IGCSE Physics 0625 · Topic 4.3.2

IGCSE Physics Series and Parallel Circuits Questions and Worked Solutions

50 topic-wise practice questions on Series and Parallel Circuits (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 questions16 Core24 Extended10 Challenge136 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: Series and Parallel Circuits

Components can be connected in series (one path) or parallel (several branches). Current, p.d. and resistance behave differently in each.

Key formulas

R = R₁ + R₂ + …	resistors in series	Core
1/R = 1/R₁ + 1/R₂	two resistors in parallel	Extended
I = I₁ + I₂	currents at a junction (parallel)	Extended
V = V₁ + V₂	p.d.s in series add up to the supply	Extended
E = E₁ + E₂ + …	e.m.f.s of cells in series	Core

Prerequisite knowledge

R = V/I (4.2.4)
Fractions and reciprocals

Exam tips & common mistakes

Calculate parallel combinations first, then add series parts
Check: parallel total must be smaller than the smallest branch
Use V = IR on each part separately

Key facts from the syllabus (4.3.2)

Current is the same at every point in a series circuit
In parallel, the current from the source is larger than the current in each branch
The combined resistance of two resistors in parallel is less than either one alone
Lamps in parallel: each gets the full p.d. and can be switched independently; if one fails the others stay on
Extended the sum of currents into a junction equals the sum out (charge is conserved)
Extended the p.d. across parallel branches is the same

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

Core[1 mark]

State how the current varies around a series circuit.

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Method: Same everywhere.

The current is the same at every point

Core[2 marks]

Resistors of 4.0 Ω, 6.0 Ω and 10 Ω are in series. Calculate the combined resistance.

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Method: Add in series.

R = 4.0 + 6.0 + 10 = 20.0 Ω

Core[3 marks]

A 12 V battery is connected to 8.0 Ω and 4.0 Ω resistors in series. Calculate the current.

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Method: R total = 12 Ω.

I = 12 ÷ 12 = 1.00 A

Core[2 marks]

Four 1.5 V cells are connected in series. Calculate the total e.m.f.

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Method: Add e.m.f.s.

E = 4 × 1.5 = 6.00 V

Core[1 mark]

In a parallel circuit, compare the current from the source with the current in each branch.

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Method: Source current is larger.

The current from the source is larger than the current in each branch

Core[1 mark]

State how the combined resistance of two resistors in parallel compares with each one.

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Method: Less than either.

It is less than either resistor on its own

Core[2 marks]

State two advantages of connecting lamps in parallel in a lighting circuit.

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Method: Independence and full p.d.

Each lamp can be switched independently; if one fails the others stay lit; each gets the full supply p.d.

Core[2 marks]

Two identical lamps are in series. One breaks. State what happens to the other and explain.

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Method: Circuit broken.

It goes out, because the circuit is broken so no current flows

Core[2 marks]

Two resistors of 15 Ω and 25 Ω are in series. Calculate the combined resistance.

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Method: Add.

R = 40.0 Ω

Core[3 marks]

A 6.0 V supply is connected to 10 Ω and 20 Ω in series. Calculate the current.

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Method: R = 30 Ω.

I = 0.200 A

Core[2 marks]

Two 1.5 V cells and one 9.0 V battery are in series, all the same way. Calculate the total e.m.f.

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Method: Add.

E = 12.0 V

Core[2 marks]

Three identical lamps are in series with a 12 V battery. State the p.d. across each.

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Method: Share equally.

4.0 V

Core[2 marks]

State what happens to the current in a series circuit when another resistor is added.

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Method: R increases.

The current decreases

Core[2 marks]

State what happens to the total current from a supply when another resistor is added in parallel.

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Method: R decreases.

The total current increases

Core[2 marks]

Explain why house lights are wired in parallel.

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Method: Independent, full voltage.

Each light gets the full mains p.d. and can be switched on or off independently

Core[2 marks]

Two lamps are in parallel. One blows. State what happens to the other.

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Method: Independent.

It stays lit

Extended[2 marks]

A current of 3.0 A enters a junction and splits into two branches. One branch carries 1.2 A. Calculate the current in the other.

