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3/23/2007 ELEC A6 DC Machine
1
Lecture Notes
ELE-A6
DC Machine
3/23/2007 ELEC A6 DC Machine
2
DIRECT CURRENT (DC)
MACHINES Fundamentals
• Generator action: An emf (voltage) is induced in a
conductor if it moves through a magnetic field.
• Motor action: A force is induced in a conductor that
has a current going through it and placed in a
magnetic field
• Any DC machine can act either as a generator or as
a motor.
3/23/2007 ELEC A6 DC Machine
3
DC Generator Fundamentals
( )e
e B v
= × ⋅
=
v B l
l sin cosα β
α - angle between the direction in which the
conductor is moving and the flux is acting.
β - smallest possible angle the conductor makes
with the direction of, the vector product, ( v × B)
and for maximum induction, β = 0. Hence, e =
Blv for most cases.
( v × B) indicates the direction of the current flow
in the conductor, or the polarity of the emf.
e = induced voltage, v = velocity of the conductor, B =
flux density and l is the length of the conductor
3/23/2007 ELEC A6 DC Machine
4
Generated Voltage in a Loop
(a coil of one turn)
• For emf to be induced, the
conductors must cut the flux
lines as they move.
Otherwise, ( v × B) = 0.
eloop = eab + ebc + ecd + eda
eloop = Blv + 0 + Blv + 0
eloop = 2 B l v
3/23/2007 ELEC A6 DC Machine
5
Generated Voltage in a Loop
(a coil of one turn)
• Note: Induced voltages are always AC.
3/23/2007 ELEC A6 DC Machine
6
Generated Voltage in a Loop
(a coil of one turn)
3/23/2007 ELEC A6 DC Machine
7
DC MACHINES
Real DC machine Construction
• Stator: Stationary part of the machine.
The stator carries a field winding that is
used to produce the required magnetic
field by DC excitation. Often know as
the field.
• Rotor: The rotor is the rotating part of
the machine. The rotor carries a
distributed winding, and is the winding
where the emf is induced. Also known
as the armature.
N S
Stator with
with polesBrush
Rotor
Field
3/23/2007 ELEC A6 DC Machine
8
DIRECT CURRENT MACHINES
DC machine Construction
• The picture shows the
stator of a large DC
machine with several
poles.
• The iron core is supported
by a cast iron frame.
Dc motor stator construction
Field poles
3/23/2007 ELEC A6 DC Machine
9
DIRECT CURRENT
MACHINES
DC machine Construction
• The rotor iron core is mounted on the
shaft.
• Coils are placed in the slots.
• The end of the coils are bent and
tied together to assure mechanical
strength.
• Note the commutator mounted on
the shaft. It consists of several
copper segments, separated by
insulation.
DC motor rotor construction
Commutator
Poles
Rotor winding
Fan
Bearing
Brushes
3/23/2007 ELEC A6 DC Machine
10
DC MACHINES
Generated EMF in a Real DC Machine
Where
Z = total number of conductors, P = total number of poles
a = P for lap winding, a = 2 for wave winding, φ = flux,
ω = speed in rad/s and n = speed in rpm.
φωφω
π
φφ mgG k
a
ZP
nkn
a
ZP
E ====
260
3/23/2007 ELEC A6 DC Machine
11
DC Motor Fundamentals
L is a vector in the direction of the flow of the current.
direction indicates the direction of force.
( )F B= ×l i
F = induced force, B = flux density, I is the current passing
in the conductor and l is the length of the conductor
)( Bl ×
3/23/2007 ELEC A6 DC Machine
12
DC Motor Fundamentals
Counter EMF:
When the motor is running, internally generated
emf, (EG = EC) opposes the applied voltage, thus:
I
V E
RA
T C
A
=
−
Where: VT = terminal voltage, Ec = counter
EMF, RA is the armature resistance and IA is
the armature current
3/23/2007 ELEC A6 DC Machine
13
The relationship between the
induced EMF and torque
{ {
E B v
T B ir
therefore
E
T
B v
B ir
v
ir
v r
E
T i
Ei T
conductor
conductor
electric
power
mechanical
power
=
=
= = =
=
=
l
l
l
l
,
, ω
ω
ω
T
EI ZP
a
I
T k I
A
m A
= =
=
ω π
φ
φ
2
Where: T is the torque,
3/23/2007 ELEC A6 DC Machine
14
Example 1
Find the induced voltage for
a road with a length of 1 m.
