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A2 Daily A Level Physics question

2026-07-12 OCR A Capacitance & RC (M6) 6.1.3(b) 6.1.3(c) 4.3.2(c)(i)

In a lab setup, a 9.0 V battery with internal resistance 1.0 Ω charges a 2200 μF capacitor through a 4.0 Ω series resistor when a switch is closed. Neglect leakage. Estimate when the capacitor voltage first reaches 6.0 V, and justify your choice.

  1. A About 11 ms, using only the 4.0 Ω external resistor for the time constant, since the battery’s internal resistance doesn’t affect the charging rate.
  2. B About 7 ms, because the battery’s internal resistance lowers the terminal voltage so the capacitor gets to 6.0 V faster.
  3. C About 22 ms, since reaching two-thirds of the final voltage needs roughly two time constants.
  4. D About 12 ms, because the charging resistance is 1.0 Ω + 4.0 Ω so τ ≈ 11 ms, and 6.0 V is slightly above 63% of 9.0 V. (correct)

Answer

The correct answer is D.

Correct: D — About 12 ms, because the charging resistance is 1.0 Ω + 4.0 Ω so τ ≈ 11 ms, and 6.0 V is slightly above 63% of 9.0 V. A The internal resistance is in series during charging, so using only 4.0 Ω underestimates τ (it should be (1.0 + 4.0) Ω). B Internal resistance increases the charging time constant; it does not make the capacitor reach 6.0 V faster, and the final voltage is still about 9.0 V when current falls to zero. C Two time constants would give about 86% of the final voltage, not roughly two-thirds, so 22 ms is too large. D Includes the correct total series resistance for τ ≈ 5.0 Ω × 2.2×10⁻³ F ≈ 11 ms, with 6.0 V (≈67% of 9.0 V) reached a little after one τ, about 12 ms.

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