1
Electric Motors and Drives - Third Edition
Solutions to Review Questions - Chapters 1 and 2
Chapter 1
1)
The magnetomotive force or MMF is simply the product of the number of turns and the current,
i.e. MMF = 250  8 = 2000 AT. The ‘AT’ stands for ampere-turns, but strictly the MMF unit
is Amperes.
2)
We are told that the magnetic circuit is made of good-quality magnetic steel, which is a coded
way of saying that the reluctance of the steel part of the magnetic circuit is negligible in
comparison with the air-gap. Under these circumstances all of the MMF provided by the coil
(2000 A) is available across the gap, and the flux density in the air-gap is given by equation 1.7
as
μ NI 4π  10 7  2000
B o

 1.26 T.
g
2 10 3
The question then asks about the flux density in the iron, implying that it is different from that
in the air-gap. But as we have seen in the sketches (e.g. Fig 1.7) the flux density remains the
same unless the cross-sectional area changes, so the answer is that the flux density in the iron is
the same, i.e. 1.26 T.
If the cross-sectional area were doubled, it would make no difference to the flux density
because, as revealed in equation 1.7, the air-gap flux density only depends on the MMF and the
length of the gap. However, with twice the area, and the same flux density, the total flux
would increase by a factor of two.
An alternative way of reaching the same conclusion would be to say that if the cross-sectional
area were doubled, the reluctance of the air-gap would be halved, so for a given MMF the flux
would double.
3)
A flux density of 1.4 Tesla means 1.4 Webers of flux per square metre, so the flux crossing an
area of 18 cm2 will be given by
  1.4  18  10 4  2.52 mWb.
4)
Because no information is given about the iron part of the magnetic circuit, we are expected to
assume that we can ignore its reluctance, and assume that the only significant reluctances are
those due to the air-gaps. This hidden message is reinforced by the use of the word ‘estimate’
rather than calculate: if the (small) reluctance of the iron parts is neglected, our values of flux
density are inevitably going to be slightly on the high side.
Taking the second part of the question first, if we denote the reluctance of the 0.5 mm air-gap
by R, the reluctance of the 1 mm gap will be 2R, because reluctance is directly proportional to
length. The two air-gaps are in series, so the total reluctance is 3R. The flux through both is
the same, and from the Magnetic Ohm’s law is given by
2

MMF
1200 400


.
Reluctance
3R
R
To find the MMF across each air-gap we apply the magnetic Ohm’s law again: so the MMF
across the 0.5 mm gap is given by
MMF  Flux  Reluctance 
400
 R  400 A.
R
Similarly the MMF across the 1 mm gap is given by
400
MMF  Flux  Reluctance 
 2 R  800 A.
R
To find the flux density we can proceed as in question 2 and apply equation 1.7 to either gap.
For the 1 mm gap, this yields
μ NI 4π 10 7  800
B o

 1.0 T.
g
1  10 3
5)
In order to double the flux density, the flux would have to double, and so therefore would the
MMF. Hence the current in the field winding must double, which means that the voltage
applied to the field winding must double also. With twice the voltage and twice the current, the
power increases by a factor of 4, to 100 W.
Two things should be mentioned here. Firstly, we did not need to know the two values of flux
density, but the fact that both are well below the saturation flux density for iron provides us
with justification for ignoring the reluctance of the iron parts. Secondly, the resistance of
copper windings increases with temperature, so we must expect the field resistance to be higher
when the power dissipation is 100 W than when it is 25 W. We cannot say how hot it will get
because we have no information about the cooling system. Again, we take refuge in the fact
that we are asked to estimate, not calculate.
6)
The new rotor diameter is 299.5 mm instead of 300mm, so radius has decreased by 0.25 mm,
meaning that the new air-gap is 2.25 mm, instead of its original 2 mm. To the uninitiated, this
might appear to be of little consequence.
