ENGNG 2024 Electrical Engineering
 E Levi, 2002
1
FUNDAMENTALS OF ELECTROMECHANICAL ENERGY
CONVERSION
1. PRELIMINARY CONSIDERATIONS
Electromechanical energy conversion is achievable in a number of ways. These
possibilities rely on different fundamental laws of electrical engineering. As the only method
that has importance on the large scale is electromechanical conversion achieved by means of
electromagnetic converters, this section is fully devoted to the analysis of basic principles
involved in electromagnetic electromechanical conversion.
Electromechanical energy conversion is achieved by devices that are usually called
electric machines. In principle, laws of electromagnetics can be used to design converters with
the linear and with the rotary motion. Converters with linear motion are called linear electric
machines, while those that rely on rotating motion are called rotating electric machines. Vast
majority of existing machinery belong to the category of rotating electric machines. These
include all the machines used to generate the electricity, as well as the most of the machines
used in industry to perform some useful work while converting electric into mechanical
energy. Linear machines are used relatively rare for somewhat specialised applications. It is
for this reason that only rotating electric machines will be dealt with here. Prefix ‘rotating’
will be omitted and the converters will be called simply electric machines, implying that
devices under consideration are characterised with rotational movement.
Operating principles of electric machines involve two basic laws of electromagnetism,
namely the law of the electromagnetic induction (Faraday’s law) and the law of force creation
in an electromagnetic field (Bio-Savart’s law).
Consider the situation shown in Fig. 1. A conductor is connected to an electric source
and it carries current I. It is placed in the magnetic field of certain flux density B (which is of
course a vector; hence the arrow above the symbol in Fig. 1). Interaction of the flux density
and the conductor current leads to the creation of an electromagnetic force
BlIF
e

constant speed. Under this condition the electromagnetic force and the mechanical force are
mutually equal, but act in the opposite directions.
Consider next Fig. 2, where the same conductor is placed in the same flux density.
However, the conductor is now not connected to the electric source; instead, the electric
circuit is closed by using, say, an external resistance. The conductor is now dragged through
the flux density using mechanical force at certain speed and this is the origin of the movement
in this case. The sequence of events now reverses. An electromotive force, given with (2), is
at first induced in the conductor. Since the circuit is closed, a current starts flowing.
Interaction of the current and the flux density causes creation of the electromagnetic force.
This force again acts in the opposite direction to the mechanical force and the equilibrium is
established when the two forces are equal but act in the opposite direction. Note that in this
case the source of motion is the supplied mechanical energy. The mechanical energy is now
converted into electrical energy and the process is called generation.
B
I
F
e
v
Fig. 1 – Illustration of motoring.
B
I
F
m
v
F
e
Fig. 2 – Illustration of generation.
It is important to note here that the process of electromechanical energy conversion is
reversible. This means that either electric energy can be converted into mechanical energy, or
mechanical energy can be converted into electric energy, by means of the same physical

e
=IlB,is created on
each of the two conductors. However, one of these two forces acts to the left, while the other
ENGNG 2024 Electrical Engineering
 E Levi, 2002
3
one acts to the right (due to opposite directions of the current flow in the two conductors).
Now, a torque is created on each of the two conductors, that equals the product of the force
and the radius. However, since forces act in opposite directions at opposite sides of the
structure, the torques will both act in anticlockwise direction, initiating the rotation of the
structure in anticlockwise direction. The total electromagnetic torque will in general be the
sum of all the individual torques acting on individual conductors.
Current in
Current out
B
F
e
F
e
Fig. 3 – Torque creation in the rotating structure.
Every electric machine consists of ferromagnetic iron cores and windings mounted on
the iron cores, these elements being of essential importance for electromechanical conversion.
An electric machine consists of a stationary element, called stator, and a rotating element
(such as the one in Fig. 3) called rotor. The winding is placed in slots of the stationary stator
and/or in slots of rotational rotor. The winding consists of an appropriate number of turns. A
turn is composed of two conductors which are placed in such a way that the induced
electromotive forces in them sum up. The current therefore flows in the opposite direction, as
illustrated in Fig. 3.
As already noted and explained, the operation of electric machines relies on Faraday's
law of electromagnetic induction and on Bio-Savar's law of electromagnetic force (torque).

