Simple one-loop direct-current generator: at 90 revolutions per second, how many times per second does the rectified dc output reach its maximum?
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A90
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B180
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C270
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D360
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E45
Answer
Correct Answer: 180
Explanation
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:
Relate mechanical speed to AC cycles: 1 cycle/rev.Apply commutation: two positive humps per cycle (half-cycles inverted).Compute peaks per second: 2 * 90 = 180 peaks per second.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