Question 1(c) Merit to Excellence

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.

After long use the internal resistance rises to 0.500 Ω. The discharged battery is charged by a 24.0 V DC supply with a 15.5 Ω resistor in series.

Calculate the terminal voltage across the car battery. Begin by calculating the circuit current while it is charging.

The 2024 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

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

1. 1 Two sources face each other in one loop. What is the net EMF driving the current?

   The 24.0 V supply pushes current into the 12.0 V battery, against its EMF. Net EMF = 24.0 − 12.0 = 12.0 V.

2. 2 What resistance is in the loop? Check the diagram for everything.

   The 15.5 Ω resistor plus the battery’s 0.500 Ω internal resistance: 16.0 Ω.

3. 3 Now the charging current.

   I = 12.0 / 16.0 = 0.750 A. Kirchhoff’s loop law gives the same: 24.0 − 15.5I − 12.0 − 0.500I = 0.

4. 4 Current is being forced into the battery. Is its terminal voltage above or below its EMF?

   Above. The charger has to supply the EMF and the drop across r as well: V_T = ε + Ir = 12.0 + (0.750 × 0.500) = 12.375 V.

5. 5 Check it from the supply’s side.

   The supply’s 24.0 V less the drop across the resistor: 24.0 − (0.750 × 15.5) = 24.0 − 11.625 = 12.375 V. The same.

What the marker wanted

• Achieved One of: the net EMF, 12.0 V, or a start on Kirchhoff’s loop law for this loop.
• Merit The charging current, 0.750 A, and the voltage across the 15.5 Ω resistor that follows from it, or the whole calculation with a single slip.
• Excellence The terminal voltage, 12.4 V.

The 11.7 V in the schedule’s Merit column is the voltage across the 15.5 Ω resistor: 0.750 × 15.5 = 11.625 V, rounded up.

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.