 ⎞ ⎛ VA = 1 ⎜1 − = 0.84 pu ⎝ 0.125 ⎟⎠ 177. Topic: AC and DC Transmission Concepts (c) Over excited synchronous motors are capacitive.
164.11 = 54.7 3
XC = C=
(1.9 kV ) 2 XC
(1.9 × 103 ) 2 = 66.36 Ω 54.7 × 103
1 1 = = 47.96 μF 2π fX C 2π 50(66.36)
182. Topic: Per-Unit Quantities (1.07) 30 × 103 = 1255.11 A Rated current = 3 × 13.8 Ifault = 4270 A
11/21/2018 11:51:55 AM
310
GATE EE Chapter-wise Solved Papers
I pu =
4270 = 3.4 pu 1255.11
I g1 =
Ea X 1 + ( X 2 || X 0′ )
P1 X 2eq = P2 X 1eq X 1eq = j 0.2 + [( j 0.1 + j 0.5) ( j 0.1 + j 0.5)] = j 0.5
X 0′ = X 0 + 3( X n + X f ) = 3Z n + 0.35
Similarly, from the circuit below j0.6
1 3.4 = 0.15(3 X n + 0.35) 0.15 + 0.15 + 3 X n + 0.35
j0.2
j0.3
Xn = 1.07 pu 183. Topic: Models and Performance of Transmission Lines and Cables (1.427) For the given transmission line
j 0.3 j0.6
≡
j0.2
y
z
x
1m j0.3
j0.3
0.3 × 0.3 = 0.075 0.6 + 0.3 + 0.3 0.6 × 0.3 = 0.15 y= 0.6 + 0.3 + 0.3 0.6 × 0.3 = 0.15 z= 0.6 + 0.3 + 0.3 x=
r = 10 mm In the first case, ⎛ D⎞ L = 2 × 10 −7 ln ⎜ ⎟ ⎝ r′ ⎠
1 ⎛ ⎞ = 2 × 10 −7 ln ⎜ −3 ⎝ 10 × 10 × 0.7788 ⎟⎠
= 0.97 mH/m
X 2 eq = j1.2 pu j1.5
In the second case,
j0.2
j1.5 j0.075
3 ⎛ ⎞ D2 ×1 L = 2 × 10 ln ⎜ −3 ⎟ ⎝ 0.7788 × 10 × 10 ⎠ −7
Therefore, P1 1.2 = = 2.4 P2 0.5
1.05 × 0.97 × 10−6 = 2 × 10 −7 ln D × 100 0.7788 3
2
3 D 2 × 100 1.05 × 0.97 × 10 −6 = ln −7 0.7788 2 × 10
On solving, we get
Vs = AVR
D = 142.7 cm = 1.427 m 184. Topic: Per-Unit Quantities (2.4) In the given power system with occurrence of fault
j0.5
j0.2 1 pu
(Using T-parameters Vs =AVr +BIr) 400 = A(420) A = 0.9524 yz A = 1+ 2
j0.1
F
Line is lossless, therefore,
j0.5
A = 1−
j0.1 1.0 pu
Ch wise GATE_EE_Ch6.indd 310
185. Topic: Models and Performance of Transmission Lines and Cables (294.59) For no load condition
(ω C )(ω L) 2
11/21/2018 11:51:58 AM
Chapter 6 • Power Systems
Z1eq = j 0.4 j 0.4 = j 0.2
β l = ω LC
Z 2eq = j 0.15
where b is the propagation constant. Therefore, A = 1 − 0.9524 = 1 −
Z 0eq = j 0.15
(β l ) 2 2
Then fault current is
(β l ) 2 2
If = 3 ×
β l = 0.3085 ⇒ β =
0.3085 l
= 3×
v 2π 30 × 105 2π = ⇒ = f β 50 β
β=
311
Vprefault Z 0 + Z1 + Z 2 1 = 6 pu j (0.2 + j 0.15 + j 0.15)
188. Topic: Models and Performance of Transmission Lines and Cables (a) Van = 220 sin(100 p t)
2π × 50 30 × 105
l = 294.59 km
ia = 10 sin(100 p t) Vbn = 220 cos(100 p t),
186. Topic: Power Factor Correction (0.316) To raise power factor to unity, we have P = 500 × 0.8 = 400 kW Q = 500 × 0.6 = 300 kVAr Therefore, 300 kVAr must be provided by motor ⎛Q ⎞ ⎛ 300 ⎞ ϕ motor = tan −1 ⎜ m ⎟ = tan −1 ⎜ = 71.56° ⎝ 100 ⎟⎠ ⎝ Pm ⎠ Power factor of motor = cos(71.56°) = 0.316
ib = 10 cos(100 p t) P = Van˙ia + Vbn˙ib = 2200 W 189. Topic: Bus Admittance Matrix (b)
y12 + y13 = Y11 = -j12
y12 + y23 = Y22 = -j15 y13 + y23 = Y33 = -j17
187. Topic: Symmetrical and Unsymmetrical Fault Analysis
Solving these equations, we get
(6) From the following circuit for two identical unloaded generators connected in parallel, we have
y13 = -j 2, y23 = -j 5, y12 = -j10 Reactance on p, q, r are p=
1 = j 0.2 pu − j5
q=
−1 = j 0.1 pu j10
r=
1 = j 0.5 pu − j2
j0.4 j0.4 +
+ 1∠0° −
1∠0° −
j0.3
j0.3
j0.15 j0.15
190. Topic: Gauss-Seidel and Newton-Raphson Load Flow Methods (14) G1 = slack bus; G2 = PQ bus; G3, G4 = PV buses; L1, L2 = PQ buses; L3, L4 = PV buses; L5, L6 = PQ buses. Now, Non-linear equations = Number of unknown variables = (2 × 10) - 4 - 2 = 14 191. Topic: Bus Admittance Matrix (d) Both YBUS and ZBUS are unsymmetrical.
Ch wise GATE_EE_Ch6.indd 311
11/21/2018 11:51:59 AM
312
GATE EE Chapter-wise Solved Papers
192. Topic: AC and DC Transmission Concepts (406.2)
196. Topic: Per-Unit Quantities (0.1)
Z = 100 ∠80°, X = 3300 W
If =
Y 1 1 = = 2 X 3300
⇒ 375 =
⇒ y = 6.06 × 104 Therefore, characteristic impedance is ZC =
2 = y
3VTh z1 + z2 + z0 + 3 zn
⇒ 0.5 + 3 zn =
100 = 406.2 Ω 6.06 × 10 −4
193. Topic: Gauss-Seidel and Newton-Raphson Load Flow Methods (61) [ J ] = [ 2 n − m − 2] ⇒ 100 = 2n − 20 − 2 122 ⇒ = n ⇒ n = 61 2
197. Topic: Power Factor Correction (1.22) Using equal area criterion, we have A1 = A2 δ2
δc
∫ (P
δ0
m0
− Pmax1 sin δ ) dδ = ∫ ( Pmax1 sin δ − Pm0 ) dδ δc
⇒ Pmax1 = =
(b) cos ϕ 1 = cos ⎡⎢ tan −1 ⎛⎜ θ ⎞⎟ ⎥⎤ ⎝ P⎠⎦ ⎣
Now,
Dca = 1m
1m = dbc
a
b Dab = 1m
⇒ cos ϕ 2 = 0.707 lag
⎛D ⎞ X L = 2π f L = 2π f × 2 × 10 −7 ln ⎜ m ⎟ ⎝ DS ⎠
195. Topic: Bus Admittance Matrix y11 ( new ) = y11 (old ) + j
1 = − j 19.95 0.05
y33 ( new ) = y33 (old ) + j
1 = − j 19.95 0.05
y13 (new) = 0 y31 (new) = 0 Therefore, modified matrix is given by
Ch wise GATE_EE_Ch6.indd 312
1 pu(1.221 − 0.7295) cos 41.8 − cos 69.95
c
QC = P[tan ϕ 1 − tan ϕ 2 ]
⎡ − j19.95 ⎢ j 20 ⎢ ⎢⎣0
cos δ 0 − cos δ 2
198. Topic: Principles of Over-Current (0.289)
⇒ ϕ 2 = 45°
Pm0 (δ 2 − δ 0 )
= 1.22
−1 ⇒ 1 = 1[tan(cos 0.707) − tan ϕ 2 ]
(c)
3 = 0.8v 3.75
⇒ zn = 0.1 pu
194. Topic: Power Factor Correction
= cos (tan-1 (1)) = 0.707 (lagging)
3 ×1 0.2 + 0.2 + 0.1 + 3 zn
0 j 20 ⎤ ⎥ j 39.9 j 20 ⎥ j 20 − j19.95⎥⎦
⎛ 1 ⎞ = 2π × 50 × 2 × 10 −7 ln ⎜ ⎝ 0.01⎟⎠ = 2.89 × 10−4 W/m = 0.289 W/km 199. Topic: Principles of Over-Current (0.80) VS − VR = I ( R cos ϕ + X L sin ϕ ) Also, I=
PR VR cos ϕ
11/21/2018 11:52:01 AM
313
Chapter 6 • Power Systems
⇒ (VS − VR ) =
For 5 cycles, t = 0.1s
PR [ R cos ϕ + X L sin ϕ ] VR cos ϕ
Pa = Pm − Pe
⎡V V − VR2 − RPR ⎤ ϕ = tan −1 ⎢ S R ⎥ X L PR ⎣ ⎦
M=
KE 1000 = = 0.111 180 f 180 × 50
Power angle is
Putting in the given values, Power factor = cos ϕ = 0.8 lagging
δ=
200. Topic: Power Factor Correction
P ⎛ t2 ⎞ 60 (0.1) 2 = × = 2.7° M ⎜⎝ 2 ⎟⎠ 0.111 2
Therefore, new power angle is
(333)
δ ′ = 10 + 2.7 = 12.7° 53 52 4% Q1
6% Q2
202. Topic: Models and Performance of Transmission Lines and Cables (b) For simple transmission line, Reodance
ix V1
V2 Pmax =
300 PB
PA 400
V1V2 (1) X
With lossy transmission line, we have Z Assume 50 Hz at full load, tan θ1 =
52 − f 52 − 50 1 = = PA 400 200
(1)
tan θ 2 =
53 − f 53 − 50 1 = = PB 300 100
(2)
Also,
53 − f 1 = PB 100
53 − f 1 = 600 − PA 100
P=
V2
V1V2 AV2 2 cos( β − δ ) − cos(β − α ) (2) X Z
where a = attenuation constant and b = propagation constant So, with lossy transmission line Pmax
1 C (t ) = 1 −
⎡ e − s1t e − s2t ⎤ − ⎢ ⎥ s2 ⎦ 2 ζ 2 − 1 ⎣ s1
wn
where s1 = ζw n − w n ζ − 1 2
s2 = ζw n + w n ζ 2 − 1 (c) Critically damped, ζ = 1 ; C (t ) = 1 − e − w n t [1 + w n t ] (d) Under damped, 0 < ζ < 1 y (t ) = 1 −
e −ζw n t 1−ζ 2
21. Rise time, Tr =
IV Root Locus Technique 27. Under unity feedback configuration, the closed-loop transfer function of the system is T ( s) =
KG ( s) KN ( s) = 1 + KG ( s) D( s) + KN ( s)
(a) The roots of the characteristic equation p(s) = D(s) + KN(s) = 0 are the poles for the closedloop system (b) The location of the roots as gain K varies is s1, s2 =
−1 ± 2
1 − 4K 2
28. The centroid, that is, the starting point of the asymptotes sA is given by
σA =
∑ Real part of poles − ∑ Real part of zeros n−m
29. Angle of asymptotes is given by [sin(w d t + f )]
⎡ 1−ζ 2 f = tan −1 ⎢ ⎢⎣ ζ
Ch wise GATE_EE_Ch7.indd 317
III Stability of Control Systems
⎤ ⎥ ; wd = wn 1 − ζ 2 ⎥⎦
⎛ 1−ζ 2 ⎞ p − tan −1 ⎜ ⎟ ⎝ ζ ⎠
wn 1 − ζ 2
θA =
180( 2k + 1) for k = 0, 1, …(n − m) n−m
30. Irrational function e−Ts can be approximated in terms of ratio of two polynomials in s using Pade’s approximation
e
− Ts
=
e e
−
Ts 2
+
Ts 2
=
1 − sT / 2 s − 2 /T =− 1 + sT / 2 s + 2 /T
11/21/2018 3:45:39 PM
318
GATE EE Chapter-wise Solved Papers
31. Angle of departure/approach fdep = net contribution due to other poles and zeros farr = (-1) × net contribution due to other poles and zeros
VII Controllers 36. Types of controllers (a) Proportional (P) controller, G(s) = K p
V Frequency Response Analysis
KI (b) Integral (I) controller, G(s) = s
32. Polar plots and type and order of polar plots
(c) Derivative (D) controller, G(s) = K + sKD
33. Nyquist stability criterion states that if the G(s)H(s) contour in the G(s)H(s) plane corresponding to Nyquist contour in the s-plane encircles the point (−1+j 0) in the anticlockwise direction as many times as the number of right half of s-plane poles of G(s)H(s), then the closedloop system is stable.
⎡ ⎤ 1 (d) PID controller, G(s) = K p ⎢1 + + Td s⎥ ⎣ TI s ⎦
37. State model for armature controlled DC motor ⎡ x 1 (t ) ⎤ ⎡ − Ra /La ⎢ x (t ) ⎥ = ⎢ K /s ⎢ 2 ⎥ ⎢ ⎢⎣ x 3 (t ) ⎥⎦ ⎢⎣ 0
VI Bode Plots 34. Basic factors of G(jw) are (a) Constant gain, K K or (b) Integral factor, jw
VIII State Space Representation
K
⎡ x1 (t ) ⎤ y = [0 0 1] ⎢⎢ x2 (t ) ⎥⎥ ⎢⎣ x3 (t ) ⎥⎦
(c) Derivative factor, K( jw) or K( jw)n (d) First-order factor in the denominator, 1 1 or m jw ⎛ jw ⎞ 1+ wc ⎜⎝1+ w ⎟⎠ c
Standard form: y = Cx + Du 39. Obtaining transfer function from state model: Given: x = Ax + Bu and y = Cx + Du
(e) First-order factor in the numerator (1 + jwT) or (1 + jwT)m.
G ( s) =
(f) Quadratic factor in the denominator 1
n
− B /J 1
0 ⎤ ⎡ x1 (t ) ⎤ ⎡1/La ⎤ 0 ⎥⎥ ⎢⎢ x2 (t ) ⎥⎥ + ⎢⎢ 0 ⎥⎥ u 0 ⎦⎥ ⎣⎢ x3 (t ) ⎦⎥ ⎢⎣ 0 ⎥⎦
38. Standard form: x = ax + bu
( jw )n
⎛ jw ⎞ ⎛ jw ⎞ + 1 + 2ζ ⎜ ⎝ w ⎟⎠ ⎜⎝ w ⎟⎠
− K b /La
Y ( s) = [C ( sI − A) −1 B + D ] U ( s)
40. Solutions of homogeneous state equation: (a) x(s) = [sI − A]−1x(0) x(s) = f (s)x(0) f (s) → Resolvant matrix (b) x(t) = f (t) x(0) f (t) → state transition matrix.
2
n
(g) Quadratic factor in the numerator: 1 + 2z (jw/wn) + (jw/wn)2
35. Frequency domain specifications 1 (a) Resonant peak, M r = 2ζ 1 − ζ 2
41. Solution for non-homogeneous equation t
x(t ) = e At x(0) + ∫ e A( t − x ) Bu( x ). dx 0
(b) Resonant frequency, w r = w n 1 − 2ζ 2
42. Condition for controllability
(c) Bandwidth, w b = w n (1 − 2ζ + 2 − 4ζ + 4 ζ ) 2
(d) Phase margin ⎛ ⎞ 2ζ γ = 180° + tan −1 ⎜ ⎟ 2 4 1/ 2 ⎝ ( −2ζ + 4ζ + 1) ⎠
Ch wise GATE_EE_Ch7.indd 318
2
4 1/ 2
[U c ] = [ B must have rank n.
AB
A2 B … An −1
B]
43. Condition for observability 2
[U o ] = [C T AT C T AT C T ... AT
n −1
C T ]T
11/21/2018 3:45:41 PM
Chapter 7 • Control Systems
319
QUESTIONS 1. A three phase semi-converter feeds the armature of a separately excited DC motor, supplying a non-zero torque. For steady state operation, the motor armature current is found to drop to zero at certain instances of time. At such instances, the voltage assumes a value that is (a) equal to the instantaneous value of the AC phase voltage (b) equal to the instantaneous value of the motor back emf (c) arbitrary (d) zero (GATE 2000: 1 Mark) 2. Feedback control systems are (a) insensitive to both forward and feedback path parameter changes (b) less sensitive to feedback path parameter changes than to forward path parameter changes (c) less sensitive to forward-path parameter changes than to feedback path parameter changes (d) equally sensitive to forward and feedback path parameter changes (GATE 2000: 1 Mark)
6. A unity feedback system has open-loop transfer 25 ⎪⎧ ⎪⎫ function G ( s) = ⎨ ⎬ . The peak overshoot in ⎩⎪ ⎡⎣ s ( s + 6 )⎤⎦ ⎭⎪ the step-input response of the system is approximately equal to: (a) 5% (b) 10% (c) 15% (d) 20% (GATE 2000: 2 Marks) 7. Maximum phase-lead (0.5s + 1) , is D (s) = (0.05s + 1) (a) (b) (c) (d)
Ch wise GATE_EE_Ch7.indd 319
w=∞
0°
w=0 (a) (b) (c) (d)
s 1 (d) 2 s( s + 1) s( s +1)
The number of roots in the right half of s-plane are: (a) zero (b) 1 (c) 2 (d) 3 (GATE 2000: 2 Marks)
52 deg at 4 rad/s 52 deg at 10 rad/s 55 deg at 12 rad/s None of the answers is correct (GATE 2000: 2 Marks)
−1.42
(GATE 2000: 1 Mark)
compensator
GH-plane
4. A linear time-invariant system initially at rest, when subjected to a unit-step input, gives a response y(t) = te−t, t > 0. The transfer function of the system is: 1 1 (a) (b) ( s + 1) 2 s( s + 1) 2
5. The characteristic equation of a feedback control system is: 2s4 + s3 + 3s2 + 5s + 10 = 0
the
8. The polar plot of a type-1, 3-pole, open-loop system is shown in in the following figure. The closed loop system is
3. A unity feedback system has open loop transfer function G(s). The steady-state error is zero for (a) step input and type – 1 G(s) (b) ramp input and type – 1 G(s) (c) step input and type – 0 G(s) (d) ramp input and type – 0 G(s) (GATE 2000: 1 Mark)
(c)
of
always stable marginally stable unstable with one pole on the right half s-plane unstable with two poles on the right half s-plane (GATE 2001: 1 Mark)
9. Given the homogeneous state-space equation ⎡ −3 1 ⎤ x = ⎢ ⎥x ⎣ 0 −2⎦
the steady state value xss = lim x(t ), given the initial t →∞
state value of x(0) = [10 −10] , is T
11/21/2018 3:45:42 PM
320
GATE EE Chapter-wise Solved Papers
⎡0⎤ (a) xss = ⎢ ⎥ ⎣0⎦ ⎡ −10 ⎤ (c) xss = ⎢ ⎥ ⎣ 10 ⎦
⎡ −3⎤ (b) xss = ⎢ ⎥ ⎣ −2⎦
t
y(t ) = ∫ ( 2 + t − t )e −3( t −t ) u(t )dt , 0
the transfer function Y(s)/U(s) is 2e −2 s s+2 (b) s+3 ( s + 3) 2
(a)
2s + 5 2s + 7 (d) s+3 ( s + 3) 2
(c)
(GATE 2001: 2 Marks)
∫ s( t − t )
⎡1 ⎤ ⎡ 2 3⎤ 14. For the system X = ⎢ X + ⎢ ⎥ u, which of the ⎥ ⎣ 0 5⎦ ⎣0 ⎦ f ollowing statements is true? (a) The system is controllable but unstable (b) The system is uncontrollable but unstable (c) The system is controllable but stable (d) The system is uncontrollable but stable (GATE 2002: 2 Marks) 15. A unity feedback system has an open loop transfer function, G ( s) =
−60dB/dec
(a)
20( s + 5) 10( s + 5) (b) 2 s( s + 2)( s + 25) ( s + 2) ( s + 25)
(c)
20( s + 5) 50( s + 5) (d) s 2 ( s + 2)( s + 25) s 2 ( s + 2)( s + 25) (GATE 2001: 2 Marks)
(b)
jw
s
25
(a)
K . The root locus plot is s2
jw
rad/s 5
u(t )dt
(GATE 2002: 1 Mark)
−40dB/dec
2
2
13. The state transition matrix for the system X = AX with initial state X(0) is (a) (sI – A)−1 (b) eAtX(0) (c) Laplace inverse of [(sI – A)−1] (d) Laplace inverse of [(sI – A)−1X(0)]
−60dB/dec
0.1
⎦
(GATE 2002: 1 Mark)
dB
54
1
0
11. The asymptotic approximation of the log-magnitude versus frequency plot of a minimum phase system with real poles and one zero is shown in the following figure. Its transfer functions is
−40dB/dec
⎣0
1
t
(d)
(GATE 2001: 1 Mark) 10. Given the relationship between the input u(t) and the output y(t) to be
∫ s(t − t ) ⎢∫ u(t )dt ⎥ dt 0
⎡∞ ⎤ (d) xss = ⎢ ⎥ ⎣∞ ⎦
⎤
⎡t
t
(c)
(c)
s
(d)
jw
s
jw
s
12. Let s(t) be the step response of a linear system with zero initial conditions; then the response of this system to an input u(t) is t
(a)
∫ s(t − t )u(t )dt 0
⎤ d ⎡ ⎢ ∫ s(t − t )u(t )dt ⎥ dt ⎣ 0 ⎦ t
(b)
Ch wise GATE_EE_Ch7.indd 320
(GATE 2002: 2 Marks) 16. The transfer function of the system described by d 2 y dy du + = + 2u with u as input and y as output is dt 2 dt dt
11/21/2018 3:45:44 PM
Chapter 7 • Control Systems
321
(a)
( s + 2) ( s + 1) (b) 2 ( s + s) ( s 2 + s)
d 2 x 1 dx 1 + + x = 10 + 5e −4t + 2e −5t dt 2 2 dt 18
(c)
2 2s (d) 2 ( s + s) ( s + s)
The natural time constants of the response of the system are (a) 2 s and 5 s (b) 3 s and 6 s (c) 4 s and 5 s (d) l/3 s and l/6 s
2
(GATE 2002: 2 Marks) ⎡1⎤ ⎡2 0⎤ X + ⎢ ⎥ u; y = [ 4 0] X , 17. For the system X = ⎢ ⎥ ⎣0 4 ⎦ ⎣1⎦ with u as unit impulse and with zero initial state, the output, y, becomes (a) 2e2t (b) 4e2t (c) 2e4t (d) 4e4t (GATE 2002: 2 Marks) 18. A control system is defined by the following mathematical relationship d2x dx + 6 + 5 x = 12(1 − e −2t ) dt dt 2 The response of the system as t → ∞ is (a) x = 6 (b) x = 2 (c) x = 2.4 (d) x = -2 (GATE 2003: 1 Mark) 19. A lead compensator used for a closed-loop controller has the following transfer function s⎞ ⎛ K ⎜1 + ⎟ ⎝ a⎠ s⎞ ⎛ ⎜⎝1 + ⎟⎠ b For such a lead compensator (a) a < b (c) a > Kb
(b) b < a (d) a < Kb (GATE 2003: 1 Mark)
20. A second-order system starts with an initial condition ⎡ 2⎤ of ⎢ ⎥ without any external input. The state transition ⎣3 ⎦ ⎡e −2t 0⎤ matrix for the system is given by ⎢ . The state −t ⎥ ⎣ 0 e ⎦ of the system at the end of 1 s is given by ⎡0.271⎤ ⎡0.135⎤ (b) ⎢ (a) ⎢ ⎥ ⎥ ⎣1.100 ⎦ ⎣0.368⎦ ⎡0.271 ⎤ ⎡0.135⎤ (c) (d) ⎢0.736 ⎥ ⎢1.100 ⎥ ⎣ ⎦ ⎣ ⎦ (GATE 2003: 1 Mark) 21. A control system with certain excitation is governed by the following mathematical equation
Ch wise GATE_EE_Ch7.indd 321
(GATE 2003: 2 Marks) Common Data for Questions 22 and 23: The block diagram shown in the following figure gives a unity feedback closedloop control system. u(t) +
3 s+15
15 s+1
y(t)
22. The steady-state error in the response of the given system to unit step input is (a) 25% (b) 0.75% (c) 6% (d) 33% (GATE 2003: 2 Marks) 23. The roots of the closed-loop characteristic equation of the system are (a) -l and −15 (b) 6 and 10 (c) −4 and −15 (d) −6 and −10 (GATE 2003: 2 Marks) 24. The following equation defines a separately excited DC motor in the form of a differential equation
d 2w B dw K 2 K w= + + Va LJ dt 2 J dt LJ The above equation may be organised in the state-space form as follows ⎡ ⎤ ⎡ d 2w ⎤ ⎢ ⎥ ⎢ 2 ⎥ dw ⎥ ⎢ dt ⎥ = P ⎢ + QVa ⎢ dt ⎥ ⎢ dw ⎥ ⎢ ⎥ ⎢⎣ dt ⎥⎦ ⎣w ⎦
where P matrix is given by ⎡ B − (a) ⎢⎢ J ⎢⎣1
−
⎡ K2 K2 ⎤ B⎤ − ⎥ ⎥ (b) ⎢− LJ ⎥ LJ J⎥ ⎢ 0 ⎦⎥ 0 1⎥⎦ ⎣⎢
0 1⎤ 0⎤ ⎡1 ⎡ ⎢ ⎥ ⎢ 2 2⎥ (c) ⎢ K B (d) ⎢− B − K ⎥ − − ⎥ ⎢⎣ LJ ⎢⎣ J J ⎥⎦ LJ ⎥⎦ (GATE 2003: 2 Marks)
11/21/2018 3:45:46 PM
322
GATE EE Chapter-wise Solved Papers
25. The loop gain GH of a closed-loop system is given by the expression
K The value of K for which s( s + 2)( s + 4)
the system just becomes unstable is (a) K = 6 (c) K = 48
(b) K = 8 (d) K = 96 (GATE 2003: 2 Marks)
K 26. The asymptotic Bode plot of the transfer function s is given in the following figure. 1+ a The error in phase angle and dB gain at a frequency of w = 0.5 a are respectively |G(jw)| dB
20 log K
29. Consider the function, F ( s) =
is Laplace transform of the function f (t). The initial value of f (t) is equal to 5 (a) 5 (b) 2 5 (c) (d) 0 3 (GATE 2004: 1 Mark)
E ( s) constant in V/rad/s, then the transfer function, θ ( s) will be
w
Kt (a) Kts2 (b) s 0.1a
10a
w
Ph°
(b) 5.7°, 3 dB (d) 5.7°, 0.97 dB (GATE 2003: 2 Marks)
27. The block diagram of a control system is shown in the following figure. The transfer function G(s) = Y(s)/U(s) of the system is
− −
Y(s) Integrator
+ 3
1 (a) s ⎞ ⎛ s⎞ ⎛ 18 ⎜ 1 + ⎟ ⎜1 + ⎟ ⎝ 12 ⎠ ⎝ 3 ⎠ (c)
−
R(s) 1 s
1 (b) ⎛ s⎞ ⎛ s⎞ 27 ⎜1 + ⎟ ⎜1 + ⎟ ⎝ 6⎠ ⎝ 9⎠
1 1 (d) s ⎞ ⎛ s⎞ ⎛ ⎛ s⎞ ⎛ s⎞ 27 ⎜1 + ⎟ ⎜1 + ⎟ 27 ⎜ 1 + ⎟ ⎜ 1 + ⎟ ⎝ 12 ⎠ ⎝ 9 ⎠ ⎝ 9 ⎠ ⎝ 3⎠
+
1 s
(a)
C(s)
+
+
(c) 12
(GATE 2003: 2 Marks)
Ch wise GATE_EE_Ch7.indd 322
(GATE 2004: 1 Mark)
32. For the block diagram shown in the following figure, the C ( s) transfer function is equal to R( s)
Integrator
2
(d) Kt
(GATE 2004: 2 Marks)
9 U(s) +
(c) Kts
31. 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
45°/decade
(a) 4.9°, 0.97 dB (c) 4.9°, 3 dB
5 where F(s) s( s 2 + 3s + 2)
30. For a tachometer, if q(t) is the rotor displacement in radians, e(t) is the output voltage and Kt is the tachometer
20 dB/decade
a
28. 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) 0° (b) 45° (c) 90° (d) 180° (GATE 2004: 1 Mark)
+
s2 + s + 1 s 2 + 1 (b) s2 s2 s2 + s + 1 1 (d) s s + s +1 (GATE 2004: 2 Marks) 2
33. The state variable description of a linear autonomous system is X = AX where X is a two-dimensional state vector and A is the system matrix given by ⎡0 2⎤ A= ⎢ ⎥ ⎣2 0⎦ . The roots of the characteristic equation are
11/21/2018 3:45:48 PM
Chapter 7 • Control Systems
(a) −2 and +2 (c) -2 and −2
(b) −j 2 and +j 2 (d) +2 and +2 (GATE 2004: 2 Marks)
39. The following figure shows the root locus plot (location of poles not given) of a third order system whose openloop transfer function is Im
34. The block diagram of a closed-loop control system is given in the following 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/s, are respectively equal to R(s) + −
K s(s+2)
2√3 √3 −3 −2 −1 −√3 −2√3
C(s)
(b) 20 and 0.2 (d) 25 and 0.2 (GATE 2004: 2 Marks)
35. The unit impulse response of a second-order underdamped system starting from the rest is given by c(t ) = 12.5e −6 t 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 (GATE 2004: 2 Marks) 36. In the system shown in the following figure, the input is x(t) = sin wt, In the steady-state, the response y(t) will be x(t) y(t) s s +1 1 sin(t − 45°) (b) sin(t + 45°) 2 2 (c) sin(t - 45°) (d) sin (t + 45°) (GATE 2004: 2 Marks) (a)
1
37. The open-loop transfer function of a unity feedback as +1 control system is given as G ( s) = 2 . s The value of a to give a phase margin of 45° is equal to (a) 0.141 (c) 0.841
1 2
3 Re
(a)
K K (b) 2 3 s s + 1) ( s
(c)
K K (d) s( s 2 + 1) s( s 2 −1)
1+sP (a) 20 and 0.3 (c) 25 and 0.3
323
(GATE 2005: 1 Mark) 40. The gain margin of a unity feedback control system with ( s +1) is the open-loop transfer function G ( s) = s2 (a) 0 (c)
(b)
1
2 2 (d) ∞ (GATE 2005: 2 Marks)
41. A unity feedback system, having an open-loop gain K (1 − s) , becomes stable when G ( s) H ( s) = (1 + s) (a) |K| > 1 (c) |K| < 1
(b) K > 1 (d) K < − 1 (GATE 2005: 2 Marks)
42. When subjected to a unit step input, the closed-loop control system shown in the following figure will have a steady-state error of −
+ R(s) 3/s −
2 s+2
Y(s)
+
(b) 0.441 (d) 1.141 (GATE 2004: 2 Marks)
(a) −1.0 (b) −0.5 (c) 0 (d) 0.5 (GATE 2005: 2 Marks)
38. A system with zero initial conditions has the closedloop transfer function s2 + 4 T ( s) = ( s + 1)( s + 4)
43. In the GH(s) plane, the Nyquist plot of the loop transfer function p e −0.25 s G ( s) H ( s) = s passes through the negative real axis at the point (a) (−0.25, j 0) (b) (−0.5, j 0) (c) (−1, j 0) (d) (−2, j 0) (GATE 2005: 2 Marks)
The system output is zero at the frequency (a) 0.5 rad/s (b) 1 rad/s (c) 2 rad/s (d) 4 rad/s (GATE 2005: 1 Mark)
Ch wise GATE_EE_Ch7.indd 323
11/21/2018 3:45:50 PM
324
GATE EE Chapter-wise Solved Papers
44. If the compensated system shown in the following figure has a phase margin of 60° at the crossover frequency of 1 rad/s, then value of the gain K is R(s) + K+0.366s − (a) 0.366 (c) 1.366
Y(s) 1 s(s+1)
(b) 0.732 (d) 2.738 (GATE 2005: 2 Marks)
Statement for Linked Answer Questions 45 and 46: A state variable system 1⎤ ⎡1 ⎤ ⎡0 X (t ) = ⎢ X (t ) + ⎢ ⎥ u(t ), ⎥ ⎣0 −3⎦ ⎣0⎦
48. A discrete real all pass system has a pole at z = 2∠30°, it therefore 1 (a) also has a pole at ∠30° 2 (b) has a constant phase response over the z-plane: arg |H(z)| = constant. (c) is stable only, if it is anticausal. (d) It has a constant phase response over the unit c ircle: arg |H(eiW)| = constant. (GATE 2006: 1 Mark) 49. The Bode magnitude plot of
H ( jw ) =
|H (jw)|dB
(a)
40
with initial condition X(0) = [-1 3]T and the unit step input u(t) has
20
45. The state transition matrix
−1
1 1 −t ⎤ ⎡ ⎡ −3t ⎤ 1 (1 − e −3t ) ⎥ ⎢ 1 3 (e − e )⎥ (a) ⎢ (b) 3 ⎢ ⎥ ⎢ ⎥ e −3t ⎦ e −t ⎦ ⎣0 ⎣0 1 −t ⎤ ⎡ 1 (e − e −3t ) ⎥ (c) ⎢ 3 ⎢ ⎥ e −3t ⎦ ⎣0
(b)
(c)
⎡t − e −3t ⎤ ⎡t − e −3t ⎤ (d) X (t ) = ⎢ X ( t ) = ⎢ ⎥ −3t ⎥ e −t ⎦ ⎣ 3e ⎦ ⎣
system with the transfer function 3( s − 2) H ( s) = 2 , the matrix A in the state space 4 s − 2s + 1 . form x = Ax + Bu is equal to
20 −1
⎡ 0 1 0⎤ ⎢ 1⎥⎥ (b) ⎢ 0 0 ⎢⎣ −1 2 −4 ⎥⎦
1 0⎤ ⎡0 ⎢ 3 −2 1⎥ (c) ⎢ ⎥ ⎢⎣ 1 −2 4 ⎥⎦
0⎤ ⎡ 1 0 ⎢ 0 0 1⎥⎥ (d) ⎢ ⎢⎣ −1 2 −4 ⎥⎦ (GATE 2006: 1 Mark)
Ch wise GATE_EE_Ch7.indd 324
log (w) 0 −20 −40
+1
+2
+3
+2
+3
|H (jw)|dB
(c)
40 20 −1
a
0⎤ ⎡ 1 0 ⎢ 0 1 0⎥ (a) ⎢ ⎥ ⎢⎣ −1 2 −4 ⎥⎦
+3
40
(GATE 2005: 2 Marks) 47. For
+2
|H (jw)|dB
46. The state transition equation ⎡t − e − t ⎤ (b) X (t ) = ⎢ −3t ⎥ ⎣ 3e ⎦
+1
−40
(GATE 2005: 2 Marks) ⎡t − e − t ⎤ X (t ) = ⎢ −t ⎥ ⎣ e ⎦
log (w) 0 −20
−t (d) ⎡ 1 (1 − e ) ⎤ ⎢ ⎥ e −t ⎦ ⎣0
(a)
10 4 (1 + jw ) is (10 + jw )(100 + jw ) 2
(d)
log (w) +1 0 −20 −40
|H (jw)|dB 40 20 −1
log (w) +1 0 −20 −40
+2
+3
(GATE 2006: 2 Marks) 50. A closed-loop system has the characteristic function (s2 − 4) (s + 1) + K (s − 1) = 0. Its root locus plot against K is
11/21/2018 3:45:52 PM
Chapter 7 • Control Systems
(a)
(3)
jw
−2
−1 +1
(b)
s +2
ω
jw
−2
−1 +1
−1
Re ω
(a) (1) only (c) all, except (3)
jw
(d)
−2
s
−1
+1
+2
+2 (GATE 2006: 2 Marks)
51. The quadratic equation F(s) = s5 − 3s4 + 5s3 − 7s2 + 4s + 20 is given. F(s) = 0 has (a) a single complex root with the remaining roots being real (b) one positive real root and four complex roots, all with positive real parts (c) one negative real root, two imaginary roots, and two roots with positive real parts (d) one positive real root, two imaginary roots, and two roots with negative real parts (GATE 2006: 2 Marks) 52. Consider the following Nyquist plots of loop transfer functions over w = 0 to w = ∞. Which of these plots represents a stable closed-loop system? (1)
Im
(2)
−1
Ch wise GATE_EE_Ch7.indd 325
Re ω
− u2
S +
−1
54. If x = Re G( jw), and y = Im G( jw) then for w → 0+, 1 the Nyquist plot for G ( s) = becomes s( s + 1)( s + 2) asymptotic to the line (a) x = 0
(b) x = −
(c) x = y - 1/6
(d) x =
3 4
y 3
(GATE 2007: 2 Marks) 900 is be compensated such s( s + 1)( s + 9) that its gain-crossover frequency becomes same as its uncompensated phase-crossover frequency and provides a 45° phase margin, to achieve this, one may use
55. The system
(a) a lag compensator that provides an attenuation of 20 dB and a phase lag of 45° at the frequency of 3 3 rad/s. (b) a lead compensator that provides an amplification of 20 dB and a phase lead of 45° at the frequency of 3 rad/s
Im ω
R ω=∞
s−1 s+2
(a) stable (b) unstable (c) conditionally stable (d) stable for input u1, but unstable for input u2 (GATE 2007: 1 Mark)
s +1
S
1 s−1
jw
−1
Re
53. The system shown in the following figure is
− −2
ω =∞
−1
(b) all, except (1) (d) (1) and (2) only (GATE 2006: 2 Marks)
s +2
Im
ω =∞
u1 +
(c)
(4)
Im
325
ω =∞ Re
(c) a lag-lead compensator that provides an amplification of 20 dB and a phase lag of 45° at the frequency of 3 rad/s.
11/21/2018 3:45:54 PM
326
GATE EE Chapter-wise Solved Papers
(d) a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45° at the frequency of 3 rad/s (GATE 2007: 2 Marks) 56. 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) = … = 0. +
S
G(z)
+
with (a)
X = c0 s + c1 , Y =
(b) X = 1, Y =
1 , Z = b0 s + b1 ( s 2 + a0 s + a1 )
(c0 s + c1 ) , Z = b0 s + b1 ( s + a0 s + a1 ) 2
(c)
X = c1 s + c0 , Y =
(b1 s + b0 ) ,Z =1 ( s + a1 s + a0 )
(d)
X = c1 s + c0 , Y =
1 , ( s 2 + a1 s + a0 )
K
2
Z = b1 s + b0 (GATE 2007: 2 Marks)
This system is stable for range of values of K ⎡ 1⎤ (a) ⎢ −1, ⎥ ⎣ 2⎦
(b) [ −1,1]
⎡ 1 ⎤ (c) ⎢ − ,1⎥ ⎣ 2 ⎦
⎡ 1 ⎤ (d) ⎢ − , 2⎥ ⎣ 2 ⎦ (GATE 2007: 2 Marks)
59. The RLC series circuit shown is supplied from a variable frequency voltage source. The admittance locus of the RLC network at terminals AB for increasing frequency w is A R w
57. If the loop gain K of a negative feedback system having K ( s + 3) a loop transfer function is to be adjusted to ( s + 8) 2 induce a sustained oscillation then (a) the frequency of this oscillation must be
4
rad/s . 3 (b) the frequency of this oscillation must be 4 rad/s. (c) the frequency of this oscillation must be 4 or 4 rad/s . 3 (d) such a K does not exist. (GATE 2007: 2 Marks)
L C B
(a)
(b)
Im
Im
Re
Re w
w (c)
(d)
Im w
Im w
Re
Re
58. The system shown in the following figure (GATE 2007: 2 Marks) b0
c0
b1
c1 S
S
1/s
a0
S
1/s
P
60. Consider the feedback control system shown in the following figure which is subjected to a unit step input. The system is stable and has the following parameters kp = 4, ki = 10, w = 500 and z = 0.7. ki s
a1
Z
1 0 can be reduced to the form + X
S
Y
+ Z
Ch wise GATE_EE_Ch7.indd 326
P
+ Σ −
+ kp +
Σ
w2 s2 + 2zws +w 2
The steady-state value of Z is (a) 1 (b) 0.25 (c) 0.1 (d) 0 (GATE 2007: 2 Marks)
11/21/2018 3:45:56 PM
Chapter 7 • Control Systems
(a) 0 (c) 1
Statement for Linked Answer Questions 61 and 62: 61. A signal is processed by a causal filter with transfer function G(s). For a distortion free output signal waveform, G(s) must (a) provide zero phase shift for all frequency (b) provide constant phase shift for all frequency (c) provide linear phase shift that is proportional to frequency. (d) provide a phase shift that is inversely proportional to frequency. (GATE 2007: 2 Marks) 62. 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 (1/3) (c) a = b (d) a = b −(1/3) (GATE 2007: 2 Marks) Linked Answer Questions 63 and 64: Consider the RLC circuit shown in the figure.
C = 10 µF
(b) 0.5 (d) 2 (GATE 2008: 2 Marks)
66. The transfer functions of two compensators are C1 =
10( s + 1) s + 10 , C2 = ( s + 10) 10( s + 1)
Which one of the following statements is correct? (a) C1 is a 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 compensators (d) Both C1 and C2 are lag compensators (GATE 2008: 2 Marks) 67. The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the following figure. G(jω) (dB) 20 0 −20
R = 10 Ω L = 1mH
ei
327
eo
-40 dB/decade −20 dB/decade ω (rad/s) (log scale)
0.1
0 dB/decade
63. For a step-input e, the overshoot in the output e0 will be (a) 0, since the system is not under-damped (b) 5% (c) 16% (d) 48% (GATE 2007: 2 Marks)
This transfer function has (a) three poles and one zero (b) two poles and one zero (c) two poles and two zeros (d) one pole and two zeros (GATE 2008: 2 Marks) 68. The following figure shows a feedback system where K > 0.
64. If the above step response is to be observed on a non- storage CRO, then it would be best to have the ei as a (a) step function (b) square wave of frequency 50 Hz (c) square wave of frequency 300 Hz (d) square wave of frequency 2.0 kHz (GATE 2007: 2 Marks) 65. The transfer function of a linear time invariant system is given as 1 G ( s) = 2 s + 3s + 2 The steady-state value of the output of this system for a unit impulse input applied at time instant t = 1 will be
Ch wise GATE_EE_Ch7.indd 327
+
S
K s(s+3)(s+10)
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 (GATE 2008: 2 Marks)
69. The transfer function of a system is given as 100 . 2 s + 20 s + 100 This system is (a) an overdamped system (b) an underdamped system
11/21/2018 3:45:57 PM
328
GATE EE Chapter-wise Solved Papers
(c) a critically damped system (d) an unstable system (GATE 2008: 2 Marks)
73. The polar plot of an open-loop stable system is shown in the following figure. The closed-loop system is Im
Statement for Linked Answer Questions 70 and 71: The state space equation of a system is described by w=∞
x = Ax + Bu y = Cx
−1.42
Re
where x is state vector, u is input, y is output and 1⎤ ⎡0 ⎤ ⎡0 A= ⎢ , B = ⎢ ⎥ , C = [1 0] ⎥ ⎣0 −2⎦ ⎣1 ⎦
w=0
70. The transfer function G(s) of this system will be (a)
s ( s + 2)
(b)
s +1 s ( s − 2)
(c)
s ( s − 2)
(d)
1 s ( s + 2)
(GATE 2008: 2 Marks)
(a) (b) (c) (d)
always stable marginally stable unstable with one pole on the RH s-plane unstable with two poles on the RH s-plane (GATE 2009: 1 Mark)
74. The first two rows of Routh’s tabulation of a third-order equation are as follows:
71. A unity feedback is provided to the above system G(s) to make it a closed-loop system as shown in the figure below. r(t) +
Y(t)
Σ
G(s)
For a unit step input r(t), the steady-state error in the output will be (a) 0 (b) 1 (c) 2 (d) ∞ (GATE 2008: 2 Marks) 72. The measurement system shown in the following figure uses three sub-systems in cascade whose gains are specified as G1, G2 and 1 . The relative small errors G3
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 = ± j 2 and one root in left half s-plane (c) two roots at s = ± j 2 and one root in right half s-plane (d) two roots at s = ± j and one root in left half s-plane (GATE 2009: 1 Mark) 75. The asymptotic approximation of the log-magnitude vs. frequency plot of a system containing only real poles and zeros is shown in the following figure. Its transfer function is 80 dB
−40dB/decade
−60dB/decade
associated with each respective subsystem G1, G2 and G3 are e1, e2 and e3. The error associated with the output is Input
G1
G2
1 G3
w rad/s Output
1 ε ⋅ε (a) ε1 + ε 2 + (b) 1 2 ε3 ε 3 (c) ε1 + ε 2 − ε 3
(d) ε1 + ε 2 + ε 3 (GATE 2009: 1 Mark)
Ch wise GATE_EE_Ch7.indd 328
0.1
2 5
25
(a)
10( s + 5) s( s + 2)( s + 25)
(c)
100( s + 5) 80( s + 5) (d) s( s + 2)( s + 25) s 2 ( s + 2)( s + 25)
(b)
1000( s + 5) s ( s + 2)( s + 25) 2
(GATE 2009: 1 Mark)
11/21/2018 3:45:59 PM
Chapter 7 • Control Systems
76. The unit-step response of a unity feedback system with open-loop transfer function G ( s) =
K ( s + 1)( s + 2)
80. As shown in the following 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
is shown in the following figure. The value of K is 1 Response
329
+
100 ± 10%
−
0.75 0.5
9/100 0.25 0 0
1
2 Time(s)
(a) 0.5 (c) 4
3
4
(b) 2 (d) 6 (GATE 2009: 2 Marks)
77. The open-loop transfer function of a unity feedback system is given by G ( s) =
(e −0.1s ) s
The gain margin of this system is (a) 11.95 dB (b) 17.67 dB (c) 21.33 dB (d) 23.9 dB (GATE 2009: 2 Marks)
(a) 10 ± 1% (c) 0 ± 5%
t
(b) 10 ± 2% (d) 10 ± 10% (GATE 2010: 1 Mark)
2 , the approximate time taken for a ( s + 1) step response to reach 98% of its final value is (a) 1 s (b) 2 s (c) 4 s (d) 8 s (GATE 2010: 1 Mark)
81. For the system
82. If the electrical circuit of Fig. (b) is an equivalent of the coupled tank system of Fig. (a), then
Common Data for Questions 78 and 79: A system is described by the following state and output equations h1
dx1 (t ) = −3 x1 (t ) + x2 (t ) + 2u(t ) dt dx2 (t ) = −2 x2 (t ) + u(t ) dt y(t ) = x1 (t )
(a) Coupled tank
B
where u(t) is input and y(t) is the output. 78. The system transfer function is
A
(a)
s + 2 (b) s+3 s 2 + 5s − 6 s 2 + 5s + 6
(c)
2s − 5 2 s + 5 (d) 2 s + 5s − 6 s + 5s + 6
D C
(b) Electrical equivalent
2
(GATE 2009: 2 Marks) 79. The state-transition matrix of the above system is ⎡ e −3t (a) ⎢ −2t −3t ⎣e + e
⎡e −3t e −2t − e −3t ⎤ 0⎤ ⎥ (b) ⎢ ⎥ e ⎦ e −2t ⎦ ⎣ 0 −2 t
⎡e −3t e −2t + e 3t ⎤ ⎡e 3t e −2t − e −3t ⎤ (c) ⎢ (d) ⎢ ⎥ −2 t ⎥ e ⎦ e −2t ⎦ ⎣ 0 ⎣ 0 (GATE 2009: 2 Marks)
Ch wise GATE_EE_Ch7.indd 329
h2
(a) (b) (c) (d)
A, B are resistances and C, D capacitances A, C are resistances and B, D capacitances A, B are capacitances and C, D resistances A, C are capacitances and B, D resistances (GATE 2010: 1 Mark)
1 [ s( s + 1)( s + 2)] plotted in the complex G( jw) plane for (0 < w < ∞) is
83. The frequency response of G ( s) =
11/21/2018 3:46:00 PM
330
GATE EE Chapter-wise Solved Papers
(a)
(b)
Im
ω=0
−3/4
of 10 and a duration of one second, as shown in the following figure is r(t)
Im
Re
Re
−3/4
ω= 0
10 t
(c)
Im
(d)
1s
Im
ω =0 Re
Re
(b) 0.1 (d) 10 (GATE 2011: 1 Mark)
88. A point z has been plotted in the complex plane, as shown in the following figure. Im
ω= 0 −1/6
−1/6
(a) 0 (c) 1
(GATE 2010: 2 Marks) ⎡0 ⎤ ⎡ −1 2⎤ ,B = ⎢ ⎥ 84. The system x = Ax + Bu with A = ⎢ ⎥ ⎣ 0 2⎦ ⎣1 ⎦ ⎡0 ⎤ ⎡ −1 2⎤ is , = A= ⎢ B ⎥ ⎢1 ⎥ ⎣ 0 2⎦ ⎣ ⎦ (a) (b) (c) (d)
stable and controllable stable but uncontrollable unstable but controllable unstable and uncontrollable (GATE 2010: 2 Marks)
85. The characteristic equation of a closed-loop system is s(s + l)(s + 3) + k(s + 2) = 0, k > 0. Which of the following statements is true? (a) Its roots are always real (b) It cannot have a breakaway point in the range -l < 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 (GATE 2010: 2 Marks)
z
·
Re
The plot of the complex number y = (a)
1.3
1.2
1.0
0.8
0.5
0.3
-G(jw )
−130°
−140°
−150°
−160°
−180°
−200°
Unit circle Re y•
(b)
Im Unit circle Re
y• (c)
The gain margin and phase margin of the system are (a) 6 dB and 30° (b) 6 dB and −30° (c) −6 dB and 30° (d) −6 dB and −30° (GATE 2011: 1 Mark)
1 is z
Im
86. The frequency response of a linear system G(jw) is provided in the tabular form below |G(jw )|
Unit circle
Im
•
y
Unit circle Re
87. 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
Ch wise GATE_EE_Ch7.indd 330
11/21/2018 3:46:01 PM
Chapter 7 • Control Systems
(d)
(d) phase shift of the closed-loop transfer function at very high frequencies (w → ∞). (GATE 2011: 2 Marks)
Im Unit circle
92. A system with transfer function Re
•
y (GATE 2011: 1 Mark)
89. An open-loop system represented by the transfer function G ( s) = (a) (b) (c) (d)
( s − 1) is ( s + 2)( s + 3)
G ( s) =
93. The state variable description of an LTI system is given by
stable and of the minimum phase type. stable and of the non-minimum phase type. unstable and of the minimum phase type. unstable and of the non-minimum phase type. (GATE 2011: 2 Marks)
2⎞ ⎛ K ⎜s+ ⎟ ⎝ 3⎠ G ( s) = 2 s ( s + 2) From the root locus, it 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 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. (GATE 2011: 2 Marks) 91. A two-loop position control system is shown in the following figure. Motor 1 + + R(s) Y (s) s (s + 1) − − Ks Tacho-generator
( s 2 + 9)( s + 2) ( s + 1)( s + 3)( s + 4)
is excited by sin (w t). 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 (GATE 2012: 1 Mark)
⎛ x 1 ⎞ ⎛ 0 a1 0⎞ ⎛ x1 ⎞ ⎛ 0⎞ ⎜ x ⎟ = ⎜ 0 0 a ⎟ ⎜ x ⎟ + ⎜ 0⎟ u 1 ⎜ 2⎟ ⎜ ⎟ ⎜ 2⎟ ⎜ ⎟ ⎝ x3 ⎠ ⎝ a3 0 0⎠ ⎝ x3 ⎠ ⎝ 1⎠ ⎛ x1 ⎞ y = (1 0 0 ) ⎜ x2 ⎟ ⎜ ⎟ ⎝ x3 ⎠
90. The open-loop transfer function G(s) of a unity feedback control system is given as
where y is the output and u is the input, The system is controllable for (a) a1 ≠ 0, a2 = 0, a3 ≠ 0 (b) a1 = 0, a2 ≠ 0, a3 ≠ 0 (c) a1 = 0, a2 ≠ 0, a3 = 0 (d) a1 ≠ 0, a2 ≠ 0, a3 = 0 (GATE 2012: 2 Marks) 94. The feedback system shown in the following figure oscillates at 2 rad/s when R(s) +
K(s + 1) −
(a) (b) (c) (d)
The gain K of the Tacho-generator influences mainly (a) peak overshoot. (b) natural frequency of oscillation. (c) phase shift of the closed-loop transfer function at very low frequencies (w → 0).
Y (s)
s3 + as2 + 2s + 1
K = 2 and a = 0.75 K = 3 and a = 0.75 K = 4 and a = 0.5 K = 2 and a = 0.5 (GATE 2012: 2 Marks)
Statement for Linked Answer Questions 95 and 96: The transfer function of a compensator is given as Gc ( s) =
Ch wise GATE_EE_Ch7.indd 331
331
s+a s+b
95. 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, 6 = 1 (GATE 2012: 2 Marks)
11/21/2018 3:46:02 PM
332
GATE EE Chapter-wise Solved Papers
96. The phase of the above lead compensator is maximum at (a) (b) 2 rad/s 3 rad/s 1 (c) (d) rad/s 6 rad/s 3 (GATE 2012: 2 Marks) 97. The transfer function V2 ( s) of the circuit shown in the V1 ( s) following figure is 100 µF +
The transfer function Y ( s) for this system is U ( s) (a) (c)
− (a) 0.5s + 1 s +1 (c) s + 2 s +1
s +1 s + 6s + 2 2
s + 1 (d) 1 s2 + 4s + 2 5s 2 + 6 s + 2 (GATE 2013: 2 Marks)
⎡ x 1 ⎤ ⎡ −2 0 ⎤ ⎡ x1 ⎤ ⎡1⎤ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ + ⎢ ⎥ u, ⎣ x2 ⎦ ⎣ 0 −1⎦ ⎣ x1 ⎦ ⎣1⎦
+ V2(s)
100 µF
(b)
Common Data for Questions 100 and 101: The state variable formulation of a system is given as
10 kΩ V1(s)
s +1 5s + 6 s + 2 2
x1 (0) = 0
⎡ x1 ⎤ and y = [1 0] ⎢ ⎥ x2 ( 0 ) = 0 ⎣ x2 ⎦
− (b) 3s + 6 s+2 (d) s + 1 s+2 (GATE 2013: 1 Mark)
98. The Bode plot of a transfer function G(s) is shown in the following figure.
100. 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. (GATE 2013: 2 Marks) 101. The response y(t) to a unit step input is
32
1 1 −2t − e 2 2
20
(c) e −2t − e − t
Gain (dB)
40
(a)
1 1 (b) 1 − e −2t − e − t 2 2 (d) 1− e − t (GATE 2013: 2 Marks)
0 −8
1
10 100 ω (rad/s)
The gain (20 log|G(s)|) 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) is
102. In the formation of Routh–Hurwitz array for a polynomial, all the elements of a row have zero values. This premature termination of the array indicates the presence of (a) only one root at the origin (b) imaginary roots (c) only positive real roots (d) only negative real roots (GATE 2014: 1 Mark)
39.8 39.8 (b) s s2 32 32 103. The root locus of a unity feedback system is shown in (c) (d) s s2 the figure (GATE 2013: 1 Mark) (a)
99. The signal flow graph for a system is given below. 1 U(s) 1
s−1
s−1 −2
Ch wise GATE_EE_Ch7.indd 332
−4
1 Y(s)
jw
K=0 K=0 −2
−1
s
11/21/2018 3:46:05 PM
Chapter 7 • Control Systems
The closed loop transfer function of the system is (a)
C ( s) K = R( s) ( s + 1)( s + 2)
(b)
C ( s) −K = R( s) ( s + 1)( s + 2) + K
(c)
C ( s) K = R( s) ( s + 1)( s + 2) − K
(d)
C ( s) K = R( s) ( s + 1)( s + 2) + K (GATE 2014: 1 Mark)
107. A single-input single-output feedback system has forward transfer function G(s) and feedback transfer function H(s). It is given that |G(s)H(s)| < 1. Which of the following is true about the stability of the system? (a) The system is always stable. (b) The system is stable if all zeros of G(s)H(s) are in left half of the s-plane. (c) The system is stable if all poles of G(s)H(s) are in left half of the s-plane. (d) It is not possible to say whether or not the system is stable from the information given. (GATE 2014: 1 Mark) 108. For the given system, it is desired that the system be stable. The minimum value of a for this condition . is
104. The closed-loop transfer function of a system is T ( s) =
333
4 ( s 2 + 0.4 s + 4)
The steady state error due to unit step input is . (GATE 2014: 1 Mark)
(s + a) R(s)
+ −
s3 + (1 + a)s2 + (a − 1)s + (1 − a)
C(s)
105. The state transition matrix for the system (GATE 2014: 2 Marks)
⎡ x 1 ⎤ ⎡1 0 ⎤ ⎡ x1 ⎤ ⎡1⎤ ⎢ x ⎥ = ⎢1 1 ⎥ ⎢ x ⎥ + ⎢1⎥ u ⎦⎣ 2⎦ ⎣ ⎦ ⎣ 2⎦ ⎣
109. The Bode magnitude plot of the transfer function
is ⎡et (a) ⎢ t ⎣e
⎡ et 0⎤ 0⎤ (b) ⎥ ⎢2 t t t⎥ e⎦ ⎣t e e ⎦
⎡ et (c) ⎢ t ⎣te
⎡e t te t ⎤ 0⎤ (d) ⎥ ⎢ t ⎥ et ⎦ ⎣0 e ⎦ (GATE 2014: 1 Mark)
106. The signal flow graph of a system is shown below. U(s) is the input and C(s) is the output. h1
U(s)
h0
1 s
1
1
1 s
G ( s) =
K (1 + 0.5s)(1 + as) is shown below: Note s⎞ ⎛ s⎞ ⎛ s ⎜1 + ⎟ (1 + bs) ⎜1 + ⎟ ⎝ 8⎠ ⎝ 36 ⎠
that -6 dB/octave = −20 dB/decade. The value of a is bK . 6 dB/ 0 dB/octave −6 dB/ octave octave 0 dB/ octave dB
1
−12 dB/ octave
C(s) −a1
0 0.01
−a0
2
4
Assuming, h1 = b1 and h0 = b0 - b1a1, the input-output transfer function, G ( s) =
C ( s) of the system is given by U ( s)
(a) G ( s) =
b0 s + b1 (b) a s + a0 G ( s) = 2 1 2 s + a0 s + a1 s + b1 s + b0
(c) G ( s) =
b1 s + b0 (d) a s + a1 G ( s) = 2 0 s + a1 s + a0 s + b0 s + b1 2
(GATE 2014: 1 Mark)
Ch wise GATE_EE_Ch7.indd 333
−6 dB/ octave
8
24
36 w (rad/s)
(GATE 2014: 2 Marks) 110. A system with the open loop transfer function G ( s) =
K s( s + 2)( s 2 + 2 s + 2)
is connected in a negative feedback configuration with a feedback gain of unity. For the closed loop system to be marginally stable, the value of K is . (GATE 2014: 2 Marks)
11/21/2018 3:46:07 PM
334
GATE EE Chapter-wise Solved Papers
111. For the transfer function G ( s) =
115. The magnitude Bode plot of a network is shown in the following figure.
5( s + 4) s( s + 0.25)( s 2 + 4 s + 25)
The values of the constant gain term and the highest corner frequency of the Bode plot respectively are (a) 3.2, 5.0 (c) 3.2, 4.0
(b) 16.0, 4.0 (d) 16.0, 5.0 (GATE 2014: 2 Marks)
112. The second order dynamic system dX = PX + Qu dt y = RX has the matrices P, Q and R as follows: ⎡0⎤ ⎡ −1 1 ⎤ P=⎢ Q=⎢ ⎥ R = [0 1] ⎥ 0 3 − ⎣ ⎦ ⎣1⎦
G(jω) dB
Slope 20 dB/decade
0
1 1 log10w 3 The maximum phase angle f m and the corresponding gain Gm respectively, are (a) -30° and 1.73 dB (b) -30° and 4.77 dB (c) +30° and 4.77 dB (d) +30° and 1.73 dB (GATE 2014: 2 Marks) 116. A continuous-time LTI system with system function H(w) has the following pole-zero plot. For this system, which of the alternatives is TRUE? w
The system has the following controllability and observability properties: (a) Controllable and observable (b) Not controllable but observable (c) Controllable but not observable (d) Not controllable and not observable (GATE 2014: 2 Marks) 113. The block diagram of a system is shown in the figure
+
R(s)
-
1 s
+
-
G(s)
s
C(s)
If the desired transfer function of the system is C ( s) s = 2 R( s) s + s + 1 then G(s) is (a) 1
(b) s
(c) 1/s
(d)
w2 w1 (0, 0)
s
|H(0)| > |H(w)|; |w | > 0 |H(w)| has multiple maxima, at w1 and w2 |H(0)| < |H(w)|; |w | > 0 |H(w)| = constant; -∞ < w < ∞ (GATE 2014: 2 Marks)
(a) (b) (c) (d)
117. A Bode magnitude plot for the transfer function G(s) of a plant is shown in the figure. Which one of the following transfer functions best describes the plant? 20 log(Gj2pf )
−s s + s2 − s − 2 3
20
(GATE 2014: 2 Marks)
0
114. Consider the system described by following state space equations
−20 f(Hz)
⎡ x 1 ⎤ ⎡ 0 1⎤ ⎡ x1 ⎤ ⎡0 ⎤ ⎢ x ⎥ = ⎢ −1 −1⎥ ⎢ x ⎥ + ⎢1 ⎥ u; ⎦⎣ 2⎦ ⎣ ⎦ ⎣ 2⎦ ⎣
⎡x ⎤ y = [1 0] ⎢ 1 ⎥ ⎣ x2 ⎦
If u is unit step input, then the steady state error of the system is (a) 0 (b) 1/2 (c) 2/3 (d) 1 (GATE 2014: 2 Marks)
Ch wise GATE_EE_Ch7.indd 334
0.1 (a)
(c)
1
10 100 1 k 10 k 100 k
1000( s + 10) 10( s + 10) (b) s + 1000 s( s + 1000) s + 1000 s + 1000 (d) 10 s( s + 10) 10( s + 10) (GATE 2015: 1 Mark)
11/21/2018 3:46:10 PM
335
Chapter 7 • Control Systems
118. For the signal flow graph shown in the following figure, which one of the following expressions is equal to the transfer function
Y ( s) X 2 ( s)
X1(s)
(c)
Im
?
Re
X1( s ) = 0
X2(s) (d) 1
Im ∞ w
G2
G1
Y(s) −1
w
−1
0
G1 G2 (a) (b) 1 + G2 (1 + G1 ) 1 + G1 (1 + G2 )
(GATE 2015: 1 Mark)
G1 G2 (d) 1+ G1G2 1+ G1G2 (GATE 2015: 1 Mark)
(c)
119. An open loop control system results in a response of e−2t(sin5t + cos5t) for a unit impulse input. The DC gain of the control system is ________. (GATE 2015: 1 Mark)
121. In the signal flow diagram given in the figure, u1 and u2 are possible inputs whereas y1 and y2 are possible outputs. When would the SISO system derived from this diagram be controllable and observable? 5 u1
∞ Im w=∞ G1(s)
G2(s)
1
y2 −1
w=0
(a) When u1 is the only input and y1 is the only output. (b) When u2 is the only input and y1 is the only output. (c) When u1 is the only input and y2 is the only output. (d) When u2 is the only input and y2 is the only output. (GATE 2015: 2 Marks) 122. The transfer function of a second order real system with a perfectly flat magnitude response of unity has a pole at (2 − j3). List all the poles and zeroes.
w=∞ Re
(b)
x2
2 1
Nyquist plot of the product of G1(s) and G2(s) is
∞
1
1
w
0
(a)
1
−2
1/s
Re
Im
y1
1 u2
Re
w
x1 1/s
120. Nyquist plots of the two functions G1(s) and G2(s) are shown in the figure.
Im
Re
w
Im
(a) (b) (c) (d)
Poles at (2 ± j3), no zeroes Poles at (±2 − j3), one zero at origin Poles at (2 − j3), (−2 + j3), zeroes at (−2 − j3), (2 + j3) Poles at (2 ± j3), zeroes at (−2 ± j3) (GATE 2015: 2 Marks)
Re 1
123. Find the transfer function system
Ch wise GATE_EE_Ch7.indd 335
Y ( s) of the following X ( s)
11/21/2018 3:46:11 PM
336
GATE EE Chapter-wise Solved Papers
(a)
+
y
G1 − X(s)
1
+ H
Y(s) 0
Y(s) +
−
5
G2
−2
G1 G2 G G2 + (b) 1 + (a) 1 − HG1 1 − HG2 1 + HG1 1 + HG2 (c)
t
+
(b) y
G1 + G2 G1 + G2 (d) 1 + H (G1 + G2 ) 1 − H (G1 + G2 )
2 1
(GATE 2015: 2 Marks) 124. The open loop poles of a third order unity feedback system are at 0, −1, −2. Let the frequency corresponding to the point where the root locus of the system transits to unstable region be K. Now suppose we introduce a zero in the open loop transfer function at −3, while keeping all the earlier open loop poles intact. Which one of the following is TRUE about the point where the root locus of the modified system transits to unstable region? (a) It corresponds to a frequency greater than K (b) It corresponds to a frequency less than K (c) It corresponds to a frequency K (d) Root locus of modified system never transits to unstable region (GATE 2015: 2 Marks)
0 (c)
For this discrete-time system, which one of the following statements is TRUE? (a) The system is not stable for h > 0 1 (b) The system is stable for h > p (c) The system is stable for 0 < h < (d) The system is stable for
1 2p
1 1 0, T > 0 s(1 + Ts)(1 + 2 s)
The closed loop system will be stable if, (a) 0 < T
0, b > 0, K < 0), the steady R( s) s + as + b state error will be a (a) 0 (b) b (c)
U2 −1
(d) -p and 0 (GATE 2017: 1 Mark)
⎡ x (t ) ⎤ y(t ) = [1 0] ⎢ 1 ⎥ ⎣ x2 (t ) ⎦
(a)
( s + 2) ( s − 2) (b) 2 ( s − 2 s − 2) ( s + s − 4)
(c)
( s − 4) ( s + 4) (d) ( s 2 + s − 4) ( s 2 − s − 4)
2
(GATE 2017: 2 Marks)
11/21/2018 3:46:18 PM
Chapter 7 • Control Systems
146. For a system having transfer function G ( s) =
−s + 1 , s +1
−s + 1 , a unit step input is applied at time s +1 t = 0. The value of the response of the system at t = 1.5 s (rounded off to three decimal places) is . (GATE 2017: 2 Marks)
G ( s) =
147. Consider a causal and stable LTI system with rational transfer function H(z), whose corresponding impulse 5 response begins at n = 0. Furthermore, H (1) = . 4 1 ⎛ ( 2k − 1) x ⎞ exp ⎜ j The poles of H(z) are pk = ⎟⎠ for ⎝ 4 2 k = 1, 2, 3, 4. The zeros of H(z) are all at z = 0. Let g[n] = jnh[n]. The value of g[8] equals . (Give the answer up to three decimal places.)
