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A2 Daily A Level Physics question

2026-05-23 OCR A DC Circuits (M4) 6.2 Capacitors 6.2.2 Charging and discharging of capacitors; time constant RC (qualitative exponential behaviour)

In a lab test, a 1000 μF capacitor is charged to 6.0 V, then allowed to discharge through a resistor while a data logger records the voltage. In trial 1 the resistor is 1.0 kΩ; in trial 2 it is 2.0 kΩ. Students measure the time taken for the voltage to drop to 3.0 V. Which statement must be true?

  1. A Trial 2 takes twice as long; t2 ≈ 2 t1, because for the same capacitor the time to reach the same fraction of the initial voltage is proportional to the resistance. (correct)
  2. B Trial 1 takes twice as long; t1 ≈ 2 t2, because the larger current in trial 1 makes the voltage fall more slowly.
  3. C Both trials take the same time, because the voltage halves by the same amount regardless of the resistor used.
  4. D Trial 2 takes four times as long; t2 ≈ 4 t1, because reducing the current leads to a squared increase in the time.

Answer

The correct answer is A.

Correct: A — Trial 2 takes twice as long; t2 ≈ 2 t1, because for the same capacitor the time to reach the same fraction of the initial voltage is proportional to the resistance. A matches the idea that doubling R (with C unchanged) doubles the characteristic time to reach any fixed fraction like 1/2, so t2/t1 = 2. B reverses the effect: a smaller resistance gives a larger current and a faster discharge, so trial 1 would be quicker, not slower. C ignores the dependence on RC; the time to reach a given fraction is not the same when R changes. D introduces a squared dependence that does not apply here; the time scales linearly with R (and C), not with R squared.

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