Chapter 5 Additional Problems
X5.1 The dc machine of Fig. X5.1 is connected as a cumulative compound dc motor. The
switch (SW) is open at all times except as stated in part (b). (a) If the machine were
operated as a dc generator with direction of rotation unchanged, would the dc
generator be cumulative compound or differential compound? (b) If the machine is
operating as a motor at steady-state conditions and the switch is closed, will the motor
speed increase or decrease? (c) If terminals 1-2 were reconnected to the source so that
Vt  Vdc , will the motor reverse direction of rotation? (d) If terminals 1-2 were
reconnected so that Vt  Vdc , is the motor field connection still cumulative compound
or has it changed to differential compound?
(a) For generator action, I a reverses, but I f remains in the same direction. Therefore,
the field connection is differential compound.
(b) With diversion of armature current through the switch, the contribution of the
series field is lost and  p decreases in value. Thus,  m increases.
(c) Both I f and I a change direction; thus,  p changes directions. Since both  p and
I a reverse, Td remains in the direction of rotation, or rotation is unchanged.
(d) Since both I f and I a reverse, the fields still add and the motor remains cumulatively compound.
X5.2 For purposes of this analysis, the series dc motor of Fig. X5.2 is to be treated as
magnetically linear with negligible rotational and ohmic losses. The load torque ( TL )
has a constant value. If the motor is operating in steady-state when the switch (SW) is
1
moved from position 1 to position 2, (a) predict the new motor current and (b) the new
motor steady-state speed, both as a percentage of the prior quantities.
(a) By [5.15],
Td  K I a2  C1
Thus, I a does not change from the prior value.
(b) By [5.10] and [5.13],
m 
Vt
Vt

K  p KC I a
Since I a is constant,
m2 2VB

2
m1 VB
The new speed is 200% of the prior speed.
X5.3 Rework Problem X5.2 except let TL  kT m2 and all else is unchanged.
(a) Using [5.15],
TL  Td  kT m2  K I a2
Thus,
 m  C1I a
(1)
Based on [5.10],
m 
Vt
Vt

K  p KC I a
(2)
2
Equate (1) and (2) to see that
I a  C2 Vt
Thus,
I a 2 C2 2VB

 2
I a1
C2 VB
The current increases to 141% of the prior value.
(b) Based on (1),
 m 2 C1I a 2 C1 2 I a1


 2
 m1 C1I a1
C1I a1
The speed also increases to 141% of the prior value.
X5.4 For purposes of analysis, the shunt dc motor of Fig. X5.3 is to be treated as magnetically linear with negligible rotational losses and ohmic losses. The load torque ( TL )
has a constant value. If the motor is operating in steady-state when the switch (SW) is
moved from position 1 to position 2, (a) predict the new motor current and (b) the new
motor speed, both as a percentage of prior quantities.
(a) For magnetic linearity,
 p  K I f 
KVt
(1)
Rf
Equating load and developed torque and using (1),
 KVt
TL  C1  Td  K  p I a  K 
 Rf

Thus,
C2  2VB  I a 2
C2 VB  I a1


 I a  C2Vt I a

2I a 2
1
I a1
Armature current decreases to half of the prior value.
3
(b) Based on [5.10] and using (1),
m 
Vt

K p
Vt
 KVt
K
 Rf





Rf
K K
Speed is independent of Vt , or speed is unchanged.
X5.5 Rework Problem X5.4 except let TL  kT m2 and all else is unchanged. The solution is
more simple if part (b) is solved before part (a).
(b) For magnetic linearity,
 p  K I f 
KVt
(1)
Rf
Based on [5.10] and using (1),
m 
Vt

K p
Vt
 KVt
K
 Rf





Rf
K K
Speed is independent of Vt , or speed is unchanged.
(a) By equating load and developed torque and using (1),
 KVt
TL  Td  kT m2  K  p I a  K 
 Rf


