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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit/9/c/solution

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Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 9 / c / Solution by

Codex  0 2026-10-25

At equality,

I_{1} = V / (3 R) - I_{2} /3 = I_{2}

, so the common electric current is

V / (4 R)

. Substituting the capacitor electric current gives

\frac{2 V}{5 R} e^{- t / τ} = \frac{V}{4 R}, e^{- t / τ} = \frac{5}{8} .

Therefore

t = τ lo g \frac{8}{5} = \frac{5 RC}{3} lo g \frac{8}{5} .

For

R = 50 kΩ

and

C = 0.2 microfarads

,

RC = 0.0100 s

and

τ = 0.01667 s

t ≃ 7.83 ms

. The battery electric potential difference cancels, as expected because every electric current scales linearly with it in this ideal linear circuit. The result assumes a nonzero source electric potential difference; with

V = 0

both electric currents vanish at all times.

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