(c) s = −5 (d) s= 5 (GATE 2017: 2 Marks) 149. The range of K for which all the roots of the equation s3 + 3s2 + 2s + K = 0 are in the left half of the complex s-plane is (a) 0 < K < 6 (b) 0 < K < 16 (c) 6 < K < 36 (d) 6 < K < 16 (GATE 2017: 2 Marks) 150. Which of the following systems has maximum peak overshoot due to a unit step input? (a)
100 100 (b) s 2 + 10 s + 100 s 2 + 15s + 100
(c)
100 100 (d) 2 s + 5s + 100 s + 20 s + 100
⎡ x1 (0) ⎤ ⎡1 ⎤ If u(t) is a unit step input and ⎢ ⎥ = ⎢ ⎥ , value of ⎣ x2 ( 0 ) ⎦ ⎣ 0 ⎦ output y(t) at t = 1 s (rounded off to three decimal place) . is (GATE 2017: 2 Marks)
152. A continuous-time input signal x(t) is an eigenfunction of an LTI system, if the output is (a) kx(t), where k is an eigenvalue (b) kejw tx(t), where k is an eigenvalue and ejw t is a complex exponential signal (c) x(t)ejw t, where ejw t is a complex exponential signal (d) kH(w), where k is an eigenvalue and H(w) is a frequency response of the system (GATE 2018: 1 Mark) 153. The series impedance matrix of a short three-phase transmission line in phase coordinates is ⎡ Zs ⎢Z ⎢ m ⎢⎣ Z m
(GATE 2017: 2 Marks) 148. The root locus of the feedback control system having the characteristic equation s2 + 6Ks + 2s + 5 = 0, where K > 0, enters into the real axis at (a) s = −1 (b) s=− 5
339
Zm Zs Zm
Zm ⎤ Z m ⎥⎥ . Zs ⎥⎦
If the positive sequence impedance is (1 + j 10) W, and the zero sequence is (4 + j 31) W then the imaginary part (up to 2 decimal places). of Zm (in W) is (GATE 2018: 1 Mark)
154. Consider a unity feedback system with forward transfer function given by G ( s) =
1 ( s + 1)( s + 2)
The steady-state error in the output of the system for a unit-step input is (up to 2 decimal places). (GATE 2018: 1 Mark)
155. A DC voltage source is connected to a series L-C circuit by turning on the switch S at time t = 0 as shown in the figure. Assume i(0) = 0, v(0) = 0. Which one of the following circular loci represents the plot of i(t) versus v(t)?
2
i(t)
(GATE 2017: 2 Marks)
S t=0 L=1 H
151. Consider the system described by the following state space representation: ⎡ x1 (t ) ⎤ ⎡0 1 ⎤ ⎡ x1 (t ) ⎤ ⎡0 ⎤ ⎢ x (t ) ⎥ = ⎢0 −2⎥ ⎢ x (t ) ⎥ + ⎢1 ⎥ u(t ) ⎦⎣ 2 ⎦ ⎣ ⎦ ⎣ 2 ⎦ ⎣
+ −
+ 5V
C=1 F
v(t) −
⎡ x (t ) ⎤ y(t ) = [1 0] ⎢ 1 ⎥ ⎣ x2 (t ) ⎦
Ch wise GATE_EE_Ch7.indd 339
11/21/2018 3:46:20 PM
340
GATE EE Chapter-wise Solved Papers
i(t) (a)
i(t) v(t)
(0,-5)
(b) (5, 0) i(t)
i(t) (c)
v(t)
157. The unit step response y(t) of a unity feedback system with K open loop transfer function G ( s) H ( s) = ( s + 1) 2 ( s + 2) is shown in the figure. The value of K is (upto 2 decimal places).
(0, 5)
(d) (-5, 0)
v(t)
v(t) (GATE 2018: 2 Marks) 156. The number of roots of the polynomial s7 + s6 + 7s5 + 14s4 + 31s3 + 73s2 + 25s + 200, in the open left half of the complex plane is (a) 3 (b) 4 (c) 5 (d) 6 (GATE 2018: 2 Marks)
1.4 y(t) 1.2 1 0.8 0.6 0.4 0.2 0 0 2
4
6
8
10
12
14
16
18
20 t(sec)
(GATE 2018: 2 Marks)
ANSWER KEY 1. (b)
2. (c) 3. (a) 4. (c)
5. (c) 6. (b)
7. (d)
8. (d)
9. (a)
10. (d)
11. (d)
12. (b) 13. (c) 14. (b)
15. (b) 16. (a)
17. (b)
18. (c)
19. (a)
20. (a)
21. (b)
22. (a) 23. (d) 24. (a)
25. (c) 26. (a)
27. (b)
28. (a)
29. (d)
30. (c)
31. (b)
32. (b) 33. (a) 34. (d)
35. (d) 36. (a)
37. (c)
38. (c)
39. (a)
40. (a)
41. (c)
42. (c) 43. (b) 44. (c)
45. (a) 46. (c)
47. (b)
48. (c)
49. (a)
50. (b)
51. (c)
52. (a) 53. (d) 54. (b)
55. (d) 56. (a)
57. (b)
58. (d)
59. (a)
60. (a)
61. (c)
62. (a) 63. (c) 64. (c)
65. (a) 66. (a)
67. (c)
68. (c)
69. (c)
70. (d)
71. (a)
72. (d) 73. (d) 74. (d)
75. (b) 76. (d)
77. (d)
78. (c)
79. (b)
80. (a)
81. (c)
82. (d) 83. (a) 84. (c)
85. (c) 86. (a)
87. (a)
88. (d)
89. (b)
90. (a)
91. (a)
92. (c) 93. (d) 94. (a)
95. (a) 96. (a)
97. (d)
98. (b)
99. (a)
100. (a)
101. (a)
102. (b) 103. (c) 104. 0
105. (c) 106. (c)
107. (a)
108. (0.618) 109. (0.75)
111. (a)
112. (c) 113. (b) 114. (a)
115. (c) 116. (d)
117. (d)
118. (b)
119. (0.241) 120. (b)
121. (b)
122. (d) 123. (c) 124. (d)
125. (a) 126. (a)
127. (a)
128. (a)
129. (b)
130. (a)
131. (0.707) 132. (a) 133. (6) 134. (a)
135. (2) 136. (c)
137. (b)
138. (a)
139. (d)
140. (1.04)
141. (d)
145. (d) 146. (0.554) 147. (0.097) 148. (b)
149. (a)
150. (c)
142. (d) 143. (a) 144. (a)
151. (1.284) 152. (a) 153. (7) 154. (0.66) 155. (b) 156. (a)
110. (5)
157. (8)
ANSWERS WITH EXPLANATION 1. Topic: Transient and Steady-State Analysis of Linear Time Invariant Systems (b) For a DC motor, when an input voltage V is applied to the armature with resistance Ra. The emf from the equivalent circuit of the machine (motor back emf) can be expressed as: Eb = V − I a Ra ⇒ V = Eb + I a Ra
Ch wise GATE_EE_Ch7.indd 340
Given that, Ia = 0, we have V = Eb 2. Topic: Feedback Principle (c) The sensitivity to the forward path and feedback path parameter changes is given as SGY =
1 1 + G ( s) H ( s)
11/21/2018 3:46:21 PM
Chapter 7 • Control Systems
SHY =
G ( s) 1 + G ( s) H ( s)
Since the gain of the forward path element is always more than that of the feedback path gain (control systems are stabilised by reducing gain and increasing bandwidth using negative feedback), SGY < SHY . Hence, less sensitive to change in G than change in H.
Unit step signal
341
3. Topic: Feedback Principle (a) The steady state error, ess is zero for step input with type-1 system and for ramp input with type-2 system. The steady-state errors of systems of different types are given below:
Type 0
Type 1
Type 2
Error Constants
1 1+ K p
0
0
Position error constant: K p = lim G ( s) s→0
1 Kv
∞
Unit ramp signal
0
Velocity error constant: K v = lim sG ( s) s→0
4. Topic: Transient and Steady-State Analysis of Linear Time Invariant Systems (c) Given that y(t) = te−t(for t > 0) Therefore, Y ( s) =
1 ( s + 1) 2
Two sign-changes occur in first column of the array. Hence, there are 2 poles on right-half of s-plane. 6. Topic: Transient and Steady-State Analysis of Linear Time Invariant Systems (b) Given that unity feedback system has open loop transfer function 25 s( s + 6 ) The closed loop transfer function will be G ( s) =
and x (t) = u(t) 1 s Therefore, the transfer function of the system will be X ( s) =
So,
Y ( s) 1/ ( s + 1) 2 s = = X ( s) 1/s ( s + 1) 2 5. Topic: Routh–Hurwitz and Nyquist Criteria (c) The given characteristic equation for feedback control system is 2 s 4 + s3 + 3s 2 + 5s + 10 = 0 Formulate the Routh−Hurwitz array as shown below s4
2
s3 s2 s1
1 3 − 10 = −7 1 −35 − 10 +45 = 7 −7
s0
Ch wise GATE_EE_Ch7.indd 341
10
3 5
10
0
6 = 0.6 2×5 Now, the percentage overshoot in the step response
x=
− px
+ +
1− x 2
× 100
− p × 0.6
−
−
Sign change
+
% MP = e
Sign change
10
G ( s) 1 + G ( s) H ( s) 25 = 2 s + 6 s + 25 Therefore, the characteristic equation of the given system is s2 + 6s + 25 = 0 Hence, w n = 25 = 5 TF =
=e =e
1− ( 0.6 )2 0.6 −p × 0.8
× 100
× 100
= 0.0948 × 100 = 9.48 ≈ 10%
11/21/2018 3:46:23 PM
342
GATE EE Chapter-wise Solved Papers
7. Topic: Lag, Lead and Lead-Lag Compensators (d) Find maximum phase lead of the compensator (0.5s + 1) 0.05s + 1) On comparing with standard
We need to find,
f (t ) = e At ⎡ s + 3 −1 ⎤ [ sI − A]−1 = ⎢ s + 2⎥⎦ ⎣ 0 1 ⎤ ⎡ 1 ⎢ ( s + 3) ( s + 2)( s + 3) ⎥ ⎥ =⎢ 1 ⎢ ⎥ ⎢ 0 ⎥ ( s ) 2 + ⎣ ⎦
D ( s) =
at s + 1 ts +1 we get, t = 0.05 and a = 10. Therefore, maximum phase lead, D ( s) =
⎛ 10 − 1⎞ = sin −1 ⎜ = 55° ⎝ 10 + 1⎟⎠
wn = =
1
t a
x(t ) = e At x(0)
⎡e −3t =⎢ ⎣ 0
0.05 10
= 6.32 rad/s
e −2t − e −3t ⎤ ⎡ 10 ⎤ ⎥⎢ ⎥ e −2t ⎦ ⎣ −10 ⎦
⎡ 20e −3t − 10e −2t ⎤ x (t ) = ⎢ ⎥ 10e −2t ⎣ ⎦
1
8. Topic: Routh–Hurwitz and Nyquist Criteria (d) For the given polar plot, Nyquist plot can be drawn as shown below.
e −2t − e −3t ⎤ ⎥ e −2t ⎦
⎡e −3t =⎢ ⎣ 0
And
e At = L−1 [ sI − A]−1
⎛ a − 1⎞ f m = sin −1 ⎜ ⎝ a + 1⎟⎠
⎡0 ⎤ xss = lim x(t ) = ⎢ ⎥ t →∞ ⎣0 ⎦
Therefore,
10. Topic: Transfer Function (d) Given that, t
y(t ) = ∫ ( 2 + t − t ) e −3( t −t ) u(t ) dt
Im
0
−1.42 −1
Re
Relation between input u(t) and output y(t) is a convolution operation between the input u(t) and the impulse response of the system. Now, ⎤ 2s + 7 ⎡t Y ( s) = L[ y(t )] = L ⎢ ∫ ( 2 + t − t )e −3( t −t ) u(t )dt ⎥ = 2 ⎣0 ⎦ ( s + 3)
The number of encirclement of point (−1 + j0), N = −2. Also, the number of right side poles in open loop is, P = 0. By Nyquist stability criterion, N = P – Z −2 = 0 – Z ⇒ Z=2 Thus, closed loop system is unstable with two poles on the right half of s-plane. 9. Topic: State Space Model (a) Since, x(t ) = e At x(0)
Ch wise GATE_EE_Ch7.indd 342
And
U ( s) = L[u(t )] = 1
Therefore, Y ( s) 2s + 7 = U ( s) ( s + 3) 2 11. Topic: Bode Plots (d) From the given Bode plot we observe that, Number of poles at origin = 2 Next pole at ω = 2 and ω = 25 and zero at ω = 5 Also, 54 = −40 log (0.1) + 20 log K
⇒K =5
11/21/2018 3:46:25 PM
Chapter 7 • Control Systems
Therefore, ⎛ 1 ⎞ 5 ⎜ 1 + s⎟ ⎝ 5 ⎠ Transfer function = 1 ⎞ ⎛ 1 ⎞⎛ s 2 ⎜ 1 + s⎟ ⎜ 1 + s⎟ ⎝ 2 ⎠ ⎝ 25 ⎠ 50( s + 5) s 2 ( s + 2)( s + 25)
=
12. Topic: Transient and Steady-State Analysis of Linear Time Invariant Systems (b) Let output y(t) of LTI system to an arbitrary input u(t) is given by y(t ) = h(t ) ∗ u(t ) where h(t) is the impulse response. If s(t) is the step response. Then, h(t ) =
15. Topic: Root Loci (b) Number of poles = 2 Number of zeros = 0 Therefore, angle of asymptotes θa = 90°, 270° Also, centroid 0 σ = =0 2 Hence, the curve in option (b) fits best. 16. Topic: Transfer Function (a) Given that,
with u as input and y as output. Taking Laplace transform on both side s2 Y(s) + s Y(s) = s U(s) + 2 U(s)
Therefore,
t ⎤ d ⎡ ⎢ ∫ s(t − t )u(t )dt ⎥ dt ⎣ 0 ⎦
13. Topic: State Space Model (c) The state transition matrix for the system X = AX, with initial state X(0) is e
At
−1
−1
= L [ sI − A]
14. Topic: State Space Model (b) Given that, ⎡ 2 3⎤ A= ⎢ ⎥ ⎣ 0 5⎦ Therefore, the characteristic equation is |sI – A| = 0 s−2 0
−3 =0 s−5
(s – 2) (s – 5) = 0 s = 2, 5 Roots lie on right half of s-plane. Hence, it is unstable. Now, for controllability, ⎡1 2⎤ Qc = ⎢ ⎥ ⎣0 0⎦ |Q| = 0 Hence, the system is uncontrollable.
Ch wise GATE_EE_Ch7.indd 343
Y ( s) s+2 = 2 U ( s) ( s + s)
Therefore,
=
d 2 y dy du + = + 2u (1) dt 2 dt dt
d s( t ) dt
d ⎡d ⎤ y(t ) = ⎢ s(t ) ⎥ ∗ u(t ) = [ s(t ) ∗ u(t )] dt ⎣ dt ⎦
343
17. Topic: State Space Model (b) With zero initial condition, x (0) = 0 And t
x(t ) = ∫ e A( t −t ) BU (t )dt
0
t ⎡ e 2( t −t ) = ∫⎢ 0 0 ⎣
0 ⎤ ⎡1⎤ ⎥ ⎢ ⎥ δ (t ) dt e ⎦ ⎣1⎦ 4 ( t −t )
⎡ e 2t ⎤ = ⎢ 4t ⎥ ⎣e ⎦
Therefore, output is ⎡ e 2t ⎤ y (t ) = [ 4 0 ] ⎢ 4 t ⎥ = 4 e 2 t ⎣e ⎦ 18. Topic: Mathematical Modeling and Representation of Systems (c) Given d2x dx + 6 + 5 x = 12(1 − e −2t ) (1) dt dt 2
By final value theorem,
lim f (t ) = lim sF ( s) RHS:
t →∞
s→0
(i) Convert Eq. (1) into frequency domain by applying Laplace transform. (ii) Then apply the limit s → 0 to find f(t)|t → ∞ (LHS)
11/21/2018 3:46:27 PM
θ n = tan −1 (w / a) − tan −1 (w / b)
344
⎡w w ⎤ ⎢ − ⎥ = tan −1 ⎢ a b2 ⎥ ⎢1+ w ⎥ ⎢⎣ ab ⎥⎦
GATE EE Chapter-wise Solved Papers
Taking LT on both sides
⎡ w b − w a⎤ ⎢ ab ⎥ = tan ⎢ ⎥ 2 ⎢ ab + w ⎥ ⎢⎣ ab ⎥⎦
1 ⎤ ⎡1 s 2 X ( s) + 6 sX ( s) + 5 X ( s) = 12 ⎢ − ⎥ ⎣ s s + 2⎦ ⎡s + 2 − s⎤ X ( s)[ s 2 + 6 s + 5] = 12 ⎢ ⎥ ⎣ s( s + 2) ⎦
−1
⎡ w ( b − a) ⎤ θ n = tan −1 ⎢ >0 2 ⎥ ⎣ ab + w ⎦
24 s( s + 2)( s 2 + 6 s + 5)
X ( s) = As LT(e − at ) =
[should be greater than zero for lead compensation] Therefore, b > a ⇒ a < b
1 s+a
On applying the limit s → 0 ⎡ ⎤ 24 24 lim sX ( s) = s ⎢ ⎥ = ( 2)(5) 2 s→0 2 6 5 s ( s + )( s + s + ) ⎣ ⎦ 24 = = 2.4 10
20. Topic: State Transient Matrix (a) Given ⎡e −2t State transition matrix, f (t ) = ⎢ ⎣ 0 ⎡ 2⎤ Initial condition X (0) = ⎢ ⎥ ⎣ 3⎦
19. Topic: Lag, Lead and Lead-Lag Compensators (a) Given that lead compensator transfer function is
We have to find X(t) at t = 1, without any external input So, X (t ) = f (t ) X ( 0 )
K (1 + s / a) (1 + s / b)
⎡e −2t =⎢ ⎣ 0
Main important criterion for a lead compensation is that phase qn should be positive.
X (t )
K (1 + jw / a) (1 + jw / b)
(ii) Find the phase angle for the given transfer function
θ n = ∠H ( jw ) =
tan −1 (w / a) tan −1 (w / b)
Note that, For a + jb
⎡ a−b ⎤ tan −1 a − tan −1 b = tan −1 ⎢ ⎥ ⎣1 + ab ⎦
d 2 x 1 dx 1 + ⋅ + x = 10 + 5e −4t + 2e −5 t (1) dt 2 2 dt 18 To find the natural time constants, we need to (i) Find the response of the system. (ii) Determine the pole nearest to the imaginary axis. Taking Laplace transform of Eq. (1) on both the sides, we get s 2 X ( s) +
Therefore,
θ n = tan −1 (w / a) − tan −1 (w / b) ⎡w w ⎤ ⎢ − ⎥ = tan −1 ⎢ a b2 ⎥ ⎢1+ w ⎥ ⎢⎣ ab ⎥⎦
t =1
⎡ 2e −2 ⎤ ⎡0.271⎤ = ⎢ −1 ⎥ = ⎢ ⎥ ⎣ 3e ⎦ ⎣1.100 ⎦
21. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (b) Given
⎛ b⎞ f = tan −1 ⎜ ⎟ ⎝ a⎠
0 ⎤ ⎡ 2⎤ ⎡ 2e −2t ⎤ ⎥⎢ ⎥ = ⎢ ⎥ e − t ⎦ ⎣ 3⎦ ⎣3e − t ⎦
At t = 1
(i) Substitute s = jw. H ( jw ) =
0⎤ ⎥ e −t ⎦
1 1 10 5 2 + sX ( s) + X ( s) = + 2 18 s s+4 s+5
s 1 ⎤ 10( s + 5)( s + 4) + 5s( s + 5) + 2 s( s + 4) ⎡ X ( s) ⎢ s 2 + + ⎥ = 2 18 ⎦ s( s + 4)( s + 5) ⎣ 10( s + 5)( s + 4) + 5s( s + 5) + 2s( s + 4) X ( s) = 1⎞ 1⎞ ⎛ ⎛ s( s + 4)( s + 5) ⎜ s + ⎟ ⎜ s + ⎟ ⎝ ⎝ ⎠ 6⎠ 3
⎡ w b − w a⎤ ⎥ ⎢ = tan −1 ⎢ ab 2 ⎥ ⎢ ab + w ⎥ ⎢⎣ ab ⎥⎦ Ch wise GATE_EE_Ch7.indd 344
11/21/2018 3:46:28 PM
Chapter 7 • Control Systems
Y ( s) G ( s) = U ( s) 1 + G ( s) H ( s) 45 ( s + 1)( s + 15) = 45 1+ ( s + 1)( s + 15) 45 = ( s + 1)( s + 15) + 45
jw
Pole plot
s 5
4
3
2
1
Therefore, poles nearest to the imaginary axis are
Therefore, the characteristic equation is ( s + 1)( s + 15) + 45 = 0
1 −1 s1 = , s2 = − 3 6
s 2 + 15s + s + 15 + 45 = 0
Therefore, natural time constants are
s 2 + 16 s + 60 = 0 ( s + 6)( s + 10) = 0 s = −6, −10
T1 = 3 s and T2 = 6 s 22. Topic: Transient and Steady-State Analysis of Linear Time Invariant Systems (a) Given that: Unity feedback system H(s) = 1 Open-loop transfer function G1 ( s) = G2 ( s) =
24. Topic: State Space Model (a) Given that the differential equation of a separately excited DC motor is d 2w B dw K 2 K w= + + Va LJ dt 2 J dt LJ State space matrix is
3 s + 15
15 s +1
⎡ d 2w ⎤ ⎡ dw ⎤ ⎢ 2 ⎥ ⎢ dt ⎥ = P ⎢ dt ⎥ + QVa ⎥ ⎢ ⎢ dw ⎥ ⎣w ⎦ ⎢⎣ dt ⎥⎦
Steady-state error, ess = lim s→0
sR( s) 1 + G ( s) H ( s)
1 R( s) = ( unit step input ) s G ( s) = G1 ( s)G2 ( s) [series combination]]
We first assign the state variables, Let dw = x1 , w = x2 , dt
45 ⎛ 3 ⎞ ⎛ 15 ⎞ =⎜ = ⎝ s + 15 ⎟⎠ ⎜⎝ s + 1⎟⎠ ( s + 1)( s + 15) s ⋅1/s ess = lim s→0 45 1+ ( s + 1)( s + 15) 1 1 1 = = = = 0.25 45 1 + 3 4 1+ 15 %ess = 0.25 × 100 = 25% 23. Topic: Transfer Function (d) To find the roots of the closed-loop transfer function, we need to Y ( s) (i) Find the closed-loop transfer function = . U ( s)
(ii) Equate the denominator (characteristic equation) to find the roots.
Ch wise GATE_EE_Ch7.indd 345
345
d 2w = x 1 dt 2
Therefore, state equation becomes,
x 1 +
K2 K B x1 + x2 = Va J LJ LJ
x 1 =
B K2 K Va − x1 − x2 LJ J LJ (1)
From the state variable assigned, x2 = w dw = x1 dt (2) x 2 = x1
x 2 =
Substituting coefficients of x1 and x2 and coefficient of Va from Eqs. (1) and (2) in the given state matrix, we get
11/21/2018 3:46:30 PM
346
GATE EE Chapter-wise Solved Papers
⎡ −B ⎡ x 1 ⎤ ⎢ ⎢ x ⎥ = ⎢ J ⎣ 2⎦ ⎢ 1 ⎣ ⎡ −B P matrix = ⎢⎢ J ⎢⎣ 1
−K 2 ⎤ ⎥ ⎡ x1 ⎤ LJ ⎥ ⎢ ⎥ + x 0 ⎥⎦ ⎣ 2 ⎦
⎡K ⎤ ⎢ LJ ⎥ V ⎢ ⎥ a ⎣ 0 ⎦
−K 2 ⎤ ⎥ LJ ⎥ 0 ⎥⎦
= 20 log K − 20 log 1 +
(0.5a) 2 a2
(ii) Phase angle at w = 0.5a ⎛w⎞ f = − tan −1 ⎜ ⎟ ⎝ a⎠
K s( s + 2)( s + 4)
To find K when system becomes unstable, we need to (i) Find the characteristic equation 1 + G ( s) H ( s) = 0 K 1+ =0 s( s + 2)( s + 4) s( s + 2)( s + 4) + K = 0 s( s 2 + 6 s + 8) + K = 0 s 3 + 6 s 2 + 8s + K = 0
w2 a2
= 20 log K − 20 log 1 + 0.52 = 20 log K − 0.96
25. Topic: Routh–Hurwitz and Nyquist Criterion (c) Given that the loop gain transfer function G(s) H(s) =
= 20 log K − 20 log 1 +
(ii) From the RH array s3
1
8
s2
6
K
s1
48 − K 6
⎛ 0.5a ⎞ = − tan −1 ⎜ = −26.56° ⎝ a ⎟⎠ For asymptotic plot: Gain (dB)
= 20 log K
w = 0.5 a
Phase angle at w = 0.5a
f = −22.5° Error in gain, 20 log K - (20 log K - 0.96) = +0.96 dB. Error in phase angle = -22.5° - (-26.56)° = 4.9° 27. Topic: Transfer Function (b) The given block diagram can be redrawn as 9 U(s)
1 s
+ −
2
3 s
12
K
0
For the system to be unstable, element in the first column should be negative
1
26. Topic: Bode Plots (a) Given that transfer function K s 1+ a The maximum error occurs at corner frequency between exact and asymptotic plot. For exact plot: (i) Gain (dB)
Ch wise GATE_EE_Ch7.indd 346
w =0.5a
2
Eliminating the feedback path 1 + −
⎛ 48 − K ⎞ ⎜⎝ ⎟ ≤0 6 ⎠ 48 − K = 0 − K = −48 ⇒ K = 48
Y(s)
1 s
+ −
1 s 3
1 1 1 s = = s = s+3 s+3 1 1+ ⋅3 s s Eliminating the feedback path 2 + −
1 s 12
1 1 s = = 1 s + 12 1 + ⋅12 s
11/21/2018 3:46:32 PM
Chapter 7 • Control Systems
s = 0; 5 = A( 2) A = 2.5 s = 1; 5 = B( −1)( −1 + 2) B = −5 s = −2; 5 = C ( −2)( −2 + 1) C = 2.5
Therefore, the block diagram is reduced as, 9 U(s) − +
Y(s) 1 s+3
2
1 s + 12
2 (s+3)(s+12)
Y(s)
Eliminating feedback path, we get 2 2 Y ( s) ( s + 3)( s + 12) = = 2 U ( s) ( s + 3)( s + 12) + 18 1+ 9 ( s + 3)( s + 12) 2 2 = 2 = 2 s + 15s + 36 + 18 s + 15s + 54 2 2 = = s s ( s + 9)( s + 6) 9 × 6 ⎛⎜1 + ⎞⎟ ⎛⎜1 + ⎞⎟ ⎝ 9⎠ ⎝ 6⎠ 1 s ⎛ ⎞ ⎛ s⎞ 27 ⎜1 + ⎟ ⎜ 1 + ⎟ ⎝ 9⎠ ⎝ 6⎠
29. Topic: Mathematical Modeling and Representation of Systems (d) Given 5
30. Topic: Transfer Function (c) Given: q(t) is the rotor displacement in radians. e(t) is the output voltage. Kt is the tachometer constant in V rad/s. In AC tachometer, e(t ) ∝ w (t ) dθ (t ) e(t ) ∝ (as w = dθ /dt ) dt dθ (t ) e(t ) = K t dt
E ( s) = sK t θ ( s) 31. Topic: Routh–Hurwitz and Nyquist Criterion (b) Given s3 - 4s2 + s + 6 = 0 By Routh–Hurwitz criterion, s3
1
1
s2
-4
6
s1
−4 + ( −6) −4
s1
2.5
s0
6
s(s 2 + 3s + 2) LT
f (t ) ← → F (s) Inverse LT
To find initial value of f(t) at t → 0, we convert the given function F(s) to f(t) by inverse Laplace transform. By partial fraction method, 5 A B C = + + s( s + 3s + 2) s s + 1 s + 2 5 = A( s + 1)( s + 2) + Bs( s + 2) + Cs( s + 1)
Ch wise GATE_EE_Ch7.indd 347
t →0
E ( s) = K t ⋅ sθ ( s)
We know that the phase margin is the amount of additional phase lag required to bring the system to verge of instability. Therefore, phase margin of the system is 0°
2
lim f (t ) = 2.5 − 5 + 2.5 = 0
Taking Laplace transform on both sides,
28. Topic: Routh–Hurwitz and Nyquist Criterion (a) Given that G(s) H(s) passes through (-1 + j 0) point.
F (s) =
2.5 5 2.5 − + s s +1 s + 2
Taking inverse LT f(t) = 2.5u(t) - 5e-t + 2.5e-2t
9
=
Therefore, F ( s) =
Eliminating the series block
U(s) − +
347
There are two changes of sign from s3 to s2 and s2 to s1. Therefore, number of poles lying on right half of s-plane = 2 Number of poles lying on left half of s-plane is Total number of poles - Right half poles = 3-2=1
11/21/2018 3:46:34 PM
348
GATE EE Chapter-wise Solved Papers
32. Topic: Transfer Function (b) From the given block diagram 2
1 R(s)
C(s) 1 s
1 s
+ +
+ +
34. Topic: Transient and Steady-State Analysis of Linear Time Invariant Systems (d) Given that Damping ratio z = 0.7 Undamped natural frequency wn = 5 rad/s. We need to obtain the closed-loop transfer function K C ( s) s ( s + 2) = K R( s) 1+ (1 + sP ) s( s + 2) K = s( s + 2) + K (1 + sP )
For 1: The output is 1. s R(s) R(s) 1 s
+ + R(s)
Characteristic equation is
1 R( s) + R( s) s ⎡ 1⎤ = R( s) ⎢1 + ⎥ ⎣ s⎦
s( s + 2) + K (1 + sP ) = 0 s 2 + 2 s + K + KsP = 0
1 R(s) (1+ 1 s s + +
[
1 s
)[
s 2 + 2ζw n + w n2 = 0 (2)
Comparing Eqs. (1) and (2), we get
w n2 = K ⇒ w n = K
R(s) 1 ⎛ 1⎞ C ( s) = R( s) ⋅ ⋅ ⎜1 + ⎟ + R( s) s ⎝ s⎠ ⎡ 1 1⎤ C ( s) = R( s) ⎢1 + + 2 ⎥ ⎣ s s ⎦ 2 C ( s) s + s + 1 = R( s) s2 33. Topic: State Transient Matrix (a) Given that X is a two-dimensional state vector ⎡0 2⎤ A is the system matrix = ⎢ ⎥ ⎣2 0⎦
5 = K ⇒ K = 25 Similarly, 2ζw n = 2 + KP 2 × 0.7 × 5 = 2 + 25 P 7 − 2 = 25 P ⇒ P = 0.2 35. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (d) Given that c(t ) = 12.5e −6 t sin 8t , t ≥ 0 where c(t) is the unit impulse response of the second- order system. Transfer function
State variable, X = AX The characteristic equation is given by, sI − A = 0 ⎡ 1 0 ⎤ ⎡ 0 2⎤ sI − A = s ⎢ ⎥ ⎥−⎢ ⎣0 1⎦ ⎣ 2 0⎦ ⎡ s −2⎤ =⎢ ⎥ s⎦ ⎣ −2 sI − A = s 2 − 4 = 0 s 2 = 4 ⇒ s = ±2
Ch wise GATE_EE_Ch7.indd 348
(1)
In general characteristic equation of a second-order system is given by,
For 2: The output is R(s)(1+ 1s )
s 2 + s( 2 + KP ) + K = 0
As
H ( s) = LT [c(t )] =
12.5 × 8 ( s + 6 ) 2 + 82
LT (e − at sin bt ) =
b ( s + a) 2 + b 2
Therefore, 100 s 2 + 36 + 12 s + 64 100 = 2 s + 12 s + 100
H ( s) =
11/21/2018 3:46:35 PM
Chapter 7 • Control Systems
Steady-state value is reached as time t tends to infinity.
Phase margin = 180° + ∠G ( jw gc )
t→∞
45° = 180° + ∠G ( jw gc ) where wgc is the gain crossover frequency. It is the frequency at which the magnitude of open-loop transfer function is unity or dB magnitude is zero. Therefore,
By final value theorem, lim y(t ) = lim s ⋅ Y ( s) t →∞
349
G ( jw gc ) = 1
s→0
= lim s ⋅ H ( s) R( s) s→0
To find G ( jw gc ) , substitute s = jw in the given transfer function. We get
100 1 ⋅ 2 s→0 s + 125 + 100 s (∵ R( s) = 1 / s; unit step input) = lim s ⋅ =1
Using w = wgc
36. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (a) For the given system x(t) → sinwt Transfer function
G ( jw gc ) = Since
Y ( s) s = H ( s) = X ( s) s +1
Equation (1) becomes
jw H ( jw ) = jw +1 Magnitude of H (jw),
w +1 2
−w gc 2
(w a) gc
2
+1
(
)
2
w gc4 = w gc a + 1
From the data given w = 1 rad/s. 1 1+1
1
=
2
Phase of H (jw) ∠H ( jw ) = 90° − tan (w / 1) −1 −1
= 90° − tan (1)
37. Topic: Bode Plots (c) Given that as + 1 as + 1 G ( s) = 2 G ( s) = 2 s s = 45° Phase margin Phase margin = 45°
w 4gc − w gc2 a 2 − 1 = 0 (2) As, phase margin = 180° + ∠G ( jw gc ) tan −1 (w gc a) − 180° + 180° = 45° tan −1 (w gc a) = 45° ⇒ w gc a = 1 (1 / a) 4 − 1 − 1 = 0 ⇒ (1 / a) 4 = 2
Therefore, output of the system = H ( jw ) x[t − ∠H ( jw )] =
Substituting in Eq. (2), we get
= 90° − 45° = 45°
Ch wise GATE_EE_Ch7.indd 349
+1
Squaring both sides, we get
(∵ a + jb = R ⇒ R = a + b 2 )
w =1
2
gc
−w gc 2 =
2
∠H ( jw )
(w a)
1=
w
H ( jw ) w =1 =
jw gc a +1 (1) jw gc2
G ( jw gc ) = 1 and ∠G ( jw gc ) = 180° − 45°
Substituting s = jw, we get
H ( jw ) =
jw a +1 jw 2
G ( jw ) =
1 2
sin(t − 45°)
a 4 = 0.5 a = 0.841 38. Topic: Transfer Function (c) Given T ( s) =
s2 + 4 ( s + 1)( s + 4)
We have to find the frequency at which the system output is zero.
11/21/2018 3:46:38 PM
350
GATE EE Chapter-wise Solved Papers
Substituting s = jw in the transfer function, we get ( jw ) 2 + 4 T ( jw ) = ( jw + 1)( jw + 4) The system output is zero, so −w 2 + 4
T ( jw ) =
jw + 1 jw + 4
Substituting s = jw in the transfer function, we get s +1 s2 jw + 1 jw + 1 G ( jw ) = = ( jw ) 2 −w 2 G ( s) =
=0
this implies 4 − w 2 = 0, w 2 = 4, w = 2 rad/s
∠G ( jw pc ) = −180° tan −1 w pc = −180°
w pc = 0 Gain margin =
(d)
K ⇒ s( s 2 + 1)
G ( J w pc )
1 G ( jw pc )
or, GM in dB = 20 log
1 G ( jw pc )
where wpc is the phase cross over frequency. It is the frequency at which the phase of open-loop transfer function is 180°, that is, ∠G ( jw pc ) = −180°
Ch wise GATE_EE_Ch7.indd 350
=
1
w pc2 + 1
=
w pc2 2 1 + w pc
=0
w pc2 41. Topic: Stability Analysis (c) Given G ( s) H ( s) =
Therefore, solution option (a) is correct.
Gain margin (GM) =
G ( J w pc )
1
0− j−0 = − j /3 3−0
40. Topic: Bode Plots (a) Given the transfer function s +1 G ( s) = 2 s We have
1
From Eq. (1), we get
From the given options:
0+ j −0 = j /3 3−0
(1)
For w = wpc, we have
where n is the number of poles and m is the number of zeros. Centroid is the point of intersection of asymptote on real axis. Sum of poles − Sum of zeros Centroid = No. of poles − No. of zeros
K (c) ⇒ 2 s( s + 1)
(i )
∠G ( jw ) = tan −1 w
±180°( 2 K + 1) n−m
(a) K ⇒ 0 − 0 = 0 3 3−0 s K −1 − 0 (b) 2 = −1 / 3 ⇒ 3−0 s ( s + 1)
Also,
39. Topic: Transfer Function (a) For a third-order system, highest power of s = 3 Number of poles = 3 Asymptote:
w2 +1 w2
G ( jw ) =
K (1 − s) (1 + s)
Stability the given system can be analysed using RouthHurwitz criterion. (i) The characteristic equation for the system is given by 1 + G ( s) H ( s) = 0 K (1 − s) 1+ =0 1+ s (1 + s) + K (1 − s) = 0 1 + s + K − Ks = 0 s(1 − K ) + 1 + K = 0 s1
1-K
s0
1+K
For a system to be stable, s1 and s0 should be positive. Therefore,
11/21/2018 3:46:40 PM
Chapter 7 • Control Systems
43. Topic: Routh–Hurwitz and Nyquist Criteria (b) Given
1− K > 0 1+ K > 0 ⇒ −1 < K < 1
p e −0.25 s To find the point where the Nyquist plot of G(s) H(s) passes through negative real axis: G ( s) H ( s) =
K 0 = tan −1 ⎜ ⎝ 10 + w 2 ⎟⎠
Phase lead f C1 = tan −1 w − tan −1
C2 =
jw + 10 ( jw + 1)
w − tan −1 w 10 ⎛ w ⎞ −w ⎟ −1 ⎜ 10 = tan ⎜ 1 + w 2 / 10 ⎟ ⎟ ⎜ ⎝ ⎠
Phase lag f C 2 = tan −1
⎛ −9w ⎞ = tan ⎜ 0 ⇒ K < 390 Therefore,
s 00 : K > 0 Therefore, 0 < K < 390
69. Topic: Transfer Function (c) Given that transfer function is 100 s 2 + 20 s + 100 Nature depends on the value of damping ratio (z ). Standard transfer function of second-order system w n2 = 2 s + 2ζw n s + w n2 Therefore,
w n2 = 100 ⇒ w n = 10 2ζw n = 20
ζ=
20 =1 20
z = 1 shows that the system is critically damped.
11/21/2018 3:46:53 PM
ess = lim s ⋅ E ( s) s→0
⎡ Control ⎤ R( s) Systems Chapter 7 • = lim s ⋅ ⎢ ⎥ s→0 ⎣1 + G ( s ) H ( s ) ⎦
70. Topic: Transfer Function (d) Given that: x = Ax + Bu y = Cx
(1) ( 2)
1⎤ ⎡0 ⎤ ⎡0 A= ⎢ B=⎢ ⎥ ⎥ ⎣1 ⎦ ⎣ 0 − 2⎦ ⎡1 0 ⎤ C = [1 0] I=⎢ ⎥ ⎣0 1⎦ Taking Laplace transform of Eq. (1), we get sX ( s) = AX ( s) + BU ( s) sX ( s) − AX ( s) = BU ( s) X ( s) [ sI − A] = BU ( s) X ( s) =
[ sI − A]−1 BU ( s)
Taking Laplace transform on Eq. (2), we get Y(s) = C X(s) Y(s) = C [sI − A]-1 BU(s) Transform function is given by Y ( s) = C ( sI − A) −1 B U ( s)
359
⎡ ⎤ 1 ⎢ ⎥ s ⎥ = lim s ⎢ s→0 ⎢1 + 1 ⋅ 1 ⎥ ⎢⎣ s( s + 2) ⎥⎦ ⎡ s( s + 2) ⎤ = lim ⎢ ⎥=0 s → 0 s( s + 2) + 1 ⎣ ⎦ 72. Topic: Block Diagrams and Signal Flow Graphs (d) Given that e1 is the error associated with G1 e2 is the error associated with G2 e3 is the error associated with 1/G3. Overall gain in series combination = G1G2
1 G3
Therefore, error as applicable for product of more than two quantities is given by ε1 + ε 2 + ε 3 73. Topic: Stability Analysis (d) Mapping the mirror image of the given plot, we have
−1
⎡ s − 1 ⎤ ⎡0⎤ = [1 0] ⎢ ⎥ ⎢ ⎥ ⎣0 s + 2⎦ ⎣1 ⎦ −1
1 ⎤ ⎡1 ⎢ s s( s + 2) ⎥ ⎡0 ⎤ ⎥ ⎢ ⎥ = [1 0] ⎢ ⎥ ⎣1 ⎦ ⎢ 1 ⎢⎣0 s + 2 ⎥⎦ ⎡ 1 ⎤ ⎢ s( s + 2 ) ⎥ ⎥ = [1 0] ⎢ ⎢ 1 ⎥ ⎢⎣ s + 2 ⎦⎥ 1 = s ( s + 2) 71. Topic: Transfer Function (a) We have H ( s) = 1 R( s) = 1/s Steady-state error is given by ess = lim s ⋅ E ( s) s→0
Ch wise GATE_EE_Ch7.indd 359
⎤ ⎡ R( s) = lim s ⋅ ⎢ ⎥ s→0 ⎣1 + G ( s ) H ( s ) ⎦ ⎤ ⎡ 1 ⎥ ⎢ s ⎥ = lim s ⎢ s→0 1 ⎢1 + ⋅1⎥ ⎢⎣ s( s + 2) ⎥⎦ ⎡ s( s + 2) ⎤ = lim =0
-1.42
Mapping the circle of R → 0 to an infinite radius R → ∞ gives
-1.42
Number of closed-loop poles in right half of s place, Z = P−N where P is the number of open loop poles and N is the number of encirclements (-1, j 0). From above sketch N = −2 and P = 0, therefore Z = 0 - (−2) ⇒Z = 2. So, the system is unstable with 2 poles on right half of the s-plane.
11/21/2018 3:46:54 PM
360
GATE EE Chapter-wise Solved Papers
74. Topic: Routh–Hurwitz and Nyquist Criteria (d) From the given Routh’s array, the auxiliary equation is,
76. Topic: Feedback Principle (d) Given that K . ( s + 1)( s + 2) H ( s) = 1 G ( s) =
2s 2 + 2 = 0 2 s 2 = −2
From the figure, Steady-state error (ess) = 1 − 0.75 = 0.25 We know that
s 2 = −1 s = ± j The characteristic equation is
ess = lim s ⋅ E (s )
2 s3 + 4 s 2 + 2 s + 4 = 0
s →0
Dividing both sides by 2, we get
R (s) 0. 25 = lim s ⋅ 1 + G( s) H (s ) s →0 1/ s 0. 25 = lim s ⋅ K s →0 1+ (s + 1) (s + 2) (s + 1) (s + 2) 0. 25 = lim s →0 K + (s + 1)(s + 2)
s3 + 2 s 2 + s + 2 = 0 To find the roots s 2 ( s + 2) + 1( s + 2) = 0 ( s + 2) ( s 2 + 1) = 0 s = −2, s = ± j
[∵ R (s) = 1/ s]
2 K + 2 2 = 0. 25K + 0. 5 1. 5 K = =6 0. 25
0. 25 = 75. Topic: Bode Plots (b) From the given data: (i) Bode plot starts with slope −40 dB/dec which implies the presence of two poles at origin. (ii) Slope change from −40 dB/dec to −60 dB/dec (−20 dB/dec added) implies that there is a pole in the transfer function. wc1 = 2 rad/s. (iii) Slope change from −60 dB/dec to −40 dB/dec (+20 dB/dec added) implies that there exists a zero in the transfer function wc2 = 5 rad/s. (iv) Slope change from −40 dB/dec to −60 dB/dec. (−20 dB/dec added) implies that there is a pole in the transfer function. wc3 = 25 rad/s. Therefore,
77. Topic: Feedback Principle (d) Given that
K ( s + 5) s ( s + 2)( s + 25)
K ( jw + 5) ( jw ) ( jw + 2)( jw + 25)
Therefore, transfer function is 1000( s + 5) s ( s + 2)( s + 25) 2
Ch wise GATE_EE_Ch7.indd 360
(2)
∠G ( jw pc ) = −180° (3)
G ( jw ) =
e −0.1 jw jw
180° ⎤ ⎡ ∠G ( jw ) = ⎢ −0.1w × − 90° p ⎥⎦ ⎣ Substituting w = wpc, we have
2
K (5) ⇒ K = 1000 0.12 × 2 × 25
G ( jw pc )
Substituting j = jw in the transfer function given in Eq. (1), we get
2
T (jw)|at w = 0 = 80. Therefore, 80 = 20 log
1
where wpc is the phase crossover frequency.
To find K, substitute s = jw. We get T ( jw ) =
(e −0.1s ) (1) s
Gain margin = 20 log
T ( s) =
G ( s) =
180° ⎤ ⎡ ∠G ( jw pc ) = ⎢ −0.1w pc × − 90° p ⎥⎦ ⎣ Substituting value from Eq. (4) in Eq. (3), we get
(4)
180° ⎤ ⎡ ⎢ −0.1w pc × p ⎥ − 90° = −180° ⎦ ⎣ ⇒ w pc = 15.7 rad/s
11/21/2018 3:46:56 PM
Chapter 7 • Control Systems
Substituting value in Eq. (2), we get gain margin as 1 15.7 = 20 log 15.7 = 23.9 dB
GM = 20 log
78. Topic: State Transient Matrix (c) Given that
dx1 (t ) = −3 x1 (t ) + x2 (t ) + 2u(t ) (1) dt
dx2 (t ) = −2 x2 (t ) + u(t ) (2) dt
y(t ) = x1 (t ) (3) Taking Laplace transform on Eqs. (1), (2) and (3), we get sX 1 ( s) = −3 X 1 ( s) + X 2 ( s) + 2U ( s)
X 1 ( s) ( s + 3) = X 2 ( s) + 2U ( s) (4)
sX 2 ( s) = −2 × 2 × 2 ( s) + U ( s) X 2 ( s) ( s + 2) = U ( s)
X 2 ( s) =
U ( s) (5) s+2
Substituting Eq. (5) in Eq. (4), we get U ( s) + 2U ( s) s+2 [1 + 2( s + 2)] X 1 ( s)( s + 3) = U ( s) s+2 U ( s)( 2 s + 5) X 1 ( s) = ( s + 2)( s + 3) X 1 ( s)( s + 3) =
From Eq. (3), Y ( s) = X 1 ( s) Y ( s) =
U ( s)( 2 s + 5) ( s + 2)( s + 3)
Transfer function is Y ( s) ( 2 s + 5) = U ( s) ( s + 2)( s + 3) 79. Topic: State Transient Matrix (b) From Eqs. (1) and (2) given in Question 78, we have x 1 (t ) = −3 x1 (t ) + x2 (t ) + 2u(t ) x 2 (t ) = −2 x2 (t ) + u(t ) Therefore, state matrix is ⎡ x 1 ⎤ ⎡ −3 1⎤ ⎡ x1 ⎤ ⎡ 2⎤ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ + ⎢ ⎥ u (t ) ⎣ x2 ⎦ ⎣0 − 2⎦ ⎣ x2 ⎦ ⎣1 ⎦ ⇒ x (t ) = Ax + Bu
Ch wise GATE_EE_Ch7.indd 361
361
State transition matrix f (t) = L−1[f (s)] f (s) = (sI − A)−1 ⎡ s 0 ⎤ ⎡ −3 1 ⎤ ⎡ s + 3 − 1 ⎤ sI − A = ⎢ ⎥=⎢ ⎥−⎢ s + 2⎥⎦ ⎣0 s ⎦ ⎣0 − 2⎦ ⎣ 0 1⎤ ⎡s + 2 1 [ sI − A]−1 = ⎢ s + 3⎥⎦ ( s + 3)( s + 2) ⎣0 Therefore, 1 s+2 ⎤ ⎡ ⎢ ( s + 3)( s + 2) ( s + 3)( s + 2) ⎥ ⎥ f ( s) = ⎢ s+3 ⎥ ⎢ ⎥ ⎢0 ( + 3 )( + 2 ) s s ⎦ ⎣ 1 ⎤ ⎡ 1 ⎢ s + 3 ( s + 2)( s + 3) ⎥ ⎥ =⎢ ⎥ ⎢ 1 ⎥⎦ ⎢⎣0 s+2 ⎡e −3t f = L−1f ( s) = ⎢ ⎣0
e −2t − e −3t ⎤ ⎥ e −2t ⎦
80. Topic: Feedback Principle (a) Given that Gain = 100 with ± 10% tolerance and 9/100 attenuator. Gain without error = 10 Gain with 10% error = 10 + 0.1= 10.1 Gain with −10% error = 10 − 0.1 = 9.9. Therefore, 10 ± 1% 81. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (c) Given that 2 s +1 Convert from s domain to time domain, we have, step response, Y ( s) =
y(t ) = 2(1 − e t )u(t ) = 0.98 × 2 ⇒ t = 3.9 s ≈ 4 s 82. Topic: Mathematical Modeling and Representation of Systems (d) From the given figure, h1 and h2 are compared as voltage across the capacitors. B and D are compared as the resistance across the two points. Therefore, A and C are capacitances and B and D are resistances.
11/21/2018 3:46:57 PM
362
GATE EE Chapter-wise Solved Papers
83. Topic: Routh–Hurwitz and Nyquist Criteria (a) Given that G ( s) =
G ( jw ) =
1 s( s + 1)( s + 2)
2
(
1 2 +1
−3/4
)
2+4
=
1 < 3/ 4 6
Im
Substituting s = jw in the transfer function, we get G ( jw ) =
1 (1) jw ( jw + 1)( jw + 2)
G ( jw ) =
1
w 1+ w
4+w
2
2
84. Topic: Stability Analysis (c) Given
and
∠G ( jw ) = −90° − tan −1 w − tan −1
Re
w 2
x = Ax + Bu
(i) At w = 0,
(ii) At w − ∞, G ( jw ) = 0 and ∠G ( jw ) = −270°
⎡ −1 2⎤ A= ⎢ ⎥ ⎣0 2 ⎦
G ( jw ) = ∞ and ∠G ( jw ) = −90°
For 2 × 2 matrix, first check for [B : AB]. ⎡ −1 2⎤ ⎡0 ⎤ ⎡ 2⎤ AB = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣ 0 2⎦ ⎣1 ⎦ ⎣ 2⎦ ⎡0 2⎤ [ B : AB] = ⎢ ⎥≠0 ⎣ 1 2⎦
−270° w=∞
−180°
⎡0⎤ B = ⎢ ⎥. ⎣1 ⎦
0°
Therefore, system is controllable. Next check for stability w=0
A − λI = 0
0 −90°
⎡ −1 2⎤ ⎡ λ 0 ⎤ A − λI = ⎢ ⎥ ⎥−⎢ ⎣ 0 2⎦ ⎣ 0 λ ⎦ ⎡ −1 − λ 2⎤ =⎢ ⎥ 0 2 − λ⎦ ⎣ = ( −1 − λ )( 2 − λ ) − 0
To find the point of intersection at the real axis, we divide and multiply Eq. (1) with complex conjugate of the expression. G ( jw ) =
1 ( jw )(1 + jw )( 2 + jw ) ( − jw )(1 − jw )( 2 − jw ) × ( − jw )(1 − jw )( 2 − jw )
=
− jw ( 2 − jw − 2 jw − w 2 ) w 2 (1 + w 2 )( 4 + w 2 )
=
−2 jw − 3w + jw w 2 (1 + w 2 )( 4 + w 2 )
G ( jw ) =
2
= −2 + λ − 2λ + λ 2 = λ 2 − λ − 2 A − λ I = 0;
λ − λ − 2 = 0 ⇒ λ = 2, −1 2
Eigen values are of opposite sign, so the system is unstable.
3
−3w jw ( 2 − w ) − 2 2 2 w (1 + w )( 4 + w ) w (1 + w 2 )( 4 + w 2 ) 2
2
2
At real axis, imaginary part |G (jw)| = 0. Therefore, w (2 − w ) = 0 2
w 2 = 2; w =
2 rad/s
At w = 2 rad/s
Ch wise GATE_EE_Ch7.indd 362
85. Topic: Root Loci (c) Given that
s( s + 1)( s + 3) + K ( s + 2) = 0; K > 0. (1) By Routh-Hurwitz criterion s( s 2 + 4 s + 3) + Ks + 2 K = 0 s3 + 4 s 2 + 3s + Ks + 2 K = 0 s 3 + 4 s 2 + s( 3 + K ) + 2 K = 0
11/21/2018 3:46:59 PM
Chapter 7 • Control Systems
To form Routh array, s3
1
3+K
s2
4
2K
s1
12 + 2 K 4
s0
2K
As N = 3, the root loci branches terminate at infinity along asymptotes, so + Re(s) = -1. 86. Topic: Bode Plots (a) Gain margin = −20 log G ( jw ) = 20 log
w = w pc
1 G ( jw pc )
Phase margin = 180° + fgc
It is given that K > 0, therefore, from s
1
where fgc is the gain cross-over frequency or angle at wgc
12 + 2k > 0. 4 Thus, there is no sign change in the first column of Routh array which indicates the absence of right-half poles. To sketch the root locus: Equation (1) can be written as, 1+
363
From the given data, the frequency at which the response crosses −180° is 0.5. G ( jw ) = 0.5, at ∠G ( jw ) = −180° Therefore GM = 20 log
K ( s + 2) =0 s( s + 1)( s + 3)
1 = 6 dB 0.5
Phase margin = 180 + fgc = 180° + ( −150°) = 30°
Standard form 1 + G ( s) H ( s) = 0 For unity feedback system H(s) = 1. Therefore, G ( s) =
87. Topic: Feedback Principle (a) For step input ess = 0.1 We know that
K ( s + 2) s( s + 1)( s + 3)
1 1+ K 1 0.1 = 1+ K ⇒K =9 ess =
Poles: Zeros n = 3 m=1 s = 0 s = −2 s = −1 s = −3
Also, G ( s) =
jw
9 K = s +1 s +1
From the figure, input −3
−2
−1
0
Number of root locus branches (n) = 3. Number of root locus branches ending at zero at infinity = n − m = 2. Number of root locus branches ending at first zero = 1. Breakaway point occurs between two open-loops poles say between s = 0 and s = −1. Therefore, ΣReal part of poles − ΣReal part of zeros n−m −2) 0 − 1 − 3 − (− = −1 = 2
Centroid =
Ch wise GATE_EE_Ch7.indd 363
r (t ) = 10 [u(t ) − u(t − 1)]
s
Taking Laplace transform ⎡1 e−s ⎤ ⎡1 − e − s ⎤ R( s) = 10 ⎢ − 10 = ⎥ ⎢ ⎥ s ⎦ ⎣s ⎣ s ⎦ Therefore, steady-state error for pulse input is, ess( p ) = lim s ⋅ E ( s) s→0
R( s) 1 + G ( s) H ( s) R( s) = lim s ⋅ s→0 s + 100 s +1 10(1 − e − s ) / s = lim s ⋅ s→0 s + 10 s +1 (∵ R( s) = 10(1 − e − s ) and H ( s) = 1) = lim s ⋅ s→0
=
10 (1 − e 0 ) =0 10 / 1
11/21/2018 3:47:01 PM
ess( p ) = lim s ⋅ E ( s) s→0
= lim s ⋅ s→0
364
R( s) 1 + G ( s) H ( s) R( s)
= lim s ⋅Solved Papers GATE EE Chapter-wise s→0 s + 100 s +1 10(1 − e − s ) / s = lim s ⋅ s→0 s + 10 s +1 (∵ R( s) = 10(1 − e − s ) and H ( s) = 1) =
10 (1 − e 0 ) =0 10 / 1
88. Topic: Bode Plots (d) Given
−2
•
z
Point z lies inside the unit circle hence the value of z < 1 y = 1/z y >
[∵ z < 1]
z − is positive real and imaginary part. Therefore, ‘y’ should have positive real and negative imaginary part 89. Topic: Stability Analysis (b) Given that
( s − 1) ( s + 2)( s + 3)
Zeros (m) =1; s = 1 Poles (n) = 2; s = −2; s = −3. (i) All the poles lie on the left half of the s-plane, therefore open-loop system is stable. (ii) One zero lies on the right half of the s-plane, therefore non-minimum phase type.
91. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (a) Solving for inner loop, transfer function is 1 1 s( s + 1) = 1 s( s + 1) + K ( s) 1+ ⋅ K ( s) s( s + 1) Overall transfer function 1 Y ( s) 1 s( s + 1) + Ks = = 1 R( s) s( s + 1) + Ks + 1 1+ s( s + 1) + Ks 1 1 Y ( s) = = 2 2 R( s) s + s + Ks + 1 s + s( K + 1) + 1 Standard second-order characteristic equation is given by s 2 + 2ζw n s + w n2 = 0 Therefore, 2ζw n = 1 + K
wn = 1 K ( s + 2 / 3) s 2 ( s + 2)
Sum of poles − Sum of zeros No. of poles − No. of zeros −2 − ( −2 / 3) = = −2 / 3 3 −1
Centroid (σ A ) =
( 2k ± 1) × 180° ; Angle of asymptote θ A = n−m k = 0, 1, , n − m
Ch wise GATE_EE_Ch7.indd 364
So, three roots with nearly equal real parts exist on the left half of the s-plane.
wn2 = 1
90. Topic: Root Loci (a) G ( s) =
−2/3
Unit circle y = 1/z
G ( s) =
180° = 90° 2 3 × 180° k = 1; θ 2 = = 270° 2 For k = 0; θ1 =
ζ=
1+ K 2
Therefore, peak overshoot = e −ζp /1 −ζ
2
92. Topic: Transfer Function (c) The steady state error shall be zero when the overall function is stable and the factor due to sinusoidal input, that is, (s2 + 9). It implies that
w s + w2 2
gets cancelled by
s 2 + w 2 = s 2 + 9 ⇒ w 2 = 9 ⇒ w = 3 rad/s
11/21/2018 3:47:02 PM
Chapter 7 • Control Systems
93. Topic: State Space Model (d) Given that ⎛ x 1 ⎞ ⎛ 0 a1 0⎞ ⎛ x1 ⎞ ⎛ 0⎞ ⎜ x ⎟ = ⎜ 0 0 a2 ⎟ ⎜ x ⎟ + ⎜ 0⎟ u; ⎟ ⎜ 2⎟ ⎜ ⎟ ⎜ 2⎟ ⎜ ⎜⎝ x ⎟⎠ ⎜ a3 0 0⎟ ⎜⎝ x ⎟⎠ ⎝1 ⎠ ⎝ ⎠ 3 3
From Routh array, the characteristic equation is s3 + as2 + s (2 + k) + (k + 1) = 0 s
x = Ax + Bu (2) On comparing Eqs. (1) and (2), ⎛ 0 a1 0⎞ ⎜ ⎟ A = ⎜ 0 0 a2 ⎟ ⎜⎝ a3 0 0⎟⎠ ⎛0 ⎞ AB = ⎜ a2 ⎟ ⎜ ⎟ ⎝0 ⎠ ⎛0
s0
For an oscillatory system, we have s1: 0 which implies a( 2 + K ) − K + 1 =0 (1) a ( K + 1) a= ( K + 2)
Auxiliary equation is, as 2 + K + 1 = 0 ⇒ s 2 = −
⎛ 0⎞ B = ⎜ 0⎟ ⎜ ⎟ ⎝1 ⎠
s 2 = −( K + 1) ⋅
⎛ 0 0 a1a2 ⎞ ⎜ ⎟ A = a2 a3 0 0⎟ ⎜ ⎝ 0 a3 a1 0⎠ 2
s= j K +2
(2)
jw = j K + 2
w = K +2 2= K +2 ⇒ K = 2
0 a1a2 ⎞ a2 0 ⎟ ⎟ 0 0 ⎟⎠
Therefore, a =
This implies that a1 ≠ 0, a2 ≠ 0 and a3 may or may not be zero. 94. Topic: Feedback Principle (a) Overall transfer function of the system is,
Ch wise GATE_EE_Ch7.indd 365
( K + 2) = −( K + 2) ( K + 1)
Substituting s = jw in Eq. (2), we get
0 a1a2 ⎞ ⎛ 0⎞ ⎛ a1a2 ⎞ 0 ⎟ ⎜ 0⎟ = ⎜ 0 ⎟ ⎟⎜ ⎟ ⎜ ⎟ 0 ⎟⎠ ⎝1 ⎠ ⎜⎝ 0 ⎟⎠
Y ( s) G ( s) = R( s) 1 + G ( s) H ( s) K ( s + 1) = 3 s + as 2 + 2 s + 1 K ( s + 1) 1+ 3 s + as 2 + 2 s + 1 K ( s + 1) = 3 s + as 2 + 2 s + 1 + K ( s + 1) K ( s + 1) = 3 2 s + as + s( 2 + K ) + ( K +1)
( K + 1) a
Substituting value of a from Eq. (1), we have
A2 B = ⎜ a2 a3 0 ⎜ ⎜⎝ 0 a a 3 1 ⎛0 U = ⎜0 ⎜ ⎜⎝1
a K +1 a( 2 + K ) − K + 1 a K +1
2
s1
U = [ B : AB : A2 B] (1) We know that
2+ K
s3 1
⎡ x1 ⎤ ⎢ ⎥ y = (1 0 0 ) ⎢ x2 ⎥ ⎢⎣ x3 ⎥⎦ For a 3 × 3 matrix to check the countability, the condition is
365
(∵ H ( s) = 1)
K +1 2 +1 3 = = = 0.75 K +2 2+2 4
95. Topic: Lag, Lead and Lead-Lag Compensators (a) Substituting s = jw in the given transfer function, we get jw + a Gc ( jw ) = jw + b
For a lead compensator f > 0 w w f = tan −1 − tan −1 (1) a b ⎛w w⎞ ⎜ − ⎟ ⎛ w ( b − a) ⎞ = tan −1 ⎜ a b2 ⎟ = tan −1 ⎜ 2 ⎝ w + ab ⎟⎠ w ⎜ 1+ ⎟ ⎝ ab ⎠ f > 0 implies that
w ( b − a) > 0 b−a> 0⇒b> a
11/21/2018 3:47:04 PM
366
GATE EE Chapter-wise Solved Papers
This condition is satisfied by options (a) and (c). However, in case of lead compensator, zero is nearer to the origin as compared to pole, therefore option (c) is not valid and the correct option is (a). 96. Topic: Lag, Lead and Lead-Lag Compensators (a) From Eq. (1) of Question 78, we have 1/a 1/b df − =0 = 2 dw ( 1 + w /b ) 2 ⎛w⎞ 1⎜ ⎟ ⎝ a⎠
⎤ ⎥= ⎥⎦ 2 1 1⎛w ⎞ + = a a ⎜⎝ b 2 ⎟⎠ 2
1 ⎡ ⎛w⎞ ⎢1 + ⎜ ⎟ b ⎢⎣ ⎝ a ⎠
2
0 − 32 = −40 (Set w 2 = w ) ⎛w⎞ log ⎜ ⎟ ⎝ 1⎠
⎤ ⎥ ⎥⎦
−32 = −40 log(w ) ⇒ w = 6.309
1 1⎛w ⎞ + b b ⎜⎝ a 2 ⎟⎠ 2
We know that
2 ⎛ 1 1⎞ w ⎛ 1 1⎞ − = ⎜⎝ ⎟ ⎜ − ⎟ a b ⎠ ab ⎝ a b ⎠
w = K 1 /N
w = ab ⇒ w = ab = 1 × 2 = 2 rad/s 2
97. Topic: Transfer Function (d) The given circuit can be represented as C1 100 µF +
+ R 10 kΩ
V1(s) I(s)
C2 100 µF
−
V2(s) −
1 1 I ( s) + RI ( s) + I ( s) C1 s C2 s
1⎤ ⎡1 V1 ( s) = I ( s) ⎢ + R + ⎥ Cs ⎦ ⎣ Cs
(1)
For loop 2: 1 I ( s) Cs V2 ( s) 1⎤ ⎡ (2) V2 ( s) = I ( s) ⎢ R + ⎥ ⇒ I ( s) = ( R + 1/Cs) Cs ⎦ ⎣ Substituting Eq. (2) in Eq. (1), we get V2 ( s) = R I ( s) +
V1 ( s) =
V2 ( s) ⋅ Cs ⎡ RCs + 2 ⎤ Rcs + 1 ⎢⎣ Cs ⎥⎦
V2 ( s) 10 × 103 × 100 × 10 −6 s + 1 s + 1 = = V1 ( s) 10 × 103 × 100 × 10 −6 s + 2 s + 2
Ch wise GATE_EE_Ch7.indd 366
6.309 = K 1/ 2 K = 39.8 39.8 G ( s) = 2 s
(∵ N = 2)
99. Topic: Transfer Function (a) Using Mason's gain formula, Y ( s) s −2 + s −1 = U ( s) 1 − [ −2 s −2 − 4 s −1 − 2 s −1 − 4] s −2 (1 + s) 1 + 2 s −2 + 6 s −1 + 4 s −2 ( s + 1) = −2 2 s ( 5 s + 6 s + 2) s +1 = 2 5s + 6 s + 2
=
For loop 1: Let C1 = C2 =100 mF = C V1 ( s) =
G2 − G1 = −40 dB/decade log w 2 − log w1 0 − 32 = −40 ⎛ w2 ⎞ log ⎜ ⎟ ⎝ w1 ⎠
1/a 1/b = 1 + (w /a) 2 1 + (w /b) 2 1 ⎡ ⎛w⎞ ⎢1 + ⎜ ⎟ a ⎢⎣ ⎝ b ⎠
98. Topic: Bode Plots (b) Any two points on same line segment of Bode plot satisfies the equation of a straight line. The slope of straight line is
100. Topic: State Space Model (a) ⎡1⎤ ⎡ −2 0 ⎤ A= ⎢ ⎥ B = ⎢1⎥ C = [1 0] 0 1 − ⎣ ⎦ ⎣⎦ To check for controllability: ⎡ −2 0 ⎤ ⎡1⎤ ⎡ −2⎤ AB = ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎣ 0 −1⎦ ⎣1⎦ ⎣ −1⎦ ⎡1 −2⎤ [ BAB ] = ⎢ ⎥≠0 ⎣1 −1⎦ Therefore, system is controllable. To check for observability:
11/21/2018 3:47:05 PM
⎡ e 2t − 1⎤ ⎥ ⎢ 2 ⎥ x(t ) = e At ⎢ t ⎢e −1 ⎥ ⎢⎣ 1 ⎥⎦ ⎡ e 2t − 1⎤ ⎡e 0⎤ ⎢ 2 ⎥ ⎥ ⎢ = ⎢ Chapter 7 −• t ⎥Control e ⎦ ⎢ e t − 1 ⎥Systems ⎣0 ⎢⎣ 1 ⎥⎦ −2 t
⎡ −2 0 ⎤ CA = [1 0] ⎢ ⎥ = [ −2 0] ⎣ 0 −1⎦ ⎡C ⎤ ⎡ 1 0 ⎤ ⎢CA⎥ = ⎢ ⎥=0 ⎣ ⎦ ⎣ −2 0 ⎦ Therefore, the system is not observable. 101. Topic: State Space Model (a) Solution for the given state equation in time domain is,
⎡1 − e − 2 t ⎤ ⎢ ⎥ 2 ⎥ =⎢ −t ⎢1 − e ⎥ ⎢⎣ 1 ⎥⎦ Therefore, y(t ) = [1 0]x(t ) ⎡1 − e −2t ⎤ = [1 0] ⎢ 2 ⎥ ⎢ ⎥ ⎢⎣1 − e − t ⎥⎦ 1 1 = − e −2t 2 2
t
x(t ) = e At x(0) + e At ∫ e − Ax BU ( x )dx 0
Given x(0) = 0. We know that e At = L−1 [ sI − A]−1 ⎡ s 0 ⎤ ⎡ −2 0 ⎤ ⎡ s + 2 0 ⎤ sI − A = ⎢ ⎥=⎢ ⎥−⎢ s + 1⎥⎦ ⎣0 s ⎦ ⎣ 0 − 1⎦ ⎣0 0⎤ ⎡s + 1 1 [ sI − A]−1 = s + 2⎥⎦ ( s + 1)( s + 2) ⎢⎣0 ⎡ 1 = ⎢s + 2 ⎢ ⎣0
⎤ ⎥ ⎥ 1 / ( s + 1) ⎦ 0
102. Topic: Routh–Hurwitz and Nyquist Criteria (b) Suppose we have s4 : A1 A2 A3; s3 : B1 B2; s2 : 0 0. A row of zero indicates that s4 row forms an auxiliary equation B1 x 3 + B2 x = 0. This gives the imaginary roots which make the system marginally stable. 103. Topic: Root Loci (c) The transfer function is G ( s) =
On inverse Laplace transform, we get ⎡e −2t 0 ⎤ L−1 [f ( s)] = ⎢ ⎥ e −t ⎦ ⎣0 ⎡e 2 x 0 ⎤ [ using x = −t ] e − Ax = ⎢ ⎥ ex ⎦ ⎣0 ⎡e 2 x ⎤ ⎡e 2 x 0 ⎤ ⎡1⎤ = e − Ax ⋅ B = ⎢ ⎢ x ⎥ ⎥ ⎢⎥ e x ⎦ ⎣1⎦ ⎣e ⎦ ⎣0 2x t t ⎛e ⎞ − Ax ∫0 e ⋅ B( x) = ∫0 ⎜⎝ e x ⎟⎠ ⋅ dx ⎡ e 2t − 1⎤ ⎢ ⎥ 2 ⎥ =⎢ t ⎢e −1 ⎥ ⎢⎣ 1 ⎥⎦
1 ( s + 1)( s + 2)
Therefore, for K = 0 C ( s) K = R( s) ( s + 1)( s + 2) − K 104. Topic: Transfer Function (0) We have T ( s) =
4 G ( s) = s + 0.4 s + 4 1 + G ( s) 2
Solving we get that open-loop transfer function as G ( s) =
4 s( s + 0.4)
Error constant ( K p ) = lim G ( s) = lim
Therefore,
s→0
⎡ e 2t − 1⎤ ⎥ ⎢ 2 ⎥ x(t ) = e At ⎢ t ⎢e −1 ⎥ ⎢⎣ 1 ⎥⎦ ⎡e =⎢ ⎣0
−2 t
0⎤ −t ⎥ e ⎦
367
Steady state error (ess) =
s→0
4 =∞ s ( s + 0.4 )
1 =0 1+ K p
105. Topic: State Transient Matrix ⎡ e 2t − 1⎤ ⎥ ⎢ ⎢ 2 ⎥ ⎢ et − 1 ⎥ ⎢⎣ 1 ⎥⎦
(c) State transition matrix is, e At = L−1 ⎡⎣( sI − A) −1 ⎤⎦ ⎡ s 0 ⎤ ⎡1 0 ⎤ ⎡ s − 1 0 ⎤ sI − A = ⎢ ⎥−⎢ ⎥=⎢ ⎥ ⎣0 s ⎦ ⎣1 1 ⎦ ⎣ −1 s − 1⎦
⎡1 − e − 2 t ⎤ ⎥ ⎢ 2 ⎥ =⎢ ⎢1 − e − t ⎥ ⎢⎣ 1 ⎥⎦ Ch wise GATE_EE_Ch7.indd 367
11/21/2018 3:47:07 PM
368
GATE EE Chapter-wise Solved Papers
[ sI − A]−1 =
⎡s − 1 0 ⎤ 1 s − 1⎥⎦ ( s − 1)( s − 1) − 0 ⎢⎣ 1
⎡ 1 ⎢ s −1 =⎢ ⎢ 1 ⎢ ( s − 1) 2 ⎣ e
At
106. Topic: Block Diagrams and Signal Flow Graphs (c) By Mason’s gain formula, the transfer function is Δ1 P1 + Δ 2 P2 + C ( s) 1 = ∑ Δ k Pk = Δ R( s) Δ k where P1 = Δ = 1+
s3 s
0 ⎤ ⎥ ⎥ 1 ⎥ s − 1⎥⎦
0⎤ ⎥ et ⎦
⎡ et =⎢ t ⎣te
From Routh–Hurwitz criteria
h1 a ; P2 = b0 ; Δ1 = 1 + 1 ; 2 s s s
a1 a0 + ; Δ2 = 1 s s2
⎛ a1 ⎞ ⎛ h1 ⎞ b0 1+ ⎟ ⎜ ⎟ + 2 b s + b0 s⎠⎝ s⎠ s C ( s) ⎜⎝ = = 2 1 a a R( s) s + a1 s + a0 1 + 1 + 20 s s
a
(1 + a )1 a (1 + a ) − 1 (1 + a )
1
s1 s0
1
For a stable system elements in the first column should be positive, therefore (1 + a) > 0 and (a 2 + a - 1) > 0. a = 0.618 and a = - 0.618 a > 0, therefore, a = 0.618
Since
109. Topic: Bode Plots (0.75) From the given magnitude plot, open loop transfer function can be written as, s⎞ ⎛ s⎞ ⎛ K ⎜1 + ⎟ ⎜1 + ⎟ ⎝ 2⎠ ⎝ 4⎠ G ( s) = (1) s ⎞⎛ s⎞ ⎛ s⎞ ⎛ s ⎜1 + ⎟ ⎜1 + ⎟ ⎜1 + ⎟ ⎝ 8 ⎠ ⎝ 24 ⎠ ⎝ 36 ⎠ Given transfer function is G ( s) =
Therefore,
1
2
K (1 + 0.5s)(1 + as) (2) s⎞ ⎛ s⎞ ⎛ s ⎜1 + ⎟ (1 + bs) ⎜1 + ⎟ ⎝ 8⎠ ⎝ 36 ⎠
From Eqs. (1) and (2), we have a = 1/4 and b = 1/24. To find K, K = (w1)n
107. Topic: Stability Analysis (a) When |G(s)H(s)| < 1, it is open loop transfer function. The system is always stable. 108. Topic: Transfer Function (0.618) The closed loop transfer function is (s + a ) C ( s) s + (1 + a ) s + (a − 1) s + (1 − a ) = (s + a ) R( s) 1+ 3 s + (1 + a ) s 2 + (a − 1) s + (1 − a ) 3
20 dB/decade w1 8
where n is the number of poles from the given plot. So, K = (8)1 = 8. Therefore, a 1/ 4 1/ 4 3 = = = = 0.75 bK (1/ 24) × 8 1/ 3 4
2
C ( s) (s + a ) = 3 2 R( s) s + (1 + a ) s + (a − 1) s + (1 − a ) + ( s + a )
110. Topic: Routh–Hurwitz and Nyquist Criteria (5) Given that G ( s) =
Characteristic equation is s3 + (1 + a ) s 2 + a s − s + 1 − a + s + a = 0 s3 + (1 + a ) s 2 + a s + 1 = 0
Ch wise GATE_EE_Ch7.indd 368
log w
Closed
loop
K s( s + 2)( s 2 + 2 s + 2)
transfer
function
is
given
by
C ( s) G ( s) = R( s) 1 + G ( s) H ( s)
11/21/2018 3:47:09 PM
Chapter 7 • Control Systems
K s( s + 2)( s 2 + 2 s + 2) K 1+ s( s + 2)( s 2 + 2 s + 2) K = 2 s( s + 2)( s + 2 s + 2) + K
Observability is given by
=
Characteristic equation, s + 4s + 6s + 4s + K = 0 By Routh–Hurwitz criteria, 4
s4 s3 s2
1 4 5
s1
⎛ 20 − 4 K ⎞ ⎜⎝ ⎟ 5 ⎠
3
6 4 K
2
20 − 4K = 0 ⇒ K = 5
|Qo| = 0, so the system is not observable. 113. Topic: Transfer Function (b) C ( s) s G ( s) = 2 = ⇒ G ( s) = s R( s) s + s + 1 1 + G ( s) H ( s) 114. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (a) We know that . x = AX + BU . y = CY + DU We need to determine C[sI - A]-1 B. From the given data,
111. Topic: Bode Plot (a) Given that G ( s) =
⎡0⎤ ⎡ 0 1⎤ A= ⎢ , B = ⎢ ⎥ , C = [1 0] ⎥ ⎣ −1 −1⎦ ⎣1⎦
5( s + 4) s( s + 0.25)( s 2 + 4 s + 25)
From state space equation, transfer function −1
⎛ s −1 ⎞ ⎛ 0⎞ = (1 0) ⎜ ⎝ 1 s + 1⎟⎠ ⎜⎝1 ⎟⎠
To find corner frequencies, convert G(s) into bode form. ⎡ s⎤ 5 × 4 ⎢1 + ⎥ ⎣ 4⎦ G ( s) = 2 4 s ⎤ ⎛ ⎛ s⎞ ⎞ ⎡ 25 1 + + s × 0.25 ⎢1 + s ⎜ ⎟ ⎜ ⎝ 5 ⎠ ⎟⎠ ⎣ 0.25 ⎥⎦ ⎝ 25
=
=
where H(s) = 1 and G ( s) =
2 4 s ⎞⎛ ⎛ s⎞ ⎞ ⎛ 1 + + s ⎜1 + s ⎜⎝ ⎟⎠ ⎟ ⎝ 0.25 ⎟⎠ ⎜⎝ 25 5 ⎠
112. Topic: State Transient Matrix (c) Given that dX = PX + Qu and y = RX dt The matrices are ⎡0⎤ ⎡ −1 1⎤ P=⎢ Q = ⎢ ⎥ R = [0 1] ⎥ ⎣ 0 −3⎦ ⎣1⎦ Controllability is given by 1⎤ ⎡0 QC = [Q PQ ] = ⎢ ⎥ where QC ≠ 0 , 1 − 3 ⎣ ⎦ so the system is controllable.
