Sketch and read a simple generator voltage graph
Interpret zeros, signs, peak magnitude and repeated cycles under a stated ideal phase and terminal convention.
Section 4.5.2 · The a.c. generator
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Read signed graph axes.
- Distinguish peak magnitude from a difference between values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the magnitude of −6 V?
What you will learn
- Interpret zeros, signs, peak magnitude and repeated cycles under a stated ideal phase and terminal convention.
- Read graph quantity, terminal sign, peak size and cycle interval before inferring any load behaviour.
Label the axes and conventions
Sketch e.m.f. vertically in volts and time horizontally with a stated unit. For the supplied constant-speed, uniform-field two-pole model, take t = 0 at a zero crossing rising toward a positive peak. The graph is sinusoidal with equal-magnitude positive and negative peaks. Swapping output terminals reverses the vertical sign, not the rotation rate.
The supplied graph has peak magnitude 6 V and period 0.040 s: zero at 0, +6 V at 0.010 s, zero at 0.020 s, −6 V at 0.030 s and zero at 0.040 s. The positive peak and negative trough have the same magnitude; −6 V is not a smaller absolute voltage than +6 V.
Read voltage without inferring current
The graph describes induced e.m.f., not mechanical force, flux or load current. Current in a supplied purely resistive closed load can follow the same sign pattern, but current magnitude requires its resistance and other circuit conditions. An open output still permits a voltage graph.
A peak-to-peak difference between +6 V and −6 V is 12 V; peak magnitude is 6 V. Do not substitute either for a direct-current voltage or claim rms voltage without a separately justified relationship. This lesson does not require an rms calculation.
Pause and explain
Output terminals are exchanged while rotation is unchanged. What happens to the voltage graph?
Worked example
A graph reaches +6 V and −6 V and repeats every 0.040 s. State peak magnitude and peak-to-peak difference.
Show the worked solution
- The absolute maximum from zero is 6 V.
- The full vertical difference is 6 − (−6) = 12 V.
- The repeat interval is the supplied period.
Answer Peak magnitude 6 V; peak-to-peak 12 V; period 0.040 s.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Sketch one supplied 6 V peak, 0.040 s sine-like cycle starting at zero and rising.
Give me a hint
Mark quarter-cycle times before drawing a smooth curve.
Compare my reasoning
- Mark 0, .010, .020, .030 and .040 s.
- Assign 0, +6, 0, −6 and 0 V.
- Connect smoothly and label axes e.m.f./V and time/s.
A labelled alternating curve through the five stated points.
Look for these in your work
- Use correct sign and times.
- Keep volts on the vertical axis.
For peaks +8 V and −8 V, find peak magnitude and peak-to-peak difference.
Give me a hint
Compare maximum distance from zero with the whole vertical span.
Compare my reasoning
- Peak magnitude is 8 V.
- Span is 8 − (−8).
- The difference is 16 V.
Peak magnitude 8 V; peak-to-peak 16 V.
Look for these in your work
- Separate both quantities.
- Use voltage units.
A supplied open-output graph has alternating ±6 V peaks. A report calls it ±6 A current and says terminal reversal doubles frequency. Correct both errors.
Give me a hint
Read the vertical quantity and identify what terminal swapping changes.
Compare my reasoning
- The labelled quantity is e.m.f. in volts.
- An open output need not have sustained load current.
- Terminal reversal changes sign, not the cycle timing.
It is an e.m.f. graph; swapping terminals reverses sign while frequency remains unchanged.
Look for these in your work
- Preserve voltage units.
- Separate polarity and timing.
Common mistakes
- Keep graph quantity and units explicit.
- Peak and peak-to-peak differ.
- Terminal swapping reverses sign, not frequency.
What can you explain now?
What does a negative e.m.f. value mean on the supplied graph?
Compare with the explanation
Opposite terminal polarity to the chosen positive convention
The sign gives orientation relative to fixed terminals, not negative energy existence.
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Make it stick
Read graph quantity, terminal sign, peak size and cycle interval before inferring any load behaviour.
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.