Introduction to Electrical Engineering  Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.) 
8. Threephase Current 

In the preceding Chapter we have shown that a sinusoidal alternating voltage is induced in a conductor loop which is rotated in a homogeneous magnetic field. Several conductor loops which are mechanically connected together can also be turned in a magnetic field at the same time. Then, in each conductor loop, a sinusoidal alternating voltage is induced. An arrangement where three conductor loops displaced by 120 to each other are used has gained great importance in practice (Fig. 8.1.).
Fig. 8.1. Principle of the
threephase oberhungtype alternator
1, 2, 3  Rotatable coils
4  North pole
5  South pole
In order to conduct the electrical energy generated in this overhungtype alternator to the consumer, sliding contacts must be provided for all three conductor loops. This disadvantage is not associated with the innerpole alternator. In This type, the three conductor loops are fixed in the stator of the alternator (an alternatingcurrent generator is also called alternator) while the magnetic field in the interior is rotated. (Fig. 8.2.). The relatively low electrical energy for the production of the magnetic field must be fed to the rotor of the generator via sliding contacts.
In each of the three coils, a sinusoidal voltage is produced when the magnetic field is rotating which  in accordance with the arrangement of the coils  exhibits a phase shift of 120°. (Consequently, a voltage maximum is always reached, when the magnetic pole is turned past the coil. For the three coils, this always occurs after a rotation of the magnetic pole through 120 °). The line diagram of these three voltages is shown in Fig. 8.5.
Fig. 8.2. Principle of the
threephase innerpole alternator
1, 2, 3  Fixed coils
4  North pole
5  South pole
Fig. 8.3. Line diagram of voltages
of threephase current
The three voltages subjected to a phaseshift of 120° are called threephase current. In order to distinguish safely between the three voltages and their three coils, the three phases are marked by L1, L2 and L3 and coloured (L1: yellow; L2: green; L3: violet) according to an IECrecommendation. ^{1)} The connections of the starts of the coils can be designated by U, V and W and the ends of the coils by X, Y and Z. For the phases L1 to L3, sometimes R, S and T are used as designation.
^{1)} IEC = International Electrotechnical Commission
When a magnetic field is rotated, inside of three coils which are displaced by 120° to each other, then three sinusoidal voltages are produced which are called threephase current. A phase shift of 120° exists between every two of the three voltages.
Questions:
1. What is the difference between overhungtype alternator and innerpole alternator?
2. What are the advantages of the innerpole alternator over the overhungtype alternator?
3. How are the three phases distinguished from each other by markings?
4. What is the phase angle between the voltages of the three phases?
The most important property of the threephase current is discussed here. For this purpose, we again start from three coils displaced by 120° from each other which are connected to a threephase current according to Fig. 8.3. (Fig. 8.4.).
Fig. 8.4. Threephase winding of a
motor
Fig. 8.5. Development of the rotating field
a) Threephase line diagram with
points of time plotted on the diagram
b) Resulting magnetic field for the
points of time plotted on
In each of the three coils, an alternating field will be brough about in accordance with the alternating voltage applied. The total magnetic field resulting from the magnetic fields of the 3 coils is subject to a closer examination below. For this purpose, we have to give some explanations regarding the representations. When the positive half wave of the alternating voltage is applied to the coil, a magnetic north pole will be formed at the end of the coil which is inside the arrangement in accordance with the sense of winding of the coil. This pole is represented by cross hatching. The density of the lines of hatching corresponds to the amplitude of the voltage present at the instant of observing. When the negative half wave is applied, the south pole is formed which is represented by longitudinal hatching. For the points of time entered into the threephase current line diagram in Fig. 8.5.a, the magnetic flux produced in the three coils at the points of time a to g is shown in Fig. 8.5.b. The position of the total magnetic field brought about by the three individual fields is represented by the position of a rod magnet whose north pole is marked black. (A representation of the coils as given in Fig. 8.4. is omitted for the sake of clearness.) The position of the total magnetic field changes with the interval of time under consideration of the threephase current line diagram. It is evident that the total magnetic field has turned through half a revolution from point of time a to point of time g. When further partial pictures would be represented, then a full revolution of the total magnetic field would be performed during the duration of a cycle of the threephase current. This shows that, in a threephase winding connected to threephase current, a rotating magnetic field is formed which is called rotating field. The majority of electric motors and their mode of operation are based on the presence of the rotating field which enables extremely simple and sturdy designs of electric motors. Within a full cycle of the threephase current one revolution of the rotating field is performed in an arrangement according to Fig. 8.5. When two times three coils are arranged at the circumference, then the rotating field will perform only half a revolution within one cycle. Such an arrangement is termed as design with two pairs of poles. There are also arrangements with a higher number of pole pairs. For the calculation of the rotational speed of the rotating field we have
n_{D} = f/p
where:
n_{D} 
rotational speed of the rotating field 
f 
frequency 
p 
number of pole pairs 
Example 8.1.
