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Title: basic electrical and electronics
Description: It's about engineering and engineering is about knowledge of everything so this is the unit for basic electrical and electronics engineering
Description: It's about engineering and engineering is about knowledge of everything so this is the unit for basic electrical and electronics engineering
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JHUNJHUNUWALA
RTU BTech 1ST SEM SOLVED PAPERS
Basic Electrical and Electronics Engineering
The hard copy of this solved paper will be distributed free in
your college
...
JHUNJHUNUWALA solved papers will be distributed
free in your college for all subjects
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TECH
SEMESTER 1ST, RTU KOTA
Solved Papers
Chapterwise Arrangement According to the Latest Syllabus
BASIC ELECTRICAL & ELECTRONICS
ENGINEERING
Exam 2016-2017
JHUNJHUNUWALA
Solved Paper
BASIC ELECTRICAL & ELECTRONICS ENGINEERING
Edition 2016
Copyright © By JHUNJHUNUWALA
Information contained in this book has been obtained by author, from sources believes
to be reliable
...
This book is published with the understanding that Jhunjhunuwala and its author are
supplying information but are not attempting to render engineering or other professional
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Unit 2
Alternating Quantities: Introduction, Generation of AC Voltages, Root
Mean Square and Average Value of Alternating Currents and Voltages,
Form Factor and Peak Factor, Phasor Representation of Alternating
Quantities, Single Phase RLC Circuits, Introduction to 3-Phase AC
System
...
AC Machines: Principle of Operation of 3-Phase Induction Motor,
3-Phase Synchronous Motor and 3- Phase Synchronous Generator
(Alternator), Applications of AC Machines
...
Digital Electronics: Boolean algebra, Binary System, Logic Gates and
Their Truth Tables
...
Instrumentation : Introduction to Transducers: Thermocouple, RTD,
Strain Gauges, Load Cell and Bimetallic Strip
...
Suggested
Suggested Readings
1
...
Basic Electrical & Electronics Engineering by V
...
Vinod
Kumar & R
...
3
...
Basic Electrical and Electronics Engineering by Muthusubrmaniam, TMH
5
...
Bobrow, Oxford
University Press
6
...
Basic Electrical & Electronics Engineering by Ravish Singh, TMH
8
...
Fundamental of Electrical Engineering
1
2
...
Circuits Theorems
38
4
...
AC Circuits
88
6
...
Transformer
122
8
...
AC Machine
154
10
...
Transistor
175
12
...
Communication Systems
209
14
...
Explain the Faraday’s law of Electromagnetic Induction
...
RTU 2015
Ans :
FARADAY’S LAWS OF ELECTROMAGNETIC INDUCTION
Faraday performed a series of experiments to demonstrate the
phenomena of electromagnetic induction
...
Alternatively, a stationary conductor when placed in a
magnetic field that changes with time develops an emf across it
...
He summed
up his conclusion into two laws, known as Faraday’s laws of
electromagnetic inductions
...
It states :
When the magnetic flux linking in a coil or conductor changes,
an emf is induced in that coil or conductor
...
The essence of the first law is that the induced emf appears
in coil or conductor subjected to a changing magnetic field
...
e \ dF
dt
Page 2
Fundamentals of Electrical Engineering
Chap 1
where F is the magnetic flux linkage of the circuit or coil
...
(i)
dt
where N is constant of proportionality
...
Therefore
d (NF)
e=
dt
or
e=N
EDDY CURRENTS
An eddy current is a swirling current set up in a conductor in response
to a changing magnetic field, due to Farad’s law of induction
...
But
the conductor may not actually be part of the coils electrical circuit,
but may be the coils iron core or some other metallic part of the
system, for example, a transformer
...
Followings are some important point about eddy currents:
1
...
2
...
3
...
2
...
Also find current I3
...
Thus, the circuit will look
like as shown below:
Writing KCL at node a
I1 = 2 + 1 = 3 A
Writing KCL at node b
I2 + 1 = 2
I2 = 1 A
Writing KCL at node c
I1 = I2 + I3
I3 = I1 - I2 = 3 - 1 = 2 A
Writing KVL in loop abcd , we get
Page 4
Fundamentals of Electrical Engineering
Chap 1
V + 2I2 + 1I1 + 5I1 = 0
V + 2 ^1 h + 1 ^3 h + 5 ^3 h = 0
V = - ^2 + 3 + 15h = - 20 V
Energy stored in capacitor
E = 1 CV 2 = 1 # 4 # 10-6 ^202h = 1 # 4 # 10-6 # 400
2
2
2
or
3
...
After 0
...
14 mA
...
RTU 2015
Ans :
Given that,
Voltage across capacitor
...
(ii)
i = c dv = c 150 ^- 20e-20t h = - 3000ce-20t
dt
At t = 0
...
14 mA , so from eq
...
14 # 10-3 = 3000ce-20^0
...
14 # 10-3 = 3000ce-1
1
...
3678
or
-3
c = 1
...
033 # 10-6 F
3000 # 0
...
033 mF
Energy stored in capacitor at t = 0
...
05 sec can be calculated by Eq
...
05h = 150 ^1 - e-1h
...
3678h = 94
...
(iii), we get
E = 1 # 1
...
83 = 4
...
Fundamentals of Electrical Engineering
Page 5
Find the voltage drop between terminal AB , CB and AD in fig
...
Assuming loop currents I1 and I2 and applying KVL
6 - 1 # I1 - 2 # I1 = 0
I1 = 6 = 2 A
3
20 - 2I2 - 3I2 = 0
I2 = 20 = 4 A
5
Voltage drop across 2W resistor
V1 = 2I1 = 2 # 2 = 4 V
Voltage drop across 3W resistor
V2 = 3I2 = 3 # 4 = 12 V
Writing KVL in loop ABCD , we get
- VAB + V2 + 10 - V1 = 0
VAB = V2 + 10 - V1 = 12 + 10 - 4 = 18 V
Writing KVL in loop CBDC , we get
- VCB + V2 + 10 = 0
Page 6
Fundamentals of Electrical Engineering
Chap 1
VCB = V2 + 10 = 12 + 10 = 22 Volt
Writing KVL in loop ADCA, we get
- VAD + 10 - V1 = 0
VAD = 10 - V1 = 10 - 4 = 6 Volt
5
...
RTU 2011
Ans :
STAR-DELTA TRANSFORMATION
There are two ways in which three resistance can be connected
across three point of a network
...
The other showing in Fig 1b is
called delta (D) or mesh connection
...
Fig 1 : Star Delta Conversion
Chap 1
Fundamentals of Electrical Engineering
Page 7
Delta to Star
If the two arrangement in Fig 1 are equivalent, the resistance
between any pair of terminal (AB, BC or CA) in the two circuits
has to be same, when the third line is left open
...
(i)
RB + RC = 1 2
R1 + R2 + R3
Similarly we can write
R (R + R3)
...
(iii)
...
(v)
...
(vii)
Star to Delta
To get the reverse transformation, we multiply eq (v) and eq (vi)
so as to get
RA RB =
2
R1 R2 R 3
(R1 + R2 + R3) 2
...
(ix)
Similarly, we can write
RB RC =
2
R1 R 2 R3
(R1 + R2 + R3) 2
Adding above three equation we get
and
RC RA =
RA RB + RB RC + RC RA =
2
2
2
R1 R2 R 3 + R 1 R2 R3 + R1 R 2 R3
2
(R1 + R2 + R3)
...
(xi)
Similarly
RA RB + RB RC + RC RA
RB
...
R2 =
R3 =
RA RB + RB RC + RC RA
RC
...
RTU 2011, RU 2001
Ans :
The given circuit is as follows
Converting 6 V source into equivalent current source, we get
Chap 1
Fundamentals of Electrical Engineering
Page 9
Combining parallel connected 3W and 6W resistances,
Transforming 2 A current source into equivalent voltage source,
Transforming 4 V source again into its equivalent current source,
Adding parallel connected current sources, we get
Page 10
Fundamentals of Electrical Engineering
Chap 1
Transforming 4 A source into voltage source,
IL =
7
...
6 A
10
4+2+4
Determine the resistance between points A and B in the network
shown in fig
...
Chap 1
Fundamentals of Electrical Engineering
Page 11
R1 =
^5 # 3h + ^3 # 2h + ^2 # 5h 15 + 6 + 10
= 31 = 6
...
33 W
3
3
^5 # 3h + ^3 # 2h + ^2 # 5h
= 31 = 15
...
2 W = 5 # 6
...
76 W
5 + 6
...
33 W | | 4 W = 10
...
88 W
10
...
5 W , 2
...
76 W
resistances into equivalent Y -network as shown
...
76 # 15
...
78 = 2
...
14
2
...
5 + 2
...
5 # 2
...
64 = 2
...
14
2
...
5 + 2
...
76 # 2
...
94 = 0
...
14
2
...
5 + 2
...
02 | | 6
...
375
= 8
...
11 + 0
...
02 + 6
...
467 + 0
...
842 W
8
...
RTU 2008
Ans :
First we transform the Y -network having 4 W , 10 W and 6 W resistors
into its equivalent D -network
...
66 W
6
6
R1 =
R2 =
^6 # 4h + ^4 # 10h + ^10 # 6h
= 124 = 31 W
4
4
^6 # 4h + ^4 # 10h + ^10 # 6h
= 124 = 12
...
4 W = 10 # 12
...
53 W
10 + 12
...
87 W
31 + 2
Now, the circuit will become as shown
Now 2 W , 4 W and 20
...
Chap 1
Fundamentals of Electrical Engineering
RA =
2#4
= 8 = 0
...
6
2 + 4 + 20
...
6 = 41
...
54 W
26
...
6
4 # 20
...
09 W
2 + 4 + 20
...
RC =
RAB = 0
...
07 | | 4
...
300 + 7
...
96
7
...
96
= 0
...
915 = 3
...
Use source conversion technique to find the voltages V0 in the circuit
as shown in Fig
...
This combination is equivalent to 15 V source only
...
State and explain Kirchhoff’s Laws
...
However,
when it is coupled with Kirchoff’s two laws, we have a sufficient,
powerful set of tools for analyzing a large variety of electric circuits
...
KCL
KCL states that
At any junction (node) in an electric circuit the total current
flowing towards that junction (or node) is equal to the total current
flowing away from the junction (or node)
...
N
/I
n
=0
n=1
In fig 1 we see that the node has two current entering, I1 = 4 A and
I5 = 5 A and three current leaving, I2 = 4 A, I3 = 3 A and I 4 = 2
A
...
SIin = SIout
I1 + I5 = I2 + I3 + I 4
4+5= 4+3+2
9 A= 9A
Page 18
Fundamentals of Electrical Engineering
Chap 1
Fig 1
KVL
KVL states that
In any closed loop in a network, the algebraic sum of the voltage
drop taken around the loop is equal to the resultant emf acting in
the loop
...
e
...
N
/V
n
=0
n=1
For a closed loop
If we consider the circuit of fig 2 we may begin at point in the lower
left-hand corner
...
We move through the voltage source, which represents
a rise in potential from point a to point b
...
Moving through c to d we have potential rise
Vs2 due to voltage source continuing through resistors R2 and R3 , we
have additional drops of V2 and V3 respectively and finally we have
potential drop Vs3 of voltage source,
...
Fig 2
*******
Page 20
Analysis Method
Chap 2
CHAPTER 2
ANALYSIS METHOD
1
...
RTU 2016
Ans :
We take reference node as shown below
...
Nodal analysis is applied as below
...
6
12
3
...
15VA + 3VA - 3VB = 0
8
...
(i)
Chap 2
Analysis Method
Page 21
Writing node equation at node ‘B ’, we get
VB - 40 + VB - 0 + VB - VA = 0
36
8
...
35VB - 774 + 2
...
5VB = 774
Solving Eq
...
38 V and VB = 33
...
6
...
45 - 27
...
7058 A
8
...
Find the values of unknown currents I1, I2, I3 and unknown resistance
R1 and R2 , as shown in Figure
...
The
bottom node has been taken as refrence
...
3
0
...
3
0
...
(i)
2V2 - 2V1 + 3V2 - 360 + 18 = 0
- 2V1 + 5V2 = 342
Solving Eq
...
(ii)
V1 = 109 V
Current,
V2 = 112 V
I1 = 110 - V1 = 110 - 109 = 10 A
0
...
1
I2 = V2 - V1 = 112 - 109 = 10 A
0
...
3
I3 = 120 - V2 = 120 - 112 = 40 A
0
...
2
Current through resistor R1
20 = V1
R1
So,
Similarly,
3
...
45 W
20
20
R2 = V2 = 112 = 3
...
Any two point of this ring are
connected through an external resistance ]R\
...
RTU 2015
Ans :
If n be the number of cells connected on one side of resistor R ,
then the number of cells connected on the other side of R is ^N -n h
...
Here, n cells are shown as equivalent to a single cell of e
...
f
...
Similarly ^N -n h cells are shown equivalent
to single cell of e
...
f
...
Chap 2
Analysis Method
Page 23
Fig 1
Applying Kirchoff’s second law to closed mesh abcda
x 6N - n@r + yR = ^N - n h E
^N - n h E - yR
y
= E -d
nR
r
N-n r
^N - n h r
Applying Kirchoff’s second law to closed mesh cdafbc
x=
Thus
...
(ii)
y
ER = E + yn
r
r dN - n n r
N
R
y< n +
F=0
N ]N - ng r
Thus
4
...
e
...
Using loop current method find the current I1 and I2 in fig
...
(i)
- 3I2 - 2 + 6 - 6 ^I2 - I1h = 0
- 3I2 + 4 - 6I2 + 6I1 = 0
4 - 9I2 + 6I1 = 0
or
- 6I1 + 9I2 = 4
Solving eq
...
...
RTU 2013, 2010
Ans :
and
as V1 and V2 respectively and
We take voltage at node
assume the bottom node as refrence node
...
(i)
12V2 - 12V1 + 3V2 + 4V2 - 96 = 0
...
(i) and (ii), we get
V1 = 300 and V2 = 336
29
29
Current through 1 W resistor
I = V2 - V1 = 336 - 300 = 36 = 1
...
Find Ix in Fig
...
RTU 2012, RU 2002
Ans :
Taking bottom datum node we assign node variable as shown in Fig
1
...
(i)
Va - Vb = 10 + 2Vc - 2Vb
or
Va + Vb - 2Vc = 10
Node b: Vb - Va + Vb - 0 + Vb - Vc = 0
2
1
1
or
or
2Vb - 2Va + Vb + 2Vb - 2Vc = 0
or
- 2Va + 5Vb - 2Vc = 0
Vc - Vb Vc - 0 2I = 0
+
+ x
1
1
...
(iii)
Vc - Vb + Vc + 2 (Vc - Vb) = 0
or
- 3Vb + 4Vc = 0
Writing (ii), (iii) and (iv) in matrix form
R
VR V R V
V
S 1 1 - 2WS aW S10W
S- 2 5 - 2WS bW = S 0 W
V
S
S 0 - 3 4WS cW S 0 W
WS W S W
V
XT X T X
T
Solving above equation we get
Vb = 8 V
and
Another Method :
Vc = 6 V
Ix = Vc - Vb = 6 - 8 = - 2 A
1
1
...
Fig 2
Now applying KVL in right side mesh we get
Ix + 2 (3Ix + 10) + 1 (3Ix ) = 0
or
or
or
7
...
given below
...
Page 28
Analysis Method
Chap 2
Writing KVL for every mesh, we get
Mesh 1:
10 - 3I1 - 4 ^I1 - I2h = 0
10 - 3I1 - 4I1 + 4I2 = 0
7I1 - 4I2 = 10
Mesh 2:
...
(ii)
- 4I3 - 2I3 - 3 ^I3 - I2h = 0
- 4I3 - 2I3 - 3I3 + 3I2 = 0
3I2 - 9I3 = 0
or
I2 = 3I3
Substituting 3I3 = I2 into eq
...
(iii)
4I1 - 9I2 + I2 = 0
4I1 - 8I2 = 0
or
I1 = 2I2
Substituting I1 = 2I2 into eq
...
(iv)
7 ^2I2h - 4I2 = 10
10I2 = 10
I2 = 1 A
So,
and
Voltage,
8
...
write the node voltage equation and determine
the current across 10 ohm and 20 ohm resistances of the network
...
Writing node equations,
10 = V1 - 0 + V1 - V2
20
6
600 = 3V1 + 10V1 - 10V2
or
13V1 - 10V2 = 600
Similarly,
V2 - V1 + V2 - 0 + V2 - 20 = 0
6
2
10
...
(i) and (ii), we get
V1 = 67
...
71 V
So, current through 20 W resistance
I20 W = V1 = 67
...
37 A
20
20
Current through 10 W resistance,
...
71 = 2
...
State and explain Node-Voltage Theorem
...
RTU 2011
Ans :
NODAL ANALYSIS
Node-Voltage Method
Node voltage method and nodal analysis are similar method
...
Nodal analysis is the most general method for the analysis of
electrical circuits
...
One of the nodes is selected as
a reference node (usually but not necessarily ground), and each of
the other node voltages is referenced to this node
...
In the node voltage method, each branch current is expressed in
terms of one or more node voltages; thus, currents are not explicitly
used into the equations
...
Fig 1 : Branch Current in terms of Node Voltage
Once each branch current is defined in terms of the node
voltages, Kirchhoff’s current law can be applied at each node:
SI = 0
Figure 2 illustrates this procedure
...
However, one of the node
voltages is the reference voltage and is therefore already known,
since it is usually assumed to be zero
...
Nodal analysis provides the minimum number
of equations required to solve the circuit, since any branch voltage
or current may be determined from knowledge of nodal voltages
...
Select a reference node (usually ground)
...
2
...
3
...
4
...
Following the procedure outlined above in the box guarantees
that the correct solution to a given circuit will be found, provided
that the nodes are properly identified and KCL is applied
consistently
...
