Why Circuit Questions Matter More Than They Look
Electricity is one of the highest-yield topics in O-Level Pure Physics. It shows up in Paper 1 MCQs and structured Paper 2 questions, and it often returns in the data-based or application question at the end of the paper. It also carries into related topics: electrical energy and power, practical electricity and electromagnetism all assume you can work confidently with current and potential difference.
Yet this is where many strong students lose marks. The usual story goes like this: “I know V = IR, but I don’t know which V and which R to use.” The student has memorised the formula but has no system for applying it. When the circuit has a resistor in series with two resistors in parallel, they plug in numbers and hope.
At Intuitional we teach circuits as a method. By the end of this article you should have one framework that handles almost every D.C. circuit question in the syllabus, plus the reasoning skills for the “explain what happens” questions that examiners like to set.
The Core Ideas: Three Rules and Nothing Else
Every D.C. circuit question at O-Level rests on three rules. Understand them properly and the rest follows.
Rule 1: Current is conserved, so it splits and recombines but is never used up
Current is the rate of flow of charge: I = Q/t. Charge is not created or destroyed in a circuit, which gives two consequences:
- In series, the current is the same through every component. There is only one path, so the same charge passes through each component every second.
- At a junction, the current entering equals the current leaving. If 1.5 A enters a junction that splits into two branches, the branch currents must add up to 1.5 A.
A common misconception is that a bulb “uses up” current, so less current comes out than goes in. It does not. What the bulb converts is energy, not charge. Keeping this clear heads off many wrong MCQ options.
Rule 2: Potential difference is shared in series and equal across parallel branches
First, the two definitions students often mix up:
- Electromotive force (e.m.f.) of a source is the work done by the source in driving a unit charge around a complete circuit.
- Potential difference (p.d.) across a component is the work done per unit charge in converting electrical energy to other forms when charge passes through that component.
Both are measured in volts, where 1 V = 1 J/C. From energy conservation:
- In series, the p.d.s across the components add up to the e.m.f. of the source (assuming negligible internal resistance). The energy given to each coulomb by the battery is shared out among the components.
- Across parallel branches, the p.d. is the same. Both branches connect the same two points, so each coulomb travelling through either branch has the same energy converted.
Rule 3: V = IR applies to one thing at a time
Resistance is defined as R = V/I. The key habit is this:
V, I and R in the equation must all refer to the same component (or the same combination of components).
You can apply V = IR to a single resistor, to a parallel pair treated as one block, or to the whole circuit. You must never mix the whole circuit’s voltage with one resistor’s resistance. Most wrong answers in circuit calculations come from exactly that mix-up.
Combining resistors
- Series: RT = R1 + R2 + …
- Parallel: 1/RT = 1/R1 + 1/R2 + …
A quick sanity check for parallel combinations: the combined resistance is always smaller than the smallest individual resistance. Adding a parallel branch gives the charge an extra path, so the combination lets more current through for the same p.d. If your parallel total is larger than one of the resistors, you have made an arithmetic slip, most often by forgetting to take the reciprocal at the end.
A useful shortcut for exactly two resistors in parallel: RT = (R1 × R2) / (R1 + R2), often remembered as “product over sum”.
Resistance of a wire
The resistance of a conductor depends on its dimensions: R = ρL/A, where ρ is the resistivity of the material, L is the length and A is the cross-sectional area. Doubling the length doubles the resistance. Doubling the diameter quadruples the area, so the resistance falls to a quarter. That diameter-versus-area trap appears in MCQs every year.
The Method: Collapse → Solve → Expand
This is the framework we drill with every student. It turns a messy circuit into a sequence of small, safe steps.
- Collapse. Replace each parallel group with a single equivalent resistor. Then add the series resistors. Keep going until the whole circuit is one resistor across the battery.
- Solve. Use V = IR on the whole circuit to find the total current from the source.
- Expand. Work backwards through your collapsing steps. At each stage use Rule 1 (same current in series) or Rule 2 (same p.d. in parallel) to carry the known quantity into the next level, then apply V = IR to that block alone.
- Check. Currents at each junction should add up, and the p.d.s around the loop should add up to the e.m.f.
The aha moment is realising that you never have to solve the whole circuit at once. At every stage you know one quantity for one block, and V = IR gives you the other.
Worked Examples
Worked Example 1: A mixed series–parallel circuit
Question. A 12 V battery of negligible internal resistance is connected to a 4.0 Ω resistor in series with a parallel combination of a 6.0 Ω resistor and a 12 Ω resistor. Find (a) the current from the battery, (b) the p.d. across the 4.0 Ω resistor, and (c) the current in each parallel branch.
