Page 214 - Solutions Manual to accompany Electric Machinery Fundamentals

Therefore, the current per path is
I 120 A
I  A   15 A
a 8

(b) Duplex lap-wound:
a  mP    8  16 paths
2
Therefore, the current per path is
I 120 A
I  A   7.5 A
a 16

(c) Simplex wave-wound:
a  2m    1  2 paths
2
Therefore, the current per path is
I 120 A
I  A   60 A
a 2

7-5. How many parallel current paths will there be in the armature of an 20-pole machine if the armature is (a)
simplex lap-wound, (b) duplex wave-wound, (c) triplex lap-wound, (d) quadruplex wave-wound?

SOLUTION
(a) Simplex lap-wound:

a  mP  (1)(20) 20 paths
(b) Duplex wave-wound:
a  2m  (2)(2)  4 paths

(c) Triplex lap-wound:

a  mP  (3)(20) 60 paths
(d) Quadruplex wave-wound:
a  2m  (2)(4) 8 paths

7-6. The power converted from one form to another within a dc motor was given by
P conv  E I  AA  
m
ind
Use the equations for E A and  ind [Equations (7-38) and (7-49)] to prove that EI =   ; that is,
m
ind
AA
prove that the electric power disappearing at the point of power conversion is exactly equal to the
mechanical power appearing at that point.
SOLUTION
P conv  E I
AA
Substituting Equation (8-38) for E
A
 P K   I
conv A
  P K   I 
conv A

But from Equation (7-49),   K I , so
ind A
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