ECE 4800: Electromechanical devices by Dr. Ashraf Ali Khan
THREE-PHASE CIRCUITS
Readings:
Fundamentals of Electric Circuits, 5th ed.,
C. K. Alexander and M. N. O. Sadiku
Chapters 12.1-12.11
3.1 Balanced three-phase voltage sources.
Three-phase four-wire electrical power system.
A three-phase system for electrical power distribution is shown above. Three voltage sources are used, and each is
120° out of phase with its neighbors. Such an implementation has the advantage that it requires less conducting
wire to deliver the same amount of power as an equivalent single-phase system. It also produces less vibrations.
There are two important things to consider in setting up three-phase systems:
i. Geometrical connection of the sources
Y-connected voltages sources…
-connected voltages sources…
ii. Relative phase sequence of the three incorporated sources
ABC (positive) phase sequence…
ACB (negative) phase sequence…
Example. What is the total voltage from the three phases?
Van + Vbn + Vcn = Vp  0 + Vp  -120 + Vp  120 = Vp (1 - 0.5 – j0.866 – 0.5 + j0.866) = 0
3.2 Balanced three-phase loads.
Often it is more convenient to work with one network
in a place of another. Line currents can be more
easily related to phase currents with wye loads, for
example, and line voltages can be more easily related
to phase voltages with delta loads.
To develop a general transformation between the two
networks, we can superimpose a wye network on a
delta network and equate the equivalent impedances
for the wye and delta terminals.
Overlaid wye and delta networks.
Wye-delta transformation…
Z1 + Z3 = Zb(Za + Zc) / (Za + Zb + Zc),
Z2 + Z3 = Za(Zb + Zc) / (Za + Zb + Zc), and
Z1 + Z2 = Zc(Za + Zb) / (Za + Zb + Zc).
(1)
(2)
(3)
Take (1) - (2),
Take (2) - (3),
Take (3) - (1),
Z1 - Z2 = Zc(Zb – Za) / (Za + Zb + Zc).
Z3 - Z1 = Zb(Za – Zc) / (Za + Zb + Zc).
Z2 - Z3 = Za(Zc – Zb) / (Za + Zb + Zc).
(4)
(5)
(6)
Take (3) + (4),
Take (2) + (6),
Take (1) + (5),
Z1 = ZbZc / (Za + Zb + Zc).
Z2 = ZaZc / (Za + Zb + Zc).
Z3 = ZaZb / (Za + Zb + Zc).
(7)
(8)
(9)
Take (7)(8)+(8)(9)+(9)(7), Z1Z2+Z2Z3+Z3Z1 =
(ZaZbZcZc+ZaZaZbZc+ZaZbZbZc)/(Za+Zb+Zc)2 = ZaZbZc/(Za+Zb+Zc).
(10)
Take (10)/(7),
Take (10)/(8),
Take (10)/(9),
(11)
(12)
(13)
Za = (Z1Z2 + Z2Z3 + Z3Z1) / Z1.
Zb = (Z1Z2 + Z2Z3 + Z3Z1) / Z2.
Zc = (Z1Z2 + Z2Z3 + Z3Z1) / Z3.
Notice that if it is balanced with Za = Zb = Zc = Z, and Z1 = Z2 = Z3 = ZY, then Z = 3 ZY.
-Y conversion:
Z1 =
For Y- conversion:
For balanced loads:
Zb Zc
Z a + Zb + Zc
Za =
Z1Z 2 + Z 2 Z 3 + Z 3 Z1
Z1
Z2 =
Za Zc
Z a + Zb + Zc
Zb =
Z1Z 2 + Z 2 Z 3 + Z 3 Z1
Z Z + Z 2 Z 3 + Z 3 Z1
Zc = 1 2
Z2
Z3
Z  = 3ZY
Z3 =
Z a Zb
Z a + Zb + Zc
ZY = Z  / 3
.
.
.
3.3 Balanced three-phase networks.
3.3.1 Wye-Wye connections.
The balanced wye-wye voltage source and load connection system (shown below) is the easiest system to analyze.
As we will see, other source and load systems can be transformed to this orientation before being analyzed. The
analysis can be carried out with source, line, and load impedances together as ZY = ZS + Zl + ZL.
The wye-wye network…
Phase voltages (voltages of sources):
Van = Vp  0, Vbn = Vp  -120, Vcn = Vp  +120.
Line voltages (line-to-line voltage difference):
Vab = Van + Vnb = Van - Vbn = Vp  0-Vp  -120=Vp(1+1/2+j√3/2)=√3Vp  30.
Vab = √3Vp  30, Vbc = √3Vp  -90, Vca = √3Vp  150.
Line currents and phase currents: (apply KVL to each phase and we obtain)
Ia = Van / ZY = Ip  0
Ib = Ip  -120,
Ic = Ip  120.
Neutral line current:
Ia + Ib + Ic = 0, so In = -(Ia + Ib + Ic) = 0, thus the neutral line can be removed.
Practical conclusions for Y-Y networks:
● Source line voltage magnitudes are √3 times the source phase voltage magnitudes and are +300 phase-shifted.
● Load phase currents are equal to the load line currents.
WYE-WYE SUMMARY:
Phase Voltages (abc):
Van = Vp  0,
Vbn = Vp  -120,
Vcn = Vp  120.
Line Voltages (abc):
