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ENGNG2024 Electrical Engineering

E Levi, 2001 1

STEADY STATE OPERATION OF DC MACHINES

1. INTRODUCTION

Electric DC machines, as indeed any other type of electric machine, can be used to

either produce electric energy from the input mechanical energy, or to convert electric energy

into output mechanical energy. These two possible operating regimes are called generation and

motoring. As already mentioned, DC machines used to be in the past the major source of DC

power. In order to produce electric power DC machines were operated as generators.

Nowadays, however, use of DC generators is becoming more and more rare. DC power is

obtained instead by means of power electronic converters. The remaining applications of DC

machines are today restricted to motoring. In this operating regime a DC machine is operated

as a DC motor: it consumes DC power, while delivering at its shaft mechanical power. Theshaft drives a certain load, that is characterised with the load torque.

A DC machine consists of stationary part, called stator, and rotating part, called rotor.

Both stator and rotor are equipped with one winding. Stator winding is supplied from a DC

voltage source and the role of this winding is to produce magnetic flux in the air gap of the

machine. This flux is stationary in space. Rotor winding is again supplied from a DC voltage

source (for motoring): DC supply is connected to the rotor winding through a special

assembly, that is composed of brushes and commutator. Brushes are stationary, while

commutator is fixed to the rotor and hence rotates together with the rotor. This assembly

enables supply of electric power from stationary power supply to rotating winding on rotor.

Principle of operation of this assembly is illustrated by means of Fig. 1. Rotor winding is shown

in a very simplified manner, as consisting of just one coil, connected to two segments of the

commutator. Motoring action is assumed and the current is therefore delivered to the rotor

winding through the stationary brushes and rotating commutator. Two positions of the rotor

winding are shown in Fig. 1. Terminal current (current brought to the brushes) and the winding

current are illustrated in Fig. 2. As can be seen from these two figures, current inside the rotor

winding is reversed (commutated) after each half-revolution of the rotor. Current inside the

rotor is therefore AC, while the terminal current is DC. Frequency of the current inside the

rotor winding equals frequency of rotation.

A

ia = Ia ia

i=Ia

B

B

ia = Ia ia

i=Ia

A

Fig. 1 - Current reversal (commutation) in rotor coil by means of the commutator.

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E Levi, 2001 2

ia Ia i Ia

0 = t 0 2 =t

Ia

Fig. 2 - Terminal current and current through rotor coil.

Note that such a situation regarding frequencies in the two windings is the only possible

one that satisfies the condition of average torque existence. Since the stator winding is supplied

with pure DC current of zero frequency, the machine can develop an average torque if and

only if the rotor winding frequency equals the frequency of rotation. This means that it is not

possible to realise an electric machine with DC currents flowing in both stator and rotorwindings. Such a situation would result in the possibility of developing an average torque at

zero speed only. At zero speed however converted power equals zero and therefore such a

machine could not do the process of electromechanical energy conversion.

Let the stator winding, which is called excitation or field winding as well, be supplied

with constant DC voltage equal to Vf . Current that flows through this winding is in steady-

state operation determined with

If = Vf /Rf (1)

Flux produced in the air gap of the machine is, neglecting saturation of the magnetic circuit,

proportional to this current. Hence

f = c1 If (2)It has to be emphasised that the excitation winding can be replaced with permanent magnets.

Many of the modern DC motors rely on permanent magnet excitation and in such a case there

is not any possibility of changing the excitation flux, since it is fixed by the magnet properties.

Rotor rotates at certain angular speed [rad/s], in the stationary flux produced by the

excitation winding. Thus, according to the basic law of electro-magnetic induction, e = B l v

(where v is linear speed, B is flux density and l is length of the conductor), there is an induced

electromotive force (emf) in the rotor winding

E= c2 f = k If (3)

Induced emf is proportional to the flux (i.e., excitation current) and to the speed of rotation,

through a constant determined with constructional features of the machine. Speed of rotation

is the so-called angular speed of rotation in [rad/s] and it is correlated with the speed n in

revolutions per minute [rpm] through ( ) = 2 60 n . As rotor winding (often called armaturewinding) is supplied from a DC voltage source and as the winding is of certain resistance, then

the voltage equilibrium equation for the armature winding (index a) is

Va = Ra Ia + E (4)

According to the basic law of electromagnetic force creation, if a conductor that carries

current I moves in the flux of flux density B, then an electromagnetic force F = B I l acts on

the conductor. As rotor winding rotates in the flux density produced by the excitation winding,

an electromagnetic torque is produced, equal to

Te = c2 f Ia = k If Ia (5)

This electromagnetic torque is the reason why the rotor rotates. If there is load connected tothe shaft of the machine, then this load opposes rotation with its torque, called load torque TL.

