The questions are on the internal resistance test. Take it under real conditions before reading the mark scheme.
Quick answer: The emf of a source is the energy it gives each coulomb of charge. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on internal resistance, from emf and lost volts to falling terminal voltage, with the equivalent answers to accept.
Marks are awarded for a correct formula, a clear substitution and an answer with its unit. Accept answers within rounding error and award method marks if an earlier slip is carried through consistently.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB B.5
Question 1 Emf and lost volts (2 marks)
The emf is the total energy per coulomb supplied by the battery. The terminal p.d. is smaller because energy per coulomb is lost inside the battery in its internal resistance (lost volts, Ir).
Marking: 1 mark for emf as energy per unit charge supplied, 1 mark for the terminal p.d. being smaller by Ir.
Question 2 Cell and resistor (3 marks)
- I = ε / (R + r) = 6.0 / (2.5 + 0.50) = 2.0 A.
- Lost volts = Ir = 2.0 × 0.50 = 1.0 V.
- V = 6.0 − 1.0 = 5.0 V.
Marking: 1 mark for each part.
Question 3 Junction (2 marks)
Current in = 4.5 + 1.2 = 5.7 A. Current out = 2.0 + 1.3 + I. So I = 5.7 − 3.3 = 2.4 A, by Kirchhoff’s first law (conservation of charge).
Marking: 1 mark for the answer, 1 mark for naming the law.
Question 4 Loop equation (3 marks)
- 15 = 2.0 × 2.0 + 2.0 × 3.0 + 2.0 × R, i.e. the emf equals the sum of the p.d.s.
- 15 = 4.0 + 6.0 + 2.0R, so 2.0R = 5.0 and R = 2.5 Ω.
- V = IR = 2.0 × 3.0 = 6.0 V.
Marking: 1 mark for each part.
Question 5 Potential divider (3 marks)
- V = 12 × 1.0 / (3.0 + 1.0) = 3.0 V.
- The lower resistor and the load are in parallel: R = 0.50 kΩ. V = 12 × 0.50 / (3.0 + 0.50) = 1.71 V.
Marking: (a) 1 mark. (b) 1 mark for the parallel resistance, 1 mark for the output.
Question 6 Power in a battery (3 marks)
- P = εI = 9.0 × 1.5 = 13.5 W.
- P = I²r = 1.5² × 0.60 = 1.35 W.
- Efficiency = (13.5 − 1.35) / 13.5 = 90 %.
Marking: 1 mark for each part.
Question 7 Short circuit (2 marks)
I = ε / r = 12 / 0.040 = 300 A. The very large current produces intense heating in the wires and battery, which can cause fire or burns.
Marking: 1 mark for the current, 1 mark for the danger.
Question 8 Falling terminal voltage (2 marks)
The lost volts Ir across the internal resistance is proportional to the current. The terminal p.d. is V = ε − Ir, so as I increases, Ir increases and V decreases while the emf stays constant.
Marking: 1 mark for lost volts proportional to current, 1 mark for V = ε − Ir.