 I a

or
 m2  CVt I a
Since speed is constant,
I a 2 Vt1 VB 1



I a1 Vt 2 2VB 2
The armature current changes to 50% of the prior value.
X5.6 The separately-excited dc motor of Fig. X5.4 was operated at no-load and the
following data were recorded:
 m  1000 /30 rad / s
I a  0.95 A
Vt  240 V
VB  150 V
The field voltage ( VB ) is unchanged, but the motor is loaded so that it supplies an
output power Ps  10 HP at 1000 rpm to a coupled mechanical load. At this load point,
determine (a) the rotational losses, (b) the armature current, (c) the terminal voltage,
and (d) the efficiency. Neglect armature reaction.
4
(a) Since the speed is unchanged, PFW can be found from the no-load data.
PFW  Vt I a  I a2 Ra   240  0.95   0.95  0.2   227.8 W
2
(b) For the no-load condition,
E  Vt  I a Ra  240   0.95 0.2  239.81 V
Then,
K p 
E
m

239.81
 2.29 V  s / rad
1000 / 30
The developed torque at the load point is
Td 
Pd
m

Ps  PFW
m

10  746   227.8  73.41 N  m
1000 / 30
Thus,
Ia 
Td
73.41

 32.06 A
K  p 2.29
(c) The load point terminal voltage is
Vt  E  I a Ra  239.81   32.06  0.2   246.22 V
(d) The total losses are
150   733.37 W
VB2
2
 227.8   32.06   0.2  
Rf
75
2
Losses  PFW  I a2 Ra 

Ps 100 
Ps  losses

10 746 100   91.05%
10 746  733.37
5
X5.7 A 230 V dc machine has an emf constant K  212.21 V  ´s / Wb  rad and Ra  0.278  .
The field is separately excited and produces 0.01 Wb per pole. I a  36 A for all parts
of the problem. (a) At what speed does the machine operate as a dc generator with
rated terminal voltage? (b) At what speed does the machine operate as a dc motor with
rated terminal voltage? (c) If an external resistance Rx  1  is added in series with the
armature circuit, at what speed does the machine operate as a dc motor with rated
terminal voltage? (d) If the motor of part (b) is known to be supplying 10 HP to a
coupled mechanical load, determine the rotational losses at the point of operation.
(a)
E  Vt  I a Ra  230   36  0.278  240 V
m 
E
240

 113.09 rad/s  1080 rpm
K  p  212.21 0.01
m 
230   36  0.278 
V I R
E
 t a a 
 103.67 rad/s  990 rpm
K p
K p
 212.21 0.01
(b)
(c)
m 
(d)
Vt  I a  Ra  Rx 
K p

230   36  0.278  1
 212.21 0.01
 86.7 rad/s  828 rpm
Pd  EI a  Vt  I a Ra  I a    230  36  0.2728 36   7919.7 W
PFW  Pd  Ps  7919.7  10  746   459.7 W
X5.8 A series dc motor that can be treated as magnetically linear is operating at 1200 rpm
with Vt  250 V and I a  50 A . It is known that Ra  Rs  0.2  . If torque is increased
until the speed is reduced to 750 rpm while Vt remains constant in value, determine
the new values for Td and I a .
Let 1 denote the 1200 rpm case and 2 denote the 750 rpm case.
E1  Vt  I a1  Ra  Rs   250  50  0.2   240 V
Based on [5.14],
K 
E1
I a1 m1

240
Vs
 0.0382
50 1200 / 30 
A  rad
E2  K I a 2m 2  Vt  I a 2  Ra  Rs 
6
Whence,
Ia2 
Vt
230

 73.01 A
K  m 2  Ra  RS  0.0382  750 / 30   0.2
By [5.15],
Td 2  K I a22   0.0382  73.01  230.62 N  m
2
X5.9 A shunt dc motor draws an input current I L  75 A while driving a coupled mechanical
load at a speed of  m  1000 / 30 rad/s . Efficiency at the point of operation is known
to be 92%. Also, Ra  0.1  and R f  200  . (a) Calculate the value of output power
to the mechanical load. (b) Determine the value of rotational losses at the point of
operation. (c) Find the value of developed torque.
(a)
Ps 