1 s + s +1 2
G ( s) 1 = 1 + G ( s) H ( s) s 2 + s + 1
s⎞ ⎛ 3.2 ⎜1 + ⎟ ⎝ 4⎠
Gain = 3.2 and wn = 5 (highest corner frequency/natural frequency).
Ch wise GATE_EE_Ch7.indd 369
⎡0 0 ⎤ P T RT ⎤⎦ = ⎢ ⎥ ⎣ 1 −3⎦
Qo = ⎡⎣ RT
K
s0 K Condition for marginally stable configuration is
369
Steady state error, ess = ess =
det A 1+ K p
=
1 s +s 2
det A 1+ K p
=
1 1 = 1 + lim G ( s) 1 + lim G ( s) s→0
s→0
1 1 = 1 + lim G ( s) 1 + lim G ( s) s→0
s→0
ess =
1 1 = =0 1+ ∞ ∞
115. Topic: Bode Plots (c) We know that ⎡1 − a ⎤ f m = sin −1 ⎢ ⎥ ⎣1 + a ⎦ From the given plot, G ( s) = K ×
3 [1/ 3 + s] (1 + 3s) =K (1 + s) ( s + 1)
11/21/2018 3:47:11 PM
370
GATE EE Chapter-wise Solved Papers
Putting K = 1, we get
1 ⎞ ⎛ k ⎜1 + s ⎝ 1000 ⎟⎠ Therefore, G ( s) = 1 ⎞ ⎛ ⎜⎝1 + s⎟⎠ 10
1 1 1 = ⇒ =1 T 3 aT So, a =
1 1 1 and w m = = 3 a 3 G ( s) w = 1 = 3
4 4 /3
20 log k = 20 ⇒ k = 10 Substituting for k in expression for G(s), we have 1 ⎞ ⎛ 10 ⎜1 + s ⎝ 1000 ⎟⎠ 10(1000 + s) = G ( s) = (10 + s) 1 ⎞ ⎛ 1000 × ⎜⎝1 + s⎟⎠ 10 10
= 3
G ( s) in dB = 20 log 3 = 4.77 dB ⎡1 − 1 / 3 ⎤ −1 f m = sin −1 ⎢ ⎥ = sin (1 / 2) = 30° ⎣1 + 1 / 3 ⎦
G ( s) =
118. Topic: Transfer Function (b) We have P1 = G2, so ∆1 = 1 − (non-touching part of loop 1) = 1 − 0 = 1 ∆ = 1 − [−G1G2 − G1] = 1 + G1G2 + G1 = 1 + G1[1 + G2]
116. Topic: Root Loci (d) The given plot can be redrawn as w
p1
z2
z1
p2
z∗1
p∗2
PΔ G2 (1) G2 Y ( s) = 1 1 = = 1 + G1 (1 + G2 ) 1 + G1 (1 + G2 ) Δ X 2 ( s)
s
p∗1
119. Topic: Transfer Function (0.241) We have
z∗2
g (t ) = e −2t (sin 5t + cos 5t )
From the given plot, the transfer function is
The output response for a unit impulse input is obtained by taking Laplace transform of the given function. Therefore,
( s − z1 )( s − z1* )( s − z2 )( s − z2* ) H ( s) = K ( s − p1 )( s − p1* )( s − p2 )( s − p2* ) Put s = jw, H ( jw ) = K
( s + 1000) 10( s + 10)
G ( s) =
w 2 + z1
2
w 2 + z1
2
w 2 + z2
2
w 2 + z2
w 2 + p1
2
w 2 + p1
2
w 2 + p2
2
w 2 + p2
2
As |G(s)|s = 0 = 0 for DC gain, we have
2
From the plot |z1| = |p2| and |z2| = |p1| Therefore, H(jw) = K.
117. Topic: Bode Plots (d) The Bode plot can be redrawn in ideal form as follows.
5 s+2 + ( s + 2 ) 2 + 5 2 ( s + 2) 2 + 5 2
G(0) =
2 5 2 5 + 2 = + = 0.241 2 2 29 29 2 +5 2 +5 2
120. Topic: Routh–Hurwitz and Nyquist Criteria (b) From the given Nyquist plot,
G1 ( s) =
1 and G2 ( s) = s s
1 ×s =1 s This value is constant, hence resulting Nyquist plot is Im
Therefore, G ( s) =
Gain 0 dB/dec −20 dB 0 dB
1 10
Ch wise GATE_EE_Ch7.indd 370
Re
1k
11/21/2018 3:47:13 PM
Chapter 7 • Control Systems
121. Topic: P, PI and PID Controllers (b) Controllability matrix M = [B AB] from the given signal flow when is W the only input 12 × (0.33 R L =diagram, ) 2 = 1u.133
The Routh array is 1
K1 + 2
s
3
3K1
s1
6 − K1 3
0
s0
3K1
0
So, for the system to be unstable, 6 − K1 ≤ 0 ⇒ K1 ≥ 6 3 After adding zero, we have G ( s) H ( s) =
The Routh array is
123. Topic: Transfer Function (c) The given system can be represented as
s3
1
K1 + 2
2
s
3
3K1
s1
2
0
s
3K1
0
0
[X(s) − HY(s)] +
jΩ
G1[X(s) − HY(s)]
−
+ H Y(s)
− G2
Y(s) + G2 [X(s) − HY(s)]
−2
+ Y(s) = G1[X(s) — HY(s)] + G2 [X(s) − HY(s)] = X(s) [G1 + G2] — HY(s) [G1 + G2] Y(s) + HY(s) (G1 + G2) = X(s) (G1 + G2) Y(s) [1 + H(G1 + G2)] = X(s) (G1 + G2) G1 + G2 Y ( s) = X ( s) 1 + H (G1 + G2 )
K1 s( s + 1)( s + 2)
125. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (a) xk +1 − xk y − yk and − xk = k +1 h h System will not be stable for h > 0. yk =
126. Topic: Transfer Function (a) We have u(t) = 1, U(s) = 1/s and
q( s) = 1 + G ( s) H ( s) = s( s + 1)( s + 2) + K1 = 0 = s3 + 3s 2 + 2 s + K1 = 0
s 0
In the Routh array, for any value of K1 ≥ 6 , system does not go in right half cycle of s = 0 plane. Thus, the root locus of the modified system with zero at s = −3, never transits to unstable region.
124. Topic: Root Loci (d) The open loop transfer function of the system is G ( s) H ( s) =
−1
Unstable
[X(s) − HY(s)]
Ch wise GATE_EE_Ch7.indd 371
K1 ( s + 3) s( s + 1)( s + 2)
q( s) = 1 + G ( s) H ( s) = s3 + 3s 2 + ( K1 + 2) s + 3K1 = 0
122. Topic: Transfer Function (d) Pole 1: (2 − j3) and Pole 2: (2 + j3) For a second order system there should be two poles and the other two are zeros at −2 + 3j and −2 − 3j.
X(s)
s3 2
⎡1 × (30⎤.33) 2 = 1.33 W R ML == 12 ⎢1 3⎥ = 0 ( not controllable) ⎣⎡1 3⎦⎤ M 0 2( not controllable) 33 R L == 12 × 3(−0⎥ .2= ⎤ ) = 1.33 W ⎡⎢⎣10only When uM is the input, ⎦ ≠ 0 (controllable). =⎢ 2 ⎡101 3−⎤21⎥⎦ ⎣ =⎤ 0 ( not controllable) M =⎡ M = ⎢⎢⎣1C 3⎥⎦ ⎥ ≠ 0 (controllable). ⎡⎣1 ⎤ 1⎦ N=⎢ ⎥ 0 − 2⎤ ⎡CA M = ⎣⎡ C ⎦⎤ ⎥ ≠ 0when (controllable ). output, The observability y1 is the only N = ⎢⎢⎣11 ⎥matrix ⎡⎣ CA ⎦ 01⎤⎦ ≠ 0 (observable). N=⎢ ⎡15C ⎤− 20⎥⎦ ⎣ ⎤ ⎡ N= N = ⎢⎢⎣ CA ⎥⎦ ⎥ ≠ 0 (observable). ⎣ 5 − 2⎦ 0⎤ ⎡1 N=⎢ ⎥ ≠ 0 (observable). ⎣ 5 − 2⎦
G1
371
G ( s) =
1 − 2s 1+ s
11/21/2018 3:47:15 PM
372
GATE EE Chapter-wise Solved Papers
⎛ 1 − 2s ⎞ ⎛ 1 ⎞ Y ( s) = U ( s)G ( s) = ⎜ ⎝ 1 + s ⎟⎠ ⎜⎝ s ⎟⎠ By partial fraction method,
129. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (b) Given that Y ( s) s = R( s) s + 2
1 − 2s A B = + s(1 + s) s (1 + s)
and y(t ) = A cos( 2t + f ), r(t) = cos2t
1 − 2s = A(1 + s) + B(s) A = 1; B = −3; ⇒ 1 − 3 s s +1 1 So, the LHS is − 3 s s +1 Taking inverse Laplace transform, we get y(t) = u(t) − 3e−t u(t) = u(t) [1 − 3e−t] y(0) = 1 − 3e0 = 1 − 3 = −2 (graph starts at −2)
G ( s) =
1 + G(s) = 0 2 s(s + 3s + 2) + K = 0 − K = s3 + 3s2 + 2s dK = 0 , ⇒ 3s2 + 6s + 2 = 0 ds Thus, s = −0.42 (or) −1.57. As K is positive, s = −0.42. 128. Topic: Transfer Function (a) Given that
x = AX + BU ; Y = CX + DU
H ( jw ) =
−1
2
= 0.707
∠H ( jw ) = 90 − 45° = 45°
130. Topic: Transfer Function 100 (a) Given that G ( s) = ( s + 1)3 G ( jw ) = =
100 (1 − 3w ) + j (3w − w 2 )
=
100[(1 − 3w 2 ) − jw (3 − w 2 ) 2 ] (1 − 3w 2 ) + w 2 (3 − w 2 ) 2
100 100 = (1 + jw )3 1 + ( jw )3 + 3( jw ) 2 + 3 jw 2
Using Im[G ( jw )] = 0 w (3 - w 2) = 0 ⇒ w = 0 or ± 3
Therefore, y (t ) =
s ( s − 2 ) ( −2 )
Ch wise GATE_EE_Ch7.indd 372
1
∠H (w ) = tan −1 w = 45°
⎡ s − 2 −1⎤ ⎡1⎤ 1 ⎤ ⎡1⎤ ⎡s = [3 0 ] ⎢ ⎥ ⎢1⎥ = [3 0] ⎢ s 2 −2 s − 2⎥⎦ ⎢⎣1⎥⎦ ⎣ ⎦ ⎣⎦ ⎣ ⎡ s + 1⎤ 1 3( s + 1) 3 0] ⎢ [ ⎥ = 2 s − 4 s − 2s + 2 ⎣ ⎦ s − 2s + 2 2
=
131. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (0.707) We have 1 1 1 H (w ) = = = 2 1+1 2 w +1
⎡x ⎤ y = [3 0 ] ⎢ 1 ⎥ ⎣ x2 ⎦ Transfer function C (sI − A)−1B
=
2 4+4
Thus, w ph = 3 rad/s
⎡ x 1 ⎤ ⎡ 2 1 ⎤ ⎡ x1 ⎤ ⎡ 4 ⎤ ⎢ x ⎥ = ⎢ −2 0 ⎥ ⎢ x ⎥ + ⎢ 4 ⎥ U ⎦⎣ 2⎦ ⎣ ⎦ ⎣ 2⎦ ⎣
⎛ ⎛ w⎞⎞ , ∠H ( jw ) = 90° − ⎜ tan −1 ⎜ ⎟ ⎟ ⎝ 2 ⎠⎠ ⎝ w +4 2
Therefore,
dx1 dx = 2 x1 + x2 + 4 ; y = 3x1; 2 = −2 x1 + 4 dt dt
w
K s( s + 1)( s + 2)
jw s ⇒ H ( jw ) = jw + 2 s+2
H ( jw ) =
127. Topic: Root Loci (a) Given transfer function
H ( s) =
Thus, A =
1 2
1 2
cos(t − 45°)
= 0.707
11/21/2018 3:47:19 PM
Chapter 7 • Control Systems
132. Topic: Bode Plots (a) The plot shows three poles, of which there is a double pole at the corner frequency near w = 8 rad/s, and a single pole near w = 0.5 rad/s. Initial pole is positive, therefore, one zero exists at s = 0. Only (a) satisfies these conditions.
136. Topic: Transfer Function (c) We know that 1 + G(s)H(s) = 0 Substituting values for the unity feedback system, we have
133. Topic: State Space Model (6) Given that
1+
⎡1⎤ ⎡1 0 ⎤ x (t ) = ⎢ x(t ); x(0) = ⎢ ⎥ ⎥ ⎣0 2⎦ ⎣1⎦
Using Routh–Hurwitz criteria
−1
s3
−1 ⎧ 0 ⎤ ⎫⎪ ⎡e t −1 ⎪ ⎡ s − 1 = L ⎨⎢ ⎬=⎢ s − 2⎥⎦ ⎪⎭ ⎣ 0 ⎪⎩ ⎣ 0
0⎤ ⎥ e 2t ⎦
⎡ et ⎤ x (t ) = f (t ) ⋅ x ( 0 ) = ⎢ 2 t ⎥ ⎣e ⎦
K ( s + 1) =0 s(1 + Ts)(1 + 2 s)
2Ts3 + (2 + T)s2 + (1 + K)s + K = 0
f (t ) = L [ sI − A] −1
y(t) = e6 + e25
(1 + K )
2T
2
(2 + T ) s1 (2 + T ) (1 + K ) − 2TK 2+T K s0 s
w=0 w=0
w=∞
It encircles (−1 + j 0) once, in clockwise direction. 135. Topic: Routh–Hurwitz and Nyquist Criteria (2) For the given polynomial equation s3 + 5.5s2 + 8.5 + 3 = 0 Let s = (r - 1) (r - 1)3 + (5.5) (r - 1)2 + 8.5(r - 1) + 3 = 0 r3 + 2.5r2 + 0.5r - 1 = 1
T +2 T −2 T +2 0 0, therefore (2 + T) (1 + K) − 2TK > 0
y(t ) = e loge 2 + e 2 loge 2 = 2 + 4 = 6
373
K = s+ For breakaway point,
4 −5 s
dK =0 ds
11/21/2018 3:47:21 PM
374
GATE EE Chapter-wise Solved Papers
where
dK 4 = 1− 0 − 2 = 0 ds s s =±2 Gain K =
K v = lim s→0
1× 2 =1 2
s ⋅ ( Ks + b) b = s( s + a − K ) a − K
Therefore, from Eq. (1), ess =
139. Topic: Routh–Hurwitz and Nyquist Criteria (d) Using Routh–Hurwitz criteria, we have K(K + 2) > 3 ⇒ (K + 3) (K - 1) > 0 ⇒K>1
142. Transfer Function (d) H ( s) =
Vo ( s) 1 − s = Vi ( s) 1 + s
140. Topic: Feedback Principle H ( jw ) =
Ke − s (1.04) G ( s) = s Now, Phase margin = 180° + f = 30°
f = ∠G ( jw ) w = w
∠H ( jw ) = −2 tan −1 w
∠H ( jw ) = 0° gc
(1)
and at w = ∞, ∠H ( jw ) = −p
Also, ∠G ( jw ) = −90° − 57.3°w
1 − jw 1 + jw
Now, at w = 0,
⇒ f = −150°
a−K b
(2)
At w = w gc ,
143. Topic: Lag, Lead and Lead-Lag Compensators (a) The pole–zero diagram is shown below
G ( jw ) = 1 k ×1 =1 w ⇒ w = w gc = K rad/s ⇒
(3)
From Eqs. (2) and (3), we have ∠G ( jw ) = −90° − 57.3° K (4)
—1
—0.1
Substituting Eq. (1) in Eq. (4), we have -150° = -90° - 57.3° K ⇒ K = 1.047 ≈ 1.04
141. Topic: Transfer Function (d) Closed loop transfer function is Ks + b s 2 + as + b Open loop transfer function is Ks + b Ks + b = 2 s + as + b − Ks − b s( s + a − K )
—100 —10
Steady-state error is given by 1 ess = (1) Kv
Ch wise GATE_EE_Ch7.indd 374
The maximum phase lead is between 0.1 and 1. Therefore, 0.1 < w < 1 144. Topic: Transfer Function (a) Y ( s) = Transfer function = U1 ( s ) =
k1 1 ⋅ LJ s 2 ⎡ R k1k2 ⎤ 1 − ⎢− − 2⎥ ⎣ LS LJs ⎦
k1 LJs 2 + RJs + k1k2
11/21/2018 3:47:24 PM
Chapter 7 • Control Systems
145. Topic: State Space Model ⎡1⎤ ⎡1 2⎤ X = ⎢ ⎥ X + ⎢ 2⎥ u; Y = [1 0]X 2 0 ⎣ ⎦ ⎣ ⎦ We know that, Transfer function = C [SI - A]-1 B + 0 Now,
Given H (1) =
(d)
⎡ ⎢ s2 [SI − A]−1 = ⎢ ⎢ ⎢⎣ s 2
5 −s−4 2 −s−4
5 . Therefore, from Eq. (2), we have 4 25 (3) k= 16
Substituting Eq. (3) in Eq. (2), we get H ( z) =
2 ⎤ 2 s − s − 4⎥ ⎥ s −1 ⎥ s 2 − s − 4 ⎥⎦
⎡ s+4 ⎤ ⎢ 2 ⎥ [SI − A]−1 . B = ⎢ s − s − 4 ⎥ 4 s − ⎢ ⎥ ⎢⎣ s 2 − s − 4 ⎥⎦
375
25 z 16 ⎛ 4 1 ⎞ ⎜⎝ z + ⎟⎠ 4
⇒ H (8) = 0.097 = g (8) 148. Topic: Root Loci (b) Characteristic equation is s2 + 6Ks + 2s + 5 = 0 G ( s) = −1 =
Therefore, Y (s ) s+4 = 2 U (s ) s − s − 4
6 Ks s + 2s + 5 2
jw 2j
146. Topic: Transfer Function (0.554) G ( s) =
Now,
−s + 1 s +1
—1
—√5
1 1 2 Y ( s) = G ( s) ⋅ = − s s s +1
—2j
K=
−t
y(t ) = u(t ) − 2e u(t ) ⇒ y(1.5) = 1 − 2e −1.5 = 1 − 0.44626 ⇒ y(1.5) = 0.554 147. Topic: Transient and Steady-State Analysis of Linear Time Invariant System (0.097) z=
1 2
e
j ( 2 k −1)
p 4
1 1 (1 + j ), z2 = ( −1 + j ), 2 2 1 1 z3 = ( −1 − j ), z4 = (1 − j ) 2 2
kz 4 H ( z) = (1) ( z − z1 )( z − z2 )( z − z3 )( z − z4 )
H ( z) =
kz 4 1 z + 4 4
Ch wise GATE_EE_Ch7.indd 375
The point at which root locus enters into the real axis is given by
(2)
dK =0 ds
dK −( s 2 + 2 s + 5) = =0 ds 6s
⇒
⇒ s=± 5
( k = 1, 2, 3, 4,...)
⇒ z1 =
−( s 2 + 2 s + 5) 6s
Therefore, s=− 5 149. Topic: Routh–Hurwitz and Nyquist Criteria (a) Given that s3 + 3s2 + 2s + K = 0 Using Routh–Hurwitz criterion, K < 6 and K > 0 ⇒0 Pout for both transformer and emitter follower.
Codes: (a) (v), (ii), (iii), (iv), (i) (b) (v), (ii), (iii), (i), (iv) (c) (ii), (i), (iii), (iv), (v)
(c) Pin < Pout for transformer and Pin = Pout for emitter follower. (d) Pin = Pout for transformer and Pin < Pout emitter follower.
(d) (v), (ii), (i), (iii), (iv)
(GATE 2009: 2 Marks)
(GATE 2009: 1 Mark) 106. 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 (GATE 2009: 1 Mark) 107. The following circuit has R = 10 kW and C = 10 mF. The input voltage is a sinusoid at 50 Hz with an rms value of 10 V under ideal conditions, the current IS from the source is
Ch wise GATE_EE_Ch9.indd 444
109. An ideal op-amp circuit and its input waveform are shown in the following figures. The output waveform of this circuit will be 3 2 1 V 0 −1 −2 −3
t4 t5
t6 t
t1 t2
t3
11/21/2018 6:12:10 PM
Chapter 9 • Analog and Digital Electronics
445
+6 V (d)
1 kΩ
− +
Vin
6
Vout
−3 V
V
2 kΩ
1 kΩ
0 t2
t4
t6
t
−3 (GATE 2009: 2 Marks)
6
(a)
110. In an 8085 microprocessor, the contents of the accumulator, after the following instructions are executed, will become XRA A MVIB F0H SUB B
V t3
t6
0 t
(a) 01 H (c) F0 H
(b) 0F H (d) 10 H (GATE 2009: 2 Marks)
−3
111. Give that the op-amp is ideal, the output voltage Vo is 2R
(b)
6 +10 V R V
Vo + t3
t6
0
+2 V
t
−3
-10 V
(a) 4 V (c) 7.5 V
(b) 6 V (d) 12.12 V (GATE 2010: 1 Mark)
(c)
112. Assuming that the diodes in the given circuit are ideal, the voltage Vo is
6
D1
V
10 kΩ
D2
10 kΩ t6 0
Vo
10 V t2
t2
15 V
t 10 kΩ
Ch wise GATE_EE_Ch9.indd 445
−3
11/21/2018 6:12:11 PM
446
GATE EE Chapter-wise Solved Papers
(a) 4 V (c) 7.5 V
(b) 5 V (d) 12.12 V (GATE 2010: 1 Mark)
113. The given 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 V
+
−
I
(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 (GATE 2010: 1 Mark) 115. The transistor circuit shown in the given figure uses a silicon transistor with VBE = 0.7 V, IC / IE and a DC current gain of 100. The value of Vo is +10 kΩ
(a)
I
10 kΩ
V
50 kΩ
(b)
I Vo V 100 kΩ
(c)
I
(a) 4.65 V (c) 6.3 V
(b) 5V (d) 7.32 V (GATE 2010: 2 Marks)
V
116. The TTL circuit shown in the figure given below is fed with the waveform X (also shown). All gates have equal propagation delay of 10 ns. The output Y of the circuit is
(d)
I
X 100 ns
V 1
0
(GATE 2010: 1 Mark) 114. 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 and SP means Stack pointer.
Ch wise GATE_EE_Ch9.indd 446
t
X Y
11/21/2018 6:12:12 PM
Chapter 9 • Analog and Digital Electronics
(a) Y
447
(c) X
1 F
0
t
Y Z
(b) Y (d) X
1 0
t F
(c) Y
Y Z
1 0
(GATE 2010: 2 Marks)
t 119. For the following circuit,
(d) Y
R
1 0
t +12 V
(GATE 2010: 2 Marks)
Statement for Linked Answer Questions 117 and 118: The following Karnaugh map represents a function F.
R
+12 V +
Vi
+
Vo
R
F
YZ
-12 V
R 00
01
11
10
0
1
1
1
0
1
0
0
1
0
-12 V
X
R
117. A form of the function F is (a) F = XY + YZ (c) F = XY + YZ
R
(b) F = XY + YZ (d) F = XY + YZ
the CORRECT transfer characteristic is
(GATE 2010: 2 Marks)
(a)
118. Which of the following circuits is a realization of the above function F ?
Vo +12 V
(a) X +6 V F
-6 V
Y
Vi
Z (b) X -12 V F
Y
Z
Ch wise GATE_EE_Ch9.indd 447
11/21/2018 6:12:14 PM
448
GATE EE Chapter-wise Solved Papers
(b)
Vo
(a) 1
(b) 0
(c) X
(d) X
+12 V
(GATE 2011: 1 Mark)
+6 V -6 V
Vi
-12 V
121. 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) half the frequency of the time base (sweep) oscillator. (GATE 2011: 1 Mark)
(c)
122. The transistor used in the following circuit has b of 30 and ICBO is negligible.
Vo +12 V
2.2 kΩ
15 kΩ 1 kΩ
D VCE = 0.7 V
-6 V
+6 V VCE = 0.2 V
Vi VZ = 5 V -12 V
−12 V
(d)
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
Vo +12 V
(GATE 2011: 2 Marks) 123. A clipper circuit is shown below. -6 V
1 kΩ
+6 V Vi
D Vi -12 V
Vo
VZ = 10 V 5V
(GATE 2011: 1 Mark) 120. The output Y of the logic circuit given below is X Y
Ch wise GATE_EE_Ch9.indd 448
Assuming the forward voltage drops of the diodes to be 0.7 V, the input-output transfer characteristics of the circuit are
11/21/2018 6:12:14 PM
Chapter 9 • Analog and Digital Electronics
(a)
: : 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 (a) POP D (b) POP H DAD H DAD D PUSH D INX H INX H INX H PUSH H (c) POP H (d) XTHL DAD D INX D PUSH H INX D INX D XTHL
Vo 4.3
4.3
Vi
(b) 10 Vo 4.3
4.3
10
449
(GATE 2011: 2 Marks)
Vi
125. A two-bit counter circuit is shown in the following figure.
(c)
Q Vo
J
O
J
O
K
Q
K
Q
5.7
−0.7
QB
Clk
−0.7
5.7
Vi
If the state QAQB of the counter at the clock time tn is “10” then the state QAQB of the counter at tn + 3 (after three clock cycles) will be (a) 00 (b) 01 (c) 10 (d) 11 (GATE 2011: 2 Marks)
(d)
10 Vo
126. In sum of products function
5.7
f ( X , Y , Z ) = Σ ( 2, 3, 4, 5)
10 Vi
(a) XY , XY
−5.7
(b) XY , XYZ , XYZ (GATE 2011: 2 Marks)
124. A portion of the main program to call a subroutine SUB in an 8085 environment is given below. : : LXI D,DISP LP: CALL SUB
Ch wise GATE_EE_Ch9.indd 449
the prime implicants are
(c) XYZ , XYZ , XY XYZ , XYZ , XYZ , XYZ (d) (GATE 2012: 1 Mark) 127. 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
11/21/2018 6:12:16 PM
450
GATE EE Chapter-wise Solved Papers
(a) 4 (c) 8
(b) 6 (d) 10 (GATE 2012: 1 Mark)
128. Consider the given circuit. A
(b) High-pass filter with f 3dB =
1 rad/s R1C
(c) Low-pass filter with f 3dB =
1 rad/s R1C
(d) High-pass filter with f 3dB =
1 rad/s ( R1 + R2 )C
Clk
(GATE 2012: 2 Marks)
B
131. The state transition diagram for the logic circuit shown is
In this circuit, the race around (a) does not occur. (b) occurs-when Clk = 0 (c) occurs when Clk = 1 and A = B = 1 (d) occurs when Clk = 1 and A = B = 0 (GATE 2012: 1 Mark)
2-1 MUX D Clk
Q Q
X1 Y X0 Select
129. The voltage gain AV of the given circuit is A
13.7 V (a) A = 1
A=0
12 kΩ A=1 C Vo
100 kΩ
Q=0
Q=1 A=0
C b = 100 10 kΩ
(b) A = 0
Vi
A=0
A=1 Q=0
Q=1 A=0
(a)
AV ≈100 AV ≈ 200 (b)
(c)
AV ≈ 20 (d) AV ≈ 10
(c) A = 0
(GATE 2012: 2 Marks)
A=1
A=0
130. The circuit shown is a Q=0
C + Input −
(d)
+5 V
R1 0
−
A=1
+ Output −
+ 0
Ch wise GATE_EE_Ch9.indd 450
1 rad/s ( R1 + R2 )C
A=1
A=0
−5 V (a) Low-pass filter with f 3dB =
Q=1 A=1
R2
Q=0
Q=1 A=0 (GATE 2012: 2 Marks)
11/21/2018 6:12:17 PM
Chapter 9 • Analog and Digital Electronics
132. In the following circuit what is the output voltage (Vout) in volts if a silicon transistor Q and an ideal op-amp are used?
1 kΩ
135. In the following circuit, the knee current of the ideal Zener diode 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
Q
+15 V
451
100 Ω
Vout
ILoad
10 V
+ 5V
+ -
RL -15 V
VZ = 5 V
(a) −15 (b) −0.7 (c) +0.7 (d) +15 (GATE 2013: 1 Mark) 133. In the feedback network shown as follows, if the feedback factor k is increased, then the
+V − in
V1 + −
A0
+ −
(a) 125 and 125 (c) 250 and 125
(b) 125 and 250 (d) 250 and 250 (GATE 2013: 2 Marks)
136. A voltage 1000 sinwt volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in Volts is
Vout 1 kΩ W
X
Y Z
Vf = kVout + −
k
+ −
(a) input impedance increases and output impedance decreases (b) input impedance increases and output impedance also increases (c) input impedance decreases and output impedance also decreases (d) input impedance decreases and output impedance increases (GATE 2013: 1 Mark) 134. 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) an AND gate (b) an OR gate (c) an XOR gate (d) a NAND gate (GATE 2013: 1 Mark)
Ch wise GATE_EE_Ch9.indd 451
+ 1 kΩ − (b) (sinwt + | sinwt |)/2 (d) 0 for all t
(a) sinwt (c) (sinwt | sinwt |)/2
(GATE 2013: 2 Marks) 137. In the following circuit, the op-amps are ideal. Then Vout in V is −2 V
1 kΩ
1 kΩ +15 V −
+15 V −
+
Vout +
1 kΩ
−15 V
−15 V
+1 V 1 kΩ 1 kΩ
11/21/2018 6:12:18 PM
452
GATE EE Chapter-wise Solved Papers
(a) 4 (c) 8
(b) 6 (d) 10
+VCC 10 kΩ
(GATE 2013: 2 Marks) 138. The clock frequency applied to the digital circuit shown in the given figure is 1 kHz. If the initial state of the output Q of the flip-flop is ‘0’, then the frequency of the output waveform Q in kHz is
Q
X O
T
Q
Clk
(a) 0.25 (c) 1
C C
Vo 10 kΩ hfe = 100
Vi
C
(a) 1 (c) 20
(b) 10 (d) 100 (GATE 2014: 1 Mark)
Q
(b) 0.5 (d) 2 (GATE 2013: 2 Marks)
141. The figure shows the circuit of a rectifier fed from a 230 V (rms), 50 Hz sinusoidal voltage source. If we want to replace the current source with a resistor so that the rms value of the current supplied by the voltage source remains unchanged, the value of the resistance (in ohms) (Assume diodes to be ideal.) is
139. In the following circuit, Q1 has negligible collectorto-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
10 A 230 V, 50 Hz
+VCC (GATE 2014: 1 Mark) R1 Z R2 X
142. The transistor in the given circuit should always be in active region. Take VCE(sat) = 0.2 V, VBE = 0.7 V. The maximum value of RC in Ω which can be used is .
Q1 Diode RC + 5V
Y (a) XY
RS = 2 kΩ
(b) XY
b = 100
+ 5V
(c) XY
(d) XY (GATE 2013: 2 Marks)
(GATE 2014: 1 Mark)
140. The magnitude of the mid-band voltage gain of the circuit shown in figure is (assuming h fe of the transistor to be 100)
143. The sinusoidal AC source in the figure has an rms value 20 V. Considering all possible values of RL, the of 2
Ch wise GATE_EE_Ch9.indd 452
11/21/2018 6:12:19 PM
Chapter 9 • Analog and Digital Electronics
minimum value of RS in Ω to avoid burn out of the Zener diode is .
453
0X/1, 10/1 11/0
RS
Q=0
Q=1
0X/1, 10/1
11/0 20 V Ö2
5V 1/4W
RL The logic gate represented by the state diagram is (a) XOR (b) OR (c) AND (d) NAND
(GATE 2014: 1 Mark)
(GATE 2014: 1 Mark)
144. An operational-amplifier circuit is shown in the following figure.
148. In 8085A microprocessor, the operation performed by the instruction LHLD 2100H is (a) (H) ← 21H, (L) ← 00H (b) (H) ← M(2100H), (L) ← M(2101H) (c) (H) ← M(2101H), (L) ← M(2100H) (d) (H) ← 00H, (L) ← 21H (GATE 2014: 1 Mark)
R
+Vsat vi
−
R
+Vsat − +
+ −Vsat
−Vsat
vo
R2
149. Given that the op-amps in the figure are ideal, the output voltage Vo is V2
R1
+ −
R R
R −
2R
The output of the circuit for a given input vi is ⎛R ⎞ ⎛ R ⎞ (a) − ⎜ 2 ⎟ vi (b) − ⎜1 + 2 ⎟ vi R1 ⎠ ⎝ R1 ⎠ ⎝ ⎛ R ⎞ (c) ⎜1 + 2 ⎟ vi 1 R1 ⎠ ⎝
(d) +Vsat or -Vsat (GATE 2014: 1 Mark)
145. A cascade of three identical modulo-5 counters has an overall modulus of (a) 5 (b) 25 (c) 125 (d) 625 (GATE 2014: 1 Mark) 146. Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter? (a) 1 0 0 0 (b) 1 0 0 1 (c) 0 0 1 1 (d) 1 1 0 0 (GATE 2014: 1 Mark) 147. A state diagram of a logic gate which exhibits a delay in the output is shown in the figure, where X is the do not care condition, and Q is the output representing the state.
Ch wise GATE_EE_Ch9.indd 453
Vo +
V1
− +
R
R R
(a) (V1 − V2 ) (b) 2(V1 − V2 ) (c) (V1 − V2 ) / 2 (d) (V1 + V2 ) (GATE 2014: 2 Marks) 150. In the figure shown, assume the op-amp to be ideal. Which of the alternatives gives the correct Bode plots V (ω ) for the transfer function o ? Vi (ω ) +VCC 1 kΩ Vi
+ 1 µF
Vo − −VCC
Rf
11/21/2018 6:12:20 PM
454
GATE EE Chapter-wise Solved Papers
151. Assuming the diodes to be ideal in the figure, for the output to be clipped, the input voltage Vi must be outside the range.
(a) f
V (w) 20 log o Vi(w)
10 kΩ 103
0
w
2 −10 1 10 10 −20 −30
10 102 103
0
w
−p/4 1 −p/2
10 kΩ Vi 1V
(a) −1 V to −2 V (c) +1 V to −2 V
(b) f
V ( w) 20 log o Vi(w) 103
0
w
2 −10 1 10 10 −20 −30
−p/2 −p/4 0 −p/4 −p/2
103 w
(b) −2 V to −4 V (d) +2 V to −4 V (GATE 2014: 2 Marks)
152. An oscillator circuit using ideal op-amp and diodes is shown in the figure. R
1 10 102
+5 V − Vo
C (c)
+
−10 1 10 −20 −30
−5 V
f
V ( w) 20 log o Vi(w) 0
Vo
2V
w
102
−p/2 −p/4 2 3 0 1 10 10 10
3 kΩ
w
1 kΩ
−p/4 −p/2
103
1 kΩ
The time duration for positive part of the cycle is Δt1 Δt1 − Δt2
f
and for negative part is Δt2. The value of e RC will be . (GATE 2014: 2 Marks)
−p/2 −p/4 0
153. Two monoshot multivibrators, one positive edge triggered (M1) and another negative edge triggered (M2), are connected as shown in the following figure.
(d) V ( w) 20 log o Vi(w) 0 2 3 −10 1 10 10 10 −20 −30
w
w
2 3 −p/4 1 10 10 10 −p/2
10 kW
(GATE 2014: 2 Marks)
Ch wise GATE_EE_Ch9.indd 454
+5 V
10 µF
M1
M2 Q1
Q2
Q1
Q2
vo
11/21/2018 6:12:22 PM
Chapter 9 • Analog and Digital Electronics
The monoshots M1 and M2 when triggered produce pulses of width T1 and T2, respectively, where T1 > T2. The steady state output voltage vo of the circuit is
(c)
455
vo
(a) vo
vi T1
T2
T1
T2
T1
1
t (b) vo T1
T1
T1
(d)
T1
vo
vi
t
−1
(c) vo T2
T1
T2
T1
T2 (GATE 2014: 2 Marks)
t (d) vo T2
T2
T2
T2
T2
155. Which of the following logic circuits is a realization of the function F whose Karnaugh map is shown in figure?
T2
AB
t (GATE 2014: 2 Marks) 154. The transfer characteristics of the op-amp circuit shown in the following figure is
0
00
01
1
1
11
10
C 1
R
1
1
(a) A
A
B C A
B C A
R +Vsat
R −
vi
R
+ −Vsat A R
R
(a)
B C A
vo −1
−
+Vsat
+
−Vsat
(a) A B C A B CC A
(a)
B
B A 1
B A
(d) (c)
C
(b) B C A C
(a)
(b)
(b)
(c) B A B A CC C
(b) vo
(b)
(a)
B A vi C C
vo
A B C (c) (b)
(d)
B (c)
(d)
(d)
C
vi B
(c)
Ch wise GATE_EE_Ch9.indd 455
B (d)
(GATE 2014: 2 Marks)
11/21/2018 6:12:23 PM
456
GATE EE Chapter-wise Solved Papers
156. An output device is interfaced with 8-bit microprocessor 8085A. The interfacing circuit is shown in figure
(c) J
Qn
T T flip-flop
Clk
AB
K
8 BDB 8 3L ´ 8L decoder A15 0 Output I2 1 A14 I1 port 2 A13 I0 A12 3 8 A11 4 E1 5 E2 Output device 6 E3 7 WR IO/M
Qn (d) Qn
J
T T flip-flop
Clk K
Qn (GATE 2014: 2 Marks) BCB
The interfacing circuit makes use of 3 line to 8 line decoder having 3 enable lines E1 , E2 , E3 . The address of the device is (a) 50H (b) 5000H (c) A0H (d) A000H (GATE 2014: 2 Marks) 157. The SOP (sum of products) form of a Boolean function is Σ(0, 1, 3, 7, 11), where inputs are A, B, C, D (A is MSB, and D is LSB). The equivalent minimised expression of the function is (a) ( B + C )( A + C )( A + B )(C + D ) (b) ( B + C )( A + C )( A + C )(C + D )
159. In an 8085 microprocessor, the following program is executed Address location
Instruction
2000H
XRA A
2001H
MVI B, 04H
2003H
MVI A, 03H
2005H
RAR
2006H
DCR B
2007H
JNZ 2005
200AH
HLT
At the end of program, register A contains (a) 60H (b) 30H (c) 06H (d) 03H (GATE 2014: 2 Marks)
(c) ( B + C )( A + C )( A + C )(C + D ) (d) ( B + C )( A + B )( A + B )(C + D ) (GATE 2014: 2 Marks) 158. A J-K flip-flop can be implemented by T flip-flops. Identify the correct implementation.
160. A 3-bit Gray counter is used to control the output of the multiplexer as shown in the figure. The initial state of the counter is (000)2. The output is pulled high. The output of the circuit follows the
(a) Qn
J
T Clk
3-bit Gray counter
T flip-flop
K
A2 A1 A0
+5 V
S0 Qn (b) Qn
J
Clk
T Clk
S1
R
4 ×1 MUX Output
T flip-flop (a) I0, 1, 1, I1, I3, 1, 1, I2
K Qn
Ch wise GATE_EE_Ch9.indd 456
I0 I1 I2 I3
E 0 1 2 3
(b) I0, 1, I1, 1, I2, 1, I3, 1
11/21/2018 6:12:25 PM
Chapter 9 • Analog and Digital Electronics
(c) 1, I0, 1, I1, I2, 1, I3,1
457
(a) 0 (b) p (c) p/2 (d) −p/2
(d) I0, I1, I2, I3, I0, I1, I2, I3 (GATE 2014: 2 Marks) 161. A hysteresis type TTL inverter is used to realise an oscillator in the circuit shown in the figure.
(GATE 2015: 1 Mark) 164. In the given circuit, the silicon transistor has b = 75 and a collector voltage VC = 9 V. Then the ratio of RB and RC is ————————.
10 kΩ 15 V RC +5 V
RB VC Vo
0.1 µF
If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in ms), for which output is LOW, is . (GATE 2014: 2 Marks)
(GATE 2015: 1 Mark) 165. In the following circuit, the input voltage Vin is 100 sin (100pt). For 100pRC = 50, the average voltage across R (in Volts) under steady-state is nearest to
162. Of the four characteristics given below, which are the major requirements for an instrumentation amplifier? P. High common mode rejection ratio Q. High input impedance R. High linearity S. High output impedance (a) P, Q and R only (c) P, Q and S only
(b) P and R only (d) Q, R and S only (GATE 2015: 1 Mark)
163. Consider the circuit shown in the following figure. In this circuit, R = 1 kW and C = 1 mF. The input voltage is sinusoidal with a frequency of 50 Hz, represented as a phasor with magnitude Vi and phase angle 0 radian as shown in the figure. The output voltage is represented as a phasor with magnitude Vo and phase angle d radian. What is the value of the output phase angle d (in radian) relative to the phase angle of the input voltage? R
Vin +
C
+
−
R C
(a) 100 (c) 200
−
(b) 31.8 (d) 63.6 (GATE 2015: 1 Mark)
166. The operational amplifier shown in the following figure is ideal. The input voltage (in Volt) is Vi = 2 sin(2p × 2000t). The amplitude of the output voltage Vo (in Volt) is ________. 0.1 µF
1 kΩ Vi
1 kΩ − +
Vo
C −
vi = Vi∠0
+ C R
Ch wise GATE_EE_Ch9.indd 457
vo = Vo∠d
(GATE 2015: 1 Mark) 167. In the following circuit, the transistor is in active mode and VC = 2 V. To get VC = 4 V, we replace RC with R ′C . Then the ratio R ′C / R C is ________.
11/21/2018 6:12:26 PM
458
GATE EE Chapter-wise Solved Papers
(a) The CB junction is forward biased and the BE junction is reverse biased. (b) The CB junction is reverse biased and the BE junction is forward biased. (c) Both the CB and BE junctions are forward biased. (d) Both the CB and BE junctions are reverse biased. (GATE 2015: 1 Mark)
+10 V RC RB
VC
(GATE 2015: 1 Mark)
170. In the 4 × 1 multiplexer, the output F is given by F = A ⊕ B. Find the required input I3 I2 I1 I0.
168. The filters F 1 and F 2 having characteristics as shown in Figures (a) and (b) are connected as shown in Figure (c).
Vo/Vi
I0 I1 4 × 1 I2 MUX I3 S1 S0
F1
Vi
F
Vo f1
A f
(a) 1010 (c) 1000
(a)
B (b) 0110 (d) 1110 (GATE 2015: 1 Mark)
171. Consider the following sum of products expression, F.
F2
F = ABC + A B C + ABC + ABC + A B C
Vo/Vi Vi
Vo
The equivalent product of sums expression is (a) F = ( A + B + C )( A + B + C )( A + B + C )
f2 f
(b) F = ( A + B + C )( A + B + C )( A + B + C )
(b)
(c) F = ( A + B + C )( A + B + C )( A + B + C ) R/2
(d) F = ( A + B + C )( A + B + C )( A + B + C ) (GATE 2015: 1 Mark)
+Vsat − +
R F1 Vi
Vo
−Vsat
R F2 (c)
172. The op-amp shown in the figure has a finite gain A = 1000 and an infinite input resistance. A step voltage Vi = 1 mV is applied at the input at time t = 0 as shown. Assuming that the operational amplifier is not saturated, the time constant (in millisecond) of the output voltage Vo is C
The cut-off frequencies of F1 and F2 are f1 and f2, respectively. If f1 < f2, the resultant circuit exhibits the characteristics of a (a) band-pass filter (b) band-stop filter (c) all pass filter (d) high-Q filter (GATE 2015: 1 Mark) 169. When a bipolar junction transistor is operating in the saturation mode, which one of the following statements is TRUE about the state of its collector-base (CB) and the base-emitter (BE) junctions?
Ch wise GATE_EE_Ch9.indd 458
R 1 kΩ V 1 mV
t = 0s
+ −
1 µF − A = 1000
+
+ Vo −
11/21/2018 6:12:27 PM
Chapter 9 • Analog and Digital Electronics
(a) 1001 (c) 11
(b) 101 (d) 1 (GATE 2015: 2 Marks)
173. The saturation voltage of the ideal op-amp shown below is ±10 V. The output voltage V o of the following circuit in the steady-state is
output. The reference voltage is +5 V. The output of the ADC, at the end of 3rd clock pulse after the start of conversion, is (a) 1010 0000 (b) 1000 0000 (c) 0000 0001 (d) 0000 0011 (GATE 2015: 2 Marks) 177. In the following sequential circuit, the initial state (before the first clock pulse) of the circuit is Q1Q0 = 00. The state (Q1Q0), immediately after the 333rd clock pulse is
1 kΩ +10 V 0.25 µF
459
− +
Vo 2 kΩ
−10 V
Q1
Q0 J 0 Q0
J1 Q1
K 0 Q0
K1 Q1
2 kΩ Clk (a) square wave of period 0.55 ms. (b) triangular wave of period 0.55 ms. (c) square wave of period 0.25 ms. (d) triangular wave of period 0.25 ms. (GATE 2015: 2 Marks) 174. The figure shows a digital circuit constructed using negative edge triggered J-K flipflops. Assume a starting state of Q2 Q1 Q0 = 000. This state Q2 Q1 Q0 = 000 will repeat after ———————— number of cycles of the clock Clk.
1 J0 Q0 Clk Clock 1 Q0 K
J1
Q1
J2
Clock
Clock
1
0
Q2
1 K1
Q1
K2
Q2
(GATE 2015: 2 Marks) 175. f(A, B, C, D) = PM(0, 1, 3, 4, 5, 7, 9, 11, 12, 13, 14, 15) is a maxterm representation of a Boolean function f (A, B, C, D) where A is the MSB and D is the LSB. The equivalent minimised representation of this function is (a) ( A + C + D )( A + B + D )
178. A Boolean function f (A, B, C, D) = Π(1, 5, 12, 15) is to be implemented using an 8 × 1 multiplexer (A is MSB). The inputs ABC are connected to the select inputs S2 S1 S0 of the multiplexer, respectively. 0 1 2 3 4 5 6 7
f(A, B, C, D)
S2 S1 S0 A B C
Which one of the following options gives the correct inputs to pins 0, 1, 2, 3, 4, 5, 6, 7 in order? (a) D, 0, D, 0, 0, 0, D , D (b) D , 1, D , 1, 1, 1, D, D (d) D , 0, D , 0, 0, 0, D, D
ACD + ABCD + ABCD
(d) ( B + C + D )( A + B + C + D )( A + B + C + D ) (GATE 2015: 2 Marks) 176. An 8-bit, unipolar successive approximation register type ADC is used to convert 3.5 V to digital equivalent
Ch wise GATE_EE_Ch9.indd 459
(b) 01 (d) 11 (GATE 2015: 2 Marks)
(c) D, 1, D, 1, 1, 1, D , D
(b) ACD + ABD (c)
(a) 00 (c) 10
(GATE 2015: 2 Marks) 179. A transistor circuit is given below. The Zener diode breakdown voltage is 5.3 V as shown. Take base of emitter voltage drop to be 0.6 V. The value of the current gain b is ________.
11/21/2018 6:12:29 PM
460
GATE EE Chapter-wise Solved Papers
183. The output expression for the Karnaugh map shown below is
10 V 220 Ω
4.7 kΩ
BC 00 01 11 10 A 0 1 0 0 1
0.5 mA 1 1 470 Ω
5.3 V
(GATE 2016: 1 Mark)
1
1
(a)
A+C A + B (b)
(c)
A + C (d) A+C (GATE 2016: 1 Mark)
180. The circuit shown below is an example of a
184. For the circuit shown below, taking the op-amp as ideal, the output voltage Vout in terms of the input voltages V1, V2 and V3 is
R2 C
9Ω
+15 V − +
Vin R1
1Ω
Vout V3 −15 V
(a) low pass filter. (c) high pass filter.
1
181. A temperature in the range of − 40 °C to 55 °C is to be measured with a resolution of 0.1°C. The minimum number of ADC bits required to get a matching dynamic range of the temperature sensor is (a) 8 (b) 10 (c) 12 (d) 14 (GATE 2016: 1 Mark)
−
1Ω
V1
(b) band pass filter. (d) notch filter. (GATE 2016: 1 Mark)
+VCC +
4Ω
Vout −VSS
V2 (a) 1.8V1 + 7.2V2 − V3
(b) 2V1 + 8V2 − 9V3
(c) 7.2V1 + 1.8V2 − V3
(d) 8V1 + 2V2 − 9V3 (GATE 2016: 2 Marks)
185. The current state QA QB of a two J-K flip-flop system is 00. Assume that the clock rise-time is much smaller than the delay of the J-K flip-flop. The next state of the system is
182. Consider the following circuit which uses a 2-to-1 multiplier as shown in the following figure. The Boolean expression for output F in terms of A and B is
5V QA
QB
J
J
O Y S
QA
F K
K
1 A (a)
CLK
B
A ⊕ B (b) A+ B
(c) A + B (d) A⊕ B (GATE 2016: 1 Mark)
Ch wise GATE_EE_Ch9.indd 460
(a) 00 (c) 11
(b) 01 (d) 10 (GATE 2016: 2 Marks)
11/21/2018 6:12:30 PM
461
Chapter 9 • Analog and Digital Electronics
186. A 2-bit flash Analog to Digital Converter (ADC) is given in the following figure. The input is 0 ≤ Vin ≤ 3 Volts. The expression for the LSB of the output B0 as a Boolean function of X2, X1, and X0 is
D1
D2
I0
3V 100 Ω 200 Ω
200 Ω
100 Ω
+ −
X2
+ −
X1
+ −
X0
B1
Digital circuit
B0
(a) D1 conducts for greater than 180° and D2 conducts for greater than 180°. (b) D2 conducts for more than 180° and D1 conducts for 180°. (c) D1 conducts for 180° and D2 conducts for 180° (d) D1 conducts for more than 180° and D2 conducts for 180°. (GATE 2017: 1 Mark) 190. The Boolean expression AB + AC + BC simplifies to
Vin
(a) BC + AC (b) AB + AC + B
(a)
X 0 [ X 2 ⊕ X1 ]
(b) X 0 [ X 2 ⊕ X 1 ]
(c)
X 0 [ X 2 ⊕ X 1 ] (d) X 0 [ X 2 ⊕ X1 ]
(c)
(GATE 2017: 1 Mark)
(GATE 2016: 2 Marks) 187. The Boolean expression ( a + b + c + d ) + (b + c ) simplifies to (a) 1 (b) a, b (c) a, b
AB + AC (d) AB + BC
(d) 0 (GATE 2016: 2 Marks)
188. For the circuit shown in the figure below, assume that diodes D1, D2 and D3 are ideal.
191. For 3-input logic circuit shown below, the output Z can be expressed as P Q
Z
R PQ + R (a) Q + R (b) P+Q+R (c) Q + R (d)
D1
(GATE 2017: 1 Mark)
R V1 + D2 V(t) = p sin(100 p t)V
D3
R
V2 −
192. The approximate transfer characteristic for the circuit shown below with an ideal operational amplifier and diode will be V55
The DC components of voltages V1 and V2, respectively, are (a) 0 V and 1 V (b) -0.5 V and 0.5 V (c) 1 V and 0.5 V (d) 1 V and 1 V
Vin
− + D −V55 V0
(GATE 2017: 1 Mark) R 189. In the circuit shown, the diodes are ideal, the inductance is small, and I0 ≠ 0. Which one of the following statements is true?
Ch wise GATE_EE_Ch9.indd 461
11/21/2018 6:12:32 PM
462
GATE EE Chapter-wise Solved Papers
(a)
VCC = 10 V
V0
(b)
4R
RB
Vin
b = 29
C
B
V0
E Vin (c)
R
V0 For this circuit, the value of Vin
(d)
RB is R
(a) 43 (c) 121
V0 Vin (GATE 2017: 2 Marks)
193. The circuit shown in the figure uses matched transistor with a thermal voltage VT = 25 mV. The base currents of the transistors are negligible. The value of the resistance R in kΩ that is required to provide 1 mA bias current for the differential amplifier block shown is . (Give the answer up to one decimal place.)
(b) 92 (d) 129 (GATE 2017: 2 Marks)
195. For the circuit shown in the following figure, assume that the op-amp is ideal. Which one of the following is TRUE? R R R
R − +
Vo
2R
12 V VS
2R Differential amplifier
(a) Vo = VS (b) Vo = 1.5 VS (c) Vo = 2.5 VS (d) Vo = 5 VS
1 µA
1 mA
(GATE 2017: 2 Marks) 196. The output expression for the Karnaugh map shown below is CD
R
−12 V
00
01
11
10
00
0
0
0
0
01
1
0
0
1
11
1
0
1
1
10
0
0
0
0
AB
(GATE 2017: 2 Marks) 194. For the circuit shown in the following figure, it is V given that VCE = CC . The transistor has b = 29 and 2 VBE = 0.7 V when the B-E junction is forward biased.
Ch wise GATE_EE_Ch9.indd 462
11/21/2018 6:12:33 PM
463
Chapter 9 • Analog and Digital Electronics
The minimum number of clock after which the output Z would again become zero is . (GATE 2017: 2 Marks)
(a) BD + BCD (b) BD + AB (c) BD + ABC (d) BD + ABC (GATE 2017: 2 Marks) 197. The logical gate implemented using the circuit shown below, where V1 and V2 are inputs (with 0 V as digital 0 and 5 V as digital 1) and VOUT is the output, is
200. A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current?
5V i0
+
1 kΩ L o v 0 a d
Vout V1
1 kΩ Q1 V2 1 kΩ Q2
− (a) NOT (c) NAND
(b) NOR (d) XOR
(a) v0 ≥ 0 and i0 < 0 (b) v0 < 0 and i0 < 0
(GATE 2017: 2 Marks) 1 digit timer counter possesses a base clock of 2 frequency 100 MHz. When measuring a particular input, the reading obtained is the same in: (i) frequency mode of operation with a gating time of one second and (ii) period mode of operation (in the × 10 ns scale). The frequency of the unknown input (reading obtained) in Hz is . (GATE 2017: 2 Marks)
(c) v0 ≥ 0 and i0 ≥ 0
198. A 10
199. For the synchronous sequential circuit shown below, the output Z is zero for the initial conditions QA QB QC = QA′ QB′ QC′ = 100 .
(d) v0 < 0 and i0 ≥ 0 (GATE 2018: 1 Mark) 201. In the logic circuit shown in the figure, Y is given by A B Y C D
D Q Q
D Q
D Q
Q
Q
QA Clock
QB
[MSB]
QC Z
[MSB] Q′A Q′B D Q D Q D Q Q
Ch wise GATE_EE_Ch9.indd 463
Q
Q
(a) (b) (c) (d)
Y = ABCD Y = (A + B) (C + D) Y=A+B+C+D Y = AB + CD (GATE 2018: 1 Mark)
Q′C 202. The op-amp shown in the figure is ideal. The input v impedance in is given by iin
11/21/2018 6:12:35 PM
464
GATE EE Chapter-wise Solved Papers
Z iin
+
+ −
Vin
D
Q
D
Q
D
Q
fOUT
Vo −
C
C
C
fIN R2
(a) Z
R1
R1 R (b) −Z 2 R2 R1 (d) − Z
(c) Z
R1 R1 + R2
(a) It can be used for dividing the input frequency by 3. (b) It can be used for dividing the input frequency by 5. (c) It can be used for dividing the input frequency by 7. (d) It cannot be reliably used as a frequency divider due to disjoint internal cycles. (GATE 2018: 2 Marks)
(GATE 2018: 1 Mark) 203. The waveform of the current drawn by a semi-converter from a sinusoidal AC voltage source is shown in the figure. If I0 = 20 A, the rms value of fundamental component of the current is A (up to 2 decimal places). Voltage and current
205. Digital input signals A, B, C with A as the MSB and C as the LSB are used to realize the Boolean function F = m0 + m2 + m3 + m5 + m7, where mi denotes the i th minterm. In addition, F has a don’t care for m1. The simplified expression for F is given by (a)
A+C AC + BC + AC (b)
(c) C + A (d) AC + BC + AC (GATE 2018: 2 Marks) 206. In the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain b = 100. The base-emitter voltage drop is a constant, VBE = 0.7 V. The value of the Thevenin equivalent resistance RTh(in W) as shown in the figure is (up to 2 decimal places).
Vm sin(w t) I0 0
wt
I0 30°
a 10 W
180° + 210°
Vm sin(wt)
I1
(GATE 2018: 1 Mark) 204. Which one of the following statements is true about the digital circuit shown in the figure?
Ch wise GATE_EE_Ch9.indd 464
10 kW
15 V
+
-
1 kW
RTh
10.7 V -
b (GATE 2018: 2 Marks)
11/21/2018 6:12:36 PM
465
Chapter 9 • Analog and Digital Electronics
ANSWER KEY 1. (d)
2. (b)
3. (b)
4. (d)
5. (a) 6. (c)
7. (b)
8. (c)
9. (d)
10. (b)
11. (c)
12. (b)
13. (c)
14. (d)
15. (a) 16. (c)
17. (a)
18. (a)
19. (b)
20. (a)
21. (*)
22. (b)
23. (b)
24. (d)
25. (c) 26. (d)
27. (d)
28. (a)
29. (c)
30. (b)
31. (b)
32. (b)
33. (d)
34. (a)
35. (c) 36. (c)
37. (c)
38. (b)
39. (d)
40. (a)
41. (a)
42. (c)
43. (d)
44. (d)
45. (*) 46. (b)
47. (b)
48. (a)
49. (d)
50. (b)
51. (c)
52. (a)
53. (d)
54. (a)
55. (c) 56. (c)
57. (b)
58. (b)
59. (c)
60. (b)
61. (d)
62. (b)
63. (d)
64. (b)
65. (d) 66. (d)
67. (a)
68. (d)
69. (a)
70. (b)
71. (c)
72. (c)
73. (c)
74. (a)
75. (a) 76. (a)
77. (a)
78. (b)
79. (d)
80. (b)
81. (d)
82. (b)
83. (b)
84. (b)
85. (c) 86. (d)
87. (b)
88. (a)
89. (d)
90. (a)
91. (a)
92. (a)
93. (a)
94. (c)
95. (d) 96. (b)
97. (a)
98. (c)
99. (d)
100. (b)
101. (b)
102. (b)
103. (a)
104. (b)
105. (b) 106. (c)
107. (a)
108. (d)
109. (d)
110. (d)
111. (b)
112. (b)
113. (c)
114. (d)
115. (a) 116. (a)
117. (b)
118. (d)
119. (d)
120. (a)
121. (c)
122. (d)
123. (c)
124. (c)
125. (c) 126. (a)
127. (b)
128. (a)
129. (d)
130. (b)
131. (d)
132. (b)
133. (c)
134. (c)
135. (b) 136. (d)
137. (c)
138. (b)
139. (b)
140. (d)
141. (23)
142. (22.33) 143. (300)
144. (d)
145. (c) 146. (d)
147. (d)
148. (c)
149. (b)
150. (a)
151. (b)
152. (6)
153. (c)
154. (c)
155. (c) 156. (b)
157. (a)
158. (b)
159. (a)
160. (a)
161. (0.635) 162. (a)
163. (d)
164. (105.13) 165. (c) 166. (1.25) 167. (0.75) 168. (b)
169. (c)
170. (b)
171. (a)
172. (a)
173. (a)
174. (6)
175. (c) 176. (a)
177. (b)
178. (b)
179. (19) 180. (a)
181. (d)
182. (d)
183. (b)
184. (d)
185. (c) 186. (a)
187. (d)
188. (b)
189. (a)
190. (a)
191. (c)
192. (a)
193. (172.69) 194. (d)
195. (c) 196. (d)
197. (b)
198. (10 ) 199. (6)
200. (c)
201. (d)
202. (b)
203. (17.39) 204. (b)
205. (b) 206. (92)
8
* None of the options is correct.
ANSWERS WITH EXPLANATION 1. Topic: Operational Amplifiers: Characteristics and Applications (d) For the given circuit, 1 kΩ − +
3. Topic: Amplifiers: Equivalent Circuit and Frequency Response (b) Cross over distortion is a type of distortion that occurs in push-pull class B amplifiers. This happens during the crossing over of the signal between devices, in this case, from upper transistor to the lower and vice versa.
5V Vo −5 V
4. Topic: Operational Amplifiers: Characteristics and Applications (d) From the given circuit 100 W
1 A
− +
Vp
+
Vo Vo I
The output voltage, Vo = −1 mA × 1 kΩ = −1 V 2. Topic: 8085 Microprocessor: Architecture (b) INTR (Interrupt request) is not a vectored interrupt.
Ch wise GATE_EE_Ch9.indd 465
Vs −
Ip = 0 90 W 1 kW I 10 W
11/21/2018 6:12:37 PM
466
GATE EE Chapter-wise Solved Papers
Therefore, on removal of this capacitor, only resistance will be there, which will cause a reduction in voltage gain.
