Download Chapter Three-phase circuits

Chapter
5
Three-phase circuits
Table of Contents 5.1 Introduction .............................................................................................................................. 3 5.2 From single‐phase to three‐phase systems .............................................................................. 3 5.2.1 Modelling three‐phase lines when induced EMFs between wires cannot be neglected ................ 7 5.3 The single‐phase equivalent of the three‐phase circuit ......................................................... 10 5.4 Power in three‐phase systems ................................................................................................ 12 5.5 Historical notes ....................................................................................................................... 14 5.5.1 Tesla’s short biography ............................................................................................................... 14 5.6 References .............................................................................................................................. 14 5.7 Proposed exercises ................................................................................................................. 15 For the teacher
In this chapter, coherently with the approach followed in chapter four, the three-phase
system is introduced modeling the lines connecting generators and loads as purely resistive
branches: i.e. neglecting cross conduction between wires (both conductive and
displacement currents) and the effect of EMF induced in meshes containing the lines due to
the alternating current circulating in the line conductors.
This is due to the fact that in this book the tricky nature of R-L models of transmission
lines is evidenced, and the effect of cross-conductor currents and induced EMFs are
considered an “advanced” topic, that is dealt with in Appendix B.
The teacher might have met books in which the conductor lines are substituted by an R-L
couple, and may have experienced some discomfort due to the fact that in three-phase
system, even more than in single phase ones, the meaning of those L’s is really
questionable.
For a practical point of view, limiting the analysis to resistive models of the lines has
negligible effects, since in low-voltage lines, probably when only ones this book’s students
will be required to analyse quantitatively, the line inductances are nearly all negligible or
quasi negligible.
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
5.3
5.1 Introduction
Every reader of this book has surely seen, in their lives several times overhead high-voltage
transmission lines, such as the ones depicted in fig. 5.1. It can seen from the pictures that in these
lines energy is transferred using three-wire lines: the left-hand pole carries three wires, the right
hand one holds four wires, but the highest one has just the purpose of protecting the line from
lightning and does not contribute to carrying electric energy.
Fig. 5.1. Two three-phase overhead lines (left: medium voltage; right: high voltage).
This is a very visible evidence of the fact that electric energy is used and transferred, when
power overcome a few kW always in systems called three-phase systems, that are based on three
wires for transferring electric power.
This chapter shows the genesis of three-phase lines and systems and gives some information
on how to make computations with these systems in the particularly important (and easy) cases of
balanced three-phase systems.
5.2 From single-phase to three-phase systems
Imagine we want to feed three load impedance Zu by means of a generator of sinusoidal
EMF., situated relatively far from the load, at distance L (fig. 5.2 a). Its electrical behaviour, if
induced EMF between wires can be neglected, can be analysed as represented in fig. 5.2 b).
Rl
+
3Iu
+ E
Rl
L
a)
Fig. 5.2. Single-phase transmission to three loads.
b)
Iu
Iu
Iu
Zu
Zu
Zu
5.4
Chapter 5: Three-phase circuits
Consider that the load requires a power S=P+jQ to be delivered at a voltage U (1).
The current flowing though the line is I=S/U and the power lost through the line is Pl=2RlI2
Now it will be shown that there is a technique that allows to transfer the same power under
the same voltage to the three loads while reducing the line losses, if the quantity of copper for the
lines is the same, or reducing the quantity of involved copper at equal line losses.
Consider the arrangement shown in fig. 5.3 The three electromotive forces have equal
amplitudes, and are out of phase with each other by 2π/2 rad. Such set of voltages is called
balanced three-phase set of voltages. It will be shown in the next chapter that these sets of
voltages are very easily created in electrical machines.
In fig. 5.3 b) a circuit modelling the system of figure 5.3a is reported, valid in case induced
EMFs between wires can be neglected.
Ea
+
+
Eb
+
+
Ec
+
+
Rl
Ia
A
IAC
IAB
Z ab
Rl
Rl
Ec

