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AQA GCSE Physics · Topic 4.2 Electricity
Series and Parallel Circuits, with Calculations
In this lesson you'll learn to: state the rules for current, potential difference and
resistance in series and parallel circuits, explain why adding a resistor in parallel
decreases the total resistance, use V = I R to work through series, parallel and combined
circuits, and explain how a sensing circuit shares the voltage.
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Prefer to watch? Scroll to the bottom for the video on this topic
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A video walkthrough is coming soon — it will appear at the bottom of this page
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Start here — have a play
Don't read anything yet. On tab 1, switch on the ammeters, then the voltmeters, and change the resistor. Do the ammeters ever disagree? On tab 2, switch on the meters and add a branch: watch what happens to the currents in the branches that were already there. Tab 3 mixes the two. Tab 4 is a sensing circuit: heat the thermistor, darken the room for the LDR, and watch the voltmeter.
1 coulomb of charge1 joule of energy
Battery
Resistor
What the meters say
1 coulomb of charge1 joule of energy
Battery
Branches
What the meters say
Circuit
1 coulomb of charge1 joule of energy
Battery
Sensor
charge1 joule of energy per coulomb
Temperature
cold
hot
Who gets the 6 V?
Charges flow in the direction of conventional current, from + to −. Cells are 1.5 V. Meters are ideal.
▲ Every reading comes from V = I R, so you can check each one with your calculator.
Seen the rules in action? Now use them.
Almost every question on this topic is a V = I R calculation with one of the series or parallel rules on top. The notes give you the rules, then work through one of each type.
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Revision notes
The rules
| Series | Parallel | |
|---|---|---|
| Current | Same everywhere: I₁ = I₂ = I₃ | Splits between the branches, then adds back up: Itotal = I₁ + I₂ + … |
| Potential difference | Shared: Vtotal = V₁ + V₂ + … | Same across every branch as across the battery |
| Resistance | Adds up: Rtotal = R₁ + R₂ | Total is less than the smallest resistor |
⭐ Why parallel lowers the total resistance
Adding a resistor in parallel gives the charge another path. The p.d. across each branch stays the same, so the old branches carry the same current as before, and the new branch adds its own current. More total current from the same p.d. means a smaller total resistance (R = V ÷ I).
Ω
Every calculation: V = I R
Use it on one component at a time, or on the whole circuit, but keep the V, I and R from the same place.
V volts · I amps · R ohms✓
Check it adds up
Series: the component voltages must add up to the battery. Parallel: the branch currents must add up to the battery current.
Sensing circuits
A fixed resistor in series with a thermistor or LDR shares the battery's p.d. The component with the bigger resistance takes the bigger share of the voltage, because the current is the same through both.
- Thermistor gets hotter → its resistance decreases → it takes a smaller share → the voltmeter across it reads less.
- LDR gets darker → its resistance increases → it takes a bigger share → the voltmeter across it reads more. That rising voltage can switch on a street light.
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How to write it in the exam
Every question below follows the same plan: decide series or parallel, apply the rule, then use V = I R. Try each one before you read the answer.
The plan for any circuit calculation
1. Series or parallel? — which rule applies here
2. Total first — total resistance, then the current from the battery
3. Then each part — V = I R for each component
4. Check — voltages add up in series, currents add up in parallel
2. Total first — total resistance, then the current from the battery
3. Then each part — V = I R for each component
4. Check — voltages add up in series, currents add up in parallel
decide & checkthe whole circuiteach component
Question 1 · series · 5 marks
A 9 V battery is connected in series with a 2 Ω resistor and a 4 Ω resistor. Calculate (a) the total resistance, (b) the current, and (c) the p.d. across each resistor.
- (a) Series, so resistances add: R = 2 + 4 = 6 Ω (1)
- (b) I = V ÷ R = 9 ÷ 6 (1) = 1.5 A (1)
- (c) Same current through each. V₁ = 1.5 × 2 = 3 V; V₂ = 1.5 × 4 = 6 V (1)
- Check: 3 + 6 = 9 V, the battery's p.d. ✓ (1)
Question 2 · parallel · 5 marks
A 12 V battery is connected to a 4 Ω resistor and a 12 Ω resistor in parallel. Calculate the current in each resistor, the current from the battery, and the total resistance.
- Parallel, so each resistor has the full 12 V across it (1)
- I₁ = 12 ÷ 4 = 3 A; I₂ = 12 ÷ 12 = 1 A (1)
- Battery current = 3 + 1 = 4 A (1)
- Total resistance = V ÷ I = 12 ÷ 4 (1) = 3 Ω (1)
Check: 3 Ω is less than the smallest resistor (4 Ω), as it must be in parallel.
Question 3 · parallel junction · 1 mark
The battery current is 2.5 A. It splits into two branches. One branch carries 0.8 A. What is the current in the other branch?
- Currents in parallel add up to the total: 2.5 − 0.8 = 1.7 A (1)
Question 4 · series and parallel · 4 marks
A 6 V battery is connected to a 2 Ω resistor in series with two 6 Ω resistors that are in parallel with each other. Calculate the current from the battery and the p.d. across the parallel pair.
- Two equal resistors in parallel share the current equally, so together they act like half of one: 6 ÷ 2 = 3 Ω (1)
- Total resistance = 2 + 3 = 5 Ω, so I = 6 ÷ 5 = 1.2 A (1)
- p.d. across the 2 Ω resistor = 1.2 × 2 = 2.4 V (1)
- p.d. across the parallel pair = 6 − 2.4 = 3.6 V (1), so each 6 Ω resistor carries 3.6 ÷ 6 = 0.6 A
Question 5 · sensing circuit · 5 marks
A 6 V battery is connected to a 1000 Ω fixed resistor in series with a thermistor. A voltmeter across the thermistor reads 2 V. (a) Calculate the resistance of the thermistor. (b) Describe what happens to the voltmeter reading when the thermistor is heated.
- (a) p.d. across the fixed resistor = 6 − 2 = 4 V (1)
- Current = 4 ÷ 1000 = 0.004 A, the same through both, because they are in series (1)
- Thermistor resistance = 2 ÷ 0.004 = 500 Ω (1)
- (b) Heating decreases the thermistor's resistance (1), so it takes a smaller share of the 6 V and the voltmeter reading decreases (1)
Shortcut: in series, the voltages are in the same ratio as the resistances. The thermistor has half the fixed resistor's p.d., so it has half its resistance: 500 Ω.
📘 Now do it in your workbook
Fill in the series and parallel rules, work through three sensing circuits (LDR, thermistor and variable resistor), then the three 6-mark calculations, including designing a circuit to double the current.
AQA GCSE Physics Workbook · ELECTRICITY · pages 9, 10 & 11The full workbook covers the whole of Topic 4.2 Electricity, with exam-style questions and worked answers.
✅ Can you do it? Tick as you go
0% complete🎉 Nice work! You've ticked off every objective for this spec point.
Don't forget to hit “Mark complete” at the bottom of the lesson.
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Prefer to watch? Here's the whole thing
Video walkthrough — coming soon
Everything above, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it.
Everything above, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it. The video is on its way.
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Video coming soon
The walkthrough for this lesson is being filmed and will appear right here. Until then, the animation and the notes above cover every mark.
Now put it into practice
Pages 9, 10 and 11 of the workbook have the series and parallel rules, three sensing circuits and three 6-mark calculations, each with full answers at the back.
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