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AS Daily A Level Physics question

2025-11-10 OCR A DC Circuits (M4) 4.3.3(c)(i)

A student makes a light sensor using a 6.0 V battery, a 1.0 kΩ fixed resistor, and an LDR in series. The output voltage V_out is measured across the LDR, with the positive terminal taken at the end nearest the battery’s positive. In bright light the LDR is 200 Ω; in darkness it is 5.0 kΩ. Which statement must be true as the room goes from bright to dark? Consider the sign of V_out and what happens when one resistance is much larger than the other.

  1. A V_out decreases by about a factor of 5, because the current falls and less voltage appears across the larger resistance.
  2. B V_out increases only slightly (well under 50%), since the 1.0 kΩ resistor still takes most of the 6.0 V.
  3. C V_out is roughly unchanged, because in series the voltage share is about equal regardless of light level.
  4. D V_out rises from about 1.0 V to about 5.0 V (≈5×), since a much larger LDR takes nearly all the 6.0 V. (correct)

Answer

The correct answer is D.

Correct: D — V_out rises from about 1.0 V to about 5.0 V (≈5×), since a much larger LDR takes nearly all the 6.0 V. In bright: V_out ≈ (200/(1000+200))×6 ≈ 1.0 V; in dark: V_out ≈ (5000/(1000+5000))×6 ≈ 5.0 V; polarity stays positive across the LDR. A This reverses the trend: the larger resistance in series takes a larger share of the supply, so V_out should increase, not decrease. B Understates the change: when the LDR dominates (5.0 kΩ vs 1.0 kΩ), most of the 6.0 V appears across it, so the increase is much more than 50%. C The voltage split is not generally equal; it is proportional to resistance, so changing the LDR from 200 Ω to 5.0 kΩ strongly alters V_out. D Matches the correct ratios and limiting-case reasoning, and keeps the sign positive as defined.

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