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E Levi, 2001 3

In any steady-state condition electromagnetic torque and load torque are equal and act in the

opposite direction. While electromagnetic torque acts in the direction of rotation, load torque

acts in the opposite direction, as illustrated in Fig. 3 for two possible directions of rotation, and

expressed with the following relation:

Te = TL (6)

Angular speed of rotation in [rad/s] is correlated with speed in [rpm] through thealready given scaling factor

n [rpm] = (60/2 ) [rad/s] (7)

Te Te

rotor rotor

TL TLforward motoring reverse motoring

Fig. 3 - Directions of electromagnetic torque, load torque and speed in motoring.

Note that constant of proportionality in expressions for induced emf and

electromagnetic torque, k, is the same as long as speed in expression for emf is given in rad/s.

Input and output powers of the motor are electrical and mechanical powers,

respectively, and are given with the following expressions:

Pin = VaIa + Vf If (8)

Pout = Te (9)

The difference of the two powers represents power loss in the machine, which includes

mechanical loss, iron loss and loss in the windings. In what follows mechanical loss and iron

loss will be frequently neglected, but copper loss will always be accounted for. Power loss in

the machine and the efficiency are given with

Ploss = Pin - Pout (10)

= Pout / Pin (11)

Equations (1)-(11) completely describe steady-state operation of a DC motor. Standard data

that are given for a DC motor on its nameplate are so-called rated values of output power,

armature voltage and current, excitation voltage and current, and speed of rotation in rpm.

Rated values will always be identified in examples with index n. Note that, as rated power isalways output power, rated power for a motor is mechanical output power.

Load torque that an electric motor drives can be either speed independent (say, crane

lifting a load) or speed dependent. Frequently met speed dependent load torques are either

linearly proportional to speed or proportional to speed squared (pumps, compressors, fans,

etc.). These three types of load torques are illustrated in Fig. 4. If the load torque is constant,

then it follows from (5) that product of excitation current and armature current is constant as

well. This is the simplest case and all the examples will assume that a DC motor drives a

constant speed independent load torque.

Depending on how excitation winding and armature winding are supplied from DC

voltage sources, various types of DC motors may be identified. Two of the types that are

nowadays in wide application are separately excited DC motor and series excited DC motor.

The third one, rarely used nowadays, is the shunt excited DC machine. In a separately excited

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E Levi, 2001 4

DC machine, two voltage DC sources are used. Shunt and series excited DC machines require

only one voltage source, with the stator and rotor winding connected in parallel and in series,

respectively. These three DC machine types are considered in more detail in what follows.

TL = k

TLTL = const.

TL = k2

Fig. 4 - Illustration of various types of load torques.

2. SEPARATELY EXCITED DC MOTOR

Excitation winding and armature winding of a separately excited DC motor are

supplied from two independent DC power sources. A separately excited DC motor is described

in steady-state operation with the following set of equations, that essentially only summarise

again (1)-(6)V E R I

V R I

c I

E c kI

T c I kI I

T T

a a a

f f f

f f

f f

e f a f a

L e

= +

=

=

= =

= =

=

1

2

2

(12)

Equivalent circuit of a separately excited DC motor is shown in Fig. 5. Mechanical, speed-

torque characteristic of the motor (speed against electromagnetic torque) can be derived from

(12) as follows:

( )

( )

V E R I kI R I I V kI R

T kI I kI V kI R

V

kI

R

kI T

I V R

V

kI

R

kIT

a a a f a a a a f a

e f a f a f a

a

f

a

fe

fn fn f

an

fn

a

fn

e

= + = +

=

= =

=

=

=

( )

( )

2

2

If both voltages have rated values then

(13)

Speed-torque characteristic, with rated voltages applied to both windings, is the so-called

natural operating characteristic and is shown in Fig. 5 with bold trace. Equation (13) enables

examination of available speed control methods for a separately excited DC motor, that are

beyond the scope here. As can be seen from the speed torque characteristic, speed slightly

decreases, in a linear manner, as load is increased. The highest value of the operating speed isunder no-load conditions (i.e., load torque equal to zero). Rated operating point is indicated in

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E Levi, 2001 5

Fig. 5 as well. Speed drop in a separately excited DC machine from no-load condition to full

(rated) load conditions is typically a few tens of rpms.

