THREE PHASE CIRCUITS: POWER DEFINITIONS AND VARIOUS COMPONENTS

8/20/2019 THREE PHASE CIRCUITS: POWER DEFINITIONS AND VARIOUS COMPONENTS

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Chapter 2

THREE PHASE CIRCUITS: POWER

DEFINITIONS AND VARIOUS

COMPONENTS

(Lectures 9-18)

2.1 Three-phase Sinusoidal Balanced System

Usage of three-phase voltage supply is very common for generation, transmission and distribution

of bulk electrical power. Almost all industrial loads are supplied by three-phase power supply for

its advantages over single phase systems such as cost and efficiency for same amount of power

usage. In principle, any number of phases can be used in polyphase electric system, however

three-phase system is simpler and giving all advantages of polyphase system. In previous section,

we have seen that instantaneous active power has a constant term V Icosφ as well pulsating termV I cos(2ωt − φ). The pulsating term does not contribute to any real power and thus increases theVA rating of the system.

In the following section, we shall study the various three-phase circuits such as balanced, un-

balanced, balanced and unbalanced harmonics and discuss their properties in details [1]–[5].

2.1.1 Balanced Three-phase Circuits

A balanced three-phase system is shown in Fig. 2.1 below.

Three-phase balanced system is expressed using following voltages and currents.

va(t) =√

2V sin(ωt)

vb(t) =√

2V sin(ωt − 120◦) (2.1)vc(t) =

√ 2V sin(ωt + 120◦)

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a

b

c

a

b

c

Fig. 2.1 A three-phase balanced circuit

and

ia(t) =√

2I sin(ωt − φ)ib(t) =

√ 2I sin(ωt − 120◦ − φ) (2.2)

ic(t) =√

2I sin(ωt + 120◦ − φ)

In (2.1) and (2.2) subscripts a, b and c are used to denote three phases which are balanced. Balancedthree-phase means that the voltage or current magnitude (V or I ) are same for all three phases andthey have a phase shift of −120o and 120o. The currents are assumed to have φ degree lag withtheir respective phase voltages. The balanced three phase system has certain interesting properties.

These will be discussed in the following section.

2.1.2 Three Phase Instantaneous Active Power

Three phase instantaneous active power in three phase system is given by,

p3φ(t) = p(t) = va(t)ia(t) + vb(t)ib(t) + vc(t)ic(t)= pa + pb + pc (2.3)

In above equation, pa(t), pb(t) and pc(t) are expressed similar to single phase system done previ-ously. These are given below.

pa(t) = V I cos φ {1− cos2ωt} − V I sin φ sin2ωt pb(t) = V I cos φ {1− cos2(ωt − 120o)} − V I sin φ sin2(ωt − 120o) (2.4) pc(t) = V I cos φ {1− cos2(ωt + 120o)} − V I sin φ sin2(ωt + 120o)

Adding three phase instantaneous powers given in (2.4), we get the three-phase instantaneous

power as below.

p(t) = 3 V I cos φ − V I cos φ{cos2ωt + cos 2(ωt − 120o) + cos 2(ωt + 120o)}− V I sin φ{sin2ωt + sin 2(ωt − 120o) + sin 2(ωt + 120o)} (2.5)

Summation of terms in curly brackets is always equal to zero. Hence,

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p3φ(t) = p(t) = 3V I cos φ. (2.6)

This is quite interesting result. It indicates for balanced three-phase system, the total instantaeous

power is equal to the real power or average active power (P ), which is constant. This is the reason

we use 3-phase system. It does not involve the pulsating or oscillating components of power as incase of single phase systems. Thus it ensures less VA rating for same amount of power transfer.

Here, total three-phase reactive power can be defined as sum of maximum value of preactive(t)terms in (2.4). Thus,

Q = Qa + Qb + Qc = 3V I sin φ. (2.7)

Is there any attempt to define instantaneous reactive power q (t) similar to p(t) such that Q isaverage value of that term q (t)?. H. Akagi et al. published paper [6], in which authors defined terminstantaneous reactive power. The definition was facilitated through αβ 0 transformation. Briefly

it is described in the next subsection.

2.1.3 Three Phase Instantatneous Reactive Power

H. Akagi et.al. [6] attempted to define instantaneous reactive power(q (t)) using αβ 0 transforma-tion. This transformation is described below.

The abc coordinates and their equivalent αβ 0 coordinates are shown in the Fig. 2.2 below.

av

bv

cv

j

60o

- c /2

- b /2

-j c

j b

O

Fig. 2.2 A abc to αβ 0 transformation

Resolving a, b, c quantities along the αβ axis we have,

vα =

2

3 (va − vb

2 − vc

2 ) (2.8)

vβ =

2

3

√ 3

2 (vb − vc) (2.9)

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Here,

23

is a scaling factor, which ensures power invariant transformation. Along with that, we

define zero sequence voltage as,

v0 = 2

3 1

2(va + vb + vc) (2.10)

Based on Eqns.(4.60)-(2.10) we can write the above equations as follows.

v0(t)vα(t)vβ(t)

= 23

1√ 2 1√ 2 1√ 21 −12 −120

√ 32

−√ 32

va(t)vb(t)vc(t)

(2.11)

v0vαvβ

= [Aoαβ]

vavbvc

The above is known as Clarke-Concordia transformation. Thus, va, vb and vc can also be expressedin terms of v0, vα and vβ by pre-multiplying (2.11) by matrix [A0αβ]

−1, we have

vavbvc

= [A0αβ]−1 v0vα

vβ

It will be interesting to learn that

[A0αβ]−1 = [Aabc] =

23

1√ 2 1√ 2 1√ 21 −12

−12

0√ 32

−√ 32

−1

[A0αβ]−1 =

23

1√ 2

1 01√ 2

−12

√ 32

1√ 2

−12

−√ 32

= [A0αβ]T = [Aabc] (2.12)

Similarly, we can write down instantaneous symmetrical transformation for currents, which is

given below.

i0iαiβ

= 23

1√ 2 1√ 2 1√ 21 −12 −120

√ 32

−√ 32

iaibic

(2.13)Now based on ’0αβ ’ transformation, the instantaneous active and reactive powers are defined asfollows. The three-phase instantaneous power p(t) is expressed as the dot product of 0αβ compo-nents of voltage and currents such as given below.

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p(t) = vα iα + vβ iβ + v0 i0

= 2

3 va − vb

2 −

vc

2 ia − ib

2 −

ic

2 +√

3

2

(vb

−vc)

√ 3

2

(ib

−ic)

+ 1√

3(va + vb + vc)

1√ 3

(ia + ib + ic)

= va ia + vb ib + vc ic (2.14)

Now what about instantaneous reactive power? Is there any concept defining instantaneous reactive

power? In 1983-84,authors H.akagi have attempted to define instantaneous reactive power using

stationary αβ 0 frame, as illustrated below. In [6], the instantaneous reactive power q (t) is definesas the cross product of two mutual perpendicular quantities, such as given below.

q (t) = vα × iβ + vβ × iαq (t) = vαiβ − vβiα

= 2

3

va − vb

2 − vc

2

√ 32

(ib − ic) −√

3

2 (vb − vc)

ia − ib

2 − ic

2

= 2

3

√ 3

2

(−vb + vc) ia +

va − vb

2 − vc

2 +

vb2 − vc

2

ib +

−va + vb

2 +

vc2

+ vb

2 − vc

2

ic

= − 1√

3[(vb − vc) ia + (vc − va) ib + (va − vb) ic]

= − [vbcia + vcaib + vabic] /√ 3 (2.15)This is also equal to the following.

q (t) = 1√

3

(ib − ic) va +

−ib

2 +

ic2 − ia + ib

2 +

ic2

vb +

−ib

2 +

ic2

+ ia − ib2 − ic

2

vc

=

1√ 3

[(ib − ic) va + (ic − ia) vb + (ia − ib) vc] (2.16)

2.1.4 Power Invariance in abc and αβ 0 Coordinates

As a check for power invariance, we shall compute the energy content of voltage signals in two

transformations. The energy associated with the abc0 system is given by (v2a + v2b + v

2c ) and the

energy associated with the αβ 0 components is given by

v20 + v2α + v

2β

. The two energies must

be equal to ensure power invariance in two transformations. It is proved below. Using, (2.11) and

