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Website : www. aceengineeringpublications.com E Engineering Publications ( Sister oncern of E Engineering cademy, Hyderabad) Hyderabad Delhi Bhopal Pune Bhubaneswar Bengaluru Lucknow Patna hennai ijayawada isakhapatnam Tirupati Kukatpally Kolkata Bits & Bytes GTE PRTE BOOKLET Electronics & ommunication Engineering (olume - ) Network Theory, Signals and Systems, ontrol Systems, nalog Electronic ircuits, Digital Electronic ircuits & Microprocessors Expected Questions with Solutions E is the leading institute for coaching in ESE, GTE & PSUs H O: Sree Sindhi Guru Sangat Sabha ssociation, # 4--3//, King Koti, bids, Hyderabad-5. Ph: 4-33448 / 9 / /, 4-475437 7 ll ndia st Ranks in ESE 43 ll ndia st Ranks in GTE

opyright E Engineering Publications 8 ll rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, digital, recording or otherwise, without the prior permission of the publishers. Published at : E Engineering Publications Sree Sindhi Guru Sangat Sabha ssociation, # 4--3//, King Koti, bids, Hyderabad 5, Telangana, ndia. Phones : 4-33449 / / www.aceenggacademy.com Email: info@aceenggpublications.com hyderabad@aceenggacademy.com uthors : Subject experts of E Engineering cademy, Hyderabad While every effort has been made to avoid any mistake or omission, the publishers do not owe any responsibility for any damage or loss to any person on account of error or omission in this publication. Mistakes if any may be brought to the notice of the publishers, for further corrections in forthcoming editions, to the following Email-id. Email : info@aceenggpublications.com First Edition : Revised Edition : 8 Printed at : Karshak rt Printers, Hyderabad. Price : `. 4/- SBN :978--9458-93-8

Dear Students, Solutions of all previous GTE Questions are already available. Every year about % of questions will have repetitive nature. However, rest of the questions are from untapped areas (never asked areas) and few from Previous Engineering Services & ivil Services Questions. Keeping this in view, possible questions are prepared in various subjects (chapter wise) along with their hints/solutions. The student is advised to practice the questions systematically so that their chances of getting high score in GTE Exam will increase. Foreword GTE in Electronics & ommunication Engineering GTE PRTE BOOKLET (Bits & Bytes) (OLUME - ) The student is advised to solve the problems without referring to the solutions. The student has to analyze the given question carefully, identify the concept on which the question is framed, recall the relevant equations, find out the desired answer, verify the answer with the final key such as (a), (b), (c), (d), then go through the hints to clarify his answer. This will help to face numerical answer questions better. The student is advised to have a standard text book ready for reference to strengthen the related concepts, if necessary. The student is advised not to write the solution steps in the space around the question. By doing so, he loses an opportunity of effective revision. s observed in the GTE Exam, number of sets may be possible, being online exams. Hence, don t skip any subject. ll are equally important. t is believed that this book is a aluable aid to the students appearing for competitive exams like ES, SRO and Other PSU s. This book can also be used by fresh Teachers in Engineering in improving their oncepts. With best wishes to all those who wish to go through the following pages. Y.. Gopala Krishna Murthy, M Tech. ME, hairman & Managing Director, E Engineering cademy, E Engineering Publications.

Electronics & ommunication Engineering GTE PRTE BOOKLET (Bits & Bytes) (OLUME - ) MN NDEX S.No. Name of the Subject Page No. Network Theory - 99 Signals & Systems - 83 3 ontrol Systems 84-4 nalog Electronic ircuits 7-359 5 Digital Electronic ircuits 3-4 Microprocessors 43-434

hapter Basic oncepts olume ( Questions) (a) P and Q (b) Q and R (c) R and P (d) R only. Figure shows the current flowing through 4. The equivalent capacitance of the network a capacitor. The charge acquired by the shown in the figure, is capacitor during the interval t = to t = sec c c is i(t), mps c/ c/ c/ t (sec) c (a) c/4 (b) c/3 (c) 5c/ (d) 3c (a) (c) (b).5 (d).5 5. Determine the input resistance R in. Which of the statements given below is /are correct? P) ll the Passive networks are always symmetrical Q) ll the active networks are always nonreciprocal R) ll the linear networks are always reciprocal (a) P and Q (b) Q and R (c) R and P (d) Q only 3. onsider the following statements: P) The Kirchoff s Laws are valid only for the linear networks. Q) The Ohm s Law is applicable only to the lumped elements. R) The Ohm s Law is valid only at a constant temperature. Which of the above statements is/are true.? (a).3 (b).343 (c).4 (d).53. E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata R in R in stages