A threephase current winding (also known as polyphase winding) with three pairs of poles is connected to a threephase current having a frequency of 50 Hz. Determine the rotational speed of the rotating field per minute!
Given:
f = 50 Hz
p =3
To be found:
n_{D} in rpm
Solution:
n_{D} = f/p
n_{D} = (50 1/s)/3
Since the speed per minute is required, the above calculation will, however, result in the revolutions per second, multiplying by 60 s/min is necessary.
n_{D} = (50 1/s · 60 s/min)/3
n_{D} = 1000 rpm
The rotating field has a speed of 1000 revolutions per minute.
When exchanging two phases shown in Fig. 8.5. in the manner represented in Fig. 8.6., the sense of rotation of the rotating field is inverted. This property of the threephase current is also of advantage to the construction of electric motors.
Fig. 8.6. Threephase winding with
the phases exchanged as compared to Fig. 8.5.
Inside a polyphase winding connected to threephase current a rotating field is formed (rotating magnetic field) whose sense of rotation can be inverted by exchanging two connections. The rotational speed of the rotating field is dependent on the frequency and the number of pole pairs. Simple designs of motors are possible due to the rotating field involved in threephase current.
Questions and problems:
1. Under which conditions is a rotating field formed?2. Demonstrate the reversal of the sense of rotation due to the exchange of two phases as shown in Fig. 8.6. with the help of a representation as given in Fig. 8.5.b!
3. What is the technical importance of the rotating field?
4. Calculate the possible rotational speeds of the rotating field for arrangements with 1 to 10 pairs of poles when the frequency is 50 Hz!
The explanations given in Section 8.1. show that threephase current involves three phases with two connections each. Consequently, six lines would be necessary as connection between generator and consumer or load. Such an open threephase system ins not used in practice. By a combination of certain conductors, which is termed as interlinking, connecting lines can be saved. In practice two of such interlinking connections are used, namely, the star connection (also known as Yconnection) and the delta connection.
The interlinking of the three phases of a threephase current into a star connection is shown in Fig. 8.7. For this purpose, the ends of the three generator coils (X, Y and Z) are connected and, in most cases, the connecting point is brought out as neutral conductor N. Then, four lines are required between generator and consumer.
Fig. 8.7. Star connection
When the load, resistors also arranged, in star connection have the same resistance values and cause an equal phase angle between current and voltage (in Fig. 8.7., pure active load, with a phase angle of 0 is represented.), then a line diagram will be brough about for the three currents which resembles that shown in Fig. 8.5. When we take a closer look at the sum of the three instantaneous values at any point of time in this line diagram, we will find that this sum is always equal to 0. Let us study this fact with particular respect to certain points of time on the basis of the line diagram shown in Fig. 8.5.a. For example, at point of time b, the phases L1 and L3 have half of the positive peak value and phase L2 the negative peak value. At point of time c, phase L3 has the value of 0, and the values of phases L1 and L2 are equal but of opposed directions. From this follows that no current is flowing in the neutral conductor R with the same phase load. In practice, equal phase load is seldom given so that a current flows in the neutral conductor which, however, is smaller than the current in the phase conductors L1, L2 and L3.
