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A2 Daily A Level Physics question

2026-02-25 OCR A Applied contexts: sensor circuits OCR-A 4.2.3 Electrical circuits — potential dividers; meter loading (qualitative); sensor applications OCR-A 4.2.2 Energy, power and resistance — LDR/thermistor behaviour (qualitative)

A lab light monitor uses a 5.0 V supply with an LDR in the upper leg and a 10 kΩ fixed resistor in the lower leg; the midpoint voltage Vout is read by a data logger whose input resistance is 10 kΩ to 0 V. Over the operating range, doubling the light intensity roughly halves the LDR’s resistance. Which statement must be true about how Vout changes when the light doubles?

  1. A It increases, but by less than the ideal-divider prediction, because the logger’s finite input resistance effectively lowers the bottom resistance and reduces sensitivity. (correct)
  2. B It increases by exactly the ideal-divider factor; the logger’s input is high so it does not load the divider appreciably.
  3. C It decreases; the lower node is dragged toward 0 V when the LDR draws more current in brighter light.
  4. D It stays approximately the same; the regulated 5 V supply and the logger input together keep the midpoint voltage nearly fixed.

Answer

The correct answer is A.

Correct: A — It increases, but by less than the ideal-divider prediction, because the logger’s finite input resistance effectively lowers the bottom resistance and reduces sensitivity. A: Halving the top (LDR) resistance moves the divider output toward +5 V, but the 10 kΩ input in parallel with the 10 kΩ lower resistor halves the effective bottom resistance, reducing both Vout and its change compared with the no-loading (Rin → ∞) case. B: Wrong because Rin is comparable to the lower resistor, so loading is significant; only if Rin were much larger would the ideal prediction hold. C: Wrong because with the LDR on the upper leg, reducing its resistance pulls the midpoint toward +5 V, not toward 0 V. D: Wrong because a regulated supply fixes the rails, not the divider node; the logger input cannot “hold” Vout constant, it merely loads the divider.

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