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AS 91526 · 2023 paper · alternating current

Question 3(a) Merit

Practises AC theory: RMS, reactance, impedance & resonance

Kate builds a series circuit with a signal generator at 200 Hz, an AC ammeter, a lamp of 15.0 Ω and a 1.20 µF capacitor.

f200 Hz
R15.0 Ω
C1.20 µF

Show that the capacitive reactance is 663 Ω, and hence determine the impedance of the circuit.

The 2023 paper, page 8 ↗ Marking schedule, page 6 ↗

Work it out

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

  1. 1 Which formula turns a capacitance and a frequency into a reactance? Put C in farads.

    XC = 1 / (2πfC), with C = 1.20 µF = 1.20 × 10−6 F.

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

    XC = 1 / (2π × 200 × 1.20 × 10−6) = 663.15 Ω, which rounds to the 663 Ω asked for.

  3. 3 The lamp’s resistance and the capacitor’s reactance are at right angles on a phasor diagram. How do they combine?

    Z = √(R² + XC²) = √(15.0² + 663.15²) = 663.32 Ω.
Answer XC = 663.15 Ω and Z = 663 Ω (663.32 Ω). The capacitor dominates; the lamp adds almost nothing.

What the marker wanted

  • Achieved Either value on its own: XC = 663.15 Ω, shown, or Z = 663.32 Ω.
  • Merit Both: the reactance shown correctly and the right impedance.

In our words, from NZQA’s 2023 marking schedule.

✕

Common mistake

Stopping once 663 Ω is shown. The question says “and hence”, and the Merit column needs the impedance as well.

How did it go?

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