# Chapter 8: Sinusoidal Steady-State in Three-Phase Circuits – Electric Circuit Analysis

## Sinusoidal Steady-State in Three-Phase Circuits

##### CHAPTER OBJECTIVES
• To define and explain balanced and unbalanced three-phase systems and to extend the concepts of active power, reactive power and complex power to such systems.
• To develop and employ relations between line and phase quantities in Y-connected and Δ-connected three-phase circuits.
• To emphasize constraints on line quantities in three-phase three-wire and four-wire systems.
• To develop a systematic procedure to analyse sinusoidal steady-state in three-phase balanced circuits using Y-equivalent and single-phase equivalent circuits.
• To show how to use phasors, phasor equivalent circuits and phasor diagrams for solving three-phase circuits under sinusoidal steady-state condition.
• To introduce and illustrate the problem of neutral-shift voltage in four-wire systems.
• To illustrate circuit analysis procedures for unbalanced three-phase circuit analysis through solved examples.
• To introduce and explain symmetrical components in detail and use it in analysis of unbalanced circuits.
• To develop power relations using sequence components.
##### INTRODUCTION

Electrical power generation, transmission and distribution employ a particular system of three sinusoidal voltages/currents with certain symmetry in voltage/current magnitude and relative phase. This system of three sinusoidal voltage/current waveforms is called a three-phase voltage/current and a circuit employing a three-phase voltage/current system is called a three-phase circuit. A three-phase circuit, like any other circuit, will have to go through a transient period subsequent to the application of source functions before it can settle down to a sinusoidal steady-state condition. We focus on analysis of three-phase circuits under sinusoidal steady-state condition in this chapter.

No new concepts are required for analysis of three-phase circuits under sinusoidal steady-state conditions. It requires only the application of all the concepts we developed in the last two chapters. But before we take up the analysis of three-phase circuits, let us see why electrical engineers thought up three-phase system in the first place.

##### 8.1 THREE-PHASE SYSTEM VERSUS SINGLE-PHASE SYSTEM

Industrial heaters are used in various material processing industries for a variety of heating applications. Typically, heater boxes that contain heating elements powered by AC supply, transfer the heat developed to a gaseous (usually air) or liquid medium which conducts the heat to the material that is to be heated up. Resistor elements made of some specially constructed wire are laid out uniformly inside the heater box to develop uniform heating of the contact surface. Many such heating elements are finally connected in parallel to form the heater resistive load. Consider such a heater which, after all parallel connection of sub-elements, is finally represented by three equal resistors of R each. These three resistors may be connected in parallel and supplied from a single-phase AC source or they may be supplied individually from different AC sources – for instance, for redundancy and reliability. Fig. 8.1-1 (a) Three equal resistors powered from a single-phase supply (b) Three equal resistors supplied from three identical sources

The circuit in Fig. 8.1-1(a) shows the three heater elements tied in parallel and supplied from a single-phase AC supply. The rms value of source is specified as VP. Let r be the resistance of the connecting cable and 3I be the phasor current in the cable. r is usually small compared to R and hence we can take the total circuit resistance as R/3 ignoring the contribution 2r. Therefore, circuit current magnitude ≈ 3VP/R A rms. Therefore, magnitude of I is ≈ VP/R A rms. The total power loss in the cable is ≈18r (VP/R)2 W.

The circuit in Fig. 8.1-1(b) shows the same three heater elements supplied from same single-phase supply – but with sources shown as three identical sources. Since the current in the connecting cable is only one-third of the earlier value we will choose a cable that has one-third the area of cross section that the cable in the circuit in Fig. 8.1-1(a) had. This implies that the cables in the circuit in Fig. 8.1-1(b) have a resistance of 3r each. Thus, the total power loss in the circuit in Fig. 8.1-1(b) is ≈6 × 3r × (VP/R)2 = 18r (VP/R)2 W again. Fig. 8.1-2 From single-phase circuit to three-phase circuit

Next, we imagine that the negative polarity terminals of all the three sources are joined together to form a common node and that that a single cable connects this node to a similarly formed node on the load side as shown in the circuit of Fig. 8.1-2(a). The return currents from the three load branches flow through this cable. Then, this cable must have three times the cross-sectional area of the other cables in circuit in Fig. 8.1-2(a). Therefore, its resistance will be r. The power loss in four cables put together will again be 18r (VP/R)2 W.

Non-uniform heating will result if the three branches get different powers delivered to them. Therefore, we wish to keep the power dissipated in the three resistors equal. Therefore, the three sources must have the same rms voltage. But that does not mean that they should be in-phase too. We are not particularly worried about the relative phase of currents flowing in the three branches of a heater resistance. However, is there any advantage in making the phase angles non zero?

We assume in circuit of Fig. 8.1-2(b) that the phase of source connected at the terminal identified Y is at an angle α with respect to the source connected at terminal marked as R. Similarly, the phase of source connected at the terminal identified as B is at an angle β with respect to the source connected at terminal marked as R. Moreover, we assume that now we are using an extra-thick cable between node N to a similar node on the load side such that it is virtually a resistance-free connection. Now, the currents flowing out from the sources have different phase; but same magnitude of VP/(R + 3r) A rms. Note that we use ‘R’ as a terminal marking and ‘R’ as symbol and value for a resistance. The cable connecting the node N to a similar node on the load side will have a current In in it. Why do not we make the magnitude of this current zero so that the cable power loss will become zero in the corresponding cable even if we reduce the size of that cable? We try to find a (α, β) pair such that magnitude of In is zero. α and β should be such that (sinα + sinβ) = 0 and (cosα + cosβ) = −1 simultaneously. First equation implies that β = α and with this constraint the second equation results in α = ±120°. Therefore, (α, β) = (–120°, 120°) or (120°, –120°) are the two choices for (α, β) pair such that the current In = 0.

If we make one of these two choices, the total power delivered to the resistors remains the same; however, the total cable loss becomes ≈9r (VP/R)2 W, i.e., only 50% of the cable loss in the case of single-phase supply.

If a cable (or any resistance for that matter) carries only zero current in a circuit, then, that cable can be removed without affecting the behaviour of any circuit variable in the circuit. This implies that we can remove the cable connecting the common point N of the sources to the common point in the load without affecting the load currents or load power. This is possible since the instantaneous currents in the three lines add up to zero at all t leaving only zero current in the common point link. Let us verify this with α = –120° β = 120°. Hence, no wire is needed between the common point of sources and common point of load. Therefore, we need only three wires of 3r Ω each in circuit of Fig. 8.1-2(b) instead of two wires of r Ω each as in circuit in Fig. 8.1-1(a). Cross-sectional area of wires in circuit of Fig. 8.1-1(a) will be three times that of the wires in circuit of Fig. 8.1-2(b). Therefore, the amount of copper (volume or weight) we use in wiring up circuit in Fig. 8.1-2(b) is only 50% of the copper that we use in wiring up circuit in Fig. 8.1-1(a).

A set of three sinusoidal quantities, all at the same frequency, with equal peak (and hence rms) values and shifted successively by 120° in phase is defined as a Balanced Three-Phase Quantity. Therefore, if x1(t), x2(t) and x3(t) is a three-phase set, then, and x1(t) + x2(t) + x3(t) = 0 for all t.

Each limb or branch of a three-phase system (source or load) is termed as a phase. R, Y and B are used to designate the line terminals of a three-phase source or load.

If the peak values are unequal and/or successive phase shifts are different from 120°, the set will be called an Unbalanced Three-Phase Quantity.

A balanced three-phase system requires only 50% of the copper and incurs only 50% of the power loss to transmit a given amount of power to load compared to a single-phase system.

The sub elements of an industrial heater designed for three-phase operation are sometimes arranged in such a way that adjacent sub-elements belong to different phases. That is, if three sub-elements are adjacent, the first one will be a part of R that gets connected to R-line, the second one will be part of R that gets connected to Y-line and the third one will be part of R that gets connected to B-line. Then, the cycle repeats. This is, obviously, done on purpose. We discuss the sum of instantaneous power delivered by the three phases of a three-phase system under balanced operation to appreciate the purpose involved.

Let us assume that the impedance per branch (i.e., phase) is Z = Z∠θ and that Y-phase voltage lags R-phase by 120° and B-phase voltage lags Y-phase voltage by 120°. Then, the instantaneous currents in three lines can be expressed as where Im = Vm/Z.

Also,     vRN(t) = Vm cos(ωt), vYN (t) = Vm cos(ωt –120°), and VBN (t) = Vm cos(ωt +120°).

where Now, the instantaneous powers delivered by the sources in each phase are expressed as The sum of instantaneous powers can now be expressed as A phase angle of –240° is same as +120° and a phase angle of +240° phase is same as –120°. Therefore, the three terms in the sums enclosed by curly brackets in the preceding equation form balanced three-phase sets. Sum of a three-phase balanced set is zero for all time instants. Therefore, If the power dissipated in the three branches of a balanced three-phase load acts on the same physical system, then, a balanced three-phase source delivers active power to the load with zero power pulsation. The double-frequency power pulsation in the case of a single-phase AC supply can never be reduced to zero.

Some instances in which the total active power from three branches of a three-phase load acts on the same physical system are industrial heaters and three-phase motors. In a three-phase heater, power dissipated in three branches will heat up the same body. In a three-phase motor, the torque developed by the branch windings act on the same mechanical shaft. This leads to steady heating and steady temperature in a heater and speed without fluctuations in the case of a motor.

The next point of superiority of three-phase system over single-phase system has to be accepted as a statement of fact. It concerns matters beyond the scope of a book on circuits. It is the fact that Electrical Machines and Power Systems Components are more efficiently designed in three-phase version rather than in single-phase version in all senses of the word ‘efficiency’ – efficiency in material utilisation, efficiency in power utilisation etc. Hence, we conclude:

A balanced three-phase system comprising balanced three-phase sources and balanced three-phase loads is superior to a single-phase system thanks to the following facts:

1. Power loss in transmission system is lower in three-phase system.
2. Copper utilisation is superior in three-phase system.
3. Power delivered to a balanced three-phase load by a balanced three-phase supply is free of pulsation.
4. Electrical equipment designed for three-phase operation is more efficient than their single-phase counter parts.
##### 8.2 THREE-PHASE SOURCES AND THREE-PHASE POWER

Three-phase sources and loads can take two forms – star connected or delta connected. The star connection is also referred to as Y-connection. The delta connection is referred to as Δ-connection and mesh connection too.

