 Solution

Let $C$ denote the capacitance omitted after the word “capacitance” in the source sentence. For the ideal series RC charging circuit, Kirchhoff's voltage law gives $RI+Q/C=V$, with $I=dQ/dt$ and $Q(0)=0$. ThereforeThe electric current falls from $V/R$ to zero, while the electric charge rises from zero to $CV$. Their initial slopes are $-V/(R\tau )$ and $V/R$, respectively.

Reducing $RC$ shortens the approach to the relevant final values. Which heights change depends on the component changed: reducing $R$ at fixed $C$ increases $I(0)$ but leaves $Q(\infty )$ unchanged; reducing $C$ at fixed $R$ leaves $I(0)$ unchanged but reduces $Q(\infty )$. A shorter time constant alone does not specify both endpoint amplitudes.