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A2 Daily A Level Physics question

2026-06-09 OCR A High level GCSE electricity, Potential dividers and sensor circuits, Meter loading (qualitative and simple quantitative) OCR-A Module 4.2.1 Charge, current and potential difference OCR-A Module 4.2.2 Energy, power and resistance OCR-A Module 4.3.1 Electrical circuits: series/parallel, Kirchhoff’s laws OCR-A Module 4.3.2 Potential dividers and sensor circuits; loading effects

A student tests a light sensor circuit powered by 5.0 V: a 10 kΩ fixed resistor is in series with an LDR to 0 V, and a voltmeter with 20 kΩ input resistance is connected across the LDR to read V_out at the junction. The LDR is 10 kΩ in the dark and 2.0 kΩ in bright light. Which statement must be true about the meter reading as the light increases? Estimate numerically.

  1. A It rises to about 4.2 V because the LDR carries more current in bright light.
  2. B It stays near 2.5 V since the meter’s resistance is much larger than the divider resistances.
  3. C It falls only slightly, from about 2.0 V to about 1.7 V, because the meter shunts little current.
  4. D It falls from about 2.0 V (dark) to about 0.8 V (bright), as the effective lower resistance drops from ≈6.7 kΩ to ≈1.8 kΩ. (correct)

Answer

The correct answer is D.

Correct: D — It falls from about 2.0 V (dark) to about 0.8 V (bright), as the effective lower resistance drops from ≈6.7 kΩ to ≈1.8 kΩ. A The reading will not rise: with the meter across the LDR (the lower element), reducing its resistance reduces its share of the 5.0 V, so an increase to ~4.2 V is opposite to the divider behaviour. B 20 kΩ is not much larger than 10 kΩ, so loading is significant; the dark case gives R_eff = (10∥20) ≈ 6.7 kΩ and V_out ≈ 5 × 6.7/(10 + 6.7) ≈ 2.0 V, not ~2.5 V. C The drop is not slight: in bright light R_eff = (2.0∥20) ≈ 1.8 kΩ, giving V_out ≈ 5 × 1.8/(10 + 1.8) ≈ 0.77 V, not ~1.7 V. D Matches the calculation: about 2.0 V in the dark to about 0.8 V in bright light once the meter’s loading is included.

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