: 4 : Network Theory n the circuit shown in figure, the value of R in is (a) 3.3 (b).3 (c) 4.3 (d) 5.3 7. The equivalent inductance across the terminals and B is H H (a) (/5) (b) (4/5) (c) (8/5) (d) (/5). n the circuit of figure, the current will be 8 4 5 8 H H B H (a) (c) 4 (b) (d) 8 (a).5 H (c) 4 7 H (b) 3.5 H (d) 8. For the circuit shown in figure, the value of R eq is R eq 5 3 5 7 H. onsider the following circuit: S 3 3 3 S n the above circuit, when S =3, = 4, what is the value of when S = 9? (a) 5 (b) (c) (d) 5 (a) (c) 3 (b) 8 (d) 4. onsider the following circuit: 5 9. n the circuit shown in the given figure, current is 4 5 3 4 What is the value of current in the 5 resistor? (a) (b) (c) 3 (d) 4 E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: 5 : Basic oncepts 3. onsider the network shown in Fig. 4 i i 4 8 3 The node voltage is (a) 3.5 (b).75 (c) 4.5 (d) 7. onsider the network shown in figure. Determine the currents i and i (a), 4 (b) 4, 5 L (c) 3, (d), 3 4. onsider the network shown in Fig. 4 8 3 i 4 i 3 The current L is given by (a) 5 (b) 5 3 (c) 5 (d) 3 Determine the currents i and i 3 (a), (b), (c) -, (d), - 8. For the circuit shown in figure, the value of i x is i x 5. The potential difference across the 3 mmete resistor as shown in figure is 4 3 (a) (c) 3 (b) (d) 4 3 (a) Zero (b) (c) 3.5 (d) 7. onsider the network shown in figure. 9. n the circuit shown, the current is 8 3 9 9 3 (a) (b) (c) 3 (d) 4 E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: : Network Theory. onsider the following network The value of x is 3. onsider the circuit shown in figure. The voltages B and y are respectively 3 x 5 5 8 x B x B y (a) (c) 4 (b) (d) 4 (a), 3 (c) 3, 3 (b), (d),. The average value of the current waveform shown in fig., i(t) 4. The current waveform shown in fig. is applied to an inductance of value H. The voltage across the inductor at t = sec is i(t) 5 (a).5 (b) 5. (c) 7.5 (d). 3 4 (a) 5 (b) 5 (c) (d) t (sec). n the circuit shown in fig., the potential difference is 3 4 5 4 5. Given =, = in the circuit shown in fig., the S is given by S R (a) 5.5 (b) (c) 4.5 (d) E Engineering Publications (a) 7 (b) 5 (c) 7 (d) None Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: 7 : Basic oncepts. onsider the following network. The value of is 9. For the circuit shown below, different time constants are given: 4 R k.5f k.5f R 7 (a) (b) 7 3 7 (c) (d) 7 3 7. onsider the following network. i The value of current i is (a).577 (b).77 (c).777 (d).877..5-3 S. -3 S 3..5-3 S 4. -3 S What are the charging and discharging times respectively? (a), (b), 3 (c), 3 (d), 4 3. For the circuit shown in fig., the value of R in is 4 M 8. onsider the following network. i M (a) k (b) k (c) k (d) M R in The power associated with source is (a).9 W (b).9 W (c) 3.9 W (d) 4.9 W 3. n the circuit shown in fig., if the current through the source is zero, then the value of R is E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: 8 : Network Theory R (a) 9 (b) (c) (d) 3. Match the following List P. KL Q. KL R. Tellegen s Theorem List. onservation of mass.. onservation of energy. 3. onservation of harge. (a) P, Q, R (b) P, Q, R (c) P3, Q, R (d) P3, Q, R 34. The voltage across a resistor is shown in figure. verage power dissipated by the resistor is v(t) - 3 (a) 5W (c) W 3 x 4 3 5 3 (b) W (d) W (t) 35. onsider the network shown in figure. The current source shown in figure. (a) bsorbs W (b) Delivers W (c) bsorbs W (d) Delivers W 3. onsider the circuit shown in figure. 33. n the circuit shown, the value of K for y = is y 5 5 a b X K X (a) (b) (c) 4 (d) 4 The source shown in figure. (a) bsorbs 5W (b) Delivers 5W (c) bsorbs W (d) Delivers W E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: 9 : Basic oncepts 37. onsider the circuit shown in figure. 4. onsider the circuit shown in figure. a 5 5 b 8 The Thevenin s equivalent across a, b is (a) (, ) (b) (, ) (c) (, ) (d) (, ) 38. onsider the circuit shown in figure. The total power absorbed in the circuit is (a) W (b) 44W (c) 4W (d) 4W 4. n inductance of H is connected across the terminals of a device whose vi characteristics is shown in fig. The time constant of the combination is 8 i The voltage source shown in figure v (a) Delivers W (b) bsorbs W (c) Delivers 4W (d) bsorbs 4W (a).5 sec (b).4 sec (c).8 sec (d) sec 39. onsider the circuit shown in figure. 8 The current source shown in figure (a) Delivers W (b) bsorbs W (c) Delivers 4W (d) bsorbs 4W E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

