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Three-Phase Systems

Introduction to symmetrical and balanced three-phase systems

Summary: an introduction to the three-phase system, the technology commonly used to transmit and use electric power. Here we discuss how to generate a three-phase system that is symmetrical in its voltages and balanced in its currents. We then look at the characteristics of wye (star) and delta connections of the phases.

Notation. We use dotted letters, for example \(\small \dot{V},\, \dot{I},\, \dot{Z}\), for phasors and complex operators. The same notation also stands for the corresponding complex numbers (e.g. \(\small \dot{Z} = R +j\,X\)). We keep the standard notation for rotating vectors (e.g. \(\small \vec{I}, \vec{V}\)).
Undotted letters denote the magnitudes of complex or vector quantities (e.g. \(\small Z = \sqrt{R^2 + X^2}\)).
For the imaginary unit, instead of the letter \(\small i = \sqrt{-1}\) used in mathematics, we use the letter \(\small j = \sqrt{-1}\). When we want to stress the dependence on another quantity such as time, we use lowercase letters, for example: \(\small v(t) = V_M\,sin(\omega t)\).

1.1 - Generating a sinusoidal alternating EMF

Let us briefly recall a few key points about generating sinusoidal alternating EMFs. For more detail, see Alternating Current 1: sinusoidal electrical quantities.
To generate an EMF, according to the law of electromagnetic induction, we need relative motion between a magnetic field and a conductor or an electric circuit⁽¹⁾. Technically, the most convenient solution is to rotate a conductor at constant angular velocity (uniform circular motion) inside a magnetic field. It makes no difference whether the conductor moves with respect to the magnetic field or the other way around. What matters for generating the induced EMF is the relative motion between magnetic field and conductor.

    The induced electromotive force is given in general by Faraday’s law:  
\[e(t) = - \frac{\Delta\,\Phi(t)}{\Delta\,t} \] or by the so-called generator formula, which suits our case better: \[e(t) = Blv \; sin(\omega t) \] The rotation of the conductor inside the magnetic field B induces in the conductor a sinusoidal alternating EMF of the form:
\[e(t) = Blv \; sin(\omega t + \psi) = U_M\,sin(\omega t + \psi) \] where the amplitude \(\small U_M\) equals \(\small B\,l\,v\) and \(\small \psi\) is the phase angle that sets the value of the function at the initial instant.
We also know that sinusoids can be represented by vectors rotating at angular velocity \(\small \omega\), equal to the angular frequency of the sinusoid.
Rotating vectors taken at the initial instant (a snapshot of the system at t = 0) and with a length equal to the RMS value of the quantity are called phasors.

1.2 - Generating a symmetrical set of three-phase voltages

The efficiency of an AC voltage generator (alternator) can be improved by moving from a single-phase to a three-phase system, which makes better use of the machine’s geometry (although this is not the only reason three-phase systems are used). For the generator to be of practical use, the active sides must be connected to the outside world. For our “teaching machine” we need a set of coaxial slip rings that rotate with the moving part (the rotor) and on which the brushes rest (sliding contacts, usually made of graphite). Each slip ring is connected to the corresponding active side, which can thus supply an external circuit.
For current to flow, however, the circuit must be closed, so the generating phases have to be connected to each other in some way, for example in a wye (star) connection. In this case, as we will see, we can also add a fourth ring to connect the conductor called “neutral” N (conductor 0) to the star point.
In an actual alternator, however, a different arrangement is used: the phases (the armature) are fixed in the stationary part of the machine (the stator), while the field system (the magnetic field) rotates (the rotor) inside the stator.
This makes it easier to take the currents and voltages produced by the generator, which can be connected directly to the outside without brushes and rings, components that are inherently fragile and need frequent maintenance.
In technical jargon, each active side, bundle of conductors or winding in which the sinusoidal EMF is produced is called a “phase” or “generating phase”. A three-phase machine has three of these phases, set 120° apart, \(\small \dfrac{2\,\pi}{3}\) radians from each other⁽²⁾.
In our teaching model the generating phases are the three active sides R, S and T. On each of them a sinusoidal alternating EMF is produced, called the phase voltage, which we denote by \(\small U_1\) for phase R, \(\small U_2\) for phase S and \(\small U_3\) for phase T.
We then have:

