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Title: Basic electrical engineering
Description: For preparation

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2 A
...
Circuits

2
...
It states that
when current carrying conductor cut the magnetic field then emf induced in the conductor
...
1
...
of turns of coil
A = Area of coil (m2)
ω=Angular velocity (radians/second)

Magnetic Flux

N

m= Maximum flux (wb)

S

Wire
Wire
Loop(Conductor)
Axis of Rotation
Loop(Conductor)
Axis of Rotation
Figure 2
...
1 Generation of EMF



When coil is along XX’ (perpendicular to the lines of flux), flux linking with coil= m
...
When coil is
making an angle  with respect to XX’ flux linking with coil,  = m cosωt [ = ωt]
...
2 Alternating Induced EMF



According to Faraday’s law of electromagnetic induction,

d
dt
(  cos t )
e   Nd m
dt
e   N m (  sin t )  

e  N

Where,

Em  N m 
N  no
...
C
...
Shown in figure 2
...

Phase
angle

N

135
D

180

225

e  Em sin t

e

90
C
B

E

45

225 270 315 360

A 0/360

F

Induced
emf

0 45 90 135 180

ωt

t  00

e0

t  900

e  Em

t  1800 e  0

H 315
G
270

t  2700

S

e   Em

t  3600 e  0
Figure 2
...
2

Definitions

 Waveform
It is defined as the graph between magnitude of alternating quantity (on Y axis) against time
(on X axis)
...
4 A
...
Waveforms

 Cycle
It is defined as one complete set of positive, negative and zero values of an alternating
quantity
...
C
...

Generally denoted by small letters
...
g
...
Generally denoted by capital letters
...
g
...

e
...
Vave = Average value of voltage
Iave = Average value of current
 RMS value
It is the equivalent dc current which when flowing through a given circuit for a given time
produces same amount of heat as produced by an alternating current when flowing through
the same circuit for the same time
...
g
...
Symbol is
f
...

 Time period
It is defined as time taken to complete one cycle
...
Unit is seconds
...
Power Factor = pf = cos,
where  is the angle between voltage and current
...
It is given by product of rms voltage and rms
current and cosine angle between voltage and current
...

Active Power= P= I2R = VI cos
...


Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

3

2 A
...
Circuits
 Reactive power
The power drawn by the circuit due to reactive component of current is called as reactive
power
...

Reactive Power = Q= I2X = VIsin
...

 Apparent power
It is the product of rms value of voltage and rms value of current
...

Apparent Power = S = VI
...

 Peak factor/ Crest factor
It is defined as the ratio of peak value (crest value or maximum value) to rms value of an
alternating quantity
...
414 for sine wave
...
Denoted by
Kf
...
11 for sine wave
...


+V

In Phase (  )

Positive Phase ()

t

V(t) = Vmsinωt

Negative Phase (-)

+V

0

0

-V

+V



t

-V

V(t) = Vmsin(ωt+

0

-V

-
t

V(t) = Vmsin(ωt-

Figure 2
...
C
...

 Lagging phase difference
A quantity which attains its zero or positive maximum value after the other quantity
...
C
...
3

Derivation of average value and RMS value of sinusoidal AC signal

 Average Value
Graphical Method

Analytical Method

Voltage

Area Under the Curve

Voltage

Vm

V4

V5 V6

Vm

V7

V3

V8

V2

V9

V1

V10

Time

Time

180 /n

Figure 2
...
of Values

Figure 2
...
 v10
10

V

m

Vave 

Sin t d t

0



Vm

  cos t 0

V
  m  cos   cos 0 

2V
 m

 0
...
C
...
8 Graphical Method for RMS Value

Vrms 

One Full Cycle

Figure 2
...
of instantaneous values
Total No
...
curve
Base of the curve
2

V

2
m

Vrms 
Vrms

v 2  v 2  v 2  v 2  v 2 
...
707 Vm

Vrms 
Vrms
Vrms

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

6

2 A
...
Circuits
2
...

Phasor is a quantity that has both “Magnitude” and “Direction”
...
10 Phasor Representation of Alternating Quantities

Phase Difference of a Sinusoidal Waveform
 The generalized mathematical expression to define these two sinusoidal quantities will be
written as:

v  Vm Sin t

i  I m sin ( t   )
Voltage (v)
+Vm
Current (i)

+Im

0

-Im
-Vm

ωt



V


LEAD
ω



I
Figure 2
...
12 Phasor Diagram of Voltage & Current



As show in the above voltage and current equations, the current, i is lagging the voltage, v by
angle 
...
2
...
2
...


Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

7

2 A
...
Circuits
2
...
2
...


Circuit Diagram

It

R

vt=Vmsinωt

Where,
vt = Instantaneous Voltage
Vm = Maximum Voltage

VR

VR = Voltage across Resistance

Figure 2
...
2
...


Waveforms and Phasor Diagram


The sinewave and vector representation of vt  Vm Sin t & it  I m sin t are given in
Fig
...
14 & 2
...


V,i

vt=Vmsinωt

ω

it=Imsinωt
0

IR
ωt

Figure 2
...
15 Phasor Diagram of Voltage & Current for Pure
Resistor

Basic Electrical Engineering (3110005)

8

2 A
...
Circuits
Power
 The instantaneous value of power drawn by this circuit is given by the product of the
instantaneous values of voltage and current
...


2
...
2
...


Circuit Diagram

it

vt=Vmsinωt

L

VL

Figure 2
...
C
...
2
...
This back emf will oppose
the instantaneous rise or fall of current through the coil, it is given by
eb  -L



di
dt

As, circuit does not contain any resistance, there is no ohmic drop and hence applied voltage
is equal and opposite to back emf
...
17 Waveform of Voltage & Current for Pure Inductor

dt

Vm   cos t 


L  

it  

V

Vm
cos t
L



it  I m sin t  90


v,i

ω

90



 Vm

 L  I m 



From the above equations it is clear that
the current lags the voltage by 900 in a
purely inductive circuit
...
18 Phasor Diagram of Voltage & Current for Pure
Inductor

Power
 The instantaneous value of power drawn by this circuit is given by the product of the
instantaneous values of voltage and current
...
C
...


2
...
2
...
c
...


Circuit Diagram

it
q+
C

vt=Vmsinωt

VC
q-

Figure 2
...
2
...
Thus, the alternating supply applied to the plates of
the capacitor, the capacitor is charged
...

dq
it 
dt
dCVm sin t
it 
dt

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

11

2 A
...
Circuits
it  CVm sin t
Vm
cos t
1 / C
V
it  m cos t
Xc

it 

it  I m sin( t  90o )


Vm
 Im )
Xc

(

From the above equations it is clear that the current leads the voltage by 900 in a purely
capacitive circuit
...
20 Waveform of Voltage & Current for Pure Capacitor

Figure 2
...

Instantaneous Power
p( t )  v  i



p( t )  Vm sin t  I m sin t  90



p( t )  Vm sin t  I m cos t
p( t )  Vm I m sin t cos t
2Vm I m sin t cos t
2
V I
 m m sin 2t
2

p( t ) 
p( t )

Average Power
2

Pave 


0

Vm I m
sin 2 t
2
d t
2

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

12

2 A
...
Circuits
2

Pave
Pave
Pave



The average power consumed by purely capacitive circuit is zero
...
8


V I   cos t 
 m m

4 
2
0
V I
 m m   cos 4  cos 0
8
0

Series Resistance-Inductance (R-L) Circuit

Consider a circuit consisting of a resistor of resistance R ohms and a purely inductive coil of
inductance L henry in series as shown in the Figure 2
...

VL

VR

L

R
it

vt=Vmsinωt
Figure 2
...

But the voltage across them will be different
...


Waveforms and Phasor Diagram
 The voltage and current waves in R-L series circuit is shown in Fig
...
23
...
23 Waveform of Voltage and Current of Series R-L Circuit






We know that in purely resistive the voltage and current both are in phase and therefore
vector VR is drawn superimposed to scale onto the current vector and in purely inductive
circuit the current I lag the voltage VL by 90o
...
This is shown in the Fig
...
24
...
Next VL is drawn 90o leading the I
...


Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

13

2 A
...
Circuits
ω

VL

VL
R

+V

I

VR

VL
=
V

I


L

R

I

VR

Figure 2
...
If supply voltage

v  Vm Sin t
i  I m sin

 t   

Vm
Z

Impedance Triangle

Power Triangle





R

VR=I*R
Figure 2
...
26 Impedance Triangle Series
R-L Circuit

XL
  tan
R
1

R 2  X L2

en
(V t Po
A) w
e

Ap
pa
r

Z

V=
I*Z

XL

VL=I*XL

Reactive Power,Q
(VAr)

r,S

Voltage Triangle

Where I m 



Real Power,P
(Watt)
Figure 2
...
C
...

