AS 91526 · 2024 paper · DC and capacitors

Question 1(d) Merit to Excellence

Stefan investigates new and used car batteries. His car has a new battery with an EMF of 12.0 V and an internal resistance of 0.0500 Ω, connected to a 7.00 Ω load.

ε	12.0 V
r	0.0500 Ω
R	7.00 Ω

A 12.0 V DC supply is connected through a switch to a 0.140 F capacitor, two lamps and a resistor. Lamp A is in series with the capacitor and Lamp B in series with the resistor, and the two branches are in parallel across the supply. The resistor and each lamp have a resistance of 10.0 Ω.

Give an in-depth explanation of the brightness of lamps A and B — as soon as the switch closes, and ten seconds later.

The 2024 paper, page 3 ↗ Marking schedule, page 2 ↗

Work it out

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

1 Trace the circuit. Which lamp shares a branch with the capacitor, and which with the resistor?
Lamp A is in series with the 0.140 F capacitor; Lamp B is in series with the 10.0 Ω resistor. The branches are in parallel, so each has the full 12.0 V across it.
2 The instant the switch closes, the capacitor is empty. What voltage is across it, and so across Lamp A?
None across the capacitor yet, so Lamp A gets all 12.0 V: I_A = 12.0 / 10.0 = 1.20 A, and P = 14.4 W.
3 What about Lamp B’s branch?
Lamp B shares its 12.0 V with the resistor: 20.0 Ω in all, so I_B = 0.600 A, 6.00 V across the lamp and P = 3.60 W. At the start Lamp A is much the brighter, with four times the power.
4 How quickly does the capacitor fill? Find the time constant of Lamp A’s branch.
τ = RC = 10.0 × 0.140 = 1.40 s. Ten seconds is about seven time constants, so the capacitor is as good as full at 12.0 V, and Lamp A’s current is down to about 1 mA (1.20 × e^−10/1.40).
5 So what do you see at ten seconds?
Lamp A is out: a full capacitor lets no more current through its branch. Lamp B is just as it was, because nothing in its branch changes with time.

Answer At switch-on Lamp A is brighter than Lamp B: the empty capacitor leaves it the whole 12.0 V (1.20 A), while Lamp B shares 12.0 V with the resistor (0.600 A). Lamp A dims as the capacitor charges (τ = 1.40 s) and is out by ten seconds; Lamp B glows steadily, unchanged, the whole time.

What the marker wanted

Achieved One observation: Lamp A lights at first, Lamp A goes out, Lamp B stays on, or Lamp B is the dimmer one.
Merit Two of these, with reasons: Lamp A bright at first while the capacitor charges; Lamp A out once it is full, because its current stops; Lamp B still on; Lamp A the brighter of the two, since it has the larger current, voltage and power (or Lamp B the dimmer, since the resistor takes half of its 12.0 V).
Excellence All of them: Lamp A bright at first and out once the capacitor is full, Lamp B on throughout, and the comparison of the two lamps with its reason.

In our words, from NZQA’s 2024 marking schedule.

Common mistake

Deciding that the current all goes one way because that path is easier. Each branch has the full 12.0 V across it and carries its own current. NZQA’s 2024 report names that “least resistance” idea as a common misconception, and says many candidates misunderstood what a capacitor does in a parallel circuit.

How did it go?

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