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AS 91526 · 2025 paper · AC inductors and capacitors

Question 3(a) Achieved

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

Show that the reactance of the inductor is 94.2 Ω.

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 Which formula turns an inductance and a frequency into a reactance?

    XL = 2πfL

  2. 2 Substitute. It is a “show that”, so keep the unrounded result.

    XL = 2π × 25.0 × 0.600 = 94.248 Ω
Answer 94.248 Ω, which rounds to the 94.2 Ω asked for.

What the marker wanted

  • Achieved The substitution written out: 2π × 25 × 0.6 (or 50π × 0.6).

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

✕

Common mistake

Going straight to 94.2 Ω. The schedule’s evidence here is the substitution itself, so the line that earns the mark is 2π × 25.0 × 0.600.

How did it go?

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