#### 8.2.1 The Y-connected Source

Two options to construct a three-phase Y-connected source are shown in Fig. 8.2-1.

In Fig. 8.2-1(a) the terminal identified by Y has a voltage phasor that lags the voltage phasor of terminal identified as R by 120°. Further, the terminal identified as B has a voltage phasor that leads the R-line by 120° – i.e., B-line voltage lags R-line voltage by 240°. Thus, similarly located points on the waveform will appear first in R-line, then in Y-line after one-third of a cycle period and finally in B-line after another one-third of a cycle period. See the waveforms in Fig. 8.2-1(a). Thus, the sequence of appearance of peaks will be …–RYB–RYB–… A three-phase source that has this sequence of appearance of similarly located waveform points is called a positive sequence three-phase source. Note that the sequence can also be written as …–YBR–YBR–… or BRY–BRY–…

The source in Fig. 8.2-1(b) illustrates a negative phase sequence source. In this case, the voltage of B-line lags that of R-line by 120° and voltage of Y-line lags that of R-line by 240°. Thus, the order of appearance of similarly located waveform points in time will be …–RBY–RBY–… (equivalently, …–BYR–BYR–… or …–YRB–YRB–… too) in this case.

There are loads that cannot distinguish between the two sequences – a three-phase industrial heater is one. There are loads that can distinguish between the sequences – a three-phase induction motor is one. If it rotates in clockwise direction for positive sequence applied voltage, it will rotate in counter-clockwise direction for negative sequence applied voltage. So will a three-phase synchronous motor. The three-phase sources we consider from this point onwards will be positive sequence sources by default.

The voltage VP used in Fig. 8.2-1 is an rms value. We will specify three-phase source in rms values in this chapter.  Fig. 8.2-1 (a) A positive sequence Y-connected three-phase source (b) A negative sequence Y-connected three-phase source

The Y-connected source has a common point at which the three phase-sources join. This point is called the ‘neutral’ point of the source. Phases in the context of three-phase source or load refer to the three branches that constitute the three-phase system. Lines refer to the three terminals – R, Y and B – that come out of the three-phase source or load and are available for external connections. Thus, in the case of Y-connected source, the phase-source is connected between a Line and the Neutral. Therefore, the phase-source voltage is same as Line-to-Neutral voltage in the case of Y-connected source. As a matter of common practice, phase-source voltage is shortened to phase voltage. Beginners often get confused between phase voltage (i.e., phase-source voltage or phase-load voltage) and line-to-neutral voltage and conclude that they are the same. They are not the same always. Phase voltage is the voltage across phase source or phase load. Phase voltages exist for any three-phase source or load. Line-to-neutral voltage will exist only if there is a neutral point – i.e., only for Y-connected source or load.

Though line-to-neutral voltage does not exist unless it is a Y-connected system, line-to-line voltages exist for all three-phase systems. VRY is the phasor voltage of the R-line with respect to the Y-line. VYB and VBR are similarly defined. Line-to-line voltage is usually shortened to line voltage.

Similarly, phase current is the phase-source current or phase-load current flowing in one of the branches that constitute the three-phase source or load as per passive sign convention. Line current is the current that flows out of or into one of the three terminals R, Y or B. Line current is usually taken to be flowing out of a source line terminal and flowing into a load line terminal. Refer to Fig. 8.2-2 that shows all the phase and line quantities in a Y-connected source. VRN, VYN and VBN are the phase voltages and IRN, IYN and IBN are the phase currents. There is a neutral point in an Y-connected source. Hence, line-to-neutral voltages exist and they are the same as phase voltages. Line currents are IR, IY and IB and are assumed to flow out of the source towards the load. In a Y-connected source, the line currents defined in this manner are equal to negative of phase currents defined as per passive sign convention.

Since the R-line is same as positive terminal of the source that feeds the R-line in a Y-connected source, this source may be referred to as the R-phase source. Similarly for other two sources too. Fig. 8.2-2 A Y-connected source with all phase and line quantities identified

We start at the Y-line terminal and traverse through the two sources, reach R-line terminal and move to Y-terminal by falling through VRY. The KVL equation we get is VRY = VRNVYN. Refer to the phasor diagram shown in Fig. 8.2-2. VRN is taken as reference and a horizontal line of length VP (the rms value of phase sources) to suitable scale is drawn to represent it. The remaining two phase voltages –VYN andVBN – are at –120° and –240° positions as shown. Now, the equation VRY =VRNVYN is implemented by rotating VYN by 180°, moving a copy of VRN to the tip of –VYN and completing the triangle. The result is the first line voltage VRY.

We observe by projecting the lengths of VRN and –VYN on to VRY that magnitude of VRY is 2 cos30° VP = √3 VP. We represent the magnitude of line voltage by VL. Hence, VL = √3 VP. We observe further that the first line voltage, VRY, leads the first phase voltage VRN by 30°.

The remaining two line voltages are also obtained by using equations VYB = VYNVBN and VBR = VBNVRN. Obviously, the three line voltages form a three-phase balanced set of voltages.

The line voltages in a balanced Y-connected source form a balanced three-phase set of voltages that are √3 times in magnitude and 30° ahead in phase with respect to phase voltages.

The line currents in Y-connected source need not be cophasal with the phase voltages. That depends on the load. Assuming that the load is a balanced one, the line currents themselves will form a three-phase balanced set of phasors that are displaced in phase by a phase-lag angle θ with respect to respective phase voltages.

Thus, IR = ILθ, IY = ILθ 120° and IB = IL–θ + 120°, where IL is the rms value of line current. These current phasors are shown in relation to phase voltage phasors in Fig. 8.2-3(b). Also shown are the phase positions of line voltages with respect to phase voltages in Fig. 8.2-3(a). We make the following observations from these phasor diagrams.

1. The line current lags the phase-source voltage by θ and it lags behind the corresponding line voltage by (30° + θ).
2. Each source contributes an active power of VpIL cosθW. Hence, total three-phase active power delivered by the three-phase source = 3 VPIL cosθ = √3 VLIL cosθ W where θ is the angle by which the ‘negative of phase current defined as per passive sign convention’ (i.e., –IRN in the case of R-line) lags behind the phase voltage.
3. Each source contributes a reactive power of VPIL sinθ W. Hence, total three-phase reactive power delivered by the three-phase source = 3 VPIL sinθ = √3 VLIL sinθ VArs, where θ is the angle by which the ‘negative of phase current defined as per passive sign convention’ (i.e.,IRN in the case of R-line) lags behind the phase voltage. Equivalently, θ is the angle by which the current delivered by a phase source lags behind the phase voltage.
4.  Therefore, the complex power delivered by the three-phase source S = √3 VLILθVA. Fig. 8.2-3 (a) Phasor diagram of phase and line voltages of a Y-connected source (b) Phasor diagram of line currents and phase voltages in Y-connected source

#### 8.2.2 The ∆-connected Source

The generators in a power system are almost invariably Y-connected. However, power systems use Y-Δ transformers and a Δ-connected three-phase source can be used to model the secondary side of such a transformer. A Δ-connected three-phase source and the phasor diagram of its voltages and currents are shown in Fig. 8.2-4.

We assume that the load on the source is balanced. VRY is taken as the reference phasor for phasor diagrams.

Note that phase voltage and line voltage are the same for this source and that there is no neutral point and hence no line-to-neutral voltage can be defined for this source. Applying KCL at the three corners of delta, we get the following equations. This system of equations has no unique solution since we can add a constant to IRY, IYB and IBR without affecting these equations. Such a constant added to the three currents will represent a circulating current within the delta-loop. Such a circulating current is possible only since three ideal sources are connected in a loop. However, in practice, all the sources in the delta will have some small impedance or other which will force the condition that (IRY + IYB +IBR)Z = 0 where Z is the small source impedance of the sources. Assuming that the circulating current component in delta is zero, let the current –IRY be IP∠–θ. Then, the remaining two phase currents will be –IYB = IP∠–θ–120° and –IBR = IΡ∠–θ + 120°, where IP is the magnitude of phase currents.

Now, IR can be constructed by IR = –IRY– (–IBR). This is shown in the phasor diagram in Fig. 8.2-4. The resulting IR = √3 IP ∠–(30° + θ) A rms. Fig. 8.2-4 A delta-connected source and its phasor diagrams

Hence, the line current magnitude is IL= √3 IP in a delta-connected three-phase source.

The line currents delivered by a balanced Δ-connected source form a balanced three-phase set of currents that are √3 times in magnitude and 30° behind in phase with respect to phase currents delivered by the phase-sources.

We make the following observations from these phasor diagrams.

1. The delivered phase current lags the phase-source voltage by θ and line current lags behind the corresponding line voltage by (30° + θ).
2. Each source contributes an active power of VLIP cosθ W. Hence, total three-phase active power delivered by the three-phase source = 3 VLIP cosθ = √3 VLIL cosθW, where θ is the angle by which the ‘negative of phase current defined as per passive sign convention’ (i.e., –IRY in the case of RY-phase source) lags behind the phase voltage.
3. Each source contributes a reactive power of VLIP sin θW. Hence, total three-phase reactive power delivered by the three-phase source = 3 VLIP sinθ = √3 VLIL sinθ VArs where θ is the angle by which the ‘negative of phase current defined as per passive sign convention’ (i.e., –IRY in the case of RY-phase source) lags behind the phase voltage. Equivalently, θ is the angle by which the current delivered by a phase-source lags behind the phase voltage.
4. Therefore, the complex power delivered by the three-phase source S = √3 VLILθVA.

Therefore, we state the following conclusion:

The three-phase complex power delivered by a three-phase source is given by

S = √3 VLILθ = [√3 VLIL cosθ + j √3 VLIL sinθ] VA

where θ is the angle by which the current phasor delivered by a phase source lags behind its voltage phasor. This relationship is independent of whether the source is Y-connected or Δ-connected. It should be noted that θ is not the angle between line voltage phasor and line current phasor.

