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AS 91526 · 2024 paper · alternating current

Question 3(a) Achieved

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.

L70.0 mH
R40.0 Ω
V RMS110 V
f60.0 Hz

Calculate the peak voltage of the power supply.

The 2024 paper, page 6 ↗ Marking schedule, page 6 ↗

Work it out

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

  1. 1 How are the RMS and peak values of a sine wave related?

    Vpeak = √2 × VRMS. Only the 2 goes under the root.

  2. 2 Substitute.

    Vpeak = √2 × 110 = 155.6 V.
Answer 156 V

What the marker wanted

  • Achieved The peak voltage, 156 V.

In our words, from NZQA’s 2024 marking schedule.

✕

Common mistake

Taking the root of 2 × 110 as a whole, which gives 14.8 V. NZQA’s 2024 report says many candidates did exactly that.

How did it go?

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