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 out-of-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).

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.

oVV VV0∠= pano120−∠= pbn

o120∠= pcn VV

• Phase voltages sum up to zero, i.e., 0=++ cnbnan VVV • Two possible combinations:

bc (+) acb (−) a positive phase sequence negative phase sequence

0

120

240

an p

bn p

cn p

V V

V V

V V

= ∠ °

= ∠ + °

= ∠ + °

0

120

240

an p

bn p

cn p

V V

V V

V V

= ∠ °

= ∠ − °

= ∠ − °

• Balanced line voltages are equal in magnitude and are out of phase with one another by 120 degrees.

o303 ∠= VV pab opbc 903 −∠= VV o1503 ∠= pca VV

• Line voltages sum up to zero, i.e., 0=++ cabcab VVV

• 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

YYan VV θ∠= o303 +∠= YYab VV θ

o120−∠= Yθ ⎯⎯ →⎯ Δ−Y Ybn VV o903 −∠= YYbc VV θ

o240−∠= YYcn VV θ o2103 −∠= YYca VV θ

Δ−Y SOURCE TRANSFORMATION

ΔΔ∠= θVVab o303

−∠= ΔΔ θVVan

o120−∠= ΔΔ θVVbc ⎯⎯ →⎯ −Δ Y o1503

−∠= ΔΔ θVVbn

o240−∠= ΔΔ θVVca o2703

−∠= ΔΔ θVVcn

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

Δ==== ZZZZZY 33211 Z Z Ycba Z Z Z= = = = 3Δ

Ex. Practice problem 12.1

Given that o30110∠=bnV V, find anV and cnV , 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 o30120∠=anV V, find the line voltages and line currents.

BALANCED Y-Δ CONNECTION

Ex. A balanced abc-sequence Y-connected source with phase o voltage 10100∠=anV 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 o20240 −∠=ABV V. If the source is connected to a Δ-connected

load of Ω, find the phase and line currents assuming the abc sequence.

o4020∠

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 o0330∠=abV 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 o35202.19 ∠=aI A, find ABI and .ABV

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 Δ- connected 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, o15240∠=abV V and .1512 Ω+= jZY Calculate the line currents.