Instantaneous power

pt  v  i

pt  Vm sin t  I m sin  t   
pt  Vm I m sin t  sin  t   
2 Vm I m sin t  s in  t   

pt 

pt 



2

Vm I m
cos - cos(2t- )
2

Thus, the instantaneous values of the power consist of two components
...
r
...
time and second component vary with time
...
C
...
9


Series Resistance-Capacitance Circuit

Consider a circuit consisting of a resistor of resistance R ohms and a purely capacitive of
capacitance farad in series as in the Fig
...
28
...
28 Circuit Diagram of Series R-C Circuit




In the series circuit, the current it flowing through R and C will be the same
...

The vector sum of voltage across resistor VR and voltage across capacitor VC will be equal to
supply voltage vt
...
29 Waveform of Voltage and Current of Series R-C Circuit





We know that in purely resistive the voltage and current in a resistive circuit both are in
phase and therefore vector VR is drawn superimposed to scale onto the current vector and in
purely capacitive circuit the current I lead the voltage VC by 90o
...
This is shown in the Fig
...
30
...
Next VC is drawn 90o lagging the I
...

VR

I
VR
I

R

I

-
V=
V

VC
C

C+

VR

ω

VC

Figure 2
...
C
...
If supply voltage

v  Vm Sin t

Voltage Triangle
VR=IR

Where, I m 

Vm
Z

Impedance Triangle

Real Power,P
(Watt)

R

A

Power Triangle

-

-

O

-

e
ow
tP
e n VA )
(

Z

IZ
V=

r
pa
Ap

-XC

VC=I(-XC)

r,S

Reactive Power,Q
(VAr)

i  I m sin  t   

D

Figure 2
...
32 Impedance Triangle
Series R-L Circuit

where, Z  R 2  X C2

Figure 2
...
f
...

Instantaneous power

pt  v  i

pt  Vm sin t  I m sin  t   
pt  Vm I m sin t  sin  t   
pt 

2 Vm I m sin t  sin  t   
2

Vm I m
cos - cos(2t   )
2
Thus, the instantaneous values of the power consist of two components
...
r
...
time and second component vary with time
...
C
...
10 Series RLC circuit


Consider a circuit consisting of a resistor of R ohm, pure inductor of inductance L henry and
a pure capacitor of capacitance C farads connected in series
...
34 Circuit Diagram of Series RLC Circuit

Phasor Diagram
VL

VR

I

Current I is taken as reference
...
35 Phasor Diagram of Series RLC Circuit

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

18

2 A
...
Circuits


Since VL and VC are in opposition to each other, there can be two cases:
(1) VL > VC
(2) VL < VC
Case-1
When, VL > VC, the phasor diagram would be
as in the figure 2
...
37
Phasor Diagram

VR

ω

-

VL-VC

V

V


I

Figure 2
...

If vt  Vm Sin t

 t   

series RL circuit Pave  VI cos 
...


it  I m Sin





tan 

IR

X



where, Z  R 2  X C  X L



R
I X L  XC
1

1

2

R2  X C  X L

 IZ

The angle  by which V leads I is given by 

V



 ( IR )2  I X C  XL

where, Z  R 2  X L  X C

tan 



Figure 2
...


Basic Electrical Engineering (3110005)

19

2 A
...
Circuits
2
...
2
...
C
...

R

L

C

VR

VL

VC

it
f

vt=Vmsinωt
Figure 2
...
Since X L and XC
are function of frequency, at a particular frequency of applied voltage, XL and XC will become
equal in magnitude and power factor become unity
...
In a series circuit since
current I remain the same throughout we can write,
IXL = IXC

i
...


VL = VC

Phasor Diagram
 Shown in the Fig
...
39 is the phasor diagram of series resonance RLC circuit
...

The supply voltage

V 




VC

VR2  (VL  VC )2

V  VR

i
...
the supply voltage will drop across the
resistor R
...
39 Phasor Diagram of Series Resonance RLC
Circuit

Resonance Frequency
At resonance frequency XL = XC




2 f r L 

1
2 f r C

Bhavesh M Jesadia -EE Department

 f r is

the resonance frequency



Basic Electrical Engineering (3110005)

20

2 A
...
Circuits


f r2 



fr 

1

 2 

2

LC

1
2 LC

Q- Factor
 The Q- factor is nothing but the voltage magnification during resonance
...

 Q- factor = Voltage magnification
V
Q  Factor  L
VS
IX L

IR
L
 r
R
2f r L

R

XL
R



 Q  Factor 

But f r 

1
2 LC

1 L
R C

Graphical Representation of Resonance
 Resistance (R) is independent of frequency
...