However, the reluctance of the air-gap has increased by 12.5%, so, assuming that we can
neglect the reluctance of the iron parts, the MMF will have to increase by 12.5% in order to
maintain the same flux density.
As we saw in question 5, this means that the power dissipated in the field windings has to
increase by a factor of (1.125)2, i.e. 1.27. Such a large increase in heating will almost certainly
be unacceptable, because unless the original cooling system had been overdesigned (in which
case the field windings were running well-below their allowable temperature rise), the
permissible rise will certainly be exceeded under the new conditions.
So what are the options? In the short run, the only course open is to tolerate the reduced flux
density, which will be 89% of its original value. At rated armature voltage the motor will then
run 12.5% above its rated speed, which can be restored by reducing the armature voltage to
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approximately 89% of its original value. With the same full-load armature current the full-load
torque and power will then be 11% below their original values.
In the longer term it might be possible to add some extra turns to the field winding (if there is
space!), but even then the field voltage will have to be raised to maintain the same field current.
7)
The sentence in brackets is intended to encourage us to pursue our suspicion that to answer the
question, we must first know the maximum allowable flux density. In section 1.3.5 it is
mentioned that flux densities in iron are not usually much above 1.5 T, so this is a reasonable
figure to use to make an estimate. On this basis the cross sectional area is given by
A
 5 10 3

 33  10 4 m 2 or 33 cm 2 .
B
1.5
8)
This is an exercise in applying equation 1.2, i.e. F  B I l. So for (a) the force is
0.8  4  0.25  0.8 N, while for (b) the total force is 20  0.8  2  0.25  8 N. The result for
(b) is ten times as great as that for (a) because the total current in the coil-side is 40 A.
9)
To estimate the torque we first need to calculate the total tangential electromagnetic force, and
then multiply by the radius at which the force acts. We are told that there are 120 conductors,
but that only 75% of the circumference is covered by the poles: this means that only 75% of the
conductors will be exposed to the radial flux density at any instant, i.e. we can assume that 90
of the 120 conductors make a contribution to the torque. We also make the very important
assumption that if the conductors lying under a N pole carry positive current, those under a Spole will be carrying negative current, so that they all contribute to torque.
The electromagnetic force on one conductor is given by F  B I l  0.4  50  0.5  10 N.
Note that we use the mean flux density under the pole-face in this calculation: some conductors
may be exposed to a slightly higher flux density than others, but we do not need the details as
long as we are given the average flux density.
The total tangential force is thus 90  10 = 900 N. We assume that the force acts at the centre
of the conductors, but all we know about the diameter of the conductors is that it must be less
than 1 cm in order to fit in the air-gap. If we take the diameter as 0.8 cm, the radius at which
the electromagnetic force acts will be 15 + 0.4 = 15.4 cm or 0.154 m. The torque is therefore
given by T  Force  radius  900  0.154  139 Nm.
10)
In the machines we have looked at in chapter 1, the radial flux density has been taken to be
constant under the pole-face, and zero elsewhere (see for example Figure 1.11). However, in
question 9 we acknowledged that there is some variation even under a pole-face. But in many
a.c. machines (discussed in later chapters) the radial flux density varies sinusoidally around the
periphery (see for example Figure 5.1). In such cases the meaning of the ‘average flux density’
(used to calculate average force, for example) is the average over a half-sinewave, i.e. over one
‘pole’. The average of a half-sinewave turns out to be 2/π times the peak.
A pedantic person might point out that since there are always an even number of poles, and that
flux is a circuital quantity, the nett flux entering the rotor surface is always balanced by an equal
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quantity of flux leaving the rotor surface. Strictly speaking, therefore, the average flux density
over the rotor surface is indeed zero, but this is of no use when we want to calculate torque!
11)
The rewound (220 V) field winding must produce the same MMF as the original winding did
when it was supplied at 110 V. If we are not sure how to proceed further, we can begin by
arguing that, to do the same job as the old winding, the new one would probably consume the
same power, in which case if the current in the 110 V field was I, the current in the 220 V one
would be I/2. By progressing further with this approach, it should soon become clear whether
or not we are on the right lines.
