Lecture 2: Polyphase Systems / Circuits
• Analysis of a balanced Y- connected system
• Derivation of expression of complex power
• Analysis of power flow in balanced Y-Y and Y-  connected system
Voltage – current relationship
B. Balanced Y- connected system
 Balanced Y connected source
 Balanced  connected load (ZAB = ZBC = ZCA = Z)
𝐼𝑎
a
Van
Vab
Vcn
n
Vca
Vbn
b
Vbc
𝑜
Van = 𝑉𝑝ℎ ∠0
Vbn = 𝑉𝑝ℎ ∠ −120𝑜
Vcn = 𝑉𝑝ℎ ∠ +120𝑜
Vab =
Vbc =
Vca =
VL =
c
A
VAB
VCA
𝐼𝑏
B
ZBC
VBC
𝐼𝑐
C
𝐼𝐵𝐶
3𝑉𝑝ℎ ∠30𝑜 = VAB
3𝑉𝑝ℎ ∠ − 90𝑜 = VBC Line voltages are same as phase voltages in  connected load
3𝑉𝑝ℎ ∠ − 210𝑜 = VCA
3𝑉𝑝ℎ
Phase currents of the load:
VAB
VBC
VCA
𝐼𝐴𝐵 =
, 𝐼𝐵𝐶 =
, 𝐼𝐶𝐴 =
𝑍𝐴𝐵
𝑍𝐵𝐶
𝑍𝐶𝐴
Line currents of the load: These currents are obtained from the
phase current applying the KCL at the nodes A, B, and C of the
load. Thus
𝐼𝑎 = 𝐼𝐴𝐵 − 𝐼𝐶𝐴 , 𝐼𝑏 = 𝐼𝐵𝐶 − 𝐼𝐴𝐵 , 𝐼𝑐 = 𝐼𝐶𝐴 − 𝐼𝐵𝐶
Voltage – current relationship
A
𝐼𝑎
VAB
VCA
B 𝐼𝑏
𝐼𝐵𝐶 = 𝐼𝐴𝐵 ∠ − 120𝑜 , 𝐼𝐶𝐴 = 𝐼𝐴𝐵 ∠ + 120𝑜 = 𝐼𝐴𝐵 ∠ − 240𝑜
𝐼𝑎 = 𝐼𝐴𝐵 − 𝐼𝐶𝐴 = 𝐼𝐴𝐵 (1 - 1 ∠ + 120𝑜 ) = 3𝐼𝐴𝐵 ∠ − 30𝑜
VBC 𝐼 𝐼𝐵𝐶
𝑐
C
𝐼𝑏 = 𝐼𝑎 ∠ − 120𝑜 , 𝐼𝑐 = 𝐼𝑎 ∠ + 120𝑜
Note:
The magnitude of the line currents of the load = 3 times the
magnitude of the phase currents of the load
𝐼𝐿 = 3𝐼𝑝ℎ = 𝐼𝑎 = 𝐼𝑏 = 𝐼𝑐
𝐼𝑝ℎ = 𝐼𝐴𝐵 = 𝐼𝐵𝐶 = 𝐼𝐶𝐴
ZBC
𝐼𝑐
𝐼𝐶𝐴
30o
𝐼𝐴𝐵
30o
𝐼𝑏
30o
𝐼𝐵𝐶
−𝐼𝐶𝐴
𝐼𝑎
Voltage – current relationship of load
 Summary
Connection
Phase voltages/Currents
Van = 𝑉𝑝ℎ ∠0𝑜
Vbn = 𝑉𝑝ℎ ∠ −120𝑜
