AS 91526 · 2024 paper · alternating current
Question 3(b) Merit
Practises Calculating the phase angle between voltage and current
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.
| L | 70.0 mH |
|---|---|
| R | 40.0 Ω |
| V RMS | 110 V |
| f | 60.0 Hz |
Calculate the phase difference between the supply voltage and the circuit current, and state which one leads. Begin by showing that the reactance of the inductor is 26.4 Ω.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Put the inductance in base units, then find its reactance.
70.0 mH = 0.0700 H. XL = 2πfL = 2π × 60.0 × 0.0700 = 26.39 Ω, which rounds to the 26.4 Ω asked for. -
2 Sketch the phasor diagram. Which way does each voltage point?
Take the current as the reference, along the horizontal. The bulb’s voltage (IR) is in phase with it; the inductor’s (IXL) points straight up, 90° ahead. The supply voltage is their sum, the diagonal. -
3 Which ratio gives the angle between the supply voltage and the current?
tan φ = XL / R = 26.39 / 40.0 = 0.660, so φ = 33.4° (0.583 rad). -
4 Which one leads?
The supply voltage: the circuit is inductive, so on the diagram it is turned ahead of the current. (A check on the size: Z = √(40.0² + 26.4²) = 47.9 Ω.)
What the marker wanted
- Achieved One of: the reactance shown with L = 0.070 H, giving 26.4 Ω; the angle, 33° (0.583 rad); that the supply voltage leads; or a correctly labelled phasor diagram.
- Merit XL = 26.4 Ω and the angle, 33° (0.583 rad), and says that the supply voltage leads.
In our words, from NZQA’s 2024 marking schedule.
✕Common mistake
Putting in 70.0 for the inductance instead of 0.0700 H, which makes the reactance a thousand times too big. The schedule insists the show-that uses L = 0.070 H, and NZQA’s 2024 report singles out unit prefixes as badly handled that year.How did it go?
Got it moves it further out. Review brings it back tomorrow.