Physics · 4 Electricity and magnetism

4.2.3 Electromotive force and potential difference

Before you start

Current as the flow of charge, measured in amps, and charge in coulombs (4.2.2).
Work done and energy transfer, both measured in joules.
That a circuit transfers energy from a source to components.

Syllabus objectives covered (7)

Define electromotive force (e.m.f.) as the electrical work done by a source in moving a unit charge around a complete circuit
Know that e.m.f. is measured in volts (V)
Define potential difference (p.d.) as the work done by a unit charge passing through a component
Know that the p.d. between two points is measured in volts (V)
Describe the use of voltmeters (analogue and digital) with different ranges
Recall and use the equation for e.m.f. E = W / Q
Recall and use the equation for p.d. V = W / Q

Part 1 of 4

Two words for two different jobs

Both e.m.f. and p.d. are measured in volts, and both are about energy per unit charge. What differs is where in the circuit you are looking.
E.m.f. is about the source giving energy out. It is the electrical work done by a source in moving a unit charge around a complete circuit.
P.d. is about a component using energy up. It is the work done by a unit charge passing through that component.
So a cell has an e.m.f.; a lamp has a p.d. across it. Picture the charge picking up energy at the cell and dropping it off at each component.
Notice the phrase 'unit charge' in both definitions, meaning per coulomb. A 1.5 V cell gives 1.5 joules of energy to every coulomb that passes through it.
Write the definitions out rather than paraphrasing. 'The energy given to the charge' is loose; 'the electrical work done by a source in moving a unit charge around a complete circuit' is the marking point.
Both are measured in volts (V), and a p.d. is always quoted between two points or across a component — never at a single point.

Part 2 of 4

The equations

Both definitions turn into the same-shaped equation, with $W$ the work done in joules and $Q$ the charge in coulombs.
$E = \frac{W}{Q} V = \frac{W}{Q}$
$E$ is the e.m.f. of the source; $V$ is the p.d. across the component. Same arithmetic, different context — use the symbol that matches what the question is about.
This tells you what a volt is: one volt is one joule per coulomb. That single sentence explains why both quantities share a unit.
The most useful rearrangement is $W = Q V$ : the energy transferred when a charge $Q$ passes through a p.d. of $V$ .
Combine it with $Q = I t$ from 4.2.2 and you get $W = I V t$ , which is exactly the electrical energy equation you meet in 4.2.5. The topics join up.
Neither equation is given to you in Paper 4. Practise writing both, and the definition of the volt, from memory.

Part 3 of 4

Energy round a circuit

Follow one coulomb of charge round a simple circuit and the difference between the two quantities becomes obvious.
It leaves the cell having been given, say, 6 J of energy. The cell's e.m.f. is 6 V.
It passes through a lamp and gives up 4 J there, which the lamp transfers to light and thermal energy. The p.d. across that lamp is 4 V.
It passes through a second lamp and gives up the remaining 2 J. The p.d. across that lamp is 2 V.
It returns to the cell with no energy left, ready to be given another 6 J. Notice that $4 + 2 = 6$ : the p.d.s across the components add up to the e.m.f. of the source.
That result is not a coincidence — it is conservation of energy applied to a circuit, and you will meet it again as a circuit rule in 4.3.2.

Part 4 of 4

Using a voltmeter

A voltmeter measures p.d., and it must be connected in parallel with the component — across it, not in the path of the current.
This follows from what a p.d. is: a difference between two points. You have to touch both points to measure a difference.
Contrast it with the ammeter, which goes in series. Ammeter in, voltmeter across — and the two are easy to swap under pressure.
A voltmeter should have a very high resistance, so that almost no current flows through it and the circuit is barely disturbed.
That is the opposite requirement from an ammeter's very low resistance, and for the same underlying reason: a meter should measure without changing what it measures.
Ranges work as they do for ammeters. Choose the range whose maximum is just above the p.d. you expect: too large and the needle barely moves, too small and it goes off scale.
For analogue meters, read the needle straight on to avoid parallax error. Digital voltmeters avoid that, and usually give more decimal places.

Worked examples

Example 1. A battery does $36$ J of work moving a charge of $4.0$ C around a complete circuit. Calculate the e.m.f. of the battery.

Show the working

The question describes work done by a source moving charge around a complete circuit, which is the definition of e.m.f. So use $E = W / Q$ .
Substitute the values.
$E = \frac{36}{4.0} = 9.0 V$
Give the unit — e.m.f. is measured in volts.
Sense-check it: 9 V is a standard battery, and 9 J per coulomb is exactly what 9 V means.

Answer: $E = 9.0$ V.

Example 2. A charge of $15$ C passes through a resistor. The p.d. across the resistor is $4.0$ V. Calculate the energy transferred in the resistor.

Show the working

Energy is wanted, so rearrange $V = W / Q$ to make $W$ the subject.
$W = Q V$
Substitute.
$W = 15 \times 4.0 = 60 J$
The unit is the joule, since this is an energy.
Interpret it: each coulomb gives up 4 J in the resistor, and 15 coulombs pass, so 60 J is transferred — mostly to thermal energy in this case.

Answer: $W = 60$ J.