We have IP = 0 VP = 10 × I And
8. Topic: A/D and D/A Converters (c) A parallel comparator is faster because A/D conversion is performed simultaneously through a set of comparators, hence is also called flash type ADC. The, integrating type ADC is slowest among of the given ADC types.
Vo = 10 I + 90 I Vo = 100 × I Therefore, feedback factor
b=
Vp Vo
=
10 × I 1 = 100 × I 10
5. Topic: Combinational and Sequential Logic Circuits (a) For the product of sum (POS) function, we prepare K-map shown as follows and then make groups of zeros.
9. Topic: Combinational and Sequential Logic Circuits (d) Truth table of NOR gate. A B Y=A+B
AB C
A+B
A+B
A+B A+B
C
1
1
f
0
C
0
0
f
0
A′
C′
So, the minimal POS function described by the K-map is A′C ′. Thus, option (a) is correct. 6. Topic: A/D and D/A Converters (c) In a dual slope ADC having an N-bit counter, when the input signal Va is being integrated, the N-bit counter is allowed to count up to a value proportional to Va. Analog Np(Va)
Switch control
A
B
Y
0
0
1
0
1
0
1
0
0
1
1
0
Given case
Truth table of Ex-NOR gate A B Y = A′B′ + AB A
B
Y
0
0
1
0
1
0
1
0
0
1
1
1
Given case
Control R
− +
+ −
X CLK
CLK
7. Topic: Amplifiers: Equivalent Circuit and Frequency Response (b) When an emitter resistance is added in common emitter (CE) amplifier, its voltage gain is reduced, but the input impedance increases. When a bypass capacitor is connected in parallel with an emitter resistance, the voltage gain of CE amplifier increases because at mid-band and high-band frequencies, the capacitor will behave as short circuit and overall gain will increases.
Ch wise GATE_EE_Ch9.indd 466
10. Topic: Multiplexer (b) The output of the given 4-to-1 MUX, f = XYI 0 + XYI1 + XYI 2 + XYI 3 Therefore, f = XY ⋅ 0 + XY ⋅1 + XY ⋅1 + XY ⋅1 = XY + XX + XY = XY + X (Y + Y ) = XY + X =X+Y
11/21/2018 6:12:38 PM
Chapter 9 • Analog and Digital Electronics
11. Topic: Operational Amplifiers: Characteristics and Applications (c) For an amplifier, the product of gain (G) and bandwidth (B) is constant. Therefore, for the given operational amplifier.
Third loop Now, accumulator has (30)H and B has (10)H Then, ADD B in accumulator ACC 1 0 1 0 0 0 0 0 CY[0] 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 RLC 1 0 1 1 0 0 0 0 CY[1] Here, carry is generated. Now program will HLT. So, NOP instruction is executed 3 times.
G1 B1 = G2 B2 B2 = =
G1 B1 G2 105 × 10 = 10 4 = 10 kHz 100
14. Topic: Sample and Hold Circuits (d) Leakage current is given by
12. Topic: Oscillators and Feedback Amplifiers (b) Given oscillator is an astable multi-vibrator, time period is given by
IL =
1 × Q and Q = CV 100 here C = 0.1 nF, V = 5 V and t = 1 µs. Therefore, 1 0.5 × × 0.1 × 10 −9 × 5 100 IL = = 2.5 × 10 −6 = 2.5 μA 1 × 10 −6
where τ = RC and b is feedback factor which is given by V+ 1 = V− 2
15. Topic: Operational Amplifiers: Characteristics and Applications (a) We know that 2π fVm Slew rate = ωVm = 2π fVm V/s = V/μs 106 Therefore,
Therefore, 1⎞ ⎛ 1+ ⎜ 2 ⎟ = 2τ ln 3 T = 2τ ln ⎜ 1⎟ ⎜1− ⎟ ⎝ 2⎠ 13. Topic: 8085 Microprocessor (c ) For the given program for 8085 Microprocessor; First loop Accumulator has (10)H and B has (10)H Then, ADD B in accumulator Accumulator has (20)H 0
RLC
0
0
1
1
0
0
0
0
0
0
0
0
Second loop Now, accumulator has (20)H and B has (10)H Then, ADD B in accumulator ACC RLC
Ch wise GATE_EE_Ch9.indd 467
0 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0
0000 0000 0000 0000
Q2 t
where Q2 = 0.5 ×
⎛1+ b ⎞ T = 2τ ln ⎜ ⎝ 1 − b ⎟⎠
b=
467
0
0
0
f = =
Slew rate × 106 2π × Vm 62.8 × 106 = 1 MHz 2π × 10
16. Topic: Characteristics of Diodes, BJT, MOSFET (c) Given that, Pinch voltage, VP = −5 V Gate-to source voltage, VGS = −3 V Transconductance, gm = 1 mA/V The transconductance of n-channel JFET is given by
gm = 1 mA/V =
−2 I DSS VP
⎛ VGS ⎞ ⎜⎝1 − V ⎟⎠
−2 I DSS −5
⎛ −3 ⎞ ⎜⎝1 − ⎟⎠ −5
P
2
⎛ 5⎞ I DSS = ⎜ ⎟ = 6.25 mA ⎝ 2⎠
11/21/2018 6:12:40 PM
468
GATE EE Chapter-wise Solved Papers
We know that transconductance is maximum when VGS = 0. Therefore, ( g m ) max =
=
−2 I DSS VP −2 × 6.25 = 2.5 mA/V −5
17. Topic: Combinational and Sequential Logic Circuits (a) Characteristic equation for D flip-flop Qn+1 = D So, frequency of signal available at Q is 10 kHz. 18. Topic: Simple Diode Circuits: Rectifiers (a) The given circuit is half-wave rectifier. So, the DC output current is given by, Vm i (t ) = π ( RD + RL )
1 1 1 0 0 ……. 0 1 1 1 0 1 ……. 1 ⇒ E000 – EFFF 21. Topic: Operational Amplifiers: Characteristics and Applications (*) Insufficient Data. 22. Topic: Combinational and Sequential Logic Circuits (b) For the given NAND logic circuit, the final output P ⋅Q
P Q
PQ
R ⋅S
R S
R ⋅S
Given forward resistance of diode is 5 W and load resistance = 45W, therefore, i (t ) =
Y = ( PQ ).( RS )
Vm V = m π (5 + 45) π (50)
19. Topic: Simple Diode Circuits: Clipping (b) In the positive half cycle, at peak input voltage, the circuit will be as shown below,
P Q ⋅ RS =P+Q+R+S
=P+Q+R+S
23. Topic: Amplifiers: Biasing (b) Given that, hFE = 99; VBE = 0.7; and we need to determine the collector current. In the given circuit, 3.3 kΩ
1 kΩ IC + 33 kΩ
+ 3.3 V 10 V −
1 kΩ Vo
12 V IB
−
0.7 V
4V L1
Therefore, Vo = 3.3 + 0.7= 4 V Now, for negative half cycle, the circuit will be
Applying KVL to the loop 1, we have
1 kΩ + + 1 kΩ Vo
10 V −
Therefore, Vo =
− 1 × 10 = 5 V 1+1
20. Topic: 8085 Microprocessor: Architecture (a) According to the given logic circuit, A15 A14 A13 A12 A11 ……. A0
Ch wise GATE_EE_Ch9.indd 468
3.3 kΩ
4 − (33 × 103 ) I B − VBE − (3.3 × 103 ) I E = 0 4 − (33 × 103 ) I − V − (33.3 × 103 ) I E = 0 4 − (33 × 103 ) I B − 0.7 −B(3.3 BE × 10 ) ( hFEE + 1) I B = 0 4 − (33 × 103 ) I B − 0.7 − (3.3 × 103 ) ( hFEE + 1) I B = 0 ⎛ IE ⎞ ⎛⎜∵ I E = 1 + hFE ⎞⎟ ⎝ I ⎠ ⎜∵ I B = 1 + hFE ⎟⎠ B 3 3⎝ 4 − (33 × 10 ) I B − 0.7 − (3.3 × 10 ) (99 + 1) I B = 0 4 − (33 × 103 ) I B −30.7 − (3.3 × 1033) (99 + 1) I B = 0 4 − 0.7 = (33 × 10 ) I B + (3.3 × 10 ) 100 I B 3 4 − 0.7 = (33 × 103 ) I3 B + (3.3 × 10 ) 100 I 3.3 = I B [33 × 10 + 3.3 × 103 × 100] B 3 3 3.3 = I B [33 × 10 + 3.3 × 10 × 100] 3.3 IB = 3 3 3.3 I B = [33 × 103 + 3.3 × 103 × 100] [33 × 10 + 3.3 × 10 × 100]
11/21/2018 6:12:41 PM
Chapter 9 • Analog and Digital Electronics
Since, IC = IBhFE, substituting values of IB and hFE we have IC =
3.3 mA (0.33 + 3.3)
24. Topic: Operational Amplifier: Characteristics and Applications (d) For the given circuit:
25. Topic: Multiplexer (c) Output of MUX is the sum of three inputs. This shows that MUX implements full adder. Therefore, Sum = P ⊕ Q ⊕ Cin Truth table is
Let V1 be the voltage at positive terminal and V2 be the voltage at negative terminal. By applying nodal method at negative terminal, we have V2 − Vin V2 − Vout + =0 R1 R1 V2 Vin V2 Vout − + − =0 R1 R1 R1 R1 2V2 Vin + Vout = R1 R1
(1)
With positive terminal voltage V1, we have. V1 − Vin V −0 + 1 =0 R (1/jωC ) V1 Vin − + V1 ( jωC ) = 0 R R V1 (1 + jωCR) Vin = R R Vin = V1 (1 + jωCR)
V1 =
469
P
Q
Cin
Output
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
Therefore, Sum = PQCin + PQCin + PQCin + PQCin (1) The exact solution is (2)
Vin (3) 1 + jω CR
Substituting Eq. (3) in Eq. (1) and using that for an ideal op-amp V1 = V2, we get ⎛ Vin ⎞ = Vin + Vout 2⎜ ⎝ 1 + jω CR⎟⎠ ⎡ ⎤ 2 − 1⎥ = Vout Vin ⎢ + ω 1 j CR ⎣ ⎦ Vout 2 − 1 − jω CR 1 − jω CR = = Vin 1 + jω CR 1 + jω CR
S = PQI 0 + PQI1 + PQI 2 + PQI 3 (2)
From Eqs. (1) and (2), we have I 0 = Cin ; I1 = Cin ; I 2 = Cin ; and I 3 = Cin So, I 0 = I 3 = Cin and I1 = I 2 = Cin 26. Topic: 8085 Microprocessor: Programming and Interfacing (d) 27. Topic: Operational Amplifier: Characteristics and Applications (d) In the given circuit, for an ideal op-amp V1 = V2 = 0 10 kΩ
Vx 10 kΩ
Therefore, 1 kΩ
θ = tan ( −ω CR) − tan (ω CR) −1
−1
= 180° − tan −1 ω CR − tan −1 ω CR = π − 2 tan −1 ω CR R
1 kΩ
V1
Vin V2
− + Vout
Hence, θ max = π (or)180°
Ch wise GATE_EE_Ch9.indd 469
11/21/2018 6:12:43 PM
470
GATE EE Chapter-wise Solved Papers
Applying nodal analysis:
Substituting values,
At the node 1, we have 10 × 103 =
0 − Vin 0 − Vx + =0 1 10
Vin Vx V + = 0 ⇒ Vin = − x 1 10 10 At the node 2, we have
= (1)
Vin = −
T1 = 0.75 T 0.693( RA + RB )C 0.75 = 0.693( RA + 2 RB )C (2)
V Vx = − out 10 120
Vout = −120 Vin
28. Topic: Amplifiers: Biasing (a) From the given diagram, VCC = 40 V From the given data bforced = 10 , therefore, transistor is in saturation. V − VCE I C = CC RC
40 − 0.2 ≈ 4A 10
29. Topic: VCOs and Timers (c) Given, that C = 10 nF, frequency f = 10 kHz and duty cycle = 0.75. From the circuit we have that the capacitor gets charged in RA + RB and discharges in RB. Therefore, Charging time T1 = 0.693 (RA + RB)C Discharging time T2 = 0.693 RA . C We know that f =
Ch wise GATE_EE_Ch9.indd 470
RA + RB 3 = RA + 2 RB 4 4 RA + 4 RB = 3RA + 6 RB 4 RA − 3RA = 6 RB − 4 RB RA = 2 RB
(2) Substituting Eq. (2) in Eq. (1), we get 2 R B + 2 RB = 14.4 × 103 4 R B = 14.4 × 103 R B = 3.6 kΩ RA = 2 × 3.6 kΩ = 7.2 kΩ 30. Topic: A/D and D/A Converters
We know that for saturation, VCE ≈ 0.2 V Therefore, IC =
Substituting C = 10 × 10−9, we get
Duty cycle D is given by
12(Vx ) − Vout = 0 Vout 12 Substituting Eq. (2) in Eq. (1), we get
1 0.693RA C + 0.693RBC + 0.693RBC
RA + 2 RB = 14.4 × 103 (1)
Vx − 0 Vx − Vout Vx − 0 + + =0 10 10 1 Vx Vx Vout + − + Vx = 0 10 10 10 Vx + Vx − Vout + 10Vx = 0 Vx =
1 0.693( RA + RB )C + 0.693RBC
1 1 = T T1 + T2
(b) For the given dual slope A/D converter, the number of bits n = 10 and clock frequency fclk = 1 MHz = 106 Hz. The maximum frequency fm of input in the given A/D converter is fm =
1 Tm
The maximum frequency and clock frequency are related as Tm = 2nTclk Therefore, fm = =
f clk 2n
⎛ 1 ⎞ ⎜⎝∵ Tclk = f ⎟⎠ clk
106 ≈ 1 kHz 210
11/21/2018 6:12:45 PM
Chapter 9 • Analog and Digital Electronics
31. Topic: Combinational and Sequential Logic Circuits (b) The K-Map with X, Y, Z for the given Boolean expression is
X
Y
Qn
Qn +1
0
0
0
1 →0
471
YZ 00
01
11
10
0
0
1
1 →1
0
0
1
0
1
0
1
0
0
1
0
1
1
1
0
1
1
1 →3
1
0
0
1 →4
1
0
1
0
1
1
0
0
1
1
0
0
X
From the K-map XYZ = 010; XYZ = 001; XYZ = 110; XYZ = 101; XYZ = 111 Therefore, the simplified function is Y Z + Y Z + XZ 32. Topic: Combinational and Sequential Logic Circuits (b) For the given shift register, X3
X2
X1
Karnaugh map YQn X
X0
Clk 1
0
1
1
X
1
1
Therefore,
X
Qn +1 = YQn [ X + X ] + XQn (Y + Y ) (∵ A + A = 1) (1)
= YQn + XQn
Replacing X = J and Y = K, we have
The truth table is Clock pulse I/P _ 1 1 0 2 0 3 0 4 1 5 0 6 1 7 1
YQn
1
Y
X = X 0 ⊕ X1 Y = X2
Qn +1 = KQn + JQn
(2)
From Eqs. (1) and (2), we get
Shift register 1010 1101 0110 0011 0001 1000 0100 1010
Thus, content of the shift register will become 1010 again on the 7th clock pulse. 33. Topic: Combinational and Sequential Logic Circuits (d) For the given J-K flip-flop, the characteristic table is
Ch wise GATE_EE_Ch9.indd 471
X
YQn
0
I/P
Z
Y Qn YQn
J = Y; K = X Alternately X
Y
Qn+1
J
K
0
0
1
1
0
0
1
Qn
0
0
1
0
Qn
1
1
1
1
0
0
1
− Clearly J matches Y and K matches X. 34. Topic: 8085 Microprocessor: Architecture (a) Given that number of memory chips = 8; number of address lines = 12 and number of data lines = 4. Therefore, total memory size is
11/21/2018 6:12:46 PM
472
GATE EE Chapter-wise Solved Papers
Therefore,
= ( 212 × 4) × 8 bits = ( 212 ⋅ 22 ) × 8 = 2 × 8 = 2 bytes = 16 kbytes 24
24
I R = I C1 + I B1 + I B2 I R = I C1 + I B1 + I B2 = I C2 + 2 I B2 = I C2 + 2 I B2 I = I C2 + 2 ⋅ IC2C2 = I C2 + 2 ⋅ b b = I C2 (1 + 2 / b) = I C2 (1 + 2 / b)
(∵ 8 bit = 1 bytes)
35. Topic: 8085 Microprocessor: Programming and Interfacing (c) The execution flow is given as follows → LX I H, 1FFE H = (1 F)H L = (FE)H → MOV B, M B = 1FFE → INR L L = L + (1)H = (FE)H + 1H = (FF)H → ADD B A=A+B → INR L L = L + (1)H = (FF)H +(1)H = 00 → MOV M, A 1F00 = A → XOR A A = A XOR A = 0 Therefore, exact address in 1F00. 36. Topic: Characteristics of Diodes, BJT, MOSFET (c) From the given figure, we have that RZ = 0.1 kW; VZ = 3.3 V and V1 = 3.5 V R1 ⇒ V1 = 3.5 V By nodal method, IZ =
V1 − VZ 3.5 − 3.3 = = 2 mA RZ 0.1 × 103
37. Topic: Amplifiers: Equivalent Circuit and Frequency Response (c) For the given circuit, IR
+3 V
As, IR =
(∵ I C1 = I C2 and I B1 = I B2 ) (∵ I C1 = I C2 and I B1 = I B2 ) (∵ b = I C /I B ) (∵ b = I C /I B )
−5 + 0.7 = −4.3 mA 103
Substituting values, we have I C2 =
+4.3 = 4.2914 ≈ 4.3 mA 2 1+ 1000
38. Topic: Oscillators and Feedback Amplifiers (b) Feedback circuit samples the output voltage. As IFB is fed to the output, input signal is in shunt connection with it. Therefore, it has shunt-shunt configuration. 39. Topic: Amplifiers: Biasing (d) 40. Topic: Combinational and Sequential Logic Circuits (a) When the input is 1, output is stable. When the input is 0, output is stable. Therefore, it is a bistable multivibrator. 41. Topic: A/D and D/A Converters (a) We know that for n-bit quantizer, number of levels = 2n. As +Vsat and −Vsat indicates two levels, 2n = 2 ⇒ n = 1 42. Topic: Simple Diode Circuits: Rectifiers (c) For the given circuit, we can see that diode D2 is always forward biased and let us assume D1 is reverse biased.
1 kΩ
1 kΩ
I IC1
1 kΩ
IB1+IB2 b = 1000 IB1
V1
IB2 5V
8V V2
−5 V VBE 1 = VBE2 = VBE
Ch wise GATE_EE_Ch9.indd 472
Let the voltages at the two terminals of D1 be V1 and V2. Since D1 is reverse biased, V1 < V2. Applying nodal analysis
11/21/2018 6:12:47 PM
Chapter 9 • Analog and Digital Electronics
V1 − 5 V1 − ( −8) + =0 1 1 V1 − 5 + V1 + 8 = 0
Vo ( s) = Vi
AV ( s) =
V2 = 0
Therefore, V1 < V2 and the assumption that D1 is reverse biased is correct. Hence, ID1 = 0 mA
ω0 = Q =
+ r RB
gmv
V −
RC E
Rin
rd =
Rin = RB || rd =
Ix =
Vx − Vy
10 × 10 = 5 kΩ 10 + 10
5×5 = 2.5 kΩ 5+5
Vx − Vy
(1)
1 × 106 Vx = V1 = V2
100 × 10
3
+
Vx − 0 =0 10 × 103
Vx − Vy + 10Vx = 0 ⇒ 11Vx = Vy (2)
Substituting from Eq.(2) in Eq.(1), we have Ix =
44. Topic: Amplifiers: Biasing (d) From the given circuit
Vx − 11Vx −10Vx = 106 106
Then impedance
VSD (sat) = VSG + VT Therefore,
10 −9 × 2 = 10000 rad/s 200 × 103 × 10 −18
46. Topic: Operational Amplifier: Characteristics and Applications (b) For ideal op-amp,
b 50 = = 5 kΩ g m 10 × 10 −3
RB = 10 kΩ || 10 kΩ =
(C1 + C2 ) RC1C2
g m rd = b
(2)
2
ω 0 C1 + C2 = Q RC1C2
VO
C
(ω 0 /Q ) A0 s s + (ω 0 /Q ) s + ω 02
Comparing Eqs. (1) and (2), we have
43. Topic: Amplifiers: Equivalent Circuit and Frequency Response (d) Given that gm = 10 ms. From the following small signal AC equivalent circuit, we have B
(1)
2
2V1 = −3 ⇒ V1 = −1.5
VS
s/R2C1 C1 + C2 1 s + s+ R1C1C2 R1 R2C1C2
473
4SG− 0+ V =T4 V VSDV(sat) SG = V VSD(sat) V ==44−−10==3 4VV SG
VSD(sat) = 4 − 1 = 3 V
Rin =
Vx Vx −106 = = = −100 kΩ I x −10Vx /106 10
47. Topic: Schmitt Trigger (b) From the given circuit, we have +5 V
We know that for PMOS in saturation region, VSD < VSD(sat), therefore 200 Ω
VSD < 3 ⇒ VS − VD < 3
Vo
4 − I D R < 3 ⇒ − I D R < −1 ID R > 1 ⇒ R >
1 10 −3
R > 1000 Ω 45. Topic: Simple Active Filters (*) From the given circuit, we have R1 = 200 kW, C1 = C2 = 1 nF. The voltage gain is given by
Ch wise GATE_EE_Ch9.indd 473
Q1
Q2
Vi = 0
1 kΩ
1.25 mA
11/21/2018 6:12:49 PM
474
GATE EE Chapter-wise Solved Papers
Q1 is in cut-off region and Q2 is in saturation region. Therefore,
50. Topic: Combinational and Sequential Logic Circuits (b)
5 − I C RC − VCE(sat) − 1.25 = 0
Clk
5 − I C RC − 0.1 − 1.25 = 0 5 − I C RC = 1.35
(1)
Vo = VCC − I C RC
Q
From Eq. (1) Vo = 5 − I C R C = 1.35 V
O/P = Clk (AND) Q
48. Topic: Combinational and Sequential Logic Circuits (a) We have Y = ( A ⋅ BC + D )( A ⋅ D + B ⋅ C ) = ( ABCD ) + ( ABC BC ) + ( AD ) + BCD = ABCD + 0 + AD + BCD = AD( BC + 1) + BCD = AD + BCD
(∵ BC + 1 = 1)
49. Topic: Combinational and Sequential Logic Circuits (d) Output is given by, Y = ( A0 ⊕ B0 ) ⋅ ( A1 ⊕ B1 ) ⋅ ( A2 ⊕ B2 ) ⋅ ( A3 ⊕ B3 ). A ⊕ B = AB + AB A⋅ B = A ⊕ B Comparing the numbers A0 A1 A2 A3 , B0 B1 B2 B3 with given options, we have For option (a) 1010,1010
Y = (1 ⊕ 1) ⋅ (0 ⊕ 0) ⋅ (1 ⊕ 1) ⋅ (0 ⊕ 0) = 0⋅0⋅0⋅0 = 1
51. Topic: 8085 Microprocessor: Programming and Interfacing (c) The execution flow for the given programme START MVI A, 14H: Move 14 H to register A [Instruction cycle = 1] RLC: Rotate accumulator left without carry Executed until accumulator becomes 0. [Instruction cycle = 6] JNZ : Jump on non- zero] Executed until accumulator becomes 0 [ Instruction cycle = 5] HALT [Instruction cycle = 1] Therefore, number of instruction cycles = 1 + 6 + 5 + 1 = 13 52. Topic: Simple Diode Circuits: Rectifiers (a) For the given circuit, let D1 be forward biased and D2 be reversed biased (open).
For option (b) 0101, 0101
= (0 ⊕ 0) ⋅ (1 ⊕ 1) ⋅ (0 ⊕ 0) ⋅ (1 ⊕ 1) = 0⋅0⋅0⋅0 = 1
For option (c) 0010, 0010
= (0 ⊕ 0) ⋅ (0 ⊕ 0) ⋅ (1 ⊕ 1) ⋅ (1 ⊕ 0) = 0 ⋅ 0 ⋅ 0 ⋅1 = 0
Alternately
For the NOR gate, output to be zero; atleast one input must be 1.
XOR output is 1 only if inputs do not match. This is possible only with option (d).
Ch wise GATE_EE_Ch9.indd 474
2 kΩ
D2
2 kΩ
1 mA = (0 ⊕ 0) ⋅ (0 ⊕ 0) ⋅ (1 ⊕ 1) ⋅ (0 ⊕ 0) =1
V1
For option (d) 0010, 0011
D1
V2 Then V1 = 1 × 2 = 2V V2 = 0 ⇒ V1 > V2 Therefore, ID2 = 0 mA
11/21/2018 6:12:51 PM
Chapter 9 • Analog and Digital Electronics
53. Topic: Characteristics of Diodes, BJT, MOSFET (d) Given that Vth = 1 V and VGS = 2 V
59. Topic: Operational Amplifier: Characteristics and Applications (c) For an inverting amplifier,
VDS(sat) = VGS − Vth = 2 − 1 = 1 V
Gain AV =
⇒VDS > VDS(sat) (Saturation region) High current flows from drain to source, therefore Vab = 0
R1 =
55. Topic: 8085 Microprocessor: Programming and Interfacing (c)
Q(t + 1) = D Q(t + 1) = Q(t ) ⊕ X = Q (t ) X + Q (t ) X
− RF R1
The voltage gain is to lie between −10 and −25, so For AV = −10
54. Topic: Characteristics of Diodes, BJT, MOSFET (a)
56. Topic: Combinational and Sequential Logic Circuits (c) For the given digital circuit, let Q(t) be the present state and Q(t + 1) the next state. For D to be flip-flop, D = Q(t) ⊕X
475
−106 = 100 kΩ −10
For AV = −25 R1 =
−106 = 40 kΩ −25
Since R1 should be large, R1 = 100 kW 60. Topic: Amplifiers: Equivalent Circuit and Frequency Response (b) Direct coupled amplifier provides gain at dc (or) very low frequency.
[∵ A ⊕ B = AB + AB ]
If X = 0; Q(t + 1) = Q(t ) Gain
If X = 1; Q(t + 1) = Q(t ) Therefore, it is a T flipflop.
Frequency 57. Topic: A/D and D/A Converters (b) Resolution =
3.5 3.5 ⇒ 2n = = 250 n 2 14 × 10 −3
Therefore, n = 8 (as 28 = 256, so closest value). Alternately ⎡ Resolution ⎤ Bit size = 2 ⎢ ⎥ ⎣ Full scale O/P ⎦
61. Topic: Characteristics of Diodes, BJT, MOSFET (d) Ideal op-amp goes into saturation region between +V and −V. The input is given to negative terminal, therefore during positive half cycle it saturates to −V and vice versa.
⎡ 14 ⎤ = 2⎢ ⎥ = 8 ⎣ 3.5 ⎦ Vout
t
58. Topic: Amplifiers: Biasing (b) In the active region
b= IB =
Ch wise GATE_EE_Ch9.indd 475
IE IB IE 1 × 10 −3 = = 10 μA 100 b
(∵ I C I E )
62. Topic: Characteristics of Diodes, BJT, MOSFET (b) Transconductance, gm =
∂id ( 2 − 1) mA = = 1 ms ∂ vgs ( 2 − 1) V
11/21/2018 6:12:52 PM
476
GATE EE Chapter-wise Solved Papers
63. Topic: Characteristics of Diodes, BJT, MOSFET (d) Small signal equivalent circuit, G
67. Topic: Simple Diode Circuits: Rectifiers (a) Let D2 be OFF, then the circuit is 1Ω
D Vo 1Ω
Vin
gmvgs
+ VD2 −
RD = 10 kΩ 10 V
1Ω
D1
D3
5A
S D1 is ON because it is connected with 10 V source, then the circuit is
Vo = − g m vgs RD Vo = − g m RDVin
(∵ vgs = Vin )
Voltage gain is given by V AV = o = − g m RD = −( 1 ms)(10 kΩ) = −10 Vin
1Ω 1Ω
10 V 64. Topic: Combinational and Sequential Logic Circuits (b) The truth table for the given output is X1
X2
Q
1
0
1
0
0
1
0
1
0
D3
5A
Since, we have assumed D2 to be OFF, VD2 ≤ 0 which is possible only if D3 is in OFF state. Therefore, D1 is ON and D2 = D3 = OFF
For the circuits given in: Option (a): If X1 = 1, Q = 0 Option (b): If X1 = 1, and X2 = 0, Q = 1 Option (c): If X1 = 0, Q = 0 Option (d): If X1 = 0, Q = 1 Hence, the circuit corresponding to the given output is that given in option (b). 65. Topic: Combinational and Sequential Logic Circuits (d) In the given circuit, Q1 and Q3 are inverter circuits; Q4 and Q5 are OR circuits and Q2 is output inverter. Therefore, Q = X1 + X 2 = X1 ⋅ X 2 66. Topic: Combinational and Sequential Logic Circuits (d) Given that X1 = 1 and X2 = 1. The output Q is unstable.
68. Topic: Operational Amplifier: Characteristics and Applications (d) During positive half cycle D1 will be reverse biased. There is no feedback to op-amp, hence it enters into saturation region. Therefore, VP = −12 V. During negatives half cycle D1 will be forward biased. Therefore, VP = Vγ + V− For virtual ground: V+ = V− = 0 and VP = Vγ = 0.7 V . Therefore, the voltage waveform is as shown in option (d). 69. Topic: Characteristics of Diodes, BJT, MOSFET (a) In the given circuit When Vi < 10 V; D1 and D2 are OFF. Then Vo = 10 V. When Vi > 10 V; D1 is ON and D2 is OFF. Then Vo = Vi Hence, the transfer characteristics for the given circuit are as shown in option (a). Vo
Q(t+1) = Q(t)
X1 = 1
1Ω
D1
Q(t) 10 X2 = 1 Q(t) Vi 10
Ch wise GATE_EE_Ch9.indd 476
11/21/2018 6:12:54 PM
Chapter 9 • Analog and Digital Electronics
70. Topic: Amplifiers: Biasing (b) Let the BJT be in the active region:
477
71. Topic: Oscillators and Feedback Amplifiers (c) Given that Vo = ±12 V
+12 V 2.2 kΩ
Open
Rth C
15 kΩ Vin
−
Q
Vo = 12 V + 100 kΩ
-12 V Then 10 kΩ
Vth − Vin Vth − ( −12) + =0 15 100 20 Vth − 20Vin + 3Vth + 36 = 0 23Vth = −36 + 20 × 5 Vth = 2.78 V Rth =
15 kΩ × 100 kΩ = 13.04 kΩ 15 kΩ + 100 kΩ
By nodal method, when Vo = +12 V Vth − 12 Vth − 0 + =0 10 10 2Vth = 12 ⇒ Vth = 6 V When Vo = −12 V
Therefore, the circuit becomes,
Vth + 12 Vth − 0 + =0 2 10 Vth = −10 V
2.2 kΩ 13.04 kΩ
b = 30
-10
Vth = 2.78 V
2.78 − Rth I B − 0.7 = 0 2.78 − (13.04 × I B ) − 0.7 = 0 I B = 0.157 mA Saturation current is, I B(sat) = I C(sat) I B(sat)
I C(sat)
b 12.2 = = 5.4 mA 2.2 5.47 mA = = 0.181 mA 30
Hence, IB(sat) > IB and the assumption that the transistor is operating in the active region is correct.
Ch wise GATE_EE_Ch9.indd 477
6
72. Topic: Combinational and Sequential Logic Circuits (c) In the given circuit: For transistor Q1 VCE = VC − VE = VC − 3 (1) For transistor Q2 VBE = VB − VE = 0.7 V ⇒ VB = 0.7 V (∵VE = 0 V ) Also, for Q2 VC = 0.7 V Substituting in Eq. (1), we have VCE = 0.7 − 3 = −2.3 Therefore, transistor Q1 is reverse ON and transistor Q2 is always ON.
11/21/2018 6:12:55 PM
478
GATE EE Chapter-wise Solved Papers
73. Topic: A/D and D/A Converters (c) For the 3-bit binary down counter, the output is 111 110 101 100 011 010 001 000 The output achieved when connected to R-2R ladder type DAC is 7 3 5 1 6 2 4 0 The correct expected sequence is 7 6 5 4 3 2 1 0 Thus, the counter and DAC are working properly but the connection from counter to DAC is not proper.
H ( s) =
1 / sC 1 = R + 1 / sC 1 + RCs
First order RC filters have a roll-off of -20 dB/decade. 78. Topic: A/D and D/A Converters (b) Here, n = 8 since it is 8-bit ADC. Thus, SNR of an ADC is given by: 6.02n + 1.76 dB = 6.02 × 8 + 1.76 = 49.92 ≈ 50 dB For frequency:
74. Topic: Multiplexer (a)
⎡ 1 ⎤ 20 log ⎢ ⎥ = 50 dB 2 ⎣ 1+ ω ⎦ 20 − log ⎡⎣1 + ω 2 ⎤⎦ = 50 dB ⇒ −10 log ω 2 = 50 dB 2 50 − 20 log ω = 50 dB ⇒ log ω = = 2.5 decades 20
F ( I 0 S1S0 ) = I 0 S1S0 + I1S1S0 + I 2 S1S0 + I 3 S1S0 F ( A, B, C ) = ABC + ABC + 1⋅ BC + 0 ⋅ BC = ABC + ABC + BC = ABC + ABC + BC ( A + A)
[∵ A + A = 1]
= ABC + ABC + ABC + ABC = Σ (1, 2, 4, 6)
79. Topic: Amplifiers: Biasing (d) By applying KVL to the circuit, we have
75. Topic: 8085 Microprocessor: Programming and Interfacing (a) From the given programme: MVI H⎫ ⎬ Stores 255 in H and L registers. MVI L⎭ DCR L [Decrements L by one]. JNZ [Checks whether value has become 0]. Therefore, DCR L is executed 255 times as when content of L becomes 0, the content of H starts decrementing by 255 times.
10 − I C (103 ) − 0.7 − 270 × 103 I B = 0
b I B + 270 I B = 9.3 mA IB =
9.3 = 0.025 mA (∵ b = 100) 270 + 100
Base current (saturation): 10 − I C (1) − VCE − I E (1) = 0 (1)
Substituting I C = I E and VCE 0 in Eq. (1), we have
76. Topic: 8085 Microprocessor: Programming and Interfacing (a) XCHG → Exchanges the content of DE register pair with H L register pair. INR M → Increments the content of memory. Therefore, the correct sequence of instruction is XCHG INR M 77. Topic: A/D and D/A Converters (a) Consider the following equivalent circuit
(∵ I C = b I B )
10 − 2 I C(sat) = 0 I C(sat) = 5 mA Therefore, I B(sat) =
I C (sat)
b 5 = = 0.5 mA 100
Since IB < IB(sat), the transistor is operating in the active region.
R C
Ch wise GATE_EE_Ch9.indd 478
80. Topic: Characteristics of Diodes, BJT, MOSFET (b) Let us assume that the transistor operating in active region. From the figure we have
11/21/2018 6:12:57 PM
Chapter 9 • Analog and Digital Electronics
83. Topic: Operational Amplifier: Characteristics and Applications (b) Switch is closed at initial position. Capacitor will charge from 0 V to 20 V. Voltage across the capacitor is,
VBE(ON) = 0.6 V VB − VE = 0.6 6.6 − VE = 0.6 ⇒ VE = 6 V Appling KCL at the emitter terminal, IE = IB + IC
VC (t ) = VC (∞) + [VC (0) − VC (∞)]e − t / RC
6 − 6.6 6 + 1 × 103 10 0.6 6 = + = 0.6 A 1000 10 =
Substituting VC (0) = 0 V and VC (∞) = 20 V, we have
VC (t ) = 20 + [0 − 20]e − t / RC = 20(1 − e − t / RC )
(1)
At the positive terminal,
VCE = VC − VE = 10 − 6 = 4 V
V+ − Vout V+ − 0 10 + = 0 ⇒ V+ = Vout 10 100 11
Therefore, power dissipated is, VCE × I C = 4 × 0.6 = 2.4 W
As the given diode in the circuit is the Zener diode, output will be the range of −5 V to +5 V. Therefore,
81. Topic: Operational Amplifier: Characteristics and Applications (d) The given circuit is a V–I converter. Thus, the output current always depends on input voltage. We know that for an ideal op-amp: V1 = V+ = V−
V+ =
10 (5) 11
(2)
From Eqs. (1) and (2), we get
(
)
10 × 5 11 10 × 5 = 20 × 11 17 = 22
20 1 − e − t / RC =
Therefore, V1 − V V1 − 0 + =0 R1 R2
1 − e − t / RC
V1 V V1 − + =0 R1 R1 R2
e − t / RC Hence,
⎡ R + R2 ⎤ V V1 ⎢ 1 ⎥= ⎣ R1 R2 ⎦ R1 ⎡ R + R2 ⎤ V1 = V ⎢ 1 ⎥ ⎣ R2 ⎦
−1
For ideal op-amp, I L = I1 =
V1 V ⎡ R2 ⎤ = ⎢ ⎥ r r ⎣ R1 + R2 ⎦
82. Topic: VCOs and Timers (b) Given that the maximum and minimum voltage is +7 V and −7 V, f = 1 kHz and duty cycle D = 50%. The given circuit is a class B amplifier, therefore, efficiency is given by
η=
π Vp 4 VCC
Here, Vp is the peak input voltage = 7 V and VCC is the supply voltage = 10 V. Substituting values, we have
η=
Ch wise GATE_EE_Ch9.indd 479
479
π 7 ⋅ = 55% 4 10
⎛ 22 ⎞ t = RC ln ⎜ ⎟ =1 × 103 × .01 × 10 −6 × 0.256 = 2.57 μs ⎝ 17 ⎠ 84. Topic: VCOs and Timers (b) In the given astable multivibrator, charging of capacitor takes places through RA + RB and discharge takes place through RB with charging (Tch) and discharging (Tdis) time given by Tch = 0.693 (RA + RB)C Tdis = 0.693 (RB)C However, in the given circuit, during charging diode will be forward biased, therefore, charging of capacitor takes place through RA only and the discharge through RB. Given RA = RB, therefore Tch = Tdis 85. Topic: Characteristics of Diodes, BJT, MOSFET (c) For the positive half cycle of the input, that is 180°, the p-side of the diode is connected to a positive input, hence the diode conducts. Further, a pure inductor lags 90° behind the given input. Hence, the diode conducts for 270° as given in the plot below
11/21/2018 6:12:58 PM
480
GATE EE Chapter-wise Solved Papers
90. Topic: 8085 Microprocessor: Programming and Interfacing (a)
Vi
E 91. Topic: Characteristics of Diodes, BJT, MOSFET (a) When the diodes are in the reverse bias condition
Conduction for 270°
Vo = 86. Topic: Combinational and Sequential Logic Circuits (d) Output is given as, Y = ( A ⊕ B ) ⊕ (C ⊕ D ) For even number of 1’s we have: (i) A B C D (ii) A 1
1
1
1
0
B
C
D
0
1
1
10 sin ω t × 10 = 5 sin ωt 10 + 10
92. Topic: A/D and D/A Converters (a) In case of single slope integrating type and successive approximation type ADCs, the time for conversion of analog to digital signal is independent of the input voltage. So, the time of conversion for 2.5 V input will also be TA and TB. 93. Topic: 8085 Microprocessor: Programming and Interfacing
0
0
0
0 For odd number of 1’s we have B C D (ii) A (i) A 0 1 1 1 0
0
0 0 B
C
D
1
1
1
1
1
0
(a) The sequence of instruction is LXI H, 2500 H (Starting address) MOV A, M (Moves the data from memory to accumulator) 94. Topic: Amplifiers: Biasing (c) Given that VDD = 0.7 V. For current mirror circuit,
IC
1
1
Therefore, the output will be 1, if number of 1’s in the input are odd. 87. Topic: Combinational and Sequential Logic Circuits (b) For the given hexadecimal number AB.CD, the binary equivalent is A B C D 1010 1011 1100 1101
5V 1 kΩ
VB Q1
Q2 IB1
IB2
Thus (AB.CD)16 = (1010 1011 1100 1101)2 Octal equivalent is, 010 101 011 110 011 010 = ( 253.632)8 2
5
3
6
3
88. Topic: 8085 Microprocessor: Programming and Interfacing (a) 89. Topic: 8085 Microprocessor: Programming and Interfacing (d) 11010110 is the 2’s complement of 00101010 = (42)10. 42/2 = 21 = (00010101)2 In 2’s complement format: 00010101 → 11101011
Ch wise GATE_EE_Ch9.indd 480
-5 V
2
Since, b of the transistors is high I C1 = I C2 and I B1 = I B2 VB = −5 + 0.7 = − 4.3 V Since, diode D1 is forward biased, I = I C2 I C1 =
0 − ( −4.3) = 4.3 mA 1 × 103
11/21/2018 6:13:00 PM
Chapter 9 • Analog and Digital Electronics
95. Topic: Characteristics of Diodes, BJT, MOSFET (d)
481
98. Topic: Simple Active Filters (c) At low frequencies,
In the positive half cycle, D1 is ON and D2 is OFF.
ω → 0,
In the negative half cycle, D1 is ON and D2 is ON. This implies that C1 charges up to +5V, so
1 =∞ ωc
VC 1 = 5 V
R2
Therefore, VC2 will charge up to −10 V in the opposite direction, so VC 2 = − 10 V
R1 96. Topic: Operational Amplifier: Characteristics and Applications (b) Let the input waveform be sinusoidal
−
Vi VA
Vo +
R3 R4
Vin t
On nodal analysis at the negative and positive terminals of the op-amp, we have
Based on the transfer characteristics, Output at rectifier P is
VA − Vi VA − Vo + =0 R1 R2
VP
Given that R1 = R2 = RA, therefore t VA Vi VA Vo − + − =0 RA RA RA RA
Output at rectifier Q is
2VA − Vi − Vo = 0
VQ
2VA = Vi + Vo
t
Similarly, VA − Vi VA − 0 + =0 R3 R4
Therefore, the net output will be sinusoidal. The output waveforms show that rectifier P is connected to the inverting terminal of op-amp and Q is connected to non-inverting terminal. 97. Topic: VCOs and Timers (a) Given that Vp-p = 5 V; Vi = 0 V. The shift in input voltage will produce a DC shift of 2.5 V to the triangular waveform and the output waveform will be:
5V
Using R3 = R4 = RB
2VA = Vi From Eqs. (1) and (2), we have
(2)
2VA = 2VA + Vo ⇒ Vo = 0 Therefore, low frequency signals will be stopped. For high frequencies,
2.5 V
ω → ∞,
Ch wise GATE_EE_Ch9.indd 481
(1)
1 = 0 ⇒ Vi = Vo ωc
11/21/2018 6:13:01 PM
482
GATE EE Chapter-wise Solved Papers
On obtaining complement of each term, we have,
R2
F = (Y + Z )( X + Z )( X + Y + Z ) 101. Topic: Combinational and Sequential Logic Circuits
R1 −
Vi R3
Vo
(b)
f =
+
TON2 1 , D= TON1 + TON2 TON1 + TON2
R4
Therefore, high frequency signals are allowed. Hence, it is a high-pass filter 99. Topic: Simple Active Filters (d) The cut-off frequency of a high-pass filter is
ω hpf =
1 2π RA C
For the given circuit (low-pass) I/P
HPF
ω lpf
LPF
102. Topic: 8085 Microprocessor: Programming and Interfacing (b) Given that SP = 2700 + 1 (Stack pointer) PC = 2100 H (Program counter) HL = 0000 H (HL register) Program execution proceeds as follows DAD SP (ADD SP to HL) HL = HL + SP HL = 0000 H + 2700 H = 2700 H PCHL (Load program counter with HL contents) PC = HL = 2700 H Therefore, content in PC = 2700 H and HL = 2700 H
O/P
1 = = 2ω hpf R 2π A ⋅ C 2
Initially, it allows high frequency signal to pass and at particular point it allows low frequency signal to pass. Therefore it is a band-pass filter and the gain vs. frequency characteristics are given by graph in option (d). 100. Topic: Combinational and Sequential Logic Circuits (b) From given logic gate diagram,
103. Topic: Characteristics of Diodes, BJT, MOSFET (a) When a forward biased diode is reverse biased, initially it conducts due to the flow of stored charges. After some time it goes into non-conducting state. So current characteristic of the diode during switching is as given. 104. Topic: Oscillators and Feedback Amplifiers (b) The equivalent circuit for the given amplifier is Ro Vi
Vout +
F = ∑ m(1, 3, 5, 6)
VE
= XYZ + XYZ + XYZ + XYZ Using K-map.
−
YZ Y Z YZ X 1 X X
YZ
+ −
AVi
YZ +
2 kΩ
1
1
Vfb 1
F = XZ + YZ + X YZ In product of sum form, F = (Y + Z )( X + Z )( X + Y + Z )
Ch wise GATE_EE_Ch9.indd 482
Ri
1 kΩ
−
From the circuit, it is clear that the feedback samples output voltage (Vfb) which adds as a negative voltage to the input. Hence, the op-amp circuit shows voltagevoltage feedback.
11/21/2018 6:13:03 PM
Chapter 9 • Analog and Digital Electronics
105. Topic: 8085 Microprocessor: Architecture (b) The increasing order of speed of data access is as given in option (b), that is, Magnetic tape CD ROM Dynamic RAM Cache memory Processor register 106. Topic: Combinational and Sequential Logic Circuits (c) Universal gates are NAND and NOR, that is, any number of logical circuits can be drawn with these gates. 107. Topic: Operational Amplifier: Characteristics and Applications (a) Given that R = 10 kW, C = 10 mF, f = 50 Hz and Vrms = 10 V. Let V1 be the voltage at positive terminal and V2 be the voltage at negative terminal. For ideal op-amp, V1 = V2 = VS. From the circuit, we have V2 − 0 V2 − Vo + = 0 ⇒ (1+ RCs)VS = Vo ( s) 1/Cs R [∵V1 = V2 = VS and Vo = Vo ( s)] Therefore,
483
Output power Pout = AVAIPin As current gain is >1, Pout > Pin 109. Topic: Operational Amplifier: Characteristics and Applications (d) The output for the given Schmitt trigger circuit for the two possible states is given by Vo(H) = 6 V Vo(H)[state =6V 1]
[state 1]
] [state 2] Vo(L) = −3 V Vo(L)[state = −32V At the non-inverting terminals, the threshold (upper) voltage is obtained as Vth − 6 Vth − 0 + =0 2 1 Vth = 2 V Similarly, at the inverting terminal, the threshold (lower) voltage is obtained as Vth1 − ( −3) Vth1 + =0 2 1 Vth1 = −1 V Voltage upper threshold limit = +2 V Voltage lower threshold limit = −1 V Thus for upper limit Vin > 2 V
Vout = +6 V
Vin < 2 V
Vout = −3 V
For lower limit
V − Vo IS = S R We have
Vin < −1 V
Vout = 6 V
Vin > −1 V
Vout = −3 V
VS (1+ RCs) = Vo ( s)
Vo
VS − Vo ( s) = RCs ⋅VS Therefore,
6 RCs ⋅VS R I S = jω C ⋅VS IS =
I S = 2π fC ⋅VS = 2π f × 10 × 10 × 10 −6
= 2π × 50 × 10 × 10 −6 × 10 = 10 π mA leading by 90° 108. Topic: Amplifiers: Biasing (d) For emitter follower Voltage gain (AV) = 1 Current gain (AI) > 1
Ch wise GATE_EE_Ch9.indd 483
(∵ s = jω ) (∵ ω = 2π f )
t -3 110. Topic: 8085 Microprocessor: Programming and Interfacing (d) When the given instructions are executed (i) XRA A: A⊕A ⇒ A = 0 (ii) MVI B, FOH: B = F00H (iii) SUB B: A=A−B
11/21/2018 6:13:04 PM
484
GATE EE Chapter-wise Solved Papers
Then 2’s complement is performed as B > A and added to A. A=0 0 0 0 0 0 0 0 B= 0 0 0 0 2s complement of ( − B ) = 0 0 0 1 0 0 0 0 A − B = A + ( − B) = 0 0 0 1 0 0 0 0 = 10 H
(i) Saves the contents of the program counter into the stack. This is executed by ((SP)) ← (PC) SP = SP − 2 (decrement) (ii) Stores address in program counter (PC) ←Addr Hence, correct sequence of operations is as given in option (d).
(2’s complements of B) 111. Topic: Operational Amplifier: Characteristics and Applications (b) For the given ideal op-amp,
115. Topic: Amplifiers: Biasing (a)
V+ = 2 V and V− = 2 V
+10 V
By nodal analysis, V− − 0 V− − Vo + =0 R 2R 2 2 − Vo + = 0 ⇒ Vo = 6 V R 2R
10 kΩ
50 kΩ 1
112. Topic: Characteristics of Diodes, BJT, MOSFET (b) In the given circuit, diode D2 is always OFF and diode D1 can be OFF or ON. The equivalent circuit is
Vo 100 Ω
D1 10 kΩ
In loop 1, Vo
10 V 10 kΩ
10 − (10 × 103 )I B − VBE − Vo = 0
(1)
Given that I C ≈ IE = 100 IB 10 Vo = × 10 = 5 V 10 + 10 113. Topic: Characteristics of Diodes, BJT, MOSFET (c) In the given composite switch, the diodes are connected in series, so current cannot be negative. The voltage across switch is zero when the current flows through the switch and can be positive or negative when current through the switch is zero. Hence, the I–V characteristic is as shown in option (c). 114. Topic: 8085 Microprocessor: Programming and Interfacing (d) The instruction CALL Addr performs two operations:
Ch wise GATE_EE_Ch9.indd 484
Vo = 100 IB 100 Therefore, IB =
Vo 10000
Substituting value of IB in Eq. (1), we have ⎛ V ⎞ 10 − (10 × 103 ) ⎜ o ⎟ − 0.7 − Vo = 0 ⎝ 10000 ⎠ Vo = 4.65 A
11/21/2018 6:13:05 PM
Chapter 9 • Analog and Digital Electronics
X
116. Topic: Combinational and Sequential Logic Circuits (a) In the given logic circuit,
X Y
X Y
X Y
485
F =X Y + YZ
Y
Y
B Z
Y = X ⊗B Each gate will have a propagation delay of 10 ns. Therefore, output waveform is
X1
t
0
B1 0
119. Topic: Operational Amplifier: Characteristics and Applications (d) The given circuit is the Schmitt trigger circuit in which there is more than one op-amp to produce square wave. (i) First half of the circuit is a differential amplifier with negative feedback Va = −Vi There is
Va = −Vi (ii) Phase shift is zero. (iii) Second op-amp has a positive feedback, so it acts as Schmitt trigger and since Va = −Vi, it is a non-inverting Schmitt trigger. Therefore,
t
12 = 6 V (upper threshold limit) 2 = −6 V (lower threshold limit)
Vth1 = Vth2
Y1 t
0
120. Topic: Combinational and Sequential Logic Circuits (a) Generally XOR operation is X X
117. Topic: Combinational and Sequential Logic Circuits
Y = XX + XX
(b) YZ 00
01
11
10
0
1
1
1
0
1
0
0
1
0
X
Y
For the given circuit, the second input is inverted, therefore Y = X X + XX = XX + XX
By grouping all pairs of 1s, we have X Y + YZ Therefore, F = X Y + YZ 118. Topic: Combinational and Sequential Logic Circuits (d) From the above solution, we have function F = X Y + YZ . Therefore, the circuit for realization is
Ch wise GATE_EE_Ch9.indd 485
= X + X =1 121. Topic: Multiplexer (c) 122. Topic: Amplifiers: Biasing (d) In the given Zener diode, the base-emitter and base-collector junctions are forward biased and hence operating in saturation condition. Therefore, we cannot use IC = bIB. So collector current is given by IC =
V − VCE 12 − 0.2 = = 5.36 mA RC 2.2 × 103
11/21/2018 6:13:07 PM
486
GATE EE Chapter-wise Solved Papers
123. Topic: Simple Diode Circuits: Clipping (c) Case (i): When −0.7 V < Vi < 5.7 V Here, the Zener diode will be in OFF state and normal diode will be in OFF state. Therefore, output will follow input.
126. Topic: Combinational and Sequential Logic Circuits (a) From the Karnaugh map, we have YZ 00 01 11 10 X 1
Case (ii): W hen Vi ≤ −0.7 V. Here, the Zener diode will be forward biased, therefore Vo = −0.7 V
124. Topic: 8085 Microprocessor: Programming and Interfacing
LP: CALL SUB LP + 3 The instructions suitable to be executed preceding this, from the given options are: (HL = LP + 3)
DAD D (HL = HL + DE) (LP + 3 +DE) = LP + DISP + 3) 125. Topic: Combinational and Sequential Logic Circuits (c) For J-K flip-flop: Qn +1 = JQn + KQn From the given figure; J = QB ; K = QB Output of J-K flip-flop is given by QA ( n +1) = QB QA + QB QA = QB
a 1
a 0
b 1
b 0
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
Therefore, number of combinations for which Y = 1 is 6. 128. Topic: Combinational and Sequential Logic Circuits (a) Race around occurs when the output of the circuit (Y1 and Y2) oscillates between 0 and 1. Considering the conditions given in the options: For option (b): Clk = 0. Output of NAND gate: A1 = B1 = 0 = 1 Output of next NAND gate: Y2 = Y1 ⋅1 = Y1
Output of t flipflop is given by
Y1 = Y2 ⋅1 = Y2
QB( n +1) = Q A
If
Clk pulse
QA
QB
QA(n + 1)
QB(n + 1)
tn
1
0
1
0
tn + 1
1
0
1
0
tn + 2
1
0
1
0
tn + 3
1
0
1
0
Ch wise GATE_EE_Ch9.indd 486
1
127. Topic: Combinational and Sequential Logic Circuits (b) If A > B; Y = 1 Let A = a1a0 and B = b1b0.
LXI D, DISP
PUSH H (Last two value of the stack will be HL value.
1 F = XY + XY
(c) In the given portion of the programme, the RET instruction should return the control to LP + DISP + 3.
POP H
1
0
Case (iii): When Vi ≥ 5.7 V ere, normal diode will be forward biased, H therefore Vo = 5.7 V
1
Y1 = 0 Y2 = Y1 = 1 (Output does not oscillate) Y2 = 0 Y1 = Y2 = 1 (Output does not oscillate) For option (c): Clk =1, A = B = 1 Output of NAND gates: A1 = B1 = 0 and Y1 = Y2 = 1 (No oscillation in the output). For option (d): Clk =1, A = B = 0 Output is: A1 = B1 = 1 (No oscillation in the output)
11/21/2018 6:13:08 PM
Chapter 9 • Analog and Digital Electronics
129. Topic: Amplifiers: Equivalent Circuit and Frequency Response (d) From the given circuit, on DC analysis we have
gm =
487
b 100 = = 0.04 rπ 2.5 × 103
By applying KCL at the output node 13.7 V v − vπ vo + g m vπ + o =0 RF RC
12 kΩ IB
Substituting RC = 12 k W, RF = 100 k W, we have gm = 0.04 s
VC 100 kΩ
⎡ 1 ⎡ 1⎤ 1⎤ vo ⎢ + ⎥ + vπ ⎢ g m − ⎥ = 0 RF ⎦ ⎣ ⎣ RC RF ⎦
+ 0.7 V −
vo (9.33 × 10 −5 ) + vπ (0.04) = 0 vo = − 428.72 V
Applying KVL at the input side,
v v v − vo vi = π + π + π RS R S rπ RF
VC − 100 I B − 0.7 = 0 VC = 100 I B + 0.7
(1)
⎡1 1 1⎤ V = vπ ⎢ + + ⎥ − C ⎣ RS rπ RF ⎦ RF v = vπ (5.1 × 10 −4 ) − o RF
Emitter current is given by IE =
13.7 − VC IC 12 × 103
(2)
We know that,
(5)
From Eqs. (4) and (5), we have I C = (1 + b ) I B
v vi −5.1 × 10 −4 vo − o = RF RS 428.72
Substituting from Eq. (2), we have 13.7 − VC 100 I B 12 × 103
(4)
Applying KCL at the input node, we have
VC − 100 I B − VBE = 0
vi = −1.16 × 10 −6 vo − 1 × 10 −5 vo 10 × 103 = −1.16 × 10 −5 vo
(3)
Solving for IB from Eqs. (1) and (3), we get IB = 0.01 mA From small signal analysis, the voltage source is transformed into equivalent current source.
Therefore, voltage gain AV =
Vo Vo = = 8.96 10 Vi 10 × 103 × −1.16 × 10 −5Vo
RF = 100 kΩ vo + v RS
vπ
RS
v rπ
gmvπ
− For shunt-shunt feedback amplifier obtained, rπ =
Ch wise GATE_EE_Ch9.indd 487
VT 25 × 10 −3 = = 2.5 kΩ I B 0.01 × 10 −3
130. Topic: Simple Active Filters (b) Replacing the elements by their Laplace equivalent and substituting s = jw
RC 1 1 ⇒ Cs Cjω R⇒R Vo ( s) ⇒ Vo ( jω )
C=
Vi ( s) ⇒ Vi ( jω )
11/21/2018 6:13:10 PM
488
GATE EE Chapter-wise Solved Papers
From the given diagram, to obtain transfer function, 0 − Vi ( jω ) 0 − Vo ( jω ) + =0 1 R2 + R1 jω C Vo ( jω ) −Vi ( jω ) = 1 R2 +R jω C 1 ⎤ ⎡ R2 Vo ( jω ) = −Vi ( jω ) ⎢ ⎥ ⎣1/jω C + R1 ⎦
So,
−Vi ( jω ) R2 =0 ∞ R ω → ∞; Vo ( jω ) = − 2 Vi ( jω ) R1
ω → 0; Vo ( jω ) =
Therefore, it is a high-pass filter. Let H ( jω ) =
V − R2 = o [transfer function ] . R1 − j1/ω C Vi
or | H ( jω ) | = R2 /R1 At 3 dB frequency, | H (∞) | = 2 | H ( jω 0 ) | R × 2 = 2 R1
R2
1
R12 +
1 ω 02 C 2
(at ω = ω 0 )
132. Topic: Operational Amplifier: Characteristics and Applications (b) We know that ⎛ RI ⎞ Vo = ηVT ln ⎜ o ⎟ ⎝ Vin ⎠ Given that R = 1 k W, V = 5 V and VT is the thermal voltage. For ideal transistor, η = 1; I 0 = I CBO . Therefore, Vout = − 0.7 V 133. Topic: Oscillators and Feedback Amplifiers (c) 134. Topic: Combinational and Sequential Logic Circuits (c) Let P1 and P2 be the switches and 0 and 1 represent OFF and ON states, respectively. P1
P2
0 0 1 1
0 1 0 1
or,
ω0 =
1 rad/s (high pass filter) R1C
131. Topic: Multiplexer (d) Let Qn be the present state and Qn+1 be the next state. It is given from figure, D = Y = AX 0 + AX 1 Qn +1 = D = AX 0 + AX 1 Qn +1 = AQn + AQn
0⎫ 1 ⎪⎪ ⎬ XOR gate 1⎪ 0⎪⎭
135. Topic: Characteristics of Diodes, BJT, MOSFET (b) Current through the 100 W resistor, I100 =
Squaring and cross-multiplying, we get 1 1 2 R12 = R12 + 2 2 ⇒ R12 = 2 2 ω0C ω0 C
O /P
100 − 5 = 0.05 A = 50 mA 100 I K = 10 mA
Therefore, I L = I100 − I K = 50 − 10 = 40 mA Resistance is given by RL =
VZ 5 = × 1000 = 125Ω I L 40
RL = ∞; maximum power rating of Zener diode occurs for R = ∞, therefore I Z = 50 mA Power rating = 50 × 5 = 250 W 136. Topic: VCOs and Timers
[∵ X 0 = Q ; X 1 = Q ]
(d) Given that: VYZ = 1000 sin ω t
If A = 0
Qn +1 = Qn
If A = 1
Qn +1 = Qn
During positive half cycle, all the diodes will be in reverse bias condition.
Ch wise GATE_EE_Ch9.indd 488
11/21/2018 6:13:13 PM
Chapter 9 • Analog and Digital Electronics
Therefore, voltage VWX = 0 During negative half cycle, all the diodes will be in forward bias condition and W and X will be shorted, therefore, VWX = 0 137. Topic: Operational Amplifier: Characteristics and Applications (c) In the first part of the circuit, ⎛ −R ⎞ ⎛ R ⎞ V1 = ⎜ F ⎟ ⋅ ( −2) + ⎜1 + F ⎟ (1). ⎝ R1 ⎠ ⎝ R2 ⎠ ⎛ 11⎞ ⎛ −1⎞ = ⎜ ⎟ ( −2) + ⎜1 + ⎟ (1) ⎝ 1⎠ ⎝ 11⎠ = 4V
489
The waveform shows that the time period of output waveform is twice that of clock input or the frequency of the output waveform f =
1 ( f clk ) = 0.5 kHz 2
139. Topic: Combinational and Sequential Logic Circuits (b) If Y = 0; Z cannot be at positive potential as diode conducts. Therefore, Y = 0 and Z = 0. If X = 1; the transistor will be ON and the output is grounded. Therefore, X = 1 and Z = 0. The truth table is
In the second part,
X
⎛R ⎞ Vout = ⎜ F ⎟ V1 ⎝ R⎠ ⎛ 1⎞ = ⎜1 + ⎟ 4 ⎝ 1⎠
Y
Z
0 0 0 1 1 0 1 1
0 1 0 0
= 2×4 = 8 V Thus, Z = XY 138. Topic: Combinational and Sequential Logic Circuits (b) From the given digital circuit, we have
(
)
X = ⎡( Q ⊕ Q ) ⋅ Q ⊕ Q ⎤ ⎣ ⎦ Q⊕Q =1
140. Topic: Amplifiers: Equivalent Circuit and Frequency Response (d) The equivalent circuit is 10 kΩ
IB
IC Vo
Q⊕Q = 0 Therefore, X = 1.0 = 0 = 1 So at T, input is always 1 because the clock is negative edge – triggered. So, the output waveform is as shown below.
VS
Clk So,
hfe
hfeIb
10 kΩ
Voltage gain = Output voltage Input voltage
We have h fe = 100, so T AV =
Vo − h fe I b (10 × 103 ) − hfe (10 × 103 ) = = VS I b (10 × 103 ) + I b hf e [10 × 103 + hfe ]
100 × 10 × 103 ( hf e is negligible in comparison) 10 × 103 = −100
=−
2T
Ch wise GATE_EE_Ch9.indd 489
AV = 100
11/21/2018 6:13:14 PM
490
GATE EE Chapter-wise Solved Papers
141. Topic: Simple Diode Circuits: Rectifiers (23) From the given circuit
10 A
where Vsat2 in the output of second op-amp. vo = ±Vsat
Therefore,
145. Topic: Combinational and Sequential Logic Circuits (c) For three identical modulo- 5 counters, overall modulus = 53 = 125. 146. Topic: Combinational and Sequential Logic Circuits
Vs
(d) From the given information, we have 8 1 1 0 1
RMS voltage VS = 230 V = Vrms I rms =
Vrms 230 ⇒R= = 23 Ω R 10
142. Topic: Amplifiers: Biasing (22.33) Given that VCE(sat) = 0.2 V , VBE = 0.7 V. Now, base current IB =
VEE − VBE 5 − 0.7 = = 2.15 mA R 2 × 103
RS(min) =
PZ 0.25 = = 0.05 A VZ 5
Vm − VZ 20 − 5 = = 300 Ω IZ 0.05
144. Topic: Operational Amplifier: Characteristics and Applications (d) Let Vsat1 be the output of the first op-amp. The second op-amp is the non-inverting amplifier. Therefore, net output voltage, ⎡ R⎤ vo = ⎢1 + ⎥ Vsat2 > Vsat1 ⎣ R⎦
Ch wise GATE_EE_Ch9.indd 490
1 0 1 1 0
= 8 = 9 = 3 = 12
147. Topic: Combinational and Sequential Logic Circuits (d) The truth table of the given state diagram is:
5 − VCE 5 − 0.2 Therefore, R C = = = 22.33 Ω IC 0.215
PZ = VZ I Z ⇒ I Z =
2 0 0 1 0
Valid states in BCD is form 0 − 9, so 1100 is the invalid state.
Current gain I b = C ⇒ I C = (100)( 2.15)(10 −3 ) = 0.215 A IB
143. Topic: VCOs and Timers ⎛ 1⎞ (300) Given that Vm = 20 V, VZ = 5 V, PZ = ⎜ ⎟ W. ⎝ 4⎠ Now,
4 0 0 0 1
A
B
X
Q
0
0
1
1
0
1
1
1
1
0
1
1
1
1
0
0
From the truth table, it is observed that
(i) If any one of the input is low, the output is high.
(ii) If both the inputs are high, the output is low. This represents the logic of NAND gate.
148. Topic: 8085 Microprocessor: Programming and Interfacing (c) The instruction LHLD 2100H loads the register pair HL with content directly from the memory to destination location. The register pair requires 2-bit of data. Since 8085 microprocessor can accommodate 1-bit data at a given address, the first-bit is available at the location mentioned in the instruction and the second bit is automatically accessed from the next location (address), that is 2101H. On execution of instruction, data is also moved similarly as (L) ←M(2100H) and (H) ←M(2101H)
11/21/2018 6:13:15 PM
491
Chapter 9 • Analog and Digital Electronics
149. Topic: Operational Amplifier: Characteristics and Applications (b) From the given circuit V2
+ − V2
If
150. Topic: Operational Amplifier: Characteristics and Applications (a) For the given circuit
R
+Vcc
1 kΩ
+ −
Vi R
Vo2
R
I1 Vo1/2 − Vo2/2 +
2R
1 µF
Vo
−Vcc
Vo Rf
V1 V1
− +
R
Vo1
R
For low pass filter, first order transfer function is given by
R
Vo ( s) 1 = Vi 0 ( s) 1 + sCR
By applying modal analysis, V2 − V1 V2 − Vo2 + = 0 (1) 2R R
Replace s = jw Vo ( jω ) 1 1 = = Vi ( jω ) 1 + ( jω )(103 )(10 −6 ) 1 + ( jω )10 −3
V1 − V2 V1 − Vo1 + = 0 (2) 2R R From Eq. (1), we have
Replace jw → s, we get
V2 V1 V2 Vo2 − + − =0 2R 2R R R Vo2 3V2 V1 = − R 2R 2R 3V − V Vo2 = 2 1 (3) 2 From Eq. (2), we have V1 V2 V1 Vo1 − + − =0 2R 2R R R Vo1 3V1 − V2 = R 2R Vo1 =
3V1 − V2 (4) 2
From Eqs. (3) and (4), we get
Vo ( s) = Vi ( s)
1 1+
s 1000
Corner frequency wc = 1000 Low frequency gain = 0 151. Topic: Simple Diode Circuits: Clipping (b) From the given circuit Vi [two diodes are OFF] 2
(i) Without clipping Vo =
(ii) When clipped Vi must be outside the range (−2 V to −4 V).
152. Topic: Operational Amplifier: Characteristics and Applications (6) When the op-amp is operated in the positive saturation region Vo = 5 V diode D1 is ON and diode D2 is OFF. Vb = Vo ×
1 = 1.25 V (By potential divider rule) 4
(V1 − V2 the ) op-amp is operated in the negative satura⎛ 3V − V ⎞ ⎛ 3V − V ⎞ 4V 4V = 2(V − V ) Vo = Vo1 − Vo2 = ⎜ 1 2 ⎟ − ⎜ 2 1 ⎟ = 1 − 2 = 4 When ⎝ 2 ⎠ ⎝ 2 ⎠ tion2region 1 2 2 2 Vo = − 5 V, diode D1 is OFF and diode D2 is ON. (V1 − V2 ) ⎛ 3V1 − V2 ⎞ ⎛ 3V2 − V1 ⎞ 4V1 4V2 1 = − − = 2(V1 − V2 ) = Vo1 − Vo2 = ⎜ =4 Vb = −Vo × = −2.5 V ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2 2 2 2
Ch wise GATE_EE_Ch9.indd 491
11/21/2018 6:13:17 PM
492
GATE EE Chapter-wise Solved Papers
M2 generates pulse width T2; T1 > T2 +5 V 1.5
Dt1
−2.5 −5 V
Dt2
T2
Let Δt1 = time duration for positive part of cycle, Δt2 = time duration for negative part of cycle. 5 = 1.25 Volt 4
Upper triggering point UTP = Vsat + ( UTP − Vsat ) e ⎛ 5 ⎞ 5 + ⎜ − − 5⎟ e ⎝ 2 ⎠ e
−
t1 τ
−
t1 RC
t −1 t
(i) When Q2 = 1; Q2 = 0; ⇒ Q2 ⋅ Q2 = 0 . Therefore, M1 does not generate pulse width T1.
