3. 9/12/2006 ELE A6 Three-phase Circuits 3
Advantages of Three-phase
Circuits
• Smooth flow of power (instantaneous power
is constant).
• Constant torque (reduced vibrations).
• The power delivery capacity tripled
(increased by 200%) by increasing the
number of conductors from 2 to 3
(increased by 50%).
4. 9/12/2006 ELE A6 Three-phase Circuits 4
Three-phase Circuits
Wye-Connected System
• The neutral point is grounded
• The three-phase voltages have
equal magnitude.
• The phase-shift between the
voltages is 120 degrees.
V
0
V =
°
∠
=
an
V
bn
V 120
V °
−
∠
=
cn
V 240
V °
−
∠
=
5. 9/12/2006 ELE A6 Three-phase Circuits 5
Three-phase Circuits
Wye-Connected System
• Line-to-line voltages are the
difference of the phase voltages
30
V
3 ∠
=
= bn
an
ab V
-
V
V
90
-
V
3 ∠
=
= cn
bn
bc V
-
V
V
150
V
3 ∠
=
= an
cn
ca V
-
V
V
6. 9/12/2006 ELE A6 Three-phase Circuits 6
Three-phase Circuits
Wye-Connected System
• Phasor diagram is used to visualize
the system voltages
• Wye system has two type of
voltages: Line-to-neutral, and
line-to-line.
• The line-to-neutral voltages are
shifted with 120o
• The line-to-line voltage leads the
line to neutral voltage with 30o
• The line-to-line voltage is times
the line-to-neutral voltage
3
7. 9/12/2006 ELE A6 Three-phase Circuits 7
Three-phase Circuits
Wye-Connected Loaded System
• The load is a balanced load and each one = Z
• Each phase voltage drives current through the
load.
• The phase current expressions are:
I
V
z
I
V
z
I
V
z
a
an
b
bn
c
cn
= = =
, ,
8. 9/12/2006 ELE A6 Three-phase Circuits 8
Three-phase Circuits
Wye-Connected Loaded System
• Since the load is balanced (Za = Zb =
Zc) then: Neutral current = 0
• This case single phase equivalent
circuit can be used (phase a, for
instance, only)
• Phase b and c are eliminated
Io
a
Van
a
Za
Ia
a
n
9. 9/12/2006 ELE A6 Three-phase Circuits 9
Wye-Connected System with balanced load
• A single-phase equivalent circuit is used
• Only phase a is drawn, because the magnitude of currents and voltages
are the same in each phase. Only the phase angles are different (-120o
phase shift)
• The supply voltage is the line to neutral voltage.
• The single phase loads are connected to neutral or ground.
Three-phase Circuits
ln
V
Load
10. 9/12/2006 ELE A6 Three-phase Circuits 10
Three-phase Circuits
Balanced Delta-Connected System
• The system has only one voltage :
the line-to-line voltage ( )
• The system has two currents :
– line current
– phase current
• The phase currents are:
LL
V
I
V
z
a
ab
=
11. 9/12/2006 ELE A6 Three-phase Circuits 11
Three-phase Circuits
Delta-Connected System
The line currents are:
• In a balanced case the line
currents are:
ca
ab
a I
I
I −
=
ab
bc
b I
I
I −
=
bc
ca
c I
I
I −
=
I I
line phase
= ∠ −
3 30
12. 9/12/2006 ELE A6 Three-phase Circuits 12
Three-phase Circuit
Delta-Connected System
• The phasor diagram is used to
visualize the system currents
• The system has two type of
currents: line and phase currents.
• The delta system has only line-to-
line voltages, that are shifted by
• The phase currents lead the line
currents by
• The line current is times the
phase current and shifted by 30
degree.
°
30
3
13. 9/12/2006 ELE A6 Three-phase Circuits 13
Three-phase Circuit
• Circuit conversions
– A delta circuit can be converted to an equivalent wye
circuit. The equation for phase a is (will not be used for
this course):
– Conversion equation for a balanced system is:
ab
ab
ca
bc
ca
a
Z
Z
Z
Z
Z
Z
+
+
=
3
ab
Z
Za =
14. 9/12/2006 ELE A6 Three-phase Circuits 14
Three-phase Circuit
Power Calculation
• The three phase power is equal the sum of the phase powers
• If the load is balanced:
• Wye system:
• Delta system:
c
b
a P
P
P
P +
+
=
( )
φ
cos
I
V
3
P
3
P phase
phase
phase =
=
LN
LL
L
phase
LN
phase V
3
V
I
I
V
V =
=
=
( ) ( )
φ
φ cos
I
V
3
cos
I
V
3
P L
LL
phase
phase =
=
phase
LL
phase
Line V
V
I
3
I =
=
( ) ( )
φ
φ cos
I
V
3
cos
I
V
3
P =
= L
LL
phase
phase
15. 9/12/2006 ELE A6 Three-phase Circuits 15
Three-phase Circuit
Power measurement
• In a four-wire system (3 phases and a neutral)
the real power is measured using three single-
phase watt-meters.
