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A2 Daily A Level Physics question

2026-07-10 OCR A Consolidation: DC circuits mixed reasoning (A2) Module 4.2.2 Energy, power and resistance Module 4.2.3 Electrical circuits (series/parallel, meters) Module 4.2.5 Potential divider and sensor circuits PAG4: Electrical circuits — use of meters; assessing meter loading

A student builds a 5.0 V potential divider using two identical 10 kΩ resistors. They measure the output by connecting a digital voltmeter (input resistance 100 kΩ) from the midpoint to ground (i.e. across the lower 10 kΩ). Which statement must be true about the meter reading compared with the ideal 2.50 V?

  1. A The reading is higher than 2.50 V by about 5% (≈2.62 V) because the meter adds resistance in series.
  2. B The reading is lower than 2.50 V by about 5% (≈2.38 V) because the meter loads the lower resistor in parallel. (correct)
  3. C The reading remains 2.50 V because a voltmeter does not affect a circuit.
  4. D The reading drops to about half (≈1.25 V) because the meter halves the lower resistance.

Answer

The correct answer is B.

Correct: B — The reading is lower than 2.50 V by about 5% (≈2.38 V) because the meter loads the lower resistor in parallel. A The meter is not in series with the divider; it is in parallel with the lower resistor, so it draws extra current and reduces the output rather than increasing it. B Parallel loading gives R_eq = (10 kΩ || 100 kΩ) ≈ 9.09 kΩ, so V_out ≈ 5 × 9.09/(10 + 9.09) ≈ 2.38 V, about 5% below 2.50 V. C Real voltmeters have finite input resistance (100 kΩ here), so they do affect the circuit; only an ideal infinite-resistance meter would leave 2.50 V unchanged. D Halving the output would require the lower resistance to be much smaller (e.g. 5 kΩ with a 10 kΩ in parallel), not the slight change from 10 kΩ to 9.09 kΩ caused by a 100 kΩ meter.

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