hapter Basic oncepts. ns: (b) Method - : q - = rea of lr triangle rea of lr triangle ()() ()( ) =.5 =.5 Method - : < t < : (, ) (, ) y (x ) y = x i = t v v() t id t t t < t < : (, ) (, ) v y (x ) y = x y = x i = t olume ( Solutions) t id t t d t t t id t t t (t )d t t t t t t v() Q t t 4t 4 t 4t 3 4t 5 485 t 4t 5.5 oulombs. ns: (d). f a symmetrical network consists of only Linear passive elements, then it is also reciprocal if a symmetrical network consists of dependent source along with passive elements then it is a Non-reciprocal network.. ll the active networks are always Nonreciprocal. 3. ns: (d) Ohms law limitations. pplicable for both lumped and distributed elements. The Ohm s Law is valid only at a constant temperature. Kirchoffs laws are valid for linear as well as non-linear networks E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: : Basic oncepts 4. ns: (b). ns: (b) R in R in /3 R in 5. ns: (b) We can write G in... ohm =.99 ohm. R in =.344 R in R in = ǁ ( R in ) R in = 3 R R in (3 R in ) = 5 3R in R in =.3 7. ns: (c) H H in R in. H B H H H H [Note: n K Tn n! where H H B n!&t Tn n Tn. This can also be used to find the sum... ] H 4/3 B E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata H

: : Network Theory L B = 8. ns: (d) 4 / 3 4 / 3 4 / 3 7 / 3 4 / 7H 3 5 7 3 4 = 3 = 3 4 ( 8 ) 5 5 =.4 R eq 5 3. ns: (b) Superposition theorem is easy to apply here. 3 5 4 with only source acting = 4 R eq = 8 8 8 = 4 = 8 with only 5 source acting = 5 = 4 With both source acting, current =. 9. ns: (d) Using mesh analysis, and writing matrix equation by inspection 3. ns: (a) 3 3 s 3 s 3 4 s acting alone, 3 3 7 3 4 3 4 4 3 = 3 s Determinant of matrix = 5 7 4 8 3 5 5, 4 E Engineering Publications because of s because of s s s = 3 Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata 3.5