Example 1:

A separately excited DC motor has the following rated data: 500 V, 100 A, 1000 rpm.

Armature resistance is 0.5 . Excitation flux is constant and equal to rated. Calculaterated torque and rated power of the motor and evaluate efficiency of the motor in the

rated operation if power loss in the excitation winding is 5 kW.

Solution:As rated voltage and rated armature current are known, then from (12) one can calculate induced

electro-motive force in rated operating conditions: E R I E V R I x

E kI kI n kI E

n

x

T kI I x

P T x x

P P

an n a an n an a an

n fn n fn n fn

n

n

en fn an

n en n

n n in

= +

= = =

= =

= = =

= = =

= = =

= = + = =

500 0 5 100 450

2

60 2

60

30 450 1000 4.3

4.3 100 430

4302

601000 45030

45030 50000 5000) 4503 55

.

/ ( )

/ / ( . /

V

Rated torque is then

Nm

Rated power is mechanical (output)power,

W

Input power is the power delivered to the armature winding (500 V times 100 A)

plus the power delivered to the excitation winding (5 kW). The efficiency in rated

operation is ratio of output to input power. Hence

0819.

Example 2:

Separately excited DC motor, 230 V, armature resistance = 0.2 , operates under

no-load conditions at 1200 rpm. Under rated conditions armature current is 40 A.

Find the rated speed and rated electromagnetic torque of the motor. Excitation flux is

constant and rated.

Solution:Speed is in this example known for no-load operating conditions. If there is not load connected to

the shaft of the motor, then load torque is zero. From (12) it follows that electromagnetic torque is

zero as well. As electromagnetic torque is proportional to the armature current, then zero torque

indicates that armature current is zero. No-load point is denoted with index 0. Hence

E V R I V

E kI n kI E

n

V

n

x

E V R I x

E kI n nE

kI

x

T kI I x

an a a an

fn fnan

n an a an

n fn n nn

fn

en fn an

0 0

0 00

0 0

2

60 2

60

2

60

30 230 1200 1 83

230 0 2 40 222

2

60 2

60

30 222 1 83 1150

1 83 40 73 2

= =

=

= = = =

= = =

= = = =

= = =

/ ( ) .

.

/ ( . )

. .

Then for rated operating conditions

V

rpm

Nm

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E Levi, 2001 6

Example 3:

A separately excited DC motor, whose rated data are 440 V, 120 A, 970 rpm, armature

resistance = 0.16 , is loaded with such a load torque that the armature current is 40

A. Find the speed and torque of the motor for this operating condition. Excitation flux

is constant and equal to rated.

Solution:Rated data for voltage and current apply to the armature. When excitation flux is constant, as the case

is here, excitation winding data are not needed and are hence not given. The motor in this question

operates in an operating point other than rated. However, initial calculations always have to deal with

rated operating point, in order to find necessary values for subsequent calculations.

For rated operating point

V E R I E V R I x

E kI kI n kI E

n

x

T kI I x

P T x x

E V R I x

E kI kI n nE

kI

x

an n a an n an a an

n fn n fn n fnn

n

en fn an

n en n

an a a

fn fn

fn

= +

= = =

= =

= = =

= = =

= = =

= = =

= =

= =

440 0 16 120 420 8

2

60 2

60

30 420 8 970 4 14

4 1 4 120 497

4972

60970 50484

440 0 16 40 433 6

2

60 2

60

30 433 6 4 14

1 1

1 1 1 11

. .

. / ( ) .

.

. .

. / ( .

V

Nm

W

In new operating point

V

) .

. .

. . .

=

= = =

= = =

100013

4 14 40 165 6

16562

60100013 173438

1 1

1 1 1

rpm

Nm

W

T kI I x

P T x x

e fn a

e

The two operating points are illustrated in Figure 5.