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Method: Currents in = currents out.

I = 3.0 − 1.2 = 1.80 A

Extended[2 marks]

Explain why the sum of the currents into a junction equals the sum of the currents out.

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Method: Charge conserved.

Charge is conserved: it cannot be created or destroyed at the junction

Extended[2 marks]

In a series circuit with a 9.0 V supply, the p.d. across one resistor is 3.5 V. Calculate the p.d. across the other (only two resistors).

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Method: p.d.s add in series.

V = 9.0 − 3.5 = 5.50 V

Extended[2 marks]

Two resistors in parallel are connected to a 6.0 V supply. State the p.d. across each.

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Method: Same p.d. in parallel.

6.0 V

Extended[2 marks]

Calculate the combined resistance of 6.0 Ω and 3.0 Ω in parallel.

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Method: 1/R = 1/R₁ + 1/R₂.

R = 2.00 Ω

Extended[2 marks]

Calculate the combined resistance of 20 Ω and 30 Ω in parallel.

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Method: Product over sum.

R = (20 × 30) ÷ 50 = 12.0 Ω

Extended[3 marks]

A 12 V supply is connected across 4.0 Ω and 12 Ω in parallel. Calculate the current in each branch and the total current.

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Method: I = V/R each.

I₁ = 3.00 A, I₂ = 1.00 A, total = 4.00 A

Extended[4 marks]

A 6.0 Ω and a 3.0 Ω resistor in parallel are connected in series with a 4.0 Ω resistor and a 12 V supply. Calculate the current from the supply.

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Method: Parallel = 2.0 Ω, total 6.0 Ω.

R = 6.00 Ω; I = 12 ÷ 6.0 = 2.00 A

Extended[3 marks]

For the circuit in the previous question, calculate the p.d. across the 4.0 Ω resistor and across the parallel pair.

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Method: V = IR.

V₄ = 2.0 × 4.0 = 8.00 V; parallel pair = 12 − 8 = 4.00 V

Extended[3 marks]

Three 12 Ω resistors are connected in parallel. Calculate the combined resistance.

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Method: 1/R = 3/12.

R = 4.00 Ω

Extended[3 marks]

A 10 Ω and a 15 Ω resistor are in series with a 10 V supply. Calculate the p.d. across each.

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Method: I = 10/25.

I = 0.400 A; V₁₀ = 4.00 V; V₁₅ = 6.00 V

Extended[3 marks]

Two identical lamps are in parallel across a 6.0 V supply, each drawing 0.30 A. Calculate the total current and the resistance of each lamp.

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Method: I adds; R = V/I.

Total I = 0.600 A; R = 6.0 ÷ 0.30 = 20.0 Ω

Extended[2 marks]

Currents of 0.40 A and 0.60 A leave a junction. Calculate the current entering it.

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Method: Sum.

I = 1.00 A

Extended[2 marks]

Calculate the combined resistance of 10 Ω and 40 Ω in parallel.

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Method: Product/sum.

R = 8.00 Ω

Extended[2 marks]

Calculate the combined resistance of two 50 Ω resistors in parallel.

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Method: Half.

R = 25.0 Ω

Extended[3 marks]

Calculate the combined resistance of 2.0 Ω, 3.0 Ω and 6.0 Ω in parallel.

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Method: Reciprocals.

R = 1.00 Ω

Extended[3 marks]

A 9.0 V supply is connected across 3.0 Ω and 6.0 Ω in parallel. Calculate the current in each branch and from the supply.

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Method: I = V/R.

3.00 A, 1.50 A; total 4.50 A

Extended[3 marks]

A 12 V supply is connected to 4.0 Ω in series with (6.0 Ω parallel with 12 Ω). Calculate the supply current.

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Method: Parallel = 4.0 Ω.

R = 8.00 Ω; I = 1.50 A

Extended[3 marks]

For the circuit in the previous question, calculate the current in the 6.0 Ω resistor.

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Method: V across pair = 6.0 V.

