IGCSE Physics 0625 · Topic 4.3.3

IGCSE Physics Action and Use of Circuit Components Questions and Worked Solutions

50 topic-wise practice questions on Action and Use of Circuit Components (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 questions14 Core26 Extended10 Challenge138 marksTextbook §4.3 · pp. 209–221

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: Action and Use of Circuit Components

This subtopic looks at how components share p.d. and how a potential divider uses two resistors (or a resistor and a sensor) to give a chosen fraction of the supply p.d.

Key formulas

R₁ ÷ R₂ = V₁ ÷ V₂	potential divider	Extended
V_out = V_in × R₂ ÷ (R₁ + R₂)	output across R₂	Extended
V = I R	p.d. increases with R at constant current	Core

Prerequisite knowledge

Series circuits (4.3.2)
Thermistors and LDRs (4.3.1)
Ratios

Exam tips & common mistakes

The output is always across the component you measure it over — read the diagram carefully
Dark → LDR resistance high → larger share of p.d.
Hot → thermistor resistance low → smaller share of p.d.

Key facts from the syllabus (4.3.3)

For a constant current, the p.d. across a conductor increases as its resistance increases
In series, the larger resistance has the larger share of the supply p.d.
Extended a variable potential divider (potentiometer) gives an output from zero to the full supply
Extended with a thermistor or LDR in a potential divider, the output changes with temperature or light — used in sensing circuits

Textbook: §4.3 Electric circuits, pp. 209–221

Core[2 marks]

For a constant current, state how the p.d. across a conductor changes as its resistance increases.

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Method: V = IR.

The p.d. increases

Core[2 marks]

Two resistors of 2.0 Ω and 8.0 Ω are in series with a constant current of 0.50 A. Calculate the p.d. across each.

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Method: V = IR.

1.00 V and 4.00 V

Core[2 marks]

In a series circuit, explain why the larger resistor has the larger p.d.

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Method: Same current, V = IR.

The current is the same through both, so V = IR is larger for the larger resistance

Core[2 marks]

A thermistor and a fixed resistor are in series across a supply. The temperature rises. State what happens to the p.d. across the thermistor.

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Method: Thermistor R falls.

It decreases, because the thermistor's resistance falls

Core[2 marks]

An LDR and a fixed resistor are in series. It gets darker. State what happens to the p.d. across the LDR.

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Method: LDR R rises.

It increases, because the LDR's resistance increases

Core[2 marks]

A variable resistor is in series with a lamp. Its resistance is increased. State what happens to the brightness of the lamp.

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Method: Less current.

The lamp gets dimmer (current decreases)

Core[2 marks]

Explain how a dimmer switch uses a variable resistor to control a lamp.

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Method: Changes resistance in series.

Changing the resistance in series changes the current through the lamp, and so its brightness

Core[2 marks]

A 3.0 Ω and a 9.0 Ω resistor are in series with a current of 0.50 A. Calculate the p.d. across each.

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Method: V = IR.

1.50 V and 4.50 V

Core[2 marks]

State which resistor in a series circuit has the larger share of the supply p.d.

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Method: Larger R.

The one with the larger resistance

Core[2 marks]

A variable resistor in series with a lamp is increased. State what happens to the p.d. across the variable resistor.

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Method: Larger share.

It increases

Core[2 marks]

An LDR and fixed resistor are in series across 6 V. The light gets brighter. State what happens to the p.d. across the fixed resistor.

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Method: LDR R falls.

It increases

Core[2 marks]

A thermistor and fixed resistor are in series. The thermistor is cooled. State what happens to the p.d. across the thermistor.

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Method: R rises.

It increases

Core[2 marks]

State why the p.d. across a component increases when its resistance increases with constant current.

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Method: V = IR.

Because V = IR, so for constant I, V is proportional to R

Core[2 marks]

A 12 Ω resistor carries a constant 0.25 A. Calculate the p.d., and the p.d. if it were replaced by 24 Ω with the same current.

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Method: V = IR.

3.00 V and 6.00 V

Extended[2 marks]

Describe the action of a variable potential divider.

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Method: Sliding contact.

A resistor with a sliding contact; moving the slider changes the fraction of the supply p.d. taken as output, from zero up to the full supply

Extended[1 mark]

State the potential divider equation for two resistors.

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Method: R₁/R₂ = V₁/V₂.

R₁ ÷ R₂ = V₁ ÷ V₂

Extended[3 marks]

A 12 V supply is across 4.0 kΩ and 8.0 kΩ in series. Calculate the p.d. across the 8.0 kΩ resistor.

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Method: Ratio 1:2.

