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1ELEC 24409: Circuit Theory 2 Dr. Kalyana Veluvolu

Three Phase Circuits


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 andresidential wiring


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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 startingcharacteristics 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 self-

starting and do not require a special design or additional starting circuitry.


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.

Single Phase, Three phase Circuits


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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.


Balanced Three-phase Voltages

Three-phase four-wire system

Neutral Wire

A Three-phase Generator

Voltages having 120° phase difference


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Balanced Three phase Voltages

a) Wye Connected Source b) Delta Connected Source

a) abc or positive sequence b) acb or negative 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

= ∠ °

= ∠ + °

= ∠ + °

Neutral Wire


Balanced Three phase Loads

a) Wye-connected load b) Delta-connected load

1 2 3

Conversion of Delta circuit to Wye or Wye to Delta.Balanced Impedance Conversion:

Y

a b c

Z Z Z Z

Z Z Z Z ∆

= = =

= = =

1Z 3 Z

3Y Y

Z Z ∆ ∆

= =

A Balanced load has equal impedances on all the phases


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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.


Balanced Wye-wye Connection A balanced Y-Y system, showing the source, line and load impedances.

Source ImpedanceLine Impedance

Load Impedance


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Balanced Wye-wye Connection

Phase voltages are: V an, V bn and V cn.

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: V ab, V bc and V ca

Magnitude of line voltages is √3 times the magnitude of phase voltages. V L= √3 V p

Line current I n add up to zero.Neutral current is zero:

I n= -( I a+ I b+ I c)= 0


Balanced Wye-wye Connection

Magnitude of line voltages is √3 times the magnitude of phase voltages. V L= √3 V p

3

0 , 120 ,

30

3 90

3 21

120

0

an p bn p cn p

ab an nb an bn

bc bn cn

ca cn an

p

p

an bn p

V

V V V V V V

V V V V V

V V V

V V V

V

V V V

= ∠ ° = ∠ − ° = ∠ + °

= + = − =

= − =

∠ °

∠ − °

= + ∠ −= − = °

Line current I n add up to zero.Neutral current is zero:

I n= -( I a+ I b+ I c)= 0


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Balanced Wye-wye Connection Phasor diagram of phase and line voltages

= 3 3 3 = 3 L ab bc ca

an bn cn p

p an bn cn

V V V V

V V V V

V V V V

= = =

= =

= = =


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.

ana

Y

V I Z

=


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Balanced Wye-delta Connection

AB AB

BC BC

CACA

V I

Z V

I Z

V I

Z

∆

∆

∆

=

=

=

Line currents are obtained from the phase currents I AB, I BC and I CA

3 303 30

3 30

a AB CA

b BC AB

c CA BC

AB

BC

CA

I I I

I I I

I

I

I I I

I

= − =

= − =

∠ − °

∠ − °

∠− −= = ° 3

L a b c

p AB BC CA

L p

I I I I

I I I I

I I

= = =

= = =

=

Three phase sources are usually Wye connected and three phase loads are Delta

connected.

There is no neutral connection for the Y- system.


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Balanced Wye-delta Connection

3

Z ∆

Single phase equivalent circuit of the balanced Wye-delta connection

Phasor diagram of phase and line currents

3

L a b c

p AB BC CA

L p

I I I I

I I I I

I I

= = =

= = =

=


Balanced Delta-delta Connection Both the source and load are Delta connected and balanced.

, ,a AB CA b BC AB c CA BC

I I I I I I I I I = − = − = −

, , BC CA AB AB BC CAV V V

I I I Z Z Z

∆ ∆ ∆

= = =


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Balanced Delta-wye Connection

30

3

pV ∠− °

Transforming a Delta connected source

to an equivalent Wye connection Single phase equivalent of Delta Wye connection


Power in a Balanced System The total instantaneous power in a balanced three phase system is constant.

