1. POLYPHASE CIRCUITS
LEARNING GOALS
Three Phase Circuits
Advantages of polyphase circuits
Three Phase Connections
Basic configurations for three phase circuits
Source/Load Connections
Delta-Wye connections
Power Relationships
Study power delivered by three phase circuits
3. Proof of Theorem
For a balanced three phase circuit the instantaneous power is constant
)
(
cos
2
3
)
( W
I
V
t
p m
m
)
240
cos(
)
240
cos(
)
120
cos(
)
120
cos(
)
cos(
cos
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
t
t
t
t
t
t
I
V
t
p
t
i
t
v
t
i
t
v
t
i
t
v
t
p
m
m
c
cn
b
bn
a
an
power
ous
Instantane
)
cos(
)
cos(
2
1
cos
cos
3cos cos(2 )
( ) cos(2 240 )
2
cos(2 480 )
m m
t
V I
p t t
t
)
120
sin(
sin
)
120
cos(
cos
)
120
cos(
)
120
sin(
sin
)
120
cos(
cos
)
120
cos(
cos
cos
0
)
120
cos(
)
120
cos(
cos
Proof
Lemma
5
.
0
)
120
cos(
0
)
120
cos(
)
120
cos(
cos
)
120
cos(
)
480
cos(
)
120
cos(
)
240
cos(
t
5. SOURCE/LOAD CONNECTIONS
BALANCED Y-Y CONNECTION
120
|
|
120
|
|
0
|
|
p
cn
p
bn
p
an
V
V
V
V
V
V
Positive sequence
phase voltages
ab
V
bc
V
ca
V
Line voltages
30
|
|
3
2
3
2
1
|
|
|
|
)
120
sin
120
(cos
1
|
|
120
|
|
0
|
|
p
p
p
p
p
p
bn
an
ab
V
j
V
V
j
V
V
V
V
V
V
210
|
|
3
90
|
|
3
p
ca
p
bc
V
V
V
V
Voltage
Line
|
|
3 p
L V
V
Y
cn
c
Y
bn
b
Y
an
a
Z
V
I
Z
V
I
Z
V
I
;
;
120
|
|
;
120
|
|
;
|
|
L
c
L
b
L
a I
I
I
I
I
I
0
n
c
b
a I
I
I
I For this balanced circuit it is enough to analyze one phase
6. Relationship between
phase and line voltages
LEARNING EXAMPLE For an abc sequence, balanced Y - Y three phase circuit
30
208
ab
V
Determine the phase voltages
Balanced Y - Y
Positive sequence
a-b-c
30
|
|
3 p
ab V
V
120
|
|
120
|
|
0
|
|
p
cn
p
bn
p
an
V
V
V
V
V
V
Positive sequence
phase voltages
)
30
30
(
3
|
|
ab
an
V
V
30
by
lags ab
an V
V
30
208
ab
V
60
120
180
120
60
120
an
bn
an
V
V
V
The phasor diagram could be rotated by any angle
7. LEARNING EXAMPLE For an abc sequence, balanced Y - Y three phase circuit
source 120( ) , 1 1 , 20 10
rms
phase line phase
V V Z j Z j
Determine line currents and load voltages
Because circuit is balanced
data on any one phase is
sufficient
0
120
Chosen
as reference
120 0
120 120
120 120
an
bn
cn
V
V
V
Abc sequence
rms
A
j
V
I an
aA
)
(
65
.
27
06
.
5
65
.
27
71
.
23
0
120
11
21
rms
A
I
rms
A
I
cC
bB
)
(
65
.
27
120
06
.
5
)
(
65
.
27
120
06
.
5
57
.
26
36
.
22
)
10
20
( aA
aA
AN I
j
I
V
rms
V
VAN )
(
08
.
1
15
.
113
rms
V
V
rms
V
V
CN
BN
)
(
92
.
118
15
.
113
)
(
08
.
121
15
.
113
8. LEARNING EXTENSION For an abc sequence, balanced Y - Y three phase circuit
voltages
line
the
Find
.
)
(
90
120 rms
V
Van
Relationship between
phase and line voltages
30
|
|
3 p
ab V
V
30
by
lags ab
an V
V
30
by
leads
an
ab V
V
rms
V
Vab )
(
120
120
3
rms
V
V
rms
V
V
ca
bc
)
(
240
120
3
)
(
0
120
3
voltages
phase
the
Find
.
)
(
0
208 rms
V
Vab
30
by
lags ab
an V
V
rms
V
Van )
(
30
3
208
rms
V
V
rms
V
V
cn
bn
)
(
90
3
208
)
(
150
3
208
9. LEARNING EXTENSION For an abc sequence, balanced Y - Y three phase circuit
load 104.02 26.6 ( ) , 1 1 , 8 3
rms
phase line phase
V V Z j Z j
Determine source phase voltages
rms
V)
(
6
.