Page 32
Analysis Method
Chap 2
Fig 3 : Circuit for Nodal Method
We can identify 4 node as shown in Fig 4 and apply nodal
analysis in following steps
1
...
2
...
Here it
must be noted that the node voltage Va is fixed at 10 V
...
3
...
Va = 10 V
Vb - Va Vb - Vc Vb = 0
+
+
2
1
1
Vc - Vb + Vc + 4 = 0
1
1
...
(ii)
...
Once the node voltage are known, current through
any branch of the network can be determined
...
A network excited only by current sources is shown in Fig
...
RTU 2009
Ans :
We take the bottom node as refrence node and apply nodal analysis
as shown below
...
(i)
Page 34
Analysis Method
Chap 2
80 = 8V2 - 8V1 + 2V2 + V2
- 8V1 + 11V2 = 80
Solving eq
...
(ii)
V2 = 400 V
17
So, current I through 2 W resistor
380 - 400
17 = - 20 = - 10 A
I = V1 - V2 = 17
2
2
17 # 2
17
or
I = - 0
...
Fig
...
RTU 2009
Ans :
Given that,
V1 = V2 = V3 = 10 V
R1 = R2 = R3 = R 4 = R5 = R6 = 5 W
Writing KVL for each mesh,
Mesh 1:
V1 - I1 R1 - ^I1 - I3h R 4 - ^I1 - I2h R5 = 0
10 - ^I1 # 5h - ^I1 - I3h 5 - ^I1 - I2h 5 = 0
Chap 2
Analysis Method
Page 35
2 - I1 - I1 + I3 - I1 + I2 = 0
3I1 - I2 - I3 = 2
...
(ii)
Mesh 3:
V3 - I3 R3 - ^I3 - I2h R6 - ^I3 - I1h R 4 = 0
10 - 5I3 - ^I3 - I2h 5 - ^I3 - I1h 5 = 0
2 - I3 - I3 + I2 - I3 + I1 = 0
- I1 - I2 + 3I3 = 2
Substracting eq
...
(i), we get
...
(iii) from eq
...
(iv)
4I2 - 4I3 = 0
or
I2 = I3
From eq
...
(v)
I1 = I2 = I3
So, from eq
...
Find the voltage across the 20 W resistance as shown in Fig
...
RU 2005
Page 36
Analysis Method
Chap 2
Ans :
NODAL METHOD:
We assign the nodes a and b as shown below and take the bottom
node as refrence
...
(i)
V2 - V1 + 2V2 - 80 = 0
- V1 + 3V2 = 80
Current in 20 W resistor,
I1 = V1 - V2
20
Substituting I1 into eq
...
(ii)
19V1 - 3V2 + 6V1 - 6V2 = 120
25V1 - 9V2 = 120
From eq
...
(iii)
Chap 2
Analysis Method
Page 37
Current through 20 W resistor
I1 = V1 - V2 = 1 ;180 - 1060 E = - 0
...
(i)
Mesh 2:
12 - 6I2 - 10 ^I2 - I1h + 20I1 = 0
12 - 16I2 + 30I1 = 0
- 30I1 + 16I2 = 12
- 15I1 + 8I2 = 6
Solving eq
...
788 A
33
I2 = - 0
...
788 A
*******
...
State and explain Thevenin’s theorem with suitable example
...
RTU 2013, 11, 07, RU 2002
or
State and explain Thevenin’s theorem
...
RTU 2014
Ans :
THEVENIN’S THEOREM
Thevenin’s theorem is very important in circuit analysis
...
A large circuit may be replaced by a single
independent voltage source and a single resistor
...
Thevenin’s theorem allows us to find current through or voltage
across any element in the network
...
Fig 1 : Thevenin Equivalent
The following sequence of steps given in outlined box leads to
solve thevenin’s equivalent across any pair of terminals in a network
...
Remove the load i
...
Chap 3
2
...
4
...
Apply any preferred method (KCL, KVL, nodal analysis,
mesh analysis etc
...
The Thevenin voltage is VTh = Voc
Steps to Obtain Thevenin Resistance:
1
...
e voltage sources are short
circuited and current sources are open circuited) and obtain
the equivalent resistance Req seen from terminals
...
e
...
2
...
Short the load terminals and find the
short circuit current Isc through it with all sources present in
the network
...
Now replace the linear circuit with its Thevenin’s equivalent
and find the voltage across or current through that branch
...
Fig 2
Removal of Load Resistance :
We remove the resistor RL = 4 W and mark open circuit voltage as
shown in fig
...
Page 40
Circuit Theorems
Chap 3
Fig 3
Thevenin Voltage :
Now we determine the open circuit voltage using voltage divider
rule as follows
Voc = 6 # 12 = 8 V
3+6
The open circuit voltage is equal to the Thevenin voltage
...
The resistor 3 W and 6 W are in parallel and this
combination is in series with 2 W resistor
...
The current through this resistor is
IL = VTH
= 8 = 1A
4+4
RTH + RL
Chap 3
Circuit Theorems
Page 41
Fig 5
2
...
RTU 2016
Ans :
We consider the effect of each source i
...
, we calculate the current
through 5 W resistor due to individual sources and then add the
currents to get total current
...
Writing KVL in the loop,
6 - 5I1 - 5Vx1 - Vx1 = 0
6 - 5I1 - 6Vx1 = 0
...
(i), we get
6 - 5I1 - 6I1 = 0
or
11I1 = 6
I1 = 6 A
11
Current due to 2 A source:
Now we consider only 2 A current source and replace 6 V source by
short circuit
...
Circuit Theorems
Page 43
Find the Thevenin’s equivalent circuit for the network shown below:
RTU 2015
Ans :
Thevenin Voltage (Open circuit voltage):
To obtain the Thevenin voltage, we remove the load resistance
RL i
...
, open circuit the load terminals
...
Writing mesh equation in mesh 1, we get
V1 - ^I1 # 1h - ^I1 - I2h 1 - aV1 = 0
V1 - I1 - I1 - aV1 = 0
2I1 = ^1 - a h V1
^1 - a h
V1
2
Writing mesh equation in mesh 2, we get
I1 =
aV1 - 1 ^I2 - I1h + bI1 - ^1 # I2h - VTh = 0
aV1 - ^0 - I1h + bI1 - 0 - VTh = 0
aV1 + I1 + bI1 - VTh = 0
VTh = aV1 + I1 ^1 + b h
Substituting I1 from eq
...
(i)
^I2 = 0h
Page 44
Circuit Theorems
Chap 3
^1 - a h
V1
2
VTh = V1 :2a + 1 - a + b - ab D
2
V1 1 + a + b - ab = V
VTh = 6
@
oc
2
VTh = aV1 + ^1 + b h
or
Thevenin resistance :
To determine Thevenin resistance, we have to obtain short circuit
current through load terminals as shown below
...
(ii)
bI1 - 1 # Isc + aV1 - 1 ^Isc - I1h = 0
aV1 + ^b + 1h I1 - 2Isc = 0
From eq
...
(iii), we get
^1 - a h V1 Isc
aV1 + ^b + 1h;
+ E - 2Isc = 0
2
2
I1 =
aV1 +
^b + 1h^1 - a h V1 ^b + 1h
Isc - 2Isc = 0
+
2
2
^b + 1hE
2
V1 :2a + b - ab + 1 - a D = Isc : 4 - b - 1D
2
2
V1 :a + b - ab + 1 - a D = Isc ;2 2
^3 - b h
V1 :1 + a + b - ab D = Isc
2
2
...
Find the current in 5 W resistance using superposition theorem
...
Current due to 2 A source only:
In this case we consider the effect of only 2 A current source and
remove the 5 V source i
...
, replace it by a short circuit
...
To remove a
current source, it will be replaced by an open circuit
...
(i)
Chap 3
Circuit Theorems
Page 47
Substituting Vx2 into eq
...
State and explain superposition theorem
...
RTU 2014
or
State and Explain superposition theorem
...
RTU 2007
Ans :
SUPERPOSITION THEOREM
It states that
The current through, or voltage across, an element in a linear
circuit (having more than one independent source) is equal to the
algebraic sum of the currents or voltages produced independently
by each source acting alone while all other sources are set to zero
...
In order to “zero” a
voltage source, It is replaced with a short circuit, since the voltage
across a short circuit is zero volts
...
An illustration of theorem is shown in the fig 3
...
1 where two
sources are connected in the circuit
...
Page 48
Circuit Theorems
Chap 3
I = I1 + I2
Steps to apply superposition theorem in a linear network are
given in following outlined box :
Steps to Apply Superposition Theorem :
1
...
2
...
3
...
4
...
Superposition theorem can not be applied to power calculations
since power in not a linear quantity, it is proportional to square of
either current or voltage
...
Fig 1 : Illustration of Superposition Theorem
To apply superposition theorem in given circuit, we consider the
effect of each source acting alone and add up all
...
To remove 6 A current source, replace it
by an open circuit and, to remove 8 V voltage source, replace it by
a short circuit as shown below
...
5 A
2+6
Current due to 6 A source only :
Now we consider only 6 A current source acting in the circuit and
remove all other sources i
...
, short circuit 12 V source and 8 V
source
...
5 A
8
2+6
Current due to 8 V source only :
Now we consider the 8 V source only and remove all other sources
i
...
, replace 12 V source by short circuit and 6 A source by open
circuit
...
5 + 4
...
Find the current in the ammeter of the 2 W resistance branch as
shown in fig
...
RTU 2013
Ans :
We obtain Thevenin equivalent across 2 W resistor (load resistance)
...
The circuit can be redrawn in a simplified manner as shown below
...
(i)
V1 - Vb = 10I2
Subtracting eq
...
(ii), we get
...
369 V
271
Thevenin Resistance :
To obtain Thevenin resistance, we remove all the sources as shown
Page 52
Circuit Theorems
Chap 3
below
...
RA =
10 # 1 = 10 W
10 + 1 + 5
16
RB =
5#1 = 5 W
10 + 1 + 5
16
10 # 5 = 50 W
10 + 1 + 5
16
Now we add following series connected resistances
10 + RA = 10 + 10 = 170 W
16
16
RC =
Chap 3
Circuit Theorems
Page 53
6 + RB = 6 + 5 = 101 W
16
16
Equivalent Thevenin resistance
RTh = Rab
= ;170 | | 101E + 50
16
16
16
= 3
...
125 = 7
...
369 = 0
...
62 mA
7
...
Compute the power dissipated in 9 W resistor by applying
superposition in circuit of figure
...
Superposition theorem can not be applied to power calculation
...
Current Due to the 32 V Source :
We remove the 4 A current source as shown in fig 1
...
Va + Va - 32 + Va = 0
6+9
12
4
or
5Va + 15 (Va - 32) + 4Va = 0
or
5Va + 15Va - 480 + 4Va = 0
24Va = 480
Va = 480 = 20 V
24
or
or
Current through 9 W is
I32 V = Va = 20 = 4 A
6+9
15 3
Current Due to the 4 A Source :
We remove the 32 V voltage source as shown in fig 2
...
Circuit is redrawn as shown in fig 3
...
Current through 9 W may
be find using current divider rule as follows :
4
I 4A =
(6 + 3) = 2 A
6+3+9
Current Due to Both Source :
Total current through 9 W resistor is
I9W = I32 V + I 4 A
= 4 + 2 = 10 A
3
3
Power dissipated in 9 W resistor is
10 2
2
P9W = I 9W R = b 3 l 9 = 100 W
8
...
by applying the superposition
theorem
...
Current due to 10 V source only :
We consider that only 10 V source is acting in the circuit and make
20 V source short circuit as shown below
...
(i)
Chap 3
Circuit Theorems
Page 57
Substituting V1 into eq
...
0638 A
Current due to 20 V voltage source only :
In this case we consider only 20 V source acting in the circuit and
remove 10 V source i
...
, replace it by short circuit
...
V1l- V2l + V1l- 20 + V1l- 0 = 0
2
3
10
15V1l- 15V2l+ 10V1l- 200 + 3V1l = 0
28V1l- 15V2l = 200
Similarly,
V2l- 20 + V2l- 20 + V2l- V1l = 0
6
2
1
...
(ii) and (iii), we get
V1l = 17
...
23 A
...
44 = 1
...
I = I1 + I2 = 1
...
0638 = 1
...
at
terminals x and y
...
RTU 2012, RU 2006
Ans :
Thevenin voltage :
To obtain Thevenin voltage, we remove the load and make load
terminals open circuit as shown
...
Chap 3
Circuit Theorems
RA =
11
8
11
8
#1
= 11 W
43
+1+3
11
8
#3
=
+1+3
RC = 11 3 # 1 =
8 +1+3
Now, further simplifying the circuit
RB =
Voc = VTh =
11
8
11
43
Page 59
110
43
+ 110
43
33 W
43
24 W
43
# 11 = 10 V
Thevenin Resistance :
To obtain Thevenin resistance, we make all sources zero
...
Consider circuit of
Fig
...
So,
Current through 9 W resistor
IL = VTh
= 10 = 1 A
1+9
RTh + RL
10
...
RTU 2010
Ans :
Chap 3
Circuit Theorems
Page 61
To apply superposition, we consider the effect of each source acting
alone in the circuit and then add the currents or voltages
...
e
...
Now, we transform the D -network compromising 8 W , 4 W and 5 W
resistors into equivalent Y -network as shown
...
870 W
17 17
10
10
= 1
...
870
5 + 1
...
870
5+
17
From current division in circuit of Fig
...
2597 # 1
...
335 A
Current due to 5 V source only :
Now, we consider only 5 V source acting in the circuit and replace
10 V source by short circuit
...
35 W
9
Page 63
Page 64
Circuit Theorems
Chap 3
5
= 1
...
35 + 1
So, total current through 1 W resistor
I1m =
I1 = I1l+ I1m = 0
...
149 = 1
...
Find the current in the 3 ohm resistor of the circuit of Fig
...
RTU 2009
Ans :
Removal of Load Resistance :
We remove the load resistance and mark open circuit voltage as
shown in fig 1
...
Therefore
VTH = Voc = Va = 36 V
Thevenin Resistance :
We remove both source as shown in fig 2
...
Thus
Fig 2
RTH = 12 | | 6 = 12 # 6 = 4 W
12 + 6
Load Current :
We draw the thevenin equivalent circuit and put load resistor as
shown in fig 3
...
Calculate the current in the 4 W resistor of Fig
...
RTU 2008
Page 66
Circuit Theorems
Chap 3
Ans :
We consider the effect of each source acting alone in the circuit and
then add all the results
...
e
...
So,
I1 = 12 = 1 A
4+8
Due to 6 A source only :
Now we consider only 6 A source and replace both the voltage
sources by short circuit
...
e
...
Chap 3
Circuit Theorems
Page 67
I3 = - 6 = - 6 = - 0
...
5 = - 3
...
Transform the current sources of circuit of Fig
...
RTU 2008
Ans :
The given circuit is as shown below
...
Obtain the current in 10 W resistor using superposition Theorem
...
Due to 50 V source only :
Here we consider only 50 V voltage source acting alone in the circuit
Chap 3
Circuit Theorems
Page 69
and remove the 5 A current source i
...
, replace it by an open circuit
...
e
...
Page 70
Circuit Theorems
Chap 3
Using current division
20
-5 =- 2 A
30 + 20 ^ h
Total current through 10 W resistor
I2 =
I = I1 + I2 = 1 - 2 = - 1 A
15
...
determine the current I through
the 3 W resistor by application of :
(i) Superposition principle
(ii) Thevenin’s theorem
RU 2001
or
Determine the current through 3 ohm resistor in the circuit as
shown in Fig
...
RU 2001
Ans :
Chap 3
Circuit Theorems
Page 71
(a) Super Position Theorem
We determine the effect of every source and then add to get the
total effect as per superposition theorem
...
Writing KCL equation for this node
we get
Va - 10 Va
V
+ + a =0
2
2 1+3
2Va - 20 + 2Va + Va = 0
or
Va = 4 V
The current through 3 W may be get by ohms law as follows
I10V = Va = 1 A
4
Current Due to 10 A Source :
We remove the 10 V voltage source as shown as shown in fig 2
...
So we simplify
this circuit as shown in fig 2
...
Thus
current through 3 W resistor is
I = I10V + I10A = 1 + 2 = 3 A
(b) Thevenin Theorem
Removal of Load Resistor :
We remove the load resistor and mark open circuit voltage as shown
in fig 3
...
Writing KCL for
this node we have
Va - 10 + Va = 10
2
2
Va - 10 + Va = 20 or Va = 15 V
The open circuit voltage or thevenin equivalent voltage is equal to
node voltage Va because there is no voltage drop across 1 W resistor
...
Fig 4
In right side of this circuit two 2 W are parallel and this combination
is in series with 1 W resistor
...
Current through the load resistor is
IL = VTH
= 15 = 3 A
2+3
RTH + RL
Fig 5
*******
Page 74
AC Circuits Basics
Chap 4
CHAPTER 4
AC CIRCUITS BASICS
1
...
The effective or rms value of alternating current is that steady
current (dc) which when flowing through a given resistance for a
given time produces the same amount of heat as produced by the
alternating current when flowing through the same resistance for
the same time
...
The
average of this power over one cycle, as follows
...
Hence the effective value of an ac wave is the
square root of the mean of the squared function
...
or
Ieff =
#
Since it is the effective value of voltage and current which is used
in everyday usage, the symbols V and I (without any subscript) are
taken to mean the effective values
...
Let us find the rms value of the sinusoidal waveform shown in
fig 1
...
Hence, the rms value
is given as
Chap 4
AC Circuits Basics
Ieff =
1
2p
#
2p
=
2
Im
4p
#
Page 75
2p
i2 d (wt) =
0
1
2p
#
0
2p
2
I m sin2 wtd (wt)
(1 - cos 2wt) d (wt)
0
2
I m wt sin 2wt 2p = Im
2 D0
4p :
2
The rms value is always greater than the average value, except for
a rectangular wave
...
