= Solution
Put $e_t=e^{-t/\tau}$. From $u=V/2-RI_1$ and $I_0=I_1+I_2$,
$$
\boxed{I_1=\frac V{4R}(1-e_t),\qquad I_2=\frac V{8R}(1+e_t),\qquad I_0=\frac V{8R}(3-e_t).}
$$
At $t=0^+$ the three values are $I_1=0$ and $I_2=I_0=V/(4R)$. At late times they are $I_1=V/(4R)$, $I_2=V/(8R)$, and $I_0=3V/(8R)$. Thus $I_1$ and $I_0$ increase monotonically while $I_2$ decreases; every transient has the same <time constant>. By $10\tau$, the exponential correction is only $e^{-10}\simeq4.54\times10^{-5}$ of its initial amplitude.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-rl-branch-currents.png]
{title=Switch-on currents in the resistive network containing one inductor}
{description=Currents are normalized by V/R. Values at switching and at equilibrium are marked for all three source-diagram current directions.}
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