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Chapter 12
Three-Phase Circuit
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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Structure of ower System
: 10~20kV
: 154, 345, 765kV
: 154, 66kV : 380V
: 120/220V: 6.6, 11.4, 22.9kV
Energy
Source
Fossil
Fuel
Hydro
Power
Power Plant
Ther
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Mecha
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Electri
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Transf
ormati
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Transf
ormer
Power
Transmission
Line
Transf
ormati
on
High
Voltage
Transmis
sion
Netwo
rk
Trans
format
ion
Distribu
tion
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mption
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ormer
Transformer
for Distribution
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toryLarge
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Mecha
nical
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Substation Substation
Power
Distribution
Line
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12.1 What is a Three-Phase Circuit?
a
b
A
B
V pÐf Z L
Single Phase vs. Three Phase
N
S
a
n
b
n
c n
N
S
a
b
magneto
a
b
A
B
V pÐ0°
Z L
c C
V pÐ-120°
n N
V pÐ+120°
Z L
Z L
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• It is a system produced by a generator consisting ofthree sources having the same amplitude andfrequency but out of phase with each other by 120°.
12.1 What is a Three-Phase Circuit?
Three sources
with 120° out
of phaseFour wired
system
neutral line
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Advantages:
1. Most of the electric power is generated anddistributed in three-phase.
2. The instantaneous power in a three-phase systemcan be constant.
3. The amount of power of the three-phase systemis more economical than that of the single-phase.
4. In fact, the amount of wire required for a three-
phase system is less than that required for anequivalent single-phase system.
12.1 What is a Three-Phase Circuit?
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12.2 Balanced Three-Phase Voltages
• Balanced three-phase voltages are defined as the voltagesequal in magnitude and are out of phase with each other by120 o.
Sinusoidal form of the balanced three-phase voltages
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12.2 Balanced Three-Phase Voltages
• Two possible configurations:
Three-phase voltage sources:
(a) Y-connected configuration; (b) Δ-connected configuration
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cnbnan
p pcn
pbn
pan
leadsturninwhichleads
V V
V
V
V,VV
V120V240V
V120V
V0V
°Ð°-Ð
°-аÐ
9
12.2 Balanced Three-Phase VoltagesThree-phase voltage configuration has two possible combinations
with different phase sequence.
In both cases,
abc or positive sequence(clockwise direction) acb or negative sequence(counterclockwise direction)
bncnan
p pbn
pcn
pan
leadsturninwhichleads
V V
V V
V,VV
V120V240V
V120VV0V
°Ð°-Ð
°-аÐ
effective or rms
0an bn cn
an bn cn
v v v
v v v
+ +
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12.2 Balanced Three-Phase Voltages
• Balanced phase voltages are equal inmagnitude and are out of phase with each otherby 120°.
• The phase sequence is the time order in whichthe voltages pass through their respectivemaximum values.
Graph in the time domain for the positive sequence
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12.2 Balanced Three-Phase Voltages
Example 1
Determine the phase sequence of the
set of voltages.
)110cos(200)230cos(200
)10cos(200
°- °-
°+
t vt v
t v
cn
bn
an
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12.2 Balanced Three-Phase Voltages
Solution:
The voltages can be expressed in phasor formas
We notice that Van leads Vcn by 120° and Vcn inturn leads Vbn by 120°.
Hence, we have an acb (negative) sequence.
V110200V
V230200V
V10200V
°-Ð
°-Ð
°Ð
cn
bn
an
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12.3 Balanced Three-Phase Connection
Balanced Load Configuration
Z1
Z2
Z3
a
b
c
n
Za
ZbZc
a
b
c
Y connected Load D connected Load
- A balanced load is one that in which the phase impedance are equal in
magnitude and phase.
1 2 3, =Y a b c Z Z Z Z Z Z Z Z D
- Y-connected Load can be interchangeable with D connected load each other.