B = 3 T
V = 1 m/s0).(
with90makes
page)the(into3
).(
o
=×
×
==×
×=
lBv
lBv
vBBv
lBveind
3/23/2007 ELEC A6 DC Machine
15
Example 2
Example: Find the
force for a conductor
with a length of 1 m.
B = 0.35 T
i = 2 A
pagetheintoisdirectionThe
7.0)35.0(*)2(*)1()( NiBlF ==×=
3/23/2007 ELEC A6 DC Machine
16
Example 3
Vn
a
ZP
EG 8.17210*30*600*
2*60
6*4*48
60
voltage.inducedthecalculatemWb,30
ispoleperfluxtheandrpm600@rotatingisarmatureThe
/slot.conductors4withslots48hasaramatureThewinding.
WaveaasconnectedarmatureanhasmachineDCpole6A
3
=== −
φ
3/23/2007 ELEC A6 DC Machine
17
Example 4
A six-pole DC machine has a flux per pole of 30 mWb.
The armature has 536 conductors connected as a lap
winding. The DC machine runs at 1050 rpm and it
delivers a rated armature current of 225 A to a load
connected to its terminals, calculate:
A) Machine constant, Km
B) Generated voltage, EG
C) Conductor current
D) Electromagnetic torque.
E) Power delivered by the machine.
3/23/2007 ELEC A6 DC Machine
18
Example 4
kW32.63)
.84.575225*03.0*31.85)
5.37
6
225
)(
:windinglapisitSince)
V4.28196.109*03.0*31.85
rad/s96.109
60
1050*2
60
2
)
31.85
6*2
536*6
2
)
===
===
===
===
===
===
aG
am
a
mG
m
IETPe
mNIkTd
A
a
I
conductorI
c
kE
n
b
a
ZP
ka
ω
φ
φω
ππ
ω
ππ
3/23/2007 ELEC A6 DC Machine
19
DC Motors
Equivalent circuit.
The equivalent circuit of DC Motors (and Generators) has two
components:
• Armature circuit: it can be represented by a voltage source and a
resistance connected in series (the armature resistance). The armature
winding has a resistance, Ra.
• The field circuit: It is represented by a winding that generates the
magnetic field and a resistance connected in series. The field winding
has resistance Rf.
3/23/2007 ELEC A6 DC Machine
20
Classification of DC Motors
• Separately Excited and Shunt Motors
- Field and armature windings are either connected separate or in
parallel.
• Series Motors
- Field and armature windings are connected in series.
• Compound Motors
- Has both shunt and series field so it combines features of series and
shunt motors.
3/23/2007 ELEC A6 DC Machine
21
Shunt DC Motors
– The armature and field windings are connected in parallel.
– Constant speed operation.
3/23/2007 ELEC A6 DC Machine
22
Shunt DC Motors
By KVL around the outer loop:
VT - IA RA - EC = 0 I
V
RF
T
F
=
nkIE
mII
nkE
fC
ff
gC
=
=∝
=
φφ
φ
so,
but
3/23/2007 ELEC A6 DC Machine
23
Starting of Shunt DC Motors
• At the starting of a DC motor, EC = 0, so:
• To limit IA, a resistance is inserted in series
with RA then removed after the development
of EC
a
T
A
R
V
I =
3/23/2007 ELEC A6 DC Machine
24
Speed Regulation
Speed regulation is the percentage change
in speed from no-load to full-load as a
function of the full load speed.
SR
n n
n
NL FL
FL
=
−
×100%
3/23/2007 ELEC A6 DC Machine
25
Power Flow and Losses in
DC Motors
Pin = VtIL
I2R Losses
Copper losses
Core Losses
(Both cores)
Vt = Terminal voltage
IL= Line current
Mechanical
Losses
Stray Losses
Pout
P(developed) = ECIa
3/23/2007 ELEC A6 DC Machine
26
Example 1
Q1) A 240 V, shunt DC motor takes an
armature current of 20 A when running at
960 rpm. The armature resistance is 0.2 Ω.
Determine the no load speed if the no load
armature current is 1 A.
