IGCSE Physics 0625 · Topic 4.2.3

IGCSE Physics Electromotive Force and Potential Difference Questions and Worked Solutions

50 topic-wise practice questions on Electromotive Force and Potential Difference (Unit 4: Electricity and Magnetism), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions18 Core22 Extended10 Challenge124 marksTextbook §4.2 · pp. 184–208

Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Electromotive Force and Potential Difference

Electromotive force (e.m.f.) is the energy a source gives to each unit of charge. Potential difference (p.d.) is the energy each unit of charge transfers in a component. Both are measured in volts with a voltmeter in parallel.

Key formulas

E = W ÷ Q	e.m.f. E (V), work done by the source W (J), charge Q (C)	Extended
V = W ÷ Q	p.d. V (V), energy transferred in the component W (J)	Extended
1 V = 1 J/C	definition of the volt	Core

Prerequisite knowledge

Charge and current (4.2.1, 4.2.2)
Energy and work (1.7)

Exam tips & common mistakes

Voltmeter in parallel, ammeter in series
Use Q = It first if only current and time are given
A reversed cell subtracts from the total e.m.f.

Key facts from the syllabus (4.2.3)

e.m.f. is the electrical work done by a source in moving a unit charge around a complete circuit
p.d. is the work done by a unit charge passing through a component
Both are measured in volts (V)
Voltmeters (analogue and digital) are connected in parallel; choose a suitable range
e.m.f.s of cells in series add (taking direction into account)

Textbook: §4.2 Electrical quantities, pp. 184–208

Core[2 marks]

Define electromotive force (e.m.f.).

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Method: Work per unit charge by source.

The electrical work done by a source in moving a unit charge around a complete circuit

Core[1 mark]

State the unit of e.m.f.

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Method: Volt.

volt (V)

Core[2 marks]

Define potential difference (p.d.).

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Method: Work per unit charge in component.

The work done by a unit charge passing through a component

Core[1 mark]

State the instrument used to measure p.d. and how it is connected.

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Method: Voltmeter in parallel.

A voltmeter, connected in parallel with the component

Core[2 marks]

Explain why a voltmeter should have a very high resistance.

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Method: Draws little current.

So it draws very little current and does not change the circuit

Core[2 marks]

A voltmeter has ranges 0–3 V and 0–15 V. A battery is about 9 V. State which range to use.

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Method: Must exceed reading.

0–15 V, since 9 V exceeds the 3 V range

Core[2 marks]

State the difference between e.m.f. and p.d.

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Method: Source vs component.

e.m.f. refers to energy given to charge by a source; p.d. refers to energy transferred by charge in a component

Core[1 mark]

State what a reading of 6 V across a lamp means.

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Method: 6 J per coulomb.

6 J of energy is transferred for each coulomb of charge passing through the lamp

Core[2 marks]

Three 1.5 V cells are connected in series, all facing the same way. Calculate the total e.m.f.

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Method: Add e.m.f.s.

E = 3 × 1.5 = 4.50 V

Core[2 marks]

Two 1.5 V cells are connected in series but one is reversed. State the total e.m.f.

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Method: Opposing e.m.f.s cancel.

0 V

Core[2 marks]

State the meaning of a cell e.m.f. of 1.5 V.

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Method: 1.5 J per C.

1.5 J of energy is given to each coulomb of charge passing through the cell

Core[2 marks]

Describe how a voltmeter is connected to measure the p.d. across a lamp.

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Method: Parallel.

Across the lamp (in parallel), with its positive terminal connected towards the positive of the supply

Core[2 marks]

A voltmeter is connected across a cell with no other circuit. State what it measures.

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Method: e.m.f.

The e.m.f. of the cell

Core[2 marks]

A battery has an e.m.f. of 12 V. State the total p.d. across the components in a simple series circuit.

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Method: Sum = e.m.f.