13 , ,3
Y Y Z Z Z Z D D
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Delta-Y Conversion
Similarly
321
213 )(
R R R
R R R
R R cb ++
+
+ 321321 )(
R R R
R R R
R R ac ++
+
+ Rbc = Rca =(2) (3)
From (1), (2), (3)
R R R
R R R
R R R
R R R
R R R
R R
R
c
b
a
++
++
++
321
13
321
32
321
21
++
++
++
R
R R R R R R R
R
R R R R R R R
R
R R R R R R R
a
accbba
c
accbba
b
accbba
3
2
1
Delta to Y
Y to Delta
Delta to Y
conversion
Y to Delta
conversion
(4)
(5)
(6)14
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12.3 Balanced Three-Phase Connection
•Four possible connections
1. Y-Y connection (Y-connected source with a
Y-connected load)
2. Y-Δ connection (Y-connected source with aΔ-connected load)
3. Δ-Δ connection
4. Δ-Y connection
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Three Line (Wire) System
, , L ab bc cav v v v
- Line Current
, , L a b c I I I I - Line Voltage
A
B
C
a
c
b
Ia
Ib
Ic
Three Phase
Signal Source
Three Phase
Load
wrt the
interconnection
Ib
Ic
ZY
ZY ZY
A
B C
Ia
N
I AN
IBN ICN
Ib
Ic
A
B C
ZD ZD
ZDIBC
I AB
ICA
Ia
1
2
, ,, ,
L AN AN CN
L AB BC CA
v v v vv v v v
- Phase Current
1
2
, ,
, ,
P AN BN CN
P AB BC CA
I I I I
I I I I
- Phase Voltage
Y-connected
Δ -connected
Y-connected
Δ -connected
In Y-connected system
phase current = line current
phase voltage ≠ line voltage
In Δ -connected system
phase current ≠ line current
phase voltage = line voltage
Need to find phase (line) current & phase (line) voltage
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12.3 Balanced Three-Phase Connection
The line voltages lead their corresponding phase voltage by 30o.
°-Ð-
°-Ð-
°Ð
++
°-Ð-°Ð-
°Ð
°-Ð
°Ð
2103VVV
903VVV
similarly,
3032
3
2
11
1200VVV
V,V,V
voltageslineorvoltagesline-to-lineThe
V120V
V120V
V0V
pancnca
pcnbnbc
p p
p pbnanab
cabcab
pcn
pbn
pan
V
V
V jV
V V
areand
V
V
V
Phasor diagram illustrating the relationship between line and phase voltages in
a balanced Y-Y system.
Van
Vbn
Vcn Vab
Vbc
Vca
Line voltage:abc sequence
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12.3 Balanced Three-Phase Connection
The voltage across the neutral line is zero, so the neutral line can be removed
without affecting the system.
0IZV
0III-I
0III
240I240VV
I
120I
120VV
I
VI
++
++
-Ð-Ð
-Ð
-Ð
nnnN
cccn
ccc
o
a
Y
o
an
Y
cnc
o
aY
o
an
Y
bn
b
Y
ana
or
Z Z
Z Z
Z
Applying KVL to each phase,
we obtain line currentsby KVL
- Only three wires are needed in the balanced three phase system
- Find line currents, which are phase currents in Y-configuration
Line current:
abc sequence
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12.3 Balanced Three-Phase Connection
An alternative way of analyzing a balanced Y-Y system is to do so on
a “per phase” basis.
From Ia, we use the phase sequence to obtain other line currents. Thus, as long
as the system is balanced Y-Y connected, we need only analyze one phase.
Y
ana
Z
VI
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12.3 Balanced Three-Phase Connection
Example 2
Calculate the line currents in the three-wire Y-Y
system shown below:
A
A
A
j j j Z
Z
ooo
ac
oo
ab
o
o
o
a
o
Y
Y
ana
2.9881.68.26181.6240II
8.14181.6120II
8.2181.68.21155.16
0110I
8.21155.1661581025
VI
Ð-Ð-Ð
-Ð-Ð
-ÐÐÐ
Ð+++-
Balanced Y-Y system ? OK. Then,line impedance
load impedance
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12.3 Balanced Three-Phase Connection
°-а-Ð
°-Ð-+
°-Ð-°Ð
°-Ð
°-а-аÐ
303II,303II,
303I866.05.01I
240101II-II
240II
240II,120II,0II
I-III-III-II ,,,
CAc BC b
AB AB
ABCA ABa
ABCA
ABCA AB BC m AB
BC CAc AB BC bCA ABa
similarly
j
Since
The line currents are obtained from the phase currents.
