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Basic Electrical Engineering: NEE-201
UNIT: 1

(NOORUL ISLAM)

DC CIRCUIT ANALYSISAND NETWORK THEOREM
1
...
For
example, the resistor can conduct current equally well in either direction of current flow
...


I

I

R1

R2

Fig
...
1(a1): Unilateral Element

Fig
...
1(a2) : Unilateral Element

R1 = R2
Unilateral Network: An Unilateral circuit refers to those type of electrical circuit where the current flow occurs only one direction
...
For
example, diode conducts current in forward direction whereas it blocks current in backward direction
...
11(a3):

Fig
...
Thus, all three passive circuit elements, i
...
,
Resistance, Inductance and capacitance are linear elements
...

Non-linear element: An element is said to be non – linear when the volt – ampere characteristic is other than a straight line, and so
the electric circuit containing them are called nonlinear circuits
...


Voltage, V

Figure: Linear

Voltage, V

Figure: Nonlinear

Active Element:
The active elements generate energy
...
In a complete circuit voltage
or current source are active elements, which deliver power in the circuit
...
Resistor, capacitor, inductor etc are passive
elements because they take energy from circuit
...

Loop
A closed path in circuit where more than two meshes can be occurred is called loop i
...
there may be many meshes in a loop, but a
mesh does not contain on one loop
...

****
Q
...

Answer: Circuit is a close loop path giving a return path for the current
...
Discuss Ohm’s law
...
This relationship is called Ohm’s Law and be stated as;
The Current “I” flowing through a conductor is directly proportional to the potential difference i
...
Voltage “V” across its ends
provided the physical conditions (i
...
temperature, strain, etc) do not change
...
e
...

Current = Potential Difference / Resistance, I = V / R
Q
...

Answer:

A voltage source can be converted to a current source by finding the current (I) of the current source

I 

V
, which is
Ri

the short circuit current of the voltage source, and placing the resistance

Ri of the voltage source in the parallel to the current source

as depicted in the fig(a)
...


A

A

Ri
=

Network

I

Ri

I=V/R

Network

V
B

B

Fig
...
b

Similarly, a current source may be converted to a voltage source by finding

V  I Ri and putting the internal parallel resistance Ri

of the current source in the series with the voltage source as depicted in fig
...

1
...

Answer: Superposition Theorem
In any linear network, containing more than one independent energy source, then the complete response (voltage or current) in
any branch of network is equal to the sum of the response due to each independent source acting one at a time with all other ideal
independent sources are made inactive (short the voltage source and open the current source)
...

(ii) Applicable only for Linear and bilateral network
...

Example:
Calculate current in R2 Ω resistor by
R1
R3
R1
R3
superposition theorem,
in Figure: (a)
...
C
...
1
...
2(b)

I x R3
R2  R3

R1

R3
I22

(ii) Considering Source ‘V’, From figure: (c),

I22 

V
R2  R3

R2

O
...


V

Therefore, Current in R2 Ω resistor

IR 2  I2  I22

Fig
...
2(c)

1
...

Answer Thevenin’s theorem: Any two terminal linear network consisting of energy sources (voltage or current) and resistances can
be replaced with an equivalent circuit consisting of a voltage source (V th) and a series resistance (Rth) connected to the load
...

Rth(Thevenin’s equivalent resistance) is the resistance of the network as viewed from the load terminals, with the
independent sources replaced by their internal resistance
...

Solution:

Thevenin’s Equivalent

IL
R2

V

a

R1

IL

Rth
Vth

R3

RL

Figure: (a)
R3

RL

R2

S
...


Rth

b
Calculation of Rth,

Figure: (b)

Rth  R3 

I1

R1  R2
R1  R2

Figure:3(c)
R

R1

R2

V

Vth

Calculation of Vth,
from figure: (d),

Figure: (d)

V
I1 
R1  R2

VTH  I1  R2 

V  R2
R1  R2

Therefore, current in load RL,

IL 

Vth
R th  R L

1
...

Answer Norton’s theorem: Any two terminal linear networks consisting of energy sources (voltage or current) and resistances can be
replaced with an equivalent circuit consisting of a Current Source (IN) and a parallel resistance (RN) connected to the load
...

RN (Norton’s equivalent resistance) is the resistance of the network as viewed from the load terminals, with the independent sources
replaced by their internal resistance
...
1
...
1
...
6(b), load current is given by,

IL 

IN  R N
RN  RL

1
...
For which type of circuits is it generally applied
...