(ii) After T2 duration, Q2 = 0 ⇒ Q2 = 1 ⇒ Q1Q2 = 1 .
Therefore, M1 generates pulse width T1.
= UTP
=
Hence, output would be
5 4
Q2(t) = vo(t)
= 0.5
(1)
Similarly, Lower triggering point LTP = 2.5 V −Vsat + ( LTP + Vsat ) e e
+
−
t2 τ
t − 2 τ
= 3
=−
T2 Q1(t)
5 2 (2)
Δt1 = 0.6931 τ
T1 Q2(t)
Δt 2 = −1.098 τ
Δt1 Δt 2 − = 1.791 τ τ
Δt1 − Δt 2 = e1.791 ≅ 6 τ
T2 Overall output waveform
153. Topic: Oscillators and Feedback Amplifiers (c) Non-stable multivibrator generates pulse width T1. ... T2 T1
Ch wise GATE_EE_Ch9.indd 492
T1
T2
where T1 > T2.
11/21/2018 6:13:19 PM
493
Chapter 9 • Analog and Digital Electronics
154. Topic: Operational Amplifier: Characteristics and Applications (c)
(i) When vi > 0; vo = 0, slope is 0 as the v oltage at negative and positive terminals is the same for ideal op-amp.
(ii) When vi < 0; vo = vi, slope is 1. Hence, the transfer characteristic of op-amp circuit is
For logic circuit in option (c) A
A
C
C
AC A C + BC BC B
For logic circuit in option (d)
vo
A
A
AC
C
A C +BC
Slope = 0 BC
vi
B
Slope = 1
From the given K-map exact logic diagram is option (c).
155. Topic: Combinational and Sequential Logic Circuits (c) From the given K-map,
156. Topic: 8085 Microprocessor: Programming and Interfacing (b) Given 3 × 8 decoder. So, Enable line E1 is active high which is output of AND gate Enable lines E2 and E3 are active low . Address will be in 16 bits as IO/M should be active low which indicates that it is memory mapped I/O interfacing.
AB 00
01
1
1
11
10
C 0 1
To enable E1 line, A12 = 1 and A11 = 0 To select output port through second line,
1
1
A15 A14 A13 0 1 0
F = CAB + CAB + CAB + CAB = CA( B + B ) + CB( A + A) = AC + BC
A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 A0 0
1
0
1
0
0
0
0
0
0
0 0
0
0 = 5000H
For circuit in option (a)
0
0
0
0
157. Topic: Combinational and Sequential Logic Circuits (a) The K-map for the function is
A A A BC B C
BC
CD 00
01
11
10
00
1
0
1
0
01
0
0
1
0
11
0
0
0
0
10
0
0
1
0
AB
For logic circuit in option (b) A A A + BC B C
Ch wise GATE_EE_Ch9.indd 493
BC
11/21/2018 6:13:21 PM
494
GATE EE Chapter-wise Solved Papers
Minimised expression is,
2005H RAR
→ Rotate accumulator right with carry
2006H DCR B
→ Decrement content of B register by one
F = ( B + C )( A + C )( A + B )(C + D ) 158. Topic: Combinational and Sequential Logic Circuits (b) Consider option (b). The present and the next state characterstic table for the J-K flip-flop is
2007H JNZ 2005H → Jump on no zero to location 2005H →
200AH HLT
Qn
J
K
Qn+1
T
0
0
0
0
0
0
0
1
0
0
0
1
0
1
1
0
1
1
1
1
1
0
0
1
0
= 0000 0001 1 First rotate accumulator right
1
0
1
0
1
= 1000 0000 1 Second rotate accumulator right
1
1
0
1
0
= 1100 0000 0 Third rotate accumulator right
1
1
1
0
1
= 0110 0000 0 Fourth rotate accumulator right
Carry A=
0000 0011 0
So, finally, A = 0110 0000 = 60H
The K-map for T flip-flop is J-K 00
01
11
10
0
0
0
1
1
1
0
1
1
0
Qn
Since B = 04 serves as the counter for the loop, A gets rotated right through carry flag 4 times.
160. Topic: Multiplexer (a) For the given counter Decimal
Binary
Gray
Output
0
000
000
I0
1
001
001
1
2
010
011
1
The characteristic equation for T flip-flop is
3
011
010
I1
T = Qn J + Qn K
4
100
110
I3
5
101
111
1
6
110
101
1
7
111
100
I2
From the given options, T = ( J + Qn )( K + Qn ) = JK + JQn + KQn + Qn Qn
= JK + JQn + KQn
(as Qn Qn = 0)
= JQn + KQn
On elimination of redundant term J-K. 159. Topic: 8085 Microprocessor: Programming and Interfacing (a) 2000H XRA A
→ [A] clears to 00H, Carry = 0, Z = 1
2001H MVI B, 04H→ [B] = 04H 2003H MVI A, 03H→ [A] = 03H
Ch wise GATE_EE_Ch9.indd 494
Output ‘1’ represents high impedance state. 161. Topic: Sample and Hold Circuits (0.635) Given that LTP = 0.9 V and UTP = 1.7 V. VC (t ) = Vmax + (Vmin − Vmax ) e − t / RC LTP = 0 + (1.7 - 0)e−t/RC 0.9 = (1.7)e − t /(10 ×10
3
× 0.1 × 10 −6 )
Therefore, t = 0.635 ms. 162. Topic: Characteristics of Diodes, BJT, MOSFET (a) Characteristics of instrumentation amplifier (i) High common mode rejection ratio (CMRR) (ii) High input impedance
11/21/2018 6:13:21 PM
Chapter 9 • Analog and Digital Electronics
(iii) (iv) (v) (vi)
High linearity (high open loop again) Very low DC offset Low drift Low noise
163. Topic: Operational Amplifier: Characteristics and Applications (d) Given that R = 1 kΩ, C = 1 mF, vi = Vi ∠0 and vo = Vo ∠δ . The value of output phase angle (d ) relative to the phase angle of the input voltage is −p/2 as Vo ∠δ = RC ωVi ∠(−90° ) where RC ωVi is a real quantity.
166. Topic: Operational Amplifier: Characteristics and Applications (1.25) Let Z1 = R1 = 1 kW. Now, 1 1 1 × 103 × j 0.1 × 10 −6 ω jω C Z2 = = 1 1 R2 + 1 × 103 + jω C j 0.1 × 10 −6 ω R2 ×
=
IC + IB = IE = I Bb + I B =
6 15 − 9 = RC RC
6 RC
IB[1 + b ] = 76 I B = 76 I B =
6 RC
6 RC
(b /VC ) (75/ 9) 8.3 = = IB = RB RB RB ⎛ 8.3 ⎞ R 6 76 × 8.3 76 × ⎜ ⎟ = ⇒ B= = 105.13 RC 6 ⎝ R B⎠ RC 165. Topic: Simple Diode Circuits: Rectifiers (c) For the sinusoidal voltage, the time period is 1 2π T= = = 20 ms 100π 50
103 j 0.1 × 10 −3 ω + 1
V = 2 sin (2p × 2000t) ⇒ w = (2p × 2000) rad/s
164. Topic: Amplifiers: Biasing (105.13) Given that b = 75 and VC = 9 V. We know that IC = b IB. Therefore,
495
Therefore, Z 2 =
103 1 + j1.25
V = −Vi ×
Z2 Z1
= −2 sin( 2π × 2000t ) ×
103 1 × 1 + j1.25 103
= −1.25 sin( 2π × 2000t − 51.3°) So, amplitude of output voltage = 1.25 V. 167. Topic: Amplifiers: Biasing (0.75) We have VC = 2 V, VC′ = 4 V . Then I C R C = 10 − VC = 10 − 2 = 8 V I C R ′C = 10 − VC′ = 10 − 4 = 6 V Therefore,
I C R ′C 6 R′ = ⇒ C = 0.75 I C RC 8 RC
168. Topic: Operational Amplifier: Characteristics and Applications (b) The output waveform if f1 < f2 is Vo/Vi
For the circuit, the time constant is
τ = RC =
1 50 = s 100π 2π
So, time constant is greater than time period. In the first half cycle, after charging, the voltage across R is 2Vm = 2 × 100 = 200 V. In the second half cycle also, the voltage across R is 2Vm = 2 × 100 = 200 V Note: The given circuit is a voltage doubler, hence voltage across two capacitors is twice the voltage across each capacitor.
Ch wise GATE_EE_Ch9.indd 495
f f1
f2
The signals are blocked at frequencies less than f1 and greater than f2. Therefore, it is a band-stop filter. 169. Topic: Characteristics of Diodes, BJT, MOSFET (c) For saturation mode, collector–base is forward biased and base–emitter is forward biased. Hence, for the saturation condition, both the junctions should be in forward bias.
11/21/2018 6:13:24 PM
496
GATE EE Chapter-wise Solved Papers
170. Topic: Multiplexer (b) Given F = A ⊕ B . Now A ⊕ B = A′ B + AB ′, so, the truth table for the multiplexer is S0
S1
F
0
0
A′ B′ → I 0 0
0
1
A′ B → I11
1
0
AB ′ → I 2 1
1
1
AB → I30
171. Topic: Combinational and Sequential Logic Circuits (a)
A 0
BC 01
1
BC 11
1 0
1
A 1 4
Therefore,
174. Topic: Combinational and Sequential Logic Circuits (6) From the given circuit, we have Flip-flop 1 functions as a MOD -2 counter. The next two flip-flops, flip from MOD (2n − 1) Johnson counter. So, overall modulus is MOD-6 counter. Therefore, state Q2Q1Q0 = 000 will repeat after 6 cycles of the Clk. 175. Topic: Combinational and Sequential Logic Circuits
BC 10
(c) The K-map for the given function is
1 1
R2 2 = = 0.5 R1 + R2 2 + 2
⎛ 1 + 0.5 ⎞ T = 2 × 1 × 103 × 0.25 × 10 −6 log ⎜ = 0.55 ms ⎝ 1 − 0.5 ⎟⎠
So, the required input is 0110.
BC BC 00 A
b=
3
2
7
6
CD CD 00 AB
CD 01
CD 11
CD 10
1 5
F = ABC + ABC + ABC + ABC + ABC
AB 00
0
0
0
1
AB 01
0
0
0
1
AB 11
0
0
0
0
AB 10
1
0
0
1
Minterm, F = Σm(0, 1, 3, 5, 7) and Maxterm, F = P (2, 4, 6) F = ( A + B + C )( A + B + C )( A + B + C ) Taking complement F = ( A + B + C )( A + B + C )( A + B + C ) 172. Topic: Operational Amplifier: Characteristics and Applications (a) Given that A = 1000, Vi = 1 mV, R = 1 kΩ, C = 1 mF. Now, we have A=
Vo Vi
or Vo = 1000 × 1 mV = 1000 mV Time constant = (1000 + 1)RC = 1001 × 103 × 10−6 = 1001 ms. 173. Topic: VCOs and Timers (a) The given circuit is a astable multivibrator. It produces square wave output. Therefore, T = 2 RC log
Ch wise GATE_EE_Ch9.indd 496
(1 + b ) (1 − b )
f ( A, B, C , D ) = ABC D + ABCD + ABCD + ABCD = AC D( B + B ) + ABD(C + C ) = AC D + ABCD + ABCD 176. Topic: A/D and D/A Converters (a) In a successive approximation register (SAR) type ADC, the analog voltage input undergoes conversion through control logic to give digital equivalent voltage output on application of clock pulses. It loads value to the output register with MSB = 1 and remaining bit = 0 and cross-checks the logic conditions as follows If Vi > VDAC, then it maintains the loaded bit. If Vi < VDAC, then it clears the loaded bit. The process will continue for 8 clock pulses and the output obtained is given by (resolution) × (Decimal equivalent of binary voltage applied). Therefore from the given values Resolution =
Vref 5 = 2n − 1 28 − 1
11/21/2018 6:13:26 PM
Chapter 9 • Analog and Digital Electronics
At the first clock pulse, Value at the output register is (10000000)2 = (128)10
179. Topic: Amplifiers: Biasing (19) For the given circuit, VBE = 0.6 V. Therefore,
VDAC = 128 × 20 × 10 −3 V = 2.56 V Given Vi = 3.5 V. So Vi > VDAC. Hence maintains the loaded bit. The output at the end of first clock pulse is (10000000)2 At the second clock pulse, Value at the output register is (11000000)2 = (192)10
5.3 V − 0.6 V = 0.01 A 470 Ω
IE =
IB = 0.5 mA = 0.0005 A IC = 0.01A − 0.0005A (0.01 − 0.0005) = 19 0.0005
b=
VDAC = 192 × 20 × 10 −3 V = 3.84 V So Vi < VDAC. Hence, clears the loaded bit. So the output at the end of second clock pulse is (10000000)2. At the third clock pulse, Value at the output register is (10100000)2 = (160)10
Alternately
IE = b + 1 ⇒ b = 19 IB
180. Topic: Simple Active Filters (a) For the given circuit
VDAC = 160 × 20 × 10 −3 V = 3.2 V
1 ⎤ ⎡ R2 ⎢ jω C R2 + jω C ⎥ Vout ⎥ = −⎢ Vin R1 ⎢⎣ ⎥⎦
So Vi > VDAC. Hence, maintains the loaded bit. So, the output at the end of third clock pulse is (10100000)2
⎡ ⎤ R2 = −⎢ ⎥ ⎣ R1 ( R2 jω C + 1) ⎦
177. Topic: Combinational and Sequential Logic Circuits (b) The truth table for the sequential circuit is K1 (Q0 )
J1 (Q ) 0
J0 (Q1 )
K0 (Q1)
Q1
Q0
0
0
0
1
1
0
0
1
1
0
1
0
1
1
1
0
0
1
1
0
0
1
0
1
0
0
333 will complete 83 4 cycles and remainder clock = 1. Therefore, cycle ends at Q1 Q0 = 00. Therefore, next state at 333rd cycle is Q1 Q0 = 01.
For a Johnson MOD-4 counter,
178. Topic: Multiplexer (b) Given that f(A, B, C, D) = Π(1, 5, 12, 15) [max term]. Therefore, f (A, B, C, D) = ∑(0, 2, 3, 4, 6, 7, 8, 9, 10, 11, 13, 14) I0
I1
I2
I3
I4
I5
I6
I7
D(0)
0
2
4
6
8
10
12
14
D(1)
1
3
5
7
9
11
13
15
D
1
D
1
1
1
D
D
Ch wise GATE_EE_Ch9.indd 497
497
Let -R2/R1 = k, R2C = t, j w = s. Then Vout k = Vin sτ + 1 This is standard low pass filter expression. 181. Topic: A/D and D/A Converters (b) Total degrees = 55° − (−40°) = 95°C and resolution of 0.1 °C is required. Therefore, number of steps = 950. From the given options, 10 bits which implies 210 = 1024 steps; will cover the entire range 182. Topic: Multiplexer (d) F = BI 0 + BI1
= B. A + BA
=A
= A⊕ B
B (XNOR operation)
183. Topic: Combinational and Sequential Logic Circuits (b) For the given Karnaugh map, two quads cover all the 1’s as shown. BC A 00 01 11 10 0 1 0 0 1 1 1
1
1
1
Therefore, A + C
11/21/2018 6:13:28 PM
498
GATE EE Chapter-wise Solved Papers
184. Topic: Operational Amplifier: Characteristics and Applications (d) For the given circuit, at the two nodes we have
However, D3 is ON for negative half cycle; shorting out V2, and full voltage appears on V1. So V1 (avg) =
1 ⎞ ⎛4 VA = ⎜ V1 + V2 ⎟ ⎝5 5 ⎠
π /2 1 ⎡ π sin 100π t dω t ⎢ 2π ⎣ ∫0 2 2π ⎤ + ∫ π sin 100 πt d ω t ⎥ π ⎦
VB = V3 Vout = −9VB + 10VA= −9V3 + 8V1 + 2V2 185. Topic: Combinational and Sequential Logic Circuits (c) For the given two J-K flip-flop system Initially QA = 0; hence QA = 1 Therefore, QB = 1 (in the next cycle). As J and K are both high, so QA toggles and becomes 1. Hence, the next state is 11.
= -0.5 V
189. Topic: Simple Diode Circuits: Rectifiers (a) Both diodes conduct for more than 180°. V0
186. Topic: A/D and D/A Converters (a) For the given ADC Input
wt
Output
X2
X1
X0
B1
0 0 0 0
0 0 1 1
0 1 0 1
0 0 1 1
B0
D1
X1X0 0 X2 00 1 ⇒ 0 0 0 1 X 1
B0 = X 2 X 1 X 0 + X 2 X 1 X 0
= X 0 ( X 2 X1 + X 2 X1 )
= X 0 ( X 2 ⊕ X1 )
B0 01 1
11 0
10 X
X
1
X
D2
190. Topic: Combinational and Sequential Logic Circuits (a) K-map for given expression: A
BC 00 01 11 10 0 0 0 1 0 = BC + AC
187. Topic: Combinational and Sequential Logic Circuits (d) The given function
1 1
0
1
1
191. Topic: Combinational and Sequential Logic Circuits (c) P
X
Q
Z
F = ( a + b + c + d ) + (b + c ) R
Y
can be simplified using De Morgan’s theorem as X = P ⋅Q
F = ( a + b + c + d ) ⋅ (b + c ) = ( a ⋅ b ⋅ c ⋅ d ) ⋅ (bc)
Y = Q⋅R
=0 188. Topic: Simple Diode Circuits: Rectifiers (b) For positive half-cycle, D1 is ON. p/2
⇒ V1 −p
Ch wise GATE_EE_Ch9.indd 498
Therefore, Z = X ⋅ Q ⋅ Y = PQ ⋅ Q ⋅ QR = PQ + Q + QR = PQ + Q + QR = Q( P + 1) + QR = Q + QR = (Q + Q )(Q + R) = Q + R
11/21/2018 6:13:30 PM
Chapter 9 • Analog and Digital Electronics
192. Topic: Operational Amplifier: Characteristics and Applications (a) When Vin > 0,
194. Topic: Amplifiers: Biasing (d) Using KVL in inner and outer loops,
VA = +V55, Don, V0 = Vin
When Vin < 0,
10 = (1 + b )IB × 4R + IBRB + 0.7 + (1 + b )IB × R ⇒ 9.3 = 150IBR + IBRB
VA = -V55, Doff, V0 = 0 Therefore,
499
(1)
and 10 = (1 + b )IB × 4R + 5V + (1 + b )IBR
V0
⇒ 5 = 150IBR
VA = Vin
VA = 0
⇒ IBRB = 4.3
(2)
Solving Eqs. (1) and (2), we get
Vin
RB = 129 R 193. Topic: Amplifiers: Equivalent Circuit and Frequency Response (172.69)
195. Topic: Operational Amplifier: Characteristics and Applications (c) Consider the figure. I3 R
R
VB
I1
12 V I2 I3
I0 = 1 µA
Rq
Q2 VBE1
VBE2
VS
I0
VA = VS/2 I3 =
VB − VA = I3R
VBE1 = VBE2 + I0R
⇒ VB = VA + I3R
I0R = VBE2 − VBE1 Now,
Ch wise GATE_EE_Ch9.indd 499
VA VS = R 2R
Now,
From the above circuit,
I1
2R 2R
−12 V
Vo +
Here, R
R=
−
VA VA
1 mA Q1
R
R
VT ln( I R /I 0 ) I0
⎛ 1 mA ⎞ 0.025 ln ⎜ ⎝ 1 μA ⎟⎠ = 1 μA = 172.69 kΩ
⇒ VB =
VS VS + = VS 2 2
Also, I2 = VS/R Therefore, I1 = I 2 + I 3 =
VS VS 3 ⎛ VS ⎞ + = R 2 R 2 ⎜⎝ R ⎟⎠
11/21/2018 6:13:31 PM
500
GATE EE Chapter-wise Solved Papers
199. Topic: Combinational and Sequential Logic Circuits (6)
Now, Vo − VB = I1 R =
3 VS 2 R
Therefore,
CLK
QA
QB
QC
QA′
QB′
QC′
Z
0
1
0
0
1
0
0
0
5 3 Vo = VS + VS = VS = 2.5 VS 2 2
1
0
1
0
1
1
0
1
2
0
0
1
1
1
1
1
196. Topic: Combinational and Sequential Logic Circuits
3
1
0
0
0
1
1
1
4
0
1
0
0
0
0
1
5
0
0
1
0
0
0
1
6
1
0
0
1
0
0
0
(d) Consider the below Karnaugh map. CD 00
01
11
10
00
0
0
0
0
01
1
0
0
1
11
1
0
1
1
AB
Therefore, the minimum number of clock is six cycles. 200. Topic: Simple Diode Circuits: Rectifiers (c) During forward bias + + ON
10
0
0
0
+
0
Therefore, the output expression is
off
i0 L o V0 [Free wheeling a diode is off] d
-
off
ON
- -
BD + ABC
Let Vin (t ) = Vm sin(ωt ), V0 (t ) = Vn sin(ωt ) [ positive]
197. Topic: Combinational and Sequential Logic Circuits (b) When V1 = 5 V, Vout = 0 When V2 = 5V, Vout = 0 Vout = 5 V only when both V1, V2 = 0. Therefore, the logical gate is NOR gate. 198. Topic: Schmitt Trigger (108) Let f = Input frequency n=N umber of cycles of repetitive signal = 100 × 106 t = Gate time = 1 s f = On 10
n 100 × 106 = = 100 MHz t 1
1 digit display ⇒ 100000000.00 Hz 2
1 1 Time period = = 8 seconds = Period of input signal f 10 ⇒ P = 10 ns
Ch wise GATE_EE_Ch9.indd 500
During reverse bias i0 + ON
off
L o a d
+
−
off
ON
[Free wheeling diode is off] V0(t) = Vm sin(wt) [positive]
−
Vin(t) Vm 0
p
q
2p
–Vm
Vm 0 a
p p+a
2p
q
SCR allows current only in one direction - anode to cathode (i0 ≥ 0).
11/21/2018 6:13:33 PM
501
Chapter 9 • Analog and Digital Electronics
201. Topic: Combinational and Sequential Logic Circuits (d)
⇒ Vin ×
Y1
A
⇒ Vin ⋅
B
( R1 + R2 ) × R2
R1 + ZI = 0 ( R1 + R2 ) in
R1 = − ZI in R2
Y ⇒
C Y2
D
Y1 = A ⋅ B = AB
Y2 = CD
Y = ( AB) ⋅ (CD )
= AB + CD
Vin R = −Z ⋅ 2 I in R1
203. Topic: Characteristics of Diodes, BJT, MOSFET (17.39 A) i0(t)
I0 0
202. Topic: Operational Amplifier: Characteristics and Applications (b) For the given op-amp
a
p
p+a
2p
q = wt
-I0 Fourier series representation on i0(t) is i0 (t ) =
Z
nα ⎞ ⎛ 4I0 ⎞ ⎛ ⎛ nα ⎞ ⎜⎝ nπ ⎟⎠ cos ⎜⎝ 2 ⎟⎠ sin ⎜⎝ nω t − 2 ⎟⎠ n =1, 3, 5 ∞
∑
RMS value of nth harmonic is iin
Vin
⎛ nα ⎞ 2 2 ⎛ nα ⎞ I 0 cos ⎜ ⎟ cos ⎜ ⎟ = ⎝ ⎠ ⎝ 2 ⎠ 2 n π 2nπ
4I0
+ Vo
+ −
RMS value of fundamental is
−
I s1 ( rms) = Vin R2
=
R1
2 2 α I 0 .cos π 2 2 2 ⎛ 30° ⎞ × 20 × cos ⎜ = 17.39 A ⎝ 2 ⎟⎠ π
204. Topic: Combinational and Sequential Logic Circuits -Vin + ZIin + V0 = 0
Vin = V0
(1)
(b) D0 = Q1 + Q2 Q0 Q1 Q2
R2 R1 + R2
0 0 0
Substituting this in Eq. (1), we get −V0 R2 + ZI in + V0 = 0 R1 + R2 ⎛ R2 ⎞ + ZI in = 0 ⇒ V0 ⎜1 − ⎝ R1 + R2 ⎟⎠ ⎛ R1 ⎞ ⇒ V0 ⎜ + ZI in = 0 ⎝ R1 + R2 ⎟⎠
Ch wise GATE_EE_Ch9.indd 501
1 1 1 0 0
0 1 1 1 0
0 1 1 1 1
1
0 0
⎫ ⎪ ⎪ ⎪ ⎬ 5 cycles; Divide by 5 ⎪ ⎪ ⎪⎭
So, frequency will be divided by 5. 205. Topic: Combinational and Sequential Logic Circuits (b) Given F = m0 + m2 + m3 + m5 + m7 and
m1 = don’t care
11/22/2018 12:26:28 PM
502
GATE EE Chapter-wise Solved Papers
BC
BC
1
X
BC
The ac model for the common base circuit is
BC
C
B A
0
1 1
1
A
1 3
2 bi0 open
rπ
f=A+C
1
10 kΩ 4
5
7
10 kΩ E
6
a I0
So, F = A + C
1 kΩ
206. Topic: Amplifiers: Biasing (92) DC model: Applying KVL at input loop, we have Ie =
b
B
10.7 − 0.7 10 = 10 kΩ 10 kΩ + 1 kΩ + 1 kΩ 1+ b 101
rπ = 2.72 Ω b
10 10 = = = 9.17 mA 1 kΩ + 0.099 k 1.09 k
re =
RTh
10 kΩ/b a I0
VT 25 mV = = 2.72 Ω I e 9.17 mA
1 kΩ RTh
Vab b
bre = rp = 0.272 kW RTh = 10 kΩ
Vab = 1 kΩ || 2.72 Ω + 100 Ω I0
= 1 kW || 102.72 W 1 kΩ
10.7 V
Ch wise GATE_EE_Ch9.indd 502
+ –
= 1 kW || 0.102 kW IE =
1 × 0.102 0.102 = = 0.092 kΩ = 92 Ω 1 + 0.102 1.102
11/21/2018 6:13:37 PM
Power Electronics
C H A P T E R 10
Syllabus Characteristics of semiconductor power devices: Diode, Thyristor, Triac, GTO, MOSFET, IGBT; DC to DC conversion: Buck, Boost and Buck-Boost converters; Single and three phase configuration of uncontrolled rectifiers, Line commutated thyristor based converters, Bidirectional ac to dc voltage source converters, Issues of line current harmonics, Power factor, Distortion factor of ac to dc converters, Single phase and three phase inverters, Sinusoidal pulse width m odulation.
CHAPTER ANALYSIS Topic Characteristics of Semiconductor Power Devices
GATE 2009
GATE 2010
GATE 2011
GATE 2012
GATE 2013
GATE 2014
GATE 2015
GATE 2016
GATE 2017
GATE 2018
3
2
3
2
4
3
3
3
1
1
2
2
2
2
DC to DC Conversion Single and Three Phase Configuration of Uncontrolled Rectifiers
1
2
Line Commutated Thyristor Based Converters
1
Bidirectional AC to DC Voltage Source Converters
1
2
2
2
1
2
1
1 1
1
Issues of Line Current Harmonics Power Factor
3
Distortion Factor of AC to DC Converters
1
Single Phase and Three Phase Inverters
2
1
1
1
1
1 2
1
2
2
3
Sinusoidal Pulse Width Modulation
3
1
1
IMPORTANT FORMULAS I. Basic p-n Diode Characteristics
II. BJTs
1. The forward current of the p-n junction
2. Collector current
where IS is the reverse saturation current, V is the applied forward voltage across the device and q is the charge of an electron, K is the Boltzman’s constant and T is the junction temperature in Kelvin.
Ch wise GATE_EE_Ch10.indd 503
I CS a + IB 1− a 1− a where ICS is the reverse saturation current of junction JCB and IC/IE is the alpha (a) parameter of the transistor or IC =
I F = I S (e qV / KT − 1)
I C = β I B + (β + 1) I CS
11/21/2018 6:19:27 PM
504
GATE EE Chapter-wise Solved Papers
a is called the large signal common 1− a emitter current gain of the transistor. where β =
IV. GTO 7. Anode current IA =
3. Conduction loss
a n I G + (i CB01 + i CB02 ) 1 − (a n + a p )
PC = I C × VCE(sat)
V. Performance Parameters
and power loss
8. Input power factor:
P = VIC
pf =
4. Switching energy loss for one cycle E=
VI C (TOFF + TOn ) 6
Mean power Mean power = Vrms I rms VS I S
For sinusoidal voltage input ⎛I ⎞ pf = ⎜ 1rms ⎟ cos ϕ 1 ⎝ I rms ⎠
and total switching loss Psw =
VI C (T OFF + T On ) ⋅ f s watts 6
9. Displacement factor = cos f1 10. Distortion factor =
III. Thyristors
11. Total harmonic distortion
5. Total anode current I +I I A = CO1 CO2 1 − (a1 + a 2 )
n
THD =
I CO = 1 − (a1 + a 2 )
where a1and a2 are current gains and ICO1 and ICO2 are reverse saturation currents of the CB junctions of Q1 and Q2, respectively. I CO I CO1 + I CO2 is the total reverse leakage current of J2 junction. In gate assisted turn-on
a I + I CO IA = 2 G 1 − (a1 + a 2 )
=
i=2
2 i
I1rms
=
2 2 − I1rms I rms
I1rms
2 2 I 2rms + I 3rms + I n2rms
I1rms
⎛I ⎞ CF(or c) = ⎜ max ⎟ ⎝ I rms ⎠ 13. Ripple factor: RF(or r) =
2 2 I rms − I avg
I avg
14. Form factor
L R = 2D C where D is the damping factor and usually taken as 0.5 2
⎛ ⎞ ⎜ 2V ⋅ D ⎟ 1 C = ⎜ max ⎟ ⎜ dv ⎟ L ⎜⎝ dt ⎟ max ⎠
Ch wise GATE_EE_Ch10.indd 504
∑I
12. Crest factor:
6. R and C from the snubber circuit
I1rms I rms
⎛I ⎞ FF(or F) = ⎜ rms ⎟ ⎝ I avg ⎠ 15. Ripple factor in terms of form factor 2
⎛I ⎞ RF = ⎜ rms ⎟ − 1 ⎝ I avg ⎠ = (FF)2 − 1
11/21/2018 6:19:29 PM
Chapter 10 • Power Electronics
VI. Converters: AC-DC Rectifiers 16. Single-phase uncontrolled half-wave rectifier (with resistive load):
18. Single-phase uncontrolled full-bridge rectifier with RLE load
(a) Output average voltage Vo (avg) =
∫ V d (ω t )
V 1 = Vm sin ω td( ω t ) = m ∫ π 2π 0
(b) Output rms voltage Vo(rms) =
π V 1 Vm2 sin 2 ω t d ω t = m ∫ π0 2
(b) rms value of the output voltage π V 1 Vm2 sin 2 ω td ( ω t ) = m ∫ 2π 0 2
VD (rms) =
19. Single-phase half-wave controlled rectifier with resistive load
(a) Average value of the output voltage
where Vm is the peak value of the input voltage.
π
1 2 V sin ω t d ω t = Vm π ∫0 m π
o
0
π
(a) Average voltage Vo(avg) =
2π
1 2π
Vo(avg) =
(c) Ripple factor of output voltage Vo (RF) =
Vo2(rms) − Vo2(avg) Vo (avg)
=
1 π2 − 4 2
Vo(rms) =
=
(a) Average output voltage is given by, Vo (avg) = Vo (avg) =
1 2π
2π
∫ Vo dω t = 0
β
1 Vm sin ω t dω t 2π ∫0
Vm (1 − cos β ) 2π
=
Vo(avg) =
2β − sin 2β 2π
Vo (rms) Vo (avg)
Vo(rms) =
2 Vo(RF) = Vo(FF) −1 =
Ch wise GATE_EE_Ch10.indd 505
π ( 2β − sin 2β ) −1 2(1 − cos β ) 2
sin 2a ⎤ ⎡ ⎢π − a + 2 ⎥ ⎣ ⎦
2 o
dω t
0
1 2π
2π
∫ V dω t o
0
Vm (cos a − cos β ) 2π
1 2π
2π
∫V
2 o
dω t
0
1
V ⎛ β − a sin 2a − sin 2β ⎞ 2 = m⎜ + ⎟⎠ 2 ⎝ π 2π
2β − sin 2 β =π 2π (1 − cos β ) 2
(d) Ripple factor of output voltage
Vm2 4π
∫V
(b) Load voltage rms value
(c) Form factor of the voltage waveform Vo (FF) =
=
1 Vm2 sin 2 ω t d (ω t ) 2π ∫0 Vm 2
2π
(a) Load average voltage
(b) rms output voltage Vo (rms) =
1 2π
20. Half-wave controlled rectifier with resistive-inductive (RL) load
β
Vm (1 + cos a ) 2π
(b) Output voltage rms value
17. Single-phase uncontrolled half-wave rectifier (with RL load):
505
(c) Average load current I o(avg) =
Vo(avg) R
=
Vm (cos a − cos β ) 2π R
11/21/2018 6:19:31 PM
506
GATE EE Chapter-wise Solved Papers
21. Single phase half-controlled converter with RLE load
(a) Average load voltage π
π
1 1 = ∫ Vo dω t = ∫ Vm sin ω t dω t π0 πa
Vo (avg)
Vm (1 + cos a ) π
=
24. Three-phase full-wave uncontrolled bridge rectifier feeding RLE load (a) Average output voltage Vo(avg) =
3 π
π /3
Vo (avg) − E
Vm sin(ϕ − a ) + sin ϕ e π − Z 1 − e tan ϕ
−
a tan ϕ
25. Three-phase half-controlled converter feeding RLE load
(a) Output voltage for a ≤ p/3 ⎡ 3 2VL ⎢ 0 π⎞ ⎛ Vo = sin ⎜ ω t + 2 ⎟ d( ω t ) ∫ ⎢ ⎝ 2π 3⎠ ⎢a − π3 ⎣
ii = −io ; for p + a ≤ wt ≤ 2p
a+
ii = 0 ; otherwise
+
=
(a) Average output voltage π +a
2
∫ V dω t = π V
m
o
cos a
(b) Output voltage rms value 1 π
π +a
∫V
2 o
d ( ωt ) =
Vm
a
3 2 VL (1 + cos a ) 2π
(b) Output voltage for a ≥ p/3 ⎡ ⎤ 3 2VL ⎢ 2π π⎞ ⎛ ⎥ Vo = sin ⎜ ω t + ⎟ d ( ω t ) ⎥ 2π ⎢ ∫π ⎝ 3⎠ ⎢a − 3 ⎥ ⎣ ⎦
a
Vo (rms) =
∫
⎤ π⎞ ⎛ ⎥ sin ⎜ ω t + ⎟ d (ω t )⎥ ⎝ 3⎠ ⎥ ⎦
π 3
0
22. Single-phase fully-controlled bridge converter with rle load
1 π
⎛ 3 3⎞ 2VL2 sin 2 ωt d ( ω t ) = ⎜1 + VL 2π ⎟⎠ ⎝ π /3
∫
where VL is the rms line to line voltage.
ii = io ; for a ≤ wt ≤ p
Vo (avg) =
3 2 V π L
2π / 3
3 π
Vo(rms) =
Input current ii =
2VL sin ω t d ω t =
(b) Output voltage rms value
R Vm = (1 + cos a − π E ) πR
∫
(b) Average load current I o (avg) =
2π / 3
2
=
3 2 VL (1 + cos a ) 2π
23. Three-phase half-wave uncontrolled rectifier
Vo(avg) =
26. Three-phase fully-controlled bridge converter feeding RLE load
(a) Average output voltage 3 2π
5π / 6
∫
Vm sin ω t d (ω t ) =
π /6
3 3 Vm 2π
3 Vo = π
(b) Output voltage rms value ⎡ 3 Vo(rms) = ⎢ ⎣ 2π
⎤ Vm2 sin 2 ω t d (ω t ) ⎥ π /6 ⎦
5π / 6
∫
(a) Output voltage
1
2
Ch wise GATE_EE_Ch10.indd 506
∫
π 3
Vo dω t =
a
3 2 V cos a π L
(b) rms voltage
1
⎡1 3 3 ⎤2 =⎢ + ⎥ Vm ⎣ 2 4π ⎦
a+
Vo (rms)
3 = π
a+
∫ a
π 3
⎤ ⎡ 3 3 V dω t = VL ⎢1 + cos 2a ⎥ 4π ⎣ ⎦
1
2
2 o
11/21/2018 6:19:32 PM
Chapter 10 • Power Electronics
VII. DC to DC Conversion
29. Boost or step-up converters
27. Choppers
(a) Case 1: Switch ON i L (t ) =
30. Buck-boost converters
i L = I o e − Rt / L
VIII. Inverters
TOFF = (1 - D)Ts
output voltage
TOn ⋅VDC = D ⋅VDC Ts
Vo =
VDC (1 − e − RT / L ) + I min e − RT / L R
32. PWM modulation index md =
where t ′ = t − TON
33. Sinusoidal PWM
(h) Average voltage
28. Buck or step-down converters
(a) Frequency modulation ratio
Vo =
mf =
f ωc = c f ref ω ref
(b) Amplitude modulation ratio or modulation index
(a) Average voltage T
1 1 Vo dt = ∫ T 0 T
TOn
Ch wise GATE_EE_Ch10.indd 507
md =
⎛ TOn ⎞ = DVs T ⎟⎠
∫ V dt = V ⎜⎝ s
0
(b) Duty ratio D=
V1 VDC
where VDC is applied input DC voltage and V1 is the peak value of fundamental component of PWM.
T Vo (avg) = On ⋅VDC = TOn f sVDC Ts
2VDC sin nω t n = 1, 3, 5 nπ
IX. Pulse Width Modulation
(f) For switch OFF period, that is, TON ≤ Ts or, 0 ≤ t - TON ≤ Ts or 0 ≤ t ′ ≤ Ts,
(g) i L (t ′ ) = I max e − Rt ′ / L
∞
∑
(e) For switch ON period, that is, 0 ≤ t ≤ TON i L (t ) =
31. Single-phase full-bridge VSI: Harmonic component of
(d) Average output voltage Vo (avg) =
Average output voltage ⎛T ⎞ ⎛ D ⎞ Vo = Vs ⎜ On ⎟ = Vs ⎜ ⎝ 1 − D ⎟⎠ ⎝ TOFF ⎠
TOn TS
(c) Duty cycle D =
Average output voltage ⎛ T ⎞ ⎛ T ⎞ ⎛ 1 ⎞ = Vs ⎜ = Vs ⎜ Vo = Vs ⎜ ⎟ ⎝ 1 − D ⎟⎠ ⎝ T − TOn ⎟⎠ ⎝ TOFF ⎠
VDC (1 − e − Rt / L ) + I o e − Rt R
(b) Case 2: Switch OFF
507
TOn TOn = T (TOn + TOFF )
s
Aref Ac
X. Speed Control of DC Motor Drives 34. Torque
T = Kaϕ I a
35. Voltage or emf
E = K a ϕω
36. Power
P = ωT = VI a − I a2 Ra
11/21/2018 6:19:34 PM
508
GATE EE Chapter-wise Solved Papers
QUESTIONS 1. A thyristorised, three phase, fully controlled converter feed a DC load that draws a constant current. Then the input AC line current to the converter has (a) an rms value equal to the DC load current (b) an average value equal to the DC load current (c) a peak value equal to the DC load current (d) a fundamental frequency component, whose rms value is equal to the DC load current (GATE 2000: 1 Mark) 2. Triangular PWM control, when applied to a three phase, BJT based voltage source inverter, introduces (a) low order harmonic voltages on the DC side (b) very high order harmonic voltages on the DC side (c) low order harmonic voltage on the AC side (d) very high order harmonic voltage on the AC side (GATE 2000: 1 Mark) 3. A diode whose terminal characteristics are related as ⎛V ⎞ iD = I S ⎜ ⎟ , where IS is the reverse saturation current ⎝ VT ⎠ and VT is the thermal voltage (= 25 mV), is biased at ID = 2 mA. Its dynamic resistance is: (a) 25 ohms (c) 50 ohms
(b) 12.5 ohms (d) 100 ohms (GATE 2000: 2 Marks)
4. In the circuit shown in the following figure, the value of the base current IB will be 5V
5 kΩ
IB + 0.7 V −
b = 80
5. A three phase voltage source inverter supplies a purely inductive three phase load. Upon Fourier analysis, the output voltage waveform is found to have an hth order harmonic of magnitude ah times that of the fundamental frequency component (ah < 1). The load current would then have an hth order harmonic of magnitude (a) zero (b) ah times the fundamental frequency component (c) hαh times the fundamental frequency component (d) ah/h times the fundamental frequency component (GATE 2000: 2 Marks) 6. A step-down chopper operates from a DC voltage source Vs, and feeds a DC motor armature with a back emf Eb. From oscilloscope traces, it is found that the current increases for time tr, falls to zero over time tf, and remains zero for time t0, in every chopping cycle. Then the average DC voltage across the freewheeling diode is (a)
(Vs tr + Eb tf ) Vs t r (b) + + t t t ( r f 0) (t r + t f + t 0 )
(c)
Vs t r + Eb (tf + t0 ) (Vs tr + Eb t0 ) (d) t + t + t ( r f 0) (t r + t f + t 0 ) (GATE 2000: 2 Marks)
7. The main reason for connecting a pulse transformer at the output stage of a thyristor triggering circuit is to (a) amplifying the power of the triggering pulse (b) provide electrical isolation (c) reduce the turn on time of the thyristor (d) avoid spurious triggering of the thyristor due to noise (GATE 2001: 1 Mark) 8. AC-to-DC circulating current dual converters are operated with the following relationship between their triggering angles (a1 and a2). (a) a1 + a2 = 180° (b) a1 + a2 = 360° (c) a1 − a2 = 180° (d) a1 + a2 = 90° (GATE 2001: 1 Mark) 9. A half-wave thyristor converter supplies a purely inductive load, as shown in the following figure. If the triggering angle of the thyristor is 120°, the extinction angle will be
−10 V (a) (b) (c) (d)
0.0 micro amperes 18.2 micro amperes 26.7 micro amperes 40.0 micro amperes
+ L − Vmsin w t (GATE 2000: 2 Marks)
Ch wise GATE_EE_Ch10.indd 508
11/21/2018 6:19:35 PM
Chapter 10 • Power Electronics
(a) 240° (c) 200°
(b) 180° (d) 120° (GATE 2001: 2 Marks)
10. A single-phase full-bridge voltage source inverter feeds a purely inductive load, as shown in the following figure, where T1, T2, T3, T4 are power transistors and D1, D2, D3, D4 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? T 1 D1
D3 T3
(b) Reactive power absorbed by the rectifier is maximum when α = 30° (c) Reactive power absorbed by the rectifier is maximum when α = 15° (d) Reactive power absorbed by the rectifier is maximum when α = 10° (GATE 2002: 2 Marks) 14. In the chopper circuit shown in the following 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? L
L
+
+
s
−
iL T4 D4
509
+
D2 T2
V1
−
+ C −
R Vo −
(a) 5 ms (c) 20 ms
(b) 10 ms (d) 2.5 ms (GATE 2001: 2 Marks)
11. 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 (GATE 2002: 1 Mark) 12. A step down chopper is operated in the continuous condition mode in steady state with a constant duty ratio D. If V0 is the magnitude of the DC output voltage and if Vs V is the magnitude of the DC input voltage, the ratio 0 is Vs given by (a) D (c)
(a) D = 1 −
Vs 2L (b) D= V0 RT
(c) D = 1 −
2L RT (d) D= RT L (GATE 2002: 2 Marks)
15. Following figure (a) shown an inverter circuit with a DC source voltage Vs. The semiconductor switches of the inverter are operated in such a manner that the pole voltages v10 and v20 are as shown in the following figure (b). What is the rms value of the pole-to-pole voltage v12?
+ Vs
V1
−
V2
(b) 1 – D
1 D (d) 1− D 1− D
0 (a)
(GATE 2002: 1 Mark) 13. A three phase thyristor bridge rectifier is used in a HDVC link. The firing angle α (as measured from the point of natural communication) is constrained to lie between 5° and 30°. If the DC side current and AC side voltage magnitudes are constant, which of the following statements is true (neglect harmonics in the ac side currents and commutation overlap in your analysis) (a) Reactive power absorbed by the rectifier is maximum when α = 5°
Ch wise GATE_EE_Ch10.indd 509
Vs V10 0
p
2p
Vs V20
q
q Tj (b)
11/21/2018 6:19:36 PM
510
GATE EE Chapter-wise Solved Papers
(a)
Vs ϕ
ϕ Vs (b) π π 2
A J3
(c) Vs
p
Vs ϕ (d) 2π π
J2 n J1
(GATE 2002: 2 Marks) 16. In the single phase diode bridge rectifier shown in the following figure, the load resistor is R = 50 Ω. The source voltage is v = 200 sin (ωt), where ω = 2π × 50 radians per second. The power dissipated in the load resistor R is
V
R
p
G
n
K (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. (GATE 2003: 1 Mark)
(a)
3200 400 W (b) W π π
(c) 400 W
(d) 800 W
19. The figure given below 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
(GATE 2002: 2 Marks) D 17. The variation of drain current with gate-to-source voltage (ID − VGS characteristic) of a MOSFET is shown in the following figure. The MOSFET is
VDS G
ID
S IS (a) IS 0
VGS
(a) an n-channel depletion mode device. (b) an n-channel enhancement mode device. (c) a p-channel depletion mode device. (d) a p-channel enhancement mode device. (GATE 2003: 1 Mark) 18. The figure given below 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
Ch wise GATE_EE_Ch10.indd 510
VDS
(b) IS
VDS
11/21/2018 6:19:37 PM
Chapter 10 • Power Electronics
(c)
511
10 V IS ID
RD
VDS
(d)
IS
VDS
(a) 2.8 mA
(b) 2.0 mA
(c) 1.4 mA
(d) 1.0 mA (GATE 2003: 2 Marks)
(GATE 2003: 1 Mark) 20. The speed/torque regimes in a DC motor and the control methods suitable for the same are given, respectively in Group II and Group I.
23. A voltage signal 10 sinwt is applied to the circuit with ideal diodes, as shown in the following figure. The maximum and minimum values of the output waveform Vout of the circuit are, respectively. 10 kW
Group I
Group II
P. Field control
1. Below base speed
+
Q. Armature control
2. Above base speed
Vin −
3. Above base torque
+
4V
4V
10 kW
−
4. Below base torque The match between the control method and the speed/ torque regime is (a) P − 1; Q − 3 (b) P − 2; Q − 1 (c) P − 2; Q − 3 (d) P − 1; Q − 4 (GATE 2003: 1 Mark) 21. A fully-controlled natural commutated three-phase bridge rectifier is operating with a firing angle a = 30°. The peak-to-peak voltage ripple expressed as a ratio of the peak output DC voltage at the output of the converter bridge is (a) 0.5
(b)
(a) +10 V and −10 V
(b) +4 V and −4 V
(c) +7 V and −4 V
(d) +4 V and −7 V (GATE 2003: 2 Marks)
24. A phase-controlled half-controlled single-phase converter is shown in the following figure. The c ontrol angle a = 30° a = 30°
VDC IDC
VAC
3 2
⎛ 3⎞ (c) ⎜1 − 2 ⎟ (d) 3 − 1 ⎠ ⎝
The output DC voltage waveshape will be as shown in VDC
(GATE 2003: 1 Mark) 22. For the n-channel enhancement MOSFET shown in the following figure, the threshold voltage Vth = 2 V. The drain current ID of the MOSFET is 4 mA when drain resistance RD is 1 kΩ. If the value of RD is increased to 4 kΩ, drain current ID will become
Ch wise GATE_EE_Ch10.indd 511
Vout
VDC
t
Figure A
t Figure B
11/21/2018 6:19:38 PM
512
GATE EE Chapter-wise Solved Papers
VDC
28. The circuit in the following figure shows a full-wave rectifier. The input voltage is 230 V (rms) single-phase AC. The peak reverse voltage acreoss the diodes D1 and D2 is
VDC
t
t
D1
Figure C
230 V, 50 Hz, AC
Figure D
(a) Figure A (c) Figure C
(b) (d)
Figure B Figure D
230 V/50-0-50 V
(GATE 2003: 2 Marks) 25. A chopper is employed to charge a battery as shown in the following 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 5A L = 20 mH VDC
D2
60 V
(a) 100 2 V
(b) 100 V
(c) 50 2 V
(d) 50 V (GATE 2004: 1 Mark)
29. The triggering circuit of a thyristor is shown in the following 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 Load
t 12 V
Chopper VDC
(a) 0.48 A (c) 2.4 A
(b) 1.2 A (d) 1 A (GATE 2003: 2 Marks)
26. An inverter has a periodic output voltage with the output waveform as shown in the following figure. 1 a 2p 0
Vb = 12 ± 4 V R
200 µs 1 ms
p −1
a
100 V
Vb (a) 10000 Ω (c) 1200 Ω
(b) 1600 Ω (d) 800 Ω (GATE 2004: 1 Mark)
30. The circuit in the following figure shows a three-phase half-wave rectifier. The source is a symmetrical, threephase four-wire system. The line-to-line voltage of the source is 100 V. The supply frequency is 400 Hz. The ripple frequency at the output is R
Y When the conduction angle a = 120°, the rms fundamental component of the output voltage is
B
(a) 0.78 V (c) 0.90 V
N
(b) 1.10 V (d) 1.27 V (GATE 2003: 2 Marks)
27. With reference to the output waveform given in Question 26, the output of the converter will be free from 5th harmonic when (a) a = 72° (b) a = 36° (c) a = 150° (d) a = 120° (GATE 2003: 2 Marks)
Ch wise GATE_EE_Ch10.indd 512
R
(a) 400 Hz (c) 1200 Hz
(b) 800 Hz (d) 2400 Hz (GATE 2004: 1 Mark)
31. A MOSFET rated for 15 A, carries a periodic current as shown in the following figure. The ON state resistance of the MOSFET is 0.15 Ω. The average ON state loss in the MOSFET is
11/21/2018 6:19:39 PM
513
Chapter 10 • Power Electronics
(a) 10 A (c) 20 A
I
(b) 30 A (d) 40 A (GATE 2004: 2 Marks)
10 A wt p
0
3p
2p
(a) 33.8 W (c) 7.5 W
(b) 15.0 W (d) 3.8 W (GATE 2004: 2 Marks)
32. The triac circuit shown in the following figure controls the AC output power to the resistive load. The peak power dissipation in the load is
a=
230Ö2 sin wt
35. 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 Ω. The converter is working from a 230 V, single-phase AC source with a firing angle of 30°. Under this operating condition, the speed of the motor will be (a) 339 rpm (c) 366 rpm
36. 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
p 4
R = 10 Ω
(a) 400 V (c) 400
(a) 3968 W (c) 7935 W
(b) 5290 W (d) 10580 W (GATE 2004: 2 Marks)
33. The following 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 S
100 V
10 W
D
(a) 1.6 A (c) 8.0 A
(b) 359 rpm (d) 386 rpm (GATE 2004: 2 Marks)
(b) 6.4 A (d) 10.0 A (GATE 2004: 2 Marks)
(b) 400 2 V
400 2 V (d) V 3 3 (GATE 2005: 1 Mark)
37. The output voltage waveform of a three-phase squarewave inverter contains (a) only even harmonics (b) both odd and even harmonics (c) only odd harmonics (d) only triple harmonics (GATE 2005: 1 Mark) 38. The following figure shows the voltage across a power semiconductor device and the current through the device during a switching transition. Is the transition a turn ON transition or a turn OFF transition? What is the energy lost during the transition? v, i V v
34. The following figure shows a chopper. The device S1 is the main switching device. S2 is the auxiliary commutation device. S1 is rated for 400 V, 60 A. S2 is rated for 400 V, 30 A. The load current is 20 A. The main device operates with a duty ratio of 0.5. The peak current through S1 is
I
i t t1
S1 2 mF
S2
200 V
D 200 mH
Ch wise GATE_EE_Ch10.indd 513
20 A
(a) Turn On,
t2
VI (t1 + t 2 ) 2
(b) Turn OFF, VI (t1 + t2)
11/21/2018 6:19:40 PM
514
GATE EE Chapter-wise Solved Papers
(c) Turn ON, VI (t1 + t2) (d) Turn OFF,
every positive half cycle of the input voltage. If peak value of the instantaneous output voltage equals 230 V, then firing angle a is close to
VI (t1 + t 2 ) 2 (GATE 2005: 2 Marks)
+ 39. An electronic switch S is required to block voltages of either polarity during its OFF state as shown in the Fig. (a). This switch is required to conduct in only one direction during its ON state as shown in Fig. (b) i 1 ±
1′
S − +
1′
1 S ±
(a) i
1′
1 S (b)
Which of the following are valid realisations of the switch S? 1′
− (a) 45° (c) 90°
(b) 135° (d) None of these (GATE 2005: 2 Marks)
42. The speed of a three-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 three-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 (a)
Q.
P. 1
R Vo
1′
S − +
1
230 V rms 50 Hz
V
1′
1
f R. 1
1′
S. 1
1′
(b) (a) Only P (c) P and R
50
V
(b) P and Q (d) R and S (GATE 2005: 2 Marks)
40. The given figure shows a step-down chopper switched at 1 kHz with a duty ratio k = 0.5. The peak-peak ripple in the load current is close to
f 50
(c)
V
200 mH + f i
5Ω
50
100 V − (a) 10 A (c) 0.125 A
V
(b) 0.5A (d) 0.25 A (GATE 2005: 2 Marks)
41. Consider a phase-controlled converter shown in the following figure. The thyristor is fired at an angle a in
Ch wise GATE_EE_Ch10.indd 514
(d)
f
50 (GATE 2006: 1 Mark)
11/21/2018 6:19:41 PM
Chapter 10 • Power Electronics
43. A single-phase half-wave uncontrolled converter circuit is shown in the following figure. A two-winding transformer is used at the input for isolation. Assuming the load current to be constant and v = Vm sinwt, the current waveform through diode D2 will be iL
D1
Vin
V
L
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 (GATE 2006: 2 Marks) 46. A voltage commutation circuit is shown in the f ollowing figure. If the turn OFF time of the SCRs is 50 ms and a safety margin of 2 is considered, then what will be the approximate minimum value of capacitor required for proper commutation?
D2
(a)
50 Ω
p
0
515
50 Ω C
2p 100 V
(b)
T1 p
0
2p
(a) 2.88 mF (c) 0.91 mF
(c) 0
p
2p
p
2p
(d) 0
(GATE 2006: 1 Mark) 44. 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 the following figure. It is found that the output voltage is free from fifth harmonic for pulse width 144°. What will be percentage of third harmonic present in the output voltage (Vo3/Vo1max)? V
2d
T2 (b) 1.44 mF (d) 0.72 mF (GATE 2006: 2 Marks)
47. A solar cell of 350 V is feeding power to an AC supply of 440 V, 50 Hz through a three-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 Ω, then each thyristor will be reverse biased for a period of (a) 125° (b) 120° (c) 60° (d) 55° (GATE 2006: 2 marks) 48. A single-phase bridge converter is used to charge a battery of 200 V having an internal resistance of 0.2 Ω as shown in the following figure. The SCRs are triggered by a constant DC signal. If SCR 2 gets open circuited, then what will be the average charging current?
3π/2 π/2 −V (a) 0.0% (c) 31.7%
π 2d
(b) 19.6% (d) 53.19% (GATE 2006: 2 Marks)
45. A three-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 60°. The load current is assumed constant at 10 A due to high load inductance.
Ch wise GATE_EE_Ch10.indd 515
200 V battery
2π
230 V 50 Hz
(a) 23.8 A (c) 11.9 A
SCR1
SCR2
SCR3
SCR4 (b) 15 A (d) 3.54 A (GATE 2006: 2 Marks)
11/21/2018 6:19:42 PM
516
GATE EE Chapter-wise Solved Papers
49. An SCR having a turn ON time of 5 ms, latching current of 50 mA and holding current of 40 mA is triggered by a short duration pulse and is used in the circuit shown in the following figure. The minimum pulse width required to turn the SCR ON will be
20 Ω 100 V
5 kΩ 0.5 H
(a) 251 ms (c) 100 ms
(b) 150 ms (d) 5 ms (GATE 2006: 2 Marks)
Statement for Linked Answer Questions 50 and 51: A voltage commutated chopper operating at 1 kHz is used to control the speed of DC motor as shown in the following figure. The load current is assumed to be constant at 10 A. M + 1 µF
A
iL
V = 250 V −
2 mH
50. The minimum time in ms for which the SCR (M) should be ON is (a) 280 (b) 140 (c) 70 (d) 0 (GATE 2006: 2 Marks) 51. The average output voltage of the chopper will be (a) 70 V (b) 47.5 V (c) 35 V (d) 0 V (GATE 2006: 2 Marks) 52. A single-phase fully-controlled thyristor bridge AC − DC converter is operating at a firing angle of 25° and an overlap angle 10° 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 (GATE 2007: 1 Mark) 53. A three-phase, fully controlled thyristor bridge converter is used as line commutated inverter to feed 50 kW power at 420 V DC to a three-phase, 415 V(line), 50 Hz
Ch wise GATE_EE_Ch10.indd 516
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 (GATE 2007: 1 Mark) 54. 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 freewheeling. (b) because the converter does not provide for freewheeling for high values of triggering angles. (c) or else the freewheeling 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. (GATE 2007: 1 Mark) 55. 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 carier devices, MOSFETs are voltage driven. (b) True, because MOSFETs have 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. (GATE 2007: 1 Mark) 56. A single-phase voltage source inverter is controlled in a single pulse-width modulated mode with a pulse width of 150° in each half cycle. Total harmonic distortion is defined as THD =
2 Vrms − Vi 2 × 100 V1
where V1 is the rms value of the fundamental component of the output voltage. The THD of output AC voltage wave form is (a) 65.65% (b) 48.42% (c) 31.83% (d) 30.49% (GATE 2007: 2 Marks) 57. A voltage source inverter is used to control the speed of a three-phase, 50 Hz, squirrel cage induction motor.
11/21/2018 6:19:42 PM
517
Chapter 10 • Power Electronics
Its slip for rated torque is 4%. The flux is maintained at rated value. If the stator resistance and rotational losses are neglected, then the f requency of the impressed voltage to obtain twice the rated torque at starting should be (a) 10 Hz (b) 5 Hz (c) 4 Hz (d) 2 Hz (GATE 2007: 2 Marks) 58. A three-phase, 440 V, 50 Hz AC mains fed thyristor bridge is feeding a 440 V DC, 15kW. 1500 rpm separately excited DC motor with a ripple free continuous 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 (GATE 2007: 2 Marks) 59. 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 Ω. The charging current is (a) 2.43 A (b) 1.65 A (c) 1.22 A (d) 1.0 A (GATE 2007: 2 Marks) 60. The circuit in the figure is a current commutated DC− DC chopper where, ThM is the main SCR and ThAux is the auxiliary SCR, The load current is constant at 10 A. ThM is ON. ThAux is triggered at t = 0. ThM is turned OFF between ThM
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. 1W R +10 V
PT
S C R
L
200 V
61. The resistance R should be (a) 4.7 kW (c) 47 W
(b) 470 W (d) 44.7 W (GATE 2007: 2 Marks)
62. The minimum approximate volt-second rating of the 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 be applied) (a) 2000 mVs (b) 200 mVs (c) 20 mVs (d) 2.0 mVs (GATE 2007: 2 Marks) 63. A 3 V DC supply with an internal resistance of 2 Ω supplies a passive non-linear r esistance characterised by the 2 . The power dissipated in the non-linrelation VNL = I nL ear resistance is (a) 1.0 W (b) 1.5 W (c) 2.5 W (d) 3.0 W (GATE 2007: 2 Marks) 64. In the single-phase voltage controller circuit shown in the following figure, for what range of triggering angle (a), the output voltage (vo) is not controllable?
ThAux Load 230 V
10 µF 25.28 µH
Vo
Vs −
(a) 0 ms < t ≤ 25 ms (b) 25 ms < t ≤ 50 ms (c) 50 ms t ≤ 75 ms (d) 75 ms < t ≤ 100 ms
50 W
+
(GATE 2007: 2 Marks)
(b) 45° < a < 135° (d) 135° < a < 180° (GATE 2008: 1 Mark)
Common Data for Questions 61 and 62: A 1:1 pulse transformer (PT) is used to trigger the SCR given in the following 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
65. A three-phase voltage source inverter is operated in 180° conduction mode. Which one of the following statements is true? (a) Both pole-voltage and line-voltage will have third harmonic components.
Ch wise GATE_EE_Ch10.indd 517
(a) 0° < a < 45° (c) 90° < a < 180°
j50 W
11/21/2018 6:19:43 PM
518
GATE EE Chapter-wise Solved Papers
(b) Pole-voltage will have third harmonic component but line-voltage will be free from third harmonic. (c) Line-voltage will have third harmonic component but pole-voltage will be free from third harmonic. (d) Both pole-voltage and line-voltage will be free from third harmonic components. (GATE 2008: 1 Mark) 66. 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 capacitors C1 and C2 will be
(a)
Vo
π
0
2π ω t
(a) Vo (b)
vC1 t=0 S
D2
+ −
0
π
4π/3
2π ωt
C1 +
5 sinwt
π/3
D1
(a) vC1 = 10 V,
vC2 = 5 V
(b) vC1 = 10 V,
vC2 = −5 V
(c) vC1 = 5 V,
vC2 = 10 V
(d) vC1 = 5 V,
vC2 = −10 V
C2
−
vC2
Rload
(b) Vo (c)
0
(GATE 2008: 2 Marks) 67. A single-phase fully-controlled bridge converter supplies a load drawing constant and ripple free load current. If triggering angle is 30°, then input power factor will be (a) 0.65 (b) 0.78 (c) 0.85 (d) 0.866 (GATE 2008: 2 Marks) 68. A single-phase half-controlled converter shown in the following figure is feeding power to highly inductive load. The converter is operating at a firing angle of 60°. If firing pulses are suddenly removed, then steady state voltage (Vo) waveform of the converter will become
Vo
π/3
π
4π/3
2π ω t
(d)
(c) Vo
0
π
2π ω t
(d)
(GATE 2008: 2 Marks)
69. A 220 V, 20 A, 1000 rpm, separately excited DC motor has an armature resistance of 2.5 Ω. The motor is controlled by a step down chopper with a frequency of 1 kHz. The input DC voltage to the chopper is 250 V. The duty cycle of the chopper for the motor to operate at a speed of 600 rpm delivering the rated torque will be (a) 0.518 (b) 0.608 (c) 0.852 (d) 0.902 (GATE 2008: 2 Marks)
Ch wise GATE_EE_Ch10.indd 518
11/21/2018 6:19:43 PM
519
Chapter 10 • Power Electronics
70. A 220 V, 1400 rpm, 40 A separately excited DC motor has an armature resistance of 0.4 Ω. The motor is fed from a single-phase circulating current dual converter with an input AC line voltage of 220 V (rms). The approximate firing angles of the dual converter for motoring operation at 50% of rated torque and 1000 rpm will be (a) 43°, 137° (b) 43°, 47° (c) 39°, 141° (d) 39°, 51° (GATE 2008: 2 Marks) 71. A single-phase voltage source inverter is feeding a purely inductive load as shown in the following figure. The inverter is operated at 50 Hz in 180° square wave mode. Assume that the load current does not have any DC component. The peak value of the inductor current io will be
total harmonic distortion (%THD) and the rms value of fundamental component of the 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 (GATE 2008: 2 Marks) 74. In the circuit shown in the following figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous. The average voltage across the load and the average current through the diode will, respectively, be
IL = 4 A
ID
L
20 V
D
S
Vo
Load
0.1 H 200 V
(a) 10 V, 2 A (c) 40 V, 2 A
io
(b) 10 V, 8 A (d) 40 V, 8 A (GATE 2008: 2 Marks)
(a) 6.37 A (c) 20 A
(b) 10 A (d) 40 A (GATE 2008: 2 Marks)
75. The following circuit has a source voltage Vs as shown in the graph. The current through the circuit is also shown. The element connected between a and b could be
72. 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 following 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 Io = 10 A will be
2W
− Vs +
Vs (volts)
io
a
230 V 50 Hz
15 10 5 0 −5 −10 −15 0
150 V
(b) 51° (d) 136° (GATE 2008: 2 Marks)
73. 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 30°. The approximate
Ch wise GATE_EE_Ch10.indd 519
R 10 kΩ
100 200 300 400 Time (ms)
Current (mA)
(a) 44° (c) 129°
1.5 1 0.5 0 −0.5 −1 −1.5 0
b
100 200 300 400 Time (ms)
11/21/2018 6:19:44 PM
GATE EE Chapter-wise Solved Papers
(a) a
(d)
b
(a) a (b) (b) a (c)
1.5 Current
520
b b
1 0.5
(c) 0
(d) a
0
b (GATE 2009: 1 Mark)
(d)
76. 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. (GATE 2009: 1 Mark) 77. 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
+
50
78. The current source inverter shown in the following figure is operated by alternately turning on thyristor pairs (T1, T2) and (T3, T4). If load is purely resistive, then theoretical maximum output frequency obtainable will be
T1
T3
0.1 µF −
+
D1 D3
10 A
10 Ω D2 −
+
0.1 µF
L = (0.1/p)H −
40
(GATE 2009: 2 Marks)
−
Vs = 10 sin 100p t
(a)
20 30 Time (ms)
D4
D +
10
T4
T2
Current
1.5 (a) 125 kHz (c) 500 kHz
1
(GATE 2009: 2 Marks)
0.5 00
10
(b) 1.5 Current
(b) 250 kHz (d) 50 kHz
20 30 Time (ms)
40
50
1
79. 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 L0 to be negligible.
0.5
+ 00
10
20 30 Time (ms)
40
50
TM L1
(c) Current
1.5
100 V 1
Ch wise GATE_EE_Ch10.indd 520
TA
L0
0.5 00
C1 − +
D1 10
20 30 Time (ms)
40
50
D0
C0
8Ω
−
11/21/2018 6:19:45 PM
Chapter 10 • Power Electronics
(a) 0.2 mF (c) 2 mF
(b) 0.02 mF (d) 20 mF
(a) 150 V (c) 300 V
521
(b) 210 V (d) 100 V
(GATE 2009: 2 Marks)
(GATE 2010: 1 Mark)
80. 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.
Statement for Linked Answer Questions 83 and 84: The LC circuit shown in the given figure has an inductance L = 1 mH and a capacitance C = 10 mF. L
A +
−
B +
−
C
−
+
i
D
is
is I
is
Vs
IV
Vs
Codes: A B (a) I II (b) II IV (c) IV III (d) IV III
Vs
Vs
C III I I II
t=0
is
III
II
C
−
+
D IV III II I
− 100 V +
83. The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t = 0. The current i through the circuit is (a) 5 cos (5 × 103t) A (b) 5 sin (104t) A (c) 10 cos (5 × l03t) A (d) 10 sin (104t) A (GATE 2010: 2 Marks)
(GATE 2009: 2 Marks) 81. The power electronic converter shown in the following 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
84. The LC circuit of Question 83 is used to c ommutate a thyristor, which is initially carrying a current of 5 A as shown in the following figure. The values and initial conditions of L and C are the same as in the given figure. The switch is closed at t = 0. If forward drop is negligible, then time taken for the device to turn OFF is L
A P Vi
L i + Vo −
B
(a) step-down chopper (buck converter) (b) half-wave rectifier (c) step-up chopper (boost converter) (d) full-wave rectifier
C t=0 100 V
20 Ω
5A
(a) 52 ms (c) 312 ms
(b) 156 ms (d) 26 ms (GATE 2010: 2 Marks)
(GATE 2010: 1 Mark) 82. The fully controlled thyristor converter in the given figure is fed from a single-phase source. When the firing angle is 0°, the DC output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60°, assuming continuous conduction?