CA BC AB p
cba L
p L
I
I
I I
III
III
where,3
Phasor diagram illustrating the
relationship between phase
and line currents in delta
connection.
An alternative way of analyzing the Y-∆ circuit is to
transform the ∆ connected load to an equivalent Y
connected load. Using Y-∆ transformation formula
3
ZZ DY
phase current
line current
Each line current lags the
corresponding phase
current by 30o.
abcsequence
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12.3 Balanced Three-Phase ConnectionExample 3
A balanced abc -sequence Y-connected source with( ) is connected to a Δ-connected load (8+j4) per phase. Calculate the phase and line currents.
SolutionUsing single-phase analysis,
Other line currents are obtained using the abc phasesequence
°Ð 10100Van
A57.1654.33
57.26981.2
10100
3/Z
VI an °-Ð
°Ð
°Ð
D
a
A43.10354.33120II
A57.13654.33120II
°Ð°+Ð
°-а-Ð
ac
ab
°-Ð 303II ABa
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• A balanced Δ- Δ system is a three-phase system with a
balanced Δ -connected source and a balanced Δ -connectedload.
12.3 Balanced Three-Phase Connection
V120V
V120V
V0V
°Ð
°-Ð
°Ð
pca
pbc
pab
V
V
V
Assuming positive sequence,
In this configuration, line voltages are same asphase voltages.
If there is no line impedance, the phase voltages
of the delta-connected source are equal to the
voltages across the impedances;
CAca Bbc ABab VV,VV,VV C
Hence, the phase currents are
DD
DD
DD
Z Z
Z Z
Z Z
caCACA
bc BC BC
ab AB A
VVI
,VV
I
,VV
I B
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• A balanced Δ- Δ system is a three-phase system with a balanced Δ -
connected source and a balanced Δ -connected load.
12.3 Balanced Three-Phase Connection
The line currents are obtained from the phase currents by
applying KCL. (same w/ balanced Y-Δ system)
°-а-Ð
°-Ð-+
°-Ð-°Ð°-Ð
303II,303II,
303I866.05.01I
240101II-II240II
I-III-III-II ,,,
CAc BC b
AB AB
ABCA ABa
ABCA
BC CAc AB BC bCA ABa
similarly
j
Since
CA BC AB p
cba L
p L
I
I
I I
III
III
where,3
Each line current lags the
corresponding phase current
by 30o.
An alternative way of analyzing the Δ- Δ circuit is to convert both the source and the load to their Y equivalents.
3ZZ DY see the next section.
Phasor diagram illustrating the
relationship between phase
and line currents.
Each line current lags thecorresponding phase
current by 30o.
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• A balanced Δ- Δ system is a three-phase system with a balanced Δ -
connected source and a balanced Δ -connected load.
12.3 Balanced Three-Phase Connection
The line currents are obtained from the phase currents by
applying KCL. (same w/ balanced Y-Δ system)
°-а-Ð
°-Ð-+
°-Ð-°Ð°-Ð
303II,303II,
303I866.05.01I
240101II-II240II
I-III-III-II ,,,
CAc BC b
AB AB
ABCA ABa
ABCA
BC CAc AB BC bCA ABa
similarly
j
Since
CA BC AB p
cba L
p L
I
I
I I
III
III
where,3
Each line current lags the
corresponding phase current
by 30o.
An alternative way of analyzing the Δ- Δ circuit is to convert both the source and the load to their Y equivalents.
3ZZ DY see the next section.
Phasor diagram illustrating the
relationship between phase
and line currents.
Each line current lags thecorresponding phase
current by 30o.
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• A balanced Δ-Y system is a three-phase system with abalanced Δ -connected source and a balanced Y-
connected load.
12.3 Balanced Three-Phase Connection
Assuming positive sequence,
V120VV,120VV,0V °Ð°-аРpca pbc pab V V V
These are also the line voltages and phase voltages.