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squares of the respective components, we have the following.

v2α =

2

3

va − vb

2 − vc

2

2v2α =

23v2a + v2b4 + v2c4 − 2vavb2 + 2vbvc4 − 2vavc2

= 2

3v2a +

v2b6

+ v2c

6 − 2vavb

3 +

vbvc3 − 2vavc

3 (2.17)

Similary we can find out square of vβ term as given below.

v2β =

√ 3

2

2

3 (vb − vc)

2=

12

v2b + v2c − 2vbvc

= v2b

2 +

v2c2 − vbvc (2.18)

Adding (2.17) and (2.18), we find that,

v2α + v2β =

2

3 v2a + v2b + v2c − vcvb − vbvc − vcva=

v2a + v

2b + v

2c

− v2a3

+ v2b

3 +

v2c3

+ 2vavb

3 +

2vbvc3

+ 2vavc

3

=

v2a + v

2b + v

2c

− 13

(va + vb + vc)2

=

v2a + v2b + v

2c

− 1√ 3

(va + vb + vc)

2(2.19)

Since v0 = 1√ 3

(va + vb + vc), the above equation, (2.19) can be written as,

v2α + v2β + v

20 = v

2a + v

2b + v

2c . (2.20)

From the above it is implies that the energy associated with the two systems remain same instant to

instant basis. In general the instantaneous power p(t) remain same in both transformations. Thisis proved below.

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Using (2.14), following can be written.

p(t) = vαiα + vβiβ + voio

p(t) =

v0vα

vβ

T

i0iα

iβ

=

[Aabc] vavb

vc

T [Aabc] iaib

ic

=

vavbvc

T [Aabc]T [Aabc] iaib

ic

=

vavbvc

T

[Aabc]−1 [Aabc]

iaibic

=

va vb vc iaib

ic

= vaia + vbib + vcic (2.21)

In the above, the following property of matrices of from (2.12), is used.

[Aabc]T [Aabc] = [Aabc]

−1 [Aabc] = I (2.22)

In above, I is identity matrix.

2.2 Instantaneous Active and Reactive Powers for Three-phase Circuits

In the previous section instantaneous active and reactive powers were defined using αβ 0 trans-formation. In this section we shall study these powers for various three-phase circuits such as

three-phase balanced, three-phase unbalanced, balanced three-phase with harmonics and unbal-

anced three-phase with harmonics. Each case will be considered and analyzed.

2.2.1 Three-Phase Balance System

For three-phase balanced system, three-phase voltages have been expressed by equation (2.1). For

these phase voltages, the line to line voltages are given as below.

vab =√

3√

2V sin(ωt + 30◦)

vbc =√

3√

2V sin(ωt − 90◦)vca =

√ 3√

2V sin(ωt + 150◦) (2.23)

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0

o

aV V

bV

bV 3

3 0 o

a b V

V

cV

Fig. 2.3 Relationship between line-to-line and phase voltage

The above relationship between phase and line to line voltages is also illustrated in Fig. 2.3. For

the above three-phase system, the instantaneous power p(t) can be expressed using (2.21) and it isequal to,

p(t) = vaia + vbib + vcic

= vαiα + vβiβ + v0i0

= 3 V I cosφ (2.24)

The instantaneous reactive power q (t) is as following.

q (t) = − 1√ 3

[√

3√

2V sin(ωt − 90o)√

2I sin (ωt − φ)

+√

3√

2V sin(ωt + 150o)√

2I sin (ωt−

120o

−φ)

+√ 3√ 2V sin(ωt + 30o)√ 2I sin (ωt + 120◦ − φ)]= −V I [cos (90◦ − φ) − cos(2ωt − 90o − φ)

+cos(90o − φ) − cos (2ωt − 30o − φ)+cos(90o − φ) − cos (2ωt + 150o − φ)]

= −V I [3 sin φ− cos (2ωt − φ + 30o) − cos (2ωt − φ + 30o + 120o)− cos (2ωt − φ + 30o − 120o)]

= −V I [3 sin φ − 0]q (t) = −3V I sin φ (2.25)

The above value of instantaneous reactive power is same as defined by Budeanu’s [1] and is given

in equation (2.7). Thus, instantaneous reactive power given in (2.15) matches with the conven-

tional definition of reactive power defined in (2.7). However the time varying part of second terms

of each phase in (2.4) has no relevance with the definition given in (2.15).

Another interpretation of line to line voltages in (2.15) is that the voltages vab, vbc and vca have

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90o phase shift with respect to voltages vc, va and vb respectively. These are expressed as below.

vab =√

3vc∠− 90ovbc =

√ 3va∠− 90o (2.26)

vca =√

3vb∠− 90o

In above equation, vc∠ − 90o implies that vc∠ − 90o lags vc by 90o. Analyzing each term in(2.15) contributes to,

vbcia =√

3va∠− 90◦. ia=√

3√

2V sin(ωt − 90◦) .√

2I sin (ωt − φ)=√

3V I 2 sin (ωt − 90◦) . sin(ωt − φ)=√

3V I [cos (90◦ − φ) − cos (2ωt − 90◦ − φ)]=√

3V I [sin φ − cos {90◦ + (2ωt − φ)}]=√

3V I [sin φ + sin (2ωt − φ)]=√

3V I [sin φ + sin 2ωt cos φ

−cos2ωt sin φ]

vbcia/√ 3 = V I [sin φ (1 − cos2ωt) + cos φ sin2ωt]Similarly,

vcaib/√

3 = V I

sin φ

1− cos

2

ωt − 2π

3

+V I cos φ. sin2

ωt − 2π

3

vabic/

√ 3 = V I

sin φ

1− cos

2

ωt +

2π

3

+V I cos φ. sin2ωt +

2π

3 (2.27)Thus, we see that the role of the coefficients of sin φ and cos φ have reversed. Now if we takeaverage value of (2.27), it is not equal to zero but V I sin φ in each phase. Thus three-phase reactivepower will be 3V I sin φ. The maximum value of second term in (2.27) represents active averagepower i.e., V I cos φ. However, this is not normally convention about the notation of the powers.But, important contribution of this definition is that average reactive power could be defined as the

average value of terms in (2.27).

2.2.2 Three-Phase Unbalance System

Three-phase unbalance system is not uncommon in power system. Three-phase unbalance may

result from single-phasing, faults, different loads in three phases. To study three-phase systemwith fundamental unbalance, the voltages and currents are expressed as following.

va =√

2V a sin (ωt − φva)vb =

√ 2V b sin (ωt − 120o − φvb) (2.28)

vc =√

2V c sin (ωt + 120o − φvc)

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and,

ia =√

2I a sin (ωt − φia)ib =

√ 2I b sin (ωt − 120o − φib) (2.29)

ic =√

2I c sin (ωt + 120o − φic)

For the above system, the three-phase instantaneous power is given by,

p3φ(t) = p(t) = vaia + vbib + vcic

=√

2V a sin (ωt − φva)sin(ωt − φia)+√

2V b sin (ωt − 120o − φvb)√

2I b sin (ωt − 120o − φib) (2.30)+√

2V c sin (ωt + 120o − φvc)

√ 2I c sin (ωt + 120

o − φic)Simplifying above expression we get,

p3φ(t) = V aI a cos φa {1 − cos(2ωt − 2φva)}

pa,active−V aI a sin φa sin (2ωt − 2φva) pa,reactive

+V bI b cos φb [1 − cos {2 (ωt − 120◦) − 2φvb}]−V bI b sin φb sin {2 (ωt − 120◦) − 2φvb}+V cI c cos φc [1 − cos {2 (ωt + 120◦) − 2φvc}]−V cI c sin φc sin {2 (ωt + 120◦) − 2φvc} (2.31)

where φa = (φia − φva)Therefore,

p3φ (t) = pa,active + pb,active + pc,active + pa,reactive + pb,reactive + pc,reactive

= pa + pb + pc + pa + pb + pc (2.32)where,

pa = P a = V aI a cos φa

pb = P b = V bI b cos φb (2.33)

pc = P c = V cI c cos φc

and

pa = −V aI a cos (2 ωt − φa − 2 φva) pb = −V bI b cos (2 ωt − 240o − φb − 2 φvb) (2.34) pc = −V cI c cos (2ωt + 240− φc − 2 φvc)Also it is noted that,

pa + pb + pc = vaia + vbib + vcic = P (2.35)