Answer
Maximum Power Transfer Theorem
A resistive load connected to a d
...
network receives maximum power when the load resistance is equal to the Thevenin’s
equivalent resistance of the network as seen from the load terminals
...
1
...
1
...

Derivation: From fig
...
20(b), The load current,

IL 

Vth
Rth  RL

The power delivered to the load RL is,

PL  I RL
2
L

Above equation can be written as,

or, PL 

PL 

Vth 2
RL
Rth  RL 2

Vth 2 
( Rth  RL ) 2
1 4 Rth 
( Rth  RL ) 2


PL will be maximum, if Rth – RL = 0, Rth = RL
...
22 Derive expression for the conversation from (a) Star to Delta Network (b) Delta to Star Network
...
A familiar case is a three terminal
network for example delta network or star network
...
However, converting delta network into star and vice - versa often simplifies the network and makes it possible to apply
series–parallel circuit techniques
...
1
...
1
...
Since both are identical therefore, the resistance between any pair of lines should be the same
In case of delta connection
In case of star connection
From figure (a)
from figure (b)
Resistance between points 1-2:

Resistance between points 2-3:

Resistance between points 3-1:

R12 ( R23  R31 )
R12  R23  R31
R23 ( R12  R31 )
R12  R23  R31
R31 ( R23  R12 )
R12  R23  R31



R1  R2




R2  R3
R3  R1

Adding equation (1), (2) and (3), we get,

2( R12  R23  R23  R31  R31  R12 )
 2( R1  R2  R3 )
R12  R23  R31

(1)

(2)

(3)

R12  R23  R23  R31  R31  R12
 R1  R2  R3
R12  R23  R31
R23  R31
Now, subtracting equation (1) from equation (4), R3 
R12  R23  R31
or,

subtracting equation (2) from equation (4),

subtracting equation (3) from equation (4),

(4)

(5)

R12  R31
R12  R23  R31
R23  R12
R2 
R12  R23  R31
R1 

(6)

(7)

Now, multiplying equations (5x6, 6x7 and 7x5) and adding all, we get:

R3 R1  R1 R2  R2 R3 

or,

2
2
2
R12  R23  R31  R12  R23  R31  R12  R23  R31
( R12  R23  R31 ) 2

R3 R1  R1 R2  R2 R3 

R12  R23  R31 ( R12  R23  R31 )
( R12  R23  R31 ) 2

R3 R1  R1 R2  R2 R3 

or,

R12  R23  R31
( R12  R23  R31 )

(8)

Now divide equations (8/5, 8/6 and 8/7), we get:

R3 R1  R1 R2  R2 R3
R3
R R  R1 R2  R2 R3
R23  3 1
R1
R R  R1 R2  R2 R3
R31  3 1
R2
R12 

(9)

(10)

(11)

1
...
Find the resistances of the equivalent star connection
...
1
...
1
...
1
...
1
...
Define Form Factor and Peak Factor
...
e
...
F
...
e
...
F
...
11 Explain series resonance in R-L-C circuit
...

Answer:
VL

I

R
900

AC

L

V

VR

900

I

C
VC

Fig
...
11(b): Phasor Diagram

Fig
...
11(a): R-L-C Series Circuit
From fig
...
11(a) Current flowing in the circuit is given by,

I 
Where,

V
R  (X L  XC )
2

X L   L  2 f L

and

XC 

(i)

1
1

 L 2 f C

V
will be maximum
...

R
1
fr 
or,
,
fr = Resonance Frequency
2  LC

X L  X C , then I 

From equation (i), If
and from above,

 Z  R2  ( X L  X C )2

2

2 f r L 

1
2 f r C

In this case, voltage across capacitor becomes equal to the voltage across inductor
...

In series R-L-C circuit, at resonance voltage across capacitor or inductor is more than the voltage across the whole circuit
...

