Why this topic matters for the exam
Electricity and D.C. circuits appears in every single O-Level Pure Physics paper, across all three components: multiple choice, structured questions, and often a circuit-based calculation worth several marks. It rewards students who are precise and punishes those who guess. The good news is that almost all of it reduces to a handful of relationships used carefully. Once a student is fluent with current, potential difference, resistance, and how circuits combine, this becomes one of the most reliable places to bank full marks.
The difference between a B3 and an A1 in this topic is rarely about understanding a hard idea. It is usually about not dropping a unit, not confusing e.m.f. with p.d., and reasoning correctly about what happens when a component is added or removed. This article fixes exactly those things.
Core concepts clearly explained
Current, charge, and potential difference
Electric current is the rate of flow of electric charge. In symbols, I = Q / t, where I is current in amperes (A), Q is charge in coulombs (C), and t is time in seconds (s). One ampere is one coulomb per second. Conventional current flows from the positive terminal to the negative terminal in the external circuit, which is opposite to the direction the electrons actually drift.
Potential difference (p.d.), also called voltage, between two points is the work done to drive one coulomb of charge between those points. In symbols, V = W / Q, measured in volts (V). One volt is one joule per coulomb. P.d. is what pushes current through a component; without a p.d. across it, no current flows.
Resistance and Ohm's law
Resistance is a measure of how strongly a component opposes the flow of current. It is defined as R = V / I, with resistance in ohms (Ω). One ohm is one volt per ampere.
Ohm's law states that the current through a metallic conductor is directly proportional to the p.d. across it, provided the temperature (and other physical conditions) remain constant. A conductor that obeys this is called ohmic, and its V–I graph is a straight line through the origin. A filament lamp is non-ohmic: as current increases, the filament heats up, its resistance rises, and the V–I graph curves. Students should be able to sketch and interpret both.
Resistivity (R = ρL / A)
Resistance also depends on the wire itself. The relationship is R = ρL / A, where ρ (rho) is the resistivity of the material in ohm-metres (Ω m), L is the length of the wire in metres, and A is the cross-sectional area in square metres. The key qualitative takeaways for the exam: a longer wire has more resistance, a thicker wire (larger A) has less resistance, and resistivity is a property of the material itself, not its shape.
e.m.f. versus p.d.
This distinction is examined almost every year. The electromotive force (e.m.f.) of a source is the work done by the source in driving one coulomb of charge around a complete circuit; it is the energy the source supplies per unit charge. Potential difference is the energy converted from electrical to other forms per unit charge in a component.
In a real cell there is internal resistance, so when current flows some energy is "lost" inside the cell. This means the p.d. measured across the terminals (the terminal p.d.) is slightly less than the e.m.f. when the cell is supplying current. When no current flows, the terminal p.d. equals the e.m.f. A clean exam answer: e.m.f. is energy supplied per unit charge by the source; p.d. is energy transferred per unit charge to a component.
Resistors in series
In a series circuit there is only one path. Therefore:
- The current is the same everywhere: I is identical through each component.
- The p.d.s across the components add up to the supply p.d.: V = V1 + V2 + ...
- The total resistance is the sum: Rtotal = R1 + R2 + ... Adding a resistor in series always increases total resistance.
Resistors in parallel
In a parallel circuit there are multiple paths sharing the same two end points. Therefore:
- The p.d. is the same across each branch.
- The currents in the branches add up to the total current: I = I1 + I2 + ...
- The total resistance is found from 1 / Rtotal = 1 / R1 + 1 / R2 + ... The combined resistance is always smaller than the smallest individual resistor. Adding a resistor in parallel always decreases total resistance.
A useful shortcut for exactly two resistors in parallel: Rtotal = (R1 × R2) / (R1 + R2). And for n identical resistors of value R in parallel, the total is simply R / n.
Electrical power and energy
Power is the rate of energy transfer. The core formula is P = VI, with power in watts (W). Combining with V = IR gives two more very useful forms:
- P = I2R (handy when you know the current and resistance)
- P = V2 / R (handy when you know the p.d. and resistance)
Electrical energy transferred is E = Pt = VIt, measured in joules (J) when time is in seconds. For household billing, energy is measured in kilowatt-hours (kWh), where 1 kWh is the energy used by a 1 kW appliance running for 1 hour.
Potential dividers
A potential divider is two (or more) resistors in series used to provide a fraction of the supply voltage. For two resistors R1 and R2 in series across a supply Vin, the p.d. across R2 is:
Vout = Vin × R2 / (R1 + R2)
The voltage splits in the same ratio as the resistances. Replacing one resistor with a thermistor (resistance falls as temperature rises) or a light-dependent resistor (LDR) (resistance falls as light intensity rises) lets the output voltage respond to the environment—the basis of many sensor circuits in the syllabus.
Worked exam-style examples
Example 1 — Series and parallel combined
A 12 V battery (assume negligible internal resistance) is connected to a 4 Ω resistor in series with a parallel combination of a 6 Ω resistor and a 3 Ω resistor. Find (a) the total resistance, (b) the current from the battery, and (c) the p.d. across the parallel section.
(a) First combine the parallel pair: 1 / RP = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so RP = 2 Ω. (Check: it is smaller than 3 Ω, the smaller branch—good.) Then add the series resistor: Rtotal = 4 + 2 = 6 Ω.
(b) I = V / Rtotal = 12 / 6 = 2 A from the battery.