 Inductive reactance (XL) is directly proportional to frequency
...


X L  2 fL
 XL  f


Capacitive reactance(XC) is inversely proportional to frequency
...

1
XC 
2 fC
1
 XC 
f



Impedance (Z) is minimum at resonance frequency
...


I

V
Z

For f  fr , I 

V
is maximum,IMAX
R

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

21

2 A
...
Circuits


Power factor is unity at resonance frequency
...
f
...
F
...
40 Graphical Representation of Series Resonance RLC Circuit

2
...
2
...

R

L

IC

IL

I= IL cosL

it
IC

V

L

C
IL
IL sinL

vt=Vmsinωt
Figure 2
...
42 Circuit Diagram of Parallel Resonance RLC
Circuit

The current IC can be resolved into its active and reactive components
...


Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

22

2 A
...
Circuits




A parallel circuit is said to be in resonance when the power factor of the circuit becomes
unity
...

In the phasor diagram shown, this will happen when IC = IL sin  and I = IL cos 
...

I C  I L sin 
V
V XL

XC ZL Z L
Z L2  X L X C
Z L2  2 f r L

R

1
L

2 f r C C



L
C
L  1  R2
2
r   2   2
CL  L
2

 2r L2 

 2f r 

1
2

fr 



2

L  1  R2
  2  2
CL  L
1
R2
 2
LC L

If the resistance of the coil is negligible,
fr 

1
2 LC

Impedance
 To find the resonance frequency, we make use of the equation I = IL cos  because, at
resonance, the supply current I will be in phase with the supply voltage V
...
Thus, impedance at the resonance is maximum
...

Z

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

23

2 A
...
Circuits
Q-Factor
 Q- factor = Current magnification

Q  Factor 


IL
I
I L sin 
sin 

I L cos 
cos 

 tan  


r L
R

2f r L
R

 Q  Factor 

But f r 

1
2 LC

1 L
R C

Graphical representation of Parallel Resonance



Conductance (G) is independent of frequency
...

Inductive Susceptance (BL) is inversely proportional to the frequency
...


BL 


1
1

,
jX L j2 fL

 BL 

1
f

Capacitive Susceptance (BC) is directly proportional to the frequency
...
F
...
43 Graphical Representation of Parallel Resonance RLC Circuit

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

24

2 A
...
Circuits


Admittance (Y) is minimum at resonance frequency
...

I  VY



Power factor is unity at resonance frequency
...
12 Comparison of Series and Parallel Resonance
Sr
...

1

Description
Impedance at resonance

Series Circuit
Minimum
Z=R
Maximum

2

Current

3

Resonance Frequency

4

Power Factor

5

Q- Factor

6

It magnifies at resonance

Bhavesh M Jesadia -EE Department

I

V
R

fr 

Maximum
Z

1
2 LC

1 L
R C

Voltage

L
RC

Minimum

I

Unity

fr 

Parallel Circuit

V
L / RC

fr 

1
2 LC

Unity

fr 

1 L
R C

Current

Basic Electrical Engineering (3110005)

25

2 A
...
Circuits
Three - Phase AC Circuits
2
...

conductors
...


In large industries and for running
heavy loads
...
14 Generation of three phase EMF
R

S

N
B

Y

Figure 2
...


Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

26

2 A
...
Circuits




Now, we consider 3 coil C1(R-phase), C2(Y-phase) and C3(B-phase), which are displaced
1200 from each other on the same axis
...
2
...

The coils are rotating in a uniform magnetic field produced by the N and S pols in the
counter clockwise direction with constant angular velocity
...
The emf induced in these three
coils will have phase difference of 1200
...
e
...

e

eR=Emsinωt

Em

eY=Emsin(ωt-120)
eB=Emsin(ωt-240)

e
ωt

0

120
240
Figure 2
...
2
...


eB

120

120

eR
12
0

eY
Figure 2
...
15 Important definitions
 Phase Voltage
It is defined as the voltage across either phase winding or load terminal
...

Phase voltage VRN, VYN and VBN are measured between R-N, Y-N, B-N for star connection and
between R-Y, Y-B, B-R in delta connection
...
C
...
It is denoted by VL
...