Phase R (phase 1) \(\qquad \qquad u_1(t) = U_{1M} \; sin(\omega t) \)

Phase S (phase 2) \(\qquad \qquad u_2(t) = U_{2M} \; sin\left(\omega t - \dfrac{2\,\pi}{3}\right) \)

Phase T (phase 3) \(\qquad \qquad u_3(t) = U_{3M} \; sin\left(\omega t - \dfrac{4\,\pi}{3}\right) = U_{3M} \; sin\left(\omega t + \dfrac{2\,\pi}{3}\right) \)

The three sinusoids are conveniently described by the three rotating vectors \(\small \vec{U}_1\), \(\small \vec{U}_2\), \(\small \vec{U}_3\), or by the corresponding phasors. The set of voltages produced in this way is called symmetrical precisely because the voltages are equal in magnitude and equally shifted (120°) from one another.

As we know, rotating vectors in the same diagram have the same frequency. The corresponding phasors can be represented by complex numbers. For example, in the initial situation shown here, calling \(\small U_i \quad i=1,2,3\) the (common) magnitude of the three phasors, equal to the RMS values \(\small U_i= \dfrac{U_{iM}}{\sqrt{2}}\), we have:

Phase R (phase 1) \(\qquad \qquad \dot{U_1} = U_{1} \; [cos(0) + j\,sin(0)] \)

Phase S (phase 2) \(\qquad \qquad \dot{U_2} = U_{2} \; \left[cos\left(-\dfrac{2\,\pi}{3}\right) + j\,sin\left(-\dfrac{2\,\pi}{3}\right)\right] \)

Phase T (phase 3) \(\qquad \qquad \dot{U_3} = U_{3} \; \left[cos\left(\dfrac{2\,\pi}{3}\right) + j\,sin\left(\dfrac{2\,\pi}{3}\right)\right] \)

Working out the calculations, and calling \(\small U\) the common magnitude of the three phasors, the formulas above give:

Phase R (phase 1) \(\qquad \qquad \dot{U_1} = U + j\,0 \)

Phase S (phase 2) \(\qquad \qquad \dot{U_2}= -\dfrac{1}{2}U - j\,\dfrac{\sqrt{3}}{2}\,U \)

Phase T (phase 3) \(\qquad \qquad \dot{U_3}= \dfrac{1}{2}U + j\,\dfrac{\sqrt{3}}{2}\,U \)

If, for example, U = 230 V, we have:

Phase R (phase 1) \(\qquad \qquad \dot{U_1} = 230 + j\,0 \quad V \)

Phase S (phase 2) \(\qquad \qquad \dot{U_2}= -115 - j\,199 \quad V \)

Phase T (phase 3) \(\qquad \qquad \dot{U_3}= 115 + j\,199 \quad V \)

Activity 1

1.3 - Connecting the phases

As we said, the three active sides, which from now on we will call phases, must be connected to each other and to the outside.
In a case like the one we are studying, the connection to the outside would be made through a set of three rings, each connected to one phase, on which slide the brushes connected to the external circuits.

Activity 2


There are two kinds of internal connection between the phases: wye (star) and delta (triangle).

1.4 - Wye (star) connection

In the wye connection three corresponding terminals of the phases are connected to the same point, called the star point. In this case the phase voltages \(\small U_1\), \(\small U_2\) and \(\small U_3\) are also called star voltages (often written \(\small E_1\), \(\small E_2\) and \(\small E_3\)). We now have two sets of voltages:

• the phase voltages \(\small U_1\), \(\small U_2\) and \(\small U_3\), already mentioned, measured between the free ends (R (1), S (2) and T (3)) of the active sides and the star point;
• the line voltages \(\small U_{12}\), \(\small U_{23}\) and \(\small U_{31}\), which we could also write as \(\small U_{RS}\), \(\small U_{ST}\) and \(\small U_{TR}\), measured between pairs of free ends of the active sides (R-S, S-T and T-R).