This series resonance circuit is called acceptor circuit
...
The practical application of such circuit is found in radio-receivers
...
707) of its maximum value at

I
V
R
I
V
I 0 
R
2R
I0 

A

B

resonance
...
2
...


f1

fr

Fig
...
11(c)
At Point A, f1 lower cut-off frequency,

At Point B, f2 upper cut-off frequency

f2

f

I0

Current at half power point =

2
2

Power dissipated in the circuit at frequency f1,

 I 
I2
PA  I R   0  R  0 R
 2 
2



Power dissipated in the circuit at frequency f2,

 I0 
I2
2
 R 0 R

PB  I B R 
 2 
2



2
A

2

Therefore, power dissipated in the circuit at resonant frequency = P0  I 0 R
2

and, PA  PB 

1 2
I0 R
2

Quality Factor
For a series resonant circuit the quality factor ( Q0 ) is a measure of the voltage magnification

Voltage Magnificat ion 

Voltage across L or C at resonance
Supply voltage at resonance

V
I L  L
Q0  L 0  0  0
V
IR
R
Q0 

or,

0 L
R



1
LC

or,



V
Q0  C 0 
V

I

0 C
IR



1
 0 CR

1 L
L

R R C

2
...


Answer: From fig
...
11(c)

V V

Zr R
Ir
V

Current in R-L-C series circuit at half power points is given by, I 
2
2R
Current in R-L-C series circuit at resonance is given by,

Ir 

(i)

and also, Current in R-L-C series circuit at half power points is given by,

V

Z

I

V

(ii)

R2  (X L  X C )2

From equation (i) and (ii), we have,

V
2R



V
R  (X L  XC )
2

or,

2 R2  R2  ( X L  X C )2

or,

2R  R 2  ( X L  X C ) 2

2

( X L  X C )2  R2

X L  XC   R

(iii)

f 1 and upper half frequency is f 2
...
13 Discuss the parallel ac circuit and what is Admittance?
Solution: A parallel circuit is one in which two or more impedances are connected in parallel across the supply voltage
...
2
...
2
...
It is denoted by Y and is measured in unit Siemens or mho
...

If B negative then it is Inductance
2
...
Derive the resonant frequency
...
2
...

Fig
...
14(b)
shows
equivalent
parallel
circuit
...

From fig
...
14(c), circuit will be in electrical resonance if
Where, I L  I R  L Sin
therefore, I R  L Sin 

or,

R X

2
L

V
C

R X
2

C

I L  IC

AC

V
IC 
,
XC

Fig
...
14:(a)
I1
I2

XL

V
2

Coil

2
L



V
XC

I

L

R
C
V

Fig
...
14:(b)

or,

2
X L X C  R2  X L ,

or,

2
XL 

L
2
 R2  X L
C

or,

L
 R2 ,
C

0 

or,

1
Therefore Resonance frequency, f 0 
2
If R  L , Then

f0 

1
2

Ic

1
R2
 2 ,
LC L

IR

V

θ

1
R2

LC L2

1
LC

IL

IR-L

Fig
...
14:(c)
2
...

Answer : The impedance offered by the parallel circuit at resonance is called dynamic impedance denoted as Z D
...
As current drawn at resonance is minimum
...
It rejects the
unwanted
frequencies
and
hence
it
is
used
as
filter
in
radio
receiver
...
2
...
16 Derive the quality factor of the parallel RLC circuit at resonance
...

therefore,

Q  factor 

Circulatin g current between L and C
Line Current

V
L
XC
 L
Z
I2
Q  factor 

 D  CR  r
V
1
XC
R
I
r C
ZD



Q

2 f r L
R

But for parallel resonance circuit,

if, R very small, then

So, Q-factor ,

Q

fr 

1
2

1
2

fr 

1
2

1
LC

1 L 1 L

LC R R C

1
R2
 2
LC L

(from fig
...
14)

( ZD

 Dynamic impedance )

Power factor,

Cos 

VR
170

 0
...
20

Q
...


Q

TruePower
VICos
PowerFacto r 

 Cos
ApparantPo wer
VI

θ
P

Fig: Power Triangle
Causes of Low Power Factor
(i) Transformer
(ii) Induction motors
(iii) Induction Generators
(iv) High Intensity discharge (HID) Lighting
Problem of Low Power Factor: All the problems of low power factor arise due t more current needed to supply a
given power
...
1 For star and delta connected system in 3- phase circuit prove that

VL  3 V ph and I L  3 I ph

Answer
Let, phase voltage of three phase valance
system is Vph and line voltage VL
...
3
...
3
...