Electrical Power Engineering adopts a convention to specify a three-phase source by specifying its line-to-line voltage in rms value. We follow this convention unless stated otherwise in the remaining portions of this chapter. Thus a 400 V, 50 Hz source means a source that has a line-to-line voltage magnitude of 400 V rms at 50 Hz. If it is Y-connected its phase voltage will be 400/√3 V rms and peak of its phase voltage will be 400√2/√3 V.

##### 8.3 ANALYSIS OF BALANCED THREE-PHASE CIRCUITS

Balanced three-phase circuits comprising a balanced three-phase source and a balanced three-phase load can be of four varieties, viz., –Y–Y, Y–Δ, Δ–Y or Δ–Δ. We develop a common analysis procedure for all the possible configurations.

#### 8.3.1 Equivalence Between a Y-connected Source and a Δ-connected Source

Let VRY = VL∠0°, VYB = VL∠–120° and VBR = VL∠120° be the three line voltages observed in a three-phase circuit and let IR = IL(θ + 30°), IY = IL(θ + 150°) and IB = IL∠–(θ–90°) be the observed line currents where the observed phase lag of first line current is expressed as 30° plus some angle designated by θ. If the source is within a black box and we are to guess whether it is a Y-connected source or a Δ-connected source, we will find that we will not be able to resolve the matter using the observed line voltage phasors and line current phasors. This is so because both configurations shown in Fig. 8.3-1 will result in these observed line voltages and line currents. Fig. 8.3-1 Illustrating equivalence between a Y-connected source and a Δ-connected source

Hence, for every set of line voltages and line currents observed in a three-phase circuit, there exist a delta-connected source and a star-connected source which will justify the observed line quantities and which are indistinguishable from outside.

Therefore, if a balanced three-phase source is really Δ-connected, it may be replaced by an equivalent Y-connected source for purposes of analysis. The details of phase-source currents in the real delta-connected source can be obtained after the circuit problem is solved using the star equivalent source.

But there is nothing new in this equivalence – we already know that any Y-connected set of impedances can be transformed into a Δ-connected impedance and vice versa. Therefore, any given three-phase balanced Δ-connected impedance can be converted into a three-phase balanced Y-connected impedance such that the terminal voltage and current behaviour remain unaffected.

The individual branch currents in the branches of delta in the real delta-connected load can be obtained after line quantities have been obtained by using its equivalent Y-connected model. Three-phase symmetry and 1/√3 factor connecting line currents and phase currents in a delta-connected system can be used for this purpose. The impedance to be used in Y-connected load is 1/3 times the impedance present in Δ-connected load. See Fig. 8.3-2. #### 8.3.3 The Single-Phase Equivalent Circuit for a Balanced Three-Phase Circuit

Any balanced three-phase circuit can be equivalenced to a Y–Y balanced three-phase circuit by replacing Δ-connected sources and loads (if any) by their Y-connected equivalents developed in the last two subsections. The equivalence is valid as far as line voltages, line currents and complex power flow are concerned.

Thus, analysis of balanced three-phase circuits reduces to a three-step procedure. The Δ-connected sources and loads are replaced by their Y-connected equivalents in the first step. The resulting balanced Y-connected circuit is solved for line voltages and line currents everywhere in the second step. The relevant line voltage values and line current values are used to determine the phase-source voltages and currents in the case of those Δ-connected sources that were replaced by star equivalents earlier and phase-load voltages and currents in the case of those Δ-connected loads that were replaced by star equivalents earlier. This will constitute the third step.

The difference between a real Y-connected source/load and Y-connected equivalent of Δ-connected source/load is that the neutral point of the first is accessible for connection, whereas the neutral point of the second is an inaccessible node. However, the neutral connection in a balanced three-phase circuit does not carry any current (even if the neutral link has non-zero impedance) and hence the voltage difference between source neutral and load neutral will remain zero whether we connect the neutral points together or not. Therefore, the circuit solution in a balanced Y-connected circuit is independent of neutral connection.

If all the neutral points in a balanced three-phase circuit are at the same potential, then, each phase-source appears directly across the net impedance connected from the respective source line terminal (R or Y or B) to the load neutral point. Thus, the circuit may be solved phase by phase independent of each other. But then, why should we solve all the three phases at all? We expect the circuit solution to be balanced set of three-phase phasors in any case. Hence, we can determine the solution for Y-line and B-line if we know the solution for phasors in R-line by using three-phase symmetry. Therefore, we need to determine only the R-line phasors. The equivalent circuit that we employ to solve for R-line phasors is called the single-phase equivalent circuit for a balanced three-phase circuit. This equivalent circuit will contain the R-line phase-source in series with all impedances that come between R-line terminal of the source and neutral point of load. The neutral points of source and load are shorted to form the reference node in a single-phase equivalent circuit even if the neutral points are connected through impedance in the actual circuit. This is so since the neutral connection in a balanced circuit will not carry any current.

Note that if a three-phase circuit is star connected everywhere, the use of ‘R-phase’ instead of ‘R-line’ does not create any confusion. Such confusion can arise in the case of delta connection. A delta-connected source has only R-line and no R-phase. However, we have now decided that all our circuits will be converted to star equivalents. Thus, use of ‘R-phase’ instead of R-line is permitted.

Example: 8.3-1

A balanced Y-connected load with phase impedance of 7.9 + j5.7 Ω at 50 Hz is supplied from a 400V, 50 Hz balanced Y-connected source through connection impedance of 0.1 + j0.3 Ω in each line. (i) Find the line current and load voltage. (ii) Find the active and reactive power delivered by the source and delivered to the load. Fig. 8.3-3 Single-phase equivalent for Example 8.3-1

Solution

It is a Y–Y connection already. The single-phase equivalent circuit required for solving the problem is shown in Fig. 8.3-3.

A three-phase source is specified by specifying line-to-line voltage rms value. Thus, the source voltage in single-phase equivalent must be 400/√3 = 230.9 V rms in magnitude. The R-phase source voltage is taken as the reference phasor with an angle of 0°. Solving the circuit in Fig. 8.3-3, we get, Active power delivered by source = 3 × 230 × 23.09 × cos(0 – (–36.9°)) = 12.75 kW

Reactive power delivered by source = 3 × 230 × 23.09 × sin(36.9°) = 9.56 kVAr

Active power delivered to load = 3 × 224.9 × 23.09 × cos(–1.1° – (–36.9°)) = 12.64 kW

Reactive power delivered to load = 3 × 224.9 × 23.09 × sin(–1.1° – (–36.9°)) = 9.11 kVAr

Magnitude of line voltage across load = √3 × 224.9 = 389.56V rms

Complete Phasor Solution

Phase voltages at source: 230∠0°, 230∠–120°, 230∠120° (V rms)

The first line voltage, i.e., VRY, is known to lead the first phase voltage, i.e., VRN, by 30°. Therefore, line voltages at source terminals: 400∠30°, 400∠–90°, 400∠150° (V rms)

Phase voltages at load: 224.9∠–1.1°, 230∠–121.1°, 230∠118.9° (V rms)

Line voltages at load terminals: 389.56∠28.9°, 389.56∠–91.1°, 389.56∠148.9° (V rms)

Line currents: 23.09∠–36.9°, 23.09∠–156.9°, 23.09∠83.1° (A rms)

Complex power delivered by source = √3 × 400 × 23.09∠(0 – (36.9°)) = 16∠36.9° kVA

Complex power in load = √3 × 389.56 × 23.09∠(–1.1– (36.9°)) = 15.54∠35.8° kVA

Example: 8.3-2

A balanced Δ-connected load with phase impedance of 23.7 + j17.1 Ω at 50 Hz is supplied from a 400 V, 50 Hz balanced Y-connected source through connection impedance of 0.1 + j0.3 Ω in each line. (i) Find the line current and load voltage. (ii) Find the load-branch currents. (iii) Find the active and reactive power delivered by the source and delivered to the load.

Solution

The circuit with all the relevant phasors identified is shown in Fig. 8.3-4. Fig. 8.3-4 A Y-connected source delivering power to a Δ-connected load through connection impedance

The delta-connected impedance is transformed into an Y-connected one by star-delta transformation equations. Since the delta impedances are equal, the impedance in star connection will be one-third of delta branch impedance, i.e., 7.9 + j5.7 Ω. The circuit after this transformation is shown in Fig. 8.3-5.

Now, the single-phase equivalent of this circuit can be identified as the same as in Example: 8.3-1 and hence the solution is same as in that example.

∴Line current (R) = 23.09∠–36.9° A rms, Load phase voltage (R) = 224.9∠–1.10 V rms and Load line voltage (RY) = 389.56∠28.9° V rms.

Load active power can be found as 3 × 224.9 × 23.09 × cos(36.9°–1.1°) or as √3 × 389.56 × 23.09 × cos(36.9°–1.1°). The value is 12.64 kW. Similarly, the load reactive power is 3 × 224.9 × 23.09 × cos(36.9°–1.1°) or √3 × 389.56 × 23.09 × cos(36.9°–1.1°). The value is 9.11 kVAr. Fig. 8.3-5 The Y-connected equivalent of circuit in Fig. 8.3-4

Applying KCL at the corners of delta in R-line and Y-line in Fig. 8.3-4, we get

IRYIBR = IR and IYB IRY = IY

Multiplying the second equation by –1 and adding the result to the first equation, we get

2IRY (IBR + IYB) = IR IY

But (IRY + IBR + IYB) = 0 since we expect these three currents to be a three-phase balanced set. Therefore, –(IBR + IYB) = IRY.

Therefore, 2IRY + IRY = IRIY IRY = (IR IY)/3.

We know that IR = 23.09–36.9° A rms. IR, IY and IB form a balanced three-phase set of phasors. Therefore, IY = 23.09–156.9° A rms.

∴ 3 IRY = (23.09∠–36.9° – 23.09∠–156.9°) A rms.

This is the difference between two phasors with 120° between them. We have determined the result of this kind of difference operation before. The result will have a magnitude that is √3 × 23.09 and will lead the first quantity in the difference by 30°.

∴ 3 IRY = 40∠–6.9° and IRY = 13.33 ∠–6.9° A rms.

This can also be obtained by observing that RY line voltage at load is 389.5628.9° V rms and IRY is the current drawn by 23.7 + j17.1Ω (29.22535.8° Ω) from this voltage.

