Welcome! Why do old fairy lights all go out when one bulb breaks, but your house lights don’t? In this lesson you’ll learn the rules for series and parallel circuits, then design the lighting for a treehouse. It takes about one hour. Work through the steps in order.
Before you start
- You need: your exercise book or paper, a pen, a ruler and a calculator.
- Write the title and date: Series and parallel circuits – Circuit Builder.
- The golden rule: write your answer first, then tap Show answer. Correct mistakes in a different colour.
- Look out for the blue 📝 boxes. Copy each one into your book neatly – these are your revision notes.
📝 Copy into your book
By the end of this lesson I will be able to:
1. state the rules for current, potential difference and resistance in series and parallel circuits
2. use V = IR in circuit calculations
3. explain why houses are wired in parallel.
Step 1: Starter (5 minutes)
A string of old-style fairy lights goes completely dark when just one bulb breaks. In your house, when one bulb breaks the others stay on. Why do you think they behave differently? Try drawing both circuits.
Show answer
The fairy lights are wired in series – one single loop. A broken bulb breaks the loop, so no current flows anywhere. House lights are wired in parallel – each light has its own branch, so if one breaks, current still flows through the others.
📝 Copy into your book
SERIES: components in one loop. If one breaks, they all stop.
PARALLEL: components on separate branches. If one breaks, the others keep working.
Step 2: Read and write (8 minutes)
First, copy these three questions into your book. Then read the text below carefully and answer them in full sentences.
- What is the same for every component in a series circuit?
- What is the same for every branch in a parallel circuit?
- What happens to the total resistance when you add resistors in parallel?
📖 Read: Series and parallel rules
In a series circuit, there is only one path, so the current is the same through every component. The potential difference of the supply is shared between the components. The total resistance is the sum of the individual resistances: Rtotal = R1 + R2. Adding more resistors in series increases the total resistance.
In a parallel circuit, there are several branches. The potential difference across each branch is the same as the supply. The total current from the supply is the sum of the currents through the separate branches. Adding a resistor in parallel gives the current another path, so the total resistance decreases. The total resistance of two resistors in parallel is always less than the smallest individual resistance.
Show answers
(a) The current.
(b) The potential difference – it’s the same as the supply p.d.
(c) It decreases – it becomes less than the smallest resistance.
Step 3: Key facts (6 minutes)
These rules are worth marks in almost every electricity paper. Copy the table carefully:
📝 Copy into your book
| Series | Parallel | |
|---|---|---|
| Current | Same everywhere | Splits between branches; total = sum of branches |
| Potential difference | Shared between components | Same across each branch |
| Total resistance | Rtotal = R1 + R2 | Less than the smallest resistor |
potential difference = current × resistance V = I R
Step 4: Circuit calculations (10 minutes)
Circuit 1 (series): a 4 Ω resistor and a 6 Ω resistor are in series with a 12 V battery.
Q1 Calculate the total resistance.
Show answer
R = 4 + 6 = 10 Ω
Q2 Calculate the current.
Show answer
I = V ÷ R = 12 ÷ 10 = 1.2 A (the same through both resistors)
Q3 Calculate the p.d. across each resistor. Check they add up to 12 V.
Show answer
4 Ω: V = 1.2 × 4 = 4.8 V
6 Ω: V = 1.2 × 6 = 7.2 V
4.8 + 7.2 = 12 V ✓
Circuit 2 (parallel): the same two resistors are now connected in parallel with the 12 V battery.
Q4 Calculate the current through each resistor and the total current from the battery.
Show answer
Each branch has the full 12 V.
4 Ω: I = 12 ÷ 4 = 3 A
6 Ω: I = 12 ÷ 6 = 2 A
Total = 3 + 2 = 5 A – much more than in series, because the total resistance is lower.
✅ Score out of 4.
Step 5: Main task – Circuit Builder (14 minutes)
You’re designing the lighting for a treehouse! You have a 12 V battery, three 12 V lamps, some switches and wires. The owners want:
- All three lamps to shine at full brightness.
- The other lamps to stay on if one lamp breaks.
- Each lamp to be switched on and off by itself.
- One main switch that turns everything off.
Your task: draw the circuit diagram with a ruler, using correct circuit symbols. Then explain how your design meets each of the four requirements.
Stuck? A 12 V lamp only shines at full brightness if it gets the full 12 V. Which type of circuit gives every branch the full supply p.d.?
Show a model design
Design: the three lamps are connected in parallel, each on its own branch with its own switch. The main switch is placed next to the battery, before the circuit splits into branches.
1. In parallel, each branch gets the full 12 V, so every lamp is at full brightness. (In series they would share the 12 V – only 4 V each – and be dim.)
2. Each lamp has its own path, so if one breaks, current still flows through the others.
3. A switch in each branch only breaks that branch.
4. The main switch is in the part of the circuit that all the current flows through, so opening it stops the current to every lamp.
✅ Check: did you use proper symbols, and explain all four requirements?
Step 6: Exam practice (12 minutes)
Look at the marks – they tell you how many points to make. Answer in full sentences, then mark yourself.
Q1 (2 marks) A 15 Ω resistor and a 25 Ω resistor are connected in series. Calculate the total resistance.
Show model answer
R = R1 + R2 = 15 + 25 (1) = 40 Ω (1)
Q2 (3 marks) Explain why the lights in a house are connected in parallel rather than in series.
Show model answer
In parallel each light gets the full supply p.d. (1), so they all work at full brightness. Each light can be switched on and off independently (1). If one light breaks, the others keep working because current can still flow through their branches (1).
Q3 (4 marks) A 3.0 Ω resistor and a 6.0 Ω resistor are connected in parallel to a 6.0 V battery. Calculate the total current from the battery.
Hint: work out the current in each branch first.
Show model answer
Each branch has 6.0 V across it (1).
3.0 Ω branch: I = 6.0 ÷ 3.0 = 2.0 A (1)
6.0 Ω branch: I = 6.0 ÷ 6.0 = 1.0 A (1)
Total = 2.0 + 1.0 = 3.0 A (1)
✅ Add up your marks out of 9 and write the score in your book.
Step 7: Exit ticket (5 minutes)
- In a series circuit, the potential difference is…
- Adding a resistor in parallel makes the total resistance smaller because…
- One thing I’m still not sure about is…
🎉 Well done – lesson complete! Your 📝 boxes are your revision notes for this topic. Want more? Try the Circuits and mains electricity questions.
