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AS Daily A Level Physics question

2026-03-19 OCR A Electricity (M4) Module 4: Electricity — 4.3.1 Potential dividers (including sensors)

A garden light uses a 9.0 V supply and a potential divider made from a 3.0 kΩ fixed resistor in series with an LDR. The output voltage V_out is measured across the LDR. In shade the LDR is 2.0 kΩ, but in bright sunlight it halves to 1.0 kΩ. What happens to V_out, and by roughly how much?

  1. A It increases to about 4.5 V because halving the LDR’s resistance gives it a larger share of the 9.0 V.
  2. B It decreases to about 1.8 V because the voltage across a resistor simply halves when its resistance halves.
  3. C It stays about 3.0 V because the supply is fixed and the two resistors still share it similarly.
  4. D It decreases from about 3.6 V to about 2.3 V since the LDR gets a smaller fraction of the total resistance. (correct)

Answer

The correct answer is D.

Correct: D — It decreases from about 3.6 V to about 2.3 V since the LDR gets a smaller fraction of the total resistance. Using V_out = 9 × R_LDR/(3.0 + R_LDR): in shade, 9 × 2.0/(3.0+2.0) = 3.6 V; in bright light, 9 × 1.0/(3.0+1.0) = 2.25 V ≈ 2.3 V, so V_out decreases by about 40%. A assumes a lower resistance takes more of the voltage, but the voltage share is proportional to resistance, so V_out should decrease, not increase. B assumes V_out halves with R, ignoring the series partner; because the total also changes, the drop is to 2.25 V, not 1.8 V. C assumes little change because the supply is fixed, but changing the resistance ratio changes how the 9.0 V is divided. D correctly applies the divider ratio: a smaller LDR resistance means a smaller fraction of the total, so the measured voltage falls from about 3.6 V to about 2.3 V.

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