Ea
Ib B
IBC
Z bc
Ic C
Z ca
Eb
Z ab=Z bc=Z ca
L
a)
b)
Fig. 5.3. Three-phase transmission to three loads.
Using this arrangement, it can be easily demonstrated that the three loads are fed with the
same active and reactive power of the arrangement of fig. 5.3 but with reduced losses in the lines.
To do this, consider figure 5.4. The voltages between wires at the source terminals can be
determined using the common rule for vector differences (figure 5.4 a); the voltages between the
corresponding phases at the load terminals will be similar to them, and are reported with a
common origin in fig. 5.4 b). Considering that by hypothesis the three load impedances are equal
to each other, the currents absorbed by the three loads will have the same amplitudes and phase
displacements with respect to the corresponding voltages. Therefore they will be of the type
shown in fig. 5.4 b). Considering KCL at nodes A, B, C, the corresponding line currents are
found by difference of the currents flowing in the load impedances, and will be of the type shown
in fig. 5.4 b). The amplitudes of the line currents will therefore be equal to the load currents,
multiplied by 3 , as shown by the construction reported in fig. 5.4c).
U ca=E c-E a


U CA
Ia
ICA
U bc
Ea
Eb
ICA
IAB
U ab
U BC


3 / 2 I AB
60°
IAB
IBC
a)
Ia / 2 
U AB
Ec
b)
c)
Fig. 5.4. Phasors relative to the three-phase transmission shown in fig. 5.3.
1
remember that when in alternating current a voltage value is quoted, in case of no other specification, it is always
intended to refer to the r.m.s. value of the sinusoidal waveform.
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
5.5
The conclusion is that the total line losses will be
P' jl  3Rl ( 3I u ) 2  9 Rl I u2
Compared with the solution of fig. 5.2:
Pjl  2 Rl (3I u ) 2  2  9Rl I u2
The ratio of the two powers is therefore
P' jl / Pjl  0.5
that demonstrates that the transmission of electric power shown in fig. 5.3, called three-phase
transmission is more effective than the single-phase one (2). Naturally, in case one wants the
power loss to be unchanged, he can reduce the total copper used for the three-phase transmission,
In this case, instead of having a reduction in power loss one would have had reduction in line
cost.
Another very important advantage of three-phase systems consists in the fact that, differently
from the single-phase transmission, the instantaneous power delivered by three-phase generators,
and transferred to three-phase loads is constant. This will be demonstrated in the next section.
The three load impedances discussed earlier, were connected between the line wires, and
formed a triangle. This was done because equality of the voltage at the load terminals was
searched in comparison with the single-phase transmission.
Indeed also star connection is possible, and very frequently used.
Consider the scheme reported in fig. 5.5. In this case the load is constituted by three starconnected impedances.
As usual, the left part of the figure represents a physical system whose mathematical model
is the circuit reported in the right part of the figure.
Obviously, that circuit, although being evidently a three-phase circuit, can be studied using
the usual rules for solving AC circuits.
In particular the nodal analysis can be uses, assuming as reference node O. The following
equation can be thus written:
Ea  VN Eb  VN Ec  VN


0
Rl  Z a Rl  Z a Rl  Z a
Or:
Ea  Eb  Ec
VN
3
(°)
Rl  Z a
Rl  Z a
Ec

Ea
-E a
Eb
It must now be observed any balanced three-phase set of quantities
(voltages, currents) has null sum, as can be immediately inferred from the
scheme aside: evidently  Ea  Eb  Ec  Ea  Eb  Ec  0
As a consequence eq. (°) yields:
2
The analysis performed considered only the amount of copper, at equal line-to-line voltages. A more detailed
analysis should have also considered the cost of insulating to each other three wires instead of two. However, it can
be seen that even with this more detailed analysis the advantages of three-phase transmission remains sound.
5.6
Chapter 5: Three-phase circuits
VN
 VN 
Rl  Z a
0
03
This shows that in the circuit of fig. nodes O and N share the same potential.
Points O and N are called star points or neutral points of the three-phase system.
Ea
+
+
Za
Rl
Ec
+
O