− 100 V +
85. A three-phase current source inverter used for the speed control of an induction motor is to be realized using MOSFET switches as shown in the following figure. Switches S1 to S6 are identical switches. Id A
+ S1
Ch wise GATE_EE_Ch10.indd 521
S5
S6
S2
Induction motor
B A
VDC −
S3
S4 B
11/21/2018 6:19:46 PM
522
GATE EE Chapter-wise Solved Papers
The proper configuration for realising switches S1 to S6 is
If the maximum value of load current is 10A, then the maximum current through the main (M) and auxiliary (A) thyristors will be
(a)
(a) iM max = 12 A and iA max = 10 A
(b)
A
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 (GATE 2011: 2 Marks) (a)
B
(c)
(b) (d)
A
B A
Statement for Linked Answer Questions 89 and 90: A solar energy installation utilises a three-phase bridge converter to feed energy into power system through a transformer of 400 V/400 V, as shown in the following figure. The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance 10 Ω. Filter choke
(c)
B
86. Circuit turn-OFF time of an SCR is defined as the time (a) taken by the SCR to turn 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. (GATE 2011: 1 Mark) 87. A lossy capacitor CX, rated for operation at 5 kV, 50 Hz is represented by an equivalent circuit with an ideal capacitor CP is parallel with a resistor RP. The value of CP is found to be 0.102 mF and the value of RP = 1.25 MΩ. Then the power loss and tand 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 (d) 20 W and 0.04 (GATE 2011: 2 Marks) 88. A voltage commutated chopper circuit, operated at 500 Hz, is shown in the following figure. M
iM
A
iA
iL = 10 A
+ 0.1 µF
1 mH
Ch wise GATE_EE_Ch10.indd 522
89. The maximum current through the battery will be (a) 14 A (b) 40 A (c) 80 A (d) 94 A (GATE 2011: 2 Marks) 90. The kVA rating of the input transformer is (a) 53.2 kVA (b) 46.0 kVA (c) 22.6 kVA (d) 19.6 kVA (GATE 2011: 2 Marks) Common Data Questions 91 and 92: The input voltage given to a converter is v i = 100 2 sin(100 π t )V The current drawn by the converter is i i = (10 2 (100 π t − π / 3) + 5 2 sin(300 π t + π / 4) +2 2 sin(500 π t − π / 6)A 91. The input power factor of the converter is (a) 0.31 (b) 0.44 (c) 0.5 (d) 0.71 (GATE 2011: 2 Marks)
Load
200 V
-
Battery
B (d) 2011: 1 Mark) (GATE
92. The active power drawn by the converter is (a) 181 W (b) 500 W (c) 707 W (d) 887 W (GATE 2011: 2 Marks)
11/21/2018 6:19:47 PM
Chapter 10 • Power Electronics
93. The average power delivered to an impedance (4 − j 3) Ω by a current 5 cos (100 pt + 100) A is (a) 44.2 W (b) 50 W (c) 62.5 W (d) 125 W (GATE 2012: 1 Mark) 94. The i-v characteristics of the diode in the circuit given below are
(a) 6.6 A (c) 5.8 A
523
(b) 5.0 A (d) 4.2 A (GATE 2012: 2 Marks)
Common Data for Questions 98 and 99: In the three-phase inverter circuit shown in the following figure, the load is balanced and the gating scheme is 180°-conduction mode. All the switching devices are ideal.
1 kW
+ −
10 V
i + v −
S1
S3
S5
+ −
R = 20 W
vph Vd
R S4 ⎧ v − 0.7 A; for v ≥ 0.7 ⎪ i = ⎨ 500 ⎪⎩0 A ; for v < 0.7 The current in the circuit is (a) 10 mA (c) 6.67 mA
(b) 9.3 mA (d) 6.2 mA (GATE 2012: 1 Mark)
95. A half-controlled single-phase bridge rectifier is supplying an RL 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
S6
S2
3-phase inverter
R 3-phase balanced load
98. The rms value of load phase voltage is (a) 106.1 V (b) 141.4 V (c) 212.2 V (d) 282.8 V (GATE 2012: 2 Marks) 99. If the DC bus voltage Vd = 300 V, the power consumed by three-phase load is (a) 1.5 kW (b) 2.0 kW (c) 2.5 kW (d) 3.0 kW (GATE 2012: 2 Marks)
100. Thyristor T in the following figure is initially OFF and is 1 ⎛ a⎞ (b) ⎜⎝1− ⎟⎠ 2 π triggered with a single pulse of width 10 ms. It is given a a ⎛ 100 ⎞ ⎛ 100 ⎞ (c) (d) μH and C = ⎜ μF that L = ⎜ ⎟ 2π π ⎝ π ⎠ ⎝ π ⎟⎠ (GATE 2012: 1 Mark) Assuming latching and holding currents of the thyristor 96. The typical ratio of latching current to holding current are both zero and the initial charge on C is zero, T conin a 20A thyristor is ducts for (a) 5.0 (b) 2.0 + (c) 1.0 (d) 0.5 L T (GATE 2012: 1 Mark) 15 V C 97. In the circuit shown in the following figure, an ideal switch S is operated at 100 kHz with a duty ratio of 50%. Given that ∆iC is 1.6 A peak-to-peak and Io is 5A DC, the peak current in S is (a) 10 ms (b) 50 ms S L (c) 100 ms (d) 200 ms + DiC Io (GATE 2013: 2 Marks) + R vo 24 V Common Data for Questions 101 and 102: In the following C D − figure, the chopper feeds a resistive load from a battery source. MOSFET Q is switched at 250 kHz, with a duty ratio of 0.4. − All elements of the circuit are assumed to be ideal. (a)
Ch wise GATE_EE_Ch10.indd 523
11/21/2018 6:19:48 PM
524
GATE EE Chapter-wise Solved Papers
Ra = 1 Ω, 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
100 µH
+ − 12 V
Q
470 µF
20 kΩ + 200 V −
101. The average source current in amperes in steady-state is (a) 3/2 (b) 5/3 (c) 5/2 (d) 15/4 (GATE 2013: 2 Marks) 102. The peak-to-peak source current ripple in amperes is (a) 0.96 (b) 0.144 (c) 0.192 (d) 0.288 (GATE 2013: 2 Marks) Statement for Linked Anser Questions 103 and 104: The voltage source inverter (VSI) shown in the following figure 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 Ω and L = 9.55 mH. Q1
D3 + vo −
+ − Q4
(b) 0.5 (d) 0.7 (GATE 2013: 2 Marks)
106. The following figure shows four electronic switches (i), (ii), (iii) and (iv). Which of the switches can block voltages of either polarity (applied between terminals ‘a’ and ‘b’) when the active device is in the OFF state? a
b (i)
L io
R
Q2 D4
D2
103. In the interval when vo < 0 and io > 0 the pair of devices which conducts the load current is (a) Q1, Q2 (b) Q3, Q4 (c) D1, D2 (d) D3, D4 (GATE 2013: 2 Marks) 104. Appropriate transition, that is, zero voltage switching (ZVS)/zero current switching (ZCS) 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 (GATE 2013: 2 Marks) 105. The separately-excited DC motor given in the figure 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,
Ch wise GATE_EE_Ch10.indd 524
(a) 0.4 (c) 0.6
a
a
a
b (iii)
b
Q3 D1
VDC
La, Ra
b (ii)
(iv)
(a) (i), (ii) and (iii) (b) (ii), (iii) and (iv) (c) (ii) and (iii) (d) (i) and (iv) (GATE 2014: 1 Mark) 107. A step-up chopper is used to feed a load at 400 V DC from a 250 V DC source. The inductor c urrent is continuous. If the ‘off’ time of the switch is 20 μs, the switching frequency of the chopper in kHz is . (GATE 2014: 1 Mark) 108. In a constant V/f control of induction motor, the ratio V/f is maintained constant from 0 to base frequency, where V is the voltage applied to the motor at fundamental frequency f. Which of the following statements relating to low frequency operation of the motor is TRUE? (a) At low frequency, the stator flux increases from its rated value. (b) At low frequency, the stator flux decreases from its rated value. (c) At low frequency, the motor saturates. (d) At low frequency, the stator flux remains unchanged at its rated value. (GATE 2014: 1 Mark)
11/21/2018 6:19:48 PM
525
Chapter 10 • Power Electronics
angle of the bridge converter is a = 30°. The input power . factor of the bridge is
109. The figure shows the circuit diagram of a rectifier. The load consists of a resistance 10 Ω and an inductance 0.05 H connected in series. Assuming ideal thyristor and ideal diode, the thyristor firing angle (in degree) needed to obtain an average load voltage of 70 V is .
iS + Load
VS
+ 325 sin(314t) V
−
Load − (GATE 2014: 2 Marks)
(GATE 2014: 2 Marks)
110. Figure (i) shows the circuit diagram of a chopper. The switch S in the circuit in figure (i) is switched such that the voltage VD across the diode has the wave shape as shown in figure (ii). The capacitance C is large so that the voltage across it is constant. If switch S and the diode are ideal, the peak to peak ripple (in A) in the inductor current is .
113. A single-phase SCR based AC regulator is feeding power to a load consisting of 5 Ω resistance and 16 mH inductance. The input supply is 230 V, 50 Hz AC. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are (a) 30° and 46 A (b) 30° and 23 A (c) 45° and 23 A (d) 45° and 32 A
S
1 mH
(GATE 2014: 2 Marks)
+ C
VD
100 V
Load
− (i)
VD
114. The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 μs is applied to the SCR. The maximum value of R in Ω to ensure successful firing of the SCR is . SCR
100 V
+ t(ms) 0
0.05
0.1 (ii)
0.15
500 Ω
100 V
R
0.2 200 mH
(GATE 2014: 2 Marks) 111. The figure shows one period of the output voltage of an inverter. a should be chosen such that 60° < a < 90°. If rms value of the fundamental component is 50 V, . then a in degree is
(GATE 2014: 2 Marks) 115. A three-phase fully controlled bridge converter is fed through star-delta transformer as shown in the figure. IR
100 V
0
100 V
100 V
a
wt (degree) 180 − a 180 360 − a 180 + a 360 −100 V −100 V −100 V (GATE 2014: 2 Marks)
112. A fully controlled converter bridge feeds a highly inductive load with ripple free load current. The input supply (VS) to the bridge is a sinusoidal source. Triggering
Ch wise GATE_EE_Ch10.indd 525
R
1:K
Io
Y
B
The converter is operated at a firing angle of 30°. Assuming the load current (Io) to be virtually constant at 1 pu and transformer to be an ideal one, the input phase current waveform is
11/21/2018 6:19:49 PM
526
GATE EE Chapter-wise Solved Papers
(a)
(a)
2/3 K
IR
1/3 K
p
0
2p
VDC 2 q
VAO
p
p−q
2p
−VDC 2
(b)
2K/3 K/3
IR
p
0
2p
(b) VDC 2 VAO
q
p−q
p
2p
q
p−q
p
2p
(c) 2K/3
IR p
0
2p
(c) VDC 2 VAO
(d) 2/3 K
IR p
0
2p (d)
VDC 2
(GATE 2014: 2 Marks) 116. A single-phase voltage source inverter shown in figure is feeding power to a load. The triggering pulses of the devices are also shown in the figure.
VAO
q p−q
p
2p
−VDC 2 (GATE 2014: 2 Marks)
S1
117. In the following chopper, the duty ratio of switch S is 0.4. If the inductor and capacitor are sufficiently large to ensure continuous inductor current and ripple free capacitor voltage, the charging current (in ampere) of the 5 V battery, under steady-state, is ————————.
S2
C iL VDC
O
A C
Load
S3
B S4
S
L 3Ω
+ 20 V − S1, S4
q
p−q
p
2p
+ −
5V
(GATE 2015: 1 Mark) S2, S3
p+ q
2p − q
If the load current is sinusoidal and is zero at 0, p, 2p..., the node voltage VAO has the waveform
Ch wise GATE_EE_Ch10.indd 526
118. Consider a HVDC link which uses thyristor based line commutated converters as shown in the following figure. For a power flow of 750 MW from system 1 to system 2, the voltages at the two ends, and the current, are given by: V1 = 500 kV, V2 = 485 kV and I = 1.5 kA. If the
11/21/2018 6:19:51 PM
527
Chapter 10 • Power Electronics
direction of power flow is to be reversed (that is, from system 2 to system 1) without changing the electrical connections, then which one of the following combinations is feasible? System 1
(b) VL =
δ VDC 2
System 2
I
1 VDC 1− δ 1 and VC = VDC 1− δ
(a) VL = 0 and VC =
δ VDC 1− δ δ and VC = VDC 1− δ (GATE 2015: 2 Marks)
(c) VL = 0 and VC = + V1 − (a) (b) (c) (d)
+ V2 −
(d) VL =
V1 = −500 kV, V2 = −485 kV and I = 1.5 kA V1 = −485 kV, V2 = −500 kV and I = 1.5 kA V1 = 500 kV, V2 = 485 kV and I = −1.5 kA V1= −500 kV, V2 = −485 kV and I = −1.5 kA (GATE 2015: 1 Mark)
119. The circuit shown is meant to supply a resistive load RL from two separate DC voltage sources. The switches S1 and S2 are controlled so that only one of them is ON at any instant. S1 is turned on for 0.2 ms and S2 is turned on for 0.3 ms in a 0.5 ms switching cycle time period. Assuming continuous conduction of the inductor current and negligible ripple on the capacitor voltage, the output voltage Vo (in Volt) across RL is ————————. S1 L S2 10 V
C + −
+ − 5V
RL
+ Vo −
(GATE 2015: 2 Marks) 120. A self-commutating switch SW, operated at duty cycle d is used to control the load voltage as shown in the figure VL
δ VDC 2
121. The single-phase full-bridge voltage source inverter (VSI), shown in figure, has an output frequency of 50 Hz. It uses unipolar pulse width modulation with switching frequency of 50 kHz and modulation index of 0.7. For Vin = 100 V DC, L = 9.55 mH, C = 63.66 mF, and R = 5 W, the amplitude of the fundamental component in the output voltage Vo (in volt) under steady state is +
+ Vin + −
Full-bridge V R VSI −
d
C
VC
RL
Under steady state operating conditions, the average voltage across the inductor and the capacitor respectively, are
Ch wise GATE_EE_Ch10.indd 527
Vo −
122. A 3-phase 50 Hz square wave (6 step) VSI feeds a 3-phase, 4-pole induction motor. The VSI line voltage has a dominant 5th harmonic component. If the operating slip of the motor with respect to fundamental component voltage is 0.04, the slip of the motor with respect to 5th harmonic component of voltage is ————————. (GATE 2015: 2 Marks) 123. A buck converter feeding a variable resistive load is shown in the figure. The switching frequency of the switch S is 100 kHz and the duty ratio is 0.6. The output voltage Vo is 36 V. Assume that all the components are ideal, and that the output voltage is ripple free. The value of R (in ohm) that will make the inductor current (iL) just continuous is ________.
D
L SW
R
C
(GATE 2015: 2 Marks)
S
VDC
L
+ 60 V
iL
5 mH + 36 V Vo
R
(GATE 2015: 2 Marks) 124. For the switching converter shown in the following figure, assume steady-state operation. Also assume that the
11/21/2018 6:19:52 PM
528
GATE EE Chapter-wise Solved Papers
components are ideal, the inductor current is always positive and continuous and switching period is TS. If the voltage VL is as shown, the duty cycle of the switch S is ________.
VL +
−
VL 30 V
TOFF
TON
Vo 0
L
t
+ + −
Vin
S C
R
−
− 20 V TS
15 V (b)
(GATE 2016: 1 Mark)
t TS
VL
−45 V (GATE 2015: 2 Marks) 125. In the given rectifier, the delay angle of the thyristor T1 measured from the positive going zero crossing of Vs is 30°. If the input voltage Vs is 100sin(100 pt) V, the average voltage across R (in Volt) under steady-state is ––––––––.
127. A steady DC current of 100 A is flowing through a power module (S, D) as shown in Figure (a). The V-I characteristics of the IGBT (S) and the diode (D) are shown in Figures (b) and (c), respectively. The conduction power loss in the power module (S, D), in watts, is ________. (a)
(b) IS(A) S
D3
T1 Vs
R
D Vo=1 V dV/dI = 0.02 Ω
+ Vo −
100 A
−
+ D4
VS(Volt)
D2 V-I Characteristic of IGBT (c)
ID(A)
(GATE 2015: 2 Marks) 126. A buck converter, as shown in Figure (a) below, is working in steady state. The output voltage and the inductor current can be assumed to be ripple free. Figure (b) shows the inductor voltage VL during a complete switching interval. Assuming all devices are ideal, the duty cycle of the buck converter is ________.
VS(Volt) V-I Characteristic of diode (GATE 2016: 1 Mark)
M + + Vg −
−
VL
D
+
C Vo
− (a)
Ch wise GATE_EE_Ch10.indd 528
Vo=0.7 V dV/dI = 0.01 Ω
R
128. A three-phase diode bridge rectifier is feeding a constant DC current of 100 A to a highly inductive load. If three-phase, 415 V, 50 Hz AC source is supplying to this bridge rectifier then the rms value of the current in each diode, in ampere, is ________. (GATE 2016: 1 Mark) 129. A buck-boost DC-DC converter, shown in the figure below, is used to convert 24 V battery voltage to 36 V
11/21/2018 6:19:53 PM
Chapter 10 • Power Electronics
DC voltage to feed a load of 72 W. It is operated at 20 kHz with an inductor of 2 mH and output capacitor of 1000 μF. All devices are considered to be ideal. The peak voltage across the solid-state switch (S ), in volt, is _______.
529
132. The voltage (vS) across and the current (iS) through a semiconductor switch during a turn-ON transition are shown in figure. The energy dissipated during the turn-ON transition, in mJ, is ____________.
vS 600 V + 24 V −
Load
− +
S 2 mH
0 36 V
t 50 A
(GATE 2016: 1 Mark)
iS 100 A 0
130. A single-phase thyristor-bridge rectifier is fed from a 230 V, 50 Hz, single-phase AC mains. If it is delivering a constant DC current of 10 A, at firing angle of 30°, then value of the power factor at AC mains is (a) 0.87 (b) 0.9 (c) 0.78 (d) 0.45 (GATE 2016: 2 Marks) 131. The switches T1 and T2 in Figure (a) are switched in a complementary fashion with sinusoidal pulse with modulation technique. The modulating voltage vm(t) = 0.8 sin(200 p t) V and the triangular carrier voltage (vC) are as shown in Figure (b). The carrier frequency is 5 kHz. The peak value of the 100 Hz component of the load current (iL), in ampere, is _________.
+ T1 VDC/2 = 250 V iL
t
T1 = 1 ms T2 = 1 ms
(GATE 2016: 2 Marks) 133. A single-phase full-bridge voltage source inverter (VSI) is fed from a 300 V battery. A pulse of 120° duration is used to trigger the appropriate devices in each halfcycle. The rms value of the fundamental component of the output voltage, in volts, is (a) 234 (b) 245 (c) 300 (d) 331 (GATE 2016: 2 Marks) 134. A full-bridge converter supplying an RLE load is shown in the following figure. The firing angle of the bridge converter is 120°. The supply voltage vm(t) = 200p sin (100 p t) V, R = 20 W, E = 800 V. The inductor L is large enough to make the output current IL a smooth DC current. Switches are lossless. The real power fed back to the source, in kW, is ________. Load
XL =16 Ω at R = 12 Ω 100 Hz +
T2 IL
VDC/2 = 250 V
T1
L
T3
−
R = 20 Ω
Bridge
(a)
Vm vC
T2
−
T4
E = 800 V +
1
0.8
t vm
(b)
(GATE 2016: 2 Marks)
Ch wise GATE_EE_Ch10.indd 529
(GATE 2016: 2 Marks) 135. A three-phase voltage source inverter (VSI) as shown in the figure is feeding a delta connected resistive load of 30 W/phase. If it is fed from a 600 V battery, with 180° conduction of solid-state devices, the power consumed by the load, in kW, is ________.
11/21/2018 6:19:53 PM
530
GATE EE Chapter-wise Solved Papers
138. For the power semiconductor devices IGBT, MOSFET, diode and thyristor, which one of the following statements is TRUE?
+
30 Ω
30 Ω 30 Ω
600 V
− (GATE 2016: 2 Marks) 136. A DC-DC boost converter, as shown in the following figure is used to boost 360V to 400 V, at a power of 4 kW. All devices are ideal. Considering continuous inductor current, the rms current in the solid state switch (S), in ampere, is ________.
(a) All the four are majority carrier devices. (b) All the four are minority carrier devices. (c) IGBT and MOSFET are majority carrier devices, whereas diode and thyristor are minority carrier devices. (d) MOSFET is majority carrier device, whereas IGBT, diode, and thyristor are minority carrier devices. (GATE 2017: 1 Mark) 139. In the converter circuit shown below, the switches are controlled such that the load voltage v0(t) is a 400 Hz square wave.
10 mH + S 1 mF
360 V
S1
S3
S4
Load + n0(t) − S2
+ 220 V −
Load + 400 V −
−
140. A 3-phase voltage source inverter is supplied from a 600 V DC source as shown in the figure below. For a star connected resistive load of 20 Ω phase, the load power for 120° device conduction in kW is .
20 Ω + 600 V −
Ω
137. A single-phase bi-directional voltage source converter (VSC) is shown in the figure below. All devices are ideal. It is used to charge a battery at 400 V with power of 5 kW from a source Vs = 220 V (rms), 50 Hz sinusoidal AC mains at unity pf. If its AC side interfacing inductor is 5 mH and the switches are operated at 20 kHz, then the phase shift (d ) between AC mains voltage (Vs) and fundamental AC rms VSC voltage (VC1), in degree, is ________.
The RMS value of the fundamental component of v0(t) in volts is . (GATE 2017: 1 Mark)
20
(GATE 2016: 2 Marks)
20
Ω
5 mH XS
+ 400 V −
(GATE 2017: 1 Mark)
(GATE 2016: 2 Marks)
141. A phase-controlled, single-phase, full-bridge converter is supplying a highly inductive DC load. The converter is fed from a 230 V, 50 Hz, AC source. The fundamental frequency in Hz of the voltage ripple on the DC side is (a) 25 (b) 50 (c) 100 (d) 300 (GATE 2017: 1 Mark)
1 mF
IS 220 V AC
IS d
VS VC1
Ch wise GATE_EE_Ch10.indd 530
ISXS
11/21/2018 6:19:54 PM
Chapter 10 • Power Electronics
142. A three-phase voltage source inverter with ideal devices operating in 180° conduction mode is feeding a balanced star-connected resistive load. The DC voltage input is Vdc. The peak of the fundamental component of the phase voltage is (a) (c)
145. The figure below shows an uncontrolled diode bridge rectifier supplied from a 220 V, 50 Hz, 1-phase ac source. The load draws a constant current I0 = 14 A. The conduction angle of the diode D1 in degrees (rounded off to two decimal places) is .
Vdc 2Vdc (b) π π 3Vdc 4Vdc (d) π π
531
LS = 10 mH
D1
D3 I0 = 14 A
220 V, 50 Hz
(GATE 2017: 1 Mark) 143. The figure below shows the circuit diagram of a controlled rectifier supplied from a 230 V, 50 Hz, 1-phase voltage source and a 10:1 ideal transformer. Assume that all devices are ideal. The firing angles of the thyristors T1 and T2 are 90° and 270°, respectively. T1
T2
10 1 230 V. 50 Hz
R
D3 D1
IDA
D2
D4
(GATE 2017: 2 Marks) 146. The figure below shows a half-bridge voltage source inverter supplying an RL-load with R = 40 W and ⎛ 0.3 ⎞ L = ⎜ ⎟ H. The desired fundamental frequency of ⎝ π ⎠ the load voltage is 50 Hz. The switch control signals of the converter are generated using sinusoidal pulse width modulation with modulation index, M = 0.6. At 50 Hz, the RL-load draws an active power of 1.44 kW. The value of DC source voltage VDC in volts is VDC
The RMS value of the current through diode D3 in amperes is . (GATE 2017: 1 Mark) 144. The input voltage VDC of the buck-boost converter shown below varies from 32 V to 72 V . Assume that all components are ideal, inductor current is continuous and output voltage is ripple free. The range of duty ratio D of the converter for which the magnitude of the steadystate output voltage remains constant at 48 V is
S1 − VDC
+ −
L
C
R
Vo +
D2
+ −
S1 R
L S2
VDC
+ −
(a) 300 2
(b) 500
(c) 500 2 (d) 1000 2 (GATE 2017: 2 Marks) 147. In the circuit shown, all elements are ideal and the switch S is operated at 10 kHz and 60% duty ratio. The capacitor is large enough so that the ripple across it is negligible and at steady state acquires a voltage as shown. The peak current in amperes drawn from the 50 V DC source is . (Give the answer up to one decimal place.) S
(a)
2 3 2 3 ≤ D ≤ (b) ≤D≤ 5 5 3 4
1 2 (c) 0 ≤ D ≤ 1 (d) ≤D≤ 3 3 (GATE 2017: 2 Marks)
Ch wise GATE_EE_Ch10.indd 531
− 50 V
75 V
0.6 mH
5Ω
+
(GATE 2017: 2 Marks)
11/21/2018 6:19:56 PM
532
GATE EE Chapter-wise Solved Papers
+
148. In the circuit shown in the figure, the diode used is ideal. The input power factor is . (Give the answer up to two decimal places.)
2Ω V0
Vm sin(wt)
100 sin (100 πt) V
10 mH
10 Ω − (GATE 2017: 2 Marks)
149. Four power semiconductor devices are shown in the figure along with their relevant terminals. The device(s) that can carry dc current continuously in the direction shown when gated approximately is (are) MT1
A
D
A G
G
G
K I
Thyristor (a) (b) (c) (d)
MT2 I
TRIAC
G
K S I
I
GTO
(GATE 2018: 2 Marks) 151. The figure shows two buck converters connected in parallel. The common input dc voltage for the converters has a value of 100 V. The converters have inductors of identical value. The load resistance if 1 Ω. The capacitor voltage has negligible ripple. Both converters operate in the continuous conduction mode. The switching frequency is 1 kHz, and the switch control signals are as shown. The circuit operates in the steady state. Assuming that the converters share the load equally, the average value of is1, the current of switch S1 (in Ampere), is (up to 2 decimal places).
MOSFET
iS1 + 100 V -
Triac only Triac and MOSFET Triac and GTO Thyristor and Triac
V given by V0 = m (3 + cos a ), where Vm = 80 π volts and 2π α is the firing angle. If the power delivered to the loss(up to 2 decless battery is 1600 W, α in degree is imal places).
Ch wise GATE_EE_Ch10.indd 532
S1
L
1W
C
S2
(GATE 2018: 1 Mark) 150. A phase controlled single phase rectifier, supplied by an AC source, feeds power to an RLE load as shown in the figure. The rectifier output voltage has an average value
+ 80 V − Battery
L
Switch control signals S1 t S2 t 0
0.5 ms
1 ms (GATE 2018: 2 Marks)
11/21/2018 6:19:57 PM
Chapter 10 • Power Electronics
533
ANSWER KEY 1. (c)
2. (b)
3. (b)
4. (b)
5. (d)
6. (c)
7. (b)
8. (a)
9. (a)
10. (a)
11. (c)
12. (a)
13. (b)
14. (c)
15. (b)
16. (c)
17. (c)
18. (d)
19. (b)
20. (b)
21. (a)
22. (a)
23. (d)
24. (b)
25. (a)
26. (a)
27. (a)
28. (a)
29. (d)
30. (c)
31. (c)
32. (d)
33. (c)
34. (d)
35. (d)
36. (b)
37. (c)
38. (a)
39. (c)
40. (c)
41. (b)
42. (d)
43. (d)
44. (b)
45. (c)
46. (a)
47. (a)
48. (c)
49. (b)
50. (b)
51. (c)
52. (a)
53. (c)
54. (b)
55. (d)
56. (c)
57. (c)
58. (a)
59. (d)
60. (b)
61. (c)
62. (a)
63. (a)
64. (a)
65. (b)
66. (d)
67. (b)
68. (a)
69. (b)
70. (c)
71. (c)
72. (c)
73. (b)
74. (c)
75. (a)
76. (c)
77. (d)
78. (c)
79. (a)
80. (c)
81. (a)
82. (a)
83. (d)
84. (a)
85. (c)
86. (c)
87. (c)
88. (a)
89. (b)
90. (c)
91. (c)
92. (b)
93. (b)
94. (d)
95. (d)
96. (b)
97. (c)
98. (b)
99. (d)
100. (c)
101. (b)
102. (c)
103. (d)
104. (b)
105. (d)
106. (c)
107. (31.25) 108. (b)
109. (69.3)
110. (2.5)
114. (6060) 115. (b)
116. (d)
117. (1)
119. (7)
120. (a)
111. (77.15) 112. (0.78) 113. (c)
118. (b)
121. (56)
122. (0.808) 123. (2500) 124. (0.75) 125. (61.56)
126. (0.4) 127. (170)
128. (57.7) 129. (60)
131. (10)
132. (75)
133. (a)
134. (6)
135. (24)
136. (3.5) 137. (9.21)
138. (d)
139. (198.06) 140. (9)
141. (c)
142. (b)
143. (0)
144. (a)
145. (224.17) 146. (c)
148. (0.7)
149. (a)
147. (40)
130. (c) 150. (90)
151. (50)
Answers with Explanation 1.
Topic: Line Commutated Thyristor Based Converters (c) The line current on AC side of a three-phase fully-controlled bridge rectifier is a quasi-square wave with peak value equal to constant load current.
2.
Topic: Sinusoidal Pulse Width Modulation (b) A very high order harmonic voltage is introduced on the DC side of a triangular PWM controlled, BJT based voltage source inverter. This contributes a disadvantage of this technique.
4. Topic: Characteristics of Semiconductor Power Devices: Diode (b) For the given circuit
+5 V
3. Topic: Characteristics of Semiconductor Power Devices: Diode (b) Dynamic or AC resistance (rd) is the resistance offered by the diode to the time varying input signal. It is given by ΔVD rd = ΔI D
5 KΩ
b = 80 + 0.7 V IB
=
VT iD 25 mV 2 mA
= 12.5 Ω
Ch wise GATE_EE_Ch10.indd 533
6.3 KΩ IE −10 V
For the given diode rd =
−
writing KVL for input loop, we have −0.7 – IE × 6.3 × 103 + 10 = 0 IE =
9.3 6.3 × 103
IE = 1.476 mA
11/21/2018 6:19:57 PM
534
GATE EE Chapter-wise Solved Papers
For CE configuration, we have, IE = (1 + β)IB ⇒ IB =
IE 1.476 = (1 + β ) (1 + 80) = 0.0182 mA
IB = 18.2 μA 5.
6.
Vm (cos a − cos ω t )(1) ωL Now, at extinction angle β, ωt = β, i = 0 Therefore, from Eq. (1), V 0 = m (cos a − cos β ) ωL
Topic: Single Phase and Three Phase Inverters (d) In a three phase voltage source inverter, feeding a purely inductive three-phase load, the load current will have h-th order harmonic of magnitude a h /h times the fundamental component. Topic: DC to DC Conversion (c) The voltage across the diode is shown below
i=
cos β = cos a Therefore, For n = 1,
β = 2nπ ± a
Also,
VD
β = 2π + 120° = 480° β = 2π – 120° = 240°
10. Topic: Single Phase and Three Phase Inverters (a) Each feedback diode conducts for one-fourth of the total time period. Therefore, diode conduction time
Vs
TD =
Eb tr
tf
to
Time-period =
Therefore, the average voltage will be Vavg =
Vs t r + Eb t0 t r + tf + t0
7. Topic: Characteristics of Semiconductor Power Devices: Thyristors (b) The thyristor is used in the applications where the voltage is very high (in kV). But to trigger this thyristor small pulses of only few volts are required. Thus, there is a huge difference in the operating power levels of trigger circuit and thyristor circuit. 8. Topic: Bidirectional AC to DC Voltage Source Converters (a) AC to DC circulating current dual converters are operated with a condition that a1 + a 2 = 180° 9. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (a) When the thysistor is ON, Vm sin ω t = L i
Therefore,
Ch wise GATE_EE_Ch10.indd 534
Vm sin ω t dt L a
1 1 = = 20 ms Frequency 50 Hz
Therefore, TD =
20 m s = 5 ms 4
11. Topic: Line Commutated Thyristor Based Converters (c) The source current, Ia ⎛ nπ ⎞ cos ⎜ ⎟ ⎝ 6⎠ n where n is odd number. For n = 3 is = 0 ( cos π /2 = 0) For n = 5 −I is = a 5 Hence, lowest harmonic component present in the source current is fifth harmonic. Therefore, Frequency = 5 × 50 = 250 Hz is ∝
12. Topic: DC to DC Conversion (a) The output voltage of a step-down chopper is given by, Vo = DVs
t
∫ di = ∫ 0
di dt
Time-period 4
Therefore,
Vo =D Vs
11/21/2018 6:19:59 PM
Chapter 10 • Power Electronics
13. Topic: Bidirectional AC to DC Voltage Source Converters (b) The reactive power consumed by the three-phase rectifier is given by Q = 3VL I L1 sin a Q ∝ sin a
Therefore,
Hence, Q will be higher for higher values of a. 14. Topic: Characteristics of Semiconductor Power Devices: Diode (c) For the given chopper circuit, the boundary condition is achieved when D = 1−
2L RT
15. Topic: Single Phase and Three Phase Inverters (b) For the given inverter circuit, the pole voltages are as shown in the figure. V10 Vs q V20 Vs
q V0
Vs
f
f
−Vs
q
From the figure, v12 = v10 − v20 Therefore, RMS value, Vrms = Vs
P=
2 Vrms R
( 200 / 2 ) 2 P= = 400 W 50
Ch wise GATE_EE_Ch10.indd 535
17. Topic: Characteristics of Semiconductor Power Devices: MOSFET (c) It is a p-channel depletion mode device. When VGS = 0 , there exist ID, therefore it is a depletion mode device. ID increases for negative values of gate voltage. 18. Topic: Characteristics of Semiconductor Power Devices: Thyristors (d) When the thyristor is turned ON, it is operating in forward conducting mode. Therefore, J1, J2 and J3 are all forward biased. 19. Topic: Characteristics of Semiconductor Power Devices: MOSFET (b) MOSFETs have three modes which are forward blocking, forward conduction and reverse conduction. When MOSFET is ON, it means we are giving signal current to it. We have two cases: (i) When IS < 0, diode is in non-conducting mode so VDS = 0. (ii) When IS > 0, diode is reversed biased so VDS = 0 When MOSFET is OFF, it means no signal current is there. So, IS = 0, but diode is forward baised so VDS > 0. The reverse conduction takes place when internal body diode conducts. So, the graph depicted in option (b) is correct. 20. Topic: Power Factor (b) Field control is above base speed and armature control is below base speed. 21. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (a) In fully controlled rectifier, a = 30°. Vp-p = Vm − Vm cos(π / 6 + a )
ϕ π
16. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (c) The rms output voltage of a single-phase bridge rectifier V Vrms = m 2 Therefore, power
535
Vp-p = Vm [1 − cos(π / 6 + 30°)] Vp-p Vm
= 0.5
22. Topic: Characteristics of Semiconductor Power Devices: MOSFET (a) We know that VD(sat) = VGS − Vth Vth = 2 V Therefore,
VD(sat) = VGS − 2
From the circuit, VDS = VGS, therefore
11/21/2018 6:20:00 PM
536
GATE EE Chapter-wise Solved Papers
23. Topic: Characteristics of Semiconductor Power Devices: Diode (d) For the given circuit, in the positive half cycle, Vin > 4V, D2 is ON and D1 is OFF and Vout is 4 V.
VDS(sat) = VDS − 2 VDS = VDS(sat) + 2 VDS > VDS(sat) In the saturation region, current becomes I D = K (VGS − Vth )
10 kW +
2
4 = K (VGS − 2) 2 (1)
Vin
− 4V +
4V
From the circuit,
+
Vout
−
10 kW
−
VGS = VDS = 10 − I D RD = 10 − ( 4)(1) = 6 V (2)
Substituting the value of VGS from Eq. (2) in Eq. (1), we get
In the negative half cycle, D1 is ON and D2 is OFF, so from the circuit 10 kΩ
4 = K ( 6 − 2) 2 4 = K ( 4) K = 1/ 4
+
2
Vin
I
When RD is increased to 4 kW, let the new drain current be I D′ . Then ′ = VGS ′ VDS
Vout
− −
Vin − 10 I + 4 − 10 I = 0 Vin + 4 = 20 I
1 I D′ = (VGS − 2) 2 (3) 4 ′ = VDS ′ = 10 − I D′ × RD′ VGS ′ = 10 − 4 I D′ VGS
Therefore, substituting for VGS in Eq. (3), we get 1 I D′ = (10 − 4 I D′ − 2) 2 4 4 I D′ = (8 − 4 I D′ ) 2 4 I D′ = 4 2 ( 2 − I D′ ) 2 2
I D′ = 4 ( 4 + I D′ − 4 I D′ ) I ′D = 16 + 4 I D′ − 16 I D′ 4 I D′ − 17 I D′ + 16 = 0 I D′ = 2.84 mA ≈ 2.8 mA
Ch wise GATE_EE_Ch10.indd 536
+
10 kΩ
′ − VTh ) 2 I D′ = K (VGS
Also,
− 4V +
4V
I=
Vin + 4 20
Maximum value in negative half cycle = −10 V. Therefore, I=
−10 + 4 3 = − mA 20 10 I=
Vin − Vout 10
−3 −10 − Vout = 10 10 Vout = − 7 V 24. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (b) Since current source connected at load side, so, current flows continuously either through T1T2 or T3T4. Since the firing angle is 30°, so T1T2 conduct for 30° < wt < 210°. Similarly T3T4 conduct at 30° + p. Since in negative cycle there is a cut-off, so output DC voltage is given by
11/21/2018 6:20:03 PM
Chapter 10 • Power Electronics
537
VDC + Vi −
− Vo +
t + Vi −
25. Topic: DC to DC Conversion (a) During turn-ON of chopper, TOn
TOn
∫ V dt = ∫ L
0
0
+
min
= L(imax − imin ) = L( ΔI )
= (60 − 12)TOn = 48TOn Therefore, volt-area, 48TOn 48 × 0.2 × 10 −3 ΔI = = = 0.48 A L 20 × 10 −3 26. Topic: Single Phase and Three Phase Inverters (a) Output voltage, Vo =
⎛ 4Vs ⎞ ⎜ ⎟ (sin nd )(sin nω t )(sin nπ / 2) n = 1, 3, 5 ⎝ nπ ⎠
Therefore, rms value of fundamental component, Vrms =
4Vs
So, D2 is reverse biased and peaks across diode VD2 = −2Vm Therefore, peak input voltage PIV = 2Vm = 2 × 50 2 = 100 2 V 29. Topic: Characteristics of Semiconductor Power Devices: Thyristors (d) Given that Vb = 12 ± 4 V
2π
Vo(rms) =
4Vs 2π
× sin 60° = 0.78 Vs
27. Topic: Issues of Line Current Harmonics (a) On removing 5th harmonic, 5d = 0, π , 2π Therefore, Pulse width, 2d = a = 0,
Therefore, Vb =16 V(max) and 8 V (min) Therefore, required value of R is
sin d × 1
When a = 120° ; 2d = 120° ⇒ d = 60°.
2π 4π , = 0°, 72°, 144° 5 5
28. Topic: Characteristics of Semiconductor Power Devices: Diode (a) For the given full-wave rectifier, the voltage Vi = 50 V(rms), therefore Vm = 50 2 V .
Ch wise GATE_EE_Ch10.indd 537
In the positive half cycle, voltage across D2 VD2 = Vi − Vo = −2Vo (Vi = Vo )
∞
∑
−
During the positive cycle, D1 conducts and during negative cycle D2 conducts, so Vo = Vi
imax
⎛ di ⎞ L ⎜ ⎟ dt = ∫ Ldi ⎝ dt ⎠ i
Voltage applied to the inductor L is
VD2
R=
Vb (min) IG
=
8 = 800 Ω 10 × 10 −3
30. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (c) There are three continuous pulses for output voltage in one cycle of input voltage. So, Ripple frequency = 3 f = 3 × 400 = 1200 Hz 31. Topic: Characteristics of Semiconductor Power Devices: MOSFET (c) Given that R = 0.15 Ω and I = 15 A. Therefore, average power loss π /ω 1 I 2 Rdt ∫ ( 2π /ω ) 0 ω = × 10 2 × 0.15 × π /ω 2π = 7.5 W
=
11/21/2018 6:20:05 PM
538
GATE EE Chapter-wise Solved Papers
32. Topic: Characteristics of Semiconductor Power Devices: TRIAC (d) Given that Vs = 230 × 2 sin ω t and R = 10 Ω. Therefore,
Peak power =
R
=
Then, speed of the motor
( 2 × 230) = 10580 W 100
Vs = 100 V Duty ratio k = 0.8 R = 10 Ω
I = I o + Vs C /L
E1 = ∫ vidt = V ∫idt
35. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (d) The single phase half-controlled rectifier can be represented as
Ra + Ea −
− Average output voltage of single-phase controlled rectifier is given by Vm (1 + cos a ) π
230 2 (1 + cos 30°) π = 193.201 V
Ch wise GATE_EE_Ch10.indd 538
(V is constant, so v = V )
1 = VIt1 2
(1)
In the time interval t2, current is constant. The energy loss is given by E2 = ∫ vidt = I ∫ vdt
( I is constant, so i = I )
1 = VIt 2 2
+
Vo
38. Topic: Characteristics of Semiconductor Power Devices: Diode (a) The transition is turn ON. In the time interval t1, voltage is constant. The power loss is given by vi, therefore energy loss
2 × 10 −6 = 20 + 200 200 × 10 −6 1 = 20 + 200 × = 40 A 10
=
36. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (b) Given that Vrms = 400 V and f = 50 Hz
37. Topic: Single Phase and Three Phase Inverters (c) A three-phase square wave symmetrical inverter contains only odd harmonics
34. Topic: Line Commutated Thyristor Based Converters (d) Peak current through S1,
V=
183.201 = 386.40 rpm ≈ 386 rpm 0.5
Vo = 2 Vrms = 400 × 2 V
kVs 0.8 × 100 = =8A R 10
Vm
N=
Instantaneous voltage, purely resistive
Average current through the diode
Single phase rectifier
Substituting values,
2
33. Topic: Line Commutated Thyristor Based Converters (c) Given that
=
E = V – I a Ra E = 193.201 − 5 × 2 = 183.201 V
Vm(peak) = 230 × 2 (Vm(peak) ) 2
We know that
(2)
Hence from Eqs. (1) and (2), we total energy loss during the transition as E = E1 + E2 1 1 = VIt1 + VIt 2 2 2 39. Topic: Characteristics of Semiconductor Power Devices: Diode (c) The given switch is required to block the voltage in both forward and reverse direction and conduct current only in the forward direction. In Figure P, the thyristor blocks voltage in both polarities until gate is triggered.
11/21/2018 6:20:06 PM
Chapter 10 • Power Electronics
In Figure R, the transistor along with diode blocks voltage in both polarities. In devices given in Figures Q and S, current flows through the diode. 40. Topic: Line Commutated Thyristor Based Converters (c) Given that,
43. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (d) In the first half cycle, θ=0 D1 is ON and D2 is OFF.
k = 0.5, f = 1 kHz,Vs = 100 and L = 200 mH.
p
0 1 T= = 10 −3 s 1 × 10 −3 Ta =
2p
44. Topic: Single Phase and Three Phase Inverters (b) In PWM inverter,
L 200 × 10 −3 = = 40 × 10 −3 s R 5
− kT / TS )(1 − e − (1− k )T / Ta ) ⎤ Ripple = Vs ⎡⎢ (1 − e ⎥ R⎣ 1 − e −T / TS ⎦
ΔI max =
539
Vs 100 = = 0.125 A 4 fL 4 × 1 × 10 −3 × 200 × 10 −3
41. Topic: Line Commutated Thyristor Based Converters (b) Given that Vrms = 230 V and f = 50 Hz. Therefore, Vm = 2 Vrms = 2 × 230 V For (a < 90°) ,
∞
4Vs sin nd sin nωt sin nπ / 2 n = 1 nπ
Vo = ∑
Pulse width, 2d = 144° For fundamental component (n = 1), 4Vs sin 72° sin ωt π Third harmonic component (n = 3), 4V Vo3 = s sin(3 × 72°) sin 3ω t 3π Vo1 =
Therefore, 4Vs sin(3 × 72°) Vo3 = 3π = 19.61% ≈ 19.6% 4Vs Vo1max sin 72° π 45. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (c) Given that V = 400 V, f = 50 Hz, a = 60° and I L = 10 A.
Vpeak = Vm sin a = 230
2 × 230 sin a = 230 V = 400 V, f = 50 Hz, a = 60° and I L = 10 A. 1 sin a = ⇒ a = 135° Input displacement factor (IDF) is given by 2 cos a = cos 60° = 0.5 42. Topic: Single Phase and Three Phase Inverters Input power factor (IPF) = Distortion factor × cos a (d) In variable frequency variable voltage (V/f ) method, Distortion factor on varying frequency, V/f should be constant. Therefore, 4 × 10 maximum flux density should also remain constant. sin 60° I (Fundamental) π × 2 = s = = 0.955 E = 4.44 f ϕ m N Is 10 × 2 / 3 E = 4.44 f Bm AN Therefore, input power factor = 0.955 × 0.5 = 0.478 V 46. Topic: Line Commutated Thyristor Based Converters (a) Given that T = 100 and R = 50. We know that T = RC ln 2. Therefore, f 50
Ch wise GATE_EE_Ch10.indd 539
C=
T 100 = = 2.88 μF R ln 2 50 × 0.693
11/21/2018 6:20:09 PM
540
GATE EE Chapter-wise Solved Papers
47. Topic: Bidirectional AC to DC Voltage Source Converters (a) Given that: E = 350 V, R = 0.5 Ω and I o = 20 A. Therefore,
set I1 as IL and I2 as IR
Vs = E − I o R
IL =
= 350 − ( 20 × 0.5) = 340 V Also, Vs =
Vs 100 = = 20 mA R2 5 × 103
IR =
R − 1t Vs 100 (1 − e L ) = (1 − e −40 t ) R1 20
Anode current I = I1 + I 2
3Vm cos a π
0.05 = 20 × 10 −3 + 5(1 − e −40 t )
3( 440) × 2 cos a π cos a = 55°
0.05 − 0.02 = 5(1 − e −40 t )
340 =
0.03 5 t = 150 μs
1 − e −40 t =
Therefore, each thyristor will be reverse biased for 180° − 55° = 125° 48. Topic: Line Commutated Thyristor Based Converters (c) Average value of current, 1 π −θ1 = (Vm sin ωt − E ) dθ 2π R ∫θ1
I avg I (avg)
50. Topic: Line Commutated Thyristor Based Converters (b) Given that I L = 10 A. The minimum time for which the main SCR should be ON for charging TOn =
1 = [2Vm cos θ − E (π − 2θ1 )] 2π R
= 3.14 × 2 × 10 −3 × 10 −6 = 140 μs
1 = [2 × ( 230 × 2 ) cos θ − 200(π − 2θ1 )] 2π × 2 ⎛ E⎞ ⎛ 200 ⎞ θ1 = sin −1 ⎜ ⎟ = sin −1 ⎜ = 38° = 0.66 rad ⎝ 2 × 230 ⎟⎠ ⎝ Vm ⎠
π = π LC ω0
51. Topic: Line Commutated Thyristor Based Converters (c) We have that TOn = 140 μs
Therefore, 1 [2 2 × 230 cos 38° − 200(π − 2(0.66))] 4π = 11.9 A
I (avg) =
Average output TOn ×V T 1 1 T = = 3 = 1 ms f 10
Vo(avg) =
49. Topic: Characteristics of Semiconductor Power Devices: Thyristors (b) Given that TOn = 5 μs, I L = 50 mA and I H = 40 mA. Minimum gate pulse width or duration is decided by the external circuit to allow the device current to rise above its latching current (IL) level I1
Vo =
52. Topic: Line Commutated Thyristor Based Converters (a) Given that a = 25°, μ = 10° and I o = 20
I2
20 W
140 × 10 −6 × 250 = 35 V 1 × 10 −3
Io =
Vm [cos a − cos(a + μ )] ω Ls
Substituting values, we have 5 kW
100 V
20 = 0.5 H
230 2 [cos 25° − cos(25° + 10°)] 2π × 50 Ls
Ls = 0.0045 H
Ch wise GATE_EE_Ch10.indd 540
11/21/2018 6:20:11 PM
Chapter 10 • Power Electronics
Vo =
2Vm cos a ω Ls I o − π π
2 × 230 × 2 × cos 25 3.14 2 × 3.14 × 50 × 4.5 × 10 −3 × 20 − 3.14 = 187.7 − 9 = 178.7 V =
V I 178.7 × 20 = 0.78 Displacement factor = o o = Vs I s 230 × 20 53. Topic: Line Commutated Thyristor Based Converters (c) Given that P = 50 KW and VDC = 420 V. P = VDC × I DC I DC =
50 × 103 = 119.05 A 420
RMS value of thyristor current =
3 = 68.73 A 54. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (b) The converter has to have a free-wheeling diode because the converter does not provide freewheeling for high values of triggering angles. 55. Topic: Characteristics of Semiconductor Power Devices: MOSFET (d) The given statement is false, because MOSFETs can be operated as excellent constant current sources in the saturation region. 56. Topic: Sinusoidal Pulse Width Modulation (c) Given total harmonic distortion THD =
2 Vrms − V12
V1
× 100; d = 150° (pulse width).
We have, Vrms = V1 = Vo(rms) =
THD =
150 ⋅Vs = 0.91Vs 180 0.4Vs 2π
sin 75° = 0.869Vs
(0.91Vs ) 2 − (0.869Vs ) 2 0.869Vs
57. Topic: Bidirectional AC to DC Voltage Source Converters (c) Given that f = 50 Hz and slip s = 4% So, voltage is 50 Hz of 4% = 2 Hz We know that Trated ∝ s To obtain twice the rated torque the frequency of impressed voltage is = 2 × 2 = 4 Hz 58. Topic: Line Commutated Thyristor Based Converters (a) Given that Vo = 440 V, P = 15 kW, f = 50 Hz and N = 1500 rpm. On neglecting losses, 3 × 2 × 440 750 × 2π cos a = K m × (1) π 60 We know that Vo = K m ω m = E
440 = K m ×
119.05
= 31.83%
541
2π N 2π × 1500 ⇒ 440 = K m × 60 60
K m = 2.8 Substituting for Km in Eq. (1), we get cos a = 0.37 Vt =
3 2 × 440 cos a π
3 2 × 440 × 0.37 π = 219.9 V =
Ia =
15000 = 34 A 440
I s(rated) = I a 2 / 3 = 27.8 A Power factor =
Vt I s 3Vs I s(rated)
= 0.354
59. Topic: Bidirectional AC to DC Voltage Source Converters (d) Given that V = 230 V, f = 50 Hz, VDC = 12 V, R = 19.04 Ω. V1 4 = V2 1 V2 =
V1 230 = = 57.5 V 4 4
For the diode to conduct VDC = Vm sin θ1 12 = (57.5) 2 sin θ1
Ch wise GATE_EE_Ch10.indd 541
11/21/2018 6:20:14 PM
542
GATE EE Chapter-wise Solved Papers
1W
Substituting values, we have VDC = Vm sin θ1 12 = (57.5) 2 sin θ1 +10 V
PT
D1
S C R
R
⇒ θ1 = 8.486 or 0.148 rad. V1
Therefore, Vm = 81.32 V
D2
1 [(2Vm cos θ1 ) − VDC (π − 2θ )] 2π R 1 = [2 × 81.32 cos θ1 − 12(π − 2θ1 )] 2π × 19.04
I avg(charging) =
= 1.005 Ω /A ≈ 1 ΩA
ThM im
The transistor is in the active state, so there is 1 V drop on the primary side, VCE = 1 V and pulse obtained on the secondary side is V1 = 9 V. Again there is a gate-cathode voltage drop VGK = 1 V and gate cathode junction voltage V2 = 9 V. The voltage drop across resistance R is
60. Topic: DC to DC Conversion: Buck, Boost and Buck-Boost Converters (b) The given circuit can be represented as 1
Io
ic 10 µF 25.28 µH
VR = V2 − V1 − VGK = 9 −1−1 = 7 V Therefore, for SCR to turn ON VR 7 R= = = 46.67 ≈ 47 Ω I G(max) 150 62. Topic: Line Commutated Thyristor Based Converters (a) Consider the following circuit for the SCR.
ThAux
Load
RL = 1 Ω + i V L − T
L + 200 V −
Using Kirchoff’s current law at point 1, we have
During ON-state, forward voltage drop across SCR, VT = 1 V. Then applying KVL around the loop, we get
im + ic = I o im = I o − ic = I p sin ω Δ t When im = 0 , ThM is turned OFF, that is, ⎛I ⎞ 1 Δt = sin −1 ⎜ o ⎟ = 25 μ s ω ⎝ Ip ⎠ Hence, ThM is off between t < T < t + Δt T=
π = π LC = 50 μs ω
Therefore, 25 μs < t < 50 μs 61. Topic: Line Commutated Thyristor Based Converters (c) The given transformer circuit can be represented as
Ch wise GATE_EE_Ch10.indd 542
200 V
+ V −CE
Average charging current,
230 V
V2
L
V −L
di − RiL − VT = 0 dt
Substituting values, we get diL − iL − 1 = 0 dt ⇒ iL = 199 (1 − e − t / 0.15 ) 200 − 150 × 10 −3
If T is the time required for ia to rise to iL then, iL = 199 (1 − e −T / 0.15 ) 250 × 10 −3 = 199 (1 − e −T / 0.15 ) T = 188.56 μs Given that voltage across PT = 10 V, therefore, voltage rating of the pulse transformer is 10 V × 188.50 μs ≈ 2000 μVs
11/21/2018 6:20:16 PM
Chapter 10 • Power Electronics
543
63. Topic: Power Factor (a) Circuit based on the given data is as below.
67. Topic: Power Factor (b) Given that a = 30°. We know that input power factor
2Ω
PF = Distortion factor (DF) × Displacement factor = DF × cos a
+
We know that
+ 3V
V0 =
−
I 02
DF =
I0 −
Therefore, PF = 0.9 × cos 30° = 0.78
From the circuit, we have
I + 2 I − 3V = 0 2 0
2 0
I 0 + 3I 0 − I 0 − 3 = 0 2
I 0 ( I 0 + 3) − 1( I 0 + 3) = 0 ⇒ I 0 = 1 A or I 0 = −3A Using I0 = 1A, we have, V0 = I 02 ⇒
V0 = 1V
Power dissipated, P0 = V0 I 0 = 1W
68. Topic: Characteristics of Semiconductor Power Devices: Thyristors (a) As the current flowing through T1 and T3 is more than holding current, T1 − T3 ⇒ On. 69. Topic: Bidirectional AC to DC Voltage Source Converters (b) For rated speed, V = E + I a Ra
64. Topic: Line Commutated Thyristor Based Converters (a) From the circuit, we have ω L 50 R + jω L = 50 + 50 j ; tan ϕ = = =1 R 50
ϕ = tan −1 1 = 45°. Therefore, firing angle should be higher than 45° which implies output voltage will be uncontrollable for 0 < α < 45°. 65. Topic: Single Phase and Three Phase Inverters (b) A three-phase voltage source inverter is operated in180° mode. Third harmonics will be absent in the line voltage due to cos( nπ / 6) = 0 in the expression VL =
4Vs π⎞ ⎛ nπ ⎞ ⎛ cos ⎜ ⎟ sin n ⎜ ωt + ⎟ ⎝ 6⎠ ⎝ 6⎠ n =1, 3, 5 nπ ∞
∑
The pole voltage is not zero, for n = 3, as is evident from the following expression VP =
∞
2Vs sin nωt n = 6 k ±1 nπ
∑
66. Topic: Characteristics of Semiconductor Power Devices: Diode (d) In the positive half cycle, D1 is ON and D2 is OFF. In the negative half cycle, D1 is ON and D2 is ON. This implies that C1 charges up to +5V, so vC1 = 5 V. Therefore, vC2 will charge up to 10 V in the opposite direction, so vC2 = −10 V.
Ch wise GATE_EE_Ch10.indd 543
I s1 2 2 I o 1 = × = 0.9 Is Io π
E = V − I a Ra = 220 − 20 × 2.5 = 170 V For 600 rpm, E2 =
E1 × N2 N1
170 × 600 = 102 V 1000 Since torque is dependent on the current, so rated =
T ∝ I.
torque
Now, when voltage is applied, Vi = E2 + I a Ra = 120 + 20 × 2.5 = 152 V Duty cycle D can be calculated as Vo = DVi D=
Vo 152 = = 0.608 Vi 250
70. Topic: Bidirectional AC to DC Voltage Source Converters (c) At rated speed and torque, E = V − I a Ra = 220 − 40 × 0.4 = 204 V Since torque is dependent on current, at 1000 rpm and 50% of rated torque,
11/21/2018 6:20:18 PM
544
GATE EE Chapter-wise Solved Papers
1 × 40 = 20 A 2 E 204 × 1000 = 145.71 V E2 = 1 × N 2 = 1400 N1 I2 =
Thus, output of the first converter 2V Vo = m cos a1 = 153.71 V π 2 × 2 × 20 cos a1 = 153.71 π a1 ≈ 39° Therefore, the firing angle of rectifier, a 2 = 180 − a1 ≈ 141°
For continuous mode, V − 2 I + 150 = 0
I=
( L = 0)
V +150 2
When I = 10 A, the equation above becomes V + 150 2 V = 20 − 150 = −130 V
10 =
We know that V=
71. Topic: Single Phase and Three Phase Inverters (c) For the given inverter the plot of current is given as follows.
2Vm cos a π
2 × 2 × 230 cos a π cos a = −0.627 −130 =
a = cos −1 ( −0.627) = 128.86° ≈ 129
200 p
2p
−200
73. Topic: Line Commutated Thyristor Based Converters (b) Given that I = 10 A and a = 30° I rms = I × 2 / 3 = 2 / 3 × 10 = 8.16 A
Peak current occurs at (π /ω ) di Vs = L dt di Vs = L di 200 = 0.dt 1di× dt Vs = L di = 0π.1dt× 1 Therefore, 200 di = × 1dt / 50 = di 2 π 100 200 = π0.1 × 1 di = × 11dt / 50 =1 2ππ × dimax = 200 × 100 = 20 A 0.1 100 1 Therefore, di = 2π × 11/ 50 =1 100 dimax = 200 × × = 20 A 0.11 100 1 dimax = 200 × × = 20 A 0.1 100 72. Topic: Line Commutated Thyristor Based Converters (c) The circuit for the given controlled converter bridge I
R = 2Ω V
=
I L = I D = 4 A and Vs = 20 V When the switch is closed, I D = 0; Vo = 0 Vs 20 = = 40 V 1 − D 14 − 0.5 Average current is
150 V
Ch wise GATE_EE_Ch10.indd 544
1 −1 DF 2
1 − 1 = 0.3105 = 31% (0.955) 2 I 0.78 × 10 = = 0.955 Distortion (DF) = a1Thyristor 74. Topic: Linefactor Commutated Based Ia 0.816 × 10 Converters (c) When switch is open, THD =
Average voltage is
L
+
−
(i) THD I ao ==0.78 1× 10−=1 7.8 A (Fundamental component) DF 2 I 0.78 × 10 1 = 0.955 factor (DF) = a1 = THD = (ii) Distortion −1 2 Ia 0.816 × 10 DF I 0.78 × 10 Distortion factor (DF) = a1 = = 0.955 Ia 0.816 × 10
0+4 = 2A 2
11/21/2018 6:20:20 PM
Chapter 10 • Power Electronics
75. Topic: Characteristics of Semiconductor Power Devices: Diode (a) When a forward biased diode is reverse biased, initially it conducts due to the flow of stored charges. After some time it goes into non-conducting state. So, current characteristic of diode during switching is as shown in option (a). 76. Topic: Distortion Factor of AC to DC Converters (c) SCR is considered to be a semi-controlled device because it can be tuned ON but not OFF with a gate pulse
79. Topic: Single Phase and Three Phase Inverters (a) Given that Duty ratio = 0.8 dv Maximum, = 50 V/ μ s dt Current, Imax = Duty ratio × Current = 0.8 × 12.5 = 10 A Also, I max = C1
77. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (d) We know that di 1 ; i L = ∫ VL dt dt L For 0 < ω t < π ,
VL = L
VL = Vs = 10 sin ω t =
C1 =
dv dt
10 = 0.2 × 10 −6 = 0.2 μF 50 × 106
80. Topic: Characteristics of Semiconductor Power Devices (c) Device Characteristics
diL dt
1 π iL = ∫ Vs dt = 10 sin 100πtdt = − cos 100π t + c L 0.1 ∫ At 100π t =
545
Is +
−
Vs
π , iL = 0 ; c = 0 2
Is
i L(peak) = 1 A +
−
Vs
For p < wt, VL = Vs = 0 Is
Vin ωt
+
Vs
−
1 iL
Is
0
10
20
30 ωt
78. Topic: Single Phase and Three Phase Inverters (c) At start: T3, T4 are conducting at T = 0 (i) T1T2 is ON T3T4 is force commutated (ii) T1T2 is force commutated T3T4 is ON
Vs +
−
81. Topic: Bidirectional AC to DC Voltage Source Converters (a) For the given converter, Vo = kVi k 0.7 V and diode be turned ON
Ch wise GATE_EE_Ch10.indd 547
547
11.4 = 3.8 > 0.7 3
v − 0.7 3.8 − 0.7 = = 6.2 mA 500 500
95. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (d) For continuous conduction of load current, the freewheeling diode conducts from p to p + a in a cycle. Thus fraction of cycle that free wheel diode conducts is a/p. 96. Topic: Characteristics of Semiconductor Power Devices: Thyristor (b) In a thyristor, IL > IH and IL is taken as two or three times that of IH. 97. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (c) I s = I o +
ΔiC = 5 + 0.8 = 5.8 A 2
98. Topic: Single Phase and Three Phase Inverters (b) The rms output line voltage, VL =
2 2 VDC = × 300 = 141.4 V 3 3
99. Topic: Single Phase and Three Phase Inverters (d) P = 3 ×
VL2 (141.4) 2 = 3× = 3 kW R 20
100. Topic: Characteristics of Semiconductor Power Devices: Thyristor (c) We know that, VC = V (1 − cos ω t) iC = I m sin ω t 1 1 = LC 100 /π 2π f = 1/100/ π 1 π 2π f = ⇒ f = 100 200
ω=
11/21/2018 6:20:25 PM
548
GATE EE Chapter-wise Solved Papers
T=
1 = 200 μs f
Conduction period = T/2 = 100 ms. This is because thyristor will conduct for one half cycle and will be open for next half cycle.
VQ
3 −Q 4
=0
before Q3 and Q4 are ON. So, zero voltage switching. 105. Topic: Characteristics of Semiconductor Power Devices: Diode
(d) Given that VR = 150 V, I aR = 20 A, La = 0.1 mH, Ra = 1 Ω 101. Topic: Characteristics of Semiconductor Power Devices: MOSFET VR = 150 V, I aR = 20 A, La = 0.1 mH, Ra = 1 Ω , T = 0.5TR (b) When Q is ON, iC = −iR
where TR is the rated torque. I = 0.5 I aR = 0.5 × 20 = 10 A
When Q is OFF, iC = iL − iR
E = V − ( I aR Ra ) = 150 − ( 20 × 1) = 130 V
Average capacitor current iC = 0 The new duty ratio D′ = 1− D I iL = in R D′ Vo Vi (12 / 0.6) = = =1A IR = 20 R D ′/R Therefore, iL =
IR 1 = = 5/ 3 A D ′ 1 − 0.4
102. Topic: Characteristics of Semiconductor Power Devices: MOSFET (c) We know that peak-to-peak source current = peak–peak induction current Vi ⋅k ×T L 1 1 T= = f 250 × 103
I L(p-p) =
I L(p-p) =
12 1 × 0.4 × = 0.192 −6 100 × 10 250 × 103
103. Topic: Bidirectional AC to DC Voltage Source Converters (d) Voltage and current will be positive when Q1 and Q2 are ON and Q3 and Q4 are OFF. As the load is inductive, IGBT will not conduct during its reverse current and so this current will flow through D3 and D4 104. Topic: Bidirectional AC to DC Voltage Source Converters (b) Since Q3 and Q4 are turned before, Q1 and Q2 this implies that D3 and D4 will be in conduction mode. Therefore
Ch wise GATE_EE_Ch10.indd 548
V = E + I a Ra = 130 + (10 )(1) = 140 V For chopper output to be 140 V 140 = D × 200 Duty cycle =
140 = 0.7 200
106. Topic: Characteristics of Semiconductor Power Devices: Diode (c) From the given switches: (i) blocks voltage in the forward direction. (ii) blocks voltage in the forward and reverse directions. (iii) blocks voltage in the forward and reverse directions. (iv) blocks voltage in the forward direction. Therefore, (ii) and (iii) can block voltages of either polarity. 107. Topic: Characteristics of Semiconductor Power Devices: Thyristor (31.25) Given that Vo = 400 V, Vs = 250 V, Toff = 20 μs. We know that V 1 ; TOFF = (1 − D )T ; Vo = s T 1− D V 250 Vo = s ⇒ (1 − D ) = ⇒ D = 0.375 1− D 400
f =
TOFF = (1 − D)T 20 × 10−6 = (1 − 0.375)T ⇒ T = 32 μs f =
1 1 = = 31250 Hz = 31.25 kHz T 32 × 10 −6
108. Topic: Power Factor (b) At low frequency, the voltage applied is low and the voltage across airgap to produce flux reduces as the stator impedance drop becomes comparable to applied voltage (at rated torque).
11/21/2018 6:20:27 PM
Chapter 10 • Power Electronics
109. Topic: Characteristics of Semiconductor Power Devices: Thyristors (69.3) Given that RL = 10 Ω, L2 = 0.05 H, Vo = 70 V. The average output voltage is given by V Vo = m (1 + cos a ) 2π 525 70 = (1 + cos a ) 2π So, a = 69.3°
549
112. Topic: Line Commutated Thyristor Based Converters (0.78) For fully controlled converter bridge: Input power factor = 0.9 cosa = 0.9 × cos 30° = 0.78 113. Topic: Line Commutated Thyristor Based Converters (c) Given that Vs = 230 V; f = 50 Hz; R = 5 Ω; L = 16 mH ⎛ XL ⎞ ⎛ 2π FL ⎞ a = tan −1 ⎜ = tan −1 ⎜ ⎟ ⎝ R⎠ ⎝ R ⎟⎠
110. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (2.5) From the circuit diagram and output waveform Vo
⎛ 2π × 50 × 16 × 10 −3 ⎞ = tan −1 ⎜ ⎟⎠ 5 ⎝
a = 45.1° The current flowing through SCR will be maximum at angle a = ϕ , γ = π .
