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Three-phase AC supply

Abdul R. Beig Department of Electrical Engineering, The Petroleum Institute, Abu Dhabi, UAE

Abstract

This chapter presents the basic concepts and analysis of the balanced three-phase circuit. A brief review of generation of balanced three-phase voltages is presented. The three-phase connection of wye- and delta-connected sources, and wye- and delta-connected loads is given. The analysis of different types of three-phase networks is presented. The calculation of active, reactive, and apparent power in a balanced three-phase circuit is presented.

Keywords

three-phase circuits

balanced three-phase source

balanced three-phase load

balanced three-phase system

wye-connected source

delta-connected source

wye-connected load

delta-connected source

active power

reactive power

apparent power

10.1. Introduction

The three-phase system is an economical way of bulk power transmission over long distances and for distribution. The three-phase system consists of a three-phase voltage source connected to a three-phase load by means of transformers and transmission lines. Two types of connections are possible, namely delta (∆) connection and star or wye (Y) connection. The load and the source can be either in delta or star. The transmission line will be in delta connection. Normally the power system is operated in balanced three-phase condition. Most of the bulk loads such as industrial loads are three-phase balanced loads in nature. However, when there is a mix of single-phase and three-phase loads such as residential loads, the load will be unbalanced. Nevertheless, an unbalanced load can be resolved into a set of more than one balanced system. Hence, the analysis of a balanced system at steady state will be enough to understand three-phase systems.

10.2. Generation of three-phase voltages

The bulk of electricity is generated using three-phase alternating current (AC) generators also known as alternators. The AC generators are driven by hydroturbines, gas turbines, steam turbines, or internal combustion engines. AC generators have three separate windings distributed around the inner periphery of the stator [1]. The three windings corresponding to three phases of the system are electrically separated by 120°. The rotor consists of field windings or electromagnets and is rotated at synchronous speed by the prime mover. The rotatory magnetic field cuts the windings in the stator and according to Faraday’s law [1,2] voltages are induced in these windings.

The three-phase windings are designed such that the induced three-phase voltages are sinusoidal, have same frequency, same peak voltage, and are phase displaced by 120°. The three-phase voltages under balanced conditions are defined as in (10.1) and the waveforms are shown in Figure 10.1.

$\begin{array}{l} v_{a} = V_{m} \sin ω t v_{b} = V_{m} \sin (ω t - \frac{2 π}{3}) v_{c} = V_{m} \sin (ω t - \frac{4 π}{3}) or \\ v_{c} = V_{m} \sin (ω t + \frac{2 π}{3}) \end{array}$ $\begin{array}{l} v_{a} = V_{m} \sin ω t v_{b} = V_{m} \sin (ω t - \frac{2 π}{3}) v_{c} = V_{m} \sin (ω t - \frac{4 π}{3}) or \\ v_{c} = V_{m} \sin (ω t + \frac{2 π}{3}) \end{array}$

(10.1)

Figure 10.1 Waveforms of induced voltage in three-phase windings.

Let us denote these three phases as a, b, and c. Phase a voltage is taken as reference, then phase b voltage lags phase “a” by 120° (or 2π/3 rad) and phase “c” voltage lags phase “a” voltage by 240° (or 4π/3 rad). The phase sequence is a–b–c. Instead of a–b–c notation, the phases can also be donated by R (red)–Y (yellow)–B (blue) color code. The earlier mentioned phase sequence is referred to as positive phase sequence. If any of the phases are interchanged (e.g., phase “b” and phase “c” are interchanged) that is phase “c” lags phase “a” by 120° and phase “b” lags phase “a” by 240° then the resulting a-c-b phase sequence is referred as negative phase sequence.

For the past the few decades, major emphasis has been given on renewable energy sources such as wind energy, photovoltaic systems, fuel cells, etc. The generation of electricity from those sources is explained in Chapters 3, 4, and 6, respectively. In these types of electric sources three-phase electricity is generated with the help of direct current (DC) to AC converters [3]. The generation of balanced three-phase sinusoidal power from DC sources is explained in Chapter 15.

10.3. Connections of three-phase circuits

The three-phase sources can be connected either in delta or wye (or star-Y). Similarly the three-phase load can be connected either in delta or wye [4,5].

10.3.1. Wye-connected balanced source

A wye-connected balanced source is shown in Figure 10.2. In Y connection, the common terminal “n” is known as the neutral terminal of the source. Z_sa, Z_sb, and Z_sc are the source impedance of the sources in phases a, b, and c. For balanced sources, Z_sa = Z_sb = Z_sc = Z_s. The voltages v_a, v_b, and v_c are the source internal voltages (or the induced emf [electromotive force]) of the three phases a, b, and c. For balanced source, all three voltages will have the same frequency and amplitude, and are phase displaced by 120° (2π/3 rad) as defined in (10.1).

Figure 10.2 Wye-connected balanced source.

10.3.1.1. Phase voltages

Phase voltages are the voltages at the terminals “a,” “b,” and “c” measured with reference to neutral point “n.”

In Figure 10.2, under open circuit conditions:

$\begin{array}{l} v_{a n} = V_{m} \sin ω t \\ v_{b n} = V_{m} \sin (ω t - \frac{2 π}{3}) v_{c n} = V_{m} \sin (ω t - \frac{4 π}{3}) \end{array}$ $\begin{array}{l} v_{a n} = V_{m} \sin ω t \\ v_{b n} = V_{m} \sin (ω t - \frac{2 π}{3}) v_{c n} = V_{m} \sin (ω t - \frac{4 π}{3}) \end{array}$

(10.2)

The phase voltages in phasor form are:

$V_{a n} = V_{\emptyset} ∠ 0 °, V_{b n} = V_{\emptyset} ∠ - 120 ° and V_{c n} = V_{\emptyset} ∠ - 240 °$ $V_{a n} = V_{\emptyset} ∠ 0 °, V_{b n} = V_{\emptyset} ∠ - 120 ° and V_{c n} = V_{\emptyset} ∠ - 240 °$

where V_∅ is the RMS value of the phase voltage and

V_{\emptyset} = \frac{V_{m}}{\sqrt{2}}

$V_{\emptyset} = \frac{V_{m}}{\sqrt{2}}$

Under balanced conditions, v_an + v_bn + v_cn = 0 and V_an + V_bn + V_cn = 0.

10.3.1.2. Line voltages

The voltage difference between two line terminals or two phases is the line voltage. Following the Kirchhoff’s voltage law, line voltages v_ab, v_bc, and v_ca can be expressed in terms of phase voltages as in (10.3):

$\begin{array}{l} v_{ab} = v_{a n} - v_{b n} = V_{m} \sin (ω t) - V_{m} \sin (ω t - \frac{2 π}{3}) = \sqrt{3} V_{m} \sin (ω t + \frac{π}{6}) \\ v_{bc} = v_{b n} - v_{c n} = V_{m} \sin (ω t - \frac{2 π}{3}) - V_{m} \sin (ω t - \frac{4 π}{3}) = \sqrt{3} V_{m} \sin (ω t - \frac{π}{2}) \\ v_{ca} = v_{c n} - v_{a n} = V_{m} \sin (ω t - \frac{4 π}{3}) - V_{m} \sin (ω t) = \sqrt{3} V_{m} \sin (ω t - \frac{7 π}{6}) \end{array}$ $\begin{array}{l} v_{ab} = v_{a n} - v_{b n} = V_{m} \sin (ω t) - V_{m} \sin (ω t - \frac{2 π}{3}) = \sqrt{3} V_{m} \sin (ω t + \frac{π}{6}) \\ v_{bc} = v_{b n} - v_{c n} = V_{m} \sin (ω t - \frac{2 π}{3}) - V_{m} \sin (ω t - \frac{4 π}{3}) = \sqrt{3} V_{m} \sin (ω t - \frac{π}{2}) \\ v_{ca} = v_{c n} - v_{a n} = V_{m} \sin (ω t - \frac{4 π}{3}) - V_{m} \sin (ω t) = \sqrt{3} V_{m} \sin (ω t - \frac{7 π}{6}) \end{array}$

(10.3)

The line voltages in phasor form are:

$V_{ab} = V_{L} ∠ 30 °, V_{bc} = V_{L} ∠ - 90 ° and V_{ca} = V_{L} ∠ - 210 °,$ $V_{ab} = V_{L} ∠ 30 °, V_{bc} = V_{L} ∠ - 90 ° and V_{ca} = V_{L} ∠ - 210 °,$

where V_L is the RMS value of the line voltage and

|V_{L}| = (\frac{\sqrt{3} V_{m}}{\sqrt{2}}) = \sqrt{3} | V_{\emptyset} |

$|V_{L}| = (\frac{\sqrt{3} V_{m}}{\sqrt{2}}) = \sqrt{3} | V_{\emptyset} |$

Note: Under balanced conditions, v_ab + v_bc + v_ca = 0 and V_ab + V_bc + V_ca = 0.

The phasor diagram of the phase voltages and line voltages in a wye-connected source is shown in Figure 10.3.

Figure 10.3 Phasor diagram of line and phase voltage vectors of wye-connected sources.