Ra IaVa = E + Ra Ia

Rf Vf = Rf If

Va E If E = k If

Vf Te = k If Ia

1

n

Te1 Ten Te = TL

Fig. 5 - Equivalent circuit and natural speed torque curve of a separately excited DC motor.

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E Levi, 2001 7

Example 4:

A separately excited DC motor has the following rated data: 500 V, 100 A, 1000 rpm.

Armature resistance is 1 . Excitation flux is constant and equal to rated.

a) Calculate rated torque and rated power of the motor.

b) Evaluate efficiency of the motor in rated operation if power loss in the excitation

winding is 5 kW.c) The motor drives a load such that the armature current is 40 A. Calculate torque and

speed for this operating regime. Determine efficiency of the motor in this operating

point.

d) Plot speed - torque curve and denote the two calculated points.

Solution:Note that parts a) and b) are the exam version of the Example 1, with minor changes and additions!

a) As rated voltage and rated armature current are known, then from (12) one can calculate induced

electro-motive force in rated operating conditions:

%7373.055/40)500050000/(40000/Hencepower.inputoutput toofratioisoperation

ratedinefficiencyThekW).(5windingexcitationthetodeliveredpowertheplus

A)100timesV(500windingarmaturethetodeliveredpowertheispowerInputb)

kW40100060

2382

power,(output)mechanicalispowerRated

Nm38210082.3

thenistorqueRated

82.3)1000/(40030

60

260

2

V4001001500

===+==

===

===

===

==

===

+=

innn

nenn

anfnen

n

n

fnnfnnfnn

anaannananan

PP

xxTP

xIkIT

x

n

EkInkIkIE

xIRVEIREV

c) The load torque is now smaller, since current is only 40 A. As torque is directly proportional to the

armature current through the constant kIfn (which is 3.82) then the new torque is 40 A/100 A times the

rated torque, i.e. 152.8 Nm. Further

%6.73736.025/4.18

kW255000405005000

kW4.18115060

28.152

lyrespectiveare,powerinputandOutput

rpm1150)82.3/(46030)2/(6060

2

V460401500

===

=+=+=

===

====

===

xIVP

TP

xxkIEnnkIE

xIRVE

aanin

eout

fnfn

aaan

d) The two operating points are as indicated in Fig. 5.

3. SHUNT EXCITED DC MOTOR

A shunt excited DC motor is the motor where the excitation winding and the armature

winding are connected in parallel and supplied from the same voltage source. The equivalent

circuit is illustrated in Fig. 6. As long as the supply voltage is constant and rated (this is the

assumption valid here; voltage variation is a mean of speed control, that is beyond the scope ofinterest at present), all the equations given for a separately excited DC motor remain valid. It is

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E Levi, 2001 8

only necessary to replace individual voltages for the excitation and the armature winding in

(12)-(13) with one and the same value V of the supply voltage. The torque-speed curve of the

shunt motor is therefore the same as the one for a separately excited DC motor and remains to

be as given in Fig. 5.

Ra

Ia

V Rf E

If

Fig. 6 Equivalent circuit of a shunt excited DC machine.

Example 5:

A shunt excited DC motor, whose rated data are 440 V, 122 A, 970 rpm, armature

resistance = 0.16 , excitation winding resistance = 220 , is loaded with such a load

torque that the terminal current is 42 A. Find the speed and torque of the motor for this

operating condition.

Solution:Note that in the case of the shunt motor the rated current is the terminal current. That is , the given

rated current is the sum of the rated armature and rated excitation winding current. Since the ratedvoltage and excitation winding resistance are known, the rated excitation current is 440/220 = 2 A and

is the same regardless of the motor loading. Thus the rated armature current is 122 2 = 120 A, and

the armature current for the new operating point is 42 2 = 40 A.

Note that the rated motor data and the armature currents are now the same as in the Example 3 for a

separately excited motor. Hence the solution is the same as well, with rated armature voltage symbol

being replaced with the rated voltage symbol.