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and, pa + pb + pc = −V aI acos(2ωt − φva − φib)−V bI bcos {2(ωt − 120) − φvb − φib}−V cI ccos {2(ωt + 120) − φvc − φic}

= 0

This implies that, we no longer get advantage of getting constant power, 3V I cos φ from interactionof three-phase voltages and currents. Now, let us analyze three phase instantaneous reactive power

q (t) as per definition given in (2.15).

q (t) = − 1√ 3

(vb − vc)ia + (vc − va)ib + (va − vb)ic

= − 2√ 3

{V bsin(ωt − 120o − φvb) − V csin(ωt + 120o − φvc)} I asin(ωt − φia)

+ {V csin(ωt + 120o − φvc) − V asin(ωt − φva)}√

2I bsin(ωt − 120o − φib) (2.36)+{V asin(ωt − 120o − φva) − V b sin(ωt − 120o − φvb)}√ 2I c sin(ωt + 120o − φic)

From the above,

√ 3 q (t) = −

V bI a {cos(φia − 120o − φvb) − cos(2ωt − 120o − φia − φvb)}

−V cI a {cos(φia + 120o − φvc) − cos(2ωt + 120o − φia − φvc)}+V cI b {cos(φib + 240o − φvc) − cos(2ωt − φib − φvc)} (2.37)−V aI b {cos(φib − 120o − φva) − cos(2ωt − 120o − φva − φib)}+V aI c {cos(φic − 120o − φva) − cos(2ωt + 120o − φva − φic)}

−V bI c {cos(φic − 240o

− φvb) − cos(2ωt − φic − φvb)}Now looking this expression,we can say that

1

T

T 0

q (t)dt = − 1√ 3

V bI a cos(φia − φvb − 120o)

−V cI a cos(φia − φvc + 120o)+V cI b cos(φib + 240

o − φvc)−V aI b cos(φib − 120o − φva)+V aI c cos(φic − 120o − φva)−V bI c cos(φic − 240o − φvb)

= q a(t) + q b(t) + q c(t)

= V aI a sin φa + V bI b sin φb + V cI c sin φc (2.38)Hence the definition of instantaneous reactive power does not match to that defined by Budeanue’s

reactive power [1] for three-phase unbalanced circuit. If only voltages or currents are distorted, the

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above holds true as given below. Let us consider that only currents are unbalanced, then

va(t) =√

2V sin(ωt)

vb(t) =√

2V sin(ωt − 120◦) (2.39)vc(t) = √ 2V sin(ωt + 120◦)

and

ia(t) =√

2I a sin(ωt − φa)ib(t) =

√ 2I b sin(ωt − 120o − φb) (2.40)

ic(t) =√

2I c sin(ωt + 120o − φc)

And the instantaneous reactive power is given by,

q (t) = − 1√ 3

[vbcia + vcaib + vabic]

=

− 1√ 3

[√

3 va∠

−π/2 ia +

√ 3 vb∠

−π/2 ib +

√ 3 vc∠

−π/2 ic]

= −[√ 2V sin(ωt − π/2)√ 2I a sin(ωt − φia)+√

2V sin(ωt − 120o − π/2)√ 2I b sin(ωt − 120o − φib)+√

2V sin(ωt + 120o + π/2)√

2I c sin(ωt + 120o − φic)]

= −[V I acos(π/2 − φia) − cos {π/2 − (2ωt − φia)}+V I bcos(π/2 − φib) − cos(2ωt − 240o − π/2− φib)+V I ccos(π/2 − φic) − cos(2ωt + 240o − π/2− φic)]

= −[(V I a sin φia + V I b sin φib + V I c sin φic)+V I a sin(2ωt − φia) + V I b sin(2ωt − 240o − φib) + V I c sin(2ωt + 240o − φic)]

Thus,

Q = 1

T

T 0

q (t)dt = −(V I a sin φia + V I b sin φib + V I c sin φic) (2.41)

Which is similar to Budeanu’s reactive power.

The oscillating term of q (t) which is equal to q (t) is given below.q (t) = V I a sin(2ωt − φia) + V I b sin(2ωt − 240o − φib) + V I c sin(2ωt + 240o − φic) (2.42)

which is not similar to what is being defined as reactive component of power in (2.4).

2.3 Symmetrical components

In the previous section, the fundamental unbalance in three phase voltage and currents have been

considered. Ideal power systems are not designed for unbalance quantities as it makes power sys-

tem components over rated and inefficient. Thus, to understand unbalance three-phase systems,

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a concept of symmetrical components introduced by C. L. Fortescue, will be discussed. In 1918,

C. L Fortescue, wrote a paper [7] presenting that an unbalanced system of n-related phasors can

be resolved into n system of balanced phasors, called the symmetrical components of the originalphasors. The n phasors of each set of components are equal in length and the angles. Although,the method is applicable to any unbalanced polyphase system, we shall discuss about three phase

systems.

For the discussion of symmetrical components, a complex operator denoted as a is defined as,

a = 1∠120o = e j2π/3 = cos 2π/3 + j sin 2π/3

= −1/2 + j√

3/2

a2 = 1∠240o = 1∠− 120o = e j4π/3 = e− j2π/3 = cos 4π/3 + j sin 4π/3= −1/2 − j

√ 3/2

a3 = 1∠360o = e j2π = 1

Also note an interesting property relating a, a2 and a3,

a + a2 + a3 = 0. (2.43)

3

1 o

a o

21 120

o

a

1 120o

a

o

Fig. 2.4 Phasor representation of a, a2 and a3

These quantities i.e., a, a2 and a3 = 1 also represent three phasors which are shifted by 120o

from each other. This is shown in Fig. 2.4.

Knowing the above and using Fortescue theorem, three unbalanced phasor of a three phase un-

balanced system can be resolved into three balanced system phasors.

1. Positive sequence components are composed of three phasors, equal in magnitude, phase shift

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of −120o and 120o between phases with phase sequence same to that of the original phasors.

2. Negative sequence components consist of three phasors equal in magnitude, phase shift of

120o and −120o between phases with phase sequence opposite to that of the original phasors.

3. Zero sequence components consist of three phasors equal in magnitude with zero phase shiftfrom each other.

These are denoted as following.

Positive sequence components: V a+, V b+, V c+

Negative sequence components: V a−, V b−, V c−

Zero sequence components: V a0, V b0, V c0

Thus, we can write,

V a = V a+ + V a− + V a0V b = V b+ + V b− + V b0 (2.44)

V c = V c+ + V c− + V c0

Graphically, these are represented in Fig. 2.5. Thus if we add the sequence components of each

phase vectorially, we shall get V a, V b and V s as per (2.44). This is illustrated in Fig. 2.6.

Va

Vc

Vb

Vb

Va

Vc

0Va0Vb

0Vc

(a) (b) (c)

Fig. 2.5 Sequence components (a) positive sequence (b) negative sequence (c) zero sequence

Now knowing all these preliminaries, we can proceed as following. Let V a+ be a reference phasor,

therefore V b+ and V c+ can be written as,

V b+ = a2V a+ = V a+∠− 120◦

V c+ = aV a+ = V a+∠120◦ (2.45)

Similarly V b− and V c− can be expressed in terms of V a− as following.

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Va

VaVa

0Va

Vc

Vc

0Vc

Vc

Vb

Vb

Vb

0Vb

o

Fig. 2.6 Unbalanced phasors as vector sum of positive, negative and zero sequence phasors

V b− = aV a− = V a−∠120◦

V c− = a2V a− = V a−∠− 120◦ (2.46)

The zero sequence components have same magnitude and phase angle and therefore these are

expressed as,

V b0 = V c0 = V a0 (2.47)

Using (2.45), (2.46) and (2.47) we have,

V a = V a0 + V a+ + V a− (2.48)

V b = V b0 + V b+ + V b−= V a0 + a

2 V a+ + a V a− (2.49)

V c = V c0 + V c+ + V c−

= V a0 + a V a+ + a2 V a− (2.50)

Equations (2.48)-(2.50) can be written in matrix form as given below.V aV bV c

=1 1 11 a2 a

1 a a2

V a0V a+V a−

(2.51)41


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Premultipling by inverse of matrix [Asabc] =

1 1 11 a2 a1 a a2

, the symmetrical components areexpressed as given below.