10.3.1.3. Phase currents

The current flowing in the source winding or the source is the phase current. Let us assume that the phase current in the source is given by (10.4)

$\begin{array}{l} i_{\emptyset a} = I_{m} \sin (ω t - θ) \\ i_{\emptyset b} = I_{m} \sin (ω t - \frac{2 π}{3} - θ) \\ i_{\emptyset c} = I_{m} \sin (ω t - \frac{4 π}{3} - θ) \end{array}$ $\begin{array}{l} i_{\emptyset a} = I_{m} \sin (ω t - θ) \\ i_{\emptyset b} = I_{m} \sin (ω t - \frac{2 π}{3} - θ) \\ i_{\emptyset c} = I_{m} \sin (ω t - \frac{4 π}{3} - θ) \end{array}$

(10.4)

where θ is the phase angle between the phase voltage and corresponding phase current. In phasor form the above can be written as:

$I_{\emptyset a} = I_{\emptyset} ∠ - θ °, I_{\emptyset b} = I_{\emptyset} ∠ - 120 ° - θ °, and I_{\emptyset c} = I_{\emptyset} ∠ - 240 ° - θ °,$ $I_{\emptyset a} = I_{\emptyset} ∠ - θ °, I_{\emptyset b} = I_{\emptyset} ∠ - 120 ° - θ °, and I_{\emptyset c} = I_{\emptyset} ∠ - 240 ° - θ °,$

where

|I_{\emptyset}| = \frac{I_{m}}{\sqrt{2}}

$|I_{\emptyset}| = \frac{I_{m}}{\sqrt{2}}$

Note : i_{\emptyset a +} i_{\emptyset b} + i_{\emptyset c} = 0

$Note : i_{\emptyset a +} i_{\emptyset b} + i_{\emptyset c} = 0$

10.3.1.4. Line currents

The current flowing out of the terminals is known as the line current. For a wye-connected source the line current is equal to the phase current:

$I_{La} = I_{\emptyset a}, I_{Lb} = I_{\emptyset b}, I_{Lc} = I_{\emptyset c}, and I_{L} = I_{\emptyset}$ $I_{La} = I_{\emptyset a}, I_{Lb} = I_{\emptyset b}, I_{Lc} = I_{\emptyset c}, and I_{L} = I_{\emptyset}$

(10.5)

Note: In a balanced wye-connected system:

• The line voltage leads the corresponding phase voltage by 30° and the magnitude of the line voltage is

\sqrt{3}

$\sqrt{3}$

times the magnitude of the phase voltage.

• The phase current is equal to the line current.

10.3.2. Delta-connected balanced source

A delta-connected balanced source is shown in Figure 10.4. Z_sa, Z_sb, and Z_sc are the source impedance of phases a, b, and c and for balanced sources, Z_sa = Z_sb = Z_sc = Z_s. The voltages v_a, v_b, and v_c are internal voltages of the source (or the induced emf) of the three phases a, b, and c. For balanced source, all three voltages will have the same frequency, equal amplitude, and phase displaced by 120° (2π/3 rad) as defined in (10.1).

Figure 10.4 Delta-connected balanced source.

10.3.2.1. Line voltages

The windings of each phase in delta connections now appear between two lines as shown in Figure 10.4. Under open-circuit conditions, the phase voltage for a delta-connected source is given by:

$\begin{array}{l} v_{ab} = v_{a} = V_{m} \sin ω t \\ v_{bc} = v_{b} = V_{m} \sin (ω t - \frac{2 π}{3}) \\ v_{ca} = v_{c} = V_{m} \sin (ω t - \frac{4 π}{3}) \end{array}$ $\begin{array}{l} v_{ab} = v_{a} = V_{m} \sin ω t \\ v_{bc} = v_{b} = V_{m} \sin (ω t - \frac{2 π}{3}) \\ v_{ca} = v_{c} = V_{m} \sin (ω t - \frac{4 π}{3}) \end{array}$

(10.6)

The line voltages in phasor form are:

$V_{ab} = V_{L} ∠ 0 °, V_{bc} = V_{L} ∠ - 120 °, and V_{ca} = V_{L} ∠ - 240 °$ $V_{ab} = V_{L} ∠ 0 °, V_{bc} = V_{L} ∠ - 120 °, and V_{ca} = V_{L} ∠ - 240 °$

where V_L is the RMS value of the line voltage and

|V_{L}| = \frac{V_{m}}{\sqrt{2}}

$|V_{L}| = \frac{V_{m}}{\sqrt{2}}$

Note:

v_{ab} + v_{bc} + v_{ca} = 0 so i_{cir} = 0

$v_{ab} + v_{bc} + v_{ca} = 0 so i_{cir} = 0$

10.3.2.2. Phase voltages

For a delta-connected source, the phase voltages are equal to line voltages:

$V_{\emptyset} = V_{L}$ $V_{\emptyset} = V_{L}$

(10.7)

10.3.2.3. Phase currents

The current flowing in the source winding or the source is the phase current. Let us assume that the phase current in the source is given by (10.8):

(10.8)

where θ is the phase angle between the phase voltage and corresponding phase current. The phase currents in phasor form are:

$\begin{array}{c} I_{\emptyset a} = I_{\emptyset} ∠ - θ °, I_{\emptyset b} = I_{\emptyset} ∠ (- 120 ° - θ) ° \\ and I_{\emptyset c} = I_{\emptyset} ∠ (- 240 ° - θ °), where |I_{\emptyset}| = \frac{I_{m}}{\sqrt{2}} \end{array} .$ $\begin{array}{c} I_{\emptyset a} = I_{\emptyset} ∠ - θ °, I_{\emptyset b} = I_{\emptyset} ∠ (- 120 ° - θ) ° \\ and I_{\emptyset c} = I_{\emptyset} ∠ (- 240 ° - θ °), where |I_{\emptyset}| = \frac{I_{m}}{\sqrt{2}} \end{array} .$

10.3.2.4. Line currents

The current flowing out of the terminals is the line current. For a delta- connected source the line current is given by (10.9).

Applying Kirchhoff’s current law at nodes a, b, and c, respectively, we get:

$\begin{array}{l} i_{La} = i_{\emptyset a} - i_{\emptyset c} = I_{m} \sin (ω t - θ) - I_{m} \sin (ω t - \frac{2 π}{3} \pm θ) = \sqrt{3} I_{m} \sin (ω t - \frac{π}{6} \pm θ) \\ i_{Lb} = i_{\emptyset b} - i_{\emptyset a} = I_{m} \sin (ω t - \frac{2 π}{3} \pm θ) - I_{m} \sin (ω t \pm θ) = \sqrt{3} I_{m} \sin (ω t - \frac{5 π}{6} \pm θ) \\ i_{Lc} = i_{\emptyset c} - i_{\emptyset b} = I_{m} \sin (ω t - \frac{4 π}{3} \pm θ) - I_{m} \sin (ω t \pm θ) = \sqrt{3} I_{m} \sin (t - \frac{3 π}{2} \pm θ) \end{array}$ $\begin{array}{l} i_{La} = i_{\emptyset a} - i_{\emptyset c} = I_{m} \sin (ω t - θ) - I_{m} \sin (ω t - \frac{2 π}{3} \pm θ) = \sqrt{3} I_{m} \sin (ω t - \frac{π}{6} \pm θ) \\ i_{Lb} = i_{\emptyset b} - i_{\emptyset a} = I_{m} \sin (ω t - \frac{2 π}{3} \pm θ) - I_{m} \sin (ω t \pm θ) = \sqrt{3} I_{m} \sin (ω t - \frac{5 π}{6} \pm θ) \\ i_{Lc} = i_{\emptyset c} - i_{\emptyset b} = I_{m} \sin (ω t - \frac{4 π}{3} \pm θ) - I_{m} \sin (ω t \pm θ) = \sqrt{3} I_{m} \sin (t - \frac{3 π}{2} \pm θ) \end{array}$

(10.9)

The line currents of a delta-connected source in phasor form are written as:

$\begin{array}{l} I_{La} = I_{L} ∠ \pm θ °, I_{Lb} = I_{L} ∠ - 120 ° \pm θ ° and I_{L c} = I_{L} ∠ - 240 ° \pm θ °, \\ where |I_{L}| = \sqrt{3} | I_{\emptyset} | = \frac{I_{m}}{\sqrt{2}} \end{array}$ $\begin{array}{l} I_{La} = I_{L} ∠ \pm θ °, I_{Lb} = I_{L} ∠ - 120 ° \pm θ ° and I_{L c} = I_{L} ∠ - 240 ° \pm θ °, \\ where |I_{L}| = \sqrt{3} | I_{\emptyset} | = \frac{I_{m}}{\sqrt{2}} \end{array}$

Note: In a balanced delta-connected system:

• The line current lags the corresponding phase current by 30° and the magnitude of the line current is

\sqrt{3}

$\sqrt{3}$

times the magnitude of the phase current.

• The phase voltage is equal to the line voltage.

The phasor diagram of voltages and currents in a delta-connected source is given in Figure 10.5.

Figure 10.5 Phasor diagram of line currents and phase currents in a delta-connected source.

