AS 91526 · 2025 paper · DC circuits

Question 1(a) Achieved

Mereana has a caravan with a simple battery-powered lighting circuit of total resistance 6.00 Ω. The battery has an EMF of 12.6 V and an internal resistance of 0.100 Ω.

R	6.00 Ω
ε	12.6 V
r	0.100 Ω

When the switch is closed the current from the battery is 2.07 A.

Show that the terminal voltage of the battery is 12.4 V.

The 2025 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

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

1 Which equation links the EMF, the current and the internal resistance to the voltage across the terminals?
V_T = ε − Ir
2 Put the values in. It is a “show that”, so keep the unrounded result.
V_T = 12.6 − (2.07 × 0.100) = 12.6 − 0.207 = 12.393 V

Answer 12.393 V, which rounds to the 12.4 V asked for.

What the marker wanted

Achieved The unrounded terminal voltage, 12.393 V. Working it out as the voltage across the 6.00 Ω lights instead (2.07 × 6.00 = 12.42 V) is accepted too.

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

Common mistake

Writing 12.4 V as your answer. The question has already told you 12.4 V, so the mark is for the unrounded value that shows you got there yourself.

How did it go?

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