This article is for Secondary 3 and 4 students taking O-Level Physics in Singapore, and for their parents.
Three rules from the SEAB O-Level Physics syllabus (6091) carry the reasoning for series and parallel circuits:
- The current at every point in a series circuit is the same.
- In a series circuit, the sum of the potential differences is equal to the potential difference across the whole circuit.
- In a parallel circuit, the sum of the currents in the separate branches is equal to the current from the source.
Work in a fixed order: redraw the circuit, mark each part as series or parallel, apply the matching rule, then check the answer against the same rule.
The scope is topics 14 (Current of Electricity) and 15 (D.C. Circuits) in Section V of syllabus 6091. Topic 13 (Static Electricity) gives the prerequisite idea of charge. Magnetism, electromagnetism and induction are outside this article. Resistivity and internal resistance are not among the topic 14 and 15 learning outcomes.
Start with the definitions
Charge is measured in coulombs. The topic 13 outcomes state that unlike charges attract and like charges repel.
Current is “the rate of flow of charge”, measured in amperes. The syllabus asks students to “recall and apply the relationship charge = current × time”, which rearranges to I = Q/t. For example, if 30 C passes a point in 10 s (hypothetical values), I = 30 ÷ 10 = 3 A.
In practice, convert units before you substitute. Two minutes becomes 120 s before you divide.
Potential difference (p.d.) is “the work done per unit charge in driving charges through the component”, measured in volts. E.m.f. is “the work done per unit charge by the source in driving charges around a complete circuit”. The e.m.f. refers to the source; the p.d. refers to a component. The syllabus also expects you to distinguish conventional current from electron flow.
Tip 1: Redraw the circuit and confirm a complete loop
A misconception documented in a teaching module is that one wire is enough to make a circuit. The teaching point in that module is that there must be a connection from each end of the battery to two points on the bulb holder.
A useful habit: trace the path with your pencil from one battery terminal, through every component, and back to the other terminal. If the trace stops, the loop is not complete.
Hypothetical example: a diagram shows a battery, a switch and a bulb joined by a single wire from the battery to the bulb. The trace stops at the bulb, so the loop is open. Add the return wire before you calculate anything.
Tip 2: In a series circuit, the current is the same everywhere
The syllabus states that “the current at every point in a series circuit is the same”. It also states that the p.d.s across the components of a series circuit add up to the p.d. across the whole circuit.
A documented misconception is that current is “used up” as it flows round the circuit. The teaching point in the same module is that “the current is the same all the way round a series circuit”. What each component uses up is energy. Its p.d. measures how much energy each coulomb gives up there.
A quick test: every ammeter placed in one series loop should read the same value.
Example (hypothetical): a 6 V supply feeds two resistors in series. The p.d. across the first is 2 V, so the p.d. across the second is 6 V − 2 V = 4 V. The two p.d.s add back to the supply: 2 V + 4 V = 6 V.
Tip 3: In a parallel circuit, branch p.d.s are equal and branch currents add
The syllabus states that “the potential difference across the separate branches of a parallel circuit is the same”, and that “the sum of the currents in the separate branches of a parallel circuit is equal to the current from the source”.
For two resistors in parallel, use 1/R = 1/R1 + 1/R2. The sum gives 1/R, so invert the result at the end to get R.
Check: the total resistance must be smaller than the smallest branch. For 6 Ω and 3 Ω, 1/R = 1/6 + 2/6 = 3/6 = 1/2, so R = 2 Ω. The answer is smaller than 3 Ω.
A second check uses branch currents: on a 6 V supply, the 6 Ω and 3 Ω branches carry 1 A and 2 A (hypothetical values). Together they give 3 A, and 6 V ÷ 2 Ω = 3 A.
Tip 4: Use R = V/I, and check whether the component is ohmic
The syllabus requires you to state that resistance = p.d. / current and to apply R = V / I. This relation is also called Ohm’s law. R = V/I applies to an ohmic conductor, which the syllabus describes as a metallic conductor at constant temperature.
