Potential difference and electromotive force

Key idea: Voltage is energy transferred per unit charge. E.m.f. describes energy supplied to charge; potential difference describes energy transferred from electrical stores.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 2 of 3

Check your understanding

By the end of this lesson, you should be able to

Use V = W/Q in calculations.
Distinguish e.m.f. and p.d. using energy rather than calling either a force.
Interpret source and component voltages in joules per coulomb.

Learn the idea

Big question: What does a volt tell you about each coulomb as it moves through a circuit?

Interpret voltage as joules per coulomb

Potential difference V = W/Q measures energy transferred per unit charge between two points. A p.d. of 6 V means 6 J is transferred for every coulomb passing through the component.

Voltage is not a substance that flows and e.m.f. is not a mechanical force. Both p.d. and e.m.f. are measured in volts because both compare energy with charge.

Check your understanding: What does 12 V across a motor mean?

The motor transfers 12 J of electrical energy for each coulomb passing through it.

Distinguish supply from transfer

E.m.f. is energy supplied per unit charge by a source. P.d. across a component is energy transferred from electrical energy per unit charge in that component.

For a real source under load, terminal p.d. can be below e.m.f. because some energy per coulomb is transferred inside the source. The complete loop still conserves energy.

Check your understanding: A 1.5 V cell has terminal p.d. 1.2 V. What happened to the other 0.3 J per coulomb?

It was transferred inside the cell, usually to internal energy.

Discharging-source model: the ideal e.m.f. and internal resistance are in series. The terminal p.d. across the load is V = ε − Ir.

Key ideas

Both e.m.f. and p.d. are measured in volts.
Name the energy transfer and the charge quantity.
Do not add source and component voltages without first tracing the circuit energy account.

Relationships to know

V = W/Q
e.m.f. ε = energy supplied / charge
p.d. = energy transferred / charge

Follow the reasoning

Worked example

Use charge to compare source and component transfers

Question: A battery supplies 480 J while 40 C passes. The external circuit receives 420 J. Find the e.m.f., terminal p.d. and energy transferred inside the battery per coulomb.

Step 1: Calculate energy supplied per charge
Why: This is the definition of e.m.f.
Working: ε = 480/40 = 12.0 J C⁻¹ = 12.0 V.
Step 2: Calculate external transfer per charge
Why: This is the terminal p.d. across the external circuit.
Working: V = 420/40 = 10.5 V.
Step 3: Account for the difference
Why: Energy supplied must equal external plus internal transfers.
Working: Internal transfer per coulomb = 12.0 − 10.5 = 1.5 J C⁻¹.

Answer: E.m.f. = 12.0 V, terminal p.d. = 10.5 V and 1.5 J per coulomb is transferred internally.

Check: Across 40 C, the internal transfer is 60 J; 420 J + 60 J = 480 J.

Now try it with support

Practise with support

A motor transfers 360 J when 45 C passes through it. Find its p.d. and state the meaning of your answer.

Hints

Use V = W/Q.
Express the interpretation in joules per coulomb.

View the guided answer

V = 360/45 = 8.0 V. The motor transfers 8.0 J of electrical energy for each coulomb passing through it.

Your turn

Practise independently

A source moves 24 C and supplies 144 J while 120 J is transferred in the external circuit. Determine its e.m.f. and terminal p.d.

Check your answer

E.m.f. = 144/24 = 6.0 V. Terminal p.d. = 120/24 = 5.0 V. The difference means 1.0 J per coulomb is transferred inside the source.

Common mistakes and exam guidance

Watch out for

Calling e.m.f. a mechanical force.
Reversing the roles of energy supplied by the source and energy transferred by a component.

In an exam

For a ‘distinguish’ question, use both phrases ‘energy supplied per unit charge’ and ‘energy transferred per unit charge’.
Give voltage units even when the calculation begins with joules and coulombs.

Put the ideas together

Exam-style practice [6 marks]

A rechargeable source moves 2.5 × 10⁴ C through a circuit. It supplies 1.50 × 10⁵ J, while a motor transfers 1.20 × 10⁵ J. Calculate the source e.m.f., motor p.d. and internal energy transfer. Explain the distinction between the two voltages.

Plan before you answer

Divide each relevant energy by the same charge.
Use conservation for the internal transfer.
Define e.m.f. and p.d. in words.

View the marking points and model answer

Marking points

Finds ε = 6.0 V.
Finds motor p.d. = 4.8 V.
Finds internal energy = 3.0 × 10⁴ J.
Or internal transfer = 1.2 J C⁻¹.
Defines e.m.f. as energy supplied per charge.
Defines p.d. as energy transferred per charge.

Model answer

ε = 1.50 × 10⁵/(2.5 × 10⁴) = 6.0 V. The motor p.d. is 1.20 × 10⁵/(2.5 × 10⁴) = 4.8 V. The internal transfer is 1.50 × 10⁵ − 1.20 × 10⁵ = 3.0 × 10⁴ J, or 1.2 J C⁻¹. E.m.f. is energy supplied by the source per charge; p.d. is energy transferred by a component per charge.

Finish from memory

Three-question recap

Define one volt.
Check
One joule of energy transferred per coulomb.
What does e.m.f. describe?
Check
Energy supplied by a source per unit charge.
Why may terminal p.d. be below e.m.f.?
Check
Some energy per charge is transferred inside the source.

Try this next

Combine energy per coulomb with charge per second to obtain electrical power.

Continue to Electrical power Return to Currents Review Current, charge flow and drift velocity

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027