Ea
Eb
+
Rl
Zb
Ec
+
Rl
Zc
N
Eb
+
Z a=Z b=Z c
L
a)
b)
Fig. 5.5. Three-phase transmission to a star-connected load.
It is easy to convince ourselves that this is a rather general conclusion:
Result: potential of neutral points in three-phase balanced systems
in three-phase a system fed by a balanced set of voltages, if all the impedances present in
the three phases are equal to each other, all the neutral points of the star-connected
impedances and voltage sources share the same potential
Therefore entire three-phase networks can be built, where several three-phase loads might be
present, e.g. constituted by impedances star or delta connected. In case of balanced networks, in
which the impedances present in the three phases are equal to each other and the sources are all
constituted by balances three-phase sets, all the star centers share the same potential. In the
example of figure 5.6, for instance, it is VN1=VN2=VN3=VO. When impedances are delta connected,
they can be replaced by three, equivalent, star-connected equivalents. The star-delta conversion
formulas are the same used in chapter 4 for generic AC circuits, that, in the case considered here,
of impedances equal to each other become simply:
Zstar=3*Zdelta.
O
Ea
+
Rl
Eb
+
Rl
Ec
+
Rl
I2a
+
R2
Z4
R2
Z4
R2
U 2a
Z1
Z1
Z1
N1
E
Z1
Z2
Z2
Z4
R2
Z2
Z 3D
a)
Z 3D
N2
Rl
+
Z4
Z 3D
Z2
Z 3s
N4
Z 3S 
Z 3D
3
Fig. 5.6. A three-phase system with several components and its unifilar representation.
b)
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
5.7
It is apparent that when large balanced three-phase networks are to be represented it is
superfluous to always replicate circuit parameters for the three phase. Therefore very often, in
these cases, the so-called unifilar representation is reported, such as that shown in fig 5.6 b),
where indication that a single line indicates indeed three elements is given by use of the three
inclined small bars
.
Three-phase systems are characterised by two kinds of voltages: line-to-line and line-toneutral voltages.
Line-to-lint voltages are the voltages that can be measured between two of the three wires,
while line-to-neutral are the voltages between a common point, called neutral point. Both are
expressed as r.m.s. values.
There is always a very simple and important relation between line and phase voltages of a
three-phase system:
U line  3  U phase
Ec

Ea
60°
Ea
Eb
U ab
Eb
3 / 2  Ea
This can be easily seen looking at the phasors of these
voltages as in the drawing aside. For instance it is
U ab  Ea  Eb ; U  U  3  E  3 U
line
ab
a
phase
The duality of line and phase voltages is so important that very commonly three-phase
systems are named after two numbers, representing the two voltages.
For instance in the European Union countries the voltage used for distributing energy at users
needing slightly more power than homes is the following:
400/230 V
The first number represents the line-to-line nominal voltage of this system, while the second
one represents the phase voltage. Note that the ratio between the two is only approximately 3 :
this was chosen to avoid decimal numbers in the representation of any of the two nominal values.
Then a three-phase system is named using just a single value, it always refers to the r.m.s.
value of the line-to-line voltages: for instances some standard voltage levels are 10 kV, 15 kV,
20 kV, 130 kV, 220 kV. All these values are line RMS voltages.
5.2.1 Modelling three-phase lines when induced EMFs between wires cannot be
neglected
Consider again the system reported in fig. 5.5a. Let now discuss what happens when the
EMFs between wires are not neglected.
To understand the situation a three-phase wire of the line must be considered such as that
reported in figure 5.7a. There are fluxes that are linked with the three different meshes constituted
by the different couples of conductors.
5.8
Chapter 5: Three-phase circuits
ab
ia
a
ib
bc
+ eab
ca
eca
b
uba
+
ab
+
bc
ca
-
eba
+
ic
-
c
a)
b)
Fig. 5.7. A three-dimensional view of the conductors constituting a transmission line (left)
and a cross-section representation to better show the mesh-coupling fluxes (right).
The following equations can be written, in general for time-varying functions (left) and for
phasors (right):
dab

eab  uab  Ra ia  Rbib  dt

dbc

ebc  ubc  Rbib  Rcic 
dt

dca

eca  uca  Rcic  Raia  dt

d ab

 Eab  U ab  Ra Ia  Rb Ib  dt

d bc

 Ebc  U bc  Rb Ib  Rc Ic 
dt

d ca

 Eca  U ca  Rc Ic  Ra Ia  dt

By detailed analysis of these equations it can be demonstrated that they can transformed in a
more useful form, that, for the phasor formulation, is:
d Ia

