UNIT 4 THREE PHASE CIRCUITS

Structure 4.1 Introduction

Objectives

4.2 The Three Phase System 4.2.1 Tbe Nature of the 3-Phase System 4.2.2 Merits of a 3-Phase System

4.3 Characteristics of 3-phase Systems 4.3.1 Balanced Sets of Voltages and Currents 4.3.2 Delta and Star Connections 4.3.3 Relation between Line and Phase Quantities

4.4 Analysis of Balanced 3-phase Circuits 4.4.1 Star Connected Load Fed from a Balanced Supply 4.4.2 Delta Connected Inad Fed from a Balanco'd Supply 4.4.3 PowerConsiderations 4.4.4 Power Factor Correction 4.4.5 Single Phase and Single Line Representations

4.5 Summary

I 4.6 Answers to SAQs / 41 INTRODUCTION I 1 A sinusoidal voltage source with 2 tenninals having a single voltage output is termed a

! single phase source. Circuits incorporating such sources are called single phase (1-phase) circuits and formed the subject of our study in Unit 3. In contrast, a polyphase system contains sources each of which has several voltage outputs with a fixed phase difference between them. The three-phase (3-phase) system is the most common example of a polyphase system.

The generation and transmission of electrical energy and its utilization in bulk form is effected through 3-phase systems. In this Unit, you will first learn the precise nature of a 3-phase system and the advantages it provides relative to a single-phase system. You will then be introduced to the terminology, classification and characteristics of 3-phase circuits. Analysis of balanced 3-phase circuits will be taken up next. The representation of a balanced 3-phase system by a single line diagram and issues related to power will also be considered. You will observe that the analysis methods used in this unit are straightforward extensions of those employed for single phase circuits.

I

Objectives After completing a study of this unit, you should be able to

describe the features of a 3-phase system,

distinguish between delta and star-connections of sources and loads,

distinguish between 3-wire and 4-wire systems,

explain the meaning of phasc sequence,

distinguish between balanced and unbalanced systems,

differentiate between phase and line quantities and calculate one set fmm the other,

analyse balanced 3-phase circuits comprising loads connected in star or delta, - represent a 3-phase balanced circuit by a single-phase circuit and inter-relate the variables in the two systems, .

make calculations relating to power, reactive power, apparent power and p.f. for a given 3-phase circuit, and

design the capacitor sizes needed for p.f. improvement in a 3- phase circuit.

Introduction to Circuits 4.2 THE THREE PHASE SYSTEM

4.2.1 The Nature of a 3-Phase System A single phase a.c. generator consists of a rotating magnet driven by a prime mover and a winding embedded in the stationary part of the machine called the stator. Figure 4.1 shows an elementary form of tbe generator with a single turn coil AA' on the stator. As the magnet rotates, the flux lines linking with the coil undergo a periodic variation and hence induce a periodic emf in the latter. The frequency of this emf is fixed by the speed of rotation of the magnet. Special steps are taken in the design and construction of the machine to make the waveform of the induced voltage sinusoidal. Thus the coilfunctions a s a single-phase a.c. voltage source with terminals A, and A2, to which a load may be connected.

Fig. 4.1 : Elementary single phase generator and its circuit representation

Figure 4.2 illustrates the construction of an elementary 3-phase generator. Here we have 3 identical coilsAA1, BB', CC' placed on the stator with a displacement of 120" from one

Fig. 4.2 : Elementary 3-phase generator and its circuit representation

another. The three emf's e,, eB, e, generated in the coils therefore have the same rms value but have a phase difference of 120" from one another as shown in Figure 4.3. A -

Fig. 4.3 : Voltages produced in a 3-phase generator (a) Waveforms (b) Phasors

3-phase generator can therefore be viewed as a composite unit comprising 3 single phase voltage sources with a fixed phase difference of 120" between any two of them. In practice, it is rare for a 3-phase generator to have all the six terminals brought out. The three coils are connected either in star or delta and only 3 or 4 terminals are brought out, as we shall see later.

A three-phase system is one which contains 3-phase sources besides 3-phase load impedances and feeder lines interconnecting them.

The three individual sections which constitute a 3-phase arrangement are referred to as Phase A, Phase B, and Phase C respectively. Another common practice is to label them as R(red), Y(ye1low) and B(b1ue) phases. We shall follow thg former convention in our work.

4.2.2 Merits of a 3-phase System Let us now look at the advantages provided by 3-phase systems relative to single-phase systems.

A 3-phase a.c. generator utilises the available space on the stator more effectively than a 1-phase generator and has 50% inore kVA rating for the same physical size. All commercial power stations therefore employ 3-phase generators as they cost less than single-phase generators for the same kVA rating.

The cost of electrical transmission and distribution lines used to carry bulk power from generating stations to receiving substations and distribute power from the substations to different load centres depends substantially on the volume of conducting material (usually aluminium) required for constructing these lines. It turns out that a 3-phase arrangement for transmission and distribution requires less conductor material for the lines and is therefore less expensive than a 1-phase system for handling the same amount of power at a given system voltage.

In power utilization, a 3-phase motor develops essentially a constant output torque whereas a single-phase motor can inherently provide only a pulsating torque. Not only is the three-phase motor consequently quieter in operation but it also provides better starting characteristics, higher efficiency of power conversion from electrical to mechanical form and better p.f. It is also in general cheaper than a 1-phase motor of the same power rating.

The foregoing economic and technical advantages have led to the universal adoption of the 3-phase system for the generation, transmission and utilization of bulk power. Small electrical loads, of typically less than 3 kW power rating, are however designed and built for single -phase operation. These include electric lights, fans, heaters and small motors needed for various domestic appliances, machine tools, pump sets and the like. The benefits that may stem from 3-phase operation of these loads are not commensurate with the additional cost of manufacturing them to be suitable for 3-phase use and of running 3-phase lines to each individual item. In practice, these small loads are fed from single phase supplies available from a 3-phase distribution system.

SAQ 1 Distinguish between a 3-phase generator and a single-phase generator.

SAQ 2 Fill up the blanks

A 3-phase generator has more than a 1-phase generator of the same physical size. A 3-phase transmission line employs less 2

than a 1-phase transmission line for the same power transmitted and the same system voltage. The torque developed by a 3-phase motor is - 3

while the torque developed by a 1-phase motor is 4

Three-Phase Circuits

introduction to Circuits

I

4.3 CHARACTERISTICS OF 3-PHASE SYSTEMS

After having been acquainted with the nature of 3-phase systems and their advantages, you will study in this section the characteristics of 3-phase sources, loads and associated systems in greater detail.