10.3.3. Wye-connected balanced load

A wye-connected load is shown in Figure 10.6. A, B, and C are the terminals of phase a, phase b, and phase c, respectively The three loads are connected at common point “N” called the neutral point of the load. Z_A, Z_B, and Z_C are the source impedance of phases a, b, and c. For a balanced load, Z_A = Z_B = Z_C = Z. If we take the phase voltage V_AN as reference, then

$V_{AN} = V_{\emptyset} ∠ 0 °, V_{BN} = V_{\emptyset} ∠ - 120 ° and V_{CN} = V_{\emptyset} ∠ - 240 ° .$ $V_{AN} = V_{\emptyset} ∠ 0 °, V_{BN} = V_{\emptyset} ∠ - 120 ° and V_{CN} = V_{\emptyset} ∠ - 240 ° .$

The line voltages are:

$\begin{array}{l} V_{AB} = V_{AN} - V_{BN} = \sqrt{3} | V_{\emptyset} | ∠ 30 ° = |V_{L}| ∠ 30 °, \\ V_{BC} = V_{BN} - V_{CN} = \sqrt{3} | V_{\emptyset} | ∠ - 90 ° = |V_{L}| ∠ - 90 °, \\ V_{CA} = V_{CN} - V_{AN} = \sqrt{3} | V_{\emptyset} | ∠ - 210 ° = |V_{L}| ∠ - 210 ° \\ and \\ |V_{L}| = \sqrt{3} |V_{\emptyset}| \end{array}$ $\begin{array}{l} V_{AB} = V_{AN} - V_{BN} = \sqrt{3} | V_{\emptyset} | ∠ 30 ° = |V_{L}| ∠ 30 °, \\ V_{BC} = V_{BN} - V_{CN} = \sqrt{3} | V_{\emptyset} | ∠ - 90 ° = |V_{L}| ∠ - 90 °, \\ V_{CA} = V_{CN} - V_{AN} = \sqrt{3} | V_{\emptyset} | ∠ - 210 ° = |V_{L}| ∠ - 210 ° \\ and \\ |V_{L}| = \sqrt{3} |V_{\emptyset}| \end{array}$

When a wye-connected balanced load is supplied by the balanced source then the phase currents in the load will also be balanced.

$I_{\emptyset A} = I_{\emptyset} ∠ \pm θ °, I_{\emptyset B} = I_{\emptyset} ∠ - 120 ° \pm θ °, and I_{\emptyset C} = I_{\emptyset} ∠ - 240 ° \pm θ °,$ $I_{\emptyset A} = I_{\emptyset} ∠ \pm θ °, I_{\emptyset B} = I_{\emptyset} ∠ - 120 ° \pm θ °, and I_{\emptyset C} = I_{\emptyset} ∠ - 240 ° \pm θ °,$

where θ is the impedance angle, same as the phase angle between the phase voltage and the load current.

For a wye-connected load, the line currents are given by:

i_{LA} = i_{\emptyset A}, i_{LB} = i_{\emptyset B}, i_{LC} = i_{\emptyset C}, and I_{L} = I_{\emptyset}

$i_{LA} = i_{\emptyset A}, i_{LB} = i_{\emptyset B}, i_{LC} = i_{\emptyset C}, and I_{L} = I_{\emptyset}$

10.3.4. Delta-connected balanced load

A delta-connected load is shown in Figure 10.7. A, B, and C are the terminals of phase a, phase b, and phase c, respectively. Z_A, Z_B, and Z_C are the respective load impedances. For balanced sources, Z_A = Z_B = Z_C = Z. When the balanced load is supplied by a balanced source, the line currents as well as the phase currents in the load are balanced.

For a delta-connected load, the line voltages are given by:

$V_{AB} = V_{L} ∠ 0 °, V_{BC} = V_{L} ∠ - 120 °, and V_{CA} = V_{L} ∠ - 240 °,$ $V_{AB} = V_{L} ∠ 0 °, V_{BC} = V_{L} ∠ - 120 °, and V_{CA} = V_{L} ∠ - 240 °,$

For a delta-connected load, V_L = V_∅.

When a delta-connected balanced load is supplied by a balanced source then the phase currents in the load will also be balanced:

where θ is the impedance angle.

For a delta-connected load, the line currents are given by:

$\begin{array}{l} I_{LA} = I_{\emptyset A} - I_{\emptyset B} = \sqrt{3} |I_{\emptyset}| ∠ - 30 \pm θ ° \\ I_{LB} = I_{\emptyset B} - I_{\emptyset C} = \sqrt{3} |I_{\emptyset}| ∠ - 150 \pm θ ° \\ I_{LC} = I_{\emptyset C} - I_{\emptyset A} = \sqrt{3} |I_{\emptyset}| ∠ - 270 \pm θ ° \\ and |I_{L}| = \sqrt{3} |I_{\emptyset}| \end{array}$ $\begin{array}{l} I_{LA} = I_{\emptyset A} - I_{\emptyset B} = \sqrt{3} |I_{\emptyset}| ∠ - 30 \pm θ ° \\ I_{LB} = I_{\emptyset B} - I_{\emptyset C} = \sqrt{3} |I_{\emptyset}| ∠ - 150 \pm θ ° \\ I_{LC} = I_{\emptyset C} - I_{\emptyset A} = \sqrt{3} |I_{\emptyset}| ∠ - 270 \pm θ ° \\ and |I_{L}| = \sqrt{3} |I_{\emptyset}| \end{array}$

10.4. Circuits with mixed connections

There are mainly four types of networks in three-phase systems and these are listed in Table 10.1.

Table 10.1

Types of three-phase networks

Type of connection	Source	Load
Wye–wye connection	Wye	Wye
Wye–delta connection	Wye	Delta
Delta–wye connection	Delta	Wye
Delta–delta connection	Delta	Delta

The delta-connected source or load can be transformed into an equivalent wye-connected source or load, respectively [4,5]. Hence, all the other three connections can be analyzed by converting them into an equivalent wye–wye network. In the following subsection analysis of a wye–wye network is presented.

10.4.1. Wye–wye network

Line currents (I_L) are the currents flowing in each line and phase current (I_∅) is the current flowing in an individual source or load. For simplicity we limit our analysis to passive loads, that is, loads consisting of R, L, and C components only.

The circuit in Figure 10.8 shows a typical wye–wye network of the power system.

Z_sa = Z_sb = Z_sc = Z_s and Z_La = Z_Lb = Z_Lc = Z_L are the impedances of the line connecting the source to load; these are equal and referred to as line impedances.

Z_A = Z_B = Z_C = Z are the load impedances connected in wye. N is the neutral point of the load.

Applying Kirchhoff’s current law at source neutral “n,” the node equation is:

$\begin{array}{l} \frac{V_{a' n} - V_{N n}}{Z_{sa} + Z_{La} + Z_{A}} - \frac{V_{b' n} - V_{Nn}}{Z_{sb} + Z_{Lb} + Z_{B}} + \frac{V_{c' n} - V_{Nn}}{Z_{sc} + Z_{Lc} + Z_{C}} = \frac{V_{Nn}}{Z_{Ln}} \\ \frac{V_{a' n} + V_{b' n} + V_{c' n}}{Z_{s} + Z_{L} + Z_{}} = \frac{V_{Nn}}{Z_{Ln}} + \frac{V_{Nn}}{Z_{s} + Z_{L} + Z_{}} \end{array}$ $\begin{array}{l} \frac{V_{a' n} - V_{N n}}{Z_{sa} + Z_{La} + Z_{A}} - \frac{V_{b' n} - V_{Nn}}{Z_{sb} + Z_{Lb} + Z_{B}} + \frac{V_{c' n} - V_{Nn}}{Z_{sc} + Z_{Lc} + Z_{C}} = \frac{V_{Nn}}{Z_{Ln}} \\ \frac{V_{a' n} + V_{b' n} + V_{c' n}}{Z_{s} + Z_{L} + Z_{}} = \frac{V_{Nn}}{Z_{Ln}} + \frac{V_{Nn}}{Z_{s} + Z_{L} + Z_{}} \end{array}$

For a balanced source

V_{a' n} + V_{b' n} + V_{c' n} = 0

$V_{a' n} + V_{b' n} + V_{c' n} = 0$

$\frac{V_{Nn}}{Z_{Ln}} + \frac{V_{Nn}}{Z_{s} + Z_{L} + Z_{}} = 0$ $\frac{V_{Nn}}{Z_{Ln}} + \frac{V_{Nn}}{Z_{s} + Z_{L} + Z_{}} = 0$

So, for a balanced three-phase circuit, V_Nn = 0.

This means that there is no voltage difference between the source neutral “n,” and the load neutral “N.” Consequently, I_N = 0, Hence, we can remove the neutral line from the balanced wye–wye configuration resulting in a three-phase –three-wire wye–wye connection or we can connect n to N resulting in a three-phase–four-wire wye–wye connection [1,4,5].

10.4.1.1. Line currents and phase currents

The line currents (I_LA, I_LB, and I_LC) are the currents flowing out of source terminals and into the load terminals through the line impedance.

In Figure 10.8, the voltages at the source in each phase are defined as:

V_{a' n} = |V_{\emptyset}| ∠ 0 °, V_{b' n} = |V_{\emptyset}| ∠ - 120 °, and V_{c' n} = |V_{\emptyset}| ∠ - 240 °

$V_{a' n} = |V_{\emptyset}| ∠ 0 °, V_{b' n} = |V_{\emptyset}| ∠ - 120 °, and V_{c' n} = |V_{\emptyset}| ∠ - 240 °$

The voltage

V_{a' n}

$V_{a' n}$

is taken as the reference vector.