Vcn = 𝑉𝑝ℎ ∠120𝑜
Y-Y
𝐼𝑎 =
Van
𝑍𝑌
𝐼𝑏 = 𝐼𝑎 ∠ −120𝑜
𝐼𝑐 = 𝐼𝑎 ∠120𝑜
Y-
VAB = 3𝑉𝑝ℎ ∠30𝑜
VBC = 3𝑉𝑝ℎ ∠ − 90𝑜
VCA = 3𝑉𝑝ℎ ∠ − 210𝑜
VL = 3𝑉𝑝ℎ , 𝑉𝑝ℎ is the source phase
voltage
VAB
𝐼𝐴𝐵 =
𝑍
VBC
𝐼𝐵𝐶 =
𝑍
VCA
𝐼𝐶𝐴 =
𝑍
Line voltages/Currents
Vab =
Vbc =
Vca =
VL =
3𝑉𝑝ℎ ∠30𝑜
3𝑉𝑝ℎ ∠ − 90𝑜
3𝑉𝑝ℎ ∠ − 210𝑜
3𝑉𝑝ℎ
Line current = Phase current
Line voltage = Phase voltage
𝐼𝑎 = 3𝐼𝐴𝐵 ∠ − 30𝑜
𝐼𝑏 = 𝐼𝑎 ∠ − 120𝑜
𝐼𝑐 = 𝐼𝑎 ∠ + 120𝑜
𝐼𝐿 = 3𝐼𝑝ℎ , 𝐼𝑝ℎ is the phase
current of load
Relationship between Y and  connected load impedances
  (ZAB = ZBC = ZCA) to Y (ZA = ZB = ZC) conversion
The impedance seen between the lines A and B in the star
connected load is ZA + ZB (series combination of ZA and ZB). In
A
the delta connected load, the impedance seen between A and B is
ZAB in parallel with the series combination of ZCA and ZBC.
ZA
ZB
𝒁𝑨 + 𝒁𝑩 = 𝒁𝑨𝑩 || 𝒁𝑩𝑪 + 𝒁𝑪𝑨
𝒁𝑨𝑩 𝒁𝑩𝑪 + 𝒁𝑪𝑨
=
𝒁𝑨𝑩 + 𝒁𝑩𝑪 + 𝒁𝑪𝑨
ZC
Similarly,
ZBC
B
C
𝒁𝑩𝑪 𝒁𝑨𝑩 + 𝒁𝑪𝑨
𝒁𝑩 + 𝒁 𝒄 =
𝒁𝑨𝑩 + 𝒁𝑩𝑪 + 𝒁𝑪𝑨
𝒁𝑪𝑨 𝒁𝑨𝑩 + 𝒁𝑩𝑪
𝒁𝒄 + 𝒁 𝑨 =
𝒁𝑨𝑩 + 𝒁𝑩𝑪 + 𝒁𝑪𝑨
𝒁𝑨𝑩 𝒁𝑪𝑨
𝒁𝑨 =
𝒁𝑨𝑩 + 𝒁𝑩𝑪 + 𝒁𝑪𝑨
𝒁𝑩 =
𝒁𝑨𝑩 𝒁𝑩𝑪
𝒁𝑨𝑩 + 𝒁𝑩𝑪 + 𝒁𝑪𝑨
𝒁𝑩𝑪 𝒁𝑪𝑨
𝒁𝐶 =
𝒁𝑨𝑩 + 𝒁𝑩𝑪 + 𝒁𝑪𝑨
Relationship between  and Y connected load impedances
 Y (ZA = ZB = ZC) to  (ZAB = ZBC = ZCA) conversion
A
𝒁𝑨𝑩
ZA
𝒁𝑨 𝒁𝑩 + 𝒁𝑩 𝒁 𝑪 + 𝒁𝑪 𝒁𝑨
=
𝒁𝑪
𝒁𝑩𝑪 =
ZB
ZC
ZBC
B
𝒁𝐶𝐴
C
𝒁𝑨 𝒁𝑩 + 𝒁𝑩 𝒁𝑪 + 𝒁𝑪 𝒁𝑨
𝒁𝑨
𝒁𝑨 𝒁𝑩 + 𝒁𝑩 𝒁𝑪 + 𝒁𝑪 𝒁𝑨
=
𝒁𝑩
 For balanced load: Y (ZA = ZB = ZC = ZY) and  (ZAB = ZBC = ZCA= Z): Z =3 ZY
Instantaneous and average power