IL  IR  IB ,

From figure3
...
3
...
12

If the three phase star connected alternator is VP, What will the line voltage, (a) When the phase are correctly
connected, (b) when the connection to one of the phase are reversed
...
a
V'L/2

Va

V’L

60 0
Z VP

VP

Vb

Z
VL

VP
VP

V’L

60 0
Z
VP

Vc

Fig
...
c,

Fig
...


line voltage between a and c, Vac=

3 V ph
'

line voltage between a-b and b- c, Vab= Vbc = V L

VL'
Cos 60 0 

2

V ph

or,

V ph 

VP

1 VL'
V ' V ph

2 or, L
2

3
...
(i) lagging (ii) leading
...
3
...
3
...
The phasor diagram for such a balanced star connected load is shown in figure
(b)
...

and currents lagging behind their phase voltages by an angle
Current through wattmeter




...
3
...
18 What are the essential requirements on an instrument?
Answer: For satisfactory operation of any indicating instrument, following systems must be present in an instrument
...


(ii) Controlling system producing controlling torque,
(iii) Damping system producing damping torque,
Deflecting toque,

Tc
...


d produced by,

(i) Magnetic effect
(ii) Thermal effect
(iii) Electrostatic effect
(iv) Induction effect
(v) Hall effect
Controlling toque,
(i)
(ii)

c produced by,

Spring control
Gravity control

Damping toque,

TD produced by,

(i)
Air friction damping
(ii)
Fluid friction damping
(iii)
Eddy current damping
3
...

Answer These instruments are used either as ammeter or
voltmeter and are suitable for d
...
works only
...

Construction: PMMC instrument consists of:
(i) A light rectangular coil of many turns of fine wire
wound on an aluminum former inside which is an iron core
...


N

S

(iii) Permanent horse shoe magnet
...
The current
is led into and out of the coil by means of spring
...
3
...

Torque Equation: If i is the current in ampere flowing through the coil of turns N and length l meters and B is the flux density in
the air gap
...


Then deflecting torque,

d  F  r  B i l N r  B i N A

d  i

Therefore,

( A  l r )

as B, N , A are fixed

Since the instrument is spring controlled,

c 

Therefore,
At steady position,

Td  Tc

Therefore,

 i

Advantages:
(i) Uniform scale
...

(iii) High efficiency
...

(v) Very accurate and reliable
...
c
...

(ii) Expensive
...

(ii) D
...
galvanometer to detect small currents
...

3
...

Answer:
This type of instrument is principally used
for the measurement of alternating currents
and voltages (it is also used in d
...


Graduated Scale

These are two types
(i)

Attraction type

(ii)

Spring

Air
damper

Repulsion type

(i) Attraction type:
In this a single soft-iron vane (or moving
iron) is mounted on the spindle and is
attracted towards the coil when operating
current flow through it
...
3
...
An oval shaped soft-iron is attached to the spindle in such a way that it
can move in and out of the coil
...

The controlling torque is provided by the spiral spring arranged at the top of the moving element
...

Working principle: The operating current flowing through the coil sets up a magnetic field and it attracts the soft-iron piece towards
it
...

Deflecting torque:
The force, F pulling the soft iron piece towards the coil is directly proportional to:
(i) Field strength H produced by the coil
...


F  mH

F  H2
Therefore, Td  H

(m  H )

2

Td  I 2

(I  H )

Since, the instrument is spring control
...
One of these vanes is fixed and the
other is free to move
...


spring

Damping is provided by air friction due to the motion of
a piston in an air chamber
...
As
the fixed vane can not move, the movable vane deflects
and causes the pointer to move from zero position

Fixed
iron

Fig
...
20(b): Repulsion type MI instrument
Deflecting torque: The force, F pulling the soft iron piece towards the coil is directly proportional to the pole strengths m1 and m2
developed in the fixed and movable vanes
...
Therefore,
At steady position,

Td  Tc

Therefore,

c 

 I2

This instrument is used in a
...
as well as d
...
, it has graduated scale
...

(ii) Used in a
...
and d
...
both
...

(iv) Reasonably accurate
...

(ii) These instruments are not as sensitive as PMMC
...

Applications:
These instruments are used for measurement of a
...
voltages and current
...
21 Explain construction and working principle of dynamometer type instrument
...
The magnetic field is produced by two air
– cored fixed coils placed on either side of the moving
coil
...
c
...
c
...
It has graduated scale and controlling
torque is provided by springs
...


Fixed
Coil

Fixed
Coil
Moving
Coil

Fig
...
21: Dynamometer type instrument

Working principle: Two fixed coils and the moving coil are so connected that the same current flows through the two coils (fixed
and moving coils)
...
The result is that moving coil moves the pointer
over the scale
...
Therefore,
At steady position,

Td  Tc

Therefore,

 I2

c 
since,

i f  im  I

This instrument is used in a
...
as well as d
...
, it has graduated scale
...
c
...
c
...

