A2 Daily A Level Physics question

2026-02-14 OCR A Medical Imaging (M6) OCR Physics A (H156/H556) Module 4.2 Energy, power and resistance OCR Physics A (H156/H556) Module 4.3 Electrical circuits — potential dividers

A light-dependent resistor (LDR) is in series with a 2.0 kΩ fixed resistor across a stable 5.0 V supply. The output voltage V_out is measured across the fixed resistor. In bright light the LDR is about 200 Ω; in dim light it increases to about 20 kΩ. When the room is darkened, what happens to V_out, and why?

A It increases from about 0.45 V to about 4.5 V, because the larger LDR resistance forces more voltage across the fixed resistor.
B It increases slightly from about 2.3 V to about 2.7 V, since the 5.0 V is shared nearly equally despite the change.
C It stays close to 5.0 V because the supply is fixed, so the output across the fixed resistor is essentially unaffected.
D It decreases from about 4.5 V to about 0.45 V, as the fixed resistor takes a smaller fraction of the 5.0 V when the LDR’s resistance becomes much larger. (correct)

Answer

The correct answer is D.

Correct: D — It decreases from about 4.5 V to about 0.45 V, as the fixed resistor takes a smaller fraction of the 5.0 V when the LDR’s resistance becomes much larger. In bright light: fraction = 2.0 kΩ / (2.0 kΩ + 0.20 kΩ) ≈ 0.91, so V_out ≈ 4.5 V; in dim light: 2.0 kΩ / (20 kΩ + 2.0 kΩ) = 2/22 ≈ 0.091, so V_out ≈ 0.45 V. A reverses the voltage division: when the LDR’s resistance increases, more of the supply voltage is dropped across the LDR, not the fixed resistor. B assumes an almost equal split; the large change in the LDR’s resistance shifts the division by about a factor of ten, not a slight amount. C confuses the fixed 5.0 V of the supply with the measured output across only the fixed resistor, which does change with the resistance ratio. D is the only option consistent with voltage division in series.

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