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ELECTRICAL ENGINEERINGESE TOPIC WISE OBJECTIVE
SOLVED PAPER-I
From (1992 – 2017)
Office : Phone : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064
Mobile : E-mail:
Web :
8130909220, 9711853908 [email protected], [email protected]
iesmasterpublication.org
Typeset at : IES Master Publication, New Delhi-110016
IES MASTER PUBLICATIONF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016Phone : 011-26522064, Mobile : 8130909220, 9711853908E-mail : [email protected], [email protected] : iesmasterpublication.org
All rights reserved.Copyright © 2017, by IES MASTER Publications. No part of this booklet may be reproduced,or distributed in any form or by any means, electronic, mechanical, photocopying, recording,or otherwise or stored in a database or retrieval system without the prior permission ofIES MASTER, New Delhi. Violates are liable to be legally prosecuted.
First Edition : 2017
PREFACE
It is an immense pleasure to present topic wise previous years solved paper of EngineeringServices Exam. This booklet has come out after long observation and detailed interaction with thestudents preparing for Engineering Services Exam and includes detailed explanation to all questions.The approach has been to provide explanation in such a way that just by going through thesolutions, students will be able to understand the basic concepts and will apply these concepts insolving other questions that might be asked in future exams.
Engineering Services Exam is a gateway to a immensly satisfying and high exposure job in engineeringsector. The exposure to challenges and opportunities of leading the diverse field of engineeringhas been the main reason for students opting for this service as compared to others. To facilitateselection into these services, availability of arithmetic solution to previous year paper is the needof the day. Towards this end this book becomes indispensable.
Mr. Kanchan Kumar ThakurDirector–IES Master
Note: Direction of all Assertion Reasoning (A–R) type of questions coveredin this booklet is as follows:
DIRECTIONS:
The following four items consist of two statements, one labelled as ‘AssertionA’ and the other labelled as ‘Reason R’. You are to examine these twostatements carefully and select the answer to these two statements carefullyand select the answer to these items using the codes given below:
(a) Both A and R are individually true and R is the correct explanation ofA
(b) Both A and R are individually true but R is not the correct explanationof A
(c) A is true but R is false
(d) A is false but R is true.
Note: Direction of all Statement-I and Statement-II type of questions coveredin this booklet is as follows:
DIRECTION:
Following items consists of two statements, one labelled as ‘Statement (I)’and the other as ‘Statement (II)’. You are to examine these two statementscarefully and select the answers to these items using the code given below:
(a) Both Statement : (I) and Statement (II) are individually true andStatement (II) is the correct explanation of Statement (I).
(b) Both Statement (I) and Statement (II) are individually true butStatement (II) is not the correct explanation of Statement (I).
(c) Statement (I) is true but Statement (II) is false
(d) Statement (I) is false but Statement (II) is true.
1. Circuit Theory ............................................................................................. 01–226
2. Control Systems ....................................................................................... 227–423
3. Electro Magnetic Field Theory .................................................................. 424–550
4. Electrical Materials ................................................................................... 551–618
5. Electrical and Electronic Measurements ................................................... 619–765
6. Computer Fundamentals .......................................................................... 766–767
7. Engineering Mathematics ......................................................................... 768–772
CONTENTS
S YL L A BU SCircuits elements. Kirchoff’s Laws. Mesh and nodal analysis. Network Theorems andapplications. Natural response and forced response. Transient response and steady stateresponse for arbitrary inputs. Properties of networks in terms of poles and zeros. Transferfunction. Resonant circuits. Three-phase circuits. Two-port networks. Elements of two-element network synthesis.
Contents1. Network Elements -------------------------------------------------------------- 01-52
2. Transient and Steady State Response------------------------------------ 53-92
3. Resonance ------------------------------------------------------------------------93-130
4. Network Theorems ----------------------------------------------------------- 131-156
5. Two-Port Network ------------------------------------------------------------ 157-180
6. Network Functions ----------------------------------------------------------- 181-199
7. Network Synthesis ----------------------------------------------------------- 200-218
8. 3-Phase Circuits, Network Graphs and Filters --------------------- 219-226
1
IES – 19921. The number of turns of a coil having a time
constant T are doubled. Then the new timeconstant will be(a) T (b) 2T(c) 4T (d) T/2
2. Current having wave form shown is flowingin a resistance of 10 ohms. The averagepower is
10A
0 1 2 3 t(sec)
(a)1000 W
1 (b)1000 W
2
(c)1000 W
3 (d) 1000 W4
3. A 24 V battery of internal resistance r = 4ohm is connected to a variable resistance R.The rate of heat dissipated in the resistor ismaximum when the current drawn from thebattery is I. The current drawn from thebattery will be (I/2) when R is equal to(a) 8 ohm (b) 12 ohm(c) 16 ohm (d) 20 ohm
4. In the figure shown, if we connect a sourceof 2 V, with internal resistance of 1 atA'A, with positive terminal at A', then thecurrent through R is
1A 1
1
R=2
A
A(a) 2 A (b) 1.66 A(c) 1 A (d) 0.625 A
5. In the circuit shown the value of I isI
10A 5A428
8
(a) 1 A (b) 2 A(c) 4 A (d) 8 A
6. When all the resistances in the circuit areof one ohm each, the equivalent resistanceacross the points A and B will be
A
C
B
D(a) 1 (b) 0.5(c) 2 (d) 1.5
7. A battery is connected to a resistancecausing a current of 0.5 A in the circuit.The current drops to 0.4 A when additionalresistance of 5 is connected in series. The
1
Circuit Theory 3
current will drop to 0.2 A when theresistance is(a) 10 (b) 15(c) 25 (d) 30
8. The current in resistor R shown in figurewill be
+– +
– 1V2V
R=1K
1K 1K
1K1A
(a) 0.2 A (b) 0.4 A(c) 0.6 A (d) 0.8 A
9. The circuit shown is a linear time invariantone and the sources are ideal. Choose fromthe answers given below, the values of voltageacross and current through 1 resistor.
