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AS Daily A Level Physics question

2026-05-23 OCR A Electrons, Waves and Photons (M4): DC circuits OCR-A Module 4.2.2: Series and parallel circuits; potential dividers OCR-A Module 4.2.3: Internal resistance and use of meters (qualitative loading)

In a lab, a 9.0 V supply is connected to two 100 kΩ resistors in series. A digital multimeter with 1.0 MΩ input resistance is connected across the upper 100 kΩ resistor to measure its potential difference. Which statement best describes the reading and why?

  1. A 4.50 V; the meter draws negligible current so the divider remains exactly equal.
  2. B 4.71 V; the extra current through the upper branch increases the drop across it.
  3. C 4.05 V; the meter effectively reduces the upper resistance to about 50 kΩ, so the drop roughly halves.
  4. D 4.28 V; the upper resistance becomes about 91 kΩ (100 kΩ || 1.0 MΩ), so the fraction across it is 91/(91+100) of 9.0 V. (correct)

Answer

The correct answer is D.

Correct: D — 4.28 V; the upper resistance becomes about 91 kΩ (100 kΩ || 1.0 MΩ), so the fraction across it is 91/(91+100) of 9.0 V. A assumes an ideal voltmeter with infinite resistance; 1.0 MΩ in parallel with 100 kΩ is not negligible, giving about a 5% reduction from 4.50 V. B is wrong because reducing the effective upper resistance decreases, not increases, the share of the supply across it. C overestimates loading; 100 kΩ in parallel with 1.0 MΩ is about 91 kΩ, not 50 kΩ, so the drop is not as low as 4.05 V. D correctly applies the parallel combination then the divider ratio to get about 4.28 V.

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