 Eab  U ab  Ra Ia  Rb Ib  Laeq dt

d Ib

 Ebc  U bc  Rb Ib  Rc Ic  Lbeq
dt

d Ic

 Eca  U ca  Rc Ic  Ra Ia  Lceq dt

The parameters Laeq, Lbeq, Lceq, can be shown to be an easy function of the line geometry,
although details are not reported here.
The system can therefore be studied by the metacircuit3 reported in fig. 5.8.
O
Ea
+
Rl
Laeq
Za
Eb
+
Rl
Lbeq
Zb
Ec
+
Rl
Lceq
N
Zc
Fig. 5.8. Metacircuit representing the system of fig. 5.5a.
when the EMFs between wires are not neglected.
3
For the concept of metacircuit the reader is suggested to consult sect 4.3.2.
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
5.9
It has been seen in the previous section that three-phase systems are particularly effective
when voltages and currents are constituted by balanced sets: this was shown to happen when all
the supply voltages are constituted by a balanced set and all the impedances on different phases
are equal to each other. It is natural to require, for perfect balancing of the system, that also
transmission lines must have symmetry: all the conductors must have the same resistance and
their spatial disposition is such that when the conductors are traversed by a balanced set of
currents, ab, bc, ca constitute a balanced set of fluxes as well. This happens when the line has
physical symmetry, i.e., considering the disposition reported in figure 5.7b, when the conductors
are positioned at the vertexes of an equilateral triangle.
When a transmission line, fed by a balanced set of voltages, feeds a balanced load (i.e. a load
constituted by three equal impedances or a more complex balanced load), and the line is
physically symmetrical, the three fluxes ab, bc, ca constitute a balanced set of fluxes.
It must be added that in normal engineering practice it is possible to assume very often, with reasonable
degree of approximation, that the three-phase system of currents flowing through a line is balanced even
though, for practical reasons, the conductors are not disposed at the vertexes of an equilateral triangle.
Under these conditions, the values of the inductances Laeq, Lbeq, Lceq are equal to each other
and their value is:
D
Leq  k ln
(5.1)
kr r
where D is the distance between two conductors of the line, r is its radius, and k and kr are
constant values. More details on the value of Leq are not supplied here, because they are well
beyond the scope of this book, and because even electrical engineers tend to read them on
manuals instead of computing them directly. Eq 5.1 is reported just to show that Leq increases
when D increases, and decreases when r increases; the dependence is a lograritmic one.
It the equality between all the quantities oeprating in the different conductors of the circuit is
to be evidenced, the drawing can be transformed obtainig that reported in fig, 5.9.
O
Ea
+
Rl
Leq
Ia
Z
Eb
+
Rl
Lbeq
Ib
Z
Ec
+
Rl
Lceq
Ic
Z
Fig. 5.9. The metacircuit reported in fig. 5.8, when the system is fully balanced.
As a final consideration of this section, it is worth while noticing that in low voltage lines (up
to 1 kV), inductances Leq are very small and designers neclect them very often. In other terms, at
the usual frequency, in low voltage lines it can often be assumed that  Leq  Rl . Only in case of
very large LV cables, having cross section of copper larger than 95 square millimeters,
consideration of Leq has some improtance.
On the other hand, in medium voltage lines (say between 1 and 35 kV)4 this line inductance
becomes important, and the corresponding reactance Xeq=Leq typically reaches the same order of
magnitue of line resistances Rl.
When much higher voltage lines are considered reactances Leq tend to become dominant
with respect to line resistances.
4
The threshold of 35 kV has set here because it is considered in the International Standard IEC 60038. Individual
countries might have slightly different thresholds: for instance in Italy this value is set to 30 kV.
5.10
Chapter 5: Three-phase circuits
5.3 The single-phase equivalent of the three-phase circuit
