S2 Science | Simple Circuits: Current and Voltage in Series and Parallel

S2 Science | Simple Circuits: Current and Voltage in Series and Parallel — Tutor Prep Guide

2026-11-15

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Students can draw a circuit but cannot predict what happens when one bulb is removed. This lesson uses two rules, one about current and one about voltage, to turn a diagram into a system they can reason about.

Objectives — Students can: (1) explain that a circuit must be closed for current to flow and name the role of the cell, switch and bulb; (2) use the rules that current is the same everywhere in series and voltage is the same across parallel branches to predict what happens when a component is removed; (3) divide voltage by current to find resistance and use current to explain bulb brightness.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Circuit components	What does each of a cell, bulb, switch and wire do?	Cell supplies energy, bulb uses it, switch controls, wires connect
Unit of current	What is the unit of current?	The ampere, A
Unit of voltage	What is the unit of voltage?	The volt, V

02 Lesson flow (45 minutes)

Time	Stage	Tutor does	Student does
0-5 min	Start with a prediction	Draw two bulbs in series and ask what happens if one is removed	Predict first, then give a reason
5-17 min	Closed circuits	Build a cell, bulb and switch circuit and operate the switch	Explain the difference between an open and a closed circuit
17-29 min	Series and parallel	Build one of each and measure current and voltage	State that series current is equal, parallel voltage is equal
29-36 min	Resistance and brightness	Use 4.5 V and 0.3 A to find 15 ohms, then compare two bulbs in series at 2.25 V each	Calculate 15 ohms and explain why the bulbs dim
36-42 min	Everyday comparison	Ask whether household appliances are in series or parallel, and why	Answer parallel, so each can be switched independently
42-45 min	Exit check	Ask for the total current of two 0.3 A bulbs in parallel	Answer 0.6 A

03 Teaching script (read it out)

Scene 1: No closed circuit, nothing happens

Start with hands, not definitions. Give each group a cell, a bulb, a switch and two wires and ask them to make the bulb light without asking questions. Most succeed immediately but cannot explain why. Only then draw the circuit diagram: current leaves the positive terminal, passes through the wire, the bulb and the switch, and returns to the negative terminal. Remove one wire and the bulb goes out, which is the definition of an open circuit. Emphasise that current is not used up along the way; the cell provides energy, the bulb converts electrical energy into light and heat, and the same current flows at every point in a single loop. The common misconception is that current is consumed, which must be corrected here.

Ask students to trace the diagram with a finger and say where current leaves and returns

Scene 2: Series and parallel: two rules cover everything

Do not start with formulas. Start with two observable rules. Series has one path, so the current is the same everywhere, while the voltage is shared: two identical bulbs across 4.5 V get 2.25 V each. Parallel has several branches, so the voltage is the same across each branch, while the current splits and rejoins: two bulbs drawing 0.3 A each give 0.6 A in the main line. Use the rules on three predictions. Remove a bulb in series and the other goes out because the path is broken. Remove a bulb in parallel and the other stays lit because its branch is intact. Add a bulb in parallel and the main current rises. These three predictions are the standard exam questions, so rehearse until students can state them unprompted.

Ask students to give the current rule in series and in parallel using the words one path and several branches

Scene 3: Resistance and brightness: calculate, do not guess

Brightness is about power, not simply more electricity. Begin with resistance: R equals V divided by I, so a bulb at 4.5 V drawing 0.3 A has 15 ohms. Now put two identical bulbs in series: each takes 2.25 V, the current falls to 0.15 A, and the total power is 4.5 times 0.15, about 0.675 W, so both are dimmer than a single bulb. Let students explain why: resistances add in series, total resistance rises, current falls. Parallel does the opposite. Two 15 ohm bulbs in parallel give an equivalent resistance of 7.5 ohms, total resistance falls, current rises, so the bulbs are brighter but the cell drains faster. This converts brightness from an adjective into a calculated result.

Ask why bulbs in series are dimmer and calculate the equivalent resistance of two 15 ohm bulbs in parallel

04 Question types and worked steps

Type	Example	Steps	Answer
Voltage sharing	Three 1.5 V cells in series supply two identical bulbs in series. What voltage is across each bulb?	1) Total voltage is 3 times 1.5, so 4.5 V. 2) Identical bulbs share it equally. 3) 4.5 divided by 2	2.25 V each
Current in parallel	Two bulbs in parallel each draw 0.3 A. What is the current in the main line?	1) Currents add in parallel. 2) 0.3 plus 0.3. 3) Give the answer in A	0.6 A
Resistance	A bulb has 4.5 V across it and 0.3 A through it. What is its resistance?	1) R equals V divided by I. 2) 4.5 divided by 0.3. 3) Give the answer in ohms	15 ohms
Predicting a change	Two bulbs are in series and one is removed. What happens to the other and why?	1) Series has one path. 2) Removing a bulb breaks it. 3) No closed path means no current	The other bulb goes out because the circuit is open

Sanity check (teach this): Three checks for circuit questions: identify series or parallel first by counting the paths; keep the two rule sets apart, since series shares voltage and parallel shares current; and check that the answer is physically sensible, for example resistance rising when current falls. When asked which bulb is brighter, always justify with current or power, never with more electricity.

05 Independent practice (3 questions)

Question	Answer	What to watch
Three 1.5 V cells in series give what total voltage?	4.5 V	Check they did not treat the cells as parallel
What is the resistance of a bulb at 4.5 V and 0.3 A?	15 ohms	Check they did not divide current by voltage
Two bulbs in series, one removed: does the other stay lit?	No, the circuit is open	Check the explanation uses one path

06 Common mistakes and what to say

Mistake	Student writes	Say this
Current being used up	Current leaves the cell and gets less after each bulb.	In series there is one path, so current is the same everywhere. What is shared is voltage.
Parallel current assumed equal	Two bulbs in parallel draw the same current, so the main line is the same too.	Parallel keeps voltage equal and adds currents: 0.3 plus 0.3 gives 0.6 A.
Brightness by intuition	Parallel is always brighter because there is more electricity.	Justify with current and power: parallel lowers equivalent resistance, current rises, so it is brighter but drains faster.
Current in an open circuit	If one bulb is removed, a small current still flows in the other.	An open circuit means no closed path, so the current is zero rather than smaller.

07 Homework (paper, section and question numbers)

Density, floating and the particle model, question 3 — Same approach: state the rule first, then predict
Burning and oxidation, question 1 — Both are systems where all conditions must be met
Respiration, circulation and excretion, question 2 — The circulatory system also has a main line and branches, so the comparison transfers

08 Exit check (last 5 minutes)

Question	Answer	Decision
What is true of current at every point in a series circuit?	It is the same everywhere	Answering that it decreases means current is still treated as used up
Two 0.3 A bulbs in parallel: what is the main current?	0.6 A	Answering 0.3 A means parallel currents were not added
What is the resistance of a bulb at 4.5 V and 0.3 A?	15 ohms	Answering 1.35 means voltage and current were multiplied

09 Tutor FAQ

Should S2 students learn all three rearrangements of Ohm law?

Teach resistance equals voltage divided by current and let students derive the others from the same bulb data. Memorising three forms adds load without understanding.

What if there is no lab equipment?

Use a circuit simulator or magnets on the board as bulbs. The value is in predicting and explaining, not in the hardware.

Why are household appliances in parallel?

So each can be switched separately and each receives the full supply voltage. Mention that the total current rises with more appliances, which is why extensions can overload.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.