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A2 Daily A Level Physics question

2026-04-24 OCR A Synoptic: RC + circuits power; measurement strategies (no ln) OCR-A 6.2 Capacitors: time constant and charge/discharge (qualitative); experimental determination of RC OCR-A 4.2 DC circuits: measuring instruments; voltmeter resistance and loading effects

Which statement must be true when a 100 μF capacitor discharges through a 100 kΩ resistor and the voltage is monitored with a handheld digital voltmeter of 1.0 MΩ input resistance by timing the drop from 8.0 V to 4.0 V? Compared with using an ideal (infinite-resistance) voltmeter, what happens to the measured half-time t_meas, and why?

  1. A A t_meas is about 10% larger, because the meter slightly increases the discharge resistance.
  2. B B t_meas is about 9% smaller, because the meter in parallel makes the effective resistance about 91 kΩ instead of 100 kΩ. (correct)
  3. C C t_meas is essentially unchanged (within 1%), because 1.0 MΩ is much bigger than 100 kΩ.
  4. D D t_meas is about 50% smaller, because the meter draws a current comparable to the discharge current.

Answer

The correct answer is B.

Correct: B — B t_meas is about 9% smaller, because the meter in parallel makes the effective resistance about 91 kΩ instead of 100 kΩ. A The meter does not increase the discharge resistance; its finite input resistance provides an additional parallel path so the effective resistance is lower, not higher. B This is correct: 100 kΩ in parallel with 1.0 MΩ gives R_eff ≈ (100k×1000k)/(1100k) ≈ 91 kΩ, so all characteristic times (including the half-time) scale down by about 91/100 ≈ 0.91, i.e. ~9% smaller. C The change is not within 1%; the 10:1 ratio gives a parallel combination about 9% lower than 100 kΩ, so the timing error is noticeable. D A 1.0 MΩ meter is not comparable to 100 kΩ; the parallel reduction is modest, not a factor of two.

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