ELECTRICAL ENGINEERING ESE TOPIC WISE OBJECTIVE SOLVED PAPER-I
From (1992 – 2017)
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Phone : 011-26522064
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IES MASTER PUBLICATION F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-26522064, Mobile : 8130909220, 9711853908 E-mail : ies_master@yahoo.co.in, info@iesmaster.org Web : iesmasterpublication.org
All rights reserved. Copyright Š 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 of IES MASTER, New Delhi. Violates are liable to be legally prosecuted. First Edition : 2017
Typeset at : IES Master Publication, New Delhi-110016
PREFACE It is an immense pleasure to present topic wise previous years solved paper of Engineering Services Exam. This booklet has come out after long observation and detailed interaction with the students 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 the solutions, students will be able to understand the basic concepts and will apply these concepts in solving other questions that might be asked in future exams. Engineering Services Exam is a gateway to a immensly satisfying and high exposure job in engineering sector. The exposure to challenges and opportunities of leading the diverse field of engineering has been the main reason for students opting for this service as compared to others. To facilitate selection into these services, availability of arithmetic solution to previous year paper is the need of the day. Towards this end this book becomes indispensable.
Mr. Kanchan Kumar Thakur Director–IES Master
Note: Direction of all Assertion Reasoning (A–R) type of questions covered in this booklet is as follows: DIRECTIONS: The following four items consist of two statements, one labelled as ‘Assertion A’ and the other labelled as ‘Reason R’. You are to examine these two statements carefully and select the answer to these two statements carefully and 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 of A (b) Both A and R are individually true but R is not the correct explanation of 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 covered in 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 statements carefully and select the answers to these items using the code given below: (a) Both Statement : (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I). (b) Both Statement (I) and Statement (II) are individually true but Statement (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.
CONTENTS 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
1 S YL L A B U S Circuits elements. Kirchoff’s Laws. Mesh and nodal analysis. Network Theorems and applications. Natural response and forced response. Transient response and steady state response for arbitrary inputs. Properties of networks in terms of poles and zeros. Transfer function. Resonant circuits. Three-phase circuits. Two-port networks. Elements of twoelement network synthesis.
Contents 1.
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 – 1992 1.
2.
1
The number of turns of a coil having a time constant T are doubled. Then the new time constant will be (a) T
(b) 2T
(c) 4T
(d) T/2
Current having wave form shown is flowing in a resistance of 10 ohms. The average power is
1
1A
A R=2 A
5.
(a) 2 A
(b) 1.66 A
(c) 1 A
(d) 0.625 A
In the circuit shown the value of I is I 28 10A
4
5A
8
10A
0
3.
4.
1
2
3
t(sec)
(a)
1000 W 1
(b)
1000 W 2
(c)
1000 W 3
(d)
1000 W 4
6.
(a) 1 A
(b) 2 A
(c) 4 A
(d) 8 A
When all the resistances in the circuit are of one ohm each, the equivalent resistance across the points A and B will be C
A 24 V battery of internal resistance r = 4 ohm is connected to a variable resistance R. The rate of heat dissipated in the resistor is maximum when the current drawn from the battery is I. The current drawn from the battery will be (I/2) when R is equal to
A
B
D
(a) 8 ohm
(b) 12 ohm
(a) 1
(b) 0.5
(c) 16 ohm
(d) 20 ohm
(c)
2
(d) 1.5
In the figure shown, if we connect a source of 2 V, with internal resistance of 1 at A'A, with positive terminal at A', then the current through R is
7.
A battery is connected to a resistance causing a current of 0.5 A in the circuit. The current drops to 0.4 A when additional resistance of 5 is connected in series. The
Circuit Theory
3
current will drop to 0.2 A when the resistance is (a) 10
(b) 15
25
(d) 30
(c) 8.
In the circuit shown in the figure, the voltage across the 2 ohm resistor is 4 6A
2 1
3
The current in resistor R shown in figure will be 1K
1K R=1K 1A
+ – 2V
9.
12.
(c) 0.6 A
(d) 0.8 A
5 + 5V V1 – –
(c) 2 V
(d) zero
Four networks are shown below in figures (1), (2), (3) and (4) 1.
The circuit shown is a linear time invariant one and the sources are ideal. Choose from the answers given below, the values of voltage across and current through 1 resistor.
+
(b) 4 V
IES – 1994 13.
(b) 0.4 A
3V
(a) 6 V
1K
+ 1V –
(a) 0.2 A
+ –
R I(t) amp
C
2.
I1 1 1A
(a) 1V, 1A
L
I(t)
R
I(t)
L
I(t) amp
C
3.