12 V (ignoring internal resistance)

Core[2 marks]

Explain the difference between an analogue voltmeter with range 0–5 V and one with 0–50 V when measuring 3.6 V.

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Method: Precision.

The 0–5 V range gives a larger deflection and a more precise reading

Core[2 marks]

State the unit of potential difference in terms of joules and coulombs.

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Method: V = J/C.

1 V = 1 J/C

Core[2 marks]

Three 2.0 V cells are connected in series. State the total e.m.f.

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Method: Add.

6.0 V

Core[2 marks]

State why a voltmeter connected in series in a circuit makes lamps go out.

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Method: Very high resistance.

Its very high resistance reduces the current almost to zero

Extended[2 marks]

A cell does 18 J of work moving 4.0 C round a circuit. Calculate its e.m.f.

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Method: E = W ÷ Q.

E = 18 ÷ 4.0 = 4.50 V

Extended[2 marks]

A charge of 5.0 C passes through a lamp with p.d. 12 V. Calculate the energy transferred.

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Method: W = VQ.

W = 12 × 5.0 = 60.0 J

Extended[2 marks]

240 J is transferred when charge passes through a heater with p.d. 230 V. Calculate the charge.

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Method: Q = W ÷ V.

Q = 240 ÷ 230 = 1.04 C

Extended[3 marks]

A 9.0 V battery drives a current of 0.20 A for 60 s. Calculate the work done by the battery.

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Method: Q = It; W = EQ.

Q = 0.20 × 60 = 12 C; W = 9.0 × 12 = 108 J

Extended[3 marks]

A component transfers 360 J when 1.5 A flows for 20 s. Calculate the p.d. across it.

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Method: Q = It, V = W/Q.

Q = 30 C; V = 360 ÷ 30 = 12.0 V

Extended[2 marks]

A 1.5 V cell moves 200 C of charge. Calculate the energy it transfers.

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Method: W = EQ.

W = 1.5 × 200 = 300 J

Extended[3 marks]

A car battery of e.m.f. 12 V supplies 1.2 × 10⁵ C. Calculate the energy supplied in MJ.

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Method: W = EQ.

W = 12 × 1.2×10⁵ = 1.4 × 10⁶ J = 1.44 MJ

Extended[2 marks]

Explain why connecting a voltmeter in series stops the circuit working properly.

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Method: Huge resistance.

Its very high resistance reduces the current almost to zero

Extended[3 marks]

A charge of 25 C gains 150 J from a battery and transfers 90 J in a lamp. Calculate the e.m.f. and the p.d. across the lamp.

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Method: E = W/Q; V = W/Q.

E = 150 ÷ 25 = 6.00 V; V = 90 ÷ 25 = 3.60 V

Extended[2 marks]

State the energy given to each coulomb by a 230 V mains supply.

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Method: 1 V = 1 J/C.

230 J

Extended[2 marks]

Calculate the e.m.f. of a source that does 54 J of work moving 6.0 C.

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Method: E = W/Q.

E = 9.00 V

Extended[2 marks]

Calculate the energy transferred when 3.0 C passes through a 230 V heater.

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Method: W = VQ.

W = 690 J

Extended[2 marks]

A p.d. of 6.0 V transfers 360 J. Calculate the charge.

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Method: Q = W/V.

Q = 60.0 C

Extended[3 marks]

A 12 V battery drives 2.0 A through a motor for 5.0 minutes. Calculate the charge and energy.

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Method: Q = It.

Q = 600 C; W = 7200 J

Extended[3 marks]

A lamp transfers 900 J when 0.50 A flows for 2.5 minutes. Calculate the p.d.

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Method: Q = 75 C.

V = 900 ÷ 75 = 12.0 V

Extended[2 marks]

State the relationship between p.d., energy and charge for a component.

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Method: V = W/Q.

V = W ÷ Q

Extended[3 marks]

A cell of e.m.f. 1.5 V delivers 2400 J before going flat. Calculate the total charge delivered.

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Method: Q = W/E.

