Question

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

Answer 1

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.