(iii) Free from hysteresis and eddy current errors
...

(ii) Expensive
...
1 What is Earthing?
Answer: The connection of Eectrical Machinery to the general mass of earth, with a conducting material of very low resistance is
called earthing or grounding
...
2 State the advantages of grounded neutral supply system
...

(ii) Spikes of over voltages are easily dissipated to the earth
...

(iii) Less stress on insulation, if there is earth fault elsewhere
...
3 Explain the necessity of earthing
...
The windings and coils inside the frame of the machine carry the
current
...


Fuse

Line

Line
I

Ibody

I

Machine

Im
V

Ri

Rm

V
Person

Rbody
Neutral

Neutral
Earth

Earth resistance RE

Fig
...
3(a) Machine is not earthed

Fig
...
3(b) Equivalent Circuit

V

From figure: 4
...

so,

Ri  

I body 

therefore,

V
  Rbody  RE

0

(ii)

so, at normal conditions, there is no current passing through the body of the person and hence there is no danger of the shock
...

so,

Ri  0

'
I body 

therefore,

from equation (ii) and (iii),

V

(iii)

0  Rbody  RE

'
I body >> I body

Hence, when the machine is not earthed, there is always a danger of the shock, under certain fault conditions
...
So, current is sufficiently high to cause a

Im

fatal shock
...


Rbody

Neutral
Earth resistance RE

The current through the body of the person can be obtained by
using the results of current division in a parallel combination
...
3(c),

Ibody
IE
RE

Since, the total leakage current divided into two parts viz, one
flowing through Rbody and another flowing through new earthing

Ri

Rm

Fig
...
3(c) Machine is not earthed

RE
Rbody  RE

and current through the earthing connection,

IE  I 
Since,

Rbody
Rbody  RE

Rbody >> RE , therefore,

I E >> I body ,

So, entire leakage current I passes through the earthing contact bypassing the body of the person
...
Not only this but the current I, is high due to which fuses blows off and thus it helps to
isolate the machine from the electric supply
4
...
Give analogy between them
...
Exciting force is m
...
f
...
Response is flux,



S

Exciting force is e
...
f
...
m
...

Re luc tan ce

1
A

3
...
Permeance =

Reluctance, R 

1
Re luc tan ce

5
...


Flux Density, B 

R

Conductance =

m
...
f
...
m
...

l

I
A

Dissimilarities
7
...


Electric current flow in the circuit
...


There is no magnetic insulator to oppose the flux
...


9
...


Current can not flow through air
...
In magnetic circuit, energy is required only for creating
the flux but not for maintaining it
...


4
...

Answer:
Flux, Φ

Key

AC

E1, N1

E2, N2

Load

Fig
...
10: Transformer
A transformer works on the principle of mutual inductance
...
It has two coils wound on the same iron core
...

When the input voltage is connected across the primary coil of N1 turns as shown in figure
...


By faraday’s law,

E1  N1

d
dt

(i)

Since the flux ф is also linking with the secondary coil
...

dt

Therefore, by faraday’s law, the mutually induced emf in secondary coil is,

E2  N 2

d
dt

(ii)

Therefore, from equations (i) and (ii),

E1 N1

E2 N 2

4
...
m
...
equation of the single phase transformer
...


N1 = Number of turns in primary winding
...

The instantaneous flux is given by,

  max Sint

Therefore, instantaneous value of sinusoidally e
...
f
...
13

Draw the equivalent circuit diagram of single phase transformer and refer to primary and refer to the secondary
parameter
...
4
...


N
'
x2 = x2  1
N
 2

2


 = Secondary reactance referred to



I1

r1

x1

Rc

V1

I 2'

IC

r2'

'
x2

V2'

Xɸ

primary side of transformer
...


N
V2' = V2  1
N
 2

Fi
Fig
...
13(b): Equivalent circuit parameter of transformer referred to
primary side


 = Secondary resistance referred to primary side of transformer
...


N
x = x1  2
N
 1
'
1

2


 =Primary reactance referred to secondary



side of transformer
...
4
...




N
V1' = V1  2
N
 1


 =Primary reactance referred to secondary side of transformer
...




N
X = X  2
N
 1


 = Magnetizing reactance of transformer referred to secondary
...

N 
 2
4
...

