AS 91526 · 2025 paper · AC inductors and capacitors

Question 3(d) Merit to Excellence

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.

L	0.600 H
R(coil)	2.40 Ω
R	68.0 Ω
V RMS	9.00 V
f	25.0 Hz

The supply frequency is reduced further, to 16.0 Hz.

Calculate the RMS current in the circuit. A phasor diagram may assist.

The 2025 paper, page 9 ↗ Marking schedule, page 7 ↗

Work it out

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

1 Work out both reactances at the new frequency.
X_L = 2π × 16.0 × 0.600 = 60.32 Ω.
X_C = 1 / (2π × 16.0 × 1.20 × 10⁻⁴) = 82.89 Ω.
2 What is the total resistance in the circuit? Do not forget the coil.
The 68.0 Ω resistor plus the coil’s own 2.40 Ω: R = 70.4 Ω.
3 Combine them. On a phasor diagram the two reactances point in opposite directions, at right angles to R.
Z = √(R² + (X_L − X_C)²) = √(70.4² + (60.32 − 82.89)²) = √(70.4² + 22.57²) = 73.93 Ω.
4 Now the current.
I_RMS = V_RMS / Z = 9.00 / 73.93 = 0.122 A.

Answer 0.122 A

What the marker wanted

Achieved Either reactance (X_L = 60.3 Ω or X_C = 82.9 Ω), or a correct phasor diagram.
Merit The impedance, 73.9 Ω, or the right method carried through from a wrong impedance.
Excellence The current, 0.122 A.

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

Common mistake

Leaving out the coil’s 2.40 Ω. That gives Z = 71.6 Ω and 0.126 A, which the schedule caps at Merit, and NZQA’s 2025 report names this exact slip.

How did it go?

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