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A2 Daily A Level Physics question

2026-02-01 OCR A High level GCSE electricity OCR-A 4.3.2 Potential dividers OCR-A 4.3.1 Series and parallel circuits OCR-A 4.2.2 Energy, power and resistance (ohmic behaviour) OCR-A Practical: meter loading (qualitative)

A 12 V supply feeds a potential divider: a fixed 4.0 kΩ resistor from the supply to an LDR to ground. The output Vout is taken across the LDR. In dim light the LDR is 8.0 kΩ; in bright light it halves to 4.0 kΩ. A voltmeter of 12 kΩ is then connected across Vout. Which statement must be true about the measured change in Vout when the light increases from dim to bright?

  1. A With the 12 kΩ meter connected, Vout falls by about 21% (≈6.5 V to ≈5.1 V), smaller than the 25% fall without the meter, because the meter in parallel lowers the LDR leg more in the dim case. (correct)
  2. B With the meter connected, Vout falls by about 25% as before, because 12 kΩ is large so the divider is effectively unchanged.
  3. C With the meter connected, Vout falls by about 30% (a larger change), because the extra current through the meter exaggerates the drop across the top resistor.
  4. D With the meter connected, Vout rises when the light increases, because the meter draws current that increases the share of voltage across the LDR.

Answer

The correct answer is A.

Correct: A — With the 12 kΩ meter connected, Vout falls by about 21% (≈6.5 V to ≈5.1 V), smaller than the 25% fall without the meter, because the meter in parallel lowers the LDR leg more in the dim case. A The loaded lower leg is 8.0 kΩ‖12 kΩ = 4.8 kΩ (dim) and 4.0 kΩ‖12 kΩ = 3.0 kΩ (bright), giving Vout ≈ 12×4.8/(4+4.8) ≈ 6.55 V to 12×3/(4+3) ≈ 5.14 V, a ~21% drop, less than the 25% (8.0 V→6.0 V) without loading. B Assumes an ideal, infinite-resistance meter; 12 kΩ is comparable to 8 kΩ and 4 kΩ, so the shunt alters the divider ratio and reduces the percentage change. C Reverses the effect of loading: the shunt compresses the range by reducing the lower-leg resistance most when it is large, so the percentage change is smaller, not larger. D Gets the direction wrong: as the LDR resistance halves, the fraction R2/(R1+R2) decreases, so Vout must fall; a meter does not impose a voltage.

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