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡 + 𝜃𝑣 = 𝑉∠𝜃𝑣 ,
𝑖(𝑡)
Sinusoidal
source
Passive
linear
network
𝑣 𝑡
𝑖 𝑡 = 𝐼𝑚 sin 𝜔𝑡 + 𝜃𝑖 = 𝐼∠𝜃𝑖
s 𝑡 = 𝑣 𝑡 𝑖 𝑡 = 𝑉𝑚 sin 𝜔𝑡 + 𝜃𝑣 𝐼𝑚 sin 𝜔𝑡 + 𝜃𝑖
1
= 2 𝑉𝑚 𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖 − cos 2𝜔𝑡 + 𝜃𝑣 + 𝜃𝑖
1
1
= 2 𝑉𝑚 𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖 − 2 𝑉𝑚 𝐼𝑚 cos 2𝜔𝑡 + 𝜃𝑣 + 𝜃𝑖
Fig: 1
𝑃 = 𝑆𝑎𝑣
𝑣 𝑡
𝜃𝑣
𝜃𝑖
𝜙
s 𝑡
𝜋
𝑇
0
1
𝑠 𝑡 𝑑𝑡 = 𝑉𝑚 𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖
2
 Instantaneous power changes with time, difficult to
measure. Hence, average power is used to measure the
power using wattmeter.
𝜔𝑡
 When s(t) is positive, power is absorbed by the load
(passive linear network).
 When s(t) is negative, power is absorbed by the source.
Fig: 2
Load (passive linear network) has energy storing
1
𝑉 𝐼
elements such as inductors and capacitors.
2 𝑚𝑚
𝑖 𝑡
0
1
=
𝑇
2𝜋
1
1
𝑉 𝐼 cos 𝜃𝑣 − 𝜃𝑖 = 𝑉𝑚 𝐼𝑚 cos𝜙
2 𝑚𝑚
2
𝜔𝑡
Instantaneous and average power
𝑃 = 𝑆𝑎𝑣
1
=
𝑇
𝑇
𝑠 𝑡 𝑑𝑡 =
0
1
1
𝑉𝑚 𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖 = 𝑅𝑒 𝑉𝑚 𝐼𝑚 𝑒 𝑗
2
2
𝜃𝑣 −𝜃𝑖
= 𝑅𝑒
𝑉𝑚
2
𝑒 𝑗𝜃𝑣
𝐼𝑚
2
𝑒 −𝑗𝜃𝑖 = 𝑅𝑒 𝑉𝐼 ∗
𝑆 = 𝑉𝐼 ∗ = 𝑉∠𝜃𝑣 𝐼∠ − 𝜃𝑖 = 𝑉𝐼∠ 𝜃𝑣 − 𝜃𝑖 = 𝑉𝐼cos𝜙 + 𝑗𝑉𝐼sin𝜙 = 𝑃 + 𝑗𝑄
 S, P and Q are the total power or apparent power or complex power, P is average power or active
power, and Q is reactive power.
 Unit of S is VA or kVA or MVA, unit of P is Watt (W) or kW or MW, and unit of Q is VAR or kVAR or
MVAR.