100
Pin 

100
Vt I L 
92
 250 75  17,250 W
100
(b)
Ia  IL 
Vt
250
 75 
 73.75 A
Rf
200
PFW  Vt I L  Ps 
Vt 2
 I a2 Ra
Rf
PFW   250  75   17, 250 
 250 2 
200
 73.75 0.1  643.6 W
(c)
Td 
Ps  PFW
m

17,250  643.6
 170.87 N  m
1000 / 30
X5.10 The magnetically linear, separately-excited dc motor of Fig. X5.5 has negligible
rotational losses and drives a constant torque coupled mechanical load. The motor is
operating at steady-state conditions. Assume that Vt I a Ra . The switch is moved
from position 1 to position 2. Determine the percentage change in motor speed and
armature current from the initial operating condition.
7
Let 1 denote the initial operating point and 2 denote the new operating point.
By [5.10],
 m1 
VB
K  p1
m2 
VB
VB
2VB


1
K  p 2 K 2  p1
K  p1


Thus, speed doubles, or the new speed is 200% of the initial speed.
Since TL is constant and TFW  0 , Td 2  Td 1 .
K  p2 Ia2  K
 12  p1  I a 2  K  p1I a1
Thus,
I a 2  2 I a1
The armature current doubles, or the new armature current is 200% of the initial
armature current.
X5.11 Rework Problem X5.10 except that the coupled mechanical load torque is described
by TL  kT  m .
The determination of speed follows the identical procedure of Problem X5.10
so that the new speed is 200% of the initial speed.
Ratio of torques with TFW  0 gives
Td 2 TL 2 kT m2


2
Td1 TL1 kT m1
8
Hence,
Td 2  2Td 1
K  p2 Ia2  K
 12  p1  I a2  2K  p1Ia1
or
I a 2  4 I a1
The new value of armature current is 400% of the initial value of armature current.
X5.12 A shunt dc motor has the OCC of Fig. 5.59. Vt  250 V and I f  1.35 A . The motor is
operating at 1200 rpm while supplying 25 HP to a coupled mechanical load. At the
point of operation, the rotational losses are 550 W. (a) Determine the value of
developed torque. (b) Calculate the value of armature current. (c) Find the value of
armature resistance. (d) Determine the value of efficiency at the point of operation.
(a)
Pd  Ps  PFW   25 746   550  19,200 W
Td 
Pd
m

19,200
 152.79 N  m
1200 / 30
(b) From Fig. 5.59 for I f  1.35 , E 235 V .
Ia 
Pd 19,200

 81.7 A
E
235
Ra 
Vt  E 250  235

 0.1836 
Ia
81.7
(c)
(d)

Ps 100%
Vt I L

 25 746100%  89.83%
 25081.7  1.35
X5.13 For the shunt dc motor of Problem X5.12, the mechanical load torque is increased so
that the speed is reduced from 1200 rpm to 1150 rpm. Vt and I f are unchanged.
Determine the values of (a) motor line current and (b) output power if it is known that
rotational losses vary as speed squared.
(a) Since field current is unchanged,
E
1150
 235  225.2 V
1200
Ia 
Vt  E 250  225.2

 135.08 A
Ra
0.1836
9
(b)
Ps  Pd  PFW  E I a  PFW
2
 1150 
Ps   225.2 135.08  
 550  30,925.1 W  41.45 HP
 1200 
X5.14 The OCC for the separately excited dc generator of Fig. X5.6 is given by Fig. 5.59.
(a) If I f  1 A ,  m  1500 / 30 rad/s , and the switch SW is open, find Vt . (b) If
I f  1.5 A ,  m  1200 / 30 rad/s , and the switch is closed, determine the value of I a .
(c) If the generator is operating as described in part (b) with PFW  400 W , calculate
the input mechanical power.
(a) From Fig. 5.59 with I f  1 A , E 190 V at 1200 rpm. Adjust to 1500 rpm and
note that I a  0 .
Vt  E 
1500
190  237.5 V
1200
(b) Enter Fig. 5.59 with I f  1.5 A to find E 242 V at 1200 rpm. By KVL,
Ia 
(c)
E
242

