Lecture 20: Three-phase circuits
Vijay Singh∗
May 2, 2003
1
Balanced Y-Y connection
ib = Ib
a
A
c
ia = Ia
C
Zy
n
Zy
N
Zy
B
b
ic = Ic
Figure 1: Balanced Y-Y connection.
2
Phase voltages and Line voltages
Vcn
120◦
Van
Phase voltages are:
Van
Vbn
Vcn
= Vp 6 0
= Vp 6 − 120◦
= Vp 6 − 240◦
= Vp 6 120◦
(1)
(2)
(3)
120◦
Vbn
Equations (1), (2), (3) represent a positive phase sequence (please see phasor diagram of Figure 2.
Figure 2: Phasor Diagram.
∗ Professor and Chairman, Department of Electrical & Computer Engineering, University of Kentucky, Lexington, KY,
USA. E-mail: vsingh@engr.uky.edu. Document prepared using LATEX and figures created using Metagraf by Ramprasad
Potluri (potluri@engr.uky.edu).
1
Vcn
Van
Note that
Van + Vbn + Vcn = 0
(4)
(please see Figure 3).
Vab , Vbc , and Vca are called line-to-line voltages or
just line voltages.
3
VAB , VBC , VCA
Van , Vbn , Vcn
...
...
. . . . . . Line Voltages
. . . . . . Phase Voltages
VAN , VBN , VCN
...
. . . . . . Phase Voltages
Vbn
Figure 3: Phasor Diagram.
Relating Phase and Line Voltages
Vnb
Vab
=
=
=
=
=
=
Vab
=
Van + Vnb
Van − Vbn (please see Figure 4)
Vp 6 0 − Vp 6 − 120◦
Vp − Vp (cos(−120◦ ) + j sin(−120◦ ))
"
Ã
√ !
√ #
3
3
3
1
= Vp
Vp − Vp − − j
+j
2
2
2
2
#
"√
√
√
3
1
Vp 3
+j
= Vp 3 [j sin 30◦ + cos 30◦ ]
2
2
√
3Vp 6 30◦
(5)
Vab
30◦
30◦
Van
120◦
Vbn
Figure 4: Phasor Diagram.
Vab
Vcn
Vca
30◦
Van
Vbn
Similarly,
Vbc
Vca
=
=
√
√
3Vp 6 − 90◦
(6)
◦
3Vp 6 − 210
(7)
√
3Vp
√
Line voltages have amplitude equal to 3 times the
phase voltages.
VL = |Vab | = |Vbc | = |Vca | =
2
Vbc
Figure 5: Phasor Diagram.
4
Line Currents
Ia
Ib
Ic
Ia + Ib + Ic
Van
Vp 6 0
=
Zy
Zy
= Ia 6 −120◦
= Ia 6 −240◦
= 0
=
(8)
(9)
(10)
(11)
Note that, in a Y-Y configuration, Line current = IL = Phase current = Iph .
5
Example
◦
Given, VAN = 4406 20 . Find Vab .
We know,
VAN
j1.8
1Ω
a
A
ZL
14 Ω
+
Zy
= Van
Zy + ZL
Van
ZY
−
So,
j12
Van
=
=
Vab
Vab
6
=
=
15 + j13.8
14 + j12
486.376 22◦
√
3(486.37)6 (22◦ + 30◦ )
842.46 52◦
4406 0
n
N
Figure 6: Problem 10.14 (old Irwin textbook).
Conversion between ∆ and Y sources
Ia = Line Current
a
V ca
V ab
Load
Ib
c
b
V bc
In a ∆ connection, Line voltage
and Phase voltage are the same
(Figure 7).
Ic
Figure 7: Delta.
3
a
In a Y connection, Line current
and Phase current are the same
(Figure 8).
Van
=
Vbn
=
Vcn
=
=
V
√L 6
3
VL
√ 6
3
VL
√ 6
3
VL
√ 6
3
Ia
Van
◦
− 30
Vcn
− 150◦
− 270◦
n
Vbn
Ib
c
b
90◦
Ic
Vab
Vbc
Vca
= VL 6 0
= VL 6 120◦
= VL 6 − 240◦ = VL 6 120◦
7
Conversion between ∆-load and Y-load
Figure 8: Y.
b
Z1
a
a
b
Za
Z2
Zb
Z3
Zc
c
c
Figure 9: Y-connected load.
7.1
Figure 10: Delta-connected load.
