= Solution
At equality, $I_1=V/(3R)-I_2/3=I_2$, so the common <electric current> is $V/(4R)$. Substituting the <capacitor> <electric current> gives
$$
\frac{2V}{5R}e^{-t/\tau}=\frac{V}{4R},\qquad e^{-t/\tau}=\frac58.
$$
Therefore
$$
\boxed{t=\tau\log\frac85=\frac{5RC}{3}\log\frac85}.
$$
For $R=50\ \mathrm{k\Omega}$ and $C=0.2\ \mathrm{microfarads}$, $RC=0.0100\ \mathrm s$ and $\tau=0.01667\ \mathrm s$, giving $\boxed{t\simeq7.83\ \mathrm{ms}}$. 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.
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