4.3.1 Balanced Sets of Voltages and Currents At any section of the circuit representing a 3-phase system, there exist three voltages and three currents which constitute the variables of interest. Three such voltages or currents are said to form a balanced set if they have equal effective values and if the phase difference hetween any two is 120'. A balanced set of three voltages c, VB and Vc is depicted in Figure 4.4.

Fig. 4.4: A balanced set of three voltages with A B C phase sequence (a) Waveforms (b) Phasars

Note that similar events in the three waveforms (e.g., positive peak values) occur in the sequence ABCABC ..... . For this reason, the three voltages are said to have the ABC phase sequence. (We could as well have called it the BCA or CAB phase sequence but, by convention, choose the natural alphabetical order). Referring to the phasor diagram in Figure 4.4(b), if one were to imagine the three phasors to rotate in the anticlockwise direction, they sweep past a stationary point in the sequenceABC. This alternative way of judging the phase sequence from a phasor diagram would be useful when the waveforms are not explicitly plotted.

There exists a second possible phase sequence for a balanced voltage set, as depicted in Figure 4.5. Here similar events in the three signals occur in the sequence ACBACB .... . This sequence is called the ACB phase sequence (it could as well have been called CBA or BAC phase sequence). Note that the three related phasors now sweep past a stationary observer in the orderACB.

Fig. 4.5: A balanced set of voltages with ACB phase sequence (a) Waveforms (b) Phasors

In normal practice, the individual phases are so labelled as to correspond to the ABC phase sequence. We shall assume this to be the phase sequence in all our further work unless the contrary is specifically indicated. What has been discussed above with respect to a set of balanced voltages holds equally well with respect to a set of 3 currents. A set of balanced 3 phase voltages or currents would then have the following expressions (with ABC phase sequence assumed).

vA = \/Z V sin (wt + 8)

iA = \/Z I sin (wt + P)

ic = \/Z I sin (wt + p + 2 ~ 1 3 )

The corresponding phasors would be - v A = V ~ ~ ; ~ = v ~ ( 8 - 2 r c / 3 ) ; V C = v ~ ( e + ~ 3 ) (4.3)

The following properties of balanced voltages (or currents) are noteworthy:

WithABC phase sequence, vA leads vB by 120°, vB leads vc by 120" and vc leads v, by 120". WithACB phase sequence, vA leads vc by 120°, vc leads vB by 120" and vB leads vA by 120".

If the set of voltages (or currents) is known to be balanced and the phase sequence is fixed, the data pertaining to one voltage or current would suffice to deduce the other two. For example, if it is known then VB = 100 L 30" and that the phase sequence is ABC, it follows that FA = 100 L 150" and Vc = 100 L -90".

The sum of three balanced quantities is identically zero in time domain.

iA + iB + iC = o (4-6) The above results can be proved through manipulation of the trigonometric expressions in Eq. (4.1) and (4.2). They can also be verified by observing that the ordinates of the three pertinent waveforms like those in Figure 4.4(a) add up to zero at every instant of time.

The equivalent results in phasor domain are

VA + VB + Vc = o - zA +IB + I C = 0 . (4-8)

To check the validity of Eq.(4.7) please refer to Figure 4.6. Since VA and have equal magnitudes and are 120" apart, their resultant VA + VB is at 60" from VA and has the same magnitude. VA + VB is therefore equal and opposite to Vc. Hence VA + VB + VC = 0. If drawn from end to end, the three directed line segments VA, VB and Vc form a closed triangle as seen in Figure 4.6(b). This is an alternative way of showing that VA, VB and Vc add up to zero.

Three-Phase Circuits

Fig. 4.6 : Demonstration of the result, +Te +Tc- 0.

Introduction to Circuits

Fig. 4.7 : Examples of sets of unbalanced voltages.

Finally, you should note that three voltages / currents are unbalanced if their effective values are not equal or their phase differences are not 120" or both. Figure 4.7 gives examples of sets of unbalanced voltages.

Example 4.1

At a certain section in a 3-phase circuit, ,VB = 120 L 60" andjc = 4 L 180". If the - - voltages and currents are balanced and the phase sequence is ABC, deduce E, Vc, IA andiB.

Solution

With ABC sequence, VB leads Vc by 120" and lags by 120". Thus - VA = 120 L 180"; Vc = 120 L -60". - Ic leads?* by 120" and lagsTB by 120".

~ e n c e , i * = 4 L 60"; iB = 4 L 300" = 4 L -60".

SAQ 3 State if the following assertions are true or false.

1 Three currents in a 3-phase system are balanced if their phasors are equal.

2 In a set of balanced 3-phase voltages withABC phase sequence leads the other two voltages & and Vc.

3 If iA + iB + iC = 0, then iA' , iB and ic form a balanced set of 3-phase currents.

SAQ 4 Taking vA and iA as in Eq. (4.1) and (4.2), write the expressions for the other quantities if phase seqnence is ACB.

SAQ 5 Taking the following to be balanced sets of voltages /currents with ABC phase sequence, fill the blanks

1 VA = fi x .............I sin (wt + 2 ............. ) ;

VB = f i x lOOsin(wt + ................ > vc = fi x .............. ? sin (wt + 45"). - - - 8 ............. 2 IA = ............... L ............. ; IB = ? L 60" ; Ic = 4 L ..............

4.3.2 Delta and Star Connections You would recall that a 3-phase generator essentially consists of 3 single-phase sources, having output voltages say, eA, e, and e, A balanced 3-phase source is one in which these three voltages form a balanced set. All commercial 3-phase generators are built in

I this manner. The three single phase sources are connected internally either in delta or in I star as shown in Figure 4.8 and terminals brought out for connection to external loads.

Note the symmetrical way of connecting the three single-phase sources. In the delta connection, A, is connected to B,, B2 is connected to C, and C, is connected to A,. In the star connection A,, B2 and C, are joined together. Such an orderly method of connections is needed to ensure the balanced condition of the voltages available between the terminals A, B and C of the 3-phase generator.

(a ) (b) (C Fig. 4.8 : Internal connections ola 3-phase voltage source

(a) Delta connection Ib) Star connection with 3 terminals (c) Star connection with 4 terminals.

In the delta connection, the effective emf of the three series connected sources around the closed circuit is eA + eB + e,. If the three voltages do not add up to zero there would be a large circulating current in the delta even with no load connected to terminals A, B, C and this is clearly an undesirable situation. However, for a balanced source this contingency does not arise as eA + eB + e, = 0 (vide Eq. (4.5)). It is this fact which makes the delta connection of a 3-phase source feasible.