W8.1734313.100060

26.165

Nm6.1654014.4

rpm13.1000)14.4/(6.43330

60

260

2

V6.4334016.0440

pointoperatingnewIn

W5048497060

2497

Nm49712014.4

14.4)970/(8.42030

60

260

2

V8.42012016.0440

111

11

1

1111

11

===

===

===

==

===

===

===

===

==

===

+=

xxTP

xIkIT

x

kI

EnnkIkIE

xIRVE

xxTP

xIkIT

x

n

EkInkIkIE

xIRVEIREV

e

afne

fn

fnfn

aan

nenn

anfnen

n

n

fnnfnnfnn

anannanann

The two operating points remain to be as illustrated in Figure 5.

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E Levi, 2001 9

4. SERIES EXCITED DC MOTOR

In a series excited DC motor, as the name suggests, excitation winding and armature

winding are connected in series and fed from a single DC power source. Equivalent circuit of a

series excited DC motor is illustrated in Fig. 7. Due to series connection of the two windings,

equations (1)-(6) now become( )V E R R I

I I I

E kI kI

T kI I kI

T T

a f

a f

f

e f a

e L

= + +

= =

= =

= =

=

2

(14)

As excitation winding current is one and the same as armature current, then operation of a

series motor with variable load torque results in variable flux in the machine. Electromagnetic

torque is therefore proportional to the current squared, while emf is proportional to the

product of current and speed. Mechanical torque-speed characteristic is derived using the sameprocedure as for a separately excited DC motor:

( ) ( )( )

( )[ ]

( )[ ]

V R R I E R R I kI I V

R R k

T kIkV

R R k

TkV

R R k

a f a f

a f

e

a f

en

a f

= + + = + +

=+ +

= =

+ +

=

+ +

22

2

2

2

For rated voltage supply

(15)

Torque-speed curve is shown in Fig. 7 as well, with bold trace (rated operating point isindicated in the Figure 7). It substantially differs from the curve valid for a separately excited

motor. Small variation of the torque produces large variation in operating speed. Note that

series DC motor must not be ever allowed to run unloaded. As follows from (15), operation

with zero electromagnetic torque results when speed approaches infinity. This means that, if a

series motor is allowed to run without load, its rotor speed will reach such a high speed that

the motor will eventually disintegrate.

Rf Ra

I Te

V VfVa E Ten

Vn

n

Fig. 7 - Equivalent circuit and torque-speed characteristics of a series excited DC motor.

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E Levi, 2001 10

It should be noted that all the calculations regarding operation of a series DC motor

remain the same as in the case of a separately excited DC motor as long as operation with the

constant load torque is considered. However, if the load torque is speed dependent, then

change of load torque implies quadratic change of armature current, while in the case of the

separately excited motor the change is linear.

Example 6:

A series DC motor has armature and excitation winding resistances of 0.2 and 0.1 ,

respectively. Rated motor date are 1000 rpm, 40 A and 450 V. Calculate rated torque,

rated power and efficiency in rated operation.

Solution:From rated data

( )

E V R R I x

E kI k E I n

T kI x

P T x x

P V I x

P P

n n a f n

n n n n n n

en n

n en n

in n n

n n in

= + = + =

=

= =

= = =

= = =

= = =

= = =

=

( ) ( . . )

/ .

.

/

/ / .

450 0 1 0 2 40 438

30 0 10456

010456 40 167.3 Nm

167.3 1000 30 17520

450 40 18000

17520 18000 0 9733

480

2 2

V

W

W

Difference between input and output power is 480 W and this must

equal loss on resistances of the two windings: (0.1+ 0.2)x40 W.2

Example 7:

A series DC motor operates under rated conditions with 161.2 Nm torque, 1000 rpmspeed, 41 A current and 420 V voltage. Resistance of excitation and armature winding

is 0.2 . Find the speed and current of the motor if the torque is now 87 Nm.

Solution:Note that this example does not involve speed control and serves the purpose of illustrating

calculations for the case when load torque changes with speed. From the given data

E V RI x

T kI T kI T

T

I

I

I I T T E V RI x

E kI E kI E

E

n I

n I

n nE I

E I

x

x

n n n

en n ee

en n

n e en

n

n n n

n n n

nn

n

= = =

= = = = = =

= = =

= = =

= = =

= = =

420 0 2 41 411 8

161 2 87 87 161 2

41 87 1 61 2 30 12420 0 2 30 12 413 98

100041398 41

4118 30121368

21 1

2 1 12

2

1 1

1 1

1 1 11 1 1

11

1

. .

