Three Phase Circuits
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
1
Three Phase Circuits
Chapter Objectives:
Be familiar with different three-phase configurations and how
to analyze them.
Know the difference between balanced and unbalanced circuits
Learn about power in a balanced three-phase system
Know how to analyze unbalanced three-phase systems
Be able to use PSpice to analyze three-phase circuits
Apply what is learnt to three-phase measurement and
residential wiring
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Three phase Circuits
An AC generator designed to develop a single sinusoidal voltage for each rotation
of the shaft (rotor) is referred to as a single-phase AC generator.
If the number of coils on the rotor is increased in a specified manner, the result is a
Polyphase AC generator, which develops more than one AC phase voltage per
rotation of the rotor
In general, three-phase systems are preferred over single-phase systems for the
transmission of power for many reasons.
1. Thinner conductors can be used to transmit the same kVA at the same voltage, which
reduces the amount of copper required (typically about 25% less).
2. The lighter lines are easier to install, and the supporting structures can be less
massive and farther apart.
3. Three-phase equipment and motors have preferred running and starting
characteristics compared to single-phase systems because of a more even flow of power
to the transducer than can be delivered with a single-phase supply.
4. In general, most larger motors are three phase because they are essentially selfstarting and do not require a special design or additional starting circuitry.
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Single Phase, Three phase Circuits
a) Single phase systems two-wire type
b) Single phase systems three-wire type.
Allows connection to both 120 V and
240 V.
Two-phase three-wire system. The AC sources
operate
at different phases.
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Dr. Kalyana Veluvolu
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Three-phase Generator
The three-phase generator has three induction coils placed 120° apart on the stator.
The three coils have an equal number of turns, the voltage induced across each coil
will have the same peak value, shape and frequency.
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Dr. Kalyana Veluvolu
Balanced Three-phase Voltages
Three-phase four-wire system
Neutral Wire
A Three-phase Generator
Voltages having 120° phase difference
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Dr. Kalyana Veluvolu
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Balanced Three phase Voltages
Neutral Wire
a) Wye Connected Source
b) Delta Connected Source
Van = V p ∠0°
Van = V p ∠0°
Vbn = V p ∠ − 120°
Vbn = V p ∠ + 120°
Vcn = V p ∠ − 240°
Vcn = V p ∠ + 240°
abcTheory
or positive
ELEC 24409: a)
Circuit
2
sequence
b)
or negative
Dr. acb
Kalyana
Veluvolu sequence
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Balanced Three phase Loads
A Balanced load has equal impedances on all the phases
a) Wye-connected load
b) Delta-connected load
Balanced Impedance Conversion:
Conversion of Delta circuit to Wye or Wye to Delta.
ZY = Z1 = Z 2 = Z 3
Z ∆ = Z a = Zb = Zc
Z∆ = 3ZY
ELEC 24409: Circuit Theory 2
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ZY = Z ∆
3
Dr. Kalyana Veluvolu
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Three phase Connections
Both the three phase source and the three phase load can be
connected either Wye or DELTA.
We have 4 possible connection types.
• Y-Y connection
• Y-∆ connection
• ∆-∆ connection
• ∆-Y connection
Balanced ∆ connected load is more common.
Y connected sources are more common.
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Dr. Kalyana Veluvolu
Balanced Wye-wye Connection
A balanced Y-Y system, showing the source, line and load impedances.
Line Impedance
Source Impedance
Load Impedance
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Balanced Wye-wye Connection
Line current In add up to zero.
Neutral current is zero:
In= -(Ia+ Ib+ Ic)= 0
Phase voltages are: Van, Vbn and Vcn.
The three conductors connected from a to A, b to B and c to C are called LINES.
The voltage from one line to another is called a LINE voltage
Line voltages are: Vab, Vbc and Vca
Magnitude of line voltages is √3 times the magnitude of phase voltages. VL= √3 Vp
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Dr. Kalyana Veluvolu
Balanced Wye-wye Connection
Line current In add up to zero.
Neutral current is zero:
In= -(Ia+ Ib+ Ic)= 0
Magnitude of line voltages is √3 times the magnitude of phase voltages. VL= √3 Vp
Van = V p ∠0°, Vbn = V p ∠ − 120°, Vcn = V p ∠ + 120°
Vab = Van + Vnb = Van − Vbn = 3V p ∠30°
Vbc = Vbn − Vcn = 3V p ∠ − 90°
Vca = Vcn − Van = Van + Vbn = 3V p ∠ − 210°
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Balanced Wye-wye Connection
Phasor diagram of phase and line voltages
VL = Vab = Vbc = Vca
= 3 Van = 3 Vbn = 3 Vcn = 3V p
V p = Van = Vbn = Vcn
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Single Phase Equivalent of Balanced Y-Y Connection
Balanced three phase circuits can be analyzed on “per phase “ basis..