3/23/2007 ELEC A6 DC Machine
27
Example 1
rpm45.975960*
236
8.239
.).(
.).(
.).(
.).(
.).(
constantisIassuming
V8.2392.0*1240load)(no
V2362.0*20240load)(full
f
==⇒=⇒
=−=
=−=⇒
+=
lnn
lFn
lnn
lFE
lnE
E
E
ERIV
c
c
c
c
caat
loadfullload,no == F.l.n.l.
3/23/2007 ELEC A6 DC Machine
28
Example 2
Q2) A 120 V shunt motor has the following
parameters: Ra = 0.4 Ω, RF = 120 Ω and
rotational (core, mechanical and stray) losses are
240 W. On full load, the line current is 19.5 A and
the motor runs at 1200 rpm, find:
• The developed power
• The output power, and
• The output torque.
3/23/2007 ELEC A6 DC Machine
29
Example 2
N.m67.14
1200
60
2
1843
)(
watt18432402083(b)
watt1.20835.18*6.112*
V6.112)4.0(*)5.18(120
A5.1815.191
120
120
)(
===
=−=−=
===
=−=−=
=−=⇒==
−=
πω
τ out
out
developedout
acdeveloped
aaTc
af
fLa
P
c
lossesrotationalPP
IEP
RIVE
IAI
IIIa
3/23/2007 ELEC A6 DC Machine
30
Separately Excited DC Motors
– The armature winding supplies the load.
– The field winding is supplied by a separate DC source whose
voltage is variable.
– Good speed control.
3/23/2007 ELEC A6 DC Machine
31
Separately Excited DC Motors
3/23/2007 ELEC A6 DC Machine
32
Series DC Motors
– The armature and field winding are connected in series.
– High starting torque.
3/23/2007 ELEC A6 DC Machine
33
Series DC Motors
By KVL around the loop:
VT - IA (RA + Rf) - EC = 0
nkIE
mII
nkE
AC
AA
gC
=
=∝
=
φφ
φ
so,
but
3/23/2007 ELEC A6 DC Machine
34
Example 3
Q1) A DC series motor is operated at full
load from a 240 V supply at a speed of
600 rpm. The EC is found to be 217.2
V at a line current of 38 A, find:
a) The armature resistance assuming the
series field resistance is 0.2 Ω.
b) Find the no-load speed given that the
no-load current is 1 A.
3/23/2007 ELEC A6 DC Machine
35
Example 3
rpm151,25
V6.2396.0*1240
:loadnoAt(b)
6.0
38
2.217240
)(*)(
2
11
22
1
2
=⇒=
=−=
Ω=
−
=+⇒
++=
n
In
In
E
E
E
RR
RRIEVa
a
a
c
c
c
fa
faact
3/23/2007 ELEC A6 DC Machine
36
Compound DC Motors
3/23/2007 ELEC A6 DC Machine
37
Speed Control of DC Motors
Speed can be controlled by varying:
1) Armature circuit resistance using an external
resistance RA Ext.
2) IF can be varied by using an external
resistance Radj in series with RF to control the
flux, hence the speed.
3) The applied voltage to the armature circuit
resistance, if the motor is separately excited
3/23/2007 ELEC A6 DC Machine
38
Torque developed by shunt
motor
( )
T
K
R
k
V
R
k
T
kV
k
T
I
RIkV
RIEV
m
A
m
T
A
m
mT
m
A
AAmT
AACT
2
forsolving
so
but
so
φφ
ω
ω
φ
φω
φ
φω
−=
+=
=
+=
+=
3/23/2007 ELEC A6 DC Machine
39
Torque developed by shunt
motor
If VT and IF (hence )
are constant, speed is
directly proportional to
the torque
3/23/2007 ELEC A6 DC Machine
40
Torque developed by series
motor
3/23/2007 ELEC A6 DC Machine
41
Comparison between
different DC Motors
3/23/2007 ELEC A6 DC Machine
42
Comparison of DC Motors
Shunt Motors: “Constant speed” motor (speed regulation is very
good). Adjustable speed, medium starting torque. (Start = 1.4 × TFL)
Applications: centrifugal pump, machine tools, blowers fans,
reciprocating pumps, etc.
Series Motors: Variable speed motor which changes speed
drastically from one load condition to another. It has a high
starting torque.
Applications: hoists, electric trains, conveyors, elevators, electric
cars, etc.
Compound motors: Variable speed motors. It has a high starting
torque and the no-load speed is controllable unlike in series
motors.