To produce the same MMF with half the current the number of turns must be 2N, where N is
the original number of turns. Because the current in the 2N turns is only I/2, the cross-sectional
area of the original conductor can be halved, leaving the current density in the wire the same as
it was in the original wire. This gives the new winding twice as many turns, but the crosssectional area of each wire is halved, so the new winding should fit in the same space as the
original one. We should now check that our initial assumption - that the power of the new one
would be the same as the old one – is correct.
Let the resistance of the original winding be R: it was supplied at 110 V, so the current I was
given by I = 110/R and the power consumption was (110)2/R.
The new winding has twice as many turns, so if it were made of the same wire its resistance
would be 2R. But the cross-sectional area of the new wire is only half of the original, so each
turn of the new coil has twice the resistance of a turn of the original wire. The total resistance
of the new winding is therefore 4R. The new winding is supplied at 220 V, so the current is
220/4R = 55/R , i.e. the new current is, as expected, half of the original. The power
consumption is 220  55/R = (110)2/R, the same as the original. So our original belief that to
produce the same MMF, the same power would be required, is seen to be correct.
The new winding provides the same MMF, and contains the same volume of copper, occupies
the same space, and dissipates the same power. We can conclude from this that what really
matters is the amount of copper and how hard we work it (i.e. the current density): the number
of turns and the cross-sectional area of wire can be chosen to suit any desired operating voltage.
12)
The discussion in question 11 related to the field windings, but the same argument can be
applied to all the windings in the machine. It should therefore be clear that there will be very
little external difference, except that the cable for the 110 V version would have to be thicker to
carry just over twice as much current as the 240 V version.
13)
The windings inevitably have resistance, say R. Hence when they carry a steady current (I),
there is a continuous power dissipation of I2 R, and this is equal to the power supplied by the
voltage source. If the field windings were made of superconducting wire having zero
resistance, the steady-state power dissipation would be zero.
The energy supplied during the transient period (when the winding is first switched on, and the
build-up of current is influenced by the inductance of the winding) is divided between that
dissipated in the resistance as heat, and that stored in the magnetic field. Once the field is
established, and the current becomes steady, no further energy is required to maintain the
magnetic field.
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14)
The direction of power flow in all of the electrical machines in this book is inherently
reversible, i.e. they can convert electrical energy into mechanical energy or vice-versa,
generally without any modification. So the answer to the question is: no difference.
15)
Mechanical power is given by the product of torque and speed, so a low-speed motor needs
more torque than a high-speed one to produce the same power. The torque of an electrical
machine (with a given sophistication of cooling system) is broadly determined by the volume of
its rotor, which in turn is closely related to its overall volume. Hence for a given power a motor
that runs say ten times as fast will be ten times smaller than its low-speed equivalent, and thus
will be cheaper to manufacture. The high-speed motor will have to be geared down to provide a
low-speed drive, but this still turns out to be cheaper than a direct-drive motor in most cases.
Chapter 2
1)
(a) Taking the bottom rail as the reference for voltages, the potential of point x oscillates
sinusoidally with an amplitude of 20 V, while the potential of point y remains constant at
+10 V. The diode with the higher anode potential will conduct, thereby reverse-biassing the other
one. The voltage at the load will therefore follow the sinewave while the voltage is greater than
+10 V, and be held at +10 V at all other times, as shown by the thick line in part (a) of the figure
below.
Fig 2A here.
(b) This part probably needs a little more thought.. If we imagine that V2 reduces gradually from
the +10 V that applied in part (a), we can see that the horizontal line in the load voltage waveform
simply moves progressively downwards until it reaches zero. As this happens, the diode D2
conducts for shorter and shorter intervals. When V2 reaches zero we obtain the waveform shown
at (b). We observe that whichever diode is conducting, its current is flowing upwards, and
returning by flowing downwards through the load.