. This may sound awkward but will be clarified
later on. In deriving (3) the use was made of the correlation between the angle travelled by the
rotor and its speed of rotation,
ENGNG 2024 Electrical Engineering
 E Levi, 2002
4
= dt
ωθ
(4)
that reduces for a constant speed of rotation to
θ
=
ω
t. Chain differentiation rule was applied
as well. The total flux of the winding is called flux linkage and is denoted with
ψ
in (3). It
depends on the flux seen by each conductor
Φ
and on the number of turns N. Flux linkage is
ψ
=N
Φ
.
Electromotive force in an electric machine is induced either due to rotation of a
winding in the flux density, or due to rotation of the flux density with respect to a stationary
winding. Change of flux linkage can be caused either by mechanical motion or by change of
current in time. This is reflected in (3) and will be elaborated in detail later on.
Let us further clarify the two operating regimes of electric machines, generating and
motoring. Generating is discussed first. Due to the action of the

T
e
Fig. 4 – Torque and speed directions in generation (left) and motoring (right).
In what follows a generalised electromechanical converter is discussed at first. The
analysis is valid for any type of electric machine; the only constraint is that there is only one
degree of freedom for mechanical motion (i.e. rotor can rotate along one axis only).
2. GENERAL MODEL OF AN ELECTRIC MACHINE
2.1 Losses and efficiency
Efficiency of an electric machine is defined in the same way as for any other device,
as ratio of the output to input power
ENGNG 2024 Electrical Engineering
 E Levi, 2002
5
11 <
+
−=
+
==
lossout
loss
lossout
out
inout
PP
P
PP
P
PP
η
(5)

Since mechanical power is a product of torque and speed, this means that the mechanical loss
torque is taken as proportional to the speed of rotation.
One important point to note is that the nature of the input and output power depends
on the role of the machine. In motoring the input power is electrical, while the output power is
mechanical. In generation it is the other way round, the input power is mechanical while the
output power is electrical (in generation, there may be some windings that take electrical
power as well, while some other windings generate electrical power). It has to be remembered
that the rated power of the machine (power for which the machine has been designed), which
is always given on the nameplate of the machine, is the
output power. Hence, in generation
the known rated power (always identified further on with an index
n) is the output electrical
power, while in motoring it is the output mechanical power.
2.2 Power flow in an electric machine
Since the role of the input and the output power is dependent on the function that the
machine performs, the two cases are treated separately. In what follows lower case symbols
are used for all the quantities, meaning that
instantaneous time-domain variables are under
consideration. The idea behind the subsequent development is to develop a general
mathematical model that is valid for any rotating electric machine. It is for this reason that the
number of windings is not specified. Instead, it is taken as being equal to
n, where this is an
arbitrary number. The electric machine is for the time being a black box. There are two doors
that enable access to the machine, electrical door and mechanical door. The power can be
either delivered to the machine, or taken away from the machine, through these two doors.
Electromechanical conversion takes place inside the box and the converted power is
p
c
.Fig.5
illustrates power flows inside an electric machine for motoring and generating. Apart from

storage
Electrical output
power
Mechanical input
power
Copper
losses in
windings
Mechanical
loss
Converted
power
Electromagnetic
energy storage
Mechanical energy
storage
Small electrical
input power
Fig. 5 – Power flow in an electric machine for motoring and generation, respectively.
As can be seen from Fig. 5, apart from input and output power and losses, there are
two internal storages of energy inside the machine. The first one is the stored electromagnetic
energy, while the second one is the stored mechanical energy. Stored mechanical energy is the
energy stored in rotating masses (kinetic energy) and it is in every aspect analogous to the
energy stored under linear movement (which is
2
2
1
mvW
mech
= ,wherem is the mass of the

()
112222
212111
22112112
2
22
2
11
2
1
2
1
2
1
iLiL
iLiL
iiiiLiLiLW
e
+=
+=
+=++=
ψ
ψ
ψψ
(10b)
Taking index
e for electrical power and index m for mechanical power in Fig. 5, one can write
the following power balance equations:
Motoring:
m

dW
pp
−++=
++=
−
(12)
Note that storages are energies, as defined in (9)-(10). Powers are time derivatives of energies
and this is taken into account in formulation of (11)-(12). In generation some windings make
take the power (
p
e2
), while other winding actually generate the power (p
e1
).
Equations (11)-(12) enable formulation of the converted power that is defined as
mec
tp
ω
= (13)
in terms of other known powers and derivation of the equation for motion of rotating masses
in terms of known parameters and inputs of the machine. This is a tedious procedure for the
generalised
n-winding converter and most of the derivations will be therefore omitted. Only
the starting equations and the final equations are presented in the next sub-section. It is to be
noted that all the powers, as well as all the other variables (currents, flux linkages) were
denoted with lower-case letters in this section. These are instantaneous time domain
quantities, and the same approach is used in the following sub-section. This enables creation
of a general mathematical model, in terms of time-domain instantaneous quantities, that is
valid for all possible existing types of electric machines with rotational movement.
2.3 Mathematical model

