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A2 Daily A Level Physics question

2026-06-22 OCR A DC circuits: meter loading; internal resistance & terminal pd (Module 4.3) OCR A Physics A Module 4.3.1: Kirchhoff’s laws OCR A Physics A Module 4.3.3: Electromotive force and internal resistance Practical circuits: meter loading and instrument resistance (qualitative)

A student measures the emf of a 1.50 V cell that has an internal resistance of 50 Ω using an analogue voltmeter with sensitivity 1 kΩ per volt set to the 1.5 V range (so the meter’s input resistance is about 1.5 kΩ). The meter is connected directly across the cell. Which statement best describes the meter loading effect on the reading, and by roughly how much?

  1. A The reading is unchanged at 1.50 V because the meter draws no current from the cell.
  2. B The reading is higher than 1.50 V because adding the meter increases the total resistance in the circuit.
  3. C The reading is lower by about 3% because the meter’s finite resistance draws a small current, causing a drop across the cell’s internal resistance. (correct)
  4. D The reading is lower by about 30% because the meter and the internal resistance split the voltage roughly equally.

Answer

The correct answer is C.

Correct: C — The reading is lower by about 3% because the meter’s finite resistance draws a small current, causing a drop across the cell’s internal resistance. A The meter is not ideal (about 1.5 kΩ here), so it draws current; the terminal pd is therefore slightly less than the emf. B Adding the meter does not raise the terminal pd; it draws current and causes an internal voltage drop, so the reading falls, not rises. C Using a simple divider: fraction lost ≈ r/(r + R_meter) = 50/(50 + 1500) ≈ 50/1550 ≈ 0.032 ≈ 3%, so the reading is about 1.45 V, slightly below 1.50 V. D 30% is far too large; 50 Ω is much smaller than 1.5 kΩ, so only about 1/31 of the emf is lost, not about a third.

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