. / .

/ . .. . .

.

. ..

V

Nm Nm

AV

4 rpm

Example 8:

A series DC motor, 220 V, 1500 rpm, 270 A has combined resistance of armature and

field winding of 0.11 Ohms. The motor drives a load that is characterised with constant loadtorque, equal to the motor rated torque. Find the rated torque and rated power of the motor.

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E Levi, 2001 11

Solution:

( )

E V R R I x

E kI kI E n

T kI xP T x x

P V I x

P P

n n a f n

n n n n n n

en n

n en n

in n n

n n in

= + = =

=

= =

= = =

= = =

= = =

= = =

( ) .

/ .

. .. / .

/ . / .

220 0 1 1 270 190

30 1 21

1 2 1 270 326 6326 6 1500 30 51 3 kW

220 270 59.4

51 3 5 9.4 0 86

2

V

Nm

kW

5. TUTORIAL QUESTIONS

Q1. A series DC motor operates under rated conditions with 161.2 Nm torque, 1000 rpmspeed, 41 A current and 420 V voltage. Combined resistance of excitation and

armature winding is 0.2 .

a) Sketch the equivalent circuit of the series DC motor, derive its torque-speed

characteristic and show graphically toque-speed curve.

b) Calculate rated power of the motor and its efficiency in rated operation.

c) Find the speed and current of the motor when the load torque is 87 Nm. Determine

output power and efficiency in this operating point.

Q2. A series DC motor, 420 V, 41 A, 1000 rpm, has armature and excitation winding

resistances of 0.12 and 0.08 , respectively.

a) Calculate rated torque and rated power of the motor.

b) The motor drives a load whose torque is proportional to the speed of rotation

squared, TL = 0.015 2. Determine operating speed, current and torque of the motor

for this load torque.

Q3. a) Sketch equivalent circuits of (i) separately and (ii) series excited DC motors and

write the basic steady-state equations for each type.

b) A separately excited DC motor has the following rated data: 500 V, 100 A, 1000

rpm. Armature resistance is 0.5 . Excitation flux is constant and equal to rated.

i) Calculate rated torque and rated power of the motor.

ii) The motor drives a load such that the armature current is 40 A. Calculate

torque and speed for this operating regime.

iii) Plot speed - torque curve and denote the two calculated points.

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E Levi, 2001 12

Q4. a) Sketch equivalent circuits of (i) separately and (ii) series excited DC motors and


b) Explain the role of commutator in DC machines using simple representation of rotor

winding with just one turn and sketch the waveform of current in the turn assumingthat current at rotor terminals is constant DC.

c) A separately excited DC motor has the following rated data: 400 V, 100 A, 950 rpm.

Armature resistance is 1 . Excitation flux is constant and equal to rated. Mechanical

losses and iron core losses can be neglected and armature voltage is constant and equal

to rated. Calculate rated torque and rated power of the machine whose data are given.

If the motor now drives a load such that the armature current is 50 A, calculate speed

of rotation, torque and output power for this operating regime. Plot to scale torque-

speed curve and denote the two calculated points.

Q5. a) A series DC motor has armature resistance of 0.12 ohms and excitation winding

resistance of 0.08 ohms. When the motor operates at 1000 rpm the current is 41 A and

the torque is 176 Nm. The motor is supplied from a constant 420 V DC source.

Calculate the current and the speed when the motor operates with torque equal to 87

Nm.

b) Sketch equivalent circuits of (i) separately and (ii) series excited DC motors and


c) Explain the role of commutator in DC machines using simple representation of rotor

winding with just one turn and sketch the waveform of current in the turn assuming

that current at rotor terminals is constant DC.

Q6. A 220-V dc shunt motor draws 10-A at 1800-rpm. The armature circuit resistance is

0.2- and the field winding resistance is 440-.

a) Calculate the torque developed by the motor under the above conditions.

b) If the field current is unchanged, determine the speed and line current when the

motor produces a torque of 20-Nm.

c) Derive an expression representing the motor speed [rpm]-torque [Nm]

characteristic and then determine the machine no-load speed.

4_2024-dcm

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