Calculate component potential difference from work and charge
Use V = W/Q for a stated component account and rearrange for work or passed charge with consistent units.
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
- Define component p.d.
- Rearrange work-per-charge relationships.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does p.d. refer to?
What you will learn
- Use V = W/Q for component p.d.
- Find component work or charge and distinguish this boundary from source work.
Use the component work account
For a stated component, V = W/Q: p.d. equals the work done as charge passes through the component divided by that charge. With W in J and Q in C, V is in volts. A lamp transferring 12 J for 3 C has a p.d. of 4 V across its terminals.
The equivalent rearrangements are W = VQ and Q = W/V for a non-zero given voltage. If V is constant or a matching work-per-charge average, 5 V across a component for 0.20 C corresponds to 1.0 J. A single varying-voltage observation does not define the whole passed-charge account.
Choose the correct boundary
Use total electrical work transferred in the named component account, not only a useful light output unless the question says these are the same. For a lamp transferring 12 J electrically while 3 C passes, of which 3 J leaves as useful light, its p.d. is 4 V, not 1 V. The smaller quotient describes the useful light share per charge.
Convert prefixes separately and state the component and its two points. A zero-charge, zero-work interval gives no unique V through division; it does not prove zero p.d. across an open component. Voltage, current and total work remain different quantities.
Pause and explain
12 J is transferred electrically in a lamp for 3 C, with 3 J useful light. What p.d. follows?
Worked example
A lamp transfers 12 J electrically when 3 C passes. Only 3 J is useful light. Find the lamp p.d.
Show the worked solution
- The p.d. account uses total electrical component work, 12 J.
- Divide by the matching 3 C.
- V = 12/3 = 4 V.
Answer 4 V, not the 1 V useful-light-per-charge quotient.
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.
Find component p.d. when 12 J transfers electrically for 3 C passing.
Give me a hint
Use the complete component work divided by matching charge.
Compare my reasoning
- V = W/Q.
- 12/3 = 4.
- The p.d. is 4 V.
4 V.
Look for these in your work
- Match work and charge.
- Use volt units.
Find work for a constant 5 V p.d. and 0.20 C, then charge for 8 J at 4 V.
Give me a hint
Use W = VQ and Q = W/V in their respective accounts.
Compare my reasoning
- W = 5 × 0.20 = 1.0 J.
- Q = 8/4 = 2 C.
- Both use their specified component boundaries.
1.0 J; 2 C.
Look for these in your work
- Choose the correct rearrangement.
- Keep the two accounts separate.
A component takes 20 J electrically for 4 C, but only 8 J is useful output. Compare its p.d. with the useful-output-per-charge value.
Give me a hint
Calculate the complete input account and the useful share separately.
Compare my reasoning
- Component p.d. = 20/4 = 5 V.
- Useful output per charge = 8/4 = 2 J/C.
- The useful quotient is not the complete component p.d.
P.d. 5 V; useful share 2 J/C, which is not the complete p.d. account.
Look for these in your work
- Use total electrical work for p.d.
- Identify the smaller useful-energy share.
Common mistakes
- Use the component's complete electrical work account.
- Useful output alone may be only one share.
- Keep volts, joules and coulombs distinct.
What can you explain now?
What is the equation for component p.d. using work and charge?
Compare with the explanation
V = W/Q.
The component work must match the passed charge and the chosen terminal boundary.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Calculate p.d., work or charge using the component's complete matching electrical account.
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.