Noting from figure 1 that the angle between, say, \(\small \dot{U_1}\) and \(\small \dot{U_2}\) is 120°, while the angles between \(\small \dot{U_1}\) and \(\small \dot{U_{12}}\) and between \(\small \dot{U_2}\) and \(\small \dot{U_{12}}\) are 30°, we can derive the relation between the magnitudes of phase and line voltages.
Calling \(\small U = U_1 = U_2 = U_3\) the common magnitude of the three phase voltages, we have:
\[ U_{12} = 2\,U\,cos\,30° = 2\,U\,\dfrac{\sqrt{3}}{2} = \sqrt{3}\,U \] or, in short:
\[ U_{line} = \sqrt{3}\,U_{phase} \] Since line voltages are often written with the letter V (usually with double subscripts) and phase voltages with E, we also give the relation: \[ V = \sqrt{3}\,E \]
Note that the subscripts follow the cyclic order 1-2, 2-3, 3-1, or R-S, S-T, T-R, to match the sequence of lags:
1 -> 2 -> 3 -> 1, R -> S -> T -> R. In fact we see R “go by” first, then S, then T, then R again, and so on: S lags R, T lags S, and so on.

The vectors representing the line voltages can be obtained as differences of the phase voltages:

\(\qquad \qquad \dot{U_{12}} = \dot{U_1}-\dot{U_2} \)
\(\qquad \qquad \dot{U_{23}}= \dot{U_2}-\dot{U_3} \)
\(\qquad \qquad \dot{U_{31}}= \dot{U_3}-\dot{U_1} \)

Phase and line voltages can also be drawn as vectors all applied at the vector star point, as in the next figure. Here it is clear that the line voltages lead the phase voltages by 30°.

To express the relation between the two sets of voltages mathematically, and represent the corresponding phasors correctly, we use the operator for a counterclockwise rotation of 30°:
\[ \rho \angle 30°= e^{j\,\small{\dfrac{\pi}{6}}} = cos\,\frac{\pi}{6} + j\,sin\,\frac{\pi}{6}\] Any vector \(\small \dot{U}_{k\,l}\) (k and l are subscripts taking the values 1, 2 and 3, permuted cyclically: 12-23-31) can then be written as:
\[ \dot{U}_{k\,l} = \sqrt{3}\,\dot{U}_{k}\, \rho \angle 30° \qquad k,l = \{1,2,3\}\] or, in the other notation:
\[ \dot{V}_{k\,l} = \sqrt{3}\,\dot{E}_{k}\, \rho \angle 30° \qquad k,l = \{1,2,3\}\]
That is: the line voltages are \(\small \sqrt{3}\) times larger in magnitude than the phase voltages and lead them by 30°.

We can think of the alternator with its phases connected in wye as a set of three voltage generators, \(\small U_1\,,U_2\,,U_3\), producing equal EMFs \(\small U\) shifted 120° from each other.
For example, if \(\small \dot{U}_1 = 230 \angle 0° \), \(\small \dot{U}_2 = 230 \angle -120° \) and \(\small \dot{U}_3 = 230 \angle +120° \), from what we have said:

\(\small \dot{U}_{12} = \sqrt{3}\,\dot{U}_1\, \rho \angle 30° = 400 \angle 30°\) V, \(\small \dot{U}_{23} = \sqrt{3}\,\dot{U}_2\, \rho \angle 30° = 400 \angle -90°\) V and \(\small \dot{U}_{31} = \sqrt{3}\,\dot{U}_3\, \rho \angle 30° = 400 \angle 150°\) V

The diagram on the right, the usual way of drawing a wye connection, should also make it clear that the line voltages \(\small \dot{U}_{k\,l}\) are obtained as the difference⁽⁴⁾ of the phase voltages with the same subscripts:
\(\small \dot{U}_{k\,l} = \dot{U}_k - \dot{U}_l\), for example \(\small \dot{U}_{1\,2} = \dot{U}_1 - \dot{U}_2\), and similarly for the others.