Let us define:

$Z_{\emptyset} = Z_{s} + Z_{L} + Z_{} = |Z_{\emptyset}| ∠ θ$ $Z_{\emptyset} = Z_{s} + Z_{L} + Z_{} = |Z_{\emptyset}| ∠ θ$

$\begin{array}{l} I_{sA} = I_{LA} = I_{A} = \frac{V_{a' n}}{Z_{sa} + Z_{La} + Z_{A}} = \frac{V_{\emptyset}}{Z_{sa} + Z_{La} + Z_{A}} = \frac{|V_{\emptyset}| ∠ (0 - \emptyset) °}{|Z_{\emptyset}|} = I_{\emptyset} ∠ - \emptyset °, \\ where |I_{\emptyset}| = \frac{|V_{\emptyset}|}{|Z_{\emptyset}|} \end{array}$ $\begin{array}{l} I_{sA} = I_{LA} = I_{A} = \frac{V_{a' n}}{Z_{sa} + Z_{La} + Z_{A}} = \frac{V_{\emptyset}}{Z_{sa} + Z_{La} + Z_{A}} = \frac{|V_{\emptyset}| ∠ (0 - \emptyset) °}{|Z_{\emptyset}|} = I_{\emptyset} ∠ - \emptyset °, \\ where |I_{\emptyset}| = \frac{|V_{\emptyset}|}{|Z_{\emptyset}|} \end{array}$

$\begin{array}{l} I_{sB} = I_{LB} = I_{B} = \frac{V_{b' n}}{Z_{sb} + Z_{Lb} + Z_{B}} = \frac{V_{\emptyset}}{Z_{sb} + Z_{Lb} + Z_{B}} = \frac{|V_{\emptyset}| ∠ 0 ° - \emptyset °}{|Z_{\emptyset}|} = I_{\emptyset} ∠ - \emptyset ° \\ I_{sC} = I_{LC} = I_{C} = \frac{V_{c' n}}{Z_{sc} + Z_{Lc} + Z_{C}} = \frac{V_{\emptyset}}{Z_{sc} + Z_{Lc} + Z_{C}} = \frac{|V_{\emptyset}| ∠ 0 ° - \emptyset °}{|Z_{\emptyset}|} = I_{\emptyset} ∠ - \emptyset ° \\ I_{A} + I_{B} + I_{C} = 0 \end{array}$ $\begin{array}{l} I_{sB} = I_{LB} = I_{B} = \frac{V_{b' n}}{Z_{sb} + Z_{Lb} + Z_{B}} = \frac{V_{\emptyset}}{Z_{sb} + Z_{Lb} + Z_{B}} = \frac{|V_{\emptyset}| ∠ 0 ° - \emptyset °}{|Z_{\emptyset}|} = I_{\emptyset} ∠ - \emptyset ° \\ I_{sC} = I_{LC} = I_{C} = \frac{V_{c' n}}{Z_{sc} + Z_{Lc} + Z_{C}} = \frac{V_{\emptyset}}{Z_{sc} + Z_{Lc} + Z_{C}} = \frac{|V_{\emptyset}| ∠ 0 ° - \emptyset °}{|Z_{\emptyset}|} = I_{\emptyset} ∠ - \emptyset ° \\ I_{A} + I_{B} + I_{C} = 0 \end{array}$

For a wye-connected load or source, the line current = phase currents.

So I_L = I_∅.

Note: For a balanced three-phase wye–wye network:

• The load currents are balanced, that is, they are equal in magnitude and phase displaced by 120°.

• There is no current in the neutral wire.

We can represent the previously mentioned system by an equivalent per phase network using phase quantities. The per phase equivalent circuit for any one phase (say, for phase a) is given in Figure 10.9.

Figure 10.9 Per-phase equivalent circuit.

10.4.1.2. Line voltages at the load terminals

In Figure 10.8, the phase voltages at load terminal are given by:

$\begin{array}{l} V_{AN} = I_{A} Z \\ V_{BN} = I_{B} Z \\ V_{CN} = I_{C} Z \end{array}$ $\begin{array}{l} V_{AN} = I_{A} Z \\ V_{BN} = I_{B} Z \\ V_{CN} = I_{C} Z \end{array}$

It can be seen that all the three-phase voltages are balanced.

$V_{AN} + V_{BN} + V_{CN} = 0$ $V_{AN} + V_{BN} + V_{CN} = 0$

The line voltages are given by:

$\begin{array}{l} V_{AB} = V_{AN} - V_{BN} \\ V_{BC} = V_{BN} - V_{CN} \\ V_{CA} = V_{CN} - V_{AN} \end{array}$ $\begin{array}{l} V_{AB} = V_{AN} - V_{BN} \\ V_{BC} = V_{BN} - V_{CN} \\ V_{CA} = V_{CN} - V_{AN} \end{array}$

It can be seen that all the three-line voltages are balanced. Hence, V_AB + V_BC + V_BC = 0 and as seen in Section 10.3.1, the line voltages will be leading the corresponding phase voltage by 30° and

|V_{L}| = \sqrt{3} |V_{\emptyset}|

$|V_{L}| = \sqrt{3} |V_{\emptyset}|$

10.4.1.3. Line voltages at the source terminals

In Figure 10.8, the phase voltages at the phase terminal are given by:

$\begin{array}{l} V_{an} = V_{a' n} - I_{sa} Z_{sa} = V_{a' n} - I_{A} Z_{s} \\ V_{bn} = V_{b' n} - I_{sb} Z_{sb} = V_{b' n} - I_{B} Z_{s} \\ V_{cn} = V_{c' n} - I_{sc} Z_{sc} = V_{c' n} - I_{C} Z_{s} \end{array}$ $\begin{array}{l} V_{an} = V_{a' n} - I_{sa} Z_{sa} = V_{a' n} - I_{A} Z_{s} \\ V_{bn} = V_{b' n} - I_{sb} Z_{sb} = V_{b' n} - I_{B} Z_{s} \\ V_{cn} = V_{c' n} - I_{sc} Z_{sc} = V_{c' n} - I_{C} Z_{s} \end{array}$

It can be seen that all the three-phase voltages are balanced:

$V_{an} + V_{bn} + V_{cn} = 0$ $V_{an} + V_{bn} + V_{cn} = 0$

The line voltages at the source terminals are given by:

$\begin{array}{l} V_{ab} = V_{an} - V_{bn} \\ V_{bc} = V_{bn} - V_{cn} \\ V_{ca} = V_{cn} - V_{an} \end{array}$ $\begin{array}{l} V_{ab} = V_{an} - V_{bn} \\ V_{bc} = V_{bn} - V_{cn} \\ V_{ca} = V_{cn} - V_{an} \end{array}$

It can be seen that all the three-line voltages are balanced.

As seen in Section 10.3.1, the line voltage is leading the corresponding phase voltage by 30° and

|V_{L}| = \sqrt{3} |V_{\emptyset}|

$|V_{L}| = \sqrt{3} |V_{\emptyset}|$

Example 10.1

A balanced three-phase wye-connected generator has an internal impedance of 0.2 + j0.5 Ω per phase and the phase to neutral voltage is 220 V. The generator feeds a balanced three-phase wye-connected load having a load impedance of 10 + j8 Ω in each phase. The impedance of the line connecting the generator to the load is 0.2 + j0.8 Ω in each line. The a-phase internal voltage of the generator is specified as the reference phasor.

1. Construct a phase equivalent circuit of the system.

2. Calculate the line currents I_LA, I_LB, and I_LC.

3. Calculate the phase voltages at the load, V_AN, V_BN, and V_CN.

4. Calculate the line voltages V_AC, V_BC, and V_CA at the load terminals.

5. Calculate the phase voltages at the generator terminals, V_an, V_bn, and V_cn.

6. Calculate the line voltages at the generator terminals, V_ab, V_bc, and V_ca.

Solution

1. a-Phase equivalent circuit (Figure 10.10):

2. Line currents I_LA, I_LB, and I_LC:

$\begin{array}{l} I_{LA} = \frac{V_{an}}{Z_{\emptyset}} = \frac{220 ∠ 0 °}{(0.2 + 0.2 + 10) + j (0.5 + 0.8 + 8)} = \frac{220 ∠ 0 °}{13.95 ∠ 41.8 °} = 15.77 ∠ - 41.8 ° A \\ I_{LB} = 15.77 ∠ (- 41.8 ° - 120 °) = 15.77 ∠ - 161.8 ° A \\ I_{LC} = 15.77 ∠ (- 41.8 ° + 120 °) = 15.77 ∠ 78.2 ° A \end{array}$ $\begin{array}{l} I_{LA} = \frac{V_{an}}{Z_{\emptyset}} = \frac{220 ∠ 0 °}{(0.2 + 0.2 + 10) + j (0.5 + 0.8 + 8)} = \frac{220 ∠ 0 °}{13.95 ∠ 41.8 °} = 15.77 ∠ - 41.8 ° A \\ I_{LB} = 15.77 ∠ (- 41.8 ° - 120 °) = 15.77 ∠ - 161.8 ° A \\ I_{LC} = 15.77 ∠ (- 41.8 ° + 120 °) = 15.77 ∠ 78.2 ° A \end{array}$

3. Phase voltages at the load, V_AN, V_BN, and V_CN:

$\begin{array}{l} V_{AN} = I_{LA} Z_{A} = (15.77 ∠ - 41.8 °) (10 + j 8) = 202 ∠ - 3.14 ° V \\ V_{BN} = 202 ∠ (- 3.14 ° - 120 °) = 202 ∠ - 123.14 ° V \\ V_{CN} = 202 ∠ (- 3.14 ° + 120 °) = 202 ∠ 116.86 ° V \end{array}$ $\begin{array}{l} V_{AN} = I_{LA} Z_{A} = (15.77 ∠ - 41.8 °) (10 + j 8) = 202 ∠ - 3.14 ° V \\ V_{BN} = 202 ∠ (- 3.14 ° - 120 °) = 202 ∠ - 123.14 ° V \\ V_{CN} = 202 ∠ (- 3.14 ° + 120 °) = 202 ∠ 116.86 ° V \end{array}$