I In the star connection, the common terminal of the 3 sources (star point) is called the

I neutral point. Here there exist two possible arrangements. Where a separate terminal is not provided for the neutral point as in Figure 4.8(b), the generator fonns part of what is

i I known as a 3-wire 3-phase system. On the other hand, the arrangement shown in Figure

4.8(c) permits connection of the generator in a 4-wire 3-phase system. The terminals I

i A, B, C are called the line terminals and N is called the neutral terminal.

A 3-phase load generally comprises three impedances in a configuration suitable for connection in a 3-phase circuit. Similar to the connections in a 3-phase generator, here also we have 3 possible connections as shown in Figure 4.9. Notice that the configuration in Figure 4.9(c) is suitable for connection only in a 4-wire 3-phase system.

r A 3-phase load is balanced ifthe three complex impedances are equal i.e.,

ZA = ZB = ZC for a star-connected load (4.9a) - -

and ZAB = ZBC = ZCA for a delta-connected load (4.9b)

Tluee-Phase Circuits

Fig. 4.9: 3-phase loads. (a) Delta connection (b)Star connection with 3 terminals (c) Star connection with 4 terminals

Introduction to Circuits Not on)y the magnitudes but also the angles of the impedances should be equal for a 3-phase load to be balanced. We can therefore use a common symbol Zy for the three

impedances in star and &for the three impedances in delta.

You may recall the star-delta conversion formulas for resistances discussed in Unit 2. The same formulas hold for complex valued impedances as well. Thus a delta-connected 3-phase load has an equivalent star-connected configuration and vice-versa. If the load in Figure 4.9(b) is to be equivalent to that in Figure 4.9(a), we require

For balanced loads, the formulas reduce t?

I A 3-phase circuit is formed through the interconnection of 3- phase sources and 3-phase loads. If all the sources are balanced and have the same phase sequence and all the loads are also balanced, then the 3-phase circuit is said to be balanced. A characteristic of a balanced 3-phase circuit is that the voltages a r~d currents at any arbitrary location are balanced. In our study we shall be concerned only with balanced systems.

SAQ 6 Fill up the blanks

The connection of sources / impedances is suitable either for 3-wire or for Cwire three-phase systems but the 2

connection of sources/impedances is suitable only for 3

three-phase systems.

SAQ 7 State if the following assertions are true or false.

(a) Three impedances &, ZB and Zc form a balanced 3-phase load if ZA + ZB + & = 0.

(b) The neutral point is not available in a 3-phase delta-connected source.

Example 4.2

A balanced 3-phase load is formed by three impedances of 60 +j90 ohms each, connected in delta. If this load is equivalent to a star-connected load having Zy in each leg of the star, calculate Zy.

Solution

4.3.3 Relations between Line and Phase Quantities In 3-phase circuits one distinguishes between line voltages and currents on one hand and phase voltages and currents on the other. Phase quantities are the internal voltages or currents associated with the single phase sources constituting a 3-phase source or the three impedances constituting the 3-phase load. Line quantities, on the other hand, are those which can be measured at the terminals. These are the voltages between and the currents in the external supply lines connected to the terminals. We now proceed to show the relationship between the two sets. We use the double subscript.notation in the following analysis. Recall that v, represents the voltage drop from x t o y and i, represents the current from x to y.

S ta r Connection

Refening to the circuits of Figure 4.10, the following constitute the line quantities and phase quantities in a star-connected sourcelload.

'Ibree-Phase Circuits Phase currents : @ ~ ) p ,Gc)p - - -

Phase voltages : VAN VBN , VCN

Line currents : (TAIL 9 GB)L , Gc)L - - -

Line voltages : VAB 9 VBC , VCA - - - ( In) L A

- -41~)~ -(iBlL

la) (b) Fig. 4.10: Phase and line quantities in a star-connected (a) source and (b) load.

We straightaway note that -

(IA )p = @A )L , (TB)p = (TB)L , Gc)p = (?c)L - - - - -

and VAB= VAhr+VNB=VAN-VBN - - - - VBc = VBN + VNC = VBN - VCN

- - - VcA = VcN + VNA = VcN - VAN (4.12)

In a balanced system, each one of the above four sets of phase and line voltages and currents is balanced. Let us designate the rms values of the phase currents and voltages as I, and Vp and the rms values of the line currents and voltages as I, and VL respectively. With this notation,

(IA)P = VB)P = (Ic)P = IP

and (IA)L = (IB)L = (Ic)L = IL. We therefore have

IL = Ip (4.13)

To find the relation between VL and Vp, let us start taking VAN as the reference phasor. Since the phase voltages form a balanced set,

- - - VAN'VpLOO; VBN=VPL-120"; VCN=VPL12O0 - -

We have VAE = v&, - VBN = VP L 0" - VP L -120"

= Vp [1 - (- 0.5 - f l / 2 ) ]

= ( f i l 2 ) Vp ( f i + j l ) = f iVp L 30"

- - - VBC = VBN - VCN = VP L -1 20" - Vp L 120"

= fivp L - 90" - - - VCA=VCN-VAN= V P L 120'- V p ~ 0 ' = f i ~ p ~ 150"

The disposition of the phasors of the phase voltages in the complex plane and the construction of the line voltages therefrom are illustrated in Figure 4.11. Note for example that is the sum of VAN and - VBN. Since the latter two have the same magnitude and are displaced from each other by 60°, their resultant has a magnitude 2 cos 60" = f i times each and an angle exactly midway between them.

Introduction to Circuits Hence cB = f ivP L 30". You are advised to make use of this graphical deduction of line

voltages from.phase voltages wherever ueeded, as it is simpler and more illustrative than purely analytical methods.

Fig. 4.11: Derivation of line voltages from phase voltages in a balanced star configuration

From the foregoing analysis, we have the important result,

VL = *BVp

Note that if the three line voltages phasors were to rotate in the anticlockwise direction, their first subscripts as well as the second subscripts appear to a stationary observer in the orderABCABC.,.. indicating that the phase sequence of the line voltages is also ABC.

The following important characteristics of a star configuration in a balanced system emerge from the foregoing derivations :

Line currents have the same effective value as phase currents. IL = Ip.

* Line voltages have fi times the effective value of phase voltages. VL = d 3 V p

* For a balanced set of phase voltages withABC phase sequence, the line voltages are also balanced, have the same phase sequence and are displaced from the phase voltage set by 30" (VAB leads VAN^^ 30").

Delta connection

For the delta connected source and load illustrated in Figure 4.12, we make the following identification of the line and phase qua~tities.

(a (b)

Fig. 4.12: Phase and line quantities in a delta-connected (a) source and (b) load.