(c) The p.d. across the 4 Ω resistor is V = IR = 2 × 4 = 8 V. The remaining p.d. is across the parallel section: VP = 12 − 8 = 4 V. (Check using RP: VP = IRP = 2 × 2 = 4 V. Consistent.) The current then splits: through the 6 Ω branch, 4 / 6 = 0.67 A; through the 3 Ω branch, 4 / 3 = 1.33 A; total 2 A as expected.
Example 2 — Power, energy, and cost
An electric kettle is rated "230 V, 2300 W". It is operated for 5 minutes. Find (a) the current it draws, (b) the resistance of its heating element, (c) the electrical energy converted in joules, and (d) the energy used in kWh.
(a) From P = VI, I = P / V = 2300 / 230 = 10 A.
(b) R = V / I = 230 / 10 = 23 Ω. (Cross-check with P = V2 / R: R = V2 / P = 2302 / 2300 = 52900 / 2300 = 23 Ω. Consistent.)
(c) Convert time to seconds: t = 5 × 60 = 300 s. Then E = Pt = 2300 × 300 = 690 000 J (6.9 × 105 J).
(d) Power in kilowatts is 2.3 kW; time in hours is 5 / 60 = 0.0833 h. Energy = 2.3 × 0.0833 = 0.19 kWh (to 2 significant figures).
Example 3 — Potential divider with a thermistor
In a potential divider, a fixed 2000 Ω resistor is in series with a thermistor across a 6.0 V supply. The output voltage is taken across the fixed resistor. At room temperature the thermistor has a resistance of 4000 Ω. (a) Find the output voltage at room temperature. (b) State and explain what happens to the output voltage when the thermistor is warmed.
(a) Using Vout = Vin × Rfixed / (Rfixed + Rthermistor): Vout = 6.0 × 2000 / (2000 + 4000) = 6.0 × 2000 / 6000 = 2.0 V.
(b) When the thermistor is warmed, its resistance decreases. The total resistance falls, so the current rises, but more importantly a larger share of the supply voltage now appears across the fixed resistor. Therefore the output voltage increases. (Sanity check: if the thermistor resistance dropped to 1000 Ω, Vout = 6.0 × 2000 / 3000 = 4.0 V—higher, as predicted.)
Common exam traps and mistakes
Ammeter in series, voltmeter in parallel
An ammeter measures current and must be connected in series with the component, so all the current passes through it. An ideal ammeter has zero resistance. A voltmeter measures p.d. and must be connected in parallel across the component. An ideal voltmeter has infinite resistance, so it draws no current and does not disturb the circuit. Connecting them the wrong way round is a classic multiple-choice trap.
Adding or removing a bulb in series versus parallel
This is the single most misunderstood idea in the topic. In a series circuit, components share one path, so:
- Adding another bulb in series increases total resistance, so the current falls and every bulb gets dimmer.
- If one bulb in series "blows" (open circuit), the whole circuit breaks and all bulbs go out.
In a parallel circuit, each branch is independent across the same p.d., so:
- Each branch keeps its own p.d. regardless of the others, so existing bulbs are unaffected in brightness when another branch is added.
- Adding more parallel branches decreases total resistance, so the current drawn from the supply increases.
- If one bulb in a parallel branch blows, the others keep working. This is why household wiring uses parallel connection.
Identical versus different resistor splits
When current reaches a parallel junction, it does not automatically split equally. It splits equally only when the branches have identical resistance. With different resistances, more current takes the path of lower resistance. In Example 1, the 3 Ω branch carried twice the current of the 6 Ω branch. Students who assume an automatic 50:50 split lose marks every time.
Other frequent slips
- Forgetting to convert time to seconds before using E = Pt. Minutes and hours must be converted, or the energy will be wrong by a large factor.
- Treating the parallel formula answer as final without inverting. Students compute 1 / Rtotal and forget the last step of taking the reciprocal.
- Confusing e.m.f. with terminal p.d. Remember terminal p.d. drops below e.m.f. only when current flows, because of internal resistance.
- Dropping units. A1 candidates write A, V, Ω, W, and J every time. Missing or wrong units cost marks even when the number is correct.
- Misreading non-ohmic graphs. For a filament lamp, resistance is V / I at that specific point—not the gradient of the curve.
How to practise and revise
This topic rewards drilling, not just reading. A practical revision routine:
- Master the definitions first. Write out, from memory, the definitions of current, p.d., resistance, e.m.f., and Ohm's law—with units. Marks for definitions are free if the wording is precise.
- Build a formula sheet. Group the three power forms (P = VI, P = I2R, P = V2 / R) and the series and parallel rules side by side. Knowing which form to reach for is half the battle.
- Redraw every circuit. Before calculating, redraw the circuit and label known currents and p.d.s. Identify which parts are series and which are parallel before touching the calculator.
- Always sanity-check. Total resistance of a parallel combination must be smaller than the smallest branch. The p.d.s in series must add up to the supply. These checks catch most arithmetic errors.
- Do past-year structured questions under timed conditions. Work through at least one full circuit question per revision session, then mark it strictly against the marking scheme, paying attention to units and significant figures.
- Explain it aloud. If a student can clearly explain why a bulb in a parallel circuit stays bright when another is added, they truly understand it—and that understanding is what the trickier questions test.
If your child can confidently move between series and parallel reasoning, apply the power and energy formulae cleanly, and avoid the well-known traps, Electricity and D.C. circuits becomes one of the most dependable A1-scoring topics in the whole paper—and that is exactly the kind of steady, exam-focused progress we build with our students at Intuitional.