Lecture 1 Basics of Electric Circuits

Lectures on Electrical
and Instrument
Engineering
Focused, To-The-Point and Practical!

Personal Introduction
Engr. Arslan Ahmed Amin is a professional Electrical and
Instrumentation Engineer serving Pakistan’s pioneer Oil and Gas
Organization, Pakistan Petroleum Limited. He obtained his Bachelor's
degree in Electrical Engineering from the prestigious University of
Engineering and Technology, Lahore in 2010 and started his
professional career with Pakistan Petroleum Limited. He has served
this organization for more than 5 years and achieved lots of
accomplishments in the development of the systems of newly installed
210 MMSCFD gas compression facility. He actively contributed his
services in commissioning, testing, maintenance and upgradation of
the E&I systems. He completed his Master’s in Business Administration
(M.B.A.) in 2014 from Virtual University of Pakistan, Lahore through
distance learning and afterwards obtained Masters (M.Sc.) in Electrical
Engineering from University of Engineering and Technology, Lahore in
2015.

 M.Sc. Electrical Engineering (University of Engineering
and Technology, Lahore)
 M.B.A. Online (Virtual University of Pakistan, Lahore)
 B.Sc. Electrical Engineering with Honors (University of
Engineering and Technology, Lahore)
Education

 05 years’ experience in industrial process controls and
electrical power systems domain with Pakistan
Petroleum Limited (PPL) in CMMS (SAP) environment.
 Experience of commissioning, testing and maintenance
of latest systems regarding Power Generation, Field
Instrumentation, Distributed Control System, Safety
Instrumented System, Gas Turbines, PLCs, Analyzers and
Utility packages.
Experience

 ‘Circuits and Electronics’ from Massachusetts Institute
of Technology (MIT) USA.
 ‘Project Management’ from Virtual University of
Pakistan (VU).
 ‘Production and Operations Management’ from Virtual
University of Pakistan (VU).
 ‘Conflict Management’ from Virtual University of
Pakistan (VU).
 ‘Crisis Management’ from Virtual University of Pakistan
(VU).
Professional Courses

 QMS 9001, ISO 14001 EMS and OHSAS 18001, ERP System, Cost of
Quality, Productivity Improvement Techniques, Process Safety
Management, Hazard Identification and Risk Assessment, HAZOP,
SIL systems, Occupational health and Safety, Permit to work
system, Safety Modules (Complete), Communication Skills, Team
Work Skills, Decision Making Skills (Organized by PPL)
 SAP System (R3P version) Maintenance Work Orders Processing,
Contracts Management, Spares and Material.
 ‘Instrumentation and Controls Fundamentals’ from OMS Institute of
Management and Technology, Lahore.
 Installation, calibration and maintenance of Fire and Gas detectors
by Det-tronics.
 Generation, Transmission and Distribution at WAPDA
Engineering Academy Faisalabad.
Professional Trainings

 Among Top 10 students in the session of 240 students in B.Sc.
Electrical Engineering.
 Received Dean’s Honor Role award in consecutive five
semesters for excellent academic performance in B.Sc.
Electrical Engineering.
 Overall Topped in F.Sc. in Board of Intermediate and
Secondary Education, Faisalabad 2006.
 Winner of Quaid-e-Azam Scholarship.
 Gold medal winner in District Science Quiz Competition
Faisalabad.
 Represented as ‘Talent of Pakistan Youth’ in China in 2007 by
Ministry of Youth, Pakistan.
Academic Achievements

Lets Start!
Lecture-1
Basics of Electric Circuits

Electric Circuit
 An electric circuit is an interconnection of
electrical elements.

Systems of UNITS
Quantity Basic Unit Symbol
Length meter m
Mass kilogram kg
Time second s
Electric current ampere A
Thermodynamic
Temperature
kelvin K
Luminous
intensity
candela cd

Electric Current
Current is the flow of
electricity, much like
the flow of water in a
pipe. It is measured in
Amperage as opposed
to gallons per minute
of water.