When V2 becomes negative, we might be tempted to think that the output voltage could become
negative for part of each cycle, but this is not possible because the current would then be trying to
flow up through the load and down through a diode. The diodes can only conduct from anode to
cathode (i.e. in the direction of the broad arrow) so we conclude that neither diode will conduct and
the waveform will be as shown in (b) regardless of the value of the negative voltage V2.
(c) Whether the diode is above or below the voltage source makes no difference since they are in
series.
2)
The reference to a motor load means that there is inductance present. If, as is most likely, we can
assume that the current is continuous, the mean d.c. voltage from a fully-controlled single-phase
converter is given by combining equations 2.3 and 2.5 to yield
6
2 2
V cos  .
 rms
(If the current is discontinuous we cannot determine the voltage without knowing the load and the
motor parameters.)
Vdc 
The maximum voltage is obtained when  = 0, giving a value of 207 V. This ignores the voltdrop across the diodes, so in practice the true figure will be nearer to 205 V. The reference to lowimpedance mains signals that we can neglect the volt-drop due to supply system impedance.
3)
The mean output voltage is given by equation 2.6, i.e.
3
Vdc  Vd 0 cos  
2 Vrms cos  .

Substituting Vrms = 415 and Vdc = 300 gives  = 57.6.
The frequency does not affect the formula for the average d.c. voltage, so there would be no
change.
4)
This is the sort of deceptively simple question that can easily trip up the unwary. Whatever
approach is taken, experience shows that it is essential to define terms clearly and proceed
systematically.
We will begin by noting that the circuit is as shown in Figure 2.7, and the output voltage
waveforms for continuous-current operation are shown in Figure 2.9. We should also recall that
devices T1 and T4 conduct for half a cycle, while T2 and T3 conduct for the other half cycle.
Let us focus attention on the waveform of voltage across T1, i.e. the potential difference between
its anode and its cathode. We can see that the anode of T1 is permanently connected to terminal A
of the mains supply, so its anode is always at the potential of terminal A. So we need to see what
happens to the potential of the cathode of T1 in order to sketch the voltage across it.
For the half-cycle when T1 and T4 are conducting (which we conventionally take as
predominantly during the ‘positive’ half-cycle of the mains), the forward volt-drop across T1 when
it is conducting will be very small, so we will assume for the purposes of sketching that we can
take the ‘on’ voltage across T1 as zero for this 180of conduction.
During the other 180conduction period, T2 and T3 are conducting (with negligible volt-drop), so
the cathode of T1 is connected - via T2 - to terminal B of the supply, while its anode remains as
before, i.e. connected to terminal A of the mains. The potential difference between the anode and
cathode of T1 is therefore the potential difference between terminals A and B, which is of course
the mains voltage, VAB. So for the second period the voltage across T1 is the mains voltage. The
complete waveform of the voltage across T1 is therefore as shown in the sketch below.
Fig 2B here.
5)
We are told that the d.c. load current is smooth at 25 A. Referring to Figure 2.7, we recall that for
one period of 180 the load current flows out from T1 and returns through T4, i.e. upwards through
the supply in Figure 2.7, while during the other period of 180 the load current flows out from T2
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and returns through T3, i.e. downwards through the supply. Since the load current is constant at 25
A, the supply current is a 25 A square-wave, as shown in the figure below.
Fig 2C here.
We should note that because we have ignored supply inductance (which causes the ‘overlap’ effect
discussed in this chapter), the current commutates instantaneously, and the current waveform is
rectangular. In practice the current takes a finite time to commutate, and the waveform is
trapezoidal, but still very much non-sinusoidal and far from ideal from the point of view of the
supply authorities, who do not welcome harmonic components of current!
To calculate the peak supply power we can see from the sketch that this occurs when the voltage
reaches its peak, so the peak power is given by
Pmax  Vmax  25  230 2  25  8.13kW.
