A device draws 0.50 A from a 9.0 V supply. Calculate the power it dissipates. [1 mark]

EnglandPhysicsSyllabus dot point

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

Combining resistors in series and parallel, the application of Kirchhoff's two laws, the conservation of charge and energy in circuits, and power dissipation given by P = VI, P = I squared R and P = V squared over R.

A focused answer to AQA A-Level Physics 3.5.1.4, covering the rules for combining resistors in series and parallel, the application of Kirchhoff's two laws, conservation of charge and energy in circuits, and the three equations for electrical power dissipation.

Generated by Claude Opus 4.810 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
Combining resistors
Kirchhoff's two laws
Series and parallel behaviour
Power and energy
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What this dot point is asking

AQA specification point 3.5.1.4 wants you to combine resistors in series and parallel, apply Kirchhoff's two laws, use the conservation of charge and energy in circuits, and calculate electrical power and energy with the three power equations.

Combining resistors

The series result follows from Kirchhoff's second law: the same current $I$ flows through each resistor, so the total voltage $V = IR_1 + IR_2 + IR_3 = I(R_1 + R_2 + R_3)$ , giving an effective resistance that is the sum.

The parallel result follows from Kirchhoff's first law: each branch has the same voltage $V$ across it, so the total current $I = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dfrac{V}{R_3} = V\left(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right)$ , and dividing by $V$ gives the reciprocal sum. The parallel combination always has a resistance smaller than the smallest individual resistor, because adding extra paths makes it easier for current to flow.

Kirchhoff's two laws

The second law is best read as a bookkeeping statement about energy per unit charge. A charge that travels once around a complete loop gains potential energy at the source (the EMF) and loses it across the components (the $IR$ drops), returning to its starting potential. The two must balance exactly.

Series and parallel behaviour

In series, the current is the same through every component, and the supply voltage is shared between them in proportion to their resistances. Remove or break one component and the whole loop stops conducting, which is why old fairy-light strings all go out when one bulb fails.

In parallel, each branch has the full supply voltage across it, and the total current is the sum of the branch currents. Branches are independent, so removing one leaves the others working, which is why household circuits are wired in parallel.

Power and energy

The three forms are algebraically identical (substitute $V = IR$ ), but each is convenient in a different situation. Use $P = I^2 R$ when the current is fixed (as in a series chain) and $P = \dfrac{V^2}{R}$ when the voltage is fixed (as across parallel branches). For example, in a series chain the largest resistor dissipates the most power, whereas across a fixed voltage the smallest resistor dissipates the most.

Combining resistors and finding branch power

Two resistors, $12 \text{ }\Omega$ and $6.0 \text{ }\Omega$ , are connected in parallel across a $9.0 \text{ V}$ supply of negligible internal resistance. Find the total current drawn and the power dissipated in the $12 \text{ }\Omega$ resistor.

step 1: Find the total resistance

$\dfrac{1}{R} = \dfrac{1}{12} + \dfrac{1}{6.0} = \dfrac{1}{12} + \dfrac{2}{12} = \dfrac{3}{12}$ , so $R = 4.0 \text{ }\Omega$ .

step 2: Find the total current

$I = \dfrac{V}{R} = \dfrac{9.0}{4.0} = 2.25 \text{ A}$ .

step 3: Use the branch voltage for the power

Each branch has the full $9.0 \text{ V}$ across it, so for the $12 \text{ }\Omega$ resistor use $P = \dfrac{V^2}{R} = \dfrac{(9.0)^2}{12}$ .

step 4: Evaluate

$P = \dfrac{81}{12} = 6.75 \text{ W} \approx 6.8 \text{ W}$ .

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Q1. Two $4.0 \text{ }\Omega$ resistors are connected in series. State the total resistance. [1 mark]

Cue. $4.0 + 4.0 = 8.0 \text{ }\Omega$ .

Q2. A device draws $0.50 \text{ A}$ from a $9.0 \text{ V}$ supply. Calculate the power it dissipates. [1 mark]

Cue. $P = VI = 9.0 \times 0.50 = 4.5 \text{ W}$ .

Q3. State which conservation law underlies Kirchhoff's second law. [1 mark]

Cue. Conservation of energy.

Exam-style practice questions

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

AQA 20185 marksA

6.0 \text{ }\Omega

resistor and a

3.0 \text{ }\Omega

resistor are connected in parallel, and this combination is connected in series with a

4.0 \text{ }\Omega

resistor across a

12 \text{ V}

supply of negligible internal resistance. Calculate the total resistance of the circuit and the total power dissipated.

Show worked answer →

First combine the parallel pair: $\dfrac{1}{R_p} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{6.0} + \dfrac{2}{6.0} = \dfrac{3}{6.0}$ , so $R_p = 2.0 \text{ }\Omega$ .

Add the series resistor: $R_{\text{total}} = 2.0 + 4.0 = 6.0 \text{ }\Omega$ .

Power from the supply: $P = \dfrac{V^2}{R} = \dfrac{(12)^2}{6.0} = 24 \text{ W}$ .

Markers reward the reciprocal sum for the parallel pair, adding the series resistor afterwards, and a correct power equation. A frequent error is treating all three resistors as one parallel group.

AQA 20224 marksState Kirchhoff's two laws and explain the conservation principle that each one expresses.

Show worked answer →

Kirchhoff's first law: the sum of the currents entering a junction equals the sum of the currents leaving it. This expresses conservation of charge, since charge cannot accumulate at a point in a steady circuit.

Kirchhoff's second law: around any closed loop, the sum of the EMFs equals the sum of the potential differences (the sum of $IR$ terms). This expresses conservation of energy, since a unit charge taken once around a loop returns to its starting potential, so the energy gained from sources equals the energy dissipated.

Markers reward correct statements of both laws and correctly pairing each with conservation of charge and conservation of energy respectively.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)