(b) 1V, 6A (c) 5V, 5A
I(t)
R
I(t)
L
I(t) amp
C
(d) None of the above 10.
The equivalent capacitance across ab will be : 0.1F
4. a
b
c 0.1F
0.1 F
d
L
0.1F
0.1F
(a)
0.2 F
(b) 0.1 F
(c)
0.5 F
(d) 0
IES – 1993 11.
RI(t) + volts –
R
d I t dt
+ –
L
+ –
C
volts L
1 I t dt C
volts
Of these networks, (a) all the four networks are equivalent (b) no two networks are equivalent
Assertion (A) : Kirchhoff’s current law is valid for an ac circuit containing R, L and C.
(c) networks shown in figures (2), (3) and (4) are equivalent
Reason (R) : The sum of rms currents at any junction of the circuit is always zero.
(d) networks shown in figures (3) and (4) are equivalent
IES Objective Solutions Topic Wise Paper-I 1992-2017
4 14.
15.
The number of 2 F , 400 V capacitors needed to obtain a capacitance value of 1.5 F rated for 1600 V is (a) 12
(b) 8
(c) 6
(d) 4
B
3
1
A
(c) 2
(d) 1
19.
B
P
RL
(c)
(c) 5 A
(d) zero
(d) P
P
RL
20.
1
(b) 6 A
V
A voltage source with an internal resistance RS, supplies power to a load RL. The power delivered to the load varies with RL as (a) (b)
RL
(a) 10 A
0
(c)
I
RL
A signal is described by s(t) = r(t – a) – r(t – b), a < b, where r(t) is a unit ramp function starting at t = 0. The signal s(t) is represented as (a) s(t) (b) s(t)
In the circuit shown in the given Fig., the current I through RL is 120
–
B
P
I 5A
+
I
I
The value of the current I flowing in the 1 ohm resistor in the circuit shown in the given figure will be + 5V –
A
(d)
R
5 II A 4
(b) 3
B
(c)
E
(a) 4
R
V B
C I
17.
+ –
III 2
A
(b)
R
A connected planar network has 4 nodes and 5 elements. The number of meshes in its dual network is D
16.
A
(a)
a
0 a
t
b
b(a.b)
(d) s(t)
s(t)
60 RL=30
420V
0 a
420V
21.
18.
(a) 2 A
(b) zero
(c) –2 A
(d) 6 A
A simple equivalent circuit of the 2–terminal network shown in figure, is A
t
0 a
+ V –
(a) equals
dS t dt
(c) is the reciprocal of S(t) B
b
(a.b) t
A unit impulse input to a linear network has a response R(t) and a unit step input to the same network has response S(t). The response R(t)
(b) equals the integral of S(t)
R I
b
(d) has no relation with S(t)
IES Objective Solutions Topic Wise Paper-I 1992-2017
24 1
VA = (5/4)V
t3 = 1000 3 0 1000 W 3
= Sol–3:
Sol–5:
(b)
Current through R is, V 5/4 I = A = = 0.625A 2 2 (b) I
From circuit diagram 4 r + 24V –
28
I 10A
5A
4
8
R I
24 24 = r+R 4+R The rate of heat dissipated in the resistor is maximum, when r = R = 4
I =
+ –
24 = 3A 4 +4 when current drawn from battery is I/2
Sol–6:
C C
(d)
1
1
Connect source of 2V with internal resistance of 1 in the given figure. The figure is
A
1 1 1
A
B
1
1
2
2
1 1
D 1
I R 2
1
Using delta to star conversion.
2V 1
2
1V
B
1 4
I R
1 2
1 2 A
A 1
B
1
1
D 1
1A
40V
(b) Given all the resistance of the circuit are 1 .
