AS 91526 · 2024 paper · DC and capacitors
Question 1(b) Merit
Practises DC circuits
Stefan investigates new and used car batteries. His car has a new battery with an EMF of 12.0 V and an internal resistance of 0.0500 Ω, connected to a 7.00 Ω load.
| ε | 12.0 V |
|---|---|
| r | 0.0500 Ω |
| R | 7.00 Ω |
Stefan watches the voltmeter, which is across the battery’s terminals, while opening and closing the switch to the load.
Explain why the readings are the same or different. Values and calculations are not required.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Where is the voltmeter connected, and what does it measure?
Across the battery’s terminals, so it reads the terminal voltage: the EMF less anything lost inside the battery. -
2 With the switch open, how much current flows? What does the voltmeter read?
None, so no voltage is lost across r and the voltmeter reads the whole EMF, 12.0 V. -
3 Once the switch is closed, what happens inside the battery?
Current flows through the 7.00 Ω load and through the internal resistance as well. Part of each coulomb’s energy becomes heat inside the battery: the lost volts, Ir. -
4 So how do the two readings compare?
The closed reading is lower, because VT = ε − Ir. (Not needed, but for a sense of size: I = 12.0 / 7.05 = 1.70 A and Ir = 0.085 V, so it reads about 11.9 V.)
What the marker wanted
- Achieved One point: 12.0 V with the switch open, under 12.0 V with it closed, voltage or energy lost across the internal resistance, or just that the two readings differ.
- Merit The full answer with the reason the closed reading is lower: voltage (energy) lost across the internal resistance once current flows, in words or as V = ε − Ir.
In our words, from NZQA’s 2024 marking schedule.
✕Common mistake
Saying the readings differ without saying why. That is one Achieved point: NZQA’s 2024 report puts spotting the difference at Achieved and explaining it at Merit.How did it go?
Got it moves it further out. Review brings it back tomorrow.