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AS Daily A Level Physics question

2026-04-22 OCR A DC Circuits (Module 4): meter loading; internal resistance; terminal pd OCR A Physics H156/H556 Module 4: 4.3.1 Circuits (practical use of meters; potential dividers) OCR A Physics H156/H556 Module 4: 4.3.3 Internal resistance and terminal pd Practical skills: meter loading (qualitative effect of finite meter resistance)

A student measures the emf E of a single cell by placing a voltmeter directly across its terminals. The cell’s internal resistance is 100 Ω. Meter A has an input resistance of 10 kΩ; Meter B has an input resistance of 1.0 kΩ. The cell is measured with A and then with B. Which statement must be true about the two readings relative to E?

  1. A Both are about 0.99E; reducing the meter resistance by 10× barely changes the result.
  2. B Meter B (1.0 kΩ) reads higher because it takes more current, so a larger drop appears across it.
  3. C Meter B reads lower; its percentage shortfall is about 9% of E, whereas Meter A’s is about 1% (roughly nine times larger). (correct)
  4. D Meter B reads lower, but by only about 1% compared with Meter A.

Answer

The correct answer is C.

Correct: C — Meter B reads lower; its percentage shortfall is about 9% of E, whereas Meter A’s is about 1% (roughly nine times larger). With finite meter resistance Rm in parallel with nothing else, the meter completes a series path with the cell’s internal resistance r, so the reading is E × Rm/(r + Rm). For A: 10 kΩ gives VA/E ≈ 10000/(100 + 10000) ≈ 0.990 (≈1% shortfall); for B: 1.0 kΩ gives VB/E ≈ 1000/(100 + 1000) ≈ 0.909 (≈9% shortfall). A The 1.0 kΩ case is not ≈0.99E; it is ≈0.91E, so the change is significant. B Lowering the meter resistance increases current but decreases the fraction of E appearing across the meter (Rm/(r + Rm)), so the reading falls, not rises. C This matches the calculated percentages and the correct direction. D The drop is not only about 1%; it is about 9% vs about 1%, a much larger change.

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