THREE-PHASE CIRCUITS
PART I
AC GENERATOR
• Single-phase AC generator - designed to generate a single
sinusoidal voltage for each rotation of the shaft (rotor).
• Polyphase AC generator - designed to generate multiple outof-phase sinusoidal voltages for each rotation of the rotor by
increasing the number of coils on the stator.
• 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.
→ the three sinusoidal voltages are out of phase by 120°
• Two types of configuration: Wye (Y) or Delta (Δ)
SINGLE-PHASE AC GENERATOR
THREE-PHASE AC GENERATOR
FARADAY’S LAW
• The physical or experimental law governing the operation of
and AC generator.
• “The electromotive force (EMF) induced in a circuit is directly
proportional to the time rate of change of magnetic flux through
the circuit.”
• The EMF can either be produced by changing B (induced EMF)
or by changing the area, e.g., by moving the wire (motional
EMF).
Æ
It is the relative movement between the coil and the magnet
that matters.
THREE-PHASE VOLTAGE SOURCES
Y-connected Source
Δ-connected Source
(uncommon)
BALANCED Y-CONNECTED VOLTAGE SOURCE
• Balanced phase voltages are equal in magnitude and are out of
phase with one another by 120 degrees.
Van = V p ∠0o
Vbn = V p ∠ − 120o
Vcn = V p ∠120o
• Phase voltages sum up to zero, i.e., Van + Vbn + Vcn = 0
• Two possible combinations:
Van = V p ∠0°
Van = V p ∠0°
Vbn = V p ∠ − 120°
Vbn = V p ∠ + 120°
Vcn = V p ∠ − 240°
Vcn = V p ∠ + 240°
abc (+)
positive phase sequence
acb (−)
negative phase sequence
• Balanced line voltages are equal in magnitude and are out of
phase with one another by 120 degrees.
Vab = 3 V p∠30o
Vbc = 3 V p ∠ − 90o
Vca = 3 V p ∠150o
• Line voltages sum up to zero, i.e., Vab + Vbc + Vca = 0
• The magnitude of line voltages is √3 times the magnitude of the
phase voltages.
• Line voltages lead their corresponding phase voltages by 30°
Y−Δ SOURCE TRANSFORMATION
Vab = 3 VY ∠θY + 30o
Van = VY ∠θY
Vbn = VY ∠θY − 120o
Vcn = VY ∠θY − 240o
Y −Δ
⎯⎯⎯→
Vbc = 3 VY ∠θY − 90o
Vca = 3 VY ∠θY − 210o
Δ−Y SOURCE TRANSFORMATION
Vab = VΔ ∠θ Δ
Vbc = VΔ ∠θ Δ − 120o
Vca = VΔ ∠θ Δ − 240o
Δ −Y
⎯⎯⎯→
Van =
VΔ
∠θ Δ − 30o
3
Vbn =
VΔ
∠θ Δ − 150o
3
Vcn =
VΔ
∠θ Δ − 270o
3
BALANCED THREE-PHASE LOAD CONFIGURATIONS
• A balanced load is one in which the phase impedances are equal
in magnitude and in phase.
• Two possible configurations: Wye or Delta
• Conversion from Y to Δ or Δ to Y
1
ZY = Z1 = Z 2 = Z 3 = Z Δ
3
Z Δ = Z a = Z b = Z c = 3Z Y
Ex. Practice problem 12.1
Given that Vbn = 110∠30o V, find Van and Vcn , assuming a
positive sequence.
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.
1)
Y-Y connection
2) Y-Δ connection
3) Δ-Δ connection
4) Δ-Y connection
• Balanced Δ connected load is more common.
• Y connected sources are more common.
BALANCED Y-Y CONNECTION
Ex. Calculate line currents in the three-wire Y-Y system shown.
Ex. Practice problem 12.2
A Y-connected balanced three-phase generator with an impedance
of 0.4+j0.3 Ω per phase is connected to a Y-connected balanced
load with an impedance of 24+j19 Ω per phase. The line joining
the generator and the load has an impedance of 0.6+j0.7 Ω per
phase. Assuming a positive sequence for the source voltages and
that Van = 120∠30o V, find the line voltages and line currents.
BALANCED Y-Δ CONNECTION
Ex. A balanced abc-sequence Y-connected source with phase
voltage Van = 100∠10o V is connected to a Δ-connected
balanced load of 8+j4 Ω per phase. Calculate the phase and the
line currents.
Ex. Practice problem 12.3
One line voltage of a balanced Y-connected source is
V AB = 240∠ − 20o V. If the source is connected to a Δ-connected
load of 20∠ 40o Ω, find the phase and line currents assuming the
abc sequence.
BALANCED Δ-Δ CONNECTION
Ex. A balanced Δ-connected load having an impedance of 20−j15
Ω is connected to a Δ-connected, positive sequence generator
having Vab = 330∠0o V. Calculate the phase currents of the
load and the line currents.
Ex. Practice problem 12.4
A positive-sequence, balanced Δ-connected source supplies a
balanced Δ-connected load. If the impedance per phase of the load
is 18+j12 Ω and I a = 19.202∠35o A, find I AB and V AB .
BALANCED Δ-Y CONNECTION
Ex. A balanced Y-connected load with a phase impedance of
40+j 25 Ω is connected to a balanced, positive sequence Δconnected source with a line voltage of 210 V. Calculate the
phase currents. Use Vab as a reference.
Ex. Practice problem 12.5
In a balanced Δ-Y circuit, Vab = 240∠15o V and ZY = 12 + j15 Ω.
Calculate the line currents.