Vab = √3Vp  30,
Vbc = Vab  -120,
Vca = Vab  120.
Line Currents:
Ia = Van / ZY,
Ib = Ia  -120,
Ic = Ia  120.
Rules:
● Source line voltage magnitudes are √3 times the source phase voltage magnitudes and are +300 phase-shifted.
● Load phase currents are equal to the line currents.
Example. Obtain the line currents in the Y-Y three-phase circuit.
3.3.2 Wye-Delta connections.
The balanced wye-delta voltage source and load connection system (shown below) has some added complexities in
comparison to the analysis in the previous section. Here, line voltages are needed to calculate the current flow
through the individual load impedances (phase currents).
The wye-delta network…
Phase voltages:
Van = Vp  0,
Vbn = Vp  −120,
Vcn = Vp  120◦.
Vbc = Vab  −120 = VBC,
Vca = Vab  120 = VCA.
IBC = VBC / Z,
ICA = VCA / Z.
Line voltages:
Vab = √3Van  30 = VAB,
Phase currents:
IAB = VAB / Z,
Line currents: (use KCL at node A)
Ia = √3IAB  −30,
=> Ia = IAB − ICA = IAB(1 − 1  120) = √3IAB  −30
Ib = √3IAB  −150,
Ic = √3IAB  90.
The same results can be found with a load transformation (delta-wye):
Transform load to a wye circuit with ZY = Z  / 3 => I a =
Van Vab  − 30
3Vab  − 30
=
=
= 3I AB  − 30
ZY
Z
3Z Y
Practical conclusions for Y- networks:
● Source line voltage magnitudes are √3 times the source phase voltage magnitudes and are +300 phase-shifted.
● The line current magnitudes are √3 times the load phase magnitudes and are -300 phase-shifted.
WYE-DELTA SUMMARY:
Phase Voltages (abc):
Van = Vp  0,
Vbn = Vp  -120,
Vcn = Vp  120.
Line Voltages (abc):
Vab = VAB = √3Vp  30,
Vbc = VBC = Vab  -120,
Vca = VCA = Vab  -120.
Line Currents:
Ia = √3IAB  -30,
Ib = Ia  -120,
Ic = Ia  120.
Rules:
● Source line voltage magnitudes are √3 times the source phase voltage magnitudes and are +300 phase-shifted.
● Load phase current magnitudes are 1/√3 times the line current magnitudes and are +300 phase-shifted wrt line.
Example. In the Y-Δ system shown, the source is a positive sequence with Van = 120  0° V and phase impedance
Z = 2 – j3 Ω . Calculate the line voltage VL and the line current IL.
Method 1: Voltage/current relationships
This method works for all cases but can become complex.
Method 2: Load transformation to calculate IL:
This method is useful if you need the line currents.
Delta-wye transform: ZY = Z / 3 => ZY = (2-j3)/3 = 0.667–j =>
Ia = Van/ZY = 120  0 / (0.667 - j) = 99.8  56.31
3.3.3 Delta-Delta connections.
The balanced delta-delta voltage source and load connection system (shown below) has some advantages in that it
has (assuming no line impedance) phase voltages appear directly across the load impedances.
The delta-delta network…
Line voltages (equal to phase voltages):
Vab = Vp  0 = VAB,
Vbc = Vp  −120 = VBC,
Vca = Vp  120 = VCA.
Phase currents:
IAB = VAB / Z,
IBC = VBC / Z,
ICA = VCA / Z.
Line currents: (from the phase currents by applying KCL at nodes A, B, and C)
Ia = IAB − ICA = IAB(1 − 1  120) = IAB(1 + 1/2 – j√3/2) = √3IAB  −30
Ia = √3IAB  −30, Ib = √3IAB  −150, Ic = √3IAB  90.
Conclusion for -  networks:
Find the line currents by multiplying the phase current magnitude by √3 and subtracting 300 from the phases.
Practical conclusions for  - networks:
● Source line voltages are equal to the source phase voltages.
● The line current magnitudes are √3 times the load phase magnitudes and are -300 phase-shifted.
DELTA-DELTA SUMMARY:
Phase Voltages (abc):
Vab = Vp  0,
Vbc = Vp  -120,
Vca = Vp  120.
Line Voltages (abc):
Vab = VAB = Vp  0,
Vbc = VBC = Vp  -120,
Vca = VCA = Vp  120.
Line Currents:
Ia = √3IAB  -30,
Ib = Ia  -120,
Ic = Ia  120.
Rules:
● Source line voltages are equal to the source phase voltages.
● Load phase current magnitudes are 1/√3 times the line current magnitudes and are +300 phase-shifted wrt line.
Example. Find the line and phase currents below given ZL = 12 + j9 Ω per phase.
3.3.4 Delta-Wye connections.