Therefore, IRY= 389.5628.9° ÷ 29.22535.8° = 13.33∠–6.9° A rms as before.

Now, the remaining two branch currents can be obtained by using three-phase symmetry as IYB = 13.33∠–126.9° A rms and IBR = 13.33∠–113.1° A rms

Example: 8.3-3

A delta-connected source with an open-circuit voltage of 200 V rms at 50 Hz is used to deliver power to a star-connected load with phase impedance of 5.9 + j7.7 Ω. Each phase source in the Δ-connected source has internal impedance of 0.3 + j0.9 Ω in series with it. (i) Find the phase voltage and line voltage across the load. (ii) Find the line currents and delta branch currents. (iii) Find the active power and reactive power delivered to the load and consumed by the internal impedance of the source.

Solution

The three-phase circuit to be analysed is shown in Fig. 8.3-6. Fig. 8.3-6 A Delta-connected source with source impedance supplying a star-connected load

The open-circuit line voltage across source terminals is given to be 200 V rms. The balanced delta connection shown in Fig. 8.3-6 will not have any current in delta when the terminals are left open since the loop voltage inside delta is zero. Therefore, the source voltages must themselves be 200 V rms. A straightforward determination of Y-equivalent of delta connected source is not possible in this case. We transform the ‘voltage source in series with impedance combinations’ into ‘current source in parallel with impedance combinations’ by using source transformation theorem. This is shown in Fig. 8.3-7. Fig. 8.3-7 Applying source transformation theorem on the delta-connected source in Fig. 8.3-6

We take the RY line quantity as the reference phasor in this example. Now, the delta-connected current source has a star equivalent and the delta-connected source impedance also has a star equivalent. Combining the current source in star equivalent and the impedance in star equivalent into a voltage source in series with impedance, we get (200/√3)∠–30° = 115.5∠–30° V rms in series with (0.3 + j0.9)/3 =0.1 + j0.3 Ω in R-phase of the final star equivalent. The complete star equivalent of the source along with the load is shown in Fig. 8.3-8.

The single-phase equivalent of this circuit is shown in Fig. 8.3-9.

1. The line current (R) is 115.5∠–30° V rms ÷ 6 + j8 Ω = 11.55 ∠–83.13° A rms and the phase voltage (RN) across load is 11.55∠–83.13° A rms × (5.9 + j7.7) Ω = 11.55∠–83.13° × 9.7∠52.54° =112∠–30.6° V rms. Fig. 8.3-8 The star equivalent of the circuit in Fig. 8.3-6

Hence, line voltage (RY) across the load = 112√3(–30.6° + 30°) = 194∠–0.6° V rms.

2.  The line current (R) is 115.5∠–30°  ÷ 6 + j8 = 11.55∠–83.13° A rms. Refer to Fig. 8.3-6. The current marked as I1 = [2000° - line voltage (RY) across load]  ÷ (0.3 + j0.9) = [2000° – 194∠–0.6°]  ÷ 0.9487 71.57° = (6 + j2)  ÷ 0.9487 71.57° = 6.68∠–51.13° A rms. Then, I2 will be 6.68∠– 171.13° A rms and I3 will be 6.6868.87° A rms Fig. 8.3-9 Single-phase equivalent of circuit in Fig. 8.3-6

3. The active power delivered to the load = √3 × 194 × 11.55 × cos(–30°–(–83.13°)) = 2.33 kW. The reactive power delivered to load = √3 × 194 × 11.55 × sin(–30°–(–83.13°)) = 3.11 kVAr.

The internal impedance of the source is 0.3 + j0.9 Ω and it carries an rms current of 6.68 A. Therefore, the active power consumed by the internal impedance of the delta-connected source = 3 × 6.682 × 0.3 = 40.2 W. The reactive power consumed by this impedance = 3 × 6.682 × 0.9 = 120.6 VAr.

##### 8.4 ANALYSIS OF UNBALANCED THREE-PHASE CIRCUITS

An unbalanced three-phase circuit is one that contains at least one source or load that does not possess three-phase symmetry. A source with the three source-function magnitudes unequal and/or the phase displacements different from 120° can make a circuit unbalanced. Similarly, a three-phase load with unequal phase impedances can make a circuit unbalanced.

The single-phase equivalent circuit technique of analysis does not work for unbalanced three-phase circuits. General circuit analysis techniques like mesh analysis or nodal analysis will have to be employed for analysing such circuits.

#### 8.4.1 Unbalanced Y–Y Circuit

Y–Y connection is typically used in the last mile in a power system, i.e., in the low-tension (LT) distribution system.

The primary distribution system usually runs at 11 kV line voltage level. Distribution transformers that are rated for 11 kV/400 V and are Δ-connected in the primary side and Y-connected in the secondary side are used to step down the primary distribution voltage to the LT distribution level. Such transformers are located at the load centre location of the area that a particular transformer is expected to serve. The neutral point of secondary side of a distribution transformer is usually earthed solidly.

4-wire LT lines emanating from the secondary of transformer distribute power to various subareas of the service area. Individual consumers are provided service by means of service drops tapped from these lines. Both single-phase loads and three-phase loads are provided power from these LT lines at various points. Thus, the load on the LT line will be unbalanced in general. Even if the single phase loads are to get distributed equally among three lines (R, Y and B), only the transformer secondary will perceive a balanced load and various sections of the LT line will see varying degree of unbalance in load. This is due to the spatially dispersed nature of loads. Thus, the neutral wire in an LT distribution line will invariably carry current – that too, different currents in different locations – even if the aggregate load is balanced.

The system in which the neutral wire is available for connecting single-phase loads and three-phase loads that require a neutral tie-up is called a four-wire system. The system in which neutral wire is not available for load connection is called a three-wire system. This could be Y-Y system with neutrals isolated, Y–Δ, Δ–Y or Δ–Δ system.

The exposed metal parts of all electrical equipment at the consumer location are tied together and earthed locally. This earth point is made available as the third pin of all three-pin power points installed in the consumer electrical system. If the earthing resistances at the transformer neutral and at the consumer location are negligibly small, then, the earth-pin at all consumer locations will be at the same potential as the transformer secondary neutral. However, the neutral-pin at the consumer locations will differ in potential from the transformer neutral potential due to the neutral wire drop. This difference will be more if there is unbalance in the load and neutral is carrying heavy currents. Moreover, this potential difference between the earth-pin and neutral-pin will be different for different consumers due to spatially distributed nature of line load. This problem of neutral – earth voltage difference is called the neutral- shift problem. Electronic equipment in general, and, computing equipment in particular, are sensitive to this neutral-shift voltage and often malfunction when it exceeds certain pre-specified levels. Damage to components can also take place due differences in neutral-shift potential that exist among various power points when different components of an interconnected computing system are powered from different power points.

Example: 8.4-1

A balanced three-phase source of 400 V, 50 Hz supplies an unbalanced load through an LT 4-wire line as in Fig. 8.4-1. Each wire in the line has impedance of 0.1 + j0.3 Ω. (a) Find (i) phase and line voltages at load location and neutral-shift voltage, (ii) wire currents, (iii) power and reactive power delivered to the load, (iv) three-phase apparent power and power factor. (b) Repeat (a) assuming that the neutral wire is a thick conductor with zero impedance. (c) Repeat (a) assuming that the neutral wire is open. Fig. 8.4-1 Unbalanced three-phase circuit in Example 8.4-1

Solution

(a) We solve the problem by finding out the phasor voltage at load neutral with respect to the ground (earth) node G by using the KCL equation at load neutral. The sum of currents flowing away from that node is set to zero. i.e., Solving for V, we get, V = –2.31–j5.49 = 5.96–112.8° V rms.

Now, we can find wire currents as The load phase voltages can be determined by multiplying line current phasors by phasor impedance of each phase. The line voltages across the load can be determined in terms of the phase voltage phasors now. The active power and reactive power delivered to the load is determined by adding up the corresponding power delivered to each phase of the load. Two different definitions of apparent power and power factor are possible in the case of unbalanced three-phase circuits. The first one is using the total complex power, S, delivered to the load. In this definition, the magnitude of S is taken as apparent power and ratio of active power to the magnitude of S is taken as average power factor of the circuit. The apparent power in the load in the circuit in this example then is (13.732 + 0.7442)°·5 = 13.75 kVA and power factor is 0.9985 lead.

The second definition accepts the sum of apparent powers of load phases as the three-phase apparent power. Power factor is taken as the ratio of active power to apparent power.. The apparent power in the load in the circuit in this example then is 15.88 kVA and power factor is 0.865. We cannot specify it as a lead or lag power factor in this case.

The second definition is more appropriate in the sense that power factor and apparent power are expected to give us an indication as to the utilisation efficiency of current in carrying active power. The first method of calculating apparent power and power factor will hide the fact that underutilization of current is taking place when one phase takes leading reactive power and another phase takes lagging reactive power. The phases may consume large reactive power and yet the complex power may come out with only real part or with small imaginary part. Therefore, the definition of apparent power and power factor based on apparent power of individual phases of load is a more meaningful one.

(b) The neutral-shift voltage in this case is zero since the source neutral and load neutral are tied together by a zero impedance link. The circuit becomes a collection of three single-phase circuits sharing a common point at neutral. Therefore, the line currents can be obtained as The load phase voltages can be determined by multiplying line current phasors by phasor impedance of each phase. The line voltages across the load can be determined in terms of the phase voltage phasors now. The active power and reactive power delivered to the load are determined by adding up the corresponding power delivered to each phase of the load. The three-phase apparent power is 15.86 kVA and average power factor is 0.864.

(c) The neutral points are unconnected and hence the system is a three-wire system in this case. The neutral-shift voltage V is found out from the node equation at the load neutral. i.e., Solving for V, we get, V = –74.55 V rms.

Observe the unacceptably large neutral-shift voltage when the source neutral and load neutral are isolated.

Now, we can find wire currents as The load phase voltages can be determined by multiplying line current phasors by phasor impedance of each phase. Observe the over-voltage in R-phase. The line voltages across the load can be determined in terms of the phase voltage phasors now. The active power and reactive power delivered to the load are determined by adding up the corresponding power delivered to each phase of the load.

P = PR + PY + PB = 15.82kW
Q = QR + QY+ QB = –0.53kVAr

Apparent power is 17.11 kVA and average power factor is 0.925.