- - - Phase currents : IAB , IBC ICA

! Phase voltages : ( VAB)P ( V B c ) p ( VCA)P -

I Line currents : IA B , ?C I

Line voltages : ( V M > L , < 6 c > L , (VcA>L

Here we straightaway note that

( F u ) p = (&)L ; ( V B C ) p (FBC)L ; ( V C A ) p = (&A)L

For the currents we have

Three-Phase Circuits

t For a balanced system, let us designate the rms values of phase and line voltages as V p and V, respectively and of phase and line currents as I, and I, respectively. We then note

I Vp = VL (4.17)

To deduce the corresponding relation for currents, let us refer to the phasor diagram of

I Figure 4.13, which has been drawn taking the phase currentiu as reference and using the relations in Eq. (4.16).

Fig. 4.13 : Deduction of line currents from phase currents in a balanced delta configuration.

I For instance,

= 2 1 p c o s 3 0 0 ~ - 3 0 0 = 4 3 - ~ p ~ - 3 0 0

Since ITA( = I iBl = IjCl = IL in a balanced configuration, we have

IL = 43-1~ (4.18) I

A balanced delta configuration then has the following important characteristics :

1 Phase voltages and line voltages have the same effective value. Vr, = Vp i

Line currents have 6 times the effective value of phase currents. Ir, = G I P

The line currents and phase currents have the same phase sequence and the set of line current phasors is displaced by 30" from the set of phase current phasors (line currenti, lags& by 30")

Introduction to Cucuits SAQ 8

Fill up the blanks:

In a 3-phase system are the voltages that can be measured between the supply lines, whereas 2

are the voltages associated with the individual legs of a 3-phase source or load.

In a star configuration the is 6 times the_ 4, whereas in the 5

configuration, the is 6 times the I

Example 4.3

The line voltage & across a star-connected load has a phasor 400 L 70". Find of theA phase voltage.

Solution -

[€VAN= ~ ~ ~ 8 , t h e n ~ = ~ ~ L 0 - 1 2 0 ~

We have

Therefore Vp = 4 0 0 1 6 = 231V, 8 = 40'

Hence = 23 1 L 40"

SAQ 9 What do you understand by

(a) a balanced 3-phase voltage,

(b) a balanced 3-phase load,

(c) a balanced 3-phase circuit?

SAQ 10 hi the balanced delta configuration of Figure 4.8(a), let the phase voltage be 100 V. If now the termi~lals of the generatorA are interchanged, what would be the net voltage acting around the loop?

SAQ 11 Two three-phase loads, connected in star and delta, are in parallel as shown, where all values are impedances in ohms. Find an equivalent delta-connected configuration.

C 0 I I

Fig. 4.14 for SAQ 11

4.4 ANALYSIS OF BALANCED 3-PHASE CIRCUITS

You would recall from the previous section that a balanced 3-phase circuit comprises only balanced sources feeding balanced loads. In this section you will learn how to analyse a few typical 3-phase balanced circuits.

The supply lines A, B, C feeding a 3-phase load in a commercial power distribution network do not emanate from an actual 3-phase generator, but from a 3-phase transformer. However for our purposes, we may visualize the transformer to function as a 3-phase source.

A 3-phase supply is conventionally designated by its line voltage since it is this voltage that can be measured at the supply terminals. Thus the term a 3-phase 400-Vsupply is intended to convey that the line voltage of the supply is 400 V. If the source is delta-connected, the phase voltage of the source is also 400 V. If however the source is star-connected the phase voltage of the source is 400 / fi - 230V. Now in a 3-phase 4-wire system, we have not only the line voltages VAB, VBC and VcA available but aiso the phase voltages VAN V,,, V,,. The 3-phase 400 V, 4-wire supply system is the standard adopted in our country and many others for power distribution. Apart from 3-phase loads, this system caters to single phase 230 V loads (e.g., lights and fans) which are connected between a line (A, B or C) and the neutral. These single-phase loads are distributed evenly between the three phases as far as possible, so that the overall load on the system is nearly balanced.

4.4.1 Star-connected Load Fed from a Balanced Supply Figure 4.15 shows a 3-phase star-connected load comprising three impedances Z L a, connected to a 3 phase star-connected source in a 4-wire system. Let the source have a line voltage of V, volts. Let us analyse the circuit, taking VAN as reference.

-

Fig. 4.15 : Star-connected load in a 4-wire system

- VL - VL v w = ~

VL L O ; V B N = z L - W 3 ; V - - L W 3

CN - fi t

Assuming that the impedances of the feeder lines are negligible, the phase voltages at the load are also the same as at the source. We then have,

Note that after having found jA, we could have straightaway deduced & and jc invoking

the properties of a balanced set of currents. Separate calculatioils forjB and jc as shown above are therefore not necessary in a balanced system. Figure 4.16 is the phasor diagram showing all the relevant quantities. The line voltages at the load are:

Three-Phase Circuits

introduction to Circuits

Fig. 4.16 : Phasor diagram for the circuit of Fig. 4.15

The current in the neutral line is jN =& +iB +& This is zero for a balanced load fed from a balanced supply. Thus the neutral wire does not carry any current. Even if it is omitted (i.e., even if the balanced 3-phase load is fed from a 3-wire source) the line currents as calculated above do not undergo any change. It would thus appear that the provisioii of a neutral wire is not necessary. However, in practice, there is no assurance that the loads would be perfectly balanced on the three phases. In the event of an unbalance in the load impedances, the neutral wire serves to maintain the balance of the load phase voltages and carries the out of balance current&, = j A +TB +j,.

Let us now consider a delta-connected source connected to the same star-connected load of Figure 4.15 in a 3-wire configuration. If a delta-connected source has the same line voltages FA,, VBC and 6, it can be deduced that the load phase voltages and hence the load currents will remain the same as in the earlier analysis. In other words, such a delta connected source is equivalent to a 3-terminal star-connected source with phase voltages given by Eq. (4.19). This equivalence comes about because both have the same line to line tenniiial voltages.

Example 4.4

Three impedances of 100 +j80 ohms each are connected in star across a balanced 400 V, 3-phase, 3-wire supply. Firid the line currents taken by the load and the voltage across each impedance. Draw a phasor diagram.

Solution

Since we are interested only in the load currents aiid voltages, it is immaterial to us whether the source is coni~ected in delta or star. We arc given that VL = 400 V . Helice Vp = 4001fi = 231 V. This is the voltage across each impedance.

Ip = V f l = 2 3 l d m = 1.8A. The line currents also have a value of 1.8 A. The phasor diagram taking VAN as reference is given in Figure 4.17(b). Note that each phase current (e.g.TAN) lags the respective phase voltage (e.g. &) by

tan- '(8011 00) = 39".

