Chapter 12 Polyphase Circuits - KOCWelearning.kocw.net/KOCW/document/2016/pusan/yimoonsuk/1.pdf · 2016. 7. 19. · 회로이론-ІI 2016년1학기 이문석 1 Chapter 12 Polyphase

회로이론-ІI 2016년1학기 이문석 1

Chapter 12Polyphase Circuits

12.1 Polyphase Systems

12.2 Single-Phase Three-Wire Systems

12.3 Three-Phase Y-Y Connection

12.4 The Delta(∆) Connection

12.5 Power Measurement in Three-Phase Systems



• A three-phase generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator).

A three-phase generator The generated voltages

• Three separate windings or coils with terminals a-a’, b-b’, c-c’ are physically placed 120° apart around the stator.



120° out of phase with each other.

Advantages of Polyphase systems (mostly three-phase system :

• Most of the electric power is generated and distributed in three-phase.• The instantaneous power in a three-phase system can be constant.• The amount of power, the three-phase system is more economical than the single-phase. • In fact, the amount of wire required for a three-phase system is less than that required for an equivalent single-phase system.

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• The notation Vab indicates the voltage between point a (+) and point b (−).

• Indicating the + and − terminals is redundant if double-subscript voltages are used.

• KVL equations can be written correctly without reference to a circuit diagram:

V V V , V V V V , V Vad ab bd ad ab bc cd ab ba= + = + + = −

Double-subscript Notation

The double subscript notation Iab means the current from a to b by the direct path

When the direct path is not obvious (e.g. between c and d) we need to use a different notation.



This is an example three-phase source with a neutral.

V V VV V 100 0 100 120100 ( 50 86.6) 173.2 30

100 3 30

ab an nb

an bn

j

= += − = ∠ °− ∠− °= − − − = ∠ °

= ∠ °

One possible representation of a three-phase system of voltages

V 100 0V 100 120V 100 240

an

bn

cn

VVV

= ∠ °= ∠− °= ∠− °



Three output terminals a, n, and b, and Van=Vbn

1V V V V V , V 2V 2Van nb an bn ab an nb= = → = − = =

Single-phase three-wire system:

The typical North American household is provided single-phase 3-wire power, where V1≅110 V and Vab=2V1.

When the A and B loads are balanced, the neutral wire from n to N carries no current:

V VI , I I IZ Z

an nbaA Bb aA Bb

p p

= = → =

0nN Bb Aa Bb aAΙ = Ι + Ι = Ι − Ι =



Example 12.1 Determine the power delivered to each of the three loads as well as the power lost in the neutral wire and each of the two lines.

1 1 2 1 3

2 2 3 2 1

3 1 3 2 3

1

2

3

115 0 50( ) 3( ) 0(20 10) 100( ) 50( ) 0

115 0 3( ) 100( ) 011.24 19.839.398 24.4710.37 21.80

j− ∠ °+ Ι + Ι − Ι + Ι − Ι =

+ Ι + Ι − Ι + Ι − Ι =− ∠ °+ Ι − Ι + Ι − Ι + Ι =

Ι = ∠− ° Α→ Ι = ∠− ° ΑΙ = ∠− ° Α

1

3

3 1

11.24 19.8310.37 158.20

0.9459 177.70

aA

bB

nN

aA bB nN

Ι = Ι = ∠− ° Α→ Ι = −Ι = ∠ ° ΑΙ = Ι − Ι = ∠− ° Α

→ Ι + Ι + Ι =

250 1 2

2100 3 2

220 10 2

(50) 206

(100) 117

(20) 1763j

P W

P W

P W+

= Ι − Ι = = Ι − Ι =

= Ι =

21

23

2

(1) 126

(1) 108

(3) 3

aA

bB

nN nN

P W

P W

P W

= Ι = = Ι =

= Ι =

Load power Line loss

115 11.24 cos(19.83 ) 1216115 10.37 cos(21.80 ) 1107

an

bn

P WP W

= × × ° = = × × ° =

Delivered PowerTransmission efficiency

2086 89.8%2086 237

= =+



Balanced three-phase sources

V V Van bn cn= = V V V 0an bn cn+ + =

V V 0 , V 120 , V 240

V 0 , V 120 , V 240

an p bn p cn p

an p bn p cn p

V Vor

V V V

= ∠ ° = ∠− ° = ∠− °

= ∠ ° = ∠ ° = ∠ °

The abc (positive) and cba (negative) sequences:

Positive sequences:Van leads Vbn by 120°Vbn leads Vcn by 120°

Negative sequences:Van leads Vcn by 120°Vcn leads Vbn by 120°



Line-to-Line Voltages

For a balanced source and balanced loads, the neutral wire current INn=0 and so the neutral wire could have any impedance, including ∞.

V V V 3 30

V V V 3 90

V V V 3 210

ab an bn p

bc bn cn p

ca cn an p

V

V

V

= − = ∠ °

= − = ∠− °

= − = ∠− °

V 3 , V V V 0L p ab bc caV and= + + =

( )VI , I I IZ

V V 120I I 120Z Z

I I 240I I I I 0

anaA aA bB cC L p

p

bn anbB aA

p p

cC aA

Nn aA bB cC

I I= = = = =

∠− °= = = ∠− °

= ∠− °→ = + + =



Example 12.2 Find the total power delivered to the loads.

V 200 0 , V 200 120 , V 200 240an bn cn= ∠ ° = ∠− ° = ∠− °

V 200 0 2 60100 60

V 200 120 2 180100 60

V 200 240 2 300100 60

anaA

p

bnbB

p

cncC

p

Z

Z

Z

∠ °Ι = = = ∠− °

∠ °

∠− °Ι = = = ∠− °

∠ °

∠− °Ι = = = ∠− °

∠ °

* of phase A Re V

200 2 cos(0 60 ) 200

Total average power is 600

av an aAP

W

W

= Ι

= × × °+ ° =



Example 12.3 A balanced three-phase system with a line voltage of 300V is supplying a balanced Y-connected load with 1200W at a leading PF of 0.8.Find the line current and the per-phase load impedance.

