AS 91526 · 2023 paper · alternating current

Question 3(d) Merit to Excellence

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.

f	200 Hz
R	15.0 Ω
C	1.20 µF

The 0.200 H inductor from part (c) is still in series with the capacitor and the lamp.

Explain how the inductor affects the impedance of the circuit, and why there is one frequency at which the impedance equals the 15.0 Ω resistance, making the lamp glow brightly.

The 2023 paper, page 9 ↗ Marking schedule, page 7 ↗

Work it out

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

1 How does the inductor’s reactance depend on the frequency?
X_L = 2πfL, so it rises in step with f. At 2000 Hz it is 2513 Ω against the capacitor’s 66 Ω: a large impedance, a small current, and the lamp is off.
2 As the frequency comes down, what happens to each reactance?
X_L falls and X_C rises, so at one frequency they are equal: 325 Hz, where both are 408 Ω.
3 Why do equal reactances cancel instead of adding?
The inductor’s voltage is 90° ahead of the current and the capacitor’s 90° behind, so they are 180° apart. Equal in size, they cancel at every instant: in Z = √(R² + (X_L − X_C)²) the bracket is zero.
4 So what is the impedance, and what does the lamp do?
Z = R = 15.0 Ω, the smallest it can be. The current is at its largest and the lamp glows brightly. Either side of 325 Hz one reactance outweighs the other, Z rises, and the lamp goes off again.

Answer The inductor adds a reactance, X_L = 2πfL, that grows with frequency, so at high f the impedance is large and the lamp is off. As f falls, X_L falls and X_C rises until they are equal (325 Hz). Their voltages are 180° out of phase and cancel, so Z = R = 15.0 Ω, its minimum; the current is at its maximum, and the lamp glows brightly.

What the marker wanted

Achieved One idea: X_L rises with frequency, X_C equals X_L but acts the opposite way, at resonance Z is at its minimum or I at its maximum, or X_L − X_C = 0.
Merit Z is at its minimum at resonance because X_L = X_C, with the reason they cancel: out of phase, shown by X_L − X_C or a phasor diagram.
Excellence Why X_L and X_C cancel (opposite, by formula or phasor diagram), AND that Z is at its minimum and the current at its maximum.

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

Common mistake

Explaining why Z = R and stopping there. The Excellence column also wants what that does to the current: at its maximum, which is why the lamp glows brightly.

How did it go?

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