(a)

Fig. 4.17 for Example 4.4 (a) Circuit diagram

(b)

(b) Phasor diagram

4.4.2 Delta Connected Load Fed from a Balanced Supply Figure 4.18 shows a balanced delta-connected load fed from a star-connected balanced source in a 3-wire system. Let the line voltages have an rms value of VL volts and let VAB be the reference. Thus

Fig. 4.18 : Delta-connected load fed from a balanced supply

The line voltages are also the phase voltages, directly appearing across the three impedances. We then have,

i We therefore have for the phase current, Ip = Vp / Z - -

NOW I ~ = ~ - I ~ ~ = I ~ L ( - C Y ) - I p L ( 2 T c / 3 - ~ )

= f i I P L ( - a - x / 6 ) I - -

i IB = TBC - IAB = 6 1 ~ L (- a - 5x16) - - -

I Ic = IcA - IBc = 6 1 p L ( ~ ~ 1 2 - a)

Note that the three line currents have an effective value

IL = fiVL/Z

The phasor diagram showing the voltages as wcll as the phase and line currents at the load is given in Figure 4.1 9. Thc phase voltages of the star-connected source have a magnitude V L l a and have a displacenlent of 30" from the pertinelit line voltages as already seen in Section 4.3.3.

Tbree-Phase Circuits

Vec Fig. 4.19: Phasor diagram Lor the circuit of Fig. 4.18

Introduction to Circuits - Thus VM - vLh/5 L - ~ 1 6

These are 11ot drawn in Figure 4.19.

Obviously, a delta conilected source with the same line voltages as the star-connected source would produce the same set of currents in the load.

Example 4.5

If the three i~npedances of Example 4.4 are reconnected in delta and fed from the same 400 V, 3-phase balanced supply, what would be the line and phase currents? Draw a phasor diagram.

Solution

So as to compare the answers with those in Example 4.4, let us take VAB to have the same phase in both.

Thus

We have

Thus

Fig. 4.20 : for Example 4.5

Thus 1, =5.4A

Note that all line currents are trebled in co~nparison with those in Example 4.4. To maintain the same current, the delta-connectcd impedances should have three times the value of star-connected impedances. This is in keeping with Eq. (4.11).

SAQ 12 What is the function of a neutral wire in a 3-phase 4-wire system? What current does it carry in a balanced system?

3-Phase Circuits SAQ 13

A balanced star-connected load draws 5 A line currents from a 1000 V 3-phase source. If the current in a particular phase leads the corresponding phase voltage by 60°, find the complex value of impedance in each phase of the load.

SAQ 14 A star-connected source feeds a delta-connected balanced load containing 60 + j90 ohms in each phase. If the line currents in the circuit have 4 A effective value, find the phase voltage of the source.

4.4.3 Power Considerations Having considered the methods of analysis of simple balanced three-phase circuits, let us now examine how the power in a 3-phase circuit may be computed.

The total powerlreactive power supplied by a 3-phase source can be computed as the sum of the powerslreactive powers supplied by the three single phase sources. In fact the complex power supplied by the 3-phase source is equal to the sum of the complex powers supplied by the three individual single phase sources. A similar statement is true of loads as well. The total powerlreactive powerlcomplex power drawn by a 3-phase load is the sum of i~~dividual values of powerlreactive powerlcomnplex power drawn by the three impedances constituting the 3-phase load.

Let PA, PB and PC be the values of power consumed in the three phases A, B, C of the balanced star connected load of Figure 4.15.

Then the total power PT consunled by the load would be :

as Vm=VBN=VcN=Vp and I A = I B = I C = I p

But since in a star-connected system Vp = VI,l f i and Ip = IL, a11 alternative expression for PT would be

P , = 3(VL1fi) I, cos a = fiVLIL cos a.

Similarly, the power consumed by the balanced delta-connected load of Figure 4.18 would be

PT = PAR + PBC + PCA = VAB IAB, cos a + VBC IBC cos a + VCA ICA cosa = 3 Vp Ip cos a ,

as VAB = VBC = VCA = VP a ~ ~ d IAB = IBC = ICA = IP'

In a delta-connected circuit, VL = Vp and IL = firP' Hence an alternative expression for PT would be

PT = 3vL(ILlf i ) cos a = f i VL I, cos a.

In summary, we have the following two equivalent expressions for total power consumed by a 3-phase load, irrespective of whether the load is connected in star or delta.

P T = 3Pph = 3Vp Ip cos a

= d3VL I, cos a ,

where Pph is the power in one phase. The formula in Eq. (4.23b) is often used since it expresses PT in terms of externally measurable quantities VL and IL. It is important to note

Introduction to Circuits that even in this expression, u is the angle of the balanced load impedances or the angle of lag of a phase current with respect to the associated phase voltage. It does not represent the angle between a line voltage and a line current.

It can be similarly shown that the total reoctivepower QT and opparentpower ST of the balanced 3-phase load are given by

Q T = 3 V p ~ p s i n a = f i ~ L I L s i n a ; S T = 3 v P I p = f i V,I, (4.24)

Thus

P T = S T c 0 s a ; Q T = S T s i n a = P T t a n a

The power factor (p.fi) of tlie balartced load is defined in a similar manner as in single phase circuits.

P.F. of a balanced load = (PTIST) = cos a , (4.26)

where a is the angle of the balanced load impedances.

When several 3-phase balanced loads are connected on a system the aggregate active power Pa is the sum of the individual three-phase powers. The aggregate reactive power Qa is the sum o l the individual three-phase reactive powers. However the aggregate apparent power Sa is not the sum of the individual apparent powers as the p.f. of each balanced load may be different. S, and the overall p.f. are given by

so=- (4.27)

P a Overall p.f. = - So

SAQ 15 Fill up the blanks

1 The p.1. of a balanced 3-phase load is cosine of the. . . . . . . . . . . . . . . . . . . . . . . . . of impedance in each phase. It is also equal to cosine of the angle between a phase voltage and the corresponding . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2. Furthermore it is the ratio of total power to . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . of the load.

Example 4.6

Find.P, QT and ST taken by the load in Example 4.4. What is the value of p.f. of the balanced load.

Solution

PT = 3Pp,, = 3Vp Ip cos a = 3 x 231 x 1.80 x 100

= 974 W (100% go2)%

Q T = 3 Q P h = 3 V p I p ~ i n a = 3 X 2 3 I X 1 . 8 0 ~ 80

= 779 VAR (1 00' + 80')"

Alternative :

Q T = f i VLIL sin a =OX 400 x 1.80 x 80

= 779 VAR (loo2 + go2)%

P.F. of the balanced load = 1 0 0 d B = 0.78 lagging

Example 4.7

A 3-phase tnotor takes 5 kW at 0.8 p.f lagging from a 3-phase 400 V supply. If the motor is represented by a balanced delta- connected set of impedances, find the impedance in each phase. What is the reactive power taken by the motor?