V a0V a+V a−

= 131 1 11 a a2

1 a2 a

V aV bV c

(2.52)= [A012]

V aV bV c

The symmetrical transformation matrices A012 and Asabc are related by the following expression.

[A012] = [Asabc]−1 = [Asabc]∗ (2.53)

From (2.52), the symmetrical components can therefore be expressed as the following.

V a0 = 1

3(V a + V b + V c)

V a+ = 1

3(V a + aV b + a

2V c) (2.54)

V a− = 1

3(V a + a

2V b + aV c)

The other component i.e., V b0, V c0, V b+, V c+, V b−, V c− can be found from V a0, V a+, V a+. Itshould be noted that quantity V a0 does not exist if sum of unbalanced phasors is zero. Since sumof line to line voltage phasors i.e., V ab+V bc +V ca = (V a

−V b)+(V b

−V c)+(V c

−V a) is always

zero, hence zero sequence voltage components are never present in the line voltage, regardless of

amount of unbalance. The sum of the three phase voltages, i.e., V a + V b + V c is not necessarilyzero and hence zero sequence voltage exists.

Similarly sequence components can be written for currents. Denoting three phase currents by

I a, I b, and I c respectively, the sequence components in matrix form are given below.I a0I a+I a−

= 13

1 1 11 a a21 a2 a

I aI bI c

(2.55)Thus,

I a0 = 1

3(I a + I b + I c)

I a+ = 1

3(I a + aI b + a

2I c)

42


17/41

I a− = 1

3(I a + a

2I b + aI c)

In three-phase, 4-wire system, the sum of line currents is equal to the neutral current (I n). thus,

I n = I a + I b + I c

= 3I a0 (2.56)

This current flows in the fourth wire called neutral wire. Again if neutral wire is absent, then zero

sequence current is always equal to zero irrespective of unbalance in phase currents. This is illus-

trated below.

a

b

c

a

b

c

(a)

a

b

c

a

b

c

(b)

Fig. 2.7 Various three phase systems (a) Three-phase three-wire system (b) Three-phase four-wire system

In 2.7(b), in may or may not be zero. However neutral voltage (V Nn ) between the system and

load neutral is always equal to zero. In 2.7(a), there is no neutral current due to the absence of theneutral wire. But in this configuration the neutral voltage, V Nn, may or may not be equal to zerodepending upon the unbalance in the system.

Example 2.1 Consider a balanced 3 φ system with following phase voltages.

V a = 100∠0o

V b = 100∠− 120oV c = 100∠120

o

Using (2.54), it can be easily seen that the zero and negative sequence components are equal to

zero, indicating that there is no unbalance in voltages. However the converse may not apply.Now consider the following phase voltages. Compute the sequence components and show that the

energy associated with the voltage components in both system remain constant.

V a = 100∠0o

V b = 150∠− 100oV c = 75∠100

o

43


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Solution Using (2.54), sequence components are computed. These are:

V a0 = 1

3(V a + V b + V c)

= 31.91∠−

50.48o V

V a+ = 13

(V a + aV b + a2V c)

= 104.16∠4.7o V

V a− = 1

3(V a + a

2V b + aV c)

= 28.96∠146.33o V

If you find energy content of two frames that is abc and 012 system, it is found to be constant.

E abc = k [V 2

a + V 2

b + V 2

c ] = 381.25 kE 0+− = 3 k [V

2a0 + V

2a+ + V

2a−] = 381.25 k

Thus, E abc = E 0+− with k as some constant of proportionality.

The invariance of power can be further shown by following proof.

S v = P + jQ = [ V a V b V c ]

I aI bI c

∗

=

V aV bV c

T I aI bI c

∗

=

[Asabc] V a0V a+

V a−

T [Asabc] I a0I a+

I a−

∗

=

V a0V a+V a−

T

[Asabc]T [Asabc]

∗

I a0I a+I a−

∗

(2.57)

The term S v is referred as vector or geometric apparent power. The difference between will begiven in the following. The transformation matrix [Asabc] has following properties.

[Asabc]T [Asabc]

∗ = 3 [I ] (2.58)

44


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The matrix, [I ], is identity matrix. Using (2.58), (2.57) can be written as the following.

S v = P + jQ =

V a0V a+V a−

T 3[I ]I a0I a+

I a−

∗

= 3V a0V a+

V a−

T I a0I a+I a−

∗S v = P + jQ = V aI

∗a + V bI

∗b + V cI

∗c

= 3 [V a0I ∗a0 + V a+I

∗a+ + V a−I

∗a−] (2.59)

Equation (2.59) indicates that power invariance holds true in both abc and 012 components. But,this is true on phasor basis. Would it be true on the time basis? In this context, concept of instanta-

neous symmetrical components will be discussed in the latter section. The equation (2.59) further

implies that,

S v = P + jQ = 3 [ (V a0I a0 cos φa0 + V a+I a+ cos φa+ + V a−I a− cos φa−)

+ j(V a0I a0 sin φa0 + V a+I a+ sin φa+ + V a−I a− sin φa−) ] (2.60)

The power terms in (2.60) accordingly form positive sequence, negative sequence and zero se-

quence powers denoted as following. The positive sequence power is given as,

P + = V a+I a+ cos φa+ + V b+I b+ cos φb+ + V c+I c+ cos φc+

= 3V a+I a+ cos φa+. (2.61)

Negative sequence power is expressed as,

P − = 3V a−

I a−

cos φa−

. (2.62)

The zero sequence power is

P 0 = 3V a0I a0 cos φa0. (2.63)

Similarly, sequence reactive power are denoted by the following expressions.

Q+ = 3V a+I a+ sin φa+

Q− = 3V a−I a− sin φa−Q0 = 3V a0I a0 sin φa0 (2.64)

Thus, following holds true for active and reactive powers.

P = P a + P b + P c = P 0 + P 1 + P 2

Q = Qa + Qb + Qc = Q0 + Q1 + Q2 (2.65)

45


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Here, positive sequence, negative sequence and zero sequence apparent powers are denoted as the

following.

S + = |S +| =

P +2 + Q+2 = 3V a+I a+

S − = |S +| =

P −2 + Q−2 = 3V a−I a−

S 0 = |S +

| = P 02 + Q02 = 3V a0I a0 (2.66)The scalar value of vector apparent power (S v) is given as following.

S v = |S a + S b + S c| = |S 0 + S + + S −|= |(P a + P b + P c) + j(Qa + Qb + Qc)| (2.67)=

P 2 + Q2

Similarly, arithematic apparent power (S A) is defined as the algebraic sum of each phase or se-quence apparent power, i.e.,

S A = |S a| + |S b| + |S c|= |P a + jQa| + |P b + jQb| + |P c + jQc| (2.68)=

P 2a + Q2a +

P 2b + Q

2b +

P 2c + Q

2c

In terms of sequence components apparent power,

S A = |S 0| + |S +| + |S −|= |P 0 + jQ0| + |P + + jQ+| + |P − + jQ−| (2.69)= P

02 + Q02 + P +2 + Q+2 + P

−2 + Q−2

Based on these two definitions of the apparent powers, the power factors are defined as the follow-

ing.

Vector apparent power = pf v = P

S v(2.70)

Arithematic apparent power = pf A = P

S A(2.71)

46


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Example 2.2 Consider a 3-phase 4 wire system supplying resistive load, shown in Fig. 2.8

below. Determine power consumed by the load and feeder losses.

'a

'c

'n

a

b

c

n

Va

Vb

Vc

RIa

In

Ib

Ic

r j x

r j x

r j x

r j x

'

Fig. 2.8 A three-phase unbalanced load

Power dissipated by the load = (√

3V )2

R =

3V 2

R

The current flowing in the line =

√ 3V

R = |V a − V b

R |

and I b = −I a

Therefore losses in the feeder =

√ 3V

R

2× r +

√ 3V

R

2× r

= 2 rR3 V 2R Now, consider another example of a 3 phase system supplying 3-phase load, consisting of three

resistors (R) in star as shown in the Fig. 2.9. Let us find out above parameters.

