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AS Daily A Level Physics question

2025-08-23 OCR A A Level Physics 4.3.3(c)(i) 4.3.1(d)

An LDR and a fixed resistor R are connected in series across a stable d.c. supply. The output voltage Vout is taken across the LDR. A student measures Vout with a digital voltmeter whose input resistance is high but finite. The LDR’s resistance is very large in darkness and very small in bright light. Which statement must be true about the measured Vout compared with the value you would predict if the voltmeter drew no current?

  1. A In the dark, the measured Vout is lower than the ideal value because the voltmeter provides a parallel path that reduces the effective resistance of the LDR. (correct)
  2. B In the dark, the measured Vout equals the supply emf because almost no current flows through the divider.
  3. C In bright light, the measured Vout is higher than the ideal value because the voltmeter adds to the LDR’s resistance.
  4. D The voltmeter reading is unaffected because voltmeters do not draw current.

Answer

The correct answer is A.

Correct: A — In the dark, the measured Vout is lower than the ideal value because the voltmeter provides a parallel path that reduces the effective resistance of the LDR. B — Wrong because Vout only tends towards the supply voltage if the measuring device draws negligible current; with finite meter resistance in parallel with a very high LDR, the output is pulled down from the ideal value. C — Wrong because the voltmeter is in parallel with the LDR, not in series, so it cannot increase the effective resistance; in bright light the LDR is low resistance and the loading is negligible, not an increase above the ideal value. D — Wrong because real voltmeters have finite resistance and do draw a small current, which can load high-resistance sources and change the reading.