=
Fig 1
(ii) Form Factor and Peak Factor
The ratio of the effective value to the average value is known as
the form factor of a waveform of any shape (sinusoidal or nonsinusoidal)
...
the peakier the wave, the greater is the form factor and vice versa
...
Hence
the former has greater form factor 1
...
Similarly,
a triangular wave is more peaky than a sine wave and has a form
factor of 1
...
(iii) Peak Factor
The peak factor or crest factor or amplitude factor of a waveform
is defined as the ratio of its peak (or maximum) value to its rms
Page 76
AC Circuits Basics
Chap 4
value
...
707Vm = 1
...
637Vm
2Vm /p
and
K p = Vm = Vm
= 2 = 1
...
For instance, when an alternating voltage is applied
across a cable or capacitor, the breakdown of insulation will
depend upon the maximum voltage
...
(iv) Phase
Phase is the fraction of the time-period or cycle that has elapsed
since it last passed from the chosen zero positive or origin
...
Phase angle is equivalent of phase expressed in radians or
p
degrees
...
Thus, phase angle, f = 2T t
...
Fig 2 : Sinusoidal Waveform
Chap 4
2
...
36 26
...
36 sin ^440t + 26
...
36 = 15
...
AC Circuits Basics
Chap 4
Find the angle by which i2 lags i1 if
i1 = 120 cos ]100pt - 30cg and
(i)
(ii)
(iii)
(iv)
i2 = - 8 cos ^100pt + 20ch
i2 = 5 sin ^100pt + 50ch
i2 = - 6 sin ^100pt - 30ch
i2 = 2 sin ^100pt - 20ch + 2 cos ^100pt - 20ch
RTU 2015, RU 2002
Ans :
For comparing two quantities the frequency, sign and the sin/cos
must be same, so for simplify convert the i1 in sine term
i1 = 120 cos ^100pt - 30ch
6sin ^90c + qh = cos q@
= 120 sin ^90c + 100pt - 30ch
= 120 sin ^100pt + 60ch
(i) i2 = - 8 cos ^100pt + 20ch
The term is cos, so converting it into sine term
= - 8 sin ^90c + 100pt + 20ch
6sin ^90c + qh = cos q@
= - 8 sin ^100pt + 110ch
The sign must be positive, so
6sin ^180c + qh = - sin q@
100pt - 70ch
= 8 sin ^100pt + 290ch = 8 sin ^
Now for angle by which i2 lags i1
i2 = 8 sin ^100pt + 110c + 180ch
angle = phase angle of i1 - phase angle of i2
= 60c - ^- 70ch = 130c
So, i2 lags i1 by 130c
(ii) i2 = 5 sin ^100pt + 50ch
So the angle by which i2 lags i1 is
= 60c - 50c = 10c
i
...
,
(iii) i2 = - 6 sin ^100pt - 30ch
First the negative sign must be removed, so
i2 lags i1 by 10c
i2 = 6 sin ^100pt - 30c + 180ch 6sin ^180c + qh = - sin q@
= 6 sin ^100pt + 150ch
The angle of which i2 lags i1 is
= 60c - 150c = - 90c
So, i2 lags i1 by - 90c or i2 leads i1 by 90c
(iv) i2 = 2 sin ^100pt - 20ch + 2 cos ^100pt - 20ch
Chap 4
AC Circuits Basics
Page 79
= 2 sin ^100pt - 20ch + 2 sin ^90c + 100pt - 20ch
= 2 sin ^100pt - 20ch + 2 sin ^100pt + 70ch
The angle by which i2 lags i1 is
= 60c - ^20ch + 60c^- 70ch
= 80c - 10c = 70c
4
...
A voltage wave has variation as shown below:
Find the average and RMS value (effective value) of voltage wave
shown in figure
...
25 V
4
4
(b) Effective or rms value
Vrms =
=
Area under squared half period
Half period
#
0
T
v2 ^ t h dt
T/2
=
v ^ t h dt = 1
2
#
0
v ^ t h dt
4 2
4
^2t h2 dt +
^8h2 dt
=1
2
#
0
=1
2
3 3
4
4 ;t E + 64 6t @3 = 1 4 ^9 - 0h + 64 ^4 - 3h
3 0
2
4 2
#
0
3
#
3
4
= 1 36 + 64 = 1 100 = 5 V
2
2
5
...
Determine the from factor of sine
wave
...
Thus,
Chap 4
AC Circuits Basics
Page 81
the average value of a sinusoidal current i shown in fig
...
p
Area under half cycle
Iav =
id (wt)
=1
p 0
Length of half cycle
#
=1
p
#
0
p
p
Im sin wtd (wt) = Im [- cos wt]0
p
= Im [- cos p + cos 0c] = 2Im = 0
...
The period of sinusoidal waveform is 2p
...
Thus,
Form factor,
K f = Irms
Iav
For a sinusoidal voltage wave form,
I / 2
= 0
...
11
K f = Irms = m
Iav
0
...
A periodic voltage wave from has been shown in figure
...
(iii) RMS Value
(iv) Average Value
...
Therefore frequency
1
f = 1 =
= 1-1
T
10
^100 # 10-3h
f = 10 cycles/sec
Equation of the wave
for 0 # t # 100 msec
v = 10
100t
Wave is unsymmetrical (Half wave areas are not equal), therefore
average value is
100
100 10
Vav = 1 # V dt = 1 #
t dt
100 0
100 0 100
=
10
10000 - 0E = 10
;
2
2
10000 #
Vav = 5 Volts
Square of the rms value
^Vrms h2 = 1
100
Vrms
7
...
773 Volts
3
Explain the generation of single phase AC voltage & derive it EMF
equation
...
This
alternator consists of permanent magnet with two poles N and S,
and a single turn rectangular coil
...
These conductor are connected to
each other on one ends A'B'
...
The coil can rotate around
its own axis in clockwise or anti-clockwise directions
...
The brushes do not rotate
...
A load is connected between
the two brushes as shown in fig 1
...
Due to rotation, the conductor A
and B cut the magnetic lines of flux produced by the permanent
magnet
...
Due to this induced
emf, current flows through the load
...
The shape of induced emf is sinusoidal as shown in
fig 2
...
Here q is the angle between the line of magnetic flux and
normal to area
...
e
...
e the plane of the coil is parallel to the field (fig 1b),
2
e = Vm sin p = Vm
2
When wt = p ,
When wt =
3p
2
,
e = Vm sin p = 0
e = Vm sin
3p
2
= - Vm
(fig 1c)
(fig 1d)
Page 84
AC Circuits Basics
When wt = 2p ,
e = Vm sin 2p = 0
Chap 4
(fig 1a)
Fig (a) 0c Position : Coil sides move parallel to flux lines
...
Fig (b) 90c Position : Coil end A is positive with respect to B
...
Fig (c) 180c Position : Coil again cutting no flux
...
Fig (d) 270c Position : Voltage polarity has reversed
...
Fig 1
Chap 4
AC Circuits Basics
Page 85
Fig 2
8
...
RTU 2013
Ans :
The given waveform is half-wave rectified sinusoidal voltage wave
form
...
(a) Average Value
Area Under 1 Cycle
Period
The expression for given waveform is
Vav =
Vm sin t for 0 < t < p
v (t) = )
0
for p < t < 2p
The area under 1 cycle is
Area =
#
2p
v (t) dt =
p
#V
sin t dt
= Vm
2p
#
0
0
0
m
Thus average value is
p
Vav =
#
0
p
#V
0
m
sin tdt
2p
p
sin qdq = [- cos q] 0
p
sin tdt = 2Vm = Vm
p
2p
Page 86
AC Circuits Basics
Chap 4
= [cos q] 0 = [1 - (- 1)] = 2
p
(b) Effective Value i
...
rms
The effective rms value is
#
Vrms =
T
0
#
=
0
v2 (t) dt
T
=
#
0
2p
v2 (t) dt
2p
p
V 2m sin2 tdt + 0
2p
= Vm
2p
#
p
sin2 tdt
0
p = Vm
= Vm
2
2
2p
p 1 - cos 2t
p
# sin2 tdt = # b 2 ldt
0
0
=1
2
#
0
p
p
(1 - cos 2t) dt = 1 :t - sin 2t D
2
2 0
= 1 6p + 0 - 0 - 0@ = p
2
2
#
0
9
...
RTU 2013
Ans :
Refer to Q
...
Find the average and RMS value of the waveform shown in Fig
...
The period of the waveform
is 1 and half period is 0
...
The mathematical expression is
i ^ t h = 10t - 5 ,
0 1 t 1 1 sec
Average value,
Iav =
#
0
T/2
i ^ t h dt
T/2
= 1
0
...
5
=
#
0
0
...
5
i ^ t h dt = 2 #
0
0
...
5
2
= 2 ;10 t - 5tE = 2 65t2 - 5t @0
...
5
0
0
2
0
= 10 8^0
...
5B = 10 60
...
5@ = - 2
...
5 A
RMS value is given by
#
0
I rms =
1
0
...
5
i2 ^ t h dt
T/2
=
#
0
0
...
5
^100t2 - 100t + 25h dt
=
1 100 t3 - 100 t2 + 25t 0
...
5 ;
0
=
1 ;100 0
...
5 2 + 25 0
...
5 3 ^ h
=
1 4
...
5 + 12
...
5 6
= 2
...
In the circuit shown in Fig
...
Draw the phases diagram
...
15 W inductor & j15 W
0
...
02
= j ^2 # p # 50 # 0
...
28 W
400 mF capacitor & 1
j wC
1
j # 2pf # 400 # 10-6
1
= - j 7
...
Impedances of the branches
&
Z1 = 10 + j 15 W = 18
...
30c W
Z2 = 5 + j 6
...
02 51
...
95 W = 12
...
48c W
Chap 5
AC Circuits
Page 89
Now, total impedance of the circuit is
ZZ
Z = Z1 + Z2 | | Z3 = Z1 + 2 3
Z2 + Z3
= 10 + j 15 +
8
...
47c # 12
...
48c
5 + j 6
...
95
= 10 + j 15 +
102
...
99c
102
...
99c
= 10 + j 15 +
15 - j 1
...
09 - 6
...
78 19
...
39 + j 2
...
39 + j 17
...
78 46
...
8 46
...
78 19
...
8 19
...
8 19
...
02 51
...
45 - 32
...
8 19
...
30 57
...
77 - 38
...
02 56
...
2 56
...
AC Circuits
Chap 5
1
A coil having a resistance of 5W and an inductance of ^ p h Henry is
100
connected in series with a capacitor of ^ p h microfarad A 200 V,
50 Hz alternating voltage source (sinusoidal) is applied across the
circuit
...
(ii) Find the current flowing through the voltage source
...
(iv) Draw the phasor diagram
...
A series RL circuit has resistance and reactance of 15 ohm and
10 ohm respectively
...
RTU 2014, 2008
Page 92
AC Circuits
Chap 5
Ans :
R = 15 W
Given that
XL = 10 W
Total impedance of the circuit
Z = - jXC | | ^R + jXL h =
- jXC ^R + jXL h
- jXC + R + jXL
- jXC ^15 + j10h
10XC - j15XC
=
15 + j ^10 - XC h
15 + j ^10 - XC h
15 - j ^10 - XC h
10XC - j15XC
Z=
#
15 + j ^10 - XC h 15 - j ^10 - XC h
150XC - j10XC ^10 - XC h - j 225XC + 15XC ^10 - XC h
=
^15h2 + ^10 - XC h2
150XC + 15XC ^10 - XC h j 710XC ^10 - XC h + 225XC A
=
^15h2 + ^10 - XC h2
^15h2 + ^10 - XC h2
For unity power factor
=
Im ^Z h = 0
10XC ^10 - XC h + 225XC = 0
XC 710 ^10 - XC h + 225A = 0
10 ^10 - XC h + 225 = 0
10 ^XC - 10h = 225
Thus
XC - 10 = 22
...
5 W
1 = 32
...
5 # 2pf
1
32
...
14 # 50
= 97
...
AC Circuits
Page 93
Two coil A and B are connected in series across a 240 volt, 50 Hz
supply
...
015 H
...
Find
the inductance of A and the resistance of B
...
RTU 2014, RU 2002
Ans :
Given circuit is shown below
We have,
Active power in the network
P = 3 kW
Reactive power,
Q = 2 kVAR
If f is the power factor angle, then from power triangle we know
that
Q
tan f =
P
tan f = 2 = 0
...
69c
Real power is given as
P = VI cos f
3000 = 240 # I # cos ^33
...
02 A
This is the current in the circuit
...
Z = V = 240 = 15
...
02
I
In phasor form
Z = 15
...
98 33
...
3 + j 8
...
3 = RA + RB
So,
RB = 13
...
3 - 5 = 8
...
015
= 4
...
86 W
XA + XB = 8
...
86 - XB = 8
...
71
XA = 4
...
15
LA = 4
...
15 = 0
...
15h2 = 6
...
3h2 + ^4
...
54 W
VA = ZA # I
VA = 6
...
02 = 97
...
54 # 15
...
29 V
5
...
(i)
A resistance of 500 W
(ii) An inductance of 2 H
(iii) A capacitance of 10 mF
A source voltage of 200 V , 50 Hz is applied, determine the total
current drawn from the supply complex power, active power,
reactive power and power factors of the circuit
...
31 W
1
XC = 1 =
2pfC
2p # 50 # 10 # 10-6
= 318
...
Total Admittance of the circuit
Y = 1+ 1 + 1
R jXL - jXC
1
1
= 1 +
+
500 j 628
...
30
= 2 # 10-3 - j ^1
...
14 # 10-3h
= ^2 - j1
...
14h # 10-3 mho
= ^2 + j1
...
55h # 10-3
1
=
2
...
77c # 10-3
Z = 395
...
77c W
Total current in the circuit
200 0c
I =V =
= 0
...
77c A
Z
395
...
77c
Complex power
S = V I * = 200 0c # 0
...
77c
= 100 - 37
...
50 cos ^37
...
50 sin ^37
...
24 W
Power factor = cos f = cos ^37
...
79
6
...
below, determine :
(i)
The value of resistance, inductance and capacitance
(ii) Power factor of the circuit
RTU 2012, RU 2006, 2001
Ans :
Let the voltage across R , L and C be VR , VL and VC respectively
...
(i), we get
2
2
2
V R + VC = V R + ^VL - VC h2
or
2
2
2
V R + VC = V R + ^VL - VC h2
2
VC = ^VL - VC h2
VC = VL - VC
or
2VC = VL
VC = VL = 40 = 20 V
2
2
Chap 5
AC Circuits
Page 97
From Eq
...
61 V
Now, we have
VR = I # R = 82
...
61
R = 82
...
52 W
5
VL = I # 2pf L = 40
5 # 2p # 50 # L = 40
40
L=
= 25
...
95 # 10-4 F = 0
...
52 + j 6XL - XC @
2pfC
V
= 16
...
52 + j ; 40 - 20 E
5
5
I
I
= 16
...
52 + j 4 W
or
Z = 17 13
...
6c
Power factor = cos f = cos ^13
...
97 (Lagging)
7
...
When there is complex impedance in a circuit, part of the energy
is alternately stored and returned by the reactive part, and part of
it is dissipated by the resistance
...
Consider a circuit having complex impedance
...
Then the power at any instant of time is
P ^ t h = v ^ t h i ^ t h = Vm cos wt Im cos ^wt + qh
From eq
...
...
(ii)
Average Power
To find the average value of any power function, we have to take
a particular time interval from t1 to t2 ; by integrating the function
from t1 to t2 and dividing the result by the time interval t2 - t1 , we
get the average power
...
(iii)
Average power
P = 1 # P ^ t h dt
t2 - t1 t
In general, the average value over one cycle is
T
...
(iv) over one
cycle, we get average power
T
Pav = 1 # 'Vm Im 7cos ^2wt + qh + cos qA dt 1
2
T 0
2
1
T
T
...
(v), the first term becomes zero, and the second term remains
...
(vi)
Pav = Vm Im cos q W
2
We can write eq
...
(vii)
Chap 5
AC Circuits
Page 99
In eq
...
Thus
Pav = Veff I eff cos q
To get average power, we have to take the product of the effective
values of both voltage and current multiplied by cosine of the phase
angle between voltage and the current
...
Hence, the average power is
2
2
Pav = 1 Vm Im = 1 I m R = I eff R
2
2
If we consider a purely reactive circuit (i
...
, purely capacitive or
purely inductive), the phase angle between voltage and current is
90c
...
If the circuit contains complex impedance, the average power is
the power dissipated in the resistive part only
...
A 100 W resistance is connected in series with a choke coil, when
a 440 V , 50 Hz single phase alternating voltage is applied to this
combination the voltage across the resistance and the choke coil are
200 V and 300 V respectively
...
(ii) The total resistance and impedance of the circuit
...
We have,
Voltage across resistance, VR = 200 V
Voltage across coil,
Vcoil = 300 V
So, current in the circuit
Page 100
AC Circuits
I =
Chap 5
VR
R
I = 200 = 2 A
100
Let
VRC = Voltage across coil resistance
VLC = Voltage across coil inductance
We can write
2
2
Vcoil = V RC + V LC
2
2
^300h2 = V RC + V LC
Total voltage in the circuit
2
^VR + VRC h2 + V LC
2
^440h2 = ^VR + VRC h2 + V LC
Substracting eq
...
(ii),
...
(ii)
2
2
2
^440h2 - ^300h2 = ^VR + VRC h2 + V LC - V RC - V LC
2
103600 = ^VR + VRC h2 - V RC
2
2
2
103600 = V R + 2VR VRC + V RC - V RC
2
103600 = V R + 2VR VRC
103600 = ^200h2 + 2 ^200h VRC
63600 = 2 # 200 # VRC
or
VRC = 63600 = 159 V
2 # 200
This is the voltage across the coil resistance
...
(i), we get
or
2
^300h2 = ^159h2 + V LC
2
V LC = ^300h2 - ^159h2 = 64719
or
VLC = 254
...
Now, we have
VRC = I # RC = 159
RC = 159 = 159
2
I
Coil resistance,
So,
RC = 79
...