+=
ψ
ψ
(15)
where
=
−
n
n
R
R
R
R
R
1
2
1=
nnnnn
n
n
n
LLLL
LLLL
LLLL
LLLL
L



3
2
1
=
n
ψ
ψ
ψ
ψ
ψ3
2
1
(16b)
Note that in any electrical machine
L
ij
= L
ji
.
Input electrical power in motoring is
viivivivp
T
nne
=+++=
2211
(17)

2
1

2
1

2
1
2
1
1)1(32231131132112
22
22
2
11
=
++++++++++=
−−
(20)
Current sign in voltage equation (15) is such that the current is positive when it flows
into the winding. Hence in generation all the windings that generate will have negative
currents since the current flow will be in the opposite direction from assumed positive current
flow.
Mechanical power and mechanical loss are governed with
ω
ω
ω
kt
tp
tp

2
1
ω
JW
m
= (23)
Whatremainstobedoneistosubstituteallthepowersandderivativesofstored
energies into the power balance equations (11)-(12). This enables, first of all, calculation of
the converted power and the electromagnetic torque. Regardless of which of the two regimes
is considered, the converted power is found to be
i
dt
Ld
ip
T
c
2
1
=
(24)
Since according to (13) converted power is
ω
ec
tp = and since one can write using chain
differentiation rule that
()()()
ωθθθ
dLddtddLddtLd ==/ , one finds the electromagnetic
torque in the form
i

ePM
Le
+=−
+=−
(26a)
On the left-hand one has the difference between the driving torque (electromagnetic torque in
motoring, prime mover torque in generation) and the opposing torque (load torque in
motoring and electromagnetic torque in generation). On the right-hand side the first term is
the acceleration/deceleration torque (that exists only during transients and is zero in steady-
state) and the second term is the torque that describes mechanical losses. This particular
torque can be always taken as part of the load (or prime mover) torque since it is mechanical
in nature. One then arrives at the equation of mechanical motion in the frequently used form
generation
motoring
dt
d
JtT
dt
d
JTt
ePM
Le
ω
ω
=−
=−
(26b)
which shows that in any steady state (at constant speed)
generation0
motoring0

generation0
motoring0,
ePM
eL
mme
tT
tT
tk
dt
d
Jtt
ω
ω
(28)
i
d
Ld
it
T
e
θ
2
1
=
dtd
θω
=
where J and k are parameters of the machine and
=
−

223221
113121
=
n
v
v
v
v
v3
2
1
=
n
i
i
i
i
i3
2
1
=
n
ψ
ψ

hypothetical electric machine varies as a sine function of time, with the period equal to the
period of rotation. The instantaneous torque does exist. But, it is positive in the first half-cycle
and negative in the second half-cycle. The average torque is zero and hence the average
converted power will be zero even if the machine runs at a constant speed. The machine will
do motoring in the first half-cycle and generating in the second half-cycle, with a net zero
converted power over one cycle. Thus it follows that, if useful electromechanical conversion
is to take place, average torque of the machine must differ from zero. Average
electromagnetic torque T
e
can only exist if the certain correlation between stator current
(voltage) frequency, rotor current (voltage) frequency and the frequency of rotation is
satisfied. It can be shown that T
e
will be of nonzero value if and only if
rs
ωωω
−= (30)
where indices s and r identify stator and rotor angular frequency. Note that DC case is
encompassed by (30). Note as well that, according to (30), it is not possible to realise useful
electromechanical energy conversion if both stator and rotor windings are supplied with DC
currents. In such a case an average torque can only exist at zero speed. But converted power
equals zero at zero speed.
On the basis of (30) is it is now possible to classify the most commonly used electric
machines into three categories:
1. Synchronous machines: rotor frequency is zero. Hence frequency of rotation
equals stator frequency.
2. Induction machines: both stator and rotor windings carry AC currents. Rotor speed
is related with the two angular frequencies as
rs
ωωω