From the star point O a fourth conductor can be taken, called the neutral N and often marked 0.
The phase voltages can therefore also be seen as the potential differences between a phase and the neutral:

\(\qquad \qquad \dot{U_{1}} = \dot{U_{1 0}} \qquad \qquad \dot{E_{1}} = \dot{E_{1 0}} \)
\(\qquad \qquad \dot{U_{2}}= \dot{U_{20}} \qquad \qquad \dot{E_{2}}= \dot{E_{20}}\)
\(\qquad \qquad \dot{U_{3}}= \dot{U_{30}} \qquad \qquad \dot{E_{3}}= \dot{E_{30}} \)

1.5 - Delta connection

The delta connection connects the end of each phase to the start of the next one, until a closed loop is formed.

    In our example the end of R is connected to the start of S, the end of S to the start of T and the end of T to the start of R.  The starts of R, S and T are connected to the outside, forming three nodes.
    In this connection there is only one set of voltages, the phase voltages, which are also line voltages:

\[ \dot{U}_{12} = \dot{U}_{RS} = \dot{U}_1 \] \[ \dot{U}_{23} = \dot{U}_{ST} =\dot{U}_2 \] \[ \dot{U}_{31} = \dot{U}_{TR} = \dot{U}_3 \]
Note that:

\(\small \dot{U}_{12} + \dot{U}_{23} + \dot{U}_{31} = \dot{U}_{11}=0\)

that is, the sum of the three phase (and line) voltages around the delta is the potential difference \(\small \dot{U}_{11}\) of a point (R, in this case) with respect to itself, and must therefore be zero.
This vector sum is indeed zero, and this holds for any set of three vectors equal in magnitude and equally shifted (120°): you can easily check it by adding two of the three vectors with the parallelogram rule (figure 4) and seeing that the result is equal and opposite to the third vector.
Figure 5 shows a very common way of drawing the delta connection. As already said, in this case there is only one set of symmetrical voltages, which are both line and phase voltages.
As we will see, the advantage of the delta connection is that it can deliver to the outside currents \(\small \sqrt{3}\) times larger than those delivered by the individual generating phases  (the generators in the diagram of fig. 5).

1.6 - The three-phase load and the currents

As we said, the generator can deliver currents, and therefore power, if an electrical load is connected to its terminals.
Let us suppose, to begin with, that the generator phases are connected in wye. Through the slip rings the machine can connect to the outside, to three impedances, which we assume identical and in turn connected in wye, forming the electrical load of the generator.

Activity 3: load and currents

1.7 - The wye connection: more details

We have already looked at several aspects of the wye connection, especially on the generator side. To sum up, we have seen that:

• There are two sets of voltages: the phase voltages \(\small \dot{U}_1,\, \dot{U}_2\,,\dot{U}_3\) (or \(\small \dot{E}_1,\, \dot{E}_2\,,\dot{E}_3\)) and the line voltages \(\small \dot{U}_{12},\, \dot{U}_{23}\,,\dot{U}_{31}\) (or \(\small \dot{V}_{12},\, \dot{V}_{23}\,,\dot{V}_{31}\); we will mostly use V and E for loads).
  Phase voltages are measured between a phase and the star point (neutral), line voltages between two phases.
• The two sets of voltages are symmetrical (a condition always sought) when they are equal in magnitude and equally shifted from one another by 120°.
• If the load is also connected in wye, there is only one set of currents, which are both phase and line currents: \(\small \dot{I}_1,\, \dot{I}_2\,,\dot{I}_3\).
• The system is balanced if, the three load impedances being identical, the currents are equal in magnitude and equally shifted from one another by 120°. This condition is highly desirable and always sought, but in practice it can be approached and almost never fully reached. In such cases we speak of (symmetrical and) unbalanced systems, which are beyond the scope of this introductory module.

