AS 91526 · 2025 paper · inductors

Question 2(a) Achieved

Mereana connects an old coil to her caravan battery. The coil is modelled as a 0.600 H inductor in series with a 2.40 Ω resistor; the 12.6 V battery has an internal resistance of 0.100 Ω.

L	0.600 H
R(coil)	2.40 Ω
ε	12.6 V
r	0.100 Ω

State the size of the induced EMF across the inductor when the switch is first closed.

The 2025 paper, page 5 ↗ Marking schedule, page 4 ↗

Work it out

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

1 At the instant the switch closes, how much current is flowing?
None yet. The inductor opposes the sudden change, so the current starts from zero.
2 With no current, how much voltage is dropped across the resistances?
V = IR = 0 across both the coil’s 2.40 Ω and the battery’s 0.100 Ω.
3 So where does the battery’s EMF end up?
All of it is across the inductor, as an induced EMF opposing the battery.

Answer 12.6 V

What the marker wanted

Achieved The right value, 12.6 V.

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

Common mistake

Answering 0 V, which is how a capacitor behaves at switch-on, not an inductor. An inductor starts with the whole EMF across it. NZQA’s 2025 report lists confusing capacitors with inductors among the common reasons for Not Achieved.

How did it go?

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