Answer:
There are mainly two kinds of losses in a transformer
...

(ii) Copper Loss or Ohmic Loss
...
e
...

Ke = Proportionality constant which depends upon the volume and resistivity of the core material, thickness of laminations and the
units employed
...

and, f = frequency of the alternating flux
...
5 to 2
...

2

Ohmic Loss: When transformer is loaded, ohmic loss ( I 1

2
R1  I 2 R2 ), occurs in both the primary and secondary winding

resistances
...


 Efficiency ,  


Where,

output power
Input power

V2 I 2 Cos 2
V2 I 2 Cos 2  Pc  Pi

Pi  Total core loss

,

Pc  Total Copper loss

2
Pc  I 2 re 2 , re 2 = equivalent resistance referred to secondary and,

Cos 2  Load Power Factor
Condition for maximum efficiency:
Since, efficiency is given by:
Where,



V2 I 2 Cos 2
2
V2 I 2 Cos 2  I 2 re 2  Pi

V2 , re 2 , Cos 2 and Pi have constant value
...

dI 2

2
(V2 I 2 Cos 2  I 2 re 2  Pi )  (V2 Cos 2 )  V2 I 2 Cos 2  (V2 Cos 2  2  I 2 re 2 )  0

or,

2
(V2 I 2 Cos 2  I 2 re 2  Pi )  I 2 (V2 Cos 2  2  I 2 re 2 )  0

or,

2
Pi  I 2 re 2

Core loss = Copper Loss

(This is condition for maximum efficiency)

4
...

Answer:
Auto-Transformers

A transformer in which part of the winding is common to both
of the primary and secondary circuits is known as an autotransformer
...

As in ordinary transformer, the transformation ratio is equal to
turn-ratio
...
e
...
It provides continuously varying voltage
...
Saving in copper requirements
...

c
...

d
...

e
...

Demerits:
a
...

b
...

4
...

Answer
a
...

b
...
e
...

c
...
In audio applications, tapped autotransformers are used to adapt speakers to constant-voltage audio distribution systems, and
for impedance matching
...

Three Phase Synchronous Machines: Principle of operation of alternator and synchronous motor and their applications
...
1
Discuss the principle of operation of three phase induction motor
...

The flux passes through air gap, sweeps past the rotor surface and
cuts the rotor conductors which as yet are stationary
...
m
...
is induced on the rotor conductor
...

(iv) Now we have current carrying rotor conductors which are
placed in the rotating magnetic field produced by the stator
...

The direction of rotation is given by Fleming left hand rule
...
5
...


(vii) The speed of rotor is always less than synchronous speed, because at synchronous speed the induced emf in rotor will be
zero and hence no rotor current and therefore no rotor torque
...
2 Discuss why single phase induction motors do not have starting torque
...

Answer: When single phase supply applied to the stator of the single phase induction motor, a pulsating magnetic field is
produced(fields builds up in one direction, falls zero and then builds up in opposite direction), Which produces pulsating torque
...
Therefore, single phase induction motor is not self starting
...
The induction motor responds to each magnetic field
separately, and the net torque in the motor is equal to the sum of torques due to each of the two magnetic fields
...

Hence at standstill, the net torque is zero
...

However, if the rotor is given an initial rotation by auxiliary
means in either direction
...
The motor will therefore keep running in the direction
of initial rotation
...
3
Discuss the starting methods of single phase induction motor
...
The main
winding and starting winding are displaced 900 and it has very
low resistance and high inductive reactance
...

Its starting torque is 1
...

Applications: This type of motor is used in washing machine,
food mixers, grinders, floor polishers, blowers, centrifugal pumps,
small drills, lathes etc
...
The main
winding and starting winding are displaced 90 0
...
The main winding
and starting winding are connected in parallel during starting and
disconnected from the supply automatically by the centrifugal
switch Cs when the motor speed reaches about 70% to 80% of
synchronous speed
...
0 to 3
...

The value of rated capacitor must higher and starting resistance
should be low to obtain a high starting torque
...


Ra

Capacitor Start Motor

Auxiliary
+ main
winding

Torque,
T

Applications: This type of motor is used in pumps and
compressors, conveyors and machine tools etc
...
The main winding and
starting winding are displaced 900
...


Squirrel Cage Rotor
Xm
AC

C
Rm

Applications:

Xa

These motors are used where quiet operation is essential, as in
offices, class room, theater etc
...