Conductors
 Free Electrons (e)
 Easily Directed
 Usually metals
 Copper
 Aluminum
 Gold
 Platinum
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- +

Semi-Conductors
 Dielectrics
 4 Valence Electrons
 Polarize with Some
Electron Flow due to
Electrical Fields
+
-

Insulators
 No Free Electrons
 No Current Flow with
Field
+
-

Why Does Current Flow?
 A voltage source provides the energy (or work) required to produce a current
 Volts = joules/Coulomb = dW/dQ
 A source takes charged particles (usually electrons) and raises their potential
so they flow out of one terminal into and through a transducer (light bulb or
motor) on their way back to the source’s other terminal

Voltage
 Voltage is a measure of the potential energy that causes a current to flow
through a transducer in a circuit
 Voltage is always measured as a difference with respect to an arbitrary
common point called ground
 Voltage is also known as electromotive force or EMF outside engineering

Voltage (Volts - V or E)
Voltage is the electrical
pressure in the system,
much like water pressure.
Electrical pressure is
measured in Volts as
opposed to Pounds per
Square Inch. (ie: 110V like
water from a tap, 4160 like
a fire hose)

Resistance (Ohms - R or Ω)
Resistance is simply the
restriction of current
flow in a circuit.
Smaller wire
(conductors) and poor
conductors have higher
resistance.

Resistance
e
e
e e
e
e
e
e
e
e
e
e
e
e
Many Collisions = Heat!
Fewer Collisions = Less Heat!

Ohm’s Law
Current, Voltage, and Resistance relate as follow:
I = E / R

A Circuit
 Current flows from the higher voltage
terminal of the source into the higher
voltage terminal of the transducer before
returning to the source
+
Source
Voltage
-
I
+ Transducer -
Voltage
The source expends
energy & the transducer
converts it into
something useful
I

Passive Devices
 A passive transducer device functions only when energized by a source in a
circuit
 Passive devices can be modeled by a resistance
 Passive devices always draw current so that the highest voltage is present on
the terminal where the current enters the passive device
+ V > 0 -
I > 0
 Notice that the voltage is
measured across the device
 Current is measured
through the device

Active Devices
 Sources expend energy and are considered active devices
 Their current normally flows out of their highest voltage terminal
 Sometimes, when there are multiple sources in a circuit, one overpowers
another, forcing the other to behave in a passive manner

Power
 The rate at which energy is transferred from an active source or used by a
passive device
 P in watts = dW/dt = joules/second
 P= V∙I = dW/dQ ∙ dQ/dt = volts ∙ amps = watts
 W = ∫ P ∙ dt – so the energy (work in joules) is equal to the area under the
power in watts plotted against time in seconds

Power
The power consumed or created is just the Voltage
multiplied by the Current
P = V x I
Eg:
If 3 amps flowing through a component
generate 12 volts across the component
the power is 3 x 12 = 36 watts

Some power calculations
Current
I
Voltage
V
Power
P
2 Amps 5 Volts
9 Amps 36 Watts
10 Watts
4 Volts

Conservation of Power
 Power is conserved in a circuit - ∑ P = 0
 We associate a positive number for power as power absorbed or used by a
passive device
 A negative power is associated with an active device delivering power
I
+
V
-
If I=1 amp
V=5 volts
Then passive
P=+5 watts
(absorbed)
If I= -1 amp
V=5 volts
Then active
P= -5 watts
(delivered)
If I= -1 amp
V= -5 volts
Then passive
P=+5 watts
(absorbed)

Example
 A battery is 11 volts and as it is charged, it increases to 12 volts, by a current
that starts at 2 amps and slowly drops to 0 amps in 10 hours (36000 seconds)
 The power is found by multiplying the current and voltage together at each
instant in time
 In this case, the battery (a source) is acting like a passive device (absorbing
energy)

