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A2 Daily A Level Physics question

2026-04-09 OCR A Fields: Gravitational, Electric & Magnetic (M5/M6) 6.2.3 Capacitors — charging/discharging; time constant (R×C); qualitative exponential behaviour 6.2.1 Capacitance — C = Q/V; steady-state voltage across a capacitor

Two lab timing circuits each connect a 9.0 V battery, a resistor R, and an initially uncharged capacitor C in series with a switch. Circuit A uses R = 20 kΩ with C = 50 µF. Circuit B uses R = 10 kΩ with C = 100 µF. The switch is closed at t = 0. At t = 1.0 s, which statement must be true, and why?

  1. A After 1.0 s, the capacitor in circuit B has the higher voltage because the smaller resistor makes it charge faster even though its capacitor is larger.
  2. B After 1.0 s, both capacitors have reached the same fraction of 9.0 V (about 0.63), because R×C is the same in both circuits. (correct)
  3. C The initial currents are the same in both circuits since an uncharged capacitor initially behaves like a wire, so their voltages rise equally at first.
  4. D After a very long time, the circuit with the larger capacitor settles to a lower final voltage because more charge is needed to reach the same voltage.

Answer

The correct answer is B.

Correct: B — After 1.0 s, both capacitors have reached the same fraction of 9.0 V (about 0.63), because R×C is the same in both circuits. A: This ignores that the larger C in B offsets the smaller R; here R×C = (20 kΩ×50 µF) = (10 kΩ×100 µF) = 1.0 s, so after 1.0 s the voltages are equal (~0.63×9.0 V ≈ 5.7 V), not higher in B. B: Correct because equal R×C means equal progress towards the final value at a given time; at one such time the charging capacitor is about 63% of the supply. C: An uncharged capacitor does act like a short initially, but the initial currents are 9.0/20 kΩ = 0.45 mA and 9.0/10 kΩ = 0.90 mA, not the same; equal early slopes arise here from equal R×C, not equal currents. D: The final voltage is fixed by the supply, so both approach 9.0 V; a larger C just means more charge at the same final voltage, not a lower voltage.

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