Ra

Capacitor Run Motor

The value of capacitor should be in the range of 2 to 3 µF
...
The main
winding and starting winding are displaced 90 0
...
The two capacitors are connected in
parallel at starting
...
In order to obtain
high starting torque capacitive reactance should be low, short time
rated and electrolyte type
...
It is
called run capacitor
...
They have a higher efficiency than motors that run on
the main winding alone
...


CS

Nr = 0

Main
winding
Speed
Torque - Speed Characteristic

Nr = Ns

Shaded Pole Motor:
A shaded pole motor is self starting single phase induction motor
...
The stator is made up
of salient poles
...
This part is called shaded coil
...
The flux links shaded coil and causes
to circulate current in the ring this current produces flux which
lag behind the flux in the unshaded portion and hence resultant
flux produces
...


Torque,
T

Auxiliary
+ main
winding

Nr = 0

Speed

Nr = Ns

Torque - Speed Characteristic

5
...

Answer

Tmax

Tmax
r4

r3
r2
r1
Nr = 0

r2

r 1 < r2 < r 3 < r4

r1 < r 2 < r3 < r 4

Speed

Nr = N s

Fig
...
4(a):Torque – Speed Characteristic

Since, for three phase Induction motor,

Torque( )  k

Where, E2 = e
...
f
...

s = slip
r2 = Rotor resistance per phase
x2 = Rotor reactance per phase
Since,

r3

Torque, T

Torque, T

r4

Starting torque

 r2 and

r1

Slip

s=0

Fig
...
4(b):Torque – Slip Characteristic
2
E2
2

r2
s

 r2 
2
   x2
 s 

s =1

Maximum Torque is constant
0 ≤ S ≤ 1(Motoring Mode)

Torque( )  k

2
sE2 r2
r22  ( sx2 ) 2

(i) At s = 0, T = 0

when

N r = Ns

\

(ii) At s ≈ 0,

so,

2
s 2 x 2 can be neglected

T s
(iii) At s ≈ 1,

then,

T

2
2
s 2 x2  r22  x2

since

2
x2  r22

1
s

(iv) At s = 1,

Nr = 0

Torque( )  k

2
E2 r2
2
r22  x2

Also from above equations, at slip (s) = 0
...

As we increase the rotor resistance, starting torque increases and at r2 = x2 , starting is equals to maximum torque
5
...

Consider a three phase synchronous motor whose stator is
wound for two poles
...
C
...
C
...
When three phase winding is excited by a 3phase A
...
supply then the flux produced by the
3-phsse
winding is always rotating type
...
If the frequency of the three phase ac
supply is f Hz and stator is wound for P number of poles, then
the speed of the rotating magnetic field is given by,

Ns 

Stator

3- Φ
A
...

Supply
D
...

Supply

Salient
Pole
Rotor

120 f
r
...
m
...
9

If rotor rotates by external means like primemover or damper winding with synchronous speed, then a strong magnetic locking
develops between stator magnetic field and rotor magnetic field
...

5
...
C
...
C
...
10(a)

NS

Fig:5
...
The stator is also
wound for two poles NS and SS
...
The stator winding produces a rotating magnetic fields which revolves round the stator at synchronous speed (Ns = 120f/P)
...
Thus we have a situation in which there exist a pair of revolving
armature poles (i
...
Ns – Ss) and a pair of stationary rotor poles (i
...
NR – SR)
...
It is clear that poles N S and NR repels each other and so do the
poles SS and SR
...
After a period of half cycle ( or 1/2f ) or 1/100
second), the polarities of the stator poles are reversed but the polarities of the rotor poles remain the same
...
Therefore, the rotor tends to move in the clockwise direction
...
Due to high inertia of the rotor, the motor fails to start
...
e
...

5
...

Answer
Advantages
(i) Synchronous motors usually operate at higher efficiency, especially the low speed unit power factor ranges
...

(iii) Synchronous motor can be constructed with wide air gaps than induction motors, which make them better mechanically
...
Also, a "capacitive" power
factor, (current phase leads voltage phase), can be obtained by increasing this current slightly, which can help achieve a
better power factor correction for the whole installation
...

(ii) Synchronous motor is not self starting
...

(iv) Synchronous motors can not be used for variable speed jobs as there is no possibility of speed regulation
...
12 Explain V – curves, inverted V – curves and applications of synchronous motor
...

If V =Applied voltage
I = Line cCurrent
CosФ = Power factor
Then the input to the 3- phase synchronous motor is given
by,
Iamin
P = 3 VI CosФ
...