Example 3. A student wishes to measure the p.d. across a lamp in a circuit powered by a $6$ V battery. Describe how the voltmeter should be connected, and explain why it must have a high resistance.

Show the working

State the connection: the voltmeter goes in parallel with the lamp, one lead on each side of it.
Explain why: p.d. is the difference between two points, so the meter must be connected to both points.
Now the resistance. If the voltmeter had a low resistance, a significant current would flow through it rather than through the lamp.
That extra path would change the currents in the circuit, so the p.d. being measured would not be the p.d. that existed before the meter was connected.
A high resistance means almost no current is diverted, so the reading is a true measure of the p.d. across the lamp.

Answer: Connect it in parallel with the lamp. It needs a high resistance so that almost no current flows through it, leaving the circuit essentially undisturbed and the reading a true value of the p.d.

Where people go wrong

Mistake: Using e.m.f. and p.d. as interchangeable words.

Instead: E.m.f. describes a source supplying energy; p.d. describes a component using it.

They share a unit but not a meaning. Definitions are examined precisely, and each carries its own marking point.

Mistake: Connecting a voltmeter in series.

Instead: A voltmeter goes in parallel, across the component.

P.d. is a difference between two points, so the meter has to be attached to both.

Mistake: Saying a voltmeter should have a low resistance, like an ammeter.

Instead: A voltmeter needs a high resistance; an ammeter needs a low one.

The voltmeter is in parallel and must divert as little current as possible; the ammeter is in series and must add as little resistance as possible.

Mistake: Writing 'the p.d. at point X'.

Instead: P.d. is always between two points or across a component.

It is a difference, so a single point cannot have one.

Mistake: Defining e.m.f. as 'the voltage of the battery'.

Instead: It is the electrical work done by the source in moving a unit charge around a complete circuit.

Restating the unit is not a definition, and 'voltage' is not the term the syllabus uses.

Check yourself

Define electromotive force.
Answer
The electrical work done by a source in moving a unit charge around a complete circuit.
Define potential difference.
Answer
The work done by a unit charge passing through a component.
What is a volt, expressed in terms of joules and coulombs?
Answer
One joule per coulomb.
A cell does $24$ J of work driving $3.0$ C round a circuit. What is its e.m.f.?
Answer
$E = W / Q = 24 / 3.0 = 8.0$ V.
How much energy is transferred when $5.0$ C passes through a p.d. of $12$ V?
Answer
$W = Q V = 5.0 \times 12 = 60$ J.
State how a voltmeter is connected and what resistance it should have, and contrast this with an ammeter.
Answer
A voltmeter is connected in parallel and has a high resistance. An ammeter is connected in series and has a low resistance.
A p.d. of about $2$ V is to be measured. Why is a 0–20 V range a poor choice?
Answer
The reading would be only a tenth of full scale, giving a small deflection that cannot be read precisely. A 0–5 V range would be better.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A cell of e.m.f. $4.5$ V is connected to two lamps in series. The p.d. across the first lamp is $2.8$ V.

(a) Define potential difference. 1 mark No calculator

Mark scheme

B1 the work done by a unit charge passing through a component

(b) Determine the p.d. across the second lamp. 2 marks

Mark scheme

C1 recognises that the p.d.s across the components sum to the e.m.f.
A1 1.7 V; unit required

(c) Calculate the energy transferred in the first lamp when $8.0$ C of charge passes through it. 2 marks

Mark scheme

C1 W = QV used
A1 22.4 J (accept 22 J); unit required

Question 2. A student sets up a circuit to measure the current through a resistor and the p.d. across it.

(a) State how the ammeter and the voltmeter should each be connected. 2 marks No calculator

Mark scheme

B1 ammeter in series with the resistor
B1 voltmeter in parallel with (across) the resistor

(b) Explain why the voltmeter must have a high resistance. 2 marks No calculator

Mark scheme

B1 so that very little current flows through the voltmeter
B1 so the circuit is not significantly altered and the reading is a true value of the p.d.

(c) The student uses an analogue voltmeter. Suggest one source of error in reading it, and how to avoid it. 2 marks No calculator

Mark scheme

B1 parallax error from viewing the needle at an angle
B1 read with the eye directly in front of the needle (accept: use a mirror scale, or use a digital meter)

Question 3. A battery does $54$ J of electrical work in driving $6.0$ C of charge around a complete circuit.

(a) Calculate the e.m.f. of the battery. 2 marks

Mark scheme

C1 E = W / Q used
A1 9.0 V; unit required

(b) Explain why e.m.f. and p.d. are both measured in volts, even though they describe different things. 2 marks No calculator

Mark scheme

B1 both are energy (work done) per unit charge
B1 one volt is one joule per coulomb, so the same unit applies to both, owtte

Words to know

Electromotive force (e.m.f.): The electrical work done by a source in moving a unit charge around a complete circuit, measured in volts.
Potential difference (p.d.): The work done by a unit charge passing through a component, measured in volts.
Volt (V): One joule per coulomb.
Voltmeter: A high-resistance meter connected in parallel to measure p.d.
Ammeter: A low-resistance meter connected in series to measure current.

All Physics lessons Next: 4.2.4 Resistance