Imaginary axis
Imaginary axis
+jQ (lagging power factor),
R-L load
S
𝜙
𝜙
P
Real axis
𝑗𝑋𝐿 (lagging power factor)
𝑍
𝜙
𝜙
R
Real axis
S
𝑍
Fig: 1
-jQ (leading power factor),
R-C load
−𝑗𝑋𝐶 (leading power factor)
Fig: 2
Instantaneous and average power
 For pure R load:
𝑖 𝑡
a. Time-domain analysis:
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡
𝑅
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡,
𝑖 𝑡 = 𝐼𝑚 sin 𝜔𝑡
s 𝑡
1
𝑉 𝐼
2 𝑚𝑚
𝑣 𝑡
𝑖 𝑡
1
𝑉 𝐼 cos
2 𝑚 𝑚
𝜃𝑖 = 𝜃𝑣 = 0
𝜋
2𝜋
𝜔𝑡
s 𝑡 = 𝑣 𝑡 𝑖 𝑡 = 𝑉𝑚 𝐼𝑚 𝑠𝑖𝑛2 𝜔𝑡 =
1
𝑃 = 𝑆𝑎𝑣 = 𝑇
𝑇
𝑠
0
1
1
𝑉 𝐼
2 𝑚 𝑚
𝑡 𝑑𝑡 = 2 𝑉𝑚 𝐼𝑚 = 𝑉𝐼,
b. Phasor analysis:
1
𝜃𝑣 − 𝜃𝑖 = 2 𝑉𝑚 𝐼𝑚 cos𝜙 =
𝐼
1
1
1 − cos 2𝜔𝑡 = 2 𝑉𝑚 𝐼𝑚 − 2 𝑉𝑚 𝐼𝑚 cos 2𝜔𝑡
𝑄=0
𝑉
𝑆 = 𝑉𝐼 ∗ = 𝑉𝐼 = 𝑉𝐼 + 𝑗0 = 𝑃 + 𝑗𝑄
1
𝑉 𝐼
2 𝑚 𝑚
Instantaneous and average power
 Pure L load:
𝑖 𝑡
a. Time-domain analysis:
s 𝑡
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡
𝐿
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡,
𝑖 𝑡 = 𝐼𝑚 sin(𝜔𝑡 − 90𝑜)
1
𝑉 𝐼
2 𝑚𝑚
𝑣 𝑡
𝑖 𝑡
𝑃=0
2𝜋
𝜋
s 𝑡 =𝑣 𝑡 𝑖 𝑡 =
1
𝑃 = 𝑆𝑎𝑣 = 𝑇
𝑇
𝑠
0
𝜔𝑡
1
𝑉 𝐼
2 𝑚 𝑚
− sin 2𝜔𝑡
𝑡 𝑑𝑡 = 0,
b. Phasor analysis:
𝐼
90𝑜
𝜃𝑣 = 0, 𝜃𝑖 = −90𝑜
𝑄 = 𝑉𝐼
𝑉
𝑆 = 𝑉𝐼 ∗ = 𝑉𝐼∠90𝑜 = 0 + 𝑗𝑉𝐼 = 𝑃 + 𝑗𝑄
Instantaneous and average power
𝑖 𝑡
 Pure C load:
a. Time-domain analysis:
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡,
