POLYPHASE CIRCUITS

Introduction:

Three-phase systems are commonly used in generation, transmission and distribution of electric

power. Power in a three-phase system is constant rather than pulsating and three-phase motors start

and run much better than single-phase motors. A three-phase system is a generator-load pair in

which the generator produces three sinusoidal voltages of equal amplitude and frequency but

differing in phase by 120 from each other. As shown below:

(a) Schematic layout of generator of 3-

phase

(b) Waveform of induced emf in R,Y,B

coils

(c) phasor Diagram

VR = Vm sin ωt

VY = Vm sin (ωt-120o)

VB = Vm sin (ωt-240o)

OR

VR = Vm 0

VY = Vm -120o

VB = Vm 120o

Algebraic or Vector sum at any time of all emfs

V = VR + VY + VB = 0

Phase sequence: Order in which emfs or induced current attain their peak value, i.e. here

it is R – Y – B or sometimes also specified as a-b-c

Fig.1

A three-phase system is shown in Fig 1. In a special case all impedances are identical

Za = Zb = Zc = Z

Such a load is called a balanced load and is described by equations

IV

ZI

V

ZI

V

Za

ab

bc

c .

Using KCL, we have

I I I IZ

V V Vn a b c a b c 1

,

where

.02

3j

2

1

2

3j

2

11V240sinj240cos120sinj120cos1V

ee1VVVV

mm

240j120j

mcba

Setting the above result into (4), we obtain

In 0 .

Since the current flowing through the fourth wire is zero, the wire can be removed (see

Fig.2)

Fig. 2

The system of connecting the voltage sources and the load branches, as depicted in Fig. 2, is called

the Y system or the star system. Point n is called the neutral point of the generator and point n’ is

called the neutral point of the load.

Each branch of the generator or load is called a phase. The wires connecting the supply to the

load are called the lines. In the Y-system shown in Fig. 2 each line current is equal to the

corresponding phase current, whereas the line-to-line voltages ( or simply line voltages ) are not

equal to the phase voltages.

Methods of interconnection of three phase circuit. (i) Y- connection or Star connection (ii) Δ – connection or Delta connection

Y- connection:

n’ n

Ia Z Va

Ic Z Vc

Ib Z Vb

Useful Definitions:

(i) Phase voltage and Phase current: Voltage of any one of phases w.r.t. neutral or

star point

(ii) Line voltage and Line current: Voltage between any two phase and current in

any line

(iii) Balanced system or load: Voltage in all three phase are equal in magnitude

and displaced from each other by equal angle

(iv) Unbalanced system or load: if one of two above condition or both is not

satisfied it is known as unbalanced system of load.

Now we consider the Y-connected generator sources (see Fig. 3).

Fig.3

The phasors of the phase voltages can be generally written as follows

.

V V V e

V Ve

V Ve

a mj

bj

cj

120

240

o

o

We determine the line voltages Vab, Vbc, Vca ( see Fig.3). Using KVL, we obtain

.e3Ve2

3

2

3V

2

3j

2

3V

2

3j

2

11VVVV

30j

a3

3tanj

22

a

aabaab

1

Thus,

V V eab aj 3 30o

.

Holds and similarly we obtain

Vca

c Vbc

Vab

n

a Va

Vc

b Vb

V V ebc bj 3 30o

V V eca cj 3 30o

.

The phasor diagram showing the phase and line voltages is shown in Fig.4.

Fig.4

Thus, the line voltages Vab, Vbc, Vca form a symmetrical set of phasors leading by 30 the set

representing the phase voltages and they are 3 times greater.

V V V Vab bc ca a 3 .

Relationship between line and phase quantities in delta connection:

(i) Voltage:

The voltage between two lines is VL. As shown in fig.

the phase coils are connected between lines, the

magnitude of phase voltage will be equal to

magnitude of line voltage VL.

ERY = EYB = EBR =VL = VPh

(ii) Current:

Balanced 3 phase delta connected load

Phasor diagram

Since the system is balanced the magnitude of the phase currents i.e. IRY, IYB, IBR are equal

but they are displayed 120o from one another as shown in figure. The currents are assumed

as lagging behind the respective phase voltages by an angle of Φ.

Similarly IR, IY and IB are currents flowing through line R, Y and B respectively. The values

of line current can be easily determined by applying KCL at terminal R, Y and B. according

to KCL; IR = IRY - IBR = IRY + (-IBR)

= √(IRY2

+ IBR2 + 2 IRY * IBR cos 60o)

But IRY = IYB = IBR = IPh

Hence IR = √3 IPh

Thus IR = IY= IB= √3 IPh

Expression of Power:

Apparent Power:

For three phase power S = 3 *VPh * IPh

Here VPh=VL and IPh= (IL/ √3)

Hence S = 3 * VL * (IL/ √3)

= √3 VL IL

Active power P:

For three phase power P = 3 *VPh * IPh* cos Φ

= √3 VL IL cos Φ

Power measurement:

Three wattmeter method:

This method is employed for measurement of power in 3 phase, 4 wire system. It can

also be employed in 3 phase 3 wire delta connected load, where power consumed by

each load is required to be determined separately.

