AC Circuits
Three-Phase Circuits
Contents
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What is a Three-Phase Circuit?
Balance Three-Phase Voltages
Balance Three-Phase Connection
Power in a Balanced System
Unbalanced Three-Phase Systems
Application – Residential Wiring
2
Single-phase system
Sinusoidal voltage sources
A simple AC generators
When a conductor rotates in a constant magnetic field, a sinusoidal wave is generated.
More pole pieces? More
coils?
When the loop is moving perpendicular to the lines
of flux, the maximum voltage is induced.
Normal household:
Single-phase three-wire
When the conductor is moving parallel
with the lines of flux, no voltage is
induced.
Single-phase two-wire
neutral
Polyphase
Three-phase circuit:
A system produced by a generator consisting of three sources having the same
amplitude and frequency but out of phase with each other by 120°.
Three sources
with 120° out
of phase
Four wired
system
4
Advantages of a three-phase circuit
• Most of the electric power is generated and distributed
in three-phase:
– Less/more than 3-phase required…taken from 3-phase
• The instantaneous power in a three-phase system can
be constant (not pulsating):
– uniform power transfer and less vibration of three-phase
machine
• The amount of power, the three-phase system is more
economical that the single-phase.
• In fact, the amount of wire required for a three-phase
system is less than that required for an equivalent
single-phase system.
5
Balance Three-Phase Voltages
•
A three-phase generator consists of a rotating magnet
(rotor) surrounded by a stationary winding (stator).
It can supply power to both singlephase and three-phase load
A three-phase generator
The generated voltages
6
Balance Three-Phase Voltages
•
Two possible configurations:
Four wires
Three wires
Phase voltage
Three-phase voltage sources: (a) Y-connected ; (b) Δ-connected
Balanced voltages:
Van  Vbn  Vcn  0
Van  Vbn  Vcn
7
Balance Three-Phase Voltages
•
Balanced phase voltages are equal in magnitude and are out of phase
with each other by 120°.
Van  V p
Van  Vbn  Vcn  V p
2
3
j
4
3
Vbn  V p e
Vcn  V p e
•
j
abc/positive
sequence
 Vpe
j
2
3
acb/negative
sequence
Van  V p
Vcn  V p e
j
2
3
j
4
3
Vbn  V p e
 Vpe
j
2
3
The phase sequence is the time order in which the voltages pass through
their respective maximum values.
abc/positive
sequence
acb/negative
sequence
•
A balanced load is one in which the phase impedances are equal in
magnitude and in phase
Transform:
Z   3Z Y
Z1  Z 2  Z 3  Z Y
Z a  Zb  Zc  Z 
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Example
Example 1
Solution:
Determine the phase sequence
of the set of voltages.
The voltages can be expressed in
phasor form as
van  200 cos(t  10)
vbn  200 cos(t  230)
vcn  200 cos(t  110)
Van  20010 V
Vbn  200  230 V
Vcn  200  110 V
We notice that Van leads Vcn by
120° and Vcn in turn leads Vbn by
120°.
Hence, we have an acb sequence.
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Balance Three-Phase Connection
Four possible connections
1. Y-Y connection (Y-connected source with a Yconnected load)
2. Y-Δ connection (Y-connected source with a Δconnected load)
3. Δ-Δ connection
4. Δ-Y connection
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Balance Y-Y system
A balanced Y-Y system is a three-phase system with a balanced y-connected
source and a balanced y-connected load.
Z Y  ZS  Zl  ZL
Line voltage:
Phasor diagram?
Vab  Van  Vnb  Van  Vbn  V p  V p e
VL  3V p , where

