Discussion
A separately excited dc motor develops 50 N·m at 30 A. At 45 A with the same flux it develops
Answer: C
Correct answer: 75 N·m
With the flux constant, \( T = k\Phi I_a \) makes torque proportional to armature current:
\[ T = 50 \times \frac{45}{30} = 75 \text{ N·m} \]The relation is linear, not square-law - the 112 N·m answer squares the ratio, which would be right only for a series machine, where the flux itself rises with the current.
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