1
Electric Motors and Drives - Third Edition
Solutions to Review Questions - Chapter 3
The equivalent circuit shown here should be in our minds when we tackle any d.c. machine
questions:Fig 3A here
In the majority of cases we will be considering steady-state operation, so the current will be
constant and therefore we can ignore the armature inductance in our calculations.
Unless told to the contrary, we will assume that the volt-drop across the brushes can be
ignored.
1)
As we are not told otherwise, we are expected to assume that the question refers to the steadystate running speed, in which case the answer is that the speed is determined by the armature
voltage. Justification, if required, would be along the lines below.
Whenever the speed is steady, the motor torque must be equal and opposite to the load torque.
Except for very tiny d.c. motors it is safe to assume that when the motor is unloaded, the
friction torque is very small, so the motor torque would also be very small. Motor torque is
proportional to armature current, so we can expect the armature current of an unloaded motor to
be very small, hence the term IR in the armature voltage equation V  E  IR is negligible and
we can say that at no-load V  E. E is the motional e.m.f. induced in the armature, and is
directly proportional to the angular velocity (speed), i.e. E  k  , hence the speed is given by
V
  , i.e. the speed is determined by the applied voltage.
k
Note that in the majority of d.c. motors the term IR will be small compared with the armature
voltage V even when the motor is on load and the current I is not small, so to a first
approximation we can say that the on-load speed will also be determined by the applied voltage,
the speed when loaded only being slightly less that that of the unloaded motor.
2)
As discussed in the answer to question 1, the steady running current must be such as to produce
a torque equal and opposite to the load torque, so in the steady state it is the load torque that
determines the armature current.
3)
The answer to question 2 indicates that the steady running current is always determined by the
load torque. When no ‘real’ load torque is applied, we are left with friction, due to bearings,
fan, and (especially in a d.c. machine) brush friction. The friction torque is therefore reflected
in the no-load current.
2
4)
The answers are that the drop in speed from no-load depends directly on the load torque and
the armature resistance.
First, let us consider the effect of load torque. For any given load, the speed will settle when the
motor torque Tm equals the load torque TL . Motor torque is proportional to armature current,
T
i.e. Tm  k I , hence the steady current is given by I  L , i.e. the steady armature current is
k
proportional to the load torque. Combining the armature voltage equation V  E  IR and the
e.m.f. equation E  k  , and substituting for I from above gives the speed as
V  R
    2 TL .
k k 
This equation shows that the no-load speed (i.e. when TL  0 ) is given by 0 
V
, and the
k
 R
drop in speed that is attributable to load is given by  2 TL . The drop in speed is therefore
k 
directly proportional to the load torque and to the armature resistance.
We note also that the drop in speed for a given load is inversely proportional to the square of
the motor constant. So if we were to reduce the field current so that, say, the flux was halved to
double the no-load speed, we would find that because k had been halved, the drop in speed for a
given load torque would be four times as great as with full flux. This matter was discussed in
chapter 3 and illustrated in Figure 3.12.
What the question means when it refers to ‘little ones slow down more than large ones’ really
means that the percentage drop in speed between no-load and full-load is usually higher in
small motors than in large ones. The reason is simply that in small machines the term IR
represents a higher fraction of the applied voltage than it does in large machines.
Alternatively we could say that the reason is that ‘the per-unit resistance is higher in small
V
machines’, meaning that the ratio ( Armature resistance  rated ) is larger in small machines
Irated
than in large ones.
5)
To reverse the direction of rotation we must reverse the direction of current in the armature or
the direction of current in the field. In a separately-excited motor it is usually easiest to reverse
the connections to the field, because the field current is less and the wires are thinner. In a
series motor, the field and armature carry the same current, so either can be reversed.
6)
If the motor is producing more than its continuously-rated torque its armature current will be
above the continuously-rated value and therefore it will overheat. If the cooling of an existing
motor is improved it should be possible to increase the continuous rating without overheating,
but other problems due to commutation and brush wear must be anticipated.
3
7)
There is no basic difference, the same machine is capable of both motoring and generating.
8)
An observer watching the ammeter might conclude that the reason for the increase in current
when load was applied was that the resistance of the motor armature circuit had reduced.
V  E V  k