Fig. 8.7. also shows that a voltage can be tapped between phase conductor and neutral conductor (U_{L1N}, U_{L2N}, U_{L3N}) which is designated by U_{Str}, and between two phase conductors (U_{L1L2}, U_{L1L3} U_{L2L3}) which is designated by U_{L}. The voltage between the phase conductors is produced by series connection of two generator coils. In order to be in a position to calculate the magnitude relations between U_{Str} and U_{L}, a similar circuit in a direct current circuit will be considered as a repetition (Fig. 8.8.).
Fig. 8.8. Determination of the
voltage between the points A and B
Two voltage sources are connected in series and the point of connections is considered as the reference point (e.g. frame or earth). In this example, point A has a positive voltage (U_{A}) with respect to the reference point and point B has a negative voltage (U_{B}) with respect to the reference point. Between the points A and B is the difference of the voltages U_{A} and U_{B} related to the reference point.
U_{AB} = U_{A}  U_{B}
When we assume that U_{A} = 2 V and U_{B} =  2 V, then we have for U_{AB}
U_{AB} = +2 V  ( 2 V)
U_{AB} = 4 V
Similar conditions are given in a star connection; however, the phase shift between the voltages sources connected in series must be taken into consideration. In order to ascertain the voltage between the phase conductors, we first have to represent the three phase voltages (also known as star voltage) with their phase shift of 120° to each other in the form of voltage vectors.
In order to be in a position to determine the voltage between the phase conductors, two voltages related to the neutral conductor as reference point in each case must be subra... l from each other by adding a further vector to the tip of a given vector but having opposite direction. As resultant vector we obtain the linetoline voltage. By comparison of magnitudes in a vector graph true to scale, we obtain as magnitude relation between phase voltage and linetoline voltage
U_{L} = 1.73 · U_{Str}
Fig. 8.9. Graphical determination of the linetoline voltage
a) Vector diagram of the phase
voltages
b) Vector diagram of the phase and
linetoline voltages
The numerical factor 1.73 is the root extracted from 3 and, consequently, the equation can be written as
_{}
where:
U_{L} 
linetoline voltage (or phasetophase voltage) 
U_{str} 
phase voltage or star voltage or voltage to neutral 
Fig. 8.10. Triangle for calculating
the magnitude relation between phase voltage and linetoline voltage
(U_{Str} = phase voltage, U_{L} = linetoline voltage)
The ralation of magnitude between U_{L }and U_{Str} can also be determined mathematically. For this purpose, Fig. 8.10. shows a triangle as a part of the representation given in Fig. 8.9. Using the trigonometric functions, we have
U_{L}/2 = U_{str} · 30°U_{L} = 2 · U_{str} · cos 30°
_{}
_{}
Example 8.2.
The phase voltage of a threephase current network is 220 V. Which linetoline voltage is available in this network?
Given:
U_{Str} = 220
To be found:
U_{L}
Solution:
_{}U_{L} = 1.73 · 220 V
U_{L} = 380 V
A linetoline voltage of 380 V is available.
Example 8.2. demonstrates a typical case. With such a network, the voltage of 220 V desired by households can be supplied as the phase voltage while a linetoline voltage of 380 V is available from the same network for industrial enterprises.
In practice, sometimes the phase relation between phase voltage and linetoline voltage is utilised. Thus, Fig. 8.9.b shows that, for example U_{L1L2} exhibits a phase shift of 90° with respect to U_{L3N}. In general, it holds that the linetoline voltage has a phase shift of 90° with respect to the phase voltage of the phase conductor not under consideration.