The analysis of the delta-wye voltage source and load network is very similar to the previous networks. Here, we
can immediately see the line voltages, though, and we must “drop these down” to give the phase voltages before
we calculate the phase currents.
These phase currents are equal to the line currents.
The delta-wye network…
Ia
Line voltages (equal to phase voltages):
Vab = Vp  0 = VAB,
Vbc = Vp  −120 = VBC,
Vca = Vp  120 = VCA.
Phase currents (equal to line currents): (apply KVL to loop aANBba)
−Vab + ZY Ia − ZY Ib = 0 => (Ia − Ib) = Vab/ZY = Vp  0/ZY
(Ia − Ib) = Ia (1-1  -120) = Ia (1+1/2+j√3/2) = √3Ia  30
√3Ia  30 = Vp  0/ZY
Therefore,
Ia =
Vp
3  − 30
ZY
Ib =
Vp
3  − 150
Ic =
ZY
Vp
3 90
ZY
The same results can be found with a source transformation (delta-wye):
Transform source to a wye circuit with Van = Vp/√3  -30 => I a =
Van V p  − 30
=
ZY
3ZY
The same results can be found with a load transformation (wye-delta):
Transform load to a delta circuit with Z  = 3ZY => I ab =
V  − 30
Vab V p 0
=
= I a = 3I ab − 30 = p
Z
3ZY
3ZY
Practical conclusions for  -Y networks:
● Source line voltages are equal to the source phase voltages.
● Load phase currents are equal to the load line currents.
DELTA-WYE SUMMARY:
Phase Voltages (abc):
Vab = Vp  0,
Vbc = Vp  -120,
Vca = Vp  120.
Line Voltages (abc):
Vab = VAB = Vp  0,
Vbc = VBC = Vp  -120,
Vca = VCA = Vp  120.
Line Currents:
Ia =
Ib = Ia  -120,
Ic = Ia  120.
Vp
3  − 30
ZY
,
Rules:
● Source line voltages are equal to the source phase voltages.
● Load phase currents are equal to the load line currents.
Example. For the balanced circuit below with Vab = 125  0° V, find the line currents Ia, Ib, and Ic.
Method 1: (voltage transformation)
Method 2: (load transformations)
Wye-delta transform: Z = 3ZY => Z = 3(24 - j15) = (72 - j45) =>
IAB = VAB/Z = VAB/Z = ∠32 => Ia = √3 IAB  -30 = 2.55  2 => Ib = 2.55  -118 => Ic = 2.55  122.
3.4 Added complexities (Self Study)
3.4.1 Line impedances.
Line impedances can often be dealt with by transforming to a wye-wye connection. This gives a series relationship
between the load and line impedances and lets the problem be solved on a per-phase basis.
Example. In the circuit below with Vab = 440  10° , Vbc = 440  250° , Vca = 440  130° V, find the line currents.
3.4.2 Phase sequence.
Positive (abc) sequence.
Van = Vp  0
Vbn = Vp  -120
Vcn = Vp  +120.
Negative (acb) sequence.
Van = Vp  0
Vbn = Vp  +120
Vcn = Vp  -120.
Negative phase sequence analyses are similar to analyses with positive phase sequences with a few exceptions:
i. Phase of sources: rotating in the clockwise direction consecutively increases by 1200.
ii. Line voltages: lag the phase voltages by 300.
iii. Line currents: lead the phase currents by 300.
Example. If Vab = 400  0 V in a balanced Y-connected three-phase generator, find the phase voltages, assuming
the phase sequence is: (a) abc and (b) acb.
3.4.4 Power analyses.
The analysis of power in three-phase loads is identical to what we covered in the previous power section (with the
exception that we now need to solve for the power in three separate loads).
cos( A + B ) = cos A cos B − sin A sin B
cos( A − B ) = cos A cos B + sin A sin B
cos A cos B = 1 / 2cos( A + B ) + cos( A − B )
Consider for this case, a wye-connected load with ZY = Z Y  and:
Vp as the phase voltage magnitude in rms units,
VL = 3V p as the line voltage magnitude in rms units,
Ip as the phase current magnitude in rms units, and
IL = Ip as the line current magnitude in rms units.
Phase voltage waveforms: v AN (t ) = 2V p cos(t ) vBN (t ) = 2V p cos(t − 120  ) vCN (t ) = 2V p cos(t + 120  ) .
Phase current waveforms: ia (t ) = 2 I p cos(t −  ) ib (t ) = 2 I p cos(t −  − 120  ) ic (t ) = 2 I p cos(t −  + 120  ) .
Total instantaneous real power: p(t ) = pa (t ) + pb (t ) + pc (t ) = v AN (t )ia (t ) + vBN (t )ib (t ) + vCN (t )ic (t )