, where the motional e.m.f. E
R
R
is proportional to the speed, . At no-load the motional e.m.f. is almost equal to the applied
voltage so very little current flows into the motor. When load is applied the speed reduces,
thereby reducing E and thus allowing more current to flow into the motor to produce the extra
torque required to balance the load. So it is the change in motional e.m.f that alters the current:
the actual armature resistance does not change.
In fact we know that the current is given by I 
9)
We can consider the no-load condition, when the motional e.m.f. E is very nearly equal to the
applied voltage V. If we reduce the flux that is cut by the armature conductors, they will have
to cut through the weakened flux faster to achieve the same e.m.f., so the weaker the field, the
higher the no-load speed.
Alternatively we can use the result from the solution to question 4, i.e. that the no-load speed is
V
given by 0  , where k is the e.m.f. constant, which is proportional to the field flux. If the
k
flux is reduced, so is k, leading to a higher no-load speed.
10)
E  V  IR  220  15(0.8)  208 Volts.
We are told that the field current is ‘suddenly’ reduced by 10%, and that the flux is proportional
to the field current. (We know that the current in an inductance (the field circuit) cannot
change instantaneously, so we suppose that what the question means is ‘very quickly, compared
with any subsequent changes that may be initiated by the reduction in flux’.)
A reduction of flux by 10% will cause the motional e.m.f. to reduce by 10%, so the new e.m.f.
220  187.2
 41 A. Note that
is 0.9  208 = 187.2 V. So the new current will be given by I 
0.8
a modest reduction of only 10% in the flux causes a dramatic increase in the armature current,
which jumps from 15 A to 41 A.
The increased current will lead to more torque, but not quite in proportion to the increase in
current because there has been a reduction in the flux. In most of the calculations in the book
the flux has remained constant, in which case the torque is proportional to the current. But the
torque depends on the product of the flux and the current, so if we denote the original flux by
0.9  41
 2.46. The surge of torque will
 , the ‘new’ and ‘original’ torques are in the ratio
  15
lead to a rapid acceleration to the new (higher) steady speed.
11)
The equivalent circuit is shown in Figure 3B.
Fig 3B here.
4
The steady field current is given by If = 110V/110  = 1 A, so the armature current Ia = 25 + 1
= 26 A.
The generated e.m.f. is given by E – IaRa = V, i.e. E = 110 + 26(0.5) = 123 Volts.
Electrical output power
.
Mechanical input power
The useful power (supplied to the battery) is 110 V  25 A = 2750 W.
For a generator, efficiency  is defined by  
The power converted from mechanical to electrical is EIa = 123  26 = 3198 W. To this must
be added the 220 W of friction and other losses to yield a total input power from the diesel
2750
 100%  80.5%.
engine of 3198 + 220 = 3418 W. Hence the efficiency is
3418
It ought to be pointed out that in this example the terminal voltage is fixed by virtue of the
connection to a battery, which we treat as an ideal voltage source. The machine generates 123
Volts and 26 A, 1 A of which is required to provide the excitation flux without which the e.m.f.
of 123 V would not exist. The machine is therefore self-excited, so it doesn’t need to be
connected to a battery to act as a generator.
But what if we simply drove the machine from the diesel engine, without first connecting it to a
d.c. source? When it is stationary, there is no motional e.m.f., so there is nothing to produce a
current in the field winding and no flux. So no matter how fast the rotor turns there will be no
generated voltage. There appears to be a paradox here: we can see from the discussion above
that it can self-excite once it is going, but it appears that it will not get going on its own.
In reality, shunt connected generators can self-excite, provided that there is sufficient residual
magnetic flux (i.e. remaining from the last time the machine was operating) to provide the
initial ‘kick-start’ to begin the process of self-excitation. Assuming that he process begins by
residual action, it is easy to see that because the field current and flux increase as the e.m.f.
builds up, and the e.m.f. is proportional to the speed, there is a strong positive-feedback effect
taking place. So then we might ask what limits the build-up of voltage, which on the face of it
might be expected to go on increasing indefinitely: will there be a stable (i.e. settled) value, and
if so what will determine it?
The answer is that a stable operating position is reached because of the non-linear relationship
between flux and field current caused by progressive saturation of the iron in the magnetic
circuit. A typical relationship between field current and generated e.m.f. (at constant speed) is
shown in Figure 3C, where the initial e.m.f. due to residual magnetism has been exaggerated for
clarity. Because e.m.f. is directly proportional to speed, this graph also represents the field
current vs. flux relationship, and it clearly shows the onset of saturation at high levels of field
current.
Fig 3C here
5
The straight-line graph relating field current and e.m.f. (i.e. I f 
E
) is also shown. The
R f  Ra
stable value of E is at the point where the two graphs intersect, and is shown by the dot.
12)
When speed is steady, the motor torque must be equal and opposite to the load (friction) torque.
In this case the friction torque is equal to ‘2A worth’ of motor torque – so to quantify the torque
we need the motor constant, k.
We recall that the motor constant not only links torque and current via Tm  k I , but also
motional e.m.f. and speed via E  k . We can use the second expression to find k, after first
finding the e.m.f. at 1800 rev/min.
The armature voltage equation yields E = 250 – 2(1) = 248 Volts, and the corresponding
2
 188.5 rad/sec. Hence
angular speed is given by 1800 
60
248
Volts
Nm
k 
 1.316
 1.316
. Hence the torque is 2 A  1.316 Nm/A = 2.63 Nm.
188.5
rad/sec
A
13)
The answer to this question depends on what happens to the excitation (flux) after the motor is
disconnected. Clearly there can be no immediate change in the speed so, if it was a permanentmagnet motor or a separately-excited one in which the flux was fully maintained after the
armature was disconnected, the motional e.m.f. would be unchanged, i.e. 248 Volts. The
armature current is zero, so there is no volt-drop across the armature resistance and the terminal
voltage will be 248 V immediately after the disconnection. There is no armature current, and
therefore no torque apart from friction, so the speed will reduce gradually and the terminal
voltage will decline at the same rate.
If the motor was series-connected the flux would be zero after disconnection so there would be
no e.m.f. and therefore no voltage at the terminals.
Things are more complicated for a shunt motor, and not recommended for the faint-hearted.
The discussion at the end of question 11 is relevant to the rest of this one.
With a shunt-connected motor, the field flux will be sustained because the motional e.m.f.
(which was previously opposing the applied voltage) will now act across the series combination
of the armature resistance Ra and the field resistance Rf . The armature current will now equal
the field current, and it will flow in the opposite direction through the armature. The voltage at
the terminals will therefore be the voltage across the field winding, which, ignoring the field
Rf
inductance, will be given by potential divider action as
 E. The negative armature
R f  Ra
current (generator action) represents ‘dynamic braking’, so the speed (and e.m.f./terminal
voltage) will therefore reduce more rapidly than if friction was the only source of torque.
14)
When I is 32 A, the em.f. is given by E  250  32(1)  218 Volts. From the answer to
question 12 we know that the motor constant is 1.316 Volts/rad/sec, so the speed at full load is
6
218/1.316 = 165.65 rad/sec or 1582 rev/min.
We are not told the rated power, but because the rated current is 32 A, the torque produced by
the motor at full-load is 32  1.316 = 42.1 Nm. We know from the answer to question 12 that
the no-load friction torque is 2.6 Nm, so if we assume that this also applies when loaded, and
ignore any extra load losses, the rated torque is given by 42.1 - 2.6 = 39.5 Nm.
15)
e.m.f
110

