AS 91526 · 2025 paper · DC circuits

Question 1(c)(i) Achieved

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

A 90.0 F capacitor is fully charged by a 4.20 V battery, then discharged for 2500 s through a 6.00 Ω resistor.

Sketch current against time for 2500 s, showing the initial current and the current after one time constant.

The 2025 paper, page 3 ↗ Marking schedule, page 2 ↗

Work it out

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

1 What sets how quickly the current dies away?
The time constant: τ = RC = 6.00 × 90.0 = 540 s.
2 What is the current at the moment the switch closes?
The capacitor starts at 4.20 V, so I₀ = V / R = 4.20 / 6.00 = 0.700 A.
3 How much of the starting current is left after one time constant?
About 37% (e⁻¹): I = 0.700 × e⁻¹ = 0.258 A. Using 0.37 gives 0.259 A. Either way, 0.26 A at t = 540 s.
4 Where does the curve start, and where is it heading by 2500 s?
Start at 0.700 A on the current axis and pass through 0.26 A at 540 s. The fall keeps slowing: about 0.095 A at 1080 s and 0.035 A at 1620 s. By 2500 s (about 4.6 time constants) it is almost at zero, but it never touches the axis.

Answer A falling exponential curve: 0.700 A at t = 0, 0.26 A at t = 540 s, levelling off towards zero by 2500 s.

What the marker wanted

Achieved Any one of: τ = 540 s, the 0.700 A starting current, or a falling curve that starts at 0.700 A.

The schedule marks (c)(i) and (c)(ii) together as one item: your graph and your explanation are read as one answer.

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

Common mistake

A curve with the right shape but no values on it. Past Achieved, the schedule wants the 0.700 A start and the 0.26 A point at 540 s actually shown.

How did it go?

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