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AS Daily A Level Physics question

2026-02-22 OCR A Electricity (Module 4) OCR-A Module 4.2 Energy, power and resistance (AS) OCR-A Module 4.3 Direct-current circuits: internal resistance and terminal pd (AS)

A 6.0 V cell with internal resistance 1.0 Ω powers a small lab heater that can be switched between 1.0 Ω and 2.0 Ω. When you switch from 1.0 Ω to 2.0 Ω, what happens to the power dissipated in the heater?

  1. A It decreases to about 0.89 of the original (≈11% drop), since the heater’s voltage rises but the current falls more. (correct)
  2. B It halves, because doubling the resistance halves the power at a fixed supply voltage.
  3. C It roughly doubles, because power increases with resistance when using P = I²R.
  4. D It is unchanged, because halving the current and doubling the resistance cancel in P = IV.

Answer

The correct answer is A.

Correct: A — It decreases to about 0.89 of the original (≈11% drop), since the heater’s voltage rises but the current falls more. Using V_load = ε·R/(R+r) and P_load = V_load²/R gives P ∝ R/(R+r)²; from R=1 Ω to 2 Ω with r=1 Ω, the ratio is (2/3)²/2 ÷ (1/2)²/1 = 8/9 ≈ 0.89. B assumes the terminal pd stays at the emf, ignoring internal resistance, so it wrongly takes P ∝ 1/R. C assumes the current is unchanged, so it misapplies P = I²R to predict an increase. D assumes the voltage across the heater exactly doubles as the current halves, which is not true here (it rises from 3 V to 4 V, not to 6 V).

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