= 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 $Q(0)=0$. Therefore
$$
\boxed{Q(t)=CV\bigl(1-e^{-t/\tau}\bigr),\qquad I(t)=\frac VR e^{-t/\tau},\qquad\tau=RC.}
$$
The <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.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-resit-rc-charging.png]
{title=RC charging with the same resistance and two different capacitances}
{description=The reference time is R times the larger capacitance. Reducing capacitance halves the time constant and final charge while keeping the initial current fixed.}
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