Find source work or transferred charge from e.m.f.
Rearrange source e.m.f. = W/Q while keeping per-charge labels, total work and open-circuit cases distinct.
Section 4.2.3 · Electromotive force and potential difference
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
- Calculate source e.m.f. = W/Q.
- Rearrange a product with consistent units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What quantity results from V multiplied by C?
What you will learn
- Use W = e.m.f. × Q and Q = W/e.m.f.
- Interpret a source voltage without treating it as a fixed total energy.
Recover a total from the per-charge value
The source equation rearranges to W = e.m.f. × Q. A stated constant 9 V e.m.f. supplying work for 2 C gives 18 J. For a varying source account, one voltage reading does not automatically give the average work per charge for all the charge passed.
To find charge, Q = W/e.m.f. for a non-zero stated e.m.f. If a 4 V source does 12 J of electrical work, the matching passed charge is 3 C. Retain the source-work boundary rather than substituting only the useful output of one component.
Retain the model conditions
A voltage label is a per-charge value, so doubling Q at the same stated e.m.f. doubles W. It does not double e.m.f. A source with non-zero e.m.f. and zero passed charge can supply zero work during that open-circuit interval without losing its e.m.f.
If internal source losses or different time-dependent operating values are relevant, use the supplied complete source account and matching per-charge value. A numerical work result does not alone identify chemical capacity, useful energy efficiency or electrical power without additional information.
Pause and explain
A 4 V source does 12 J of electrical work. What charge corresponds to this account?
Worked example
A source has constant e.m.f. 9 V and 2 C passes in its stated source account. Find the electrical work.
Show the worked solution
- Use W = e.m.f. × Q.
- Multiply 9 J/C by 2 C.
- W = 18 J.
Answer 18 J of source electrical work.
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.
Find source work for constant e.m.f. 9 V and passed charge 2 C.
Give me a hint
Multiply the source work per coulomb by the charge amount.
Compare my reasoning
- W = e.m.f. × Q.
- 9 × 2 = 18.
- The result is source work in joules.
18 J.
Look for these in your work
- Use the matching charge.
- State the source account.
Find source work at 6 V for 0.50 C and charge for 12 J at 4 V.
Give me a hint
Use a product for work and a quotient for charge.
Compare my reasoning
- W = 6 × 0.50 = 3 J.
- Q = 12/4 = 3 C.
- The results name different quantities.
3 J; 3 C.
Look for these in your work
- Select the right rearrangement.
- Keep work and charge units distinct.
A 9 V source has no charge pass during an open interval. Find the interval work under this model and assess a claim that its e.m.f. must therefore be zero.
Give me a hint
Use zero Q in the product without replacing a known source value.
Compare my reasoning
- W = 9 × 0 = 0 J.
- The source's stated e.m.f. is still 9 V.
- Zero transfer during the interval does not erase the source property.
0 J during the interval; zero e.m.f. does not follow.
Look for these in your work
- Separate interval work and source property.
- Avoid dividing an uninformative 0/0 account.
Common mistakes
- A per-charge value needs a matching charge amount.
- Use source work rather than one useful output.
- One instantaneous voltage need not represent a varying whole account.
What can you explain now?
At a fixed e.m.f., what happens to source work when the passed charge doubles?
Compare with the explanation
It doubles.
The fixed work-per-charge value multiplies the doubled charge amount.
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Make it stick
Find source work or charge and retain the distinction between an interval transfer and a source per-charge value.
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.