: 3 : Basic oncepts But = = 4 (given) 4 =.5 = 3.5 When s = 9, 9.5 = = 3.5.5 = 5. ns: (a) 5 5 5 4 5 5 = 5 = 7 5 = 5 = 5 5 = 5 = 5 olts 5 5 5 3. ns: (a) i 4 5 4 4 i 5 8 3 4 3 8 88 4 3 7 4 3 = But, = 4 = 4 3 = 9 = 5 5 8 9 = 4 8 = 4 i i = = 4 4. ns: (c) 4 4 3 8 88 4 3 7 4 3 = But, = 4 = 4 3 = 9 = 5 i 4 i 3 8 3 E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: 4 : Network Theory 5 8 9 = 4 8 = i, i 3 8 8 5. ns: (a) No current flows in 3 resistor. ns: (a) pply KL at Node 9 3 3 9 3 9 = 3.5 4 3 3 i pply KL in above circuit (53) = = = = 5 = 3 L 5 3 8. ns: (c) R eq = = 3 9. ns: (b) By KL 3 = =. ns: (d) 3 5 5 7. ns: (c) L 5 5 (5) =53 Nodal 5 5 5 = 4 x = 4 E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: 5 : Basic oncepts. ns: (c) i(t) 5.54( ) = ( ) = 4.5 volts. {Note: Placing the Polarity marking for( ) is our choice. f we choose to mark negative at terminal and positive at terminal ; we would get( )= 4.5 ) i(t) is periodic a period in t of sec. i (t) ave = = / t(sec) t 5 t t 5 t t t t 5dt 5t = 7.5 5 t t t t 5t t 5dt 5t t t. ns: (c) n the figure, all resistances are in ohms and all source values are in volts. The circuit is effectively in parts: currents and voltages in different elements can be easily calculated, and are shown figure. The battery does not affect these in any way. ( - ) with positive at terminal marked. 4.5 8.5H () 3 4 5 5.5 4 4 =.5 3. ns: (c) 8 y = 3 ( x ) y = x = y = 3 x B y = 3 = B B = 3 4. ns: (c) di = L dt i(t) = mt Slope m = 5 i(t) = 5t = = di(t) dt B i(t) t sec = 5 x = y E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: : Network Theory 5. ns: (c) ssume, in source branch & in, branch s s 4 R = pply KL at Node = = = pply KL to above loop s 4 = s = 7. ns: (a) Nodal 4 By KL = = 7 R 7. ns: (a) Nodal L = 3.5 = 5. () = L = = Nodal 4 = 3.. () =.79 & =.9 i =.577 8. ns: (a) Nodal 3.5 = 5. () Nodal 4 = 3.. () =.79 & =.9 i =.577 P = =.9 W 9. ns: (c) Time constant while charging.5 F.5 3 = sec Time constant while discharging:.5 F in series with.5 F gives.5 F. orresponding time constant 3 =.5.5 sec 3. ns: (c) By applying Nodal at 4 M M = R k 3. ns: (c) oltage across R (with polarity as shown) = E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: 7 : Basic oncepts Using KL; = = = R R R = 3. ns: (c) KL onservation of charge Tellegen s theorem onservation of energy KL onservation of energy 33. ns: (b) x x x x y x x x y = 3 x = y y = x = x y x y k x y k = k = k x 34. ns: (b) (t) P avg 35. ns: (b) /3 T P = P = = W Pdt / 3 /3 dt 3 3 Writing KL = = P = = W (delivered) 3. ns: (d) 5 5 5 = = = = / 3 i 5 v v 5 t dt v E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata

: 8 : Network Theory i = 5 5 = P = = W (delivered) 37. ns: (c) R eq = = pply superposition theorem Thevenin s equivalent th =, R th = 38. ns: (c) i 8 i i 8 = 3 = 8 = 8 8 = 3 = R eq = i 8 3 i 8 i = 3 = 5 P 8 = 8 5 = 4 W (delivered) 39. ns: (c) P = ( ) = 4 W (delivered) 4. ns: (b) Total power absorbed = Total power delivered = 4 4 = 44 W 4. ns: (b) i i 5 Time constant = L R.4sec 5 E Engineering Publications Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru hennai ijayawada izag Tirupati Kukatpally Kolkata