This document discusses three-phase circuits. It covers the advantages of three-phase circuits, basic configurations including wye and delta connections, power relationships, and methods for power factor correction using capacitors. Power in a balanced three-phase circuit remains constant over time. Transformations can be used to change a wye connection to a delta or vice versa. Capacitors can be added to a three-phase load to increase the power factor by supplying reactive power.

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-Y connections
Power Relationships
Study power delivered by three phase circuits
Power Factor Correction
Improving power factor for three phase circuits

2. THREE PHASE CIRCUITS
)
)(
240
cos(
)
(
)
)(
120
cos(
)
(
)
)(
cos(
)
(
V
t
V
t
v
V
t
V
t
v
V
t
V
t
v
m
c
m
bn
m
an










Voltages
Phase
ous
Instantane
2
120

m
V
Vcn
Vbn
Van
0 120 240
3 phase
voltage

3. _
_
_
+
+
+
n
a
b
c
c
b
a
V0
V-120
V-240
Wye Connected
Source

4. a
b
c
a
b
c
+
+
+
_
_
_
Delta
Source
Delta Source
Vab = | Vab |  0
Vbc = Vab  -120
Vca = Vab  -240

5. Zl
Zl
Zl
ZL
ZL ZL
a
n
b
c
A
B C
N
Wye – Wye System

6. Zl
Zl
Zl
a
b
c
A
B C
ZL
Z
L
Z
L
+
+
+
_
_
_
Delta – Delta System

7. Zl
Zl
Zl
a
b
c
A
B C
+
+
+
_
_
_
ZL
ZL
ZL
Delta – Wye System

8. _
_
_
+
+
+
n
a
b
c
c
b
a
V0
V-120
V-240
A
B C
Z Z
Z
IaA ICA
IBC
IAB

9. Van
Vbn
Vcn
- Vbn
Vab
Vab = Van - Vbn
Vab = 3 Van 30o
30o

10.  

























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

11. IAB
IBC
ICA
- ICA
IaA
IaA = 3 IAB -30o

12. )
240
cos(
)
(
)
120
cos(
)
(
)
cos(
)
(
















t
I
t
i
t
I
t
i
t
I
t
i
m
c
m
b
m
a
Currents
Phase
Balanced
)
(
)
(
)
(
)
(
)
(
)
(
)
( t
i
t
v
t
i
t
v
t
i
t
v
t
p c
cn
b
bn
a
an 


power
ous
Instantane
Theorem
For a balanced three phase circuit the instantaneous power is constant
)
(
cos
2
3
)
( W
I
V
t
p m
m


)
)(
240
cos(
)
(
)
)(
120
cos(
)
(
)
)(
cos(
)
(
V
t
V
t
v
V
t
V
t
v
V
t
V
t
v
m
c
m
bn
m
an










Voltages
Phase
ous
Instantane
INSTANTANEOUS POWER

13. 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



14. Y




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



15. POWER FACTOR CORRECTION
Balanced
load
Low pf
lagging
Similar to single phase case.
Use capacitors to increase the
power factor
Keep clear about total/phase
power, line/phase voltages
added
be
to
Power
Reactive
old
new Q
Q
Q 


To use capacitors this value
should be negative
2
1
sin
cos pf
pf f
f 


 

2
1
tan
pf
pf
f



0

 Q
lagging
f
P
Q 
tan


16. LEARNING EXAMPLE
leading
pf
rms
kV
V
Hz
f line 94
.
0
.
5
.
34
|
|
,
60 

 :
Required
626
.
0
1
sin
cos 2




 pf
pf f
f 

MW
P
MVA
Q
old
old
72
.
18
02
.
15
|
|


2
1
tan
pf
pf
f



0

 old
Q
lagging
f
f
f
pf
S
Q
S
P
jQ
P
S



cos
sin
|
|
cos
|
|





f
P
Q 
tan

MVA
Q
leading
pf
MW
P
new
new
old
8
.
6
94
.
0
72
.
18








MVA
Q
MVA
Q
273
.
7
82
.
21
02
.
15
8
.
6








capacitor
per
rms
kV
V
Y
3
5
.
34


 capacitor
connection
2
3
6
3
10
5
.
34
60
2
10
273
.
7 






 






 C

F
C 
6
.
48


17. LEARNING BY DESIGN
Proposed new store
rms
A
170
at
rated
wire
#4ACSR
kVA
j
S
420
560
9
.
36
700
1





kVA
j
kVA
S
866
500
60
1000
2





kVA
j
kVA
S
349
720
8
.
25
800
3





kVA
kVA
j
Stotal 



 57
.
42
2417
1635
1780
rms
A
V
S
I
line
total
line 1
.
101
10
8
.
13
3
10
417
.
2
3
|
|
|
| 3
6






 Wire is OK
new
new
old
Q
pf
P




old
new Q
Q
Q 


)
(
tan new
f
P 
 kVA
28
.
758

kVA
72
.
876


2
capacitor
per
|
| CV
Q 
  
  3
/
10
8
.
13
60
2
3
/
10
72
.
876
2
3
3






C F

2
.
12
