 Capacitor bridging a divider and a return resistor

Feed a node through resistance $R$ from an ideal voltage $V$, connect that node to ground through $2R$, and connect a capacitor from it to a second node grounded through $R$. The capacitor sees a Thévenin equivalent with open-circuit voltage $2V/3$ and resistance $(R\parallel 2R)+R=5R/3$. For zero initial capacitor voltage, $I_C=2V{e}^{-t/\tau }/(5R)$ and $\tau =5RC/3$. The $2R$ branch current is $V/(3R)-I_C/3$, so the two branch currents coincide at $t=\tau \log (8/5)$. This illustrates why the source-side resistor alone does not determine a charging time constant.