100 V
1/ 2
0.05 I2
I rms
0.1
∆I
2 ⎤ ⎡ 1 a + π ⎛ Vm ⎞ ω a =⎢ sin( t − ) dω t⎥ ⎜ ⎟ ∫ ⎠ ⎥⎦ ⎢⎣ 2π a ⎝ 2
=
I1
Vm 2 × 230 = = 2.3 A 2 Z 2 52 + 5.04 2
T(ms) TON
TOFF
Peak to peak ripple in inductor current is 0.05 ⎞ ⎛ 100 − ⎜100 × ⎟ ⎝ Vs − Vo 0.1 ⎠ ⋅ Tm = (0.05 × 10 −3 ) ΔI = L 1 × 10 −3 = 2.5 A 111. Topic: Single Phase and Three Phase Inverters (77.15) Given that Vs = 100 V. The rms value of b V01 = 1 (a0 = 0) 2 p/2 a 4Vs ∫ sin q dq − ∫ sin q d q b1 = p 0 a 4Vs (− cos q )a0 − ( − cos q )ap / 2 = p 4Vs = [− cos a + cos 0 + cos p / 2 − cos a ] p 4Vs = [1 − 2 cos a ] p The rms value of V01 =
4Vs 2p
(1 − 2 cos a ) = 50
400 400 = 50 Vs =] 100] 50s = [100 (1 − 2 cos (1 −a2)cos a ) = [V 2p 2p Therefore, a = 77.15°
Ch wise GATE_EE_Ch10.indd 549
114. Topic: Distortion Factor of AC to DC Converters (6060) Given that IL = 50 mA, t = 50 μs. When SCR is ON, I = I1 + I2 =
V V (1 − e − t / τ ) + R1 R2
(1)
(for given width of gate pulse) Time constant of RL circuit,
τ=
L 200 × 10 −3 = = 0.4 × 10 −3 R 500
Substituting values in Eq. (1), we have −50 ×10 ⎞ 100 100 ⎛ 0.4 ×10 −3 1 − e ⎜ ⎟+ 500 ⎝ ⎠ R2 −6
40 × 10 −3 =
Solving, we get R2 = 6060 Ω. 115. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (b) The given circuit can be represented as 1 B + R −
2 Y
3 When R has maximum voltage, and lines 1 and 3 conduct, then current through R =
I o × 2R 2I o = 3R 3
11/21/2018 6:20:28 PM
550
GATE EE Chapter-wise Solved Papers
and current through B and Y =
Io 3
When line 2 conducts, then current through R = 2I o . 3 So, the input phase current waveform is
Io 3
119. Topic: Bidirectional AC to DC Voltage Source Converters (7) When S1 is ON
and current through B and Y =
10 V
2K/3 K/3
0.2 ms
2p
p
0
When S2 is ON
IR 116. Topic: Single Phase and Three Phase Inverters (d) The node voltage is,
5V
VDC 2 0
0.2 q
p−q
p
0.3 ms 2p
VDC 2
10 V
As the load current is sinusoidal, it implies continuous conduction. When S1 is turned off from 0 to q, the diode parallel to S3 conducts and VAO = −VDC / 2 . From q to
5V
0.2
(p - q), S1 conducts VAO = VDC / 2 and from (p - q) to (q), diode conducts VAO = −VDC / 2. 117. Topic: Characteristics of Semiconductor Power Devices: MOSFET (1) Given that duty ratio, D = 0.4; supply voltage, Vs = 20 V. Therefore, Vo = DVs = 0.4 × 20 = 8 V Charging current, I o =
Vo − E 8 − 5 3 = = =1 A R 3 3
118. Topic: Line Commutated Thyristor Based Converters (b) Initially, as V1 > V2, we have V1 − V2 R To reverse the power flow, polarity of voltage is changed but direction of the current is kept same, so I=
Vo =
For current to be positive, V1 > V2 ⇒ -485 > -500 ⇒ V1 = −485 kV, V2 = −500 kV and I = 1.5 kA
0.5
10 × 0.2 + 5 × 0.3 =7V 0.5
120. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (a) The average voltage across inductor is always zero, so VL = 0. The average voltage across capacitor is equal to the output voltage and as the converter is boost converter, so VC = VD =
1 × VDC 1− δ
121. Topic: Single Phase and Three Phase Inverters (56) Modulation index m = 0.7, Vm = 100 V, C = 63.66 mF, L = 9.55 mH and R = 5 Ω. m=
V −V I= 1 2 R
Ch wise GATE_EE_Ch10.indd 550
0.5
Vo =
d ⇒ d = 0.7 × p = 126° π
2VDC d 2 × 100 126° sin = sin = 56 V π π 2 2
11/21/2018 6:20:30 PM
Chapter 10 • Power Electronics
122. Topic: Single Phase and Three Phase Inverters (0.808) The slip for 5th harmonic voltage =
(1500 × 5 − 1440) = 0.808 1500 × 5
128. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers 2π (57.7) Each diode conducts for radians 3
R=
2 fL 2 × 100 × 10 × 5 × 10 = 1− D 1 − 0.6
= 2500 Ω
124. Topic: Characteristics of Semiconductor Power Devices: MOSFET (0.75) Given that VS = 15 V, therefore, VS − Vo = −45 V, Vo = VS + 45 = 60 V. So,
=
15 ⇒ 1 − D = 0.25 ⇒ D = 0.75 1− D
125. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (61.56) Given that a = 30°, Vs = 100 (sin 100 pt) V ⇒ Vm = 100 V. So, Vavg
V 100 = m [3 + cos a ] = [3 + cos 30°] = 61.56 V 2π 2π
126. Topic: Characteristics of Semiconductor Power Devices: MOSFET (0.4) For a Buck converter, VL (avg) = 0. So, 30 (TON) - 20 (TOFF ) = 0 But,
TOn T = D and OFF = 1 − D TOn T
Therefore, 30 D = 20 (1 − D ) ⇒ D = 0.4 127. Topic: Characteristics of Semiconductor Power Devices: IGBT (170) For the given current direction, the diode will be conducting. Therefore, from the given V–I characteristics of the diode, the diode drop can be calculated as VD = 0.7 + 100 × 0.01 = 1.7 V. The conduction loss Pc = 1.7 × 100 = 170 W.
Ch wise GATE_EE_Ch10.indd 551
∫I
2 0
dω t
0
100 3
= 57.7 A
129. Topic: DC to DC Conversion: Buck, Boost and Buck-Boost Converters (60) When switch S is OFF diode is ON. −
+ 24 V −
V Vo = S 1− D 60 =
2π
I 1 2π = 0 2π 3 3
= I0
In a buck converter, for the inductor current iL to be continous, the required resistance is −3
1 2π
I D(rms) =
123. Topic: Characteristics of Semiconductor Power Devices: Diode (2500) We have f = 100 kHz, Vo = 36 V, L = 5 mH, D = 0.6 and Vin = 60 V
3
551
36 V +
Therefore, V = 24 + 36 = 60 V 130. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (c) Input pf = pf of AC mains ⇒
2 2 2 2 cos a = cos 30° = 0.78 π π
131. Topic: Characteristics of Semiconductor Power Devices: IGBT (10) We have, v ma = m = 0.8 vC (V01 ) peak = ma ( I 01 ) peak = =
VDC = 0.8 × 250 = 200 V 2
(V01 ) peak Z1
=
200 122 + 16 2
200 R + (ω L) 2 2
= 10 A
132. Topic: Distortion Factor of AC to DC Converters T1
T2
0
T1
(75) Energy = ∫ V ⋅ idt + ∫ V ⋅ idt
11/21/2018 6:20:32 PM
552
GATE EE Chapter-wise Solved Papers
⎡1 ⎤ ⎡1 ⎤ = V ⎢ IT1 ⎥ + I ⎢ VT2 ⎥ ⎣2 ⎦ ⎣2 ⎦ ⎡150 ⎤ ⎡1 ⎤ = 600 ⎢ × 1 × 10 −6 ⎥ + 100 ⎢ × 600 × 1 × 10 −6 ⎥ 2 2 ⎣ ⎦ ⎣ ⎦ = 75 mJ 133. Topic: Single Phase and Three Phase Inverters (a) We know that for full-bridge VSI V01(rms) =
2 2 V sin d π S
Given that, 2d = 120°, d = 60°. Therefore, V01(rms) =
2 2 3 ⋅ 300 ⋅ = 234 V π 2
134. Topic: Single Phase and Three Phase Inverters (6) For the given converter, the output voltage and current can be obtained as follows Vo =
2Vm ⎛ 200π ⎞ ⎛ 2π ⎞ cos a = 2 ⎜ cos ⎜ ⎟ = − 200 V ⎟ ⎝ π ⎠ ⎝ 3⎠ π EI o = I o2 R + Vo I o 800 I o = I ( 20) + 200 I o 2 o
Power = Vs I s 360 × I s = 4000 ⇒ I s = 11.11 A ⎛T ⎞ I switch = I s ⎜ On ⎟ ⎝ T ⎠
1/ 2
= I s D = 11.1 0.1 = 3.5 A
137. Topic: Bidirectional AC to DC Voltage Source Converters (9.21) For the given single-phase, bidirectional VSC, P = Vs I s cos θ 5 × 103 = 220 I s × 1 ⇒ I s = 22.72 A tan δ =
I s X s 22.72 × 2π × 50 × 5 × 10 −3 = Vs 220
δ = 9.21° 138. Topic: Characteristics of Semiconductor Power Devices: IGBT (d) MOSFET is majority carrier device, whereas IGBT, diode, and thyristor are minority carrier devices. 139. Topic: Line Commutated Thyristor Based Converters (198.06) n0(t)
Io = 30 A Power fed to source = VoIo = 200 × 30 = 6 kW 135. Topic: Single Phase and Three Phase Inverters (24) For the given three-phase VSI
P=3
2 ph
V
R
=
VL =
2 VS = Vph 3
Vph =
2 × 600 3
3×
Vrms = =
2 × 600 × 600 3 = 24 kW 30
136. Topic: DC to DC Conversion: Buck, Boost and Buck-Boost Converters (3.5) For the boost converter Vo 1 = Vs 1 − D Therefore, duty factor is 400 1 = ⇒ D = 0.1 360 1 − D
Ch wise GATE_EE_Ch10.indd 552
p
2p
0
wt
4Vs 2π 4 × 220 2π
= 198.06 V 140. Topic: Single Phase and Three Phase Inverters (9) Vphase 300 V
−300 V
2p 3
p
2p
wt
11/21/2018 6:20:34 PM
553
Chapter 10 • Power Electronics
Vrms =
145. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers
1 2π (300) 2 ⋅ = 244.94 V π 3
(224.17) ΔVd0 = 4 fLS ⋅ I 0
Therefore, the load power is given by
= 4 × 50 × (10 × 10-3) × 14
3 × ( 244.94) 2 3V 2 /R = ≈ 9 kW 20
= 28 V Now,
141. Topic: DC to DC Conversion: Buck, Boost and Buck-Boost Converters (c)
Vd0 =
Vm (cos a − cos(a + μ )) π
⇒ 28 =
V0
220 2 (1 − cos μ ) (Since a = 0) π
⇒ cos μ = 0.717 ⇒ m = 44.17° n = 2, 4, 6 wt
Therefore, Conduction angle = 180 + 44.17 = 224.17° 146. Topic: Single Phase and Three Phase Inverters (c)
All even harmonics are present. Therefore, 2fs = 2 × 50 = 100 Hz 142. Topic: Single Phase and Three Phase Inverters (b) Three-phase VSI 180° mode is
V01peak = V01 rms =
VR p
Vdc/3
2p
wt
VR n = ⇒VR1 =
6 Vdc / 3 sin n ω t nπ 2 Vdc sin ω t π
143. Topic: Single and Three Phase Configuration of Uncontrolled Rectifiers (0) The RMS value of the current through diode D3 is 0A as D2 is OFF and it will not turn ON for R load. 144. Topic: DC to DC Conversion: Buck, Boost and Buck-Boost Converters D Vs D (32) ⇒ 48 = ⇒ D = 3/ 5 1− D 1− D
Similarly, when Vs = 72 V, Vo = 48 V, D = 2/5 Therefore, 2 3 ≤D≤ 5 5
Ch wise GATE_EE_Ch10.indd 553
MVs 2 2
=
0.6 Vs 2 2 2
0.3 ⎞ ⎛ Z1 = R 2 + (ω L) 2 = 40 2 + ⎜ 2π × 50 × ⎟ = 50 ⎝ π ⎠ ⎛ ωL⎞ ⎛ 30 ⎞ = tan −1 ⎜ ⎟ = 36.869° f 1 = tan −1 ⎜ ⎝ R ⎟⎠ ⎝ 40 ⎠
Now,
(a) Vo =
MVs 0.6Vs = , 2 2
Active power = V01 I 01 cos f1 =
V012 cos f1 Z1
2
1 ⎛ 0.6 ⎞ ⇒ 1440 = ⎜ V × × 0.674 ⎝ 2 2 S ⎟⎠ 50 ⇒ 1440 =
0.045 × 0.674 × Vs2 50
⇒ Vs ≈ 1540 V Vdc =
VS = 770 V ≈ 500 2 2
147. Topic: Sinusoidal Pulse Width Modulation (40) Buck–boost converter is Vo =
DVS 0.6 × 50 = = 75 1 − D 1 − 0.6
11/21/2018 6:20:36 PM
554
GATE EE Chapter-wise Solved Papers
(2) When MT1 is +ve and MT2 is -ve
Vo I S D 3 = = = VS I o 1 − D 2 Io =
+
Vo = 15A ⇒ I S = 22.5A R
I
Large capacitor ⇒ iC = 0 at steady state -
I L = I S + I 0 = 37.5 A DVS 0.6 × 50 ΔI L = = = 5A fL 10 × 103 × 0.6 × 10 −3
(3) When MT1 is -ve and MT2 is +ve - G
Therefore,
I ΔI L = 40 A 2
(iL ) peak = I L + 148. Topic: Power Factor (0.70)
+ GTO: GTO conducts only from anode to cathode A
V0 I
100 sin 100 pf
G
p PF =
K
Vm / 2 Vm / 2
= 0.707
MOSFET D D
149. Topic: Characteristics of Semiconductor Power Devices (a) SCR: conducts only from anode to cathode
G
G S S
(1) When D is +ve and S is -ve, diode is OFF
A
D(+ve) I G S(-ve) I
K Triac (1) Triac is combination of two anti parallel SCRs
(2) When D is -ve and S is +ve, diode is ON
MT1 D
I
G MT2
Ch wise GATE_EE_Ch10.indd 554
S
11/21/2018 6:20:38 PM
Chapter 10 • Power Electronics
150. Topic: Single Phase and Three Phase Inverters (90) Power delivered to lossless battery = EI 0 = 1600 W 80 I 0 = 1600 I 0 = 20 A From given 1− ϕ rectifier, V0 = E + I 0 R Vm [3 + cos a ] = E + I 0 R 2π 80π [3 + cos a ] = 80 + 20.2 2π 40[3 + cos a ] = 120
555
151. Topic: Bidirectional AC to DC Voltage Source Converters (50) S1 and S2 are complimentary. Load is always connected to source. Therefore,
V0 = Vs = 100 V I0 =
(1)
V0 100 = = 100 A R 1
Pin = P0 Vs × Is = V0I0 Þ⇒ Is = I0 = 100 A [from Eq. (1)] Þ⇒ I s1 =
I s 100 = = 50 A 2 2
3 + cos a = 3 cos a = 0
a = 90°
Ch wise GATE_EE_Ch10.indd 555
11/21/2018 6:20:39 PM
Ch wise GATE_EE_Ch10.indd 556
11/21/2018 6:20:39 PM
aPPENDIX
Solved GATE (EE) 2019 QUESTIONS
Chapter 1: Engineering Mathematics 1.
s+3 The inverse Laplace transform of H ( s) = 2 for s + 2s + 1 t ≥ 0 is (a) 3te−t + e−t (b) 3e−t (c) 2te−t + e−t (d) 4te−t + e−t
Topic: Differential Equations (b) The equation ∂2u 2 ⎛ ∂2u ∂2u ⎞ −c ⎜ 2 + 2 ⎟ = 0 ∂t 2 ∂y ⎠ ⎝ ∂x represents the Wave equation. Note:
(1 Mark)
1 ∂2u (Wave Equation) c 2 ∂t 2 ∇ 2ϕ = 0 (Laplace Equation)
∇2 E =
Topic: Transform Theory s+3 ,t ≥0 s + 2s + 1 ⎡ s+3 ⎤ ⎡ s+3 ⎤ L−1 [ H ( s)] = L−1 ⎢ 2 = L−1 ⎢ 2 ⎥ ⎥ ⎣ s + 2s + 1⎦ ⎣ ( s + 1) ⎦
(c) Given H ( s) =
2
2 ⎤ ⎡ s +1+ 2 ⎤ ⎡ 1 + = L−1 ⎢ = L−1 ⎢ 2 ⎥ s + 1 (s s + 1 ) ( s + 1) 2 ⎥⎦ ⎣ ⎦ ⎣ ⎡ 1 ⎤ ⎡ 1 ⎤ = L−1 ⎢ + 2 L−1 ⎢ 2 ⎥ ⎥ ⎣ s + 1⎦ ⎣ ( s + 1) ⎦ = e − t (1) + 2e − t (t )
⎡ −1 ⎡ 1 ⎤ ⎤ ⎢∵ L ⎢ 2 ⎥ = t ⎥ ⎣s ⎦ ⎦ ⎣
Hence L-1[H(s)] = 2te-t + e-t 2.
M is a 2 × 2 matrix with eigenvalues 4 and 9. The eigenvalues of M 2 are (a) 4 and 9 (b) 2 and 3 (c) −2 and −3 (d) 16 and 81 (1 Mark)
Topic: Linear Algebra (d) 4 and 9 are the eigenvalues of matrix M, which is a 2 × 2 matrix. The eigenvalues of M 2 are (4)2 = 16 and (9)2 = 81. 3.
The partial difference equation ∂2u 2 ⎛ ∂2u ∂2u ⎞ −c ⎜ 2 + 2 ⎟ = 0 ∂t 2 ∂y ⎠ ⎝ ∂x where c ≠ 0 is known as (a) Heat equation (b) Wave equation (c) Poisson’s equation (d) Laplace equation (1 Mark)
GATE_EE_2019.indd 1
∇ 2V = −
ρ (Poisson’s Equation) ε
∂u − α∇ 2 u = 0 (Heat Equation) ∂t 4.
Which one of the following functions is analytic in the region |z| ≤ 1? (a)
z2 −1 z
(b)
z2 −1 z+2
(c)
z2 −1 z − 0.5
(d)
z2 −1 z + j 0.5 (1 Mark)
Topic: Complex Variables (b) Given region is |z| ≤ 1, which represents the region inside and on the unit circle |z| = 1. The functions given in the options (a), (c), (d) are not analytic functions in the region |z| ≤ 1, since the singular points -j(0.5), 0.5, 0 lie inside |z| = 1. z2 −1 . This implies Z = -2 is the singular Let f ( z ) = z+2 2 point. However, this lies outside |z| = 1. Therefore z − 1 z+2 is analytic in the region |z| ≤ 1. 5. The mean square of a zero-mean random process is kT , where k is Boltzmann’s constant, T is the absolute C temperature, and C is a capacitance. The standard deviation of the random process is
6/24/2019 10:36:29 AM
A2
GATE EE Chapter-wise Solved papers
kT C C kT
(a) (c)
(b)
kT C
(d)
kT C
7.
(1 Mark)
⎡0 1 1 ⎤ The rank of the matrix M = ⎢⎢1 0 1 ⎥⎥ is ⎢⎣1 1 0 ⎥⎦
(3)
⎡1 0 1 ⎤ ⎢0 1 1 ⎥ ⎢ ⎥ ⎢⎣1 1 0 ⎥⎦
kT kT −0 = C C
Standard deviation = V ( X ) =
R2 → R2 - R1 ⎡1 1 0 ⎤ ⎢0 −1 1 ⎥ ⎢ ⎥ ⎢⎣0 1 1 ⎥⎦ R3 → R3 - R1
kT C
6. If f = 2x3 + 3y2 + 4z, the value of line integral C
grand f ⋅ dr
evaluated over contour C formed
⎡1 0 1 ⎤ ⎢0 1 1 ⎥ ⎢ ⎥ ⎢⎣0 1 −1⎥⎦
by the segments (−3, −3, 2) → (2, −3, 2) → (2, 6, 2) → (2, 6, −1) is . (1 Mark) Topic: Calculus (139) f = 2x3 + 3y2 + 4z ∂f ∂f ∂f grad f = i + j + k ∂x ∂y ∂z ∂ (2 x3 ) ∂ (3 y 2 ) ∂ (4 z) +j +k ∂x ∂y ∂z 2 = 6x i + 6y j + 4k i j k Curl grad f = ∂ /∂x ∂ /∂y ∂ /∂z 6x2 So,
∫ grad c
6y
4
f . dr = ∫ 6 x 2 dx + 6 ydy + 4 dz c
( 2 , −3, 2 )
∫
=
6 x 3 dx + 6 ydy + 4 dz
( −3, −3, 2 ) ( 2,6 , 2 )
∫
+
( 2 , 6 , −1)
∫
6 x dx + 6 ydy + 4 dz 2
( 2,6 , 2 )
= 70 + 81 - 12 = 139
GATE_EE_2019.indd 2
(Echelon form)
Consider a 2 × 2 matrix M = [v1 v2], where v1 and v2 ⎡ uT ⎤ are the column vectors. Suppose M −1 = ⎢ 1T ⎥ where ⎣ u2 ⎦ T T u1 and u2 are the row vectors. Consider the following statements. Statement 1: u1T v1 = 1 and u2T v2 = 1 Statement 2: u1T v2 = 0 and u2T v1 = 0 Which of the following option is CORRECT? (a) Statement 1 is true and Statement 2 is false. (b) Statement 2 is true and Statement 1 is false. (c) Both the statements are true. (d) Both the statements are false.
6 x 2 dx + 6 ydy + 4 dz
( 2 , −3, 2 )
+
Non-zero rows are 3. Therefore, rank of matrix = 3 8.
=i
⎡0 1 0 ⎤ M = ⎢⎢1 0 1 ⎥⎥ ⎢⎣1 1 0 ⎥⎦
Interchanging row 1 and row 2
E(X) = 0 V(X) = E(X2) - [E(X)]2
∫
(1 Mark)
Topic: Linear Algebra
Topic: Probability and Statistics (d) Given kT E( X 2 ) = C and
=
.
(2 Marks) Topic: Linear Algebra (c) Let ⎡a M 2× 2 = ⎢ 11 ⎣ a21
a12 ⎤ ⎡a ⎤ ⎡a ⎤ , where v1 = ⎢ 11 ⎥ , v2 = ⎢ 12 ⎥ ⎥ a22 ⎦ ⎣ a21 ⎦ ⎣ a22 ⎦
6/24/2019 10:36:35 AM
appendix • Solved GATE (EE) 2019
10. A periodic function f(t), with a period of 2π, is represented as its Fourier series, ⎧ A sin t , 0 ≤ t ≤ π f (t ) = ⎨ π ≤ t < 2π ⎩ 0,
Then M −1 =
⎡ a22 1 ⎢ a11a22 − a21a12 ⎣ −a21
−a12 ⎤ a11 ⎥⎦
where u1T =
1 [ a22 a11a22 − a21a12
−a12 ]
u2T =
1 [ −a21 a11a22 − a21a12
a11 ]
−a12 ⎤ ⎡ a11 ⎤ =1 a11a22 − a21a12 ⎥⎦ ⎢⎣ a21 ⎥⎦ a11 ⎤ ⎡ a12 ⎤ =1 a11a22 − a21a12 ⎥⎦ ⎢⎣ a22 ⎥⎦
Topic: Calculus
a0 =
a11 ⎤ ⎡ a11 ⎤ =0 a11a22 − a21a12 ⎥⎦ ⎢⎣ a21 ⎥⎦
ated counter-clockwise is (a) +8jπ (c) −4jπ
z3 + z 2 + 8 ∫ z + 2 dz evaluz =5 (b) −8jπ (d) +4jπ
(2 Marks)
Topic: Complex Variables z3 + z 2 + 8 z+2 So, singular point of F(z) is Z = -2, which lies inside C |z| = 5. Using Cauchy’s integral formula, we have (a) Let F ( z ) =
C
C
z3 + z 2 + 8 z3 + z 2 + 8 dz = ∫ dz z+2 z − ( −2) C
= 2π j[23 + 22 + 8]z = −2 ⎡ By Cauchy’s formula ⎤ ⎢ f ( z) ⎥ ⎢ ⎥ dz jf z = 2 π ( ) o ∫C z − zo ⎢⎣ ⎥⎦ ⇒
∫ C
GATE_EE_2019.indd 3
∞
∞
n −1
n =1
f (t ) = a0 + ∑ an cos nt + ∑ bn sin nt
(d)
−a12 ⎤ ⎡ a12 ⎤ =0 a11a22 − a21a12 ⎥⎦ ⎢⎣ a22 ⎥⎦
The closed loop line integral
∫ F ( z )dz = ∫
∞
where
So Statement II is also true. Hence, both the statements are correct. 9.
∞
series coefficients a1 and b1 of f(t) are A A (a) a1 = ; b1 = 0 (b) a1 = ; b1 = 0 π 2 A A (c) a1 = 0; b1 = (d) a1 = 0; b1 = π 2 (2 Marks)
So Statement I is true. Now a22 ⎡ u1T v2 = ⎢ ⎣ a11a22 − a21a12 −a21 ⎡ u2T v1 = ⎢ ⎣ a11a22 − a21a12
f (t ) = a0 ∑ n =1 an cos nt + ∑ n =1 bn sin nt , the Fourier
If
Therefore a22 ⎡ u1T v1 = ⎢ ⎣ a11a22 − a21a12 −a21 ⎡ u2T v2 = ⎢ ⎣ a11a22 − a21a12
A3
z3 + z 2 + 8 dz = 2π j ( −8 + 4 + 8) = 8π j z+2
a1 =
2π
2 2π
∫
f ( x )dx (T0 = 2π)
∫
f (t ) cos ω0 t dt
0 2π
2 2π
0
2π ⎛ ⎞ ⎜ ω0 = T = 1 ⎟ 0 ⎝ ⎠
2 = f (t ) cos t dt 2π ∫ π A = sin 2t dt = 0 2π ∫0 b1 = =
2π
2 T0
∫
f (t ) sin ω0 . t dt
0
2 2π
2π
∫
f (t ) sin t dt
0
π
=
1 A sin t . sin t dt π ∫0
=
A A 2 sin 2 t dt = [1 − cos 2t ] dt ∫ 2π 0 2π ∫0
=
A 2π
π
π
π
A ⎡ sin 2t ⎤ ⎢t − 2 ⎥ = 2 ⎣ ⎦0
11. If A = 2xi + 3yj + 4zk and u = x2 + y2 + z2, then div(uA) at (1, 1, 1) is . (2 Marks) Topic: Calculus (45) ∇ . (uA) = (∇u ) . A + u(∇ . A) u = x2 + y2 + z2 (Scalar) A = 2 × i + 3yj + 4zk (Vector) ∇u =
∂u ∂u ∂u i+ j+ k ∂x ∂y ∂z
6/24/2019 10:36:41 AM
A4
GATE EE Chapter-wise Solved papers
∇u =
Topic: Network Graph, KCL, KVL, Node and Mesh Analysis (1.4) Apply Nodal Analysis at node V. We get
∂x 2 ∂y 2 ∂z 2 i+ j+ k ∂x ∂y ∂z
= 2xi + 2yj + 2zk
V − 20 V − 5I −2+ =0 2 3
(∇u ) . A = ( 2 × i + 2 yj + 2 zk ) . ( 2 × i + 3 yj + 4 zk ) = 4x2 + 6y2 + 8z2 ∇.(uA) = u(∇. A) + (∇u ). A = (x2 + y2 + z2) (2 + 3 + 4) + 4x2 + 6y2 + 8z2
where I =
20 − V . So 2 I 2Ω
= 9(x2 + y2 + z2) + (4x2 + 6y2 + 8z2)
V
3Ω
At (1, 1, 1), (uA) = ∇(uA) = 9(12 + 12 + 12 ) + ( 4 *12 + 6 *12 ) dw = (9 × 3) + (4 + 6 + 8)
20 V
+ −
2A
= 45 12. The probability of a resistor being defective is 0.02. There are 50 such resistors in a circuit. The probability of two or more defective resistors in the circuit (round off to two decimal places) is .
V − 20 7V − 100 + =2 2 6 ⇒ 10V = 172 ⇒ V = 17.2 20 − 17.2 2.8 = = 1.4 A I= 2 2
(2 Marks) Topic: Calculus (0.26) The probability of defective resistance = 0.02. Total number of resistors N = 50.
λ = Np = 50 × 0.02 = 1
14. The current I flowing in the circuit shown below in amperes is .
Probability of two or more defective resistors in the circuit
I 50 Ω
P(x ≥ 2) = 1 - P(x < 2) = 1 − [ P ( x = 0) + P ( x = 1)]
5I
25 Ω
40 Ω
20 Ω
20 Ω
200 V ⎡ e −λ λ 0 e −λ λ1 ⎤ = 1− ⎢ + ⎥ 1! ⎦ ⎣ 0!
160 V
= 1 − e − λ (1 + 1) = 1−
{λ = 1}
2 = 0.26 e
Chapter 2: Electric Circuits 13. The current I flowing in the circuit shown below in amperes (round off to one decimal place) is A. I
2Ω
3Ω
Topic: Network Graph, KCL, KVL, Node and Mesh Analysis (0) Applying Millman’s theorem 200 160 100 80 + − − 40 25 2 E = 50 1 1 1 1 + + + 50 40 25 20 =0V 0 E I= = =0A Req + 20 Req + 20
Req 20 V
2A
80 V (2 Marks)
= 1 − e − λ (1 + λ )
100 V
I
+ 5I — Veq = E
20 Ω
(1 Mark)
GATE_EE_2019.indd 4
6/24/2019 10:36:47 AM
appendix • Solved GATE (EE) 2019
Chapter 3: Electromagnetic Fields
=
15. A co-axial cylindrical capacitor show in Figure (i) has dielectric with relative permittivity εr1 = 2. When one-fourth portion of the dielectric is replaced with another dielectric of relative permittivity εr2, as shown in Figure (ii), the capacitance is doubled. The value of εr2 is
The given condition is C2 = 2C1 Putting the values in Eqs. (1) and (2) we get ⎡ ⎤ ⎢ 4πε × l ⎥ πε 0 l ⎡ ε r 2 ⎤ 0 ⎥ 3+ = 2⎢ 2 ⎥⎦ ⎛ R ⎞ ⎢⎣ ⎢ ⎛ R⎞ ⎥ ln ⎜ ⎟ ln ⎢ ⎜⎝ r ⎟⎠ ⎥ ⎝r⎠ ⎣ ⎦ εr2 3+ =8 2 εr2 = 5 2 εr = 10
R
r
r
e r1 = 2
e r1 = 2
Figure (i)
Figure (ii)
⇒ ⇒ ⇒
(1 Mark) Topic: Capacitance of Simple Configurations (10) The capacitance of co-axial cylindrical capacitor is C=
2πε r1ε 0 l ⎛R⎞ ln ⎜ ⎟ ⎝r⎠
where r = inner diameter of capacitor, R = outer diameter of capacitor, l = length of cylinder. Given εr1 = 2. Hence,
2πε 0 2l C1 = (1) ⎛R⎞ ln ⎜ ⎟ ⎝r⎠ C2
C1 R
r
16. A 0.1 μF capacitor charged to 100 V is discharged through a 1 kΩ resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1 V is . (2 Marks) Topic: Capacitance of Simple Configurations (0.4) Voltage rise across capacitor in time t V (t ) = V0 e
GATE_EE_2019.indd 5
t τ
τ = RC = 1000 × 0.1× 10 −7 = 10 −4 sec V (t ) = 100e
−
t 10 −4 4
1 = 100e −10 t
⇒
4
0.01 = e −10 t t = 0.46 ms
⇒ ⇒
17. The magnetic circuit shown below has uniform cross-sectional area and air gap of 0.2 cm. The mean path length of the core is 40 cm. Assume that leakage and fringing fluxes are negligible. 10 cm
I 10 cm
3π π × ε 0 × ε r1 × l × ε0 × εr2 × l 2 C2 = + 2 ⎛R⎞ ⎛R⎞ ln ⎜ ⎟ ln ⎜ ⎟ r ⎝ ⎠ ⎝r⎠
−
Given V0 = 100 V. Now
e r1 = 2 When one-fourth portion of capacitor is filled with another dielectric (εr2), then the total capacitance = C1 + C2 where
πε 0 l ⎡ ε r 2 ⎤ 3+ (2) 2 ⎥⎦ ⎛R⎞⎢ ln ⎜ ⎟ ⎣ ⎝r⎠ {∵ ε r1 = 2 given}
e r2 R
A5
0.2 cm
6/24/2019 10:36:51 AM
A6
GATE EE Chapter-wise Solved papers
When the core relative permeability is assumed to be infinite, the magnetic flux density computed in the air gap is 1 tesla. With same Ampere-turns, if the core relative permeability is assumed to be 1000 (linear), the flux density in tesla (round off to three decimal places) calculated in the air gap is .
If we see options (a), (c), (d), in these u(t) is multiplied which means that their impulse responses are zero for negative value of time, hence they are causal. If we check option (b) h(t) = 1 + e-at u(t) we can see h(t) = 1, t < 0, hence it represents a noncausal system.
(2 Marks) Topic: Magnetic Circuits, Self and Mutual Inductance of Simple Configurations (0.834) Length of air gap a = 0.2 cm Length of magnetic core = 40 - 0.2 = 39.8 cm
or
MMF = Reluctance × Flux = Rϕ ϕ NI NI ϕ= , B= = R A RA 1 B∝ R
19. The output response of a system is denoted as y(t), and its 10 Laplace transform is given by Y ( s) = . 2 s( s + s + 100 2 ) The steady state value of y(t) is (a) (c)
1
(b) 10 2
10 2 1
(d) 100 2
100 2
(1 Mark)
When the relative permeability of core, μ r = ∞ Ra = R1 + R2 = 0 +
Topic: Laplace Transform and z-Transform (a) To find steady-state value, we need to find y(∞)
0.2 μ0 A
y(∞) = lim sY ( s)
When the relative permeability of core, μr = 1000 μ0 39.8 0.2 Rb = R1 + R2 = + 1000 μ0 A μ0 A
= lim
Chapter 4: Signals and Systems 18. The symbols a and T represent positive quantities, and u(t) is the unit step function. Which one of the following impulse response is NOT the output of a causal linear time-invariant system?
10 100 2
=
1 10 2
20. A 5 kVA, 50 V/100 V, single-phase transformer has a secondary terminal voltage of 95 V when loaded. The regulation of the transformer is (a) 4.5% (b) 9% (c) 5% (d) 1% (1 Mark) Topic: Single Phase Transformer: Equivalent Circuit, Phasor Diagram, Open Circuit and Short Circuit Tests, Regulation and Efficiency (c) % Voltage regulation
(b) e − a ( t +T ) u(t ) (d) e − a ( t −T ) u(t )
=
(1 Mark)
Topic: Linear Time Invariant and Causal Systems (c) If an LTI system is causal, we should have the condition h(t) = 0, t < 0
GATE_EE_2019.indd 6
10 s s ( s 2 + s + 100 2 )
Chapter 5: Electrical Machines
0.2 μ0 A
Ra 0.2 = = 0.834 = Rb 0.2398 0.2398 μ0 A
(c) 1+ e − at u(t )
=
B1 Rb = B2 Ra
(a) e + at u(t )
s →0
0.2398 = μ0 A
⇒ B2 =
s →0
=
VNo Load − VFull Load VNo Load
× 100
100 − 95 × 100 = 5% 100
21. The parameter of an equivalent circuit of a three-phase induction motor affected by reducing the rms value of the supply voltage at the rate frequency is
6/24/2019 10:36:57 AM
appendix • Solved GATE (EE) 2019
(a) (b) (c) (d)
rotor resistance rotor leakage reactance magnetizing reactance stator resistance (1 Mark)
Topic: Three Phase Induction Motors: Principle of Operation, Types, Performance, Torque-Speed Characteristics, No-Load and Blocked Rotor Tests, Equivalent Circuit, Starting and Speed Control (c) The magnetizing reactance depends on the air gap v flux. The air gap flux is proportional to . f 22. A three-phase synchronous motor draws 200 A from the line at unity power factor at rated load. Considering the same line voltage and load, the line current at a power factor of 0.5 leading is (a) 100 A (b) 200 A (c) 300 A (d) 400 A (1 Mark) Topic: Synchronous Machines: Cylindrical and Salient Pole Machines, Performance, Regulation and Parallel Operation of Generators, Starting of Synchronous Motor, Characteristics (d) Power, P = 3 VI cos ϕ , for the same voltage and load P1 = VI1 cos ϕ1 , P2 = VI 2 cos ϕ2 V I1 cos ϕ1 = V I 2 cos ϕ2 I1 cos ϕ1 200 × 1 = = 400 A cos ϕ2 0.5
23. The total impedance of the secondary winding, leads, and burden of a 5 A CT is 0.01 Ω. If the fault current is 20 times the rated primary current of the CT, the VA output of the CT is . (1 Mark) Topic: Single Phase Transformer: Equivalent Circuit, Phasor Diagram, Open Circuit and Short Circuit Tests, Regulation and Efficiency (100) Impedance of CT’s secondary = 0.01 Ω (given) If fault current is 20 times on primary side, the secondary current of CT would also be 20 times. That is, 20 × 5 = 100 A. The VA output of CT = I2R = 1002 × 0.01 = 100 VA
GATE_EE_2019.indd 7
24. A single-phase transformer of rating 25 kVA, supplies a 12 kW load at power factor of 0.6 lagging. The additional load at unit power factor in kW (round off to two decimal places) that may be added before this transformer exceeds its rated kVA is . (2 Marks) Topic: Single Phase Transformer: Equivalent Circuit, Phasor Diagram, Open Circuit and Short Circuit Tests, Regulation and Efficiency (7.21) Connected load = 12 kW at 0.6 PF lag. P = VI cos ϕ
or
VI =
P 12 = = 20 kVA cos ϕ 0.6
Connected reactive power Q = VI sin ϕ = 20 × sin 53.13 Q = 16 kVAr The transformer rating given as S = 25 kVA S = P2 + Q2 25 = ( P + Padditional ) 2 + Q 2 252 = P (12 + Padditional ) 2 + 16 2 Padditional = 7.21 kW
Now P1 = P2 (given condition). So
⇒ I2 =
A7
25. A 220 V DC shunt motor takes 3 A at no-load. It draws 25 A when running at full-load at 1500 rpm. The armature and shunt field resistances are 0.5 Ω and 220 Ω, respectively. The no-load speed in rpm (round off to two decimal places) is . (2 Marks) Topic: Three Phase Induction Motors: Principle of Operation, Types, Performance, Torque-Speed Characteristics, No-Load and Blocked Rotor Tests, Equivalent Circuit, Starting and Speed Control (1579.33) Full-load speed NFL = 1500 rpm Full-load current IFL = 25 A 220 =1A 220 Full-load armature current Field current I F =
I aFL = I FL − I F = 24 A EbFL = V − I aFL Ra = 220 - 24 × 0.5 = 208 V
6/24/2019 10:37:01 AM
A8
GATE EE Chapter-wise Solved papers
Topic: Starting of Synchronous Motor
For no-load condition INL = 3 A
(245.36)
V ph = 3VL
If = 1 A
VL =
IaNL = 3 - 1 = 2 A
3 Synchronous impedance
EbNL = 220 - 2 × 0.5 = 219 V We know that Eb ∝ N . So
220 3
= 127.01 V
I = 10 ∠38.86 A Internal voltage per phase
or No-load speed 219 × 1500 = 1579.3269 = 1579.33 rpm 208
26. A delta-connected, 3.8 kW, 400 V(line), three-phase, 4-pole, 50-Hz squirrel-cage induction motor has the following equivalent circuit parameter per phase referred to the stator: R1 = 5.39 Ω, R2 = 5.72 Ω, X1 = X2 = 8.22 Ω. Neglect shunt branch in the equivalent circuit. The starting line current in amperes (round off to two decimal places) when it is connected to a 100 V (line), 10 Hz, three-phase AC source is . (2 Marks) Topic: Three Phase Induction Motors: Principle of Operation, Types, Performance, Torque-Speed Characteristics, No-Load and Blocked Rotor Tests, Equivalent Circuit, Starting and Speed Control (14.949) X1 = X2 = 8.22 Ω, VL = Vph = 400 V (at 50 Hz) R1 = 5.39 Ω, R2 = 5.72 Ω Now at 10 Hz, 8.22 V ph = 100 V, X 1 = X 2 = = 1.644 Ω 5 VPh V I Ph = P = 2 Z ( R1 + R2 ) + ( X 1 + X 2 ) 2 100 = (5.39 + 5.72) 2 + (1.644 + 1.644) 100 = (11.11) 2 + (3.288) 2 = 8.3611 For delta connection, I L = 3I ph I L = 3 × 8.3611 = 14.949 A 27. A 220 V (line) three-phase, Y-connected, synchronous motor has a synchronous impedance of (0.25 + j2.5) Ω/phase. The motor draws the rated current of 10 A at 0.8 pf leading. The rms value of line-to line internal voltage in volts (round off to two decimal places) is . (2 Marks)
GATE_EE_2019.indd 8
=
Z = (0.25 + j2.5) Ω
EbNL N FL = EbFL N NL N NL =
V ph
Eb =
220 3
− (0.25 + j 2.5) × 10 ∠36.86
= 141.658 ∠8.728° V
(Phase to phase)
So, Line voltage = 3 × 141.658 = 245.36 28. A 30 kV, 50 Hz, 50 MVA generator has the positive, negative, and zero sequence reactances of 0.25 pu, 0.15 pu, and 0.05 pu, respectively. The neutral of the generator is grounded with a reactance so that the fault current for a bolted LG fault and that of a bolted threephase fault at the generator terminal are equal. The value of grounding reactance in ohms (round off to one decimal place) is . (2 Marks) Topic: Regulation and Parallel Operation of Generators (1.8) X1 = 0.25 pu X2 = 0.15 pu X0 = 0.05 pu ILG = ILLL 3 Vpu ( X 0 + X1 + X 2 + 3 X n )
=
X pu X1
⇒
3 ×1 1 = (0.05 + 0.25 + 0.15 + 3 X n ) 0.25
⇒
3 1 = (3 X n + 0.45) 0.25
Xn = 0.1 pu
⇒ or
X n = 0.1× Z base = 0.1× = 0.1×
( KVbase ) 2 MVAbase
30 2 = 1.8 Ω 50
6/24/2019 10:37:07 AM
appendix • Solved GATE (EE) 2019
Chapter 6: Power Systems
(1 Mark) Topic: Bus Admittance Matrix (0.1) The line admittance between bus 1 and 2 = −y12 = −j20. There are two lines in parallel. So, the admittance of each line will be −j yper line = 20 = − j10 2 Therefore, 1
Magnitude of the series reactance of each line in pu is 1 1 = = 0.1 pu j10 10 30. A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase and the capacitance is 0.01 μF/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end. The receiving end line voltage in kV (round off to two decimal places) will be . (2 Marks) Topic: Models and Performance of Transmission Lines and Cables (418.84) Supply voltage VS = 400 kV Length of line L = 300 km Inductance of line L = 1 mH/km/phase Capacitance of line C = 0.01 μF/km/phase Finding the parameters 1 1 V= = −3 LC 1× 10 × 0.01× 10 −6 = 3.16 × 105 km / sec
β=
GATE_EE_2019.indd 9
VS 400 = = 418.84 kV A 0.955 31. In the single machine infinite bus system shown below, the generator is delivering the real power of 0.8 pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees (round off to one decimal place) is .
2π fl 2π × 50 × 300 = = 0.29 V 3.16 × 105
XL1 = 0.4 pu
Xt = 0.2 pu G
XL2 = 0.4 pu
V = 1∠0° ∞
Xd′ = 0.25 pu (2 Marks) Topic: Bus Admittance Matrix (20.5) XL1 and XL2 are in parallel. Therefore
yper line
j 1 = = = j10 − j10 − j × j10
β2 (0.29) 2 = 1− = 0.955 2 2 So, receiving end voltage is given by A = 1−
29. The Ybus matrix of a two-bus power system having two identical parallel lines connected between them in pu is ⎡ − j8 j 20 ⎤ given as Ybus = ⎢ ⎥. ⎣ j 20 − j8⎦ The magnitude of the series reactance of each line in pu (round off up to one decimal place) is .
Impedance of each line =
A9
1 1 1 + = 0.4 0.4 0.2 or
XL1 and XL2 = 0.2 pu
Total reactance of system is XT = 0.25 + 0.2 + 0.2 = 0.65 pu P = Vpu × I pu × cos ϕ ⇒
0.8 = 1 × Ipu × 0.8 Ipu = 1 pu E = V + j Ipu XT = 1 + 1 ∠ −36.86 × j 0.65 = 1.484 ∠20.51° pu
Therefore, power angle of generator = 20.51°. 32. In a 132 kV system, the series inductance up to the point of circuit breaker location is 50 mH. The shunt capacitance at the circuit breaker terminal is 0.05 μF. The critical value of resistance in ohms required to be connected across the circuit breaker contacts which will give no transient oscillation is . (2 Marks) Topic: Circuit Breakers, System Stability Concepts, Equal Area Criterion (500) The value of inductance connected in series L = 50 × 10−3 H
6/24/2019 10:37:10 AM
A10
GATE EE Chapter-wise Solved papers
Value of capacitance connected in shunt
In G(s) plane, the Nyquist plot of G(s) passes through the negative real axis at the point (a) (−0.5, j0) (b) (−0.75, j0) (c) (−1.25, j0) (d) (−1.5, j0)
C = 0.05 × 10 F Critical value of resistance −6
R=
1 L 1 50 × 10 −3 = = 500 Ω 2 C 2 0.05 × 10 −6
(1 Mark) Topic: Transfer Function
π e −0.25 s s For the Nyquist plot (a) G ( s) =
Chapter 7: Control Systems 33. A system transfer function is a s 2 + b s + c1 H ( s) = 1 2 1 a2 s + b2 s + c2
∠G ( jω ) = −180° −180° = −90° − 0.25 ×
If a1 = b1 = 0, and all other coefficients are positive, the transfer function represents a (a) low-pass filter (b) high-pass filter (c) band-pass filter (d) notch filter
−90° = −025 ×
ω pc =
(1 Mark)
π
180 ω pc
π
4π −90 × π = = 2π −180 × 0.25 2
Magnitude at this frequency is
Topic: Transfer Function (a) H ( s) =
180 ω pc
a1 s 2 + b1 s + c1 a2 s 2 + b2 s + c2
G ( 2π ) =
or c1 H ( s) = a2 s 2 + b2 s + c2
π e −0.25 j ( 2π ) 1 = = 0.5 2 j 2π
At negative real axis the co-ordinate becomes (-0.5, j0).
{∵ a1 = b1 = 0}
Putting s = 0 we get
c H (0) = 1 = Constant c2 Hence as low frequency s → 0 ⇒ w → 0. Putting s = ∞ we get H (∞) = 0 Hence as high frequency s → 0 ⇒ w → ∞. So, the system passes low frequency and blocks high frequency. This represents a low-pass filter.
35. The characteristic equation of a linear time-invariant (LTI) system is given by Δ(s) = s4 + 3s3 + 3s2 + s + k = 0 The system BIBO stable if 12 9 8 (c) 0 < k < 9
(a) 0 < k
3 (d) k > 6 (1 Mark)
Topic: Routh-Hurwitz and Nyquist Criteria (c) Finding Routh’s array
Go in
Cut-off frequency
Frequency 34. The open-loop transfer function of a unity feedback system is given by G ( s) =
GATE_EE_2019.indd 10
π e −0.25 s s
S4
1
3
k
S3
3
1
0
S2
8 3
k
0
S1
⎛8 ⎞ 8 ⎜ 3 − 3k ⎟ 3 ⎝ ⎠
0
0
S0
k
0
0
6/24/2019 10:37:16 AM
appendix • Solved GATE (EE) 2019
For stability, all elements of first column should we positive. So ⎛8 ⎞ ⎜ 3 − 3k ⎟ ⎝ ⎠ > 0 and k > 0 8 3 Solving ⎛8 ⎞ ⎜ − 3k ⎟ ⎝3 ⎠ > 0 ⇒ 8 > 3k ⇒ 3k < 8 ⇒ k < 8 8 3 3 9 3 8 So we have (k > 0) and ⎛⎜ k < 8 ⎟⎞ . So 0 < k < 9 9⎠ ⎝ 36. The asymptotic Bode magnitude plot of a minimum phase transfer function G(s) is shown below. |G( jw)| (dB)
So in total the transfer function has 3 poles, hence at w = ∞. The net phase contributed by 3 poles is −3 x −270°C or . 2 Hence Statement I is false and Statement II is true. 37. The transfer function of a phase lead compensator is 1 ⎞ ⎛ 3⎜ s + 3T ⎟⎠ ⎝ given by D( s) = . 1⎞ ⎛ s + ⎜ T⎟ ⎝ ⎠ The frequency (in rad/sec), at which 3D( jω ) is maximum, is (a)
3 T2
(b)
1 3T 2
(c)
3T
(d)
3T 2 (2 Marks)
60 —20 dB/decade Topic: Transfer Function
40 —40 dB/decade
20 0 1
10
20 w (log scale)
1 ⎞ 3(3sT + 1) ⎛ 3⎜ s + 3T ⎟⎠ ⎝ 3T (b) D( s) = = ( sT + 1) 1⎞ ⎛ ⎜s+ T ⎟ T ⎝ ⎠ ⎛ 1 + 3sT ⎞ =⎜ ⎟ ⎝ 1 + sT ⎠
—60 dB/decade Consider the following two statements Statement I: Transfer function G(s) has three poles and one zero.
Frequency at which ∠D( jω ) is maximum
Statement II: At very high frequency (ω → ∞), the 3π phase angle ∠G ( jω ) = − . 2 Which one of the following option is correct? (b) Statement I is false and Statement II is true. (c) Both the statements are true. (d) Both the statements are false. (a) Statement I is true and Statement II is false.
(2 Marks) Topic: Bode Plots, Root Loci, Stability Analysis (b) From the given bode-plot, we can say that At origin, there is a pole at origin, since the initial slope is -20 db/dec. At w = 1, the change in slope is -40 - (-20) = -20 db/sec, so it implies one pole at w = 1. At w = 20, the change in slope is -60 - (-40) = -20 db/dec, so it implies one pole at w = 20.
GATE_EE_2019.indd 11
A11
ωmax = =
1 T α 1 T 3
1 ⎧ ⎫ = 3⎬ ⎨∵ α = 1/ 3 ⎩ ⎭
=
1 3T 2
38. Consider a state-variable model of a system ⎡ x 1 ⎤ ⎡ 0 ⎢ x ⎥ = ⎢ ⎣ 2 ⎦ ⎣ −α
1 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ r ⎢ ⎥+ −2 β ⎥⎦ ⎣ x2 ⎦ ⎢⎣α ⎥⎦
⎡x ⎤ y = [1 0 ] ⎢ 1 ⎥ ⎣ x2 ⎦ where y is the output and r is the input. The damping ratio ξ and the undamped natural frequency ωn (rad/sec) of the system are given by (a) ξ = (c) ξ =
β α
; ωn = α
α ; ωn = β β
(b) ξ = α ; ωn =
β α
(d) ξ = β ; ωn = α (2 Marks)
6/24/2019 10:37:24 AM
A12
GATE EE Chapter-wise Solved papers
Topic: State Space Model, State Transition Matrix (a) From the given state space model, we can say that 1 ⎤ ⎡0⎤ ⎡ 0 A= ⎢ , B = ⎢ ⎥ , C = [1 0 ] , D = ? ⎥ ⎣ −α −2 β ⎦ ⎣α ⎦ In order to calculate ξ, wn, we need the transfer function of the system, which is given by T ( s) = C ( sI − A) −1 B ⎡ ⎛ s 0⎞ = [1 0] ⎢⎜ ⎟ ⎣⎝ 0 s ⎠
⎛ 0 −⎜ ⎝ −α
t ⎤ − V ⎡ −V i(t ) = ⎢ m sin(θ − ϕ ) ⎥ e τ + m sin(ωt − ϕ ) Z ⎣ Z ⎦
⎡ −V ⎤ −t m =⎢ sin(θ − ϕ ) ⎥ e τ ⎢⎣ R 2 + X L2 ⎥⎦ Vm sin(ωt − ϕ ) + 2 R + X L2
−1
1 ⎞ ⎤ ⎡0⎤ ⎥ −2β ⎟⎠ ⎦ ⎢⎣α ⎥⎦
⎛ ωL ⎞ and ϕ = tan −1 ⎜ ⎟ ⎝ R ⎠ For maximum DC offset current,
−1
−1 ⎤ ⎡ 0 ⎤ ⎡s = [1 0 ] ⎢ ⎥ ⎢ ⎥ ⎣α s + 2 β ⎦ ⎣α ⎦ ⎡ ⎡ s + 2β 1 = [1 0 ] ⎢ ⎢ ⎣ s( s + 2 β ) + α ⎣ −α
sin(q - f) = ±1
q - f = ±90° ωL ⎞ θ = ± 90° + tan −1 ⎛⎜ ⎟ = ± 90° + 45° ⎝ R ⎠ = −45° or 135° Alternately, we can solve this using power electronics concept. DC offset maximum at θ = ±90° + (pf angle)
1⎤ ⎤ ⎡ 0 ⎤ ⎥ s ⎥⎦ ⎦ ⎢⎣α ⎥⎦
⎡α ⎤ 0] ⎢ ⎥ 1 ⎡ ⎤⎡α ⎤ ⎣α s ⎦ = 2 = [1 0 ] ⎢ 2 ⎢ ⎥ ⎥ ⎣ s + 2 β s + α ⎦ ⎣α s ⎦ s + 2 β s + α α = 2 s + 2β s + α
⇒
[1
Comparing with the standard second order control system ωn 2 s + 2 ξωn s + ωn2 We get ωn = α ; 2 ξ α = 2 β ⇒ ξ =
β α
Chapter 8: Electrical and Electronic Measurements 39. In the circuit shown below, the switch is closed at t = 0. The value of θ in degrees which will give the maximum value of DC offset of the current at the time of switching is R = 3.77 Ω L = 10 mH
t=0
v(t) = 150 sin (377t + q)
(a) 60 (c) 90
= ±90° + 45° = -45° or 135° According to the options given we have θ = -45°
40. Five alternators each rated 5 MVA, 13.2 kV with 25% of reactance on its own base are connected in parallel to a busbar. The short-circuit level in MVA at the busbar is . (1 Mark) Topic: Measurement of Voltage, Current, Power, Energy and Power Factor (100) Reactance of alternator is 25% or 0.25 pu Equivalent reactance of five alternators in parallel 0.25 = = 0.05 pu 5 1 Short-circuit current = I SC = = 20 pu 0.05 Short-circuit MVA = 20 × 5 = 100 MVA
0.25
(b) −45 (d) −30
A1
A2
0.25 A3
0.25 A4
0.25
0.25
A5
(1 Mark) Topic: Measurement of Voltage, Current, Power, Energy and Power Factor (b) When the switch is closed at t = 0, the expression for i(t) is given as
GATE_EE_2019.indd 12
41. The line currents of a three-phase four wire system are square waves with amplitude of 100 A. These three currents are phase shifted by 120° with respect to each other. The rms value of neutral current is
6/24/2019 10:37:30 AM
appendix • Solved GATE (EE) 2019
(a) 0 A
(b)
100
= -170 sin(377t - 30°)
A
= 170 cos(377t + 60°)
3 (d) 300 A
(c) 100 A
(2 Marks) Topic: Measurement of Voltage, Current, Power, Energy and Power Factor (c) IN = Ia + Ic + Ib I N rms = =
1 π /3
π /3
1 π /3
π /3
∫
dθ
=
0
= 100 A
100 p
8 2
cos (30°)
43. A moving coil instrument having a resistance of 10 Ω, gives a full-scale deflection when the current is 10 mA. What should be the value of the series resistance, so that it can be used as a voltmeter for measuring potential difference up to 100 V? (a) 9 Ω (b) 99 Ω (c) 990 Ω (d) 9990 Ω
Ia
2p 3
2
×
= 588.89 W
IN rms = I = 100 A
p 3
−170
= 680 cos(30°)
Hence,
—100 Ic 100
To calculate the phase difference between v(t) and i(t), both of them should be either +ve cos or +ve sin. So ⎛ −170 ⎞ ⎛ 8 ⎞ × =⎜ × cos(60° − 30°) ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
0
2
π⎞ ⎛ i(t ) = 8 cos ⎜ 377t + ⎟ A 6⎠ ⎝ = 8 cos(377t + 30°)
Pavg = Vrms I rms cos(θ v − θi )
I N2 dθ
∫ 100
A13
4p 3
5p 3
2p
(2 Marks) Topic: Measurement of Voltage, Current, Power, Energy and Power Factor (d) Voltage across instrument is
100 Ib
Vm = Current through meter × Resistance of meter = 10 × 10−3 × 10 = 0.1 V
IN
Series resistance required to extend the range of instrument.
42. The voltage across and the current through a load are expressed as follows π⎞ ⎛ v(t ) = −170 sin ⎜ 377t − ⎟V 6⎠ ⎝
π⎞ ⎛ i(t ) = 8 cos ⎜ 377t + ⎟ A 6⎠ ⎝ The average power in watts (round off one decimal place) consumed by the load is (2 Marks) Topic: Measurement of Voltage, Current, Power, Energy and Power Factor π⎞ ⎛ (588.89) Given v(t ) = −170 sin ⎜ 377t − ⎟ V 6⎠ ⎝
GATE_EE_2019.indd 13
⎡V ⎤ Rse = Rm ⎢ − 1⎥ V ⎣ m ⎦ ⎡100 ⎤ = 10 ⎢ − 1⎥ ⎣ 0.1 ⎦ = 9990 Ω
Chapter 9: Analog and Digital Electronics 44. Given, Vgs is the gate-source voltage, Vds is the drain source voltage, and Vth is the threshold voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in saturation are (a) Vgs < Vth; Vds ≥ Vgs - Vth (b) Vgs > Vth; Vds ≥ Vgs - Vth
6/24/2019 10:37:35 AM
A14
GATE EE Chapter-wise Solved papers
(c) Vgs > Vth; Vds ≤ Vgs - Vth
Vx − Vout Vx − 0.01 + =0 100 10
(d) Vgs < Vth; Vds ≤ Vgs - Vth (1 Mark)
Vx - Vout + 10 Vx - 0.1 = 0
Topic: Characteristics of Diodes, BJT, MOSFET (b) For NMOS transistor to be in saturation Vgs should be greater than Vth, that is, Vgs > Vth and Vds ≥ Vgs - Vth. Note: Cutoff: Vgs < Vth
⎛ 1 ⎞ Vout = 11 Vx - 0.1 = 11⎜ ⎟ - 1 ⎝ 22 ⎠ = 0.4 V = 400 mV I2
Linear: Vgs > Vth, Vds < Vgs - Vth
100 kΩ
I1 10 kΩ
Saturation: Vgs > Vth, Vds ≥ Vgs - Vth
V1 10 mV
45. A current controlled current source (CCCS) has an input impedance of 10 Ω and output impedance of 100 kΩ. When this CCCS is used in a negative feedback closed loop with a loop gain of 9, the closed loop output impedance is (a) 10 Ω (b) 100 Ω (c) 100 kΩ (d) 1000 kΩ (1 Mark) Topic: Simple Active Filters, VCOs and Timers (d) Output impedance = 100 kΩ Loop gain = Aβ = 9
— Vx Vout Vy
V2 50 mV
+
10 kΩ 100 kΩ
47. The output expression for the Karnaugh map shown below is PQ RS
00
01
11
10
00
0
1
1
0
01
1
1
1
1
= 100 kΩ (1 + 9)
11
1
1
1
1
= 1000 kΩ
10
0
0
0
0
Closed loop output impedance Z0F = Z0(1 + Aβ)
46. In the circuit below, the operational amplifier is ideal. If V1 = 10 mV and V2 = 50 mV, the output voltage (Vout) is
(b) QR + S
(a) QR + S
(d) QR + S
(c) QR + S
100 kΩ
(2 Marks) Topic: Combinational and Sequential Logic Circuits (a) The output is
10 kΩ v1
—
v2
+
RQ + S or QR + S
Vout
10 kΩ RQ
100 kΩ PQ 00
01
11
10
00
0
1
1
0
01
1
1
1
1
RS (a) 100 mV (c) 500 mV
(b) 400 mV (d) 600 mV
S (2 Marks)
Topic: Operational Amplifiers: Characteristics and Applications 100 1 (b) Vx = × 0.05 = V 100 + 10 22
GATE_EE_2019.indd 14
11
1
1
1
1
10
0
0
0
0
48. In the circuit shown below, X and Y are digital inputs, and Z is a digital output. The equivalent circuit is a
6/24/2019 10:37:39 AM
appendix • Solved GATE (EE) 2019
A15
Topic: Single Phase and Three Phase Inverters (112.88) Given unipolar PWM and one pulse per half cycle. This means it is a single pulse modulation.
X Y Z
V0 =
⎡ 4Vs
∞
∑ ⎢⎣ nπ sin
n =1, 3
nπ ⎤ sin nd ⎥ sin n ωt 2 ⎦
RMS value of fundamental component (a) NAND gate (c) XOR gate
(b) NOR gate (d) XNOR gate (2 Marks)
π ⎫ 1 ⎧ 4V V0 = ⎨ s × sin × sin d ⎬ 2 π ⎩ ⎭ 2 For 75% of DC voltage
Topic: Combinational and Sequential Logic Circuits (c) X
XY
X
⇒
sin d =
Y Z = XY + XY
Y X
4Vs 1 × 1× sin d × π 2
0.75 Vs =
⇒
0.75 × π × 2 = 0.833 4
d = 56.408°
Pulse width = 2 × d = 2 × 56.408 = 112.88
XY
Y So, Z = XY + XY . This is the expression for XOR gate.
Chapter 10: Power Electronics
51. The enhancement type MOSFET in the circuit below operates according to the square law. μnCox = 100 μA/V2, the threshold voltage (VT) is 500 mV. Ignore channel length modulation. The output voltage Vout is
49. A six-pulse thyristor bridge rectifier is connected to a balanced three-phase, 50 Hz AC source. Assuming that the DC output current of the rectifier is constant, the lowest harmonic component in the AC input current is (a) 100 Hz (b) 150 Hz (c) 250 Hz (d) 300 Hz
VDD = 2 V
5 µA Vout
(1 Mark) Topic: Issues of Line Current Harmonics (c) Six-pulse thyristor bridge rectifier is connected to balanced three-phase supply, with constant current. The input supply current will have quasi-square wave. Thus, the current will not have the harmonics which is multiple of 3, i.e., 3, 9, ..., etc. Hence, the lowest harmonic component is the 5th harmonic. The lowest harmonic frequency = 5 × 50 = 250 Hz. 50. The output voltage of a single-phase full bridge voltage source inverter is controlled by unipolar PWM with one pulse per half cycle. For the fundamental rms component of output voltage to be 75% of DC voltage, the required pulse width in degrees (round off up to one decimal place) is . (1 Mark)
GATE_EE_2019.indd 15
W = L
(a) 100 mV (c) 600 mV
(b) 500 mV (d) 2 V (2 Marks)
Topic: Characteristics of Semiconductor Power Devices: Diode, Thyristor, Triac, GTO, MOSFET, IGBT (c) For the MOSFET in saturation region Given
ID =
1 ⎛W μnCox ⎜ 2 ⎝L
⎞ 2 ⎟ (VGS − VT ) (1) ⎠
ID = 5 × 10−6 A μnCox = 100 × 10−6
6/24/2019 10:37:41 AM
A16
GATE EE Chapter-wise Solved papers
W 10 μm = = 10 L 1 μm
⇒
VT = 500 × 10−3 V = 0.5 V Substituting the values in Eq. (1) we get 1 5 × 10 −6 = × 100 × 10 −6 × 10(VGS − 0.5) 2 2 ⇒ VGS = 0.6 V 52. A DC-DC buck converter operates in continuous conduction mode. It has 48 V input voltage, and it feeds a resistive load of 24 Ω. The switching frequency of the converter is 250 Hz. If switch-on duration is 1 ms, the load power is (a) 6 W (b) 12 W (c) 24 W (d) 48 W (2 Marks) Topic: D C to DC Conversion: Buck, Boost and BuckBoost Converters (c) Given R = 24 Ω, Vin = 48 V, f = 250 Hz, Ton = 1 ms. Time period T = Duty cycle =
1 1 = = 4 ms f 250
Ton 1 = T 4
As load is resistive and it does not say that the current is ripple-free, hence, the load power can be written as Pout
(V (load ) =
0 ( rms )
=
)
2
R
( =
0.25 × 48) 24
2
53. In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The convertor operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in μH) is . (2 Marks) Topic: D C to DC Conversion: Buck, Boost and BuckBoost Converters (24) Given P = 120 W, Vin = 24 V, Vout = 48 V. Vin 1− D
D = 0.5 P = Vout × Iout
⇒
120 = 48 × Iout
⇒
120 = 2.54 A 48 Vin × Iin = Vout × Iout I out =
⇒
I in =
120 =5A 24
The inductor operates at the boundary between continuous and discontinuous conduction modes. ΔI I in = I L = L 2 {IL = inductor current} ΔI L = L=
DVin = 2 × 5 = 10 fL 0.5 × 24 = 24 μH 50 × 103 × 10
54. A fully-controlled three-phase bridge converter is working from a 415 V, 50 Hz, AC supply. It is supplying constant current of 100 A at 400 V to a DC load. Assume large inductive smoothing and neglect overlap. The rms value of the AC line current in amperes (round off to two decimal places) is . (2 Marks)
2
24 × 24 (0.5 × 48) = = 24 W 24 24
Vout =
Vin 24 = Vout 48
⇒
VGS = Vout = 600 mV.
So
GATE_EE_2019.indd 16
1− D =
Topic: Bidirectional AC to DC Voltage Source Converters (81.65) In the three-phase bridge converter, the rms value of AC current is given as
I rms = I
2 2 = 100 3 3
= 81.649 A ≈ 81.65 A 55. A single-phase fully-controlled thyristor converter is used to obtain an average voltage of 180 V with 10 A constant current to feed a DC load. It is fed from single-phase AC supply of 230 V, 50 Hz. Neglect the source impedance. The power factor (round off to two decimal places) of AC mains is . (2 Marks) Topic: Line Commutated Thyristor Based Converters (0.782) Output voltage Vout = 180 V Supply voltage VS = 230 V
6/24/2019 10:37:45 AM
appendix • Solved GATE (EE) 2019
Maximum supply voltage
PF =
Vmax = 230 × 2 = 325.26 V Vout =
2 Vm cos ϕ π
⇒
2 × 325.26 180 = cos α π
⇒
cos ϕ = 0.869
GATE_EE_2019.indd 17
where Thus
i
A17
IS1 (cos ϕ 1) I0
⎛ 1.414 ⎞ IS1 = ⎜ 2× ⎟ I0 3.14 ⎠ ⎝ ⎛ 1.414 ⎞ PF = 2 × ⎜ × 0.869 ⎝ 3.14 ⎟⎠ = 0.782
6/24/2019 10:37:47 AM
GATE_EE_2019.indd 18
6/24/2019 10:37:47 AM
aPPENDIX
Solved GATE (EE) 2020 QUESTIONS
Chapter 1: Engineering Mathematics 1.
ax3 + bx2 + cx + d is a polynomial on real x over real coefficients a, b, c, d where in a ≠ 0. Which of the following statements is true? (a) d can be chosen to ensure that x = 0 is a root for any given set a, b, c. (b) No choice of coefficients can make all roots identical. (c) a, b, c, d can be chosen to ensure that all roots are complex. (d) c alone cannot ensure that all roots are real.
Topic: Calculus (d) We have lim
α →∞
sin pθ sin qθ dθ
α
1 ⎡ sin( p − q)θ sin( p + q)θ ⎤ = lim − α →∞ 2α ⎢ ( p + q) ⎥⎦ 0 ⎣ ( p − q) = lim
α →∞
1 ⎡ sin( p − q)α sin( p + q)α ⎤ − ⎥=0 2α ⎢⎣ ( p − q) ( p + q) ⎦
When α → ∞,
(a) Let F ( x ) = ax + bx + cx + d 2
F ( x ) = ax 3 + bx 2 + cx + 0 = 0, a ≠ 0
α
−α
α 1 × 2 ∫ sin pθ sin qθ dθ 0 α →∞ 2α 1 [cos( p − q)θ − cos( p + q)θ ]dθ = lim α → ∞ 2α
Topic: Numerical Methods Now, for x = 0 to be one of the root of polynomial, d can be chosen as zero. Therefore,
∫
= lim
(1 Mark) 3
1 2α
3.
sin( p − q)α sin( p + q)α and →0 2α ( p − q) 2α ( p + q)
The value of the following complex integral, with C representing the unit circle centered at origin in the counterclockwise sense, is: z2 +1 ∫ 2 dz C z − 2z
3 2 ⇒ ax + bx + cx = 0
⇒ x( ax 2 + bx + c) = 0 Hence, x = 0 is the root for any values of a, b and c only when d = 0. 2.
Which of the following is true for all possible non-zero choices of integers m, n; m ≠ n, or all possible non-zero choices of real numbers p, q; p ≠ q, as applicable? (a)
1 π
(b)
1 2π
(c)
1 2π
∫
α →∞
Topic: Complex Variables (c) Given that, I = ∫ C
π
sin mθ sin nθ dθ = 0
0
(d) lim
(a) 8πi (b) −8πi (c) −πi (d) πi (1 Mark)
∫
π /2
∫
π
−π / 2
−π
Now, the poles of f ( z ) =
sin pθ sin qθ dθ = 0
∫
α
−α
Since z = 2 lies outside, the unit circle, therefore residue of f(z) at z = 0 is Residue at ( z = 0) = lim z ⋅ z →0
sin pθ sin qθ dθ = 0 (1 Mark)
Solved Chapter Wise GATE EE 2020 Paper.indd 1
z2 +1 are given by z2 − 2z
z 2 − 2 z = 0 ⇒ z ( z − 2) = 0 ⇒ z = 0, 2
sin pθ sin qθ dθ = 0
1 2α
z2 dz , z = 1 z2 − 2z
= lim z →0
z2 +1 ( z 2 − 2z)
z 2 + 1 −1 = ( z − 2) 2
8/10/2020 10:07:36 AM
B2
GATE EE Chapter-wise Solved papers
We know that, I =∫ C
z +1 dz = 2π i (Sum of residues) z2 − 2z
(b) 7 and 1. (d) 1 and 1/4. (2 Marks)
Topic: Calculus (a) Given that, y = 3 x 2 + 3 x + 1
⎛ −1 ⎞ = 2π i ⎜ ⎟ = −π i ⎝ 2 ⎠ 4.