Step 1: Collapse
Parallel pair: 1/RP = 1/6.0 + 1/12 = 2/12 + 1/12 = 3/12, so RP = 4.0 Ω.
Check: 4.0 Ω is smaller than 6.0 Ω, the smallest resistor in the group. ✓
Series with the 4.0 Ω resistor: RT = 4.0 + 4.0 = 8.0 Ω.
Step 2: Solve
Whole circuit: I = V/RT = 12 / 8.0 = 1.5 A.
Step 3: Expand
The 4.0 Ω resistor is in series with the battery, so it carries the full 1.5 A (Rule 1).
p.d. across 4.0 Ω: V = IR = 1.5 × 4.0 = 6.0 V.
The parallel block also carries 1.5 A and has an equivalent resistance of 4.0 Ω, so the p.d. across it is 1.5 × 4.0 = 6.0 V. As a check, 6.0 + 6.0 = 12 V = e.m.f. ✓
Both branches have 6.0 V across them (Rule 2):
- 6.0 Ω branch: I = 6.0 / 6.0 = 1.0 A
- 12 Ω branch: I = 6.0 / 12 = 0.50 A
Step 4: Check
1.0 + 0.50 = 1.5 A, which matches the current entering the junction. ✓
Also notice that the smaller resistance carries the larger current. When two branches share the same p.d., current divides in inverse proportion to resistance. Examiners often ask for this reasoning in words.
Worked Example 2: Adding a branch, an “explain what happens” question
Question. A second 12 Ω resistor is now connected in parallel with the two resistors in Example 1. State and explain what happens to (a) the current from the battery and (b) the current in the 6.0 Ω resistor.
Many students say that the 6.0 Ω resistor is “unaffected because it is in parallel”. That is true only if nothing else is in series with the parallel block. Here the 4.0 Ω resistor is in series with it, so the answer changes. Let’s calculate first, then write the explanation.
Collapse
1/RP = 1/6.0 + 1/12 + 1/12 = 2/12 + 1/12 + 1/12 = 4/12, so RP = 3.0 Ω.
RT = 4.0 + 3.0 = 7.0 Ω.
Solve
I = 12 / 7.0 = 1.7 A (to 2 s.f.). This is up from 1.5 A.
Expand
p.d. across 4.0 Ω = (12/7.0) × 4.0 ≈ 6.9 V.
p.d. across the parallel block = 12 − 6.9 ≈ 5.1 V (down from 6.0 V).
Current in 6.0 Ω = 5.1 / 6.0 ≈ 0.86 A (down from 1.0 A).
The written explanation (what earns the marks)
- Adding a resistor in parallel decreases the effective resistance of the parallel combination, so the total resistance of the circuit decreases.
- Since the e.m.f. is unchanged, the current from the battery increases (I = V/R).
- The larger current through the 4.0 Ω resistor means a larger p.d. across it.
- The p.d.s must still add up to 12 V, so the p.d. across the parallel combination decreases.
- So the current in the 6.0 Ω resistor decreases.
Look at the structure: each sentence links one cause to one effect, and each link uses one of the three rules. This chain style of reasoning scores full marks and works for any “what happens if” question.
Worked Example 3: A potential divider with a thermistor
Question. A 9.0 V supply is connected across a fixed 2.0 kΩ resistor in series with a thermistor. At 20 °C the thermistor has a resistance of 4.0 kΩ. The output voltage Vout is taken across the fixed resistor.
(a) Calculate Vout at 20 °C.
(b) The temperature rises and the thermistor’s resistance falls to 1.0 kΩ. Calculate the new Vout.
(c) Explain why Vout changes.
The potential divider shortcut
In a series pair, the supply p.d. is shared in proportion to resistance (the current is the same in both, so V ∝ R):
V1 = [R1 / (R1 + R2)] × Vsupply
(a) Vout = [2.0 / (2.0 + 4.0)] × 9.0 = (1/3) × 9.0 = 3.0 V.
(b) Vout = [2.0 / (2.0 + 1.0)] × 9.0 = (2/3) × 9.0 = 6.0 V.
You can also use Collapse → Solve → Expand. In part (a), RT = 6.0 kΩ, I = 9.0 / 6000 = 1.5 mA, and Vout = 1.5 × 10−3 × 2000 = 3.0 V. Same answer. The ratio method is simply faster once you trust it.
