4.30625 · Section 4: Electricity and Magnetism

Electric circuits

Key learnings

Resistance, Ohm's law and wire factors
Resistance $R = V/I$ = ratio of the p.d. across a component to the current through it; unit ohm (Ω). "Opposition to current" is too vague and scores zero. Ohm's law: for a metal conductor at constant temperature, p.d. is directly proportional to current — its I–V graph is a straight line through the origin (an ohmic conductor). Wire resistance: a longer wire has more; a thicker wire (larger area) has less; higher-resistivity material has more.
Series and parallel circuit rules
Series (one path): current the same everywhere ( $I_1 = I_2$ ); p.d.s add to the supply ( $V_1 + V_2 = V_S$ ); resistances add ( $R_T = R_1 + R_2$ ). Parallel (branches): p.d. the same across each branch ( $V_1 = V_2 = V_S$ ); branch currents add ( $I_T = I_1 + I_2$ ); $1/R_T = 1/R_1 + 1/R_2$ , always less than the smallest branch. (E) derive it from $I_T = V/R_1 + V/R_2 = V(1/R_1 + 1/R_2)$ together with $I_T = V/R_T$ .
Electrical power and energy
Power (rate of energy transfer; unit watt, W): $P = IV = I^2 R = V^2/R$ . Energy: $E = IVt = Pt$ (joule, J), with $t$ in seconds — convert minutes by ×60. e.g. $P = IV = 0.4 \times 9.0 = 3.6$ W. An LDR's resistance falls as light intensity rises, and an NTC thermistor's falls as temperature rises; both are used in potential-divider sensing circuits.

Common mistakes

Drawn from real examiner reports.

Resistance is V ÷ I, not V × I

Many candidates write $R = V \times I$ instead of $R = V/I$ . With $V = 4.8$ V and $I = 0.4$ A: wrong $R = 4.8 \times 0.4 = 1.92\ \Omega$ ; correct $R = 4.8/0.4 = 12\ \Omega$ . Resistance is the ratio V ÷ I, not the product — write $R = V/I$ before substituting to keep the method mark, and sense-check the size of the answer.

June 2024 Paper 32 Q7(a): "The most common error was using an incorrectly rearranged form of the equation V = I × R with the most common of these being R = V × I = 4.8 × 0.4 = 1.92." Examiner report explicitly named this as the most frequent calculation error on that question.

Energy is IVt, not IV; time in seconds

Two linked errors on energy questions. (1) Using $E = IV$ (that is power) instead of $E = IVt = Pt$ — power is the rate; energy = power × time, so $E = IV$ gives watts, not joules. (2) Not converting time: $E = IVt$ needs $t$ in seconds, so "5 minutes" is $t = 300$ s. Leaving minutes makes the answer 60 times too small.

June 2024 Paper 32 Q8(a): "Many candidates did not recall the equation for energy transferred, E = I × t × V. The most common error was the use of the equation for electrical power. Those who did recall the correct equation often did not convert the time to seconds."

(E) Resistance is V/I, not the I–V gradient

(E) On an I–V characteristic, resistance is not the gradient. The gradient is $\Delta I/\Delta V$ = conductance (equal to $1/R$ ), not resistance. To find $R$ at a point, read $V$ and $I$ there and compute $R = V/I$ . For a non-ohmic component (e.g. a filament lamp) the gradient changes along the curve, so the gradient at a point is not $V/I$ at that point.

November 2023 Paper 23 Q27: "Most common choice was option B. Candidates who chose this may have thought that resistance is the gradient of the current-potential difference characteristic as this is a common misconception. Resistance is actually found by dividing the potential difference by the current." June 2024 Paper 23 Q27: identical misconception reproduced.

Do not use the series formula in parallel

Adding resistances ( $R_T = R_1 + R_2$ ) is the series rule. In parallel, use $1/R_T = 1/R_1 + 1/R_2$ , which gives a total smaller than the smallest branch. Writing $R_T = 6 + 12 = 18\ \Omega$ for two parallel resistors is a flagged error — the self-check fails at once, since a parallel total of $18\ \Omega$ exceeds the $6\ \Omega$ branch.

(E) A filament lamp is non-ohmic

(E) Do not treat a filament lamp as ohmic. As the current rises, the filament heats up, its atoms vibrate more and electrons collide more, so its resistance increases — each extra volt drives less extra current and the I–V graph curves (flattens). Only an ohmic conductor at constant temperature gives a straight line through the origin.

Every parallel branch gets the full supply

In a parallel circuit every branch has the full supply p.d. — do not divide the supply between the branches (a flagged error). This is why household lamps are wired in parallel: each gets the full 230 V and works independently, so one failed lamp does not switch off the others. In series, by contrast, the p.d.s share out and one break stops the whole circuit.

Exam tips

State the circuit type first
Before applying any rule, write whether the circuit is series or parallel. This commits you to the correct rule set (same current vs same p.d.) and is itself credited on "describe/explain" questions where you must justify the behaviour.
Write the equation, finish with the unit
Write the equation in symbols first — $R = V/I$ , $P = IV$ , $E = IVt$ — then substitute; this earns the method mark even if the arithmetic slips. Finish with the correct unit (Ω, W, J) and give a sensible number of significant figures.
Convert units before substituting
Check units before you calculate: time in seconds (not minutes), current in amperes (not milliamps), p.d. in volts. A single unconverted quantity — for example a current left in mA — throws the whole answer out.
Self-check series and parallel answers
Finish with a sanity check. In series, the component p.d.s must add up to the supply. In parallel, the combined resistance must be less than the smallest branch. If either check fails, you have applied the wrong rule.

Revision notes