No load

Field current, If

Fig
...
12(a): V - Curves
Inverted V- Curves:
The curve obtained by plotting magnitude of the armature
current and field current is known as inverted V-Curves
...
0

Full load
Half load
No load

CosΦ

Applications:
(i) Synchronous motors are used to regulate the voltage at
the end of transmission line
...

(iii) Synchronous motors are used in houses and substations
in parallel to the bus bars to improve the power factor
...


Field current, If

Fig
...
12(b): Inverted V - Curves

5
...

Answer:
DC Generator: It is an electrical device which converts mechanical input power into dc Electrical Power
...
It states that “whenever the number of magnetic lines of force i
...

The magnitude of induced e
...
f in a conductor is proportional to the rate of change of flux associated with the conductor
...
m
...
This is mathematically given by;

e

d
dt

DC Motor: It is an electrical device which converts dc electrical input power into mechanical Power
...

The magnitude of force experienced by the a motor is given by,

F  BiL
Where, B= Flux density due to the flux produced by the field winding
...

i = Magnitude of current passing through the coductor
...
e
...

5
...

Answer Let,
Z = Number of conductors or coil sides in series per phase
...

P = Number of rotor poles
...
p
...

The time taken in one revolution (dt) = 60/N second, and
each conductor is cut by Pф webers in one revolution, i
...
dф = Pф
Therefore, Average e
...
f
...

dt
P
PN
 e

60
60
N
e

Since, there are Z conductors in series per phase
...
m
...
/phase,
or,

Eav =

PZ 120 f

60
P
Eav  2 f Z

Eav 

PN
Z
60
Since,

N

120 f
P

Since, r
...
s
...
m
...
per phase = Average value/phase x form factor



Erms  Eav  1
...
22 fZ Volts

If

K  is the winding factor of the armature winding, then,
E
...
F
...
22 f  Z K  Volts

5
...

Answer:
Let,
Z = Number of conductors or coil sides in armature winding of dc machine
...


P = Number of rotor poles
...
p
...

The time taken in one revolution (dt) = 60/N second, and each conductor is cut by Pф webers in one revolution of
armature, i
...
dф = Pф
Therefore, Average e
...
f
...

dt
P
PN
 e

60
60
N
e

Let, there are ‘A’ parallel paths in the armature winding
...
m
...
induced in one path,
or,

E  Eav 

PZN
60 A

In DC generator, induced e
...
f
...
m
...


Eg 

PZN
60 A

In, DC motor, induced e
...
f
...
m
...


Eb 

PZN
60 A

5
...

Answer: Torque Equation:
Let T is the average electromagnetic torque developed by the armature
...
p
...
then Mechanical power developed by the armature,

P  T  2nT = E I a
m
Therefore,

T 

E

But,
Therefore,
Where,

EI a


ZPn

A
PZI a
0
...
159 
A

Therefore,

T    Ia
Hence the torque developed by a dc motor is directly proportional to the flux per pole and armature current
...
17

Draw the characteristics of DC motor and application (series motor, shunt motor)
...
c
...
Following
are the three important characteristics of a d
...
motor
...
5
...
c
...
5
...
c
...
c
...
c
...


T  Ia

Hence, T - Ia characteristic is a straight line
passing through the origin
...
m
...
Eb in a shunt motor are almost
Constant under normal conditions
...

(iii) Speed and Torque characteristic (N - T):
The speed N of a dc motor is
given by;

Eb


T    Ia

and

Speed, N

N

Armature Current, Ia

From above two equations,

N

Eb I a
T

Torque, T

Since, under normal conditions, speed of shunt
Motor is constant
...


Characteristics of Series Motors:
(i) Torque and Armature Current characteristic (T - Ia):

T    Ia

For d
...
series motor, flux ф

TI

 Ia

2
a

Torque, T

We know that in a d
...
motor,

Hence, T - Ia characteristic is parabola
passing through the origin
...
m
...
Eb in a series dc motor is

almost constant under normal conditions
...
Therefore, speed of a series motor as
given in the figure
...
18 What are the applications of dc series motor and dc shunt motor?
Answer:
Direct current motors are very commonly used as variable – speed drives and in applications where severe torque variations occur
...

Compound motors
These motors are used where constant high starting torque and fairy constant speed is required, for example, presses, shears,
conveyor, elevators, rolling mills etc

---