Question 2(d) Merit to Excellence

Mereana connects an old coil to her caravan battery. The coil is modelled as a 0.600 H inductor in series with a 2.40 Ω resistor; the 12.6 V battery has an internal resistance of 0.100 Ω.

Maximum energy is stored in the inductor when a steady current flows. The inductor is replaced by a capacitor storing an equal amount of energy.

Calculate the capacitance required, when fully charged by the 12.6 V battery.

The 2025 paper, page 6 ↗ Marking schedule, page 4 ↗

Work it out

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

1. 1 Once the current is steady, what limits it?

   Only resistance, because a steady current induces nothing: I = ε / R = 12.6 / 2.50 = 5.04 A.

2. 2 How much energy does the inductor hold at that current?

   E = ½LI² = ½ × 0.600 × 5.04² = 7.62 J.

3. 3 Which formula gives a capacitor’s energy from its capacitance and voltage? What voltage does it reach?

   E = ½CV². Fully charged, with no current flowing, the capacitor reaches the battery’s full 12.6 V.

4. 4 Set the two energies equal and solve for C.

   C = 2E / V² = (2 × 7.620) / 12.6² = 0.0960 F.

What the marker wanted

• Achieved The steady current, 5.04 A.
• Merit The energy stored in the inductor, 7.62 J.
• Excellence The capacitance, 0.096 F. The schedule also accepts 0.104 F, which is what you get if you leave out the battery’s 0.100 Ω and carry that through consistently.

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

How did it go?

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