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, $\frac{x-v}{4R}+\frac{x}{4R}+\frac{x}{36R}+\frac{x-v}{6R}=0\Rightarrow x=\frac{3}{5}v$

$\therefore 4IR=\frac{2}{5}v or v=\frac{4\times 5IR}{2}=10IR$

 In steady state: $\frac{x_1-v}{4R}+\frac{x_1}{4R}=0\Rightarrow x_1=\frac{v}{2}, \frac{x_2-v}{6R}+\frac{x_2}{36R}=0\Rightarrow x_2=\frac{6v}{7}$

Charge on capacitor  $=\frac{25}{7}\text{CIR}$