4 W
XL = 254
...
4 = 127
...
2
127
...
40 H
(ii) Total resistance of circuit
RT = R + RC = 100 + 79
...
5
Total impedance of circuit
Z = RT + jXL = 179
...
2 W
or
Z = ^179
...
2h2 = 220 W
(iii) Power absorbed by coil
PC = I 2 # RC = ^2 h2 # 79
...
5 = 718 W
9
...
02 H is connected in
series with another coil of resistance 6 W and inductance 15 mH
across a 230 V, 50 Hz supply
...
(ii) Voltage drop across each coil
...
RTU 2010
Ans :
First coil,
Resistance,
R1 = 10 W
Inductance,
L1 = 0
...
02
XL1 = 6
...
283 W
Second coil,
Resistance,
R2 = 6 W
Inductance,
L2 = 15 mH = 0
...
015
XL2 = 4
...
712 W
(i) Impedance of the circuit
Z = Z1 + Z2 = 10 + j 6
...
712
= 16 + j10
...
41 34
...
85 A
19
...
85 ^10h2 + ^6
...
95 V
Voltage drop in second coil,
V2 = I # Z2 = 11
...
712h2
V2 = 90
...
85h2 ^10 + 6h
= 2246
...
In a series RC circuit, the value of R = 10 W and C = 25 nano farad
...
5 volt
...
RTU 2009
Ans :
We have
R = 10 W
1
XC = 1 =
2pfC
2p # 50 # 25 # 10-9
= 127
...
5
I # XC = 2
...
5
= 0
...
32 # 103
Voltage across resistance
I =
VR = I # R
Resistance,
Reactance,
or
R = 10 W
1
XC = 1 =
2pfC
2 # p # 50 # 106 # 25 # 10-9
XC = 0
...
5 V
I # XC = 2
...
5 = 19
...
127
So, voltage across resistance
VR = I # R = 19
...
8 V
Total voltage across series combination
2
2
V = V R + VC =
^196
...
5h2 = 196
...
A 50 Hz sinusoidal voltage V = 311 sin wt is applied to RL series
circuit
...
02 Henry
...
(ii) Obtain expression for instantaneous current
...
02 H
XL = 2pf L = 2p # 50 # 0
...
28 W
Impedance of the circuit
Z = R + jXL = 5 + j 6
...
47c W
(i) Current in the circuit
311 0c
I =V =
= 38
...
47c A
Z
8 51
...
875 = 27
...
47c
(ii) Instantaneous current,
I = 38
...
47c
i ^ t h = 38
...
47ch A
*******
Chap 6
Three Phase Circuits
Page 105
CHAPTER 6
THREE PHASE CIRCUITS
1
...
RTU 2012, 2010
or
Explain two wattmeter method of measuring power and power
factor
...
RTU 2007, RU 2006, 2002
Ans :
Principle of Working
The two-wattmeter method is the most commonly used method
for three-phase power measurement
...
Fig 1 : Two-wattmeter method for measuring three-phase power
Page 106
Three Phase Circuits
Chap 6
Notice that the current coil of each wattmeter measures the line
current, while the respective voltage coil is connected between the
line and the third line and measures the line voltage
...
Although the
individual wattermeters no longer read the power taken by any
particular phase, the algebraic sum of the two wattmeter readings
equals the total average power absorbed by the load, regardless of
whether it is wye- or delta-connected, balanced or unbalanced
...
(i)
PT = P1 + P2
We will show here that the method works for a balanced threephase system
...
Our objective
is to apply the two-wattmeter method to find the average power
absorbed by the load
...
Due to the load impedance, each
voltage coil leads its current coil by f , so that the power factor
is cos f
...
Thus, the total phase difference between the
phase current Ia and line voltage Vab is f + 30c, and the average
power read by wattmeter W1 is
P1 = Re 6 ab I a@
V *
= Vab Ia cos ^f = 30ch = VL IL cos ^f + 30ch
...
(iii)
V
We now use the trigonometric identities
cos ^A + B h = cos A cos B - sin A sin B
...
(ii) and (iii):
P1 + P2 = VL IL 7cos ^f + 30ch + cos ^f - 30chA
= VL IL ^cos f cos 30c - sin f sin 30c
+ cos f cos 30c + sin f sin 30ch
...
Eq (v) shows that the sum of the wattmeter readings gives the total
average power,
PT = P1 + P2
...
(vii) with Eq
...
(ix)
QT = 3 ^P2 - P1h
From Eqs
...
(x)
ST = PT + QT
Dividing Eq
...
(vi) gives the tangent of the power factor
angle as
Q
...
Thus, the
two-wattmeter method not only provides the total real and reactive
powers, it can also be used to compute the power factor
...
(vi), (ix) and (xi), we conclude that:
1
...
2
...
3
...
Although these results are derived from a balanced wye-connected
load, they are equally valid for a balanced delta-connected load
...
We use the three-wattmeter method
to measure the real power in a three-phase four-wire system
...
Draw the explain phasor diagram to show relationship between
phase/line voltages and currents for a star-connected lagging
load connected to a balanced 3 phase voltage source
...
RTU 2015
Ans :
STAR OR WYE (Y) CONNECTIONS
In star or wye (Y) connection, the similar ends (either start or
finish) of three windings are connected to a common point called
star or neutral point
...
Fig 1 : Star Connection
As shown in fig 1, the finish terminals R', Y ' and B ' of the three
Chap 6
Three Phase Circuits
Page 109
windings are connected to form a star or neutral point
...
The current flowing through each phases is called phase
current I ph and current flowing through each conductor is called
line current IL
...
Relation between Phase Voltages and Line Voltages
The star connection is shown in fig 2a where we have shown all
voltage and current
...
Their phasor are shown in fig
2b and 2c
...
So, we can write
VRN = VYN = VBN = Vph
The three phasor representing the set of phase are
VRN = Vph 0c
VYN = Vph - 120c
and
VBN = Vph - 240c
Now,
VRY = VRN - VYN
= Vph 0c - Vph - 120c
= Vph - Vph (cos 120c - j sin 120c)
= Vph - Vph c- 1 - j 3 m
2
2
= Vph c 3 + j 3 m = Vph 3 30c
2
2
Thus, the magnitude of VRY is is 3 Vph and the phase angle with
respect to the reference phasor VRN is f = 30c
Similarly, we can get
VYB = 3 Vph - 90c and VBR = 3 Vph 150c
Thus, the magnitude of the line voltage VL for star connection is
given as
VL =
3 Vph
Relations Between Phase Current and Line Current
From fig
...
Thus
where,
IR = IRN , IY = IYN and IB = IBN
IRN = IYN = IBN = I ph (phase current)
and
IR = IY = IB = IL (Line current)
Hence, in star connections line current is equal to phase current
...
All the phase currents are displaced by
120c with respect to each other, ‘f ’ is the phase angle between
phase voltage and phase current (lagging load is assumed)
...
3 that the angle between the line
Chap 6
Three Phase Circuits
Page 111
(phase) current and the corresponding line voltage is ^30 + fh c for
a lagging load
...
Fig 3
3
...
If the supply voltage is balanced 440 volt, find
the true power drawn by the load, the power factor and line current
...
991 or f = - 71
...
51ch = 0
...
317
11 # 103
3 # 440 # 0
...
53 A
So, value of line current is 45
...
4
...
8
...
9 lag
...
8
For a star connected system, total average power is given by
P=
So,
3 VL IL cos f
3
50 # 10 =
3 # 1000 # IL # 0
...
8
IL = 36
...
I ph = IL = 36
...
08 # 0
...
86 A
Reactive component of current
= IL sin f = IL 1 - cos2 f = 36
...
8h2
= 36
...
6 = 21
...
9 new output power would be
Pnew =
or
5
...
08 # 0
...
24 kW
Derive with neat circuit and phasor diagram, the relationships
between the phase and line voltages and currents for a balanced
three phase delta- connected load, and hence deduce an expression
for the total three phase power in terms of the line voltage, the line
Chap 6
Three Phase Circuits
Page 113
current and the power factor
...
The three line conductors are run from three
junctions of the mesh
...
The three conductors R,Y and B are run from the three junction
called line conductor
...
Similarly, voltage across each phase is called phase voltage Vph
and voltage across two line conductors is called line voltages VL
...
Let IRR', IYY' and IBB' , be the rms value of the
phase currents in the three windings of the generator
...
Therefore, we can write
(say)
IRR' = IYY' = IBB' = I ph
The three phasor representing the set of phase currents can be
written as
IRR' = I ph 0c ,
IYY' = I ph - 120c
IBB' = I ph - 240c = I ph 120c
From 2a it can be seen that the phase current IRR' flows away
from the line conductor R , whereas the phase current IBB' flows
towards it
...
Similarly, we can get
IY =
3 I ph - 150c and IB =
3 I ph - 270c
Thus, the magnitude of the line current IL for delta connection is
given as
IL =
3 I ph
Relation Between Phase Voltage and Line voltages
From fig 2a, it is clear that voltage across terminals RR' is the same
as across terminals R and Y
...
POWER IN 3-PHASE SYSTEM
Consider one phase only
...
The average active power consumed by this load is
given by
P1 = Vph I ph cos f
where, f is the phase angle of the load
...
Hence, the total power consumed is
P = 3P1 = 3Vph I ph cos f
For a star-connected system, we have VL =
...
Hence,
3 I ph
Page 116
Three Phase Circuits
P = 3VL (IL / 3 ) cos f =
Chap 6
3 VL IL cos f
Thus, it follows that, for any balanced load (current in either Y
or D ), the total power is given as
P = 3 VL IL cos f
While using above equation, it is important to note that f is the
angle of the load impedance per phase and not the angle between
VL and IL
...
A delta load of ZAB = 52 45c W , ZBC = 52 - 30c W and ZCA = 10 0c W
are connected to a 230 volt, 3-phase source
...
RTU 2009
Ans :
The circuit is shown as below
...
42 - 45c = 3
...
12 A
52 45c
230 120c
IBC =
= 4
...
42 A
52 - 30c
230 120c
ICA =
= 23 120c = - 11
...
92 A
10 0c
IAB =
IA = IAB - ICA = 3
...
12 + 11
...
92 = 14
...
04
IB = IBC - IAB = - j 4
...
12 + j 3
...
12 + j1
...
5 + j19
...
42 = - 11
...
34
Chap 6
7
...
The power factor of
the motor is 0
...
If the two wattmeter method is used to
measure the three phase power supplied to the motor
...
47
f = cos-1 ^0
...
96c
So,
PT = P1 + P2 =
Real power,
So,
P1 + P2 =
3 VL IL cos f
3 # 450 # 24 # 0
...
88 W
Reactive power,
QT = P2 - P1 =
...
96ch
P2 - P1 =
So,
3 # 450 # 24 # 0
...
82
Adding eq
...
(i)
...
88 + 16498
...
7
P2 = 25290
...
35
2
P2 = 12
...
79 - 12
...
85 kW
8
...
54 + j3
...
Determine the line currents and draw the phasor diagram
...
Thus,
V
110 0c
Phase current,
I ph = ph =
Z ph
5 45c
= 21 - 45c A
Line current,
IL =
3 I ph =
3 621 - 45c@
IL = 36
...
The phasor diagram is shown below
...
Three Phase Circuits
Page 119
Three 20 W resistor are connected in (i) star (ii) delta across a
415 V and currents and the power taken from main in each case
...
RTU 2007, RU 2005
Ans :
(i) Star Connection
Phase voltage
Phase current
VL = 415
VPh = 415 = 239
...
6 = 11
...
(ii) Delta connection
Phase voltage
Phase current
VPh = 415
VPh = VL (for delta connection)
IPh = VPh = VL = 415 = 20
...
94 Amp
Power Taken from Source in Case
(i) Star Connection
[ f = 0c (power factor)]
P=
So
3 VL IL cos f
P=
3 # 415 # 11
...
24 W = 8
...
91 # 1
= 25833
...
83 kW
When One of the Resistor is Disconnected
(i) Star Connection
The circuit no longer remains a 3-phase circuit but consists of two
20 W resistors in series across a 415 V supply
...
375 Amp
20 + 20
Power absorbed in both resistors
= 415 # 10
...
31 kW
Hence by disconnecting one resistor, the power consumption is
reduced by half
...
75 Amp
...
75h2 # 20
= 17
...
*******
Page 122
Transformer
Chap 7
CHAPTER 7
TRANSFORMER
1
...
RTU 2016
or
Explain working principle and emf equation of a single phase
transformer
...
RU 2006
Ans :
A transformer is an electrical device, having no moving parts,
which by electromagnetic induction transfer electric energy form
one circuit to another circuit at the same frequency, usually with
changed value of voltage and current
...
Can the transformer operate on dc ?
The answer is no
...
As there is no change in the
flux linkage with the secondary winding, the induced emf in the
secondary winding is zero
...
Most of these fluxes is linked with the other coil and produces
mutually induced emf in that coil
...
The first coil, in which electrical energy is fed from
AC supply is called primary and the second coil from which the
power is obtained is called secondary
...
The thick
line denotes the iron core
...
When N2 > N1 , the transformer is called a step up transformer
...
EMF Equation
Consider a sinusoidal varying voltage V1 applied to the primary
windings of the transformer shown in fig 1
...
As per the law of electromagnetic
induction, the induced emf in a winding of N turns is given by
E = - N dF = - N d (F m sin wt)
dt
dt
Page 124
Transformer
Chap 7
= - NwF m cos wt = - NwF m sin (wt - p )
2
Thus the peak value of induced emf is Em = wNF m
...
(i)
E = Em = wNF m =
= 4
...
11
Average Value
This equation, known as emf equation of transformer, can be used
to find the emf induced in any winding (primary and secondary)
linking with flux F
...
44f F m # N1
Similarly rms value of emf induced in secondary,
...
(iii)
E2 = 4
...
So emf induced E1 in primary voltage is
equal to applied voltage V1 in primary and terminal voltage V2 in
secondary is equal to the emf induced E2 in secondary
...
It is denoted by letter K
...
Using eq (ii) and eq (iii) we can write
E1 = 4
...
44f N2 F m
Then, the transformation ratio can be written as
...
By selecting
K properly, the transfer of voltage can be done from any value to
any other desired value
...
There can be two cases :
(1) When K 2 1 i
...
N2 > N1 , the transformer is known as step-
Chap 7
Transformer
(2)
2
...
In this case voltage towards secondary side is
higher
...
e
...
In this case voltage towards primary is
higher
...
cm
...
(b) The no
...
30 A
6600
I1 is the full load primary current
...
of primary turns
Page 126
Transformer
Chap 7
N1 = N2 # V1 = 80 # 6600
400
V2
(iii)
N1 = 1320
Let F m is the peak value of flux
...
44 f F m N1
6600 = 4
...
44 # 50 # 1320
F m = 0
...
So,
Bm = F m = 0
...
81 Tesla
(iv) As obtained in part (iii), maximum flux value in the core is
F m = 0
...
Explain the following terms: Per unit voltage regulation
RTU 2015
Ans :
VOLTAGE REGULATION
With the increase in the load on a transformer, there is a change in
its secondary terminal voltage
...
It increases if the power factor is leading
...
Let V2 (0) be the secondary terminal voltage at no load, and V2 be
the secondary terminal voltage at full load
...
We can compare this
change in voltage with respect to either the no-load voltage or the
full-load voltage, and it can be expressed as either per unit basis or
percentage basis
...
% regulation up =
4
...
RTU 2015
Ans :
In a transformer it is observed that, all the flux linked with primary
winding does not get linked with secondary winding
...
This leakage
flux does not link with both the windings, and hence it does not
contribute to transfer of energy from primary winding to secondary
winding
...
Hence,
leakage flux produces an effect equivalent to an inductive coil in
series with each winding
...
5
...
It has 50 turn on its secondary
...
4
50
N2
(ii) No
...
KVA rating = V2 I2 = 220 # 180 = 39
...
5 kV, 400 Hz step down transformer is to be operated at 60
Hz
...
(ii) Transformation ratio in both frequency situations
...
So, highest
safe input voltage
Vsafe = 200 # 60
400
Vsafe = 30 V
The 27
...
Therefore,
k = V2 = 200 = 7
...
5
V1
Transformation ratio with 60 Hz frequency
k = Vsafe = 30 = 1
...
5
V2
(ii)
*******
Page 130
DC Machines
Chap 8
CHAPTER 8
DC MACHINES
1
...
RTU 2016
Ans :
APPLICATIONS OF DC MACHINES
Applications of electrical machines depends on factors like cost,
maintenance, nature of power supply required, ease of speed control,
etc
...
Applications of DC Generators
DC generators are commonly used in dynamometers (for measuring
torque etc
...
Applications of separately excited dc generators are
1
...
It works as control generator in Ward-Leonard system of
speed control
3
...
Application of self excited dc generators are as follows
1
...
2
...
3
...
Applications of DC Motors
DC machines are widely used in variable-speed devices and where
severe torque variations occur
...
Applications of shunt, series and compound motors are discussed
separately below
...
2
...
2
...
g
...
DC series motors are used where high starting torque is
required
...
Series motors acquire very high speed at no-load or at very
light load, hence they should not be used for a belt drive
...
For example, rolling mill
drive, punching press, planning or miling machine etc
...
The effect of armature flux on the main
field flux is called as armature reaction
...
But, there is an axis (or a plane) along which
armature conductors move parallel to the flux lines and, hence,
they do not cut the flux lines at the moment
...
Brushes are always placed along MNA because reversal
of current in the armature conductors takes place along this axis
...
Armature Reaction
The effect of armature reaction is well illustrated in the figure below
...
In this case, magnetic flux lines due to the field
poles are uniform and symmetrical to the polar axis
...
N
...
) coincides with the ‘Geometric Neutral Axis’
(G
...
A
...
Now, in case the machine is running, both the fluxes (flux due
to the armature conductors and flux due to the field winding) will
be present at a time
...
