4.2.2 Series and Parallel Circuits, with Calculations — Physics with Kate

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4.2.2 · sensing circuits 4.2.1.4

AQA GCSE Physics · Topic 4.2 Electricity

2.7 · 2.8 · 2.17 · 2.18 · 2.19

Edexcel International GCSE Physics · Section 2 Electricity · 2 (c) Energy and voltage in circuits

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.

▶ Prefer to watch? Scroll to the bottom for the video on this topic ▶ 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

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.

Read the notes ↓

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Revision notes

The rules

	Series	Parallel
Current	Same everywhere: I₁ = I₂ = I₃	Splits between the branches, then adds back up: I_total = I₁ + I₂ + …
Potential difference	Shared: V_total = V₁ + V₂ + …	Same across every branch as across the battery
Resistance	Adds up: R_total = 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).

Edexcel onlySeries or parallel? Advantages and disadvantages

Edexcel iGCSE (2.7) asks you to compare the two ways of wiring lamps. Every point comes from the rules above: in series the current has one path and the voltage is shared; in parallel every branch has the full voltage and its own path.

	Advantages	Disadvantages
Series	Simple circuit, fewer wires · all the bulbs work from one switch · each bulb takes a share of the voltage, so lower-voltage bulbs can be used and they are less likely to overheat	If one bulb breaks, they all go out · bulbs can't be switched on and off separately · adding more bulbs makes each one dimmer
Parallel	Each bulb gets the full voltage, so it stays bright · if one bulb breaks, the others stay on · each bulb can have its own switch	More wire, more complex · the cell supplies a larger total current, so it goes flat sooner

Edexcel only⭐ Why your house is wired in parallel

Every appliance gets the full 230 V, each one has its own switch, and one broken lamp doesn't turn off everything else. Decorative lights are often in series: one switch, and each tiny bulb only needs a small share of the mains voltage.

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

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.

These are the exam-style questions from pages 14 and 15 of the Edexcel workbook.

Page 14 · decorative lamps · 13 marks

Lamp X runs from the mains. It has seven identical bulbs connected in series. Each bulb has a working resistance of 390 Ω and the voltage across each bulb is 33 V.

(a) A disadvantage of series: if one bulb breaks, all the bulbs go out (or they can't be controlled separately) (1)
(b) An advantage: less likely to overheat / simpler circuit with fewer wires / lower-voltage bulbs can be used / one switch controls them all (1)
(c)(i) voltage = current × resistance, V = I R (1)
(c)(ii) 33 = I × 390 (1); I = 33 ÷ 390 (1) = 0.085 A (1)
(c)(iii) E = I × V × t (1); 2.5 hours = 2.5 × 3600 = 9000 s (1); use all 7 bulbs: V = 7 × 33 = 231 V (1); E = 0.085 × 231 × 9000 ≈ 180 000 J (1)
(d) One bulb in Lamp Y is brighter (1): with five bulbs sharing the mains, each has a larger voltage (about 46 V), Lamp Y has a lower total resistance so a bigger current flows, and more energy is transferred to each bulb every second (2)

Where marks go missing: in (c)(iii), using 33 V (one bulb) instead of 231 V (the whole lamp), or leaving the time in hours.

Page 15 · Question 1 · LDR sensing circuit · 6 marks

An LDR is in series with a 1000 Ω fixed resistor and a 6 V battery. The output across the LDR switches on a security light when it gets dark.

(a) Light decreases → LDR resistance increases, so the current decreases (1) and the p.d. across the LDR increases (1)
(b) p.d. across the fixed resistor = 6 − 4 = 2 V, so I = 2 ÷ 1000 = 0.002 A (1); R = 4 ÷ 0.002 = 2000 Ω, because the same current flows through both in series (1)
(c) Total R = 5000 + 1000 = 6000 Ω (1); I = 6 ÷ 6000 = 0.0010 A (1)

Same method for Q2 and Q3: thermistor (b) 1000 Ω, (c) 0.0020 A · variable resistor (b) 600 Ω, (c) 0.0050 A.

📘 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.

Fill in the rules and make up meter readings, draw series and parallel circuits, complete the advantages and disadvantages table, answer the 13-mark decorative lamps question, then three 6-mark sensing-circuit calculations.

AQA GCSE Physics Workbook · ELECTRICITY · pages 9, 10 & 11 Edexcel iGCSE Physics Workbook · ELECTRICITY · pages 11, 13, 14 & 15

⬇ Free: pages 9, 10 & 11 (PDF) 📄 Full workbook PDF · £2.99 Paperback on Amazon ⬇ Free: pages 11, 13, 14 & 15 (PDF) 📄 Full workbook PDF · £2.99 Paperback on Amazon

The full workbook covers the whole Electricity topic, with exam-style questions and worked answers.

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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.

🎬

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.

Pages 11, 13, 14 and 15 of the Edexcel workbook are free: the rules, the advantages and disadvantages table, the decorative lamps question and three sensing-circuit calculations.

⬇ Free: pages 9, 10 & 11 (PDF) 📄 Full workbook PDF · £2.99 Paperback on Amazon ⬇ Free: pages 11, 13, 14 & 15 (PDF) 📄 Full workbook PDF · £2.99 Paperback on Amazon