+– 5V
5
1A1V1+–
I1
(a) 1V, 1A(b) 1V, 6A(c) 5V, 5A(d) None of the above
10. The equivalent capacitance across ab will be :0.1 F
0.1 F
0.1 F 0.1 F 0.1 F
a bcd
(a) 0.2 F (b) 0.1 F(c) 0.5 F (d) 0
IES – 199311. Assertion (A) : Kirchhoff’s current law is
valid for an ac circuit containing R, L and C.Reason (R) : The sum of rms currents atany junction of the circuit is always zero.
12. In the circuit shown in the figure, the voltageacross the 2 ohm resistor is
+–
4 2
1 3 3V6A
(a) 6 V (b) 4 V(c) 2 V (d) zero
IES – 199413. Four networks are shown below in figures
(1), (2), (3) and (4)1.
R
L
C
I(t)amp
2.R
L
CI(t)amp
I(t)
I(t)
3.R
L
CI(t)amp
I(t)
I(t)
4. RRI(t)volts
+–
+–
+–
L
C
volts
d I tLdt
volts
1L I t dtC
Of these networks,(a) all the four networks are equivalent(b) no two networks are equivalent(c) networks shown in figures (2), (3) and
(4) are equivalent(d) networks shown in figures (3) and (4)
are equivalent
4 IES Objective Solutions Topic Wise Paper-I 1992-2017
14. The number of 2 F , 400 V capacitors neededto obtain a capacitance value of 1.5 F ratedfor 1600 V is(a) 12 (b) 8(c) 6 (d) 4
15. A connected planar network has 4 nodes and5 elements. The number of meshes in itsdual network is
A
B C
D E
I II
III
1
2
3 4
5
(a) 4 (b) 3(c) 2 (d) 1
16. The value of the current I flowing in the 1ohm resistor in the circuit shown in thegiven figure will be
+–
5V 5A 1
I
(a) 10 A (b) 6 A(c) 5 A (d) zero
17. In the circuit shown in the given Fig., thecurrent I through RL is
120 60R =30L
I
420V 420V
(a) 2 A (b) zero(c) –2 A (d) 6 A
18. A simple equivalent circuit of the2–terminal network shown in figure, is
A
B
R
VI+–
(a) A
B
R
V+–
(b) A
B
R
(c) A
B
R
I
(d) A
B
VI+–
19. A voltage source with an internal resistanceRS, supplies power to a load RL. The powerdelivered to the load varies with RL as(a)
P
RL
(b)P
RL
(c)P
RL
(d)P
RL
20. A signal is described by s(t) = r(t – a) –r(t – b), a < b, where r(t) is a unit rampfunction starting at t = 0. The signal s(t) isrepresented as(a) s(t)
0 a b t
(b) s(t)
0 a b(a.b)
(c) s(t)
0 a b t
(d) s(t)
0 a bt(a.b)
21. A unit impulse input to a linear networkhas a response R(t) and a unit step input tothe same network has response S(t). Theresponse R(t)
(a) equals dS t
dt(b) equals the integral of S(t)(c) is the reciprocal of S(t)(d) has no relation with S(t)
2 4 IES Objective Solutions Topic Wise Paper-I 1992-2017
AV = (5/4)V Current through R is,
I = AV2 = 5 / 4
2 = 0.625ASol–5: (b)
8428
I
40V40V+– +
–
I
28
84 5A10A
In the circuit, applying KVL,–40 + 4I+ 28I + 8I – 40 = 040 I = 80
I = 8040 = 2A
Sol–6: (b)Given all the resistance of the circuitare 1 .
1 1
11
22A B
C
D
11 1
11
A B
C
D1
1
Using delta to star conversion.