Energy
 The energy is the area under the power curve
 Area of triangle = .5 ∙ base ∙ height
 W=area= .5 ∙ 36000 sec. ∙ 22 watts = 396000 J.
 W=area= .5 ∙ 10 hr. ∙ .022 Kw. = 110 Kw.∙hr
 So 1 Kw.∙hr = 3600 J.
 Since 1 Kw.∙hr costs about $0.10, the battery costs $11.00 to charge

AC and DC Current
•DC Current has a constant value
•AC Current has a value that changes sinusoidally
Notice that AC current
changes in value and
direction
No net charge is
transferred

AC v DC
• DC can be produced chemically or mechanically;
AC must be produced mechanically
• DC can be easily stored; AC cannot
• AC is easier, and thus cheaper, to produce
• AC can easily be transformed to other voltages
• AC can be transmitted more economically

Polarity
• Some components (like a bulb) can be connected either
way round – they will still work

Polarity
• Some components (like a diode) can be connected either
way round – they work one way but not the other

Circuit Elements
 Ideal Independent Source: provides a specified
voltage or current that is completely independent of
other circuit variables
 Ideal Independent Voltage Source:

Circuit Elements
 Ideal independent current source

Circuit Elements
 Ideal dependent voltage source
 Ideal dependent current source

Equivalent electrical circuit
Vbatt
(b)
+
–
Ihead
Ibatt
Itail Istart Ifan Ilocks Idash

Electrical vehicle battery
pack
DC-AC converter
(electric drive)
12 V12 V12 V12 V12 V
AC motor
(a)
Vbatt1 Vbatt2 Vbattn

Various representations of an
electrical system
HeadlightCar
battery
+ –
R
i
i
+
–
v
Source
Load
(a) Conceptual
representation
Power flow
(b) Symbolic (circuit)
representation
(c) Physical
representation
+_
i
+
–
vVS

Volt-ampere characteristic of
a tungsten light bulb
0.1
0.2
0.3
0.5
0.4
–0.5
–0.4
–0.3
–0.2
0–20–30–40–50–60 –10 5040302010 60
–0.1
i (amps)
v (volts)
Variable
voltage
source
Current
meter
+
–
v
i

The resistance element
i
R v
+
–
A
l 1/R
i
v
i-vcharacteristicCircuit symbolPhysical resistors
with resistanceR.
Typical materials are
carbon, metal film.
R =
l
A

Resistor color code
b 4 b 3 b 2 b 1
Color bands
black
brown
red
orange
yellow
green
0
1
2
3
4
5
blue
violet
gray
white
silver
gold
6
7
8
9
10%
5%
Resistor value = ( b 1 b 2 ) 10b3;
b4 = % tolerance in actual value

The current
1.5 V
+_
R
v+ –
v– +
+
–
v
i
R
i flows through each of
the four series elements. Thus, by
KVL,
1.5 = v1+v2+ v3
R 1
R 2
R 3
R n
R N
R EQ
Nseries resistors are equivalent to
a single resistor equal to the sum of
the individual resistances.

Parallel circuits
+
–
v
KCL applied at this node
The voltage v appears across each parallel
element; by KCL, iS = i1 +i 2+i 3
N resistors in parallel are equivalent to a single equivalent
resistor with resistance equal to the inverse of the sum of
the inverse resistances.
RN REQR1 R2 R3 Rn
i1 i2 i3
iS R1 R2 R3

Wheatstone bridge circuits
c
R2
R3R1
vS a
+
_
(a)
Rx
vbva b
d
c
R2
R3R1
+_
(b)
Rx
vbva b
d
avS

A force-measuring instrument
R2
R3R1
vS
R4
vbva
d
c
+
–
ia ib
h
w
Beam cross section
R2 , R3 bonded
to bottom surface
F