4. Line voltages V_AB, V_BC, and V_CA at the load terminals:

$\begin{array}{l} V_{AB} = \sqrt{3} V_{AN} ∠ 30 ° = \sqrt{3} \times 202 ∠ (- 3.14 ° + 30 °) = 349.9 ∠ 26.86 ° V \\ V_{BC} = \sqrt{3} V_{BN} ∠ 30 ° = \sqrt{3} \times 202 ∠ (- 123.14 ° + 30 °) = 349.9 ∠ - 93.14 ° V \\ V_{CA} = \sqrt{3} V_{CN} ∠ 30 ° = \sqrt{3} \times 202 ∠ (116.86 ° + 30 °) = 349.9 ∠ 146.86 ° V \end{array}$ $\begin{array}{l} V_{AB} = \sqrt{3} V_{AN} ∠ 30 ° = \sqrt{3} \times 202 ∠ (- 3.14 ° + 30 °) = 349.9 ∠ 26.86 ° V \\ V_{BC} = \sqrt{3} V_{BN} ∠ 30 ° = \sqrt{3} \times 202 ∠ (- 123.14 ° + 30 °) = 349.9 ∠ - 93.14 ° V \\ V_{CA} = \sqrt{3} V_{CN} ∠ 30 ° = \sqrt{3} \times 202 ∠ (116.86 ° + 30 °) = 349.9 ∠ 146.86 ° V \end{array}$

5. Phase voltages at the generator terminals, V_an, V_bn, and V_cn.

Phase current in the source = I_LA:

$\begin{array}{l} V_{an} = 220 ∠ 0 ° - I_{sa} Z_{s} = 220 ∠ 0 ° - (15.77 ∠ - 41.8 °) (0.2 + j 0.5) = 212.43 ∠ - 1.02 ° V \\ V_{bn} = 212.43 ∠ (- 1.02 ° - 120 °) = 212.43 ∠ - 121.02 ° V \\ V_{cn} = 212.43 ∠ (- 1.02 ° + 120 °) = 212.43 ∠ 118.98 ° V \end{array}$ $\begin{array}{l} V_{an} = 220 ∠ 0 ° - I_{sa} Z_{s} = 220 ∠ 0 ° - (15.77 ∠ - 41.8 °) (0.2 + j 0.5) = 212.43 ∠ - 1.02 ° V \\ V_{bn} = 212.43 ∠ (- 1.02 ° - 120 °) = 212.43 ∠ - 121.02 ° V \\ V_{cn} = 212.43 ∠ (- 1.02 ° + 120 °) = 212.43 ∠ 118.98 ° V \end{array}$

6. Line voltages at the generator terminals, V_ab, V_bc, and V_ca:

$\begin{array}{l} V_{ab} = \sqrt{3} V_{an} ∠ 30 ° = \sqrt{3} \times 212.43 ∠ (- 1.02 ° + 30 °) = 367.94 ∠ 28.98 ° V \\ V_{bc} = \sqrt{3} V_{bn} ∠ 30 ° = \sqrt{3} \times 212.43 ∠ (- 121.02 ° + 30 °) = 367.94 ∠ - 91.02 ° V \\ V_{bc} = \sqrt{3} V_{bn} ∠ 30 ° = \sqrt{3} \times 212.43 ∠ (118.98 ° + 30 °) = 367.94 ∠ 148.98 ° V \end{array}$ $\begin{array}{l} V_{ab} = \sqrt{3} V_{an} ∠ 30 ° = \sqrt{3} \times 212.43 ∠ (- 1.02 ° + 30 °) = 367.94 ∠ 28.98 ° V \\ V_{bc} = \sqrt{3} V_{bn} ∠ 30 ° = \sqrt{3} \times 212.43 ∠ (- 121.02 ° + 30 °) = 367.94 ∠ - 91.02 ° V \\ V_{bc} = \sqrt{3} V_{bn} ∠ 30 ° = \sqrt{3} \times 212.43 ∠ (118.98 ° + 30 °) = 367.94 ∠ 148.98 ° V \end{array}$

Figure 10.10 Per-phase circuit for Example 10.1.

10.4.2. Delta to wye conversion

A delta-connected load (or source) can be converted to a wye-connected load (or source) and vice versa. This transformation is particularly useful to convert all the other types of configurations to get equivalent wye–wye configurations.

Assume that Z_D is the load connected in delta as shown in Figure 10.11. The supply voltage is

V_{AB} = V ∠ 0 °, V_{BC} = V ∠ - 120 °, and V_{CA} = V ∠ - 240 °

$V_{AB} = V ∠ 0 °, V_{BC} = V ∠ - 120 °, and V_{CA} = V ∠ - 240 °$

Figure 10.11 Delta-connected load and its wye equivalent.

Let us use the principle of superposition by considering only one source at a time. First consider source V_AB:

$V_{AB} = V ∠ 0 °; V_{BC} = 0 (open circuit); V_{CA} = 0 (open circuit)$ $V_{AB} = V ∠ 0 °; V_{BC} = 0 (open circuit); V_{CA} = 0 (open circuit)$

The current due to the delta-connected load is:

$I_{LA} = \frac{V_{AB}}{Z_{D} | | 2 Z_{D}}$ $I_{LA} = \frac{V_{AB}}{Z_{D} | | 2 Z_{D}}$

Let Z_Y be the equivalent wye-connected load in each phase supplied from the same source.

The current due to the wye-connected load is:

$I_{LA} = \frac{V_{AB}}{Z_{Y} + Z_{Y}}$ $I_{LA} = \frac{V_{AB}}{Z_{Y} + Z_{Y}}$

Both the currents should be equal, hence:

$\begin{array}{l} \frac{V_{AB}}{(Z_{D} \times 2 Z_{D}) / 3 Z_{D}} = \frac{V_{AB}}{2 Z_{Y}} \\ \frac{3}{2 Z_{D}} = \frac{1}{2 Z_{Y}} \\ 3 Z_{Y} = Z_{D} \\ Z_{Y} = \frac{Z_{D}}{3} \end{array}$ $\begin{array}{l} \frac{V_{AB}}{(Z_{D} \times 2 Z_{D}) / 3 Z_{D}} = \frac{V_{AB}}{2 Z_{Y}} \\ \frac{3}{2 Z_{D}} = \frac{1}{2 Z_{Y}} \\ 3 Z_{Y} = Z_{D} \\ Z_{Y} = \frac{Z_{D}}{3} \end{array}$

(10.10)

10.4.3. Wye-connected source/delta-connected load

The wye-connected source and delta-connected load are shown in Figure 10.12. The delta-connected load can be replaced by an equivalent wye-connected load using delta to wye conversion. In Figure 10.12, for a balanced load the load in each phase, Z_A = Z_B = Z_C = Z_D.

Figure 10.12 Wye-connected source and delta-connected load.

Using equation (10.9), the previously mentioned load can be replaced by an equivalent wye-connected load of

Z_{Y} = \frac{Z_{D}}{3}

$Z_{Y} = \frac{Z_{D}}{3}$

Now the resulting wye–wye network can be solved using the procedure given in Section 10.4.1.

The magnitude load current in each phase in a delta-connected load is given by

I_{\emptyset} = \frac{I_{L}}{\sqrt{3}}

$I_{\emptyset} = \frac{I_{L}}{\sqrt{3}}$

and the phase current will be leading the line current by 30°.

Example 10.2

A balanced three-phase wye-connected generator has an internal impedance of 0.2 + j0.5 Ω/Ø and an internal voltage 220 V per phase. The generator feeds a delta-connected load having an impedance of 30 + j24 Ω in each phase. The impedance of the line connecting the generator to the load is 0.1 + j0.9 Ω in each line. The a-phase internal voltage of the generator is specified as the reference phasor.

1. Construct the a-phase equivalent circuit of the system.

2. Calculate the line currents I_LA, I_LB, and I_LC.

3. Calculate the phase voltages at the load, V_AN, V_BN, and V_CN.

4. Calculate the phase currents of the load.

5. Calculate the line voltages at the generator terminals, V_ab, V_bc, and V_ca.

Solution

1. Construct the a-phase equivalent circuit of the system (Figure 10.13):

$Z_{Y} = \frac{Z_{D}}{3} = \frac{30 + j 24}{3} = 10 + j 8$ $Z_{Y} = \frac{Z_{D}}{3} = \frac{30 + j 24}{3} = 10 + j 8$

2. Line currents I_LA, I_LB, and I_LC:

$\begin{array}{l} I_{LA} = \frac{V_{an}}{Z} = \frac{220 ∠ 0 °}{(0.2 + 0.1 + 10) + j (0.5 + 0.9 + 8)} = \frac{220 ∠ 0 °}{14 ∠ 42.38 °} = 15.78 ∠ - 42.38 ° A \\ I_{LB} = 15.78 ∠ (- 42.38 ° - 120 °) = 15.78 ∠ - 162.38 ° A \\ I_{LC} = 15.78 ∠ (- 42.38 ° + 120 °) = 15.78 ∠ 77.62 ° A \end{array}$ $\begin{array}{l} I_{LA} = \frac{V_{an}}{Z} = \frac{220 ∠ 0 °}{(0.2 + 0.1 + 10) + j (0.5 + 0.9 + 8)} = \frac{220 ∠ 0 °}{14 ∠ 42.38 °} = 15.78 ∠ - 42.38 ° A \\ I_{LB} = 15.78 ∠ (- 42.38 ° - 120 °) = 15.78 ∠ - 162.38 ° A \\ I_{LC} = 15.78 ∠ (- 42.38 ° + 120 °) = 15.78 ∠ 77.62 ° A \end{array}$