26
02
.
104
8
3
j
Currents are not required. Use inverse
voltage divider
AN
an V
j
j
j
V
3
8
)
1
1
(
)
3
8
(
41
.
3
15
.
1
73
5
84
3
8
3
8
3
8
4
9 j
j
j
j
j
30
120
an
V
Positive sequence
a-b-c
150
120
90
120
cn
bn
V
V
10. DELTA CONNECTED SOURCES
Relationship between
phase and line voltages
30
|
|
3 p
ab V
V
30
by
lags ab
an V
V
120
120
0
L
ca
L
bc
L
ab
V
V
V
V
V
V
90
3
150
3
30
3
L
cn
L
bn
L
an
V
V
V
V
V
V
150
120
90
120
30
120
cn
bn
an
V
V
V
180
208
60
208
60
208
ca
bc
ab
V
V
V
Example
Convert to an equivalent Y connection
11. LEARNING EXAMPLE Determine line currents and line voltages at the loads
Source is Delta connected.
Convert to equivalent Y
120
120
0
L
ca
L
bc
L
ab
V
V
V
V
V
V
90
3
150
3
30
3
L
cn
L
bn
L
an
V
V
V
V
V
V
Analyze one phase
rms
A
j
IaA )
(
14
.
49
38
.
9
2
.
4
1
.
12
30
)
3
/
208
(
rms
V
j
VAN )
(
71
.
30
65
.
118
19
.
49
38
.
9
)
4
12
(
Determine the other phases using the balance
rms
A
I
rms
A
I
cC
bB
)
(
86
.
71
38
.
9
)
(
14
.
169
38
.
9
30
|
|
3 p
ab V
V 3 118.65 0.71
AB
V
3 118.65 120.71
3 118.65 119.29
BC
CA
V
V
12. LEARNING EXTENSION Compute the magnitude of the line voltage at the load
Source is Delta connected.
Convert to equivalent Y
120
120
0
L
ca
L
bc
L
ab
V
V
V
V
V
V
90
3
150
3
30
3
L
cn
L
bn
L
an
V
V
V
V
V
V
Analyze one phase
30
120
1
.
4
1
.
10
4
10
j
j
VAN
10
1
.
0
j
Only interested in magnitudes!
rms
V
VAN )
(
57
.
118
90
.
10
77
.
10
120
|
|
30
|
|
3 p
ab V
V rms
V
VAB )
(
4
.
205
|
|
13. DELTA-CONNECTED LOAD
Method 1: Solve directly
120
|
|
120
|
|
0
|
|
p
cn
p
bn
p
an
V
V
V
V
V
V
Positive sequence
phase voltages
210
|
|
3
90
|
|
3
30
|
|
3
p
ca
p
bc
p
ab
V
V
V
V
V
V
120
|
|
120
|
|
|
|
I
Z
V
I
I
Z
V
I
I
Z
V
I
CA
CA
BC
BC
AB
AB
currents
phase
Load
BC
CA
cC
AB
BC
bB
CA
AB
aA
I
I
I
I
I
I
I
I
I
currents
Line
Method 2: We can also convert the delta
connected load into a Y connected one.
The same formulas derived for resistive
circuits are applicable to impedances
3
Z
ZY
case
Balanced
Z
L
AB
aA
L
aA
Y
an
aA Z
V
I
I
Z
V
I
3
/
|
|
3
/
|
|
|
|
|
|
Z
L
Z
Z
|
|
Z
30
30
|
|
3
|
|
line
line I
I
Line-phase current
relationship
15. Y
b
a
ab R
R
R
)
(
|| 3
1
2 R
R
R
Rab
3
2
1
3
1
2 )
(
R
R
R
R
R
R
R
R b
a
3
2
1
2
1
3 )
(
R
R
R
R
R
R
R
R c
b
3
2
1
3
2
1 )
(
R
R
R
R
R
R
R
R a
c
SUBTRACT THE FIRST TWO THEN ADD
TO THE THIRD TO GET Ra
a
b
b
a
R
R
R
R
R
R
R
R 1
3
3
1
c
b
c
b
R
R
R
R
R
R
R
R 1
2
1
2
REPLACE IN THE THIRD AND SOLVE FOR R1
Y
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
c
b
a
3
2
1
1
3
3
2
1
3
2
3
2
1
2
1
Y
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
a
c
c
b
b
a
c
a
c
c
b
b
a
b
a
c
c
b
b
a
3
2
1
tions
Transforma
Y
OF
REVIEW
3
3
2
1
R
R
R
R
R
R Y
Y
16. LEARNING EXAMPLE
02
.
37
52
.
12
54
.
7
10 j
Z
Delta-connected load consists of 10-Ohm resistance in series
with 20-mH inductance. Source is Y-connected, abc sequence,
120-V rms, 60Hz. Determine all line and phase currents
rms
V
Van )
(
30
120
54
.