 0.70 Volts/rad/sec = 0.70 Nm/A.
speed 1500  2
60
when I = 10 A, Torque = 7 Nm.
a) k 
b) The gravitational force on the mass is given by F  mg  5  9.81  49.05N. Hence the
torque exerted at the motor shaft is 49.05  0.8 = 39.24 Nm.
The motor must exert an equal and opposite torque to achieve equilibrium, so the motor
current is given by 39.24/0.70 = 56.06 A.
The stability question can be addressed by considering that, with the arm horizontal and zero
nett torque, we make a small change to one of the parameters and see if the system takes up a
new equilibrium. If we slightly reduced the current in the motor, the load torque would then
exceed the motor torque and the weight would move downwards. But as it did so the torque it
exerts reduces because the line of action of the force moves closer to the axis of the motor. So
when it has moved down to the point where the load torque again equals the motor torque, it
will find a stable equilibrium.
However, if we slightly increase the current, the motor torque will be greater than the load
torque and the weight will begin to move upwards. In so doing its line of force moves closer to
the axis and the torque it exerts gets less. The amount by which the motor torque exceeds the
load torque therefore increases with movement, and we have an unstable equilibrium.
So there isn’t a simple answer to the question ‘is it stable’, because the stability depends on how
the equilibrium is disturbed.
c) This is another straightforward exercise using the armature voltage equation. First we need
2 

to find the bck e.m.f. which is given by E  k   0.70  1430 
  104.8 Volts. Then
60 