We look at one phase, say phase a and analyze the single phase equivalent circuit.
Because the circuıit is balanced, we can easily obtain other phase values using their
phase relationships.
Ia =
ELEC 24409: Circuit Theory 2
Van
ZY
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ELEC 24409: Circuit Theory 2
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Dr. Kalyana Veluvolu
Balanced Wye-delta Connection
Three phase sources are usually Wye connected and three phase loads are Delta
connected.
There is
no neutral connection for the Y-∆ system.
I AB =
VAB
Z∆
I BC =
VBC
Z∆
I CA =
VCA
Z∆
Line currents are obtained from the phase currents IAB, IBC and ICA
I a = I AB − I CA = I AB 3∠ − 30°
I L = I a = Ib = Ic
I b = I BC − I AB = I BC 3∠ − 30°
I p = I AB = I BC = I CA
I c = I CA − I BC = I CA 3∠ − 30°
ELEC 24409: Circuit Theory 2
I L = 3I p
Dr. Kalyana Veluvolu
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Balanced Wye-delta Connection
Phasor diagram of phase and line currents
I L = I a = Ib = Ic
I p = I AB = I BC = I CA
I L = 3I p
Single phase equivalent circuit of the balanced Wye-delta connection
Z∆
3
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Balanced Delta-delta Connection
Both the source and load are Delta connected and balanced.
I AB =
V
V
VAB
, I BC = BC , I CA = CA
Z∆
Z∆
Z∆
I a = I AB − I CA , I b = I BC − I AB , I c = I CA − I BC
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Balanced Delta-wye Connection
Transforming a Delta connected source
to an equivalent Wye connection
Single phase equivalent of Delta Wye connection
Vp∠−30°
3
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Power in a Balanced System
The total instantaneous power in a balanced three phase system is constant.
v AN = 2V p cos(ωt ) vBN = 2V p cos(ωt − 120°) vCN = 2V p cos(ωt + 120°)
ia = 2 I p cos(ωt − θ ) ib = 2 I p cos(ωt − θ − 120°) ib = 2 I p cos(ωt − θ + 120°)
p = pa + pb + pc = v AN ia + vBN ib + vCN ic
p = 2V p I p [ cos(ωt ) cos(ωt − θ ) + cos(ωt − 120°) cos(ωt − θ − 120°) + cos(ωt + 120°) cos(ωt − θ + 120°)]
1
cos A cos B = [cos( A + B ) + cos( A − B )] Using the identity and simplifying
2
p = 3V p I p cosθ
The instantenous power is not function of time.
The total power behaves similar to DC power.
This result is true whether the load is Y or ∆ connected.
The average power per phase Pp = p .
3
Pp = p
ELEC 24409: Circuit Theory 2
3
= V p I p cos θ
Dr. Kalyana Veluvolu
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Power in a Balanced System
The complex power per phase is Sp. The total complex power for all phases is S.
p = 3V p I p cos θ
1
p = V p I p cos θ
3
Sp = Pp + jQp = Vp I p∗
Pp =
Qp =
1
p = V p I p sin θ
S p = Vp I p
3
Complex power for each phase
P = Pa + Pb + Pc = 3Pp = 3V p I p cos θ = 3VL I L cos θ
Q = 3Q p = 3V p I p sin θ = 3VL I L sin θ
2
∗
S=3Sp = 3Vp I p = 3I p Z p =
3Vp 2
Zp
∗
Total complex power
S = P + jQ = 3VL I L ∠θ
Vp , I p , VL and I L are all rms values, θ is the load impedance angle
ELEC 24409: Circuit Theory 2
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Dr. Kalyana Veluvolu
Power in a Balanced System
∗
2
S=3Sp = 3Vp I p = 3I p Z p =
3Vp 2
Z p∗
Toal complex power
S = P + jQ = 3VL I L ∠θ
Vp , I p ,VL and I L are all rms values, θ is the load impedance angle
Notice the values of Vp, VL, Ip, IL for different load connections.
VL = 3 V p
VL = V p
IL = I p
IL = 3 I p
Ip
Vp
Ip
VL
Vp
VL
VL
Vp
Ip
VL
ELEC 24409: Circuit TheoryY
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VL
connected load.