(a) 7 and 1/4. (c) −2 and −1/2.
2
Consider the initial value problem below. The value of y at x = ln 2, (rounded off to 3 decimal places) is . dy = 2 x − y, dx
So,
dy d2 y dy = 6 x + 3, = 6 and =0 dx dx dx 2
For maximum and minimum value,
y ( 0) = 1
6x + 3 = 0 (1 Mark)
⇒x=−
Topic: D ifferential Equations (0.886) The given equation is
Hence, x =
dy = 2 x − y , y ( 0) = 1 dx ⇒
−1 is a point of minima. Therefore, 2 2
y
dy + y = 2x dx
x =−
This is a linear differential equation of first order.
Now, the solution of the equation is given by
Now,
⇒ ye = ∫ 2 xe dx ⇒ ye x = 2[ xe x − ∫ e x dx ] + C ⇒ ye x = 2[ xe x − e x ] + C ⇒ ye x = 2e x ( x − 1) + C ⇒ y = 2( x − 1) + Ce − x Given, y(0) = 1, therefore 1 = 2(0 − 1) + C ⇒ C = 3
= 12 − 6 + 1 = 7 y x = 0 = 3(0) + 3(0) + 1 = 1 Hence, maximum and minimum values of y are 7 and 1/4 respectively.
x
⇒ ye x = 2 ∫ xe x dx
y x =−2 = 3( −2) 2 + 3( −2) + 1
y(IF) = ∫ Q(IF)dx x
1 2
⎛ 1⎞ ⎛ 1⎞ = 3⎜ − ⎟ + 3⎜ − ⎟ +1 2 ⎝ ⎠ ⎝ 2⎠ 1 3 3 = − +1 = 4 2 4
Since x ∈ [−2, 0], hence absolute maximum and minimum values need to consider.
dy + Py = Qx dx Pdx 1dx So, integrating factor, IF = e ∫ = e ∫ = e x
1 2
6.
The number of purely real elements in a lower triangular representation of the given 3 × 3 matrix, obtained through the given decomposition is . ⎡ 2 3 3 ⎤ ⎡ a11 ⎢3 2 1⎥ = ⎢a ⎢ ⎥ ⎢ 12 ⎢⎣ 3 1 7 ⎥⎦ ⎢⎣ a13 (a) 5 (c) 8
0 a22 a23
0 ⎤ ⎡ a11 0 ⎥⎥ ⎢⎢ a12 a33 ⎥⎦ ⎢⎣ a13
0 a22 a23
= 2(ln 2 − 1) + 3e − ln 2 3 = 2(ln 2 − 1) + 2 = 0.8862 5.
For real numbers, x and y, with y = 3x2 + x + 1, the maximum and minimum value of y for x ∈ [−2, 0] are respectively, .
(2 Marks) Topic: Linear Algebra (c) Given that, ⎡2 ⎢3 ⎢ ⎢⎣ 3 ⎡2 ⇒ ⎢⎢ 3 ⎢⎣ 3
3 3 ⎤ ⎡ a11 0 0 ⎤ ⎡ a11 0 2 1 ⎥⎥ = ⎢⎢ a12 a22 0 ⎥⎥ ⎢⎢ a12 a22 1 7 ⎥⎦ ⎢⎣ a13 a23 a33 ⎥⎦ ⎢⎣ a13 a23 3 3 ⎤ ⎡ a11 0 0 ⎤ ⎡ a11 a12 ⎢ ⎥ 2 1 ⎥ = ⎢ a12 a22 0 ⎥⎥ ⎢⎢ 0 a22 0 1 7 ⎥⎦ ⎢⎣ a13 a23 a33 ⎥⎦ ⎢⎣ 0
2 ⎡ 2 3 3 ⎤ ⎡ a11 ⎢ ⇒ ⎢⎢ 3 2 1 ⎥⎥ = ⎢ a12 a11 ⎢⎣ 3 1 7 ⎥⎦ ⎢⎣ a13 a11
Solved Chapter Wise GATE EE 2020 Paper.indd 2
T
(b) 6 (d) 9
Therefore, y = 2( x − 1) + 3e − x For x = ln 2, we get y = 2(ln 2 − 1) + 3e − ln 2
0⎤ 0 ⎥⎥ a33 ⎥⎦
a11a12 2 a122 + a22 a13 a12 + a23 a22
0⎤ 0 ⎥⎥ a33 ⎥⎦
T
a13 ⎤ a23 ⎥⎥ a33 ⎥⎦
⎤ a11a13 ⎥ a12 a13 + a22 a23 ⎥ 2 2 ⎥ a132 + a23 + a33 ⎦ 8/10/2020 10:07:38 AM
⎡2 ⎢3 ⎢ ⎢⎣ 3 ⎡2 ⇒ ⎢⎢ 3 ⎢⎣ 3
3 3 ⎤ ⎡ a11 0 0 ⎤ ⎡ a11 0 ⎥ ⎢ 2 1 ⎥ = ⎢ a12 a22 0 ⎥⎥ ⎢⎢ a12 a22 1 7 ⎥⎦ ⎢⎣ a13 a23 a33 ⎥⎦ ⎢⎣ a13 a23 3 3 ⎤ ⎡ a11 0 0 ⎤ ⎡ a11 a12 2 1 ⎥⎥ = ⎢⎢ a12 a22 0 ⎥⎥ ⎢⎢ 0 a22 0 1 7 ⎥⎦ ⎢⎣ a13 a23 a33 ⎥⎦ ⎢⎣ 0
2 ⎡ 2 3 3 ⎤ ⎡ a11 ⎢ ⎢ ⎥ ⇒ ⎢ 3 2 1 ⎥ = ⎢ a12 a11 ⎢⎣ 3 1 7 ⎥⎦ ⎢⎣ a13 a11
So,
0⎤ 0 ⎥⎥ a33 ⎥⎦
T
a13 ⎤ a23 ⎥⎥ a33 ⎥⎦
⎤ a11a13 ⎥ a12 a13 + a22 a23 ⎥ 2 2 ⎥ a132 + a23 + a33 ⎦
a11a12 2 a122 + a22 a13 a12 + a23 a22
a112 = 2 ⇒ a11 = ± 2 , a11a12 = 3 ⇒ a12 = ± a11a13 = 3 ⇒ a13 = ±
3
appendix • Solved GATE (EE) 2020
Topic: Calculus (-3) The given vector function is F = yaˆ x − xaˆ y We know that dl = ( dxaˆ x + dyaˆ y ) iˆ
So, the line integral of given vector function is ∫ F ⋅ dl = ∫ ( yaˆ x − xaˆ y ) ⋅ (dxaˆ x + dyaˆ y )
,
2 3
iˆ
C
,
2
C
= ∫ ( y dx − x dy ) C
Now, 2 a122 + a22 =2⇒
−5 ⇒ a22 = 2 a13 a12 + a23 a22 = 1 ⇒
Given curve C is y = x2 ⇒ dy = 2 x dx
9 2 + a22 =2 2
2 ⇒ a22 =
Therefore,
5 j, 2
2
−1
C
( x 2 dx − 2 x 2 dx ) 2
⎡ x3 ⎤ = ∫ − x dx = − ⎢ ⎥ −1 ⎣ 3 ⎦ −1 ⎡ 8 1 ⎤ −9 = −⎢ + ⎥ = = −3 ⎣3 3⎦ 3 2
7 9 j + a23 × a22 = 1 ⇒ a23 = 2 10 −11
j 10 Since a23 is imaginary in both the cases and a13 is real. 2 2 + a33 = 7, we get a33 = real So, from a132 + a23 ⇒
∫ F ⋅ dl = ∫
If we consider a12 and a13 different, then a23 =
8.
9 49 74 2 − + a33 = 7 ⇒ a33 = ± 2 10 10
9 121 2 371 − + a33 = 7 ⇒ a33 = ± 2 100 100 Hence, the number of purely real elements in a lower triangular representation of the given matrix is 8. Let ax and ay be unit vectors along x and y directions, respectively. A vector function is given by F = ax y − a y x The line integral of the above function
∫ F ⋅ dl
C
along the curve C, which follows the parabola y = x2 as shown in the following figure is (rounded off to 2 decimal places).
2
Chapter 2: Electric Circuits The Thevenin equivalent voltage, VTH, in V (rounded off to 2 decimal places) of the network shown below, is .
⇒
7.
B3
2Ω
3Ω +
4V + −
3Ω
5A
VTH − (1 Mark)
Topic: Thevenin’s Theorem, Norton’s Theorem, Superposition Theorem, Maximum Power Transfer Theorem (14.00) We have 4V
2 Ω VTH 3 Ω +
y 4V + −
4
3Ω
5A
VTH −
Applying KCL, we have
C 1 −1
2
x (2 Marks)
Solved Chapter Wise GATE EE 2020 Paper.indd 3
VTH − 4 =5 2 ⇒ VTH = 5 × 2 + 4 = 14 V
8/10/2020 10:07:39 AM
B4 9.
GATE EE Chapter-wise Solved papers
A resistor and a capacitor are connected in series to a 10 V DC supply through a switch. The switch is closed at t = 0, and the capacitor voltage is found to cross 0 V at t = 0.4τ , where τ is the circuit time constant. The absolute value of percentage change required in the initial capacitor voltage if the zero crossing has to happen at t = 0.2τ is (rounded off to 2 decimal places).
Topic: Biot-Savart’s Law, Ampere’s Law, Curl, Faraday’s Law (c) The given vector function is F = (5 y − k1 z )aˆ x + (3 z + k2 x )aˆ y ( k3 y − 4 x )aˆ z For a conservative field, ∇ × F = 0
(2 Marks)
∂ ∂ ∂ =0 ∂x ∂y ∂z (5 y − k1 z ) (3 z + k2 x ) ( k3 y − 4 x )
iˆ
aˆ x
Topic: Transient Response of Dc and Ac Networks (54.98) We have,
t=0
+ VC −
⇒ ( k3 − 3)aˆ x − ( k1 − 4)aˆ y + ( k2 − 5)aˆ z = 0 k1 − 4 = 0 ⇒ k1 = 4
VC (t ) = VC (∞) + [VC (0) − VC (∞)]e − t /τ (1)
Now, VC (t ) = 0 at t = 0.4τ in first case, therefore 0 = 10 + [VC1 − 10]e −0.4 where VC1 is the initial voltage of capacitor in the first case. VC1 = ( −10 + 10e −0.4 ) / e −0.4 = −4.918 V Again, VC (t ) = 0 at t = 0.2τ in second case, therefore 0 = 10 + [VC2 − 10]e −0.2 VC2 = ( −10 + 10e −0.2 ) /e −0.2 = −2.214 V where VC2 is the initial voltage of capacitor in the second case. Hence, absolute value of percentage change in initial 4.918 − 2.214 voltage = × 100 4.918 = 54.98%
Chapter 3: Electromagnetic Fields 10. The vector function expressed by F = aˆ x (5 y − k1 z ) + aˆ y (3 z + k2 x ) + aˆ z ( k3 y − 4 x ) represents a conservative field, where ax, ay, az are unit vectors along x, y and z directions, respectively. The values of constants k1, k2, k3 are given by: (a) k1 = 3, k2 = 3, k3 = 7 (b) k1 = 3, k2 = 8, k3 = 5 (c) k1 = 4, k2 = 5, k3 = 3 (d) k1 = 0, k2 = 0, k3 = 0 (2 Marks)
Solved Chapter Wise GATE EE 2020 Paper.indd 4
Therefore, k3 − 3 = 0 ⇒ k3 = 3
C
The capacitor voltage VC(t) is given by
aˆ x
⇒ aˆ x ( k3 − 3) − aˆ y ( −4 + k1 ) + aˆ z ( k2 − 5) = 0
R
10 V
aˆ y
k2 − 5 = 0 ⇒ k2 = 5
11. The static electric field inside a dielectric medium with relative permittivity, εr = 2.25, expressed in cylindrical coordinate system is given by the following expression ⎛3⎞ E = ar 2r + aφ ⎜ ⎟ + az 6 ⎝r⎠ where ar, aϕ, az are unit vectors along r, ϕ and z directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, ε0, in SI units is given by: (a) 3ε0 (b) 4ε0 (c) 5ε0 (d) 9ε0 (2 Marks) Topic: Coulomb’s Law, Electric Field Intensity, Electric Flux Density, Gauss’s Law (d) The electric field density of the medium is given by D =εE 3 ⎛ ⎞ = ε 0ε r ⎜ 2raˆr + aˆϕ + 6 aˆ z ⎟ r ⎝ ⎠ 3 ⎛ ⎞ = 2.25ε 0 ⎜ 2raˆr + aˆϕ + 6 aˆ z ⎟ r ⎝ ⎠ The volume charge density is given by ρV = ∇ ⋅ D =
1 ∂ 1 ∂Dϕ ∂Dz ( r Dr ) + + r ∂r r ∂ϕ ∂z
=
1 ∂ 1 ∂ ⎛ 6.75ε 0 ( 4.5 ε 0 r 2 ) + ⎜ r ∂r r ∂ϕ ⎝ r
⎞ ∂ ⎟ + z (13.5ε 0 ) ⎠ ∂
1 = ( 4.5 × 2 × ε 0 × r ) + 0 + 0 = 9ε 0 r
8/10/2020 10:07:42 AM
B5
appendix • Solved GATE (EE) 2020
12. Let aˆr , aˆϕ and aˆ z be unit vectors along r , ϕ and z directions, respectively in the cylindrical coordinate system. For the electric flux density given by D = ( aˆr 15 + aˆϕ 2r − aˆ z 3rz ) Coulomb/m2, the total electric flux, in Coulomb, emanating from the volume enclosed by a solid cylinder of radius 3 m and height 5 m oriented along the z-axis with its base at the origin is: (a) 54π (b) 90π (c) 108π (d) 180π iˆ
iˆ
Topic: Self and Mutual Inductance of Simple Configurations (138.63) Magnetic flux density due to infinite long wire is μ I B = 0 aˆϕ 2π Pr So, magnetic flux crossing square loop is ϕ = ∫∫ B ⋅ dS
μ0 I aˆϕ ( dr dz )aˆϕ 2π r μ I 2 dr 1 = 0 ∫ dz 2π r =1 r ∫z = 0 μ I μ I = 0 (ln r ) rr ==12 [ z ]1z = 0 = 0 (ln 2) 2π 2π
(2 Marks)
= ∫∫
Topic: C oulomb’s Law, Electric Field Intensity, Electric Flux Density, Gauss’s Law (d) Given that, D = 15aˆr + 2raˆϕ − 3rzaˆ z Current density is given by ρV = ∇ ⋅ D
Hence, the mutual inductance is
1 ∂ 1 ∂ ∂ ( rDr ) + ( Dϕ ) + ( Dz ) = r ∂r r ∂ϕ ∂z 1 ∂ 1 ∂ ∂ (15r ) + ( 2r ) + ( −3rz ) = ∂z r ∂r r ∂ϕ 15 15 = + 0 − 3r = − 3r r r Now, total electric flux crossing closed surface is ψ = ∫ D ⋅ dS = ∫∫∫ (∇ ⋅ D )dV S
V
⎛ 15 ⎞ = ∫∫∫ ⎜ − 3r ⎟ r dr dϕ dz r ⎝ ⎠ = 15∫
3
0
2π
∫ ∫
5
ϕ =0 z =0
dr dϕ dz − 3∫
3
0
2π
∫ ∫
5
ϕ =0 z =0
r 2 dr dϕ dz
3
⎡ r3 ⎤ = 15 × 3 × 2π × 5 − 3 × ⎢ ⎥ × 2π × 5 ⎣ 3 ⎦0 = 450π − 27 × 2π × 5 = 180 π C 13. A conducting square loop of side length 1 m is placed at a distance of 1 m from a long straight wire carrying a current I = 2 A as shown below. The mutual inductance, in nH (rounded off to 2 decimal places), between the conducting loop and the long wire is . z
ϕ μ0 ln 2 = I 2 × 2π 4π × 10 −7 × ln 2 = = 1.386 × 10 −7 H = 138.63 nH 2π
M=
Chapter 4: Signals and Systems 14. Which of the following statements is true about the two sided Laplace transform? (a) It exists for every signal that may or may not have a Fourier transform. (b) It has no poles for any bounded signal that is nonzero only inside a finite time interval. (c) The number of finite poles and finite zeroes must be equal. (d) If a signal can be expressed as a weighted sum of shifted one sided exponentials, then its Laplace Transform will have no poles. (1 Mark) Topic: Applications of Fourier Transform, Laplace Transform and z-Transform (b) Two sided Laplace transform has no poles for any bounded signal that is non-zero only inside a finite time interval. n
⎛1⎞ 15. Consider a signal x[n] = ⎜ ⎟ 1[n], where 1[n] = 0 if ⎝2⎠ n < 0. and 1[n] = 1 if n ≥ 0. The z-transform of x[n − k],
I = 2A
a=1m d=1m
k > 0 is a=1m (2 Marks)
Solved Chapter Wise GATE EE 2020 Paper.indd 5
z −k with region of convergence being 1 1 − z −1 2
8/10/2020 10:07:43 AM
B6
GATE EE Chapter-wise Solved papers
(a)
z >2 z < 2 (b)
(c)
z
2 2 (1 Mark)
Topic: Applications of Fourier Transform, Laplace Transform and z-Transform
(c)
y A ≠ kx A ; yR = kxR
(d) y A ≠ kx A ; yR ≠ kxR (1 Mark) Topic: Shifting and Scaling Operations (a), (b) The average value of x(t) is given by xA =
n
⎛1⎞ (d) Given that, x[n] = ⎜ ⎟ 1[n] ⎝2⎠ Since,
yA = k
1 T
∫ x(t )dt = kx
A
T
The rms value of x(t) is given by
It is right sided causal sequence 1[n] = u[n] unit step sequence
xR =
n
⎛1⎞ So, x[n] = ⎜ ⎟ u[n] ⎝2⎠ By z-transformation, we get X ( z) =
∫ x(t )dt
T
Average value of y(t) is given by
for n ≥ 0 for n < 0
⎧1 1[n] = ⎨ ⎩0
1 T
1 T
∫x
2
(t )dt
T
The rms value of y(t) is given by
1 1 ROC z > 1 −1 2 1− z 2
yR =
x( n − k ) = − z X ( z ) 1 = z −k . 1 1 − z −1 2
1 ROC z > 2
For every causal sequence the ROC is always exterior of the circle of radius R. ROC > | R | ROC
∫k
2
x 2 (t )dt
T
Case I: If k is real and positive, then
From property of z-transformation, we have −k
1 T
yR = kxR Case II: When k is imaginary or complex or real and negative, then yR ≠ kxR Hence, both the options (a) and (b) are correct in the given condition. 17. Suppose for input x(t) a linear time-invariant system with impulse response h(t) produces output y(t), so that x(t ) * h(t ) = y(t ), Further, if x(t ) * h(t ) = z (t ), which of the following statements is true? (a) For all t ∈ ( −∞, ∞), z (t ) ≤ y(t ) (b) For some but not all t ∈ ( −∞, ∞), z (t ) ≤ y(t )
1/2
(c) For all t ∈ ( −∞, ∞), z (t ) ≥ y(t ) (d) For some but not all t ∈ ( −∞, ∞), z (t ) ≥ y(t ) (2 Marks)
16. xR and xA are, respectively, the rms and average values of x(t) = x(t − T), and similarly, yR and yA are, respectively, the rms and average values of y(t) = kx(t). k, T are independent of t. Which of the following is true? (a)
y A = kx A ; yR = kxR
(b) y A = kx A ; yR ≠ kxR
Solved Chapter Wise GATE EE 2020 Paper.indd 6
Topic: Linear Time Invariant and Causal Systems (c) Given that x(t ) * h(t ) = y(t ) and z (t ) = x(t ) * h(t ) ∞
So, y(t ) = ∫ x(τ )h(t − τ )dτ −∞ ∞
and z (t ) = ∫ x(τ ) h(t − τ ) d −∞
8/10/2020 10:07:45 AM
appendix • Solved GATE (EE) 2020
B7
Topic: Linear Time Invariant and Causal Systems (c) The given difference equation of the system is
Case 1: Let h(t)
y[n] − ay[n − 1] = b0 x[n] − b1 x[n − 1] x(t)
Taking z-transform on both sides, we get
1
Y ( z ) − az −1Y ( z ) = b0 X ( z ) − b1 z −1 X ( z ) 0
1
and
⇒ H ( z) =
t 0
t
1
b0 b z −1 Y ( z) = − 1 −1 −1 X ( z ) (1 − az ) 1 − az
The system transfer function has one pole at z = a. So,
−1
If z > a
⇒ h( n) = 0, n < 0
Then, If z < a
z(t) y(t) 1
2
Therefore, the system is not necessarily causal. It is clear from the expression of h1(n) and h2(n), the impulse response of system is non-zero at infinitely many instants.
and 0
t
1
2
t
−1 Case 2: Let
Chapter 5: Electrical Machines
x(t)
h(t)
1
1
t and 0
h2 ( n) = −b0 a n u( −n − 1) + b1a n −1u( −n) ⇒ h( n) ≠ 0, n < 0
1 0
h1 ( n) = b0 a n u( n) − b1a n −1u( n − 1)
1
t 0
1
Then, y(t) = z(t) 1
0
1
2
t
Therefore, for all t ∈ ( −∞, ∞), z (t ) ≥ y(t ) 18. Which of the following options is true for a linear time-invariant discrete time system that obeys the difference equation: y[n] − ay[n − 1] = b0 x[n] − b1 x[n − 1] (a) y[n] is unaffected by the values of x[n - k]; k > 2. (b) The system is necessarily causal. (c) The system impulse response is non-zero at infinitely many instants. (d) When x[n] = 0, n < 0. the function y[n]; n > 0 is solely determined by the function x[n]. (2 Marks)
Solved Chapter Wise GATE EE 2020 Paper.indd 7
19. A three-phase cylindrical rotor synchronous generator has a synchronous reactance Xs and a negligible armature resistance. The magnitude of per phase terminal voltage is VA and the magnitude of per phase induced emf is EA. Considering the following two statements, P and Q. P: For any three-phase balanced leading load connected across the terminals of this synchronous generator, VA is always more than EA Q: For any three-phase balanced lagging load connected across the terminals of this synchronous generator, VA is always less than EA which of the following options is correct? (a) P is false and Q is true. (b) P is true and Q is false. (c) P is false and Q is false. (d) P is true and Q is true. (1 Mark) Topic: Synchronous Machines: Cylindrical and Salient Pole Machines, Performance, Regulation and Parallel Operation of Generators, Starting of Synchronous Motor, Characteristics (a) A cylindrical rotor synchronous generator (with ra = 0) has always positive voltage regulation for lagging power factor loads, that means, EA > VA. Whereas, for leading power factor loads, it has positive, zero and negative voltage regulation, that means, EA > VA, EA = VA and
8/10/2020 10:07:47 AM
B8
GATE EE Chapter-wise Solved papers
EA < VA are possible. Leading load can lead to negative or zero voltage regulation. Negative voltage regulation can be questionable. Hence, statement P is false and statement Q is true. 20. A three-phase. 50 Hz, 4-pole induction motor runs at no-load with a slip of 1%. With full load, the slip increases to 5%. The % speed regulation of the motor (rounded off to 2 decimal places) is .
Therefore, PC = Af + Bf 2 Case I: 450 = A(50) + B(50) 2 ⇒ A + 50 B = 9 (1) Case II: 320 = A( 40) + B( 40) 2 A + 40 B = 8 (2) From Eqs. (1) and (2), we get ( A + 50 B ) − ( A + 40 B) = 9 − 8 ⇒ 10 B = 1 1 ⇒B= 10
(1 Mark) Topic: Three Phase Induction Motors: Principle of Operation, Types, Performance, Torque-Speed Characteristics, No-Load and Blocked Rotor Tests, Equivalent Circuit, Starting and Speed Control (4.21) Given that, f = 50 Hz, P = 4, S = 1% = 0.01. Synchronous speed of motor is given by 120 f NS = P 120 × 50 = = 1500 rpm 4 Now, speed of motor at no-load, N0 = NS (1 - S) = 1500 (1 - 0.01) = 1485 rpm Speed of motor at full load when S = 5% = 0.05, N = 1500 (1 - 0.05) = 1425 rpm Therefore, the speed regulation of the motor is N0 − N × 100% N 1485 − 1425 = × 100% = 4.21% 1425
A = 8 − 40 ×
and
1 =4 10
Therefore, at 100 V and 25 Hz, PC = 4( 25) + = 100 + 62.50 = 162.50 W
22. A single 50 Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to be increased by 10 MW, as a result of which, system frequency dropped to 49.75 Hz. Further increase in load in the system resulted in a frequency of 49.25 Hz. At this condition, the power in MW supplied by the generator is (rounded off to 2 decimal places).
S.R. =
21. A single-phase, 4 kVA. 200 V/100 V, 50 Hz transformer with laminated CRGO steel core has rated no-load loss of 450 W. When the high-voltage winding is excited with 160 V, 40 Hz sinusoidal AC supply, the no-load losses are found to be 320 W. When the high-voltage winding of the same transformer is supplied from a 100 V, 25 Hz sinusoidal AC source, the no-load losses will be W (rounded off to 2 decimal places). (1 Mark) Topic: Single Phase Transformer: Equivalent Circuit, Phasor Diagram, Open Circuit and Short Circuit Tests, Regulation and Efficiency (162.50) Given that,
(1 Mark) Topic: Synchronous Machines: Cylindrical and Salient Pole Machines, Performance, Regulation and Parallel Operation of Generators, Starting of Synchronous Motor, Characteristics (130) Let the full load frequency is 50 Hz. Let the power supplied by the generator at frequency 49.25 Hz is P. Therefore, f(Hz)
50 θ
49.75 49.25
θ
V1 = 200 V,V2 =100 V, f = 50 Hz, PC = 450 W,V1′ = 160 V, f ′ = 40 Hz, PC′ = 320 W,100
Hz, PC = 450 W,V1′ = 160 V, f ′ = 40 Hz, PC′ = 320 W, V1′′ = 100 V and f ′′ = 25 Hz. We know that, V 200 160 100 = = = = 4 = Constant f 50 40 25
Solved Chapter Wise GATE EE 2020 Paper.indd 8
1 ( 25) 2 = 10
P (MW) 110
P
As the slope of Δf/ΔP is constant, therefore 50 − 49.75 49.75 − 49.25 = 100 − 110 110 − P 0.25 0.50 ⇒ = P − 110 10 ⇒ P − 110 = 20 ⇒ P = 20 + 110 = 130 MW 8/10/2020 10:07:48 AM
appendix • Solved GATE (EE) 2020
50 − 49.75 49.75 − 49.25 = 100 − 110 110 − P 0.25 0.50 ⇒ = P − 110 10 ⇒ P − 110 = 20 ⇒ P = 20 + 110 = 130 MW
⇒
(2 Marks) Topic: D C Machines: Separately Excited, Series and Shunt, Motoring and Generating Mode of Operation and their Characteristics, Starting and Speed Control of DC Motors (c) Given that, Rsh = 100 Ω, Ra = 0.2 Ω, Ia = 5 A, each brush has voltage drop 1 V and V = 250 V. Ish
1200 × ϕNL 247 = 238.5 N load × 0.95ϕNL [Since, ϕload = (1 − 0.5)ϕNL = 0.95 ϕNL ]
23. A 250 V DC shunt motor has an armature resistance of 0.2 Ω and a field resistance of 100 Ω. When the motor is operated on no-load at rated voltage, it draws an armature current of 5 A and runs at 1200 rpm. When a load is coupled to the motor, it draws total line current of 50 A at rated voltage, with a 5% reduction in the air-gap flux due to armature reaction. Voltage drop across the brushes can be taken as 1 V per brush under all operating conditions. The speed of the motor, in rpm, under this loaded condition, is closest to: (a) 1200 (b) 1000 (c) 1220 (d) 900
IL
1200 × 238.5 247 × 0.95 = 1219.68 1220 rpm
⇒ N Load =
24. Consider a permanent magnet DC (PMDC) motor which is initially at rest. At t = 0, a DC voltage of 5 V is applied to the motor. Its speed monotonically increases from 0 rad/s to 6.32 rad/s in 0.5 s and finally settles to 10 rad/s. Assuming that the armature inductance of the motor is negligible, the transfer function for the motor is (a) (c)
10 2 (b) 0.5s + 1 0.5s + 1 2 10 (d) s + 0.5 s + 0.5 (2 Marks)
Topic: DC Machines: Separately Excited, Series and Shunt, Motoring and Generating Mode of Operation and their Characteristics, Starting and Speed Control of DC Motors (b) At t = 0, a 5V DC voltage is applied to the motor, therefore V ( s) =
Ia
Rsh
Eb Ra
V
ω ( s) K = V ( s) 1 + sT
EbNL = V − I a Ra − brush voltage = 250 – 5 × 0.2 – 2 × 1 = 247 V At loaded condition, we have
250 = 2.5 A 100
I a = 50 − 2.5 = 47.5 A
Ebload = V − I a Ra − brust voltage = 250 – 47.5 × 0.2 – 2 × 1 = 238.5 V We know that Eb ∝ N ϕ So,
EbNL Eb load
Solved Chapter Wise GATE EE 2020 Paper.indd 9
5 [Since V(t) = 5u(t)] s
The speed of motor increases monotonically from 0 to 6.32 rad/s in 0.5 s, that means, mean time constant of motor is 0.5 s. Now, the transfer function of motor is
At no load, back emf is
I sh =
B9
N ×ϕ = NL NL N load × ϕload
⇒
ω ( s) K = (1) V ( s) 1 + 0.5s
The steady state value of speed of motor is 10 rad/s. So lim ω (t ) = lim sω ( s) t →∞
s →0
= lim s ⋅ s →0
⇒ 10 = lim s →0
5 K × (1 + 0.5s) s
5K (1 + 0.5s)
⇒ 5 K = 10 ⇒K =2 Hence, the transfer function for the motor is
ω ( s) 2 = V ( s) 0.5s + 1
8/10/2020 10:07:50 AM
B10
GATE EE Chapter-wise Solved papers
25. The figure below shows the per-phase Open Circuit Characteristics (measured in V) and Short Circuit Characteristics (measured in A) of a 14 kVA, 400 V, 50 Hz, 4-pole, 3-phase, delta connected alternator, driven at 1500 rpm. The field current, If is measured in A. Readings taken are marked as respective (x, y) coordinates in the figure. Ratio of the unsaturated and saturated synchronous impedances ( Z s ( unsat ) /Z s ( sat ) ) of the alternator is closest to: Voc
Isc
OCC (8, 400)
(2, 210)
and saturated impedance is
Zsat =
400 40 = (3) 20 2
From Eqs. (2) and (3), we get Z unsat 81/ 2 = = 2.025 Zsat 40 / 2 26. A cylindrical rotor synchronous generator with constant real power output and constant terminal voltage is supplying 100 A current to a 0.9 lagging power factor load. An ideal reactor is now connected in parallel with the load, as a result of which the total lagging reactive power requirement of the load is twice the previous value while the real power remains unchanged. The armature current is now A (rounded off to 2 decimal places). (2 Marks)
(1, 110) (4, 20) (0, 10)
Topic: Synchronous Machines: Cylindrical and Salient Pole Machines, Performance, Regulation and Parallel Operation of Generators, Starting of Synchronous Motor, Characteristics (125.35) Given that,
SCC
(0, 0)
If
(a) 2.100 (c) 2.000
(b) 2.025 (d) 1.000 (2 Marks)
Topic: Single Phase Transformer: Equivalent Circuit, Phasor Diagram, Open Circuit and Short Circuit Tests, Regulation and Efficiency (b) We have Voc
I a1 = 100 A, cos ϕ1 = 0.9 ⇒ ϕ1 = cos −1 (0.9) = 25.84°, Q2 = 2Q1 and P1 = P2. The real power of generator is constant. So, V1 I a1 cos ϕ1 = V2 I a2 cos ϕ2 Since, V1 = V2, terminal voltage remain constant. Then, I a1 cos ϕ1 = I a2 cos ϕ2
Isc (8, 810)
tan ϕ1 =
Q1 = tan( 25.84°) = 0.482 P1
and
tan ϕ2 =
2Q1 = 2 × 0.482 = 0.964 P1
(8, 400)
(2, 210) (1, 110) (0, 10)
Now,
⇒ cos ϕ2 = 0.718
(4, 20)
Therefore, I a1 cos ϕ2 = I a2 cos ϕ2
(0, 0) If
At 400 V, If = 8 A So, air gap line equation will be like y = mx + c
⇒ V = 100 I f + 10 (1)
At If = 8 A, unsaturated voltage Vunsat = 100 × 8 + 10 = 810 So, unsaturated impedance is
Z unsat =
Solved Chapter Wise GATE EE 2020 Paper.indd 10
810 81 = (2) 20 2
⇒ I a2 =
I a1 cos ϕ1 cos ϕ2
=
100 × 0.9 = 125.35 A 0.718
27. Windings ‘A’, ‘B’ and ‘C’ have 20 turns each and are wound on the same iron core as shown, along with winding ‘X’ which has 2 turns. The figure shows the sense (clockwise/anti-clockwise) of each of the windings only and does not relied the exact number of turns. If windings ‘A’, ‘B’ and ‘C’ are supplied with balanced 3-phase voltages at 50 Hz and there is no core saturation, the no-load RMS voltage (in V, rounded off to 2 decimal places) across winding ‘X’ is .
8/10/2020 10:07:52 AM
appendix • Solved GATE (EE) 2020
B11
(1.02) Given that, Vt = (1 + j 0) pu ,
230 ∠0° V +
E f = (1 + j 0.7) pu, X d = j 0.7 pu and X d′′ = j 0.2 pu
A X
230∠−120° V +
For a three-phase solid fault current is given by
B
If =
230 ∠120° V +
E f − Vt Xd
=
(1 + 0.7 j ) − (1 + 0 j ) j 0.7
= 1 pu
C
Now, the subtransient internal emf is given by E ′′f = Vt + jX d′′ I f = 1 + j 0.2 × 1 = 1 + j 0.2 V
(2 Marks) Topic: Three Phase Transformers: Connections, Parallel Operation (46) Consider the following figure: 230 ∠0° V
E ′′f = 12 + (0.2) 2 = 1.02 V
And
Chapter 6: Power Systems A 20
X
20
230∠−120° V B 20 230 ∠120° V
C
Voltage across winding X is given by 2 2 2 230∠0° + 230∠− 120°V + 230∠120° 20 20 20 = −23∠0° + 23∠− 120° + 23∠ ∠120°
29. A lossless transmission line with 0.2 pu reactance per phase uniformly distributed along the length of the line, connecting a generator bus to a load bus, is protected up to 80% of its length by a distance relay placed at the generator bus. The generator terminal voltage is 1 pu. There is no generation at the load bus. The threshold pu current for operation of the distance relay for a solid three phase-to-ground fault on the transmission line is closest to: (a) 1.00 (b) 3.61 (c) 5.00 (d) 6.25 (1 Mark)
VX = −
= 23∠180° + 23∠− 120° + 23∠120° = 23 cos ∠180° + j 23 sin 180° + 23 cos 120° − j 23 sin 120° + 23 cos 120° + j 23 sin 120°
Vt = 1 pu.
= 46∠180° V Hence, no-load RMS voltage across winding is 46 V. 28. A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and subtransient reactance of 0.2 pu. It is operating at (1 + j0) pu terminal voltage with an internal emf of (1 + j0.7) pu. Following a three-phase solid short circuit fault at the terminal of the generator, the magnitude of the subtransient internal emf (rounded off to 2 decimal places) is pu. (2 Marks) Topic: S ynchronous Machines: Cylindrical and Salient Pole Machines, Performance, Regulation and Parallel Operation of Generators, Starting of Synchronous Motor, Characteristics
Solved Chapter Wise GATE EE 2020 Paper.indd 11
Topic: Models and Performance of Transmission Lines and Cables (d) Given, 80% of the line is protected by distance relay, so reactance of the line as seen by the relay, x = 0.8 × 0.2 = 0.16 pu. The generator terminal voltage, Assume Vt = 1.0 pu on per phase basis. The solid three phase fault current is given by If =
Vt 1 = = 6.25 pu x 0.16
30. Out of the following options, the most relevant information needed to specify the real power (P) at the PV buses in a load flow analysis is (a) solution of economic load dispatch (b) rated power output of the generator (c) rated voltage of the generator (d) base power of the generator (1 Mark)
8/10/2020 10:07:53 AM
B12
GATE EE Chapter-wise Solved papers
Topic: G auss- Seidel and Newton-Raphson Load Flow Methods (a) The most relevant information needed to specify the real power (P) at the PV buses in a load flow analysis is solution of economic load dispatch.
20%. Real power flow through the line under both the conditions is zero. The new value of the voltage magnitude, V1 in pu (rounded off to 2 decimal places), at bus 1 is . V1
31. Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of which there are ideal transformers with turns ratios as shown. Bus admittance matrix for the system is: Vi
Vj Y
Bus i
⎡ −ti t jY (a) ⎢ 2 ⎢⎣ ti Y ⎡ ti t jY (b) ⎢ 2 ⎢⎣ −ti Y ⎡ ti2Y (c) ⎢ ⎢⎣ −ti t jY
Bus j
tj : 1
1 : ti
Bus2 (2 Marks)
Topic: Models and Performance of Transmission Lines and Cables (1.12) Given that, V1 = 1.1 pu, V2 = 1 pu. V1
V2 Q12
Bus1
−t 2j Y ⎤ ⎥ ti t jY ⎥⎦
Bus2
As real power flowing through line is zero, then P=
−ti t jY ⎤ ⎥ t 2j Y ⎥⎦ −(ti − t j ) 2 Y ⎤ ⎥ ti t j Y ⎥⎦ (2 Marks)
1 : ti
Now,
Q12 =
V1V2 V2 cos δ − 2 X X
Q12 =
1.1× 1 1 0.1 cos 0° − = (1) X X X
V1 = V1′. Then, tj : 1
Ii
V1′V2 V2 cos 0° − 2 X X ′ V 1 Q12′ = 1 − (2) X X Q12′ =
Y
Vi
V1V2 sin δ = 0 ⇒ δ = 0° X
Given, Q12 is increased by 20%, so let Q12′ = 1.2Q12 and
Topic: B us Admittance Matrix (c) Consider the following figure:
Vj
Ij
From Eqs. (1) and (2), we get
Total current is given by I = Y (tiVi − V j t j )
I i = ti I = ti2YVi − ti t jYV j (1)
I j = −t j I = − I i t jYVi + t 2j YV j
(2)
From Eqs. (1) and (2), we get ⎡ I i ⎤ ⎡ t y −ti t j y ⎤ ⎡Vi ⎤ ⎥⎢ ⎥ ⎢I ⎥ = ⎢ 2 ⎣ j ⎦ ⎢⎣ −ti t j y t j y ⎥⎦ ⎣V j ⎦ 2 i
32. Bus 1 with voltage magnitude V1 = 1.1 pu is sending reactive power Q12 towards bus 2 with voltage magnitude V2 = 1 pu through a lossless transmission line of reactance X. Keeping the voltage at bus 2 fixed at 1 pu, magnitude of voltage at bus 1 is changed, so that the reactive power Q12 sent from bus 1 is increased by
Solved Chapter Wise GATE EE 2020 Paper.indd 12
Q12
Bus1
t 2j Y ⎤ ⎥ −ti t jY ⎥⎦
⎡ ti t jY (d) ⎢ 2 ⎢⎣ −(ti − t j ) Y
V2
V1′ 1.12 = ⇒ V1′ = 1.12 pu X X
Chapter 7: Control Systems 33. Consider a linear time-invariant system whose input r(t) and output y(t) are related by the following differential equation: d 2 y (t ) + 4 y (t ) = 6 r (t ) dt 2 The poles of this system are at (a) +2j, −2j (b) +2, −2 (c) +4, −4 (d) +4j, −4j (1 Mark)
8/10/2020 10:07:55 AM
appendix • Solved GATE (EE) 2020
Topic: M athematical Modeling and Representation of Systems (a) The given differential equation of the system is
B13
Simplifying the signal flow graph, we get 1
d 1 − cd
a
d 2 y (t ) + 4 y (t ) = 6 r (t ) dt 2
1
e
Taking Laplace’s transformation, we get ad 1 − cd
s 2Y ( s) + 4Y ( s) = 6 R( s) ⇒ Y ( s)[ s + 4] = 6 R( s) Y ( s) 6 ⇒ = 2 R( s) ( s + 4) 2
e 35. Consider a negative unity feedback system with forK ward path transfer function G ( s) = , ( s + a)( s − b)( s + c)
So, the poles of the system are s2 + 4 = 0 ⇒ s = ± 2 j 34. Which of the options is an equivalent representation of the signal flow graph shown here? c 1
a
1 d e
(a)
(b)
1
a(d + c)
1
e (a + c)d
(d)
where K, a, b, c are positive real numbers. For a Nyquist path enclosing the entire imaginary axis and right half of the s-plane in the clockwise direction, the Nyquist plot of (1 + G(s)), encircles the origin of (1 + G(s))-plane once in the clockwise direction and never passes through this origin for a certain value of K. Then, the number G ( s) of poles of lying in the open right half of the 1+ G ( s) s-plane is
(1 Mark) Topic: Routh-Hurwitz and Nyquist Criteria (2) According to Nyquist Criteria, we have N = P−Z where N is number of encirclement of origin in 1 + G(s) H(s) plane in anticlockwise direction, P is number of poles in right hand side of open loop transfer function and Z is the number of poles in right hand side for closed loop transfer function. Given, number of encirclement in clockwise direction is one. So, N = −1 and from G(s) it is clear that it has one right sided pole, then P = 1. So, N = P−Z
1
1
d a 1−cd
1
1
e c a 1−cd
1
⇒ −1 = 1 − Z ⇒ Z = 2
e (1 Mark) Topic: B lock Diagrams and Signal Flow Graphs (c) Given signal flow graph is c 1
.
1
e (c)
1
1
a
36. Which of the following options is correct for the system shown below?
R(s)
+
−
1 s2
1 s+1
Y(s)
1 d e
Solved Chapter Wise GATE EE 2020 Paper.indd 13
20 s + 20
8/10/2020 10:07:56 AM
B14
GATE EE Chapter-wise Solved papers
(a) (b) (c) (d)
4th order and stable 3rd order and stable 4th order and unstable 3rd order and unstable
(a) 9 (c) 7
(2 Marks) (2 Marks)
+
−
1 s2
1 s+3
Topic: Routh-Hurwitz and Nyquist Criteria (b) The characteristic equation of the system is 1 + G ( s) H ( s) = 0
Topic: Routh-Hurwitz and Nyquist Criteria (c) Consider the following figure:
R(s)
(b) 8 (d) 6
s2 + s + 1 =0 s3 + 2 s 2 + 2 s + K ⇒ s3 + 2 s 2 + 2 s + K + s 2 + s + 1 = 0 ⇒ 1+
Y(s)
⇒ s3 + 3s 2 + 3s + ( K + 1) = 0 By using Routh-Hurwitz criteria, we have s3 s2
20 s + 20
s
R(s)
+
1 +1)
−
Y(s)
s2(s
s0
For roots to be on imaginary axis K +1− 9 =0 3 ⇒ K +1− 9 = 0 ⇒ K =8
20 s + 20 Transfer function is given by 1 2 Y ( s) s ( s + 1) = 1 20 R( s) ⋅ 1+ 2 + 20) s ( s ( s + 1) ( s + 20) = 2 s ( s + 1)( s + 20) + 20 ( s + 20) = 2 2 s ( s + 21s + 20) + 20 ( s + 20) = 4 s + 21s3 + 20 s 2 + 20 It is clear that the order of the system is four. The characteristics equation of the system is
38. The causal realization of a system transfer function H(s) having poles at (2, −1), (−2, 1) and zeroes at (2, 1), (−2, −1) will be (a) stable, real, allpass (b) unstable, complex, allpass (c) unstable, real, highpass (d) stable, complex, lowpass (2 Marks) Topic: Transfer Function (b) Given that, pole locations of the system = (2, −1) and (−2, 1); and zero locations of the system = (2, 1) and (−2, 1). Therefore, pole zero plot of given transfer function is jω
1 + G ( s) H ( s) = 0 s 4 + 21s3 + 20 s 2 + 20 = 0 Since the term of ‘s’ is missing, therefore the system is unstable. 37. Consider a negative unity feedback system with the fors2 + s + 1 , where K s + 2s 2 + 2s + K is a positive real number. The value of K for which the system will have some of its poles on the imaginary axis is .
ward path transfer function
Solved Chapter Wise GATE EE 2020 Paper.indd 14
3
1 3 3 ( K + 1) K +1− 9 3 K +1
+1 −2
2
σ
−1 Since given pole-zero plot is symmetrical about origin, hence it is an all pass system. The system is unstable because for stability of causal system all poles should lie in the left half of s-plane.
8/10/2020 10:07:57 AM
appendix • Solved GATE (EE) 2020
Now,
H ( s) =
[ s − ( 2 + j )][ s − ( −2 − j )] [ s − ( 2 − j )][ s − ( −2 + j )]
=
[ s − ( 2 + j )][ s + ( 2 + j )] [ s − ( 2 − j )][ s + ( 2 − j )]
=
s2 − (2 + j )2 s2 − (4 − 1 + 4 j ) = s2 − (2 − j )2 s2 − (4 − 1 − 4 j )
=
s −3− 4 j s2 − 3 + 4 j 2
B15
Let ω 2 = t, therefore t 2 − 201t + 9600 = 0 201 ± ( 201) 2 − 4 × 1× 9600 2 ⇒ t = 122.36, 7813 ⇒t =
Hence, ω 2 = 122.36, 78.13 ⇒ ω = 11.06, 8.84 rad/s
Hence, the transfer function of the system is complex.
Hence, the smallest positive frequency at unity gain is 8.84 rad/s.
39. A stable real linear time-invariant system with single pole at p, has a transfer function H ( s) =
s 2 + 100 with a s− p
DC gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to: (a) 8.84 (b) 11.08 (c) 78.13 (d) 122.87 (2 Marks)
Chapter 8: Electrical and Electronic Measurements 40. Currents through ammeters A2 and A3 in the figure are 1∠10° and 1∠70°, respectively. The reading of the ammeter A1 (rounded off to 3 decimal places) is A.
I2
Topic: Stability Analysis (a) The given transfer function of the system is H ( s) =
A2
I1
s 2 + 100 s− p
A1
Substituting s = jω, we get H ( jω ) = H ( jω ) =
A3
I3
( − jω ) 2 + 100 ( jω − p) −ω 2 + 100
ω 2 + p2
(1 Mark) Topic: Measurement of Voltage, Current, Power, Energy and Power Factor (1.731) Given that, I 2 = 1∠10° and I 3 = 1∠70°.
At ω = 0 (DC gain), we have
I2 = 1∠10°
100 p ⇒ p = ± 20 5=±
A2
I1 A1
The given system is stable, therefore the value of p = −20 H ( s) =
Hence,
H ( jω ) =
Now,
⇒1=
s 2 + 100 s + 20 −ω + 100 2
ω 2 + 400 −ω 2 + 100
ω 2 + 400 2 ⇒ ω + 400 = ( −ω 2 + 100) 2 ⇒ ω 2 + 400 = ω 4 + 10000 − 200ω 2 ⇒ (ω 2 ) 2 − 201ω 2 + 9600 = 0
Solved Chapter Wise GATE EE 2020 Paper.indd 15
A3 I3 = 1∠70° Applying KCL, we get I1 = 1∠10° + 1∠70° = 1(cos 10° + j sin 10°) + 1(cos 70° + j sin 70°) = 1(cos 10° + cos 70°) + j (sin 70° + sin 10°) = (1.3268) + j (1.1133) = 1.3268 + j1.1133 A = 1.731∠40° A
8/10/2020 10:07:59 AM
B16
GATE EE Chapter-wise Solved papers
41. A benchtop DC power supply acts as an ideal 4 A current source as long as its terminal voltage is below 10 V. Beyond this point, it begins to behave as an ideal 10 V voltage source for all load currents going down to 0 A. When connected to an ideal rheostat, find the load resistance value at which maximum power is transferred, and the corresponding load voltage and current. (a) Short, ∞ A, 10 V (b) Open, 4 A, 0 V (c) 2.5 Ω, 4 A, 10 V (d) 2.5 Ω, 4 A, 5 V (2 Marks)
R thermistor
+
+ 3V
−
Vout
− R1 R3 R2 +
Topic: M easurement of Voltage, Current, Power, Energy and Power Factor (c) Given that, I = 4 A and Vmax = 10 V.
0.1 V
−
(2 Marks)
I I + V −
Supply
Topic: Oscilloscopes, Error Analysis (0.08) Given that, R1 = 1 kΩ, R2 = 1.3 kΩ, R3 = 2.6 kΩ, α = − (4 ± 0.25)% /°C and T = 150 °C.
4A R
2.5
Ω
R thermistor
Since supply acts as 4 A current source, then V = 4R and P = 42R Since Vmax = 10 V, then Rmin =
+
R + 3V
Vout
−
−
R1 R3
10 = 2.5 Ω, Pmax = 16 × 2.5 = 40 W 4
R2
Now, if supply acts as 10 V source, then
+
10 V
−
0.1 V
Rthermistor = 2(1 + α T ) kΩ, α = −( 4 ± 0.25)% / °C 2.5 We have, I =
Ω
R
10 10 2 10 2 ,P= , Pmax = = 40 W R R 2.5
Hence, for maximum power transfer, R = 2.5 Ω, V = 10 V and I = 4 A 42. The temperature of the coolant oil bath for a transformer is monitored using the circuit shown. It contains a thermistor with a temperature-dependent resistance, Rthermistor = 2(1 + α T ) kΩ, where T is the temperature in °C, The temperature coefficient α , is − (4 ± 0.25)% /°C. Circuit parameters: R1 = 1 kΩ, R2 = 1.3 kΩ, R3 = 2.6 kΩ. The error in the output signal (in V. rounded off to 2 decimal places) at 150 °C is .
Solved Chapter Wise GATE EE 2020 Paper.indd 16
= −(0.04 ± 0.0025) / °C So, α max = −0.0425/ °C and α min = −0.375/ °C At α = −0.04, Rthermistor = 2(1 - 0.04 × 150) = −10 kΩ ⎡ R3 ⎤ R3 ⎢1 + ⎥ − × 0.1 ⎣ R2 ⎦ R2 1 ⎡ 2⎤ = 3× 1 + − 2 × 0.1 = −1.2 V (1 − 10) ⎢⎣ 1 ⎥⎦
Vout = 3 ×
R1 R1 + Rthermistor
Case 1: Considering α max = −0.0425/ °C, we have Rthermistor = 2[1 + ( −0.0425 × 150)] = −10.75 kΩ Vout = 3 ×
1 [1 + 2] − 2 × 0.1 = −1.12 V (1 − 10.75)
8/10/2020 10:08:01 AM
appendix • Solved GATE (EE) 2020
Case 2: Considering α min = −0.375/ °C, we have Rthermistor = 2[1 + (−0.0375 × 150)] = −9.25 kΩ Now, Vout = 3 ×
1 [1 + 2] − 2 × 0.1 = −1.29 V (1 − 9.25)
So, Vout = −1.2 ± 0.08
Chapter 9: Analog and Digital Electronics 43. A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source supplies a series combination of finite resistance, R, and a very large inductance, L. The two most dominant frequency components in the source current are: (a) 50 Hz, 0 Hz (b) 50 Hz, 100 Hz (c) 50 Hz, 150 Hz (d) 150 Hz, 250 Hz (1 Mark) Topic: S imple Diode Circuits: Clipping, Clamping, Rectifiers (c) For full bridge rectifier, High inductive load
Io
Io = load current
Io
ISource
−Io
2d
We have, I Source =
4I0 nπ sin nd sin (sin nωt + ϕn ) n 2 π n =1, 3, 5 ∞
∑
π . Square wave with time period 2 2p i Fundamental , 3rd harmonic will be dominant. Here, 2d = π and d =
I Source =
4I0 nπ
∞
∑
n =1, 3, 5
sin 2
nπ sin( nωt + ϕn ) 2
(since ISource ≠ 0) Hence, the most dominant frequency component will be f and 3f, that means, 50 Hz and 150 Hz. 44. A common-source amplifier with a drain resistance, RD = 4.7 kΩ, is powered using a 10 V power supply. Assuming that the transconductance, gm, is 520 μA/V, the voltage gain of the amplifier is closest to:
Solved Chapter Wise GATE EE 2020 Paper.indd 17
B17
(a) −2.44 (b) −1.22 (c) 1.22 (d) 2.44 (1 Mark) Topic: Amplifiers: Biasing, Equivalent Circuit and Frequency Response (a) Given, drain resistance RD = 4.7 kΩ and transconductance, gm = 520 μA/V The voltage gain for common source amplifier is given by AV = − g m RD = −(520 μ A/V ) × ( 4.7 kΩ) = −(520 × 10 −6 × 4.7 × 103 ) = −2.44 45. A sequence detector is designed to detect precisely 3 digital inputs, with overlapping sequences detectable. For the sequence (1, 0, 1) and input data (1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0), what is the output of this detector? (a) 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0 (b) 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0 (c) 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0 (d) 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0 (1 Mark) Topic: Combinational and Sequential Logic Circuits (b) Given overlapping sequence input data = 11010011010110 The output of detector is 1, when 101 is detected. So, 110 → 0 101 → 1 010 → 0 100 → 0 001 → 0 011 → 0 110 → 0 101 → 1 010 → 0 101 → 1 011 → 0 110 → 0 Hence, the output of the detector is 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0. 46. The cross-section of a metal-oxide-semiconductor structure is shown schematically. Starting from an uncharged condition, a bias of +3 V is applied to the gate contact with respect to the body contact. The charge inside the silicon dioxide layer is then measured to be +Q. The total charge contained within the dashed box shown, upon application of bias, expressed as a multiple of Q (absolute value in Coulombs, rounded off to the nearest integer) is .
8/10/2020 10:08:01 AM
B18
GATE EE Chapter-wise Solved papers
R1
I
GATE Silicon Dioxide
+ D
V
R2
I1
Si
−
BODY DASHED
I
OX (1 Mark)
Topic: C haracteristics of Diodes, BJT, MOSFET (0) Charge in each layer is shown in the following figure:
GATE SiO2
+Q +Q
For maximum power transfer, V should be maximum and I1 should pass through the load. Since I1 < 0, then diode is reversed biased and behave as open circuit. R1 + I1
R2
3V
−
−Q
Si Body
−Q
Hence, overall charge in dashed box = Q + Q - Q - Q =0 47. Consider the diode circuit shown below. The diode, D obeys the current-voltage characteristic ⎛ ⎛V ⎞ ⎞ I D = I S ⎜ exp ⎜ D ⎟ − 1⎟ , where n > 1, VT > 0. VD ⎜ ⎟ ⎝ nVT ⎠ ⎠ ⎝ is the voltage across the diode and ID is the current through it. The circuit is biased so that voltage, V > 0 and current, I < 0. If you had to design this circuit to transfer maximum power from the current source (I1) to a resistive load (not shown) at the output, what values of R1 and R2 would you choose? R1
If R2 is large, whole current I1 will try to pass through it and less value of R1 reduce the voltage drop across it. Hence, for maximum power transfer, R2 should be large and R1 should be small. 48. A non-ideal diode is biased with a voltage of −0.03 V, and a diode current of I1 is measured. The thermal voltage is 26 mV and the ideality factor for the diode is 15/13. The voltage, in V, at which the measured current increases to 1.5I1 is closest to: (a) −0.02 (b) −0.09 (c) −1.50 (d) −4.50 (2 Marks) Topic: Characteristics of Diodes, BJT, MOSFET (b) Given that, η = 15/13, V = −0.03 V and VT = 26 mV. The equation of diode current is given by I = I S ( eV /ηVT − 1)
I +
D I (a) Large R1 and large R2. (b) Small R1 and small R2. (c) Large R1 and small R2. (d) Small R1 and large R2.
(e
−0.03 (15 /13)× 26 mV
−
1.5 I1 = I S
(e
VD (15 / 23)× 26 mV
) (1)
−1
) (2)
−1
From Eqs. (1) and (2), we get VD
(2 Marks)
Topic: S imple Diode Circuits: Clipping, Clamping, Rectifiers (d) Consider the following figure:
Solved Chapter Wise GATE EE 2020 Paper.indd 18
I1 = I S
When I1 = 1.5I1, then
V
R2
I1
Vo
1.5 I1 (e 0.03 − 1) = −0.03 −1 I1 0.03 (e )
15 ⎤ ⎡ ⎢Since 13 × 26 mV = 0.03 V ⎥ ⎣ ⎦
(e −1 − 1) × 1.5 = eVD / 0.03 − 1 ⇒ eVD / 0.03 = 0.055 Taking natural logarithm on both sides, we get
8/10/2020 10:08:03 AM
B19
appendix • Solved GATE (EE) 2020
VD = ln(0.055) 0.03 ⇒ VD = −0.087 V 49. A non-ideal Si-based pn junction diode is tested by sweeping the bias applied across its terminals from −5 V to +5 V. The effective thermal voltage, VT , for the diode is measured to be (29 ± 2) mV. The resolution of the voltage source in the measurement range is 1 mV. The percentage uncertainty (rounded off to 2 decimal places) in the measured current at a bias voltage of 0.02 V is . (2 Marks) Topic: Characteristics of Diodes, BJT, MOSFET (4.75) Current across diode is given by VD
I D = I 0 (eηVT − 1)
At the end of this program, the memory location 2003H contains the number in decimal (base 10) form . (2 Marks) Topic: 8085 Microprocessor: Architecture, Programming and Interfacing (210) We have LXI H, 2001H = H load 20 and L load 01 MVI A, 21H = A load 21H INX H = H load 20 and L load 02 ADD M = A load 21 H + B1H = D2H INX H = H load 20 and L load 03 MOV M, A = 2003 load D2H HLT = Stop Therefore, the data in memory location 2003 H is (D2H)16 → (11010010)2 → (1 × 27 + 1 × 26 + 1 × 24 + 1 × 21)10 → (128 + 64 + 16 + 2)10 → (210)10
VD
Chapter 10: Power Electronics
⇒ I D = I 0 eηVT (Neglecting the value 1) Now differentiate with respect to VT ⎛ V ∂I D = − I 0 ⎜ − D2 ∂VT ⎝ ηVT
V
⎞ ηV0T ⎟e ⎠ V
0 V = − D 2 ⋅ I 0 eηVT ηVT
⇒ ⇒
51. Thyristor T1 is triggered at an angle α (in degree), and T2 at angle 180° + α , in each cycle of the sinusoidal input voltage. Assume both thyristors to be ideal. To control the load power over the range 0 to 2 kW, the minimum range of variation in α is: T1
ΔI D V = − D 2 × ΔVT ID ηVT ΔI D V = D 2 × ΔVT ID ηVT =
0.02 × 2 × 10 −3 (By taking η = 1) 1× 292 × 10 −6
40 Percentage uncertainty = 2 × 100 = 4.75% 29 50. An 8085 microprocessor accesses two memory locations (2001H) and (2002H), that contain 8-bit numbers 98H and B1H, respectively. The following program is executed: LXI H, 2001H MVI A, 21H INX H ADD M INX H MOV M, A HLT
Solved Chapter Wise GATE EE 2020 Paper.indd 19
+ 200 V 50 Hz
T2
10∠−60°Ω
−
(a) 0° to 60° (c) 60° to 120°
(b) 0° to 120° (d) 60° to 180° (1 Mark)
Topic: Characteristics of Semiconductor Power Devices: Diode, Thyristor, Triac, GTO, MOSFET, IGBT (c) Source voltages from thyristor T1 and T2 are given in the following figure: Vs
−60°
ωt 120°
8/10/2020 10:08:04 AM
B20
GATE EE Chapter-wise Solved papers
Since, load is capacitance. The output voltage for any α is given in the following figure Vo
Topic: Characteristics of Semiconductor Power Devices: Diode, Thyristor, Triac, GTO, MOSFET, IGBT (3) We have, IGBT IS I L
1 α
π
π +α
Vsource
2π
+ −
ID
Lpar ID
1/ 2
So, Vrms =
Vm ⎡ 1 ⎤ ⎢π − α + 2 sin 2α ⎥ 2π ⎣ ⎦ 2
2 rms
Rload Lload
D 2
4
2 rms
V V ⎛V ⎞ Then, Po = ⎜ rms ⎟ × R = ×5 = 100 20 ⎝ z ⎠ For calculating power range, maximum power can occur at maximum voltage, that is, For α = 0°, Vor =
200 2 = 200, Po = = 2 kW 20 2
Vm
For minimum power of zero the current should be zero, that is, when α = 120°, as current becomes negative for positive cycle and the thyristor will not conduct and thus load current is zero and, P = 0 So, the range of α should be 0° to 120°. 52. A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode, D, represented approximately as a piecewise linear plot of current vs time at diode turn-off, Lpar is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1, 2, 3 or 4) at which the IGBT experiences the highest current stress is .
3 For inductive load, load current can be assumed to be constant. Applying KCL, we have IS = IL − ID IS is maximum, where ID is minimum, that means, IGBT experiences highest current stress at point 3. 53. A single-phase inverter is fed from a 100 V DC source and is controlled using a quasi-square wave modulation scheme to produce an output waveform, v(t), as shown. The angle σ is adjusted to entirely eliminate the 3rd harmonic component from the output voltage. Under this condition, for v(t), the magnitude of the 5th harmonic component as a percentage of the magnitude of the fundamental component is (rounded off to 2 decimal places). v(t) 100 V
σ
IGBT 1
Vsource + −
0 Rload
Lpar
Diode Current
−100 V
σ
σ π/2
π
3π/2
2π
σ
σ
σ
σ 5π/2 3π
7π/2 4π
σ
σ
ωt
Lload
D
(1 Mark) 4
2
Time
3 (1 Mark)
Solved Chapter Wise GATE EE 2020 Paper.indd 20
Topic: Sinusoidal Pulse Width Modulation (20) From single pulse width modulation scheme, the output voltage is given by v (t ) =
4 VS nπ sin nd sin 2 n =1, 3, 5 nπ ∞
∑
8/10/2020 10:08:05 AM
di dt appendix di 50 • Solved GATE (EE) 2020 ⇒ = dt L tON 50 50 ⇒i= dt = × tON ∫ 0 10 mH 10 mH 50 0.6 = × = 0.3 A 10 × 10 −3 10 × 103 50 = L
For eliminating 3rd harmonic,
π 3 So, the ratio of the 5th harmonic component and the fundamental component is 3d = π ⇒ d =
4 VS 5π ⎛ 5π ⎞ sin ⎜ sin ⎟ v5 (t ) 1 5π 2 ⎝ 3 ⎠ = =− 4 VS π v1 (t ) 5 ⎛π ⎞ sin ⎜ ⎟ sin 2 π ⎝3⎠
B21
During tOFF time, inductor will discharge through inverted 50 V battery
10 mH
50 V
Therefore, percentage of ratio of the magnitudes is V05 1 = × 100 = 20% V01 5 54. In the DC-DC converter circuit shown, switch Q is switched at a frequency of 10 kHz with a duty ratio of 0.6. All components of the circuit are ideal, and the initial current in the inductor is zero. Energy stored in the inductor in mJ (rounded off to 2 decimal places) at the end of 10 complete switching cycles is . D
10 mH
1 Vdt L∫ 1 ⇒ i = iL0 + ∫ Vdt L tOFF 1 ⇒ i = 3+ ( −50)dt ∫ 0 10 mH i = iL0 +
50 × tOFF 10 × 10 −3 50 0.4 ⇒ i = 3− × = 0.1 A 10 × 10 −3 10 × 103
⇒ i = 3−
Hence, current in the inductor in one complete cycle = 0.1 A After 10 complete cycles, current in the inductor = 0.1 × 10 = 1 A Therefore, energy stored in inductor is
50 V
50 V
Q 1 2 Li 2 1 = × 10 × 10 −3 × 1 = 5 mJ 2
E= (2 Marks) Topic: D C to DC Conversion: Buck, Boost and Buck-Boost Converters (5) Given, circuit frequency, f = 10 kHz and duty ratio = 0.6 Now, during tON time, tON = α t = 0.6 /f , so tOFF = 1 - tON = (1 - 0.6)/f, we have
i
50 V
10 mH
di dt di 50 ⇒ = dt L tON 50 50 ⇒i= dt = × tON ∫ 0 10 mH 10 mH 50 0.6 = × = 0.3 A −3 10 × 10 10 × 103 50 = L
Solved Chapter Wise GATE EE 2020 Paper.indd 21
55. A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is . (2 Marks) Topic: Line Commutated Thyristor Based Converters (4.80) Given that, R = 10 Ω, L = very high and V = 230 V, f = 50 Hz, Li = 2.28 mH and α = 45° Now, ΔVd0 =
Vd0 2
[cos α − cos(α + μ )] (1)
8/10/2020 10:08:06 AM
B22
GATE EE Chapter-wise Solved papers
where ΔVd0 is average reduction in output voltage, Vd0 is 2Vm /π for single phase full bridge fully controlled rectifier. Also, ΔVd0 = 4 fLI 0 for single phase fully controlled rectifier So, I 0 R =
2Vm cos α − 4 fLI 0 π
2 × 230 2 cos 45° − 4 × 50 × 2.28 × 10 −3 × I 0 π 460 ⇒ 10 I 0 = − 0.2 × 2.28 I 0 π 460 ⇒ 10.456 I 0 = π 460 = 14.01 A ⇒ I0 = π × 10.456 ⇒ 10 I 0 =
Solved Chapter Wise GATE EE 2020 Paper.indd 22
So,
ΔVd0 =
Vd0 2
[cos α − cos(α + μ )]
2Vm [cos 45° − cos( 45° + μ )] π ×2 230 2 ⎡ 1 ⎤ ⇒ 4 × 50 × 2.28 ×10 −3 ×14.01 = − cos( 45° + μ ) ⎥ ⎢ π ⎣ 2 ⎦ ⇒ 4 fLI 0 =
⇒ 6.385 = 103.536[0.707 − cos( 45° + μ )] ⇒ 0.0616 = 0.707 − cos( 45° + μ ) ⇒ cos( 45° + μ ) = 0.6453 ⇒ 45° + μ = cos −1 (0.6453) ⇒ 45° + μ = 49.80° ⇒ μ = 4.80°
8/10/2020 10:08:06 AM
GATE The book comprises previous years’ (2000 to 2020) GATE questions in Electrical Engineering covered chapter-wise with detailed solutions. Each chapter begins with a detailed year-wise analysis of topics on which questions are based and is followed by listing of important formulas and concepts for that chapter. The book is designed to make the students well-versed with the pattern of examination, level of questions asked and the concept distribution of questions and thus bring greater focus to their preparation. It aims to be a must-have resource for an essential step in their preparation, that is, solving and practicing previous years’ papers.
Salient Features Chapter-wise coverage of GATE Electrical Engineering previous years’ questions (2000 to 2018). Chapter-wise 2019 and 2020 GATE papers available as appendix
l
Chapter on Engineering Mathematics included for complete coverage of the technical section of the GATE paper
l
Detailed topic-wise analysis of questions for each chapter
l
Important Formulas and Concepts summarized for easy recall in each chapter
l
All questions marked for level of difficulty (i.e. 1 Mark or 2 Marks)
l
Detailed solutions for all questions, tagged topic-wise
l
Packaged with unique scratch code for free 7-day subscription for online GATE tests
Customer Care +91 120 6291100 [email protected] www.wileyindia.com www.wiley.com
eISBN 978-81-265-9535-8
9 788126 595358
Chatterjee, Lather, Gupta
Wiley India Pvt. Ltd.
CHAPTER-WISE SOLVED PAPERS (2000-2020)
l
GATE
ELECTRICAL ENGINEERING
CHAPTER-WISE SOLVED PAPERS
(2000-2020)
Inside
The Graduate Aptitude Test in Engineering (GATE) is an All-India level examination for engineering graduates aspiring to pursue Master’s or Ph.D. programmes in India. The Public Sector Undertakings (PSU’s) also use GATE used as a recruiting examination. The examination is highly competitive, and the GATE score plays an important role in fulfilling the academic and professional aspirations of the students. This book is aimed at supporting the efforts of GATE aspirants in achieving a high GATE score.
ELECTRICAL ENGINEERING
About the Book
Dr. Debashis Chatterjee Dr. J. S. Lather Dr. Lalita Gupta
l l l
Detailed Exam Analysis Chapter-wise and Topic-wise. Questions from previous years’ (2000 – 2020) papers. Unique scratch code that provides access to w Free online test with analytics. w 7-Day free subscription for topic-wise GATE tests. w Instant correction report with remedial action.