3. Phase voltages at the load, V_AN, V_BN, and V_CN:

$\begin{array}{l} V_{AN} = I_{aA} Z_{Y} = (15.78 ∠ - 42.38 °) (10 + j 8) = 202 ∠ - 3.72 ° V \\ V_{BN} = 202 ∠ (- 3.72 ° - 120 °) = 202 ∠ - 123.72 ° V \\ V_{CN} = 202 ∠ (- 3.72 ° + 120 °) = 202 ∠ 116.38 ° V \end{array}$ $\begin{array}{l} V_{AN} = I_{aA} Z_{Y} = (15.78 ∠ - 42.38 °) (10 + j 8) = 202 ∠ - 3.72 ° V \\ V_{BN} = 202 ∠ (- 3.72 ° - 120 °) = 202 ∠ - 123.72 ° V \\ V_{CN} = 202 ∠ (- 3.72 ° + 120 °) = 202 ∠ 116.38 ° V \end{array}$

4. Phase currents of the load I_AB, I_BC, and I_CA:

$\begin{array}{l} I_{\emptyset A} = \frac{I_{aA} ∠ 30 °}{\sqrt{3}} = \frac{15.78 ∠ (- 42.38 ° + 30 °)}{\sqrt{3}} = 9.11 ∠ - 12.38 ° A \\ I_{\emptyset B} = 9.11 ∠ (- 12.38 ° - 120 °) = 9.11 ∠ - 132.38 ° A \\ I_{\emptyset C} = 9.11 ∠ (- 12.38 ° + 120 °) = 9.11 ∠ 107.62 ° A \end{array}$ $\begin{array}{l} I_{\emptyset A} = \frac{I_{aA} ∠ 30 °}{\sqrt{3}} = \frac{15.78 ∠ (- 42.38 ° + 30 °)}{\sqrt{3}} = 9.11 ∠ - 12.38 ° A \\ I_{\emptyset B} = 9.11 ∠ (- 12.38 ° - 120 °) = 9.11 ∠ - 132.38 ° A \\ I_{\emptyset C} = 9.11 ∠ (- 12.38 ° + 120 °) = 9.11 ∠ 107.62 ° A \end{array}$

5. Line voltages at the generator terminals V_ab, V_bc, and V_ca.

We start by calculating the phase voltages at the generator terminals, V_an, V_bn, and V_cn:

$\begin{array}{l} V_{an} = 220 ∠ 0 ° - I_{LA} Z_{an} = 220 ∠ 0 ° - (15.77 ∠ - 42.38 °) (0.2 + j 0.5) = 212.38 ∠ - 0.998 ° V \\ V_{ab} = \sqrt{3} V_{an} ∠ 30 ° = \sqrt{3} \times 212.38 ∠ (- 0.998 ° + 30 °) = 367.86 ∠ 29.002 ° V \\ V_{bc} = 367.86 ∠ (29.002 ° - 120 °) = 367.86 ∠ - 90.998 ° V \\ V_{ca} = 367.86 ∠ (29.002 ° + 120 °) = 367.86 ∠ 149.002 ° V \end{array}$ $\begin{array}{l} V_{an} = 220 ∠ 0 ° - I_{LA} Z_{an} = 220 ∠ 0 ° - (15.77 ∠ - 42.38 °) (0.2 + j 0.5) = 212.38 ∠ - 0.998 ° V \\ V_{ab} = \sqrt{3} V_{an} ∠ 30 ° = \sqrt{3} \times 212.38 ∠ (- 0.998 ° + 30 °) = 367.86 ∠ 29.002 ° V \\ V_{bc} = 367.86 ∠ (29.002 ° - 120 °) = 367.86 ∠ - 90.998 ° V \\ V_{ca} = 367.86 ∠ (29.002 ° + 120 °) = 367.86 ∠ 149.002 ° V \end{array}$

Figure 10.13 Single-phase equivalent circuit for Example 10.2.

10.4.4. Delta–wye network

The delta–wye network is shown in Figure 10.14. The delta-connected source can be represented by the equivalent wye-connected source. The resulting equivalent circuit can be solved similar to the wye–wye network.

10.4.5. Delta–delta network

The delta–delta network is shown in Figure 10.15. The delta-connected source and delta-connected load can be represented by the equivalent wye-connected source and wye-connected load, respectively. The resulting equivalent circuit can be solved similar to the wye –wye network.

10.5. Power calculation of balanced three-phase circuit

10.5.1. Balanced wye loads

Consider a wye-connected balanced load (refer to Figure 10.6) with load impedance in each phase,

Z = |Z| ∠ \pm θ

$Z = |Z| ∠ \pm θ$

Phase voltage across the load is

V_{\emptyset} = V_{\emptyset} ∠ 0 °

$V_{\emptyset} = V_{\emptyset} ∠ 0 °$

Current in the load in each phase is

I_{\emptyset} = \frac{V_{\emptyset}}{Z}

$I_{\emptyset} = \frac{V_{\emptyset}}{Z}$

$Active power per phase : P_{\emptyset} = V_{\emptyset} I_{\emptyset} \cos θ$ $Active power per phase : P_{\emptyset} = V_{\emptyset} I_{\emptyset} \cos θ$

(10.11)

$Reactive power per phase : Q_{\emptyset} = V_{\emptyset} I_{\emptyset} \sin θ$ $Reactive power per phase : Q_{\emptyset} = V_{\emptyset} I_{\emptyset} \sin θ$

(10.12)

$Apparent power per phase : S_{\emptyset} = P_{\emptyset} + j Q_{\emptyset}, |S_{\emptyset}| = V_{\emptyset} I_{\emptyset}$ $Apparent power per phase : S_{\emptyset} = P_{\emptyset} + j Q_{\emptyset}, |S_{\emptyset}| = V_{\emptyset} I_{\emptyset}$

(10.13)

In a wye-connected system:

$V_{L} = \sqrt 3 V_{\emptyset} and I_{L} = I_{\emptyset}$ $V_{L} = \sqrt 3 V_{\emptyset} and I_{L} = I_{\emptyset}$

Total active power in a three-phase circuit:

$P = 3 P_{\emptyset} = 3 V_{\emptyset} I_{\emptyset} \cos θ = \sqrt 3 V_{L} I_{L} \cos θ$ $P = 3 P_{\emptyset} = 3 V_{\emptyset} I_{\emptyset} \cos θ = \sqrt 3 V_{L} I_{L} \cos θ$

(10.14)

Total reactive power in a three-phase circuit:

$Q = 3 Q_{\emptyset} = 3 V_{\emptyset} I_{\emptyset} \sin θ = \sqrt 3 V_{L} I_{L} \sin θ$ $Q = 3 Q_{\emptyset} = 3 V_{\emptyset} I_{\emptyset} \sin θ = \sqrt 3 V_{L} I_{L} \sin θ$

(10.15)

Total apparent power in three phases:

$S = P + j Q, |S| = 3 V_{\emptyset} I_{\emptyset} = \sqrt 3 V_{L} I_{L}$ $S = P + j Q, |S| = 3 V_{\emptyset} I_{\emptyset} = \sqrt 3 V_{L} I_{L}$

(10.16)

10.5.2. Balanced delta loads

Consider a delta-connected balanced load (refer to Figure 10.7) with load impedance in each phase,

Z = |Z| ∠ \pm θ

$Z = |Z| ∠ \pm θ$

Active and reactive power consumed by the load across nodes A and B is:

$P_{AB} = V_{AB} I_{\emptyset A} \cos θ Q_{AB} = V_{AB} I_{\emptyset A} \sin θ$ $P_{AB} = V_{AB} I_{\emptyset A} \cos θ Q_{AB} = V_{AB} I_{\emptyset A} \sin θ$

Active and reactive power consumed by the load across nodes B and C is:

$P_{BC} = V_{BC} I_{\emptyset B} \cos θ Q_{BC} = V_{BC} I_{\emptyset B} \sin θ$ $P_{BC} = V_{BC} I_{\emptyset B} \cos θ Q_{BC} = V_{BC} I_{\emptyset B} \sin θ$

Active and reactive power consumed by the load across nodes C and A is:

$P_{CA} = V_{CA} I_{\emptyset C} \cos θ Q_{CA} = V_{CA} I_{\emptyset C} \sin θ$ $P_{CA} = V_{CA} I_{\emptyset C} \cos θ Q_{CA} = V_{CA} I_{\emptyset C} \sin θ$

$\begin{array}{l} |V_{AB}| = |V_{BC}| = |V_{CA}| = |V_{\emptyset}| \\ |I_{\emptyset A}| = |I_{\emptyset B}| = |I_{\emptyset C}| = |I_{\emptyset}| \end{array}$ $\begin{array}{l} |V_{AB}| = |V_{BC}| = |V_{CA}| = |V_{\emptyset}| \\ |I_{\emptyset A}| = |I_{\emptyset B}| = |I_{\emptyset C}| = |I_{\emptyset}| \end{array}$

The total active and reactive power consumed by the three-phase delta-connected load is:

$\begin{array}{l} P = P_{AB} + P_{BC} + P_{CA} = 3 V_{\emptyset} I_{\emptyset} \cos θ \\ Q = S_{AB} + S_{BC} + S_{CA} = 3 V_{\emptyset} I_{\emptyset} \cos θ \end{array}$ $\begin{array}{l} P = P_{AB} + P_{BC} + P_{CA} = 3 V_{\emptyset} I_{\emptyset} \cos θ \\ Q = S_{AB} + S_{BC} + S_{CA} = 3 V_{\emptyset} I_{\emptyset} \cos θ \end{array}$

