Series and parallel circuits | Pearson Edexcel GCSE Physics Notes

Start here

The key idea

In series, current is the same everywhere. In parallel, potential difference is the same across each branch.

Series And Parallel Circuits

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

Series circuits

In a series circuit all components are connected in one loop. The current is the same everywhere. d., so adding more resistors in series increases the total resistance.

Parallel circuits

In a parallel circuit components are connected across separate branches. d.The total current is the sum of the currents in the branches.Adding more resistors in parallel decreases the total resistance, because there are more paths for the current to flow.

Why resistance falls in parallel

Adding a resistor in parallel provides an additional route for charge to flow, increasing the total current for the same potential difference.d. / current and the current rises, the total resistance falls.The total resistance of resistors in parallel is always less than the smallest individual resistance.

Household wiring

Home appliances are wired in parallel so each receives the full mains potential difference of 230 V and can be switched on and off independently.If they were in series, switching one off would break the circuit and stop all of them, and each would receive only a share of the voltage.

Key terms

Definitions to learn

Series circuit

A circuit with one path; current is the same throughout.

Parallel circuit

A circuit with branches; p.d. is the same across each branch.

Total resistance (series)

The sum of all individual resistances.

Branch

One of the separate paths in a parallel circuit.

Supply p.d.

The total potential difference provided by the cell or mains.

Worked example

Two resistors of 4 ohms and 7 ohms are connected in series. Find the total resistance.

In a series circuit, add the individual resistances.

Calculate 4 + 7.

Final answer

11 ohms

Exam habit

State the circuit rule before applying it. Series: same current everywhere, pd adds up.Parallel: same pd across each branch, currents add up. Never apply a series rule to a parallel branch.

Watch out

Do not add resistor values in parallel using the series rule.

Examiner tips

How to score full marks

1In series the current is the same; in parallel the potential difference is the same — state which you are using.
2Adding resistors in series increases total resistance; adding them in parallel decreases it.
3Total parallel resistance is smaller than the smallest single resistor — use this to check answers.

Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A 12 V supply is connected across two parallel branches. What is the potential difference across each branch?

Mark scheme

1.Use the parallel potential-difference rule.

12 V

2A current of 3.5 A splits into two parallel branches. One branch carries 1.2 A. Find the current in the other branch.

Mark scheme

1.Currents entering a junction equal currents leaving it.

2.3 A

3Explain why adding another resistor in parallel decreases the total resistance.

Mark scheme

1.Consider the number of paths available for charge.

An extra branch provides another path for charge, so more current flows for the same potential difference.

4In a series circuit, what is true about the current at different points?[1 mark]

Mark scheme

1.Recall the series current rule.

The current is the same at every point in the circuit (1)

5Two resistors of 4 ohms and 6 ohms are connected in series. Calculate the total resistance.[2 marks]

Mark scheme

1.Use R total = R1 + R2.
2.Add 4 and 6.

R total = R1 + R2 (1) = 4 + 6 = 10 ohms (1)

6A 12 V battery is connected to two resistors in series. The p.d. across the first resistor is 8 V. State the p.d. across the second resistor and the rule you used.[2 marks]

Mark scheme

1.Recall that series p.d.s add to the supply.
2.Subtract 8 from 12.

p.d.s in series add up to the supply p.d., so second p.d. = 12 - 8 = 4 V (1); rule: total p.d. is shared between components (1)

7Two identical lamps are connected in parallel to a 6 V supply. State the potential difference across each lamp and explain why.[2 marks]

Mark scheme

1.Recall the parallel p.d. rule.

Each lamp has 6 V across it (1), because in a parallel circuit the p.d. across each branch equals the supply p.d. (1)

8Explain why connecting two resistors in parallel gives a lower total resistance than either resistor alone, and why household appliances are wired in parallel.[3 marks]

Mark scheme

1.Link the extra branch to more paths for current.
2.Link more current to lower resistance.
3.Give two practical reasons for parallel wiring.

A second branch gives charge an extra path, so for the same p.d. the total current is larger (1); since resistance = p.d. / current, a larger current means a lower total resistance (1); appliances are wired in parallel so each gets the full mains voltage and can be switched independently without affecting the others (1)

9Three resistors of 2 ohms, 5 ohms and 3 ohms are connected in series to a 10 V supply. Calculate the total resistance and the current in the circuit.[3 marks]

Mark scheme

1.Add the resistances in series.
2.Use V = I x R to find current.
3.Rearrange to I = V / R.

