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AS Daily A Level Physics question

2026-07-04 OCR A Electricity: DC circuits & sensors (M4) Module 4.2 Energy, power and resistance — I–V characteristics of components Module 4.3 Electrical circuits — Potential dividers and sensors (LDR/thermistor)

A garden light uses an LDR in a potential divider to sense daylight. A 4.0 kΩ fixed resistor is connected from +6.0 V to the output node; the LDR is connected from the output node to 0 V. In shade the LDR is 8.0 kΩ. When the light level doubles, assume the LDR resistance halves. What is the new output voltage?

  1. A 5.0 V
  2. B 4.0 V
  3. C 3.0 V (correct)
  4. D 2.0 V

Answer

The correct answer is C.

Correct: C — 3.0 V. With the LDR at the bottom, Vout = Vs × Rbottom/(Rtop + Rbottom): in shade Vout = 6 × 8/(4 + 8) = 4.0 V; when light doubles, LDR halves to 4 kΩ so Vout = 6 × 4/(4 + 4) = 3.0 V, a 25% decrease. A 5.0 V — Incorrect: reducing the lower resistance pulls the divider point closer to 0 V, not closer to the supply. B 4.0 V — Incorrect: the resistance ratio changes from 8:4 to 4:4, so the output cannot stay the same. C 3.0 V — Correct: calculation using the resistor ratio gives 3.0 V. D 2.0 V — Incorrect: halving the lower resistor does not halve the output here; the top resistor is unchanged so the new ratio gives 3.0 V, not 2.0 V.

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