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AS Daily A Level Physics question

2026-03-31 OCR A DC Circuits (M4) Module 4.3.1 Electrical circuits — application of current and potential difference rules Module 4.2.2 Energy, power and resistance — potential dividers; meter loading (qualitative)

In a lab, a student makes a 12.0 V potential divider using two identical 10 kΩ resistors to obtain about 6.0 V at the midpoint. They measure the midpoint to ground with an analogue voltmeter whose input resistance on the chosen range is 20 kΩ. Which statement must be true about the meter reading, and why?

  1. A About 6.0 V, because the voltmeter draws negligible current.
  2. B About 4.8 V, because the meter in parallel with the lower resistor reduces the lower leg’s resistance to two-thirds, shifting the divider ratio. (correct)
  3. C About 8.0 V, because adding the meter increases the total resistance so the midpoint voltage rises.
  4. D About 7.2 V, because the meter plus the lower resistor can be treated as 15 kΩ, then 12 × 15/(10 + 15).

Answer

The correct answer is B.

Correct: B — About 4.8 V, because the meter in parallel with the lower resistor reduces the lower leg’s resistance to two-thirds, shifting the divider ratio. A is wrong because assuming negligible current ignores that 20 kΩ is comparable to 10 kΩ, so the meter loads the divider and pulls the output down. B is correct: 10 kΩ ∥ 20 kΩ ≈ 6.7 kΩ, so Vout ≈ 12 × 6.7/(10 + 6.7) ≈ 4.8 V. C is wrong because loading the lower leg reduces its resistance and therefore lowers (not raises) the midpoint voltage. D is wrong because treating 10 kΩ ∥ 20 kΩ as 15 kΩ (a simple average) is incorrect; that error gives 7.2 V instead of the correct ~4.8 V.

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