In the circuit shown, $I_0$  is a constant current generator. Take  $I_0=4A$.  The switch S is kept closed for a long time and opened at time t = 0. The potential difference $(V_A-V_B)$ at time t = ln2 $\mu$ s is given by
 

First steady state voltage at t = 0 is with 0.5 $\Omega$  in parallel with 1 $\Omega$

$\Rightarrow V_i=4\left(\frac{1}{3}\right)=\frac{4}{3}volt$

Final steady state voltage at  $t\to \infty$ is with only 1 $\Omega$  in parallel
$\Rightarrow V_f=4volt$ 
Time constant after t = 0 is $RC=\left(1\Omega \right)\left(0.5\mu F\right)=0.5\mu S$
$$\begin{gathered} V=4-\frac{8}{3}{e}^{-t/\tau } \\ givent=2\tau \ln 2,V=4-\frac{8}{3}\left(\frac{1}{4}\right) \\ =4-\frac{2}{3}=\frac{10}{3}volt \end{gathered}$$