j
 1
 3
3
1
  3Vp 
  3Vpe 6
 Vp 1   j

j
 2
2 
2 
 2

Line current (KVL):
The same as
the phase
current in Y-Y
2
j 
3
Ia 
2
j 
3
Van
V
V e
, I b  bn  an
ZY
ZY
ZY
 Iae
2
j 
3
I a  I b  I c  0, I n  -I a  I b  I c   0,
, Ic  Iae
V p  Van  Vbn  Vcn
4
j 
3
VnN  0
VL  Vab  Vbc  Vca
Voltage across the neutral line is zero:
the neutral line can be removed (earth
acting as the neutral line).
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Balanced Y-Y connection
Example 2
Calculate the line currents in the three-wire Y-Y system
shown below:
Ans
I a  6.81  21.8 A
Ib  6.81  141.8 A
I c  6.8198.2 A
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Balance Y- Connection
• A balanced Y-Δ system is a three-phase system with a
balanced y-connected source and a balanced Δ-connected load.
Line current
Line voltage = the voltage across the load
VAB  Vab
Phase current I  VAB
AB
Phase current
Line current
ZΔ
I a  I AB  I CA  I AB  I AB e
4
j 
3
 3I AB e
1
j 
6
I L  3I p , where
I L  I a  Ib  Ic
I p  I AB  I BC  ICA
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Balance Y- Connection: Example
Example 3
A balanced abc-sequence Y-connected source with Van  10010 V is
connected to a Δ-connected load (8+j4) per phase. Calculate the
phase and line currents.
Solution
Using single-phase analysis,
Ia 
Van
10010

 33.54  16.57 A
Z / 3 2.98126.57
Other line currents are obtained using the abc phase sequence
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Balance - Connection
• A balanced Δ-Δ system is a three-phase system with a
balanced Δ -connected source and a balanced Δ -connected load.
Line voltage = the phase voltage
VAB  Vab
Phase current
I AB 
VAB Vab

ZΔ
ZΔ
Line current
I a  I AB  I CA  I AB  I AB e
 3I AB e
4
j 
3
1
j 
6
I L  3I p
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Balance - Connection: Example
Example 4
A balanced Δ-connected load having an impedance 20-j15  is connected to
a Δ-connected positive-sequence generator having (Vab  3300 V ).
Calculate the phase currents of the load and the line currents.
Ans:
The phase currents
I AB  13.236.87 A; I BC  13.2  81.13 A; I AB  13.2156.87 A
The line currents
I a  22.866.87 A; Ib  22.86  113.13 A; I c  22.86126.87 A
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Balance -Y Connection
• A balanced Δ-Y system is a three-phase system with a
balanced Y-connected source and a balanced Y-connected load.
Can you work out the relationship between line current and phase current?
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Balance -Y Connection: Example
Example 5
A balanced Y-connected load with a phase impedance 40+j25  is
supplied by a balanced, positive-sequence Δ-connected source with a
line voltage of 210 V. Calculate the phase currents. Use Vab as
reference.
Answer
The phase currents
I AN  2.57  62 A;
I BN  2.57  178 A;
ICN  2.5758 A;
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Balanced three-phase: Summary of phase and line V/I
Power in a Balanced System
•
Instantaneous power (Y-load, ZY = Z)
2 
2 


v AN  2V p cos t , vBN  2V p cos t 
, vCN  2V p cos t 

3 
3 


2 
2 


ia  2 I p cost   , ib  2 I p cos t   
, ic  2 I p cos t   

3 
3 


p  pa  pb  pc  v AN ia  vBN ib  vCN ic  3V p I p cos 
•
Time independent
For one phase : S  P  jQ  Vp I*p , P  V p I p cos  , Q  V p I p sin θ
Comparing the power loss in (a) a single-phase system, and (b) a three-phase system
PL2
P'loss  2 R 2 , single - phase
VL
PL2
P'loss  R' 2 , three - phase
VL
• If same power loss is tolerated in both system, three-phase system use
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only 75% of materials of a single-phase system
Unbalanced Three-Phase Systems
• An unbalanced system is due to unbalanced voltage sources
or an unbalanced load.
Ia 
VAN
V
V
, I b  BN , I c  CN ,
ZA
ZB
ZC
I n  I a  I b  I c 
• To calculate power in an unbalanced three-phase system requires that
we find the power in each phase.
• The total power is not simply three times the power in one phase but
the sum of the powers in the three phases.
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Unbalanced Three-Phase Systems: Example
Example 6
Determine the total average power, reactive power, and complex power
at the source and at the load
Ans
At the source:
Ss = -(2087 + j834.6) VA
Pa = -2087W
Pr = -834.6VAR
At the load:
SL = (1392 + j1113) VA
Pa = 1392W
Pr = 1113VAR
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Application – Residential Wiring
A 120/240 household power system
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Application – Residential Wiring (2)
Single-phase three-wire residential wiring
A typical wiring diagram of a room
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