Power supplied to load = 3

V

R

2× R = 3V

2

R

Losses in the feeder = 3 V

R2

×r =

r

R3 V 2

R Thus, it is interesting to see that power dissipated in the unbalanced system is twice the power loss

in balanced circuit. This leads to conclusion that power factor in phases would become less than

unity, while for balanced circuit, the power factor is unity. Power analysis of unbalanced circuit

shown in Fig. 2.8 is given below.

47


22/41

'a

'c

'n

a

b

c

nVa

R

Ia

r j x

r j x

r j x

r j x

R

R

'b

Vb

Vc

Ib

Ic

In

Fig. 2.9 A three-phase balanced load

The current in phase-a, I a = V a − V b

R =

V abR

=

√ 3 V aR

∠30◦

The current in phase-b, I b = −I a =√

3 V

R ∠(30− 180)o

=

√ 3 V

R ∠− 150o

The current in phase-c and neutral are zero, I c = I n = 0

The phase voltages are: V a = V ∠0o, V b = V ∠− 120o, V c = V ∠120o.

The phase active and reactive and apparent powers are as following.

P a = V aI a cos φa = V I cos 30◦ =

√ 3

2

V I

Qa = V aI a sin φa = V I sin 30◦ =

1

2 V I

S a = V aI a = V I

P b = V bI b cos φb = V I cos(−30)◦ =√

3

2 V I

Qb = V bI b sin φb = V I sin(−30)◦ = −12

V I

S b = V bI b = V I

P c = Qc = S c = 0

Thus total active power P = P a + P b + P c = 2 ×√

3

2 V I = √ 3 V I =

√ 3 V

√ 3 V

R

P = 3 V 2

RTotal reactive power Q = Qa + Qb + Qc = 0

48


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The vector apparent power, S v =

P 2 + Q2 = 3 V 2/R = P

The arithmetic apparent power, S A = S a + S b + S c = 2 V I = (2/√

3) P

From the values of S v and S A, it implies that,

pf v =

P

S v =

P

P = 1

pf A = P

S A=

P

(2/√

3) P =

√ 3

2 = 0.866

This difference between the arthmetic and vector power factors will be more due to the unbalances

in the load.

For balance load S A = S V , therefore, pf A = pf V = 1.0. Thus for three-phase electrical cir-cuits, the following holds true.

pf A

≤ pf V (2.72)

2.3.1 Effective Apparent Power

For unbalanced three-phase circuits, their is one more definition of apparent power, which is known

as effective apparent power. The concept assumes that a virtual balanced circuit that has the same

power output and losses as the actual unbalanced circuit. This equivalence leads to the definition

of effective line current I e and effective line to neutral voltage V e.

The equivalent three-phase unbalanced and balanced circuits with same power output and losses

are shown in Fig. 2.10. From these figures, to maintain same losses,

'a

'c

'n

n

Va a RIa

r j x

r j x

r j x

r j x

'bVb

Vc

Ib

Ic

In

Vn

b R

c R

'a

'c

'n

Vea

Iea

r j x

r j x

r j x

r j x

'b

e R

e R

e RIeb

Iec

I 0n

n

Veb

Vec

(a) (b)

Fig. 2.10 (a) Three-phase with unbalanced voltage and currents (b) Effective equivalent three-phase system

rI 2a + rI 2b + rI

2c + rI

2n = 3rI

2e

The above equation implies the effective rms current in each phase is given as following.

I e =

(I 2a + I

2b + I

2c + I

2n)

3 (2.73)

49


24/41

For the original circuit shown in Fig. 2.8, the effective current I e is computed using above equationand is given below.

I e =

(I 2a + I

2b )

3 since, I c = 0and I n = 0

= 2 I 2a

3 =

2 (√ 3V /R)23

=

√ 2V

R

To account same power output in circuits shown above, the following identity is used with Re = Rin Fig. 2.10.

V 2aR

+ V 2b

R +

V 2cR

+ V 2ab + V

2bc + V

2ca

3R =

3V 2eR

+ 9V 2e

3R (2.74)

From (2.74), the effective rms value of voltage is expressed as,

V e =

1

18{3 (V 2a + V 2b + V 2c ) + V 2ab + V 2bc + V 2ca} (2.75)

Assuming, 3 (V 2a + V 2b + V

2c ) ≈ V 2ab + V 2bc + V 2ca, equation (2.75) can be written as,

V e =

V 2a + V

2b + V

2c

3 = V (2.76)

Therefore, the effective apparent power (S e), using the values of V e and I e, is given by,

S e = 3 V e I e = 3√

2 V 2

RThus the effective power factor based on the definition of effective apparent power (S e), for thecircuit shown in Fig. 2.8 is given by,

pf e = P

S e=

3 V 2/R

3√

2 V 2/R=

1√ 2

= 0.707

Thus, we observe that,

S V ≤ S A ≤ S e, pf e (0.707) ≤ pf A (0.866) ≤ pf V (1.0).

When the system is balanced,

V a = V b = V c = V en = V e,

I a = I b = I c = I e,

I n = 0,

and S V = S A = S e.

50


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2.3.2 Positive Sequence Powers and Unbalance Power

The unbalance power S u can be expressed in terms of fundamental positive sequence powers P +,

Q+ and S + as given below.

S u = S 2e − S +2 (2.77)where S + = 3 V +I + and S +

2= P +

2+ Q+

2.

2.4 Three-phase Non-sinusoidal Balanced System

A three-phase nonsinusoidal system is represented by following set of equaitons.

va(t) =√

2V 1 sin(wt − α1) +√

2∞

n=2V n sin(nwt − αn)

vb(t) =√

2V 1 sin(wt − 120◦ − α1) +√

2∞n=2

V n sin(n(wt − 120◦) − αn) (2.78)

vc(t) =√

2V 1 sin(wt + 120◦ − α1) +

√ 2

∞n=2

V n sin(n(wt + 120◦) − αn)

Similarly, the line currents can be expressed as,

ia(t) =√

2I 1 sin(wt − β 1) +√

2∞

n=2I n sin(nwt − β n)

ib(t) =√

2I 1 sin(wt − 120◦ − β 1) + √ 2∞n=2

I n sin(n(wt − 120◦ ) − β n) (2.79)

ic(t) =√

2I 1 sin(wt + 120◦ − β 1) +

√ 2

∞n=2

I n sin(n(wt + 120◦ ) − β n)

In this case,

S a = S b = S c,

P a = P b = P c, (2.80)

Qa = Qb = Qc,Da = Db = Dc.

In above the terms Da, Db and Dc are known as distortion powers in phase-a,b,c respectively. Thedefinition of The distortion power, D, is given in Section 1.4.5. The above equation suggests thatsuch a system has potential to produce significant additional power loss in neutral wire and ground

path.

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2.4.1 Neutral Current

The neutral current for three-phase balanced system with harmonics can be given by the following

equation.

in = ia + ib + ic

= √ 2 [ I a1 sin (wt − β 1) + I a2 sin (2wt − β 2) + I a3 sin (3wt − β 3)+I a1 sin (wt − 120o − β 1) + I a2 sin (2wt − 240o − β 2) + I a3 sin (3wt − 360o − β 3)+I a1 sin (wt + 120

o − β 1) + I a2 sin (2wt + 240o − β 2) + I a3 sin (3wt + 360o − β 3)

+I a4 sin (4wt − β 4) + I a5 sin (5wt − β 5) + I a6 sin (6wt − β 6)+I a4 sin (wt − 4 × 120o − β 4) + I a5 sin (5wt − 5 × 120o − β 5) + I a6 sin (6wt − 6 × 120o − β+I a4 sin (wt + 4 × 120o − β 4) + I a5 sin (5wt + 5 × 120o − β 5) + I a6 sin (6wt + 6 × 120o − β

(2

+I a7 sin (7wt − β 7) + I a8 sin (8wt − β 8) + I a9 sin (9wt − β 9)+I a7 sin (7wt − 7 × 120

o

− β 7) + I a8 sin (8wt − 8 × 120o

− β 8) + I a9 sin (9wt − 9 × 120o

−+I a7 sin (7wt + 7 × 120o − β 7) + I a8 sin (8wt + 8 × 120o − β 8) + I a9 sin (9wt + 9 × 120o − β

From the above equation, we observe that, the triplen harmonics are added up in the neutral current.

