POLYPHASE CIRCUITS

LEARNING GOALS

Three Phase CircuitsAdvantages of polyphase circuits

Three Phase ConnectionsBasic configurations for three phase circuits

Source/Load ConnectionsDelta-Wye connections

THREE PHASE CIRCUITS

))(240cos()(

))(120cos()(

))(cos()(

VtVtv

VtVtv

VtVtv

mc

mbn

man

°−=°−=

=

ωωω

VoltagesPhase ousInstantane

ai

bi

ci

)240cos()(

)120cos()(

)cos()(

°−−=°−−=

−=

θωθωθω

tIti

tIti

tIti

mc

mb

ma

Currents Phase Balanced

2120=mV

)()()()()()()( titvtitvtitvtp ccnbbnaan ++=power ousInstantane

TheoremFor a balanced three phase circuit the instantaneous power is constant

)(cos2

3)( WIV

tp mm θ=

Proof of TheoremFor a balanced three phase circuit the instantaneous power is constant

)(cos2

3)( WIV

tp mm θ=

−−−+−−−+

−=

++=

)240cos()240cos(

)120cos()120cos(

)cos(cos

)(

)()()()()()()(

θωωθωω

θωω

tt

tt

tt

IVtp

titvtitvtitvtp

mm

ccnbbnaan

power ousInstantane

[ ])cos()cos(2

1coscos βαβαβα ++−=

3cos cos(2 )

( ) cos(2 240 )2

cos(2 480 )

m m

tV I

p t t

t

θ ω θω θω θ

+ −= + − −

+ − −

)120sin(sin)120cos(cos)120cos(

)120sin(sin)120cos(cos)120cos(

coscos

0)120cos()120cos(cos

φφφφφφ

φφ

φφφ

−=++=−

=

=++−+Proof

Lemma5.0)120cos( −=

0)120cos()120cos(cos =++−+ φφφ

2

cos( 240) cos( 120)

cos( 480) cos( 120)

tφ ω θφ φφ φ

= −− = +− = −

THREE-PHASE CONNECTIONS

Positive sequencea-b-c

Y-connectedloads Delta connected loads

SOURCE/LOAD CONNECTIONS

BALANCED Y-Y CONNECTION

°∠=

°−∠=

°∠=

120||

120||

0||

pcn

pbn

pan

VV

VV

VV

Positive sequencephase voltages

abV

bcV

caV

Line voltages

( )| | 0 | | 120

| | 1 (cos120 sin120)

1 3| | | |

2 2

3 | | 30

ab an bn

p p

p

p p

p

V V V

V V

V j

V V j

V

= −= ∠ °− ∠ − °

= − −

= − − − ÷ ÷

= ∠ °

°−∠=

°−∠=

210||3

90||3

pca

pbc

VV

VV

VoltageLine == ||3 pL VVY

cnc

Y

bnb

Y

ana Z

VI

Z

VI

Z

VI === ;;