In delta-connected system,

V_{L} = V_{\emptyset} and I_{L} = \sqrt 3 I_{\emptyset}

$V_{L} = V_{\emptyset} and I_{L} = \sqrt 3 I_{\emptyset}$

$P = 3 V_{\emptyset} I_{\emptyset} \cos θ = \sqrt 3 V_{L} I_{L} \cos θ$ $P = 3 V_{\emptyset} I_{\emptyset} \cos θ = \sqrt 3 V_{L} I_{L} \cos θ$

(10.17)

$Q = 3 V_{\emptyset} I_{\emptyset} \sin θ = \sqrt 3 V_{L} I_{L} \sin θ$ $Q = 3 V_{\emptyset} I_{\emptyset} \sin θ = \sqrt 3 V_{L} I_{L} \sin θ$

(10.18)

$S = P + j Q |S| = 3 V_{\emptyset} I_{\emptyset} = \sqrt 3 V_{L} I_{L}$ $S = P + j Q |S| = 3 V_{\emptyset} I_{\emptyset} = \sqrt 3 V_{L} I_{L}$

(10.19)

Example 10.3

Consider the wye–wye network in Example 10.1:

1. Calculate the active power, reactive power, and apparent power in each phase of the load.

2. Calculate the total active power, reactive power, and apparent power consumed by the load.

3. Calculate the total active power, reactive power, and apparent power supplied by the source.

Solution

From Example 10.1:

Load current

I_{A} = 15.77 ∠ - 41.8 ° A

$I_{A} = 15.77 ∠ - 41.8 ° A$

Phase voltage at load terminals is

V_{AN} = 202 ∠ - 3.14 ° V

$V_{AN} = 202 ∠ - 3.14 ° V$

1. Calculate the active power, reactive power, and apparent power in each phase of the load:

$\begin{array}{l} P_{\emptyset} = V_{\emptyset} I_{\emptyset} \cos θ = 202 \times 15.77 \times \cos (41.8 ° - 3.14 °) = 202 \times 15.77 \times \cos (38.66 °) = 2.487 kW \\ Q_{\emptyset} = V_{\emptyset} I_{\emptyset} \sin φ = 202 \times 15.77 \times \sin (41.8 ° - 3.14 °) = 202 \times 15.77 \times sin (38.66 °) = 1.989 kVAR \\ S_{\emptyset} = V_{\emptyset} I_{\emptyset} = 202 \times 15.77 = 3.185 kVA \end{array}$ $\begin{array}{l} P_{\emptyset} = V_{\emptyset} I_{\emptyset} \cos θ = 202 \times 15.77 \times \cos (41.8 ° - 3.14 °) = 202 \times 15.77 \times \cos (38.66 °) = 2.487 kW \\ Q_{\emptyset} = V_{\emptyset} I_{\emptyset} \sin φ = 202 \times 15.77 \times \sin (41.8 ° - 3.14 °) = 202 \times 15.77 \times sin (38.66 °) = 1.989 kVAR \\ S_{\emptyset} = V_{\emptyset} I_{\emptyset} = 202 \times 15.77 = 3.185 kVA \end{array}$

2. Calculate the total active power, reactive power, and apparent power consumed by the load:

$\begin{array}{l} P = 3 P_{\emptyset} = 3 \times 2.487 = 7.461 kW \\ Q = 3 S_{\emptyset} = 3 \times 1.989 = 5.967 kVAR \\ S = 3 S_{\emptyset} = 3 \times 3.185 = 9.555 kVA \end{array}$ $\begin{array}{l} P = 3 P_{\emptyset} = 3 \times 2.487 = 7.461 kW \\ Q = 3 S_{\emptyset} = 3 \times 1.989 = 5.967 kVAR \\ S = 3 S_{\emptyset} = 3 \times 3.185 = 9.555 kVA \end{array}$

3. Calculate the total active power, reactive power, and apparent power supplied by the source.

Source current =

I_{LA} = 15.77 ∠ - 41.8 ° A

$I_{LA} = 15.77 ∠ - 41.8 ° A$

Phase voltage at source

V_{an} = 220 ∠ 0 ° V

$V_{an} = 220 ∠ 0 ° V$

$\begin{array}{l} P = 3 V_{\emptyset} I_{\emptyset} \cos θ = 3 \times 220 \times 15.77 \times \cos (41.8 °) = 7.759 kW \\ Q = 3 V_{\emptyset} I_{\emptyset} \sin θ = 3 \times 220 \times 15.77 \times \sin (41.8 °) = 6.937 kVAR \\ S = 3 V_{\emptyset} I_{\emptyset} = 3 \times 220 \times 15.77 = 10.408 kVA \end{array}$ $\begin{array}{l} P = 3 V_{\emptyset} I_{\emptyset} \cos θ = 3 \times 220 \times 15.77 \times \cos (41.8 °) = 7.759 kW \\ Q = 3 V_{\emptyset} I_{\emptyset} \sin θ = 3 \times 220 \times 15.77 \times \sin (41.8 °) = 6.937 kVAR \\ S = 3 V_{\emptyset} I_{\emptyset} = 3 \times 220 \times 15.77 = 10.408 kVA \end{array}$

Example 10.4

Consider the wye–delta network in Example 10.2:

1. Calculate the active power, reactive power, and apparent power in each phase of the load.

2. Calculate the total active power, reactive power, and apparent power consumed by the load.

3. Calculate the total active power, reactive power, and apparent power supplied by the source.

Solution

From Example 10.2,

1. Calculate the active power, reactive power, and apparent power in each phase of the equivalent wye connected load

I_{φ} = 15.78 ∠ - 42.38 ° A

$I_{φ} = 15.78 ∠ - 42.38 ° A$

V_{φ} = 202 ∠ - 3.72 ° V

$V_{φ} = 202 ∠ - 3.72 ° V$

For delta-connected load.

Phase voltage = voltage across load,

V_{AB} = \sqrt{3} V_{AN} ∠ 30 ° = \sqrt{3} \times 202 \times ∠ (- 3.72 + 30) ° = 349.87 ∠ 26.28 °

$V_{AB} = \sqrt{3} V_{AN} ∠ 30 ° = \sqrt{3} \times 202 \times ∠ (- 3.72 + 30) ° = 349.87 ∠ 26.28 °$

Load current per phase

$I_{AB} = \frac{I_{φ}}{\sqrt{3}} ∠ (θ + 30) A$ $I_{AB} = \frac{I_{φ}}{\sqrt{3}} ∠ (θ + 30) A$

$I_{AB} = 9.11 ∠ - 12.38 ° A$ $I_{AB} = 9.11 ∠ - 12.38 ° A$

$P_{φ} = V_{φ} I_{φ} \cos θ$ $P_{φ} = V_{φ} I_{φ} \cos θ$

$= 349.87 \times 9.11 \times \cos (26.28 ° + 12.38 °) = 349.87 \times 9.11 \times c o s (38.66 °) = 2.488 kW$ $= 349.87 \times 9.11 \times \cos (26.28 ° + 12.38 °) = 349.87 \times 9.11 \times c o s (38.66 °) = 2.488 kW$

$Q_{φ} = V_{φ} I_{φ} \sin θ$ $Q_{φ} = V_{φ} I_{φ} \sin θ$

$= 349.87 \times 9.11 \times \sin (26.28 ° + 12.38 °) = 1.990 kVAR .$ $= 349.87 \times 9.11 \times \sin (26.28 ° + 12.38 °) = 1.990 kVAR .$

$S_{φ} = V_{φ} I_{φ} = 348.87 \times 9.11 = 3.186 kVA$ $S_{φ} = V_{φ} I_{φ} = 348.87 \times 9.11 = 3.186 kVA$

2. Calculate the total active power, reactive power, and apparent power consumed by the load.

$P = 3 P_{φ} = 3 \times 2.487 = 7.461 kW$ $P = 3 P_{φ} = 3 \times 2.487 = 7.461 kW$

$Q = 3 S_{φ} = 3 \times 1.901 = 5.703 kVAR$ $Q = 3 S_{φ} = 3 \times 1.901 = 5.703 kVAR$

$S = 3 S_{φ} = 3 \times 3.186 = 9.558 kVA$ $S = 3 S_{φ} = 3 \times 3.186 = 9.558 kVA$

3. Calculate the total active power, reactive power, and apparent power consumed by the source.

Voltage per phase at source

V_{φ} = 220 ∠ 0 ° V

$V_{φ} = 220 ∠ 0 ° V$

Phase current in source

I_{φ} = I_{A} = 15.78 ∠ - 42.38 °

$I_{φ} = I_{A} = 15.78 ∠ - 42.38 °$

$P = 3 V_{φ} I_{φ} \cos = 3 \times 220 \times 15.78 \times cos (42.38 °) = 7.693 kW$ $P = 3 V_{φ} I_{φ} \cos = 3 \times 220 \times 15.78 \times cos (42.38 °) = 7.693 kW$

$Q = 3 V_{φ} I_{φ} \sin θ = 3 \times 220 \times 15.78 \times sin (42.38 °) = 7.020 kVAR$ $Q = 3 V_{φ} I_{φ} \sin θ = 3 \times 220 \times 15.78 \times sin (42.38 °) = 7.020 kVAR$

$S = 3 V_{φ} I_{φ} = 3 \times 220 \times 15.78 = 10.414 kVA$ $S = 3 V_{φ} I_{φ} = 3 \times 220 \times 15.78 = 10.414 kVA$

10.6. Advantages and disadvantages of three-phase supply

10.6.1. Advantages

• Power transmission will be three times that of a single-phase supply.