300Single phase(per-phase) power is 400W, and the phase voltage is 3

300400 0.83

2.89 ( )

p p p

p L

V I PF I

I I A rms

= × × = × ×

→ = =

1cos (0.8)=36.9 , leading PF : current leads voltage

V 300 3Z 60 36.9I 2.89 36.9

anp

aA

− °

= = = ∠− ° Ω∠ °



Example 12.4 A balanced 600W lighting load is added (in parallel) to the system of Example 12.3. Determine the new line current.

1 1

2 2

300200 I cos(0 ) I 1.1553

300400 I 0.8 I 2.89 ( )3

A

A rms

= × × ° → =

= × × → =

1 2

1 2

If assume the angle of phase voltage is 0

I 1.155 0 , I 2.89 36.9I I I 3.87 26.6

300 3.87 cos(26.6 ) 6003

L

pP W

°

= ∠ ° = ∠ °→ = + = ∠ °

= × × ° =


12.4 The Delta (∆) Connection

A delta-connected load is also commonly used (note the absence of the neutral wire).

V V V

3 3 V 3 V 3 VL ab bc ca

p an bn cn

V

V

= = =

= = = =

V V VI , I , IZ Z Z

ab bc caAB BC CA

p p p

= = =

Line currentI I I , I I I , I I I

I I I 3aA AB CA bB BC AB cC CA BC

L aA bB cC L pI I I

= − = − = −

= = = → =

Phase currentI I Ip AB BC CAI = = =





Example 12.5 A balanced three-phase system with a line voltage of 300 V is supplying a balanced Y-connected load with 1200 W at a lagging PF of 0.8. Find the line current and the per-phase load impedance

Single phase power is 400W400 300 0.8

1.667 3 1.667 2.89p p p

p L

V I PF I

I A I A

= × × = × ×

→ = → = × =

1cos (0.8)=36.9V V 300Z 180 36.9I I 1.667 36.9

ab ABp

AB AB

− °

= = = = ∠ ° Ω∠− °

Average power :1 cos( ) cos( ) cos( )2 m m eff eff rms rmsP V I V I V Iθ φ θ φ θ φ= − = − = −


12.5 Power Measurement in Three-phase systems

Wattmeter in a single-phase system

The wattmeter is a four-terminal device that measures power delivered to the network if connected as shown: the torque applied to the moving system is proportional to the instantaneous product of the currents flowing in the two coils.

• Current coil : very low resistance, Icurrent-coil is flowing into network

• Potential coil : relatively high resistance (voltage coil), Vpotential-coil is Vnetwork

• For an upscale reading, a positive current is flowing into the (+) end of the

current coil while (+) terminal of the potential coil is positive with respect to

the unmarked end.



Practice 12.9 Determine the wattmeter readings when (+) terminal of the wattmeter is connected to x, y, and z points.

100 2I (150 130) 4I (6 12)II=15.81 108.4

j j j= + + + + −→ ∠− °

PI

(a) Probe at point , we measure the potential,(6 12)I=(5.367 26.57 )(15.81 108.4 )

84.85 135 (84.85)(15.81)cos( 135 108.4 ) 1199.5This power is absorbed by the 6 resistor.

xj

P W− ∠− ° ∠− °

= ∠− ° → = − °+ ° =Ω

P

(b) Probe at point , we measure the potential,84.85 135 4I=84.85 135 63.24 108.4 144.2 123.7

(144.2)(15.81)cos( 123.7 108.4 ) 2199 W: absorbed by the 4 and 6 resistors.

y

P∠− °+ ∠− °+ ∠− ° = ∠− °

→ = − °+ ° = Ω Ω

(c) Probe at point , we measure the potential,100 2I=70.70 8.115

(70.70)(15.81)cos(8.115 108.4 ) 499 W: absorbed by the 100V source

zj

P− ∠ °

→ = °+ ° = −



Wattmeter in a three-phase system

• Theoretically correct configuration, but useless in practice because the neutral of the Y is not always accessible and the phases of the ∆ are not available.

The sum of the powers measured by the wattmeters is the total power delivered.

The common node x is arbitrary.



One wattmeter can be eliminated if the point x is moved to a line (as shown, B).



Example 12.7 The balanced load is fed by a balanced three-phase system having Vab=230∠0° V rms and positive phase sequence. Find the reading of each wattmeter and the total power drawn by the load.

V 230 0 , V 230 120 ,V 230 120 , V 230 60

ab bc

ca ac

= ∠ ° = ∠− °= ∠ ° = ∠− °

1

2

1 2

V 230 3 304 15 4 15

8.554 105.1

V cos( V )(230)(8.554)cos( 60 105.1 )1389

V cos( V )(230)(8.554)cos( 120 134.9 )

512.5

876.5

anaA

ac aA ac aA

bc bB bc bB

j j

P ang ang

WP ang ang

W

P P P W

∠− °Ι = =

+ += ∠− °

= Ι − Ι

= − °+ °=

= Ι − Ι

= − °− °= −

= + =

Check ! Total power is 3 V cos( V )

2303 8.554 cos( 30 105.1 )3

876.3

an aA an aAang ang

W

Ι − Ι

= × × × − °+ °

=

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Chapter 12 Polyphase Circuits - KOCWelearning.kocw.net/KOCW/document/2016/pusan/yimoonsuk/1.pdf · 2016. 7. 19. · 회로이론-ІI 2016년1학기 이문석 1 Chapter 12 Polyphase