Ip
Ip
Vp
Ip
Vp
VL
∆ connected
Dr. Kalyana Veluvolu
Vp
load.
22
Power in a Balanced System
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Single versus Three phase systems
Three phase systems uses lesser amount of wire than single phase systems for the
same line voltage VL and same power delivered.
a) Single phase system
b) Three phase system
Wire Material for Single phase 2(π r 2l ) 2r 2 2
=
=
= (2) = 1.33
Wire Material for Three phase 3(π r '2l ) 3r '2 3
If same power loss is tolerated in both system, three-phase system use
only 75% of materials of a single-phase system
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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ELEC 24409: Circuit Theory 2
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Dr. Kalyana Veluvolu
VL=840 V (Rms)
IL
Capacitors for pf
Correction
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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IL =
ELEC 24409: Circuit Theory 2
S
73650
=
= 50.68A
3 VL
3 840
Dr. Kalyana
Veluvolu
Without
Pf
Correction
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Unbalanced Three Phase Systems
An unbalanced system is due to unbalanced voltage sources or unbalanced load.
In a unbalanced system the neutral current is NOT zero.
Unbalanced three phase Y connected load.
Line currents DO NOT add up to zero.
In= -(Ia+ Ib+ Ic) ≠ 0
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Three Phase Power Measurement
Two-meter method for measuring three-phase power
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Residential Wiring
Single phase three-wire residential wiring
ELEC 24409: Circuit Theory 2
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Dr. Kalyana Veluvolu
Problem 12.10 Determine the current in the neutral line.
UNBALANCED LOAD
NEUTRAL CURRENT IS NOT
ZERO
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Problem 12.12 Solve for the line currents in the Y-∆
∆ circuit. Take Z∆ =
60∠
∠45°Ω
°Ω.
°Ω
SINGLE PHASE EQUIVALENT
CIRCUIT
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Problem 12.22 Find the line currents Ia, Ib, and Ic in the three-phase network
below. Take Z∆ = 12 - j15Ω, ZY = 4 + j6 Ω, and Zl = 2 Ω.
ONE DELTA AND ONE Y
CONNECTED LOAD IS
CONNECTED
TWO Loads are parallel if they are
converted to same type.
Delta connected load is converted to
Y connection.
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Problem 12.26 For the balanced circuit below, Vab = 125∠0° V. Find the line
currents IaA, IbB, and IcC.
BALANCED Y CONNECTED LOAD.
Source voltage given is line to line, obtain
the line to neutral voltage.
ELEC 24409: Circuit Theory 2
Problem 12.47
Dr. Kalyana Veluvolu
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The following three parallel-connected three-phase loads are fed by a balanced three-
phase source.
Load 1: 250 kVA, 0.8 pf lagging Load 2: 300 kVA, 0.95 pf leading
Load 3: 450 kVA, unity pf
If the line voltage is 13.8 kV, calculate the line current and the power factor of the source. Assume that the
line impedance is zero.
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Problem 12.81
A professional center is supplied by a balanced three-phase source. The center has
four plants, each a balanced three-phase load as follows:
Load 1: 150 kVA at 0.8 pf leading Load 2: 100 kW at unity pf
Load 3: 200 kVA at 0.6 pf lagging Load 4: 80 kW and 95 kVAR (inductive)
If the line impedance is 0.02 + j0.05 Ω per phase and the line voltage at the loads is 480 V, find the
magnitude of the line voltage at the source.
ELEC 24409: Circuit Theory 2
Dr. Kalyana Veluvolu
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Problem 12.84
The Figure displays a three-phase delta-connected motor load which is connected to
a line voltage of 440 V and draws 4 kVA at a power factor of 72 percent lagging. In addition, a single 1.8
kVAR capacitor is connected between lines a and b, while a 800-W lighting load is connected between line c
and neutral. Assuming the abc sequence and taking Van = Vp ∠0°, find the magnitude and phase angle of
currents Ia, Ib, Ic, and In.
Total load is UNBALANCED. LINE
CURRENTS Ia, Ib, Ic ARE NOT
ERQUAL
Single phase, 800 W lighting load connected to phase C
only. Pf for lighting loads is unity.
I L = I1 =
ELEC 24409: Circuit Theory 2
S
3 VL
Dr. Kalyana Veluvolu
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440 V
I1 =
ELEC 24409: Circuit Theory 2
S
= 5.249 ∠ (θ + 30 ° )
3 VL
Dr. Kalyana Veluvolu
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