Question

A DC shunt machine develops an AC EMF of 250 V at 1500 rpm. Find the torque developed for an armature current of 50 A.
1. 59.6 N-m
2. 79.6 N-m
3. 69.6 N-m
4. 49.6 N-m

Answer 1

Correct Answer - Option 2 : 79.6 N-m

Concept:

Torque:

• When armature conductors of a DC motor carry current in the presence of stator field flux, a mechanical torque is developed between the armature and the stator.
•  Torque is given by the product of the force and the radius at which this force acts.

Power:

• Mechanical power developed by the motor is given by P_m = E_b × I_a
• E_b is EMF developed and I_a is armature current
• In terms of torque and speed mechanical power developed is given by P_m = torque × speed.
• Speed is in radians per second (ω).
• \({\rm{\omega }} = 2\;\pi \frac{N}{{60}}\) radians per sec.

Calculations:

\({\rm{\omega }} = \frac{{2\; \times \;{\rm{\pi }}\; \times \;1500}}{{60}}\)

= 157.08 radians/sec.

Power = generated emf × armature current

= 250 × 50

= 12500 watts.

Torque = power/speed.

= \(\frac{{12500}}{{157.08}}\)

= 79.577 Nm.