lecture 1 basics of electric circuits
TRANSCRIPT
Lectures on Electrical
and Instrument
Engineering
Focused, To-The-Point and Practical!
Personal IntroductionEngr. 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 Instituteof Technology (MIT) USA.
‘Project Management’ from Virtual University ofPakistan (VU).
‘Production and Operations Management’ from VirtualUniversity of Pakistan (VU).
‘Conflict Management’ from Virtual University ofPakistan (VU).
‘Crisis Management’ from Virtual University of Pakistan(VU).
Professional Courses
QMS 9001, ISO 14001 EMS and OHSAS 18001, ERP System, Cost ofQuality, Productivity Improvement Techniques, Process SafetyManagement, Hazard Identification and Risk Assessment, HAZOP,SIL systems, Occupational health and Safety, Permit to worksystem, Safety Modules (Complete), Communication Skills, TeamWork 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 ofManagement and Technology, Lahore.
Installation, calibration and maintenance of Fire and Gas detectorsby Det-tronics.
Generation, Transmission and Distribution at WAPDAEngineering 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 fivesemesters for excellent academic performance in B.Sc.Electrical Engineering.
Overall Topped in F.Sc. in Board of Intermediate andSecondary Education, Faisalabad 2006.
Winner of Quaid-e-Azam Scholarship.
Gold medal winner in District Science Quiz CompetitionFaisalabad.
Represented as ‘Talent of Pakistan Youth’ in China in 2007 byMinistry 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
The SI Prefixes
Charge and Current
Electric Current
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)
Voltage
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
ee
e ee
ee
e
e
ee
ee
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
+
SourceVoltage
-
I
+ Transducer -Voltage
The source expendsenergy & the transducerconverts 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 ismeasured 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
Passive Sign Convention
PSC: Example I
PSC: Example II
PSC: Example III
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
Automotive circuits
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
So
urc
e
Lo
ad
(a) Conceptualrepresentation
Power flow
(b) Symbolic (circuit)representation
(c) Physicalrepresentation
+_
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)
Variablevoltagesource
Currentmeter
+
–
v
i
The resistance element
i
R v
+
–
A
l 1/R
i
v
i-v characteristicCircuit symbolPhysical resistorswith resistance R.
Typical materials arecarbon, metal film.
R =l
A
Resistor color code
b 4 b 3 b 2 b 1
Color bands
blackbrownredorangeyellowgreen
012345
bluevioletgraywhitesilvergold
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– +
+
–
vi
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
N series 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
v S a+_
(a)
Rx
v bva b
d
c
R2
R3R1+_
(b)
R x
v bva b
d
avS
A force-measuring instrument
R2
R3R1
vS
R4
vbva
d
c
+
–
ia ib
h
w
Beam cross section
R2 , R3 bondedto bottom surface
F
Practical voltage source
R L
rSi S
+_vS
+
–
v L
Practical
voltagesource
iS =vS
rS + R L
lim iS =vS
rSRL 0
r S i S 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.
i S
+
–
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 seriescircuit
R2
R1
+_vS
A
A
Symbol forideal ammeter
Circuit for the measurementof the current i
i i
Measurement of voltage
R2
R1
+_vS
A seriescircuit
R1
+_ VV
Idealvoltmeter
Circuit for the measurementof the voltage v2
i
v2
+
–i
R2v2
+
–v2
+
–
vS
Models for practical ammeter
and voltmeter
rm
A
Practicalammeter
V
Practicalvoltmeter
rm
Measurement of power
i
R1
+_
Internal wattmeter connections
R2v 2
+
–
vS
iR1
+_
Measurement of the powerdissipated in the resistor R2:
P2 = v2 i
vS
W
R 2v2
+
–
V
A
Definition of a branch
a
rm
A
Practicalammeter
Idealresistor
Rv
A battery
A branch
Branchvoltage
Branchcurrent
+
–
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 loopsin the same circuit may in-clude some of the same ele-ments or branches.
iSR1 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)
Frequency
0 90 180 270 360
Inductance
Stores electromagnetic energy
in its magnetic field
mH
dt
diLV
t
idvL
i0
)0()(1
2
2
1LiW
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
fLX L 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
fCX C
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
fCX C
2
1fLX L 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
GT220 KV
Step down
transformer
Distribution
Power plant Transmission
systemDistribution 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
Generation & Distribution
Benefits of high voltage transmission
• Less voltage drop: good voltage regulation
• Less power loss: high transmission
efficiency
• Smaller conductor: lower costs
Generation & Distribution
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
TV
v t dtT
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
jV
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 1Capacitor ( ) (0)
C
Z = Impedance
R = Resistance
X = Reactance
XZ = =arctan( )
t
v t Ri t V RI
di tv t L V j LI
dt
i t dt v V Ij C
R jX Z
R XR
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
VI
Z
t
w
w
w
Complex Power
max
max
max max
( ) ( ) ( )
v(t) = cos( )
(t) = cos( )
1cos 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( )
1cos( )
2
cos( )
= =
V I V I
T
avg
V I
V I
V I
p t V I t
P p t dtT
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
QS
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
100117.6
0.85
P S
Q S
kWS
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
2Resistor Resistor
2Inductor Inductor L
2
Capacitor Capacitor C
Capacitor
Resistors only consume real power
P
Inductors only consume reactive power
Q
Capacitors only generate reactive power
1Q
Q
C
I R
I X
jI X X
j C Cw w
2
Capacitor
C
V
X
Example
*
40000 0400 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
VI
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 powersettings, 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
VI
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
VabVca
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
IcaIc
IabIbc
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
Three Phase Transmission Line
Wye Connection Line Voltages
Van
Vcn
Vbn
VabVca
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
IcaIc
IabIbc
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 = 1020W
13.8 0
13.8 0
13.8 0
ab
bc
ca
V kV
V kV
V kV
13.8 0138 20
138 140 138 0
ab
bc ca
kVI 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 cos 20 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 Z3
2) Δ-connected sources can be replaced by
Y-connected sources with V3 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 assources. The sequence order we’ve been using(phase b lags phase a and phase c lags phase a) isknown as “positive” sequence; later in the coursewe’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 = 10volts 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” = 10 through a second transmission line which also has an impedance of j0.1per phase.
Find
1. The load voltage Va’b’
2. The total power supplied by each
generator, SY andS
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
3j 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 VA0.1
3 5.1 4.7 VA0.1
a aa a a
a agen a
V VV I V j
j
V VS V j
j
124
Delta-Star Transformation
125
Start-Delta Transformation
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