Q = 1600 C

Extended[2 marks]

Explain why the sum of the p.d.s around a series circuit equals the e.m.f.

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Method: Energy conservation.

Energy given to each coulomb by the source equals the energy transferred by it in the components

Extended[3 marks]

In a series circuit with a 9.0 V battery, a resistor has 5.0 V across it. Calculate the energy transferred in the other component when 4.0 C flows.

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Method: V = 4.0 V.

W = 4.0 × 4.0 = 16.0 J

Extended[2 marks]

Calculate the energy given to each electron by a 6.0 V battery (e = 1.6 × 10⁻¹⁹ C).

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Method: W = VQ.

W = 9.6 × 10⁻¹⁹ J

Extended[2 marks]

Explain why e.m.f. and p.d. have the same unit.

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Method: Both J/C.

Both are energy transferred per unit charge

Extended[3 marks]

A solar cell has e.m.f. 0.50 V. Calculate the number needed in series to make a 6.0 V supply and the energy per coulomb provided.

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Method: Divide.

12 cells; 6.0 J per coulomb

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

ChallengeCore[4 marks]

Explain the difference between e.m.f. and p.d., using a circuit with a cell and two lamps as an example.

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Method: Source vs component.

e.m.f. is energy given per coulomb by the cell; p.d. is energy transferred per coulomb in each lamp; the p.d.s across the lamps add up to the e.m.f.

ChallengeCore[4 marks]

Four 1.5 V cells are connected in series, but one is accidentally reversed. Calculate the total e.m.f. and explain.

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Method: Reversed cell subtracts.

4.5 − 1.5 = 3.0 V: the reversed cell opposes the others

ChallengeCore[3 marks]

Explain why a voltmeter has a very high resistance and is connected in parallel.

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Method: Draws little current.

It must not change the current in the component it measures; in parallel with high resistance it draws almost no current

ChallengeExtended[4 marks]

A 9.0 V battery drives 0.30 A through a circuit for 10 minutes. Calculate the charge and the energy transferred by the battery.

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Method: Q = It; W = EQ.

Q = 180 C; W = 1620 J

ChallengeExtended[4 marks]

In a series circuit with a 12 V battery, lamp A has 7.5 V across it. 25 C flows. Calculate the energy transferred in lamp A and in the rest of the circuit.

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Method: W = VQ.

A: 187.5 J; rest: 4.5 × 25 = 112.5 J

ChallengeExtended[4 marks]

A charge of 600 C passes through a heater, transferring 1.38 × 10⁵ J. Calculate the p.d., and state the energy transferred by each electron (e = 1.6 × 10⁻¹⁹ C).

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Method: V = W/Q.

V = 230 V; energy per electron = 3.68 × 10⁻¹⁷ J

ChallengeExtended[4 marks]

A torch uses two 1.5 V cells and the bulb draws 0.25 A. The cells deliver 2.7 × 10⁴ J before going flat. Calculate how long the torch works.

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Method: t = W/(EI).

t = 27000 ÷ (3.0 × 0.25) = 36000 s = 10 h

ChallengeExtended[4 marks]

A solar cell provides an e.m.f. of 0.60 V. Calculate the number needed in series to charge a 3.6 V battery and the energy given to 500 C of charge.

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Method: Cells in series add.

6 cells; W = 3.6 × 500 = 1800 J

ChallengeExtended[4 marks]

A car battery of e.m.f. 12 V delivers 150 A to the starter motor for 1.5 s. Calculate the charge and energy delivered.

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Method: Q = It.

Q = 225 C; W = 2700 J

ChallengeExtended[3 marks]

Explain, using energy per coulomb, why a 12 V lamp is brighter than a 6 V lamp when each carries the same current.

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Method: More energy per coulomb.

Each coulomb transfers twice as much energy in the 12 V lamp and the same charge passes per second, so the power is double

All IGCSE Physics topics ← 4.2.2 Electric Current 4.2.4 Resistance →

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