 23.27 A
Ra  RL 0.4  10
Pin  PFW  EI a  400   242  23.27   6031.3 W
X5.15 A shunt dc motor is fed from a 250 V dc bus. The motor is operating at 1200 rpm
supplying 30 HP to a coupled mechanical load. From an ammeter reading, the line
10
current to the motor is 102 A. It is known that Ra  0.2  and R f  125  . Determine
(a) the value of rotational losses and (b) the efficiency.
(a)
Ia  I L 
Vf
Rf
 102 
250
 100 A
125
E  Vt  I a Ra  250  100  0.2   230 V
Pd  EI a   230100   23,000 W
PFW  Pd  Ps  23,000   30  746   620 W
(b)

Ps 100%
Vt I L

 30 746100%  87.76%
 250102
X5.16 A shunt dc motor nameplate has the data 230 V, 1350 rpm, 10 HP, and 37.5 A. In
addition, it is known that Ra  0.35  , rated speed rotational losses are 519 W , and
field current is 0.75 A at rated conditions. (a) Determine developed torque, the
CEMF, and efficiency at rated conditions. (b) If the motor is operating at rated
conditions when the load torque is reduced until the line current drops to 20 A,
calculate the new speed and developed torque.
(a)
Td 
Ps  PFW
m

10  746   519  56.44 N  m
1350 / 30
I a  I L  I f  37.5  0.75  36.75 A
E  Vt  I a Ra  230   36.75 0.35  217.14 V

10 746 100%  86.49%
Ps
100% 


Vt I L
 230 37.5
(b) Since I f is unchanged,  p is unchanged. Let R denote rated condition.
I a  I L  I f  20  0.75  19.25 A
E  Vt  I a Ra  230  19.25 0.35  223.26 V
nm 
E
223.26
nmR 
1350  1388.1 rpm
ER
217.14
Td 
Ia
19.25
TdR 
 56.44  29.56 N  m
I aR
36.75
11
X5.17 In test of large dc machines, a suitable load may not be available for full power
operation. Frequently, a back-to-back test using two identical machines is used, as
shown schematically by Fig. X5.7 where one machine acts as a motor (M) to drive the
other machine as a generator (G). The shafts of the two machines are mechanically
coupled. An auxiliary dc source ( VB ) is used to supply all losses (except field winding
losses). Assume that rotational losses are negligible so that the losses supplied by VB


2
2
are I B2 RB  I aG
 I am
Ra . Both fields are separately excited; however,  G  1.05  M .
It is known that EG  750 V when  m  100 rad/s . It turns out that VB must vary
directly with speed to maintain operation. Determine the value of the ratio VB /  m for
this particular set-up.
K G 
EG
m

750
 7.5
100
The motor and generator developed torques are equal.
K G I aG  K 1.05   M  I aG  K  M I aM
or
I aM  1.05 I aG
(1)
By KVL,
EG  EM  K Gm  K  mm   I aG  I aM  Ra
(2)
Substitute  G  1.05  M , (1), and Ra  0.05  into (2) to find
 m  0.287 I aG
(3)
By KCL and subsequent substitution of (1),
I B  I aM  I aG  1.05 I aG  I aG  0.05 I aG
12
(4)
Since VB supplies the losses other than field losses,


2
2
VB I B  I B2 RB  I aG
 I aM
Ra
(5)
Use (1), (4), and Ra  0.05 in (5) to yield
VB  2.1075 I aG
(6)
Ratio (6) and (3) to give
VB
m

2.1075 I aG
Vs
 7.3432
0.287 I aG
rad
13