Applications: Rolling mills, sudden temporary loads, heavy
machine tools, punches, etc
3/23/2007 ELEC A6 DC Machine
43
Home work problems
Q no.1
3/23/2007 ELEC A6 DC Machine
44
Home work problems
Q no.2

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Dc machines

  • 1. 3/23/2007 ELEC A6 DC Machine 1 Lecture Notes ELE-A6 DC Machine
  • 2. 3/23/2007 ELEC A6 DC Machine 2 DIRECT CURRENT (DC) MACHINES Fundamentals • Generator action: An emf (voltage) is induced in a conductor if it moves through a magnetic field. • Motor action: A force is induced in a conductor that has a current going through it and placed in a magnetic field • Any DC machine can act either as a generator or as a motor.
  • 3. 3/23/2007 ELEC A6 DC Machine 3 DC Generator Fundamentals ( )e e B v = × ⋅ = v B l l sin cosα β α - angle between the direction in which the conductor is moving and the flux is acting. β - smallest possible angle the conductor makes with the direction of, the vector product, ( v × B) and for maximum induction, β = 0. Hence, e = Blv for most cases. ( v × B) indicates the direction of the current flow in the conductor, or the polarity of the emf. e = induced voltage, v = velocity of the conductor, B = flux density and l is the length of the conductor
  • 4. 3/23/2007 ELEC A6 DC Machine 4 Generated Voltage in a Loop (a coil of one turn) • For emf to be induced, the conductors must cut the flux lines as they move. Otherwise, ( v × B) = 0. eloop = eab + ebc + ecd + eda eloop = Blv + 0 + Blv + 0 eloop = 2 B l v
  • 5. 3/23/2007 ELEC A6 DC Machine 5 Generated Voltage in a Loop (a coil of one turn) • Note: Induced voltages are always AC.
  • 6. 3/23/2007 ELEC A6 DC Machine 6 Generated Voltage in a Loop (a coil of one turn)
  • 7. 3/23/2007 ELEC A6 DC Machine 7 DC MACHINES Real DC machine Construction • Stator: Stationary part of the machine. The stator carries a field winding that is used to produce the required magnetic field by DC excitation. Often know as the field. • Rotor: The rotor is the rotating part of the machine. The rotor carries a distributed winding, and is the winding where the emf is induced. Also known as the armature. N S Stator with with polesBrush Rotor Field
  • 8. 3/23/2007 ELEC A6 DC Machine 8 DIRECT CURRENT MACHINES DC machine Construction • The picture shows the stator of a large DC machine with several poles. • The iron core is supported by a cast iron frame. Dc motor stator construction Field poles
  • 9. 3/23/2007 ELEC A6 DC Machine 9 DIRECT CURRENT MACHINES DC machine Construction • The rotor iron core is mounted on the shaft. • Coils are placed in the slots. • The end of the coils are bent and tied together to assure mechanical strength. • Note the commutator mounted on the shaft. It consists of several copper segments, separated by insulation. DC motor rotor construction Commutator Poles Rotor winding Fan Bearing Brushes
  • 10. 3/23/2007 ELEC A6 DC Machine 10 DC MACHINES Generated EMF in a Real DC Machine Where Z = total number of conductors, P = total number of poles a = P for lap winding, a = 2 for wave winding, φ = flux, ω = speed in rad/s and n = speed in rpm. φωφω π φφ mgG k a ZP nkn a ZP E ==== 260
  • 11. 3/23/2007 ELEC A6 DC Machine 11 DC Motor Fundamentals L is a vector in the direction of the flow of the current. direction indicates the direction of force. ( )F B= ×l i F = induced force, B = flux density, I is the current passing in the conductor and l is the length of the conductor )( Bl ×
  • 12. 3/23/2007 ELEC A6 DC Machine 12 DC Motor Fundamentals Counter EMF: When the motor is running, internally generated emf, (EG = EC) opposes the applied voltage, thus: I V E RA T C A = − Where: VT = terminal voltage, Ec = counter EMF, RA is the armature resistance and IA is the armature current
