Question 2(d) Merit to Excellence

Harry is experimenting with coils and magnets. Each part gives its own values.

A 0.600 H inductor and a lamp of resistance 2.00 Ω are connected in series to a 12.0 V battery through a switch.

Calculate the current through the lamp after 0.600 s. Begin by calculating the time constant.

The 2024 paper, page 5 ↗ Marking schedule, page 4 ↗

Work it out

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

1. 1 Once the current settles, what limits it?

   Only the lamp’s resistance, since a steady current induces nothing: I_max = 12.0 / 2.00 = 6.00 A.

2. 2 What is the time constant of an inductor with a resistance?

   τ = L / R = 0.600 / 2.00 = 0.300 s.

3. 3 How many time constants is 0.600 s?

   Two. In each time constant the current closes about 63% of the gap still left to its final value.

4. 4 What is the current once two time constants have passed?

   I = I_max(1 − e^−t/τ) = 6.00 × (1 − e⁻²) = 6.00 × 0.8647 = 5.19 A.

5. 5 Check it one time constant at a time.

   After one τ: 63% of 6.00 A = 3.78 A, leaving 2.22 A to go. After two: 3.78 + (0.63 × 2.22) = 5.18 A. The small difference comes from rounding e⁻¹ to 37%.

What the marker wanted

• Achieved One of: the final current, 6.00 A; the time constant, 0.300 s; or that one time constant means a 63% change.
• Merit All three: 6.00 A, 0.300 s and the current at t = τ (3.78 A with 63%), or the right method with one small slip.
• Excellence The current after 0.600 s: 5.18 A in the Excellence column (63% taken twice), 5.19 A in the schedule’s own working.

The schedule’s Merit column prints the current at t = τ as 3.38 A, which appears to be a misprint for 3.78 A (63% of 6.00 A).

In our words, from NZQA’s 2024 marking schedule.

Common mistake

How did it go?

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