For what seen in the previous chapter, in a balanced three-phase system, that occurs in phase
a occurs in phase b also, only with a phase displacement of 120° and then occurs in phase c, with
another phase displacement of 120 degrees.
So it is sufficient, to understand the full behaviour of the three-phase system, to evaluate
what happens on one of the three phase: all voltages and current phasors in the other phases will
have the same amplitudes and will be just rotated by 120 or 240 degrees.
By convention, the phase considered is the first of the three: phase a, if phases are named a,
b, c, phase 1 if they are named 1, 2, 3, phase r if the naming convention considers as phase names
r, s, t.
The circuit constituted by only the first phase of a three-phase system is called the singlephase equivalent of the three-phase circuit.
Just to make this approach more explicit, consider again the circuit shown in fig. 5.6. Its
behaviour can be easily evaluated considering the single-phase equivalent shown in fig. 5.10.
Rl
R2
I2a
+
+
Z1
Ea
Z2
U 2a
Z 3s
O N1  N2  N4
Z4
O
Fig. 5.10. Single-phase equivalent of the circuit of fig. 5.6.
The process for determining the single-phase equivalent of a three-phase circuit should be
now clear: since the potential of all neutral points are the same, they are connected to each other,
creating the lower common node of the circuit. The upper nodes of the single equivalent represent
the corresponding nodes in phase a of the three-phase circuit. To make things clearer, in both
figures voltage U2a and current I2a are shown.
The same process can be used when EMFs between wires of transmission lines be neglected.
For instance, the metacircuit reported in figure 5.9 can be studied by its single-phase equivalent,
reported in figure 5.11. The currents and voltages flowing in this single-phase equivalent are the
ones flowing in conductor a of the actual three-phase system.
Rl
Leq
I
+
+
E
Z
U
-
Fig. 5.11. Single-phase equivalent odf the circuit of fig. 5.9.
Example 1
Determine I1, I1, I2, I3 and UAB in the following circuit, where:
e1(t)=2380sin(314t+62°) (balanced system of voltages)
e2(t)=2216sin(314t+30°) (balanced system of voltages)
Z1=1+2j
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
5.11
Z2=1.5+2.5j
Z3=120+40j
Z1
I2
A
e1(t)
+
I1
e2(t)
Z2
+
I1
Z1
Z2
B
+
+
+
Z2
Z1
+
I3
Z3
Z3
Z3
Let’s first determine the phasors of sources:
E1=38062°
E=21630°
Now let’s transform all delta connections to equivalent star connections:
I1
I2
A
Z1
E2
Z2
+
E1
+
Z1
Z2
B
+
+
Z2
Z1
+
+
I3
Z3
Z3
Z3
where:
E1=E1/3e-j30°= 219.432° = 186.1+j116.7
Z3=Z3/3=40+j13.33
Since all the neutral points have the same potential (balanced generators and loads), the
following single-phase equivalent circuit can be used:
Z1
A
I2
E2
Z2
+
I1
E1
+
I3
Z3
To solve this circuit, let us apply Thévenin’s theorem left side the red section:
5.12
Chapter 5: Three-phase circuits
I2
Zth
E2
Z2
+
Eth
+
+
U
-
where:
E Z
Eth= 1 3 =183.1+j105.4
Z1  Z 3
Zth=Z1||Z3=1.032+j1.890.
Hence:
E  E th
I2= 2
=0.824-j0.417
Z th  Z 2
U=E2-I2Z2=184.8+j106.6=213.329.97°
I3=U/Z3=4.957+j1.012
I1=(E-U)/Z1=4.133+j1.429=4.37319.08°
To calculate the phase-to-phase voltage UAB and the phase current I1, U and I1 must be
modified in amplitude and angle, according to what discussed in sect. 5.2.
UAB=U3*ej30°=184.9+j319.9=369.559.97°
I1=I1/3*ej30°=1.654+j1.908=2.52549.08°
5.4 Power in three-phase systems
Consider the three-phase circuit reported in fig. 5.5 and fig. 5.12. How the three loads are
generic linear boxes, and the electric quantities are indicated as time functions. Since the
electromotive forces are sinusoidal also the currents are sinusoidal; because the three loads are
equal (they share the same constitutive equation u=f(i)) and the three electromotive forces are
constituted by a three-phase balanced system of voltages, also the currents constitute a threephase balanced system.
ea
+
Rl
ia
+
va
O
eb
+
Rl
ib
+
vb
ec
+
Rl
ic
+
vc
A
+ ua N
B
+ ub -
C
+ uc p(t)
ea (t )  Eˆ sin t