12
i
d
Ld
it
T
e
θ
2
1
=
<
>−
=+=+
generation0
motoring0,
ePM
eL
mme
tT
tT
tk
dt
d
Jtt
ω
ω
(28)
remains to be valid in the same form. However, (29) reduces to
=====

ii
d
dL
i
d
dL
it
sr
rs
r
r
s
se
++=
22
2
1
2
1
(33)
The first two components of the torque expression will have non-zero values only if the
winding self-inductance is a function of the rotor position. The third torque component is the
one due to the interaction of the stator and rotor winding and this component will exist in all
machines that do have windings on stator and rotor. The component of the torque due to
interaction of the stator and rotor winding is called fundamental torque component. The
torque component that exists only if self-inductances of the windings are functions of the
rotor position is called reluctance torque component. In general both torque components will
contribute to the average torque so that
lfundamenta
e

component.
Functional dependence of a self-inductance on rotor position is beyond the scope of
interest here. However, it is necessary to explain how the mutual inductance between the
stator and the rotor winding depends on rotor position. Consider Fig. 7, where magnetic axes
of the two windings are identified with symbols s and r. Magnetic axis of a winding is the
axis along which a particular winding produces flux. Stator magnetic axis is obviously
stationary, while rotor magnetic axis rotates with rotor. Let the maximum value of the mutual
inductance between the two winding be M. Flux linkage of the stator and the rotor winding
can be expressed as
ssrrrr
rsrsss
iLiL
iLiL
+=
+=
ψ
ψ
(35)
For the sake of explanation, let us assume that rotor current is constant DC and let us
investigate the contribution of this rotor current to the flux linkage in stator winding. The
maximum value of the contribution of the rotor current to the flux linkage in stator winding,
MI
r
, is shown along the magnetic axis of the rotor winding. Its projection on stator winding
magnetic axis is L
sr
I
r
. Table II lists values of the contribution for various rotor positions.
When the two axes are aligned (zero angle) the contribution is of maximum value. When the

sr
= (36)
The mutual inductance between a stationary and a rotating winding is therefore always
a function of the rotor position. Simple sine or cosine functional dependence suffices for
ENGNG 2024 Electrical Engineering
 E Levi, 2002
14
machines with uniform air-gap. If the machine is with salient poles on rotor, the expression
for the mutual inductance becomes more complicated.
As the next step, let us consider the induced emfs in the two windings of the machine
in Fig. 7. By definition
dtde
dtde
rr
ss
ψ
ψ
=−
=−
(37)
Substitution of (35) into (37) yields to
ω
θθ
ω
θθ
θ
θθ
+++=−
+++=−
+++=−

d
dL
i
dt
di
L
dt
di
Le
dt
dL
i
dt
dL
i
dt
di
L
dt
di
Le
sr
s
r
r
s
sr
r
rr
sr

ω
θ
++=−
++=−
d
dL
i
dt
di
L
dt
di
Le
d
dL
i
dt
di
L
dt
di
Le
sr
s
s
sr
r
rr
sr
r

s
ss
(39b)
Example 1:
A two winding system has the following inductances: stator winding self inductance =
0.8 H, rotor winding self-inductance = 0.2 H and mutual inductance between the stator
winding and the rotor winding = 0.4 cos
θ
[H]. The rotor revolves at constant angular
velocity of 40 rad/s and the initial value of the mechanical co-ordinate at zero time
instant equals zero. Determine the instantaneous value of induced electromotive force
in open-circuited rotor winding if the current that flows through the stator winding is
equal to 10cos(100t) A.
Solution:
In this example rotor current and its derivative are zero since the rotor winding is open circuited.
Moreover, stator and rotor self-inductances are constant. Hence the induced emf follows from (39b) as
ω
θ
+=−
d
dL
i
dt
di
Le
sr
s
s
srr
Derivatives of the stator current and the mutual inductance are

100cos40sin16040cos100sin400
+=
−=+=
−++=
+=
+=
βαβα
βαβα
Example 2:
A two-winding system has stator inductance of 0.1 H, rotor inductance of 0.04 H and
mutual inductance of 0.05 cos
θ
[H].
a) Rotor rotates at 200 rad/s and stator current is known to be 10sin200t. Find the
induced emf in rotor winding if it is open-circuited. The initial value of mechanical co-
ordinate at zero time is zero.
b) Current 10sin200t flows through both windings, which are connected in series. Find
the speeds at which average torque will exist; find the average torque values and
determine the values of the load angle which yield maximum values of the average
torque.
Solution:
a) Rotor current and its derivative are again zero. Hence once more
ω
θ
+=−
d
dL
i
dt
di

te
tttte
ω
b) Condition of average torque existence yields speeds at which the average torque will have non-zero
values:
()
=
==
±=
]rad/s[400
][rad/s0
]/[200
ω
ωω
ω
ω
ω
srad
rs
rs
Note that both self-inductances are constant. Hence the torque contains only the fundamental torque
component. Instantaneous torque is
() ()
[]
()
)sin()400cos1(5.2sin200sin5
sin05.0200sin10
2
2
2