Perhaps the easiest way to obtain the average power is to take advantage of the fact that we are told
to ignore the losses in the devices. This means that the average a.c. input power is equal to the
mean d.c. power, which is easy to calculate in this instance because the current is constant. When
the current is constant, the mean power is simply the mean d.c. voltage (which we can obtain from
equations 2.3 and 2.5) multiplied by the current.
Hence the mean power is given by Pav  Vdc  I dc 
2 2

230 cos 45  25  3.66 kW.
It is important to note that the mean power can only be obtained by multiplying the mean voltage
by the current if the current is constant. If both the voltage and current vary with time, it is
necessary to integrate the instantaneous power (i.e. the product of instantaneous voltage and
instantaneous current) to obtain the total energy per cycle, and then divide by the period to obtain
the mean power.
An alternative approach to find the mean power directly from the a.c. side exploits the fact that if,
as here, the voltage is sinusoidal, the average power can be obtained by finding the r.m.s. value of
the fundamental-frequency component of the current, I1; then taking the product Vr m s I1cosφ ,
where  is the phase angle between the fundamental components of current and voltage.
The amplitude of the fundamental component of a square wave of 25 A can readily be shown to be
4
31.83
 25  31.83 A , so the r.m.s. of the fundamental component of current is
 22.51 A.

2
Hence the average power is given by 230  22.51  cos 45  3.66 kW, as above.
6)
The average load voltage is 20 V and the source voltage is 100 V, so it follows that the transistor is
‘on’ for one-fifth of the time. The load (motor) current is constant at 10 A, so the waveforms of
source and load are as shown below.
Fig 2D here.
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The average input current is 1 A, so the input power is 100 V  1 A = 100 W. The average load
voltage is 20 V, so the load power is 20 V  5 A = 100 W.
To draw a parallel with a transformer (see chapter 7) we focus on the average values of the input
and output voltages and currents, in which case the chopper appears to function as an ideal stepdown transformer with a turns ratio of 5 to 1: the secondary voltage is one-fifth of the primary
(input) voltage, while the secondary current is five times the primary current. And like an ideal
transformer, the input and output powers are equal.
7)
The circuit and the load voltage and current waveforms are shown in Figure 2.4. The current is
described as being ‘almost constant’, which means that we treat it as constant for the purposes of
calculation.
i) The average load voltage is given by the product of the resistance and the average current, i.e. 8
 5 = 40 V. The voltage waveform itself is rectangular, being 150 V when the transistor is ‘on’
and zero when the transistor is ‘off’ and the current is freewheeling through the diode. If we
denote the on time (the mark) by Ton and the off time (the space) by Toff , the average voltage is
Ton
Ton
T
 40, 
 0.267,  on  0.364.
given by 150 
Ton  Toff
Ton  Toff
Toff
ii) Because the load current is constant, we can obtain the average load power from the product of
the mean load voltage and the current, i.e. 40  5 = 200 W.
iii) We are told to treat all the devices as ideal, so there are no losses and the input power must
equal the output power, i.e. 200 W. As a check, we can easily calculate the average source
power because the input voltage is constant (at 150 V), so the average power is simply the
product of 150 V and the average source current.
The source current is 5 A when the transistor is on, and zero when it is off. So the average
Ton
 5 0.267   1.33 A. Hence the mean source power is
source current is 5 
Ton  Toff
150  1.33 = 200 W, as above.
8)
Taking the possibilities in the order given:
As there is no tendency for current to flow upwards through the switch there is no need to prevent
it - and in any event a MOSFET contains an intrinsic diode that is reverse-biased during normal
operation, but would allow reverse current to flow if necessary.
Inductors do not need protection from high voltages: any ‘high voltages’ present are likely to be
generated by high rates of change of current through inductors anyway.