(
) (
) (
) (
2
p(t ) = V I cos + cos(2t −  ) + cos + cos(2t −  − 240 ) + cos + cos(2t −  + 240 )
2
p(t ) = V I 3 cos  + cos(2t −  ) + cos(2t −  − 240 ) + cos(2t −  + 240 )
p(t ) = 2V p I p cos(t )cos(t −  ) + cos t − 120  cos t −  − 120  + cos t + 120  cos t −  + 120 

)

p p


p p
3 cos + cos(2t −  ) +

p(t ) = V p I p 



 
cos(2t −  ) cos 240 + sin (2t −  )sin 240 + cos(2t −  ) cos 240 − sin (2t −  )sin 240 

 1 
p(t ) = V p I p 3 cos + cos(2t −  ) + 2 cos(2t −  ) −  = 3V p I p cos
 2 

Real power: P = 3V L I L cos  => either Y or  => constant in time.
(
)
(
)
(
)
(
)
Reactive power: Q = 3V L I L sin  => either Y or  => constant in time.
Total balanced three-phase average power:
P = 3VL I L cos 
.
Total balanced three-phase reactive power:
Q = 3VL I L sin 
.
Total balanced three-phase apparent power:
S = 3VL I L
.
Total balanced three-phase complex power:
S = P + jQ = 3VL I L 
.
Example. Determine the total average power, reactive power, and complex power at the source and at the load.
Example: A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the
line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor.
Example: Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in Fig. 12.22(a). Load 1 draws
30 kW at a power factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging.
Assuming the abc sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load,
(b) the line currents.
Example. Three-phase Wattmeter readings: The total power absorbed by a three-phase load can be measured
with two Wattmeters. This is a common configuration for three-phase measurements, as it allows for readings of
power in balanced wye and delta configurations. Wattmeter 1 measures the voltage of line a with respect to line b
and the current in line a. Wattmeter 2 measures the voltage of line c with respect to line b and the current in line c.
If two Wattmeters are used to measure the power of a balanced wye-connected load as described above, write
expressions for the total average power, total reactive power, total apparent power, and power factor as a function
of the Wattmeter 1 reading, P1, and the Wattmeter 2 reading, P2.
Given I a = I L 0 :
Phase voltage is VAN = I L ZY  .
Line voltage is VAB = 3I L ZY ( + 30  ) = VL ( + 30  ) .


Wattmeter 1 power is P1 = Re VAB I a = ReVL I L ( + 30  ) = VL I L cos( + 30  ).
*
Given I c = I L 120  : Phase voltage is VCN = I L ZY ( + 120  ) .
Line voltage is VBC = 3I L ZY ( + 30  − 120  ) = VL ( − 90  ) = −VCB => VCB = VL ( + 90  ) .


Wattmeter 2 power is P2 = Re VCB I c = ReVL I L ( − 30  ) = VL I L cos( − 30  ) .
cos( A + B ) = cos A cos B − sin A sin B
Given sums and diffs: with
cos( A − B ) = cos A cos B + sin A sin B
 (
) (
)
cos cos 30 − sin  sin 30 + cos cos 30
*
P1 + P2 = VL I L cos  + 30 + cos  − 30


P1 + P2 = VL I L
Meter reading sums give the total real power.
P2 − P1 = VL I L cos  − 30 − cos  + 30
P2 − P1 = VL I L
 (
) (
)
cos cos30 + sin  sin 30 − cos cos30





+ sin  sin 30 = 3VL I L cos = P

+ sin  sin 30 = VL I L sin  = Q / 3
Meter reading difference P2 – P1 times 3 gives the total reactive power Q.
(P1 + P2 )2 + 3(P2 − P1 )2 .
= costan −1 (Q / P ) = costan −1 (3(P2 − P1 ) / (P1 + P2 )).
Apparent power is S = P 2 + Q 2 =
Power factor is pf pf
THREE-PHASE CIRCUITS SUMMARY
Three-phase voltages and currents (for balanced positive abc sequences).
Connection
Phase voltages/currents
Line voltages/current