Interpret average current across changing intervals
Combine interval charge amounts before dividing by total time, including a supplied signed reversal.
Section 4.2.2 · Electric current
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
- Use Q = It for constant intervals.
- Add signed amounts and durations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What charge passes at 3 A for 2 s?
What you will learn
- Calculate a time-weighted interval average.
- Distinguish signed net passage from total movement in a reversing current.
Weight each rate by its duration
For piecewise constant current, calculate the charge in each interval and divide their total by the whole duration. If 1 A flows for 4 s and 3 A for 2 s in the same direction, the charges are 4 C and 6 C. Their total 10 C in 6 s gives an average of 10/6 A, about 1.67 A.
The arithmetic mean of the two rates is 2 A, but that is wrong here because the lower rate lasted twice as long. Use total charge divided by total time rather than averaging labels on a graph. A zero-current interval still contributes time to the denominator.
Use a stated sign reference
For a reversal, positive and negative current values represent opposite charge-passage directions relative to the chosen conventional reference. +2 A for 3 s and −1 A for 2 s give signed passages +6 C and −2 C, so the net is +4 C over 5 s and the signed average +0.80 A.
The sum of the magnitudes of those passages is 8 C, a different quantity from the signed net 4 C. Neither a zero signed average nor a smaller net passage means no charge moved. Keep the direction reference and the requested 'net' or 'total magnitude' wording explicit.
Pause and explain
1 A lasts 4 s and 3 A lasts 2 s. What is the supported average?
Worked example
A current is 1 A for 4 s and 3 A for 2 s, in one direction. Find the whole-interval average.
Show the worked solution
- Charges are 1 × 4 = 4 C and 3 × 2 = 6 C.
- Total charge is 10 C and total time 6 s.
- Average current = 10/6 A.
Answer 10/6 A ≈ 1.67 A.
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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.
Find the average for 1 A lasting 4 s then 3 A lasting 2 s in one direction.
Give me a hint
Calculate each charge and divide their sum by total time.
Compare my reasoning
- Q₁ = 4 C and Q₂ = 6 C.
- Total Q = 10 C and t = 6 s.
- Average I = 10/6 A.
10/6 A ≈ 1.67 A.
Look for these in your work
- Use duration weights.
- State the whole interval.
2 A flows for 3 s, followed by zero current for 2 s. Find the 5 s average.
Give me a hint
The inactive interval adds time but no passed charge.
Compare my reasoning
- Passed charge = 2 × 3 = 6 C.
- Total duration = 3 + 2 = 5 s.
- Average current = 6/5 = 1.2 A.
1.2 A.
Look for these in your work
- Include the zero-current interval.
- Divide by all 5 seconds.
On a stated reference, current is +2 A for 3 s then −1 A for 2 s. Find net charge, signed average and sum of passage magnitudes.
Give me a hint
Retain signs for the net, then separately add absolute passage amounts.
Compare my reasoning
- Signed charges are +6 C and −2 C, giving +4 C.
- Signed average = +4/5 = +0.80 A.
- Magnitude sum = 6 + 2 = 8 C.
Net +4 C; signed average +0.80 A; magnitude sum 8 C.
Look for these in your work
- Separate signed and magnitude accounts.
- Use the same 5 s denominator.
Common mistakes
- Weight rates by their durations.
- Include zero-current time.
- Signed net passage differs from a sum of magnitudes.
What can you explain now?
When is the simple arithmetic mean of two constant rates valid for their combined interval?
Compare with the explanation
When their durations are equal.
Equal durations give the two rates equal time weights in the total-charge calculation.
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Make it stick
Combine interval charge amounts and durations while distinguishing signed net and total magnitude passages.
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.