Practical voltage source
R L
rS
i S
+_vS
+
–
v L
Practical
voltage
source
iS =
vS
rS + RL
lim iS = vS
rSRL 0
r S iS max
vS
+
–
vL
The maximum (short circuit)
current which can be supplied
by a practical voltage source is
iS max = vS
rS
+_

Practical current source
R Li S
+
–
v Sr S
A model for practical current
sources consists of an ideal source
in parallel with an internal
resistance.
iS
+
–
v Sr S
Maximum output
voltage for practical
current source with
open-circuit load:
vS max = iS rS

Measurement of current
R2
R1
+_vS
A series
circuit
R2
R1
+_vS
A
A
Symbol for
ideal ammeter
Circuit for the measurement
of the current i
i i

Measurement of voltage
R2
R1
+_vS
A series
circuit
R1
+_ VV
Ideal
voltmeter
Circuit for the measurement
of the voltage v2
i
v2
+
– i
R2v2
+
–
v2
+
–
vS

Models for practical ammeter
and voltmeter
rm
A
Practical
ammeter
V
Practical
voltmeter
rm

Measurement of power
i
R1
+
_
Internal wattmeter connections
R2v2
+
–
vS
iR1
+
_
Measurement of the power
dissipated in the resistorR2:
P2 = v2 i
vS
W
R2v2
+
–
V
A

Definition of a branch
a
rm
A
Practical
ammeter
Ideal
resistor
Rv
A battery
A branch
Branch
voltage
Branch
current
+
–
b
i
Examples of circuit branches

Definition of a node
Examples of nodes in practical circuits
Node a
Node b
vS iS
Node c Node a
Node b
Node

Definition of a loop
Loop 1 Loop 2
Loop 3
vS
R
1-loop circuit 3-loop circuit
(How many nodes in
this circuit?)
Note how two different loops
in the same circuit may in-
clude some of the same ele-
ments or branches.
iS
R1 R2

Magnetics
NorthSouth
Magnetic Flux
Magnet

Current Flow in Conductor
- +
Current Flowing in a Conductor

Generated Field Around
Conductor
+

Magnetic Field With Coil
+
-
+
-
North Magnetic Pole
South Magnetic Pole

Interaction with Medium
NorthSouth
Magnetic Flux
MagnetMetal NS

Electrical Properties
 Frequency
 Inductance (L)
 Mutual
 Inductive Reactance (XL)
 Capacitance (C)
 Capacitive Reactance (XC)
 Phase Angle/Power Factor
 Impedance (Z)

Inductance
 Stores electromagnetic energy
in its magnetic field
 mH
dt
di
LV 
 
t
idv
L
i
0
)0()(
1
 2
2
1
LiW 
I lags V

Mutual Inductance
 When 2 coils in close
proximity, a changing
current in one coil will
induce a voltage in a
second coil
0 90 180 270 360
N1 = 5 Turns
100 Volts
N2 = 5 Turns
100 Volts

Inductive Reactance XL
 Inductive Reactance is
the AC Resistance of a
coil
 Presented as a
resistance in Ohms
 Frequency and
Inductance Dependant
fLXL 2

Capacitance
 Stores energy in an
electric field
 Dielectric between 2
plates
 The charged condition
is maintained until a
discharge path is
present
 Causes current to lead
voltage
+
-

Capacitive Reactance XC
fC
XC
2
1


Phase Angle / Power Factor
 In a coil or motor,
current lags behind
voltage
 This is represented as
an angle or a fraction
of ‘unity’
 Adding C can improve
PF
IV
0 90 180 270 360

Impedance Z
fC
XC
2
1
fLXL 2
DC
Resistance
Complex
AC
Resistance
22
)( CL XXRZ 

Summary
 Atomic Structure and Electron Movement
 Conductors, Semi-Conductors, Insulators
 Basic Electricity: Current, Voltage and Resistance
 Electrical and Magnetic Fields
 Alternating Current Electricity: L, C, XL, XC, Z