apply the armature voltage equation V  E  IR to obtain IR  110  104.8  5.2 Volts.
Hence since I = 25 A, R = 0.2 .
To drive a current of 56 A through 0.2  when the motor is at rest (i.e. E = 0) requires a voltage
of 56  0.2 = 11.5 Volts.
d) The machine is now acting as a generator, supplying power to a system at 110 V. The
generated e.m.f. E is greater than the system voltage by IR.
7
If the power supplied to the system is 3,500 W at 110 V, the current is 3500/110 = 31.82 A.
Hence the generated e.m.f. is given by E  110  31.82(0.2)  116.4 Volts. The speed is given
E 116.4
 166.28 rad / sec  1588 rev/min.
by   
k
0.70
The corresponding torque is given by 31.82  0.70 = 22.27 Nm.
The electromechanical power is EI = 116.4  31.82 = 3704 W, to which we must add 200 W to
find the mechanical input power, and 100 W for the input power to the field, making a total of
4004 W. The useful output power, supplied to the system, is 3500W, so the efficiency is
(3500/4004) 100%, i.e. 87.4%.
16)
The issue of maximum possible output power from a d.c.motor is discussed in section 3.4.6
where it is explained that the maximum mechanical output power is obtained when the back
V E
e.m.f. is half of the applied voltage, i.e. E = V/2. The armature current is given by I 
,
R
V2
so when E = V/2, the current is V/2R, and the mechanical power (EI) is given by Pmax 
.
4R
(The same result can be obtained by applying the maximum power transfer theorem of circuit
theory.)
For the toy motor the maximum possible power output is therefore 122/32 = 4.5 W. If this
power were to be developed at the speed of 5000 rev/min, the torque would be given by
Power
4 .5

 8.6 mNm .
Speed 5000  2
60
As this is less than half of the torque claimed by the manufacturer, the claim is clearly not
justified.
T 
17)
Another straightforward exercise in the basic equations, but with parameters of a much smaller
motor than hitherto.
The e.m.f. at no load is given by E  V  IR  6  0.070(2.9)  5.8 Volts, and this e.m.f is
developed at 8000 rev/min or 837.76 rad/sec. The motor constant is therefore given by
E
5.8
k 

 6.92 10 3 Volts/rad/ sec  6.92 10 3 Nm/A .
 837.76
At no-load, the motor torque is equal to the friction torque, so the friction torque is
6.92  10 3  0.070  0.484 mNm .
When the stationary motor is switched directly onto a 6 V supply, the current will – assuming
that we can ignore the armature inductance - be given by I = 6/2.9 = 2.07 A. Hence the starting
torque is given by T  k I  6.92  10 3  2.07  0.0143 Nm.
8
Acceleration is given by Accelertio n 
18)
Torque
0.0143