• Higher reliability: if one or two of the phases fail, the healthy phase can supply power.

• Two voltage levels: possible to connect single-phase and three-phase with two voltage levels (line and phase).

10.6.2. Disadvantages

• Three lines are required; expensive compared to single phase.

• Difficult to maintain exact balanced load when single-phase and three-phase loads are used.

10.7. Summary

• In a balanced three-phase system, all the three phases or line voltages will have the same frequency, equal voltage magnitude, and are phase displaced by 120°.

• In a balanced three-phase system, all the three phases or line currents will have the same frequency, equal voltage magnitude, and are phase displaced by 120°.

• In a wye-connected system the magnitude of the line voltage will be

\sqrt{3}

$\sqrt{3}$

times the magnitude of the phase voltage and the line voltage will lead the corresponding phase voltage by 30°. The line current is equal to the phase current.

• In a wye-connected system the magnitude of the line current will be

\sqrt{3}

$\sqrt{3}$

times the magnitude of the phase current and the line current will lag the corresponding phase current by 30°. The line voltage is equal to the phase voltage.

• In a balanced three-phase system the sum of all three-phase (line) voltages is equal to zero and the sum of all three phase (line) currents is equal to zero.

• A balanced wye–wye circuit can be reduced to a single-phase equivalent single-phase circuit and can be analyzed similar to a single-phase circuit using per phase quantities.

• A delta-connected source or load can be converted to an equivalent wye-connected source or load. So the wye–delta, delta–wye, and delta–delta network can be converted to an equivalent wye–wye network.

Problems

1. For each set of voltages, state whether or not the voltages form a balanced three-phase set. If the set is not balanced, explain why.

\begin{array}{l} V_{a} = 156 \cos 314 t V, \\ V_{b} = 156 \cos (314 t - 120 °) V, \\ V_{c} = 156 \cos (314 t + 120 °) V . \end{array}

$\begin{array}{l} V_{a} = 156 \cos 314 t V, \\ V_{b} = 156 \cos (314 t - 120 °) V, \\ V_{c} = 156 \cos (314 t + 120 °) V . \end{array}$

\begin{array}{l} V_{a} = 600 \sqrt{2} \sin 377 t V, \\ V_{b} = 600 \sqrt{2} \sin (377 - 240 °) V, \\ V_{c} = 600 \sqrt{2} \sin (377 + 240 °) V . \end{array}

$\begin{array}{l} V_{a} = 600 \sqrt{2} \sin 377 t V, \\ V_{b} = 600 \sqrt{2} \sin (377 - 240 °) V, \\ V_{c} = 600 \sqrt{2} \sin (377 + 240 °) V . \end{array}$

\begin{array}{l} V_{a} = 1796 \sin (ω t + 10 °) V, \\ V_{b} = 1796 \cos (ω t - 110 °) V, \\ V_{c} = 2796 \cos (ω t + 130 °) V . \end{array}

$\begin{array}{l} V_{a} = 1796 \sin (ω t + 10 °) V, \\ V_{b} = 1796 \cos (ω t - 110 °) V, \\ V_{c} = 2796 \cos (ω t + 130 °) V . \end{array}$

2. The time domain expressions for the three line-to-neutral voltages at the terminals of a wye-connected load are:

\begin{array}{l} V_{AN} = 1796 \cos ω t V, \\ V_{BN} = 1796 \cos (ω t + 120 °) V, \\ V_{CN} = 1796 \cos (ω t - 120 °) V . \end{array}

$\begin{array}{l} V_{AN} = 1796 \cos ω t V, \\ V_{BN} = 1796 \cos (ω t + 120 °) V, \\ V_{CN} = 1796 \cos (ω t - 120 °) V . \end{array}$

Write the time domain expressions for the three line-to-line voltages v_AB, v_BC, and v_CA.

3. The magnitude of the phase voltage of an ideal balanced three-phase wye-connected source is 2200 V. The source is connected to a balanced wye-connected load by a distribution line that has an impedance of 2 + j16 per line. The load impedance is 20 + j40 Ω in each phase. The phase sequence of the source is a–b–c. Use the a-phase voltage of the source as the reference. Specify the magnitude and phase angle of the following quantities:

a. The three line currents.

b. The three line voltages at the source.

c. The three phase voltages at the load.

d. The three line voltages at the load.

4. A balanced delta-connected load is powered from a wye-connected source. The load impedance is 36 + j10 Ω in each phase. The line impedance of each line connecting the source to the load is 0.1 + j1 Ω. The line voltage at the terminals of the load is 33 kV.

a. Calculate the three phase currents of the load.

b. Calculate the three line currents.

c. Calculate the three line voltages at the source.

5. In a balanced three-phase system, the source is balanced wye with line voltage = 190 V. The load is a balanced wye in parallel with a balanced delta. The phase impedance in each phase of the wye-connected load is 4 + j3 Ω and the impedance of the delta is 3 – j9 Ω per phase. The line impedance is 1.4 + j0.8 Ω per line.

a. Calculate the current in the wye-connected load.

b. Calculate the current in the delta-connected load.

c. Calculate the current in the source.

6. The delta-connected source is connected to a wye-connected load. The load impedance per phase is 100 + j200 Ω, and the line impedance is 1.2 + j12 Ω. The line voltage of the source is 6600 V. Ignore the internal impedance of the source.

a. Construct a single-phase equivalent circuit of the system.

b. Determine the magnitude of the line voltage at the terminals of the load.

c. Determine the magnitude of the phase current in the delta-connected source.

7. A three-phase delta-connected generator has an internal impedance of 0.6 + j4.8 Ω per phase with an open-circuit terminal voltage of 33,000 V. The generator feeds a delta-connected load through a transmission line with an impedance of 0.8 + j6.4 Ω. The impedance of the load is 2877 – j864 Ω per phase.

a. Calculate the magnitude of the line current.

b. Calculate the magnitude of the line voltage at the terminals of the load.

c. Calculate the magnitude of the line voltage at the terminals of the source.

d. Calculate the magnitude of phase current in the load.

e. Calculate the magnitude of phase current in the source.

8. The output of the balanced positive-sequence three-phase source is 100 kVA and power factor is 0.8 lead. The line voltage at the source is 400 V. The impedance of each line connecting the source to the load is 0.05 + j0.2 Ω.

a. Find the magnitude of the line voltage at the load.

b. Find the total complex power at the terminals of the load.

9. A balanced wye-connected load is supplied from a balanced three-phase voltage source. The three-phase voltage source is 220 V. Load impedance is 8 + j6 Ω at 50 Hz. Neglect the source internal impedance and impedance of the line connecting the source and load. Find the current in the load and source.

10. The current in a balanced wye-connected load is 8 A. The source is wye connected with a phase voltage of 230 V. The load power factor is 0.8 lagging. Neglect the source internal impedance and impedance of the line connecting the source and load. Find the resistance and reactance of the load.

11. The p and Q in a three-phase wye-connected system are 1000 W and 1000 VAR, respectively. The source voltage is 220 V. Neglect the source internal impedance and impedance of the line connecting the source and load. Find the impedance per phase and current.

12. Show that the total instantaneous power in a balanced three-phase circuit is constant and equal to 1.5V_mI_m cos θ_∅, where V_m and I_m represent the maximum amplitudes of the phase voltage and phase current, respectively.

13. A balanced three-phase source is supplying 90 kVA at 0.8 lagging to two balanced wye-connected parallel loads. The distribution line connecting the source to the load has negligible impedance. Load 1 is purely resistive and absorbs 60 kW. Find the per-phase impedance of load 2 if the line voltage is 415.69 V and the impedance components are in series.

14. The total apparent power supplied in a balanced three-phase wye–delta system is 10 kVA. The line voltage is 190 V. if the line impedance is negligible and the power factor angle of the load is 25°, determine the impedance of the load.

15. In a balanced three-phase system, the source has an a–b–c sequence, is wye connected, and V_an = 110 ∠ 20° V. The source feeds two loads, both of which are wye connected. The impedance of load 1 is 8 + j6 Ω per phase. The complex power for the a-phase of load 2 is 600 ∠ 36° VA. Find the total complex power supplied by the source.

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Table of Contents for 10: Three-phase AC supply

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Abstract

Keywords

10.1. Introduction

10.2. Generation of three-phase voltages

10.3. Connections of three-phase circuits

10.3.1. Wye-connected balanced source

10.3.1.1. Phase voltages

10.3.1.2. Line voltages

10.3.1.3. Phase currents

10.3.1.4. Line currents

10.3.2. Delta-connected balanced source

10.3.2.1. Line voltages

10.3.2.2. Phase voltages

10.3.2.3. Phase currents

10.3.2.4. Line currents

10.3.3. Wye-connected balanced load

10.3.4. Delta-connected balanced load

10.4. Circuits with mixed connections

10.4.1. Wye–wye network

10.4.1.1. Line currents and phase currents

10.4.1.2. Line voltages at the load terminals

10.4.1.3. Line voltages at the source terminals

10.4.2. Delta to wye conversion

10.4.3. Wye-connected source/delta-connected load

10.4.4. Delta–wye network

10.4.5. Delta–delta network

10.5. Power calculation of balanced three-phase circuit

10.5.1. Balanced wye loads

10.5.2. Balanced delta loads

10.6. Advantages and disadvantages of three-phase supply

10.6.1. Advantages

10.6.2. Disadvantages

10.7. Summary

Table of Contents for
10: Three-phase AC supply