AS 91526 · 2025 paper · inductors

Question 2(b) Merit

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

Calculate the size of the induced EMF across the inductor 0.240 s after the switch is 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 What resistance limits the current? Check the circuit diagram for everything in the loop.
The coil’s 2.40 Ω plus the battery’s 0.100 Ω internal resistance: R = 2.50 Ω.
2 What is the time constant of a circuit with an inductor and a resistance?
τ = L / R = 0.600 / 2.50 = 0.240 s.
3 Compare 0.240 s with τ. What does that tell you?
0.240 s is exactly one time constant, so the induced EMF has fallen to about 37% (e⁻¹) of its starting 12.6 V.
4 Finish it.
ε = 12.6 × e⁻¹ = 4.64 V.

Answer 4.64 V. The schedule’s 4.66 V comes from rounding e⁻¹ to 37% (12.6 × 0.37).

What the marker wanted

Achieved The time constant, 0.240 s, or a decayed EMF worked out from the wrong time constant.
Merit The EMF after 0.240 s. The schedule’s value is 4.66 V, worked with 37%.

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

Common mistake

Leaving out the battery’s 0.100 Ω. That gives τ = 0.250 s and an EMF of 4.82 V, which the schedule only credits at Achieved.

How did it go?

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