  • 13. 3/23/2007 ELEC A6 DC Machine 13 The relationship between the induced EMF and torque { { E B v T B ir therefore E T B v B ir v ir v r E T i Ei T conductor conductor electric power mechanical power = = = = = = = l l l l , , ω ω ω T EI ZP a I T k I A m A = = = ω π φ φ 2 Where: T is the torque,
  • 14. 3/23/2007 ELEC A6 DC Machine 14 Example 1 Find the induced voltage for a road with a length of 1 m. B = 3 T V = 1 m/s0).( with90makes page)the(into3 ).( o =× × ==× ×= lBv lBv vBBv lBveind
  • 15. 3/23/2007 ELEC A6 DC Machine 15 Example 2 Example: Find the force for a conductor with a length of 1 m. B = 0.35 T i = 2 A pagetheintoisdirectionThe 7.0)35.0(*)2(*)1()( NiBlF ==×=
  • 16. 3/23/2007 ELEC A6 DC Machine 16 Example 3 Vn a ZP EG 8.17210*30*600* 2*60 6*4*48 60 voltage.inducedthecalculatemWb,30 ispoleperfluxtheandrpm600@rotatingisarmatureThe /slot.conductors4withslots48hasaramatureThewinding. WaveaasconnectedarmatureanhasmachineDCpole6A 3 === − φ
  • 17. 3/23/2007 ELEC A6 DC Machine 17 Example 4 A six-pole DC machine has a flux per pole of 30 mWb. The armature has 536 conductors connected as a lap winding. The DC machine runs at 1050 rpm and it delivers a rated armature current of 225 A to a load connected to its terminals, calculate: A) Machine constant, Km B) Generated voltage, EG C) Conductor current D) Electromagnetic torque. E) Power delivered by the machine.
  • 18. 3/23/2007 ELEC A6 DC Machine 18 Example 4 kW32.63) .84.575225*03.0*31.85) 5.37 6 225 )( :windinglapisitSince) V4.28196.109*03.0*31.85 rad/s96.109 60 1050*2 60 2 ) 31.85 6*2 536*6 2 ) === === === === === === aG am a mG m IETPe mNIkTd A a I conductorI c kE n b a ZP ka ω φ φω ππ ω ππ
  • 19. 3/23/2007 ELEC A6 DC Machine 19 DC Motors Equivalent circuit. The equivalent circuit of DC Motors (and Generators) has two components: • Armature circuit: it can be represented by a voltage source and a resistance connected in series (the armature resistance). The armature winding has a resistance, Ra. • The field circuit: It is represented by a winding that generates the magnetic field and a resistance connected in series. The field winding has resistance Rf.
  • 20. 3/23/2007 ELEC A6 DC Machine 20 Classification of DC Motors • Separately Excited and Shunt Motors - Field and armature windings are either connected separate or in parallel. • Series Motors - Field and armature windings are connected in series. • Compound Motors - Has both shunt and series field so it combines features of series and shunt motors.
  • 21. 3/23/2007 ELEC A6 DC Machine 21 Shunt DC Motors – The armature and field windings are connected in parallel. – Constant speed operation.
  • 22. 3/23/2007 ELEC A6 DC Machine 22 Shunt DC Motors By KVL around the outer loop: VT - IA RA - EC = 0 I V RF T F = nkIE mII nkE fC ff gC = =∝ = φφ φ so, but
  • 23. 3/23/2007 ELEC A6 DC Machine 23 Starting of Shunt DC Motors • At the starting of a DC motor, EC = 0, so: • To limit IA, a resistance is inserted in series with RA then removed after the development of EC a T A R V I =
  • 24. 3/23/2007 ELEC A6 DC Machine 24 Speed Regulation Speed regulation is the percentage change in speed from no-load to full-load as a function of the full load speed. SR n n n NL FL FL = − ×100%
  • 25. 3/23/2007 ELEC A6 DC Machine 25 Power Flow and Losses in DC Motors Pin = VtIL I2R Losses Copper losses Core Losses (Both cores) Vt = Terminal voltage IL= Line current Mechanical Losses Stray Losses Pout P(developed) = ECIa
  • 26. 3/23/2007 ELEC A6 DC Machine 26 Example 1 Q1) A 240 V, shunt DC motor takes an armature current of 20 A when running at 960 rpm. The armature resistance is 0.2 Ω. Determine the no load speed if the no load armature current is 1 A.
  • 27. 3/23/2007 ELEC A6 DC Machine 27 Example 1 rpm45.975960* 236 8.239 .).( .).( .).( .).( .).( constantisIassuming V8.2392.0*1240load)(no V2362.0*20240load)(full f ==⇒=⇒ =−= =−=⇒ += lnn lFn lnn lFE lnE E E ERIV c c c c caat loadfullload,no == F.l.n.l.
