Discussion
The emf of a dc generator is doubled by
Answer: A
Correct answer: doubling either the flux per pole or the speed
From \( E = P\Phi ZN/60A \), the emf is directly proportional to both the flux per pole and the speed, so doubling either doubles it. Doubling the number of parallel paths would halve it, since \( A \) is in the denominator. Armature resistance and load current do not enter the generated emf at all, though they determine the terminal voltage through the \( I_aR_a \) drop.
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