= Solution
<Kirchhoff's current law> at the left <capacitor> terminal gives $(V-V_a)/R=V_a/(2R)+I_2$. Since $I_1=V_a/(2R)$,
$$
\boxed{I_1(t)=\frac{V}{3R}-\frac13I_2(t)=\frac{V}{3R}-\frac{2V}{15R}e^{-t/\tau}}.
$$
Thus $I_1$ rises from $V/(5R)$ to $V/(3R)$, whereas $I_2$ falls from $2V/(5R)$ to zero. Their derivatives have opposite signs, and both approach their limiting values on the same time scale $\tau=5RC/3$. The graph uses <electric currents> in units of $V/R$ and time in units of $\tau$.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit-capacitor-branch-currents.png]
{title=The capacitor's Thevenin equivalent and the two original branch currents}
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