AS 91526 · 2025 paper · DC circuits

Question 1(d) Merit to Excellence

Mereana has a caravan with a simple battery-powered lighting circuit of total resistance 6.00 Ω. The battery has an EMF of 12.6 V and an internal resistance of 0.100 Ω.

R	6.00 Ω
ε	12.6 V
r	0.100 Ω

Six 90.0 F capacitors are charged individually to 4.20 V. Three are connected in series with a 6.00 Ω resistor (Circuit A); three in parallel with a 6.00 Ω resistor (Circuit B).

For each circuit, explain how the current varies over time compared with the single capacitor in part (c).

The 2025 paper, page 4 ↗ Marking schedule, page 3 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Three equal capacitors in series: what is their total capacitance, and what voltage does the chain start at?
In series, capacitance goes down: 1/C = 3 / 90.0, so C = 30.0 F. The three 4.20 V charges add, so Circuit A starts at 12.6 V.
2 Now the same two questions for three in parallel.
In parallel, capacitance adds: C = 3 × 90.0 = 270 F. The voltage stays at 4.20 V, the same as one capacitor.
3 Work out the starting current and the time constant of each, and put them beside the single capacitor from (c).
Single: I₀ = 0.700 A, τ = 540 s.
Circuit A: I₀ = 12.6 / 6.00 = 2.10 A, τ = 6.00 × 30.0 = 180 s.
Circuit B: I₀ = 4.20 / 6.00 = 0.700 A, τ = 6.00 × 270 = 1620 s.
4 Turn those numbers into a description of each current over time.
Circuit A’s current starts three times higher and dies away three times faster. Circuit B’s starts at the same 0.700 A but falls three times more slowly, so it keeps flowing for much longer.

Answer Circuit A: 30.0 F charged to 12.6 V, so the current starts higher (2.10 A) and falls faster (τ = 180 s). Circuit B: 270 F at 4.20 V, so it starts at the same 0.700 A and falls more slowly (τ = 1620 s).

What the marker wanted

Achieved Any one comparison: total capacitance (down in series, up in parallel), the time constant, the starting current, or how fast the current falls.
Merit Links the capacitance change to the time-constant change for both circuits, or explains that Circuit A starts with a bigger current because it starts at a higher voltage.
Excellence All four, for both circuits: capacitance, time constant, starting current, and how quickly the current falls.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

Leaving Circuit A’s starting current at 0.700 A. In series the three 4.20 V capacitors add to 12.6 V, so it starts at three times the current, and explaining why is one of the ways to reach Merit.

How did it go?

Got it moves it further out. Review brings it back tomorrow.