In the delta connection shown in Fig. 8.11., one starts from the consideration that the sum of the individual voltages is always 0 at any time. Therefore, no balance current can flow within the generator in delta connection. In this circuit, only the linetoline voltage occurs. But in the junctions, a current division is obtained. For the upper junction we have
I_{L1} = I_{WZ}  I_{UX}
When the three linetoline currents have the same intensity, then the vector representation given in Fig. 8.12. is obtained. For the relation of magnitude between phase current and linetoline current a relation is given similar to the relation of magnitude between phase voltage and linetoline voltage. With the same phase load we have
where:
I_{L} 
linetoline current (or phasetophase current) 
I_{str} 
phase current 
_{}
Fig. 8.11. Vector representation of
the currents
Fig. 8.12. Delta connection
In threephase systems, two methods of interlinking are possible  the star connection and the delta connection. In the star connection, voltages can be tapped both between the phase conductors and between the neutral conductor and the phase conductors. In the delta connection, only the linetoline (phasetophase) voltage is available, however, the difference between linetoline current and phase current must be taken into consideration.
Questions and problems:
1. Sketch the star connection and the delta connection of the three coils of a generator!2. Determine the relations of magnitude between phase voltage (voltage to neutral) and linetoline (phasetophase) voltage by means of sketch true to scale! (Note: Select a length of 50 MM for the phase voltage!)
3. Calculate the voltage to neutral with a phaseto phase voltage of 220 V!
4. What is the value of the current flowing in the coils of a generator in delta connection when the current in the phase conductors is 34.6 A?
For calculating the power of threephase systems, the same relations are applicable as for the calculation of the power of alternating current systems. In accordance with the phase angle involved, a distinction is also made between effective power, reactive power and apparent power.
The star connection of three equal resistors is shown in Fig. 8.13.
For the total power, we have
P = 3 · U_{Str} · I_{L} · cos j
When the power is to be determined, on the basis of the linetoline voltage U_{L}, the following holds when using equation u.2.
_{}
_{}
Fig. 8.13. Star connection of three
resistors
When three equal resistors are connected in delta (Fig. 8.14.), the total power is written as
P = 3 · U_{L} · I_{Str} · cos j
Fig. 8.14. Delta connection of three
resistors
When the linetoline current is used, the following holds when using equation 8.3.
_{}
_{}
where:
U_{L} 
linetoline voltage 
I_{L} 
linetoline current 
cos j 
power factor 
A comparison of the equations 8.4. and 8.5. shows that, independent of the given type of connection, the same equations for calculating the power are given.
When the phase load is unequal, the total power is obtained in the form of the sum of the powers in the three phases to be determined individually.
Example 8.3.
Three resistors of 800 W each have to be interposed in a threephase network of 380 V one time in star connection and another time in delta connection. Calculate the effective power involved in each case.
Given:
U_{L} = 380 V
R = 800 W
cos j = 1
To be found:
effective power P for star connection and for delta connection
Solution:
star connection of the three resistors
_{}
In star connection, only the phase voltage drops at the three resistors. Hence, for the current I_{L} we have
_{}
This expression is entered in the initial equation
_{}P = (380 V)^{2}/800 W
P = 180.5 W
delta connection of the three resistors
_{}P = 3 · U_{L} · I_{Str} · cos j
Since the full linetoline voltage is applied to each resistor, we have for the phase current
I_{Str} = U_{L}/R
This expression is entered in the initial equation
P = 3 · U_{L}^{2}/R · cos j
P = 3 · (380 V)^{2}/800 W
P = 541.5 W
At the three resistors, a total power of 180.5 W is obtained, in star connection and of 541.5 W in delta connection.
In practice frequently advantage is taken of the possibility of obtaining different powers by changing the type of connection of the various loads. For example, for threephase motors, there are special switching devices which enable the changing over from star connection to delta connection and vice versa.
The power of a threephase system can be determined from the sum of the individual powers in the three phases in ease of unequal phase loads or from the relation given in equation 8.4. in case of equal loads irrespective of the type of connection.
Questions and problems:
1. Determine the effective power of three resistors in star connection of 200 W if the latter are connected to a threephase network with a linetoline voltage of 220 V!2. Determine the effective power when the three resistors of problem 1. are in delta connection!
3. Three unequal effective resistances (80 W, 200 W, 500 W) have to be connected in star and in delta arrangement to a threephase network with a linetoline voltage of 380 V. Determine the total effective power!