Converting Y-load to ∆-load
To establish equivalence,
Zab
=
Zbc
=
Zca
=
Z1 (Z2 + Z3 )
Z1 + Z2 + Z3
Z3 (Z1 + Z2 )
Zb + Zc =
Z3 + Z1 + Z2
Z2 (Z1 + Z3 )
Zc + Za =
Z1 + Z2 + Z3
Za + Zb =
(12)
(13)
(14)
(12)+(13)+(14) gives:
2(Za + Zb + Zc ) =
2(Z1 Z2 + Z2 Z3 + Z3 Z1 )
Z1 + Z2 + Z3
4
(15)
Z1
a
b
Za
(15)−(13) gives:
Za =
Z1 Z2
Z1 + Z2 + Z3
(16)
Zb =
Z1 Z3
Z1 + Z2 + Z3
(17)
Zc
Z2
Zb
Z3
(15)−(14) gives:
c
(15)−(12) gives:
Figure
Zc =
7.2
Z2 Z3
Z1 + Z2 + Z3
11:
Illustration of
the relationship given by equations (16), (17), (18).
(18)
Converting ∆-load to Y-load
Solving equations (16), (17), (18) for Z1 , Z2 , Z3 gives:
7.3
Z1
=
Z2
=
Z3
=
Za Zb + Zb Zc + Zc Za
Zc
Za Zb + Zb Zc + Zc Za
Zb
Za Zb + Zb Zc + Zc Za
Za
(19)
(20)
(21)
Balanced loads
A
|IaA | = IL
B
Z∆
A
Zy
Zy
Ib
B
N
Z∆
Z∆
Zy
C
|IcC | = IL
Ic
C
Figure 12: I aA =
V an
.
ZY
Figure 13: |IcC | = IL =
√
3I∆ .
For balanced loads, Z1 = Z2 = Z3 = Z∆ . Therefore, using equations (16), (17), (18) gives:
Za = Zb = Zc = ZY =
Thus,
ZY =
5
Z∆
3
(Z∆ )2
Z∆
=
3Z∆
3
Phase voltage
Vp 6 ϕ
Y
∆
√
= Vp 36 (ϕ + 30◦ )
Phase current
Ip = IL 6 θ
I∆ =
Line voltage
VL =
Line current
IL 6 θ
IL 6 θ
Load impedance
ZY 6 (ϕ − θ)
Z∆ = 3ZY 6 (ϕ − θ)
√
3Vp 6 (ϕ + 30◦ )
IL 6
√
3
(θ + 30◦ )
VL 6 (ϕ + 30◦ ) =
√
3Vp 6 (ϕ + 30◦ )
I L = IL 6 θ
a
+
VL =
√
IL 6
+
θ
a
IL 6 θ
Vp 6 θ
Zy
◦
3Vp 6 (θ + 30 )
Z∆
Zy
−
Z∆
n
−
Zy
b
c
c
b
Z∆
Figure 14: Y.
8
Figure 15: Delta.
Example
a
0.5 Ω
j0.2
A
Given that 6 (Van ) = 30◦ and PT = 1836.54 W.
What is Van ?
V an
= Van 6 30◦ = I aA (16.5 + j10.2)
= 19.4IaA 6 (θaA + 31.72)
◦
+
Van
16 Ω
−
◦
But, 6 (Van ) = 30 . So, θaA = −1.71 , and Van =
19.4IaA . Also,
j10
PL
= 612.18 = (IaA )2 RY = (IaA )2 16
3
n
So, IaA = 6.19 A (rms). Therefore,
N
Figure 16: Problem 10.15 (from old Irwin text-
Van = 120◦ and V an = 1206 30◦
book).
6
9
Power relationships
Consider Figure 14 (Y-connection) or Figure 15 (∆-connection).
VL
Pp
Qp
30◦
= Real power in each phase
= Vp Ip cos θ
= Reactive power in each phase
= Vp Ip sin θ
Vp
θ
(22)
Ip
(23)
Equations (22) and (23) are valid for Y or ∆ connection.
Figure 17: Phasor Diagram.
From Figure 14, we see that:
In a Y-connection:
VL
Vp = √ and Ip = IL
3
Therefore,
Pp
=
Qp
=
VL I L
√ cos θ
3
VL I L
√ sin θ
3
(24)
(25)
In a ∆-connection (see Figure 15):
IL
Vp = VL and Ip = √
3
Therefore, again
ST
Pp
=
Qp
=
VL I L
√ cos θ
3
VL I L
√ sin θ
3
=
=
Total power = Power in 3 phases
3(Pp + jQp )
=
PT + jQT
or,
ST
|ST |
=
√
3VL IL (cos θ + j sin θ)
q
√
=
PT2 + Q2T = 3VL IL
Thus,
ST =
√
3VL IL 6 θ
7
(26)
(27)
(28)