 E Levi, 2002
16
For the speed of 400 [rad/s]:
2/for[Nm]25.1
[Nm]-1.25sin
1.25sin-)-0t1.25sin(80)-0t-2.5sin(40
)-sin(400t2.5cos400t)-0t-2.5sin(40
)400sin()400cos1(5.2
1
max
0
πδ
δ
δδδ
δδ
δ
−==
=
+=
+=
−−−=
=
e
e
e
e
e
T
ee
T

==
tL
LL
LLL
ss
rsrs
rss
where0
sin9.0cos9.0=
H0.95=H1H1
21
21
21
Rotor winding current is constant DC, of 10 A. Stator windings are fed with two-phase
system of currents, such that
titi
ssss
ωω
sin10cos10
21
==
a) Sketch a cross-sectional view of the machine and identify the type of the machine.
b) Develop the expression for the instantaneous and average torque produced by the
machine under the assumption that the condition of average torque existence is
satisfied. Calculate the average torque for load angle equal to 30 degrees and explain
its nature.
c) Sketch the dependence of average torque on load angle δ, identify motoring and
generating part and define the region of stable operation.
Solution:
In this example a so-called two-phase machine is considered. The example will show that with the

θθ
θθ
θ
θ
sin90
)sin(90
)sin(cos)cos(sin90
cos9.0sin10sin9.0cos10
cos9.0
sin9.0
2
1
2211
=≡
=
+−=
−−−=
+−=
=
−=
+=
ee
s
se
sse
ssre
rs
rs
rsrsrsrse
Tt

FNI t
FNI t
am
bm
cm
=
=−
=−
cos
cos /
cos /
ω
ωπ
ωπ
23
43
(41)
One observes that in terms of spatial dependence, all the three individual phase magneto-
motive forces are stationary and they act along the defined magnetic axis of the winding.
From (41) one notices that each of the three m.m.f.’s is varying in time. The values of the
ENGNG 2024 Electrical Engineering
 E Levi, 2002
18
three phase m.m.f.’s in the given instant of time correspond to those met in any three phase
system.
a
c
b
F
a

=++ = =
=+−+−
aa a a
aa
2
2
3
2
4
3
2
23 43
π
π
ωωπ ωπ
,
cos cos / cos /
(42)
The expression for the resultant magneto-motive force is most easily found if one recalls the
well-known correlation
()
cos . exp( ) exp( )
δδδ
=+−05 jj. Hence
()() () ()
( )
()
()
FNIee ae ae ae ae
FNIeeaaeaaeaaeaae

43
22 22
2
22 3 4
22
ω ω ωπ ωπ ωπ ωπ
ωω ω ω ω ω
ωω
/// /
**
**
=
()
()
() ()
()
aa
FNIe e
FNIe
res m
jt jt
res m
jt
2
1
2
30
3
2
+=

t=135° F
res
ωt=45°
ω
t=0
°
Re (a)
1.5NI
m
Fig. 9 - Resultant field in the three-phase machine for sinusoidal supply conditions.
When the machine is synchronous, rotor winding carries DC excitation current and a
field is produced by this current. This field is stationary with respect to rotor. Since the rotor
rotates at synchronous speed, then, looking in from stationary stator, this rotor field rotates at
synchronous speed. This is always the case in any multi-phase AC machine: regardless of
whether the rotor rotates synchronously or asynchronously, all the fields in the machine rotate
at synchronous speed.
Since the resulting m.m.f. is responsible for the resulting flux density and ultimately
resulting flux, this means that apart from the rotating m.m.f., there is a rotating flux density
wave and a rotating flux in the machine as well. The term rotating field in general denotes any
of the three.
Example 4:
Consider a three-phase machine with cylindrical cross section of stator and rotor.
Rotor carries a single winding, supplied with DC current. Mutual inductances between
stator windings and rotor winding and three phase stator currents are given with:
θ
cosML
ar
= )3/2cos(
πθ
−= ML