All other harmonics except triplen harmonics do not contribute to the neutral current, due to their

balanced nature. Therefore the neutral current is given by,

in = ia + ib + ic =∞

n=3,6,..

3√

2I n sin(nwt − β n). (2.82)

The RMS value of the current in neutral wire is therefore given by,

I n = 3 ∞n=3,6,..

I 2n1/2 . (2.83)

Due to dominant triplen harmonics in electrical loads such as UPS, rectifiers and other power

electronic based loads, the current rating of the neutral wire may be comparable to the phase wires.

It is worth to mention here that all harmonics in three-phase balanced systems can be catego-

rized in three groups i.e., (3n + 1), (3n + 2) and 3n (for n = 1, 2, 3,...) called positive, nega-tive and zero sequence harmonics respectively. This means that balanced fundamental, 4th, 7th

10th,... form positive sequence only. Balanced 2nd, 5th, 8th, 11th,... form negative sequence only

and the balanced triplen harmonics i.e. 3rd, 6th, 9th,... form zero sequence only. But in case of

unbalanced three-phase systems with harmonics, (3n + 1) harmonics may start forming negative

and zero sequence components. Similarly, (3n + 2) may start forming positive and zero sequencecomponents and 3n may start forming positive and negative sequence components.

2.4.2 Line to Line Voltage

For the three-phase balanced system with harmonics, the line-to-line voltages are denoted as vab,vbc and vca. Let us consider, line-to-line voltage between phases a and b. It is given as following.

52


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vab(t) = va(t) − vb(t)

=∞

n=1√

2V n sin(n ωt − αn) −∞

n=1√

2V n sin(n (ωt − 120o) − αn)

=∞n=1

√ 2V n sin(n ωt − αn) −

∞n=1

√ 2V n sin((n ωt − αn) − n × 120o)

=∞n=1

√ 2V n [sin(n ωt − αn) − sin(n ωt − αn) cos(n × 120o)

+ cos(n ωt − αn) sin(n × 120o)]

=∞

n=3,6,9...

√ 2V n [sin(n ωt − αn) − sin(n ωt − αn) (−1/2)

+ cos(n ωt−

αn) (±√

3/2)=

√ 2

∞n=3,6,9...

V n

(3/2) sin(n ωt − αn) + (±

√ 3/2) cos(n ωt − αn)

=

√ 3√

2∞

n=3,6,9...V n

(√

3/2) sin(n ωt − αn) + (±1/2) cos(n ωt − αn)

(2.84)

Let√

3/2 = rn cos φn and ±1/2 = rn sin φn. This impliles rn = 1 and φn = ±30o. Using this,equation (2.84) can be written as follows.

vab(t) = √ 3√ 2∞

n=3,6,9...V n [sin(n ωt − αn ± 30o)] . (2.85)

In equations (2.84) and (2.85), vab = 0 for n = 3, 6, 9, . . . and for n = 1, 2, 4, 5, 7, . . ., the ± signof 1/2 or sign of 300 changes alternatively. Thus it is observed that triplen harmonics are missingin the line to line voltages, inspite of their presence in phase voltages for balanced three-phase

system with harmonics. Thus the following identity hold true for this system,

V LL ≤√

3 V Ln (2.86)

Above equation further implies that,

√ 3 V LL I ≤ 3 V Ln I. (2.87)

In above equation, I refers the rms value of the phase current. For above case, I a = I b = I c = I and I n = 3

∞n=3,6,9... I n

2. Therefore, effective rms current, I e is given by the following.

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I e =

3 I 2 + 3

∞n=3,6,9... I n

2

3

= I 2 +∞

n=3,6,9...I n

2

(2.88)

≥ I

2.4.3 Apparent Power with Budeanu Resolution: Balanced Distortion Case

The apparent power is given as,

S = 3V lnI =

P 2 + Q2B + D2B

= P 2 + Q2 + D2 (2.89)where,P = P 1 + P H = P 1 + P 2 + P 3 + ....

= 3V 1I 1 cos φ1 + 3∞n=1

V nI n cos φn

(2.90)

where, φn = β n − αn. Similarly,Q = QB = QB1 + QBH

= Q1 + QH (2.91)

Where Q in (2.89) is called as Budeanu’s reactive power (VAr) or simply reactive power which isdetailed below.

Q = Q1 + QH = Q1 + Q2 + Q3 + ....

= 3V 1I 1 sin φ1 + 3∞n=1

V nI n sin φn (2.92)

2.4.4 Effective Apparent Power for Balanced Non-sinusoidal System

The effective apparent power S e for the above system is given by,

S e = 3V eI e (2.93)For a three-phase, three-wire balanced system, the effective apparent power is found after cal-

culating effective voltage and current as given below.

V e =

(V 2ab + V 2bc + V

2ca)/9

= V ll/√

3 (2.94)

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I e =

(I 2a + I 2b + I

2c )/3

= I (2.95)

Therefore

S e = S = √ 3V llI (2.96)For a four-wire system, V e is same is given (2.94) and I e is given by (2.88). Therefore, theeffective apparent power is given below.

√ 3V llI ≤ 3 V lnI e (2.97)

The above implies that,

S e ≥ S A. (2.98)

Therefore, it can be further concluded that,

pf e (= P /S e) ≤ pf A (= P /S A). (2.99)

2.5 Unbalanced and Non-sinusoidal Three-phase System

In this system, we shall consider most general case i.e., three-phase system with voltage and current

quantities which are unbalanced and non-sinusoidal. These voltages and currents are expressed as

following.

va(t) =

∞n=1

√ 2V an sin(n ωt − αan)

vb(t) =∞n=1

√ 2V bn sin {n (ωt − 120o) − αbn} (2.100)

vc(t) =∞n=1

√ 2V cn sin {n (ωt + 120o) − αcn}

Similarly, currents can be expressed as,

ia

(t) =∞

n=1√

2I an

sin(n ωt−

β an

)

ib(t) =∞n=1

√ 2I bn sin {n (ωt − 120o) − β bn} (2.101)

ic(t) =∞n=1

√ 2I cn sin {n (ωt + 120o) − β cn}

55


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For the above voltages and currents in three-phase system, instantaneous power is given as follow-

ing.

p(t) = va(t)ia(t) + vb(t)ib(t) + vc(t)ic(t)

= pa(t) + pb(t) + pc(t)

= ∞

n=1

√ 2V an sin(nωt − αan)

∞n=1

√ 2I an sin(nωt − β an)

(2.102)

+

∞n=1

√ 2V bn sin {n(ωt − 120o) − αbn}

∞n=1

√ 2I bn sin {n(ωt − 120o) − β bn}

+

∞n=1

√ 2V cn sin {n(ωt + 120o) − αcn}

∞n=1

√ 2I cn sin {n(ωt + 120o) − β cn}

In (2.102), each phase power can be found using expressions derived in Section 1.4 of Unit 1. Thedirect result is written as following.

pa(t) =∞n=1

V anI an cos φan {1− cos(2nωt − 2αan)} −∞n=1

V anI an sin φan cos(2nωt − 2αan)

+

∞n=1


∞m=1,m=n

√ 2I am sin(mωt − β am)

=

∞

n=1

P an {1− cos(2nωt − 2αan)} −∞

n=1

Qan cos(2nωt − 2αan)

+

∞n=1


∞m=1,m=n

√ 2I am sin(mωt − β am)

(2.103)

In the above equation, φan = (β an − αan). Similarly, for phases b and c, the instantaneouspower is expressed as below.

pb(t) =∞n=1

P bn [1 − cos {2n(ωt − 120o) − 2αbn}] −∞n=1

Qbn cos {2n(ωt − 120o) − 2αbn}

+

∞n=1

√ 2V bn sin {n(ωt − 120o) − αbn}

∞m=1,m=n

√ 2I bm sin {m(ωt − 120o) − β bm}

(2.104)

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and

pc(t) =∞n=1

P cn [1 − cos {2n(ωt + 120o) − 2αcn}] −∞n=1

Qcn cos {2n(ωt + 120o) − 2αcn}

+ ∞

n=1

√ 2V cn sin {n(ωt + 120o

) − αcn} ∞

m=1,m=n√ 2I cm sin {m(ωt + 120

o

) − β cm}(2.105)

From equations (2.103), (2.104) and (2.105), the real powers in three phases are given as follows.