This effect is called as armature
reaction in DC machines
...
Armature reaction weakens the main flux
...
2
...
N
...
gets shifted (M
...
A
...
Brushes should be placed on M
...
A
...
So, due to armature reaction, it is hard to determine the
exact position of M
...
A
...
On the other hand, for a loaded dc motor, MNA will
be shifted in the direction opposite to that of the rotation
3
...
They have the polarity of succeeding pole (coming next in sequence
of rotation) in generator action and proceeding (which has passed
behind in rotation sequence) pole in motor action
...
Since inter poles are connected in series with armature, the
change in direction of current in armature changes direction of inter
pole
...
m
...
It also provides commutation voltage for the
coil undergoing commutation such that the commutation voltage
completely neutralizes the reactance voltage (L di/dt)
...
4
...
C
...
RTU 2015
Ans :
Voltage build up in a shunt generator depends upon field circuit
resistance
...
(3
...
If the field circuit
resistance is increased to R2 (tine OB), the generator will build up
a voltage OL, slightly less than OM
...
When the field
resistance line becomes tangent (line OC) to O
...
C
...
If the field circuit resistance is increased beyond
this point (say line OD), the generator will fail to excite
...
C
...
) is
called critical field resistance RC for the shunt generator
...
It should be noted that shunt generator will build up voltage only if
field circuit resistance is less than critical field resistance
...
Explain the following term: Commutation
...
The current generated in the armature conductors of a dc
generator are alternating
...
As conductors move out of the influence of the north pole
and enter south pole, the currents in them are reversed
...
During the period
of short circuit of an armature coil by a brush, the current in the
coil must be reversed and also brought up to its full value in the
reversed direction
...
The inductive nature of the coil opposes the reversal
of current from (+ I ) to (- I )
...
The sudden
reversal of current as the brush leaves the segment may form an arc
causing sparking at the commutator and the brush
...
Resistive Commutation (Linear Commutation)
2
...
Accelerated Commutation
6
...
RTU 2015
Ans :
Page 136
DC Machines
Chap 8
Consider a shunt wound motor shown in Figure
...
Therefore,
driving torque acts on the armature which begins to rotate
...
The applied voltage V has to force current through
the armature against the back emf Eb
...
It follows,
therefore, that energy conversion in a dc motor is only possible due
to the production of back emf Eb
...
The armature
current Ia is given by:
Ia = V - Eb
Ra
V = Eb + Ia Ra
...
VIa = electric power supplied to armature (armature input)
Eb Ia = Power developed by armature (armature output)
2
I a Ra = electric power wasted in armature (armature Cu loss)
Thus, out of the armature input, a small portion (about 5%) is
2
wasted as I a Ra and the remaining portion Eb Ia is converted into
mechanical power within the armature
...
DC Machines
Page 137
Explain why the speed-torque characteristic of a DC shunt motor is
non-linear in high-torque region
...
(i)
It is seen from Equation (i) that with increase of Te , the speed
Page 138
DC Machines
Chap 8
drops
...
Since with increase of Te , f is reduced, Te /f2 increases at
a faster rate and the speed drops more rapidly with the increase of
torque in a shunt motor as shown in Figure
...
If effect of AR is neglected, then (Ka f) 2 in Equation
(i) remains constant
...
8
...
The resistance of motor armature and field windings are 0
...
Determine the back emf generated in
the motor when it runs on full load
...
Fig 1
Shunt field current,
Ish = V = 250 = 1
...
25 = 28
...
75 # 0
...
875 = 247
...
Explain the principle of DC machine and construction of DC
Chap 8
DC Machines
Page 139
machine ?
RTU 2014, 2007
or
Explain the working principle of DC motor and compare its types
...
RTU 2011
or
Describe the essential parts of DC machines regarding their
construction
...
RTU 2010, 2008
or
Ans :
WORKING PRINCIPAL OF DC MACHINE
A very basic model of dc machine is shown in fig 1 which can work
as dc generator and motor
...
Loop can rotate along
its axis
...
Induced emf is alternating in nature and direction of it is
governed by Fleming Right Hand rule
...
This emf appear across
brushes which are stationary
...
Direction of this force is governed by
Fleming Left Rule
...
So due to these couple of magnetic forces
a torque is produced (F # L ) which turns the loop
...
The
conductor that comes under the south pole should experience a
force always upward or downward and the conductor that comes
under the north pole should experience a force always downward
or upward
...
CONSTRUCTION OF DC MACHINE
A dc machine consist of two parts namely stator and rotor
...
Fig 2 shows the basic structure of
a four-pole dc machine
...
Fig 2 : Construction of DC Machine
Stator
It is stationary part and is designed mainly for producing magnetic
flux
...
It is outermost cylindrical part which serves
two purposes
...
Material used
for yoke are basically low reluctance material such as cast iron, cast
steel or forged steel
...
(2) Poles
Pole structure is shown in fig 3
...
The function of poles are
(1) Pole core carries a field winding which is necessary to produce
the flux
...
(3) Pole shoe enlarges the area to armature core to come across
the flux, which is necessary to produce larger induced emf
...
The pole cores are fixed inside the yoke, usually by bolts
...
As it require definite
shape and size, laminated construction is used
...
It carries the current due to which the pole core behaves as an
electromagnet
...
Anamelled copper wire is used for the construction of coils as it
is good conductor and bend easily
...
It is
inner cylindrical part having armature and commutator - brush
arrangement :
(1) Armature
It is further divided into two part namely armature core and
armature winding
...
It consists of slots
on its periphery and the air ducts to permit the air flow through
armature which serves cooling purpose
...
As it has to provide a low reluctance path to the flux, it is made
up of magnetic material like cast iron or cast steel
...
Armature Winding :
Armature winding is nothing but the interconnection of the
armature conductors, placed in the slots provided on the armature
core periphery
...
As armature winding carries entire current which depends on
external load, it has to be made up of conducting material like
copper
...
It serve the
following purpose :
(1) It connects the rotating armature conductor to the stationary
external circuit through brushes
...
Commutator has split slip ring with many segments
...
The conductor in the armature
Chap 8
DC Machines
Page 143
are grouped into parallel and series combination according to the
requirement of voltage and current and then they are connected to
the armature segment
...
To establish an electrical connection between the rotating
commutator and the external stationary circuit, brushes are used to
take out the electrical power
...
The
brushes are held in particular position around the commutator by
brush holders
...
For larger machines roller bearing are used at the driving end,
and ball bearing may be used at the non driving end, i
...
at the
commutator end
...
Sleeve bearing, with ring lubrication are used
for motors when very silent running is required
...
(5) Shaft
The rotating parts such as armature core, commutator, cooling fan
etc are keyed to the shaft
...
The shaft is made of mild steel with a maximum
breaking strength
...
A 8 pole lap wound armature has 40 slots with 12 conductor per
slots, generate a voltage of 500 volts, determine the speed at which
it is running if flux per pole is 50 m Wb
...
Derive the EMF equation of DC machine
...
Let
P = Number of poles,
F = Flux per pole in Webers (Wb),
Z = Total no
...
The flux cut by
N
each conductors per second is FP # 60
...
There are total Z conductor with A parallel paths
...
Thus total emf generated between the terminals
is
E = FP N # Z Volts
60
A
Chap 8
DC Machines
Page 145
For a given machine the number of poles (P) and number of
armature conductors per parallel path ( Z ) are constant, therefore,
A
Generated emf,
where K = PZ
E = KFN
60A
or
E \ FN
E \ Fw
or
2pN
60
where w =
, the angular velocity When the machine is
operating as a generator, this induced emf is called generated
emf Eg , whereas in case of a machine operating as a motor it is
called the counter or back emf Eb
...
12
...
Calculate the generated emf with the same flux and the speed at
1500 RPM
...
5 V
13
...
Page 146
DC Machines
Chap 8
R
...
U
...
R
...
U
...
The current can be supplied
to the field winding in two different ways
...
Separately excited dc
...
Self excited dc generators
These are further classified as :
(i)
Shunt wound dc generators
(ii) Series wound dc generators
(iii) Compound dc generators
...
Separately Excited DC Generators
Separately excited DC Generator is shown in fig 1
...
There is no
physical connection between the armature and field windings
...
Fig 1 : Separately Excited DC Generator
The prime mover is a machine such as a diesel engine, which
is mechanically coupled to the armature winding to rotate
mechanically
...
Thus mechanical energy converted into electrical energy
...
Under normal
operating condition field pole possess the residual magnetism that
induce a small voltage into the armature winding, which supplies
a small field current
...
Due to this cumulative process, the
rated field current will be supplied and the generator generates the
rated armature current
...
Accordingly, the self-excited generators may be classified as follows
...
The field winding is connected across the armature winding
forming a parallel or shunt circuit
...
A very small current Ish flows through it because
this winding has many turns of fine wire having very high resistance
Rsh (of the order of 100 ohms)
...
Fig 2 : Shunt Wound Generator
Armature current,
Ia = Ish + IL
Shunt field current,
I sh = V/Rsh
Page 148
DC Machines
Chap 8
V = Eg - Ia Ra
Terminal voltage,
Power developed
= Eg Ia
Power delivered
= VIL = Eg Ia
(ii) Series Wound Generator
The conventional diagram of series wound generator is shown in
fig 3
...
Therefore, full line current IL or
armature current Ia flows through it
...
Fig 3 : Shunt Wound Generator
Armature current,
Ia = Ish
Terminal voltage,
V = Eg - Ia (Ra + Rse)
Power developed
= Eg Ia
Power delivered
= VIL = Eg Ia
(iii) Compound Wound Generator
In a compound wound generator, there are two independent field
winding on each pole
...
The two type of compound
wound generate are
Long Shunt Compound Wound
The conventional diagram of long shunt compound generator is
shown in fig 4
...
Chap 8
DC Machines
Page 149
Fig 4 : Long Shunt Compound Wound Generator
Armature Current
Ia = Ise = Ish + IL
Shunt field current,
Ish = V/Rsh
Terminal voltage,
V = Eg - Ia Ra - Ise Rse
Power Developed
= Eg Ia
Power Delivered
= VIL
Short Shunt Compound Wound
The conventional diagram of short shunt compound generator
is shown in fig 5
...
Armature Current
Ia = Ish + IL
Fig 5 : Short Shunt Compound Wound Generator
Series field current,
Shunt field current,
Terminal voltage,
Power developed
Ise = IL
E - Ia Ra
Ish = g
Rsh
V = Eg - Ia Ra - Ise Rse
= Eg Ia
Page 150
DC Machines
Chap 8
= VIL
Power Delivered
14
...
When the
generator is loaded, the terminal voltage is 240 volt
...
04 ohm and the field circuit resistance is 24 ohm
...
04 W
Field circuit resistance, Rsh = 24 W
Shunt field current,
Ish = V = 240 = 10 A
24
Rsh
Terminal voltage is given as
V = Eg - Ia Ra
240 = 254 - Ia # 0
...
04
Load current,
IL = Ia - Ish = 350 - 10 = 340 A
15
...
5 ohm
resistance at terminal voltage of 250 volt
...
24 ohm and field resistance is 250 ohm
...
RTU 2011
Ans :
We have:
P = 8 Poles
Z = 778 ;
N = 500 rpm ;
RL = 12
...
24 ohm ;
Chap 8
DC Machines
Page 151
Rsh = 250 ohm
IL = V = 250 = 20 A
12
...
24h = 250 + 5
...
04 V
f Z N P
E= # # #
60 # A
f 778 # 500 # 8
255
...
8 = f # 3112000
f = 30604
...
00983 wb = 9
...
A long shunt compound generator delivers a load current of 50 A
at 500 V and has armature, series field and shunt field resistances
of 0
...
03 W and 250 W respectively
...
Ans :
Given that,
Load current,
IL = 50 A
Terminal voltage,
V = 500 V
Armature resistance,
Ra = 0
...
03 W
Shunt field resistance, Rsh = 250 W
Terminal voltage is given by
V = Eg - Ia Ra - Ise Rse
or generated emf
Eg = V + Ia Ra + Ise Rse
Shunt field current,
Ish = V = 500 = 2 A
250
Rsh
We know that,
...
(i), we get
Eg = 500 + 52 ^0
...
03h
= 500 + 52 ^0
...
16 = 504
...
A 4 pole, 900 rpm dc machine has a terminal voltage of 220 V and
an induced voltage of 240 V at rated speed
...
2 W
...
We know that,
Eg - V = Ia Ra
240 - 220 = Ia # 0
...
2
Eg = FNZ # P
60
A
For wave wound A = 2 , so
Induced emf,
-3
240 = 10 # 10 # 900 # Z # 4
60
2
Z = 800
So, number of armature conductors is 800
...
A 6 pole dc machine has 300 conductors and each conductor is
capable 80 A without excessive temperature rise
...
015 wb and the machine is driven at 1800 r
...
m
...
Ans :
We have
P = 6;
Z = 300 ;
N = 1800 rpm ;
Chap 8
DC Machines
Page 153
f = 0
...
e
...
` Total, current,
Now,
Ia = Current per conductor # Parallel paths
= 80 # 2 = 160 A
PfNZ
E=
= 6 # 0
...
e
...
` Total current,
Now,
Ia = Current per conductor # Parallel paths
= 80 # 6 = 480 A
PfNZ
E=
= 6 # 0
...
Explain the principle of operation of 3-phase induction motor
...
When the three-phase supply is connected across the stator
windings, a rotating magnetic field of constant magnitude but
rotating at synchronous speed is produced
...
The lines of the force of the stator field cut the rotor conductors
and an alternating emf is induced in these conductors because
of the rotating magnetic field
...
Since the rotor circuit is short-circuited, currents start flowing
in the rotor conductors
...
The interaction of stator and rotor field produces a torque
which tends to move the rotor in the same direction as the
rotating field
...
The rotor moves in the direction of stator field
...
As per Lenz’s law, the rotor field
try to oppose the cause producing them
...
Hence, to reduce this relative speed the rotor
starts running in the same direction as that of stator field and
try to catch it
...
The rotor can not attain the speed of the stator field which is
equal to the synchronous speed because if the rotor is moving
at synchronous speed, there is no relative motion between
the stator field and the rotor
...
This
would cause the rotor speed to decrease
...
Therefore, the electromagnetic
torque continues to increase
...
AC Machines
Page 155
Explain the following terms:
(i)
Cogging
(ii) Skewing
RTU 2015
Ans :
(i) Cogging
The phenomenon of Magnetic Locking between the stator and the
rotor teeth is called Cogging or Teeth Locking
...
This condition arises when the number of stator
and rotor slots are either equal or have an integral ratio
...
This characteristic of the induction motor is
called cogging
...
2
...
Skewing of the rotor slots, that means the stack of the rotor
is arranged in such a way that it angled with the axis of the
rotation
...
The slots are
not made parallel to each other but are bit skewed as the skewing
prevents magnetic locking (cogging) of stator and rotor teeth and
makes the working of motor more smooth and quieter
...
It reduces locking tendency of the rotor, i
...
the tendency
of rotor teeth to remain under stator teeth due to magnetic
attraction
...
It increases the effective transformation ratio between stator
and rotor
...
It increases rotor resistance due to increased length of the
rotor conductor
...
Explain the principal and working of 3-phase induction motor and
Page 156
AC Machines
Chap 9
explain type of 3-phase induction motor
...
1
TYPES OF INDUCTION MOTOR
According to rotor type, there are two different types of induction
motor
...
(i) Squirrel Cage Induction Motor:
A squirrel cage rotor consists of copper bars placed in the rotor
slots
...
This design is referred to as
cage rotors because the entire construction resembles a squirrel
cage
...
Fig 1 : Schematic of a cage rotor
The induction motor with squirrel cage rotor has excellent
running performance, but a low starting torque
...
A wound rotor has a
complete set of three phase windings similar as the windings on the
Chap 9
AC Machines
Page 157
stator
...
The ends of the three rotor wires are tied to slip rings on the
rotor’s shaft
...
This arrangement makes it possible to connect additional
resistance in the rotor windings to give high starting torque
...
A wound rotor is shown in fig 2
...
Describe the principle of operation of 3-phase synchronous generator
...
Basically, the operating principal of alternators is quite
similar to dc generators
...
Fig 1 shows a schematic diagram of a three-phase alternator
with stationary armature and rotating field system
...
When the field system is rotated by a prime mover, the stator
conductors are cut by the magnetic flux and therefore an emf
is induced in the stator conductors
...
3
...
The direction of induced emf can be found by Fleming’s right
hand rule and frequency is given by,
f = NP
120
where,
4
...
Fig 2 shows a star connected stator winding and a dc rotor
winding
...
The conductors in stator are
arranged in such a way that phase angle between any two phase
is 120c electrical angle as shown in fig 3
...
Therefore, all the phase are having
equal induced emf
...
Fig 4 : Generated 3-phase alternating voltage waveforms
Page 160
5
...
RTU 2013
Ans :
PRINCIPAL OF OPERATION OF SYNCHRONOUS MOTOR
Let the synchronous motor has two rotor poles NR and SR
...
The motor
has direct voltage applied to the rotor winding and a three-phase
voltage applied to the stator winding
...
The stator winding produces a rotating magnetic field
rotating at synchronous speed NS = 120 f/P
...
The direct current sets up a two pole field which is stationary
so long as the rotor is not turning
...
3
...
Fig 1
4
...
10
...
Under this situation, the poles NS
and NR repel each other and so do the poles SS and SR
...
AC Machines
Page 161
Hence, the rotor has a tendency to move in the anti-clockwise
direction
...
Fig 2
6
...
Now SS and NR attract each other and so do NS and SR
...
Since the stator poles change their polarities rapidly, they tend
to pull the rotor first in one direction and then after a period
of half-cycle in the other direction
...
Therefore, it is said that a synchronous motor has no self
starting torque
...
This could be explained through following points:
1
...
2
...
Therefore
the torque on the rotor is clockwise as shown in fig 3
...
After a time corresponding to half period, the stator poles
reverse their polarities but at the same time rotor poles also
shift their polarities as shown in fig 4
...
