Question

At 1200 rpm the induced emf of a dc machine is 200 V. For an armature current of 15 A the electromagnetic torque produced would be:
1. 23.8 N-m
2. 238 N-m
3. 2000 N-m
4. 3000 N-m

Answer 1

Correct Answer - Option 1 : 23.8 N-m

Concept:

The electromagnetic torque produced in a DC machine is given by

\(T = \frac{P}{\omega } = \frac{{E{I_a}}}{\omega }\) 

where ω = speed in rad/sec

E is induced emf

I_a is armature current

Calculation:

Given that, E = 200V, I = 15A, N = 1200 rpm

\(T = \frac{{EI}}{\omega } = \frac{{EI}}{{\frac{{2\; \times \;\pi\; \times\; N}}{{60}}}}\)

\(= \frac{{200\; \times \;15}}{{2\; \times \;\pi\; \times \;\frac{{1200}}{{60}}}} = 23.87Nm\)