2 cos( ) 2 cos( 120 ) 2 cos( 120 )

2 cos( ) 2 cos( 120 ) 2 cos( 120 )

2 cos( )cos( ) cos( 120 )cos( 120 ) cos( 120 )co

AN p BN p CN p

a p b p b p

a b c AN a BN b CN c

p p

v V t v V t v V t

i I t i I t i I t

p p p p v i v i v i

p V I t t t t t

ω ω ω

ω θ ω θ ω θ

ω ω θ ω ω θ ω

= = − ° = + °

= − = − − ° = − + °

= + + = + +

= − + − ° − − ° + + °[ ]s( 120 )

1

cos cos [cos( ) cos( )] Using the identity and simplif

The instantenous power is not function of time.

The total power behav

n2

e

yi g

3 cos p p

t

A B A B A B

p V I

ω θ

θ

− + °

= + + −

=

s similar to DC power.

This result is true whether the load is Y or connected.

The average power per phase .3

cos3

p p p

p

p

pP I

P

V θ

=

= =

∆


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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 p

p p p p p

2

p p p p

Complex power for each phas

3 cos

1 1= cos = sin

3 3S V I

3 3 cos 3 cos

3 3 sin 3 sin

S=3S 3V

e

I 3

p p

p p p p p p p

a b c p p p L L

p p p L L

p

p V I

P p V I Q p V I S V I

P jQ

P P P P P V I V I

Q Q V I V I

I Z

θ

θ θ

θ θ

θ θ

∗

∗

=

= = =

= + =

= + + = = =

= = =

= = =

p L

2

p

p , , and are

Total

all rm

complex powe

s values, is the load impedance angl

3

3

r

e L

p

L L

V

Z

P jQ

I

V

V I

I

V θ

θ

∗

= + = ∠S



Notice the values of V p, V L, I p, I L for different load connections.

2

p2

p p p

p

p

p L, , and are all rms values, is the load impe

To

da

3S=3S 3V I 3

nce

al c

an

3

omplex ower

gle

p

L

p

p

L L

V

V I Z

Z

P

V

V I

I

jQ

I θ

θ

∗

∗= = =

= + = ∠S

V L

V L

V L

V p V p

V p I p

I p I p

V L

V p

I p

V L

V L V p

V p

I p

I p

Y connected load. ∆ connected load.

3 L p L p

V V I I = = 3 L p L pV V I I = =


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Single versus Three phase systems Three phase systems uses lesser amount of wire than single phase systems for thesame line voltage V L and same power delivered.

a) Single phase system b) Three phase system

2 2

'2 '2

Wire Material for Single phase 2( ) 2 2(2) 1.33

Wire Material for Three phase 3( ) 3 3

r l r

r l r

π

π = = = =

If same power loss is tolerated in both system, three-phase system use

only 75% of materials of a single-phase system


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VL=840 V (Rms)

Capacitors for pf

Correction

IL


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7365050.68A

3 3 840

Without Pf Correction

L

L

S I

V = = =


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.

I n= -( I a+ I b+ I c) ≠ 0


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Residential Wiring

Single phase three-wire residential wiring


Problem 12.10Determine the current in the neutral line.

UNBALANCED LOAD

NEUTRAL CURRENT IS NOT

ZERO


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Problem 12.12 Solve for the line currents in the Y-∆∆∆∆ circuit. Take Z∆∆∆∆

=

60∠∠∠∠45°Ω°Ω°Ω°Ω.

SINGLE PHASE EQUIVALENT

CIRCUIT


Problem 12.22Find the line currents Ia, Ib, and Ic in the three-phase networkbelow. Take Z

∆= 12 - j15Ω, Z

Y = 4 + j6 Ω, and Z

l= 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.


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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.


Problem 12.47 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.


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Problem 12.81 A professional center is supplied by a balanced three-phase source. The center hasfour 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.


Problem 12.84The

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

= V p ∠0°, find the magnitude and phase angle of

currents I a, I

b, I

c, and I

n.

Total load is UNBALANCED. LINE

CURRENTS I a, I

b, I

cARE NOT

ERQUAL

Single phase, 800 W lighting load connected to phase C

only. Pf for lighting loads is unity.

13

L

L

S I I

V = =


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440 V

1 5.249 ( 30 )3 L

S I

V θ = = ∠ + °

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