 1.04  105 rad/sec 2 .
6
Inertia 0.138  10
In a linear system work is force times distance: in a rotary system force is replaced by torque
and linear distance becomes rotary distance, i.e. angle. So in a rotary system, work is torque
times angle.
Mechanical power is the rate of doing work, i.e. work/time. So in a rotary system mechanical
power is torque times angle over time. But, assuming that speed is constant, angle over time is
angular velocity, and power is thus given by
Power  Torque  Angular velocity, i.e. P  T .
We have the equations T  k I and E  k . Hence P  T   k I 
19)
E
 EI .
k
The 24 V version clearly has to produce an output power of 30 W at 5000 rev/min, and if it is to
be ‘equivalent’ it must have the same input power. It would have the same dimensions and
magnetic circuit, but clearly the stator would have to be rewound.
The current in the 24 V motor will be half of that in the 12V version, for the same power. So if
the 24 V version has twice as many turns in each coil on the rotor, it will produce the same
torque. However the cross-sectional area of the new conductor can be half that of the 12 V
motor because it only has to carry half the current. So new coils with twice as many turns but
half the cross-section will fit in the original slots. The motional e.m.f. will be doubled, but the
current is halved so the mechanical power is the same, as required.
The new armature has twice as many turns of half the cross-sectional area, so the resistance of
the new armature is 4 times that of the original, i.e. 4 0.8 = 3.2 .
As a check, we can calculate the armature copper loss for both machines to ensure that they are
the same. If the rated current of the 12 V motor is I, the copper loss is I 2  0.8  0.8 I 2 . The
2
I
copper loss in the 24 V motor is    3.2  0.8 I 2 .
2
So the parameters of the new motor are 24 V, 30 W, 5000 rev/min, and armature resistance of
3.2 .
20)
a) When the motor is at rest the back e.m.f. is zero so if rated voltage (V) is applied the current
will be V/Ra, where Ra is the armature resistance. In large d.c. motors the current V/Ra is very
much greater than the rated current. The motor will almost certainly be supplied from a
thyristor converter, in which the thyristors would not be able to withstand such a large current.
So the control scheme would automatically limit the voltage applied to the motor in order to
restrict the current to an acceptable level.
9
b) The torque required to maintain a steady speed when a motor is unloaded is very small. The
torque produced by the motor is proportional to the current, so the no-load current is very small.
V E
The current is given by I 
, where V is the applied voltage, R is the armature resistance
R
and E is the motional or back e.m.f induced in the motor. As explained above, the no-load
current is very small, which indicates that the back e.m.f. E is almost equal to the applied
voltage. The motional e.m.f. is proportional to the speed, so the no-load speed is almost
proportional to the applied voltage.
c) When the motor is running at a steady speed, the torque it produces is equal to the load
torque. When the load torque increases the previous state of equilibrium is disturbed because
the load torque now exceeds the motor torque, so the nett torque is negative and the sytem
decelerates. The motional e.m.f. (E) is proportional to speed, so E reduces.
V E
, where V is the applied voltage, R is the armature
R
resistance and E is the back e.m.f. As E reduces, the current increase, and so does the motor
torque. The nett decelerating torque then reduces, so the deceleration is reduced but will
continue until the speed has fallen to the point where the motor torque equals the load torque, at
which point equilibrium will be restored, but at a new (lower) speed. The smaller the armature
resistance, the less the speed has to drop in order for the current to reach the new load level.
The armature current is given by I 
d) The voltage equation for a field winding is v  ri  L
di
, so the instantaneous power is
dt
di
. The first term in the power equation represents the loss of heat
dt
due to the field winding resistance, while the second term represents the rate of change of stored
energy in the magnetic field.
given by vi  i 2 R f  L f i
di
is zero, indicating that once the
dt
2
Rf )
magnetic field has been established the energy stored remains constant. The first term ( I dc
represents the heat loss per second due to resistance (‘copper loss’) and this has to be supplied
continuously, even though none of it appears as mechanical output power. If the resistance
could be made zero (e.g. with a superconducting winding) the power input would be zero once
the field current had been established.
Under d.c. conditions the second term is zero because
e) If the supply to the field is pure d.c., then apart from the very short periods when the field
flux is changing, the flux in the magnetic circuit is constant, so there is no danger of induced
eddy currents in the body of the pole and therefore no need for it to be laminated.
When the supply is form a thryristor converter, however, there will be an additional alternating
component of flux in the poles, which must therefore be laminated to minimise eddy-current
losses.
21)
The immediate effect of reducing the field current is that the flux reduces in more-or-less the
same ratio, thereby causing a fall in the motional e.m.f, E, which is proportional to the air-gap
10
V E
, where V is the applied
R
voltage and R is the armature resistance. R is small, so a modest reduction in E causes a
disproportionately large increase in armature current.
flux and the speed. The armature current is given by I 
The torque produced by the motor depends on the product of the armature current and the airgap flux. The former has increased a lot, while the latter has decreased a little, so there is a
large increase in motor torque, leading to acceleration. As the speed rises, so does the back
e.m.f. The current therefore decreases as the speed rises until the torque produced by the motor
is equal to the friction torque, at which point the speed becomes steady.
Sketches showing the variation in current and speed are shown in Figure 3D.
Fig 3D here.
22)
By making the field flux proportional to the armature current (at least up to rated current), the
separately-excited motor becomes like a series-connected motor. The torque-speed curves at
various values of armature voltage are therefore as shown in Figure 3.15b.
The fact that the separately-excited motor can be made to emulate either a shunt or a series
motor by appropriate independent control of the field current underlines the flexibility made
possible by the availability of power-electronic supplies.
23)
This is another example of the maximum power transfer theorem, and as explained in the
V2
solution to question 16 the maximum possible mechanical power is given by Pmax 
.
4R
Hence the maximum power from a 12 V motor with an armature resistance of 1  is 144/4(1)
= 36 W.
The maximum power condition occurs when the motional or back e.m.f. is equal to half of the
V E
V

