What are series circuits?

A larger resistor in a series chain takes a larger share of the voltage, because the same current through a bigger resistance produces a bigger V = IR.

What are parallel circuits?

Adding more parallel branches gives the current more paths, so the total current rises and the combined resistance falls.

Find the combined resistance of 1.0\ k and 1.0\ k in parallel. [2 marks]

A 330\ and a 470\ resistor are in series. Find the total resistance. [1 mark]

WJEC Eduqas Eduqas 2020-style practice: A 9.0\ V supply is connected to a 220\ resistor in series with a parallel combination of a 300\ and a 600\ resistor. Calculate the total resistance of the circuit and the current drawn from the supply.

Combine the parallel pair first using reciprocals: 1R_p = 1300 + 1600 = 2 + 1600 = 3600, so R_p = 200\. (Check: smaller than the smaller branch, 300\.) Add the series resistor: R_total = 220 + 200 = 420\. Current from the supply by Ohm's law: I = VR = 9.0420 = 0.0214\ A = 21\ mA. Markers reward R_p = 200\, the total 420\ and the current 21\ mA. The usual error is adding the parallel resistances directly to get 900\.

EnglandElectronicsSyllabus dot point

How do current, voltage and resistance behave in series and parallel circuits?

Series and parallel circuits: the rules for current and voltage in each, and combining resistors in series and in parallel.

An Eduqas GCSE Electronics answer on series and parallel circuits: how current and voltage divide in each, the rules that current is shared in parallel and voltage is shared in series, and combining resistors in series (add) and parallel (reciprocals).

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Eduqas wants you to know how current and voltage behave in series and in parallel circuits, and to combine resistors connected either way. These rules let you work out the current and voltage anywhere in a network, which is the basis of analysing every circuit in the course.

The answer

Series circuits

Parallel circuits

Combining resistors

Examples in context

The series and parallel rules underlie every circuit in this course. The potential divider is a series pair where the output is the voltage shared onto the lower resistor. A loaded sensor circuit puts the load resistance in parallel with part of the divider. LED indicators are often placed in parallel so each has the full supply across its own current-limiting resistor. Recognising the series and parallel groups in a circuit, and reducing them, is the first step in almost every calculation.

Try this

Q1. Two resistors are in series. State what is the same through both and what is shared between them. [2 marks]

Cue. The current is the same through both; the supply voltage is shared between them.

Q2. Find the combined resistance of $1.0\ \text{k}\Omega$ and $1.0\ \text{k}\Omega$ in parallel. [2 marks]

Cue. Equal resistors in parallel halve: $R = \frac{1000}{2} = 500\ \Omega$ .

Q3. A $330\ \Omega$ and a $470\ \Omega$ resistor are in series. Find the total resistance. [1 mark]

Cue. Series add: $330 + 470 = 800\ \Omega$ .

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20204 marksA

9.0\ \text{V}

supply is connected to a

220\ \Omega

resistor in series with a parallel combination of a

300\ \Omega

and a

600\ \Omega

resistor. Calculate the total resistance of the circuit and the current drawn from the supply.

Show worked answer →

Combine the parallel pair first using reciprocals: $\dfrac{1}{R_p} = \dfrac{1}{300} + \dfrac{1}{600} = \dfrac{2 + 1}{600} = \dfrac{3}{600}$ , so $R_p = 200\ \Omega$ . (Check: smaller than the smaller branch, $300\ \Omega$ .)

Add the series resistor: $R_\text{total} = 220 + 200 = 420\ \Omega$ .

Current from the supply by Ohm's law: $I = \dfrac{V}{R} = \dfrac{9.0}{420} = 0.0214\ \text{A} = 21\ \text{mA}$ .

Markers reward $R_p = 200\ \Omega$ , the total $420\ \Omega$ and the current $21\ \text{mA}$ . The usual error is adding the parallel resistances directly to get $900\ \Omega$ .

Eduqas 20212 marksIn a series circuit a

3.0\ \text{V}

p.d. appears across resistor A and a

5.0\ \text{V}

p.d. across resistor B, with nothing else in the circuit. State the supply voltage and explain your reasoning.

Show worked answer →

Supply voltage (1 mark): $3.0 + 5.0 = 8.0\ \text{V}$ .

Reasoning (1 mark): in a series circuit the supply voltage is shared between the components, so the individual potential differences add up to the supply voltage (this is conservation of energy: each coulomb gives up all the energy the supply gave it).

Markers reward the value $8.0\ \text{V}$ and the statement that series voltages add to the supply.

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Sources & how we know this

WJEC Eduqas GCSE (9-1) Electronics specification (C490) — WJEC Eduqas (2017)