R
IR(line)

IR(line)

(p
h)

VB

(p
h)

IR(ph)

IB

V

(
BN

)
ph

ph
)

VBR(line)

IY(ph)

N(
ph
)

IY(

VRY(line)

h)
(p

VY

)
(ph

h)
(p

IR

VRY(line)
N

IB

R

VR

VRN(ph)

1

IY(line)

VBR(line)

3

VY(ph)

Y

IY(line)

Y

2

VYB(line)
IB(line)
B

VYB(line)
IB(line)
Figure 2
...
48 Three Phase Delta Connection System

 Phase current
It is defined as the current flowing through each phase winding or load
...

Phase current IR(ph), IY(ph) and IB(Ph) measured in each phase of star and delta connection
...

 Line current
It is defined as the current flowing through each line conductor
...

Line current IR(line), IY(line), and IB((line) are measured in each line of star and delta connection
...
It is customary to denoted the 3 phases by the three colours
...
e
...

 Balance System
A system is said to be balance if the voltages and currents in all phase are equal in magnitude
and displaced from each other by equal angles
...

 Balance load
In this type the load in all phase are equal in magnitude
...

 Unbalance load
In this type the load in all phase have unequal power factor and currents
...
C
...
16 Relation between line and phase values for voltage and current in
case of balanced delta connection
...

Circuit Diagram

IR(line)

1

(p
h)

VR

IR

h)
(p

IB

VBR

h)
(p

VRY

VB

(p
h)

R

IY(ph)
IY(line)

Y

VY(ph)

2

3

VYB
B

IB(line)
Figure 2
...

VRY  VR( ph)

VYB  VY ( ph )
VBR  VB( ph)
VL  Vph
Line voltage = Phase Voltage
Relation between line and phase current
 For delta connection,

IR  line  =IR  ph  IB ph
IY line  =IY ph  IR  ph
IB line   IB ph  IY ph


i
...
current in each line is vector difference of two of the phase currents
...
C
...
50 Phasor Diagram of Three Phase Delta Connection



So, considering the parallelogram formed by IR and IB
...
C
...
17 Relation between line and phase values for voltage and current in
case of balanced star connection
...

Circuit Diagram

R

VRY

N

VBR

I

h)
Y(p

V

Y

IR(ph)

VR(ph)

IR(line)

h)
Y(p

VB

(ph
)

IB(

ph
)

IY(line)
VYB

B

IB(line)
Figure 2
...

IR( line )  IR( ph)
IY ( line )  IY ( ph)
I B( line )  IB( ph )
 IL  Iph

Line Current = Phase Current
Relation between line and phase voltage
 For delta connection,

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

31

2 A
...
Circuits
VRY =VR  ph  VY ph
VYB =VY ph  VB ph
VBR =VB ph  VR  ph

i
...
line voltage is vector difference of two of the phase voltages
...
52 Phasor Diagram of Three Phase Star Connection

From parallelogram,
VRY = VR  ph2  VY ph2  2VR  ph VY ph cos
 VL  Vph2  Vph2  2Vph Vph cos60

 2

 VL  Vph2  Vph2  2Vph2  1
 VL  3Vph2
 VL  3Vph



Similarly, VYB  VBR  3 Vph



Thus, in star connection Line voltage = 3 Phase voltage

Power

P  VphI ph cos  VphI ph cos  VphI ph cos
P  3VphI ph cos
V 
P  3 L  I L cos
 3
 P  3VL I L cos

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

32

2 A
...
Circuits
2
...


Circuit Diagram
I R(line) M

R

L

C

V

Z1

VRY

Z3

Z2
I Y(line)
Y
VBY
B

C

V

IB(lline)M

L

Figure 2
...
2
...


VBY

-VY
VB

VRY

0

I

30



B
0

30



I



IY

VR

R

VY
Figure 2
...
C
...
The phase current lag behind their respective phase voltages
by an angle 
...

Reading of wattmeter, W1  VRY IR cos1  VL IL cos 30   
Reading of wattmeter, W2  VBY IB cos2  VL IL cos 30   
Total power, P = W1+W2

 P  VL I L cos 30     VL I L cos 30   
 VL IL cos 30     cos 30    

=VL IL cos30cos  sin30sin   cos30cos   sin30sin  
 VL IL 2cos30cos 
  3

 VL I L 2
cos




  2 

 3VL I L cos


Thus, the sum of the readings of the two wattmeter is equal to the power absorbed in a 3phase balanced system
...
C
...
2
...

p
...
5

600

0

3
VL IL
2

One zero and second total
power

cos   0

900

1
 VL IL
2

1
VL I L
2

Both equal but opposite

Remark

*******************

Bhavesh M Jesadia -EE Department

Basic Electrical Engineering (3110005)

35


Title: Basic electrical engineering
Description: For preparation