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
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
- 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.
Sources for this lesson
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.
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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.