ωθθωωθ
θωθωθω
θθθ
sin5.1
)sin(5.1
)]sin()120sin()sin(
)240sin()sin()[sin(5.0
)240sin()240cos()120sin()120cos(sincos
rmee
s
smre
sss
sssmre
sssrme
cr
rc
br
rb
ar
rae
IMITt
t
tIMIt
ttt
tttIMIt
tttIMIt
d
dL
Ii
d

−= ML
br
, )3/4cos(
πθ
−= ML
cr
.
Solution:
Since rotor current is constant and since the stator currents are zero, induced emfs in the stator windings
are:
)3/4sin()()3/4sin(
)3/2sin()()3/2sin(
sin)(sin
πωωπθωω
θ
πωωπθωω
θ
ωωθωω
θ
−=−=−=
−=−=−=
==−=
tMIMI
d
dL
Ie
tMIMI
d
dL
Ie

,
and , respectively. The windings are fed from voltage
sources of known voltages vv
v
12
3
,
,
and rotor inertia and friction coefficient are J and
k. The machine runs at certain speed
ω
. Write the time domain matrix equations and
equations in developed form for the following: i) voltage equilibrium and induced
electromotive forces; ii) mechanical equilibrium; iii) power balance; iv) converted
power and electromagnetic torque.
Q2. a) State the complete time-domain mathematical model of a generalised electro-
mechanical converter with n windings and define all the matrices of the model.
b) State the general condition of average torque existence in a two-winding structure
(defineallthesymbolsused).
c) A two-winding system has stator inductance of 0.1 [H], rotor inductance of 0.04 [H]
and mutual inductance of 0.05 cos
ϑ
[H]. If the rotor rotates at 300 rad/s and stator
ENGNG 2024 Electrical Engineering
 E Levi, 2002
21
current is known to be 10 sin 300t, find the induced electromotive force in rotor
winding if it is open-circuited (the initial value of mechanical co-ordinate at zero time
is zero).
Q3. a) State the complete time-domain mathematical model of a generalised

t
12
10 10==cos
s
in
ω
ω
Sketch cross-sectional view of the machine and identify the type of the machine.
Calculate the average torque for load angle equal to 45 degrees and explain its nature.
Sketch the dependence of average torque on load angle
δ
, identify motoring and
generating part and define the region of stable operation.
Q4. An electromechanical converter has two windings only, both mounted on stator and
displaced in space by 90 degrees. The winding self-inductances and mutual inductance
can be given with
.where
2sin
2
2cos
22
2cos
22
12
22
11
δωθ
θ
θ
θ

m
sinω
s
t.
a) Sketch the cross-sectional view of the machine and identify the type of the machine.
b) Derive the expression for instantaneous torque developed in the converter and the
expression for average torque assuming that condition of average torque existence is
satisfied.
c) Calculate the average torque and explain the origin of the torque in this machine if
LLI
dqm
=== =215Aand30H, H,
δ
.
Q5. a) State the complete time-domain mathematical model of a generalised
electromechanical converter with n windings and define all the matrices and vectors of
the model.
b) Give graphical representation of the power flow in an electric machine for motoring
and generation and define all the powers in this representation in terms of terminal
quantities and parameters (use matrix form).
ENGNG 2024 Electrical Engineering
 E Levi, 2002
22
c) State the general condition of average torque existence in a two-winding structure
and classify the electrical machines with respect to the way in which this condition is
satisfied.
d) An electromechanical converter has a three winding structure, with two windings
on stator and one winding on rotor. The two stator windings are displaced in space by
90 degrees. The winding self-inductances and mutual inductances are equal to:
δωθ

Sketch the dependence of average torque on load angle
δ
, identify motoring and
generating part and define the region of stable operation.
Q6. a) State the complete time-domain mathematical model of a generalised electro-
mechanical converter with n windings and define all the matrices and vectors of the
model.
b) A two-winding system has stator inductance of 0.15 [H], rotor inductance of 0.05
[H] and mutual inductance of 0.1 cos
ϑ
[H]. If the rotor rotates at 250 rad/s and stator
current is known to be 15 sin 250t, find the induced electromotive force in rotor
winding if it is open-circuited (the initial value of mechanical co-ordinate at zero time
is zero).
c) The same system of part b) is again considered. However, current 15 sin 250t
flows now through both windings that are connected in series. Find the speeds at
which average torque will exist, determine corresponding average torque values and
the values of load angle, which yield maximum average torque values.