(c) The explanation
- As temperature increases, the resistance of the thermistor decreases.
- The fixed resistor now has a larger share of the total resistance.
- Since the supply p.d. is shared in proportion to resistance, the fixed resistor takes a larger share of the 9.0 V, so Vout increases.
The same reasoning applies to a light-dependent resistor (LDR), whose resistance decreases as light intensity increases. Exam questions often ask you to place the sensor so that an output rises when it gets dark or hot. To do that, ask which component’s share of the resistance goes up under that condition, and take the output across that component.
Worked Example 4: Reading I–V graphs correctly
Question. A filament lamp has a current of 0.20 A when the p.d. across it is 2.0 V, and 0.30 A when the p.d. is 6.0 V. Calculate its resistance at each point and explain the change.
At 2.0 V: R = V/I = 2.0 / 0.20 = 10 Ω.
At 6.0 V: R = 6.0 / 0.30 = 20 Ω.
Explanation: As the current increases, the filament gets hotter. The metal ions in the filament vibrate more vigorously, so the electrons collide with them more often. This increases the resistance, which is why the lamp’s I–V graph curves and becomes less steep at higher p.d.
By contrast, an ohmic conductor (such as a metal wire at constant temperature) has a straight-line I–V graph through the origin, because its resistance is constant.
Exam Traps and How to Avoid Them
Trap 1: Mixing quantities in V = IR
Writing I = 12 / 6.0 for the 6.0 Ω resistor in Example 1 gives 2.0 A, which is wrong. The 12 V is across the whole circuit, not across that resistor. Fix: before every V = IR, say to yourself which component or block the equation is about, and check that all three quantities belong to it.
Trap 2: Using the gradient for resistance
On a curved I–V graph, resistance at a point is V/I at that point, not the gradient of the tangent. Also look at the axes: if I is on the y-axis and V is on the x-axis, a steeper line means lower resistance. Many students read this backwards.
Trap 3: Forgetting the reciprocal
For 1/RP = 3/12, the answer is RP = 4.0 Ω, not 0.25 Ω. Use the “smaller than the smallest” sanity check every time.
Trap 4: Unit prefixes
Potential divider and sensor questions use kΩ and mA. The ratio method is safe as long as both resistances are in the same unit. When calculating a current, convert to base units first (2.0 kΩ = 2000 Ω).
Trap 5: Meter placement and ideal meters
An ammeter goes in series and ideally has zero resistance. A voltmeter goes in parallel across the component and ideally has infinite resistance. If a question says a voltmeter is placed in series, very little current flows, and the voltmeter reads close to the full e.m.f.
Trap 6: “Current is used up”
Any answer that says current decreases after passing through a bulb is wrong. Current is the same everywhere in a series loop. What decreases around the loop is potential, because energy is transferred to the components.
Trap 7: Vague explanations
“The resistance changes so the voltage changes” scores little. Examiners want direction (increases or decreases) and a reason tied to a rule at each step, as in Worked Examples 2 and 3.
How to Practise This Topic Effectively
- Annotate every circuit. Before calculating, redraw the circuit and write the known values next to each component. As you expand, fill in I and V for every component. A fully labelled diagram is itself a check.
- Practise collapsing without numbers. Take circuit diagrams from past papers and just write the collapsing order, for example “(R2 ∥ R3) then + R1”. This trains the structural skill separately from the arithmetic.
- Always run the two checks. Junction currents should add up and loop p.d.s should add up to the e.m.f. This takes ten seconds and catches most errors before the examiner does.
- Write explanation chains. For every “what happens if” question, write numbered cause-and-effect steps like the ones above. After a few, you will notice that every explanation reuses the same three rules.
- Do mixed sets, not topic blocks. Mix series, parallel, potential divider and I–V graph questions in one sitting, so you practise recognising which tool the question needs, the way you will have to in the exam.
- Link to power. Once you are confident, extend each worked example by finding the power in each component with P = VI, P = I2R or P = V2/R. This connects straight into the practical electricity questions.
The Big Picture
Circuit questions look varied, but they all come down to the same idea: three rules, applied one block at a time. Current is conserved, p.d. is shared in series and equal in parallel, and V = IR works on one thing at a time. Collapse the circuit, solve the total, expand back out, and check.
At Intuitional, our small-group Secondary Physics classes in Teck Whye are built around frameworks like this. Students learn why each step works rather than memorising patterns, so they can handle a circuit they have never seen before. If your child finds electricity confusing, or understands it in class but freezes in exams, a systematic method is usually what changes that.