Voltage supplies are very unlikely to be subject to any restriction on the rate of change of current,
but this answer may have come about as a result of confusion over the need to limit the rate of rise
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of current in some semiconductor devices. Where such restriction applies, an inductor would serve
to limit the rate of rise of current, not a diode.
Limiting the voltage across the MOSFET is the real reason for the diode. There will be a
maximum allowable voltage across the MOSFET (i.e. between Drain and Source), and if this is
exceeded the device will break down. When no current is flowing in the load (Figure Q8), the
voltage across the MOSFET is equal to the supply voltage, so clearly the device must be capable of
withstanding supply voltage. However, the real danger arises because of the load inductance, as
discussed below.
di
, i.e.
dt
the voltage is determined by the rate of change of current, or vice-versa. So if we attempt to open a
switch in a circuit containing an inductor through which current is flowing, thereby forcing a very
di
rapid rate of change of current (i.e.
tends to infinity) a very high voltage will be developed
dt
across the terminals.
The voltage across, and the current through, an inductor are related by the equation v  L
For example suppose that there was no diode in Figure Q8, and, with the MOSFET switched on, a
steady current of 2 A was flowing through an inductance of 50 mH. If we then switched off the
MOSFET such that the current was reduced to zero at a uniform rate in 1 sec, the voltage across
the inductor would be 100 kV! We know from Kirchoff’s voltage law that the sum of the voltages
around a circuit must equal the supply voltage, so it follows that almost all of the 100 kV would
appear across the MOSFET, leading to its immediate destruction.
1 2
L i . If we
2
attempt to reduce the current to zero instantaneously, we are trying to destroy energy in zero time,
which corresponds to infinite power, which is clearly impossible. The alternative is to provide a
mechanism whereby the stored energy in the inductor can be released in a more measured fashion,
and this is where the ‘freewheel’ diode comes in. When the switch opens, the diode provides an
alternative path for the current that is flowing down through the load, so the current can continue
by flowing in the closed path shown at (b) in Figure 2.4. At first the current is undiminished, so
the stored energy is unchanged, but because heat is dissipated in the resistance of the load, the
current decays exponentially as the energy stored reduces. The term ‘freewheeling’ arises by
analogy with riding a bicycle, where, having built-up kinetic energy by hard footwork, we can rest
the pedals and let the stored energy sustain our motion until friction brings us to rest.
The real problem lies in the energy stored in the inductance, which is given by E 
The last answer suggesting that the diode is there to dissipate stored energy is partially correct
in that some energy is indeed dissipated in a real diode (though not in an ideal one). But most
of the stored energy will be dissipated in the load resistance, and the resistance of the inductor
itself.
9)
We should identify the right answer (100.7 V) and explain why before speculating on the
origins of the remainder.
As explained in the answer to question 8, the freewheel diode conducts when the MOSFET
switches off, the load current flowing upwards through the diode. To find the voltage across the
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MOSFET in this condition we need to find the potential of the anode of the diode with respect
to the ground reference (i.e. the bottom of the supply). We know that the forward volt-drop
across the diode is 0.7 V, so the potential of the anode is 0.7 V higher than the potential of the
cathode. The cathode is connected to the top of the supply, so its potential with respect to
ground is 100 V. Hence the potential difference (voltage) across the MOSFET is 100.7 V. This
is therefore the maximum voltage that the MOSFET has to withstand.
The figure of 99.3 V clearly comes about because of confusion over the sign of the voltage
across the diode. The suggestion ‘zero’ presumably arises because it is believed that since there
is no current through the MOSFET during freewheeling, its voltage must also be zero. The
voltage across the load (resistance and inductance) is 0.7 V during freewheeling, because the
load is (almost) short-circuited by the diode. And the suggestion that the voltage depends on
the inductance is understandable but wrong because while the value of inductance determines
the stored energy and therefore the duration of freewheeling, the voltage across the MOSFET
will be 100.7 V regardless of the inductance.
- End of Solutions for Chapters 1 and 2 -