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AS Daily A Level Physics question

2026-04-25 OCR A DC circuits (Module 4) — Internal resistance and terminal p.d. OCR-A H156 Module 4: 4.2.3 Circuits — e.m.f., internal resistance and terminal p.d.; combining sources

A small torch bulb has resistance 3.0 Ω when lit. It is powered first by one 1.5 V cell with internal resistance 0.50 Ω, then by two identical cells in series. Which statement must be true about the terminal p.d. across the bulb when the second cell is added?

  1. A It increases by about 75%, from roughly 1.3 V to about 2.25 V. (correct)
  2. B It roughly doubles, from about 1.3 V to about 2.6 V.
  3. C It increases by less than 50%, to about 1.9 V, because internal resistance also doubles.
  4. D It barely changes (around 1.3 V) because the bulb’s resistance is much larger than the internal resistance.

Answer

The correct answer is A.

Correct: A — It increases by about 75%, from roughly 1.3 V to about 2.25 V. With one cell: total series resistance is 3.0 + 0.50 = 3.50 Ω, so Vbulb ≈ 1.5 × (3.0/3.5) ≈ 1.29 V; with two cells: total is 3.0 + 1.00 = 4.00 Ω, so Vbulb ≈ 3.0 × (3.0/4.0) = 2.25 V, an increase by a factor ≈ 1.75 (about 75%). B is wrong because doubling the e.m.f. does not double the bulb p.d. when internal resistance is not zero. C is wrong because although internal resistance doubles, the share of the e.m.f. across the bulb is 3/4 of 3.0 V, giving 2.25 V, which is more than a 50% increase. D is wrong because the internal resistance (0.50 Ω per cell) is not negligible compared with 3.0 Ω; adding the second cell significantly raises the bulb p.d.

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