R = 12 . Sol–4:
40V
80 = 2A 40
I =
24 3 = 4+R 2
8 – +
In the circuit, applying KVL, –40 + 4I+ 28I + 8I – 40 = 0 40 I = 80
I=
I 24 = 2 4+R
28
4
1
1 1 4 2
1 2
2V
1
KCL at node A, VA – 1 VA – 2 VA + + = 0 2 1 2
2VA – 1 + 2VA – 4 = 0
A
B
1 4
1 1 4 2 1
RAB = 1 1 = 0.5
Circuit Theory Sol–7:
25
(d)
Sol–8:
(a) 1k
Rth
2V
1k R 1k 1A
I = 0.5A
Vth
A 1k + –
Vth I = R = 0.5 th
Vth = 0.5 Rth
1k
1V
2V
1k 1k
1k + –
+ –
1V
1000V
KCL at node A,
...(1)
VA – 2 VA – 1 VA – 1000 + + = 0 1K 1K 2K
When additional resistance of 5 is connected in series, current drops to 0.4A, then circuit is
1006 V 5 Current through resistor R will be
VA =
Rth Vth
0.4A
VA – 1 1K
I =
5
1006 – 5 5 ×1000 = 0.201A 0.2A
=
0.4 =
Vth R th + 5
Sol–9:
(c) 5
0.4 Rth + 2 = Vth = 0.5 Rth
Rth = 20
I1
+ 5V –
(from equation (1))
V1 1 –
From circuit diagram,
When connected resistance R in series, current drops to 0.2A
V1 = 5V I1 =
Rth
0.2A
V1 5 = = 5A 1 1
Sol–10: (b) 0.1F
Vth
R
a
0.2 =
0.1F
Vth R th + R
0.1F 0.1F c d
0.2 Rth + 0.2R = Vth = 0.5 Rth 0.1F
(from equation (1)) R =
1A
+
3 3 R th = × 20 = 30 2 2
c
b
0.1F 0.1F
a
b 0.1F
0.1F d
IES Objective Solutions Topic Wise Paper-I 1992-2017
26
Given circuit is a balanced bridge, so
Sol–15: (b) Dual network of the given network is
0.1F 0.1F
A
I
a
II
b B
D
C 1
3
0.1F 0.1F
Cab = 0.05 0.05 = 0.1 F Sol–11: (c) KVL and KCL are applicable to any lumped electric circuit at any time ‘t’. Generally, the sum of the rms currents at any junction of the circuit is not zero. It depends upon the nature of the elements connected at the junction. Sol–12: (c)
E 2
III
Total no. of meshes in dual network is = 3. Sol–16: (c) I + – 5V
1
5A
Voltage across 1 resistor = 5V
A 4 6A
2 1
3
+ 3V –
Current through 1 resistor =
I= KCL at node A,
V R
5 = 5A 1
Sol–17: (d)
V V –3 6 = A + A 1 2 3VA – 3 = 12 VA = 5V Voltage across the 2 resistor is = VA – 3 = 5 – 3 = 2V Sol–13: (a) All four networks are equivalent because voltage across R, L and C are VR = i(t)R, Ldi t 1 VL = , Vc = i t dt respectively dt c are same for all four networks. Sol–14: (a) To increase the voltage rating, capacitors are to be connected in series and to increases the capacitance, capcitors are to be connected in parallel.
A I 60 R 30 L
120 420V
420V
KCL at node A, VA – 420 VA – 420 VA + + = 0 120 60 30
VA – 420 + 2VA – 840 + 4VA = 0 VA = 180V Current through RL 30 resistor
I =
VA 180 = = 6A 30 30
Sol–18: (a)
1.5 F 1600V
All capacitor 2 F, 400V
Total no. of capacitors required = 12
A R I
A Rth
+ V –
Vth B
B
Circuit Theory
27
For Rth :-
Unit impulse input response R(t) A
Unit step input response S(t)
R
Rth = R
Impulse input, d step input (t ) = dt d u t = dt
Vth = V
Response of unit impulse
Rth
B
and Sol–19: (c)
= Response of
RS I Vth
RL
Vth I = R +R s L
V
=
d [Response of u(t)] dt
=
d S t dt
Sol–22: (d)
Power through load, RL = I2RL
d [u(t)] dt
i(t) = 3t, 0 < t < 3, R = 10
2
th PL = R + R RL L s
Power (P) = i 2 t .R
at RL = 0, PL = 0
= (3t)2 × 10
Vth2 at Rs = RL, PLmax = 4 RL
= 90t2 Power dissipated in the resistor (R) is T
3
1 1 2 = T P dt = 3 90t dt 0 0
at RL = , PL = 0 Hence, correct ans. is (c) Sol–20: (c)
3
=
s(t) = r(t–a) – r(t–b), a<b = s1(t) – s2(t) s 1(t)
90 t3 = 270W 30 3 0
Sol–23: (d) Unit impulse function t is defined as
r(t–a)
0 t 0 t = t = 0
t
a s 2(t)
and
r(t–b)
t dt
=1
t
b s(t)=s1(t)–s2(t)
a
Sol–21: (a) Given :
b
Sol–24: (d) The equivalent inductance of two series connected coils is
t
Leq = L1 + L2 ± 2M where,
M = mutual inductance 16 = Leq = L1 + L2 + 2M (Series aiding) ...(i)