A B
1
1
14 1
2
A B
12
14
121
2
14 1
2
14
RAB = 1 1 = 0.5
=13
0
t10003
=1000 W
3Sol–3: (b)
From circuit diagram
R
I4r
24V+–
I = 24
r + R = 24
4 + RThe rate of heat dissipated in the resistoris maximum, when r = R = 4
I = 244 + 4 = 3A
when current drawn from battery isI/2
I2 =
244 + R
32 =
244 + R
R = 12 .Sol–4: (d)
Connect source of 2V with internalresistance of 1 in the given figure.The figure is
11
1 21A R2V
1
2R11
1V 2V
A
I
I
KCL at node A,
A A AV – 1 V – 2 V+ +2 1 2 = 0
A A2V –1 + 2V – 4 = 0
Circuit Theory 2 5
Sol–7: (d)
Rth
Vth I = 0.5A
I = th
th
VR = 0.5
Vth = 0.5 Rth ...(1)
When additional resistance of 5 isconnected in series, current drops to0.4A, then circuit is
Rth
Vth 5
0.4A
0.4 = th
th
VR + 5
0.4 Rth + 2 = Vth = 0.5 Rth
(from equation (1))
Rth = 20
When connected resistance R in series,current drops to 0.2A
Rth
Vth R
0.2A
0.2 = th
th
VR + R
0.2 Rth + 0.2R = Vth = 0.5 Rth
(from equation (1))
R = th3 R2 = 3 × 20
2 = 30
Sol–8: (a)
+– 2V +
–+–1V 1000V
A
1k1k1k
1k
1k1k 1k
1k
2V R
1V1A
KCL at node A,
A A AV – 2 V – 1 V – 1000+ +1K 1K 2K = 0
VA =1006 V
5 Current through resistor R will be
I = AV –11K
=1006 – 55 ×1000
= 0.201A 0.2ASol–9: (c)
+
–5V
V1+–
I11A
5
1
From circuit diagram,V1 = 5V
I1 = 1V1 = 5
1 = 5A
Sol–10: (b)
0.1 F
0.1 F
0.1 F
0.1 F
0.1 F
0.1 F
0.1 F0.1 F
0.1 F
a bc
a b
d
c
d
2 6 IES Objective Solutions Topic Wise Paper-I 1992-2017
Given circuit is a balanced bridge, so0.1 F
0.1 F
0.1 F
0.1 F
a b
Cab = 0.05 0.05 = 0.1 FSol–11: (c)
KVL and KCL are applicable to anylumped electric circuit at any time ‘t’.Generally, the sum of the rms currentsat any junction of the circuit is not zero.It depends upon the nature of theelements connected at the junction.
Sol–12: (c)
+
–3V6A
A4 2
1 3
KCL at node A,
6 = A AV V – 3+1 2
3VA – 3 = 12VA = 5V
Voltage across the 2 resistor is= VA – 3 = 5 – 3 = 2V
Sol–13: (a)All four networks are equivalent becausevoltage across R, L and C are VR = i(t)R,
VL = Ldi t
dt , Vc = 1 i t dtc respectively
are same for all four networks.Sol–14: (a)
To increase the voltage rating, capacitorsare to be connected in series and toincreases the capacitance, capcitors areto be connected in parallel.
1.5 F1600V
All capacitor 2 F, 400V
Total no. of capacitors required = 12
Sol–15: (b)Dual network of the given network is
A
B C13 2
ED
III
III
Total no. of meshes in dual network is= 3.
Sol–16: (c)
+– 15A5V
I
Voltage across 1 resistor = 5V
Current through 1 resistor = VR
I = 51 = 5A
Sol–17: (d)A
I
3060120
420V420V
RL
KCL at node A,
A A AV – 420 V – 420 V+ +120 60 30 = 0
VA – 420 + 2VA – 840 + 4VA = 0VA = 180V
Current through RL 30 resistor
I = AV30 =
18030 = 6A
Sol–18: (a)
V+–
I
RA
B
Rth
A
B
Vth
Circuit Theory 2 7
For Rth :-
R
A
B
Rth
Rth = Rand Vth = V
Sol–19: (c)
Vth RL
I
RS
I = th
s L
VR + R
Power through load, RL = I2RL
PL = 2
thL
s L
V RR + R
at RL = 0, PL = 0
at Rs = RL, PLmax = 2
th
L
V4R
at RL = , PL = 0Hence, correct ans. is (c)
Sol–20: (c)s(t) = r(t–a) – r(t–b), a<b
= s1(t) – s2(t)s (t)1
a t
r(t–a)
s (t)2
b t
r(t–b)
s(t)=s (t)21(t)–s
b ta
Sol–21: (a)Given :
Unit impulse input response R(t)Unit step input response S(t)
Impulse input,
( ) t = d step inputdt
= d u tdt
Response of unit impulse
= Response of ddt [u(t)]
=ddt [Response of u(t)]
= d S tdt
Sol–22: (d)
i(t) = 3t, 0 < t < 3, R = 10
Power (P) = 2i t .R= (3t)2 × 10= 90t2
Power dissipated in the resistor (R) is
=T
0
1 P dtT =
32
0
1 90t dt3
=33
0
90 t30 3
= 270W
Sol–23: (d)Unit impulse function t is definedas
t = 0 t 0
t = 0
and t dt
= 1
Sol–24: (d)The equivalent inductance of two seriesconnected coils is
Leq = 1 2L + L ± 2Mwhere, M = mutual inductance
16 = Leq = 1 2L + L + 2M(Series aiding) ...(i)
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