Hence a continuous
unidirectional torque acts on the rotor
...
AC Machines
Page 163
If now the external mover of the rotor is removed, the rotor
would continue to rotate in the clockwise direction under the
influence of torque
...
Hence, a synchronous motor always runs at synchronous
speed
...
Find the running speed of a 4-pole induction motor working on 50
Hz supply having 4% slip
...
04h = 1440 rpm
*******
Page 164
Semiconductor Diode
Chap 10
CHAPTER 10
SEMICONDUCTOR DIODE
1
...
Solar cells are large area p -n junction
diodes operated at zero-bias voltage
...
It is made
of two layers doped with n -type and p -type material
...
The thicker one of these two layers is
called base layer, and the thinner layer usually on the top part of
the device is called emitter layer
...
This type of solar cell is said to
have a p on n polarity
...
After absorption of photon energy from solar radiation
electron-hole pairs are generated in both p - and n -side of the
junction as shown in fig 2
...
If the diffusion lengths are long enough, the generated electrons
and holes of a junction will be able to reach the space charge
region by diffusion as shown in fig 3
...
Fig 3 : Conduction in photovoltaic cell
3
...
Thus, minority carriers electrons in the p -region move
to n -region and the minority carriers holes in the n -region
move to p -region
...
Semiconductor Diode
Chap 10
If the p -n diode is open circuited, the accumulation of
electrons and holes on the two sides of the junction produces
an open-circuit voltage Voc as shown in fig 4
...
Write short on: Rectifiers
...
Since a
semiconductor p -n junction diode offers a small resistance when
forward biased and large resistance when reverse biased, we make
use of it as a rectifier
...
A pure
sine wave alternates between positive and negative values (hence
it is an ac waveform) and has a zero average value over one cycle
...
The output waveform of a rectifier circuit
is described to be Pulsating dc
...
It is dc because it does not alternate
between positive and negative polarities
...
For
the sake of analysis, we often split a pulsating dc waveform into two
components, viz
...
Rectifier Circuits
Rectifier is the first stage in a regulated power supply
implementation
...
This classification is based on whether the diode(s) in
rectifier circuit conduct over half cycle of ac input or full cycle of ac
input
...
They are
(a) Full wave center tapped circuit
(b) Full wave bridge circuit
3
...
RTU 2015
Ans :
Capacitor (C) Filter
This is the simplest and cheapest filter
...
The capacitance offers a low-reactance path to the ac components
of current
...
All the dc current passes
through the load
...
During the positive half cycle, when the rectifier output
voltage is increasing, the capacitor charges to peak value Vm
as shown in fig 2
...
Page 168
Semiconductor Diode
Chap 10
Fig 2 : The approximate load-voltage waveform vo in a full wave
capacitor-filtered rectifier
2
...
The capacitor continues
to discharge until the source voltage(dotted curve) becomes
more than the capacitor voltage
...
At point C , again the input voltage exceeds the capacitor
voltage
...
4
...
The waveform can be approximated
by straight-line segments for ^wCRL >> 1h
If the total capacitor discharge voltage(ripple voltage) is Vr ,
then, the average dc output voltage is given by
Vdc = Vm - Vr
2
The ripple Vr is triangular in shape
...
Write short note on P-N Junction diode and Zener diode
...
The p -n junction has two terminals called
electrodes, one each from p -region and n -region
...
e
...
To connect the n and p -regions to the external terminals, a
metal is applied to the heavily doped n and p type semiconductor
regions
...
Such an ohmic contact has
two important properties,
1
...
2
...
Thus ohmic contacts are used to connect n and p type regions to
the electrodes
...
1a shows schematic arrangement of p -n junction
diode while the Fig
...
The p -region acts as anode while the n -region acts as cathode
...
Fig 1
V-I Characteristics of P-N Junction Diode
For a typical P-N junction diode, the V-I characteristics is shown
in Fig 2
Page 170
Semiconductor Diode
Chap 10
Fig 2 : V-I characteristics of a typical PN-junction diode
Under forward-bias condition, we find that the diode current
is very small for the first few tenths of a volt
...
But the diode
does not conduct well until the applied external voltage overcomes
the barrier potential
...
Now,
even a small increase in the applied voltage produces a sharp increase
in current
...
We
can say that the diode cuts in the circuit or turns on at this voltage
...
A very small current flows through the diode
due to the minority carriers crossing the junction
...
Hence, this current is called reverse saturation current,
Is
...
An ordinary PN-junction diode should not be operated in the
breakdown region; otherwise it may get damaged
...
The
zener diodes are fabricated with precise breakdown voltages, by
controlling the doping level during manufacturing
...
Chap 10
Semiconductor Diode
Page 171
Fig 3 Zener diode
The Fig
...
The d
...
voltage
can be applied to the zener diode so as to make it forward biased or
reverse biased
...
3b and 3c
...
V-I Characteristics of Zener Diode
The forward characteristic of a zener diode is similar to that of an
ordinary forward biased p -n junction diode
...
As soon as the applied reverse voltage reaches the zener
voltage of a diode, the reverse current abruptly increases to a very
high value
...
Fig 4 : V-I Characteristic of Zener Diode
Page 172
Semiconductor Diode
Chap 10
From the reverse characteristic, it is clear that as the reverse voltage
is increased, the reverse current remains extremely small up to a
point known as knee of the curve
...
The current
and voltage at knee point are designated as IZK and VZ respectively
...
The maximum value current can be passed through a zener
without exceeding its power dissipation capability is designated as
IZM
...
Zener Diode Equivalent Circuit
An ideal zener diode maintains a constant terminal voltage across
it, if the applied voltage exceeds the reverse breakdown voltage
...
The dc source indicates
that the effect of reverse breakdown is a constant voltage across the
zener terminals
...
A change in zener current TI Z produces a
small change in zener voltage T Z
...
Semiconductor Diode
Page 173
Describe the action of the following filter circuit:
(i)
Shunt capacitor filter
...
(iii) Choke input LC filter
...
It consists of a large value
capacitor C in shunt with the load resistor RL , as shown in fig 1
...
For dc, this is an open circuit
...
Only a small part of the ac component passes
through the load, thus the ripple voltage in RL is considerably
reduced
Fig 1 : Full-wave rectifier with capacitor filter
(ii) Series inductor filter
For high load currents inductive filter is used
...
An inductor has the fundamental property
of opposing any change in current flowing through it
...
Fig 2 : Full-wave rectifier with series inductor filter
Page 174
Semiconductor Diode
Chap 10
Whenever the current through an inductor changes a back emf is
induced in the inductor which prevents the current from changing
...
(iii) Choke input LC filter
An LC filter combines the features of both the series inductor filter
and shunt capacitor filter
...
Thus, if we combine the above two filter the ripple factor
becomes almost independent of load resistance
...
Fig 3 : Full-wave rectifier with choke-input (LC) filter
*******
Chap 11
Transistor
Page 175
CHAPTER 11
TRANSISTOR
1
...
RTU 2016
or
How a transistor works as an amplifier? Explain its amplification
phenomenon
...
As an example we consider an
npn bipolar transistor in a common-emitter configuration as shown
in fig 1
...
The voltage
sources vi represents a time-varying input voltage which has to be
amplified
...
The sinusoidal voltage vi causes a sinusoidal component of base
current superimposed on a dc quiescent value(IBQ )
...
The time-varying collector
current induces a time-varying voltages across the RC resistor which
is superimposed on a dc value (VRQ )
...
Hence, the circuit is known as a voltage
amplifier
...
Draw the circuit diagram of a single stage CE transistor amplifier
...
RTU 2015
Ans :
PRACTICAL CIRCUIT OF TRANSISTOR AMPLIFIER
A practical circuit of transistor amplifier is common emitter
configuration is shown in Fig
...
Fig 1
The various circuit elements and their functions are individually
described as follows:
(i)
Biasing Circuit : The resistors R1 , R2 and RE form the biasing
and stabilization circuit
...
When a number of stages are used the RL
represents the input resistance of the next stage
...
c
...
Its value is approximately 12 mF
...
c
...
Due to the presence of CC
the output across the resistor RL is free from the collector’s
Page 178
(iv)
(v)
3
...
c
...
In the absence of CC , RC will come in parallel
with the resistance R1 of the biasing network of next stage
and thereby congaing the biasing conditions of next stage
...
This is an electrolytic capacitor of
value approximately 12 mF
...
Thus the capacitor allows only a
...
signal to
flow but isolate the signal source from R2
...
Its value is approximately 100 mF
...
c
...
In the absence of
this capacitor, the voltage developed across RE will feed back
to the input side thereby reducing the output voltage
...
c
...
Define a and b of a transistor
...
RTU 2014, 2013, 2010
Ans :
Current Gain ( a dc ) in CB configuration
The current gain of the transistor is defined by the ratio of output
current and input current given as following
I
a dc = C
IE
a dc is called common-base current gain
...
The value of a dc is less
than, but very close to unity
...
98
...
Typical values of
b dc ranges from 50 to 400
...
Both the gains are
related with each other in following way
...
This ratio
is called ac or dynamic beta ( b ac or simply b ) or ac alpha ( a ac or
simply a )
DIC
a ac = a =
DIE V = const
CB
DIC
b ac = b =
DI B
VCE = const
Just as b dc is related to a dc , so is b related to a
...
1 = 1+ 1
a
b
or
b=
Chap 11
a or a = b
1-a
1+b
Sketch and explain input & output characteristics of a transistor
connected in common emitter configuration
...
RU 2001
Ans :
TRANSISTOR CONFIGURATIONS
The study of transistor circuits is possible using two port network
analysis, which requires two input and two output terminals
...
Three configuration are possible based on which terminal is common
or grounded
...
Common Base Configuration (CB)
2
...
Common Collector Configuration or Emitter Follower (CC)
Common Emitter Configuration
In CE configuration, the emitter is made common to the input
and the output
...
Voltage source VBB and VCC are applied in the way to make
the BE junction forward biased and the CB junction reverse biased
...
Chap 11
Transistor
Page 181
Fig 1: Common emitter configuration
Input Characteristics
In CE configuration input characteristics are obtained by plotting
the base-emitter voltage VBE with base current IB for different
values of collector-emitter voltage VCE
...
A family of curves for input characteristics are
shown in figure 2
...
After the cut-in voltage the base current
increases rapidly with a small increment in base-emitter voltage
Page 182
Transistor
Chap 11
VBE
...
ri = TVBE
TIB V is constant
CE
It is observed that the value of base current does not increase as
rapidly as that of input characteristic of a CB configuration
...
For a fixed value of VBE , base current IB decreases as VCE is
increased because a larger value of VCE results a large reverse bias
at collector base junction
...
This
is a small effect and could be ignored
...
It relates the output current IC , to the collector
to emitter voltage VCE , for various values of input current, IB
...
Fig 3: Output characteristics in CE configuration
A study of these output characteristics revels following points:
Chap 11
(1)
(2)
(3)
Transistor
Page 183
In the active region, IC increases slowly as VCE increases
...
When VCE falls below a few tenths of a volt, IC decreases
rapidly as VCE decreases
...
In this condition, both junctions
of the transistor are forward-biased
...
In cut-off region collector current is not zero when IB is zero
...
This leakage
current is the current which would flow between the collector
and emitter, if the third terminal (base) left open
...
DVCE
r0 =
DIC I = const
B
where DVCE and DIC are small changes in collector-base junction
voltage and collector current around a given point on the
characteristic curve (for a given IB )
...
Current Gain ( b dc ) in CE configuration
The dc current gain of the transistor is defined by the ratio of
output current and input current given as following
I
b dc = C
IE
where b dc is called common-emitter current gain
...
5
...
98 mA,
IE = 3
...
01 mA
...
98 = a # 3 + 0
...
98 - 0
...
97
or
a = 2
...
99
3
Now, for common emitter configuration, we have
or
IB = 30 mA
b = a = 0
...
99
For CE configuration, collector current is given by
IC = bIB + ^b + 1h ICO
= ^99 # 30 # 10-3h + ^99 + 1h^0
...
97 + 1 = 3
...
A transistor is connected in CE configuration shown in Fig
...
Calculate
the emitter current
...
9756
41
b+1
Chap 11
Transistor
Page 185
IC = 2 = 1 mA
2k
We know that,
IC = aIE
So, Emitter current
I
IE = C = 1 mA = 1
...
9756
7
...
1 mA
and 5
...
The collector leakage current with emitter
open ICEO is found to be 180 mA
...
RTU 2011
Ans :
We have,
IE = 5
...
1 - 5 = 0
...
1h + 180 # 10-3
5 = b ^0
...
180
b = 5 - 0
...
2
0
...
65 mA
1 + 48
...
A transistor is connected in common emitter (CE) configuration
in which collector supply is 8 volt and the voltage drop across
resistance RC connected in the collector circuit is 0
...
The value of RC = 800 ohm
...
96 , determine :
(i)
Collector-emitter voltage
(ii) Base-current
RTU 2009
Ans :
Collector supply voltage,VCC = 8 V
Page 186
Transistor
Collector resistance,
RC = 800 W
Voltage across RC ,
Chap 11
VR = 0
...
96
(i) Collector-emitter voltage ^VCE h is given by
VCC = VR + VCE
or
VCE = VCC - VR = 8 - 0
...
5 V
(ii) Base current is obtained as following
VR = IC RC = 0
...
5 = 0
...
625 mA
800
RC
b=
IB =
9
...
96 = 24
1-a
1 - 0
...
625 mA = 26
...
Which connection is
most common and why ?
RTU 2009
or
Give the comparison of CE, CB, CC configuration of BJT
...
Table : Comparison of Different Configuration
Sr
...
Characteristic
Parameter
Common-Base
(C-B)
CommonEmitter
(C-E)
1
...
Output dynamic
resistance
Very high
High
Low
3
...
98)
High (100)
Very high
(>100)
(< 100 W)
(< 1 MW)
(< 1 kW)
(< 45 kW)
Commoncollector
(C-C)
Very high (
750 kW )
(50 W)
Chap 11
Transistor
Page 187
Sr
...
Characteristic
Parameter
Common-Base
(C-B)
CommonEmitter
(C-E)
Commoncollector
(C-C)
4
...
Voltage gain
About 150
About 500
Less than 1
6
...
Phase relation
between input
and output
In phase
Out of phase (
180c)
In phase
8
...
The CE configuration is the only configuration which
provides both voltage and current gain greater than unity
...
( Since the
voltage gain in CC and Current gain in CB are less than unity, so
their power gain is also small)
In an amplifier stage we require a high input resistance and
low output resistance
...
In CC configuration input resistance is high and output resistance
is low as we desired, but it has a small voltage gain which is less
than unity
...
10
...
RTU 2008
Ans :
Refer to Q
...
A silicon transistor has ICBO = 1 mA at 27cC
...
If the IE = 1 mA , find the
IB at 57cC
...
99
...
99 ^1 h + 32 # 10-3
= 0
...
032 = 1
...
022 = - 0
...
Explain why a transistor cannot be achieved by connecting two
back to back diodes
...
In ebers-moll model two separate ideal diodes are connected back to
back with saturation currents IES and ICS
...
The current
sources account for the minority carrier transport across the base
...
When two isolated diodes are connected back to
back the width of the base region becomes larger than the diffusion
length and so all minority carriers will recombine in the base and
no minority carrier will be able to reach the collector
...
i,e
...
In inverted application also the common base current gain
becomes zero
...
e
...
As a result transistor
action ceases
...
TERMINALS OF A TRANSISTOR
BJT is a three terminal device, those terminal are described as
following
Emitter
As the name states, one of the outer region of transistor which emits
charge carriers, is called emitter
...
Emitter supplies majority carriers (electrons in case of
npn transistor or holes in case of pnp tansistor) to the base
...
It has small
thickness so that it may pass the injected charge carrier to the
collector
...
Collector
One of the outer region of BJT which collects or absorbs the
charge carriers (either electron or holes) is called the collector
...
e doping of collector is higher than base but
less than emitter)
...
13
...
Each junction of the transistor may
be independently forward or reverse biased
...
Forward Active Mode
When the emitter-base junction of the transistor is forward biased
and the collector-base junction is reverse biased, the transistor
operates in forward active mode
...
This bias configuration is
shown in fig 1 for npn and pnp transistors
...
Transistor has a large current in saturation
mode
...
Fig 2: Transistors biased in saturation mode
Cut-off Mode
When both the emitter-base and collector-base junctions are reverse
Chap 11
Transistor
Page 191
biased, transistor operates in cut-off mode
...
The transistor is operated in
this mode, when it is to be used as an open switch
...
Fig 3 : Transistors biased in cut-off mode
14
...
(i)
With emitter open on application of a voltage of 10 V between
collector and base results in a collector current of 0
...
(ii) With base open, an application of a voltage of 10 V between
collector and emitter results in a current of 25 mA
...
2 mA
...
25 mA
and
ICEO = 25 mA
We have the relation,
ICEO = ^b + 1h ICBO
25 = ^b + 1h 0
...
99
99 + 1
b+1
Given collector current,
or
IC = 1
...
2 = ^99 # IB h + ^99 + 1h^0
...
2 = 99IB + 0
...
175 mA
IB = 1
...
86 mA
99
15
...
determine the value of IC and VCB if it is a germanium transistor
...
Ans :
For a germanium transistor VEB = 0
...
10 = 2000IE + VEB
IE = 10 - VEB = 10 - 0
...
85 mA
Next, applying Kirchoff’s voltage law to the collector base circuit
20 = 1500IC + VCB
Thus
VCB = 20 - 1500IC
Chap 11
Transistor
Page 193
IC = IE
if neglecting base current, therefore
VCB = 20 - 1500 # 4
...
725 V
16
...
the voltage drop across the load resistance is 1
...
Determine
the value of base current if a of the transistor is 0
...
Ans :
Collector current is given by
IC = 1
...
2 mA
a = 0
...
2 mA = 1
...
