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AS 91526 · 2023 paper · transformers and inductors

Question 2(d) Merit to Excellence

Practises Electromagnetic induction: eddy currents & transformers

Kate’s school has a demonstration transformer. She connects the 12 000-turn primary coil to the mains supply at 240 V RMS. The secondary has 600 turns.

N(p)12 000
V(p)240 V RMS
N(s)600

A power station generates 50 kW. The transmission line carries it at 220 kV and has a resistance of 4.00 Ω per kilometre.

Calculate the power lost as heat over 300 km, and comment on whether it is significant compared with transmitting at 25 kV.

The 2023 paper, page 7 ↗ Marking schedule, page 5 ↗

Work it out

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

  1. 1 What is the line’s resistance over 300 km?

    R = 4.00 × 300 = 1200 Ω.

  2. 2 At 220 kV, what current carries 50 kW?

    P = VI, so I = 50 000 / 220 000 = 0.227 A.

  3. 3 Which formula gives the power turned to heat in the line?

    Ploss = I²R = 0.2273² × 1200 = 61.98 W: about 0.12% of the 50 kW, so not significant.

  4. 4 Now the same at 25 kV.

    I = 50 000 / 25 000 = 2.00 A, and Ploss = 2.00² × 1200 = 4800 W, nearly 10% of the power sent.

  5. 5 Why is the difference so large?

    At the lower voltage the same power needs 8.8 times the current, and the loss goes as I², so it is 8.8² ≈ 77 times bigger. That is why the voltage is stepped up for transmission.
Answer At 220 kV the loss is 62.0 W (61.98 W, from 0.227 A): not significant against 50 kW. At 25 kV it is 4800 W (from 2.00 A), about 77 times more and nearly 10% of the power, because the current is 8.8 times bigger and the loss goes as I².

What the marker wanted

  • Achieved One of: the current at 220 kV, 0.227 A; the current at 25 kV, 2.0 A; or that the 220 kV loss is small beside 50 kW.
  • Merit Both losses, 61.98 W and 4800 W, or the whole answer with one error or omission.
  • Excellence Both losses, and a comparison that accounts for them: the lower voltage needs a bigger current, so more is lost as heat. Or: as V goes up I comes down, and the loss goes as I², with that comparison.

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

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Common mistake

Working out both losses and leaving it there. The Excellence column wants the reason too: a lower voltage means a bigger current, and the heat lost grows as I². NZQA’s 2023 report puts using the calculations to explain why the voltage is stepped up at Excellence.

How did it go?

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