A 4-pole wave wound DC motor drawing an armature current of 20

a-4-pole-wave-wound-dc-motor-drawing-an-armature-current-of-20

A 4-pole wave wound DC motor drawing an armature current of 20 A has provided with 360 armature conductors. If the flux per pole is 0.015 Wb then the torque developed by the armature of motor is _______

A. a. 10.23 N-m

B. b. 34.37 N-m

C. c. 17.17 N-m

D. d. 19.08 N-m

This question was posed to me by my college professor while I was bunking the class.

My question is taken from EMF and Torque Production topic in division Circuit Model, Emf and Torque of DC Machines

The correct option is B. b. 34.37 N-m

For explanation I would say: DC Machine torque equation: T = ka*∅*Ia. Here, ka= ZP/(2πA), Z= total armature conductors, P= No. of poles, A= No. of parallel paths. For a wave winding A=2. So, substituting all the values in the torque equation we get torque equal to 34.37 N-m.