= Solution
Label the <capacitor>'s left terminal $a$ and right terminal $b$, with the battery's negative node at zero. The source diagram has $2R$ from $a$ to ground, $R$ from $b$ to ground, and a source-side <resistor> $R$ feeding $a$ from <electric potential difference> $V$. The <capacitor> <electric current> $I_2$ flows from $a$ to $b$; $I_1$ flows through $2R$.
With the <capacitor> removed, $V_a=2V/3$ and $V_b=0$, so the <Thévenin equivalent> seen by the <capacitor> has $V_{\rm th}=2V/3$. Suppress the ideal <electric potential difference> source to find its <electrical resistance>: the left terminal sees $R\parallel2R=2R/3$ to ground and the right sees $R$, giving $R_{\rm th}=5R/3$ between the terminals.
The initially uncharged <capacitor> <electric potential difference> $V_C=V_a-V_b$ therefore obeys
$$
R_{\rm th}C\frac{dV_C}{dt}+V_C=V_{\rm th},\qquad V_C(0)=0.
$$
Solving gives $V_C=(2V/3)(1-e^{-t/\tau})$, with $\tau=R_{\rm th}C$. The <capacitor> <electric current> is $I_2=C\dot V_C$, hence
$$
\boxed{I_2(t)=\frac{2V}{5R}e^{-t/\tau},\qquad A=\frac{2V}{5R},\qquad\tau=\frac{5RC}{3}}.
$$
At $t=0^+$, the short-like uncharged <capacitor> makes $V_a=V_b=2V/5$, so $I_2=2V/(5R)$ directly from the right <resistor>. At long times the <capacitor> carries no direct <electric current>. These endpoint checks agree with the exponential solution.
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