applied voltage, the corresponding current being given by
. The armature
R
2R
V
resistance of large motors is very small, so at rated voltage a current of
would be very high,
2R
and well beyond the capability of the converter supplying the motor.
24)
In a universal motor the field and armature are connected in series, so the torque is positive
regardless of the direction of current. If the motor is supplied at, say, 50 Hz, the current
reverses 50 times per second but is always in the same direction. The fact that the torque
pulsates 100 times per second is not usually a problem because the rotor inertia acts as a filter
and the resulting speed pulsations are very small.
25)
At the fundamental level it is true that in principle any electrical machine with rated voltage V
and rated current I could be rewound to operate at voltage kV and rated current I/k , and that the
rewound motor would contain the same amounts of active materials (copper and iron) and have
the same performance (in particular the same power (VI) as the original.
11
However, in the case of the low-voltage d.c. motor, there are several additional factors which
complicate matters.
The first relates to the size of the commutator. For a given power, the current is inversely
proportional to the voltage, so a low voltage motor obviously has a higher current than a highvoltage one. The area of brush in contact with the commutator is determined by the current it
has to carry, so the lower the voltage, the bigger the brushgear/ commutator. In handtools space
and weight are at a premium so the high-voltage motor is at an advantage.
The second matter stems form the fact that the voltage/current characteristic of the carbon
brushes is non-linear, so that under normal operation the volt-drop across the brushes contains a
more-or-less fixed component that is of the order of 1 volt, regardless of current. In a 110 V
motor the loss of one volt is not serious: but in battery-powered tools the supply voltage is only
a few volts, in which case the loss of one volt is serious, but becomes less so the higher the
supply voltage. It is therefore desirable to avoid low voltages from the point of view of
efficient use of energy.
The third factor relates to the properties of the semiconductor switches used in the chopper
drive that provides speed control. The on-state volt-drop in transistors and diodes is (rather like
the brush-drop referred to above) largely independent of the current, so that the on-state power
loss is more-or-less proportional to the current. So when efficiency is important it is preferable
to handle a given power at a high voltage and low current, rather than at a low voltage and high
current.
Taken together these factors indicate that for a given output power the designer should aim to
minimise the current, so that the higher the power, the higher the voltage.
26)
A shunt motor will indeed run in the same direction regardless of the polarity of its d.c. supply.
But applying a.c. at mains frequency (50 Hz or 60 Hz) would not be satisfactory, mainly
because of the high inductive reactance of the field winding. The impedance of the field
winding at frequency  consists of its resistance ( R f ) and its inductive reactance (  L f ). The
reactance will usually be much higher than the resistance, so for the same applied voltage, the
field current will be much less that under d.c. conditions, and it will be out of phase, lagging the
voltage by approximately 90°. The flux will therefore be very small, and will be zero at the
instant when the voltage is maximum.
In contrast the armature inductance is much smaller so the armature current will be more nearly
in phase with the applied voltage. However this means that the peak armature current during
each cycle occurs when the flux is zero, so the resultant torque is very small. All in all, not a
good idea!
27)
a) We can find the machine constant from the data given in the first paragraph. When the
machine is on open circuit there is no volt-drop across the armature resistances and the terminal
voltage is therefore the same as the induced e.m.f. Hence using the relationship E  k  ,
220
k 
 1.40 Volts/rad/ sec  1.40 Nm/A.
2
1500 
60
12
The question is all about steady-state conditions, so we must expect to make use of the fact that
if a linear (or rotary) system is not accelerating, the resultant force (or torque) must be zero. We
can make us e of this knowledge to find the tension in the rope (Fig 3Q.27a), which we need to
know in order to work out the torque exerted by the load.
The two forces acting on the mass m are the gravitational force (downwards), equal to mg, and
the tension in the rope (F) upwards. Since the descent velocity is to be constant, the nett force
must be zero, i.e. F = mg = 14.27  9.81 = 140 N.
At the drum, this (downwards) tension acts at a radius of 10 cm, so the torque exerted by the
load is 140  0.1 = 14 Nm. We are not told anything about friction torque so all we can do is to
assume it is negligible, so the total load torque is 14 Nm.
The linear speed of the rope at the drum is given as 15 m/sec, the circumference of the drum is
0.2π, and the speed of rotation of the drum and machine is therefore 15/0.2π rev/sec or 150
rad/sec.
Because the speed is steady there is no acceleration, and the machine torque must be equal and
opposite to the load torque, i.e. the machine torque must be 14 Nm at a speed of 150 rad/sec.
We keep referring to the ‘machine’ rather than the ‘motor’ because in this application we are
using the machine to restrain the descending load, not to drive it down. We need the machine
torque to act in the opposite direction to the load torque, which it will do automatically if we
complete the armature circuit with a resistance (as in Fig 3Q.27b), thereby allowing the
generated e.m.f. to drive a current in the same direction as the e.m.f. (In contrast, if for some
reason we wanted to operate as a motor rotating in the same direction, i.e. to drive down the
load, we would apply a voltage greater than the e.m.f. and force the current to flow in the
opposite direction to the e.m.f., yielding a driving torque rather than a braking torque.)
Now that we know the speed is 150 rad/sec we can calculate the generated e.m.f. as
E  k   1.4  150  210 Volts . The machine torque has to be 14 Nm, so, using Tm  k I , the
armature current must be 10 A.
The total resistance in the armature circuit is therefore given by 210/10 = 21 . The armature
resistance itself is 0.5 , so the external resistance required is 20.5 .
b) The power dissipated in the external resistor is given by I 2 R  100  20.5  2050 W , while
the power dissipated in the machine armature is 100  0.5 = 50 W.
The electrical generated power is provided from its mechanical input power, which is derived
from the steady reduction in potential energy of the lowering mass. We can check the power by
considering the power (force times speed) of the falling mass, i.e.
Pmech  Force  speed  140  15  2100 W.
In this question we have ignored all but the armature copper loss in order to simplify the
calculations, but nevertheless the situation is representative of many real-life applications, such
13
as dynamic braking of railway vehicles where kinetic energy is dumped in large resitors, often
mounted on the roof to assist cooling.
28)
The running data gives us enough to work out the machine constant, as follows.
E  V  IR  200  10(1)  190 Volts, at 3000 rev/min. Hence using
E
190
k , k
 0.605 Volt/rad/s ec  0.605 Nm/A .
2

