Page 650 - Basic Electrical Engineering

Figure 7.28

The armature winding is wave wound. The number of parallel paths is 2.

That is, all the armature conductors are connected in such a way that half the
armature current flows through each path. Thus, current flowing though each

armature conductor will be I /2 i.e., = 7.8 Amps. E is the EMF induced
a

in the armature. A voltage drop of I R takes place in the armature winding
a a
when it is supplying current. The remaining voltage, V is available across the

load terminals.

Thus,

E − I R = V
a
a
or, E = V + I R = 220 + 15.6 × 0.1 = 218.44 V
a
a

Example 7.5 A four-pole, 12 kW, 240 V dc generator has its armature coils
wave connected. If the same machine is lap connected, all other things

remaining constant, calculate the voltage, current, and power rating of the
generator.

Solution:

In a wave winding all the armature coils are arranged in two parallel paths.
The current-carrying capacity of each conductor, therefore, will be half of the

total armature current.