Generation & Distribution
Generator
10.6 KV
GT
220 KV
Step down
transformer
Distribution
Power plant Transmission
system
Distribution system

• AC generators (“alternators”) generate
electricity
• Electricity generated at 9-13 KV
• Power generated from 67.5 to 1000 MW
• Power stations: generating transformers
(GTs) to increase voltage to 132-400 KV
• Substations: step-down transformers to
reduce voltage before distribution

Benefits of high voltage transmission
• Less voltage drop: good voltage regulation
• Less power loss: high transmission
efficiency
• Smaller conductor: lower costs

83
Single phase AC circuit:
• Two wires connected
to electricity source
• Direction of current
changes many times
per second
Phase of Electricity
3-phases of an electric system
Three phase systems:
• 3 lines with electricity from 3 circuits
• One neutral line
• 3 waveforms offset in time: 50-60 cycles/second

Star connection
Phase of Electricity
Delta connection

Review of Phasors
 Goal of phasor analysis is to simplify the
analysis of constant frequency ac systems
 v(t) = Vmax cos(wt + qv)
 i(t) = Imax cos(wt + qI)
Root Mean Square (RMS) voltage of sinusoid
2 max
0
1
( )
2
T
V
v t dt
T


Phasor Representation
j
( )
Euler's Identity: e cos sin
Phasor notation is developed by rewriting
using Euler's identity
( ) 2 cos( )
( ) 2 Re V
V
j t
j
v t V t
v t V e
q
w q
q q
w q

 
 
   

Then drop the constant terms
( ) Re 2
V cos sin
I cos sin
V
V
j
V
jj t
V V
I I
V V e V
v t Ve e
V j V
I j I
q
qw
q
q q
q q
  

 
 

Advantages of Phasor Analysis
0
2 2
Resistor ( ) ( )
( )
Inductor ( )
1 1
Capacitor ( ) (0)
C
Z = Impedance
R = Resistance
X = Reactance
X
Z = =arctan( )
t
v t Ri t V RI
di t
v t L V j LI
dt
i t dt v V I
j C
R jX Z
R X
R
w
w


 
 
 
   


Device Time Analysis Phasor

RL Circuit Example
2 2
( ) 2 100cos( 30 )
60Hz
R 4 3
4 3 5 36.9
100 30
5 36.9
20 6.9 Amps
i(t) 20 2 cos( 6.9 )
V t t
f
X L
Z
V
I
Z
t
w
w

w
  

   
    
 
 
 
   
  

Complex Power
max
max
max max
( ) ( ) ( )
v(t) = cos( )
(t) = cos( )
1
cos cos [cos( ) cos( )]
2
1
( ) [cos( )
2
cos(2 )]
V
I
V I
V I
p t v t i t
V t
i I t
p t V I
t
w q
w q
     
q q
w q q



   
  
 
Power

max max
0
max max
1
( ) [cos( ) cos(2 )]
2
1
( )
1
cos( )
2
cos( )
= =
V I V I
T
avg
V I
V I
V I
p t V I t
P p t dt
T
V I
V I
q q w q q
q q
q q
 q q
    

 
 


Power Factor
Average P
Angle
ower

P
Q
S
Power Triangle Inductive Load, lagging Power
Factor.

P
Q
S
Power Triangle Capacitive load, Leading
Power Factor

*
[cos( ) sin( )
P = Real Power (W, kW, MW)
Q = Reactive Power (var, kvar, Mvar)
S = Complex power (VA, kVA, MVA)
V I V IS V I j
P
I
jQ
V
q q q q   
 


Power Factor (pf) = Cosø
If current leads voltage then pf is leading
If current lags voltage then pf is lagging

1
Relationships between real, reactive and complex power
cos
sin
Example: A load draws 100 kW with a leading pf of 0.85.
What are (power factor angle), Q and S?
-cos 0.85 31.8
100
117.6
0.85
P S
Q S
kW
S