P a =∞n=1

V anI an cos φan

P b =∞

n=1V bnI bn cos φbn (2.106)

P c =∞n=1

V cnI cn cos φcn

Similarly, the reactive powers in three phases are given as following.

Qa =∞n=1

V anI an sin φan

Qb =∞

n=1V bnI bn sin φbn (2.107)

Qc =∞n=1

V cnI cn sin φcn

Therefore, the total active and reactive powers are computed by summing the phase powers using

equations (2.106) and (2.107), which are given below.

P = P a + P b + P c =∞n=1

(V anI an cos φan + V bnI bn cos φbn + V cnI cn cos φcn)

= V a1I a1 cos φa1 + V b1I b1 cos φb1 + V c1I c1 cos φc1

+∞n=2

(V anI an cos φan + V bnI bn cos φbn + V cnI cn cos φcn)

= P a1 + P b1 + P c1 +∞n=2

(P an + P bn + P cn)

= P 1 + P H (2.108)

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and,

Q = Qa + Qb + Qc =∞n=1

(V anI an sin φan + V bnI bn sin φbn + V cnI cn sin φcn)

= V a1I a1 sin φa1 + V b1I b1 sin φb1 + V c1I c1 sin φc1

+∞n=2

(V anI an sin φan + V bnI bn sin φbn + V cnI cn sin φcn)

= Qa1 + Qb1 + Qc1 +∞n=2

(Qan + Qbn + Qcn)

= Q1 + QH (2.109)

2.5.1 Arithmetic and Vector Apparent Power with Budeanu’s Resolution

Using Budeanu’s resolution, the arithmetic apparent power for phase-a, b and c are expressed as

following.

S a =

P 2a + Q2a + D

2a

S b =

P 2b + Q2b + D

2b (2.110)

S c =

P 2c + Q2c + D

2c

The three-phase arithmetic apparent power is arithmetic sum of S a, S b and S c in the above equation.This is given below.

S A = S a + S b + S c (2.111)

The three-phase vector apparent power is given as following.

S v =

P 2 + Q2 + D2 (2.112)

Where P and Q are given in (2.108) and (2.109) respectively. The total distortion power D is givenas following.

D = Da + Db + Dc (2.113)

Based on above definitions of the apparent powers, the arithmetic and vector power factors are

given below.

pf A = P

S A

pf v = P

S v(2.114)

From equations (2.111), (2.112) and (2.114), it can be inferred that

S A ≥ S v pf A ≤ pf v (2.115)

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2.5.2 Effective Apparent Power

Effective apparent power (S e=3V eI e) for the three-phase unbalanced systems with harmonics canbe found by computing V e and I e as following. The effective rms current (I e) can be resolved intotwo parts i.e., effective fundamental and effective harmonic components as given below.

I e = I 2e1 + I 2eH (2.116)Similarly,

V e =

V 2e1 + V 2eH (2.117)

For three-phase four-wire system,

I e =

I 2a + I

2b + I

2c + I

2n

3 (2.118)

= I 2a1 + I

2a2 + ... + I

2b1 + I

2b2 + ... + I

2c1 + I

2c2 + ... + I

2n1 + I

2n2 + ...

3

=

I 2a1 + I

2b1 + I

2c1 + I

2n1 + ... + I

2a2 + I

2b2 + I

2c2 + I

2n2 + ...

3

=

I 2a1 + I

2b1 + I

2c1 + I

2n1

3 +

I 2a2 + I 2a3 + ... + I

2b2 + I

2b3 + ... + I

2c2 + I

2c3 + ... + I

2n2 + I

2n3...

3

I e =

I 2e1 + I 2eH

In the above equation,

I e1 = I 2a1 + I

2b1 + I

2c1 + I

2n1

3

I eH =

I 2aH + I

2bH + I

2cH + I

2nH

3 (2.119)

Similarly, the effective rms voltage V e is given as following.

V e =

1

18[3(V 2a + V

2b + V

2c ) + (V

2ab + V

2bc + V

2ca)]

=

V 2e1 + V 2eH (2.120)

Where

V e1 = 1

18[3(V 2a1 + V

2b1 + V

2c1) + (V

2ab1 + V

2bc1 + V

2ca1)]

V eH =

1

18[3(V 2aH + V

2bH + V

2cH ) + (V

2abH + V

2bcH + V

2caH )] (2.121)

For three-phase three-wire system, I n = 0 = I n1 = I nH .

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I e1 =

I 2a1 + I

2b1 + I

2c1

3

I eH =

I 2aH + I

2bH + I

2cH

3 (2.122)

Similarly

V e1 =

V 2ab1 + V

2bc1 + V

2ca1

9

V eH =

V 2abH + V

2bcH + V

2caH

9 (2.123)

The expression for effective apparent power S e is given as following.

S e = 3 V eI e

= 3

V 2e1 + V 2eH

I 2e1 + I

2eH

= 9 V 2e1I 2e1 + (9V 2e1I 2eH + 9V 2eH I 2e1 + 9V 2eH I 2eH )=

S 2e1 + S

2eN (2.124)

In the above equation,

S e1 = 3 V e1I e1 (2.125)

S eN =

S 2e − S 2e1=

D2eV + D

2eI + S

2eH

= 3 I 2e1V 2eH + V 2e1I 2eH + V 2eH I 2eH (2.126)In equation (2.126), distortion powers DeI , DeV and harmonic apparent power S eH are given asfollowing.

DeI = 3V e1I eH

DeV = 3V eH I e1 (2.127)

S eH = 3V eH I eH

By defining above effective voltage and current quantities, the effective total harmonic distortion

(T HDe) are expressed below.

T HDeV =

V eH

V e1

T HDeI = I eH

I e1(2.128)

Substituting V eH and I eH in (2.126),

S eN = S e1

T HD2e1 + T HD

2eV + T HD

2eI T HD

2eV . (2.129)

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In above equation,

DeI = S e1T HDI

DeV = S e1T HDV (2.130)

S eH = S e1(T HDI )(T HDV ).

Using (2.124) and (2.129), the effective apparent power is given as below.

S e =

S 2e1 + S 2eN = S e1

1 + T HD2eV + T HD

2eI + T HD

2eV T HD

2eI (2.131)

Based on above equation, the effective power factor is therefore given as,

pf e = P

S e=

P 1 + P H

S e1

1 + T HD2eV + T HD2eI + T HD

2eV T HD

2eI

= (1 + P H /P 1)

1 + T HD2eV + T HD

2eI + T HD

2eV T HD

2eI

P 1S e1

= (1 + P H /P 1) 1 + T HD2eV + T HD

2eI + T HD

2eV T HD

2eI

pf e1 (2.132)

Practically, the THDs in voltage are far less than those of currents THDs, therefore T HDeV


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f f r jx

av

bv

cv

nv

aV

bV

cV

nV

L X

L R

a I

b I

c I

L O A D

n I

Fig. 2.11 An unbalanced three-phase circuit

b. Losses in the system.

c. The active and reactive powers in each phase and total three-phase active and reactive powers.

d. Arithmetic, vector and effective apparent powers and power factors based on them.

Solution:

a. Computation of currents

va (t) = 220√

2sin(ωt)

vb (t) = 220√

2sin(ωt − 120◦)vc (t) = 220

√ 2sin(ωt + 120◦)

vab (t) = 220√ 6sin(ωt + 30◦)Therefore,

I a = 220

√ 3∠30

13∠90◦ = 29.31∠−60◦ A

I b = −I a = −29.311∠−60◦ = 29.31∠120◦ AI c =

220∠120◦

12 = 18.33∠120◦ A.

Thus, the instantaneous expressions of phase currents can be given as following.

ia(t) = 41.45sin(ωt − 60◦)ib(t) = −ia(t) = −41.45sin(ωt − 60◦) = 41.45sin(ωt + 120◦)ic(t) = 25.93sin(ωt + 120

◦)

b. Computation of losses

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The losses occur due to resistance of the feeder impedance. These are computed as below.