3000 
60
When the motor is running unloaded the current is negligible and the back e.m.f. is (almost)
equal to the applied voltage. So at 100 V, the speed of the unloaded motor is 100/0.605 = 165.3
rad/sec or 1578 rev/min.
After disconnection, the armature of the spinning machine is simply shorted through a resistor
(as in Fig 3Q.27b), the total circuit resistance being 5 . In the armature voltage equation
E
V  E  IR this condition corresponds to V = 0, so the current is given by I  
, where
RT
RT is the total resistance (5 ). The negative sign indicates that the current is in the opposite
direction to the normal motoring current, so the torque will be negative and will result in
deceleration. As the speed falls the stored rotary kinetic energy is converted to electrical form
and then dissipated as heat in the 5  resistance.
To find out how the speed varies, we need to make use of Newton’s law for a rotary system (i.e.
the relationship between the torque(s), the inertia and the angular acceleration), which is:d
Tnett  Tm  Tload  J
,
dt
d
where J is the inertia and
is the rate of change of angular velocity, ie. the angular
dt
acceleration. In this exercise we are told to ignore friction so the load torque is zero, and the
only torque is that produced by the machine, i.e. Tm.
 E 
k
k2
   k
The machine torque is given by Tm  k I  k  

.
R
R
R
T
T
 T
2
k
d
dω  k 2 
   0.
J
, or
 
The equation of motion thus becomes 

RT 
dt
dt
 JRT 
This is the first order differential equation governing the angular speed, . Equations of this
sort are very common in science and engineering. We can judge broadly how the speed varies
d
 0 ) is given by  = 0,
with time by first noting that the steady-state value of  (i.e. when
dt
i.e. the obvious result that the machine finally comes to rest. Secondly, we note that since the
two terms in the equation must always add to zero, and the second term is positive, we can
d
deduce that the deceleration or rate of change of speed (
) is always negative (hardly
dt
surprising!); but more importantly, the deceleration reduces in proportion to the speed. So
14
immediately after the disconnection, when the speed is high, the machine will decelerate
rapidly, but the slower it runs the more gradually it slows down. (See Figure 3E.)
The general solution of the governing differential equation is the form
t
JRT
.
k2
We know that at t = 0, the angular speed is 165.3 rad/sec, so o  165.3 . Substituting for J, R,
and k we find that the time-constant  = 2.73 seconds. (The significance of the time-constant is
shown in Figure 3E; if the initial gradient were to be continued, the final state would be reached
in one time-constant.)
  o e  , where τ 
The speed-time plot is sketched in Figure 3E, with the speed in rev/min rather than rad/sec. To
find the time taken to reach 100 rev/min we solve the equation 100 
t  7.53 seconds.
t
2
1578 e .73 ,
which yields
Fig 3E here.
29)
This is probably the most challenging question, although there is nothing really new in it.
The initial set-up has both motors running light at their rated voltage, the modest power being
supplied by the 520 V d.c. source.
Each motor draws 18.5 A, so we can calculate the back e.m.f. as
E  V  IR  520  18.5(0.05)  519.075 Volts . We will find that because of the delicate
balance between V and E, it is important to retain several places of decimals in some of the
subsequent calculations.
The armature copper loss is given by I o2 R  18.52  0.05  17 W.
The electromechanically-converted power is used to overcome windage and friction, and is
given by Pmech  EI o  519.118.5  9.603 kW.
The no-load speed is given as 1040.2 rev/min, and we know that the mechanical power is 9.603
9603
 88.2 Nm, for each machine .
kW. Hence the no-load torque is given by To 
2
1040.2 
60
We can also use the no-load data to calculate the motor constant from
E
519.075
k 

