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
5. SOURCE/LOAD CONNECTIONS
BALANCED Y-Y CONNECTION
°∠=
°−∠=
°∠=
120||
120||
0||
pcn
pbn
pan
VV
VV
VV
Positive sequence
phase voltages
abV
bcV
caV
Line voltages
( )
| | 0 | | 120
| | 1 (cos120 sin120)
1 3
| | | |
2 2
3 | | 30
ab an bn
p p
p
p p
p
V V V
V V
V j
V V j
V
= −
= ∠ °− ∠ − °
= − −
= − − − ÷ ÷
= ∠ °
°−∠=
°−∠=
210||3
90||3
pca
pbc
VV
VV
VoltageLine== ||3 pL VV
Y
cn
c
Y
bn
b
Y
an
a
Z
V
I
Z
V
I
Z
V
I === ;;
°+∠=°−∠=°∠= 120||;120||;|| θθθ LcLbLa IIIIII
0==++ ncba IIII 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
°−∠= 30208abV
Determine the phase voltages
Balanced Y - Y
Positive sequence
a-b-c
°∠= 30||3 pab VV
°∠=
°−∠=
°∠=
120||
120||
0||
pcn
pbn
pan
VV
VV
VV
Positive sequence
phase voltages
)3030(
3
||
°−°−∠=∴ ab
an
V
V
°30bylags aban VV
°−∠= 30208abV
°∠=
°−∠=
°−∠=
60120
180120
60120
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 10rmsphase line phaseV V Z j Z j= = + Ω = + Ω
Determine line currents and load voltages
Because circuit is balanced
data on any one phase is
sufficient
°∠0120
Chosen
as reference
120 0
120 120
120 120
an
bn
cn
V
V
V
= ∠ °
= ∠− °
= ∠ °
Abc sequence
rmsA
j
V
I an
aA
)(65.2706.5
65.2771.23
0120
1121
°−∠=
°∠
°∠
=
+
=
rmsAI
rmsAI
cC
bB
)(65.2712006.5
)(65.2712006.5
°−∠=
°−−∠=
°∠×=+×= 57.2636.22)1020( aAaAAN IjIV
rmsVVAN )(08.115.113 °−∠=
rmsVV
rmsVV
CN
BN
)(92.11815.113
)(08.12115.113
°∠=
°−∠=
8. DELTA CONNECTED SOURCES
Relationship between
phase and line voltages
°∠= 30||3 pab VV
°30bylags aban VV
⇒
°∠=
°−∠=
°∠=
120
120
0
Lca
Lbc
Lab
VV
VV
VV
°∠=
°−∠=
°−∠=
90
3
150
3
30
3
L
cn
L
bn
L
an
V
V
V
V
V
V
°∠=
°−∠=
°∠=
150120
90120
30120
cn
bn
an
V
V
V
⇒
°∠=
°−∠=
°∠=
180208
60208
60208
ca
bc
ab
V
V
V
Example
Convert to an equivalent Y connection
9. LEARNING EXAMPLE Determine line currents and line voltages at the loads
Source is Delta connected.
Convert to equivalent Y
⇒
°∠=
°−∠=
°∠=
120
120
0
Lca
Lbc
Lab
VV
VV
VV
°∠=
°−∠=
°−∠=
90
3
150
3
30
3
L
cn
L
bn
L
an
V
V
V
V
V
V
Analyze one phase
rmsA
j
IaA )(14.4938.9
2.41.12
30)3/208(
°−∠=
+
°−∠
=
rmsVjVAN )(71.3065.11819.4938.9)412( °−∠=°−∠×+=
Determine the other phases using the balance
rmsAI
rmsAI
cC
bB
)(86.7138.9
)(14.16938.9
°−∠=
°−∠=
3 118.65 0.71ABV = × ∠− °
3 118.65 120.71
3 118.65 119.29
BC
CA
V
V
= × ∠− °
= × ∠ °
10. DELTA-CONNECTED LOAD
Method 1: Solve directly
°∠=
°−∠=
°∠=
120||
120||
0||
pcn
pbn
pan
VV
VV
VV
Positive sequence
phase voltages
°−∠=
°−∠=
°∠=
210||3
90||3
30||3
pca
pbc
pab
VV
VV
VV
°+∠==
°−∠==
∠==
∆∆
∆
∆∆
∆
∆∆
∆
120||
120||
||
θ
θ
θ
I
Z
V
I
I
Z
V
I
I
Z
V
I
CA
CA
BC
BC
AB
AB
currentsphaseLoad
BCCAcC
ABBCbB
CAABaA
III
III
III
−=
−=
−=
currentsLine
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
ZYcaseBalanced
−=
=
⇒∠== ∆
ZL
AB
aA
LaA
Y
an
aA Z
V
I
I
Z
V
I
θθ
θ 3/||
3/||
||
||
ZLZZ θ∠=∆ ||
Zθθ −°=∆ 30
°−=
=
∆
∆
30
||3||
θθline
line II
Line-phase current
relationship
12. Y→∆
∆ → Y
baab RRR +=
)(|| 312 RRRRab +=
321
312 )(
RRR
RRR
RR ba
++
+
=+
321
213 )(
RRR
RRR
RR cb
++
+
=+
321
321 )(
RRR
RRR
RR ac
++
+
=+
SUBTRACT THE FIRST TWO THEN ADD
TO THE THIRD TO GET Ra
a
b
b
a
R
RR
R
R
R
R
R 1
3
3
1
=⇒=
c
b
c
b
R
RR
R
R
R
R
R 1
2
1
2
=⇒=
REPLACE IN THE THIRD AND SOLVE FOR R1
Y
RRR
RR
R
RRR
RR
R
RRR
RR
R
c
b
a
→∆
++
=
++
=
++
=
321
13
321
32
321
21
∆−
++
=
++
=
++
=
Y
R
RRRRRR
R
R
RRRRRR
R
R
RRRRRR
R
a
accbba
c
accbba
b
accbba
3
2
1
tionsTransforma
Y
OFREVIEW
↔∆
3
321
∆
∆ =⇒===
R
RRRRR Y
13. LEARNING EXAMPLE
°∠=Ω+=∆ 02.3752.1254.710 jZ
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
rmsVVan )(30120 °∠=
Ω=××= 54.7020.0602πinductanceZ
rmsA
jZ
V
I AB
AB )(98.2260.16
54.710
603120
°∠=
+
°∠
==
∆
°−=
=
∆
∆
30
||3||
θθline
line II
Line-phase current
relationship
iprelationsh
voltagephase-Line
°+=
=
∆
∆
30
||3||
phase
phaseVV
θθ rmsAIaA )(02.775.28 °−∠=
rmsAI
rmsAI
CA
BC
)(98.14260.16
)(02.9760.16
°∠=
°−∠=
rmsAI
rmsAI
cC
bB
)(98.11275.28
)(02.12775.28
°∠=
°−∠=
°∠=⇒ 02.3717.4YZ
Alternatively, determine first the line currents
and then the delta currents
Polyphase
