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AS 91526 · 2025 paper · AC inductors and capacitors

Question 3(b) Merit

Practises AC theory: RMS, reactance, impedance & resonance

The 0.600 H coil (an ideal inductor plus a 2.40 Ω resistor) is connected in series with a 68.0 Ω resistor to a variable-frequency AC supply set to 25.0 Hz and 9.00 V RMS.

L0.600 H
R(coil)2.40 Ω
R68.0 Ω
V RMS9.00 V
f25.0 Hz

An iron nail is inserted inside the coil.

Explain what effect this has on the RMS voltage across the 68.0 Ω resistor.

The 2025 paper, page 7 ↗ Marking schedule, page 6 ↗

Work it out

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

  1. 1 What does an iron core do to a coil?

    Iron concentrates the magnetic flux, so the inductance L goes up.

  2. 2 How does a bigger L change the coil’s opposition to AC?

    XL = 2πfL, so the reactance rises.

  3. 3 What happens to the whole circuit’s opposition, and so to the current?

    The impedance Z rises. The supply voltage is fixed, so the RMS current I = V / Z falls.

  4. 4 Now bring it back to the 68.0 Ω resistor.

    VR = IR with R fixed, so a smaller current gives a smaller RMS voltage across the resistor. The coil takes a bigger share of the supply voltage instead.
Answer The RMS voltage across the 68.0 Ω resistor decreases: L ↑ → XL ↑ → Z ↑ → I ↓ → VR ↓.

What the marker wanted

  • Achieved Any one link: the flux or L up, XL up, VL up, Z up, I down, or VR down.
  • Merit The whole chain, through to the resistor’s voltage falling.

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

✕

Common mistake

Stopping at “the reactance increases”. One link earns Achieved; Merit needs every step through to the resistor’s voltage.

How did it go?

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