R total = 2 + 5 + 3 = 10 ohms (1); I = V / R = 10 / 10 (1) = 1.0 A (1)

10A 6 V supply is connected to a series circuit containing a 4 ohm resistor (R1) and an unknown resistor (R2). The current in the circuit is 0.5 A. Calculate the resistance of R2 and the p.d. across each resistor.[3 marks]

Mark scheme

1.Total resistance R = V / I = 6 / 0.5.
2.R2 = R total - R1.
3.p.d. across R1 = I x R1; p.d. across R2 = I x R2.

R total = 6 / 0.5 = 12 ohms (1); R2 = 12 - 4 = 8 ohms (1); V1 = 0.5 × 4 = 2 V; V2 = 0.5 × 8 = 4 V (check: 2 + 4 = 6 V) (1)

11In a parallel circuit a 9 V supply drives current through a 3 ohm branch and a 9 ohm branch. Calculate the current in each branch and the total current from the supply.[3 marks]

Mark scheme

1.Same p.d. (9 V) across both branches.
2.I1 = 9 / 3; I2 = 9 / 9.
3.Total current = I1 + I2.

I1 = 9 / 3 = 3.0 A (1); I2 = 9 / 9 = 1.0 A (1); total current = 3.0 + 1.0 = 4.0 A (1)

12A student adds a third lamp in parallel to a circuit that already has two lamps in parallel. Explain the effect on (a) the brightness of the original two lamps and (b) the total current from the battery.[2 marks]

Mark scheme

1.P.d. across each lamp stays the same when another is added in parallel.
2.Same p.d. means same current in each lamp, so same brightness.
3.Total current increases because a new branch is added.

(a) The p.d. across each of the original lamps remains equal to the supply voltage, so the current through each is unchanged and they stay the same brightness (1); (b) the third lamp provides an extra branch, adding a new current, so the total current from the battery increases (1)

13A series circuit contains a cell of e.m.f. 9 V and internal resistance 1 ohm connected to an external resistor of 8 ohms. Calculate the current, the terminal voltage of the cell, and the power dissipated in the external resistor.[3 marks]

Mark scheme

1.Total resistance = internal + external = 1 + 8 = 9 ohms.
2.Current I = e.m.f. / R total = 9 / 9.
3.Terminal voltage = e.m.f. - I x internal resistance.
4.Power in external resistor = I² x R external.

I = 9 / 9 = 1.0 A (1); terminal voltage = 9 - (1.0 × 1) = 8 V (1); P = I² x R = 1.0² × 8 = 8 W (1)

14Describe an experiment to determine the resistance of an unknown component using a variable power supply, an ammeter and a voltmeter. Include the circuit arrangement, how to collect data, how to process it, and any precautions to ensure a fair and accurate result.[4 marks]

Mark scheme

1.Describe circuit: component in series with ammeter; voltmeter in parallel with component.
2.Vary the supply voltage; record I and V for several settings.
3.Calculate R = V / I for each pair, or plot V against I and find gradient.
4.Precautions: allow component to cool between readings; use a rheostat to fine-tune voltage.

Connect the unknown component in series with an ammeter and in parallel with a voltmeter; use a variable power supply or a cell with a rheostat in series (1); set the voltage to several different values, recording the ammeter reading (I) and voltmeter reading (V) each time (1); calculate R = V / I for each pair of readings and take a mean, or plot V on the y-axis against I on the x-axis and find the gradient, which equals the resistance if the component is ohmic (1); to ensure fair and accurate results, allow time for the component to return to room temperature between readings, as heating changes resistance; also take repeat readings at each voltage to identify and eliminate anomalous results (1)

15A circuit contains two resistors: R1 = 6 ohms in series with the parallel combination of R2 = 6 ohms and R3 = 12 ohms. The supply is 12 V. Calculate: (a) the resistance of the parallel combination, (b) the total circuit resistance, (c) the total current, (d) the p.d. across the parallel combination, and (e) the current through R3.[5 marks]

Mark scheme

1.Parallel combination: 1/R = $\frac{1}{6}$ + $\frac{1}{1}2$ .
2.R total = R1 + parallel combination.
3.I total = V / R total.
4.V across parallel = I total x R parallel.
5.I through R3 = V parallel / R3.

(a) 1/R =

\frac{1}{6}

\frac{1}{1}2

\frac{2}{1}2

\frac{1}{1}2

\frac{3}{1}2

\frac{1}{4}

, so R = 4 ohms (1); (b) R total = 6 + 4 = 10 ohms (1); (c) I total = 12 / 10 = 1.2 A (1); (d) V parallel = 1.2 × 4 = 4.8 V (1); (e) I3 = 4.8 / 12 = 0.4 A (1)

Official exam-board sources

Specification Past papers

Browse all Physics topics