2.24PH1 · Section 2: Electricity

Series and parallel circuits

Key learnings

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Series: same I, shared V, R adds
Series circuit rules:
1. Current is the same everywhere: $I_1 = I_2 = I_T$ — one path for charge.
2. Voltages add to the supply: $V_1 + V_2 = V_S$ ; the supply is shared, not duplicated.
3. Resistances add: $R_T = R_1 + R_2$ , so more series components → more resistance → less current.
Mini-example: 8 Ω + 4 Ω on 12 V → $R_T = 12$ Ω, $I = 12/12 = 1.0$ A, so the p.d. across the 4 Ω is $1.0 × 4 = 4.0$ V (not 12 V).
•
Parallel: same V, currents add, R smaller
Parallel circuit rules:
1. Voltage across each branch equals the supply: $V_1 = V_2 = V_S$ — each branch sits directly across the terminals.
2. Currents in the branches add: $I_1 + I_2 = I_T$ .
3. Resistance obeys $\dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ ; the result is always less than the smallest branch.
Two-resistor shortcut: $R_T = \dfrac{R_1 R_2}{R_1 + R_2}$ . Mini-example: 6 Ω and 12 Ω → $R_T = 4$ Ω (less than 6 Ω).
Resistance R = V/I (Ω); domestic lighting
Resistance = ratio of the p.d. across a component to the current through it: $R = V/I$ ; unit ohm (Ω). The mark scheme needs both (1) ratio of p.d. to current and (2) the unit Ω — "opposition to current" scores zero.

Domestic lighting is wired in parallel (spec 2.7–2.8): each lamp gets the full 230 V so runs at full brightness, and if one filament breaks the others stay on in their own branches. In series, one break would stop every lamp.

Common mistakes

Drawn from real examiner reports.

Full supply V across each series part

In series the supply is shared, not repeated across every component. With two resistors on 12 V the p.d. splits (e.g. 8 V and 4 V). Correct route: $R_T = R_1 + R_2$ , then $I = V_S/R_T$ , then $V_1 = IR_1$ and $V_2 = IR_2$ . Writing " $V_2 = 12$ V" applies the parallel rule to a series circuit. Self-check: $V_1 + V_2$ must equal $V_S$ .

November 2024 Paper 1P Q9(b)(iv): most candidates attempted to calculate the current in a series circuit by applying the full supply EMF across each resistor independently — the parallel voltage rule misapplied to a series circuit. Only the most able candidates answered this correctly.

Adding parallel R like series (too big)

The parallel formula needs reciprocals: $1/R_T = 1/R_1 + 1/R_2$ , then invert. Writing $R_T = R_1 + R_2$ (the series form) gives a value larger than either branch — the opposite of correct. Self-check: combined parallel resistance is always less than the smallest branch. Example: 6 Ω ∥ 12 Ω → $R_T = 4$ Ω (4 < 6 ✓); 6 + 12 = 18 Ω fails at once.

November 2024 Paper 2P examiner report Summary: examiners recommended that students "build or simulate circuits noting the effect on currents and voltages" — suggesting that many candidates lacked the intuitive grasp that adding a parallel branch decreases total resistance rather than increasing it.

Resistance is not "opposition to current"

Defining resistance as "opposition to current" earns no marks. The mark scheme wants the ratio form: $R = V/I$ — the ratio of the p.d. across a component to the current through it — plus the unit ohm (Ω). Quote both elements. The everyday phrase is too vague and lacks the quantity and unit examiners require.

Forgetting to convert mA to A

Every circuit formula ( $V = IR$ , $P = IV$ , $P = I^2R$ ) needs current in amperes. If a value is in milliamps, divide by 1000 first: 250 mA = 0.25 A. Substituting 250 into $P = I^2R$ gives an answer a million times too large. Convert mA → A (and mV → V, kΩ → Ω) before plugging in, and put the unit on the final answer.

Ammeter in series, voltmeter in parallel

An ammeter measures current, so it goes in series in the wire; its near-zero resistance would short a component if placed in parallel. A voltmeter measures p.d., so it goes in parallel across the component; its very high resistance would block the current if placed in series. Swapping them is a fatal wiring error that loses the diagram marks.

Extra parallel branch dims the others

Adding a lamp in parallel does not dim the existing lamps. Each branch keeps the full supply p.d., so the original lamp's voltage and current ( $I = V/R$ ) are unchanged — same brightness. What rises is the total current drawn from the supply, because the combined resistance falls. Do not confuse total supply current with the current in one branch.

Exam tips

Name the circuit type, then its rule
Write "series" or "parallel" first — it commits you to the right rules. Then state the rule before substituting: series $R_T = R_1 + R_2$ , $I$ same, $V_1 + V_2 = V_S$ ; parallel $1/R_T = 1/R_1 + 1/R_2$ , $V$ same. Naming it earns method marks even if arithmetic slips.
Find total resistance first
Work the whole circuit before the parts: find the combined/total resistance, then use $I = V_S/R_T$ for the supply current, then apply $V = IR$ to each component for its p.d. or current. Finding a component voltage before the total current is the usual slip.
Self-check with the conservation rule
Before boxing an answer, use the check your circuit type gives: in series the component p.d.s add to the supply ( $V_1 + V_2 = V_S$ ); in parallel the branch currents add to the total ( $I_T = I_1 + I_2$ ). If the check fails, there is an earlier error.
Advantages of parallel — give both marks
For "why are lamps connected in parallel" state both marking points: (1) each lamp gets the full supply voltage so is at full brightness, and (2) lamps work independently — if one fails the rest stay on. One point alone caps the marks; name the parallel arrangement explicitly.

Revision notes