V = 12 × 8 ÷ 12 = 8.00 V

Extended[3 marks]

A 9.0 V supply is across R₁ = 1.0 kΩ and R₂ = 2.0 kΩ. Calculate V₁ and V₂.

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Method: Share in ratio.

V₁ = 3.00 V, V₂ = 6.00 V

Extended[3 marks]

A potential divider has R₁ = 3.0 kΩ. The output across R₂ is 2.0 V and across R₁ is 4.0 V. Calculate R₂.

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Method: R₂ = R₁ × V₂/V₁.

R₂ = 3.0 × 2.0 ÷ 4.0 = 1.50 kΩ

Extended[3 marks]

A thermistor (resistance 2.0 kΩ at 20 °C) is in series with a 6.0 kΩ resistor across 8.0 V. Calculate the p.d. across the thermistor.

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Method: Ratio 2:6.

V = 8.0 × 2 ÷ 8 = 2.00 V

Extended[3 marks]

In the previous circuit, the temperature rises and the thermistor's resistance falls to 1.0 kΩ. Calculate the new p.d. across it.

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Method: Ratio 1:6.

V = 8.0 × 1 ÷ 7 = 1.14 V

Extended[3 marks]

An LDR has resistance 10 kΩ in the dark and 1.0 kΩ in the light. It is in series with a 10 kΩ resistor across 6.0 V. Calculate the p.d. across the LDR in the dark and in the light.

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Method: V = 6 × R_LDR / (R_LDR + 10).

Dark: 3.00 V; light: 0.545 V

Extended[3 marks]

Explain how an LDR in a potential divider can switch on a street lamp at night.

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Method: Output rises in dark.

In the dark the LDR's resistance rises, so the p.d. across it rises; when it passes a set value it triggers a switch (e.g. relay or transistor) that turns on the lamp

Extended[3 marks]

Describe how a thermistor in a potential divider can be used in a fire alarm.

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Method: Output changes with temperature.

As temperature rises, thermistor resistance falls, so the p.d. across the fixed resistor rises; this output triggers the alarm

Extended[2 marks]

A potential divider of two equal resistors is connected across 10 V. State the output p.d. across one.

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Method: Equal share.

5.0 V

Extended[3 marks]

A potential divider must give 3.0 V from a 12 V supply. R₂ (output) = 1.0 kΩ. Calculate R₁.

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Method: V₁ = 9.0 V.

R₁ = 1.0 × 9.0 ÷ 3.0 = 3.00 kΩ

Extended[2 marks]

Explain why a potential divider output changes when a voltmeter of low resistance is connected across it.

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Method: Parallel lowers resistance.

The voltmeter in parallel lowers the effective resistance, changing the share of p.d.

Extended[3 marks]

A potential divider uses 10 kΩ and 30 kΩ across 8.0 V. Calculate the p.d. across each.

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Method: Ratio 1:3.

2.00 V and 6.00 V

Extended[3 marks]

A potential divider must give 4.0 V across R₂ from a 10 V supply. R₁ = 3.0 kΩ. Calculate R₂.

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Method: V₁ = 6.0 V.

R₂ = 3.0 × 4.0 ÷ 6.0 = 2.00 kΩ

Extended[3 marks]

An LDR (resistance 500 Ω in light) is in series with a 4.5 kΩ resistor across 5.0 V. Calculate the p.d. across the LDR.

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Method: Share.

V = 5.0 × 0.5 ÷ 5.0 = 0.500 V

Extended[3 marks]

In the dark the LDR's resistance rises to 45 kΩ. Calculate the new p.d. across it.

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Method: Share.

V = 5.0 × 45 ÷ 49.5 = 4.55 V

Extended[2 marks]

Explain why the output of a potential divider depends on the ratio, not the actual sizes, of the resistors.

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Method: Same current.

The same current flows through both, so each p.d. is proportional to its resistance

Extended[3 marks]

A variable potential divider (potentiometer) of total resistance 1.0 kΩ is connected across 12 V. The slider is 1/4 of the way from the bottom. Calculate the output p.d.

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Method: Proportion.

3.00 V

Extended[3 marks]

Describe how a thermistor and a fixed resistor could switch on a fan when a room gets hot.

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Method: Output across fixed resistor.

As temperature rises the thermistor's resistance falls, so the p.d. across the fixed resistor rises; this output switches a relay that turns on the fan

Extended[2 marks]

State the effect of swapping the positions of the LDR and fixed resistor in a light-sensing potential divider.

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Method: Reverses response.

The output now rises in the light instead of in the dark (the response is reversed)

Extended[3 marks]

A potential divider has R₁ = 2.2 kΩ and R₂ = 4.7 kΩ across 9.0 V. Calculate the output across R₂.

