Read generator periods and compare rotation changes
Read a complete repeat interval and distinguish time-cycle changes from unspecified amplitude changes.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Use frequency as cycles per second.
- Convert milliseconds to seconds.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many seconds are 40 ms?
What you will learn
- Read a complete repeat interval and distinguish time-cycle changes from unspecified amplitude changes.
- Read equivalent cycle phases and convert units, while keeping pole count and amplitude assumptions explicit.
A full cycle repeats the same phase
For the supplied simple two-pole generator, one revolution gives one e.m.f. cycle. The period T is the time between equivalent phases, such as adjacent positive peaks. A positive peak to the next negative trough spans half a period, not a whole cycle. Frequency is f = 1/T with T in seconds.
If positive peaks occur at 10 ms and 50 ms, T = 40 ms = 0.040 s and f = 25 Hz. At constant speed, quarter-cycle points are 10 ms apart. Use explicit time units; treating forty milliseconds as forty seconds gives a very different frequency.
Hold the amplitude conditions separately
Doubling the rotation rate in the same ideal two-pole arrangement halves the period and doubles frequency. Faster change can also increase induced e.m.f. magnitude with geometry and field held fixed. However, a timing statement alone does not specify a numerical new peak for an unspecified real generator or changed field conditions.
The supplied comparison graph uses the same peak amplitude only to isolate timing. It is a graph-reading illustration, not a claim that speed doubling always leaves physical generator amplitude unchanged. Multiple pole-pair machines require their specified electrical-to-mechanical cycle relationship.
Sources for this lesson
Pause and explain
A positive peak to the next trough takes 20 ms in the supplied sine-like graph. What is the period?
Worked example
Adjacent positive peaks occur at 10 ms and 50 ms. Find period and frequency.
Show the worked solution
- T = 50 − 10 = 40 ms.
- Convert T to 0.040 s.
- f = 1/.040 = 25 Hz.
Answer Period 40 ms; frequency 25 Hz.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Positive peaks occur at 10 ms and 50 ms. Calculate T and f.
Give me a hint
Use equivalent phases and seconds in 1/T.
Compare my reasoning
- The interval is 40 ms.
- It equals .040 s.
- Frequency is 25 Hz.
40 ms and 25 Hz.
Look for these in your work
- Use adjacent equivalent phases.
- Convert the units.
A supplied two-pole generator changes from 25 to 50 revolutions per second. Find electrical frequencies and periods.
Give me a hint
Use one cycle per revolution for this stated arrangement.
Compare my reasoning
- Frequencies are 25 Hz and 50 Hz.
- Their periods are 1/25 and 1/50 s.
- Periods are .040 s and .020 s.
25 Hz/.040 s before; 50 Hz/.020 s after.
Look for these in your work
- Keep the two-pole assumption.
- Use seconds for period.
Two illustration curves have equal peaks but periods .040 s and .020 s. What do they establish, and what do they not prove about a physical generator?
Give me a hint
Separate drawn timing from the unprovided amplitude conditions.
Compare my reasoning
- The second frequency is twice the first.
- The drawing assigns equal peak amplitudes.
- It does not prove speed doubling leaves a real generator peak unchanged.
The second cycle is twice as fast; equal drawn peaks are an illustration condition, not a universal amplitude law.
Look for these in your work
- Read both periods.
- Retain the graph’s limited claim.
Common mistakes
- Adjacent unlike extrema span half a cycle.
- Convert milliseconds to seconds.
- A timing-only graph does not establish physical amplitude.
What can you explain now?
Which interval gives a complete period directly?
Compare with the explanation
One positive peak to the next positive peak
Equivalent adjacent phases repeat the full cycle.
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Make it stick
Read equivalent cycle phases and convert units, while keeping pole count and amplitude assumptions explicit.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.