# Extra Examples about DC Motors ```Example 1: A 230V dc shunt motor drives a load at 900RPM and drawing a current of 30A. The resistance
of armature circuit is 0.4Ω. The torque of the load is proportional to the speed. Calculate the resistance to be
connected in series with the armature to reduce the speed to 600RPM. Ignore armature reaction.
(ex.6.7p.158)
SOLUTION
Case I:
Supply voltage:
Armature current:
Speed:
Back emf:
Case II:
Supply voltage:
Armature current:
Speed:
Back emf:
V T = 230 V
I a1 = 30 A
(since no information is provided about the field circuit, otherwise this
current should be considered as the total current withdrawn from the
source, I T )
n1 = 900 RPM
E b 1 =V T − I a1R a = 230 − 30 &times; 0.4 = 218 V
E b 1 = k b φ n1 = k b φ 900
V T = 230 V
I a2 = ?
n 2 = 600 RPM
E b 2 =V T − I a 2 ( R a + R ) = 230 − I a 2 ( 0.4 + R )
E b 2 = k b φ n 2 = k b φ 600
Since the load is proportional to the speed ( τ ∝ n ) and ( τ ∝ I a ) then ( n ∝ I a ).
I φ n
n
600
Hence, a 2 = 2
or
I a 2 = I a1 &times; 2 = 30 &times;
= 20A
n1
900
I a1φ n1
E b 2 =V T − I a 2 (0.4 + R ) = 230 − 20(0.4 + R ) = 222 − 20R
⇒
E
n
222 − 20R 600
=
Since 2 = b 2 (magnetic flux is constant)
⇒
n1 E b 1
198
900
or
R = 3.833Ω
Example 2: A 250V dc shunt motor has an armature current of 20A when running at 1000RPM against full
load torque. The armature resistance is 0.5Ω. What resistance must be inserted in series with the armature to
reduce the speed to 500RPM at the same load torque, and what will be the speed if the load torque is halved
with this inserted resistance? (ex.6.8p.159)
SOLUTION
Case I:
Supply voltage:
Armature current:
Speed:
Back emf:
Case II:
Supply voltage:
Armature current:
Speed:
Back emf:
V T = 250 V
I a1 = 20 A
n1 = 1000 RPM
E b 1 =V T − I a1R a = 250 − 20 &times; 0.5 = 240 V
E b 1 = k b φ n1 = k b φ 1000
V T = 250 V
I a 2 = 20 A
n 2 = 500 RPM
E b 2 =V T − I a 2 ( R a + R ) = 250 − 20 ( 0.5 + R ) = 240 − 20R
E b 2 = k b φ n 2 = k b φ 500
Since
n2 Eb 2
(magnetic flux is constant)
=
n1 E b 1
Case III
Supply voltage:
Armature current:
V T = 250 V
I a 3 = 10 A
Speed:
n3 = ?
Back emf:
⇒
240 − 20R
500
=
240
1000
or
R = 6Ω
(since the load torque is halved:
τ Load = k m I a1φ
1
1 I a3
τ Load = k m I a 3φ
⇒
=
)
2
2 I a1
E b 3 =V T − I a 3 ( R a + R ) = 250 − 10 ( 0.5 + 6 ) = 185 V
E b 3 = k bφ n3
Since
n3 E b 3
(magnetic flux is constant)
=
n1 E b 1
⇒
n
185
= 3
240 1000
or
n 3 = 771 RPM
and
Example 3: A 250V dc shunt motor has an armature resistance of 0.5Ω and a field resistance of 250Ω.
When driving a constant load at 600RPM, the armature takes 20A. If it is desired to raise the speed from
600RPM to 800RPM, what resistance must be inserted in the shunt field circuit so that the change of speed
occurs? Assume the magnetizing curve is a straight line. (ex.6.1p.151)
SOLUTION
Case I:
Supply voltage:
Armature current:
Speed:
Back emf:
V T = 250 V
I a1 = 20 A
n1 = 600 RPM
E b 1 =V T − I a1R a = 250 − 20 &times; 0.5 = 240 V
E b 1 = k b φ1 n1 = k b φ1 600
Case II:
Now let the magnetic flux and armature current be φ 2 and I a 2 , respectively.
Supply voltage:
V T = 250 V
Armature current:
I a2 = ?
Speed:
n 2 = 800 RPM
Back emf:
E b 2 =V T − I a 2 R a = 250 − 0.5I a 2
E b 2 = k b φ2 n 2 = k b φ2 800
∴ τ Load = k m I a 2φ 2 = k m I a1φ 1
and
n 2 E b 2 φ1
=
&times;
n1 E b 1 φ2
since I a 2
⇒
φ
= 20 1
φ2
I a 2 = I a1
or
φ1
φ
= 20 1
φ2
φ2
800 250 − 0.5I a 2 φ1
=
&times;
φ2
600
240
or
or
⎛
φ ⎞ φ
320 = ⎜ 250 − 10 1 ⎟ &times; 1
φ2 ⎠ φ2
⎝
2
⇒
⎛ φ1 ⎞
φ1
⎜ ⎟ − 25 + 32 = 0
φ2
⎝φ 2 ⎠
φ 1 25 &plusmn; 252 − 4 &times; 32
=
= 23.65
φ2
2
or
⇒
we have a second order equation
1.35
The value 23.65 is rejected as it will not give the required increase in speed. Since magnetizing curve is a
straight line
φ1 I sh 1 V T / R sh 1
=
=
φ2 I sh 2 V T / R sh 2
R sh 2 φ1
= = 1.352
⇒
R sh 1 φ2
∴
since in general
or
V T = I sh R sh
and
φ = c I sh
where c = const .
R sh 2 = 1.353 &times; R sh 1 = 1.353 &times; 250 = 338Ω
Hence, the resistance required to be inserted in the shunt field is R = R sh 2 − R sh 1 = 338 − 250 = 88Ω .
```