986
We have,
IE = IC + IB
or
IB = IE - IC = 1
...
2 = 0
...
The collector current of an npn transistor is 4
...
If the value
of b is 120 and the base current is 40 mA
...
Ans :
Given that,
IC = 4
...
95 mA + 40 mA
= 4
...
04 mA = 4
...
95 =
120
4
...
95 = 120 # 4
...
95 = 4
...
95 - 4
...
24 # 10-3 mA = 1
...
A transistor has b = 150
...
Ans :
We have b = 150, IE = 12 mA
b
We know
a=
= 100 = 0
...
993 # 12 mA = 11
...
916 = 0
...
A transistor has IB = 105 mA and IC = 20
...
Find
(a)
b of the transistor
(b) a of the transistor
(c) emitter current
(d) Now if IB changes by + 27 mA and IC changes by + 0
...
Ans :
(a) b of the transistor
We know
b=
(b) a of the transistor
IC
= 20
...
2
IB
105 mA
Chap 11
Transistor
a=
Page 195
b
= 195
...
994
195
...
5 # 10-3
= 20
...
65 mA
Then
IB = 105 + 27 = 132 mA
and
IC = 20
...
65 = 21
...
5 mA = 160
...
What is the region of operation for the transistor in figure?
Ans :
This is a pnp transistor so we consider the junction voltages VEB
(emitter-base voltage) and VCB (collector-base voltage) as following
emitter base junction is reverse biased
VEB < 0
also the collector base junction is reverse biased
...
*******
Page 196
Digital Electronics
Chap 12
CHAPTER 12
DIGITAL ELECTRONICS
1
...
RTU 2014
Ans :
LOGIC GATES AND TRUTH TABLES
The basic elements that make up a digital system is called logic
gates
...
The most common logic
gates are OR, AND, NOT, NAND and NOR gates
...
The input-output relationship of the binary variables for each
gate can be represented in a tabular form called truth table
...
The OR gate has two or more inputs and only one output
...
The output
is LOW(0) only when all the inputs are LOW(0)
...
The output Y is written as
Chap 12
Digital Electronics
Page 197
Y = A+B
This is read as A OR B
...
The truth table shown below summarizes the
operation of the OR gate
...
The lamp is on when either A or
switch B is closed
...
Fig 2 : OR function represented by switches
AND Gate
The AND gate performs logical multiplication, commonly known as
AND function
...
An AND gate produces a HIGH(1) output only when all
the inputs are HIGH(1)
...
Page 198
Digital Electronics
Chap 12
Fig 3 : Logic symbol of a 2-input AND gate
Figure 3 shows a logic symbol for a 2-input AND gate with
inputs A and B and output Y
...
The AND function is similar to
multiplication in linear algebra and thus sometimes it is called as
logical product
...
The truth table shown below summarizes the
operation of the AND gate
...
Two switches A and B are connected in
series with a voltage source and lamp
...
If any of them is open, then lamp
will be off
...
It has one input and
one output
...
e
...
If the input value is ‘1’, the output
will be ‘0’
...
Fig 5 : Logic symbol of a NOT gate
Figure 5 shows a logic symbol for a NOT gate with inputs A
and output Y
...
Table below is
the truth table for a NOT gate
...
When the switch is closed(ON), the lamp is bypassed
and will be off
...
Fig 6 : NOT function represented by a switch circuit
Page 200
Digital Electronics
Chap 12
NAND Gate
The term NAND implies NOT-AND
...
The standard logic symbol for a
2-input NAND gate is as shown in fig 7
...
This circle represents the NOT function
...
The NAND gate produces a LOW(0) output when
all the inputs are HIGH(1)
...
Table : Truth Table for a 2-input NAND gate
Input
Output
A
B
Y = AB
0
0
1
0
1
1
1
0
1
1
1
0
NOR Gate
NOR is a contradiction of NOT-OR gates
...
The standard logic symbol for a
2-input NOR gate is as shown in fig 8
...
This bubble represents the NOT
Chap 12
Digital Electronics
Page 201
function
...
The NOR gate produces a LOW(0) output when any of the
inputs is HIGH(1)
...
Table : Truth Table for a 2-input NOR gate
Input
A
0
0
1
1
Output
B
0
1
0
1
Y = A+B
1
0
0
0
Exclusive-OR Gate
The exclusive-OR (abbreviated as Ex-OR) gate has two or more
inputs with one output
...
The output of an ExOR gate is at HIGH(1), if one and only one input is at HIGH(1)
...
Table : Table for a 2-input Ex-OR gate
Input
Output
A
B
Y = A5B
0
0
1
1
0
1
0
1
0
1
1
0
Page 202
Digital Electronics
Chap 12
The truth table shows that output is HIGH(1) if either input
A or B is HIGH exclusively, and low when both are 1 and 0
simultaneously
...
When
there is an even number of HIGH(1) inputs, such as two, four or six,
the output will be LOW(0)
...
In other words, the exclusive-NOR (abbreviated as ExNOR) gate, is an Ex-OR gate followed by inverter
...
Fig 10 : Logic symbol of 2-input Ex-NOR gate
The output of a 2-input Ex-NOR gate is HIGH when both
the inputs have same logic state, and its output is LOW when the
inputs are different
...
Table : Truth Table for a 2-input Ex-NOR gate
Input
Output
A
B
Y = A9B
0
0
1
0
1
0
1
0
0
1
1
1
For more than two inputs the output is HIGH(1), if inputs have
even number of 1s
...
Chap 12
2
...
110h2 = ^Z h10 find Z
Page 203
RTU 2016
Ans :
(i) Decimal-to-Binary Conversion
We apply repeated division method
...
The last
remainder is the MSB
...
Hence,
^1057h10 = ^1000100001h2
(ii) Decimal-to-octal conversion
To convert the given decimal integer number to octal, successively
divide the given number by 8 till the quotient is 0
...
The remainders read from bottom to top
give the equivalent octal integer number
...
d1 d0
...
d-m ) is given by
]N g10 = ^dn - 1 # 2n - 1h + ^dn - 2 # 2n - 2h +
...
+ ^d-m # 2-m h
^11011
...
5 + 0
...
75
3
...
Derive the boolean expression to represent the given circuit
...
F = ABCD + ABCD + AB C D + ABC D Implement the function
by XOR gate only
...
Digital Electronics
Chap 12
Give the truth table for the given combinational circuit
...
Digital Electronics
Page 207
List the major advantages of digital electronics over analog
electronics
RTU 2011
Ans :
ADVANTAGES OF DIGITAL SYSTEMS
Following are some of the advantages of digital systems over analog
systems:
Digital systems are easier to design
Since all the modern digital circuits use only two voltage levels,
HIGH and LOW, hence they are easier to design
...
In analog systems, signals have numerical significance;
so, their design is more complex
...
Greater accuracy and precision
Digital systems are much more accurate and precise than analog
systems, because digital systems can be expanded to handle more
digits simply by adding more switching circuits
...
Digital systems are less affected by noise
Unwanted electrical signals are called noise
...
In digital systems, noise is not critical as
long as it is so large that we can not distinguishing a HIGH from
a LOW
...
If we want to
change the system operation, we can do it easily by modifying the
program
...
Page 208
Digital Electronics
Chap 12
More digital circuitry can be fabricated on integrated circuit (IC)
devices
The fabrication of digital ICs is simpler and economical compared
to the analog ICs
...
Digital Systems are more reliable
Digital systems are more reliable than analog systems
...
Derive the mathematical expression of modulation index for frequency
modulated wave
...
RTU 2016
Ans :
MODULATION INDEX FOR FM
In frequency modulation instantaneous frequency is given as
w i = w c + K f m (t)
If we take modulating signal as
m (t) = Vm cos w m t
w i = w c + K f Vm cos w m t
This expression shows that the deviation in carrier frequency is
Tw = K f Vm cos w m t
Maximum deviation
Tw
max
= K f Vm
Now, modulation index for FM is defined as the ratio of
maximum frequency deviation and modulating frequency
K V
Maximum Frequency Deviation
mf =
= f m
wm
Modulating Frequency
It can be seen that unit of K f is hertz/volt or radian/volt
...
Now, general equation of FM
K V
f FM = Vc cos
f FM = Vc cos 6w c t + m f sin w m t @
Figures 1a, b, c and d show an unmodulated carrier, the
message(modulating) signal, frequency and phase modulated waves
...
1
...
2
...
Modulation index in FM
(m f ) can have values from
1 to any value
...
Side bands
AM spectra has only two
sidebands
...
Chap 13
Communication Systems
Page 211
Bandwidth is wider on
account of a number of
sideband pairs (200 kHz)
and is independent of the
modulating signal
...
5
...
In FM, all transmitted
power is useful
...
6
...
Area of reception is smaller
than AM
...
Frequency
range
AM broadcasts operate
in the medium and high
frequency (MF and HF)
ranges
...
8
...
Noise suppression is better
than in AM
...
Cost
AM is less complex and
less expensive
...
10
...
Adjacent channel
interference is less in FM
...
Modulating voltage
amplitude determines RF
carrier frequency
...
12
...
Bandwidth
Bandwidth is small (10
kHz) and is proportional
to the highest frequency in
the modulating signal
...
A sinusoidal carrier wave of frequency 2 MHz and amplitude 20 mV
is amplitude modulated by a sinusoidal wave of frequency 5 kHz
...
Consider
modulation index as 0
...
RTU 2016
Ans :
Given that,
Carrier frequency,
Amplitude of carrier wave,
fc = 2 MHz
Vc = 20 mV
Page 212
Communication Systems
Chap 13
Frequency of modulating signal, fm = 5 kHz
Modulation index,
ma = 0
...
8 # 20 = 8 mV
2
2
Upper side band frequency
fUSB = fc + fm = 2 # 103 + 5 = 2005 kHz
Lower sideband frequency,
fLSB = fc - fm = 2 # 103 - 5 = 1995 kHz
3
...
Draw comparison between amplitude
and frequency modulation schemes by taking important features
one by one
...
RTU 2013
Ans :
NEED OF MODULATION
Modulation is needed in communication system to achieve the
following advantage
To Reduce Height of Antenna
In radio communication, the signal is transmitted by an antenna
at the transmitter
...
The wavelength of original signal is large so required height of
antenna is also large which is practically impossible
...
This antenna height can be achieved
practically
...
Modulation provides
Chap 13
Communication Systems
Page 213
a high frequency shift to the baseband signal, so it is easier to
transmit the signals over longer distances
...
This happens
because their baseband spectrum is identical or overlapping
...
This technique of transmitting more signals
simultaneously over a single channel after these have been translated
to different frequency ranges by the process of modulation is known
as frequency division multiplexing
...
Since signal to
noise ratio(SNR) in the receiver depends on the signal bandwidth,
so it can be improved by proper control of bandwidth at modulating
stage
...
This frequency may be changed many times by
successive modulation
...
They will interfere with audio
signals
...
The carrier is a powerful signal (as its amplitude is very high in
comparison to message signal)
...
COMPARISON OF FM AND AM
Refer to Q
...
Communication Systems
Chap 13
Discuss
following
types
of
important
communication
methods
briefly
and
stating
their
important
applications
along
with
range
of
frequency
used
...
RTU 2015
Ans :
Microwave Line of Sight Communication:
Line of sight (LoS) is a type of propagation that can transmit and
receive data only where transmit and receive station are in view of
each other without any sort of an obstacle between them
...
For long distance communication, line-of-sight microwave radio
transmission using antennas is used as an alternative to the coaxial
cable or other guided media
...
It is rather easier to use antennas for direct
line-of-sight transmission
...
1
...
If the height of the
antennas are higher, the distance between the transmitter and the
receiver is also more
...
The major factors affecting this method
of transmission are rain, fog, thunderstorm, and other atmospheric
variations
...
Important applications along with range of frequency used for
line of sight are described below :
Chap 13
Frequency Range
Communication Systems
Frequency Band
Page 215
Application
Very
high VHF
frequency : 30 to
300 MHz
Ultra
high UHF
frequency : 300
MHz to 1 GHz
Mobile communication, and TV Line-ofsight required
...
Line-of-sight required
...
Mobile
communication,
commercial FM radio
...
Very short-range communication
...
MMW
TV,
and
Discuss the following type of important communication methods
briefly and stating their important applications along with range of
frequency used
...
RTU 2015
or
Discuss the configuration and properties of satellite communication
...
RTU 2012, 2011, 2008
Ans :
SATELLITE COMMUNICATION
A satellite is a radio repeater, also called transponder, placed in the
sky
...
Typical frequencies in
a satellite system are 6/4 GHz and 14/12 GHz, where 6 GHz and
14 GHz represent uplink frequencies and 4 and 12 GHz represent
downlink frequencies
...
The satellite receives the signal from the transmitting earth station,
frequency converts, amplifies and retransmits the signal towards the
receiving earth station
...
The IF modulator converts the input baseband
signal to FM, the up-converter translates the IF to an appropriate
RF carrier frequency
...
Fig 2 Earth Station Transmitter (Uplink)
Transponder
The transponder as shown in figure 3 consists of a receiving antenna
which operates at uplink frequency, a band-pass filter (BPF), an
input low noise amplifier (LNA), a frequency mixer, an output low
power amplifier (TWT) and a transmitting antenna which operates
at downlink frequency
...
The frequency of the oscillator is equal to the
frequency shift required, so it is equal to the difference of uplink
and downlink frequency
...
Fig 3 Satellite Transponder
Downlink Model (Earth Station Receiver)
A simplified block diagram of a satellite earth receiving station is
shown in figure 4
...
The output
of this is fed to a down-converter, which converts RF to IF
...
Fig 4 Earth Station Receiver (Downlink)
Advantages of Satellite Communication
(1) Since microwave frequencies are used for transmission large
bandwidth is available
...
(3) Satellite communication links are unaffected by doppler’s
Page 218
(4)
(5)
Communication Systems
Chap 13
effect
...
Satellite systems can deliver signal to any place on earth which
means that in remote area where microwave stations can not
have direct access, satellites can be used for communication
Disadvantages of Satellite Communication
(1) It is not easy to access satellites, therefore if any problem
rises in a satellite link it would become difficult to repair it
...
(3) There is an excessive time delay between a transmitted and
received signal
...
What do you mean by modulation? Discuss the amplitude
modulation in details
...
RTU 2012, 2011, 2007
Ans :
MODULATION
Modulation implies to change
...
It needs to modify in order to transmit it
efficiently through a channel
...
The message signal is called modulating signal or baseband signal
...
Depending on which parameter is being modified
there are three types of modulation:
(1) Amplitude Modulation
(2) Frequency Modulation
(3) Phase Modulation
AMPLITUDE MODULATION
If the amplitude of carrier is varied according to instantaneous value
Chap 13
Communication Systems
Page 219
of message signal (i
...
modulating signal or base band signal)
...
Let the carrier wave represented by
vc (t) = Vc cos w c t
where vc (t) is the instantaneous amplitude, Vc is peak amplitude
and w c is the carrier frequency
...
Then
m (t) = Vm cos w m t
where m (t) is instantaneous amplitude, Vm is peak amplitude
and w m is the frequency of modulating(message) signal
...
Then the general expression for the
instantaneous value of modulated is given by
f AM (t) = vc (t) + m (t) cos w c t
Substituting the value of vc (t), the above expression becomes as
f AM (t) = [Vc + m (t)] cos w c t = 6 c + Vm cos w m t @ cos w c t
V
Vm cos w t cos w t
= Vc ;1 + b l
m E
c
Vc
f AM (t) = Vc 61 + ma cos w m t @ cos w c t
...
The modulating signal, carrier signal and the
amplitude modulated wave are shown in fig 1, fig 2 and fig 3
respectively
...
Envelope of AM wave
The AM wave has a time varying amplitude which is called envelope
of the AM wave
...
The expression for AM wave as derived
earlier
f AM (t) = [Vc + m (t)] cos w c t = E (t) cos w c t
where E (t) = Vc + m (t) is the envelope of the AM wave
...
ma = Vm
Vc
Fig 4 : AM modulated wave
Figure 4 shows an amplitude modulated output waveform
...
5 : No modulation
Vc = Vmax - Vm
= Vmax - Vmax - Vmin
2
= Vmax + Vmin
2
Substituting the value of Vm and Vc , we get the another
expression for modulating index
...
Case I : If ma = 0 : It means that no amplitude modulation takes
place and equation (i) simply equal to Vc cos w c t representing
the carrier wave, shown in fig 5
...
A modulation factor between 0
...
4 is acceptable
to avoid distortion and for convenient detection
...
5
...
It means that the percentage of the modulation
is 100% (fig 7)
...
Case IV : If ma > 1
...
Any
greater depth of modulation results in a distortion of the
envelope and loss of information
...
Fig 8 : Over modulation
Analysis of AM wave
Now, proceeding with equation
f AM (t) = Vc 61 + ma cos w m t @ cos w c t
or
= Vc cos w c t + ma cos w c t cos w m t
f AM (t) = Vc cos w c t + ma Vc 8cos ^w c + w m h t + cos ^w c - w m h tB
2
= Vc cos w c t + ma Vc cos ^w c + w m h t + ma Vc cos ^w c - w m h t
2
2
Thus the amplitude modulated expressed above is sum of
following components
(1) First term of equation is unmodulated carrier signal
...
Communication Systems
Page 223
Second term of equation represents a signal that has the a
frequency which is (w m) higher than that of carrier frequency
with maximum value of amplitude 1 ma Vc
...
The third term of the equation is same as second term but
now with frequency w m less than that of carrier wave so it is
called lower side band
...
RTU 2013
Ans :
AMPLITUDE MODULATION
If the amplitude of carrier is varied according to instantaneous value
of message signal (i
...
modulating signal or base band signal)
...
Let the carrier wave represented by
vc (t) = Vc cos w c t
where vc (t) is the instantaneous amplitude, Vc is peak amplitude
and w c is the carrier frequency
...
Then
m (t) = Vm cos w m t
where m (t) is instantaneous amplitude, Vm is peak amplitude
and w m is the frequency of modulating(message) signal
...