We can obtain the line currents in many ways.
Method1.
Apply KVL to loop aANBba, writing
Y
o
pba
o pabbaY
bY aY ab
Z V I I V I I Z
or I Z I Z
00V
0V-
Ð-Ð-
-+
Since, ,120 sequenceabcin I I oab -Ð
oaa
o
a ba
303I2
3 j
2
11I12011III Ð
++-Ð--
o
ac
o
ab
Y
o
p
a I I I I Z
V I 120,120,
303/Ð-Ð
-Ð
2 3 l d h h
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Balanced Δ-Y system
12.3 Balanced Three-Phase ConnectionMethod2.
To replace the delta-connected source with its equivalent Y-
connected source.
☞ Line voltages of a Y-connected source lead their correspondingphase voltage by 30o. Therefore, we obtain each phase voltage of
the equivalent Y-connected source by and shifting its phase -30o.3
°+а-а-Ð 903
V,1503
V,303
V p
cn
p
bn
p
an
V V V
single phase equivalent circuit
Y
p
a Z
V 303/I
°-Ð
°-Ð 303
V p
an
V
V0V °Ð pab V
Recalling
V120VV,120VV,0V °Ð°-аРpca pbc pab V V V positive
sequence
Same result
from the KVL
method
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12.4 Power in a Balanced System
• Let’s examine the instantaneous power observed by the load.
☞ In the time domain, for a Y-connected load, the phase voltages are
o pCN o p BN p AN t V vt V vt V v 120cos2,120cos2,cos2 +- necessary because V p has been defined as the rms value of the phase voltage.
, Ð Z Z If Y
o pco pb pa t I it I it I i 120cos2,120cos2,cos2 +----
☞ The total instantaneous power in the load is the sum of the instantaneous powers in the three phases
oooo
p p
cCN b BN a AN cba
t t t t t t I V
iviviv p p p p
120cos120cos120cos120coscoscos2 +-++---+-
++++
☞ Applying the trigonometric identity B A B A B A -++ coscos2
1coscos
12 4 P i B l d S t
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12.4 Power in a Balanced System
• Let’s examine the instantaneous power observed by the load.
cos3cos2
12coscos3
)2(
240sinsin240coscos240sinsin240coscoscoscos3
2402cos2402cos2coscos3
p p p p
oooo
p p
oo
p p
I V I V
t where
I V
t t t I V p
-++
-
-++++
+-+--+-+
Total instantaneous power in a balanced three-phase system is constant !
The average power per phase P p for either Y-connected or ∆-connected load is p/3 or cos p p p I V P
and the reactive power per phase is sin p p p I V Q
The apparent power per phase is p p p I V S , the complex power per phase is*IVS p p p p p jQ P +
phase voltage and phase current, respectively.The total average/reactive power:
)3:,3:(
sin33sin3,cos3cos33
p L p L p L p L
L L p p p L L p p pcba
V V but I I load connected V V but I I load connected Y
I V Q I V Q I V I V P P P P P
-D-
++
The complex power: Ð+ L L p p p p p jQ P IV3Z3II3VS3S2*
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12.5 Unbalanced Three-Phase Systems
• An unbalanced system is due to unbalanced voltage sources or an
unbalanced load.
0)III(I
,Z
VI,
Z
VI ,
Z
VI
++-
cban
C
CN c
B
BN b
A
AN a
• To calculate power in an unbalanced three-phase system requires that we find the
power in each phase.
• The total power is not simply three times the power in one phase but the sum of
the powers in the three phases.
☞ Unbalanced three-phase Y-connected load
(1) The source voltages are not equal in magnitude and/or differ in phase by angles are unequal, or (2)
load impedances are unequal.
Neutral line can not be removed.
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12.5 Unbalanced Three-Phase SystemsExample 7
Consider the ∆- ∆ system shown below. Take Z1=8+6j ,Z2=4.2-2.2j , Z3=10+0j . (a) Find the phase current I AB, IBC,
and ICA. (b) Calculate the line currents IaA, IbB, and IcC.
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Problems
12.8
12.10
12.13
12.18
12.21
12.25
12.31
12.40
12.42
12.54