• In a three-wire system (three phases without
neutral) the power is measured using only two
single-phase watt-meters.
- The watt-meters are supplied by the line
current and the line-to-line voltage.
16. 9/12/2006 ELE A6 Three-phase Circuits 16
Three-phase Circuit
Power measurement
• The total power is
the algebraic sum of
the two watt-meters
reading.
Load Watt meter 1
Wattmeter 2
a
b
c
)
cos(
3
)
cos(
)
cos(
2
1
2
1
θ
θ
θ
θ
θ
L
L
T
Ic
vcb
c
cb
Ia
vab
a
ab
I
V
W
W
P
I
V
W
I
V
W
=
+
=
−
=
−
=
17. 9/12/2006 ELE A6 Three-phase Circuits 17
Example 1
A 345 kV, three phase transmission line delivers 500 MVA, 0.866 power
factor lagging, to a three phase load connected to its receiving end
terminals. Assume the load is Y connected and the voltage at the receiving
end is 345 kV, find:
• The load impedance per phase.
• The line and phase currents.
• The total real and reactive power.
Three-phase Circuits
18. 9/12/2006 ELE A6 Three-phase Circuits 18
MVAR
9
.
249
)
sin(
3
MW
433
)
cos(
3
)
(
30
7
.
836
)
(
119
206
30
238
30
7
.
836
)
866
.
0
(
cos
7
.
836
7
.
836
345
3
500
3
I
,
0
3
345
)
(
1
=
=
=
=
−
∠
=
=
+
=
∠
=
−
∠
=
−
∠
=
=
=
=
∠
=
=
−
θ
θ
φ
φ
φ
φ
φ
φ
φ
φ
L
L
L
L
L
L
I
V
Q
I
V
P
c
I
I
b
j
Z
I
A
kV
MVA
V
S
V
kV
V
I
V
Z
a
19. 9/12/2006 ELE A6 Three-phase Circuits 19
Example 2
Repeat example 2 assuming the load is Delta connected.
Three-phase Circuits
20. 9/12/2006 ELE A6 Three-phase Circuits 20
MVAR
9
.
249
)
sin(
3
MW
433
)
cos(
3
)
(
60
7
.
836
30
3
)
(
357
3
.
618
30
714
30
1
.
483
)
866
.
0
(
cos
1
.
483
1
.
483
345
*
3
500
3
I
,
0
345
)
(
1
=
=
=
=
−
∠
=
−
∠
=
+
=
∠
=
=
−
∠
=
−
∠
=
=
=
=
∠
=
=
−
θ
θ
φ
φ
φ
φ
φ
φ
φ
L
L
L
L
L
L
L
I
V
Q
I
V
P
c
I
I
b
j
I
V
Z
I
A
kV
MVA
V
S
V
kV
V
V
a
21. 9/12/2006 ELE A6 Three-phase Circuits 21
Power factor correction
• Power factor (p.f.) correction is the process of
making P.f. = 1.
• In order to correct the power factor in any system,
a reactive (either inductive or capacitive) will be
added to the load.
• If the load is inductive, then a capacitance is
added.
– Note: Correcting the P.f. WILL NOT affect the active
power, why?
22. 9/12/2006 ELE A6 Three-phase Circuits 22
Example 3
A 3-phase load draws 120 kW at a power factor of 0.85 lagging from a
440 V bus. In parallel with this load, a three phase capacitor bank that
is rated 50 kVAR is inserted, find:
• The line current without the capacitor bank.
• The line current with the capacitor bank.
• The P.F. without the capacitor bank.
• The P.F. with the capacitor bank.
Three-phase Circuits
23. 9/12/2006 ELE A6 Three-phase Circuits 23
98
.
0
)
49
.
11
(
c
capacitor)
(with
.F.
)
(
0.85
capacitor)
(no
.F.
)
(
49
.
11
6
.
160
90
6
.
65
6
.
65
440
3
10
*
50
3
1
)
sin(
load
capacitor
pure
have
we
Since
)
sin(
3
)
(
,
8
.
31
2
.
185
)
85
.
0
(
cos
25
.
185
25
.
185
)
85
.
0
(
440
3
10
*
120
I
)
cos(
3
)
(
)
(
)
(
)
(
3
)
(
)
(
1
3
=
=
=
−
∠
=
+
=
∠
=
=
=
=
=
=
−
∠
=
−
∠
=
=
=
=
−
os
P
d
P
c
I
I
I
I
A
I
I
V
Q
I
V
Q
b
I
A
I
V
P
a
capacitor
L
L
new
L
capacitor
L
capacitor
L
capacitor
L
L
capacitor
L
L
capacitor
L
L
L
L
θ
θ
θ