Note that, in all the cases, the three line voltages add up to zero. The sum of line voltages in any three-phase system is zero since these three voltages are voltages in a loop formed by R to Y, Y to B and B to R.

This example demonstrated the need for a strong neutral tie in unbalanced four-wire circuits to maintain the phase voltages at load terminal at near-nominal values.

#### 8.4.2 Circulating Current in Unbalanced Delta-connected Sources

A delta-connected source comprises three sinusoidal sources in a closed loop. If the sources are ideal independent voltage sources they have to add up to zero at all instants. Otherwise, there will be infinitely large circulating current in the closed loop. The sources will have some nonzero impedance in practice and this source impedance will limit the circulating current arising out of residual voltage in the loop. However, if the source impedance is small (and, it should be small for a good source), the circulating current will consume the current carrying capacity of the source and leave only a portion of its capacity to deliver useful power to the external load. Moreover, large circulating current in a delta loop produces large dissipation within the sources.

Therefore, the phase-source voltage phasors in a delta-connected source should add up to zero preferably. This happens naturally in the case of a balanced source. But it need not happen always in the case of unbalanced source though it can for certain combinations of magnitudes and phase angles. Essentially, the magnitudes and phase angles of the three phase-sources must be such that the phasors form a closed triangle in complex plane. This is difficult to arrange in a synchronous generator. That is why generators are always Y-connected.

It is rarely that we encounter a delta-connected source. One situation where it can appear is when a transformer secondary is delta-connected. We can use a delta-connected voltage source, with each arm of delta containing a source in series with the transformer series impedance, as a circuit model for the secondary of the transformer. Exact symmetry in a three-phase equipment is hard to achieve in practice. For instance, there can be small differences in the number of turns employed in the three phase-windings of the transformer. This may produce unbalanced voltages in the three secondary windings. Further, those voltage phasors may not ‘close’ into a triangle. Circulating current in the secondary winding of the transformer is the result.

The next example illustrates this problem as well as the application of mesh analysis and node analysis in solving unbalanced three-phase circuits.

Example: 8.4-2

A delta-connected unbalanced source delivers power to a balanced Y-connected resistive load as shown in Fig. 8.4-2. Solve the circuit completely by using mesh analysis. Fig. 8.4-2 Unbalanced delta-connected source supplying a Y-connected load in Example 8.4-2

Solution

The mesh currents are assigned as shown in Fig. 8.4-2.The mesh equations are written in matrix form by inspection. Solving for I1, I2 and I3,

I1 = 14.9877.64° A rms, I2 = 11.12∠–29.2° A rms and I3 = 10.68∠–90.48° A rms.

Now, the line currents are IR = I2 = 11.12∠–29.2° A rms, IY = I3 I2 = 11.116∠–151.77° A rms and IB =I3 = 10.6889.52° A rms.

Current delivered by 4000° V rms source = I2 I1 = 21.08∠–72.04° A rms

Current delivered by 370–120° V rms source = I3 I1 = 25.53∠–97.42° A rms

Current delivered by 370120° V rms source = – I1 = 14.98∠–102.36° A rms

Power delivered to load = [11.122 + 11.122 + 10.682] × 20 = 7227.4 W

Power delivered by 4000° V rms source = 400 × 21.08 × cos(72.04°) = 2600.3 W

Power delivered by 370∠–120° V rms source = 370 × 25.53 × cos(–120° + 97.42°) = 8772 W

Power delivered by 370120° V rms source = 370 × 14.98 × cos(120° + 102.36°) = –4095.57 W

Notice the marked deviation in phase-source currents from the levels we expect from the observed line currents. We would have expected about 6 – 7 A rms in the sources considering the fact that the line currents are around 11 A rms. The circulating current due to the residual delta-loop voltage of 30 V rms makes the delta currents much larger than what they need to be in view of load power. This results in excess power being delivered by some branches, forcing the remaining branch to absorb power rather than deliver it. This is why the 370120° V rms source had to absorb a large quantity of power.

Of course, the delta-connected source used in this example can be of no practical use. A delta-connected source is of practical utility only if the residual voltage within the delta-loop can be kept at zero or a value quite low compared to source voltage.

The line voltages appearing across the load can be calculated by subtracting the voltage drop in the source impedance of each phase-source from its source voltage value. The result of this calculation is VRY = 390∠–0.48° V rms, VYB = 375.3∠–121.8° V rms and VBR = 375.3–120.83° V rms.

##### 8.5 SYMMETRICAL COMPONENTS

Analysis of balanced three-phase circuits is considerably simpler than analysis of unbalanced three-phase circuits. Three-phase symmetry exhibited by a balanced circuit makes it possible to solve the circuit employing a single-phase equivalent circuit. This prompts us to ask the question – is it possible to bring back the symmetry enjoyed by a balanced circuit into an unbalanced three-phase circuit by some means?

We realise that an unbalanced three-phase circuit can result from unbalanced three-phase sources acting on balanced loads or balanced sources acting on unbalanced loads or unbalanced three-phase sources acting on unbalanced loads. Let us tackle the first case – unbalanced three-phase sources acting on balanced loads.

#### 8.5.1 Three-Phase Circuits with Unbalanced Sources and Balanced Loads

In this case, maybe we can bring three-phase symmetry back into the circuit if we can somehow express the unbalanced three-phase source voltages/currents as a superposition of balanced sets of three-phase voltages/currents. Is that possible?

Specialised version of a general theorem called Fortesque’s Theorem assures us that it is possible. If a set of unbalanced three-phase source functions can be expressed as a sum of balanced sets of three-phase source functions, then, the solution of unbalanced three-phase circuit can be obtained as a superposition of solution of the circuit for various balanced sets of three-phase source functions. If all the loads are balanced, each of the circuit problems that needs to be solved for applying superposition principle, will be a balanced circuit problem. Thus, symmetry can be restored to unbalanced three-phase circuits this way, provided the unbalance is only due to sources and not due to loads. The sets of three-phase balanced source components and possible single-phase components of an unbalanced source are called its Symmetrical Components.

There are three symmetrical components for an unbalanced three-phase source function. Each symmetrical component is a set of three source functions. The first set – called the positive sequence component – is a balanced three-phase set of source functions that has positive phase sequence. The second set – called the negative sequence component – is a balanced three-phase set of source functions that has negative phase sequence. The third set – called the zero sequence component – is a set of three cophasal (i.e., of same phase) single-phase source functions. It is not a three-phase set at all.

Symmetrical components are denoted by the first phasor element in each set. Thus, positive sequence component is denoted by R-phase quantity or R-line quantity of the balanced three-phase source function of positive phase sequence in phasor form. Similarly, negative sequence component is denoted by R-phase quantity or R-line quantity of the balanced three-phase source function of negative phase sequence in phasor form. Further, zero sequence component is denoted by one of the three cophasal single-phase source functions.

For instance, let 200∠–10°, 100∠–50° and 25∠–30° be the rms values of positive, negative and zero sequence components of some unbalanced phase voltage set. Then, 200∠–10° stands for a three-phase balanced voltage set (200∠–10°, 200∠–130°, 200110°) in RN, YN and BN phase voltages, respectively. The 100∠–50° negative sequence component stands for (100∠–50°, 10070°, 100∠–170°) in RN, YN and BN phase voltages, respectively. Note the phase sequence of the bracketed quantity. The zero sequence component of 25∠–30° stands for (25∠–30°, 25∠–30°, 25∠–30°) in RN, YN and BN phase voltages, respectively. Then, by Fortesque’s Symmetrical Components Theorem, the phase voltages are given by The phase voltage set is an unbalanced one and we have expressed it as a sum of three components – one balanced positive sequence three-phase voltage, one balanced negative sequence three-phase voltage and one single-phase voltage.

We could have expressed these equations in another format too. Let us define a constant complex number a as a = 1120°. When a phasor is multiplied by this operator, its magnitude stays unaffected, but it gets rotated in counter clockwise direction by 120°. Then, a2 = 1240° = 1∠–120° is the operator that can rotate a phasor by 120° in clockwise direction. Note that 1 + a + a2 = 0, a2 = a*, a3 = 10° and a4 = a. Now, the above equations can be expressed in terms of this operator a as Now, let us introduce X + , X_ and X0 as the values of positive, negative and zero sequence components (note that they are phasors) of some voltage or current instead of the three numbers – 200∠–10°, 100∠–50° and 25∠–30° – we used till now. Then, the unbalanced quantities are given in terms of symmetrical components (i.e., X + , X- and X0) by We can express this equation set in matrix form as below. XR, XY and XB can represent any phase voltage, phase current, line voltage or line current phasors in an unbalanced system. The 3 × 3 matrix in Eqn. 8.5-1 is called the Symmetric Transformation Matrix. The inverse equation required for determining sequence components from phase quantities is as below. The reader is urged to verify the matrix inversion involved.

The circuit interpretation of symmetrical components is given in Fig. 8.5-1. Here, an unbalanced star-connected voltage source is resolved into its symmetrical components. Symmetrical Components theorem assures us that the source in Fig. 8.5-1(a) can be viewed as the composite source in Fig. 8.5-1(b) in which each source limb contains three sources in series – one each from each sequence component set. Further, the superposition principle assures us that applying the source in Fig. 8.5-1(b) is the same as applying the three source sets shown in Fig. 8.5-1(c) individually. Fig. 8.5-1 Interpretation of symmetrical components

#### 8.5.2 The Zero Sequence Component

This component needs special attention. We get an expression for zero sequence component from Eqn. 8.5-2 as Zero sequence component is the average of the three three-phase quantities.

Line voltages in any three-phase circuit, balanced or unbalanced, will add up to zero by KVL. Therefore, line voltages in a three-phase system cannot have a zero sequence content anywhere in the system.

Line currents in a three-phase three-wire system will have to add up to zero by KCL. Therefore, line currents in a three-phase three-wire system cannot have a zero sequence content anywhere in the system.

Phase voltages in a three-phase system need not add up to zero. Hence, phase voltages can have zero sequence content.

Line currents in a four-wire system need not add up to zero. The sum IR + IY + IB can flow through the fourth wire (neutral wire) in the return direction. Therefore, three-phase four-wire systems can have zero sequence content in their line currents.