V = 1.5 × 4.0 = 6.0 V; I = 6.0 ÷ 6.0 = 1.00 A

Extended[3 marks]

A 20 Ω and an unknown resistor are in series across 12 V. The current is 0.40 A. Calculate the unknown resistance.

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Method: R total = 30 Ω.

R = 30 − 20 = 10.0 Ω

Extended[3 marks]

Two resistors in parallel have combined resistance 4.0 Ω. One is 12 Ω. Calculate the other.

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Method: 1/R₂ = 1/4 − 1/12.

R₂ = 6.00 Ω

Extended[2 marks]

Two lamps in series have p.d.s of 2.5 V and 3.5 V. Calculate the supply p.d.

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Method: Sum.

6.00 V

Extended[2 marks]

A lamp and a resistor are in parallel across 4.5 V. State the p.d. across the resistor.

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Method: Same in parallel.

4.5 V

Extended[3 marks]

Four 8.0 Ω resistors are connected as two parallel pairs in series. Calculate the total resistance.

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Method: Each pair 4.0 Ω.

R = 8.00 Ω

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]

Three resistors of 4.0 Ω, 6.0 Ω and 10 Ω are in series with a 12 V battery. Calculate the current and the p.d. across each resistor.

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Method: R = 20 Ω.

I = 0.600 A; 2.40 V, 3.60 V, 6.00 V

ChallengeCore[4 marks]

Explain, with reasons, three advantages of wiring house lights in parallel rather than in series.

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Method: Independence etc.

Each lamp has the full mains p.d. (full brightness); each can be switched independently; if one fails the others stay on

ChallengeCore[3 marks]

A circuit has two identical lamps in series with a cell. A third identical lamp is added in series. Explain the effect on the brightness of the lamps.

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Method: Higher R, smaller I.

Total resistance increases, current decreases, so all lamps are dimmer (each gets a third of the p.d.)

ChallengeExtended[5 marks]

A 12 V supply is connected to a 2.0 Ω resistor in series with a parallel combination of 3.0 Ω and 6.0 Ω. Calculate (a) the total resistance, (b) the supply current, (c) the current in the 3.0 Ω resistor.

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Method: Parallel = 2.0 Ω.

(a) 4.00 Ω (b) 3.00 A (c) p.d. across pair = 6.0 V; I = 2.00 A

ChallengeExtended[5 marks]

Resistors of 12 Ω, 6.0 Ω and 4.0 Ω are in parallel across 12 V. Calculate the current in each, the total current and the combined resistance.

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Method: I = V/R.

1.00 A, 2.00 A, 3.00 A; total 6.00 A; R = 2.00 Ω

ChallengeExtended[4 marks]

Two resistors in parallel draw 3.0 A from a 6.0 V supply. One resistor is 3.0 Ω. Calculate the other.

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Method: Branch currents.

I₁ = 2.0 A; I₂ = 1.0 A; R₂ = 6.00 Ω

ChallengeExtended[5 marks]

A 9.0 V battery supplies a circuit: 6.0 Ω in series with (R parallel with 12 Ω). The supply current is 1.0 A. Calculate R.

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Method: Total R = 9.0 Ω.

Parallel pair = 3.0 Ω; 1/R = 1/3 − 1/12; R = 4.00 Ω

ChallengeExtended[4 marks]

Explain, using conservation of charge, why the current entering a junction equals the current leaving it, and use it: 2.5 A enters; branches carry 0.80 A and I. Find I.

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Method: Charge conserved.

Charge cannot accumulate at a junction; I = 1.70 A

ChallengeExtended[4 marks]

Explain why adding more resistors in parallel decreases the total resistance but increases the total current drawn from a battery.

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Method: More paths.

Each new branch provides an extra path for charge, so more current flows for the same p.d. — total resistance is lower

ChallengeExtended[5 marks]

Four identical 10 Ω resistors are arranged: two in series, and this pair in parallel with the other two in series. Calculate the combined resistance and the current from a 5.0 V supply.

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Method: Two 20 Ω branches.

R = 10.0 Ω; I = 0.500 A

All IGCSE Physics topics ← 4.3.1 Circuit Diagrams and Circuit Components 4.3.3 Action and Use of Circuit Components →

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