7
020
.
0
60
2
inductance
Z
rms
A
j
Z
V
I AB
AB )
(
98
.
22
60
.
16
54
.
7
10
60
3
120
30
|
|
3
|
|
line
line I
I
Line-phase current
relationship
ip
relationsh
voltage
phase
-
Line
30
|
|
3
|
|
phase
phase
V
V
rms
A
IaA )
(
02
.
7
75
.
28
rms
A
I
rms
A
I
CA
BC
)
(
98
.
142
60
.
16
)
(
02
.
97
60
.
16
rms
A
I
rms
A
I
cC
bB
)
(
98
.
112
75
.
28
)
(
02
.
127
75
.
28
02
.
37
17
.
4
Y
Z
Alternatively, determine first the line currents
and then the delta currents
18. LEARNING EXAMPLE
lagging
20
angle
factor
power
W
P
rms
V
V
total
line
1200
)
(
208
|
|
Determine the magnitude of the line
currents and the value of load impedance
per phase in the delta
f
line
line
total
f
line
line
total
I
V
Q
I
V
P
sin
|
||
|
3
cos
|
||
|
3
f
line
line
total I
V
P
cos
3
|
||
|
3
rms
A
Iline )
(
54
.
3
|
|
30
|
|
3
|
|
line
line I
I
Line-phase current
relationship
rms
A
I )
(
05
.
2
|
|
46
.
101
|
|
|
|
|
|
I
V
Z line
20
46
.
101
Z
line
V
line
I
Power factor angle
f
- Impedance angle
line
V
19. LEARNING EXAMPLE For an abc sequence, balanced Y - Y three phase circuit
10
20
,
1
1
,
)
(
120
|
| j
Z
j
Z
V
V phase
line
rms
phase
source
Because circuit is balanced
data on any one phase is
sufficient
0
120
Chosen
as reference
120
120
120
120
0
120
an
bn
an
V
V
V
Abc sequence
rms
A
j
V
I an
aA
)
(
65
.
27
06
.
5
65
.
27
71
.
23
0
120
11
21
Determine real and reactive power per phase at the load and total real, reactive and
complex power at the source
57
.
26
36
.
22
)
10
20
( aA
aA
AN I
j
I
V
rms
V
VAN )
(
08
.
1
15
.
113
65
.
27
06
.
5
08
.
1
15
.
113
*
aA
AN
phase I
V
S
rms
VA
j
Sphase )
(
09
.
256
512
57
.
26
54
.
572
65
.
27
06
.
5
0
120
*
aA
an I
V
S phase
source
VA
j
S
78
.
281
86
.
537
65
.
27
2
.
607
phase
source
)
(
78
.
281
3
)
(
86
.
537
3
VA
Q
W
P
source
total
source
total
)
(
6
.
1821
|
|
)
(
2
.
845
6
.
1613
VA
S
VA
j
Q
P
S
source
total
source
total
source
total
source
total
20. LEARNING EXTENSION A Y -Y balanced three-phase circuit has a line voltage of
208-Vrms. The total real power absorbed by the load is 12kW
at pf=0.8 lagging. Determine the per-phase impedance
of the load
ip
relationsh
voltage
phase
-
Line
30
|
|
3
|
|
phase
phase
V
V
*
3 phase
phase
Total I
V
S
rms
V
Vphase )
(
120
3
208
|
|
*
2
*
|
|
3
3
phase
phase
phase
phase
phase
total
Z
V
Z
V
V
S
line
V
line
I
Power factor angle
Impedance angle
87
.
36
cos
8
.
0 f
f
pf
f
f
f
pf
S
Q
S
P
jQ
P
S
cos
sin
|
|
cos
|
|
kVA
pf
P
S total
total 15
|
|
88
.
2
|
|
|
|
3
|
|
2
total
phase
phase
S
V
Z
87
.
36
88
.
2
pahse
Z
f
21.
30
120
1
.
4
1
.
10
4
10
j
j
VAN
Source is Delta connected.
Convert to equivalent Y
Analyze one phase
10
1
.
0
j
rms
V
VAN )
(
57
.
118
90
.
10
77
.
10
120
|
|
LEARNING EXTENSION Determine real, reactive and complex power at both load
and source
*
2
* |
|
3
3
phase
AN
aA
AN
load
Z
V
I
V
S
*
2
* |
|
3
3
phase
total
an
aA
an
source
Z
V
I
V
S
2
2
2
2
4
10
)
4
10
(
|
57
.
118
|
3
4
10
|
57
.
118
|
3
j
j
2
2
2
2
)
1
.
4
(
)
1
.
10
(
)
1
.
4
1
.
10
(
|
120
|
3
1
.
4
1
.
10
|
120
|
3
j
j
)
496
1
.
1224
(
3
)
8
.
484
0
.
212
,
1
(
3
j
S
j
S
source
load
Polyphase