 4.7652 V/rad/sec  4.7652 Nm/A .
 1040.2  2
60
For the sake of future reference we will now look at the figures under full-load conditions, even
though the question does not ask specifically for this.
15
At full-load, the speed is given as 1000 rev/min. The full-load e.m.f. is therefore given by
2
E  4.7652  1000 
 499.0 Volts.
60
520  499
 420 A. This confirms the full-load
0.05
current given in the data. The power converted into mechanical form is given by
E f l I f l  499  420  209,580 W. If we assume that the windage and friction are the same
as at no-load (9.6 kW) we see that the useful output power is 200 kW, as per the original data.
The armature current is then given by I f l 
Now we move to the real test set-up, where the flux in machine number 1 is reduced, thereby
reducing its e.m.f., until its armature current is at its full-load value, i.e. 420 A. We are told to
assume that the voltage source is ideal, so regardless of what happens inside the two machines,
their terminal voltage will be held at rated value, i.e. 520 V. The aim is that the motor torque
provided by machine 1 will supply mechanical power to machine 2, which will generate and
supply most of the power needed by machine 1.
If motor 1 was not mechanically coupled, we would expect the speed to rise when we reduced
the field flux. But when machine 2 is generating it applies a negative (load) torque to machine
1, and therefore tends to prevent the speed rising. So it is not at all obvious at what speed things
will settle, and we will have to set up and solve the system equations in order to find out.
One thing we do know is that machine 1 has its rated armature voltage (520 V) and its rated
armature current (420 A), and therefore it must have its normal full-load e.m.f. of 499.0 Volts.
We know that motional e.m.f. is proportional to flux and speed of rotation, so if we denote the
normal (rated) flux as  , we know that at 1000 rev/min and with flux  , the e.m.f. is 499
Volts. If we denote the reduced flux in machine 1 as  ' , we know that at the steady speed of
the machines under test (say N) the e.m.f. is also 499 Volts. Hence the speed N is given by

1000
'
, where  
.
N   '  1000  , or N  '  1000 



An equivalent circuit sketch (Figure 3F) will help us to set up the remainder of the equations.
Fig 3F here.
The things we do not know are:- the e.m.f. E2, and the armature current I2, of the second
machine; the current from the supply (Is); the speed, N; the fluxes of both machines (but we do
have a relationship between them, equation 1).
Let’s look at the easy equations first, beginning with Kirchoff’s current law applied to motor 1:I s  I 2  420
The armature voltage equation for machine 2 yields
E2  0.05I 2  520
The induced e.m.f. in machine 2 is 499 V at 1000 rev/min, hence at speed N, E2 is given by
16
E2  499 
N
1000
Now we have to consider the mechanical set-up, and in particular consider the torque balance.
We will have to make the assumption that the windage and friction losses are the same during
this load test as they were when the machines ran at no-load. i.e. we will assume that a torque of
88.2 Nm per machine will be required to keep the two machines running. There is no other
mechanical load, so the nett torque of the two machines together must be 2  88.2 = 176.4 Nm.
If the flux in machine 1 was at its full value, its torque would be given by kI; but the flux has
been reduced from  to ’,so the torque per ampere is reduced in this ratio and the motoring
'
 4.7652  420    2001.38 .
torque of machine 1 is thus given by T1  k I

Machine 2 has full flux and its (-ve) generating torque is thus given by T2  k I  4.7652I 2 .
The torque balance equation is thus
T1  T2  176.4, i.e. 2001.38  4.7652I 2  176.4
At this stage we should draw together the five system equations:N 
1000

(1)
I S  I 2  420
( 2)
E2  0.05 I 2  520
(3)
E2  499 
2001.38   4.7652 I 2  176.4
N
1000
(4)
(5)
Reassuringly, there are five equations and five unknowns, which we can therefore solve to yield
the following values: N = 1077.436 rev/min; E2 = 537.64 Volts; I2 = 352.81 A;  = 0.9281; IS
= 67.19 A.
Now let us look at the outcome of all these calculations to see if it all makes sense.
Firstly, we recall that the aim of the exercise was to test machine 1 at as near as possible to full
load conditions without needing a mechanical load that could absorb 200 kW, and to minimise
the power drawn from the supply.
We have seen that the armature of machine 1 is working at full power (520 V, 420 A) and that
its mechanical power output is 200 kW, so the armature is certainly fully stretched. However
we have had to reduce the field flux by 7% from its rated value, and the speed is 7% above base
speed, so we are running at full power but with 93% torque and 107% speed, so conditions are
not precisely the same as at full load, base speed.
17
To achieve exactly rated conditions in machine 1 we would have had to increase the flux in
machine 2 instead of reducing the flux in machine 1, but that would have meant that the field
winding and the magnetic circuit of machine 2 were overloaded and therefore we could only do
this for a short period. Alternatively, we could make use of a differential gearbox to impose a
small speed difference between the machines, but this clearly pushes up the cost of the test.
As far as minimising the power drawn from the supply is concerned the arrangement is very
effective. Of the total input power of 218.4 kW to machine 1, 84% is supplied by machine 2
and only 35 kW is drawn from the supply.
- End of Solutions for Chapter 3 -