Solution

PT = 3Vp Ip cos a

5000 = 3 x 400 x Ip x 0.8 3 Ip = 5.21 A

Zp = VJIp = 40015.21 = 76.78 Q - Zp = 76.78 cos a + j76.78 sin a = 61.42 + j46.07 Q

QT = 3Vp IP sin a = PT tall a = 5000 tiin(cos- ' 0.8) = 3750 VAR

Find PT , QT and ST taken by the load in Example 4.5.

SAQ 17 Three impedances each of 100 +j100 ohms are connected in delta across a balanced 400 V, 3-phase supply. Find the total power conwmed by the load and its p.f.

Example 4.8

If three capacitors of C farads each are connected in delta across a 3-phase balanced supply of line voltage VL, find the power, reactive power and apparent power taken by the three capacitors.

Solution

In delta cotu~ection VL = Vp

cos a = 0 and sin a = - 1 for purely capacitive impedances.

PT = 3VPIPcos a = 0

QT = 3VP IP sin a = 3VL (wCVL) (- 1) = - 3wCvL2 VAR

2 ST = I QT I = 3wCVL VA

Note that the reactive power taken by the capacitors is taken as negative by convention.

4.4.4 Power Factor Correction The necessity for and the priilciple of power factor correction have bee11 explained in Unit 3. We now examine this process in relation to 3-phase circuits. In fact it is mainly 3-phase installations that call for power factor correction since they consume bulk quantities of electrical power and since the tariffs for supply of electricity to such installations penalise low power factor operation. As most industrial loads operate at lagging power factors, the process of power factor improvement amounts to connecting a 3-phase load comprising capacitors in parallel with the existing load. Such a capacitive load takes the form of several small capacitors of practical size connected in parallel in each phase, the entire assembly being referred to as a 3-phase bank of capacitors.

It has already been remarked in Section 4.4.3 that when several 3-phase loads are in parallel, the powers and reactive powers of individual loads can be added to obtain the aggregate values of power and reactive power. This principle can be used as the key to the calculation of the required capacitor values for power factor improvement. .

3-Phase Circuits

introduction to Circuits As an illustration, let us consider an electrical installation having two 3-phase 11 kV 50 Hz loads of 100 kW and 400 kW operating at 0.95 p.f. lagging and 0.8 p.f. laggil., respectively. Let it be required to raise the overall p.f. of the installation to 0.98 lagging through connection of a capacitor bank in parallel with the existing loads. The envisaged scheme is illustrated in Figure 4.21, where it is assumed that loads 1 and 2 are connected

T A 1 I

Load 1 Load 2 Capacitor Bank

Fig. 4.21: Use of p.f. correcting capacitors

in star a of the a1

nd d ctual

The work for deducing the required value of C can be organised as follows. 1 1. Load 1 (100 kW at 0.95 p.f. lagging) 100 100 ta n(cos-' 0.95)

= 32.87

2. Load 2 (400 kW at 0.8 p.f.lagging) 400 400 tan(codi 0.8) = 3 00

3. Capacitor bank - Qc

In the above it is assumed that the capacitors are pure and hence loss-free. Since the aggregate p.f. is to be 0.98 lagging, (332.87 + Q,) = 500 tan (cos- ' 0.98) = 101.53. Therefore Qc = - 231.34 kVAR

The above gives the kVAR rating of the required 3-phase capacitor bank operating at 11 kV. Qc can be related to capacitance values by Q, = - 3wcvL2 (vide Exarnple 4.8).

Therefore 3 x (loon) x C x (11 x 1 d l 2 = 231.34 x lo3 =+ c = 2.03 pF.

The apparent power take11 by the illstallatio~l before and after the connection of capacitors can be calculated as follows:

S = [(400 + 1 0 0 ) ~ + (300 + 32.87)2]" = 601 KVA

S' = [(400 + 100)~ ,t 101.53~]' = 510 KVA.

It is seen that for the same active power consumed, the kVA requirement of the installation is reduced by 15% through the use of p.f. correcti~ig capacitors.

Capacitors needed for p.f. correctio~i can be con~iected either in delta or in star. However as the required C value for given line voltage and kVAR is smaller for a delta connection, it is this con~iectio~i which is usually preferred.

SAQ 18 A delta-connected set of 3 capacitors is to be connected in parallel with a motor taking 10 kW at 0.8 p.f. lagging froin 3-phase 50 Hz, 400 V mains so as to raise the overall p.f. to 1 .O. Find the value of each capacitor.

4.4.5 Single Phase and Single-line Representations

Let us consider a balanced 3-phase system. By using the delta- star equivalence, where required, we can reduce the system to one which contains only star-corurected sources and star-connected loads and which is of course balanced. You would have noticed from our earlier work that the analysis of such circuits is no more involved than the analysis of a single phase circuit. If the currents and voltages in one-phase of the system (say A phase) are calculated, then those in the other phases can be readily deduced by adding appropriate phase shifts. To avoid unnecessary repetition and to sinlplify the circuit and phasor diagrams, we may therefore represent the entire 3-phase circuit by a single phase circuit incorporating only one phase (say A phase) of all sources, loads and supply lines. The results of analysis of such a single-phase circuit can be readily extended later to the original 3-phase circuit, to determine the currents and voltages of other phases if needed.

Figure 4.22(a) give the single-phase representation of the 3-phase circuit of Figure 4.21. Here only theA phase of the equivalent Y-connected system is shown. Let us calculate the currents in this circuit and draw the phasor diagram

t - Let v, = 1100o/f i L 0

Noting that Z1 consumes one-third of the total power of load 1 of Figure 4.21,

- loox lo3 1 I' = 3

x-x- 11000 0.95

L - cos- ' 0.95 = 5.52 L - 18.2"

Fig. 4.22: Single-phase representation of the circuit of Fig. 4.20 (a) Circuit diagram (b) Phasor diagram

Similarly from specifications of load 2,

= 20.98 - j15.76 - Ic = jIC - - - - I, = I, + I, +Ic = 26.22 - j(17.48 - I,)

If the ove-rall p.f. is to be 0.98 lagging

The phasor diagram is given in Figure 4.22(b).

If the p.f. correcting capacitors are to be delta-connected, their impedance is to be 3-times that in an equivalent star configuration. Since the impedance is inversely proportional to capacitance value, the required capacitancelphase in the delta configuration is 2.03 pF, which agrees with the analysis in Section 4.4.4.