Losses = rf (I 2a + I

2b + I

2c + I

2n)

= 0.02(29.312 + 29.312 + 18.332 + 18.332) = 47.80 W

c. Computation of various powers

Phase-a active and reactive power:

S a = V a I ∗a = 220∠0

◦ × 29.31∠60◦ = 3224.21 + j5584.49implies that, P a = 3224.1 W, Qa = 5584.30 VAr

Similarly,

S b = V b I ∗b = 220∠−120◦ × 29.31∠60◦ = −3224.21 + j5584.49

implies that, P b = −3224.1 W, Qb = 5584.30 VArFor phase-c,

S c = V c I ∗c = 220∠120

◦ × 18.33∠−120◦ = 4032.6 + j0implies that, P c = 4032.6 W, Qc = 0 VAr

Total three-phase active and reactive powers are given by,

P 3− phase = P a + P b + P c = 3224.1 − 3224.1 + 4032.6 = 4032.6 WQ3

− phase = Qa + Qb + Qc = 5584.30 + 5584.30 + 0 = 11168.60 VAr.

d. Various apparent powers and power factors

The arithmetic, vector and effective apparent powers are computed as below.

S A = |S a| + |S b| + |S c|= 6448.12 + 6448.12 + 4032.6 = 16928.84 VA

S v = |S a + S b + S c|= |4032.6 + j11168.6| = |11874.32∠70.14| = 11874.32 VA

S e = 3V eI e = 3 × 220 ×

I 2a + I 2b + I

2c + I

2n

3

= 3 × 220 ×

29.312 + 29.312 + 18.332 + 18.332

3 = 3 × 220 × 28.22

= 18629.19 VA

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Based on the above apparent powers, the arithmetic, vector and effective apparent power factors

are computed as below.

pf A = P 3− phase

S A=

4032.6

16928.84 = 0.2382

pf v = P 3− phaseS v = 4032.611874.32 = 0.3396

pf e = P 3− phase

S e=

4032.6

18629.19 = 0.2165

In the above computation, the effective voltage and current are found as given in the following.

V e =

V 2a + V

2b + V

2c

3 = 220 V

I e = I 2a + I 2b + I 2c + I 2n3 = 28.226 AExample 2.4 A 3-phase, 3-wire system is shown in Fig. 2.12. The 3-phase voltages are balanced

sinusoids with RMS value of 230 V. The 3-phase loads connected in star are given as following.

Z a = 5 + j12 Ω, Z b = 6 + j8 Ω and Z c = 12 − j5 Ω.

Compute the following.

a. Line currents, i.e., I la, I lb and I lc and their instantaneous expressions.

b. Load active and reactive powers and power factor of each phase.

c. Compute various apparent powers and power factors based on them.

N

a

c b Z

la I

lb I

lc I

saV

scV

sbV

Fig. 2.12 A star connected three-phase unbalanced load

Solution:

a. Computation of currents

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Given that Z a = 5 + j 12 Ω, Z b = 6 + j 8 Ω, Z c = 12 − j 5 Ω.

V sa = 230∠0◦ V

V sb = 230∠−120◦ VV sc = 230∠120

◦ V

V nN = 11Z a

+ 1Z b + 1Z c

V saZ a

+ V sb

Z b+ V

sc

Z c

=

11

5+ j12 + 16+ j8 +

112− j5

230∠0◦

5 + j12 +

230∠−120◦6 + 8 j

+ 230∠120◦

12 − j5

= 1

0.2013∠−37.09◦ 31.23∠−164.50◦

= −94.22− j123.18 = 155.09∠−127.41◦ V

Now the line currents are computed as below.

I al = V sa − V nN

Z a=

230∠0◦ − 155.09∠−127.41◦5 + j12

= 26.67∠−46.56◦ A

I bl = V sb − V nN

Z b=

230∠−120◦ − 155.09∠−127.41◦6 + j8

= 7.88∠−158.43◦ A

I cl = V sc − V nN

Z c=

230∠120◦ − 155.09∠−127.41◦12 − j5 = 24.85∠116.3

◦ A

Thus, the instantaneous expressions of line currents can be given as following.

ial (t) = 37.72sin(ωt − 46.56◦)ibl (t) = 11.14sin(ωt − 158.43◦)icl (t) = 35.14sin(ωt + 116.3

◦)

b. Computation of load active and reactive powers

S a = V aI ∗a = 230∠0

◦ × 26.67∠46.56◦ = 4218.03 + j4456.8S b = V bI

∗b = 230∠−120◦ × 7.88∠158.43◦ = 1419.82 + j1126.06

S c = V cI

∗c = 230

∠120◦ × 24.85∠−116.3◦ = 5703.43 + j368.11

implies that,

P a = 4218.03 W, Qa = 4456.8 VAr

P b = 1419.82 W, Qb = 1126.06 VAr

P c = 5703.43 W, Qc = 368.11 VAr

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Total three-phase active and reactive powers are given by,

P 3− phase = P a + P b + P c = 4218.03 + 1419.82 + 5703.43 = 11341.29 W

Q3− phase = Qa + Qb + Qc = 4456.8 + 1126.06 + 368.11 = 5950.99 VAr.

The power factors for phases a, b and c are given as follows.

pf a = P a

|S a|=

4218.03√ 4218.032 + 4456.82

= 4218.03

6136.3 = 0.6873 (lag)

pf b = P b

|S b|=

1419.82

1419.822 + 1126.062 =

1419.82

1812.16 = 0.7835 (lag)

pf c = P c

|S c|=

5703.43

5703.432 + 368.112 =

5703.43

5715.30 = 0.9979 (lag)

c. Computation of various apparent powers and power factors

The arithmetic, vector and effective apparent powers are computed as below.

S A = |S a| + |S b| + |S c|= 6136.3 + 1812.16 + 5715.30 = 13663.82 VA

S v = |S a + S b + S c|= |11341.29 + j5909.92| = 12807.78 VA

S e = 3V eI e = 3 × 230 ×

I 2la + I 2lb + I

2lc + I

2ln

3

= 3 × 220 × 26.672 + 7.882 + 24.852 + 023 = 3 × 230 × 21.53= 14859.7 VA

The arithmetic, vector and effective apparent power factors are computed as below.

pf A = P 3− phase

S A=

11341.29

13663.82 = 0.8300

pf v = P 3− phase

S v=

11341.29

12807.78 = 0.8855

pf e = P 3− phase

S e=

11341.29

14859.7 = 0.7632

References

[1] IEEE Group, “IEEE trial-use standard definitions for the measurement of electric power quan-

tities under sinusoidal, nonsinusoidal, balanced, or unbalanced conditions,” 2000.

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[2] E. Watanabe, R. Stephan, and M. Aredes, “New concepts of instantaneous active and reactive

powers in electrical systems with generic loads,” IEEE Transactions on Power Delivery, vol. 8,

no. 2, pp. 697–703, 1993.

[3] T. Furuhashi, S. Okuma, and Y. Uchikawa, “A study on the theory of instantaneous reactive

power,” IEEE Transactions on Industrial Electronics, vol. 37, no. 1, pp. 86–90, 1990.

[4] A. Ferrero and G. Superti-Furga, “A new approach to the definition of power components in

three-phase systems under nonsinusoidal conditions,” IEEE Transactions on Instrumentation

and Measurement , vol. 40, no. 3, pp. 568–577, 1991.

[5] J. Willems, “A new interpretation of the akagi-nabae power components for nonsinusoidal

three-phase situations,” IEEE Transactions on Instrumentation and Measurement , vol. 41,

no. 4, pp. 523–527, 1992.

[6] H. Akagi, Y. Kanazawa, and A. Nabae, “Instantaneous reactive power compensators compris-

ing switching devices without energy storage components,” IEEE Transactions on Industry

Applications, no. 3, pp. 625–630, 1984.

[7] C. L. Fortesque, “Method of symmetrical co-ordinates applied to the solution of polyphase

networks,” AIEE , 1918.

THREE PHASE CIRCUITS: POWER DEFINITIONS AND VARIOUS COMPONENTS

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