#### 8.5.3 Active Power in Sequence Components

Let the phase voltages across a balanced three-phase load circuit be VRN, VYN and VBN where N is the neutral point in the load itself or in its Y-equivalent. Let IR, IY and IB be the line currents flowing into the load. Further, let V0, V + and V- be the sequence components of phase voltages and I0, I + and Ibe the sequence components of line currents.

Then, the total active power that flows into the load, Let us define four column vectors as below. Then, the power equation can be expressed as P = Re[VtrybI*ryb] where the superscripts t indicates matrix transpose operation and * indicates complex conjugation operation.

Eqn. 8.5-1 expresses the three-phase quantities in terms of its sequence components. It is reproduced below. Using this equation, we write the column vector Vryb in terms of the column vector V0+– and write the column vector Iryb in terms of the column vector I0 + as below. Then, Therefore, AtA* is three times identity matrix of 3 × 3 order. Hence, Thus, we observe that (i) active power carried by sequence components obey superposition principle and that (ii) positive sequence component of voltage can send active power only through positive sequence component of current; negative sequence component of voltage can send active power only through negative sequence component of current and zero sequence component of voltage can send active power only through zero sequence component of current.

Balanced impedance will have negative sequence current in it only if there is negative sequence voltage applied to it. This is so since balanced impedance can produce only negative sequence voltage drop when a negative sequence current flows through it.

That is, balanced impedance is incapable of bringing about a sequence conversion involving translation of positive sequence current into negative sequence voltage and vice versa.This is called sequence-decoupling property of balanced three-phase impedances.

#### 8.5.4 Three-Phase Circuits with Balanced Sources and Unbalanced Loads

Three-phase sources, at least at generation and transmission level, are more or less balanced. They may show considerable unbalance at LT distribution level. However, three-phase circuit unbalance is usually the result of unbalanced loads rather than unbalanced sources. Symmetrical components can be used for analysing such circuits too. However, we do not take up such analysis since it is somewhat beyond the scope of an introductory text like this one. Hence, we close our discussion on symmetrical components after covering two salient points in this subsection.

The first point is that an unbalanced load does not possess sequence-decoupling property. Rather, it creates sequence coupling. Consider a balanced three-phase voltage source supplying an unbalanced three-phase load. Since the voltage source is balanced, it has no negative and zero sequence components. It is a purely positive sequence voltage source. However, the load will draw unbalanced currents since it is unbalanced. We know that unbalanced current set will contain negative sequence component. Thus, a positive sequence voltage source has produced a negative sequence current in the circuit. Moreover, the voltage drop produced across the load impedance due to the negative sequence current must be a positive sequence voltage since there is only positive sequence component in the source voltage to meet the KVL requirement. Therefore, negative sequence current succeeds in producing a positive sequence voltage drop in the case of an unbalanced load. Therefore, an unbalanced load produces sequence coupling between positive sequence components and negative sequence components.

The second observation we want to make is regarding active power. We observed in the last subsection that a negative sequence current flowing through a positive sequence voltage could not carry active power. Hence, the negative sequence current drawn by an unbalanced three-phase load from a balanced three-phase source is a wasteful current that ties up equipment capacity and increases system losses without carrying productive power. We had met with another such current before – the reactive current component. Thus, both the reactive component of positive sequence current and the entire negative sequence current drawn by an unbalanced three-phase load from a balanced supply are unproductive in effect and detrimental to the system.

The most energy-efficient way to draw active power from a balanced three-phase source is to draw it through balanced three-phase currents at unity power factor.

Symmetrical components afford us the required information to evaluate the effectiveness of current in carrying power in unbalanced loading of the system equipment. We can solve the unbalanced circuit problem by applying mesh or nodal analysis and obtain the symmetrical components of currents to evaluate how bad the current utilisation is.

Thus, with this introductory coverage on symmetrical components, we can now apply sequence components (i) to solve circuits with unbalanced sources and balanced loads and (ii) to evaluate the effectiveness of current flow in carrying active power in circuits with balanced sources and unbalanced loads. There is much more to symmetrical components. Power System Engineers go deep into symmetrical components in analysing power systems under fault condition.

Example: 8.5-1

An unbalanced 4-wire system is shown in Fig. 8.5-2. (i) Find the symmetrical components of the source phase voltages. (ii) Determine the line voltages at the source and verify that line voltage does not contain zero sequence component. (iii) Determine the line currents and neutral current by using symmetrical components. (iv) Find the load neutral voltage with respect to earth. (v) Find the phase voltages and line voltages across the load by symmetrical components. (vi) Find the total power delivered by source and delivered to load using symmetrical components. Fig. 8.5-2 Unbalanced three-phase circuit for Example 8.5-1

Solution

1. 2. Line voltages are obtained as below. Therefore, zero sequence component in line voltage is zero. Fig. 8.5-3 Single-phase equivalent circuit of the circuit in Fig. 8.5-2 for positive sequence component Fig. 8.5-4 Single-phase equivalent circuit of the circuit in Fig. 8.5-2 for negative sequence component

3. The circuit is solved for the three symmetrical components by applying superposition principle. We apply the positive sequence voltage source component first and obtain the positive sequence component of line currents and neutral current. We note that the connection impedances and load impedances are balanced. Therefore, we can use single-phase equivalent circuit for obtaining currents. The single-phase equivalent circuit is shown in Fig. 8.5-3. The source neutral and load neutral are at the same potential and hence the neutral impedance of 0.1 + j0.3 Ω will not appear in the single-phase equivalent circuit for positive sequence component.

Therefore, I + = 216.917.25° ÷ (8 + j6) = 21.69–29.62° A rms.

The single-phase equivalent circuit for negative sequence input is shown in Fig. 8.5-4. Passive balanced impedance cannot differentiate between positive phase sequence and negative phase sequence. However, this is not true in the case of all electrical equipment. AC motors can distinguish between the two and hence the equivalent circuits of motors will be different for the two phase sequences.

The load circuit is balanced. Negative sequence component is a balanced component. Therefore, neutral potential at source and load will be the same and the neutral impedance will not appear in the single-phase equivalent circuit for negative sequence input too.

Therefore, I = 39.25∠–37.58° ÷ (8 + j6) = 3.925∠–74.45° A rms.

The circuit to be solved with zero sequence input is shown in Fig. 8.5-5.

Note that all the three voltage sources have the same phase and hence all the three lines R,Y and B will carry cophasal currents in the same direction. Therefore, the neutral return current will be 3I0. Applying KVL in any one mesh, we get,

[(8 + j6) + 3 × (0.1 + j0.3)] × I0 = 16.64–168.12° ⇒ I0 = 1.542152.14° A rms

Note that the effective value of neutral line impedance in limiting the zero sequence current is three times its actual value due to 3I0 flowing in it.

Now, the line currents can be obtained by applying symmetrical component transformation equation. Fig. 8.5-5 Circuit with only zero sequence component of input voltage acting in the circuit in Fig. 8.5-2 The result will be, IR = 23.106∠–36.62° A rms, IY = 18.86∠–156.75° A rms and IB = 23.984∠–102.78° A rms.

Neutral current can arise only from zero sequence component of source voltage since the remaining two components are balanced three-phase components and the load circuit is balanced. Hence, neutral current = 3I0 = 4.63152.14° A rms.

4. Load neutral voltage with respect to earth = neutral current × neutral impedance, since the source neutral is earthed. The value is = 4.63152.14° × (0.1 + j0.3) = 1.463 ∠–136.3° V rms.
5. The symmetrical components of load phase voltages can be obtained as The values are Vl0 = 15.018∠–172.05° V rms, V1+=211.3056.2° V rms and V1= 38.237∠– 38.64° V rms. Now, the load phase voltages can be determined by symmetrical components transformation. The values obtained on substitution of sequence component values are,

VlRN = 225.09∠–0.81° V rms, VlYN =183.73∠–120.93° V rms and VlBN =233.65138.54° V rms.

The load line voltages are found as 6. The values of sequence components at source end are

Vs0 = 16.64−168.12° V rms; Vs+ = 216.917.25° V rms; Vs = 39.25− 37.58° V rms
and I0 = 1.542152.14° A rms ; I + = 21.69–29.62° A rms ; I = 3.925–74.45° A rms.

∴ Power delivered through zero sequence components

= 16.64 × 1.542 × cos(–168.12°–152.14°) = 19.73 W per phase

∴ Power delivered through positive sequence components

= 216.91 × 21.69 × cos(7.25° + 29.62°) = 3763.9 W per phase

∴ Power delivered through negative sequence components

= 39.25 × 3.925 × cos(–37.58° + 74.45°) = 123.3 W per phase

∴ Total power delivered by the source = 3 × (19.73 + 3763.9 + 123.3) W = 11.72 kW.

The values of sequence components at load end are,

Vl0 = 15.018–172.05° V rms, Vl + = 211.305.6.2° V rms and Vl = 38.237–38.64° V rms and I0 = 1.542152.14° A rms; I + = 21.69–29.62° A rms ; I = 3.925–74.45° A rms.

∴ Power delivered to the load through zero sequence components

= 15.018 × 1.542 × cos(–172.05° –152.14°) = 18.8 W per phase

∴ Power delivered to the load through positive sequence components

= 211.305 × 21.69 × cos(6.2° + 29.62°) = 3716.9 W per phase

∴ Power delivered to the load through negative sequence components

= 38.237 × 3.925 × cos(–38.64° + 74.45°) = 121.7 W per phase

∴ Total power delivered to the load = 3 × (18.8 + 3716.9 + 121.7) W = 11.572 kW

Example: 8.5-2

Single-phasing of a three-phase load leads to an extreme case of unbalanced operation in practice. An undetected open in the wiring path in one phase or a fuse blowing in a phase that leads to this kind of operation of three-phase loads in industries. It is so common that specific measures in the form of a single-phasing prevention relay is incorporated in the control gear of high-power induction motor drives in the Industry.

The circuit in Fig. 8.5-6 shows a balanced three-phase induction motor running from a balanced supply. The balanced operation of this circuit was studied in Example: 8.3-2. Assume that the line goes open in B-phase leading to single-phased operation of the motor. Find (i) line currents and sequence components of line currents, (ii) currents in the three motor windings, (iii) active and reactive power delivered to the motor and (iv) apparent power and power factor of the source. Fig. 8.5-6 Circuit for Example 8.5-2

Solution

This circuit was analysed in Example: 8.3-2 for balanced operation. The following results were obtained.