  • 28. 3/23/2007 ELEC A6 DC Machine 28 Example 2 Q2) A 120 V shunt motor has the following parameters: Ra = 0.4 Ω, RF = 120 Ω and rotational (core, mechanical and stray) losses are 240 W. On full load, the line current is 19.5 A and the motor runs at 1200 rpm, find: • The developed power • The output power, and • The output torque.
  • 29. 3/23/2007 ELEC A6 DC Machine 29 Example 2 N.m67.14 1200 60 2 1843 )( watt18432402083(b) watt1.20835.18*6.112* V6.112)4.0(*)5.18(120 A5.1815.191 120 120 )( === =−=−= === =−=−= =−=⇒== −= πω τ out out developedout acdeveloped aaTc af fLa P c lossesrotationalPP IEP RIVE IAI IIIa
  • 30. 3/23/2007 ELEC A6 DC Machine 30 Separately Excited DC Motors – The armature winding supplies the load. – The field winding is supplied by a separate DC source whose voltage is variable. – Good speed control.
  • 31. 3/23/2007 ELEC A6 DC Machine 31 Separately Excited DC Motors
  • 32. 3/23/2007 ELEC A6 DC Machine 32 Series DC Motors – The armature and field winding are connected in series. – High starting torque.
  • 33. 3/23/2007 ELEC A6 DC Machine 33 Series DC Motors By KVL around the loop: VT - IA (RA + Rf) - EC = 0 nkIE mII nkE AC AA gC = =∝ = φφ φ so, but
  • 34. 3/23/2007 ELEC A6 DC Machine 34 Example 3 Q1) A DC series motor is operated at full load from a 240 V supply at a speed of 600 rpm. The EC is found to be 217.2 V at a line current of 38 A, find: a) The armature resistance assuming the series field resistance is 0.2 Ω. b) Find the no-load speed given that the no-load current is 1 A.
  • 35. 3/23/2007 ELEC A6 DC Machine 35 Example 3 rpm151,25 V6.2396.0*1240 :loadnoAt(b) 6.0 38 2.217240 )(*)( 2 11 22 1 2 =⇒= =−= Ω= − =+⇒ ++= n In In E E E RR RRIEVa a a c c c fa faact
  • 36. 3/23/2007 ELEC A6 DC Machine 36 Compound DC Motors
  • 37. 3/23/2007 ELEC A6 DC Machine 37 Speed Control of DC Motors Speed can be controlled by varying: 1) Armature circuit resistance using an external resistance RA Ext. 2) IF can be varied by using an external resistance Radj in series with RF to control the flux, hence the speed. 3) The applied voltage to the armature circuit resistance, if the motor is separately excited
  • 38. 3/23/2007 ELEC A6 DC Machine 38 Torque developed by shunt motor ( ) T K R k V R k T kV k T I RIkV RIEV m A m T A m mT m A AAmT AACT 2 forsolving so but so φφ ω ω φ φω φ φω −= += = += +=
  • 39. 3/23/2007 ELEC A6 DC Machine 39 Torque developed by shunt motor If VT and IF (hence ) are constant, speed is directly proportional to the torque
  • 40. 3/23/2007 ELEC A6 DC Machine 40 Torque developed by series motor
  • 41. 3/23/2007 ELEC A6 DC Machine 41 Comparison between different DC Motors
  • 42. 3/23/2007 ELEC A6 DC Machine 42 Comparison of DC Motors Shunt Motors: “Constant speed” motor (speed regulation is very good). Adjustable speed, medium starting torque. (Start = 1.4 × TFL) Applications: centrifugal pump, machine tools, blowers fans, reciprocating pumps, etc. Series Motors: Variable speed motor which changes speed drastically from one load condition to another. It has a high starting torque. Applications: hoists, electric trains, conveyors, elevators, electric cars, etc. Compound motors: Variable speed motors. It has a high starting torque and the no-load speed is controllable unlike in series motors. Applications: Rolling mills, sudden temporary loads, heavy machine tools, punches, etc
  • 43. 3/23/2007 ELEC A6 DC Machine 43 Home work problems Q no.1
  • 44. 3/23/2007 ELEC A6 DC Machine 44 Home work problems Q no.2