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Method: V₂ = V × R₂/(R₁ + R₂).

V₂ = 9.0 × 4.7 ÷ 6.9 = 6.13 V

Extended[3 marks]

A thermistor has 6.0 kΩ at 25 °C. It is in series with a 2.0 kΩ resistor across 12 V. Calculate the p.d. across the fixed resistor.

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Method: Share.

V = 12 × 2 ÷ 8 = 3.00 V

Extended[2 marks]

Describe the output of a potential divider made from two equal resistors.

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Method: Half supply.

Half the supply p.d.

Extended[2 marks]

Explain why a potential divider is useful for providing a variable p.d. from a fixed battery.

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Method: Adjustable fraction.

Moving a slider changes the ratio of resistances, so any p.d. from zero to the full supply can be obtained

Extended[3 marks]

A light sensor circuit needs 2.0 V output across an LDR of 1.0 kΩ from a 6.0 V supply. Calculate the series resistor needed.

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Method: V₁ = 4.0 V.

R = 1.0 × 4.0 ÷ 2.0 = 2.00 kΩ

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]

A lamp and a variable resistor are in series with a 6.0 V battery and the current is constant at 0.20 A. The variable resistor is set to 10 Ω, then 20 Ω (assume the current stays 0.20 A). Calculate the p.d. across it each time and explain why the lamp is dimmer in practice.

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Method: V = IR.

2.00 V and 4.00 V; in practice the larger resistance reduces the current and leaves less p.d. for the lamp

ChallengeCore[3 marks]

For a constant current through a wire, explain why doubling its resistance doubles the p.d. across it.

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Method: V = IR.

V = IR, so with I constant, V is proportional to R

ChallengeExtended[5 marks]

A thermistor (5.0 kΩ at 20 °C, 1.0 kΩ at 80 °C) is in series with a 4.0 kΩ resistor across 9.0 V. Calculate the p.d. across the fixed resistor at both temperatures and explain how this could trigger a fire alarm.

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Method: Potential divider.

20 °C: 9.0 × 4 ÷ 9 = 4.00 V; 80 °C: 9.0 × 4 ÷ 5 = 7.20 V — the rising output can switch on an alarm circuit

ChallengeExtended[5 marks]

An LDR has 100 kΩ in darkness and 2.0 kΩ in light. Choose a fixed resistor so the p.d. across the LDR is 4.5 V in darkness from a 6.0 V supply, then calculate the p.d. across the LDR in light.

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Method: Ratio 4.5 : 1.5.

R = 33.3 kΩ; light: 6.0 × 2 ÷ (2 + 33.3) = 0.340 V

ChallengeExtended[4 marks]

A potentiometer of 10 kΩ is connected across 12 V. The slider is moved from the bottom to 30% of the way up, then 75%. Calculate the output each time and describe the use of such a circuit.

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Method: Proportional.

3.6 V and 9.0 V; it provides a continuously variable p.d. (e.g. volume control)

ChallengeExtended[4 marks]

A potential divider has R₁ = 3.3 kΩ and R₂ = 6.8 kΩ across 5.0 V. Calculate the output across R₂ and the current through the divider.

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Method: V₂ = V × R₂/(R₁ + R₂).

V₂ = 3.37 V; I = 0.495 mA

ChallengeExtended[4 marks]

Explain why the output of a potential divider falls if a low-resistance device is connected across R₂.

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Method: Parallel combination.

The device in parallel with R₂ reduces the effective resistance of that part, so it takes a smaller share of the supply p.d.

ChallengeExtended[4 marks]

Design a potential divider using a thermistor and a 2.2 kΩ resistor to give an output that rises as temperature rises. State where the output is taken and explain.

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Method: Output across fixed resistor.

Output across the fixed resistor: as temperature rises, thermistor resistance falls, so the fixed resistor's share of p.d. rises

ChallengeExtended[4 marks]

A 12 V supply is across two resistors. The p.d. across R₁ is 8.0 V and R₂ = 1.5 kΩ. Calculate R₁ and the current.

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Method: Ratio.

R₁ = 1.5 × 8.0 ÷ 4.0 = 3.00 kΩ; I = 4.0 ÷ 1500 = 2.67 mA

ChallengeExtended[4 marks]

An automatic night light uses an LDR potential divider and a relay. Explain the sequence of events as it gets dark.

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Method: Rising output.

Darkness raises LDR resistance; the p.d. across it rises; above a set value the relay coil is energised, its contacts close and the lamp switches on

All IGCSE Physics topics ← 4.3.2 Series and Parallel Circuits 4.4 Electrical Safety →

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