°+∠=°−∠=°∠= 120||;120||;|| θθθ LcLbLa IIIIII

0==++ ncba IIII For this balanced circuit it is enough to analyze one phase

Relationship betweenphase and line voltages

LEARNING EXAMPLE For an abc sequence, balanced Y - Y three phase circuit

°−∠= 30208abV

Determine the phase voltages

Balanced Y - Y

Positive sequencea-b-c

°∠= 30||3 pab VV

°∠=

°−∠=

°∠=

120||

120||

0||

pcn

pbn

pan

VV

VV

VV

Positive sequencephase voltages

)3030(3

|| °−°−∠=∴ aban

VV

°30by lags aban VV

°−∠= 30208abV

°∠=°−∠=

°−∠=

60120

180120

60120

an

bn

an

V

V

V

The phasor diagram could be rotated by any angle

LEARNING EXAMPLE For an abc sequence, balanced Y - Y three phase circuit

source 120( ) , 1 1 , 20 10rmsphase line phaseV V Z j Z j= = + Ω = + Ω

Determine line currents and load voltages

Because circuit is balanceddata on any one phase issufficient

°∠0120Chosenas reference

120 0

120 120

120 120

an

bn

cn

V

V

V

= ∠ °= ∠ − °= ∠ °

Abc sequence

rmsA

j

VI anaA

)(65.2706.5

65.2771.23

0120

1121

°−∠=°∠

°∠=+

=

rmsAI

rmsAI

cC

bB

)(65.2712006.5

)(65.2712006.5

°−∠=°−−∠=

°∠×=+×= 57.2636.22)1020( aAaAAN IjIV

rmsVVAN )(08.115.113 °−∠=

rmsVV

rmsVV

CN

BN

)(92.11815.113

)(08.12115.113

°∠=°−∠=

DELTA CONNECTED SOURCES

Relationship betweenphase and line voltages

°∠= 30||3 pab VV

°30by lags aban VV

⇒

°∠=°−∠=

°∠=

120

120

0

Lca

Lbc

Lab

VV

VV

VV

°∠=

°−∠=

°−∠=

903

1503

303

Lcn

Lbn

Lan

VV

VV

VV

°∠=°−∠=

°∠=

150120

90120

30120

cn

bn

an

V

V

V

⇒

°∠=°−∠=

°∠=

180208

60208

60208

ca

bc

ab

V

V

VExample

Convert to an equivalent Y connection

LEARNING EXAMPLE Determine line currents and line voltages at the loads

Source is Delta connected.Convert to equivalent Y

⇒

°∠=°−∠=

°∠=

120

120

0

Lca

Lbc

Lab

VV

VV

VV

°∠=

°−∠=

°−∠=

903

1503

303

Lcn

Lbn

Lan

VV

VV

VV

Analyze one phase

rmsAj

IaA )(14.4938.92.41.12

30)3/208( °−∠=+

°−∠=

rmsVjVAN )(71.3065.11819.4938.9)412( °−∠=°−∠×+=

Determine the other phases using the balance

rmsAI

rmsAI

cC

bB

)(86.7138.9

)(14.16938.9

°−∠=°−∠=

3 118.65 0.71ABV = × ∠ − °

3 118.65 120.71

3 118.65 119.29BC

CA

V

V

= × ∠ − °= × ∠ °

DELTA-CONNECTED LOAD

Method 1: Solve directly

°∠=

°−∠=

°∠=

120||

120||

0||

pcn

pbn

pan

VV

VV

VV

Positive sequencephase voltages

°−∠=

°−∠=

°∠=

210||3

90||3

30||3

pca

pbc

pab

VV

VV

VV

°+∠==

°−∠==

∠==

∆∆∆

∆∆∆

∆∆∆

120||

120||

||

θ

θ

θ

IZ

VI

IZ

VI

IZ

VI

CACA

BCBC

ABAB

currents phase Load

BCCAcC

ABBCbB

CAABaA

III

III

III

−=−=−=

currents Line

Method 2: We can also convert the delta connected load into a Y connected one. The same formulas derived for resistive circuits are applicable to impedances

3∆= ZZY case Balanced

−=

=⇒∠== ∆

ZL

ABaA

LaAY

anaA Z

VI

IZ

VI

θθθ 3/||

3/||||

||

ZLZZ θ∠=∆ ||

Zθθ −°=∆ 30

°−==

∆

∆

30

||3||

θθ lineline II

Line-phase currentrelationship

θ

°−==

∆

∆

30

||3||

θθ lineline II

Line-phase currentrelationship

iprelationsh voltagephase-Line

°+=

=

∆

∆

30

||3||

phase

phaseVV

θθ

LEARNING EXTENSION

currents phase the Find .4012 °∠=aAI

°∠=°−∠=

°∠=

19093.6

5093.6

7093.6

CA

BC

AB

I

I

I

Y→∆

∆ → Y baab RRR +=

)(|| 312 RRRRab +=

321

312 )(

RRR

RRRRR ba ++

+=+

321

213 )(

RRR

RRRRR cb ++

+=+

321

321 )(

RRR

RRRRR ac ++

+=+

SUBTRACT THE FIRST TWO THEN ADDTO THE THIRD TO GET Ra

a

b

b

a

R

RRR

R

R

R

R 13

3

1 =⇒=c

b

c

b

R

RRR

R

R

R

R 12

1

2 =⇒=

REPLACE IN THE THIRD AND SOLVE FOR R1

Y

RRR

RRR

RRR

RRR

RRR

RRR

c

b

a

→∆++

=

++=

++=

321

13

321

32

321

21

∆−

++=

++=

++=

Y

R

RRRRRRR

R

RRRRRRR

R

RRRRRRR

a

accbba

c

accbba

b

accbba

3

2

1

tionsTransformaY

OF REVIEW↔∆

3321∆

∆ =⇒=== RRRRRR Y

LEARNING EXAMPLE

°∠=Ω+=∆ 02.3752.1254.710 jZ

Delta-connected load consists of 10-Ohm resistance in serieswith 20-mH inductance. Source is Y-connected, abc sequence,120-V rms, 60Hz. Determine all line and phase currents

rmsVVan )(30120 °∠=

Ω=××= 54.7020.0602πinductanceZ

rmsAjZ

VI ABAB )(98.2260.16

54.710

603120 °∠=+

°∠==∆

°−==

∆

∆

30

||3||

θθ lineline II

Line-phase currentrelationship

iprelationsh voltagephase-Line

°+=

=

∆

∆

30

||3||

phase

phaseVV

θθ rmsAIaA )(02.775.28 °−∠=

rmsAI

rmsAI

CA

BC

)(98.14260.16

)(02.9760.16

°∠=°−∠=

rmsAI

rmsAI

cC

bB

)(98.11275.28

)(02.12775.28

°∠=°−∠=

°∠=⇒ 02.3717.4YZ

Alternatively, determine first the line currentsand then the delta currents

Polyphase