4.4.4Electrons, Waves and Photons
E.m.f and Potential Difference — OCR A Level Physics A H556 Revision Notes & Topical Questions
5 min read 5 parts Lesson 37 of 62 Unit quiz
Part 1 of 5The Key Difference
E.m.f. and p.d. share a unit but describe opposite energy conversions.
E.m.f.
- Energy transferred to each coulomb from chemical, mechanical or other forms into electrical energy as charge passes through the source.
P.d.
- Energy transferred from electrical energy into other forms such as heat or light, per coulomb passing through a component.
Direction of transfer
- E.m.f. involves charge gaining energy at the source, whereas p.d. involves charge losing energy in the load.
Unit
- Both quantities are measured in volts, meaning joules per coulomb.
Part 2 of 5Comparison Table
Comparison Table
Source
- E.m.f. belongs to a source, such as a cell, dynamo or solar cell; p.d. exists across any component.
Open circuit
- With no current, the p.d. across the terminals equals the e.m.f., but there is no p.d. across a connected resistor.
Closed circuit
- With current flowing, terminal p.d. is less than the e.m.f. by the lost volts if internal resistance is present.
Sum rule
- In a closed loop, the sum of e.m.f.s equals the sum of p.d.s, which is Kirchhoff's second law.
| Feature | E.m.f. | Potential difference |
|---|---|---|
| Energy conversion | Other forms → electrical | Electrical → other forms |
| Where found | Across a source | Across a component |
| Affected by current | Not affected | Changes with current and resistance |
Part 3 of 5Applying the Ideas
1
Measure the e.m.f. by connecting a high-resistance voltmeter across the cell with no load connected.
2
Connect a load and note that the voltmeter now reads the smaller terminal p.d.
3
The difference between the two readings is the lost volts, equal to the current multiplied by r.
4
Doubling the current doubles the lost volts for the same cell, so the terminal p.d. falls further.
5
Explain a dimming headlamp when a starter motor draws a large current: the lost volts in the battery rise.
Part 4 of 5Worked Example
Question
- A car battery of e.m.f.
- 12 V and internal resistance 0.40 Ω supplies a current of 5.0 A.
- Find the terminal p.d.
- Lost volts = Ir = 5.0 × 0.40 = 2.0 V.
- Terminal p.d. = ε − Ir = 12 − 2.0 = 10 V.
Answer Terminal p.d. = 10 V.
Part 5 of 5Exam Checklist & Tips
4.4.4 – Key Points for the Exam
- E.m.f. The electrical energy gained per coulomb of charge passing through a source.
- Potential difference The electrical energy transferred to other forms per coulomb across a component.
- Terminal p.d. The p.d. across a source's terminals while it supplies current.
- Open circuit A circuit with a gap so that no current flows.
Common Mistakes
- Writing that e.m.f. and p.d. are the same because both are measured in volts.
- Claiming the terminal p.d. is always equal to the e.m.f.
- Forgetting that lost volts depend on the current, so terminal p.d. changes with load.
- Describing e.m.f. as a force that pushes charge.
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