Two wattmeter method:

The connection diagram for the measurement of power in a three-phase circuit using

two wattmeters is given in Fig. This is irrespective of the circuit connection – star or

delta. The circuit may be taken as unbalanced one, balanced type being only a special

case. Please note the connection of the two wattmeters. The current coils of the

wattmeters, 1 & 2, are in series with the two phases, R &B , with the pressure or voltage

coils being connected across Y R - and Y B - respectively. Y is the third phase, in which

no current coil is connected.

Measurement of power by Two wattmeter method:

One wattmeter method:

The circuit diagram for measuring power for a balanced three-phase load is shown in

Fig. The only assumption made is that, either the neutral point on the load or source

side is available. The wattmeter measures the power consumed for one phase only. The

total power is three times the above reading, as the circuit is balanced.

Measurement of power by single wattmeter method:

Comments on two wattmeter method:

a) When the balanced load is only resistive, i.e. power factor is 1( cos Φ = 0 ), the readings

of the two wattmeters are equal and positive.

b) When power factor is 0.5 lagging, one of the wattmeter W2 gives zero reading and whole

the power is measured by wattmeter W1.

c) When power factor is more than 0.5 but less than unity, both wattmeter reads positive, W1

reads more while W2 reads less than W1.

d) When power factor is less than 0.5 but more than zero, wattmeter W2 gives negative

whereas wattmeter W1 gives positive.

e) When power factor is zero, the readings of two wattmeters are equal in magnitude but in

opposite direction.

Per-Phase Analysis

In balanced three-phase circuits the currents and voltages in each phase are equal in magnitude and

displaced from each other by 120. This characteristic results in a simplified procedure to analyze balanced

three-phase circuits. In this procedure it is necessary only to compute results in one phase and subsequently

predict results in the other phases by using the relationship that exists among quantities in the other phases.

Example 1

Three balanced three-phase loads are connected in parallel. Load 1 is Y-connected with an impedance of

150 + j50 / ; load 2 is -connected with an impedance of 900 + j600 / ; and load 3 is 95.04 kVA at

0.6 pf leading. The loads are fed from a distribution line with an impedance of 3 + j24 / . The magnitude

of the line-to-neutral voltage at the load end of the line is 4.8 kV.

a) Calculate the total complex power at the sending end of the line.

b) What percent of the average power at the sending end of the line is delivered to the load?

Solution

The per-phase equivalent circuit is first constructed. For load 1, a Y-connected balanced load, the per-

phase impedance is 150 + j50 . For load 2, a -connected balanced load, the per-phase impedance is the

equivalent Y-connected load which is Z/3 = 300 + j200 . We will represent load 3 in terms of the

complex power it absorbs per-phase. This is given by

VAj

jS

344,25008,19

8.06.03

95040/3

The voltage across the per-phase equivalents of these loads has been specified as 4800 V which is the line-

to-neutral voltage at the load end.

The per-phase equivalent circuit is shown below in Figure bellow:

Fig.: Per-Phase Equivalent Circuit

Jj50 Jj200

Van

Iline-to-line

+ +Aa

Nn

3

150 300

Jj24

S3/4800 0 V

Iline

)(949.143725.45

)(7046.118369.43

28.596.33846.70769.116.98.28

4800

344,25008,19

200300

4800

50150

4800

rmsA

rmsAj

jjj

j

jj

I

In the above step, the total current I is obtained by summing the individual currents through the three

loads. For loads 1&2, we use the expression Z

VI , and for load 3 the current is determined using

VI

S.

In the distribution line,

VARXIQ

WRIP

effloss

effloss

34.224,148243725.4533

04.528,1833725.4533

22

22

In each load,

)(02.338,1062003846.70769.113

02.507,1593003846.70769.113

)(240,138506.98.283

720,4141506.98.283

2

2

2

2

2

1

2

1

absVARjQ

WjP

absVARjQ

WjP

)(546,168251,636

032,768.0040,95

024,576.0040,95

3

3

endloadVAjS

VARQ

WP

total

Check :

148.9710004.779,649

251,631%

34.770,31604.779,649

34.224,14804.528,18546,168251,631

546,168251,63174046.118369.4348003

P

VAj

VAjjS

VAjjS

sending

total

Example 2

A balanced 230 volt (rms) three phase source is furnishing 6 kVA at 0.83 pf lagging to two - connected

parallel loads. One load is a purely resistive load drawing 2 kW. Determine the phase impedance of the

second load.

Solution

The total complex power absorbed by the load is given by

VAjS 5577.083.0106 3

Note it is specified in the problem that

VAjS 58.33464980

5577.083.0cossinsin

83.0cos

1

Load 1 absorbs VAjS 020001

As a result, Load 2 must absorb VAjSS 58.334629801

The power absorbed by Load 2 per phase is VAjjS 5266.111533.99338.334629803

1/2

Z

VS LL

2

/2

Hence,

449.26552.235266.111533.993

2302

jj

Z

449.26552.23 jZ

************