In the circuit shown below, the capacitor is initially uncharged. Immediately after the key K is closed, the reading of the ammeter is I

Applying KCL, we get

$\begin{array}{l}\frac{\mathrm{x}-\mathrm{v}}{4\mathrm{R}}+\frac{\mathrm{x}}{4\mathrm{R}}+\frac{\mathrm{x}}{36\mathrm{R}}+\frac{\mathrm{x}-\mathrm{v}}{6\mathrm{R}}=0\\ \mathrm{x}=\frac{3}{5}\mathrm{v}\\ \therefore 4\mathrm{IR}=\frac{2}{5}\mathrm{v}\\ \mathrm{v}=\frac{4\times 5\mathrm{IR}}{2}=10\mathrm{IR}\end{array}$
In steady state:
$\begin{array}{l}\frac{{\mathrm{x}}_1-\mathrm{v}}{4\mathrm{R}}+\frac{{\mathrm{x}}_1}{4\mathrm{R}}=0\Rightarrow {\mathrm{x}}_1=\frac{\mathrm{v}}{2}\\ \frac{{\mathrm{x}}_2-\mathrm{v}}{6\mathrm{R}}+\frac{{\mathrm{x}}_2}{36\mathrm{R}}=0\Rightarrow {\mathrm{x}}_2=\frac{6\mathrm{v}}{7}\\ \text{ charge on capacitor }=\frac{25}{7}\mathrm{CIR}\end{array}$