The currents through the load impedances produce voltage drops, which are the phase voltages of the load: \(\small \dot{E}_1 = \dot{Z}\,\dot{I}_1,\, \dot{E}_2 = \dot{Z}\,\dot{I}_2,\,\dot{E}_3 = \dot{Z}\,\dot{I}_3 \).
The line voltages are set by the generator and, if the line is ideal (zero line impedance), they are exactly the voltages at the generator terminals. In that case the phase voltages of the load are also equal to those of the generator.
With an unbalanced load the line voltages do not change, while the phase voltages change so that, while still satisfying the relation above, their differences still give the (imposed) line voltages.

Connecting the two star points with the neutral conductor gives a four-wire system.
If the system is balanced, no current flows in the neutral, since the sum of the currents at the star-point node is the sum of three equal vectors equally shifted, which we have already seen to be zero.
In unbalanced systems, on the other hand, current will generally flow in the neutral conductor.
With the neutral in place, not only the line voltages but also the phase voltages are imposed.

1.8 - The delta connection: more details

The delta connection is perhaps less intuitive than the wye connection, but just as important. Now the generator is connected in delta and the load in wye, but you can also have a delta-connected load.
Here too, let us sum up the features of this connection seen so far:

• There is only one set of voltages: the phase voltages \(\small \dot{U}_1,\, \dot{U}_2\,,\dot{U}_3\), which also act as line voltages: \(\small \dot{U}_{12},\, \dot{U}_{23}\,,\dot{U}_{31}\) (or \(\small \dot{V}_{12},\, \dot{V}_{23}\,,\dot{V}_{31}\)).
  They are measured between two phases of the line (or between the output terminals of the generator). Here too the voltages are symmetrical.
• There are, however, two sets of currents: phase currents \(\small \dot{I}_1,\, \dot{I}_2\,,\dot{I}_3\) and line currents \(\small \dot{I}_{L1},\, \dot{I}_{L2}\,,\dot{I}_{L3}\)
• The relation between line and phase currents follows from the fact that the line currents are differences of phase currents. The reasoning works on the generator delta as well as on the load (the current directions are opposite in the two cases). Take figure 7 as reference. For each of the nodes A, B and C we can write an equation from Kirchhoff’s current law:
  
  Node A: \(\qquad \qquad \dot{I}_{L1} = \dot{I}_1-\dot{I}_3 \)
  Node B: \(\qquad \qquad \dot{I}_{L2}= \dot{I}_2-\dot{I}_1 \)
  Node C: \(\qquad \qquad \dot{I}_{L3}= \dot{I}_3-\dot{I}_2 \)
  
  Since the phase currents are equal in magnitude and shifted by 120° (and so are the line currents), the situation is the same as the one we saw for phase and line voltages in the wye connection.
  So the magnitude of the line currents is \(\small \sqrt{3}\) times that of the phase currents, and the set of line currents lags the set of phase currents by 30°,
  that is:
  \[ \dot{I}_{L\,k} = \sqrt{3}\,\dot{I}_k \, \rho \angle -30° \qquad k = \{1,2,3\}\]

