AS 91526 · 2025 paper · DC circuits

Question 1(b) Merit

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

As the battery is used over time its internal resistance increases.

Explain the effect this has on the terminal voltage.

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 As the battery ages, what stays the same and what changes?
The EMF stays at 12.6 V. Only the internal resistance r goes up.
2 Once current flows, where does the EMF go?
It is shared between the voltage lost inside the battery and the terminal voltage across the lights: ε = V_T + Ir.
3 What happens to that split when r gets bigger?
More of the EMF is lost inside the battery, so less is left at the terminals. In one line: V_T = εR / (R + r), and R / (R + r) gets smaller as r grows.

Answer The terminal voltage decreases, because more of the unchanged EMF is lost across the larger internal resistance.

What the marker wanted

Achieved Says the terminal voltage goes down, or that more voltage is lost inside the battery.
Merit Gives the reason: the terminal voltage falls because more voltage is lost across the bigger internal resistance.

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

Common mistake

Saying the EMF drops. In this question the EMF stays the same and the terminal voltage is what falls. NZQA’s 2025 report lists mixing up EMF and terminal voltage among the common reasons for Not Achieved.

How did it go?

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