SSE JE Electrical Engineering 28 Oct 2020 Shift 1 Papers

Question : 186 of 200

Marks: +1, -0

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.

Solution:

\ω={2 \π \N }/{60}

radians per sec.

={2×\π×1500}/60

= 157.08 radians/sec.
Power = generated emf × armature current = 250 × 50 = 12500 watts.

\text "Torque "= \text "power"/\text "speed" =12500/157.08

= 79.577 Nm.

Go to Question:

Prev Question

Next Question