Simple one-loop direct-current generator: at 90 revolutions per second, how many times per second does the rectified dc output reach its maximum?

Introduction / Context:
A basic single-loop generator produces an alternating voltage per mechanical revolution. When a commutator rectifies this to direct current (pulsating dc), the waveform retains two peaks per revolution because each half-cycle of the original AC appears as a positive hump. Recognizing this helps interpret ripple frequency in simple dc machines.

Given Data / Assumptions:

• One-loop generator with a two-segment commutator (basic dc generator model).
• Mechanical speed is 90 revolutions per second (rev/s).
• We count positive maxima after rectification.

Concept / Approach:
Before rectification, one mechanical revolution produces one complete sinusoidal cycle of induced voltage. After commutation, the negative half-cycle is inverted, so both half-cycles appear as positive pulses. That yields two maxima per mechanical revolution in the pulsating dc output.

Step-by-Step Solution:

Verification / Alternative check:
Scope a model dc generator: for a single loop and two-segment commutator, you will observe two ripples per revolution. Adding coils/commutator segments increases pulse number and smooths ripple, matching machine theory.

Why Other Options Are Wrong:

• 90: Would be correct for the unrectified AC peak per period, not the rectified dc output with two peaks per revolution.
• 270 and 360: These imply three or four peaks per revolution, which do not arise with a single loop and a two-segment commutator.

Common Pitfalls:

• Confusing mechanical revolutions with electrical ripples after commutation.
• Assuming multi-coil effects; the question states a basic one-loop generator.

Final Answer:
180