 


   
  kVA
117.6sin( 31.8 ) 62.0 kVarQ     

Conservation of Power
At every node (bus) in the system
 Sum of real power into node must equal zero
 Sum of reactive power into node must equal zero
This is a direct consequence of Kirchoff’s
current law, which states that the total
current into each node must equal zero.
 Conservation of power follows since S = VI*

Conversation of Power
Example
Earlier we found
I = 20-6.9 amps
*
*
R
2
*
L
2
100 30 20 6.9 2000 36.9 VA
36.9 pf = 0.8 lagging
S 4 20 6.9 20 6.9
1600
S 3 20 6.9 20 6.9
1200var
R
L
S V I
V I
W I R
V I j
I X

        
 
       
 
       
 

Power Consumption in
Devices
2
Resistor Resistor
2
Inductor Inductor L
2
Capacitor Capacitor C
Capacitor
Resistors only consume real power
P
Inductors only consume reactive power
Q
Capacitors only generate reactive power
1
Q
Q
C
I R
I X
j
I X X
j C Cw w



   

2
Capacitor
C
V
X


Example
*
40000 0
400 0 Amps
100 0
40000 0 (5 40) 400 0
42000 16000 44.9 20.8 kV
S 44.9 20.8 400 0
17.98 20.8 MVA 16.8 6.4 MVA
V
I
V j
j
V I
j
 
   
  
      
    
     
    
First solve
basic circuit

Example
Now add additional
reactive power load
and resolve
70.7 0.7 lagging
564 45 Amps
59.7 13.6 kV
S 33.7 58.6 MVA 17.6 28.8 MVA
LoadZ pf
I
V
j
  
   
  
    

Balanced 3 Phase () Systems
A balanced 3 phase () system has
 three voltage sources with equal magnitude, but with an
angle shift of 120
 equal loads on each phase
 equal impedance on the lines connecting the generators
to the loads
Bulk power systems are almost exclusively 3
Single phase is used primarily only in low voltage, low power
settings, such as residential and some commercial

Balanced 3 -- No Neutral Current
* * * *
(1 0 1 1
3
n a b c
n
an an bn bn cn cn an an
I I I I
V
I
Z
S V I V I V I V I
  
         
   

Advantages of 3 Power
Can transmit more power for same amount of
wire (twice as much as single phase)
Torque produced by 3 machines is constant
Three phase machines use less material for
same power rating
Three phase machines start more easily than
single phase machines

Three Phase - Wye Connection
There are two ways to connect
3 systems
 Wye (Y)
 Delta ()
an
bn
cn
Wye Connection Voltages
V
V
V
V
V
V



  
   
   
+

Wye Connection Line Voltages
Van
Vcn
Vbn
Vab
Vca
Vbc
-Vbn
(1 1 120
3 30
3 90
3 120
ab an bn
bc
ca
V V V V
V
V V
V V
 



       
   
   
   
Line to line
voltages are
also balanced

Wye Connection
Define voltage/current across/through device
to be phase voltage/current
Define voltage/current across/through lines to
be line voltage/current
6
3
3 1 30 3
3
j
Line Phase Phase
Line Phase
Phase Phase
V V V e
I I
S V I


   



Delta Connection
Ica
Ic
Iab
Ibc
Ia
Ib
a
b
a
3
For the Delta
phase voltages equal
line voltages
For currents
I
3
I
I
3
ab ca
ab
bc ab
ca bc
Phase Phase
I I
I
I I
I I
S V I
 
   
 
 


Wye Connection Line Voltages
Van
Vcn
Vbn
Vab
Vca
Vbc
-Vbn
(1 1 120
3 30
3 90
3 150
ab an bn
bc
ca
V V V V
V
V V
V V
 



       
   
   
   
Line to line
voltages are
also balanced

Wye Connection Line Voltage
Define voltage/current across/through device to be phase
voltage/current
Define voltage/current across/through lines to be line
voltage/current
6
3
3 1 30 3
3
j
Line Phase Phase
Line Phase
Phase Phase
V V V e
I I
S V I


