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A2 Daily A Level Physics question

2026-06-10 OCR A Electric circuits & sensors (Module 4) OCR-A 4.2.5 Potential dividers and sensors (LDR/thermistor) OCR-A 4.2.2 Energy, power and resistance (ohmic behaviour)

A light sensor is made by placing an LDR in series with a fixed resistor across a 5.0 V supply. The output voltage V_out is taken across the fixed resistor (from the junction to ground). As the room gets brighter, what happens to V_out? Assume ideal wires and that the LDR’s resistance decreases with increasing light.

  1. A Increases, because the LDR’s resistance falls so a larger fraction of the supply appears across the fixed resistor. (correct)
  2. B Decreases, because more current flows so the fixed resistor drops less voltage.
  3. C No significant change, because the supply voltage is fixed.
  4. D First increases then decreases, because the LDR is non‑ohmic so V_out peaks at medium light.

Answer

The correct answer is A.

Correct: A — Increases, because the LDR’s resistance falls so a larger fraction of the supply appears across the fixed resistor. A increases since reducing the top resistance in a divider raises the share of the supply across the lower fixed resistor. B is wrong because greater current through an ohmic fixed resistor gives a larger, not smaller, voltage across it. C is wrong because although the supply is fixed, the divider ratio changes when the LDR’s resistance changes, so V_out changes. D is wrong because a two‑resistor divider changes monotonically with resistance; there is no peak with increasing light.

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