3-Phase Circuits

Lntroduction to Circuits In the equivalent star-connected 3-phase balanced system a neutral connection between the source and the load is not required. Even if it is present in a 4-wire system it does not carry any current. Hence the single-phase representation of Figure 4.22 can be further simplified to the one shown in Figure 4.23, in which the neutral connection is dispensed with. This is the single-line representation of the 3-phase circuit of Figure 4.21. This is only a simplified way of representation. The actual analysis is done on per-phase basis as indicated earlier. The single-line representation is widely used in power system work, to simplify the presentation of data relating to sources, feeder lines and loads in a complex system. It is a convention to indicate the line voltages and values of total 3-phase power in such representations.

11 kV Sourcc 100 kW 0.95 p.t 400 kW 0.8 p.f. Delta connected lagging Star lagging Delta Capacitor bank connected load connected load

Fig. 4.23: Single-line representation of the balanced circuit of Fig. 4.21.

SAQ 19 Fill up the blanks :

The single-phase representation of 3-phase circuits is useful only for. ............ ........... ' 3-phase systems. In this representation all sources and loads are assumed to be connected in . ................................. 2. If the current in a particular element in this representation is 6 L20° A, then the three-phase

............. currents in the original 3-phase system at the same location would be 3 ........................... . . . . . . . . . , ........................ and A.

Example 4.9 Three equal impedances each of 120 - j60 SZ are connected in delta across a 400 V, 3-phase circuit. At the same point, three other equal impedances of 60 + j80 SZ are connected in star across the same supply. Determine (a) the line current @) total power supplied (c) p.f. of the combined circuit.

Solution

In the single phase representation of an equivalent Yconnected system, shown in Figure 4.24,

Fig. 4.24 for Example 4.9

Taking the source voltage to be the reference,

Line current = 6.02 A

Total power supplied = 3 [ ~ e ( G * ) ] = 3Re[(400/1@) x (6 - j0.46)]

= 3 x (400 /n ) x 6 = 4157 W

P.F. = cos 4.4" = 0.997 leading.

Example 4.10 Show that the total instantaneous power taken by a balanced load in a balanced system is a constant and is free from pulsations.

Solution

Let the load p.f. be cos a and let the three phase voltages and three phase currents be

vA = QVp sin wt ; vg = f i V p sin (wt - W 3 )

v, = fi Vp sin (wt + N 3 )

iA = nIP sin (wt - a ) ; iB = f i ~ ~ sin (wt - a - W 3 )

ic = nip sin (wt - a + N 3 )

Instantaneous power in the 3-phases :

pA = vAiA = 2Vp Ip sin wt sin (wt - a )

= Vp Ip [cosa - cos(2wt - a ) ]

p~ = vBiB = 2Vp Ip sin (wt - 2x13) sin (wt - a - N 3 )

= Vp Ip [COS a - cos (2wt - a - 4n/3)]

Similarly

pc = vcic = Vp Ip [cos a - cos (2wt - a + 4n/3)]

Total instantaneous power = p A + p~ + p c

= 3Vdpcos a ,

the other three terms cancelling out as they form a balanced set.

Thus the total power is a constant and does not fluctuate. This is why a 3-phase , motor draws a constant value of power from the supply and develops a constant

torque when running at a steady speed. I

SAQ 20 Three equal impedances are connected in star and fed from a 300 V, 3-phase delta-connected source. If the current in each phase of the source is 5 A, find the magnitude of the load impedances.

3-Phase Circuits

SAQ 21 Three impedances each of 40 - j60 ohms are connected in star across a 400 V, 3-phase supply. Find the load p.f. and the power and reactive power drawn by the load.

SAQ 22 What should be the value of three equal impedances connected in delta, which would draw the same powrr at the same p.f. from the 400 V supply as in SAQ 217

Introduction to Circuits SAQ 23

A 3-phase balanced delta-connected load comprising 100 +jX ohms in each phase draws a power of 7.5 kW from a balanced 600 V supply. Find the line current and p.f. of the load.

SAQ 24 A star-connected 3-phase inductioil motor takes a line current of 30 A at 0.8 p.f. lagging from 440 V, 50 Hz, 3-phase supply. A 3- phase delta-connected bank of capacitors is used to raise the overall p.f. to unity. Calculate the kVA rating of the capacitor bank and the value of capacitance in each phase.

SAQ 25 The electrical load in a factory comprises (i) 10 kW of lighting load at unity p.f. (ii) 100 kW of nlotor load at 0.8 p.f. lagging and (iii) 40 kW of other loads at 0.9 p.f. lagging. Calculate (a) the overall p.f. of the load and (b) kVAR rating of capacitors to bring the overall p.f. to unity.

4.5 SUNIMARY

In this uilit; you were first introduced to the basic features of a 3-phase systein and of a 3-phase generator. You then learnt the superior features of a 3-phase systein in comparison with a single-phase system and gained an appreciation of why the former is universally adopled in commercial power sysrems.

You were then introduced to the concept of balailced 3-phase voltages and currents and the concept of phase sequence. You noted that in a set of 3 balanced quantities with a given phase sequence, symmetry permits us to deduce any two quantities - in either time or phasor domain - from a knowledge of the third. You also understood the distinctions between star and delta connections of sources and loads, between 3-wire and 4-wire 3-phase systems, between phase and line quantities and between balanced and unbalanced 3-phase systems. You noted that for balanced systems

the phase and line currents are equal in a star-connected unit but the line voltage is fi times the phase voltage;

the phase and line voltages are equal in a delta-connected unit but the line current is fi times the phase current.

Analysis of simple balanced 3-phase circuits containing either a star or a della-connected load was than considered. You noted that the neutral wire in a 4-wire balanced system does not carry any current and that in anunbalanced systetn it serves to maintain voltage balance while carrying the out of balance current in the line conductors. You observed

that a delta-connected balanced load call be replaced by an equivalent star-connected load with one-third the value of impedances and similarly a delta-connected source can be replaced by an equivalent star-connected source maintaining the same line to line voltages. You saw that a 3-phase circuit in which all sources and loads are star-connected is adequately represerlted by a 1-phase circuit which mirrors one of the three phases. From an analysis of this I-phase circuit, all the relevant quantities in the original 3-phase circuit are easily deduced.

