← Electrical Systems: past papers
AS 91526 · 2024 paper · alternating current
Question 3(c) Merit to Excellence
Practises AC theory: RMS, reactance, impedance & resonance
Rachel connects a 70.0 mH inductor with a soft iron core and a 40.0 Ω bulb in series to a 110 V RMS, 60.0 Hz supply.
| L | 70.0 mH |
|---|---|
| R | 40.0 Ω |
| V RMS | 110 V |
| f | 60.0 Hz |
Rachel slowly removes the iron core from the inductor.
Explain what she would observe in the brightness of the lamp.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 What does the soft iron core do for the inductor?
It concentrates the flux, which raises the inductance. Taking it out lowers L. -
2 How does a smaller L change the reactance, and so the impedance?
XL = 2πfL falls, and Z = √(R² + XL²) falls with it. -
3 What happens to the current, and to the lamp’s share of the voltage?
I = V / Z rises. With a smaller back EMF across the inductor, more of the 110 V is left across the lamp. -
4 What decides how bright a lamp is?
The power it converts: P = I²R (or VI, or V²/R). A bigger current through the same 40.0 Ω means more power, so a brighter lamp.
What the marker wanted
- Achieved One idea: L, XL or Z goes down, the current goes up, there is less back EMF, the field or flux is weaker, the lamp is brighter, or brightness depends on power.
- Merit The chain from L down to a brighter lamp, or less back EMF leaving more voltage across the lamp so that it is brighter.
- Excellence The chain, and why more current (or voltage) makes the lamp brighter, using a power formula.
In our words, from NZQA’s 2024 marking schedule.
✕Common mistake
Ending the chain at “the current increases”. NZQA’s 2024 report puts explaining the current at Merit and linking the inductance to the lamp’s brightness at Excellence, and the schedule wants a power formula for that last link.How did it go?
Got it moves it further out. Review brings it back tomorrow.