eb (t )  Eˆ sin t  2 / 3   

ˆ
ec (t )  E sin t  2 / 3   
ia (t )  I sin t   

ia (t )  I sin t  2 / 3     


ia (t )  I sin t  2 / 3     
ua=f(ia) ub=f(ib) uc=f(ic)
p(t)=vaia+vbib+vcic
Fig. 5.12. Circuit showing computation of instantaneous three-phase power.
In a way similar to what done in chapters 3 and 4, we can evaluate the power p(t) transferred
to charge crossing the curved line of fig. 5.12.
The charges flowing through the three phases are moved by the electric field present inside
the conductor, whose potential values, written with reference to node O are v1(t), v2(t), v3(t).
Therefore the power transferred to the load is the sum of the powers transferred though each
wire and it is:
p (t )  p a (t )  pb (t )  pc (t )  va ia  vb ib  vc ic  u a ia  ub ib  u c ic
(5.2)
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
5.13
the latter equality being justified by the fact that potentials of node N and O are the same.
Consider the expression for the instantaneous power entering a single linear circuit branch:
p  UI cos  (1  cos 2 t )  UI sin  sin 2 t  UI cos   f (2 t )
where U and I RMS values of are branch voltage and current respectively and
f ( x )  UI cos  cos x  UI sin  sin x .
f(x) has the following property:
f ( x)  f ( x  4 / 3   )  f ( x  4 / 3   ) 
UI cos   cos x  cos( x  4 / 3   )  cos( x  4 / 3   )  
UI sin  sin x  sin( x  4 / 3   )  sin( x  4 / 3   )  
UI cos   cos x  cos( x  2 / 3   )  cos( x  2 / 3   )  
UI sin  sin x  sin( x  2 / 3   )  sin( x  2 / 3   )   0
because the sum of three sinusoidal functions displaced from each other 2/3  is always zero.
Therefore it is:
p (t )  3UI cos   f (2 t )  f  2( t  2 / 3   )   f  2( t  2 / 3   )  
 3UI cos   f (2 t )  f  2 t  4 / 3   )   f  2 t  4 / 3   )  
(5.3)
 3UI cos 
Eq. (5.3) is a very important result:
Result: Instantaneous power in three-phase balanced systems
The instantaneous flowing in three-phase balances systems is constant
This is different from what happens in single-phase circuits, and is therefore another big
advantage of three-phase circuits over single-phase ones.
It will seen in the next chapters that this implies very uniform transfer of power to
mechanical shafts when three-phase machines are involved, since when the machine shafts rotate
at constant speed, whey receive constant torque, therefore with no vibration induced.
As far as active, reactive and apparent powers are concerned, their values can be obtained
trivially considering that a three-phase phasor circuit is still an AC circuit:
Ptrif  3UI cos  Qtrif  3UI sin 
S trif  U a I a*  U c Ib*  U c Ic*
S trif  3UI
(5.4)
It is remembered again that in (5.4) U is the RMS value of the line-to-neutral voltages. The
power transferred though a three-phase circuit are often expressed in terms of the line to line
voltage Ul,. From the fundamental relation U l  3U immediately:
Ptrif  3U l I cos 
Qtrif  3U l I sin 
Strif  3U l I
Example 2
Consider the circuit of Example 1 and determine the active and reactive power:
- delivered by the two three-phase generators
- absorbed by the two lines
- absorbed by the load
Assuming the frequency to be 50 Hz, determine the capacity of the capacitors (before
star connected and then delta connected) required to correct the load power factor to 1.
Remembering that the three-phase powers can be obtained multiplying by 3 the powers
calculated in the single-phase circuit, we obtain:
5.14
Chapter 5: Three-phase circuits
Sgen1=3E1I1*=2805 W+j643.6 var
Sgen2=3E2I2*=327.2 W+j501.3 var
Sline1=3I12Z1=57.37 W+j114.7 var
Sline2=3I22Z2=3.840 W+j6.400 var
Sload=3I32Z3=3071 W+j1024 var
Please note that the Boucherot’s theorem is respected:
Sgen1+Sgen2=Sline1+Sline2+Sload
The power factor correction can be carried out on the single-phase circuit, directly
obtaining the capacity of each star-connected capacitor:
L3
13.33 / (2 50)

