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A2 Daily A Level Physics question

2026-06-28 OCR A Synoptic: measurement strategies for RC transients (no ln) OCR-A H556 Module 6.2.2 Capacitors: charging/discharging and time constant OCR-A H556 Module 4.2.2 Electrical circuits: resistance combinations and power OCR-A H556 Module 1.2.1 Practical skills: measurement strategies and instrument loading

In a lab test, a student discharges a capacitor through a 100 kΩ resistor and estimates the circuit’s RC time by timing how long the capacitor’s voltage falls from 5.0 V to 2.5 V. An oscilloscope (input resistance 1.0 MΩ) is left connected across the capacitor during the timing. Compared with the true time constant without the scope attached, what is the size and sign of the error in the student’s estimate?

  1. A Approximately 1% too small (underestimates τ), as 1.0 MΩ is far bigger than 100 kΩ.
  2. B Approximately 9% too large (overestimates τ), because the added input makes discharge slower.
  3. C Approximately 9% too small (underestimates τ), since the 1.0 MΩ is in parallel with 100 kΩ. (correct)
  4. D Approximately 50% too small (underestimates τ), as the scope path largely bypasses the resistor.

Answer

The correct answer is C.

Correct: C — Approximately 9% too small (underestimates τ), since the 1.0 MΩ is in parallel with 100 kΩ. The scope’s 1.0 MΩ in parallel with 100 kΩ gives an effective discharge resistance of (100 kΩ ∥ 1.0 MΩ) ≈ 90.9 kΩ, so the time scale (and the half-value time) is about 90.9% of the true value — an underestimate by ≈9%. A The reduction is not as tiny as ~1%; the parallel combination lowers 100 kΩ to about 90.9 kΩ, which is ~9% smaller. B The estimate is not too large: adding a parallel path speeds the discharge, reducing the time, not increasing it. C This matches the correct parallel-resistance reasoning: 90.9/100 ≈ 0.909, i.e. about 9% low. D A 50% drop would require a comparable parallel resistance; 1.0 MΩ is much larger than 100 kΩ, so the effect is modest, not halving.

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