Page 658 - Basic Electrical Engineering

current is 80 A, the flux per pole is 21 mWb and armature resistance is 0.1 Ω.

Solution:

The back EMF induced in the armature of the motor is E .
b

For motor

V − E = I R a
a
b
or, E = V − I R a
b
a
= 500 − 80 × 0.1

= 492 V (ii)

Equating (i) and (ii)

Example 7.12 A dc machine induces an EMF of 240 V at 1500 rpm. Find

the developed torque for an armature current of 25 A.

Solution:

Power developed, P = E × I a

= 240 × 25

= 6000 W

Again, P = T × ω where ω is the angular velocity in radians per second.