Then the general expression for the
instantaneous value of modulated is given by
f AM (t) = vc (t) + m (t) cos w c t
Page 224
Communication Systems
Chap 13
Substituting the value of vc (t), the above expression becomes as
f AM (t) = [Vc + m (t)] cos w c t
= 6 c + Vm cos w m t @ cos w c t
V
= Vc ;1 + bVm l cos w m tE cos w c t
Vc
f AM (t) = Vc 61 + ma cos w m t @ cos w c t
...
The modulating signal, carrier signal and the
amplitude modulated wave are shown in fig 1, fig 2 and fig 3
respectively
...
Frequency Spectrum of AM wave
When a carrier is modulated by a message signal by mixing their
frequencies, two new frequencies equal to the sum and difference of
the combining frequencies are produced
...
Sidebands
carries complete information
...
Chap 13
Communication Systems
Page 225
Taking the fourier transform of the equation (i) on both sides
(which is beyond the scope of this book), we get
f AM (w) = Vc [d (w + w c) + d (w - w c)]
2
+ ma Vc [d (w + w c + w m) + d (w - w c - w m)]
4
+ ma Vc [d (w + w c - w m) + d (w - w c + w m)]
4
Fig 4 : Modulating signal spectrum
Fig 5 : Carrier signal spectrum
Fig 6 : AM signal spectrum
These spectrum are shown in figures 4, 5, and 6
8
...
(i)
v ^ t h = Vc cos ^w c t + m f sin w m t h
Comparing eq
...
(ii)
w c = 6 # 108
2pfc = 6 # 108
8
fc = 6 # 10 = 0
...
4 MHz
(ii) Modulating frequency
w m = 1250
2pfm = 1250
fm = 1250 = 198
...
94 = 994
...
Give the block diagram of a superheterodyne receiver and explain
function of each component
...
T
...
2012, 2010
Ans :
SUPERHETERODYNE RECEIVER
The radio receiver used in an AM system is called superheterodyne
AM receiver and is illustrated in Fig
...
The principal of superheterodyne receiver is that all incoming
radio frequencies are converted into a single intermediate frequency
fIF by the heterodyning(mixing) process
...
Since all the incoming radio frequencies
are converted into a fixed IF frequency so all the succeeding stages
have to operate on a fixed frequency
...
The RF Stage
The RF section is basically a tunable filter and an amplifier
that picks up the desired station by tuning the filter to the right
frequency band
...
This
frequency is above the incoming carrier frequency by an amount of
the desired intermediate frequency fIF
...
Frequency Mixer
This mixes the amplified RF signal ( fc ) and the output of the local
oscillator ( fLO )
...
This difference frequency is called the
intermediate frequency
fIF = fLO - fc
Page 228
Communication Systems
Chap 13
The mixer generates the following frequency components
(1) Difference of both frequencies fLO - fc
(2) Sum of both frequencies fLO + fc
Further only desired frequency fLO - fc is selected by IF amplifier
and the component fLO + fc will be blocked
...
IF section can effectively suppress
adjacent-channel interference because of its high selectivity
...
Detector
The detector is basically a demodulator which extract the audio
signal from the output of IF amplifier
...
10
...
^A + B h
RTU 2008
Ans :
Modulation Index of AM Wave
Refer to Q
...
Describe the basic principle of modulation and what is its need ?
Present the mathematical analysis of amplitude modulated wave
...
3
Chap 13
Communication Systems
Page 229
MATHEMATICAL ANALYSIS OF AMPLITUDE MODULATED
WAVE
Refer to Q
...
Explain frequency modulation
...
There are two types of
angle modulation
(1) Phase Modulation
(2) Frequency Modulation
Concept of Instantaneous Frequency
The angle modulated wave is mathematically expressed as,
f (t) = Vc cos [q (t)]
...
Vc = peak amplitude of the carrier
...
q (t) = is the generalized angle and a function of time
...
So, the instantaneous frequency for a frequency modulated wave
is given by
Page 230
Communication Systems
Chap 13
w i = w c + K f m (t)
In this case the angle q (t) is obtained by following expression
q (t) =
# w dt = w t + K # m (t) dt
i
c
f
Putting the value of q (t) in equation (i), we get the frequency
modulated wave given by following expression
f FM = Vc cos :w c t + K f
# m (t) dtD
Let the modulated signal is of the sinosudal form
m (t) = cos w m t
Now the expression for frequency modulated wave becomes as
f FM = Vc cos :w c t + K f
= Vc cos
#E
m
cos w m t dtD
K f Em
sin w m tF
wm
Frequency Deviation and Modulation Index for FM
In frequency modulation instantaneous frequency is given as
w i = w c + K f m (t)
If we take modulating signal as
m (t) = Vm cos w m t
w i = w c + K f Vm cos w m t
This expression shows that the deviation in carrier frequency is
Tw = K f Vm cos w m t
Maximum deviation
Tw
max
= K f Vm
Now, modulation index for FM is defined as the ratio of
maximum frequency deviation and modulating frequency
K V
Maximum Frequency Deviation
mf =
= f m
wm
Modulating Frequency
It can be seen that unit of K f is hertz/volt or radian/volt
...
Now, general equation of FM
K V
f FM = Vc cos
f FM = Vc cos 6w c t + m f sin w m t @
Figures 1a, b, c and d show an unmodulated carrier, the
message(modulating) signal, frequency and phase modulated waves
...
Dw << 1
wm
Then
BW = 2w m
Which is equivalent to AM
...
BW = 2Dw
Approximate Bandwidth for PM
BW - 2Dw = 2K p Vm w m
Phase modulation index
m p = K p Vm
13
...
Also find modulation index of AM
...
6
DE-MODULATION OF AM WAVE
Demodulation is the process of recovering the original message signal
from a modulated wave
...
The circuits used for demodulation are also know as detectors or
demodulators
...
Here we study the linear detectors only
...
A diode operating in linear
region of its characteristics can extract the envelope of an AM wave
...
Chap 13
Communication Systems
Page 233
Fig 1 : Amplitude detector (Linear diode detector)
The envelope detector is essentially a rectifier circuit with a
capacitor across it as shown in figure 1
...
So in the
positive half cycle of the rectified wave, the capacitor will
charge up to the peak value of carrier voltage
...
(3) If time constant of the circuit is properly chosen, the voltage
across the capacitor falls slow during the negative half cycle
and by that time next positive cycle is appeared
...
e
...
wc
(4) Further in positive cycle capacitor charges and this process
continues
...
(6) The output voltage as shown in figure 2 is consist of DC
voltage superimposed on the message signal and a small
ripple
...
Page 234
Communication Systems
Chap 13
Fig 2 : Output of diode detector
14
...
Ans :
Modulating signal,
m ^ t h = 15 sin 2p ^2000t h
Carrier signal,
vc = 60 sin 2p ^100000t h
Modulating frequency,
fm = 2000 Hz = 2 kHz
Carrier frequency,
fc = 100000 Hz = 100 kHz
Maximum amplitude of modulating signal
Vm = 15 V
Maximum amplitude of carrier signal
Vc = 60 V
(a) Modulation index
ma = Vm = 15 = 0
...
25 # 100 = 25 %
(c) Frequency of signal, fm = 2 kHz
frequency of carrier, fc = 100 kHz
(d) Frequency spectrum of modulated wave amplitude of lower and
upper side bands,
ma Vc = 0
...
5 V
2
2
Side band frequencies,
Chap 13
Communication Systems
fUSB = fc + fm = 100 + 2 = 102 kHz
fLSB = fc - fm = 100 - 2 = 98 kHz
Frequency spectrum
*******
Page 235
Page 236
Instrumentation
Chap 14
CHAPTER 14
INSTRUMENTATION
1
...
This property is used for measurement of temperature
...
wtn h
where R0 is the resistance at temperature t = 0cC and a , b , g
...
The number of terms necessary depends on the material, the
desired accuracy, and the range of operation
...
Theory of RTD
The expression for small temperature variation is modified as
Rt = R0 ^1 + kt h
where k is known as fundamental constant
...
Then the resistance R at any
temperature of t p c C is given by the expression
...
or
tp =
Requirements of Resistance Material Used in RTD
Platinum, nickel and copper are the most commonly used resistance
materials in RTD
...
High temperature coefficient of resistance such that there is
a large change in resistance for a relatively small change in
temperature i
...
, larger sensitivity
...
The material should have a high value of resistivity so that
minimum volume of material is used for the construction of
RTD
...
The resistance of materials should have a continuous and
linear relationship with temperature for ease in measurement
...
Stability of the electrical characteristics of the material and
resistance for good repeatability
...
Sufficient mechanical strength so that adequate ruggedness in
construction is obtained
...
Write short note on: Strain Gauges
...
RTU 2015
Ans :
STRAIN GAUGES
The strain gauge is a transducer used for measuring mechanical
surface strain and is one of the most extensively used electrical
transducers
...
This change in resistance can be measured accurately
with a wheat-stone bridge
...
Operating Principle of Resistance Strain Gauge
The working of strain gauge is based on the fact that when stress
is applied on the metal conductor its resistance changes owing
to change in length and cross-sectional area of the conductor as
depicted in fig 1
...
Resistance of an unstrained conductor is given by an expression
...
These quantities can be related with each
other by differentiating eq (i) with respect to stress s i
...
,
dR = d r l = r dl - rl dA + l dr
...
(iii)
r
R
l
A
Under strained conditions, cross-sectional area of the conductor
changes, so the diameter of the conductor is also changed
...
(iv)
D
A
Now change in diameter of wire can be related with change in length
of wire by the Poisson’s ratio n of the wire material as given below
...
(v)
D
l
From eq (iv) and (v) we have
TA = - 2n T
l
l
A
l
Substituting TA = - 2 n T into eq (iii), we get
l
A
TR = T + 2 nT + Tr
l
l
...
TR/R
G=
T/l
l
Now, from equation (vi) we can write
Tr/r
(Gauge factor)
G = 1 + 2n +
T/l
l
The gauge factor G indicates the strain sensitivity of the gauge in
terms of the change in resistance per unit resistance per unit strain
...
Applications of Strain Gauges
Electrical strain gauges are widely used because of ease in
instrumentation, high accuracy and excellent reliability
...
1
...
These are mainly useful
where a minimum disturbance due to presence of transducer
is required as in wind tunnel measurements
...
Strain gauge transducers are extensively used for analyzing
the dynamic strains in complex structures, such as the stress
and strain in bridges, automobiles, roads etc
...
Strain gauge transducers are also used for measurement of
force by the strain produced in load rings
...
Write short note on: Bimetallic strip
RTU 2016
Ans :
BIMETALLIC THERMOMETERS
Bimetallic thermometers are widely used in process industries for
temperature measurements
...
(b) The temperature coefficient of expansion is not the same
for all metals, hence their rates of expansion are different
...
Working Principal
A bimetallic thermometer consists of a bimetallic strip which is
Chap 14
Instrumentation
Page 241
constructed by bonding together two thin strips of two different
metals
...
The working of bimetallic strip can
be understand through following points:
1
...
This occurs because of the difference in
the coefficients of thermal expansion of two metals
...
The complete strip will bend into a uniform circular arc, in the
direction of the metal which expands the least, as illustrated
in fig 2, because the strips are welded together along one
entire edge
...
With one end fixed, the bimetal strip deflects in proportion to
the temperature change and square of its length, and inversely
with its thickness
...
Instrumentation
Chap 14
Thus, if one end of the bimetallic strip is firmly clamped
while the other end is kept free, as shown in fig 2, the extent
of bending can be used for indicating the strip temperature
...
Mathematical Expression
Let us consider the case of a bimetallic strip in the form of a
cantilever of length L as shown in fig 3
...
The thickness of each metal forming the strip is t/2
...
Using above expression we can measure the temperature
...
Bimetallic devices are used as sensing and control elements in
temperature control systems
...
They are used as thermostats which controls the on-off
switches of heating furnaces and automotive chokes
...
Bimetallic strips are also used as circuit breakers where the
Chap 14
4
...
Instrumentation
Page 243
overload currents cause the strip to bend and break the circuit
connection
They are widely used in refineries, oil burners, hot solder
tanks, hot wire heaters etc
...
RTU 2014
Ans :
Like electronic circuit, ICs can also be categorized as digital IC and
analog IC based on their applications
...
The output signal level depends upon the input signal
level and the output signal level is a linear function of input signal
level
...
Op amps, voltage
regulators, comparators and timers are also well-known examples of
linear ICs or analog ICs
...
These ICs operate with binary data such
as either 0 or 1
...
The main components of an IC are transistors
...
As the technology is improving day by day, the number
of transistors incorporated in a single IC chip is also increasing
...
Namely,
1
...
2
...
Page 244
3
...
5
...
Very Large Scale Integration (VLSI) where the number of
transistors incorporated in a single IC chip is from 20,000 to
10,00,000
...
Depending upon the active devices used in ICs, it can be further
classified as bipolar ICs and unipolar ICs
...
5
...
RTU 2015
Ans :
THERMOCOUPLES
Thermocouples are the most simple and most commonly used
devices for measurement of temperature
...
If the junction is heated, a voltage is generated at the junction
which is nearly proportional to the temperature
...
This principle is used to convert heat energy
to electrical energy at the junction of two conductors as shown in fig 1
...
The emf produced is proportional to the
temperature and hence to the rms value of the current
...
The emf produced in a thermocouple circuit is given by :
E = a ^Tqh + b ^Tqh2
where Tq is the difference in temperature between the hot
thermocouple junction and the reference junction of the thermocouple
which is usually 0cC
...
a is usually
very large as compared with b and therefore emf thermocouple is
E - a ^Tqh or Tq - E/a
...
Since emf is a function of temperature difference Tq , the instrument
can be calibrated to read the temperature
...
Thermocouples are cheaper than the resistance thermometers
...
Thermocouples can be constructed in small sizes
...
Thermocouples are used for measurement of temperature
upto 1400c C
...
Most of the thermocouples are rugged in construction and
can withstand high shock
...
Thermocouples follow the temperature changes with a small
time lag and as such are suitable for recording comparatively
rapid changes in temperature
...
They have a lower accuracy than those of resistance
temperature detector (RTD)
...
2
...
3
...
Maximum accuracy of
measurement is assured only when compensating wires are of
the same material as the thermocouple wires
...
Applications of Thermocouples
1
...
2
...
6
...
RTU 2015
Ans :
Transducers and Inverse Transducers
Transducer is a device that converts a non-electrical quantity into an
electrical quantity
...
For example
a piezoelectric crystal and translational and angular moving coil
elements can be employed as inverse transducers
...
Differentiate active and passive transducers
...
Active Transducers
The transducers, which develop their own voltage or current
without any auxiliary source is called active transducer
...
The energy required for production of an output signal is
obtained from the physical phenomenon being measured
...
Chap 14
Instrumentation
Page 247
Passive Transducers
The transducers which derive the power required for the energy
conversion from an external power source are called passive
transducer
...
They may absorb some energy from the physical phenomenon
under study
...
8
...
RTU 2014
Ans :
ELECTRIC TRANSDUCERS
A broad definition of an electric transducer is as follows,
An electrical transducer is a sensing device by which a physical,
mechanical or optical quantity to be measured is converted into an
electrical signal (current, voltage or frequency)
...
1
...
2
...
3
...
4
...
In
fact, when dealing with electrical or electronic signals, the
inertia effects are because of electrons, which have negligible
mass
...
The output can be indicated and recorded remotely at a
distance from the sensing medium
...
The output can be modified, modulated or amplified easily as
per requirement of the indicating or controlling equipment
...
The size and shape of electrical transducers can be suitably
designed to obtain the optimum weight and volume
...
(1) Primary and Secondary Transducer
On the basis of methods of applications, transducers may be
classified into primary and secondary transducers
...
g
...
Secondary Transducer
If the input signal is sensed first by some detector or sensor and
then its output being of some form other than input signal is given
as input to a transducer for conversion into electrical form, then
such a transducer falls in the category of secondary transducers
...
Active Transducers
The transducers, which develop their own voltage or current
without any auxiliary source is called active transducer
...
The energy required for production of an output signal is
obtained from the physical phenomenon being measured
...
Passive Transducers
The transducers which derive the power required for the energy
conversion from an external power source are called passive
transducer
...
They may absorb some energy from the physical phenomenon
under study
...
(3) Analog and Digital Transducers
On the basis of nature of output signal, transducers may be classified
into analog and digital transducers
...
Examples: Strain gauge, a thermocouple, a thermistor or an LVDT
Digital Transducers
These transducers convert the input physical phenomenon into an
electrical output which may be in form of pulse (digital)
...
An inverse transducer is a device that converts
an electrical quantity into a non-electrical quantity
...
9
...
RTU 2013
Ans :
LOAD CELLS
Load cell is defined as,
A load cell is an electromechanical device that converts weight
or force into an electrical signal and is widely used for measurement
of static and dynamic forces
...
This combination of strain
gauge-elastic member is used for weighting, called a load cell
...
It consists of a steel
cylinder, on which four identical strain gauges are mounted The
gauges R1 and R 4 are along the direction of applied load and the
gauges R2 and R3 are attached circumferentially to gauges R1 and
R 4
...
1
...
2
...
4
...
Now, when a compressive load is applied to the unit, the
vertical gauges (R1 and R 4 ) undergo compression(i
...
negative
strain) and therefore there is decrease in resistance
...
e
...
Fig 1 : Strain gauge load cell
The two strain are not equal in this case
...
The
change in the output voltage due to applied load is given by the
following expression
dV0 = 2 ^1 + n hb dR n l
R 4
The voltage is measure of the applied load
...
Crane load monitoring
2
...
Draw-bar and tool-force dynamometers
Title: basic electrical and electronics
Description: It's about engineering and engineering is about knowledge of everything so this is the unit for basic electrical and electronics engineering
Description: It's about engineering and engineering is about knowledge of everything so this is the unit for basic electrical and electronics engineering