YOU saw that the power in a three phase balanced load irrespective of whether it is star or delta-connected, is equal to 3VpIp cos a or 6 V L I L cos a. Here cos a is defined as the p.f. of the balanced load and is equal to the cosine of the angle of the three equal impedances constituting the 3-phase load. You learnt that reactive power can be similarly computed as 3VJp sin a or f i ~ ~ ~ ~ sin a. You noted that with several 3-phase loads in parallel across thc same source, the respective powers and reactive powers (and not the apparent powers) can be superposed. Finally you saw how a 3-phase capacitor bank can be used to reduce the overall reactive power a ~ ~ d thereby improve the p.f. of an installation. You also learnt how to calculate the kVAR rating and value of capacitors needed for a specified a~llount of p.f. improvement.

3.Phase Circuits

SAQ 1 : A single-phase generator has two tenninsls and produces a single output voltage between the two terminal. A 3-phase generator is a composite unit comprisi~~g 3 single phase generators, each generator producing a voltage which has fixed phase differences with the other two. The three 1-phase generators are connected internally in star or delta and the resulting 3-phase generator has 3 or 4 ex tend tenninals.

SAQ 2 : (1) kVA rating (2) conductor material (3) constant (4) pulsating

SAQ 3 :

(1) False (2) False. Vc leads c' by 120" (angle of lag / lead is limited to 180") (3) False. The converse of the statement above Eq. (4.5) is not necessarily true. For example, iA, iB and ic witb iA = - iB and ic = 0 do not form a balanced set.

SAQ 4 :

vB = a ~ s i n ( w t + e + ~ 3 )

vc = fi V sin (wt + 8 -,27c/3)

i, = \/zI sin (wt + (3 + b / 3 )

i, = \rzl sin (wt + (3 - h l 3 )

SAQ 5 :

(1) 100 (2) 285" (3) 165" (4) 100 (5) 4 (6) 180" (7) 4 (8) -60" SAQ 6 :

(1) star (2) delta (3) 3-wire

SAQ 7 :

(a) False (b) True.

SAQ 8 : (1) line voltages (2) phase voltages (3) line voltage (4) phase voltage (5) delta (6) line current (7) phase current.

SAQ 9 :

(a) A balanced 3-phase voltage source is constituted by the symmeti~al connection of three single-phase sources with equal effective values of voltage and 120" phase difference between any two voltages.

(b) A 3-phase load connected in deltalstar is balanced if the three impedanr constituting it have equal complcx values of impedance.

1 Introduction to Circuits (c) A 3-phase circuit is balanccd if it is conqtituted by balanced sources of the

same phase sequence and balanced loads.

SAQ 10 :

The net voltage would be 200 V as shown below: - Eo= -G +& + & = - 100LOO + 100L - 120" + 100L +120°

= 100 L 180" + 100 L 180" = 200 L 180" (vide Figure).

Figure for answer to SAQ 10

This voltage would drive a large circulating current in the loop, limited only by the internal impedances of the sources and can cause damage to the generator.

SAQ 11 : Converting the star into a delta, we have - ZA, = 32, = 300 + j300 ohms

Now we have two delta connected loads in parallel as shown in thc figure.

Figure for answer to SAQ 11

Across each pair of lines, FA, and ZA2 are in parallel.

They can be combined into

The equivalent load has three 150 Q-resistors connected in delta.

SAQ 12 :

The neutral wire permits the connection of single phase loads designed to operate at a voltage of 1 / a times the line voltage. The neutral wire serves to balance the phase voltages at the load even if the load impedances are unbalanced. The current in the neutral wire in a balanced system is zero.

SAQ 13 :

VL= lOOOV; V p = 1000/&V; I p = I L = 5 A

I The phase current lags the corresponding phase voltage by the impedance angle. 3-Phase Circuits

Therefore, the impedance angle here is negative and

2 = (200/G) L - 60" = (200/\r5) (cos60° - j sin60°) = 57.7 - j 100 CZ

SAQ 14 :

In a delta-connected load, Ip = I f l= 4 / f l A

Phase voltage of the load = ( 4 m ) x = 250 V

This is also the line voltage of the system as the load is deltacoimected.

Phase voltage of the star-connected source = 2501G = 144 V

Alternative method:

The equivalent star-connected load has an impedance of 20 t j30 ohms in each phase. This also draws the same line current from the source. Hence phase voltage of

the star-connected load is 4- = 144 V, which is also the phase voltage of the source.

SAQ 15 :

(1) angle (2) phase current (3) total apparent power.

SAQ 16 :

PT = fi x 400 x 5.4 x 100/(1002 + 80')' = 2921 W

QT fi x 400 x 5.4 x 80/(1002 + 8d)' = 2337 VAR

S T = G x 400 x 5.4 = 3741 VA

SAQ 17 :

P.F. = 100d- = 0.707 lagging

Ip = 400- = 2.83A

PT = 3 Vp Ip cos a = 3 x 400 x 2.83 x 0.707 = 2400 W

SAQ 18 :

Q taken by motor = 10000 tan (cos"0.8) = 7500 VAR (lagging)

Q of capacitors = (-)7500 VAR (leading) = (-)3 WCV;

C = 7500/3 x 100n x 4002 F = 49.7 C L ~ .

SAQ 19 :

I (1) balanced (2) star (3) = 6 L 20" (4) & = 6 L - 100" (5) 6 = 6 L 140"

SAQ 20 :

(Ip)soume = 5 A (ZJsource = 5fl A = (IP),,

(VL),~~,, = (V~Ilard = 300 v ; (V~)lOod = 3001fi v (vp)l0od 300 1 dT= 2o Magnitude of load impedances = - - (Ip)lOad - 5 6

SAQ 21:

Load p.f. = 40n/402 + 60' = 0.555 leading

Q = G V d L sin a = - x 400 x 3.20 4 1 - (0.555)~ = - 1846 VAR

Since the load is capacitive, the reactive power is negative.

Introduction to Circuits

SAQ 22 :

SAQ 23 :

Power in each phase = 2500 W = lp2 x 100 * I p = 5 A

Load p.f. = 7500

3 x m o x 5 \ / 5 = 0.833

The p.f. is lagging or leading depending on whetherX is positive or negative.

SAQ 24 :

Q,,, = fi x 440 x 30 v'i - ( ~ . 8 j ~ - 13,X 8 VAR

The capacitors should draw -13718 VAR

kVA rating of capacitor bank = 13.718 kVA

SAQ 25 : Pa,,,,,= 10+ 100+40= 150 kW

- d q i $ m Qaggmgate = 0 + 100 0.8 + 40 = 94.37 kVAR 0.9

1 94 37 Overall p.f. = cos [tan- (--)I - 0.85 lagging 150

To raise overall p.f. to 1.0, the capacitors should draw a reactive power of -94.37 kVAR.

kVAR rating of capacitors required = 94.37 kVAR