The syllabus asks you to sketch and interpret I-V graphs for an ohmic conductor at constant temperature, a filament lamp and a semiconductor diode. A filament lamp and a diode are non-ohmic, so one fixed value of R does not describe them across all conditions.
A documented misconception is that current “starts from one end of a battery and flows through each component of a circuit in turn”. The teaching point is that “current flows instantly in all parts of a circuit when there is a complete circuit”.
Before you use a single value of R, confirm that the component is an ohmic conductor at the stated temperature. For a filament lamp or a diode, use the V and I values the question gives at the point it asks about.
Hypothetical case: 6 V across a component gives 0.5 A through it, so R = 6 ÷ 0.5 = 12 Ω. This value applies at that p.d. and current.
Tip 5: Use the share of the p.d. in a potential divider
The syllabus asks you to “describe the action of negative temperature coefficient (NTC) thermistors and light-dependent resistors and explain their use as input transducers in potential dividers”.
Use the resistance the question gives for the thermistor or LDR at the stated condition. The current is the same in each series part, so a component’s share of the supply p.d. is its resistance divided by the total resistance.
Worked example (hypothetical): a 10 Ω fixed resistor is in series with a component of 20 Ω, on a 6 V supply. The total resistance is 30 Ω, so the current is 6 ÷ 30 = 0.2 A. The component has a p.d. of 0.2 × 20 = 4 V, and the fixed resistor has 0.2 × 10 = 2 V. The two p.d.s add back to 6 V: 4 V + 2 V.
Tip 6: Connect the ammeter in series and the voltmeter across the component
The syllabus names ammeters and voltmeters in circuit diagrams (topic 15(a)), and asks students to select and explain the use of appropriate measuring instruments (Section I 1(e)). The placement below is editorial explanation, not syllabus wording.
Place the ammeter in series, so it measures the current through the component. Place the voltmeter across the component, in parallel with it, so it reads the p.d. across that component.
A voltmeter reading across a resistor should agree with R = V/I using the current you measured.
Hypothetical example: the ammeter reads 2 A through a 3 Ω resistor. The voltmeter across the resistor should read 2 × 3 = 6 V.
Tip 7: Predict with a simulation, then check with real meters
One way to use a simulation: write down a predicted reading before you run it, then check the prediction. Use real meters as well, for practice with connections and readings.
SERC, a Carleton College resource, cites US college studies of PhET simulations and states that “PhET simulations are likely to be less effective for achieving goals such as laboratory skills and troubleshooting equipment.” The underlying studies were not checked for this article, and they do not involve O-Level students.
Hypothetical case: for a 6 V supply and a 3 Ω resistor, predict 2 A before running the simulation, then compare.
Exam habits
The syllabus lists command words for knowledge: “define, state, name, describe, explain or outline”. It lists command words for problem-solving: “predict, deduce, suggest, calculate or determine”.
Suggested habit: underline the command word before you write. A “state” question needs the rule. A “calculate” or “determine” question needs the working and the unit.
Paper 1 is 40 multiple-choice items, 1 hour, 30%. Paper 2 is structured and free response, 80 marks, 1 hour 45 minutes, 50%. Paper 3 is practical, 40 marks, 1 hour 50 minutes, 20%. These weightings cover the whole syllabus, not electricity alone. Across the theory papers, approximately 55% of the assessment objective weight is handling information and solving problems, which is also a whole-syllabus figure.
For parents
These questions are for practice at home, not a measure of progress.
- Ask your child to state the rule for current in a series circuit, and to point to two places in a drawn circuit where the current is the same.
- Ask why household lights stay on when one bulb fails.
- Ask how much charge passes in 2 minutes if the current is 0.5 A. The answer is 60 C, using 120 s.
Where to go next
Check each answer against the rule that matches the circuit; for guided practice, the programme page is https://www.exploreracademy.sg/programmes-o-level-physics/.
