Physics 9702 · Paper 4

Series & Parallel

The combination rules are the resistor rules swapped — and the derivation of the series rule has been set twice in nine months, reasoning marks included.

The rules and why

Series: one path, so every plate pair holds the SAME charge Q, and the p.d. divides. Parallel: same p.d. across both, and the charge divides. Each rule is the other one swapped.

A 2025 four-marker was simply the table (9702/41/O/N/25 Q6(a)): in series the capacitors store the same charge ( $Q_S = Q_1 = Q_2$ ) and the p.d.s add; in parallel the p.d.s are the same and the charges add.
Why the same charge in series: the plate pair in the middle is an isolated island — charge pushed off one plate must arrive on its neighbour, so every capacitor in the chain carries the same $\pm Q$ . This single sentence is the M1 that starts the derivation.

Worked example

Deriving 1/C = 1/C₁ + 1/C₂ (set twice recently)

Derive the series formula, with reasoning (9702/42/M/J/24 Q6(a)) (9702/42/F/M/25 Q5(a)).

Series capacitors store equal charge $Q$ (isolated middle section).
The p.d.s add: $V = V_1 + V_2$ .
Substitute $V = Q/C$ for each: $\dfrac{Q}{C} = \dfrac{Q}{C_1} + \dfrac{Q}{C_2}$ , cancel $Q$ : $\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$ .

Answer

\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}\ \text{(series)};\qquad C = C_1 + C_2\ \text{(parallel)}

Networks: collapse them stage by stage, exactly as with resistors but with the rules swapped. A $45\ \mu\text{F}$ in series with a (10 ∥ 20) pair: parallel first $\rightarrow 30\ \mu\text{F}$ , then $(1/45 + 1/30)^{-1} = 18\ \mu\text{F}$ (9702/42/F/M/24 Q4(a)). One capacitor in parallel with two of the same value in series gives $C + C/2 = \tfrac{3}{2}C$ (9702/42/F/M/25 Q5(b)).
Sanity checks: series always gives less than the smallest member; parallel always gives the plain sum.

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