   



Delta Connection
Ica
Ic
Iab
Ibc
Ia
Ib
3
For the Delta
phase voltages equal
line voltages
For currents
I
3
I
I
3
a ab ca
ab
b bc ab
a ca bc
Phase Phase
I I
I
I I
I I
S V I
 
   
 
 


Three Phase Example
Assume a -connected load is supplied from a 3 13.8 kV
(L-L) source with Z = 1020W
13.8 0
13.8 0
13.8 0
ab
bc
ca
V kV
V kV
V kV
  
   
  
13.8 0
138 20
138 140 138 0
ab
bc ca
kV
I amps
I amps I amps
 
    
 
      

*
138 20 138 0
239 50 amps
239 170 amps 239 0 amps
3 3 13.8 0 kV 138 amps
5.7 MVA
5.37 1.95 MVA
pf cos20 lagging
a ab ca
b c
ab ab
I I I
I I
S V I
j
        
   
      
       
 
 
   

Delta-Wye Transformation
Y
phase
To simplify analysis of balanced 3 systems:
1) Δ-connected loads can be replaced by
1
Y-connected loads with Z
3
2) Δ-connected sources can be replaced by
Y-connected sources with V
3 30
Line
Z
V



 

Per Phase Analysis
Per phase analysis allows analysis of balanced 3 systems
with the same effort as for a single phase system
Balanced 3 Theorem: For a balanced 3 system with
All loads and sources Y connected
No mutual Inductance between phases

Per Phase Analysis
Then
 All neutrals are at the same potential
 All phases are COMPLETELY decoupled
 All system values are the same sequence as
sources. The sequence order we’ve been using
(phase b lags phase a and phase c lags phase a) is
known as “positive” sequence; later in the course
we’ll discuss negative and zero sequence systems.

Per Phase Analysis Procedure
To do per phase analysis
1. Convert all  load/sources to equivalent Y’s
2. Solve phase “a” independent of the other phases
3. Total system power S = 3 Va Ia
*
4. If desired, phase “b” and “c” values can be
determined by inspection (i.e., ±120° degree phase
shifts)
5. If necessary, go back to original circuit to
determine line-line values or internal  values.

Per Phase Example
Assume a 3, Y-connected generator with Van = 10
volts supplies a -connected load with Z = -j through a
transmission line with impedance of j0.1 per phase.
The load is also connected to a
-connected generator with Va”b” = 10 through a second
transmission line which also has an impedance of j0.1
per phase.
Find
1. The load voltage Va’b’
2. The total power supplied by each
generator, SY and
S

Per Phase Example
First convert the delta load and source to equivalent
Y values and draw just the "a" phase circuit

Per Phase Example
' ' '
a a a
To solve the circuit, write the KCL equation at a'
1
(V 1 0)( 10 ) V (3 ) (V j
3
j j           

Per Phase Example
' ' '
a a a
'
a
' '
a b
' '
c ab
To solve the circuit, write the KCL equation at a'
1
(V 1 0)( 10 ) V (3 ) (V j
3
10
(10 60 ) V (10 3 10 )
3
V 0.9 volts V 0.9 volts
V 0.9 volts V 1.56
j j
j j j j
           
     
     
    volts

Per Phase Example
*'
*
ygen
*" '
"
S 3 5.1 3.5 VA
0.1
3 5.1 4.7 VA
0.1
a a
a a a
a a
gen a
V V
V I V j
j
V V
S V j
j

 
    
 
 
    
 

125
Start-Delta Transformation

THANK YOU!
For Contact
Email: arslan_engineer61@yahoo.com
LinkedIn: https://pk.linkedin.com/pub/arslan-ahmed-amin-p-e-b-sc-ee-m-sc-ee-m-b-
a/24/853/68

Lecture 1 Basics of Electric Circuits

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