=23.88 F.
2
R  X 3
402  13.332
Qsin gle phase Z3
1024 / 3
Also: C=
=23.88 F

2
2 50  213.32
 U
The capacity of the three delta-connected capacitors can be easily calculated considering
that in a balanced load:
Z=3Zthus C=C/3=7.962 F.
C=
2
3
5.5 Historical notes
5.5.1 Tesla’s short biography
Nicolas Tesla (Croatia, 1865 - New York, USA 1943) is a scientist and engineer, that gave
an outstanding contribution to the development and diffusion of electrification in the U.S. in the
early years of the XX century.
Born in Austria-Hungary, he moved in the United states in 1884. There, he participated to the
so-called “war of the currents” between those, lead by Edison, that were promoting DC current
for commercial electricity distribution and those promoting AC, lead by Westinghouse.
One of the big advantages of AC in constituted by the existence of polyphase systems,
especially three-phase systems, developed by Tesla, and whose patents were bought from
Westinghouse.
Tesla first worked under Edison then, because he felt himself underestimated, moved to
Westinghouse’s, and then could deploy all his ideas and knowledge on AC and polyphase AC
systems.
Tesla is also credited to have discovered the rotating magnetic field, that will introduced and
discussed in the next chapter, independently of Galileo Ferraris, that made the same discovery in
the same period.
The S.I. unit of magnetic flux density is the tesla, a name given in honour of him.
5.6 References
M. Ceraolo - D. Poli: Fundamentals of Electrical Engineering
5.15
5.7 Proposed exercises
5.1. Determine Ia, Ib and Ic in the following circuit,
where:
Ea=10020° (balanced system of voltages)
Zl=2+3j
Z=30+10j
Ea
+
Eb
+
Zl
Ia
Ib
Zl
Z=1+3j
Zl=18+12j
Z2=5+j
+
Zl
Ic
Z
Eb
Z
Ec
Z
I2
A
Z2
B
+
Z
Z2
I1
Z
Z1
5.2. Determine Ia, Ib, Ic and Iab in the following
circuit, where:
Eab=38030° (balanced system of voltages)
Zl=5+10j
Z=18+6j
Zl
+
Eab
Iab
Ib
Z
Z
Z1
5.5. Determine I1, I2, I3 and UAB in the following
circuit, where:
Ea=2200° (balanced system of voltages)
Eb=22030° (balanced system of voltages)
Zl=1+2j
Z2=2+3j
Z3=15+9j
I1
E1
+
Z1
Z1
I2
A
+
E2
Z2
+
Eca
Zl
Ia
Z2
Ia
Z
+
Ec
Ea
+
+
Zl
Ic
Z
Z1
Zl
+
Zl
Eca
Ib
Iab
+
Z
+
Ebc
Zl
Ic
I3
Z3
Z3
Z3
Ia
Z
Eab
Z2
Z1
+
+
5.3. Determine Ia, Ib, Ic and Iab in the following
circuit, where:
Eab=2200° (balanced system of voltages)
Zl=2+4j
Z=12+3j
Z2
B
+
+
Ebc
Z
5.4. Determine Ia, I1, I2 and UAB in the following
circuit, where:
Ea=22015° (balanced system of voltages)
5.6. Using the method of nodal voltage, determine
Ia, Ib and Ic in the following unbalanced system,
where:
Ea=2200°, Eb=210-120°, Ec=215115°
Z=1+3j, Za=20, Zb=5+25j, Zc=30-8j
Then calculate the current INO flowing in a resistor
R=10 , positioned between O and N (use
Thévenin’s theorem).
4.16
Chapter 4: Techniques for solving AC circuits
Ea
Z
Eb
Z
Ia
Za
+
O
Ib
Zb
+
Ec
Z
Ic
N
Zc
+
5.7. Using the method of nodal voltage, determine
Ia, Ib, Ic and I’ in the following unbalanced system,
where:
Ea=12730°, Eb=120-90°, Ec=130145°
Z=1+j, Za=10+5j, Zb=8j, Zc=12j, Z’=15
Ea
Z
Ia
Za
+
Eb
Z
Ec
Z
Ib
Zb
+
Ic
Zc
+
Z’
I’
Note: To solve the following problems, assume
that the voltage phasors assigned in the previous
exercises were expressed considering the voltage
RMS values.
5.8. Consider exercise 5.1. Calculate the active
and reactive power absorbed by the three-phase
load and by the line; determine the active and
reactive power delivered by the generating system
and verify the Boucherot’s theorem.
5.9. Consider exercise 5.2. Calculate the active
and reactive power absorbed by the three-phase
load and by the line; determine the active and
reactive power delivered by the generating system
and verify the Boucherot’s theorem.
5.10. Consider exercise 5.3. Calculate the active
and reactive power absorbed by the three-phase
load and by the line; determine the active and
reactive power delivered by the generating system
and verify Boucherot’s theorem.
5.11. Calculate the active and reactive power
delivered by the two three-phase generators of
exercise 5.5 and the active power losses of the two
lines.
5.12. Verify Boucherot’s theorem in exercise 5.7.
5.13. Calculate the power factor of the load of
exercise 5.4 (Z1 and Z2). Determine the capacity of
the three star-connected capacitors required to
correct the power factor to 1. Repeat if the
capacitors are delta connected.