AS 91526 · 2023 paper · transformers and inductors
Question 2(c) Merit
Practises Inductors
Kate’s school has a demonstration transformer. She connects the 12 000-turn primary coil to the mains supply at 240 V RMS. The secondary has 600 turns.
| N(p) | 12 000 |
|---|---|
| V(p) | 240 V RMS |
| N(s) | 600 |
The 12 000-turn primary is connected instead to a 12 V DC battery and a 12 V car headlamp bulb, the cores still linked with iron.
Explain why the headlamp bulb only comes on after a slight delay.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 When the switch closes, what starts to happen to the current in the primary coil?
It starts to rise from zero. -
2 What does a rising current do in a 12 000-turn coil on iron?
It builds a changing flux, and that changing flux induces an EMF in the same coil. By Lenz’s law this back EMF opposes the rise, pushing against the battery. -
3 Why does that make the current rise slowly, rather than just less?
The back EMF is ε = −L ΔI/Δt, and it can never be bigger than the battery’s 12 V (less what the lamp takes). That caps ΔI/Δt: the current can only climb so fast. 12 000 turns on iron make L huge, so the cap is low. -
4 Say the same thing with a time constant.
τ = L / R. A very large L means a long time constant, so the current takes a noticeable time to reach its full value, and the lamp only comes on properly once it has.
What the marker wanted
- Achieved Mentions the back (opposing) EMF, or that a large inductance lengthens the time constant, τ = L/R, so the current takes longer to build up.
- Merit Links the back EMF to a limit on how fast the current can change.
In our words, from NZQA’s 2023 marking schedule.
✕Common mistake
Calling it an induced current that opposes the battery. It is an EMF, a voltage: the schedule does not credit “induced current” here, and NZQA’s 2023 report singles out this slip.How did it go?
Got it moves it further out. Review brings it back tomorrow.