𝑣 𝑡 = 𝑉𝑚 sin 𝜔𝑡
s 𝑡
C
𝑖 𝑡 = 𝐼𝑚 sin(𝜔𝑡 + 90𝑜)
1
𝑉 𝐼
2 𝑚𝑚
𝑣 𝑡
𝑖 𝑡
𝜃𝑣 = 0, 𝜃𝑖 = +90𝑜
𝜋
𝑃=0
2𝜋
𝜔𝑡
s 𝑡 =𝑣 𝑡 𝑖 𝑡 =
1
𝑃 = 𝑆𝑎𝑣 = 𝑇
𝑇
𝑠
0
1
𝑉 𝐼 sin
2 𝑚 𝑚
𝑡 𝑑𝑡 = 0,
2𝜔𝑡
𝑄 = 𝑉𝐼
𝐼
b. Phasor analysis:
90𝑜
𝑉
𝑆 = 𝑉𝐼 ∗ = 𝑉𝐼∠ − 90𝑜 = 0 − 𝑗𝑉𝐼 = 𝑃 + 𝑗𝑄
Power in three phase system
𝑣𝑎 𝑡 = 𝑉𝑚 sin 𝜔𝑡 = 𝑉∠0𝑜, 𝑖𝑎 𝑡 = 𝐼𝑚 sin 𝜔𝑡 − 𝜙 = 𝐼∠ − 𝜙
sa 𝑡 = 𝑣𝑎 𝑡 𝑖𝑎 𝑡 = 𝑉𝑚 sin 𝜔𝑡 𝐼𝑚 sin 𝜔𝑡 − 𝜙
1
= 2 𝑉𝑚 𝐼𝑚 cos 𝜙 − cos 2𝜔𝑡 − 𝜙
1
1
= 2 𝑉𝑚 𝐼𝑚 cos 𝜙 − 2 𝑉𝑚 𝐼𝑚 cos 2𝜔𝑡 − 𝜙
sa 𝑡
1
𝑉 𝐼
2 𝑚𝑚
𝑣𝑎 𝑡
𝑖𝑎 𝑡
𝜙
𝜋
2𝜋
sb 𝑡 = 𝑣𝑏 𝑡 𝑖𝑏 𝑡 = 𝑉𝑚 sin 𝜔𝑡 − 120𝑜 𝐼𝑚 sin 𝜔𝑡 − 𝜙 − 120𝑜
1
1
= 2 𝑉𝑚 𝐼𝑚 cos 𝜙 − 2 𝑉𝑚 𝐼𝑚 cos 2𝜔𝑡 − 𝜙 − 120𝑜
sc 𝑡 = 𝑣𝑐 𝑡 𝑖𝑐 𝑡 = 𝑉𝑚 sin 𝜔𝑡 − 240𝑜 𝐼𝑚 sin 𝜔𝑡 − 𝜙 − 240𝑜
1
1
= 2 𝑉𝑚 𝐼𝑚 cos 𝜙 − 2 𝑉𝑚 𝐼𝑚 cos 2𝜔𝑡 − 𝜙 − 240𝑜
1
1
𝑉𝑚 𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖 = 𝑉𝑚 𝐼𝑚 cos𝜙
2
2
𝜔𝑡
𝑃 = 𝑆𝑎𝑣
1 𝑇
=
𝑠 𝑡 + 𝑠𝑏 𝑡 + 𝑠𝑐 𝑡 𝑑𝑡
𝑇 0 𝑎
1
= 3 𝑉𝑚 𝐼𝑚 cos 𝜙
2
= 3𝑉𝑝ℎ 𝐼𝑝ℎ cos 𝜙
Power in three phase system
∗
Complex power per phase: 𝑆𝑝ℎ = 𝑃𝑝ℎ + 𝑗𝑄𝑝ℎ = 𝑉𝑝ℎ 𝐼𝑝ℎ
∗
Complex power for three phase: 𝑆 = 3𝑆𝑝ℎ = 𝑃 + 𝑗𝑄 = 3𝑉𝑝ℎ 𝐼𝑝ℎ
 For Y-connected load: 𝑃 = 𝑆𝑎𝑣 = 3𝑉𝑝ℎ 𝐼𝑝ℎ cos 𝜙 = 3
𝑉𝐿
𝐼
3 𝐿
∗
2
𝑉𝑝ℎ
𝑉𝑝ℎ
= 3𝑉𝑝ℎ ∗ = 3 ∗ = 3𝑉𝐿 𝐼𝐿 ∠𝜙
𝑍𝑝ℎ
𝑍𝑝ℎ
cos 𝜙 = 3𝑉𝐿 𝐼𝐿 cos 𝜙
𝑄 = 3𝑉𝐿 𝐼𝐿 sin 𝜙
 For -connected load: 𝑃 = 𝑆𝑎𝑣 = 3𝑉𝑝ℎ 𝐼𝑝ℎ cos 𝜙 = 3𝑉𝐿
𝑄 = 3𝑉𝐿 𝐼𝐿 sin 𝜙
𝐼𝐿
cos 𝜙
3
= 3𝑉𝐿 𝐼𝐿 cos 𝜙