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

At equality, $I_1=V/(3R)-I_2/3=I_2$, so the common electric current is $V/(4R)$. Substituting the capacitor electric current givesThereforeFor $R=50\mathrm{k\Omega }$ and $C=0.2\mathrm{microfarads}$, $RC=0.0100\mathrm{s}$ and $\tau =0.01667\mathrm{s}$, giving $\boxed{t\simeq 7.83\mathrm{ms}}$. 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.