Line current (R) = 23.09∠–36.9° A rms, load phase voltage (R) = 224.9∠–1.10 V rms and load line voltage (RY) = 389.5628.9° V rms. The delta branch currents had an rms value of 13.33 A.

Load active power = 12.64 kW, load reactive power = 9.11 kVAr and source apparent power = 16 kVA. Fig. 8.5-7 The Y-connected equivalent circuit of circuit in Fig. 8.5-6 with B-line open

We replace the balanced delta-connected load impedance by a balanced star-connected impedance first to get a Y-connected equivalent circuit for the circuit in Fig. 8.5-6. This Y-connected equivalent circuit with the break in B-line is shown in Fig. 8.5-7. Note that there is only one current variable in the circuit now and that current phasor is marked as I.

1. The current phasor I can be obtained by writing the KVL equation in the only mesh that is present in the circuit.

I = [230.90° – 230.9∠–120°] ÷ [16 + j12] = 40030° ÷2036.87° = 20∠–6.87° A rms

Therefore, the line currents are IR = 20∠–6.87° A rms, IY = –20∠–6.87° A rms and IB = 0.

The symmetrical components of line current is found by using the equation The sequence components are I0 = 0, I +=11.545∠–36.87° A rms and I = 11.54523.13° A rms. Observe that the positive sequence and negative sequence components of current have equal magnitude.

2. The current I divide in two paths in the delta-connected windings. All the windings have equal impedances. Therefore, the current is shared in 2:1 ratio in RY winding and other two windings in series. The winding currents are 13.33 A rms, 6.67 A rms and 6.67 A rms.
3. Active power delivered to motor = Active power delivered by source – power dissipated in connection impedance = 400 × 20 × cos(30°–(–6.87°)) – 202 × 0.1 × 2 W = 6.32 kW

Reactive power delivered to motor = Q delivered by source – Q absorbed by connection impedance = 400 × 20 × sin(30°–(–6.87°)) – 202 × 0.3 × 2 VAr = 4.56 kVAr.

4. Apparent power of source = 230.9 × 20 + 230.9 × 20 + 230.9 × 0 VA = 9.236 kVA Power factor = 6.4/9.236 = 0.693

Example: 8.5-3

Convert the delta-connected unbalanced voltage source with source impedance in Fig. 8.5-8 into its Y-connected equivalent. Fig. 8.5-8 Delta-connected unbalanced source in Example 8.5-3

Solution

Let the three source voltage phasors be identified as V1, V2 and V3. Let the source impedance be identified as Z. We convert the delta branches into current sources in parallel with impedance form as shown in Fig. 8.5-9.

The delta-connected unbalanced current source is now resolved in terms of its sequence components as shown in Fig. 8.5-10 where I + = V+/Z, I = V/Z and I0 = V0/Z.

The first two delta-connected current sources are three-phase balanced sources and have Y-equivalents. Their Y-equivalent is obtained by finding out the first line current delivered by each. Fig. 8.5-9 Voltage source to current source transformation applied to the source in Fig. 8.5-8

The positive sequence sources deliver I + aI + into R-line. The star equivalent must have this as the source function in R-phase. Therefore, the star equivalent will have a positive sequence current source of value of The negative sequence sources deliver I a2I- into R-line. The star equivalent must have this as the source function in R-phase. Therefore, the star equivalent will have a positive sequence current source of value of The zero sequence current inside the delta circulates within it and cannot come out. Therefore, the Y-equivalent has no zero sequence current source. This is consistent with the fact that the line currents in a three-wire system cannot have zero sequence content.

Therefore, the Y-equivalent of the circuit in Fig. 8.5-10 is as shown in circuit Fig. 8.5-11(a). Note that the delta-connected impedance is also converted into equivalent star. Fig. 8.5-10 Resolution of three-phase current source in Fig. 8.5-9 in terms of its sequence components Fig. 8.5-11 Y-equivalent of circuit in Fig. 8.5-10

Now, all the three three-phase Y-connected subcircuits in circuit Fig. 8.5-11(a) are balanced circuits and hence the neutrals will be at the same potential. Therefore, they can be joined together. Then, the resulting parallel connection of two current sources and impedance in each phase can be converted into two voltage sources in series with the same impedance. The last step is to substitute for I + and I in terms of sequence components of delta phase-source voltages. The resulting Y-connected equivalent circuit is shown in circuit Fig. 8.5-11(b).

Now, we have determined the positive sequence and negative sequence components of phase voltages of the Y-connected equivalent. They are and respectively. The zero sequence component is zero. Note that there is non-zero zero sequence content in delta-connected sources. Substituting for V1, V2, V3 and a, we get V0 = 19.62∠–69.76° V rms, V+ =190.0412.6° V rms, and V =36.051∠–39.69° V rms.

The star equivalent phase voltage are obtained by ##### 8.6 SUMMARY
• A set of three sinusoidal quantities, all at the same frequency, with equal peak (and hence rms) values and shifted successively by 120° in phase is defined as a Balanced Three-Phase Quantity. Therefore, if x1(t), x2(t) and x3(t) is a three-phase set, then, and x1(t) + x2(t) + x3(t) = 0 for all t.

If the peak values are unequal and/or successive phase shifts are different from 120°, the set will be called an Unbalanced Three-Phase Quantity. A three-phase source is specified by specifying line-to-line voltage rms values.

• Each limb or branch of a three-phase system (source or load) is termed as a phase. R, Y and B are used to designate the line terminals of a three-phase source or load.
• A balanced three-phase system comprising balanced three-phase sources and balanced three-phase loads is superior to a single-phase system due to the following facts: (i) Power loss in transmission system is lower in three-phase system (ii) Copper utilisation is superior in three-phase system. (iii) Power delivered to a balanced three-phase load by a balanced three-phase supply is free of pulsation. (iv) Electrical equipment designed for three-phase operation is more efficient than their single-phase counter parts.
• The sum of line voltages in any three-phase system is zero since these three voltages are voltages in a loop formed by R to Y, Y to B and B to R.The sum of line currents in any three-phase three-wire system is zero since these three currents have to satisfy KCL.
• The line voltages in a balanced Y-connected source form a balanced three-phase set of voltages that are √3 times in magnitude and 30° ahead in phase with respect to phase voltages.
• The line currents delivered by a balanced Δ-connected source form a balanced three-phase set of currents that are √3 times in magnitude and 30° behind in phase with respect to phase currents delivered by the phase sources.
• The three-phase complex power delivered by a three-phase source is given by

S = √3 VLILθ = [√3 VLIL cosθ + j √3 VLIL sinθ ] VA where θ is the angle by which the current phasor delivered by a phase-source lags behind its voltage phasor. This relationship is independent of whether the source is Y-connected or Δ-connected. θ is not the angle between line voltage phasor and line current phasor.

• Any balanced three-phase circuit can be equivalenced to a Y-Y balanced three-phase circuit by replacing Δ-connected sources and loads (if any) by their Y-connected equivalents. The equivalence is valid as far as line voltages, line currents and complex power flow are concerned.
• A balanced Y-Y circuit can be solved by solving its R-phase and employing three-phase symmetry to arrive at the solution for Y-phase and B-phase. The R-phase circuit is solved by using the single-phase equivalent circuit.
• The sets of three-phase balanced source components and possible single-phase components of an unbalanced source are called its Symmetrical Components.
• There are three symmetrical components for an unbalanced three-phase source function. Each symmetrical component is a set of three source functions. The first set – called the positive sequence component – is a balanced three-phase set of source functions that has positive phase sequence. The second set – called the negative sequence component – is a balanced three-phase set of source functions that has negative phase sequence. The third set – called the zero sequence component – is a set of three cophasal (i.e., of same phase) single-phase source functions. It is not a three-phase set at all.
• The zero sequence component is the average of the three three-phase quantities. Line voltages in a three-phase system cannot have a zero sequence content anywhere in the system. Line currents in a three-phase three-wire system cannot have a zero sequence content anywhere in the system.
• Active power carried by sequence components obeys superposition principle.
• Positive sequence component of voltage can send active power only through positive sequence component of current; negative sequence component of voltage can send active power only through negative sequence component of current and zero sequence component of voltage can send active power only through zero sequence component of current.
• Balanced impedance is incapable of bringing about a sequence conversion involving translation of positive sequence current into negative sequence voltage and vice versa. An unbalanced load produces sequence coupling between positive sequence components and negative sequence components
• The most energy-efficient way to draw active power from a balanced three-phase source is to draw it through balanced three-phase currents at unity power factor.
##### 8.7 PROBLEMS
1. Three voltage phasors 1000° V, 100∠–120° V and 110120° V cannot be the line voltages of a three-phase source. Explain why.
2. The line currents flowing into a load are found to be 10∠–90° A, 10∠–200° A and 1030° A. Explain why the load cannot be a delta-connected one.
3. The line currents flowing into a load are found to be 10∠–30° A, 10∠–150° A and 990° A. Show that the three-phase system has to be a 4-wire system and that the load has to be star-connected.
4. The line currents flowing into a load from a balanced supply are found to be 10∠0° A, 5√2∠–135° A and 5√2135° A. Is it possible to decide whether the load is delta-connected or star connected from this data?
5. The phase-voltage phasors of a Y-connected three-phase voltage source are 1000° V, 100∠–90° V and 10090° V Calculate the line voltages and explain why line voltage magnitude is not equal to 100√3 V.
6. Three sinusoidal current sources 100° A, 10∠–120° A and 8120° A are used to form a three-phase 3-wire current source. Explain why this current source has to be a delta-connected one.
7. A balanced three-phase voltage source of 400 V, 50 Hz delivers 10 A at unity power factor to a balanced three-phase load through a connection impedance of 1 + j0 Ω. Is the line voltage magnitude across the load equal to 400 V – 10 V = 390 V? If not, why and what is the correct value?
8.  Find the symmetrical components of a set of three currents 10∠–35° A, 10∠–35° A and 10∠–35° A.
9.  An electrodynamometer type wattmeter is a measuring instrument that measures average power. It has two coils called current coil and pressure coil. The current coil terminals are usually marked M and L and the pressure coil terminals are usually marked V and COM. The meter measures the average of the product v(t) × i(t) where v(t) is the voltage of COM with respect to V and i(t) is the current flowing from M to L in the meter. In the case of sinusoidal steady-state, the meter measures Vrms Irms cosɸ where Vrms is the rms value of voltage applied across pressure coil, Irms is the rms value of current flowing through current coil and ɸ is the angle between these two phasors. Two such wattmeters can be used to measure active power in a three-wire system irrespective of the nature of source and load connections (i.e., star or delta). The wattmeter connection to do this is shown in Fig. 8.7-1. Fig. 8.7-1 Two-wattmeter connection