1.9 - Power in symmetrical and balanced three-phase systems

As we know, power is an additive quantity. This means that in a three-phase system the total power delivered or absorbed is the sum of the powers generated or absorbed by the individual phases.
Figure 8 shows the quantities involved when the load is connected in wye. Phase quantities are those that concern the individual phases: for example, the phase current is the one that actually flows through each phase impedance, the phase voltages are those measured across the same impedances, and so on. Line quantities are those on the line or, for voltages, measured between two line conductors. Using the phase quantities we can treat the three-phase system as the “sum” of three single-phase systems. Following this line of reasoning, each phase absorbs the active power:
\[ P_1 = E_1\,I_1\,cos\,\varphi_1 \qquad P_2 = E_2\,I_2\,cos\,\varphi_2 \qquad P_3 = E_3\,I_3\,cos\,\varphi_3 \] and the reactive power:
\[ Q_1 = E_1\,I_1\,sin\,\varphi_1 \qquad Q_2 = E_2\,I_2\,sin\,\varphi_2 \qquad Q_3 = E_3\,I_3\,sin\,\varphi_3 \] Since the system is symmetrical and balanced we can replace the quantities with their common values \(\small E_i = E \quad I_i = I\; (\text{phase } I)\quad \varphi_i = \varphi \quad (i= 1,2,3) \quad\) and compute the total active power as the sum of the three powers, so the total active and reactive power are: \[ P = 3\,E\,I\,cos\,\varphi \qquad \qquad Q = 3\,E\,I\,sin\,\varphi\] But since \(\small V = \sqrt{3}\,E\), we also have: \[ P = \sqrt{3}\,V\,I\,cos\,\varphi \qquad \qquad Q = \sqrt{3}\,V\,I\,sin\,\varphi\] For the delta, figure 9 likewise shows the phase and line quantities and how they are related. Following the criterion stated above, the voltages \(\small V_{ij} \quad( i,j=1,2,3)\) are both line and phase voltages. The angle \(\small \varphi \), which is always a phase quantity, is measured between phase voltage and phase current, as shown in the figure. In this case the powers are:
\[ P_1 = V_{12}\,I_1\,cos\,\varphi_1\] \[P_2 = V_{23}\,I_2\,cos\,\varphi_2 \] \[P_3 = V_{31}\,I_3\,cos\,\varphi_3 \] \[ Q_1 = V_{12}\,I_1\,sin\,\varphi_1 \qquad Q_2 = V_{23}\,I_2\,sin\,\varphi_2 \qquad Q_3 = V_{31}\,I_3\,sin\,\varphi_3 \] Here too we give the quantities their common values, noting that \(\small I_f\) is the magnitude of the phase currents \(\small I_{i} \quad (i=1,2,3)\), while \(\small I\) is the magnitude of the line currents. We get:
\[ P = 3\,V\,I_f\,cos\,\varphi \qquad \qquad Q = 3\,V\,I_f\,sin\,\varphi\] Since \(\small I_f = \dfrac{I}{\sqrt{3}}\), we also have:
\[ P = \sqrt{3}\,V\,I\,cos\,\varphi \qquad \qquad Q = \sqrt{3}\,V\,I\,sin\,\varphi\] So the same formula can be used to compute power for both delta and wye, as long as we are careful about what we put into it: voltages and currents are line quantities in both cases, while the angle is a phase quantity.


Rev. 02/08/2021 · English translation 10/2026


Books by the author and further reading

• Three-Phase Systems and the author’s other books
• Complex numbers: definitions and basic operations (in Italian)
• Euler’s formula and identity. Complex powers and logarithms (in Italian)

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⁽¹⁾ In fact, electric fields, and therefore potential differences, can be created by magnetic fields that change in time, whether or not conductors and/or electric circuits are present.

⁽²⁾ The angular distance between two phases is called the mechanical angle. It is 120° if the machine is two-pole (one pole pair), that is, if the field B is produced by a single pair of poles (North, South), as in our teaching model. The mechanical angle therefore depends on the number of poles of the alternator (for example, in a 4-pole machine, with two North-South pairs, the mechanical angle is 60°). The electrical angle, on the other hand, is always 120°: it is the angle between the phasors of the corresponding voltages.


⁽³⁾ A three-phase load supplied by a symmetrical set of three-phase voltages is called balanced if the currents through it are equal in magnitude and equally shifted (120°). This happens, for example, when the three load impedances are identical.

⁽⁴⁾ This is, of course, a vector (or geometric) difference, since the voltages are phasors (and therefore vectors) with magnitude and phase. Simply subtracting the magnitudes would give wrong results.

⁽⁵⁾ You may want to see: Alternating Current 2: series RLC
and, for power: Power in sinusoidal AC circuits (both in Italian)