Physics Numerical: What Does 'Power Delivered by the Battery' Mean?
A PGT Physics teacher is moderating a 5-mark numerical item in a school-level mock. The item reads: 'A battery of emf 12 V and internal resistance 1 Ω is connected to an external resistor of 5 Ω. Calculate the power delivered by the battery.' The keyed answer is 20 W. The marking scheme gives 2 marks for current, 2 for power, 1 for unit. One student writes: I = 12/(5+1) = 2 A; P = EI = 12 × 2 = 24 W. Another writes: terminal voltage = 12 − 2×1 = 10 V; P = I²R = 4×5 = 20 W. A third writes 28.8 W using P = V²/R with V = 12 V. The first student challenges the key, arguing that 'power delivered by the battery' means the total power produced by the battery, not the power in the external resistor. The teacher must decide whether 24 W is acceptable, how to award part marks to the third student, and how to rewrite the item for the next cycle. The item is one of 30 in a 60-mark paper; 180 students took it.
1. Calculate current, terminal voltage, total power, internal dissipation, and external power. Is 24 W a defensible reading of the exact wording? Explain.
Read the worked answer
Current I = emf/(R + r) = 12/(5 + 1) = 2 A. Terminal voltage V = emf − Ir = 12 − 2(1) = 10 V. Total power from chemical conversion = emf × I = 12 × 2 = 24 W. Internal dissipation = I²r = 4 W. Power in the external resistor = I²R = 4 × 5 = 20 W, which also equals V × I = 10 × 2.
The exact wording 'power delivered by the battery' is ambiguous. In circuit analysis, 'power delivered by the source' often means power delivered to the external circuit, 20 W. But 'power of the battery' or 'total power delivered by the battery' can mean the chemical power, 24 W. The student's 24 W is a defensible reading if the answer shows P = EI, because it correctly distinguishes emf from terminal voltage. The key should not be defended as the only possible answer.
2. How should the teacher mark the three responses? Propose a marking scheme with part marks.
Read the worked answer
Student 2 (20 W via terminal voltage) gets full 5 marks: correct current, correct terminal voltage, correct external power, unit.
Student 1 (24 W via EI) should get full 5 if the item is treated as ambiguous; if the institute insists on 20 W, award 3 or 4 marks for correct current and correct total-power calculation, with a deduction only for not answering the intended external-power quantity. The physics is not wrong; the interpretation differs.
Student 3 (28.8 W) used P = V²/R with V = 12 V. That treats the battery as ideal and ignores the internal resistance, so the terminal voltage is wrong. Award at most 1 mark for knowing the power formula; no marks for substitution or answer. A fair scheme: current 2, terminal voltage 1, external power 1, unit 1; or current 2, power 2, unit 1 with explicit alternative for total power.
3. Rewrite the stem to make a single answer defensible, and state the new key and marking scheme.
Read the worked answer
Rewrite: 'A battery of emf 12 V and internal resistance 1 Ω is connected to an external resistor of 5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, and (c) the power dissipated in the external resistor. (d) How much power is dissipated inside the battery?' New key: (a) 2 A; (b) 10 V; (c) 20 W; (d) 4 W. Marking: 1 mark each for (a) and (b), 2 for (c), 1 for (d). This removes the ambiguity, tests the internal-resistance concept, and gives a complete energy check: 24 W total = 20 W external + 4 W internal. The school should also pre-test numerical items for wording ambiguity and include a marking note for alternative valid methods.