1. Let the reference phasor be the R-phase phase voltage in the equivalent Y-connected system and let the current IR be IL ∠–θ. Then, show that, under balanced operation, (i) the wattmeter in R-line reads W1 = √3 VLIL cos(30° + θ) W, (ii) the wattmeter in B-line reads W2 = √3 VLIL cos(30° – θ) W and (iii) the total active power in the three-phase system is = W1 + W2 W. Draw the required phasor diagrams.
2. What is the range of load power factor for which the first wattmeter in two-wattmeter method of power measurement in Fig. 8.7-1 will try to read negative value? What is the corrective action needed to take readings? How is the correction incorporated in the expression for total three-phase power?
3. Wattmeters used in two-wattmeter method are often constructed with reversing switch that will reverse the voltage applied to pressure coil. What is the need for this reversing switch?
10. A 400 V rms Δ-connected source shown in Fig. 8.7-2 has I1 = 10∠–30° A rms, I2 = 10∠–120° and I3 = 10√2105° A rms. (i) Find the line currents delivered, (ii) Find the three-phase active power and reactive power delivered (iii) Find a delta-connected load that will result in this loading. Fig. 8.7-2

11. A star-connected source with zero source impedance delivers VRY = 200∠–20° V rms, VYB = 210∠–130° V rms. The R-phase phase voltage VRN is 1150° V rms. Find the third line voltage and the remaining two phase voltages. Draw the phasor diagram showing phase voltages and line voltages.
12. A delta-connected voltage source with zero source impedance delivers IR = 20∠–30° A rms, IB = 15130° A rms. The source that is directly between R-line and Y-line is found to deliver 110° A rms. Find Y-line current and the remaining two phase-source currents. Draw the phasor diagram showing all the currents.
13. A star-connected source with zero source impedance and isolated neutral delivers VRY = 400∠– 25° V rms, VYB = 390∠–135° V rms. The Y-phase phase voltage VRN is 2300° V rms. The line currents delivered by the source in R-line and B-line are IR = 10–30° A rms, IB = 12130° A rms. Find the three-phase active power and reactive power delivered by the source. Draw the phasor diagram showing phase voltages, line voltages and line currents.
14. A delta-connected voltage source with zero source impedance delivers IR = 5∠–40° A rms, IY = 15∠–100° A rms. The source that is directly between R-line and Y-line is found to deliver 110° A rms. VRY and VYB are found to be 200∠–0°V rms and 200∠–110° V rms, respectively. (i) Find the three-phase active power and reactive power delivered by the source. (ii) Find a delta-connected impedance that will cause this loading in the source. (iii) Draw the phasor diagram showing phase currents, line currents and line voltages.
15. A star-connected balanced voltage source of 400 V rms, 50 Hz rating delivers 6–j3 kVA of complex power to a passive balanced Y-connected load. Use the R-phase phase voltage as reference phasor. (i) Find the line currents. (ii) Determine the load impedance. (iii) Draw the phasor diagrams showing all phase voltages, line voltages and line currents. (iii) What will be the readings in two wattmeters in two-wattmeter method of power measurement as in Fig. 8.7-1?
16. A star-connected balanced voltage source of 400 V rms, 50 Hz rating with a source impedance of 1 + j1 Ω in series with the source in each phase delivers power to a delta-connected balanced load impedance with a branch impedance of 20 + j20 Ω at 50 Hz. Find the three-phase complex power delivered to the load.
17. A delta-connected balanced voltage source of 200 V rms, 60 Hz rating with zero source impedance delivers power to a delta-connected balanced load impedance with a branch impedance of 8–j6 Ω through an impedance-less connection. Find the line currents, load-branch currents and three-phase complex power delivered to the load without transforming the circuit into its Y-Y equivalent. Draw the phasor diagram for the circuit taking VRY as the reference phasor.
18. A delta-connected balanced voltage source of 200 V rms, 60 Hz rating with zero source impedance delivers power to a delta-connected balanced load impedance with a branch impedance of 8 + j6 Ω through an connection impedance of 1 + j1 Ω in each line. Find the line currents, load-branch currents and three-phase complex power delivered to the load by employing single-phase equivalent circuit method. Draw the phasor diagram for the circuit taking VRY as the reference phasor.
19. A delta-connected balanced voltage source of 400 V rms, 50 Hz rating with zero source impedance delivers power to a star-connected balanced load impedance with a branch impedance of 11 – j17 Ω through an connection impedance of 1 + j1 Ω in each line. Find the line currents, load-branch currents and three-phase complex power delivered to the load by using star-delta transformation of impedances.
20. A star-connected balanced voltage source of 400 V rms, 50 Hz rating with zero source impedance delivers power to a star-connected balanced load impedance with a branch impedance of 10 – j10 Ω in parallel with a balanced delta-connected impedance with a branch impedance of 30 + j60 Ω through an connection impedance of 1 + j1 Ω in each line. Find the line currents, load-branch currents in both loads, three-phase complex power delivered to each load, the complex power delivered by source and source power factor by using single-phase equivalent circuit method.
21. A star-connected balanced voltage source of 200 V rms, 60 Hz rating has an internal resistance of 1 Ω in series with an internal inductance of 3 mH in each phase. It supplies power to an unbalanced Y-connected load with 10 Ω, 15 Ω and 8 Ω in R, Y and B phases, respectively through a 3-wire connection with a connection impedance of 0.8 Ω + 4 mH inductance in each wire. The neutral of the source is firmly earthed. (i) Find the neutral potential with respect to earth at the load, (ii) Find line currents, line and phase voltages across the load and (iii) The active power delivered to each phase of the load and the total active power.
22. Repeat Problem 21 if it is a 4-wire connection with same connection impedance in the neutral wire. Find the neutral wire current too.
23. A star-connected unbalanced voltage source with VRN = 1150° V rms, VYN = 115∠–110° V rms and VBN = 132125° V rms delivers power to a balanced delta-connected load with a branch-impedance of 23.4 + j17.1 Ω through a 3-wire line with a line impedance of 0.2 + j0.3 Ω. Determine (i) line currents, (ii) line voltages across the load, (iii) active power and reactive power delivered by each phase of the source and the three-phase values and (iv) three-phase active power and reactive power delivered to the load by using star-delta transformation on load impedance and mesh analysis.
24. A delta-connected unbalanced voltage source uses 1150° V rms, 110∠–110° V rms and 122125° V in the three branches R–Y, Y–B and B–R, respectively. Each branch source has 1 Ω + 3 mH inductance in series with it. The frequency of the system is 60 Hz. This source is delivering power to an unbalanced delta-connected load that has 10 Ω resistor, 10 Ω resistor in series with 30 mH inductor and 10 Ω resistor in series with 1000 µF capacitor between R–Y, Y–B and B–R, respectively. Determine (i) line currents, (ii) source and load branch currents, (iii) line voltages across the load, (iv) active power and reactive power delivered by each phase of the source and the three-phase values and (v) three-phase active power and reactive power delivered to the load by using star-delta transformation on load impedance and mesh analysis.
25. The positive sequence component of line current drawn by a balanced delta-connected load of branch impedance 8 + j6 Ω is 20∠–30° A rms. The negative sequence component of line voltage across the load is 15∠–90° V rms. (i) Find the line voltages across the load and line currents. (ii) Find the three-phase active power and reactive power delivered to the load from three-phase quantities. (iii) Find the three-phase active power and reactive power delivered to the load from sequence components.
26. An unbalanced star-connected voltage source delivers power to a balanced star-connected load with 8 – j6 Ω in each phase over a 3-wire impedance-less connection. The load draws a line current that has 100° A rms positive sequence component and 2∠–90° A negative sequence component. The source neutral is earthed solidly and the load neutral is found to be at 10∠–150° V rms with respect to earth. (i) Find the phase voltages and line voltages across the source and load (ii) Find the active power and reactive power delivered by the source and delivered to the load.
27. An unbalanced star-connected voltage source delivers power to a balanced star-connected load with 7.9 + j5.7 Ω in each phase over a 3-wire connection with a wire-impedance of 0.1 + j0.3 Ω in each wire. The load draws a line current that has 100° A rms positive sequence component and 2∠–90° A rms negative sequence component. The source neutral is earthed solidly and the load neutral is found to be at 1050° V rms with respect to earth. (i) Find the phase voltages and line voltages across the source and load (ii) Find the active power and reactive power delivered by the source and delivered to the load.
28. An unbalanced star-connected voltage source delivers power to a balanced star-connected load with 7.9 + j5.7 Ω in each phase over a 4-wire connection with a wire-impedance of 0.1 + j0.3 Ω in each wire. The load draws a line current that has 100° A rms positive sequence component and 2∠–90° A rms negative sequence component. The neutral return current is found to be at 1.5∠– 100° A rms. (i) Find the phase voltages and line voltages across the source and load (ii) Find the active power and reactive power delivered by the source and delivered to the load.
29. The magnitudes of line voltages at a certain point in a three-phase system are measured as 400 V rms, 380 V rms and 410 V rms between R–Y, Y–B and B–R, respectively. Find the symmetrical components of the line voltage. [Hint: The equation VRY + VYB + VBR = 0 implies that line voltage phasors in a three-phase system will have to form a closed triangle.]
30. The line voltages of an ideal three-phase unbalanced source are measured to be 420 V rms, 390 V rms and 380 V rms between R–Y, Y–B and B–R, respectively. This source drives a balanced RLC load through a 3-wire connection with wire-impedance of 1 + j3 Ω in each wire. If the magnitude of positive sequence component of line voltage across load is found to be 380 V rms find the magnitude of negative sequence component of line voltage across the load.